rec.puzzles Archive (Instructions), part 01 of 35
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> Instructions <==
Instructions for Accessing rec.puzzles Archive
INTRODUCTION
Below is a list of puzzles, categorized by subject area. Each puzzle
includes a solution, compiled from various sources, which is supposed
to be definitive.
To request a puzzle, send a message to archive...@questrel.com like:
return_address your_name@your_site.your_domain
send requested_puzzle_name
For example, if your net address is "mic...@disneyland.com", to request
"decision/allais.p", send the message:
return_address mic...@disneyland.com
send allais
To request multiple puzzles, use several "send" lines in a message.
Please refrain from requesting the entire archive via email. Use FTP.
FTP
The archive has been posted to news.answers and rec.answers, which are
archived in the periodic posting archive on rtfm.mit.edu in the
anonymous ftp directory /pub/usenet.
Other archives are:
ftp.cs.ruu.nl [131.211.80.17] in the anonymous ftp
directory /pub/NEWS.ANSWERS (also accessible via mail
server requests to mail-...@cs.ruu.nl)
cnam.cnam.fr [192.33.159.6] in the anonymous ftp directory /pub/FAQ
ftp.uu.net [137.39.1.9 or 192.48.96.9] in the anonymous ftp
directory /usenet
ftp.win.tue.nl [131.155.70.100] in the anonymous ftp directory
/pub/usenet/news.answers
grasp1.univ-lyon1.fr [134.214.100.25] in the anonymous ftp
directory /pub/faq (also accessible via mail server
requests to list...@grasp1.univ-lyon1.fr), which is
best used by EASInet sites and sites in France that do
not have better connectivity to cnam.cnam.fr (e.g.
Lyon, Grenoble)
Note that the periodic posting archives on rtfm.mit.edu are also
accessible via Prospero and WAIS (the database name is "usenet" on port
210).
CREDIT
The archive is NOT the original work of the editor (just in case you were
wondering :^).
In keeping with the general net practice on FAQ's, I do not as a rule assign
credit for solutions. There are many reasons for this:
1. The archive is about the answers to the questions, not about assigning
credit.
2. Many people, in providing free answers to the net, do not have the time
to cite their sources.
3. I cut and paste freely from several people's solutions in most cases
to come up with as complete an answer as possible.
4. I use sources other than postings.
5. I am neither qualified nor motivated to assign credit.
However, I do whenever possible put bibliographies in archive entries, and
I see the inclusion of the net addresses of interested parties as a
logical extension of this practice. In particular, if you wrote a
program to solve a problem and posted the source code of the program,
you are presumed to be interested in corresponding with others about
the problem. So, please let me know the entries you would like to be
listed in and I will be happy to oblige.
Address corrections or comments to archive...@questrel.com.
INDEX
==> bicycle (analysis) <==
A boy, a girl and a dog go for a 10 mile walk. The boy and girl can
==> boy.girl.dog (analysis) <==
A boy, a girl and a dog are standing together on a long, straight road.
==> bugs (analysis) <==
Four bugs are placed at the corners of a square. Each bug walks always
==> c.infinity (analysis) <==
What function is zero at zero, strictly positive elsewhere, infinitely
==> cache (analysis) <==
Cache and Ferry (How far can a truck go in a desert?)
==> calculate.pi (analysis) <==
How can I calculate many digits of pi?
==> cats.and.rats (analysis) <==
If 6 cats can kill 6 rats in 6 minutes, how many cats does it take to
==> dog (analysis) <==
A body of soldiers form a 50m-by-50m square ABCD on the parade ground.
==> e.and.pi (analysis) <==
Without finding their numerical values, which is greater, e^(pi) or (pi)^e?
==> functional/distributed (analysis) <==
Find all f: R -> R, f not identically zero, such that
==> functional/linear (analysis) <==
Suppose f is non-decreasing with
==> integral (analysis) <==
If f is integrable on (0,inf) and differentiable at 0, and a > 0, and:
==> irrational.stamp (analysis) <==
You have an ink stamp which is so amazingly precise that, when inked
==> minimum.time (analysis) <==
N people can walk or drive in a two-seater to go from city A to city B. What
==> particle (analysis) <==
What is the longest time that a particle can take in travelling between two
==> period (analysis) <==
What is the least possible integral period of the sum of functions
==> rubberband (analysis) <==
A bug walks down a rubber band which is attached to a wall at one end and a car
==> sequence (analysis) <==
Show that in the sequence: x, 2x, 3x, .... (n-1)x (x can be any real number)
==> snow (analysis) <==
Snow starts falling before noon on a cold December day. At noon a
==> tower (analysis) <==
R = N ^ (N ^ (N ^ ...)). What is the maximum N>0 that will yield a finite R?
==> 7-11 (arithmetic/part1) <==
A customer at a 7-11 store selected four items to buy, and was told
==> arithmetic.progression (arithmetic/part1) <==
Is there an arithmetic progression of 20 or more primes?
==> clock/day.of.week (arithmetic/part1) <==
It's restful sitting in Tom's cosy den, talking quietly and sipping
==> clock/palindromic (arithmetic/part1) <==
How many times per day does a digital clock display a palindromic number?
==> clock/reversible (arithmetic/part1) <==
How many times per day can the hour and minute hands on an analog clock switch
==> clock/right.angle (arithmetic/part1) <==
How many times per day do the hour and minute hands of a clock form a
==> clock/thirds (arithmetic/part1) <==
Do the 3 hands on a clock ever divide the face of the clock into 3
==> consecutive.composites (arithmetic/part1) <==
Are there 10,000 consecutive non-prime numbers?
==> consecutive.product (arithmetic/part1) <==
Prove that the product of three or more consecutive positive integers cannot
==> consecutive.sums (arithmetic/part1) <==
Find all series of consecutive positive integers whose sum is exactly 10,000.
==> conway (arithmetic/part1) <==
Describe the sequence a(1)=a(2)=1, a(n) = a(a(n-1)) + a(n-a(n-1)) for n>2.
==> digits/6.and.7 (arithmetic/part1) <==
Does every number which is not divisible by 5 have a multiple whose
==> digits/all.ones (arithmetic/part1) <==
Prove that some multiple of any integer ending in 3 contains all 1s.
==> digits/arabian (arithmetic/part1) <==
What is the Arabian Nights factorial, the number x such that x! has 1001
==> digits/circular (arithmetic/part1) <==
What 6 digit number, with 6 different digits, when multiplied by all integers
==> digits/divisible (arithmetic/part1) <==
Find the least number using 0-9 exactly once that is evenly divisible by each
==> digits/equations/123456789 (arithmetic/part1) <==
In how many ways can "." be replaced with "+", "-", or "" (concatenate) in
==> digits/equations/1992 (arithmetic/part1) <==
1 = -1+9-9+2. Extend this list to 2 through 100 on the left side of
==> digits/equations/24 (arithmetic/part1) <==
Form an expression that evaluates to 24 that contains two 3's, two 7's,
==> digits/equations/383 (arithmetic/part1) <==
Make 383 out of 1,2,25,50,75,100 using +,-,*,/.
==> digits/equations/find (arithmetic/part1) <==
Write a program for finding expressions built out of given numbers and using
==> digits/extreme.products (arithmetic/part1) <==
What are the extremal products of three three-digit numbers using digits 1-9?
==> digits/labels (arithmetic/part1) <==
You have an arbitrary number of model kits (which you assemble for
==> digits/least.significant/factorial (arithmetic/part1) <==
What is the least significant non-zero digit in the decimal expansion of n!?
==> digits/least.significant/tower.of.power (arithmetic/part1) <==
What are the least significant digits of 9^(8^(7^(6^(5^(4^(3^(2^1))))))) ?
==> digits/most.significant/googol (arithmetic/part1) <==
What digits does googol! start with?
==> digits/most.significant/powers (arithmetic/part1) <==
What is the probability that 2^N begins with the digits 603245?
==> digits/nine.digits (arithmetic/part1) <==
Form a number using 0-9 once with its first n digits divisible by n.
==> digits/palindrome (arithmetic/part1) <==
Does the series formed by adding a number to its reversal always end in
==> digits/palintiples (arithmetic/part1) <==
Find all numbers that are multiples of their reversals.
==> digits/power.two (arithmetic/part1) <==
Prove that for any 9-digit number (base 10) there is an integral power
==> digits/prime/101 (arithmetic/part1) <==
How many primes are in the sequence 101, 10101, 1010101, ...?
==> digits/prime/all.prefix (arithmetic/part1) <==
What is the longest prime whose every proper prefix is a prime?
==> digits/prime/change.one (arithmetic/part1) <==
What is the smallest number that cannot be made prime by changing a single
==> digits/prime/prefix.one (arithmetic/part1) <==
2 is prime, but 12, 22, ..., 92 are not. Similarly, 5 is prime
==> digits/reverse (arithmetic/part1) <==
Is there an integer that has its digits reversed after dividing it by 2?
==> digits/rotate (arithmetic/part1) <==
Find integers where multiplying them by single digits rotates their digits
==> digits/sesqui (arithmetic/part1) <==
Find the least number where moving the first digit to the end multiplies by 1.5.
==> digits/squares/change.leading (arithmetic/part1) <==
What squares remain squares when their leading digits are incremented?
==> digits/squares/length.22 (arithmetic/part1) <==
Is it possible to form two numbers A and B from 22 digits such that
==> digits/squares/length.9 (arithmetic/part1) <==
Is it possible to make a number and its square, using the digits from 1
==> digits/squares/three.digits (arithmetic/part2) <==
What squares consist entirely of three digits (e.g., 1, 4, and 9)?
==> digits/squares/twin (arithmetic/part2) <==
Let a twin be a number formed by writing the same number twice,
==> digits/sum.of.digits (arithmetic/part2) <==
Find sod ( sod ( sod (4444 ^ 4444 ) ) ).
==> digits/zeros/million (arithmetic/part2) <==
How many zeros occur in the numbers from 1 to 1,000,000?
==> digits/zeros/trailing (arithmetic/part2) <==
How many trailing zeros are in the decimal expansion of n!?
==> magic.squares (arithmetic/part2) <==
Are there large squares, containing only consecutive integers, all of whose
==> pell (arithmetic/part2) <==
Find integer solutions to x^2 - 92y^2 = 1.
==> subset (arithmetic/part2) <==
Prove that all sets of n integers contain a subset whose sum is divisible by n.
==> sum.of.cubes (arithmetic/part2) <==
Find two fractions whose cubes total 6.
==> sums.of.powers (arithmetic/part2) <==
Partition 1,2,3,...,16 into two equal sets, such that the sums of the
==> tests.for.divisibility/eleven (arithmetic/part2) <==
What is the test to see if a number is divisible by eleven?
==> tests.for.divisibility/nine (arithmetic/part2) <==
What is the test to see if a number is divisible by nine?
==> tests.for.divisibility/seven (arithmetic/part2) <==
What is the test to see if a number is divisible by seven?
==> tests.for.divisibility/three (arithmetic/part2) <==
What is the test to see if a number is divisible by three?
==> alphabet.blocks (combinatorics) <==
What is the minimum number of dice painted with one letter on all six sides
==> coinage/combinations (combinatorics) <==
Assuming you have enough coins of 1, 5, 10, 25 and 50 cents, how many
==> coinage/dimes (combinatorics) <==
"Dad wants one-cent, two-cent, three-cent, five-cent, and ten-cent
==> coinage/impossible (combinatorics) <==
What is the smallest number of coins that you can't make a dollar with?
==> color (combinatorics) <==
An urn contains n balls of different colors. Randomly select a pair, repaint
==> full (combinatorics) <==
Consider a string that contains all substrings of length n. For example,
==> gossip (combinatorics) <==
n people each know a different piece of gossip. They can telephone each other
==> grid.dissection (combinatorics) <==
How many (possibly overlapping) squares are in an mxn grid? Assume that all
==> permutation (combinatorics) <==
Compute the nth permutation of k numbers (or objects).
==> subsets (combinatorics) <==
Out of the set of integers 1,...,100 you are given ten different
==> transitions (combinatorics) <==
How many n-bit binary strings (0/1) have exactly k transitions
==> contests/games.magazine (competition/part1) <==
What are the best answers to various contests run by _Games_ magazine?
==> contests/national.puzzle/npc.1993 (competition/part1) <==
What are the solutions to the Games magazine 1993 National Puzzle Contest?
==> games/bridge (competition/part1) <==
Are there any programs for solving double-dummy Bridge?
==> games/chess/knight.control (competition/part1) <==
How many knights does it take to attack or control the board?
==> games/chess/knight.most (competition/part1) <==
What is the maximum number of knights that can be put on n x n chessboard
==> games/chess/knight.tour (competition/part1) <==
For what size boards are knight tours possible?
==> games/chess/mutual.stalemate (competition/part1) <==
What's the minimal number of pieces in a legal mutual stalemate?
==> games/chess/queen.control (competition/part1) <==
How many queens does it take to attack or control the board?
==> games/chess/queen.most (competition/part1) <==
How many non-mutually-attacking queens can be placed on various sized boards?
==> games/chess/queens (competition/part1) <==
How many ways can eight queens be placed so that they control the board?
==> games/chess/rook.paths (competition/part1) <==
How many non-overlapping paths can a rook take from one corner to the opposite
==> games/chess/size.of.game.tree (competition/part1) <==
How many different positions are there in the game tree of chess?
==> games/cigarettes (competition/part1) <==
The game of cigarettes is played as follows:
==> games/connect.four (competition/part1) <==
Is there a winning strategy for Connect Four?
==> games/craps (competition/part1) <==
What are the odds in craps?
==> games/crosswords (competition/part1) <==
Are there programs to make crosswords? What are the rules for cluing cryptic
==> games/cube (competition/part2) <==
What are some games involving cubes?
==> games/go-moku (competition/part2) <==
For a game of k in a row on an n x n board, for what values of k and n is
==> games/hi-q (competition/part2) <==
What is the quickest solution of the game Hi-Q (also called Solitaire)?
==> games/jeopardy (competition/part2) <==
What are the highest, lowest, and most different scores contestants
==> games/nim (competition/part2) <==
Place 10 piles of 10 $1 bills in a row. A valid move is to reduce
==> games/online/online.scrabble (competition/part2) <==
How can I play Scrabble online on the Internet?
==> games/online/unlimited.adventures (competition/part2) <==
Where can I find information about unlimited adventures?
==> games/othello (competition/part2) <==
How good are computers at Othello?
==> games/pc/best (competition/part2) <==
What are the best PC games?
==> games/pc/reviews (competition/part2) <==
Are reviews of PC games available online?
==> games/pc/solutions (competition/part2) <==
What are the solutions to various popular PC games?
==> games/poker.face.up (competition/part2) <==
In Face-Up Poker, two players each select five cards from a face-up deck,
==> games/risk (competition/part2) <==
What are the odds when tossing dice in Risk?
==> games/rubiks/rubiks.clock (competition/part2) <==
How do you quickly solve Rubik's clock?
==> games/rubiks/rubiks.cube (competition/part3) <==
What is known about bounds on solving Rubik's cube?
==> games/rubiks/rubiks.magic (competition/part3) <==
How do you solve Rubik's Magic?
==> games/scrabble (competition/part3) <==
What are some exceptional Scrabble Brand Crossword Game (TM) games?
==> games/set (competition/part3) <==
What is the size of the largest collection of cards from which NO "set"
==> games/soma (competition/part3) <==
What is the solution to Soma Cubes?
==> games/square-1 (competition/part3) <==
Does anyone have any hints on how to solve the Square-1 puzzle?
==> games/think.and.jump (competition/part3) <==
THINK & JUMP: FIRST THINK, THEN JUMP UNTIL YOU
==> games/tictactoe (competition/part3) <==
In random tic-tac-toe, what is the probability that the first mover wins?
==> tests/analogies/long (competition/part3) <==
1. Host : Guest :: Cynophobia : ?
==> tests/analogies/pomfrit (competition/part3) <==
1. NATURAL: ARTIFICIAL :: ANKYLOSIS: ?
==> tests/analogies/quest (competition/part3) <==
1. Mother: Maternal :: Stepmother: ?
==> tests/math/putnam/putnam.1967 (competition/part3) <==
==> tests/math/putnam/putnam.1987 (competition/part4) <==
WILLIAM LOWELL PUTNAM MATHEMATICAL COMPETITION
==> tests/math/putnam/putnam.1988 (competition/part4) <==
Problem A-1: Let R be the region consisting of the points (x,y) of the
==> tests/math/putnam/putnam.1990 (competition/part4) <==
Problem A-1
==> tests/math/putnam/putnam.1992 (competition/part5) <==
Problem A1
==> Beale (cryptology) <==
What are the Beale ciphers?
==> Feynman (cryptology) <==
What are the Feynman ciphers?
==> Voynich (cryptology) <==
What are the Voynich ciphers?
==> swiss.colony (cryptology) <==
What are the 1987 Swiss Colony ciphers?
==> vcrplus (cryptology) <==
What is the code used by VCR+?
==> allais (decision) <==
The Allais Paradox involves the choice between two alternatives:
==> division (decision) <==
N-Person Fair Division
==> dowry (decision) <==
Sultan's Dowry
==> envelope (decision) <==
Someone has prepared two envelopes containing money. One contains twice as
==> exchange (decision) <==
At one time, the Canadian and US dollars were discounted by 10 cents on
==> high.or.low (decision) <==
I pick two numbers, randomly, and tell you one of them. You are supposed
==> monty.hall (decision) <==
You are a participant on "Let's Make a Deal." Monty Hall shows you
==> newcomb (decision) <==
Newcomb's Problem
==> prisoners (decision) <==
Three prisoners on death row are told that one of them has been chosen
==> red (decision) <==
I show you a shuffled deck of standard playing cards, one card at a
==> rotating.table (decision) <==
Four glasses are placed upside down in the four corners of a square
==> stpetersburg (decision) <==
What should you be willing to pay to play a game in which the payoff is
==> truel (decision) <==
A, B, and C are to fight a three-cornered pistol duel. All know that
==> K3,3 (geometry/part1) <==
Can three houses be connected to three utilities without the pipes crossing?
==> bear (geometry/part1) <==
If a hunter goes out his front door, goes 50 miles south, then goes 50
==> bisector (geometry/part1) <==
Prove if two angle bisectors of a triangle are equal, then the triangle is
==> calendar (geometry/part1) <==
Build a calendar from two sets of cubes. On the first set, spell the
==> circles.and.triangles (geometry/part1) <==
Find the radius of the inscribed and circumscribed circles for a triangle.
==> coloring/cheese.cube (geometry/part1) <==
A cube of cheese is divided into 27 subcubes. A mouse starts at one
==> coloring/triominoes (geometry/part1) <==
There is a chess board (of course with 64 squares). You are given 21
==> construction/4.triangles.6.lines (geometry/part1) <==
Can you construct 4 equilateral triangles with 6 toothpicks?
==> construction/5.lines.with.4.points (geometry/part1) <==
Arrange 10 points so that they form 5 rows of 4 each.
==> construction/square.with.compass (geometry/part1) <==
Construct a square with only a compass and a straight edge.
==> corner (geometry/part1) <==
A hallway of width A turns through 90 degrees into a hallway of width
==> cover.earth (geometry/part1) <==
A thin membrane covers the surface of the (spherical) earth. One
==> cycle.polynomial (geometry/part1) <==
What are the cycle polynomials for the Platonic solids?
==> dissections/disk (geometry/part1) <==
Can a disk be cut into similar pieces without point symmetry about the
==> dissections/hexagon (geometry/part1) <==
Divide the hexagon into:
==> dissections/largest.circle (geometry/part1) <==
What is the largest circle that can be assembled from two semicircles cut from
==> dissections/square.70 (geometry/part1) <==
Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 square be dissected into
==> dissections/square.five (geometry/part1) <==
Can you dissect a square into 5 parts of equal area with just a straight edge?
==> dissections/tesseract (geometry/part1) <==
If you suspend a cube by one corner and slice it in half with a
==> duck.and.fox (geometry/part1) <==
A duck is swimming about in a circular pond. A ravenous fox (who cannot
==> earth.band (geometry/part1) <==
How much will a band around the equator rise above the surface if it is
==> fence (geometry/part1) <==
A farmer wishes to enclose the maximum possible area with 100 meters of fence.
==> ham.sandwich (geometry/part1) <==
Consider a ham sandwich, consisting of two pieces of bread and one of
==> hike (geometry/part1) <==
You are hiking in a half-planar woods, exactly 1 mile from the edge,
==> hole.in.sphere (geometry/part1) <==
Old Boniface he took his cheer,
==> hypercube (geometry/part1) <==
How many vertices, edges, faces, etc. does a hypercube have?
==> kissing.number (geometry/part1) <==
How many n-dimensional unit spheres can be packed around one unit sphere?
==> konigsberg (geometry/part1) <==
Can you draw a line through each edge on the diagram below without crossing
==> ladders (geometry/part1) <==
Two ladders form a rough X in an alley. The ladders are 11 and 13 meters
==> lattice/area (geometry/part1) <==
Prove that the area of a triangle formed by three lattice points is integer/2.
==> lattice/equilateral (geometry/part1) <==
Can an equlateral triangle have vertices at integer lattice points?
==> manhole.cover (geometry/part1) <==
Why is a manhole cover round?
==> pentomino (geometry/part1) <==
Arrange pentominos in 3x20, 4x15, 5x12, 6x10, 2x3x10, 2x5x6 and 3x4x5 forms.
==> points.in.sphere (geometry/part1) <==
What is the expected distance between two random points inside a sphere?
==> points.on.sphere (geometry/part1) <==
What are the odds that n random points on a sphere lie in the same hemisphere?
==> revolutions (geometry/part1) <==
A circle with radius 1 rolls without slipping once around a circle with radius
==> rotation (geometry/part1) <==
What is the smallest rotation that returns an object to its original state?
==> shephard.piano (geometry/part1) <==
What's the maximum area shape that will fit around a right-angle corner?
==> smuggler (geometry/part1) <==
Somewhere on the high sees smuggler S is attempting, without much
==> spiral (geometry/part1) <==
How far must one travel to reach the North Pole if one starts from the
==> table.in.corner (geometry/part1) <==
Put a round table into a (perpendicular) corner so that the table top
==> tetrahedron (geometry/part1) <==
Suppose you have a sphere of radius R and you have four planes that are
==> tiling/count.1x2 (geometry/part1) <==
Count the ways to tile an MxN rectangle with 1x2 dominos.
==> tiling/rational.sides (geometry/part1) <==
A rectangular region R is divided into rectangular areas. Show that if
==> tiling/rectangles.with.squares (geometry/part2) <==
Given two sorts of squares, (axa) and (bxb), what rectangles can be tiled?
==> tiling/scaling (geometry/part2) <==
A given rectangle can be entirely covered (i.e. concealed) by an
==> tiling/seven.cubes (geometry/part2) <==
Consider 7 cubes of equal size arranged as follows. Place 5 cubes so
==> topology/fixed.point (geometry/part2) <==
A man hikes up a mountain, and starts hiking at 2:00 in the afternoon
==> touching.blocks (geometry/part2) <==
Can six 1x2x4 blocks be arranged so that each block touches n others, for all n?
==> trigonometry/euclidean.numbers (geometry/part2) <==
For what numbers x is sin(x) expressible using only integers, +, -, *, / and
==> trigonometry/inequality (geometry/part2) <==
Show that (sin x)^(sin x) < (cos x)^(cos x) when 0 < x < pi/4.
==> group.01 (group) <==
AEFHIKLMNTVWXYZ BCDGJOPQRSU
==> group.01a (group) <==
147 0235689
==> group.02 (group) <==
ABEHIKMNOTXZ CDFGJLPQRSUVWY
==> group.03 (group) <==
BEJQXYZ DFGHLPRU KSTV CO AIW MN
==> group.04 (group) <==
BDO P ACGIJLMNQRSUVWZ EFTY HKX
==> group.05 (group) <==
CEFGHIJKLMNSTUVWXYZ ADOPQR B
==> group.06 (group) <==
BCEGKMQSW DFHIJLNOPRTUVXYZ
==> group.07 (group) <==
CDEFLOPTZ ABGHIJKMNQRSUVWXY
==> group.08 (group) <==
COS ABDEFGHIJKLMNPQRTUVWXYZ
==> group.09 (group) <==
CDILMVX ABEFGHJKNOPQRSTUWYZ
==> group.10 (group) <==
AHIMOTUVWXY BCDEFGJKLNPQRSZ
==> group.11 (group) <==
BCDIJLMNOPQRSUVWZ AEFGHKTXY
==> group.12 (group) <==
COPSUVWXZ ABDEFGHIJKLMNQRTY
==> group.13 (group) <==
BCDEHIKOX AFGJLMNPQRSTUVWYZ
==> group.14 (group) <==
EHIS MOT ABCDFGJKLNPQRUVWXYZ
==> group.15 (group) <==
HIOX ABCDEFGJKLMNPQRSTUVWYZ
==> group.16 (group) <==
IJ ABCDEFGHKLMNOPQRSTUVWXYZ
==> group.17 (group) <==
J BDFHIKLT GPQY ACEMNORSUVWXZ
==> handshake (induction) <==
A married couple organizes a party. They only invite other married
==> hanoi (induction) <==
Is there an algorithm for solving the Hanoi tower puzzle for any number
==> n-sphere (induction) <==
With what odds do three random points on an n-sphere form an acute triangle?
==> paradox (induction) <==
Is there a non-trivial property that holds for the first 10,000 positive
==> party (induction) <==
You're at a party. Any two (different) people at the party have exactly one
==> roll (induction) <==
An ordinary die is thrown until the running total of the throws first
==> takeover (induction) <==
After graduating from college, you have taken an important managing position
==> close.antonyms (language/part1) <==
What words are similar to their antonyms in other langauges?
==> dutch/dutch.record (language/part1) <==
What are some Dutch words with unusual properties?
==> english/equations (language/part1) <==
1 = E. on a C.
==> english/etymology/acronym (language/part1) <==
What acronyms have become common words or are otherwise interesting?
==> english/etymology/fossil (language/part1) <==
What are some examples of idioms that include obsolete words?
==> english/etymology/portmanteau (language/part1) <==
What are some words formed by combining together parts of other words?
==> english/frequency (language/part1) <==
In the English language, what are the most frequently appearing:
==> english/idioms (language/part1) <==
List some idioms that say the opposite of what they mean.
==> english/less.ness (language/part1) <==
Find a word that forms two other words, unrelated in meaning, when "less"
==> english/letter.rebus (language/part1) <==
Define the letters of the alphabet using self-referential common phrases (e.g.,
==> english/malaprop (language/part1) <==
List some phrases with the same meaning that differ by one sound.
==> english/piglatin (language/part1) <==
What words in pig latin also are words?
==> english/pleonasm (language/part1) <==
What are some redundant terms that occur frequently (like "ABM missile")?
==> english/plurals/collision (language/part1) <==
Two words, spelled and pronounced differently, have plurals spelled
==> english/plurals/doubtful.number (language/part1) <==
A little word of doubtful number,
==> english/plurals/drop.terminal (language/part1) <==
What words have their plurals formed by dropping the final letter?
==> english/plurals/endings (language/part1) <==
List a plural ending with each letter of the alphabet.
==> english/plurals/man (language/part1) <==
Words ending with "man" make their plurals by adding "s".
==> english/plurals/switch.first (language/part1) <==
What plural is formed by switching the first two letters?
==> english/potable.color (language/part1) <==
Find words that are both beverages and colors.
==> english/pronunciation/autonym (language/part1) <==
What is the longest word whose phonetic and normal spellings are the same?
==> english/pronunciation/homograph/different.pronunciation (language/part1) <==
What sequence of letters has the most different pronunciations?
==> english/pronunciation/homograph/homographs (language/part1) <==
List some homographs (words spelled the same but pronounced differently)
==> english/pronunciation/homophone/homophones.alphabet (language/part1) <==
Homophones can be confusing when used to exemplify a letter. For example,
==> english/pronunciation/homophone/homophones.letter (language/part1) <==
For each letter, list homophones that differ by that letter.
==> english/pronunciation/homophone/homophones.most (language/part1) <==
What words have four or more spellings that sound alike?
==> english/pronunciation/homophone/trivial (language/part2) <==
Consider the free non-abelian group on the twenty-six letters of the
==> english/pronunciation/oronym (language/part2) <==
List some oronyms (phrases or sentences that can be read in two ways
==> english/pronunciation/phonetic.letters (language/part2) <==
What does "FUNEX" mean?
==> english/pronunciation/rhyme (language/part2) <==
What English words are hard to rhyme?
==> english/pronunciation/silent.letter (language/part2) <==
For each letter, what word contains that letter silent?
==> english/pronunciation/silent.most (language/part2) <==
What word has the most silent letters in a row?
==> english/pronunciation/syllable (language/part2) <==
What words have an exceptional number of letters per syllable?
==> english/pronunciation/telegrams (language/part2) <==
Since telegrams cost by the word, phonetically similar messages can be cheaper.
==> english/puns (language/part2) <==
Where can I find a collection of puns?
==> english/rare.trigraphs (language/part2) <==
What trigraphs (three-letter combinations) occur in only one word?
==> english/self.ref/self.ref.letters (language/part2) <==
Construct a true sentence of the form: "This sentence contains _ a's, _ b's,
==> english/self.ref/self.ref.numbers (language/part2) <==
What true sentence has the form: "There are _ 0's, _ 1's, _ 2's, ...,
==> english/self.ref/self.ref.words (language/part2) <==
What sentence describes its own word, syllable and letter count?
==> english/sentences/behead (language/part2) <==
Is there a sentence that remains a sentence when all its words are beheaded?
==> english/sentences/charades (language/part2) <==
A ....... surgeon was ....... to operate because he had .......
==> english/sentences/emphasis (language/part2) <==
List some sentences that change meaning when the emphasis is moved.
==> english/sentences/pangram (language/part2) <==
A "pangram" is a sentence containing all 26 letters.
==> english/sentences/repeated.words (language/part2) <==
What is a sentence with the same word several times repeated? Do not use
==> english/sentences/sentence (language/part2) <==
Find a sentence with words beginning with the letters of the alphabet, in order.
==> english/sentences/snowball (language/part2) <==
Construct the longest coherent sentence you can such that the nth word
==> english/sentences/weird (language/part2) <==
Make a sentence containing only words that violate the "i before e" rule.
==> english/sentences/word.boundaries (language/part2) <==
List some sentences that can be radically altered by changing word boundaries
==> english/spelling/gry (language/part2) <==
Find three completely different words ending in "gry."
==> english/spelling/j.ending (language/part2) <==
What words and names end in j?
==> english/spelling/lipograms (language/part2) <==
What books have been written without specific letters, vowels, etc.?
==> english/spelling/longest (language/part2) <==
What is the longest word in the English language?
==> english/spelling/most (language/part2) <==
What word has the most variant spellings?
==> english/spelling/near.palindrome (language/part2) <==
What are some long near palindromes, i.e., words that except for one
==> english/spelling/operations.on.words/deletion (language/part2) <==
What exceptional words turn into other words by deletion of letters?
==> english/spelling/operations.on.words/insertion.and.deletion (language/part2) <==
What exceptional words turn into other words by both insertion and
==> english/spelling/operations.on.words/insertion (language/part2) <==
What exceptional words turn into other words by insertion of letters?
==> english/spelling/operations.on.words/movement (language/part2) <==
What exceptional words turn into other words by movement of letters?
==> english/spelling/operations.on.words/substitution (language/part2) <==
What exceptional words turn into other words by substitution of letters?
==> english/spelling/operations.on.words/transposition (language/part3) <==
What exceptional words turn into other words by transposition of letters?
==> english/spelling/operations.on.words/words.within.words (language/part3) <==
What exceptional words contain other words?
==> english/spelling/palindromes (language/part3) <==
What are some long palindromes?
==> english/spelling/sets.of.words/ladder (language/part3) <==
Find the shortest word ladders stretching between the following pairs:
==> english/spelling/sets.of.words/nots.and.crosses (language/part3) <==
What is the most number of letters that can be fit into a three by three grid
==> english/spelling/sets.of.words/perfect.ladder (language/part3) <==
A "perfect" ladder comprises five-letter words where every letter is
==> english/spelling/sets.of.words/squares (language/part3) <==
What are some exceptional word squares (square crosswords with no blanks)?
==> english/spelling/sets.of.words/variogram (language/part3) <==
What is the largest known variogram (word square where repeated letters count
==> english/spelling/sets.of.words/word.torture (language/part3) <==
What is the longest word all of whose contiguous subsequences are words?
==> english/spelling/single.words (language/part3) <==
What words have exceptional lengths, patterns, etc.?
==> english/spoonerisms (language/part3) <==
List some exceptional spoonerisms.
==> english/synonyms/ambiguous (language/part3) <==
What word in the English language is the most ambiguous?
==> english/synonyms/antonym (language/part3) <==
What words, when a single letter is added, reverse their meanings?
==> english/synonyms/contradictory.proverbs (language/part3) <==
What are some proverbs that contradict one another?
==> english/synonyms/contranym (language/part3) <==
What words are their own antonym?
==> english/synonyms/double.synonyms (language/part3) <==
What words have two different synonymous meanings?
==> finnish/finnish.plural (language/part3) <==
What Finnish word is the anagram of its plural?
==> finnish/finnish.record (language/part4) <==
What are some Finnish words with unusual properties?
==> french/french.palindromes (language/part4) <==
List some French palindromes.
==> french/french.record (language/part4) <==
What are some French words with unusual properties?
==> german/german.palindromes (language/part4) <==
List some German palindromes.
==> german/german.record (language/part4) <==
What are some German words with unusual properties?
==> italian/italian.record (language/part5) <==
What are some Italian words with unusual properties?
==> multi.palindromes (language/part5) <==
List some multi-lingual palindromes.
==> norwegian/norwegian.record (language/part5) <==
What are some Norwegian words with unusual properties?
==> repeated.word (language/part5) <==
In any language, construct a sentence by repeating one word four times.
==> swedish/swedish.record (language/part5) <==
What are some Swedish words with unusual properties?
==> synonymous.reversals (language/part5) <==
What words are synonymous with their reversals in other langauges?
==> vowels.repeated (language/part5) <==
In any language, what word contains the same vowel repeated four times in a row?
==> 29 (logic/part1) <==
Three people check into a hotel. They pay $30 to the manager and go
==> ages (logic/part1) <==
1) Ten years from now Tim will be twice as old as Jane was when Mary was
==> attribute (logic/part1) <==
All the items in the first list share a particular attribute. The second
==> bookworm (logic/part1) <==
A bookworm eats from the first page of an encyclopedia to the last
==> boxes (logic/part1) <==
Which Box Contains the Gold?
==> camel (logic/part1) <==
An Arab sheikh tells his two sons to race their camels to a distant
==> centrifuge (logic/part1) <==
You are a biochemist, working with a 12-slot centrifuge. This is a gadget
==> chain (logic/part1) <==
What is the least number of links you can cut in a chain of 21 links to be able
==> children (logic/part1) <==
A man walks into a bar, orders a drink, and starts chatting with the
==> condoms (logic/part1) <==
How can a man have mutually safe sex with three women with only two condoms?
==> dell (logic/part1) <==
How can I solve logic puzzles (e.g., as published by Dell) automatically?
==> elimination (logic/part1) <==
97 baseball teams participate in an annual state tournament.
==> flip (logic/part1) <==
How can a toss be called over the phone (without requiring trust)?
==> flowers (logic/part1) <==
How many flowers do I have if all of them are roses except two, all of
==> friends (logic/part1) <==
Any group of 6 or more contains either 3 mutual friends or 3 mutual strangers.
==> hofstadter (logic/part1) <==
In first-order logic, find a predicate P(x) which means "x is a power of 10."
==> hundred (logic/part1) <==
A sheet of paper has statements numbered from 1 to 100. Statement n says
==> inverter (logic/part1) <==
Can a digital logic circuit with two inverters invert N independent inputs?
==> josephine (logic/part1) <==
The recent expedition to the lost city of Atlantis discovered scrolls
==> locks.and.boxes (logic/part1) <==
You want to send a valuable object to a friend. You have a box which
==> min.max (logic/part1) <==
In a rectangular array of people, which will be taller, the tallest of the
==> mixing (logic/part1) <==
Start with a half cup of tea and a half cup of coffee. Take one tablespoon
==> monty.52 (logic/part1) <==
Monty and Waldo play a game with N closed boxes. Monty hides a
==> number (logic/part1) <==
Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce
==> riddle (logic/part2) <==
Who makes it, has no need of it. Who buys it, has no use for it. Who
==> river.crossing (logic/part2) <==
Three humans, one big monkey and two small monkeys are to cross a river:
==> ropes (logic/part2) <==
Two fifty foot ropes are suspended from a forty foot ceiling, about
==> same.street (logic/part2) <==
Sally and Sue have a strong desire to date Sam. They all live on the
==> self.ref (logic/part2) <==
Find a number ABCDEFGHIJ such that A is the count of how many 0's are in the
==> situation.puzzles (logic/part3) <==
Jed's List of Situation Puzzles
==> smullyan/black.hat (logic/part4) <==
Three logicians, A, B, and C, are wearing hats, which they know are either
==> smullyan/fork.three.men (logic/part4) <==
Three men stand at a fork in the road. One fork leads to Someplaceorother;
==> smullyan/fork.two.men (logic/part4) <==
Two men stand at a fork in the road. One fork leads to Someplaceorother; the
==> smullyan/integers (logic/part4) <==
Two logicians place cards on their foreheads so that what is written on the
==> smullyan/painted.heads (logic/part4) <==
While three logicians were sleeping under a tree, a malicious child painted
==> smullyan/priest (logic/part5) <==
In a small town there are N married couples in which one of the pair
==> smullyan/stamps (logic/part5) <==
The moderator takes a set of 8 stamps, 4 red and 4 green, known to the
==> supertasks (logic/part5) <==
You have an empty urn, and an infinite number of labeled balls. Each
==> timezone (logic/part5) <==
Two people are talking long distance on the phone; one is in an East-
==> unexpected (logic/part5) <==
Swedish civil defense authorities announced that a civil defense drill would
==> verger (logic/part5) <==
A very bright and sunny Day
==> weighing/balance (logic/part5) <==
You are given 12 identical-looking coins, one of which is counterfeit
==> weighing/box (logic/part5) <==
You have ten boxes; each contains nine balls. The balls in one box
==> weighing/find.median (logic/part5) <==
What is the least number of pairwise comparisons needed to find the median of
==> weighing/gummy.bears (logic/part5) <==
Real gummy drop bears have a mass of 10 grams, while imitation gummy
==> weighing/optimal.weights (logic/part5) <==
What is the smallest set of weights that allow you to weigh on a
==> weighing/weighings (logic/part5) <==
Some of the supervisors of Scandalvania's n mints are producing bogus coins.
==> zoo (logic/part5) <==
I took some nephews and nieces to the Zoo, and we halted at a cage marked
==> balloon (physics) <==
A helium-filled balloon is tied to the floor of a car that makes a
==> brick (physics) <==
What is the maximum overhang you can create with an infinite supply of bricks?
==> bubbles (physics) <==
In a universe with the same physical laws, but which is mostly water
==> cannonball (physics) <==
A person in a boat drops a cannonball overboard; does the water level change?
==> magnets (physics) <==
You have two bars of iron. One is magnetized along its length, the
==> milk.and.coffee (physics) <==
You are just served a hot cup of coffee and want it to be as hot as
==> mirror (physics) <==
Why does a mirror appear to invert the left-right directions, but not up-down?
==> monkey (physics) <==
Hanging over a pulley there is a rope, with a weight at one end.
==> pole.in.barn (physics) <==
Accelerate a pole of length l to a constant speed of 90% of the speed of
==> resistors (physics) <==
What is the resistance between various pairs of vertices on a lattice
==> sail (physics) <==
A sailor is in a sailboat on a river. The current is 3 knots with respect
==> shoot.sun (physics) <==
If you are standing at the equator at sunrise, where must you point a laser
==> skid (physics) <==
What is the fastest way to make a 90 degree turn on a slippery road?
==> spheres (physics) <==
Two spheres are the same size and weight, but one is hollow. They are
==> wind (physics) <==
Is a round-trip by airplane longer or shorter if there is wind blowing?
==> pickover.01 (pickover/part1) <==
Title: Cliff Puzzle 1: Can you beat the numbers game?
==> pickover.02 (pickover/part1) <==
Title: Cliff Puzzle 2: Grid of the Gods
==> pickover.03 (pickover/part1) <==
Title: Cliff Puzzle 3: Too many 3's
==> pickover.04 (pickover/part1) <==
Title: Cliff Puzzle 4: Time in a Bottle
==> pickover.05 (pickover/part1) <==
Title: Cliff Puzzle 5: Mystery Sequence
==> pickover.06 (pickover/part1) <==
Title: Cliff Puzzle 6: Star Chambers
==> pickover.07 (pickover/part2) <==
Title: Cliff Puzzle 7: 3x3 Recursion
==> pickover.08 (pickover/part2) <==
Title: Cliff Puzzle 8: Squares and Squares and Squares ....
==> pickover.09 (pickover/part2) <==
Title: Cliff Puzzle 9: 3-Atoms and Growth
==> pickover.10 (pickover/part2) <==
Title: Cliff Puzzle 10: The Ark Series
==> pickover.11 (pickover/part2) <==
Title: Cliff Puzzle 11: The Leviathan Number
==> pickover.12 (pickover/part3) <==
Title: Cliff Puzzle 12: Slides in Hell
==> pickover.13 (pickover/part3) <==
Title: Cliff Puzzle 13: Ladders to Heaven
==> pickover.14 (pickover/part3) <==
Title: Cliff Puzzle 14: Geography Genuflection
==> pickover.15 (pickover/part3) <==
Title: Cliff Puzzle 15: Cherries in Wine Glasses
==> pickover.16 (pickover/part3) <==
Title: Cliff Puzzle 16: Undulating Squares
==> pickover.17 (pickover/part3) <==
Title: Cliff Puzzle 17: Weird Recursive Sequence
==> pickover.18 (pickover/part3) <==
Title: Cliff Puzzle 18: Difficult Nested Roots
==> amoeba (probability) <==
A jar begins with one amoeba. Every minute, every amoeba
==> apriori (probability) <==
An urn contains one hundred white and black balls. You sample one hundred
==> bayes (probability) <==
One urn contains black marbles, and the other contains white or black
==> birthday/line (probability) <==
At a movie theater, the manager announces that they will give a free ticket
==> birthday/same.day (probability) <==
How many people must be at a party before you have even odds or better
==> cab (probability) <==
A cab was involved in a hit and run accident at night. Two cab companies,
==> coupon (probability) <==
There is a free gift in my breakfast cereal. The manufacturers say that
==> darts (probability) <==
Peter throws two darts at a dartboard, aiming for the center. The
==> derangement (probability) <==
12 men leave their hats with the hat check. If the hats are randomly
==> family (probability) <==
Suppose that it is equally likely for a pregnancy to deliver
==> flips/once.in.run (probability) <==
What are the odds that a run of one H or T (i.e., THT or HTH) will occur
==> flips/twice.in.run (probability) <==
What is the probability in n flips of a fair coin that there will be two
==> flips/unfair (probability) <==
Generate even odds from an unfair coin. For example, if you
==> flips/waiting.time (probability) <==
Compute the expected waiting time for a sequence of coin flips, or the
==> flush (probability) <==
Which set contains proportionately more flushes than the set of all
==> hospital (probability) <==
A town has two hospitals, one big and one small. Every day the big
==> icos (probability) <==
The "house" rolls two 20-sided dice and the "player" rolls one
==> intervals (probability) <==
Given two random points x and y on the interval 0..1, what is the average
==> killers.and.pacifists (probability) <==
You enter a town that has K killers and P pacifists. When a
==> leading.digit (probability) <==
What is the probability that the ratio of two random reals starts with a 1?
==> lights (probability) <==
Waldo and Basil are exactly m blocks west and n blocks north from
==> lottery (probability) <==
There n tickets in the lottery, k winners and m allowing you to pick another
==> oldest.girl (probability) <==
You meet a stranger on the street, and ask how many children he has. He
==> particle.in.box (probability) <==
A particle is bouncing randomly in a two-dimensional box. How far does it
==> pi (probability) <==
Are the digits of pi random (i.e., can you make money betting on them)?
==> random.walk (probability) <==
Waldo has lost his car keys! He's not using a very efficient search;
==> reactor (probability) <==
There is a reactor in which a reaction is to take place. This reaction
==> roulette (probability) <==
You are in a game of Russian roulette, but this time the gun (a 6
==> transitivity (probability) <==
Can you number dice so that die A beats die B beats die C beats die A?
==> icecubes (real-life) <==
You have an old-fashioned refrigerator with a small freezer compartment
==> microwave (real-life) <==
Every morning when I warm my milk for breakfast, I put one cup of milk
==> books/bloopers (references) <==
What are some errors made in puzzle books?
==> books/masquerade (references) <==
What is the solution to _Masquerade_ by Kit Williams?
==> books/maze (references) <==
What is the solution to _Maze_ by Christopher Manson?
==> books/treasure (references) <==
What is the solution to _Treasure_ by Dr. Crypton?
==> books/unnamed (references) <==
What is the solution to the unnamed book by Kit Williams?
==> faq (references) <==
Where should I look if I can't find the answer here?
==> magazines (references) <==
What magazines and journals contain puzzles?
==> organizations (references) <==
What organizations exist for puzzle lovers?
==> series.00 (series) <==
Are "complete this series" problems well defined?
==> series.01 (series) <==
M, N, B, D, P ?
==> series.02 (series) <==
H, H, L, B, B, C, N, O, F ?
==> series.03 (series) <==
W, A, J, M, M, A, J?
==> series.03a (series) <==
G, J, T, J, J, J, A, M, W, J, J, Z, M, F, J, ?
==> series.03b (series) <==
A, J, B, C, G, T, C, V, J, T, D, F, K, B, H, ?
==> series.03c (series) <==
M, A, M, D, E, L, R, H, ?
==> series.04 (series) <==
A, E, H, I, K, L, ?
==> series.05 (series) <==
A B C D E F G H?
==> series.06 (series) <==
Z, O, T, T, F, F, S, S, E, N?
==> series.06a (series) <==
F, S, T, F, F, S, ?
==> series.07 (series) <==
1, 1 1, 2 1, 1 2 1 1, ...
==> series.08a (series) <==
G, L, M, B, C, L, M, C, F, S, ?
==> series.08b (series) <==
A, V, R, R, C, C, L, L, L, E, ?
==> series.09a (series) <==
S, M, S, S, S, C, P, P, P, ?
==> series.09b (series) <==
M, S, C, P, P, P, S, S, S, ?
==> series.10 (series) <==
D, P, N, G, C, M, M, S, ?
==> series.11 (series) <==
R O Y G B ?
==> series.12 (series) <==
A, T, G, C, L, ?
==> series.13 (series) <==
M, V, E, M, J, S, ?
==> series.14 (series) <==
A, B, D, O, P, ?
==> series.14a (series) <==
A, B, D, E, G, O, P, ?
==> series.15 (series) <==
A, E, F, H, I, ?
==> series.16 (series) <==
A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, X, Y?
==> series.17 (series) <==
T, P, O, F, O, F, N, T, S, F, T, F, E, N, S, N?
==> series.18 (series) <==
10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, ___ , 100, 121, 10000
==> series.19 (series) <==
0 01 01011 0101101011011 0101101011011010110101101101011011 etc.
==> series.20 (series) <==
1 2 5 16 64 312 1812 12288
==> series.21 (series) <==
5, 6, 5, 6, 5, 5, 7, 5, ?
==> series.22 (series) <==
3 1 1 0 3 7 5 5 2 ?
==> series.23 (series) <==
22 22 30 13 13 16 16 28 28 11 ?
==> series.24 (series) <==
What is the next letter in the sequence: W, I, T, N, L, I, T?
==> series.25 (series) <==
1 3 4 9 10 12 13 27 28 30 31 36 37 39 40 ?
==> series.26 (series) <==
1 3 2 6 7 5 4 12 13 15 14 10 11 9 8 24 25 27 26 ?
==> series.27 (series) <==
0 1 1 2 1 2 1 3 2 2 1 3 1 2 2 4 1 3 1 3 2 2 1 4 2 ?
==> series.28 (series) <==
0 2 3 4 5 5 7 6 6 7 11 7 13 9 8 8 17 8 19 9 10 13 23 9 10 ?
==> series.29 (series) <==
1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 ?
==> series.30 (series) <==
I I T Y W I M W Y B M A D
==> series.31 (series) <==
6 2 5 5 4 5 6 3 7
==> series.32 (series) <==
0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1
==> series.33 (series) <==
2 12 360 75600
==> series.34 (series) <==
3 5 4 4 3 5 5 4 3
==> series.35 (series) <==
1 2 3 2 1 2 3 4 2 1 2 3 4 2 2 3
==> series.36 (series) <==
ETIANMSURWDKGO
==> series.37 (series) <==
10^3 10^9 10^27 10^2 0 4 8 3
==> area.codes (trivia) <==
When looking at a map of the distribution of telephone area codes for
==> body.parts (trivia) <==
Name ten body parts that are spelled with three letters. No slang words.
==> coincidence (trivia) <==
Name some amazing coincidences.
==> eskimo.snow (trivia) <==
How many words do the Eskimo have for snow?
==> federal.reserve (trivia) <==
What is the pattern to this list:
==> jokes.self-referential (trivia) <==
What are some self-referential jokes?
==> memory.tricks (trivia) <==
When asked to name a color, many people answer "red." What are some other
==> quotations (trivia) <==
Where can I find the source for a quotation?
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> arithmetic/digits/squares/three.digits.p <==
What squares consist entirely of three digits (e.g., 1, 4, and 9)?
==> arithmetic/digits/squares/three.digits.s <==
The full set of solutions up to 10**12 is
1 -> 1
2 -> 4
3 -> 9
7 -> 49
12 -> 144
21 -> 441
38 -> 1444
107 -> 11449
212 -> 44944
31488 -> 9914 94144
70107 -> 49149 91449
3 87288 -> 14 99919 94944
956 10729 -> 9 14141 14499 11441
4466 53271 -> 199 49914 44949 99441
31487 17107 -> 9914 41941 99144 49449
2 10810 79479 -> 4 44411 91199 99149 11441
If the algorithm is used in the form I presented it before, generating
the whole set P_n before starting on P_{n+1}, the store requirements
begin to become embarassing. For n>8 I switched to a depth-first
strategy, generating all the elements in P_i (i=9..12) congruent to
a particular x in P_8 for each x in turn. This means the solutions
don't come out in any particular order, of course. CPU time was 16.2
seconds (IBM 3084).
In article <1990Feb6.0...@sun.soe.clarkson.edu>, Steven
Stadnicki suggests alternate triples of digits, in particular {1,4,6}
(with many solutions) and {2,4,8} (with few). I ran my program on
these as well, up to 10**12 again:
1 -> 1
2 -> 4
4 -> 16
8 -> 64
12 -> 144
21 -> 441
38 -> 1444
108 -> 11664
119 -> 14161
121 -> 14641
129 -> 16641
204 -> 41616
408 -> 1 66464
804 -> 6 46416
2538 -> 64 41444
3408 -> 116 14464
6642 -> 441 16164
12908 -> 1666 16464
25771 -> 6641 44441
78196 -> 61146 14416
81619 -> 66616 61161
3 33858 -> 11 14611 64164
2040 00408 -> 41 61616 64641 66464
6681 64962 -> 446 44441 64444 61444
8131 18358 -> 661 16146 41166 16164
40182 85038 -> 16146 61464 66146 61444 (Steven's last soln.)
1 20068 50738 -> 1 44164 46464 46111 44644
1 26941 38988 -> 1 61141 16464 66616 64144
1 27069 43631 -> 1 61466 41644 14114 64161
4 01822 24262 -> 16 14611 14664 16614 44644
4 05784 63021 -> 16 46611 66114 66644 46441
78 51539 12392 -> 6164 66666 14446 44111 61664
and
2 -> 4
22 -> 484
168 -> 28224
478 -> 2 28484
2878 -> 82 82884 (Steven's last soln.)
2109 12978 -> 44 48428 42888 28484
(so the answer to Steven's "Are there any more at all?" is "Yes".)
The CPU times were 42.9 seconds for {1,4,6}, 18.7 for {2,4,8}. This
corresponds to an interesting point: the abundance of solutions for
{1,4,6} is associated with abnormally large sets P_n (|P_8| = 16088
for {1,4,6} compared to |P_8| = 5904 for {1,4,9}) but the deficiency
of solutions for {2,4,8} is *not* associated with small P_n's (|P_8|
= 6816 for {2,4,8}). Can anyone wave a hand convincingly to explain
why the solutions for {2,4,8} are so sparse?
I suspect we are now getting to the point where an improved algorithm
is called for. The time to determine all the n-digit solutions (i.e.
2n-digit squares) using this last-significant-digit-first is essentially
constant * 3**n. Dean Hickerson in <90036.1...@PSUVM.BITNET>, and
Ilan Vardi in <1990Feb5.2...@Neon.Stanford.EDU>, suggest using
a most-significant-digit-first strategy, based on the fact that the
first n digits of the square determine the (integral) square root; this
also has a running time constant * 3**n. Can one attack both ends at
once and do better?
Chris Thompson
JANET: ce...@uk.ac.cam.phx
Internet: cet1%phx.ca...@nsfnet-relay.ac.uk
Hey guys, what about
648070211589107021 ^ 2 = 419994999149149944149149944191494441
This was found by David Applegate and myself (about 5 minutes on a DEC 3100,
program in C).
This is the largest square less than 10^42 with the 149-property; checking
took a bit more than an hour of CPU time.
As somebody suggested, we used a combined most-significant/least-significant
digits attack. First we make a table of p-digit prefixes (most significant
p digits) that could begin a root whose square has the 149 property in its
first p digits. We organize this table into buckets by the least
significant q digits of the prefixes. Then we enumerate the s digit
suffixes whose squares have the 149 property in their last s digits. For
each such suffix, we look in the table for those prefixes whose last q
digits match the first q of the suffix. For each match, we consider the p +
s - q digit number formed by overlapping the prefix and the suffix by q
digits. The squares of these overlap numbers must contain all the squares
with the 149 property.
The time expended is O(3^p) to generate the prefix table, O(3^s) to
enumerate the suffixes, and O(3^(p+s) / 10^q) to check the overlaps (being
very rough and ignoring the polynomial factors) By judiciously chosing p, q,
and s, we can fix things so that each bucket of the table has around O(1)
entries: set q = p log10(3). Setting p = s, we end up looking for squares
whose roots have n = 2 - log10(3) digits, with an algorithm that takes time
O( 3 ^ [n / (2 - log10(3)]) ), roughly time O(3^[.66n]). Compared to the
O(3^n) performance of either single-ended algorithm, this lets us check 50%
more digits in the same amount of time (ignoring polynomial factors). Of
course, the space cost of the combined-ends method is high.
-- Guy and Dave
--
Guy Jacobson School of Computer Science
Carnegie Mellon arpanet : g...@cs.cmu.edu
Pittsburgh, PA 15213 csnet : Guy.Jacobson%a.cs.cmu.edu@csnet-relay
(412) 268-3056 uucp : ...!{seismo, ucbvax, harvard}!cs.cmu.edu!guy
Here is an algorithm which takes O(sqrt(n)log(n)) steps to find all perfect
squares < n whose only digits are 1, 4 and 9.
This doesn't sound too great *but* it doesn't use a lot of memory and only
requires addition and <. Also, the actual run time will depend on where the
first non-{1,4,9} digit appears in each square.
set n = 1
set odd = 1
while(n < MAXVAL)
{
if(all digits of n are in {1,4,9})
{
print n
}
add 2 to odd
add odd to n
}
This works because (X+1)^2 - x^2 = 2x+1.
That is, if you start with 0 and add successive odd
numbers to it you get 0+1=1, 1+3=4, 4+5=9, 9+7=16 etc.
I've started the algorithm at 1 for convenience.
The "O" value comes from looking at at most all digits
(log(n)) of all perfect squares < n (sqrt(n) of them)
at most a constant number of times.
I didn't save the articles with algorithms claiming to be
O(3^log(n)) so I don't know if their calculations needed
to (or did) account for multiplication or sqrt() of large
numbers. O(3^log(n)) sounds reasonable so I'm going to
assume they did unless I hear otherwise.
Any comments? Please email if you just want to refresh my memory
on the other algorithms.
Andrew Charles
ac...@ihuxy.ATT.COMM
==> arithmetic/digits/squares/twin.p <==
Let a twin be a number formed by writing the same number twice,
for instance, 81708170 or 132132. What is the smallest square twin?
==> arithmetic/digits/squares/twin.s <==
1322314049613223140496 = 36363636364 ^ 2.
The key to solving this puzzle is looking at the basic form of these
"twin" numbers, which is some number k = 1 + 10^n multiplied by some number
10^(n-1) <= a < 10^n. If ak is a perfect square, k must have some
repeated factor, since a<k. Searching the possible values of k for one
with a repeated factor eventually turns up the number 1 + 10^11 = 11^2
* 826446281. So, we set a=826446281 and ak = 9090909091^2 =
82644628100826446281, but this needs leading zeros to fit the pattern.
So, we multiply by a suitable small square (in this case 16) to get the
above answer.
==> arithmetic/digits/sum.of.digits.p <==
Find sod ( sod ( sod (4444 ^ 4444 ) ) ).
==> arithmetic/digits/sum.of.digits.s <==
let X = 4444^4444
sod(X) <= 9 * (# of digits) < 145900
sod(sod(X)) <= sod(99999) = 45
sod(sod(sod(X))) <= sod(39) = 12
but sod(sod(sod(X))) = 7 (mod 9)
thus sod(sod(sod(X))) = 7
==> arithmetic/digits/zeros/million.p <==
How many zeros occur in the numbers from 1 to 1,000,000?
==> arithmetic/digits/zeros/million.s <==
In the numbers from 10^(n-1) through 10^n - 1, there are 9 * 10^(n-1)
numbers of n digits each, so 9(n-1)10^(n-1) non-leading digits, of
which one tenth, or 9(n-1)10^(n-2), are zeroes. When we change the
range to 10^(n-1) + 1 through 10^n, we remove 10^(n-1) and put in
10^n, gaining one zero, so
p(n) = p(n-1) + 9(n-1)10^(n-2) + 1 with p(1)=1.
Solving the recurrence yields the closed form
p(n) = n(10^(n-1)+1) - (10^n-1)/9.
For n=6, there are 488,895 zeroes, 600,001 ones, and 600,000 of all other
digits.
==> arithmetic/digits/zeros/trailing.p <==
How many trailing zeros are in the decimal expansion of n!?
==> arithmetic/digits/zeros/trailing.s <==
The general answer to the question
"what power of p divides x!" where p is prime
is (x-d)/(p-1) where d is the sum of the digits of (x written in base p).
So where p=5, 10 is written as 20 and is divisible by 5^2 (2 = (10-2)/4);
x to base 10: 100 1000 10000 100000 1000000
x to base 5: 400 13000 310000 11200000 224000000
d : 4 4 4 4 8
trailing 0s in x! 24 249 2499 24999 249998
==> arithmetic/magic.squares.p <==
Are there large squares, containing only consecutive integers, all of whose
rows, columns and diagonals have the same sum? How about cubes?
==> arithmetic/magic.squares.s <==
These are called magic squares. A magic square of order n (integers
from 1 to n*n) has only one possible sum: (n*n+1)*n/2.
Odd and even order squares must be constructed by different approaches.
For odd orders, the most common algorithm is a recursive scheme
devised by de la Loubere about 300 years ago. For even orders, one
procedure is the Devedec algorithm, which treats even orders not
divisible by 4 slightly differently from those which are divisible by
4 (doubly even).
For squares with odd-length sides, the following algorithm builds a magic
square:
Put 1 in the middle box in the upper row. From then on, if it's
possible to put the next number one box diagonally up and to the right
(wrapping around if the edge of the grid is reached), do so, otherwise,
put it directly below the last one.
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
...or even
47 58 69 80 1 12 23 34 45
57 68 79 9 11 22 33 44 46
67 78 8 10 21 32 43 54 56
77 7 18 20 31 42 53 55 66
6 17 19 30 41 52 63 65 76
16 27 29 40 51 62 64 75 5
26 28 39 50 61 72 74 4 15
36 38 49 60 71 73 3 14 25
37 48 59 70 81 2 13 24 35
See archive entry knight.tour for magic squares that are knight's tours.
To get a 4x4 square, write the numbers in order across each row, filling
the square...
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
then use the following pattern as a mask:
. X X .
X . . X
X . . X
. X X .
Everywhere there is an X, complement the number (subtract it from
n*n+1). For the 4x4 you get:
1 15 14 4
12 6 7 9
8 10 11 5
13 3 2 16
For n even (n>4):
Make an initial magic square by writing an n/2 magic square four times
(the same one each time). Now, although this square adds up right we
have the numbers 1 to n*n/4 written four times each. To fix this,
simply add to it n*n/4 times one of the following magic squares:
if n/2 is odd (example: n/2=7),
3 3 3 0 0 0 0 2 2 2 2 2 1 1 (there are int(n/4) 3s, int(n/4-1) 1s on each
3 3 3 0 0 0 0 2 2 2 2 2 1 1 row)
3 3 3 0 0 0 0 2 2 2 2 2 1 1
0 3 3 3 0 0 0 2 2 2 2 2 1 1 (this is row int(n/4)+1. It starts with just
3 3 3 0 0 0 0 2 2 2 2 2 1 1 the one 0)
3 3 3 0 0 0 0 2 2 2 2 2 1 1
3 3 3 0 0 0 0 2 2 2 2 2 1 1
0 0 0 3 3 3 3 1 1 1 1 1 2 2 (the lower half is the same as the upper half
0 0 0 3 3 3 3 1 1 1 1 1 2 2 with 3<->0 and 1<->2 swapped. This guarantees
0 0 0 3 3 3 3 1 1 1 1 1 2 2 that each number 1-n*n will appear in the
3 0 0 0 3 3 3 1 1 1 1 1 2 2 completed square)
0 0 0 3 3 3 3 1 1 1 1 1 2 2
0 0 0 3 3 3 3 1 1 1 1 1 2 2
0 0 0 3 3 3 3 1 1 1 1 1 2 2
if n/2 is even (example: n/2=4),
0 0 3 3 2 2 1 1 (there are n/4 of each number on each row)
0 0 3 3 2 2 1 1
0 0 3 3 2 2 1 1
0 0 3 3 2 2 1 1
3 3 0 0 1 1 2 2
3 3 0 0 1 1 2 2
3 3 0 0 1 1 2 2
3 3 0 0 1 1 2 2
References:
"Magic Squares and Cubes"
W.S. Andrews
The Open Court Publishing Co.
Chicago, 1908
"Mathematical Recreations"
M. Kraitchik
Dover
New York, 1953
==> arithmetic/pell.p <==
Find integer solutions to x^2 - 92y^2 = 1.
==> arithmetic/pell.s <==
x=1 y=0
x=1151 y=120
x=2649601 y=276240
etc.
Each successive solution is about 2300 times the previous
solution; they are every 8th partial fraction (x=numerator,
y=denominator) of the continued fraction for sqrt(92) =
[9, 1,1,2,4,2,1,1,18, 1,1,2,4,2,1,1,18, 1,1,2,4,2,1,1,18, ...]
Once you have the smallest positive solution (x1,y1) you
don't need to "search" for the rest. You can obtain the nth positive
solution (xn,yn) by the formula
(x1 + y1 sqrt(92))^n = xn + yn sqrt(92).
See Niven & Zuckerman's An Introduction to the Theory of Numbers.
Look in the index under Pell's equation.
==> arithmetic/subset.p <==
Prove that all sets of n integers contain a subset whose sum is divisible by n.
==> arithmetic/subset.s <==
Consider the set of remainders of the partial sums a(1) + ... + a(i).
Since there are n such sums, either one has remainder zero (and we're
thru) or 2 coincide, say the i'th and j'th. In this case, a(i+1) +
... + a(j) is divisible by n. (note this is a stronger result since
the subsequence constructed is of *adjacent* terms.) Consider a(1)
(mod n), a(1)+a(2) (mod n), ..., a(1)+...+a(n) (mod n). Either at
some point we have a(1)+...+a(m) = 0 (mod n) or else by the pigeonhole
principle some value (mod n) will have been duplicated. We win either
way.
==> arithmetic/sum.of.cubes.p <==
Find two fractions whose cubes total 6.
==> arithmetic/sum.of.cubes.s <==
Restated:
Find X, Y, minimum Z (all positive integers) where
(X/Z)^3 + (Y/Z)^3 = 6
Again, a generalized solution would be nice.
You are asking for the smallest z s.t. x^3 + y^3 = 6*z^3 and x,y,z in Z+.
In general, questions like these are extremely difficult; if you're
interested take a look at books covering Diophantine equations
(especially Baker's work on effective methods of computing solutions).
Dudeney mentions this problem in connection with #20 in _The Canterbury
Puzzles_; the smallest answer is (17/21)^3 + (37/21)^3 = 6.
For the interest of the readers of this group I'll quote:
"Given a known case for the expression of a number as the sum or
difference of two cubes, we can, by formula, derive from it an infinite
number of other cases alternately positive and negative. Thus Fermat,
starting from the known case 1^3 + 2^3 = 9 (which we will call a
fundamental case), first obtained a negative solution in bigger
figures, and from this his positive solution in bigger figures still.
But there is an infinite number of fundamentals, and I found by trial
a negative fundamental solution in smaller figures than his derived
negative solution, from which I obtained the result shown above. That
is the simple explanation."
In the above paragraph Dudeney is explaining how he derived (*by hand*)
that (415280564497/348671682660)^3 + (676702467503/348671682660)^3 = 9.
He continues:
"We can say of any number up to 100 whether it is possible or not to
express it as the sum of two cubes, except 66. Students should read
the Introduction to Lucas's _Theorie des Nombres_, p. xxx."
"Some years ago I published a solution for the case 6 = (17/21)^3 +
(37/21)^3, of which Legendre gave at some length a 'proof' of
impossibility; but I have since found that Lucas anticipated me in
a communication to Sylvester."
==> arithmetic/sums.of.powers.p <==
Partition 1,2,3,...,16 into two equal sets, such that the sums of the
numbers in each set are equal, as are the sums of their squares and cubes.
==> arithmetic/sums.of.powers.s <==
The solution is A = {1,4,6,7,10,11,13,16}
B = {2,3,5,8,9,12,14,15}
Let X+k be a set formed by adding k to all elements of X.
Then A+k and B+k have the property of satisfying i,ii and iii.
That means, any 16 numbers in A.P can be partioned in such a way to
satisfy i,ii and iii.
How do we arrive at the above solution without using a computer?
By the preceding discussion,
A1 = A-1 = {0,3,5,6,9,10,12,15}
B1 = B-1 = {1,2,4,7,8,11,13,14}
have the property of satisfying i,ii and iii.
Notice that all numbers of A1 have even number of 1's in binary and
all numbers of B1 have odd number of 1's in binary. Essentially the
above partition is a partition based on parity.
Observation:
Partition of (n+1) bit numbers based on parity into two
groups A and B (each consisting of 2^n elements) satisfies
sum of kth powers of elements of A = sum of kth powers of elements of B
for k=1,2,...,n. (The above puzzle is a special case n=3).
The above observation also holds for any radix. In radix r we will have
r groups.
Infact,
Any r^(n+1) terms in A.P can be partitioned into r groups (each of
r^n elements) such that sum of kth powers of all r groups is same (k=1,2,...,n)
Finding such groups with minimal number of elements (less than r^n) appears to
be more difficult!
ALL THIS APPEARS TO BE INTERESTING. IS IT A CONSEQUENCE OF A SIMPLE THEOREM OF
NUMBER THEORY? HOW DO I PROVE THIS?
Thanks in advance for any clues,
-- ram...@svl.cdc.com (Mr. Ramana (Indian analyst))
==> arithmetic/tests.for.divisibility/eleven.p <==
What is the test to see if a number is divisible by eleven?
==> arithmetic/tests.for.divisibility/eleven.s <==
If the alternating sum of the digits is divisible by eleven, so is the number.
For example, 1639 leads to 9 - 3 + 6 - 1 = 11, so 1639 is divisible by 11.
Proof:
Every integer n can be expressed as
n = a1*(10^k) + a2*(10^k-1)+ .....+ a_k+1
where a1, a2, a3, ...a_k+1 are integers between 0 and 9.
10 is congruent to -1 mod(11).
Thus if (-1^k)*a1 + (-1^k-1)*a2 + ...+ (a_k+1) is congruent to 0mod(11) then
n is divisible by 11.
==> arithmetic/tests.for.divisibility/nine.p <==
What is the test to see if a number is divisible by nine?
==> arithmetic/tests.for.divisibility/nine.s <==
If the sum of the digits is divisible by nine, so is the number.
Proof:
Every integer n can be expressed as
n = a1*(10^k) + a2*(10^k-1)+ .....+ a_k+1
where a1, a2, a3, ...a_k+1 are integers between 0 and 9.
Note that 10 is congruent to 1 (mod 9). Thus 10^k is congruent to 1 (mod 9) for
every k >= 0.
Thus n is congruent to (a1+a2+a3+....+a_k+1) mod(9).
Hence (a1+a2+...+a_k+1) is divisible by 9 iff n is divisible by 9.
==> arithmetic/tests.for.divisibility/seven.p <==
What is the test to see if a number is divisible by seven?
==> arithmetic/tests.for.divisibility/seven.s <==
Take the last digit (n mod 10) and double it.
Take the rest of the digits (n div 10) and subtract the doubled last
digit from it.
The resulting number is divisible by 7 iff the original number
is divisible by 7.
Example: Take 53445
Subtract (53445 mod 10) * 2 from (53445 div 10)
- 5 * 2 + 5344
= 5334
533 - 2 * 4 = 525
52 - 5 * 2 = 42 which is divisible by 7
so 53445 is divisible by 7.
==> arithmetic/tests.for.divisibility/three.p <==
What is the test to see if a number is divisible by three?
==> arithmetic/tests.for.divisibility/three.s <==
A number is divisible by three iff the sum of its digits is divisible by three.
First, prove 10^N = 1 mod 3 for all integers N >= 0.
1 = 1 mod 3. 10 = 1 mod 3. 10^N = 10^(N-1) * 10 = 10^(N-1) mod 3.
QED by induction.
Now let D[0] be the units digit of N, D[1] the tens digit, etc.
Now N = Summation From k=0 to k=inf of D[k]*10^k.
Therefore N mod 3 = Summation from k=0 to k=inf of D[k] mod 3. QED
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> combinatorics/alphabet.blocks.p <==
What is the minimum number of dice painted with one letter on all six sides
such that all permutations without repetitions of n letters can be formed
by placing n dice together in a line?
==> combinatorics/alphabet.blocks.s <==
n= 2 3 4 5 6
(8,4) (9,7) (9,3) (10,7) (11,7)
aijklm abcde? acdefg abcde? abkmuz
bijklm fghij? bhijkl fghij? bcpwy?
cnopqr klmno? cmnopq klmno? cdlnvz
dnopqr pqrst? dhnrvy pqrst? deqxu?
estuvw uvwxy? eiosvw uvwxy? efmowz
fstuvw afkpu? fjptwx afkpu? fgryv?
gxyz?? bglqv? gkquxy bglqv? ghnkx?
hxyz?? chmrwz lmrstu chmrwz hisuw?
dinsxz zab??? dinsxz ijoly?
ejotyz jatvx?
pqrstz
I think I can prove that there is no solution with 11 dice with 9
don't cares or with 10 dice, but I haven't checked all the details, so
I might have made a mistake. In any case, that leaves open the case
of 11 dice with 8 don't cares; my guess is that it is not possible.
-- John Rickard (jric...@eoe.co.uk)
==> combinatorics/coinage/combinations.p <==
Assuming you have enough coins of 1, 5, 10, 25 and 50 cents, how many
ways are there to make change for a dollar?
==> combinatorics/coinage/combinations.s <==
292. The table is shown below:
Amount 00 05 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 100
Coins
.01 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
.05 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
.10 1 2 4 6 9 12 16 20 25 30 36 42 49 56 64 72 81 90 100 110 121
.25 1 2 4 6 9 13 18 24 31 39 49 60 73 87 103 121 141 163 187 214 242
.50 1 2 4 6 9 13 18 24 31 39 49 62 77 93 112 134 159 187 218 253 292
The meaning of each entry is as follows:
If you wish to make change for 50 cents using only pennies, nickels and dimes,
go to the .10 row and the 50 column to obtain 36 ways to do this.
To calculate each entry, you start with the pennies. There is exactly one
way to make change for every amount. Then calculate the .05 row by adding
the number of ways to make change for the amount using pennies plus the number
of ways to make change for five cents less using nickels and pennies. This
continues on for all denominations of coins.
An example, to get change for 75 cents using all coins up to a .50, add the
number of ways to make change using only .25 and down (121) and the number of
ways to make change for 25 cents using coins up to .50 (13). This yields the
answer of 134.
==> combinatorics/coinage/dimes.p <==
"Dad wants one-cent, two-cent, three-cent, five-cent, and ten-cent
stamps. He said to get four each of two sorts and three each of the
others, but I've forgotten which. He gave me exactly enough to buy
them; just these dimes." How many stamps of each type does Dad want?
A dime is worth ten cents. -- J.A.H. Hunter
==> combinatorics/coinage/dimes.s <==
The easy way to solve this is to sell her three each, for
3x(1+2+3+5+10) = 63 cents. Two more stamps must be bought, and they
must make seven cents (since 17 is too much), so the fourth stamps are
a two and a five.
==> combinatorics/coinage/impossible.p <==
What is the smallest number of coins that you can't make a dollar with?
I.e., for what N does there not exist a set of N coins adding up to a dollar?
It is possible to make a dollar with 1 current U.S. coin (a Susan B. Anthony),
2 coins (2 fifty cent pieces), 3 coins (2 quarters and a fifty cent piece),
etc. It is not possible to make exactly a dollar with 101 coins.
==> combinatorics/coinage/impossible.s <==
The answer is 77:
a) 5c = 1 or 5;
b) 10c = 1 or 2 a's (1,2,6,10)
c) 25c = 1 or 2 b's + 1 a
d) 50c = 1 or 2 c's
e) $1 = 1 or 2 d's
total penny nickel dime quarter half
5 1 2 1 1
6 3 1 1 1
7 5 1 1
8 4 3 1
9 6 2 1
10 8 1 1
11 10 1
12 7 4 1
13 9 3 1
14 11 2 1
15 13 1 1
16 15 1
17 14 3
18 16 2
19 18 1
20 20
21 5 13 3
22 5 15 2
23 5 17 1
24 5 19
25 10 12 3
26 10 14 2
27 10 16 1
28 10 18
29 15 11 3
30 15 13 2
31 15 15 1
32 15 17
33 20 10 3
34 20 12 2
35 20 14 1
36 20 16
37 25 9 3
38 25 11 2
39 25 13 1
40 25 15
41 30 8 3
42 30 10 2
43 30 12 1
44 30 14
45 35 7 3
46 35 9 2
47 35 11 1
48 35 13
49 40 6 3
50 40 8 2
51 40 10 1
52 40 12
53 45 5 3
54 45 7 2
55 45 9 1
56 45 11
57 50 4 3
58 50 6 2
59 50 8 1
60 50 10
61 55 3 3
62 55 5 2
63 55 7 1
64 55 9
65 60 2 3
66 60 4 2
67 60 6 1
68 60 8
69 65 1 3
70 65 3 2
71 65 5 1
72 65 7
73 70 3
74 70 2 2
75 70 4 1
76 70 6
77 can't be done
78 75 1 2
79 75 3 1
80 75 5
81 can't be done
82 80 2
83 80 2 1
84 80 4
85 can't be done
86 can't be done
87 85 1 1
88 85 3
89 can't be done
90 can't be done
91 90 1
92 90 2
93-95 can't be done
96 95 1
97-99 can't be done
100 100
==> combinatorics/color.p <==
An urn contains n balls of different colors. Randomly select a pair, repaint
the first to match the second, and replace the pair in the urn. What is the
expected time until the balls are all the same color?
==> combinatorics/color.s <==
(n-1)^2.
If the color classes have sizes k1, k2, ..., km, then the expected number of
steps from here is (dropping the subscript on k):
2 k(k-1) (j-1) (k-j)
(n-1) - SUM ( ------ + SUM --------------- )
classes, 2 1<j<k (n-j)
class.size=k
The verification goes roughly as follows. Defining phi(k) as (k(k-1)/2 +
sum[j]...), we first show that phi(k+1) + phi(k-1) - 2*phi(k) = (n-1)/(n-k)
except when k=n; the k(k-1)/2 contributes 1, the term j=k contributes
(j-1)/(n-j)=(k-1)/(n-k), and the other summands j<k contribute nothing.
Then we say that the expected change in phi(k) on a given color class is
k*(n-k)/(n*(n-1)) times (phi(k+1) + phi(k-1) - 2*phi(k)), since with
probability k*(n-k)/(n*(n-1)) the class goes to size k+1 and with the same
probability it goes to size k-1. This expected change comes out to k/n.
Summing over the color classes (and remembering the minus sign), the
expected change in the "cost from here" on one step is -1, except when we're
already monochromatic, where the handy exception k=n kicks in.
One can rewrite the contribution from k as
(n-1) SUM (k-j)/(n-j)
0<j<k
which incorporates both the k(k-1)/2 and the previous sum over j.
That makes the proof a little cleaner.
==> combinatorics/full.p <==
Consider a string that contains all substrings of length n. For example,
for binary strings with n=2, a shortest string is 00110 -- it contains 00,
01, 10 and 11 as substrings. Find the shortest such strings for all n.
==> combinatorics/full.s <==
Knuth, Volume 2 Seminumerical Algorithms, section 3.2.2 discusses this problem.
He cites the following results:
Shortest length: m^n + n - 1, where m = number of symbols in the language.
Algorithms:
[Exercise 7, W. Mantel, 1897]
The binary sequence is the LSB of X computed by the MIX program:
LDA X
JANZ *+2
LDA A
ADD X
JNOV *+3
JAZ *+2
XOR A
STA X
[Exercise 10, M. H. Martin, 1934]
Set x[1] = x[2] = ... = x[n] = 0. Set x[i+1] = largest value < n such that
substring of n digits ending at x[i+1] does not occur earlier in string.
Terminate when this is not possible.
If we instead consider the strings as circular, we have a well known
problem whose solution is given by any hamiltonian cycle in the de
Bruijn (or Good) graph of dimension K. (Or equivalently an eulerian
circuit in the de Bruijn graph of dimension K-1) As a string of length
2^K is produced, it must be optimal, and any shortest sequence must be
an eulerian circuit in a dB graph.
The de Bruijn graph Tn has as its vertex set the binary n-strings.
Directed edges join n-strings that may be derived by deleting the left
most digit and appending a 0 or 1 to the right end. de Bruijn + van
Ardenne-Ehrenfest (in 1951) counted the number of eulerian circuits in
Tn. There are 2^(2^(n-1)-n) of them.
Some examples:
K=2 1100
K=3 11101000
K=4 1111001011010000
The solution to the above problem (non-circular strings) can be found
by duplicating the first K-1 digits of the solution string at the end
of the string. These are not the only solutions, but they
are of minimum length: 2^K + K-1.
We can obtain a lower bound for the optimal sequence for the general case as
follows:
Consider first the simpler case of breaking into an answer machine which
accepts d+1 digits, values 0 to n-1. We wish to find the minimal universal
code that will allow us access to any such answering machine.
Let us construct a digraph G = (V,E), where the n^d vertices are labelled
with a d sequence of digits. Notation: let [v_{i,1},v_{i,2},...,v_{i,d}]
denote the labelling on node v_i. An edge e = (v_i, v_j) is in E iff for k
in 1, ..., d-1: v_{i,k+1} = v_{j,k}, i.e., the last d-1 digits in the
labelling of the initial vertex of e is identical with the first d-1 digits
in the labelling of the terminal vertex of e. We associate with each edge a
value, t(e) = v_{j,d}, the last digit in the labelling of the terminal
vertex.
The intuition goes as follows: we are going to perform a Euler circuit of
the digraph, where the label on the current vertex gives the last d digits
in the output sequence so far. If we make a transition on edge e, we output
the tone/digit t(e) as the next output value, thus preserving the invariant
on the labelling.
How do we know that a Euler circuit exists? Simple: a connected digraph
has an Euler circuit iff for all vertices v: indegree(v) = outdegree(v).
This property is trivially true for this digraph.
So, in order to generate a universal code for the AM, we simply output 0^d
(to satisfy the precondition for being in vertex [0,...,0]), and perform an
Euler circuit starting at node [0,...,0].
Now, the total length of the universal sequence is just the number of edges
traversed in the Euler circuit plus the initial precondition sequence, or
n^d * n + d (number of vertices times the out-degree) or n^{d+1} + d. That
this is a minimal sequence is obvious.
Next, let us consider the machine AM' where the security code is of the form
[0...n-1]^d [0...m-1], i.e., d digits ranging from 0 to n-1, followed by a
terminal digit ranging from 0 to m-1, m < n.
We build a digraph G = (V, E) similar to the construction above, except for
the following: an edge e is in E iff t(e) in 0 to m-1. This digraph is
clearly non-Eulerian. In particular, there are two classes of vertices:
(1) v is of the form [0...n-1]^{d-1} [0...m-1] (``fat'' vertices)
and
(2) v is of the form [0...n-1]^{d-1} [m...n-1] (``thin'' vertices)
Observations: there are (n^{d-1} * m) fat vertices, and (n^{d-1} * (n-m))
thin vertices. All vertices have out-degree of m. Fat vertices have
in-degrees of n, and thin vertices have in-degrees of 0. Color all the
edges blue.
The question now becomes: can we put a bound on how many new (red) edges
must we add to G in order to make a blue edge covering path possible?
(Instead of thinking of edges being traversed multiple times in the blue
edge covering path, we allow multiple edges between vertices and allow each
edge to be traversed once.) Note that, in this procedure, we add edges only
if it is allowed (the vertex labelling constraint). We will first obtain a
lower bound on the length of a blue covering circuit, and then transform it
into a bound for arbitrary blue covering paths.
Clearly, we must add at least (n-m)*(n^{d-1}*m) edges incident from the fat
vertices. [ We need (n-m) new out-going edges for each of (n^{d-1}*m)
vertices to bring the out-degree up to the in-degree. ]
Let us partition our vertices into sets. Denote the range [0..m-1] by S,
the range [m..n-1] by L, and the range [0..n-1] by X.
Let S_0 = { v: v = [X^{d-1}S] }. S_0 is just the set of fat vertices.
Define in(S_0) = number of edges from vertices not in S to vertices in S.
Define out(S_0) in the corresponding fashion, and let excess(S_0) =
in(S_0)-out(S_0). Clearly, excess(S_0) = n^{d-1}m(n-m) from the argument
above. Generalizing the requirement for Eulerian digraphs, we see that we
must add excess(S_0) edges from S_0 if the blue edges connected to/within
S_0 are to be covered by some circuit (edges may not be travered multiple
times -- we add parallel edges to handle that case). In particular, edges
from S_0 will be incident on vertices of the form [X^{d-2}SX]. Furthermore,
they can not be [X^{d-2}SS] since that is a subset of S_0 and adding those
edges will not help excess(S_0). [Now, these edges may be needed if we are
to have a circuit, but we do not consider them since they do not help
excess(S_0).] So, we are forced to add excess(S_0) edges from S_0 to S_1 = {
v: v = [X^{d-2}SL] }. Color these newly added edges red.
Let us define in(S_1), out(S_1) and excess(S_1) as above for the modified
digraph, i.e., including the red excess(S_0) edges that we just added.
Clearly, in(S_1) = out(S_0) = n^{d-1}m(n-m), and out(S_1) = m*|S_1| =
m*n^{d-2}m(n-m), so excess(S_1) = n^{d-2}m(n-m)^2. Consider S_0 union S_1.
We must add excess(S_1) edges to S_0 union S_1 to make it possible for the
digraph to be covered by a circuit, and these edges must go from {S_0 union
S_1} to S_2 = { v: v = [X^{d-3}SL^2] } by a similar argument as before.
Repeating this partitioning process, eventually we get to S_{d-1} = { v: v =
[SL^{d-1}] }, where union of S_0 to S_{d-1} will need edges to S_d = { v: v
= [L^d] }, where this process terminates. Note that at this time,
excess(union of S_0 to S_{d-1}) = m(n-m)^d, but in(S_d) = 0 and out(S_d) =
m(n-m)^d, and the process terminates.
What have we shown? Adding up blue edges and the red edges gives us a lower
bound on the total number of edges in a blue-edges covering circuit (not
necessarily Eulerian) in the complete digraph. This comes out to be
n^{d+1}-(n-m)^{d+1} edges.
Next, we note that if we had an optimal path covering all the blue edges, we
can transform it into a circuit by adding d edges. So, a minimal path can
be no more than d edges shorter than the minimal circuit covering all blue
edges. [Otherwise, we add d extra edges to make it into a shorter circuit.]
So the shortest blue covering path through the digraph is at least
n^{d+1}-{n-m}^{d+1}-d. With an initial pre-condition sequence of length d
(to establish the transition invariant), the shortest universal answering
machine sequence is of length at least n^{d+1}-(n-m)^{d+1}.
While this has not been that constructive, it is easy to see that we can
achieve this bound. If we looked at the vertices in each of the S_i's, we
just add exactly the edges to S_{i+1} and no more. The resultant digraph
would be Eulerian, and to find the minimal path we need only start at the
vertex labelled [{n-1}^d], find the Euler circuit, and omit the last d edges
from the tour.
==> combinatorics/gossip.p <==
n people each know a different piece of gossip. They can telephone each other
and exchange all the information they know (so that after the call they both
know anything that either of them knew before the call). What is the smallest
number of calls needed so that everyone knows everything?
==> combinatorics/gossip.s <==
1 for n=2
3 for n=3
2n-4 for n>=4
This can be achieved as follows: choose four people (A, B, C, and D) as
the "core group". Each person outside the core group phones a member of
the core group (it doesn't matter which); this takes n-4 calls. Now the
core group makes 4 calls: A-B, C-D, A-C, and B-D. At this point, each
member of the core group knows everything. Now, each person outside the
core group calls anybody who knows everything; this again requires n-4
calls, for a total of 2n-4.
The solution to the "gossip problem" has been published several times:
1. R. Tidjeman, "On a telephone problem", Nieuw Arch. Wisk. 3
(1971), 188-192.
2. B. Baker and R. Shostak, "Gossips and telephones", Discrete
Math. 2 (1972), 191-193.
3. A. Hajnal, E. C. Milner, and E. Szemeredi, "A cure for the
telephone disease", Canad Math. Bull 15 (1976), 447-450.
4. Kleitman and Shearer, Disc. Math. 30 (1980), 151-156.
5. R. T. Bumby, "A problem with telephones", Siam J. Disc. Meth. 2
(1981), 13-18.
==> combinatorics/grid.dissection.p <==
How many (possibly overlapping) squares are in an mxn grid? Assume that all
the squares have their edges parallel to the edges of the grid.
==> combinatorics/grid.dissection.s <==
Given an n*m grid with n > m.
Orient the grid so n is its width. Divide the grid into two portions,
an m*m square on the left and an (n-m)*m rectangle on the right.
Count the squares that have their upper right-hand corners in the
m*m square. There are m^2 of size 1*1, (m-1)^2 of size 2*2, ...
up to 1^2 of size m*m. Now look at the n-m columns of lattice points
in the rectangle on the right, in which we find upper right-hand
corners of squares not yet counted. For each column we count m new
1*1 squares, m-1 new 2*2 squares, ... up to 1 new m*m square.
Combining all these counts in summations:
m m
total = sum i^2 + (n - m) sum i
i=1 i=1
(2m + 1)(m + 1)m (n - m)(m + 1)m
= ---------------- + ---------------
6 2
= (3n - m + 1)(m + 1)m/6
-- David Karr
==> combinatorics/permutation.p <==
Compute the nth permutation of k numbers (or objects).
==> combinatorics/permutation.s <==
#include <stdio.h>
/*
adapted from 'Notation as a Tool of Thought', by K.E.Iverson
Radix Representation of Permutations
*/
/* direct from radix; of given order */
dfr(short direct[],short radix[],long order)
{
if (order)
{
direct[0] = radix[0];
dfr (direct+1, radix+1, order-1);
while (--order)
direct[order] += direct[order] >= direct[0];
}
}
/* radix representation; of given order and given index */
rr(short radix[], long order, long index)
{
int i;
for (i=1; i<=order; i++)
{
radix[order-i] = index % i;
index = index/i;
}
}
show(short perm[],long order)
{
while(order--)
printf("%hd ",*perm++);
printf("\n");
}
short parity(short radix[],long order)
{
long p=0;
while(order--)
p+=*radix++;
return p%2;
}
void usage(char *name)
{
fprintf(stderr,"usage: %s order number_of_permutation\n",name);
exit(-1);
}
main(int argc, char *argv[])
{
#define MAX_ORDER 512
short radix[MAX_ORDER], direct[MAX_ORDER];
long order, nth;
if (argc!=3) usage(argv[0]);
order = atol(argv[1]);
nth = atol(argv[2]);
rr(radix, order, nth-1); /* where 0 is the first permuatation */
dfr(direct, radix, order);
printf("radix "); show(radix,order);
printf("direct "); show(direct,order);
printf("parity %d\n",parity(radix,order));
}
--
J. Henri Schueler, H&h Software, Toronto +1 416 698 9075
j...@ipsa.reuter.com
==> combinatorics/subsets.p <==
Out of the set of integers 1,...,100 you are given ten different
integers. From this set, A, of ten integers you can always find two
disjoint non-empty subsets, S & T, such that the sum of elements in S
equals the sum of elements in T. Note: S union T need not be all ten
elements of A. Prove this.
==> combinatorics/subsets.s <==
There are 2^10 = 1,024 subsets of the 10 integers, but there can be only 901
possible sums, the number of integers between the minimum and maximum sums.
With more subsets than possible sums, there must exist at least one sum that
corresponds to at least two subsets. Call two subsets with equal sums A and B.
Let C = A intersect B; define S = A - C, T = B - C. Then S is disjoint from T,
and sum(S) = sum(A-C) = sum(A) - sum(C) = sum(B) - sum(C) = sum(B-C) = sum(T).
QED
Addendum: 9 integers suffice. This was part of my Westinghouse project
in 1981 (the above problem was in Martin Gardner's Scientific American
column not long before). The argument is along the same lines, but
a bit more complicated; for starters you only work with the subsets
consisting of 3, 4, 5, or 6 of the 9 elements.
Let M(n) be the smallest integer such that there exists an n-element
set {a1,a2,a3,...,an=M(n)} of positive integers all 2^n of whose
subsums are distinct. The pigeonhole argument of subsets.s shows that
M(n)>2^n/n, and it is known that M(n)>c*2^n/sqrt(n) for some c>0.
It is still an unsolved problem (with an Erdos bounty) whether
there is a positive constant c such that M(n)>c*2^n for all n.
--Noam D. Elkies (elk...@zariski.harvard.edu)
Dept. of Mathematics, Harvard University
==> combinatorics/transitions.p <==
How many n-bit binary strings (0/1) have exactly k transitions
(an adjacent pair of dissimilar bits, i.e., a 01 or a 10)?
==> combinatorics/transitions.s <==
A transition can occur at an adjacent pair (i,i+1) where 1<=i<i+1<=n.
Since there are k transitions, there are C(n-1,k) total number of ways
that transitions can occur. But the string may start with a 1 or a 0
(after which its transitions uniquely determine the string). So there
are a total of 2C(n-1,k) such strings.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> competition/contests/games.magazine.p <==
What are the best answers to various contests run by _Games_ magazine?
==> competition/contests/games.magazine.s <==
Contest Date Archive Entry
---------------------- -------------- -----------------------------
Equations ? 1981 equations
National Puzzle 1993 1993 npc.1993
Nots and Crosses ? 1992 nots.and.crosses
Perfect Ladder July 1993 perfect.ladder
Telegrams ? telegrams
==> competition/contests/national.puzzle/npc.1993.p <==
What are the solutions to the Games magazine 1993 National Puzzle Contest?
==> competition/contests/national.puzzle/npc.1993.s <==
1. 6, 10, 11, and 12 are in one group, and everything else is in the other.
2. 20
3. The upper-right segment of the first 8.
4. 6
5. d (ball point pen, analog watch, mattress, pogo stick): springs
6. a (Fire Engine, strawberry, lobster, lips): red
7. h (Icicle, paint, nose, faucet): drip
8. f (piggy bank, basketball, jack o' lantern, drum): hollow
9. g (horseshoe, goal post, stethoscope, fish hook): shaped like letters
10. e (flying bat, Xmas tree ornament, framed picture, trapeze artist): hang
11. b (candle, owl, vampire, pajamas): all associated with night
12. 4, 7, 2, 10, 5, 8, 1, 9, 6, 3
13. 152954
14. LIMA PERU
15. 44
16. 160
17. A
18. Flo Lisa Amy Joyce Kim.
19. Top: AJKQ, 2nd Row: JKAQ, 3rd Row: KQAJ, 4th Row: JQAK
20. Joan Miro, Paavo Nurmi, Blaise Pascal
21. Top: 1, 8, 4 Middle: 6, 9, 3 Bottom: 2, 7, 5
22. d and g
23. Brenda: 87, Carrie: 85, Dick: 86, Harry: 84, Laura: 88, Tom: 83
24. If you number the columns 1-6 and letter the rows a-f, the first group
consists of 1a, 2a, 3a, 4a, 1b, 2b, 2c, 3c, 2d. Other groups are
similarly shaped, rotated around the center point of the grid.
25. 2, 6, 5
26. Top: R O M Middle: Q T A Bottom: U E D S
27. 3 X 58 = 174 = 6 X 29
28. 5 (the number of enclosed areas in the letters of each month name)
29. too hard to describe -- easier than it looked
30. 80
31. 16
32. 4 (ADBECF ADBFCE ADFCBE BFDCAE)
33. 8 (delete 3,5,7,9,12,14,17,18)
34. CONKP VALEY GSRCW TUIBI LANZS
==> competition/games/bridge.p <==
Are there any programs for solving double-dummy Bridge?
==> competition/games/bridge.s <==
I'll enclose my Double-Dummy solver for bridge. I like this program
because it contains a wildly unstructured "goto" -- which I claim is the
most readable way to write the program.
Included are two test problems for the bridge-solver: a 6-card
ending and a complete 13-card position. The former should be very fast;
the latter about 20 minutes on Sparcstation-2. Each is *very*
challenging for humans.
Regards, James
=============== clip; chmod +x; execute =============
#!/bin/sh
cat > bridge.c << 'EOF'
/*
* DOUBLE_DUMMY
* Copyright (c) 1990 by
* James D. Allen
* 19785 Stanton Ave
* Castro Valley, CA
* Permission granted to use for any non-commercial purpose
* as long as this notice is not removed.
*
* This program will solve double-dummy bridge problems.
* The algorithm is trivial: brute-force alpha-beta search (also known
* as "backtracking"). The alpha-beta is trivial since we do not
* consider overtricks or extra undertricks.
* The control flow is neat; this is a rare exception where software is
* more readable with a "goto". (Although I've tersified this to
* the point where it is perhaps unreadable anyway :-)
*/
#define NUMP 4 /* The Players: N, E, S, W */
#define NORTH 0
#define IS_CONT(x) (!((x) & 1)) /* Is x on N/S team? */
#define LHO(x) (((x) + 1) % NUMP)
#define RHO(x) (((x) + NUMP - 1) % NUMP)
char *Playername[] = { "North", "East", "South", "West" };
#define NUMS 4 /* The Suits: S, H, D, C */
char Suitname[] = "SHDC";
char *Fullname[] = { "Spades\t", "Hearts\t", "Diamonds", "Clubs\t" };
/*
* Rank indices are 2 (Ace), 3 (King), ... 14 (Deuce)
* There is also a CARD Index which can be converted to from Rank and Suit.
*/
#define CARD(Suit, Rank) (((Suit) << 4) | (Rank))
#define SUIT(Card) ((Card) >> 4)
#define RANK(Card) ((Card) & 0xf)
char Rankname[] = "??AKQJT98765432";
#define INDEX(s, c) ((char *)index(s, c) - (s))
/* A "SuitSet" is used to cope with more than one card at once: */
typedef unsigned short SuitSet;
#define MASK(Card) (1 << RANK(Card))
#define REMOVE(Set, Card) ((Set) &= ~(MASK(Card)))
/* And a CardSet copes with one SuitSet for each suit: */
typedef struct cards {
SuitSet cc[NUMS];
} CardSet;
/* Everything we wish to remember about a trick: */
struct status {
CardSet st_hold[NUMP]; /* The players' holdings */
CardSet st_lgl[NUMP]; /* The players' remaining legal plays */
short st_play[NUMP]; /* What the players played */
SuitSet st_trick; /* Led-suit Cards eligible to win trick */
SuitSet st_trump; /* Trump Cards eligible to win trick */
short st_leader; /* Who led to the trick */
short st_suitled; /* Which suit was led */
}
Status[14]; /* Status of 13 tricks and a red zone" */
#define Hold Statp->st_hold
#define Resid (Statp+1)->st_hold
#define Lgl Statp->st_lgl
#define Play Statp->st_play
#define Trick Statp->st_trick
#define Trtrick Statp->st_trump
#define Leader Statp->st_leader
#define Suitled Statp->st_suitled
/* Pick a card from the set and return its index */
pick(set)
SuitSet set;
{
return set & 0xff ? set & 1 ? 0 : set & 2 ? 1 : set & 4 ? 2
: set & 8 ? 3 : set & 16 ? 4 : set & 32 ? 5
: set & 64 ? 6 : 7 : pick(set >> 8) + 8;
}
#define highcard(set) pick(set) /* Pick happens to return the best card */
main()
{
register struct status *Statp = Status; /* Point to current status */
int tnum; /* trick number */
int won; /* Total tricks won by North/South */
int nc; /* cards on trick */
int ohsize; /* original size of hands */
int mask;
int trump;
int player; /* player */
int pwin; /* player who won trick */
int suit; /* suit to play */
int wincard; /* card which won the trick */
int need; /* Total tricks needed by North/South */
int cardx; /* Index of a card under consideration */
int success; /* Was the trick or contract won by North/South ? */
int last_t; /* Decisive trick number */
char asciicard[60];
SuitSet inplay; /* cards still in play for suit */
SuitSet pr_set; /* Temp for printing */
/* Read in the problem */
printf("Enter trump suit (0-S,1-H,2-D,3-C,4-NT): ");
scanf("%d", &trump);
printf("Enter how many tricks remain to be played: ");
scanf("%d", &ohsize);
printf("Enter how many tricks North/South need to win: ");
scanf("%d", &need);
printf("Enter who is on lead now (0=North,etc.): ");
scanf("%d", &pwin);
printf("Enter the %d cards beginning with North:\n", NUMP * ohsize);
for (player = NORTH; player < NUMP; player++) {
for (tnum = 0; tnum < ohsize; tnum++) {
scanf("%s", asciicard);
cardx = CARD(INDEX(Suitname, asciicard[1]),
INDEX(Rankname, asciicard[0]));
Hold[player].cc[SUIT(cardx)] |= MASK(cardx);
}
}
/* Handle the opening lead */
printf("Enter the directed opening lead or XX if none:\n");
Lgl[pwin] = Hold[pwin];
scanf("%s", asciicard);
if (asciicard[0] == 'X') {
strcpy(asciicard, "anything");
} else {
cardx = CARD(INDEX(Suitname, asciicard[1]),
INDEX(Rankname, asciicard[0]));
for (suit = 0; suit < NUMS; suit++)
if (suit != SUIT(cardx))
Lgl[pwin].cc[suit] = 0;
else if (!(Lgl[pwin].cc[suit] &= MASK(cardx))) {
printf("He can't lead card he doesn't have\n");
exit(1);
}
}
/* Print the problem */
for (player = NORTH; player < NUMP; player++) {
printf("\n---- %s Hand ----:\n", Playername[player]);
for (suit = 0; suit < NUMS; suit++) {
printf("\t%s\t", Fullname[suit]);
for (pr_set = Hold[player].cc[suit]; pr_set;
REMOVE(pr_set, pick(pr_set)))
printf("%c ", Rankname[RANK(pick(pr_set))]);
printf("\n");
}
}
printf("\n%s%s Trumps; %s leads %s; N-S want %d tricks; E-W want %d\n",
trump < NUMS ? Fullname[trump] : "",
trump < NUMS ? " are" : "No",
Playername[pwin], asciicard, need, ohsize + 1 - need);
/* Loop to play tricks forward until the outcome is conclusive */
for (tnum = won = success = 0;
success ? ++won < need : won + ohsize >= need + tnum;
tnum++, Statp++, success = IS_CONT(pwin)) {
for (nc = 0, player = Leader = pwin; nc < NUMP;
nc++, player = LHO(player)) {
/* Generate legal plays except opening lead */
if (nc || tnum)
Lgl[player] = Hold[player];
/* Must follow suit unless void */
if (nc && Lgl[player].cc[Suitled])
for (suit = 0; suit < NUMS; suit++)
if (suit != Suitled)
Lgl[player].cc[suit] = 0;
goto choose_suit; /* this goto is easily eliminated */
/* Comes back right away after choosing "suit" */
make_play:
Play[player] = cardx =
CARD(suit, pick(Lgl[player].cc[suit]));
if (nc == 0) {
Suitled = suit;
Trick = Trtrick = 0;
}
/* Set the play into "Trick" if it is win-eligible */
if (suit == Suitled)
Trick |= MASK(cardx);
if (suit == trump)
Trtrick |= MASK(cardx);
/* Remove card played from player's holding */
Resid[player] = Hold[player];
REMOVE(Resid[player].cc[suit], cardx);
}
/* Finish processing the trick ... who won? */
if (Trtrick)
wincard = CARD(trump, highcard(Trtrick));
else
wincard = CARD(Suitled, highcard(Trick));
for (pwin = 0; Play[pwin] != wincard; pwin++)
;
}
/* Loop to back up and let the players try alternatives */
for (last_t = tnum--, Statp--; tnum >= 0; tnum--, Statp--) {
won -= IS_CONT(pwin);
pwin = Leader;
for (player = RHO(Leader), nc = NUMP-1; nc >= 0;
player = RHO(player), nc--) {
/* What was the play? */
cardx = Play[player];
suit = SUIT(cardx);
/* Retract the played card */
if (suit == Suitled)
REMOVE(Trick, cardx);
if (suit == trump)
REMOVE(Trtrick, cardx);
/* We also want to remove any redundant adjacent plays */
inplay = Hold[0].cc[suit] | Hold[1].cc[suit]
| Hold[2].cc[suit] | Hold[3].cc[suit];
for (mask = MASK(cardx); mask <= 0x8000; mask <<= 1)
if (Lgl[player].cc[suit] & mask)
Lgl[player].cc[suit] &= ~mask;
else if (inplay & mask)
break;
/* If the card was played by a loser, try again */
if (success ? !(IS_CONT(player)) : IS_CONT(player)) {
choose_suit:
/* Pick a suit if any untried plays remain */
for (suit = 0; suit < NUMS; suit++)
if (Lgl[player].cc[suit])
/* This goto is really nice!! */
goto make_play;
}
}
}
/*
* We're done. We know the answer.
* We can't remember all the variations; fortunately the
* succeeders played correctly in the last variation examined,
* so we'll just print it.
*/
printf("Contract %s, for example:\n",
success ? "made" : "defeated");
for (tnum = 0, Statp = Status; tnum < last_t; tnum++, Statp++) {
printf("Trick %d:", tnum + 1);
for (player = 0; player < Leader; player++)
printf("\t");
for (nc = 0; nc < NUMP; nc++, player = LHO(player))
printf("\t%c of %c",
Rankname[RANK(Play[player])],
Suitname[SUIT(Play[player])]);
printf("\n");
}
return 0;
}
EOF
cc -O -o bridge bridge.c
bridge << EOF
4 6 5 2
AS JS 4S QD 8D 2C
KS QS 9H 8H AD 2D
AH 2H KD 9D 7D AC
TS 3S 2S TH TD TC
XX
EOF
bridge << EOF
1 13 13 3
3C 3H 2H AD KD 2D AS KS 7S 6S 5S 4S 3S
8C 7C 6C 5C 4C QH TH 8H 7H 8D 7D 6D 2S
AC QC JC 9C AH KH JH 9H 6H 5H 5D 4D 3D
KC TC 2C 4H QD JD TD 9D QS JS TS 9S 8S
QS
EOF
==> competition/games/chess/knight.control.p <==
How many knights does it take to attack or control the board?
==> competition/games/chess/knight.control.s <==
Fourteen knights are required to attack every square:
1 2 3 4 5 6 7 8
___ ___ ___ ___ ___ ___ ___ ___
h | | | | | | | | |
--- --- --- --- --- --- --- ---
g | | | N | N | N | N | | |
--- --- --- --- --- --- --- ---
f | | | | | | | | |
--- --- --- --- --- --- --- ---
e | | N | N | | | N | N | |
--- --- --- --- --- --- --- ---
d | | | | | | | | |
--- --- --- --- --- --- --- ---
c | | N | N | N | N | N | N | |
--- --- --- --- --- --- --- ---
b | | | | | | | | |
--- --- --- --- --- --- --- ---
a | | | | | | | | |
--- --- --- --- --- --- --- ---
Three knights are needed to attack h1, g2, and a8; two more for b1, a2,
and b3, and another two for h7, g8, and f7.
The only alternative pattern is:
1 2 3 4 5 6 7 8
___ ___ ___ ___ ___ ___ ___ ___
h | | | | | | | | |
--- --- --- --- --- --- --- ---
g | | | N | | | N | | |
--- --- --- --- --- --- --- ---
f | | | N | N | N | N | | |
--- --- --- --- --- --- --- ---
e | | | | | | | | |
--- --- --- --- --- --- --- ---
d | | | N | N | N | N | | |
--- --- --- --- --- --- --- ---
c | | N | N | | | N | N | |
--- --- --- --- --- --- --- ---
b | | | | | | | | |
--- --- --- --- --- --- --- ---
a | | | | | | | | |
--- --- --- --- --- --- --- ---
Twelve knights are needed to control (attack or occupy) the board:
1 2 3 4 5 6 7 8
___ ___ ___ ___ ___ ___ ___ ___
a | | | | | | | | |
--- --- --- --- --- --- --- ---
b | | | N | | | | | |
--- --- --- --- --- --- --- ---
c | | | N | N | | N | N | |
--- --- --- --- --- --- --- ---
d | | | | | | N | | |
--- --- --- --- --- --- --- ---
e | | | N | | | | | |
--- --- --- --- --- --- --- ---
f | | N | N | | N | N | | |
--- --- --- --- --- --- --- ---
g | | | | | | N | | |
--- --- --- --- --- --- --- ---
h | | | | | | | | |
--- --- --- --- --- --- --- ---
Each knight can control at most one of the twelve squares a1, b1, b2,
h1, g1, g2, a8, b8, b7, h8, g8, g7. This position is unique up to
reflection.
References
Martin Gardner, _Mathematical Magic Show_.
==> competition/games/chess/knight.most.p <==
What is the maximum number of knights that can be put on n x n chessboard
without threatening each other?
==> competition/games/chess/knight.most.s <==
n^2/2 for n even >= 4.
Divide the board in parts of 2x4 squares. The squares within
each part are paired as follows:
-----
|A|B|
-----
|C|D|
-----
|B|A|
-----
|D|C|
-----
Clearly, not more than one square per pair can be occupied by a knight.
==> competition/games/chess/knight.tour.p <==
For what size boards are knight tours possible?
==> competition/games/chess/knight.tour.s <==
A tour exists for boards of size 1x1, 3x4, 3xN with N >= 7, 4xN with N >= 5,
and MxN with N >= M >= 5. In other words, for all rectangles except 1xN
(excluding the trivial 1x1), 2xN, 3x3, 3x5, 3x6, 4x4.
With the exception of 3x8 and 4xN, any even-sized board which allows a tour
will also allow a closed (reentrant) tour.
On an odd-sided board, there is one more square of one color than
of the other. Every time a knight moves, it moves to a square of
the other color than the one it is on. Therefore, on an odd-sided
board, it must end the last move but one of the complete, reentrant
tour on a square of the same color as that on which it started.
It is then impossible to make the last move, for that move would end
on a square of the same color as it begins on.
Here is a solution for the 7x7 board (which is not reentrant).
------------------------------------
| 17 | 6 | 33 | 42 | 15 | 4 | 25 |
------------------------------------
| 32 | 47 | 16 | 5 | 26 | 35 | 14 |
------------------------------------
| 7 | 18 | 43 | 34 | 41 | 24 | 3 |
------------------------------------
| 46 | 31 | 48 | 27 | 44 | 13 | 36 |
------------------------------------
| 19 | 8 | 45 | 40 | 49 | 2 | 23 |
------------------------------------
| 30 | 39 | 10 | 21 | 28 | 37 | 12 |
------------------------------------
| 9 | 20 | 29 | 38 | 11 | 22 | 1 |
------------------------------------
Here is a solution for the 5x5 board (which is not reentrant).
--------------------------
| 5 | 10 | 15 | 20 | 3 |
--------------------------
| 16 | 21 | 4 | 9 | 14 |
--------------------------
| 11 | 6 | 25 | 2 | 19 |
--------------------------
| 22 | 17 | 8 | 13 | 24 |
--------------------------
| 7 | 12 | 23 | 18 | 1 |
--------------------------
Here is a reentrant 2x4x4 tour:
0 11 16 3 15 4 1 22
19 26 9 24 8 23 14 27
10 5 30 17 31 12 21 2
29 18 25 6 20 7 28 13
A reentrant 4x4x4 tour can be constructed by splicing two copies.
It shouldn't be much more work now to completely solve the problem of which 3D
rectangular boards allow tours.
Warnsdorff's rule: at each stage of the knight's tour, choose the
square with the fewest remaining exits:
1 12 23 44 3 14 25
22 43 2 13 24 35 4
11 28 45 40 47 26 15
42 21 48 27 34 5 36
29 10 41 46 39 16 33
20 49 8 31 18 37 6
9 30 19 38 7 32 17
Mr. Beverley published in the Philosophical Magazine in 1848 this
knight's tour that is also a magic square:
1 30 47 52 5 28 43 54
48 51 2 29 44 53 6 27
31 46 49 4 25 8 55 42
50 3 32 45 56 41 26 7
33 62 15 20 9 24 39 58
16 19 34 61 40 57 10 23
63 14 17 36 21 12 59 38
18 35 64 13 60 37 22 11
References:
``The Construction of Magic Knight Tours,'' by T. H. Willcocks,
J. Rec. Math. 1:225-233 (1968).
"Games Ancient and Oriental and How to Play Them"
by Edward Falkener published by Dover in 1961 (first published 1892)
"Mathematical Magic Show", Martin Gardner, ch. 14
==> competition/games/chess/mutual.stalemate.p <==
What's the minimal number of pieces in a legal mutual stalemate?
==> competition/games/chess/mutual.stalemate.s <==
6. Here are three mutual stalemate positions. Black is lower case
in the diagrams.
W Kh8 e6 f7 h7 B Kf8 e7
--+--+--+--+--+
| | k| | K|
--+--+--+--+--+
| p| P| | P|
--+--+--+--+--+
| P| | | |
--+--+--+--+--+
| | | | |
W Kb1 B Ka3 b2 b3 b4 a4
| | |
+--+--+--
| p| p|
+--+--+--
| k| p|
+--+--+--
| | p|
+--+--+--
| | K|
+--+--+--
W Kf1 B Kh1 Bg1 f2 f3 h2
| | | |
--+--+--+--+
| p| | |
--+--+--+--+
| p| | p|
--+--+--+--+
| K| b| k|
--+--+--+--+
==> competition/games/chess/queen.control.p <==
How many queens does it take to attack or control the board?
==> competition/games/chess/queen.control.s <==
Five queens are required to attack every square:
1 2 3 4 5 6 7 8
___ ___ ___ ___ ___ ___ ___ ___
h | | | | | | | | |
--- --- --- --- --- --- --- ---
g | | | | Q | | | | |
--- --- --- --- --- --- --- ---
f | | | | Q | | | | |
--- --- --- --- --- --- --- ---
e | | | | Q | | | | |
--- --- --- --- --- --- --- ---
d | | | | Q | | | | |
--- --- --- --- --- --- --- ---
c | | | | | | | | |
--- --- --- --- --- --- --- ---
b | | | | | | | | |
--- --- --- --- --- --- --- ---
a | | | | Q | | | | |
--- --- --- --- --- --- --- ---
There are other positions with five queens.
==> competition/games/chess/queen.most.p <==
How many non-mutually-attacking queens can be placed on various sized boards?
==> competition/games/chess/queen.most.s <==
On an regular chess board, at most eight queens of one color can be
placed so that there are no mutual attacks.
Here is one configuration:
-----------------
|Q| | | | | | | |
-----------------
| | |Q| | | | | |
-----------------
| | | | |Q| | | |
-----------------
| | | | | | |Q| |
-----------------
| |Q| | | | | | |
-----------------
| | | | | | | |Q|
-----------------
| | | | | |Q| | |
-----------------
| | | |Q| | | | |
-----------------
On an nxn board, if n is not divisible by 2 or 3, n^2 queens can be placed
such that no two queens of the same color attack each other.
The proof is via a straightforward construction. For n=1, the result
is obviously true, so assume n>1 and n is not divisible by 2 or 3 (thus n>=5).
Assume we are given n queens in each of these n "colors" (numbers):
0 1 2 ... n-1
The proof is by construction. The construction is easier to see then to
describe, we do both. Here is what it looks like:
0 1 2 3 4 ... n-2 n-1
n-2 n-1 0 1 2 ... n-4 n-3
n-4 n-3 n-2 n-1 0 ... n-6 n-5
...(move one row down => sub 2 (mod n); one col right => add 1 (mod n))
2 3 4 5 6 ... 0 1
People who know how a knight moves in chess will note the repetitive knight
move theme connecting queens of the same color (especially after joining
opposite edges of the board).
Now to describe this: place in each cell a queen whose "color" (number) is:
j - 2*i + 1 (mod n),
where i is the # of the row, and j is the # of the column.
Then no 2 queens of the same color are in the same:
row, column, or diagonal.
Actually, we will prove something stronger, namely that no 2 queens of the
same color are on the same row, column, or "hyperdiagonal". (The concept, if
not the term, hyperdiagonal, goes back to 19th century.) There are n
hyperdiagonals of negative slope (one of them being a main diagonal) and n
hyperdiagonals of positive slope (one of them being the other main diagonal).
Definition: for k in 0..n-1:
the k-th negative hyperdiagonal consists of cells (i,j),
where i-j=k (mod n)
the k-th positive hyperdiagonal consists of cells (i,j),
where i+j=k (mod n)
Then 0-th negative hyperdiagonal is simply the NW-SE main diagonal.
Then "1-th" positive hyperdiagonal is simply the SW-NE main diagonal.
The other 2*(n-1) hyperdiagonals appear as 2 disconnected diagonal
fragments of cells, but if you join opposite edges of an nxn
board to each other, forming a sphere, these 2 fragments
become linearly connected forming a great circle.
Now to the proof:
1) First note that the above color assignment does indeed uniquely define the
color of a queen in each of the n^2 cells.
2) no row contains 2 queens of the same color:
cells (i, a) and (i, b) contain queens of the same color =>
a-2i-1 = b-2i-1 (mod n) =>
a = b (mod n) =>
a = b (since a,b are within 1..n)
3) no column contains 2 queens of the same color:
cells (a, j) and (b, j) contain queens of the same color =>
j-2a-1 = j-2b-1 (mod n) =>
2a = 2b (mod n) =>
a = b (mod n) (since n and 2 have no common factor) =>
a = b (since a,b are within 1..n)
4) no negative hyperdiagonal contains 2 queens of the same color:
cells (a, j) and (b, k) on the same negative hyperdiagonal contain
queens of the same color =>
a-j = b-k (mod n) AND j-2a-1 = k-2b-1 (mod n) =>
2a-2j = 2b-2k (mod n) AND j-2a = k-2b (mod n) =>
(adding corresponding sides:)
-j = -k (mod n) =>
j = k.
And thus a = b, as well (see first equality, 5th line up)
5) no positive hyperdiagonal contains 2 queens of the same color:
cells (a, j) and (b, k) on the same positive hyperdiagonal contain
queens of the same color =>
a+j = b+k (mod n) AND j-2a-1 = k-2b-1 (mod n) =>
2a+2j = 2b+2k (mod n) AND j-2a = k-2b (mod n) =>
(adding corresponding sides:)
3j = 3k (mod n) =>
j = k (mod n) (since n and 3 have no common factor) =>
j = k. Likewise a = b.
As special cases with the 0th negative hyperdiagonal and 1st positive
hyperdiagonal, no 2 queens on the same main diagonal are colored the same.
Now is this condition, than n be not divisible by 2 and 3 also *necessary*?
Mike Konrad
m...@sei.cmu.edu
******
. . . o . This is a solution for the 5-queen problem
o . . . . at the torus. It has the 90 degree rotation symmetry.
. . o . .
. . . . o
. o . . .
According to T. Klove, The modular n-queen problem II,
Discrete Math. 36 (1981) 33
it is unknown how to construct symmetric (90 rot) solutions for
n = 1 or 5 (mod 12) and n has prime factors = 3 (mod 4).
He solved the smallest cases 49 and 77.
Try the next cases 121 and 133 or find a general solution.
A further reference for modular or reflected queen problems is:
Martin Gardner, Fractal Music, Hypercards and More ..., Freeman (1991)
********
For the 3-D N-queens problem the answer is four, at (1,1,2), (1,3,3),
(2,3,1), and (3,2,3).
You can't have more because with four, you must have at least two in
at least one of the three horizontal slices of the cube. For the
two-or-more-queen slice you must solve the n-queens problem for a 3x3
planar board, which allows you to place at most 2 queens, and one must
be in a corner. The two queens cover all but one spot in the adjacent
slice, so if the two-queen slice is the middle one we're already
limited to no more than 4 queens. But if we put the 2-queen slice at
the bottom or top, then the corner queen has line of sight to all
corners of the opposite slice, so it can contain at most one queen,
and so can the middle slice.
If they sit in a 4x4x4 cube, the maximum is 7. Here is a sample.
1. 4 4 3
2. 2 3 4
3. 1 2 2
4. 2 4 2
5. 3 2 1
6. 1 1 4
7. 3 1 3
If they sit in a 5x5x5 cube, the maximum is 13. Here is a sample.
1. 4 5 4
2. 3 5 1
3. 5 4 2
4. 3 1 2
5. 2 1 4
6. 2 5 5
7. 4 1 5
8. 1 5 2
9. 5 2 1
10. 2 3 1
11. 1 3 5
12. 1 1 1
13. 5 1 3
==> competition/games/chess/queens.p <==
How many ways can eight queens be placed so that they control the board?
==> competition/games/chess/queens.s <==
92. The following program uses a backtracking algorithm to count positions:
#include <stdio.h>
static int count = 0;
void try(int row, int left, int right) {
int poss, place;
if (row == 0xFF) ++count;
else {
poss = ~(row|left|right) & 0xFF;
while (poss != 0) {
place = poss & -poss;
try(row|place, (left|place)<<1, (right|place)>>1);
poss &= ~place;
}
}
}
void main() {
try(0,0,0);
printf("There are %d solutions.\n", count);
}
--
Tony Lezard IS to...@mantis.co.uk OR tony%mantis...@uknet.ac.uk
OR EVEN ar...@phx.cam.ac.uk if all else fails.
==> competition/games/chess/rook.paths.p <==
How many non-overlapping paths can a rook take from one corner to the opposite
on an MxN chess board?
==> competition/games/chess/rook.paths.s <==
For a 1 x 1 chessboard, there are 1 unique paths.
For a 1 x 2 chessboard, there are 1 unique paths.
For a 1 x 3 chessboard, there are 1 unique paths.
For a 1 x 4 chessboard, there are 1 unique paths.
For a 1 x 5 chessboard, there are 1 unique paths.
For a 1 x 6 chessboard, there are 1 unique paths.
For a 1 x 7 chessboard, there are 1 unique paths.
For a 1 x 8 chessboard, there are 1 unique paths.
For a 2 x 2 chessboard, there are 2 unique paths.
For a 2 x 3 chessboard, there are 4 unique paths.
For a 2 x 4 chessboard, there are 8 unique paths.
For a 2 x 5 chessboard, there are 16 unique paths.
For a 2 x 6 chessboard, there are 32 unique paths.
For a 2 x 7 chessboard, there are 64 unique paths.
For a 2 x 8 chessboard, there are 128 unique paths.
For a 3 x 3 chessboard, there are 12 unique paths.
For a 3 x 4 chessboard, there are 38 unique paths.
For a 3 x 5 chessboard, there are 125 unique paths.
For a 3 x 6 chessboard, there are 414 unique paths.
For a 3 x 7 chessboard, there are 1369 unique paths.
For a 3 x 8 chessboard, there are 4522 unique paths.
For a 4 x 4 chessboard, there are 184 unique paths.
For a 4 x 5 chessboard, there are 976 unique paths.
For a 4 x 6 chessboard, there are 5382 unique paths.
For a 4 x 7 chessboard, there are 29739 unique paths.
For a 4 x 8 chessboard, there are 163496 unique paths.
For a 5 x 5 chessboard, there are 8512 unique paths.
For a 5 x 6 chessboard, there are 79384 unique paths.
For a 5 x 7 chessboard, there are 752061 unique paths.
/***********************
* RookPaths.c
* By: David Blume
* d...@wdl1.wdl.loral.com (Predecrement David)
*
* How many unique ways can a rook move from the bottom left corner
* of a m * n chess board to the top right right?
*
* Contraints: The rook may not passover a square it has already visited.
* What if we don't allow Down & Left moves? (much easier)
*
* This software is provided *as is.* It is not guaranteed to work on
* any platform at any time anywhere. :-) Enjoy.
***********************/
#include <stdio.h>
#define kColumns 8 /* The maximum number of columns */
#define kRows 8 /* The maximum number of rows */
/* The following rule is for you to play with. */
#define kAllowDownLeft /* Whether or not to allow the rook to move D or L */
/* Visual feedback defines... */
#undef kShowTraversals /* Whether or nor to show each successful graph */
#define kListResults /* Whether or not to list each n * m result */
#define kShowMatrix /* Display the final results in a matrix? */
char gmatrix[kColumns][kRows]; /* the working matrix */
long result[kColumns][kRows]; /* the result array */
/*********************
* traverse
*
* This is the recursive function
*********************/
traverse (short c, short r, short i, short j )
{
/* made it to the top left! increase result, retract move, and return */
if (i == c && j == r) {
#ifdef kShowTraversals
short ti, tj;
gmatrix[i][j] = 5;
for (ti = c; ti >= 0; ti--) {
for (tj = 0; tj <= r; tj++) {
if (gmatrix[ti][tj] == 0)
printf(". ");
else if (gmatrix[ti][tj] == 1)
printf("D ");
else if (gmatrix[ti][tj] == 2)
printf("R ");
else if (gmatrix[ti][tj] == 3)
printf("L ");
else if (gmatrix[ti][tj] == 4)
printf("U ");
else if (gmatrix[ti][tj] == 5)
printf("* ");
}
printf("\n");
}
printf("\n");
#endif
result[i][j] = result[i][j] + 1;
gmatrix[i][j] = 0;
return;
}
/* try to move, left up down right */
#ifdef kAllowDownLeft
if (i != 0 && gmatrix[i-1][j] == 0) { /* left turn */
gmatrix[i][j] = 1;
traverse(c, r, i-1, j);
}
#endif
if (j != r && gmatrix[i][j+1] == 0) { /* turn up */
gmatrix[i][j] = 2;
traverse(c, r, i, j+1);
}
#ifdef kAllowDownLeft
if (j != 0 && gmatrix[i][j-1] == 0) { /* turn down */
gmatrix[i][j] = 3;
traverse(c, r, i, j-1);
}
#endif
if (i != c && gmatrix[i+1][j] == 0) { /* turn right */
gmatrix[i][j] = 4;
traverse(c, r, i+1, j);
}
/* remove the marking on this square */
gmatrix[i][j] = 0;
}
main()
{
short i, j;
/* initialize the matrix to 0 */
for (i = 0; i < kColumns; i++) {
for (j = 0; j < kRows; j++) {
gmatrix[i][j] = 0;
}
}
/* call the recursive function */
for (i = 0; i < kColumns; i++) {
for (j = 0; j < kRows; j++) {
if (j >= i) {
result[i][j] = 0;
traverse (i, j, 0, 0);
#ifdef kListResults
printf("For a %d x %d chessboard, there are %d unique paths.\n",
i+1, j+1, result[i][j]); fflush(stdout);
#endif
}
}
}
/* print out the results */
#ifdef kShowMatrix
printf("\n");
for (i = 0; i < kColumns; i++) {
for (j = 0; j < kRows; j++) {
short min = (i < j) ? i : j ;
short max = (i < j) ? j : i ;
printf("%6d", result[min][max]);
}
printf("\n");
}
#endif
}
==> competition/games/chess/size.of.game.tree.p <==
How many different positions are there in the game tree of chess?
==> competition/games/chess/size.of.game.tree.s <==
Consider the following assignment of bit strings to square states:
Square State Bit String
------ ----- --- ------
Empty 0
White Pawn 100
Black Pawn 101
White Rook 11111
Black Rook 11110
White Knight 11101
Black Knight 11100
White Bishop 11011
Black Bishop 11010
White Queen 110011
Black Queen 110010
White King 110001
Black King 110000
Record a position by listing the bit string for each of the 64 squares.
For a position with all the pieces still on the board, this will take
164 bits. As pieces are captured, the number of bits needed goes down.
As pawns promote, the number of bits go up. For positions where a King
and Rook are in position to castle if castling is legal, we will need
a bit to indicate if in fact castling is legal. Same for positions
where an en-passant capture may be possible. I'm going to ignore these
on the grounds that a more clever encoding of a position than the one
that I am proposing could probably save as many bits as I need for these
considerations, and thus conjecture that 164 bits is enough to encode a
chess position.
This gives an upper bound of 2^164 positions, or 2.3x10^49 positions.
Jurg Nievergelt, of ETH Zurich, quoted the number 2^70 (or about 10^21) in
e-mail, and referred to his paper "Information content of chess positions",
ACM SIGART Newsletter 62, 13-14, April 1977, to be reprinted in "Machine
Intelligence" (ed Michie), to appear 1990.
Note that this latest estimate, 10^21, is not too intractable:
10^7 computers running at 10^7 positions per second could scan those
in 10^7 seconds, which is less than 6 months.
In fact, suppose there is a winning strategy in chess for white.
Suppose further that the strategy starts from a strong book opening,
proceeds through middle game with only moves that Deep Thought (DT)
would pick using the singular extension technique, and finally ends in
an endgame that DT can analyze completely. The book opening might take
you ten moves into the game and DT has demonstrated its ability to
analyze mates-in-20, so how many nodes would DT really have to visit?
I suggest that by using external storage such a optical WORM memory,
you could easily build up a transposition table for such a midgame. If
DT did not find a mate, you could progressively expand the width of the
search window and add to the table until it did. Of course there would
be no guarantee of success, but the table built would be useful
regardless. Also, you could change the book opening and add to the
table. This project could continue indefinitely until finally it must
solve the game (possibly using denser and denser storage media as
technology advances).
What do you think?
-------
I think you are a little bit too optimistic about the feasibility. Solving
mate-in-19 when the moves are forcing is one thing, but solving mate-in-19
when the moves are not forcing is another. Of course, human beings are no
better at the latter task. But to solve the game in the way you described
would seem to require the ability to handle the latter task. Anyway, we
cannot really think about doing the sort of thing you described; DT is just a
poor man's chess machine project (relatively speaking).
--Hsu
i dont think that you understand the numbers involved.
the size of the tree is still VERY large compared to all
the advances that you cite. (speed of DT, size of worms,
endgame projects, etc) even starting a project will probably
be a waste of time since the next advance will overtake it
rather than augment it. (if you start on a journey to the
stars today, you will be met there by humans)
ken
==> competition/games/cigarettes.p <==
The game of cigarettes is played as follows:
Two players take turns placing a cigarette on a circular table. The cigarettes
can be placed upright (on end) or lying flat, but not so that it touches any
other cigarette on the table. This continues until one person loses by not
having a valid position on the table to place a cigarette.
Is there a way for either of the players to guarantee a win?
==> competition/games/cigarettes.s <==
The first person wins by placing a cigarette at the center of the table,
and then placing each of his cigarettes in a position symmetric (with
respect to the center) to the place the second player just moved. If the
second player could move, then symmetrically, so can the first player.
==> competition/games/connect.four.p <==
Is there a winning strategy for Connect Four?
==> competition/games/connect.four.s <==
An AI program has solved Connect Four for the standard 7 x 6 board.
The conclusion: White wins, was confirmed by the brute force check made by
James D. Allen, which has been published in rec.games.programmer.
The program called VICTOR consists of a pure knowledge-based evaluation
function which can give three values to a position:
1 won by white,
0 still unclear.
-1 at least a draw for Black,
This evaluation function is based on 9 strategic rules concerning the game,
which all nine have been (mathematically) proven to be correct.
This means that a claim made about the game-theoretical value of a position
by VICTOR, is correct, although no search tree is built.
If the result 1 or -1 is given, the program outputs a set of rules applied,
indicating the way the result can be achieved.
This way one evaluation can be used to play the game to the end without any
extra calculation (unless the position was still unclear, of course).
Using the evaluation function alone, it has been shown that Black can at least
draw the game on any 6 x (2n) board. VICTOR found an easy strategy for
these boardsizes, which can be taught to anyone within 5 minutes. Nevertheless,
this strategy had not been encountered before by any humans, as far as I know.
For 7 x (2n) boards a similar strategy was found, in case White does not
start the game in the middle column. In these cases Black can therefore at
least draw the game.
Furthermore, VICTOR needed only to check a few dozen positions to show
that Black can at least draw the game on the 7 x 4 board.
Evaluation of a position on a 7 x 4 or 7 x 6 board costs between 0.01 and 10
CPU seconds on a Sun4.
For the 7 x 6 board too many positions were unclear. For that reason a
combination of Conspiracy-Number Search and Depth First Search was used
to determine the game-theoretical value. This took several hundreds of hours
on a Sun4.
The main reason for the large amount of search needed, was the fact that in
many variations, the win for White was very difficult to achieve.
This caused many positions to be unclear for the evaluation function.
Using the results of the search, a database will be constructed
of roughly 500.000 positions with their game-theoretical value.
Using this datebase, VICTOR can play against humans or other programs,
winning all the time (playing White). The average move takes less
than a second of calculation (search in the database or evaluation
of the position by the evaluation function).
Some variations are given below (columns and rows are numbered as is customary
in chess):
1. d1, .. The only winning move.
After 1. .., a1 wins 2. e1. Other second moves for White has not been
checked yet.
After 1. .., b1 wins 2. f1. Other second moves for White has not been
checked yet.
After 1. .., c1 wins 2. f1. Only 2 g1 has not been checked yet. All other
second moves for White give Black at least a draw.
After 1. .., d2 wins 2. d3. All other second moves for White give black
at least a draw.
A nice example of the difficulty White has to win:
1. d1, d2
2. d3, d4
3. d5, b1
4. b2!
The first three moves for White are forced, while alternatives at the
fourth moves of White are not checked yet.
A variation which took much time to check and eventually turned out
to be at least a draw for Black, was:
1. d1, c1
2. c2?, .. f1 wins, while c2 does not.
2. .., c3 Only move which gives Black the draw.
3. c4, .. White's best chance.
3. .., g1!! Only 3 .., d2 has not been checked completely, while all
other third moves for Black have been shown to lose.
The project has been described in my 'doctoraalscriptie' (Master thesis)
which has been supervised by Prof.Dr H.J. van den Herik of the
Rijksuniversiteit Limburg (The Netherlands).
I will give more details if requested.
Victor Allis.
Vrije Universiteit van Amsterdam.
The Netherlands.
vic...@cs.vu.nl
==> competition/games/craps.p <==
What are the odds in craps?
==> competition/games/craps.s <==
The game of craps:
There is a person who rolls the two dice, and then there is the house.
1) On the first roll, if a 7 or 11 comes up, the roller wins.
If a 2, 3, or 12 comes up the house wins.
Anything else is a POINT, and more rolling is necessary, as per rule 2.
2) If a POINT appears on the first roll, keep rolling the dice.
At each roll, if the POINT appears again, the roller wins.
At each roll, if a 7 comes up, the house wins.
Keep rolling until the POINT or a 7 comes up.
Then there are the players, and they are allowed to place their bets with
either the roller or with the house.
-----
My computations:
On the first roll, P.roller.trial(1) = 2/9, and P.house.trial(1) = 1/9.
Let P(x) stand for the probability of a 4,5,6,8,9,10 appearing.
Then on the second and onwards rolls, the probability is:
Roller:
--- (i - 2)
P.roller.trial(i) = \ P(x) * ((5/6 - P(x)) * P(x)
(i > 1) /
---
x = 4,5,6,8,9,10
House:
--- (i - 2)
P.house.trial(i) = \ P(x) * ((5/6 - P(x)) * 1/6
(i > 1) /
---
x = 4,5,6,8,9,10
Reasoning (roller): For the roller to win on the ith trial, a POINT
should have appeared on the first trial (the first P(x) term), and the
same POINT should appear on the ith trial (the last P(x) term). All the in
between trials should come up with a number other than 7 or the POINT
(hence the (5/6 - P(x)) term).
Similar reasoning holds for the house.
The numbers are:
P.roller.trial(i) (i > 1) =
(i-1) (i-1) (i-1)
1/72 * (27/36) + 2/81 * (26/36) + 25/648 * (25/36)
P.house.trial(i) (i > 1) =
(i-1) (i-1) (i-1)
2/72 * (27/36) + 3/81 * (26/36) + 30/648 * (25/36)
-------------------------------------------------
The total probability comes to:
P.roller = 2/9 + (1/18 + 4/45 + 25/198) = 0.4929292929292929..
P.house = 1/9 + (1/9 + 2/15 + 15/99) = 0.5070707070707070..
which is not even.
===========================================================================
==
Avinash Chopde (with standard disclaimer)
a...@unhcs.unh.edu, a...@unh.unh.edu {.....}!uunet!unh!abc
==> competition/games/crosswords.p <==
Are there programs to make crosswords? What are the rules for cluing cryptic
crosswords? Is there an on-line competition for cryptic cluers?
==> competition/games/crosswords.s <==
You need to read the rec.puzzles.crosswords FAQL.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> competition/games/cube.p <==
What are some games involving cubes?
==> competition/games/cube.s <==
Johan Myrberger's list of 3x3x3 cube puzzles (version 930222)
Comments, corrections and contributions are welcome!
MAIL: myrb...@e.kth.se
Snailmail: Johan Myrberger
Hokens gata 8 B
S-116 46 STOCKHOLM
SWEDEN
A: Block puzzles
A.1 The Soma Cube
______ ______ ______ ______
|\ \ |\ \ |\ \ |\ \
| \_____\ | \_____\ | \_____\ | \_____\
| | |____ _____| | | | | |____ | | |____
|\| | \ |\ \| | |\| | \ |\| | \
| *_____|_____\ | \_____*_____| | *_____|_____\ | *_____|_____\
| |\ \ | | |\ \ | | | |\ \ | | | |
\| \_____\ | \| \_____\ | \| | \_____\ \| | |
* | |___| * | |___| *_____| | | *_____|_____|
\| | \| | \| |
*_____| *_____| *_____|
______ ______ ____________
|\ \ |\ \ |\ \ \
| \_____\ | \_____\ | \_____\_____\
| | |__________ _____| | |____ _____| | | |
|\| | \ \ |\ \| | \ |\ \| | |
| *_____|_____\_____\ | \_____*_____|_____\ | \_____*_____|_____|
| | | | | | | | | | | | | |
\| | | | \| | | | \| | |
*_____|_____|_____| *_____|_____|_____| *_____|_____|
A.2 Half Hour Puzzle
______ ______ ______
|\ \ |\ \ |\ \
| \_____\ | \_____\ | \_____\
| | |__________ _____| | |____ | | |__________
|\| | \ \ |\ \| | \ |\| | \ \
| *_____|_____\_____\ | \_____*_____|_____\ | *_____|_____\_____\
| | | | | | | | | | | | |\ \ |
\| | | | \| | | | \| | \_____\ |
*_____|_____|_____| *_____|_____|_____| *_____| | |___|
\| |
*_____|
______ ______ ______
|\ \ |\ \ |\ \
| \_____\ | \_____\ | \_____\
_____| | | _____| | | | | |
|\ \| | |\ \| | |\| |
| \_____*_____| | \_____*_____|______ ___|_*_____|______
| |\ \ | | | |\ \ \ |\ \ \ \
\| \_____\ | \| | \_____\_____\ | \_____\_____\_____\
* | |___| *_____| | | | | | | | |
\| | \| | | \| | | |
*_____| *_____|_____| *_____|_____|_____|
A.3 Steinhaus's dissected cube
______ ______ ______
|\ \ |\ \ |\ \
| \_____\ | \_____\ | \_____\
| | |__________ _____| | | | | |____
|\| | \ \ |\ \| | |\| | \
| *_____|_____\_____\ | \_____*_____| | *_____|_____\
| | | | | | |\ \ | | | |\ \
\| | | | \| \_____\ | \| | \_____\
*_____|_____|_____| * | |___| *_____| | |
\| | \| |
*_____| *_____|
____________ ______ ______
|\ \ \ |\ \ |\ \
| \_____\_____\ | \_____\ | \_____\
| | | | | | | ___________| | |
\| | | |\| | |\ \ \| |
*_____|_____|______ _________|_*_____| | \_____\_____*_____|
\ |\ \ \ |\ \ \ \ | | |\ \ |
\| \_____\_____\ | \_____\_____\_____\ \| | \_____\ |
* | | | | | | | | *_____| | |___|
\| | | \| | | | \| |
*_____|_____| *_____|_____|_____| *_____|
A.4
______
|\ \
| \_____\
| | |____ Nine of these make a 3x3x3 cube.
|\| | \
| *_____|_____\
| | | |
\| | |
*_____|_____|
A.5
______ ____________
|\ \ |\ \ \
| \_____\ | \_____\_____\
____________ | | |____ | | | |
|\ \ \ |\| | \ |\| | |
| \_____\_____\ | *_____|_____\ | *_____|_____|
| | | | | | | | | | | |
\| | | \| | | \| | |
*_____|_____| *_____|_____| *_____|_____|
______ ______
|\ \ |\ \
| \_____\ | \_____\
______ ______ | | |____ | | |__________
|\ \ |\ \ |\| | \ |\| | \ \
| \_____\ | \_____\ | *_____|_____\ | *_____|_____\_____\
| | |___| | | | | | |____ | | | | |
|\| | \| | |\| | | \ |\| | | |
| *_____|_____*_____| | *_____|_____|_____\ | *_____|_____|_____|
| | | | | | | | | | | | | | |
\| | | | \| | | | \| | | |
*_____|_____|_____| *_____|_____|_____| *_____|_____|_____|
A.6
______ ______ ______ ______
|\ \ |\ \ |\ \ |\ \
| \_____\ | \_____\ | \_____\ | \_____\
| | |____ _____| | | | | |____ | | |____
|\| | \ |\ \| | |\| | \ |\| | \
| *_____|_____\ | \_____*_____| | *_____|_____\ | *_____|_____\
| |\ \ | | |\ \ | | | |\ \ | | | |
\| \_____\ | \| \_____\ | \| | \_____\ \| | |
* | |___| * | |___| *_____| | | *_____|_____|
\| | \| | \| |
*_____| *_____| *_____|
______ ____________ ____________
|\ \ |\ \ \ |\ \ \
| \_____\ | \_____\_____\ | \_____\_____\
_____| | |____ _____| | | | _____| | | |
|\ \| | \ |\ \| | | |\ \| | |
| \_____*_____|_____\ | \_____*_____|_____| | \_____*_____|_____|
| | | | | | | | | | | | |
\| | | | \| | | \| | |
*_____|_____|_____| *_____|_____| *_____|_____|
A.7
____________
|\ \ \
| \_____\_____\
| | | |
|\| | | Six of these and three unit cubes make a 3x3x3 cube.
| *_____|_____|
| | | |
\| | |
*_____|_____|
A.8 Oskar's
____________ ______
|\ \ \ |\ \
| \_____\_____\ | \_____\
_____| | | | | | |__________ __________________
|\ \| | | |\| | \ \ |\ \ \ \
| \_____*_____|_____| x 5 | *_____|_____\_____\ | *_____\_____\_____\
| | | | | | | | | | | | | |
\| | | \| | | | \| | | |
*_____|_____| *_____|_____|_____| *_____|_____|_____|
A.9 Trikub
____________ ______ ______
|\ \ \ |\ \ |\ \
| \_____\_____\ | \_____\ | \_____\
| | | | | | |__________ _____| | |____
|\| | | |\| | \ \ |\ \| | \
| *_____|_____| | *_____|_____\_____\ | \_____*_____|_____\
| | | | | | | | | | | | | |
\| | | \| | | | \| | | |
*_____|_____| *_____|_____|_____| *_____|_____|_____|
______ ______ ____________
|\ \ |\ \ |\ \ \
| \_____\ | \_____\ | \_____\_____\
| | |____ | | |____ _____| | | |
|\| | \ |\| | \ |\ \| | |
| *_____|_____\ | *_____|_____\ | \_____*_____|_____|
| |\ \ | | | |\ \ | | | |
\| \_____\ | \| | \_____\ \| | |
* | |___| *_____| | | *_____|_____|
\| | \| |
*_____| *_____|
and three single cubes in a different colour.
The object is to build 3x3x3 cubes with the three single cubes in various
positions.
E.g: * * * as center * * * as edge * *(3) as * *(2) as
* S * * * * *(2)* space *(2)* center
* * * * * S (1)* * diagonal (2)* * diagonal
(The other two variations with the single cubes in a row are impossible)
A.10
______ ______ ______
|\ \ |\ \ |\ \
| \_____\ | \_____\ | \_____\
_____| | | | | |____ | | |____
|\ \| | |\| | \ |\| | \
| \_____*_____| | *_____|_____\ ___|_*_____|_____\
| |\ \ | | | |\ \ |\ \ \ |
\| \_____\ | \| | \_____\ | \_____\_____\ |
* | |___| *_____| | | | | | |___|
\| | \| | \| | |
*_____| *_____| *_____|_____|
______ ______ ______
|\ \ |\ \ |\ \
| \_____\ | \_____\ | \_____\
| | |__________ _____| | |____ | | |____
|\| | \ \ |\ \| | \ |\| | \
| *_____|_____\_____\ | \_____*_____|_____\ | *_____|_____\______
| |\ \ | | | | | | | | | |\ \ \
\| \_____\ | | \| | | | \| | \_____\_____\
* | |___|_____| *_____|_____|_____| *_____| | | |
\| | \| | |
*_____| *_____|_____|
B: Coloured blocks puzzles
B.1 Kolor Kraze
Thirteen pieces.
Each subcube is coloured with one of nine colours as shown below.
The object is to form a cube with nine colours on each face.
______
|\ \
| \_____\
| | | ______ ______ ______ ______ ______ ______
|\| 1 | |\ \ |\ \ |\ \ |\ \ |\ \ |\ \
| *_____| | \_____\ | \_____\ | \_____\ | \_____\ | \_____\ | \_____\
| | | | | | | | | | | | | | | | | | | | |
|\| 2 | |\| 2 | |\| 2 | |\| 4 | |\| 4 | |\| 7 | |\| 9 |
| *_____| | *_____| | *_____| | *_____| | *_____| | *_____| | *_____|
| | | | | | | | | | | | | | | | | | | | |
\| 3 | \| 3 | \| 1 | \| 1 | \| 5 | \| 5 | \| 5 |
*_____| *_____| *_____| *_____| *_____| *_____| *_____|
______ ______ ______ ______ ______ ______
|\ \ |\ \ |\ \ |\ \ |\ \ |\ \
| \_____\ | \_____\ | \_____\ | \_____\ | \_____\ | \_____\
| | | | | | | | | | | | | | | | | |
|\| 9 | |\| 9 | |\| 3 | |\| 6 | |\| 6 | |\| 6 |
| *_____| | *_____| | *_____| | *_____| | *_____| | *_____|
| | | | | | | | | | | | | | | | | |
\| 7 | \| 8 | \| 8 | \| 8 | \| 7 | \| 4 |
*_____| *_____| *_____| *_____| *_____| *_____|
B.2
Given nine red, nine blue and nine yellow cubes.
Form a 3x3x3 cube in which all three colours appears in each of the 27
orthogonal rows.
B.3
Given nine red, nine blue and nine yellow cubes.
Form a 3x3x3 cube so that every row of three (the 27 orthogonal rows, the 18
diagonal rows on the nine square cross-sections and the 4 space diagonals)
contains neither three cubes of like colour nor three of three different
colours.
B.4
Nine pieces, three of each type.
Each subcube is coloured with one of three colours as shown below.
The object is to build a 3x3x3 cube in which all three colours appears in each
of the 27 orthogonal rows. (As in B.2)
______ ______ ______
|\ \ |\ \ |\ \
| \_____\ | \_____\ | \_____\
| | |____ | | |____ | | |____
|\| A | \ x 3 |\| B | \ x 3 |\| A | \ x 3
| *_____|_____\ | *_____|_____\ | *_____|_____\
| | | | | | | | | | | |
\| B | C | \| A | C | \| C | B |
*_____|_____| *_____|_____| *_____|_____|
C: Strings of cubes
C.1 Pululahua's dice
27 cubes are joined by an elastic thread through the centers of the cubes
as shown below.
The object is to fold the structure to a 3x3x3 cube.
____________________________________
|\ \ \ \ \ \ \
| \_____\_____\_____\_____\_____\_____\
| | | | | | | |
|\| :77|77777|77: | :77|77777|77: |
| *__:__|_____|__:__|__:__|_____|__:__|
| | : |___| | : | : |___| | : |
|\| : | \| 777|777 | \| : |
| *__:__|_____*_____|_____|_____*__:__|
| | : | | |___| | | : |____
\| 777|77777|77: | \| :77|777 | \
*_____|_____|__:__|_____*__:__|_____|_____\
| | : | | : | | |
|\| : | + | 777|77777|77: |
| *__:__|__:__|_____|_____|__:__|
| | : | : | | | : |
\| + | : | :77|77777|777 |
*_____|__:__|__:__|_____|_____|
| | : | : |
\| 777|777 |
*_____|_____|
C.1.X The C.1 puzzle type exploited by Glenn A. Iba (quoted)
INTRODUCTION
"Chain Cube" Puzzles consist of 27 unit cubies
with a string running sequentially through them. The
string always enters and exits a cubie through the center
of a face. The typical cubie has one entry and one exit
(the ends of the chain only have 1, since the string terminates
there). There are two ways for the string to pass through
any single cubie:
1. The string enters and exits non-adjacent faces
(i.e. passes straight through the cubie)
2. It enters and exits through adjacent faces
(i.e. makes a "right angle" turn through
the cubie)
Thus a chain is defined by its sequence of straight steps and
right angle turns. Reversing the sequence (of course) specifies
the same chain since the chain can be "read" starting from either
end. Before making a turn, it is possible to "pivot" the next
cubie to be placed, so there are (in general) 4 choices of
how to make a "Turn" in 3-space.
The object is to fold the chain into a 3x3x3 cube.
It is possible to prove that any solvable sequence must
have at least 2 straight steps. [The smallest odd-dimensioned
box that can be packed by a chain of all Turns and no Straights
is 3x5x7. Not a 3x3x3 puzzle, but an interesting challenge.
The 5x5x5 can be done too, but its not the smallest in volume].
With the aid of a computer search program I've produced
a catalog of the number of solutions for all (solvable) sequences.
Here is an example sequence that has a unique solution (up to reflections
and rotations):
(2 1 1 2 2 1 1 1 1 1 1 1 1 1 1 1 1 2 2 1 1)
the notation is a kind of "run length" coding,
where the chain takes the given number of steps in a straight line,
then make a right-angle turn. Equivalently, replace
1 by Turn,
2 by Straight followed by a Turn.
The above sequence was actually a physical puzzle I saw at
Roy's house, so I recorded the sequence, and verified (by hand and computer)
that the solution is unique.
There are always 26 steps in a chain, so the "sum" of the
1's and 2's in a pattern will always be 26. For purposes
of taxonomizing, the "level" of a string pattern is taken
to be the number of 2's occuring in its specification.
COUNTS OF SOLVABLE AND UNIQUELY SOLVABLE STRING PATTERNS
(recall that Level refers to the number of 2's in the chain spec)
Level Solvable Uniquely
Patterns Solvable
0 0 0
1 0 0
2 24 0
3 235 15
4 1037 144
5 2563 589
6 3444 1053
7 2674 1078
8 1159 556
9 303 187
10 46 34
11 2 2
12 0 0
13 0 0
_______ ______
Total Patterns: 11487 3658
SOME SAMPLE UNIQUELY SOLVABLE CHAINS
In the following the format is:
( #solutions palindrome? #solutions-by-start-type chain-pattern-as string )
where
#solutions is the total number of solutions up to reflections and rotations
palindrome? is T or NIL according to whether or not the chain is a palindrome
#solutions by-start-type lists the 3 separate counts for the number of
solutions starting the chain of in the 3 distinct possible ways.
chain-pattern-as-string is simply the chain sequence
My intuition is that the lower level chains are harder to solve,
because there are fewer straight steps, and staight steps are generally
more constraining. Another way to view this, is that there are more
choices of pivoting for turns because there are more turns in the chains
at lower levels.
Here are the uniquely solvable chains for level 3:
(note that non-palindrome chains only appear once --
I picked a "canonical" ordering)
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;;; Level 3 ( 3 straight steps) -- 15 uniquely solvable patterns
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
(1 NIL (1 0 0) "21121111112111111111111")
(1 NIL (1 0 0) "21121111111111111121111")
(1 NIL (1 0 0) "21111112112111111111111")
(1 NIL (1 0 0) "21111111211111111111112")
(1 NIL (1 0 0) "12121111111111112111111")
(1 NIL (1 0 0) "11211211112111111111111")
(1 NIL (1 0 0) "11112121111111211111111")
(1 NIL (1 0 0) "11112112112111111111111")
(1 NIL (1 0 0) "11112112111111211111111")
(1 NIL (1 0 0) "11112111121121111111111")
(1 NIL (1 0 0) "11112111111211211111111")
(1 NIL (1 0 0) "11112111111112121111111")
(1 NIL (1 0 0) "11111121122111111111111")
(1 NIL (1 0 0) "11111112122111111111111")
(1 NIL (1 0 0) "11111111221121111111111")
C.2 Magic Interlocking Cube
(Glenn A. Iba quoted)
This chain problem is marketed as "Magic Interlocking Cube --
the Ultimate Cube Puzzle". It has colored cubies, each cubie having
6 distinctly colored faces (Red, Orange, Yellow, Green, Blue, and White).
The object is to fold the chain into a 3x3x3 cube with each face
being all one color (like a solved Rubik's cube). The string for
the chain is actually a flexible rubber band, and enters a cubie
through a (straight) slot that cuts across 3 faces, and exits
through another slot that cuts the other 3 faces. Here is a rough
attempt to picture a cubie:
(the x's mark the slots cut for the rubber band to enter/exit)
__________
/ /|
xxxxxxxxxxx |
/ / x |
/_________/ x |
| | x |
| | |
| | /
| x | /
| x | /
| x |/
-----x-----
Laid out flat the cubie faces would look like this:
_________
| |
| |
| x |
| x |
|____x____|_________ _________ _________
| x | | | |
| x | | | |
| x | x x x x x x x x x x x |
| x | | | |
|____x____|_________|_________|_________|
| x |
| x |
| x |
| |
|_________|
The structure of the slots gives 3 choices of entry face, and 3 choices
of exit face for each cube.
It's complicated to specify the topology and coloring but here goes:
Imagine the chain stretched out in a straight line from left to right.
Let the rubber band go straight through each cubie, entering and
exiting in the "middle" of each slot.
It turns out that the cubies are colored so that opposite faces are
always colored by the following pairs:
Red-Orange
Yellow-White
Green-Blue
So I will specify only the Top, Front, and Left colors of each
cubie in the chain. Then I'll specify the slot structure.
Color sequences from left to right (colors are R,O,Y,G,B,and W):
Top: RRRRRRRRRRRRRRRRRRRRRRRRRRR
Front: YYYYYYYYYYYYWWWYYYYYYYYYYYY
Left: BBBBBGBBBGGGGGGGGGBBGGGGBBB
For the slots, all the full cuts are hidden, so only
the "half-slots" appear.
Here is the sequence of "half slots" for the Top (Red) faces:
(again left to right)
Slots: ><><><><<><><<<><><>>>>><>>
Where
> = slot goes to left
< = slot goes to right
To be clearer,
> (Left):
_______
| |
| |
xxxxx |
| |
|_______|
< (Right):
_______
| |
| |
| xxxxx
| |
|_______|
Knowing one slot of a cubie determines all the other slots.
I don't remember whether the solution is unique. In fact I'm not
certain whether I actually ever solved it. I think I did, but I don't
have a clear recollection.
D: Blocks with pins
D.1 Holzwurm (Torsten Sillke quoted)
Inventer: Dieter Matthes
Distribution:
Pyramo-Spiele-Puzzle
Silvia Heinz
Sendbuehl 1
D-8351 Bernried
tel: +49-9905-1613
Pieces: 9 tricubes
Each piece has one hole (H) which goes through the entire cube.
The following puctures show the tricubes from above. The faces
where you see a hole are marked with 'H'. If you see a hole at
the top then there is a hole at the bottom too. Each peace has
a worm (W) one one face. You have to match the holes and the
worms. As a worm fills a hole completely, you can not put two
worms at both ends of the hole of the same cube.
__H__ _____ _____
| | | | | |
| | | |W | |
|_____|_____ |_____|_____ |_____|_____
| | | | | | | | |
| | |W | | |H | H | |W
|_____|_____| |_____|_____| |_____|_____|
__H__ _____ _____
| | | | | |
| | | | | W |
|_____|_____ |_____|_____ |_____|_____
| | | | | | | | |
| | | | W | H | | | H |
|_____|_____| |_____|_____| |_____|_____|
W
__W__ _____ _____
| | | | | |
| | H| |H | |
|_____|_____ |_____|_____ |_____|_____
| | | | | | | | |
| | H | | | | H| | W |
|_____|_____| |_____|_____| |_____|_____|
W
Aim: build a 3*3*3 cube without a worm looking outside.
take note, it is no matching problem, as
| |
worm> H| |H <worm
| |
is not allowed.
E: Other
E.1 Rubik's cube
E.2 Magic cube
Make a magic cube with the numbers 1 - 27.
E.3 ==> geometry/coloring/cheese.cube.p <==
A cube of cheese is divided into 27 subcubes. A mouse starts at one
corner and eats through every subcube. Can it finish in the middle?
==> geometry/coloring/cheese.cube.s <==
Give the subcubes a checkerboard-like coloring so that no two adjacent
subcubes have the same color. If the corner subcubes are black, the
cube will have 14 black subcubes and 13 white ones. The mouse always
alternates colors and so must end in a black subcube. But the center
subcube is white, so the mouse can't end there.
E.4
Cut the 3*3*3 cube into single cubes. At each slice you can
rearrange the blocks. Can you do it with fewer than 6 cuts?
==> competition/games/go-moku.p <==
For a game of k in a row on an n x n board, for what values of k and n is
there a win? Is (the largest such) k eventually constant or does it increase
with n?
==> competition/games/go-moku.s <==
Berlekamp, Conway, and Guy's _Winning_Ways_ reports proof that the
maximum k is between 4 and 7 inclusive, and it appears to be 5 or 6.
They report:
. 4-in-a-row is a draw on a 5x5 board (C. Y. Lee), but not on a 4x30
board (C. Lustenberger).
. N-in-a-row is shown to be a draw on a NxN board for N>4, using a
general pairing technique devised by A. W. Hales and R. I. Jewett.
. 9-in-a-row is a draw even on an infinite board, a 1954 result of H. O.
Pollak and C. E. Shannon.
. More recently, the pseudonymous group T. G. L. Zetters showed that
8-in-a-row is a draw on an infinite board, and have made some
progress on showing infinite 7-in-a-row to be a draw.
Go-moku is 5-in-a-row played on a 19x19 go board. It is apparently a
win for the first player, and so the Japanese have introduced several
'handicaps' for the first player (e.g., he must win with _exactly_
five: 6-in-a-row doesn't count), but apparently the game is still a win
for the first player. None of these apparent results have been
proven.
==> competition/games/hi-q.p <==
What is the quickest solution of the game Hi-Q (also called Solitaire)?
For those of you who aren't sure what the game looks like:
32 movable pegs ("+") are arranged on the following board such that
only the middle position is empty ("-"). Just to be complete: the board
consists of only these 33 positions.
1 2 3 4 5 6 7
1 + + +
2 + + +
3 + + + + + + +
4 + + + - + + +
5 + + + + + + +
6 + + +
7 + + +
A piece moves on this board by jumping over one of its immediate
neighbor (horizontally or vertically) into an empty space opposite.
The peg that was jumped over, is hit and removed from the board. A
move can contain multiple hits if you use the same peg to make the
hits.
You have to end with one peg exactly in the middle position (44).
==> competition/games/hi-q.s <==
1: 46*44
2: 65*45
3: 57*55
4: 54*56
5: 52*54
6: 73*53
7: 43*63
8: 75*73*53
9: 35*55
10: 15*35
11: 23*43*63*65*45*25
12: 37*57*55*53
13: 31*33
14: 34*32
15: 51*31*33
16: 13*15*35
17: 36*34*32*52*54*34
18: 24*44
Found by Ernest Bergholt in 1912 and was proved to be minimal by John Beasley
in 1964.
References
The Ins and Outs of Peg Solitaire
John D Beasley
Oxford U press, 1985
ISBN 0-19-853203-2
Winning Ways, Vol. 2, Ch. 23
Berlekamp, E.R.
Academic Press, 1982
ISBN 01-12-091102-7
==> competition/games/jeopardy.p <==
What are the highest, lowest, and most different scores contestants
can achieve during a single game of Jeopardy?
==> competition/games/jeopardy.s <==
highest: $283,200.00, lowest: -$29,000.00, biggest difference: $281,600.00
(1) Our theoretical contestant has an itchy trigger finger, and rings in with
an answer before either of his/her opponents.
(2) The daily doubles (1 in the Jeopardy! round, 2 in the Double Jeopardy!
round) all appear under an answer in the $100 or $200 rows.
(3) All answers given by our contestant are (will be?) correct.
Therefore:
Round 1 (Jeopardy!): Max. score per category: $1500.
For 6 categories - $100 for the DD, that's $8900.
Our hero bets the farm and wins - score: $17,800.
Round 2 (Double Jeopardy!):
Max. score per category: $3000.
Assume that the DDs are found last, in order.
For 6 categories - $400 for both DDs, that's $17,600.
Added to his/her winnings in Round 1, that's $35,400.
After the 1st DD, where the whole thing is wagered,
the contestant's score is $70,800. Then the whole
amount is wagered again, yielding a total of $141,600.
Round 3 (Final Jeopardy!):
Our (very greedy! :) hero now bets the whole thing, to
see just how much s/he can actually win. Assuming that
his/her answer is right, the final amount would be
$283,200.
But the contestant can only take home $100,000; the rest is donated to
charity.
To calculate the lowest possible socre:
-1500 x 6 = -9000 + 100 = -8900.
On the Daily Double that appears in the 100 slot, you bet the maximum
allowed, 500, and lose. So after the first round, you are at -9400.
-3000 x 6 = -18000 + 400 = -17600
On the two Daily Doubles in the 200 slots, bet the maximum allowed, 1000. So
after the second round you are at -9400 + -19600 = -29000. This is the
lowest score you can achieve in Jeopardy before the Final Jeopardy round.
The caveat here is that you *must* be the person sitting in the left-most
seat (either a returning champion or the luckiest of the three people who
come in after a five-time champion "retires") at the beginning of the game,
because otherwise you will not have control of the board when the first
Daily Double comes along.
The scenario for the maximum difference is the same as the highest
score, except that on every question that isn't a daily double, the
worst contestant rings in ahead of the best one, and makes a wrong
guess, after which the best contestant rings in and gets it right.
However, since contestants with negative scores are disqualified before
Final Jeopardy!, it is arguable that the negative score ceases to exist
at that point. This also applies to zero scores. In that case,
someone else would have to qualify for Final Jeopardy! for the maximum
difference to exist, taking one $100 or $200 question away from the
best player. In that case the best player would score 8*$200 lower, so
the maximum difference would be $281,600.00.
==> competition/games/nim.p <==
Place 10 piles of 10 $1 bills in a row. A valid move is to reduce
the last i>0 piles by the same amount j>0 for some i and j; a pile
reduced to nothing is considered to have been removed. The loser
is the player who picks up the last dollar, and they must forfeit
half of what they picked up to the winner.
1) Who is the winner in Waldo Nim, the first or the second player?
2) How much more money than the loser can the winner obtain with best
play on both parties?
==> competition/games/nim.s <==
For the particular game described we only need to consider positions for
which the following condition holds for each pile:
(number of bills in pile k) + k >= (number of piles) + 1
A GOOD position is defined as one in which this condition holds,
with equality applying only to one pile P, and all piles following P
having the same number of bills as P.
( So the initial position is GOOD, the special pile being the first. )
I now claim that if I leave you a GOOD position, and you make any move,
I can move back to a GOOD position.
Suppose there are n piles and the special pile is numbered (n-p+1)
(so that the last p piles each contain p bills).
(1) You take p bills from p or more piles;
(a) If p = n, you have just taken the last bill and lost.
(b) Otherwise I reduce pile (n-p) (which is now the last) to 1 bill.
(2) You take p bills from r(<p) piles;
I take r bills from (p-r) piles.
(3) You take q(<p) bills from p or more piles;
I take (p-q) bills from q piles.
(4) You take q(<p) bills from r(<p) piles;
(a) q+r>p; I take (p-q) bills from (q+r-p) piles
(b) q+r<=p; I take (p-q) bills from (q+r) piles
Verifying that each of the resulting positions is GOOD is tedious
but straightforward. It is left as an exercise for the reader.
-- RobH
==> competition/games/online/online.scrabble.p <==
How can I play Scrabble online on the Internet?
==> competition/games/online/online.scrabble.s <==
Announcing ScrabbleMOO, a server running at 134.53.14.110, port 7777
(nextsrv.cas.muohio.edu 7777). The server software is version 1.7.0
of the LambdaMOO server code.
To reach it, you can use "telnet 134.53.14.110 7777", and sign on. You
will have a unique name and password on the server, and directions are
provided in the opening screen on how to accomplish signing on. The
first time, you will need to type "create YourName YourPassword", and
each time thereafter, "connect YourName YourPassword".
There are currently 5 Scrabble boards set up, with global individual
high score and game-cumulative high score lists. Games can be saved,
and restored at a later time. There are complete command instructions
at each board (via the command "instructions"); usage is simple and
intuitive. There are commands to undo turns, exchange tiles, and pass,
and there are a variety of options available to change the way the
board and rack are displayed.
We do not yet have a dictionary for challenges installed on-line, and
that is coming very soon. I am seriously contemplating using the
OSPD.shar wordlist that Ross Beresford listed in a recent Usenet
article. It seems to have the full wordlist from the 1st edition
of the OSPD, plus longer words from some other source. I have
personal wordlists updating the OSPD to the 2nd edition, for words
up to 4 letters long, and will have the longer words in the near
future.
Usage of a certain dictionary for challenges is not enforced, and
really can't be. Many of the regular players there have their
personal copy of the OSPD. It's all informal, and for fun. Players
agree what dictionary to use on a game-by-game basis, though the
OSPD is encouraged. There are even commands to enable kibitzing,
if watching rather than playing is what you're into.
Come by and try it out. We have all skill levels of players, and
we welcome more!
==> competition/games/online/unlimited.adventures.p <==
Where can I find information about unlimited adventures?
==> competition/games/online/unlimited.adventures.s <==
ccosun.caltech.edu -- pub/adnd/inbound/UA
wuarchive.wustl.edu -- pub/msdos_uploads/games/UA
==> competition/games/othello.p <==
How good are computers at Othello?
==> competition/games/othello.s <==
("Othello" is a registered trademark of the Anjar Company Inc.)
As of 1992, the best Othello programs may have reached or surpassed the
best human players [2][3]. As early as 1980 Jonathon Cerf, then World
Othello Champion, remarked:
"In my opinion the top programs [...] are now equal (if not superior)
to the best human players." [1]
However, Othello's game theoretic value, unlike checkers, will likely
remain unknown for quite some time. Barring some unforeseen shortcut or
bankroll, a perfect Othello playing program would need to search in the
neighborhood of 50 plies. Today, even a general 30 ply search to end the
game, i.e. 30 remaining empty squares, is beyond reach.
Furthermore, the game of Othello does not lend itself to endgame database
techniques that have proven so effective in checkers, and in certain chess
endgames.
Progress of the best Othello computer programs:
1980
"Iago" (by Rosenbloom) [2]
1990
"Bill 3.0" (by Lee and Mahajan) [3] uses:
1. sophisticated searching and timing algorithms, e.g. iterative
deepening, hash/killer tables, zero-window search.
2. lookup tables to encode positional evaluation knowledge.
3. Bayesian learning for the evaluation function.
The average middle game search depth is 8 plies.
Exhaustive endgame search within tournament-play time constraints, is
usually possible with 13 to 15 empty squares remaining.
"Bill 3.0" defeated Brian Rose, the highest rated American Othello
player, by a score of 56-8.
1992
At the 4th AST Computer Olympiad [4][5], the top three programs were:
Othel du Nord (France)
Aida (The Netherlands)
Jacp'Oth (France)
References
----------
[1] Othello Quarterly 3(1) (1981) 12-16
[2] P.S. Rosenbloom, A World Championship-Level Othello Program,
"Artificial Intelligence" 19 (1982) 279-320
[3] Kai-Fu Lee and Sanjoy Mahajan, The Development of a World Class
Othello Program, "Artificial Intelligence" 43 (1990) 21-36
[4] D.M. Breuker and J. Gnodde, The AST 4th Computer Olympiad,
"International Computer Chess Association Journal 15-3 (1992) 152-153
[5] Jos Uiterwijk, The AST 4th Conference on Computer Games,
"International Computer Chess Association Journal 15-3 (1992) 158-161
Myron P. Souris
EDS/Unigraphics
St. Louis, Missouri
sou...@ug.eds.com
==> competition/games/pc/best.p <==
What are the best PC games?
==> competition/games/pc/best.s <==
Read "net pc games top 100" in newsgroup comp.sys.ibm.pc.games.announce.
==> competition/games/pc/reviews.p <==
Are reviews of PC games available online?
==> competition/games/pc/reviews.s <==
Presenting... the Game Bytes Issues Index! (Issues 1-8)
Game Bytes has covered well over 100 games in the past several issues.
Using this index, you can look up the particular games you're interested
in, find out what issues of Game Bytes cover them, and download those
issues. Also included is a list of the interviews GB has done to date -
- the interviews from several issues ago still contain a lot of current
material.
The easiest way to use the games index is to employ the search command
of your favorite word processor to find a distinctive string, such as
"Ultima","Perfect", or "Lemmings". The list is alphabetized; series
have been listed together rather than by individual subtitle.
All issues of Game Bytes to date are available by anonymous FTP at
ftp.ulowell.edu in the /msdos/Games/GameByte directory and are
mirrored on other FTP sites as well. Contact Ross Erickson,
ro...@kaos.b11.ingr.com, if you need assistance acquiring Game
Bytes or have other questions.
Game Bytes Interview List, Issues 1 - 7, Chronological Order
-----------------------------------------------------------------
Issue Person(s) Company Sample Games
----- --------- ------- ------------
2 Richard Garriott Origin Ultima series
3 Chris Roberts Origin Wing Commander, Strike C.
4 ID Software team Apogee* Wolfenstein 3D, Commander Keen
5 Damon Slye Dynamix Red Baron, Aces of the Pacific
5 Scott Miller Apogee Wolf3D, C. Keen, Duke Nukem
6 Bob Bates (Part 1) Legend Spellcasting 101
7 Bob Bates (Part 2) "" ""
8 Looking Glass Tech Origin Underworld 1 and 2
* distributing/producing company
Game Bytes Reviews Index, Issues 1 - 8, Alphabetical by Title
---------------------------------------------------------------------
Title Review Preview Tips
----- ------ ------- ----
A-Train 3
A.T.A.C. 5
Aces of the Pacific 3 1 8
Action Stations! 8
Air Combat 5
Air Force Commander 8
Alien 3 (Sega Genesis) 7
Amazon 8 6
Axelay (Super Nintendo) 8
B-17 Flying Fortress 6 4
B.A.T. II: The Koshan Conspiracy 7
Battlecruiser 3000 A.D. 8
Birds of Prey 7 4
Carrier Strike 6
Carriers at War 6
Castle Wolfenstein 3-D 2
Challenge of the Five Realms 4
Chessmaster 3000 2
Civilization 5
Comanche: Maximum Overkill 6
Conflict: Korea 6
Conquered Kingdoms 7
Conquests of the Longbow 3
Contra 3: The Alien Wars (Super Nintendo) 5
Crisis in the Kremlin 6
D/Generation 2
Dark Sun: Shattered Lands 6
Darklands 7 3 7
Darkseed 5
Dune 3
Dungeon Master 7
Dynamix Football 3
Earl Weaver Baseball 2 4
Ecoquest: The Search for Cetus 2 5
Eric the Unready 8
Eye of the Beholder 2 1
Eye of the Beholder 3 8
F-117A Stealth Fighter 3
F-15 Strike Eagle III 5
Falcon 3.0 1 5,8
Falcon 3.0: Operation Flying Tiger 6
Flight Simulator 4.0 Scenery 8
Front Page Sports: Football 8 6
Galactix 6
Gateway 4
Global Conquest 3
Gods 6
Gravis Gamepad 4
Great Naval Battles 8
Greens! 2
Gunship 2000 2
Hardball 3 4,5
Hardball 3 Statistical Utilities 7
Harpoon 1.3 Designer Series / IOPG 6
Heaven and Earth 4
Heimdall 7
Hong Kong Mahjong 3
Indiana Jones and the Fate of Atlantis 5
Jack Nicklaus Golf: Signature Edition 2
Joe and Mac (SNES) 2
Johnny Castaway 8
King's Quest VI: Heir Today, Gone Tomorrow 6
Laura Bow 2: The Dagger of Amon Ra 4 3
Legends of Valor 8
Les Manley: Lost in L.A. 1
Links 386 Pro 5 1
Links Courses: Troon North 2
Loom -- CD-ROM version 5
Lord of the Rings II: The Two Towers 7 3
Lost Treasures of Infocom 5
Lure of the Temptress 8
Mantis: XF5700 Experimental Space Fighter 7 4
Martian Memorandum 5
Micro League Baseball 4 6
Might and Magic: Clouds of Xeen 8
Mike Ditka's Ultimate Football 6
Monkey Island 2: LeChuck's Revenge 5
NCAA Basketball (Super Nintendo) 8
NCAA: The Road to the Final Four 3
NFL Pro League 7
NHLPA Hockey '93 (Sega Genesis) 7
Nova 9 2
Oh No! More Lemmings 3
Out of This World 6
Pirates! Gold 2
Planet's Edge 3
Pools of Darkness 2
Powermonger 5
Prince of Persia 4
Prophecy of the Shadow 7
Pursue the Pennant 4.0 4
Quest for Glory I (VGA edition) 7
Quest for Glory III: The Wages of War 7
Rampart 4
Rampart (SNES) 7
RBI Baseball 4 (Sega Genesis) 7
Red Baron Mission Builder 8 4
Rex Nebular and the Cosmic Gender Bender 8 5
Risk for Windows 1
Robosport for Windows 8
Rules of Engagement 7
Secret Weapons of the Luftwaffe 4
Sega CD-ROM (Sega Genesis) 8
Sherlock Holmes, Consulting Detective Vol.I 7
Shining in the Darkness (Sega Genesis) 4
Siege 6
SimAnt 4
Solitaire's Journey 5
Sonic the Hedgehog 2 8
Space Megaforce (SNES) 7
Space Quest V: The Next Mutation 3
Speedball 2 5
Spellcasting 301: Spring Break 8 8
Spellcraft: Aspects of Valor 3
Splatterhouse 2 (Sega Genesis) 5
S.S.I. Goldbox summary 8
Star Control 2 8
Star Legions 6
Star Trek: 25th Anniversary 1
Street Fighter 2 8
Strike Commander 3
Stunt Island 8 7
Summer Challenge 8 5
Super Hi-Impact Football (Sega Genesis) 8
Super Star Wars (SNES) 7
Super Tetris 3
Take-a-Break Pinball 6
Tegel's Mercenaries 6
Terminator 2029: Cybergen 5
The 7th Guest 5
The Castle of Dr. Brain 5
The Incredible Machine 7
The Legend of Kyrandia 7
The Lost Admiral 6
The Magic Candle II: The Four and Forty 5
The Miracle 3
The Mystical Quest (SNES) 7
The Perfect General 3
Theatre of War 6
Thrustmaster 4
Thunderhawk 2
TimeQuest 2
Tony La Russa's Ultimate Baseball II 8
Turbo Science 7
Ultima 1, 2, and 3 (First Trilogy) 7
Ultima 7: Forge of Virtue 6 4
Ultima 7: The Black Gate 3 1 5,6
Ultima Underworld: The Stygian Abyss 3 7
Ultima Underworld 2: Labyrinth of Worlds 8
V for Victory: Utah Beach 7
Veil of Darkness 8
WaxWorks 7
Wayne Gretzky Hockey III 5
Wing Commander 2 1
Wing Commander 2: Special Operations 2 4
Winter Challenge 5
Wizardry 6: Bane of the Cosmic Forge 1
Wizardry 7: Crusaders of the Dark Savant 8 5
Wordtris 4
World Circuit 7
X-Wing: Star Wars Space Combat Simulator 7
==> competition/games/pc/solutions.p <==
What are the solutions to various popular PC games?
==> competition/games/pc/solutions.s <==
Solutions, hints, etc. for many games exist at:
pub/game_solutions directory on sun0.urz.uni-heidelberg.de
pub/games/solutions directory on risc.ua.edu (130.160.4.7)
pub/msdos/romulus directory on ftp.uwp.edu (131.210.1.4)
==> competition/games/poker.face.up.p <==
In Face-Up Poker, two players each select five cards from a face-up deck,
bet, discard and draw. Is there a winning strategy for this game? What if
the players select cards alternately?
==> competition/games/poker.face.up.s <==
If the first player draws four aces, the second player draws four
kings. If the first player keeps the four aces on the draw, the second
player draws a king-high straight flush, and if the first player
pitches the aces to draw a straight flush, the second player can always
make a higher straight flush.
Instead, the winning strategy is for the first player to draw four
tens. The second player cannot draw a royal flush, and in order to
prevent the first player from getting one, the second player must draw
at least one card higher than the ten from each suit, which means he
can't do better than four-of-a-kind. Then the first player wins by
drawing a straight flush from any suit.
If the cards are dealt alternately as in real poker, the second player
can always tie with proper strategy. The second player mirrors the
first player's selections in rank and color. For example, if the first
player picks up a red queen, the second player picks up a red queen.
When they are done playing, their hands will be identical except one
will have spades and hearts where the other has clubs and diamonds, and
vice versa. Since suits aren't ranked in poker, the hands are tied.
It is unknown if there is a winning strategy if the replacement cards
are dealt together as in real poker, as opposed to alternately.
==> competition/games/risk.p <==
What are the odds when tossing dice in Risk?
==> competition/games/risk.s <==
Odds calculated with program by David Karr (ka...@cs.cornell.edu):
Attacker rolls 3 dice, defender rolls 2 dice:
Attacker Defender Probability
loses loses
0 2 2890/7776 = 0.3716563786
1 1 2611/7776 = 0.3357767490
2 0 2275/7776 = 0.2925668724
Attacker rolls 3 dice, defender rolls 1 dice:
Attacker Defender Probability
loses loses
0 1 855/1296 = 0.6597222222
1 0 441/1296 = 0.3402777778
Attacker rolls 2 dice, defender rolls 2 dice:
Attacker Defender Probability
loses loses
0 2 295/1296 = 0.2276234568
1 1 420/1296 = 0.3240740741
2 0 581/1296 = 0.4483024691
Attacker rolls 2 dice, defender rolls 1 dice:
Attacker Defender Probability
loses loses
0 1 125/216 = 0.5787037037
1 0 91/216 = 0.4212962963
Attacker rolls 1 dice, defender rolls 2 dice:
Attacker Defender Probability
loses loses
0 1 55/216 = 0.2546296296
1 0 161/216 = 0.7453703704
Attacker rolls 1 dice, defender rolls 1 dice:
Attacker Defender Probability
loses loses
0 1 15/36 = 0.4166666667
1 0 21/36 = 0.5833333333
---------------------8<------snip here--------8<--------------------
/*
* riskdice.c -- prints Risk dice odds
*
* This program calculates probabilities for one roll of the dice in Risk.
* For each possible number of dice that the attacker might roll, for each
* possible number of dice that the defender might roll, this program
* lists all the possible outcomes (number of armies lost by attacker
* and defender) and the probability of each outcome.
*
* Copyright 1993 by David A. Karr.
*/
#define MAX_ATTACK 3 /* max # of dice attacker may roll */
#define MAX_DEFEND 2 /* max # of dice defender may roll */
#define MAX_DICE MAX_ATTACK + MAX_DEFEND
void main()
{
int a; /* number of dice rolled by attacker */
int d; /* number of dice rolled by defender */
void calc();
for (a = MAX_ATTACK; a > 0; --a) {
for (d = MAX_DEFEND; d > 0; --d) {
calc( a, d );
}
}
}
void calc( a_dice, d_dice )
/*
* Purpose: Print odds for the given numbers of dice rolled
*/
int a_dice; /* number of dice rolled by attacker */
int d_dice; /* number of dice rolled by defender */
{
int num_dice; /* total number of dice rolled */
int num_armies; /* # armies that will be lost by both sides, total */
int kill_count[MAX_DEFEND + 1];
/* entry [i] counts # of times attacker loses i armies */
int roll[MAX_DICE + 1]; /* holds one roll of the dice */
int a_roll[MAX_ATTACK]; /* holds attacker's dice */
int d_roll[MAX_DEFEND]; /* holds defender's dice */
int n; /* cursor into the arrays */
int num_lost; /* # of armies lost by the attacker */
int cases; /* total # of events counted */
void dsort();
/*
* The method is pure brute force. roll[] is set successively to
* all possible rolls of the total number of dice; for each roll
* the number of armies lost by the attacker (the outcome) is
* computed and the event is counted.
* Since all the counted events are equiprobable, the count of each
* outcome merely needs to be scaled down by the total count to
* obtain the probability of that outcome.
*/
/* The number of armies at stake is min(a_dice, d_dice) */
num_armies = a_dice < d_dice ? a_dice : d_dice;
/* initialize event counters */
for (n = 0; n <= num_armies; ++n)
kill_count[n] = 0;
/*
* The roll[] array is treated as a funny odometer whose wheels each
* go from 1 to 6. Each roll of the dice appears in roll[0] through
* roll[num_dice - 1], starting with all 1s and counting up to all 6s.
* roll[num_dice] is used to detect when the other digits have
* finished a complete cycle (it is tripped when they go back to 1s).
*/
num_dice = a_dice + d_dice;
for (n = 0; n <= num_dice; ++n)
roll[n] = 1;
while (roll[num_dice] == 1) {
/* examine a new possible roll of the dice */
/*
* copy attacker's and defender's dice so as not to disturb
* the "odometer" reading.
*/
for (n = 0; n < a_dice; ++n)
a_roll[n] = roll[n];
for (n = 0; n < d_dice; ++n)
d_roll[n] = roll[n + a_dice];
/*
* sort attacker's and defender's dice, highest first.
*/
dsort(a_roll, a_dice);
dsort(d_roll, d_dice);
/*
* compare attacker's and defender's dice, count attacker's loss
*/
num_lost = 0;
for (n = 0; n < num_armies; ++n)
if (d_roll[n] >= a_roll[n])
++num_lost;
++kill_count[num_lost];
/*
* Find next roll values (bump the "odometer" up one tick).
*/
n = 0;
roll[0] += 1;
while (roll[n] > 6) {
/* place [n] rolled over */
roll[n] = 1;
/* Carry 1 into the next place (which may in turn roll over) */
++n;
roll[n] += 1;
}
}
cases = 0;
for (n = 0; n <= num_armies; ++n)
cases += kill_count[n];
printf( "Attacker rolls %d dice, defender rolls %d dice:\n\n",
a_dice, d_dice );
printf( "Attacker Defender Probability\n" );
printf( " loses loses\n" );
for (n = 0; n <= num_armies; ++n)
printf( "%5d %5d %5d/%d = %12.10lf\n",
n, num_armies - n, kill_count[n], cases,
((double) kill_count[n]) / ((double) cases) );
printf( "\n\n" );
}
void dsort( array, length )
/*
* Sort the given array in descending order.
*/
int *array;
int length; /* number of slots in the array */
{
int level, n, tmp;
/* Use bubble sort since the array will be tiny in this application */
for (level = 0; level < length - 1; ++level) {
/*
* Slots [0] through [level - 1] are already "stable."
* Bubble up the value that belongs in the [level] slot.
*/
for (n = length - 1; n > level; --n) {
if (array[n - 1] < array[n]) {
/* swap them */
tmp = array[n - 1];
array[n - 1] = array[n];
array[n] = tmp;
}
}
}
}
==> competition/games/rubiks/rubiks.clock.p <==
How do you quickly solve Rubik's clock?
==> competition/games/rubiks/rubiks.clock.s <==
Solution to Rubik's Clock
The solution to Rubik's Clock is very simple and the clock can be
"worked" in 10-20 seconds once the solution is known.
In this description of how to solve the clock I will describe
the different clocks as if they were on a map (e.g. N,NE,E,SE,S,SW,W,NW);
this leaves the middle clock which I will just call M.
To work the Rubik's clock choose one side to start from; it does
not matter from which side you start. Your initial goal
will be to align the N,S,E,W and M clocks. Use the following algorithm
to do this:
[1] Start with all buttons in the OUT position.
[2] Choose a N,S,E,W clock that does not already have the
same time as M (i.e. not aligned with M).
[3] Push in the closest two buttons to the clock you chose in [2].
[4] Using the knobs that are farest away from the clock you chose in
[2] rotate the knob until M and the clock you chose are aligned.
The time on the clocks at this point does not matter.
[5] Go back to [1] until N,S,E,W and M are in alignment.
[6] At this point N,S,E,W and M should all have the same time.
Make sure all buttons are out and rotate any knob
until N,S,E,W and M are pointing to 12 oclock.
Now turn the puzzle over and repeat steps [1]-[6] for this side. DO NOT
turn any knobs other than the ones described in [1]-[6]. If you have
done this correctly then on both sides of the puzzle N,S,E,W and M will
all be pointing to 12.
Now to align NE,SE,SW,NW. To finish the puzzle you only need to work from
one side. Choose a side and use the following algorithm to align the
corners:
[1] Start with all buttons OUT on the side you're working from.
[2] Choose a corner that is not aligned.
[3] Press the button closest to that corner in.
[4] Using any knob except for that corner's knob rotate all the
clocks until they are in line with the corner clock.
(Here "all the clocks" means N,S,E,W,M and any other clock
that you have already aligned)
There is no need at this point to return the clocks to 12
although if it is less confusing you can. Remember to
return all buttons to their up position before you do so.
[5] Return to [1] until all clocks are aligned.
[6] With all buttons up rotate all the clocks to 12.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> arithmetic/7-11.p <==
A customer at a 7-11 store selected four items to buy, and was told
that the cost was $7.11. He was curious that the cost was the same
as the store name, so he inquired as to how the figure was derived.
The clerk said that he had simply multiplied the prices of the four
individual items. The customer protested that the four prices
should have been ADDED, not MULTIPLIED. The clerk said that that
was OK with him, but, the result was still the same: exactly $7.11.
What were the prices of the four items?
==> arithmetic/7-11.s <==
The prices are: $1.20, $1.25, $1.50, and $3.16
$7.11 is not the only number which works. Here are the first 160 such
numbers, preceded by a count of distinct solutions for that price.
Note that $7.11 has a single, unique solution.
1 - $6.44 1 - $7.83 2 - $9.20 3 - $10.89
1 - $6.51 1 - $7.86 1 - $9.23 1 - $10.95
1 - $6.60 3 - $7.92 1 - $9.24 2 - $11.00
1 - $6.63 1 - $8.00 1 - $9.27 1 - $11.07
1 - $6.65 1 - $8.01 2 - $9.35 1 - $11.13
1 - $6.72 1 - $8.03 3 - $9.36 1 - $11.16
2 - $6.75 5 - $8.10 1 - $9.38 1 - $11.22
1 - $6.78 1 - $8.12 5 - $9.45 2 - $11.25
1 - $6.80 1 - $8.16 2 - $9.48 2 - $11.27
2 - $6.84 2 - $8.19 1 - $9.54 1 - $11.30
1 - $6.86 1 - $8.22 1 - $9.57 1 - $11.36
1 - $6.89 1 - $8.25 1 - $9.59 1 - $11.40
2 - $6.93 3 - $8.28 2 - $9.60 2 - $11.43
1 - $7.02 3 - $8.33 1 - $9.62 2 - $11.52
1 - $7.05 1 - $8.36 2 - $9.63 2 - $11.55
2 - $7.07 1 - $8.37 1 - $9.66 2 - $11.61
1 - $7.08 2 - $8.40 1 - $9.68 1 - $11.69
1 - $7.11 1 - $8.45 2 - $9.69 1 - $11.70
1 - $7.13 2 - $8.46 1 - $9.78 1 - $11.88
2 - $7.14 1 - $8.52 2 - $9.80 1 - $11.90
3 - $7.20 5 - $8.55 1 - $9.81 1 - $11.99
1 - $7.25 1 - $8.60 1 - $9.87 1 - $12.06
1 - $7.26 4 - $8.64 4 - $9.90 1 - $12.15
2 - $7.28 1 - $8.67 1 - $9.92 1 - $12.18
1 - $7.29 1 - $8.69 2 - $9.99 1 - $12.24
3 - $7.35 1 - $8.73 1 - $10.01 1 - $12.30
1 - $7.37 2 - $8.75 1 - $10.05 1 - $12.32
1 - $7.47 1 - $8.76 2 - $10.08 1 - $12.35
1 - $7.50 1 - $8.78 1 - $10.17 2 - $12.42
1 - $7.52 5 - $8.82 1 - $10.20 1 - $12.51
4 - $7.56 1 - $8.85 2 - $10.26 1 - $12.65
1 - $7.62 1 - $8.88 3 - $10.29 2 - $12.69
4 - $7.65 2 - $8.91 3 - $10.35 1 - $12.75
1 - $7.67 1 - $8.94 2 - $10.44 1 - $12.92
2 - $7.70 1 - $8.96 1 - $10.53 1 - $12.96
3 - $7.74 3 - $9.00 1 - $10.56 1 - $13.23
1 - $7.77 1 - $9.02 1 - $10.64 1 - $13.41
1 - $7.79 2 - $9.03 2 - $10.71 1 - $13.56
2 - $7.80 1 - $9.12 3 - $10.80 1 - $14.49
1 - $7.82 2 - $9.18 1 - $10.85 1 - $15.18
There are plenty of solutions for five summands. Here are a few:
$8.28 -- at least two solutions
$8.47 -- at least two solutions
$8.82 -- at least two solutions
--
Mark Johnson ma...@microunity.com (408) 734-8100
There may be many approximate solutions, for example: $1.01, $1.15, $2.41,
and $2.54. These sum to $7.11 but the product is 7.1100061.
==> arithmetic/arithmetic.progression.p <==
Is there an arithmetic progression of 20 or more primes?
==> arithmetic/arithmetic.progression.s <==
There is an arithmetic progression of 21 primes:
142072321123 + 1419763024680 i, 0 <= i < 21.
It was discovered on 30 November 1990, by programs running in the background
on a network of Sun 3 workstations in the Department of Computer Science,
University of Queensland, Australia.
See: Andrew Moran and Paul Pritchard, The design of a background job
on a local area network, Procs. 14th Australian Computer Science Conference,
1991, to appear.
==> arithmetic/clock/day.of.week.p <==
It's restful sitting in Tom's cosy den, talking quietly and sipping
a glass of his Madeira.
I was there one Sunday and we had the usual business of his clock.
When the radio gave the time at the hour, the Ormolu antique was
exactly 3 minutes slow.
"It loses 7 minutes every hour", my old friend told me, as he had done
so many times before. "No more and no less, but I've gotten used to
it that way."
When I spent a second evening with him later that same month, I remarked
on the fact that the clock was dead right by radio time at the hour.
It was rather late in the evening, but Tom assured me that his treasure
had not been adjusted nor fixed since my last visit.
What day of the week was the second visit?
From "Mathematical Diversions" by Hunter + Madachy
==> arithmetic/clock/day.of.week.s <==
The answer is 17 days and 3 hours later, which would have been a Wednesday.
This is the only other time in the same month when the two would agree at all.
In 17 days the slow clock loses 17*24*7 minutes = 2856 minutes,
or 47 hours and 36 minutes. In 3 hours more it loses 21 minutes, so
it has lost a total of 47 hours and 57 minutes. Modulo 12 hours, it
has *gained* 3 minutes so as to make up the 3 minutes it was slow on
Sunday. It is now (fortnight plus 3 days) exactly accurate.
Since the clock was not adjusted since the last visit, it's also
possible that the radio time shifted by one hour due to a change to or
from summer daylight saving time. However, it turns out that the only
additional possibilities that need to be considered are those of 4 days
15 hours later, when the clock would have lost 12 hours 57 minutes, and
29 days 15 hours later, when the clock would have lost 3 days 10 hours
57 minutes. Without even considering the rules for when in the month the
clock is changed, these possible solutions are ruled out because we know
that both visits were in the evening ("I spent a second evening with him").
and they involve times in a different part of the day.
==> arithmetic/clock/palindromic.p <==
How many times per day does a digital clock display a palindromic number?
==> arithmetic/clock/palindromic.s <==
The problem is underspecified. Digital clocks may run from
(a) 1:00 to 12:59
(b) 01:00 to 12:59
(c) 0:00 to 23:59
(d) 00:00 to 23:59
(e-h) any of the above with seconds appended, :00 to :59.
I agree that not all of these are common, but I have seen some uncommon
ones. For that matter, I've seen ones not on the above list -- the Toronto
subway stations used to contain digital clocks that ran from 0:00 to 12:59
(pm), then from 1:00 (pm) to 11:59 (pm), while a computer that I used to
use had the inverse anomaly -- its time of day command displayed times
from 12:00:00 to 12:59:59 (am), then 01:00:00 to 23:59:59!
I get the following results for the 8 cases.
CASE AM+PM HOURS pals/hr TOTAL OVERALL TOTAL
(a) yes 1-9 6 54
10-12 1 3 114
(b) yes 01-05 1 5
06-09 0 0
10-12 1 3 16
(c) no 0-9 6 60
10-15 1 6
16-19 0 0
20-23 1 4 70
(d) no 00-05 1 6
06-09 0 0
10-15 1 6
16-19 0 0
20-23 1 4 16
(e) yes 1-9 60 540
10-12 6 18 1116
(f) yes 01-05 6 30
06-09 0 0
10-12 6 18 96
(g) no 0-9 60 600
10-15 6 36
16-19 0 0
20-23 6 24 660
(h) no 00-05 6 36
06-09 0 0
10-15 6 36
16-19 0 0
20-23 6 24 96
--Mark Brader (m...@sq.com)
==> arithmetic/clock/reversible.p <==
How many times per day can the hour and minute hands on an analog clock switch
roles and still signify a valid time, ignoring the second hand?
==> arithmetic/clock/reversible.s <==
Have 12 clocks C1, C2 ... C12 show 1:00, 2:00, ..., 12:00. Have a
clock C0 show 12:00
Now turn C0 around 12 hours, simultaneously turning C1-C12 so their
hour hands always coincide with the minute hand of C0, i.e., as C0
spans 12 hours, C1-C12 will span 1 hour, but for each possible placing
of the hour hand, all 12 possible 'true' placings of the minute hand
will be represented by one of the 12 clocks.
Each time the hour hand of C0 coincides with the minute hand of a
C1-C12 clock we have a reversible valid time. This happens regularly
12 times each C0 hour, but the first and last time is equal (12:00), so
the number of reversible true times is 12*12-1 = 143 spaced regularly
in the 12-hour interval, ie. each 5 min 2.0979+ sec
--
stein....@tf.tele.no [X.400] stein....@nta.no [internet]
==> arithmetic/clock/right.angle.p <==
How many times per day do the hour and minute hands of a clock form a
right angle?
==> arithmetic/clock/right.angle.s <==
44. Twice each hour equals 48, less one between 2:00 and 4:00 and one between
8:00 and 10:00 for both A.M. and P.M.
==> arithmetic/clock/thirds.p <==
Do the 3 hands on a clock ever divide the face of the clock into 3
equal segments, i.e. 120 degrees between each hand?
==> arithmetic/clock/thirds.s <==
First let us assume that our clock has 60 divisions. We will show that
any time the hour hand and the minute hand are 20 divisions (120 degrees)
apart, the second hand cannot be an integral number of divisions from the
other hands, unless it is straight up (on the minute).
Let us use h for hours, m for minutes, and s for seconds.
We will use =n to mean congruent mod n, thus 12 =5 7.
We know that m =60 12h, that is, the minute hand moves 12 times as fast
as the hour hand, and wraps around at 60.
We also have s =60 60m. This simplifies to s/60 =1 m, which goes to
s/60 = frac(m) (assuming s is in the range 0 <= s < 60), which goes to
s = 60 frac(m). Thus, if m is 5.5, s is 30.
Now let us assume the minute hand is 20 divisions ahead of the hour hand.
So m =60 h + 20, thus 12h =60 h + 20, 11h =60 20, and, finally,
h =60/11 20/11 (read 'h is congruent mod 60/11 to 20/11').
So all values of m are k + n/11 for some integral k and integral n,
0 <= n < 11. s is therefore 60n/11. If s is to be an integral number of
units from m and h, we must have 60n =11 n. But 60 and 11 are relatively
prime, so this holds only for n = 0. But if n = 0, m is integral, so
s is 0.
Now assume, instead, that the minute hand is 20 divisions behind the hour hand.
So m =60 h - 20, 12h =60 h - 20, 11h =60 -20, h =60/11 -20/11.
So m is still k + n/11. Thus s must be 0.
But if s is 0, h must be 20 or 40. But this translates to 4 o'clock or
8 o'clock, at both of which the minute hand is at 0, along with the second
hand.
Thus the 3 hands can never be 120 degrees apart, Q.E.D.
This assumes, of course, that the second hand is synchronized with the
minute hand. This is not true on some inexpensive analog watches. This
also assumes that the watch is not broken (:^)).
==> arithmetic/consecutive.composites.p <==
Are there 10,000 consecutive non-prime numbers?
==> arithmetic/consecutive.composites.s <==
9973!+2 through 9973!+10006 are composite.
==> arithmetic/consecutive.product.p <==
Prove that the product of three or more consecutive positive integers cannot
be a perfect square.
==> arithmetic/consecutive.product.s <==
Three consecutive numbers:
If a and b are relatively prime, and ab is a square,
then a and b are squares. (This is left as an exercise.)
Suppose (n - 1)n(n + 1) = k^2, where n > 1.
Then n(n^2 - 1) = k^2. But n and (n^2 - 1) are relatively prime.
Therefore n^2 - 1 is a perfect square, which is a contradiction.
Four consecutive numbers:
n(n + 1)(n + 2)(n + 3) = (n^2 + 3n + 1)^2 - 1
Five consecutive numbers:
Assume the product is a integer square, call it m.
The prime factorization of m must have even numbers of each prime factor.
For each prime factor, p, of m, p >= 5, p^2k must divide one of the
consecutive naturals in the product. (Otherwise, the difference between two
of the naturals in the product would be a positive multiple of a prime >= 5.
But in this problem, the greatest difference is 4.) So we need only consider
the primes 2 and 3.
Each of the consecutive naturals is one of:
1) a perfect square
2) 2 times a perfect square
3) 3 times a perfect square
4) 6 times a perfect square.
By the shoe box principle, two of the five consecutive numbers must fall into
the same category.
If there are two perfect squares, then their difference being less than five
limits their values to be 1 and 4. (0 is not a natural number, so 0 and 1
and 0 and 4 cannot be the perfect squares.) But 1*2*3*4*5=120!=x*x where x
is an integer.
If there are two numbers that are 2 times a perfect square, then their
difference being less than five implies that the perfect squares (which are
multiplied by 2) are less than 3 apart, and no two natural squares differ by
only 1 or 2.
A similar argument holds for two numbers which are 3 times a perfect square.
We cannot have the case that two of the 5 consecutive numbers are multiples
(much less square multiples) of 6, since their difference would be >= 6, and
our span of five consecutive numbers is only 4.
Therefore the assumption that m is a perfect square does not hold.
QED.
In general the equation:
y^2 = x(x+1)(x+2)...(x+n), n > 3
has only the solution corresponding to y = 0.
This is a theorem of Rigge [O. Rigge, ``Uber ein diophantisches Problem'',
IX Skan. Math. Kong. Helsingfors (1938)] and Erdos [P. Erdos, ``Note on
products of consecutive integers,'' J. London Math. Soc. #14 (1939),
pages 194-198].
A proof can be found on page 276 of [L. Mordell, ``Diophantine
Equations'', Academic Press 1969].
==> arithmetic/consecutive.sums.p <==
Find all series of consecutive positive integers whose sum is exactly 10,000.
==> arithmetic/consecutive.sums.s <==
Generalize to find X (and I) such that
(X + X+1 + X+2 + ... + X+I) = T
for any integer T.
You are asking for all (X,I) s.t. (2X+I)(I+1) = 2T. The problem is
(very) slightly easier if we don't restrict X to being positive, so
we'll solve this first.
Note that 2X+I and I+1 must have different parities, so the answer
to the relaxed question is N = 2*(o_1+1)*(o_2+1)*...*(o_n+1), where
2T = 2^o_0*3^o_1*...*p_n^o_n (the prime factorization); this is easily
seen to be the number of ways we can break 2T up into two positive
factors of differing parity (with order).
In particular, 20000 = 2^5*5^4, hence there are 2*(4+1) = 10 solutions
for T = 10000. These are (2X+I,I+1):
(32*1,5^4) (32*5,5^3) (32*5^2,5^2) (32*5^3,5) (32*5^4,1)
(5^4,32*1) (5^3,32*5) (5^2,32*5^2) (5,32*5^3) (1,32*5^4)
And they give rise to the solutions (X,I):
(-296,624) (28,124) (388,24) (1998,4) (10000,0)
(297,31) (-27,179) (-387,799) (-1997,3999) (-9999,19999)
If you require that X>0 note that this is true iff 2X+I > I+1 and
hence the number of solutions to this problem is N/2 (due to the
symmetry of the above ordered pairs).
==> arithmetic/conway.p <==
Describe the sequence a(1)=a(2)=1, a(n) = a(a(n-1)) + a(n-a(n-1)) for n>2.
==> arithmetic/conway.s <==
This sequence and its remarkable properties were discovered, apparently
independently, by Douglas Hofstadter (circa 1975), John Conway (in the
early 1980's), B.W. Connoly, and others. Since Conway discovered many of
the deeper properties, and is the one responsible for popularizing the
sequence, it seems appropriate to name the sequence after him.
Some properties of a(n):
-- a(n+1) - a(n) = 0 or 1
-- a(2^n) = 2^(n-1)
-- n/2 <= a(n) <= n
-- A(n)= 2a(n) - n has zeros at the powers of 2 and
positive "humps" in between
-- A(2^n + t) = A(2^(n+1) - t) for t between 0 and 2^n,
i.e. the humps are symmetric
-- a(2n) <= 2a(n)
-- a(n)/n --> 1/2 as n --> infinity
-- a(n) and A(n) are self-similar, in the sense that their values on the
(N+1)'st hump can be obtained by a "magnification" process applied
to the values on the N'th hump. They are *not* chaotic sequences,
since their asymptotics and combinatorial structure are very regular
and rigid.
In a lecture at Bell Labs in 1988, Conway asked for the rate at
which a(n)/n converges to 1/2. In particular, he asked for
N(epsilon), the last value of n such that |a(n)/n - 1/2|
exceeds epsilon, in the case epsilon=1/20. This problem was
solved by Colin Mallows of Bell Labs: he found that N(1/20)=1489
and N(1/40)=6083008742. These values are reported in his article
in the American Mathematical Monthly, January 1991, p. 5.
Conway's sequence has a deep combinatorial structure, and is intimately
connected with all of the following: variants of Pascal's triangle, the
Gaussian distribution, combinatorial operations on finite sets,
coin-pushing games, and Fibonacci and Catalan numbers. All of this is
explained in my joint paper with Ravi Vakil; anyone who would like to
receive a copy in LaTex format should contact me at the e-mail address
listed below.
One byproduct of our work is an algorithm to determine N(epsilon) for
given positive epsilon. Here are some particular values:
e Last n such that |a(n)/n - 1/2| > e
---- ----------------------------------------------------------
1/20 1489 (found by Mallows in 1988)
1/30 758765
1/40 6083008742 (found by Mallows in 1988)
1/50 809308036481621
1/60 1684539346496977501739
1/70 55738373698123373661810220400
1/80 15088841875190938484828948428612052839
1/90 127565909103887972767169084026274554426122918035
1/100 8826608001127077619581589939550531021943059906967127007025
These values were computed by the Mathematica program listed below; the
algorithm is explained in our paper. To use the program, load it into
Mathematica and type neps[t] for some numerical value of t (which
should probably be smaller than 1/5 or so). The program will output N(t),
e.g. neps[1/20] = 1489. These numbers grow very fast: N(epsilon) will be
about 2^((0.138015/epsilon)^2), so N(1/1000) will have about 5735 digits.
The program requires very little memory space, but its runtime appears to
grow rapidly as epsilon decreases (on a Sun 3, it took about one second to
find the first value listed, and several minutes to find the last).
neps[eps_] := Block[{W}, L = 1 + Floor[(0.138015/eps)^2]; e = eps*2;
W = processvector[L]; While[Length[W] > 0,
nmax = 1 + Last[W][[4]]; L++; W = processvector[L]]; nmax]
processvector[L_] :=
Block[{V}, V = startvector[L]; While[notfound[V], V = newbody[V]]; V]
startvector[L_] := Select[initialvector[L], gt]
initialvector[L_] :=
Table[{L, b, Binomial[L - 1, b - 1],
2^(L + 1) - Sum[Binomial[L, c], {c, b, L}]}, {b, 0, L - 1}]
c[0] = 0
c[n_] := c[n] = Sum[Binomial[2*a, a]/(a + 1), {a, 0, n - 1}]
gt[x_] := U[x] > e
U[x_] := (x[[3]] + M[x[[1]], x[[2]]])/(x[[4]] + incp[x[[1]], x[[2]]])
M[n_, n_] = -1
M[n_, k_] := Binomial[n - 1, K[n, k]] - Binomial[n - 1, k - 1] + c[K[n, k]]
K[n_, k_] := Min[k, n - k]
incp[n_, k_] := Max[M[n, k], 1]
notfound[V_] :=
Length[V] > 0 && Last[V][[2]] > 0 && Last[V][[1]] > Last[V][[2]]
newbody[V_] := Join[Drop[V, -1], newtail[V]]
newtail[V_] := Select[{vleft[Last[V]], vright[Last[V]]}, gt]
vleft[x_] := {x[[1]] - 1, x[[2]] - 1, x[[3]], x[[4]]}
vright[x_] := {x[[1]] - 1, x[[2]], x[[3]] + S[x[[1]] - 1, x[[2]] - 1],
x[[4]] + Binomial[x[[1]] - 1, x[[2]] - 1]}
S[n_, k_] := Binomial[n - 1, k] - Binomial[n - 1, k - 1]
-Tal Kubo ku...@math.harvard.edu or ku...@zariski.harvard.edu
==> arithmetic/digits/6.and.7.p <==
Does every number which is not divisible by 5 have a multiple whose
only digits are 6 and 7?
==> arithmetic/digits/6.and.7.s <==
Yes. My proof follows:
Claim 1: For every k, there is a k-digit number whose only digits
are 6 and 7, which is divisible by 2^k.
The proof is by induction. Suppose N is a k-digit number
satisfying the above condition. Then either N = 0 (mod 2^(k+1))
or N = 2^k (mod 2^(k+1)). Note that 6(10^k) = 0 (mod 2^(k+1)),
and 7(10^k) = 2^k (mod 2^(k+1)). So, either 6*10^k + N or
7*10^k + N is divisible by 2^(k+1).
Claim 2: If m and 10 are relatively prime, then for any r,
there is a number N whose only digits are 6 and 7 such that
N = r (mod m).
Proof: Let K be the (m^2)-digit number whose only digit is 6.
There is an s, 0 <= s < m, so that K + s = r (mod m).
Let N = K + 10^(m - 1) + 10^(2m - 2) + . . . + 10^(sm - s).
Since 10^(im - i) = 1 (mod m), N = K + s (mod m) = r (mod m).
Clearly, every digit of N is either 6 or 7.
Claim 3: If n is not divisible by 5, then there is a number N whose
only digits are 6 and 7, so that N is divisible by n.
Proof: We can write n = (2^k)m, with gcd(m,10)=1.
Use claim 1 to find a k-digit number M, whose only digits are 6 and 7,
which is divisible by 2^k. Choose an integer r so that
(10^k)r + M = 0 (mod m). Use claim 2 to find a number K whose
only digits are 6 and 7, so that K = r (mod m). Let N = 10^k K + M.
Then N = 0 (mod m) and N = 0 (mod 2^k), so N is divisible by n.
Finally, the only digits of N are 6 and 7, so we are done.
--
David Radcliffe radc...@csd4.csd.uwm.edu
==> arithmetic/digits/all.ones.p <==
Prove that some multiple of any integer ending in 3 contains all 1s.
==> arithmetic/digits/all.ones.s <==
Let n be our integer; one such desired multiple is then
(10^(phi(n))-1)/9. All we need is that (n,10) = 1, and if the last
digit is 3 this must be the case. A different proof using the
pigeonhole principle is to consider the sequence 1, 11, 111, ..., (10^n
- 1)/9. We must have at some point that either some member of our
sequence = 0 (mod n) or else some value (mod n) is duplicated. Assume
the latter, with x_a and x_b, x_b>x_a, possesing the duplicated
remainders. We then have that x_b - x-a = 0 (mod n). Let m be the
highest power of 10 dividing x_b - x_a. Now since (10,n) = 1, we can
divide by 10^m and get that (x_b - x_a)/10^m = 0 (n). But (x_b -
x_a)/10^m is a number containing only the digit 1.
Q.E.D.
==> arithmetic/digits/arabian.p <==
What is the Arabian Nights factorial, the number x such that x! has 1001
digits? How about the prime x such that x! has exactly 1001 zeroes on
the tail end. (Bonus question, what is the 'rightmost' non-zero digit in x!?)
==> arithmetic/digits/arabian.s <==
The first answer is 450!.
Determining the number of zeroes at the end of x! is relatively easy once
you realize that each such zero comes from a factor of 10 in the product
1 * 2 * 3 * ... * x
Each factor of 10, in turn, comes from a factor of 5 and a factor of 2.
Since there are many more factors of 2 than factors of 5, the number of 5s
determines the number of zeroes at the end of the factorial.
The number of 5s in the set of numbers 1 .. x (and therefore the number
of zeroes at the end of x!) is:
z(x) = int(x/5) + int(x/25) + int(x/125) + int(x/625) + ...
This series terminates when the powers of 5 exceed x.
I know of no simple way to invert the above formula (i.e., to find x for
a given z(x)), but I can approximate it by noting that, except for the "int"
function,
5*z(x) - x = z(x)
which gives:
x = 4*z(x) (approximately).
The given problem asked, "For what prime x is z(x)=1001". By the above forumla,
this is approximately 4*1001=4004. However, 4004! has only
800 + 160 + 32 + 6 + 1 = 999 zeroes at the end of it.
The numbers 4005! through 4009! all have 1000 zeroes at their end and
the numbers 4010! through 4014! all have 1001 zeroes at their end.
Since the problem asked for a prime x, and 4011 = 3*7*191, the only solution
is x=4013.
The problem of determining the rightmost nonzero digit in x! is somewhat more
difficult. If we took the numbers 1, 2, ... , x and removed all factors of 5
(and an equal number of factors of 2), the remaining numbers multiplied
together modulo 10 would be the answer. Note that since there are still many
factors of 2 left, the rightmost nonzero digit must be 2, 4, 6, or 8 (x > 1).
Letting r(x) be the rightmost nonzero digit in x!, an expression for r(x) is:
r(x) = (r(int(x/5)) * w * r(x mod 10)) mod 10, x >= 10.
where w is 4 if int(x/10) is odd and 6 if it is even.
The values of r(x) for 0 <= x <= 9 are 1, 1, 2, 6, 4, 2, 2, 4, 2, and 8.
The way to see this is true is to take the numbers 1, 2, ..., x in groups
of 10. In each group, remove 2 factors of 10. For example, from the
set 1, 2, ..., 10, choose a factor of 2 from 2 and 6 and a factor of 5 from
5 and 10. This leaves 1, 1, 3, 4, 1, 3, 7, 8, 9, 2. Next, separate all the
factors that came from multiples of 5. The rightmost nonzero digit of x!
can now (hopefully) be seen to be:
r(x) = (r(int(x/5)) * w * r(x mod 10)) mod 10
where w is the rightmost digit in the number formed by multiplying the numbers
1, 2, 3, ..., 10*int(x/10) after the factors of 10 and the factors left over
by the multiples of 5 have been removed. In the example with x = 10, this
would be (1 * 1 * 3 * 4 * 3 * 7 * 8 * 9) mod 10 = 4. The "r(x mod 10)" term
takes care of the numbers from 10*int(x/10)+1 up to x.
The "w" term can be seen to be 4 or 6 depending on whether int(x/10) is odd or
even since, after removing 10*n+5 and 10*n+10 and a factor of 2 each from
10*n+2 and 10*n+6 from the group of numbers 10*n+1 through 10*n+10, the
remaining factors (mod 10) always equals 4 and 4^t mod 10 = 4 if t is odd and
6 when t is even (t != 0).
So, finally, the rightmost nonzero digit in 4013! is found as follows:
r(4013) = (r(802) * 4 * 6) mod 10
r(802) = (r(160) * 6 * 2) mod 10
r(160) = (r(32) * 6 * 1) mod 10
r(32) = (r(6) * 4 * 2) mod 10
Using a table of r(x) for 0 <= x <= 9, r(6) = 2. Then,
r(32) = (2 * 4 * 2) mod 10 = 6
r(160) = (6 * 6 * 1) mod 10 = 6
r(802) = (6 * 6 * 2) mod 10 = 2
r(4013) = (2 * 4 * 6) mod 10 = 8
Thus, the rightmost nonzero digit in 4013! is 8.
==> arithmetic/digits/circular.p <==
What 6 digit number, with 6 different digits, when multiplied by all integers
up to 6, circulates its digits through all 6 possible positions, as follows:
ABCDEF * 1 = ABCDEF
ABCDEF * 3 = BCDEFA
ABCDEF * 2 = CDEFAB
ABCDEF * 6 = DEFABC
ABCDEF * 4 = EFABCD
ABCDEF * 5 = FABCDE
==> arithmetic/digits/circular.s <==
ABCDEF=142857 (the digits of the expansion of 1/7).
==> arithmetic/digits/divisible.p <==
Find the least number using 0-9 exactly once that is evenly divisible by each
of these digits.
==> arithmetic/digits/divisible.s <==
Since the sum of the digits is 45, any permutation of the digits gives a
multiple of 9. To get a multiple of both 2 and 5, the last digit must
be 0, and thus to get a multiple of 8 (and 4), the tens digit must be
even, and the hundreds digit must be odd if the tens digit is 2 or 6,
and even otherwise. The number will also be divisible by 6, since it is
divisible by 2 and 3, so 7 is all we need to check. First, we will look
for a number whose first five digits are 12345; now, 1234500000 has a
remainder of 6 when divided by 7, so we have to arrange the remaining
digits to get a remainder of 1. The possible arrangements, in
increasing order, are
78960, remainder 0
79680, remainder 6
87960, remainder 5
89760, remainder 6
97680, remainder 2
98760, remainder 4
That didn't work, so try numbers starting with 12346; this is impossible
because the tens digit must be 8, and the hundreds digit cannot be even.
Now try 12347, and 1234700000 has remainder 2. The last five digits can
be
58960, remainder 6
59680, remainder 5, so this works, and the number is
1234759680.
==> arithmetic/digits/equations/123456789.p <==
In how many ways can "." be replaced with "+", "-", or "" (concatenate) in
.1.2.3.4.5.6.7.8.9=1 to form a correct equation?
==> arithmetic/digits/equations/123456789.s <==
1-2 3+4 5+6 7-8 9 = 1
+1-2 3+4 5+6 7-8 9 = 1
1+2 3+4-5+6 7-8 9 = 1
+1+2 3+4-5+6 7-8 9 = 1
-1+2 3-4+5+6 7-8 9 = 1
1+2 3-4 5-6 7+8 9 = 1
+1+2 3-4 5-6 7+8 9 = 1
1-2 3-4+5-6 7+8 9 = 1
+1-2 3-4+5-6 7+8 9 = 1
1-2-3-4 5+6 7-8-9 = 1
+1-2-3-4 5+6 7-8-9 = 1
1+2-3 4+5 6-7-8-9 = 1
+1+2-3 4+5 6-7-8-9 = 1
-1+2 3+4+5-6-7-8-9 = 1
-1 2+3 4-5-6+7-8-9 = 1
1+2+3+4-5+6+7-8-9 = 1
+1+2+3+4-5+6+7-8-9 = 1
-1+2+3-4+5+6+7-8-9 = 1
1-2-3+4+5+6+7-8-9 = 1
+1-2-3+4+5+6+7-8-9 = 1
1+2 3+4 5-6 7+8-9 = 1
+1+2 3+4 5-6 7+8-9 = 1
1+2 3-4-5-6-7+8-9 = 1
+1+2 3-4-5-6-7+8-9 = 1
1+2+3+4+5-6-7+8-9 = 1
+1+2+3+4+5-6-7+8-9 = 1
-1+2+3+4-5+6-7+8-9 = 1
1-2+3-4+5+6-7+8-9 = 1
+1-2+3-4+5+6-7+8-9 = 1
-1-2-3+4+5+6-7+8-9 = 1
1-2+3+4-5-6+7+8-9 = 1
+1-2+3+4-5-6+7+8-9 = 1
1+2-3-4+5-6+7+8-9 = 1
+1+2-3-4+5-6+7+8-9 = 1
-1-2+3-4+5-6+7+8-9 = 1
-1+2-3-4-5+6+7+8-9 = 1
-1+2 3+4 5-6 7-8+9 = 1
1-2 3-4 5+6 7-8+9 = 1
+1-2 3-4 5+6 7-8+9 = 1
-1+2 3-4-5-6-7-8+9 = 1
-1+2+3+4+5-6-7-8+9 = 1
1-2+3+4-5+6-7-8+9 = 1
+1-2+3+4-5+6-7-8+9 = 1
1+2-3-4+5+6-7-8+9 = 1
+1+2-3-4+5+6-7-8+9 = 1
-1-2+3-4+5+6-7-8+9 = 1
1+2-3+4-5-6+7-8+9 = 1
+1+2-3+4-5-6+7-8+9 = 1
-1-2+3+4-5-6+7-8+9 = 1
-1+2-3-4+5-6+7-8+9 = 1
1-2-3-4-5+6+7-8+9 = 1
+1-2-3-4-5+6+7-8+9 = 1
1-2 3+4+5+6+7-8+9 = 1
+1-2 3+4+5+6+7-8+9 = 1
1+2+3+4 5-6 7+8+9 = 1
+1+2+3+4 5-6 7+8+9 = 1
1 2+3 4+5-6 7+8+9 = 1
+1 2+3 4+5-6 7+8+9 = 1
1+2+3-4-5-6-7+8+9 = 1
+1+2+3-4-5-6-7+8+9 = 1
-1+2-3+4-5-6-7+8+9 = 1
1-2-3-4+5-6-7+8+9 = 1
+1-2-3-4+5-6-7+8+9 = 1
-1-2-3-4-5+6-7+8+9 = 1
-1-2 3+4+5+6-7+8+9 = 1
1-2+3 4-5 6+7+8+9 = 1
+1-2+3 4-5 6+7+8+9 = 1
1 2-3 4+5-6+7+8+9 = 1
+1 2-3 4+5-6+7+8+9 = 1
Total solutions = 69 (26 of which have a leading "+", which is redundant)
69/19683 = 0.35 %
for those who care (it's not very elegant but it did the trick):
#include <stdio.h>
#include <math.h>
main (argc,argv)
int argc;
char *argv[];
{
int sresult, result, operator[10],thisop;
char buf[256],ops[3];
int i,j,tot=0,temp;
ops[0] = ' ';
ops[1] = '-';
ops[2] = '+';
for (i=1; i<10; i++) operator[i] = 0;
for (j=0; j < 19683; j++) {
result = 0;
sresult = 0;
thisop = 1;
for (i=1; i<10; i++) {
switch (operator[i]) {
case 0:
sresult = sresult * 10 + i;
break;
case 1:
result = result + sresult * thisop;
sresult = i;
thisop = -1;
break;
case 2:
result = result + sresult * thisop;
sresult = i;
thisop = 1;
break;
}
}
result = result + sresult * thisop;
if (result == 1) {
tot++;
for (i=1;i<10;i++)
printf("%c%d",ops[operator[i]],i);
printf(" = %d\n",result);
}
temp = 0;
operator[1] += 1;
for (i=1;i<10;i++) {
operator[i] += temp;
if (operator[i] > 2) { operator[i] = 0; temp = 1;}
else temp = 0;
}
}
printf("Total solutions = %d\n" , tot);
}
cw...@media.mit.edu (Christopher Wren)
==> arithmetic/digits/equations/1992.p <==
1 = -1+9-9+2. Extend this list to 2 through 100 on the left side of
the equals sign.
==> arithmetic/digits/equations/1992.s <==
1 = -1+9-9+2
2 = 1*9-9+2
3 = 1+9-9+2
4 = 1+9/9+2
5 = 1+9-sqrt(9)-2
6 = 1^9+sqrt(9)+2
7 = -1+sqrt(9)+sqrt(9)+2
8 = 19-9-2
9 = (1/9)*9^2
10= 1+(9+9)/2
11= 1+9+sqrt(9)-2
12= 19-9+2
13= (1+sqrt(9))!-9-2
14= 1+9+sqrt(9)!-2
15= -1+9+9-2
16= -1+9+sqrt(9)!+2
17= 1+9+9-2
18= 1+9+sqrt(9)!+2
19= -1+9+9+2
20= (19-9)*2
21= 1+9+9+2
22= (-1+sqrt(9))*(9-2)
23= (1+sqrt(9))!-sqrt(9)+2
24= -1+9*sqrt(9)-2
25= 1*9*sqrt(9)-2
26= 19+9-2
27= 1*9+9*2
28= 1+9+9*2
29= 1*9*sqrt(9)+2
30= 19+9+2
31= (1+sqrt(9))!+9-2
32= -1+sqrt(9)*(9+2)
33= 1*sqrt(9)*(9+2)
34= (-1+9+9)*2
35= -1+(9+9)*2
36= 1^9*sqrt(9)!^2
37= 19+9*2
38= 1*sqrt(9)!*sqrt(9)!+2
39= 1+sqrt(9)!*sqrt(9)!+2
40= (1+sqrt(9)!)*sqrt(9)!-2
41= -1+sqrt(9)!*(9-2)
42= (1+sqrt(9))!+9*2
43= 1+sqrt(9)!*(9-2)
44= -1+9*(sqrt(9)+2)
45= 1*9*(sqrt(9)+2)
46= 1+9*(sqrt(9)+2)
47= (-1+sqrt(9)!)*9+2
48= 1*sqrt(9)!*(sqrt(9)!+2)
49= (1+sqrt(9)!)*(9-2)
50= (-1+9)*sqrt(9)!+2
51= -1+9*sqrt(9)!-2
52= 1*9*sqrt(9)!-2
53= -1+9*sqrt(9)*2
54= 1*9*sqrt(9)*2
55= 1+9*sqrt(9)*2
56= 1*9*sqrt(9)!+2
57= 1+9*sqrt(9)!+2
58= (1+9)*sqrt(9)!-2
59= 19*sqrt(9)+2
60= (1+9)*sqrt(9)*2
61= (1+sqrt(9)!)*9-2
62= -1+9*(9-2)
63= 1*9*(9-2)
64= 1+9*(9-2)
65= (1+sqrt(9)!)*9+2
66= 1*sqrt(9)!*(9+2)
67= 1+sqrt(9)!*(9+2)
68= -(1+sqrt(9))!+92
69= (1+sqrt(9))!+(9/.2)
70= (1+9)*(9-2)
71= -1-9+9^2
72= (1+sqrt(9))*9*2
73= -19+92
74= (-1+9)*9+2
75= -1*sqrt(9)!+9^2
76= 1-sqrt(9)!+9^2
77= (1+sqrt(9)!)*(9+2)
78= -1+9*9-2
79= 1*9*9-2
80= 1+9*9-2
81= 1*9*sqrt(9)^2
82= -1+9*9+2
83= 1*9*9+2
84= 1+9*9+2
85= -1-sqrt(9)!+92
86= -1*sqrt(9)!+92
87= 1-sqrt(9)!+92
88= (1+9)*9-2
89= -1*sqrt(9)+92
90= 1-sqrt(9)+92
91= -1^9+92
92= (1+9)*9+2
93= 1^9+92
94= -1+sqrt(9)+92
95= 19*(sqrt(9)+2)
96= -1+99-2
97= 1*99-2
98= 1+99-2
99= 1*9*(9+2)
100= -1+99+2
==> arithmetic/digits/equations/24.p <==
Form an expression that evaluates to 24 that contains two 3's, two 7's,
and zero or more of the operators +, -, *, and /, and parentheses. What
about two 4's and two 7's, or three 5's and one 1, or two 3's and two 8's?
==> arithmetic/digits/equations/24.s <==
7*(3+3/7)
7*(4-4/7)
5*(5-1/5)
8/(3-8/3)
==> arithmetic/digits/equations/383.p <==
Make 383 out of 1,2,25,50,75,100 using +,-,*,/.
==> arithmetic/digits/equations/383.s <==
You can get 383 with ((2+50)/25+1)*100+75.
Of course, if you expect / as in C, the above expression is just 375.
But then you can get 383 with (25*50-100)/(1+2). Pity there's no way
to use the 75.
If we had a language that rounded instead of truncating, we could use
((1+75+100)*50)/(25-2) or (2*75*(25+100))/(50-1).
I imagine your problem lies in not dividing things that aren't
divisible.
Dan Hoey
Ho...@AIC.NRL.Navy.Mil
==> arithmetic/digits/equations/find.p <==
Write a program for finding expressions built out of given numbers and using
given operators that evaluate to a given value, or listing all possible values.
==> arithmetic/digits/equations/find.s <==
As set up, it requires recompilation for different sets of numbers;
it's currently set up for 8,8,3,3; to try in other numbers, stick 'em
in the 'val' array. To find all target numbers for which the equation
is valid, uncomment the 't' loop and 'target = t', and extend the range
to be checked... you might want to turn off VERBOSE. I don't bother
with eliminating symmetries if equal vals are given (like 8 8 3 3), so
I normally use it like
numop 24 | sort | uniq
As it stands, this gives the output:
8 / (3 - (8 / 3)) = 24.0
8 / (3 - (8 / 3)) = 24.0
8 / (3 - (8 / 3)) = 24.0
8 / (3 - (8 / 3)) = 24.0
As you can see, there are five different kinds of binary trees with
exactly four leaf nodes. The program tries all four operators in each
place, and all four values in each of the leaves, guaranteeing that each
is used only once... a fairly quick operation. A small extract from
'numop 1' shows the five different shapes of trees:
((3 * 8) / 3) / 8 = 1.0
(3 * (8 / 3)) / 8 = 1.0
(3 - 3) + (8 / 8) = 1.0
3 * ((8 / 3) / 8) = 1.0
3 * (8 / (3 * 8)) = 1.0
Probably FUDGE ought to be set a little lower, for more confidence that
the equality isn't fortuitous. Extensions to other binary operators are
obvious; unary operators and more values are not. For a more general
problem I'd go recursive, use exact rational arithmetic, and have a fine
old time.
Enjoy...
Jim Gillogly <uunet!rand.org!James_Gillogly>
21 Wedmath S.R. 1993, 10:58
----------------------------------------------------------------
/* numop: using elementary operations on 4 numbers, find a
* desired result; e.g. 24.
*
* Don't worry about symmetries resulting in multiple correct answers.
*
* 11 Aug 93, SCRYER
*/
#include <stdio.h>
#define VERBOSE
#define MUL 0
#define DIV 1
#define ADD 2
#define SUB 3
#define FUDGE 0.01
float val[4] = {8, 8, 3, 3};
float eval(), atof(), fabs();
char nameop();
int divzero;
main(argc, argv)
int argc;
char *argv[];
{
int op1, op2, op3;
int iv1, iv2, iv3, iv4;
int used[4];
int i;
float target;
float e1, e2, e3;
int t, winner;
if (argc != 2)
{
fprintf(stderr, "Usage: numop <target>\n");
exit(1);
}
target = atof(argv[1]);
/* for (t = -1000; t < 1000; t++) */
{
/* target = t;*/
winner = 0;
for (i = 0; i < 4; i++) used[i] = 0;
for (op1 = 0; op1 < 4; op1++)
for (op2 = 0; op2 < 4; op2++)
for (op3 = 0; op3 < 4; op3++)
for (iv1 = 0; iv1 < 4; iv1++)
{
used[iv1] = 1;
for (iv2 = 0; iv2 < 4; iv2++)
{
if (used[iv2]) continue;
used[iv2] = 1;
for (iv3 = 0; iv3 < 4; iv3++)
{
if (used[iv3]) continue;
used[iv3] = 1;
for (iv4 = 0; iv4 < 4; iv4++)
{
if (used[iv4]) continue;
/* Case 1 */
divzero = 0;
e3 = eval(op3, val[iv3], val[iv4]);
e2 = eval(op2, val[iv1], val[iv2]);
e1 = eval(op1, e2, e3); /* (u + v) * (w - x) */
if (fabs(e1 - target) < FUDGE && ! divzero)
#ifdef VERBOSE
printf("(%.0f %c %.0f) %c (%.0f %c %.0f) = %.1f\n",
val[iv1], nameop(op2), val[iv2], nameop(op1),
val[iv3], nameop(op3), val[iv4], e1);
#else
winner = 1;
#endif
/* Case 2 */
divzero = 0;
e3 = eval(op3, val[iv1], val[iv2]);
e2 = eval(op2, e3, val[iv3]);
e1 = eval(op1, e2, val[iv4]); /* ((u + v) * w) - x */
if (fabs(e1 - target) < FUDGE && ! divzero)
#ifdef VERBOSE
printf("((%.0f %c %.0f) %c %.0f) %c %.0f = %.1f\n",
val[iv1], nameop(op3), val[iv2], nameop(op2), val[iv3], nameop(op1), val[iv4], e1);
#else
winner = 1;
#endif
/* Case 3 */
divzero = 0;
e3 = eval(op3, val[iv2], val[iv3]);
e2 = eval(op2, val[iv1], e3);
e1 = eval(op1, e2, val[iv4]); /* (u + (v * w)) - x */
if (fabs(e1 - target) < FUDGE && ! divzero)
#ifdef VERBOSE
printf("(%.0f %c (%.0f %c %.0f)) %c %.0f = %.1f\n",
val[iv1], nameop(op2), val[iv2], nameop(op3), val[iv3],
nameop(op1), val[iv4], e1);
#else
winner = 1;
#endif
/* Case 4 */
divzero = 0;
e3 = eval(op3, val[iv2], val[iv3]);
e2 = eval(op2, e3, val[iv4]);
e1 = eval(op1, val[iv1], e2); /* u + ((v * w) - x) */
if (fabs(e1 - target) < FUDGE && ! divzero)
#ifdef VERBOSE
printf("%.0f %c ((%.0f %c %.0f) %c %.0f) = %.1f\n",
val[iv1], nameop(op1), val[iv2], nameop(op3), val[iv3],
nameop(op2), val[iv4], e1);
#else
winner = 1;
#endif
/* Case 5 */ /* u + (v * (w - x)) */
divzero = 0;
e3 = eval(op3, val[iv3], val[iv4]);
e2 = eval(op2, val[iv2], e3);
e1 = eval(op1, val[iv1], e2);
if (fabs(e1 - target) < FUDGE && ! divzero)
#ifdef VERBOSE
printf("%.0f %c (%.0f %c (%.0f %c %.0f)) = %.1f\n",
val[iv1], nameop(op1), val[iv2], nameop(op2), val[iv3],
nameop(op3), val[iv4], e1);
#else
winner = 1;
#endif
}
used[iv3] = 0;
}
used[iv2] = 0;
}
used[iv1] = 0;
}
#ifndef VERBOSE
if (winner) printf("%d\n", t), fflush(stdout);
#endif
}
}
char nameop(op)
int op;
{
switch(op)
{
case MUL: return '*';
case DIV: return '/';
case ADD: return '+';
case SUB: return '-';
}
return '?';
}
float eval(op, val1, val2)
int op;
float val1, val2;
{
switch(op)
{
case MUL: return val1 * val2;
case DIV:
if (val2 == 0.)
{
divzero = 1;
#ifdef EXTREMELYVERBOSE
fprintf(stderr, "Division by zero.\n");
#endif
}
return val2 == 0.? 0. : val1 / val2;
case ADD: return val1 + val2;
case SUB: return val1 - val2;
}
return 0.;
}
==> arithmetic/digits/extreme.products.p <==
What are the extremal products of three three-digit numbers using digits 1-9?
==> arithmetic/digits/extreme.products.s <==
There is a simple procedure which applies to these types of problems (and
which can be proven with help from the arithmetic-geometric inequality).
For the first part we use the "first large then equal" procedure.
This means that are three numbers will be 7**, 8**, and 9**. Now
the digits 4,5,6 get distributed so as to make our three number as
close to each other as possible, i.e. 76*, 85*, 94*. The same goes
for the remaining three digits, and we get 763, 852, 941.
For the second part we use the "first small then different" procedure.
Our three numbers will be of the form 1**, 2**, 3**. We now place
the three digits so as to make our three numbers as unequal as possible;
this gives 14*, 25*, 36*. Finishing, we get 147, 258, 369.
Now, *prove* that these procedures work for generalizations of this
problem.
==> arithmetic/digits/labels.p <==
You have an arbitrary number of model kits (which you assemble for
fun and profit). Each kit comes with twenty (20) stickers, two of which
are labeled "0", two are labeled "1", ..., two are labeled "9".
You decide to stick a serial number on each model you assemble starting
with one. What is the first number you cannot stick. You may stockpile
unused numbers on already assembled models, but you may not crack open
a new model to get at its stickers. You complete assembling the current
model before starting the next.
==> arithmetic/digits/labels.s <==
The method I used for this problem involved first coming up with a
formula that says how many times a digit has been used in all n models.
n = k*10^i + m for some k,m with 0 <k <10, m < 10^i
f(d,n) = (number of d's used getting to k*10^i from digits 0 to (i-1)) +
(number of d's used by #'s 10^i to n from digit i) + f(d,m)
f(d,n) = i*k*10^(i-1) + (if (d < k) 10^i else if (d == k) m+1 else 0) + f(d,m)
This doesn't count 0's, which should be ok as they are not used as often
as other digits. From the formula, it is clear that f(1,n) is never
less than f(d,n) for 1<d<10.
So I just calculated f(1,n) for various n (with some help from bc).
I quickly discovered that for n = 2*10^15, f(1,n) = 2*n. After further
trials I determined that for n = 1999919999999981, f(1,n) = 2*n + 1.
This appears to be the smallest n with f(1,n) > 2*n.
==> arithmetic/digits/least.significant/factorial.p <==
What is the least significant non-zero digit in the decimal expansion of n!?
==> arithmetic/digits/least.significant/factorial.s <==
Reduce mod 10 the numbers 2..n and then cancel out the
required factors of 10. The final step then involves
computing 2^i*3^j*7^k mod 10 for suitable i,j and k.
A small program that performs this calculation is appended. Like the
other solutions, it takes O(log n) arithmetic operations.
-kym
===
#include<stdio.h>
#include<assert.h>
int p[6][4]={
/*2*/ 2, 4, 8, 6,
/*3*/ 3, 9, 7, 1,
/*4*/ 4, 6, 4, 6,
/*5*/ 5, 5, 5, 5,
/*6*/ 6, 6, 6, 6,
/*7*/ 7, 9, 3, 1,
};
main(){
int i;
int n;
for(n=2;n<1000;n++){
i=lsdfact(n);
printf("%d\n",i);
}
exit(0);
}
lsdfact(n){
int a[10];
int i;
int n5;
int tmp;
for(i=0;i<=9;i++)a[i]=alpha(i,n);
n5=0;
/* NOTE: order is important in following */
l5:;
while(tmp=a[5]){ /* cancel factors of 5 */
n5+=tmp;
a[1]+=(tmp+4)/5;
a[3]+=(tmp+3)/5;
a[5]=(tmp+2)/5;
a[7]+=(tmp+1)/5;
a[9]+=(tmp+0)/5;
}
l10:;
if(tmp=a[0]){
a[0]=0; /* cancel all factors of 10 */
for(i=0;i<=9;i++)a[i]+=alpha(i,tmp);
}
if(a[5]) goto l5;
if(a[0]) goto l10;
/* n5 == number of 5's cancelled;
must now cancel same number of factors of 2 */
i=ipow(2,a[2]+2*a[4]+a[6]+3*a[8]-n5)*
ipow(3,a[3]+a[6]+2*a[9])*
ipow(7,a[7]);
assert(i%10); /* must not be zero */
return i%10;
}
alpha(d,n){
/* number of decimal numbers in [1,n] ending in digit d */
int tmp;
tmp=(n+10-d)/10;
if(d==0)tmp--; /* forget 0 */
return tmp;
}
ipow(x,y){
/* x^y mod 10 */
if(y==0) return 1;
if(y==1) return x;
return p[x-2][(y-1)%4];
}
==> arithmetic/digits/least.significant/tower.of.power.p <==
What are the least significant digits of 9^(8^(7^(6^(5^(4^(3^(2^1))))))) ?
==> arithmetic/digits/least.significant/tower.of.power.s <==
9^11 = 9 (mod 100), so we need to find 8^...^1 (mod 10).
8^5 = 8 (mod 10), so we need to find 7^...^1 (mod 4).
7^3 = 7 (mod 4), so we need to find 6^...^1 (mod 2), but
this is clearly 0, so 7^...^1 = 1 (mod 4) ==>
8^...^1 = 8 (mod 10) ==> 9^...^1 = 9^8 (mod 100) = 21 (mod 100).
==> arithmetic/digits/most.significant/googol.p <==
What digits does googol! start with?
==> arithmetic/digits/most.significant/googol.s <==
I'm not sure how to calculate the first googol of digits of log10(e), but
here's the first 150(approximately) of them...
0.43429448190325182765112891891660508229439700580366656611445378316586464920
8870774729224949338431748318706106744766303733641679287158963906569221064663
We need to deal with the digits immediately after the decimal point in
googol*log10(e), which are .187061
frac[log(googol!)] = frac[halflog2pi + 50 + googol(100-log10(e))]
= frac{halflog2pi + frac[googol(100-log10(e))]}
= frac[.399090 + (1- .187061)]
= .212029
10 ** .212029 = 1.629405
Which means that googol! starts with 1629
==> arithmetic/digits/most.significant/powers.p <==
What is the probability that 2^N begins with the digits 603245?
==> arithmetic/digits/most.significant/powers.s <==
2^N begins with 603245 iff 603246*10^m > 2^N >= 603245*10^m for some
positive integer m ==> m+log(603246) > N*log(2) >= m+log(603245);
so 2^N begins with 603245 iff frac(log(603246)) > frac(N*log(2))
>= frac(log(603245)). If we are using natural density then N*log(2)
is uniformly distributed mod 1 since log(2) is irrational, hence the
probability is frac(log(603246)) - frac(log(603245)) =
frac(log(603246)-log(603245)) = frac(log(603246/603245)).
A neat observation is that since it is known p_n*c, where p_n is the
nth prime and c is irrational, is uniformly distributed mod 1, we get
the same answer if we replace 2^N with 2^{p_n}.
--
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
==> arithmetic/digits/nine.digits.p <==
Form a number using 0-9 once with its first n digits divisible by n.
==> arithmetic/digits/nine.digits.s <==
First, reduce the sample set. For each digit of ABCDEFGHI, such that the last
digit, (current digit), is the same as a multiple of N :
A: Any number 1-9
B: Even numbers 2,4,6,8 (divisible by 2).
C: Any number 1-9 (21,12,3,24,15,6,27,18,9).
D: Even numbers 2,4,6,8 (divisible by 4, every other even).
E: 5 (divisible by 5 and 0 not allowed).
F: Even numbers (12,24,6,18)
G: Any number 1-9 (21,42,63,14,35,56,7,28,49).
H: Even numbers (32,24,16,8)
I: Any number 1-9 (81,72,63,54,45,36,27,18,9)
Since E must be 5, I can eliminate it everywhere else.
Since I will use up all the even digits, (2,4,6,8) filling in those spots
that must be even. Any number becomes all odds, except 5.
A: 1,3,7,9
B: 2,4,6,8
C: 1,3,7,9
D: 2,4,6,8
E: 5
F: 2,4,6,8
G: 1,3,7,9
H: 2,4,6,8
I: 1,3,7,9
We have that 2C+D=0 (mod 4), and since C is odd,
this implies that D is 2 or 6; similarly we find that H is 2 or 6 ==>
{B,F} = {4,8}. D+5+F=0 (mod 3) ==> if D=2 then F=8, if D=6 then F=4.
We have two cases.
Assume our number is of the form A4C258G6I0. Now the case n=8 ==>
G=1,9; case n=3 ==> A+1+C=0 (mod 3) ==> {A,C}={1,7} ==> G=9, I=3.
The two numbers remaining fail for n=7.
Assume our number is of the form A8C654G2I0. The case n=8 ==> G=3,7.
If G=3, we need to check to see which of 1896543, 9816543, 7896543,
and 9876543 are divisible by 7; none are.
If G=7, we need to check to see which of 1896547, 9816547, 1836547,
and 3816547 are divisible by 7; only the last one is, which yields
the solution 3816547290.
==> arithmetic/digits/palindrome.p <==
Does the series formed by adding a number to its reversal always end in
a palindrome?
==> arithmetic/digits/palindrome.s <==
This is not known.
If you start with 196, after 9480000 iterations you get a 3924257-digit
non-palindromic number. However, there is no known proof that you will
never get a palindrome.
The statement is provably false for binary numbers. Roland Sprague has
shown that 10110 starts a series that never goes palindromic.
==> arithmetic/digits/palintiples.p <==
Find all numbers that are multiples of their reversals.
==> arithmetic/digits/palintiples.s <==
We are asked to find numbers that are integer multiples of their
reversals, which I call palintiples. Of course, all the palindromic
numbers are a trivial example, but if we disregard the unit multiples,
the field is narrowed considerably.
Rouse Ball (_Mathematical_recreations_and_essays_) originated the
problem, and G. H. Hardy (_A_mathematician's_apology_) used the result
that 9801 and 8712 are the only four-digit palintiples as an example
of a theorem that is not ``serious''. Martin Beech (_The_mathema-
tical_gazette_, Vol 74, #467, pp 50-51, March '90) observed that
989*01 and 879*12 are palintiples, an observation he ``confirmed'' on
a hand calculator, and conjectured that these are all that exist.
I confirm that Beech's numbers are palintiples, I will show that they
are not all of the palintiples. I will show that the palintiples do
not form a regular language. And then I will prove that I have found
all the palintiples, by describing the them with a generalized form
of regular expression. The results become more interesting in other
bases.
First, I have a more reasonable method of confirming that these
numbers are palintiples:
Proof: First, letting "9*" and "0*" refer an arbitrary string of
nines and a string of zeroes of the same length, I note that
879*12 = 879*00 + 12 = (880*00 - 100) + 12 = 880*00 - 88
219*78 = 219*00 + 78 = (220*00 - 100) + 78 = 220*00 - 22
989*01 = 989*00 + 1 = (990*00 - 100) + 1 = 990*00 - 99
109*89 = 109*00 + 89 = (110*00 - 100) + 89 = 110*00 - 11
It is obvious that 4x(220*00 - 22) = 880*00 - 88 and that
9x(110*00 - 11) = 990*00 - 99. QED.
Now, to show that these palintiples are not all that exist, let us
take the (infinite) language L[4] = (879*12 + 0*), and let Pal(L[4])
refer to the set of palindromes over the alphabet L[4]. It is
immediate that the numbers in Pal(L[4]) are palintiples. For
instance,
8712 000 87912 879999912 879999912 87912 000 8712
= 4 x 2178 000 21978 219999978 219999978 21978 000 2178
(where I have inserted spaces to enhance readability) is a palintiple.
Similarly, taking L[9] = (989*01 + 0*), the numbers in Pal(L[9]) are
palintiples. We exclude numbers starting with zeroes.
The reason these do not form a regular language is that the
sub-palintiples on the left end of the number must be the same (in
reverse order) as the sub-palintiples on the right end of the number:
8712 8712 87999912 = 4 x 2178 2178 21999978
is not a palintiple, because 8712 8712 87999912 is not the reverse of
2178 2178 21999978. The pumping lemma can be used to prove that
Pal(L[4])+Pal(L[9]) is not a regular language, just as in the familiar
proof that the palindromes over a non-singleton alphabet do not form a
regular language.
Now to characterize all the palintiples, let N be a palintiple,
N=CxR(N), where R(.) signifies reversal, and C>1 is an integer. (I
use "x" for multiplication, to avoid confusion with the Kleene star
"*", which signifies the concatenated closure.) If D is a digit of N,
let D' refer to the corresponding digit of R(N). Since N=CxR(N),
D+10T = CxD'+S, where S is the carry in to the position occupied by D'
when R(N) is multiplied by C, and T is the carry out of that position.
Similarly, D'+10T'=CxD+S', where S', T' are carries in and out of the
position occupied by D when R(N) is multiplied by C.
Since D and D' are so closely related, I will use the symbol D:D' to
refer to a digit D on the left side of a string with a corresponding
digit D' on the right side of the string. More formally, an
expression "x[1]:y[1] x[2]:y[2] ... x[n]:y[n] w" will refer to a
string "x[1] x[2] ... x[n] w y[n] ... y[2] y[1]", where the x[i] and
y[i] are digits and w is a string of zero or one digits. So 989901
may be written as 9:1 8:0 9:9 and 87912 may be written as 8:2 7:1 9.
Thus Pal(L[4])+Pal(L[9]) (omitting numbers with leading zeroes) can be
represented as
(8:2 7:1 9:9* 1:7 2:8 0:0*)*
(0:0* + 0 + 8:2 7:1 ( 9:9* + 9:9* 9))
+ (9:1 8:0 9:9* 0:8 1:9 0:0*)*
(0:0* + 0 + 9:1 8:0 ( 9:9* + 9:9* 9)). (1)
For each pair of digits D:D', there are a very limited--and often
empty--set of quadruples S,T,S',T' of digits that satisfy the
equations
D +10T =CxD'+S
D'+10T'=CxD +S', (2)
yet such a quadruple must exist for "D:D'" to appear in a palintiple
with multiplier C. Furthermore, the S and T' of one D:D' must be T
and S', respectively, of the next pair of digits that appear. This
enables us to construct a finite state machine to recognize those
palintiples. The states [X#Y] refer to a pair of carries in D and D',
and we allow a transition from state [T#S'] to state [S#T'] on input
symbol D:D' exactly when equations (2) are satisfied. Special
transitions for a single-digit input symbol (the central digit of
odd-length palintiples) and the criteria for the initial and the
accepting states are left as exercises. The finite state machines
thus formed are
State Symbol New Symbol New Symbol New
Accept? State State State
--> [0#0] Y 8:2 [0#3] 0:0 [0#0] 0 [A]
[0#3] N 7:1 [3#3]
[3#3] Y 1:7 [3#0] 9:9 [3#3] 9 [A]
[3#0] N 2:8 [0#0]
[A] Y
for constant C=4, and
State Symbol New Symbol New Symbol New
Accept? State State State
--> [0#0] Y 1:9 [0#8] 0:0 [0#0] 0 [A]
[0#8] N 8:0 [8#8]
[8#8] Y 0:8 [8#0] 9:9 [8#8] 9 [A]
[8#0] N 9:1 [0#0]
[A] Y
for constant C=9, and the finite state machines for other constants
accept only strings of zeroes. It is not hard to verify that the
proposed regular expression (1) represents the union of the languages
accepted by these machines, omitting the empty string and strings
beginning with zero.
I have written a computer program that constructs finite state
machines for recognizing palintiples for various bases and constants.
I found that base 10 is actually an unusually boring base for this
problem. For instance, the machine for base 8, constant C=5 is
State Symbol New Symbol New Symbol New
Accept? State State State
--> [0#0] Y 0:0 [0#0] 5:1 [0#3] 0 [A]
[0#3] N 1:0 [1#1] 6:1 [1#4]
[1#1] Y 0:1 [3#0] 5:2 [3#3]
[3#0] N 1:5 [0#0] 6:6 [0#3] 6 [A]
[3#3] Y 2:5 [1#1] 7:6 [1#4]
[1#4] N 1:1 [4#1] 6:2 [4#4] 1 [A]
[4#4] Y 2:6 [4#1] 7:7 [4#4] 7 [A]
[4#1] N 1:6 [3#0] 6:7 [3#3]
[A] Y
for which I invite masochists to write the regular expression. If
anyone wants more, I should remark that the base 29 machine for
constant C=18 has 71 states!
By the way, I did not find any way of predicting the size or form of
the machines for the various bases, except that the machines for C=B-1
all seem to be isomorphic to each other. If anyone investigates the
general behavior, I would be most happy to hear about it.
Dan Hoey
Ho...@AIC.NRL.Navy.Mil
May, 1992
[ A preliminary version of this message appeared in April, 1991. ]
================================================================
Dan
==> arithmetic/digits/power.two.p <==
Prove that for any 9-digit number (base 10) there is an integral power
of 2 whose first 9 digits are that number.
==> arithmetic/digits/power.two.s <==
Let v = log to base 10 of 2.
Then v is irrational.
Let w = log to base 10 of these 9 digits.
Since v is irrational, given epsilon > 0, there exists some natural number
n such that
{w} < {nv} < {w} + epsilon
({x} is the fractional part of x.) Let us pick n for when
epsilon = log 1.00000000000000000000001.
Then 2^n does the job.
==> arithmetic/digits/prime/101.p <==
How many primes are in the sequence 101, 10101, 1010101, ...?
==> arithmetic/digits/prime/101.s <==
Note that the sequence
101 , 10101, 1010101, ....
can be viewed as
100**1 +1, 100**2 + 100**1 + 1, 100**3 + 100**2 + 100**1 +1 ....
that is,
the k-th term in the sequence is
100**k + 100**(k-1) + 100**(k-2) + ...+ 100**(1) + 1
= (100)**(k+1) - 1
----------------
11 * 9
= (10)**(2k+2) - 1
----------------
11 * 9
= ((10)**(k+1) - 1)*((10)**(k+1) +1)
---------------------------------
11*9
thus either 11 and 9 divide the numerator. Either they both divide the
same factor in the numerator or different factors in the numerator. In
any case, after dividing, they leave the numerators as a product of two
integers. Only in the case of k = 1, one of the integers is 1. Thus
there is exactly one prime in the above sequence: 101.
==> arithmetic/digits/prime/all.prefix.p <==
What is the longest prime whose every proper prefix is a prime?
==> arithmetic/digits/prime/all.prefix.s <==
23399339, 29399999, 37337999, 59393339, 73939133
==> arithmetic/digits/prime/change.one.p <==
What is the smallest number that cannot be made prime by changing a single
digit? Are there infinitely many such numbers?
==> arithmetic/digits/prime/change.one.s <==
200. Obviously, you would have to change the last digit, but 201, 203,
207, and 209 are all composite. For any smaller number, you can change
the last digit, and get
2,11,23,31,41,53,61,71,83,97,101,113,127,131,149,151,163,173,181, or 191.
200+2310n gives an infinite family, because changing the last
digit to 1 or 7 gives a number divisible by 3; to 3, a number divisible
by 7; to 9, a number divisible by 11.
==> arithmetic/digits/prime/prefix.one.p <==
2 is prime, but 12, 22, ..., 92 are not. Similarly, 5 is prime
whereas 15, 25, ..., 95 are not. What is the next prime number
which is composite when any digit is prefixed?
==> arithmetic/digits/prime/prefix.one.s <==
149
==> arithmetic/digits/reverse.p <==
Is there an integer that has its digits reversed after dividing it by 2?
==> arithmetic/digits/reverse.s <==
Assume there's such a positive integer x such that x/2=y and y is the
reverse of x.
Then x=2y. Let x = a...b, then y = b...a, and:
b...a (y)
x 2
--------
a...b (x)
From the last digit b of x, we have b = 2a (mod 10), the possible
values for b are 2, 4, 6, 8 and hence possible values for (a, b) are
(1,2), (6,2), (2,4), (7,4), (3,6), (8,6), (4,8), (9,8).
From the first digit a of x, we have a = 2b or a = 2b+1. None of the
above pairs satisfy this condition. A contradiction.
Hence there's no such integer.
==> arithmetic/digits/rotate.p <==
Find integers where multiplying them by single digits rotates their digits
one position, so that the last digit become the first digit.
==> arithmetic/digits/rotate.s <==
2 105263157894736842
3 1034482758620689655172413793
4 102564 153846 179487 205128 230769
5 142857 102040816326530612244897959183673469387755
6 1016949152542372881355932203389830508474576271186440677966
1186440677966101694915254237288135593220338983050847457627
1355932203389830508474576271186440677966101694915254237288
1525423728813559322033898305084745762711864406779661016949
7 1014492753623188405797 1159420289855072463768 1304347826086956521739
8 1012658227848 1139240506329
9 10112359550561797752808988764044943820224719
In base B, suppose you have an N-digit answer A whose digits are
rotated when multiplied by K. If D is the low-order digit of A, we
have
(A-D)/B + D B^(N-1) = K A .
Solving this for A we have
D (B^N - 1)
A = ----------- .
B K - 1
In order for A >= B^(N-1) we must have D >= K. Now we have to find N
such that B^N-1 is divisible by R=(BK-1)/gcd(BK-1,D). This always has
a minimal solution N0(R,B)<R, and the set of all solutions is the set
of multiples of N0(R,B). N0(R,B) is the length of the repeating part
of the fraction 1/R in base B.
N0(ST,B)=N0(S,B)N0(T,B) when (S,T)=1, and for prime powers, N0(P^X,B)
divides (P-1)P^(X-1). Determining which divisor is a little more
complicated but well-known (cf. Hardy & Wright).
So given B and K, there is one minimal solution for each
D=K,K+1,...,B-1, and you get all the solutions by taking repetitions
of the minimal solutions.
==> arithmetic/digits/sesqui.p <==
Find the least number where moving the first digit to the end multiplies by 1.5.
==> arithmetic/digits/sesqui.s <==
Let's represent this number as a*10^n+b, where 1<=a<=9 and
b < 10^n. Then the condition to be satisfied is:
3/2(a*10^n+b) = 10b+a
3(a*10^n+b) = 20b+2a
3a*10^n+3b = 20b+2a
(3*10^n-2)a = 17b
b = a*(3*10^n-2)/17
So we must have 3*10^n-2 = 0 (mod 17) (since a is less than 10, it
cannot contribute the needed prime 17 to the factorization of 17b).
(Also, assuming large n, we must have a at most 5 so that b < 10^n will
be satisfied, but note that we can choose a=1). Now,
3*10^n-2 = 0 (mod 17)
3*10^n = 2 (mod 17)
10^n = 12 (mod 17)
A quick check shows that the smallest n which satisfies this is 15
(the fact that one exists was assured to us because 17 is prime). So,
setting n=15 and a=1 (obviously) gives us b=176470588235294, so the
number we are looking for is
1176470588235294
and, by the way, we can set a=2 to give us the second smallest such
number,
2352941176470588
Other things we can infer about these numbers is that there are 5 of
them less than 10^16, 5 more less than 10^33, etc.
==> arithmetic/digits/squares/change.leading.p <==
What squares remain squares when their leading digits are incremented?
==> arithmetic/digits/squares/change.leading.s <==
Omitting solutions that are obtained from smaller solutions by
multiplying by powers of 10, the squares of these numbers satisfy the
condition:
1. (105,145), (3375,4625), (14025,17225), (326625,454625),
(10846875,14753125), (43708125,53948125), ...
2. (45,55), (144375,175625), (463171875,560828125), ...
7. (2824483699753370361328125,2996282391593370361328125), ...
Here is how to find them. We have (y+x)*(y-x) = 10^n, and so we must have
{y+x, y-x} as {5^m*10^a, 2^m*10^b} in some order. It is also necessary (and
sufficient) that y/x lies in the interval [sqrt(3/2),sqrt(2)], or equivalently
that (y+x)/(y-x) lies in [3+sqrt(8),5+sqrt(24)] = [5.82842...,9.89897...].
Thus we need to make (5/2)^m*10^(a-b), or its reciprocal, in this range.
For each m there is clearly at most one power of 10 that will do. m=2,a=b
gives (105,145); m=3,b=a+2 gives (3375,4625), and so on.
There are infinitely many non-equivalent solutions, because log(5/2) / log(10)
is irrational.
One can use exactly the same argument to find squares whose initial 2 can
be replaced by a 3, of course, except that the range of (y+x)/(y-x) changes.
==> arithmetic/digits/squares/length.22.p <==
Is it possible to form two numbers A and B from 22 digits such that
A = B^2? Of course, leading digits must be non-zero.
==> arithmetic/digits/squares/length.22.s <==
No, the number of digits of A^2 must be of the form 3n or 3n-1.
==> arithmetic/digits/squares/length.9.p <==
Is it possible to make a number and its square, using the digits from 1
through 9 exactly once?
==> arithmetic/digits/squares/length.9.s <==
Yes, there are two such pairs: (567, 567^2=321489) and (854,854^2=729316).
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> analysis/bicycle.p <==
A boy, a girl and a dog go for a 10 mile walk. The boy and girl can
walk at 2 mph and the dog can trot at 4 mph. They also have a bicycle
which only one of them (including the dog!) can use at a time. When
riding, the boy and girl can travel at 12 mph while the dog can pedal
at 16 mph. What is the shortest time in which all three can complete
the trip?
==> analysis/bicycle.s <==
First note that there's no apparent way to benefit from letting either the
boy or girl ride the bike longer than the other. Any solution which gets the
boy there faster, must involve him using the bike (forward) more; similarly
for the girl. Thus the bike must go backwards more for it to remain within
the 10-mile route. Thus the dog won't make it there in time. So the solution
assumes they ride the bike for the same amount of time.
Also note that there's no apparent way to benefit from letting any of the three
arrive at the finish ahead of the others. If they do, they can probably take
time out to help the others. So the solution assumes they all finish at the
same time.
The boy starts off on the bike, and travels 5.4 miles. At this
point, he drops the bike and completes the rest of the trip on foot. The
dog eventually reaches the bike, and takes it *backward* .8 miles (so the
girl gets to it sooner) and then returns to trotting. Finally, the girl makes
it to the bike and rides it to the end. The answer is 2.75 hours.
The puzzle is in Vasek Chvatal, Linear Programming, W. H. Freeman & Co.
The generalized problem (n people, 1 bike, different walking and riding speeds)
is known as "The Bicycle Problem". A couple references are
Masuda, S. (1970). "The bicycle problem," University of California, Berkeley:
Operations Research Center Technical Report ORC 70-35.
Chvatal, V. (1983). "On the bicycle problem," Discrete Applied Mathematics 5:
pp. 165 - 173.
As for the linear program which gives the lower bound of 2.75 hours, let
t[person, mode, direction] by the amount of time "person" (boy, girl or dog)
is travelling by "mode" (walk or bike) in "direction" (forward or backwards).
Define Time[person] to be the total time spent by person doing each of these
four activities. The objective is to minimize the maximum of T[person], for
person = boy, girl, dog, e.g.
minimize T
subject to T >= T[boy], T >= T[girl], T >= T[dog].
Now just think of all the other linear constraints on the variables t[x,y,z],
such as everyone has to travel 10 miles, etc. In all, there are 8 contraints
in 18 variables (including slack variables). Solving this program yields the
lower bound.
==> analysis/boy.girl.dog.p <==
A boy, a girl and a dog are standing together on a long, straight road.
Simulataneously, they all start walking in the same direction:
The boy at 4 mph, the girl at 3 mph, and the dog trots back and forth
between them at 10 mph. Assume all reversals of direction instantaneous.
In one hour, where is the dog and in which direction is he facing?
==> analysis/boy.girl.dog.s <==
The dog's position and direction are indeterminate, other than that the
dog must be between the boy and girl (endpoints included). To see this,
simply time reverse the problem. No matter where the dog starts out,
the three of them wind up together in one hour.
This argument is not quite adequate. It is possible to construct problems
where the orientation changes an infinite number of times initially, but for
which there can be a definite result. This would be the case if the positions
at time t are uniformly continuous in the positions at time s, s small.
But suppose that at time a the dog is with the girl. Then the boy is at
4a, and the time it takes the dog to reach the boy is a/6, because
the relative speed is 6 mph. So the time b at which the dog reaches the
boy is proportional to a. A similar argument shows that the time the
dog next reaches the girl is b + b/13, and is hence proportional to b.
This makes the position of the dog at time (t > a) a periodic function of
the logarithm of a, and thus does not approach a limit as a -> 0.
==> analysis/bugs.p <==
Four bugs are placed at the corners of a square. Each bug walks always
directly toward the next bug in the clockwise direction. How far do
the bugs walk before they meet?
==> analysis/bugs.s <==
Since the bugs start out walking perpendicularly, and there is nothing
in the problem to alter this symmetry, the bugs are always walking
perpendicularly. Since each bug is walking perpendicularly to the line
separating it from the bug chasing it, the gap is closing at the speed
of the chasing bug. Therefore, each bug walks a distance equal to the
side of the square before it meets the next bug.
In order to conveniently express the equation of the bugs' motion,
use standard polar coordinates, and let the first bug's position at
any instant be (r(t), theta(t)). Assume the initial square is
centered at the origin.
Then by symmetry the bugs are always at four corners of some square
centered at the origin, and if they meet they meet at the origin.
Also, each bug is always walking in a direction pi/4 (45 degrees) away
from the radial line to the origin. This means that in the limit as
the time step goes to zero, the bug travels sec(pi/4) = sqrt(2) units
along its path for every unit of progress made good toward the center.
Since the corners are initially d/sqrt(2) distance from the center,
each bug travels distance d before they meet, assuming they meet at
all.
Since a bug's path always crosses radial lines at angle psi = pi/4,
the path is a logarithmic spiral with angle psi = pi/4 and equation
r(t) = e^(a*theta(t) + b)
Moreover since the bugs walk clockwise, both r(t) and theta(t)
decrease as t increases, in other words r increases as theta
increases, hence a is positive. Also, psi = pi/4 gives us
d(r)/d(theta) = r
(this is easiest to see by drawing the path for a small time step
delta-t and taking the limit as delta-t goes to 0). The solution is
r(t) = e^(theta(t) + b)
(that is, a = 1). As we know, this spiral makes infinitely many
"wraps" around the origin as the radius approaches zero, but it does
have finite length from any point inward and its limit point is the
origin, where the bugs will meet (unless one wants to quibble about
the behavior at the exact limit point).
How much closer is the bug to the origin after its first complete cycle
around the origin? Recall that r(0) = d/sqrt(2). As the bug walks
clockwise around the origin, after one full circuit its angle decreases
from theta(0) to theta(t1), where t1 (the time at which full circuit
occurs) is defined by
theta(t1) = theta(0) - 2*pi
Hence
r(0) = e^(theta(0) + b)
r(t1) = e^(theta(0) - 2*pi + b)
r(t1)/r(0) = e^(-2*pi)
The quantity we want is
r(0) - r(t1) = r(0)*(1 - e^(-2*pi))
= d * (1 - e^(-2*pi))/sqrt(2)
==> analysis/c.infinity.p <==
What function is zero at zero, strictly positive elsewhere, infinitely
differentiable at zero and has all zero derivatives at zero?
==> analysis/c.infinity.s <==
exp(-1/x^2)
There are infinitely many other such functions.
This tells us why Taylor Series are a more limited device than they might be.
We form a Taylor series by looking at the derivatives of a function at a given
point; but this example shows us that the derivatives at a point may tell us
almost nothing about its behavior away from that point.
==> analysis/cache.p <==
Cache and Ferry (How far can a truck go in a desert?)
A pick-up truck is in the desert beside N 50-gallon gas drums, all full.
The truck's gas tank holds 10 gallons and is empty. The truck can carry
one drum, whether full or empty, in its bed. It gets 10 miles to the gallon.
How far away from the starting point can you drive the truck?
==> analysis/cache.s <==
If the truck can siphon gas out of its tank and leave it in the cache,
the answer is:
{ 1/1 + 1/3 + ... + 1/(2 * N - 1) } x 500 miles.
A much harder problem occurs when the truck can siphon gas, but if it does,
it must siphon the gas into one of the initial barrels.
Now, remarkably, for initial gas values of 50, 100, 150, 200, 250, 300,
and 350 gallons, the two problems give identical optimal distances, viz.
gal dist
--- ----
50 500
100 733.3333
150 860
200 948.5714
250 1016.8254
300 1072.3810
350 1117.8355
However, for the 8 barrel case (400 gallons), the unlimited cache optimal
distance is 1157.6957, but limited cache is only 1157.2294.
What happened is that the unlimited cache optimal solution has started to
siphon out more than 50 gallons (60-80/13 to be exact) of gas on trips
to the first depot. With a limited cache, the truck can only leave a
maximum of 50 gallons (one barrel worth), and does not have a big
enough gas tank (only ten gallons) to carry around the excess.
The limited cache problem is the same as the "Desert Fox" problem
described, but not solved, by Dewdney, July '87 "Scientific American".
Dewdney's Oct. '87 Sci. Am. article gives for N=2, the optimal distance
of 733.33 miles.
In the Nov. issue, Dewdney lists the optimal distance of 860 miles for
N=3, and gives a better, but not optimal, general distance formula.
Westbrook, in Vol 74, #467, pp 49-50, March '90 "Mathematical Gazette",
gives an even better formula, for which he incorrectly claims optimality:
For N = 2,3,4,5,6:
Dist = (600/1 + 600/3 + ... + 600/(2N-3)) + (600-100N)/(2N-1)
For N > 6:
Dist = (600/1 + 600/3 + ... + 600/9) + (500/11 + ... + 500/(2N-3))
The following shows that Westbrook's formula is not optimal for N=8:
Ferry 7 drums forward 33.3333 miles (356.6667 gallons remain)
Ferry 6 drums forward 51.5151 miles (300.0000 gallons remain)
Ferry 5 drums forward 66.6667 miles (240.0000 gallons remain)
Ferry 4 drums forward 85.7143 miles (180.0000 gallons remain)
Ferry 3 drums forward 120.0000 miles (120.0000 gallons remain)
Ferry 2 drums forward 200.0000 miles ( 60.0000 gallons remain)
Ferry 1 drums forward 600.0000 miles
---------------
Total distance = 1157.2294 miles
(Westbrook's formula = 1156.2970 miles)
["Ferrying n drums forward x miles" involves (2*n-1) trips,
each of distance x.]
Other attainable values I've found:
N Distance
--- --------- (Ferry distances for each N are omitted for brevity.)
5 1016.8254
7 1117.8355
11 1249.2749
13 1296.8939
17 1372.8577
19 1404.1136 (The N <= 19 distances could be optimal.)
31 1541.1550 (I doubt that this N = 31 distance is optimal.)
139 1955.5702 (Attainable and probably optimal.)
So...where's MY formula?
I haven't found one, and believe me, I've looked.
I would be most grateful if someone would end my misery by mailing me
a formula, a literature reference, or even an efficient algorithm that
computes the optimal distance.
If you do come up with the solution, you might want to first check it
against the attainable distances listed above, before sending it out.
(Not because you might be wrong, but just as a mere formality to check
your work.)
[Warning: the Mathematician General has determined that
this problem is as addicting as Twinkies.]
Myron P. Souris
EDS/Unigraphics
sou...@ug.eds.com
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
The following output comes from some hack programs that I've used to
empirically verify some proofs I've been working on.
Initial barrels: 12 (600 gallons)
Attainable distance= 1274.175211
Barrels Distance Gas
Moved covered left
>From depot 1: 10 63.1579 480.0000
>From depot 2: 8 50.0000 405.0000
>From depot 3: 7 37.5000 356.2500
>From depot 4: 6 51.1364 300.0000
>From depot 5: 5 66.6667 240.0000
>From depot 6: 4 85.7143 180.0000
>From depot 7: 3 120.0000 120.0000
>From depot 8: 2 200.0000 60.0000
>From depot 9: 1 600.0000 0.0000
Initial barrels: 40 (2000 gallons)
Attainable distance= 1611.591484
Barrels Distance Gas
Moved covered left
>From depot 1: 40 2.5316 1980.0000
>From depot 2: 33 50.0000 1655.0000
>From depot 3: 28 50.0000 1380.0000
>From depot 4: 23 53.3333 1140.0000
>From depot 5: 19 50.0000 955.0000
>From depot 6: 16 56.4516 780.0000
>From depot 7: 13 50.0000 655.0000
>From depot 8: 11 54.7619 540.0000
>From depot 9: 9 50.0000 455.0000
>From depot 10: 8 32.1429 406.7857
>From depot 11: 7 38.9881 356.1012
>From depot 12: 6 51.0011 300.0000
>From depot 13: 5 66.6667 240.0000
>From depot 14: 4 85.7143 180.0000
>From depot 15: 3 120.0000 120.0000
>From depot 16: 2 200.0000 60.0000
>From depot 17: 1 600.0000 0.0000
==> analysis/calculate.pi.p <==
How can I calculate many digits of pi?
==> analysis/calculate.pi.s <==
long a=10000,
b,
c=2800,
d,e,
f[2801],
g;
main()
{
for (;b-c;) f[b++]=a/5;
for (;
d=0,g=c*2;
c-=14, printf("%.4d",e+d/a), e=d%a)
for (b=c;
d+=f[b]*a,f[b]=d%--g,d/=g--,--b;
d*=b);
}
==> analysis/cats.and.rats.p <==
If 6 cats can kill 6 rats in 6 minutes, how many cats does it take to
kill one rat in one minute?
==> analysis/cats.and.rats.s <==
The following piece by Lewis Carroll first appeared in ``The Monthly
Packet'' of February 1880 and is reprinted in _The_Magic_of_Lewis_Carroll_,
edited by John Fisher, Bramhall House, 1973.
/Larry Denenberg
la...@bbn.com
la...@harvard.edu
Cats and Rats
If 6 cats kill 6 rats in 6 minutes, how many will be needed to kill 100
rats in 50 minutes?
This is a good example of a phenomenon that often occurs in working
problems in double proportion; the answer looks all right at first, but,
when we come to test it, we find that, owing to peculiar circumstances in
the case, the solution is either impossible or else indefinite, and needing
further data. The 'peculiar circumstance' here is that fractional cats or
rats are excluded from consideration, and in consequence of this the
solution is, as we shall see, indefinite.
The solution, by the ordinary rules of Double Proportion, is as follows:
6 rats : 100 rats \
> :: 6 cats : ans.
50 min. : 6 min. /
.
. . ans. = (100)(6)(6)/(50)(6) = 12
But when we come to trace the history of this sanguinary scene through all
its horrid details, we find that at the end of 48 minutes 96 rats are dead,
and that there remain 4 live rats and 2 minutes to kill them in: the
question is, can this be done?
Now there are at least *four* different ways in which the original feat,
of 6 cats killing 6 rats in 6 minutes, may be achieved. For the sake of
clearness let us tabulate them:
A. All 6 cats are needed to kill a rat; and this they do in one minute,
the other rats standing meekly by, waiting for their turn.
B. 3 cats are needed to kill a rat, and they do it in 2 minutes.
C. 2 cats are needed, and do it in 3 minutes.
D. Each cat kills a rat all by itself, and take 6 minutes to do it.
In cases A and B it is clear that the 12 cats (who are assumed to come
quite fresh from their 48 minutes of slaughter) can finish the affair in
the required time; but, in case C, it can only be done by supposing that 2
cats could kill two-thirds of a rat in 2 minutes; and in case D, by
supposing that a cat could kill one-third of a rat in two minutes. Neither
supposition is warranted by the data; nor could the fractional rats (even
if endowed with equal vitality) be fairly assigned to the different cats.
For my part, if I were a cat in case D, and did not find my claws in good
working order, I should certainly prefer to have my one-third-rat cut off
from the tail end.
In cases C and D, then, it is clear that we must provide extra cat-power.
In case C *less* than 2 extra cats would be of no use. If 2 were supplied,
and if they began killing their 4 rats at the beginning of the time, they
would finish them in 12 minutes, and have 36 minutes to spare, during which
they might weep, like Alexander, because there were not 12 more rats to
kill. In case D, one extra cat would suffice; it would kill its 4 rats in
24 minutes, and have 24 minutes to spare, during which it could have killed
another 4. But in neither case could any use be made of the last 2
minutes, except to half-kill rats---a barbarity we need not take into
consideration.
To sum up our results. If the 6 cats kill the 6 rats by method A or B,
the answer is 12; if by method C, 14; if by method D, 13.
This, then, is an instance of a solution made `indefinite' by the
circumstances of the case. If an instance of the `impossible' be desired,
take the following: `If a cat can kill a rat in a minute, how many would be
needed to kill it in the thousandth part of a second?' The *mathematical*
answer, of course, is `60,000,' and no doubt less than this would *not*
suffice; but would 60,000 suffice? I doubt it very much. I fancy that at
least 50,000 of the cats would never even see the rat, or have any idea of
what was going on.
Or take this: `If a cat can kill a rat in a minute, how long would it be
killing 60,000 rats?' Ah, how long, indeed! My private opinion is that
the rats would kill the cat.
==> analysis/dog.p <==
A body of soldiers form a 50m-by-50m square ABCD on the parade ground.
In a unit of time, they march forward 50m in formation to take up the
position DCEF. The army's mascot, a small dog, is standing next to its
handler at location A. When the
B----C----E soldiers start marching, the dog
| | | forward--> begins to run around the moving
A----D----F body in a clockwise direction,
keeping as close to it as possible.
When one unit of time has elapsed, the dog has made one complete
circuit and has got back to its handler, who is now at location D. (We
can assume the dog runs at a constant speed and does not delay when
turning the corners.)
How far does the dog travel?
==> analysis/dog.s <==
Let L be the side of the square, 50m, and let D be the distance the
dog travels.
Let v1 be the soldiers' marching speed and v2 be the speed of the dog.
Then v1 = L / (1 time unit) and v2 = v1*D/L.
Let t1, t2, t3, t4 be the time the dog takes to traverse each side of
the square, in order. Find t1 through t4 in terms of L and D and solve
t1+t2+t3+t4 = 1 time unit.
While the dog runs along the back edge of the square in time t1, the
soldiers advance a distance d=t1*v1, so the dog has to cover a distance
sqrt(L^2 + (t1*v1)^2), which takes a time t1=sqrt(L^2 + (t1*v1)^2)/v2.
Solving for t1 gives t1=L/sqrt(v2^2 - v1^2).
The rest of the times are t2 = L/(v2-v1), t3 = t1, and t4 = L/(v2+v1).
In t1+t2+t3+t4, eliminate v2 by using v2=v1*D/L and eliminate v1 by
using v1=L/(1 time unit), obtaining
2 L (D + sqrt(D^2-L^2)) / (D^2 - L^2) = 1
which can be turned into
D^4 - 4LD^3 - 2L^2D^2 + 4L^3D + 5L^4 = 0
which has a root D = 4.18113L = 209.056m.
==> analysis/e.and.pi.p <==
Without finding their numerical values, which is greater, e^(pi) or (pi)^e?
==> analysis/e.and.pi.s <==
e^(pi). Put x = pi/e - 1 in the inequality e^x > 1+x (x>0).
==> analysis/functional/distributed.p <==
Find all f: R -> R, f not identically zero, such that
(*) f( (x+y)/(x-y) ) = ( f(x)+f(y) )/( f(x)-f(y) ).
==> analysis/functional/distributed.s <==
1) Assuming f finite everywhere, (*) ==> x<>y ==> f(x)<>f(y)
2) Exchanging x and y in (*) we see that f(-x) = -f(x).
3) a <> 0 ==> f((a-a)/(a+a)) = (f(a)-f(a))/(f(a)+f(a)) ==> f(0) = 0.
4) a <> 0 ==> f((a+0)/(a-0)) = f(a)/f(a) ==> f(1) = 1.
5) x<>y, y<>0 ==> f(x/y) =
f( ((x+y)/(x-y) + (x-y)/(x-y)) / ((x+y)/(x-y) - (x-y)/(x-y)) = f(x)/f(y)
==> f(xy) = f(x)f(y) by replacing x with xy and by noting that
f(x*1) = f(x)*1 and f(x*0) = f(x)*f(0).
6) f(x*x) = f(x)*f(x) ==> f(x) > 0 if x>0.
7) Let a=1+\/2, b=1-\/2; a,b satisfy (x+1)/(x-1) = x ==>
f(x) = (f(x)+1)/(f(x)-1) ==> f(a)=a, f(b)=b. f(1/\/2) = f((a+b)/(a-b))
= (a+b)/(a-b) = 1/\/2 ==> f(2) = 2.
8) By induction and the relation f((n+1)/(n-1)) = (f(n)+1)/(f(n)-1)
we get that f(n)=n for all integer n. #5 now implies that f fixes
the rationals.
9) If x>y>0 (*) ==> f(x) - f(y) = f(x+y)/f((x+y)/(x-y)) > 0 by #6.
Thus f is order-preserving.
Since f fixes the rationals *and* f is order-preserving, f must be the
identity function.
This was E2176 in _The American Mathematical Monthly_ (the proposer was
R. S. Luthar).
==> analysis/functional/linear.p <==
Suppose f is non-decreasing with
f(x+y) = f(x) + f(y) + C for all real x, y.
Prove: there is a constant A such that f(x) = Ax - C for all x.
(Note: continuity of f is not assumed in advance.)
==> analysis/functional/linear.s <==
By induction f(mx) = m(f(x)+C)-C. Let x=1/n, m=n and find that
f(1/n) = (1/n)(f(1)+C)-C. Now let x=1/n and find that f(m/n) =
(m/n)(f(1)+C)-C. f(-x+x) = f(-x) + f(x) + C ==> f(-x) = -2C - f(x)
(since f(0) = -C) ==> f(-m/n) = -(m/n)(f(1)+C)-C. Since f is
monotonic ==> f(x) = x*(f(1)+C)-C for all real x (Squeeze Theorem).
==> analysis/integral.p <==
If f is integrable on (0,inf) and differentiable at 0, and a > 0, and:
inf f(x)
Int ---------------- dx is defined
0 x
show:
inf ( f(x) - f(ax) )
Int ---------------- dx = f(0) ln(a)
0 x
==> analysis/integral.s <==
First, note that if f(0) is 0, then by substituting u=ax in
the integral of f(x)/x, our integral is the difference of two
equal integrals and so is 0 (the integrals are finite because f is
0 at 0 and differentiable there. Note I make no requirement of
continuity).
Second, note that if f is the characteristic function of the
interval [0, 1]--- i.e.
1, 0<=x<=1
f (x) =
0 otherwise
then a little arithmetic reduces our integral to that of
1/x from 1/a to 1 (assuming a>1; if a <= 1 the reasoning is similar),
which is ln(a) = f(0)ln(a) as required. Call this function g.
Finally, note that the operator which takes the function f to the
value of our integral is linear, and that every function meeting the
hypotheses (incidentally, I should have said `differentiable from the right',
or else replaced the characteristic function of [0,1] above by that of
(-infinity, 1]; but it really doesn't matter) is a linear combination of
one which is 0 at 0 and g, to wit
f(x) = f(0)g(x) + (f(x) - g(x)f(0)).
==> analysis/irrational.stamp.p <==
You have an ink stamp which is so amazingly precise that, when inked
and pressed down on the plane, it makes every circle of irrational
radius (centered at the center of the stamp) black.
Question: Can one use the stamp three times and make every point
in the plane black? [assume plane was white to begin with, and
ignore the fact that no such stamp is physically possible]
==> analysis/irrational.stamp.s <==
Yes. Center the stamp at (0,0), (1,0), and (0,pi).
Suppose there is a point (x,y) which is not covered.
Then there are rational numbers a,b,c satisfying the following equations:
(1) x^2 + y^2 = a
(2) (x-1)^2 + y^2 = b
(3) x^2 + (y-pi)^2 = c
Subtract (2) from (1) and solve for x. Thus x is rational.
From equation (2), y is algebraic. But equation (3) implies
that y is transcendental, contradiction.
==> analysis/minimum.time.p <==
N people can walk or drive in a two-seater to go from city A to city B. What
is the minimum time required to do so?
==> analysis/minimum.time.s <==
Let the speed of the car be v, the speed of a walking person be u, and
the distance between cities by L. We want to minimize T, the time t
at which all persons are at displacement x=L (city B), when they all
start at displacement x=0 (city A) at time t=0.
I'll assume that the solution has everyone starting out from city A at
the same time t=0 arriving in city B at the same time t=T so nobody is
standing around idly waiting. Let's plot everyone's movements on a
graph showing coordinates (t,x). Then at time t just after 0, (N-2)
walkers are on the line L0 through (0,0) with slope dx/dt = u, and 2
in the car are on a line through (0,0) with slope v, and at t just
before T, (N-2) walkers are on the line L1 through (T,L) with slope u,
and 2 in the car are on a line through (T,L) with slope v. Obviously
L1 lies "above" L0 (greater x coordinate given the same t coordinate).
In between t=0 and t=T, the car zigzags between L0 and L1 along lines
of slope v and -v, picking up people from L0 and dropping them off at
L1. I will not prove that this is the optimal strategy; in fact you
can make an infinite number of variations on it which all come up with
the same elapsed time.
Now examine the graph again. Say the car travels distance r between
picking someone up and dropping that person off, and distance s back
to pick up the next person. The car makes (N-1) trips forward and
(N-2) trips back to pick up and ferry everyone, so its total travel is
vT = (N - 1)r + (N - 2)s = (N - 2)(r + s) + r
Moreover the car makes (r-s) net displacement on each round trip, plus
r displacement on the extra forward trip, so
L = (N - 2)(r - s) + r
Note that a person walks distance (r-s) in the time it takes the car to
go (r+s), so
r - s = (u/v)(r + s)
A little algebraic manipulation of this equation shows us that
r - s = r * 2u/(v + u)
r + s = r * 2v/(v + u)
Plug this into the equation for L, and we get the first important
piece of information, how far the car should drive before dropping off
the passenger (once you know this, you tell it to the driver and this
guarantees the people get to B in minimum time):
L = r + (N - 2) r * 2u/(v + u)
= r * (v + u + (N - 2)*2u)/(v + u)
v + u
r = ------------ L
2uN + v - 3u
We can also find out what the elapsed time T will be:
vT = (N - 2)r*2v/(v + u) + r
= r * ((N - 2)*2v + v + u)/(v + u)
1 2vN - 3v + u
T = - * ------------ r
v v + u
Therefore
2vN - (3v - u)
T = -------------- (L/v)
2uN + v - 3u
-- David Karr (ka...@cs.cornell.edu)
==> analysis/particle.p <==
What is the longest time that a particle can take in travelling between two
points if it never increases its acceleration along the way and reaches the
second point with speed V?
==> analysis/particle.s <==
Assumptions:
1. x(0) = 0; x(T) = X
2. v(0) = 0; v(T) = V
3. d(a)/dt <= 0
Solution:
a(t) = constant = A = V^2/2X which implies T = 2X/V.
Proof:
Consider assumptions as they apply to f(t) = A * t - v(t):
1. integral from 0 to T of f = 0
2. f(0) = f(T) = 0
3. d^2(f)/dt^2 <= 0
From the mean value theorem, f(t) = 0. QED.
==> analysis/period.p <==
What is the least possible integral period of the sum of functions
of periods 3 and 6?
==> analysis/period.s <==
Period 2. Clearly, the sum of periodic functions of periods 2 and
three is 6. So take the function which is the sum of that function of
period six and the negative of the function of period three and you
have a function of period 2.
This proves that a period-2 solution exists, but not that it is minimal.
Since we're talking about integers, the only lower possibility is 1.
But the sum or difference of a period-1 and a period-3 function must have
period 3, not 6, therefore 1 is indeed impossible.
==> analysis/rubberband.p <==
A bug walks down a rubber band which is attached to a wall at one end and a car
moving away from the wall at the other end. The car is moving at 1 m/sec while
the bug is only moving at 1 cm/sec. Assuming the rubber band is uniformly and
infinitely elastic, will the bug ever reach the car?
==> analysis/rubberband.s <==
Let w = speed of bug and N = ratio of car speed/bug speed = 100. Paint N+1
equally spaced stripes on the rubberband. When the bug is standing on one
stripe, the next stripe is moving away from him at a speed slightly < w
(relative to him). Since he is walking at w, clearly the bug can reach
the next stripe. But once he reaches that stripe, the next one is only
receeding at < w. So he walks on down to the car, one stripe at a time.
The bug starts gaining on the car when he is at the next to last stripe.
==> analysis/sequence.p <==
Show that in the sequence: x, 2x, 3x, .... (n-1)x (x can be any real number)
there is at least one number which is within 1/n of an integer.
==> analysis/sequence.s <==
Throw 0 into the sequence; there are now n numbers, so some pair must
have fractional parts within 1/n of each other; their difference is
then within 1/n of an integer.
==> analysis/snow.p <==
Snow starts falling before noon on a cold December day. At noon a
snow plow starts plowing a street. It travels 1 mile in the first hour,
and 1/2 mile in the second hour. What time did the snow start
falling??
You may assume that the plow's rate of travel is inversely proportional
to the height of the snow, and that the snow falls at a uniform rate.
==> analysis/snow.s <==
11:22:55.077 am.
Method:
Let b = the depth of the snow at noon, a = the rate of increase in the
depth. Then the depth at time t (where noon is t=0) is at+b, the
snowfall started at t_0=-b/a, and the snowplow's rate of progress is
ds/dt = k/(at+b).
If the snowplow starts at s=0 then s(t) = (k/a) log(1+at/b). Note that
s(2 hours) = 1.5 s(1 hour), or log(1+2A/b) = 1.5 log(1+A/b), where
A = (1 hour)*a. Letting x = A/b we have (1+2x)^2 = (1+x)^3. Solve for
x and t_0 = -(1 hour)/x.
The exact answer is 11:(90-30 Sqrt[5]).
_American Mathematics Monthly_, April 1937, page 245
E 275. Proposed by J. A. Benner, Lafayette College, Easton. Pa.
The solution appears, appropriately, in the December 1937 issue,
pp. 666-667. Also solved by William Douglas, C. E. Springer,
E. P. Starke, W. J. Taylor, and the proposer.
See R.P. Agnew, "Differential Equations," 2nd edition, p. 39 ff.
==> analysis/tower.p <==
R = N ^ (N ^ (N ^ ...)). What is the maximum N>0 that will yield a finite R?
==> analysis/tower.s <==
ANSWER: e^(1/e)
Let N be the number in question and R the result of the process. Then
R can be defined recursively by the equation:
(1) R = N^R
Taking the logarithm of both sides of (1):
(2) ln(R) = R * ln(N)
Dividing (2) by R and rearranging:
(3) ln(N) = ln(R) / R
Exponentiating (3):
(4) N = R^(1/R)
We wish to find the maximum value of N with respect to R. Find the
derivative of N with respect to R and set it equal to zero:
(5) d(N)/d(R) = (1 - ln(R)) / R^2 = 0
For finite values of R, (5) is satisfied by R = e. This is a maximum of
N if the second derivative of N at R = e is less than zero.
(6) d2(N)/d2(R) | R=e = (2 * ln(R) - 3) / R^3 | R=e = -1 / e^3 < 0
The solution therefore is (4) at R = e:
(7) Nmax = e^(1/e)
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> cryptology/Beale.p <==
What are the Beale ciphers?
==> cryptology/Beale.s <==
The Beale ciphers are one of the greatest unsolved puzzles of all
time. About 100 years ago, a fellow by the name of Beale supposedly
buried two wagons-full of silver-coin filled pots in Bedford County,
near Roanoke. There are local rumors about the treasure being buried
near Bedford Lake.
He wrote three encoded letters telling what was buried, where it was
buried, and who it belonged to. He entrusted these three letters to a
friend and went west. He was never heard from again.
Several years later, someone examined the letters and was able to break
the code used in the second letter. The code used the text from the
Declaration of Independence. A number in the letter indicated which
word in the document was to be used. The first letter of that word
replaced the number. For example, if the first three words of the
document were "We hold these truths", the number 3 in the letter would
represent the letter t.
One of the remaining letters supposedly contains directions on how to find
the treasure. To date, no one has solved the code. It is believed that
both of the remaining letters are encoded using either the same document
in a different way, or another very public document.
For those interested, write to:
The Beale Cypher Association
P.O. Box 975
Beaver Falls, PA 15010
Item #904 is the 1885 pamphlet version ($5.00). #152 is the
Cryptologia article by Gillogly that argues the hoax side ($2.00).
A year's membership is $25, and includes 4 newsletters.
TEXT for part 1
The Locality of the Vault.
71,194,38,1701,89,76,11,83,1629,48,94,63,132,16,111,95,84,341
975,14,40,64,27,81,139,213,63,90,1120,8,15,3,126,2018,40,74
758,485,604,230,436,664,582,150,251,284,308,231,124,211,486,225
401,370,11,101,305,139,189,17,33,88,208,193,145,1,94,73,416
918,263,28,500,538,356,117,136,219,27,176,130,10,460,25,485,18
436,65,84,200,283,118,320,138,36,416,280,15,71,224,961,44,16,401
39,88,61,304,12,21,24,283,134,92,63,246,486,682,7,219,184,360,780
18,64,463,474,131,160,79,73,440,95,18,64,581,34,69,128,367,460,17
81,12,103,820,62,110,97,103,862,70,60,1317,471,540,208,121,890
346,36,150,59,568,614,13,120,63,219,812,2160,1780,99,35,18,21,136
872,15,28,170,88,4,30,44,112,18,147,436,195,320,37,122,113,6,140
8,120,305,42,58,461,44,106,301,13,408,680,93,86,116,530,82,568,9
102,38,416,89,71,216,728,965,818,2,38,121,195,14,326,148,234,18
55,131,234,361,824,5,81,623,48,961,19,26,33,10,1101,365,92,88,181
275,346,201,206,86,36,219,324,829,840,64,326,19,48,122,85,216,284
919,861,326,985,233,64,68,232,431,960,50,29,81,216,321,603,14,612
81,360,36,51,62,194,78,60,200,314,676,112,4,28,18,61,136,247,819
921,1060,464,895,10,6,66,119,38,41,49,602,423,962,302,294,875,78
14,23,111,109,62,31,501,823,216,280,34,24,150,1000,162,286,19,21
17,340,19,242,31,86,234,140,607,115,33,191,67,104,86,52,88,16,80
121,67,95,122,216,548,96,11,201,77,364,218,65,667,890,236,154,211
10,98,34,119,56,216,119,71,218,1164,1496,1817,51,39,210,36,3,19
540,232,22,141,617,84,290,80,46,207,411,150,29,38,46,172,85,194
39,261,543,897,624,18,212,416,127,931,19,4,63,96,12,101,418,16,140
230,460,538,19,27,88,612,1431,90,716,275,74,83,11,426,89,72,84
1300,1706,814,221,132,40,102,34,868,975,1101,84,16,79,23,16,81,122
324,403,912,227,936,447,55,86,34,43,212,107,96,314,264,1065,323
428,601,203,124,95,216,814,2906,654,820,2,301,112,176,213,71,87,96
202,35,10,2,41,17,84,221,736,820,214,11,60,760
TEXT for part 2
(no title exists for this part)
115,73,24,807,37,52,49,17,31,62,647,22,7,15,140,47,29,107,79,84
56,239,10,26,811,5,196,308,85,52,160,136,59,211,36,9,46,316,554
122,106,95,53,58,2,42,7,35,122,53,31,82,77,250,196,56,96,118,71
140,287,28,353,37,1005,65,147,807,24,3,8,12,47,43,59,807,45,316
101,41,78,154,1005,122,138,191,16,77,49,102,57,72,34,73,85,35,371
59,196,81,92,191,106,273,60,394,620,270,220,106,388,287,63,3,6
191,122,43,234,400,106,290,314,47,48,81,96,26,115,92,158,191,110
77,85,197,46,10,113,140,353,48,120,106,2,607,61,420,811,29,125,14
20,37,105,28,248,16,159,7,35,19,301,125,110,486,287,98,117,511,62
51,220,37,113,140,807,138,540,8,44,287,388,117,18,79,344,34,20,59
511,548,107,603,220,7,66,154,41,20,50,6,575,122,154,248,110,61,52,33
30,5,38,8,14,84,57,540,217,115,71,29,84,63,43,131,29,138,47,73,239
540,52,53,79,118,51,44,63,196,12,239,112,3,49,79,353,105,56,371,557
211,505,125,360,133,143,101,15,284,540,252,14,205,140,344,26,811,138
115,48,73,34,205,316,607,63,220,7,52,150,44,52,16,40,37,158,807,37
121,12,95,10,15,35,12,131,62,115,102,807,49,53,135,138,30,31,62,67,41
85,63,10,106,807,138,8,113,20,32,33,37,353,287,140,47,85,50,37,49,47
64,6,7,71,33,4,43,47,63,1,27,600,208,230,15,191,246,85,94,511,2,270
20,39,7,33,44,22,40,7,10,3,811,106,44,486,230,353,211,200,31,10,38
140,297,61,603,320,302,666,287,2,44,33,32,511,548,10,6,250,557,246
53,37,52,83,47,320,38,33,807,7,44,30,31,250,10,15,35,106,160,113,31
102,406,230,540,320,29,66,33,101,807,138,301,316,353,320,220,37,52
28,540,320,33,8,48,107,50,811,7,2,113,73,16,125,11,110,67,102,807,33
59,81,158,38,43,581,138,19,85,400,38,43,77,14,27,8,47,138,63,140,44
35,22,177,106,250,314,217,2,10,7,1005,4,20,25,44,48,7,26,46,110,230
807,191,34,112,147,44,110,121,125,96,41,51,50,140,56,47,152,540
63,807,28,42,250,138,582,98,643,32,107,140,112,26,85,138,540,53,20
125,371,38,36,10,52,118,136,102,420,150,112,71,14,20,7,24,18,12,807
37,67,110,62,33,21,95,220,511,102,811,30,83,84,305,620,15,2,108,220
106,353,105,106,60,275,72,8,50,205,185,112,125,540,65,106,807,188,96,110
16,73,32,807,150,409,400,50,154,285,96,106,316,270,205,101,811,400,8
44,37,52,40,241,34,205,38,16,46,47,85,24,44,15,64,73,138,807,85,78,110
33,420,505,53,37,38,22,31,10,110,106,101,140,15,38,3,5,44,7,98,287
135,150,96,33,84,125,807,191,96,511,118,440,370,643,466,106,41,107
603,220,275,30,150,105,49,53,287,250,208,134,7,53,12,47,85,63,138,110
21,112,140,485,486,505,14,73,84,575,1005,150,200,16,42,5,4,25,42
8,16,811,125,160,32,205,603,807,81,96,405,41,600,136,14,20,28,26
353,302,246,8,131,160,140,84,440,42,16,811,40,67,101,102,194,138
205,51,63,241,540,122,8,10,63,140,47,48,140,288
CLEAR for part 2, made human readable.
I have deposited in the county of Bedford about four miles from
Bufords in an excavation or vault six feet below the surface
of the ground the following articles belonging jointly to
the parties whose names are given in number three herewith.
The first deposit consisted of ten hundred and fourteen pounds
of gold and thirty eight hundred and twelve pounds of silver
deposited Nov eighteen nineteen. The second was made Dec
eighteen twenty one and consisted of nineteen hundred and seven
pounds of gold and twelve hundred and eighty eight of silver,
also jewels obtained in St. Louis in exchange to save transportation
and valued at thirteen [t]housand dollars. The above
is securely packed i[n] [i]ron pots with iron cov[e]rs. Th[e] vault
is roughly lined with stone and the vessels rest on solid stone
and are covered [w]ith others. Paper number one describes th[e]
exact locality of the va[u]lt so that no difficulty will be had
in finding it.
CLEAR for part 2, using only the first 480 words of the
Declaration of Independence, then blanks filled in by
inspection. ALL mistakes shown were caused by sloppy
encryption.
0----5----10---15---20---25---30---35---40---45---
0 ihavedepositedinthecountyofbedfordaboutfourmilesfr
50 ombufordsinanexcavationorvaultsixfeetbelowthesurfa
100 ceofthegroundthefollowingarticlesbelongingjointlyt
150 othepartieswhosenamesaregiveninnumberthreeherewith
200 thefirstdepositconsistcdoftenhundredandfourteenpou
250 ndsofgoldandthirtyeighthundredandtwelvepoundsofsil
300 verdepositednoveighteennineteenthesecondwasmadedec
350 eighteentwentyoneandconsistedofnineteenhundredands
400 evenpoundsofgoldandtwelvehundredandeightyeightofsi
450 lveralsojewelsobtainedinstlouisinexchangetosavetra
500 nsportationandvaluedatthirteenrhousanddollarstheab
550 oveissecurelypackeditronpotswithironcovtrsthtvault
600 isroughlylinedwithstoneandthevesselsrestonsolidsto
650 neandarecovereduithotherspapernumberonedescribesth
700 cexactlocalityofthevarltsothatnodifficultywillbeha
750 dinfindingit
TEXT for part 3
Names and Residences.
317,8,92,73,112,89,67,318,28,96,107,41,631,78,146,397,118,98
114,246,348,116,74,88,12,65,32,14,81,19,76,121,216,85,33,66,15
108,68,77,43,24,122,96,117,36,211,301,15,44,11,46,89,18,136,68
317,28,90,82,304,71,43,221,198,176,310,319,81,99,264,380,56,37
319,2,44,53,28,44,75,98,102,37,85,107,117,64,88,136,48,154,99,175
89,315,326,78,96,214,218,311,43,89,51,90,75,128,96,33,28,103,84
65,26,41,246,84,270,98,116,32,59,74,66,69,240,15,8,121,20,77,80
31,11,106,81,191,224,328,18,75,52,82,117,201,39,23,217,27,21,84
35,54,109,128,49,77,88,1,81,217,64,55,83,116,251,269,311,96,54,32
120,18,132,102,219,211,84,150,219,275,312,64,10,106,87,75,47,21
29,37,81,44,18,126,115,132,160,181,203,76,81,299,314,337,351,96,11
28,97,318,238,106,24,93,3,19,17,26,60,73,88,14,126,138,234,286
297,321,365,264,19,22,84,56,107,98,123,111,214,136,7,33,45,40,13
28,46,42,107,196,227,344,198,203,247,116,19,8,212,230,31,6,328
65,48,52,59,41,122,33,117,11,18,25,71,36,45,83,76,89,92,31,65,70
83,96,27,33,44,50,61,24,112,136,149,176,180,194,143,171,205,296
87,12,44,51,89,98,34,41,208,173,66,9,35,16,95,8,113,175,90,56
203,19,177,183,206,157,200,218,260,291,305,618,951,320,18,124,78
65,19,32,124,48,53,57,84,96,207,244,66,82,119,71,11,86,77,213,54
82,316,245,303,86,97,106,212,18,37,15,81,89,16,7,81,39,96,14,43
216,118,29,55,109,136,172,213,64,8,227,304,611,221,364,819,375
128,296,1,18,53,76,10,15,23,19,71,84,120,134,66,73,89,96,230,48
77,26,101,127,936,218,439,178,171,61,226,313,215,102,18,167,262
114,218,66,59,48,27,19,13,82,48,162,119,34,127,139,34,128,129,74
63,120,11,54,61,73,92,180,66,75,101,124,265,89,96,126,274,896,917
434,461,235,890,312,413,328,381,96,105,217,66,118,22,77,64,42,12
7,55,24,83,67,97,109,121,135,181,203,219,228,256,21,34,77,319,374
382,675,684,717,864,203,4,18,92,16,63,82,22,46,55,69,74,112,134
186,175,119,213,416,312,343,264,119,186,218,343,417,845,951,124
209,49,617,856,924,936,72,19,28,11,35,42,40,66,85,94,112,65,82
115,119,233,244,186,172,112,85,6,56,38,44,85,72,32,47,63,96,124
217,314,319,221,644,817,821,934,922,416,975,10,22,18,46,137,181
101,39,86,103,116,138,164,212,218,296,815,380,412,460,495,675,820
952
Evidence in favor of a hoax-
. Too many players.
. Inflated quantities of treasure.
. Many discrepancies exist in all documents.
. The Declaration of Independence is too hokey a key.
. Part 3 (list of 30 names) considered too little text.
. W.F. Friedman couldn't crack it.
. Why even encrypt parts 1 & 3?
. Why use multi-part text, and why different keys for each part?
. Difficult to keep treasure in ground if 30 men know where it was buried.
. Who'd leave it with other than your own family?
. The Inn Keeper waited an extra 10 years before opening box with
ciphers in it? Who would do this, curiousity runs too deep in
humans?
. Why did anybody waste time deciphering paper 2, which had no title?
1 & 3 had titles! These should have been deciphered first?
. Why not just one single letter?
. Statistical analysis show 1&3 similar in very obscure ways, that
2 differs. Did somebody else encipher it? And why?
Check length of keytexts, and forward/backward next word
displacement selections.
. Who could cross the entire country with that much gold and only
10 men and survive back then?
. Practically everybody who visited New Mexico before 1821, left
by way of the Pearly Gates, as the Spanish got almost every
tourist:-)
References:
"The Beale Treasure: A History of a Mystery", by Peter Viemeister,
Bedord, VA: Hamilton's, 1987. ISBN: 0-9608598-3-7. 230 pages.
"The Codebreakers", by David Kahn, pg 771, CCN 63-16109.
1967.
"Gold in the Blue Ridge, The True Story of the Beale Treasure",
by P.B. Innis & Walter Dean Innis, Devon Publ. Co., Wash, D.C.
1973.
"Signature Simulation and Certain Cryptographic Codes", Hammer,
Communications of the ACM, 14 (1), January 1971, pp. 3-14.
"How did TJB Encode B2?", Hammer, Cryptologia, 3 (1), Jan. 1979, pp. 9-15.
"Second Order Homophonic Ciphers", Hammer, Cryptologia, 12 (1) Jan. 1988,
pp 11-20.
==> cryptology/Feynman.p <==
What are the Feynman ciphers?
==> cryptology/Feynman.s <==
When I was a graduate student at Caltech, Professor Feynman showed me three
samples of code that he had been challenged with by a fellow scientist at
Los Alamos and which he had not been able to crack. I also was unable to
crack them. I posted them to Usenet and Jack C. Morrison of JPL cracked
the first one. It is a simple transposition cipher: split the text into
5-column pieces, then read from lower right upward. What results are the
opening lines of Chaucer's Canterbury Tales in Middle English.
1. Easier
MEOTAIHSIBRTEWDGLGKNLANEA
INOEEPEYSTNPEUOOEHRONLTIR
OSDHEOTNPHGAAETOHSZOTTENT
KEPADLYPHEODOWCFORRRNLCUE
EEEOPGMRLHNNDFTOENEALKEHH
EATTHNMESCNSHIRAETDAHLHEM
TETRFSWEDOEOENEGFHETAEDGH
RLNNGOAAEOCMTURRSLTDIDORE
HNHEHNAYVTIERHEENECTRNVIO
UOEHOTRNWSAYIFSNSHOEMRTRR
EUAUUHOHOOHCDCHTEEISEVRLS
KLIHIIAPCHRHSIHPSNWTOIISI
SHHNWEMTIEYAFELNRENLEERYI
PHBEROTEVPHNTYATIERTIHEEA
WTWVHTASETHHSDNGEIEAYNHHH
NNHTW
2. Harder
XUKEXWSLZJUAXUNKIGWFSOZRAWURO
RKXAOSLHROBXBTKCMUWDVPTFBLMKE
FVWMUXTVTWUIDDJVZKBRMCWOIWYDX
MLUFPVSHAGSVWUFWORCWUIDUJCNVT
TBERTUNOJUZHVTWKORSVRZSVVFSQX
OCMUWPYTRLGBMCYPOJCLRIYTVFCCM
UWUFPOXCNMCIWMSKPXEDLYIQKDJWI
WCJUMVRCJUMVRKXWURKPSEEIWZVXU
LEIOETOOFWKBIUXPXUGOWLFPWUSCH
3. New Message
WURVFXGJYTHEIZXSQXOBGSV
RUDOOJXATBKTARVIXPYTMYA
BMVUFXPXKUJVPLSDVTGNGOS
IGLWURPKFCVGELLRNNGLPYT
FVTPXAJOSCWRODORWNWSICL
FKEMOTGJYCRRAOJVNTODVMN
SQIVICRBICRUDCSKXYPDMDR
OJUZICRVFWXIFPXIVVIEPYT
DOIAVRBOOXWRAKPSZXTZKVR
OSWCRCFVEESOLWKTOBXAUXV
B
Chris Cole
Peregrine Systems
uunet!peregrine!chris
==> cryptology/Voynich.p <==
What are the Voynich ciphers?
==> cryptology/Voynich.s <==
The Voynich Manuscript is a manuscript that first surfaced in the court of
Rudolf II (Holy Roman Emperor), who bought it for some large number of
gold pieces (600?). Rudolf was interested in the occult, and the strange
characters and bizarre illustrations suggested that it had some deep
mystical/magical significance. After Rudolf's court broke up, the
manuscript was sent to (if memory serves) Athanasius Kircher, with nobody
on the list having been able to read it. It ended up in a chest of other
manuscripts in the Villa Mondragone [?] in Italy, and was discovered there
by Wilfred Voynich, a collector, in about 1910 or so. He took it to a
linguist who wasn't a cryptanalyst, who identified it as a work by the
12th century monk Roger Bacon and produced extended bogus decryptions based
on shorthand characters he saw in it. A great deal of effort by the best
cryptanalysts in the country hasn't resulted in any breakthrough. William
F. Friedman (arguably the best) thought it was written in an artificial
language. I believe the manuscript is currently in the Beinecke Rare
Book Collection at [Harvard?].
Mary D'Imperio's paper is scholarly and detailed, and provides an
excellent starting point for anyone who is interested in the subject.
David Kahn's "The Codebreakers" has enough detail to tell you if you're
interested; it also has one or more plates showing the script and some
illustrations. I believe D'Imperio's monograph has been reprinted by
Aegean Park Press. A number of people have published their own ideas
about it, including Brumbaugh, without anybody agreeing. A recent
publication from Aegean Park Press offers another decryption; I haven't
seen that one.
If you want *my* guess, it's a hoax made up by Edmund Kelley and an
unnamed co-conspirator and sold to Rudolf through the reputation of John
Dee (Queen Elizabeth I's astrologer).
--
Jim Gillogly
{hplabs, ihnp4}!sdcrdcf!randvax!jim
j...@rand-unix.arpa
I read "Labyrinths of Reason" by William Poundstone recently. I'm
posting this to so many newsgroups in part to recommend this book, which,
while of a popular nature, gives a good analysis of a wide variety of
paradoxes and philosophical quandaries, and is a great read.
Anyway, it mentions something called the Voynich manuscript, which is
now at Yale University's Beinecke Rare Book and Manuscript Library.
It's a real pity that I didn't know about this manuscript and go see it
when I was at Yale.
The Voynich manuscript is apparently very old. It is a 232-page illuminated
manuscript written in a cipher that has never been cracked. (That's
what Poundstone says - but see my hypothesis below.) If I may quote
Poundstone's charming description, "Its author, subject matter, and
meaning are unfathomed mysteries. No one even knows what language the
text would be in if you deciphered it. Fanciful picutres of nude women,
peculiar inventions, and nonexistent flora and fauna tantalize the
would-be decipherer. Color sketches in the exacting style of a
medieval herbal depict blossoms and spices that never spring from earth
and constellations found in no sky. Plans for weird, otherworldly
plumbing show nymphets frolicking in sitz baths connected with
elbow-macaroni pipes. The manuscript has the eerie quality of a
perfectly sensible book from an alternate universe."
There is a picture of one page in Poundstone's book. It's written in a
flowing script using "approximately 21 curlicued symbols," some of which
are close to the Roman alphabet, but others of which supposedly resemble
Cyrillic, Glagolitic, and Ethiopian. There is one tiny note in Middle
High German, not necessarily by the original author, talking about the
Herbal of Matthiolaus. Some astrology charts in the manuscript have the
months labeled in Spanish. "What appears to be a cipher table on the
first page has long faded into illegibility," and on the other hand, some
scholars have guessed that a barely legible inscription on the *last*
page is a key!
It is said to have "languished for a long time at the Jesuit College of
Mondragone in Frascati, Italy. Then in 1912 it was purchased by Wilfred
M. Voynich, a Polish-born scientist and bibliophile... Voynich was the
son-in-law of George Boole, the logician..." A letter written in 1666
claims that Holy Roman Emperor Rudolf II of Bohemia (1552-1612) bought
the manuscript for 600 gold ducats. He may have bought it from Dr.
John Dee, the famous astrologer. Rudolf thought the manuscript was
written by Roger Bacon! [Wouldn't it more likely have been written by
Dee, out to make a fast ducat?]
"Many of the most talented military code breakers of this century have
tried to decipher it as a show of prowess. Herbert Yardley, the
American code expert who solved the German cipher in WW1 and who cracked
a Japanese diplomatic cipher without knowing the Japanese language,
failed with the Voynich manuscript. So did John Manly, who unscrambled
the Waberski cipher, and William Friedman, who defeated the Japanese
"purple code" of the 1940's. Computers have been drafted into the
effort in recent years, to no avail."
Poundstone goes on to describe a kook, Newbold, who was apparently driven
batty in his attempt to crack the manuscript. He then mentions that one
Leo Levitov also claimed in 1987 to crack the cipher, saying that it was
the text of a 12th-century cult of Isis worshipers, and that it
describes a method of euthanasia by opening a vein in a warm bathtub,
among other morbid matters. According to Levitov's translation the text
begins:
"ones treat the dying each the man lying deathly ill the one person who
aches Isis each that dies treats the person"
Poundstone rejects this translation.
According to Poundstone, a William Bennett (see below) has analysed the
text with a computer and finds that its entropy is less than any known
European language, and closer to those of Polynesian languages.
My wild hypothesis, on the basis solely of the evidence above, is this.
Perhaps the text was meant to be RANDOM. Of course humans are lousy at
generating random sequences. So I'm wondering how attempted random
sequences (written in a weird alphabet) would compare statistically with
the Voynich manuscript.
Anyway, the only source Poundstone seems to cite, other than the
manuscript itself, is Leo Levitov's "Solution of the Voynich Manuscript,
A Liturgical Manual for the Endura Rite of the Cathari Heresy, the Cult
of Isis," Laguna Hills, Calif., Aegean Park Press, 1987, and William
Ralph Bennett Jr.'s "Scientific and Engineering Problem-Solving with the
Computer," Englewood Cliffs, New Jersey, Prentice-Hall 1976.
I will check the Bennett book; the other sounds hard to get ahold of! I
would LOVE any further information about this bizarre puzzle. If anyone
knows Bennett and can get samples of the Voynich manuscript in
electronic form, I would LOVE to get my hands on it.
Also, I would appreciate any information on:
Voynich
The Jesuit College of Mondragone
Rudolf II
The letter by Rudolf II (where is it? what does it say?)
The attempts of Yardley, Friedman and Manly
The Herbal of Matthiolaus
and, just for the heck of it, the "Waberski cipher" and the "purple
code"!
This whole business sounds like a quagmire into which angels would fear
to tread, but a fool like me finds it fascinating.
-- sender's name lost (!?)
To counter a few hypotheses that were suggested here:
The Voynich Manuscript is certainly not strictly a polyalphabetic cipher
like Vigenere or Beaufort or (the one usually called) Porta, because of
the frequent repetitions of "words" at intervals that couldn't be
multiples of any key length. I suppose one could imagine that it's an
interrupted key Vig or something, but common elements appearing at places
other than the beginnings of words would seem to rule that out. The I.C.
is too high for a digraphic system like (an anachronistic) Playfair in any
European language.
One of the most interesting Voynich discoveries was made by Prescott Currier,
who discovered that the two different "hands" (visually distinct handwriting)
used different "dialects": that is, the frequencies for pages written in
one hand are different from those written in the other. I confirmed this
observation by running some correlation coefficients on the digraph matrices
for the two kinds of pages.
W. F. Friedman ("The Man Who Broke Purple") thought the Voynich was
written in some artificial language. If it's not a hoax, I don't see any
evidence to suggest he's wrong. My personal theory (yeah, I've offered
too many of those lately) is that it was constructed by Edward Kelley,
John Dee's scryer, with somebody else's help (to explain the second
handwriting) -- perhaps Dee himself, although he's always struck me as a
credulous dupe of Kelley rather than a co-conspirator (cf the Angelic
language stuff).
The best source I know for the Voynich is Mary D'Imperio's monograph
"The Voynich Manuscript: An Elegant Enigma", which is available from
Aegean Park Press.
--
Jim Gillogly
j...@rand.org
Here's an update on the Voynich manuscript. This will concentrate on
sources for information on the Voynich; later I will write a survey of
what I have found out so far. I begin with some references to the
case, kindly sent to me by Karl Kluge (the first three) and Micheal Roe
<M....@cs.ucl.ac.uk> (the rest).
TITLE Thirty-five manuscripts : including the St. Blasien psalter, the
Llangattock hours, the Gotha missal, the Roger Bacon (Voynich)
cipher ms.
Catalogue ; 100
35 manuscripts.
CITATION New York, N.Y. : H.P. Kraus, [1962] 86 p., lxvii p. of plates, [1]
leaf of plates : ill. (some col.), facsims. ; 36 cm.
NOTES "30 years, 1932-1962" ([28] p.) in pocket. Includes indexes.
SUBJECT Manuscripts Catalogs.
Illumination of books and manuscripts Catalogs.
AUTHOR Brumbaugh, Robert Sherrick, 1918-
TITLE The most mysterious manuscript : the Voynich "Roger Bacon" cipher
manuscript / edited by Robert S. Brumbaugh.
CITATION Carbondale : Southern Illinois University Press, c1978. xii, 175 p.
: ill. ; 22 cm.
SUBJECT Bacon, Roger, 1214?-1294.
Ciphers.
AUTHOR D'Imperio, M. E.
TITLE The Voynich manuscript : an elegant enigma / M. E. D'Imperio.
CITATION Fort George E. Mead, Md. : National Security Agency/Central Security
Service, 1978. ix, 140 p. : ill. ; 27 cm.
NOTES Includes index. Bibliography: p. 124-131.
SUBJECT Voynich manuscript. [NOTE: see alternate publisher below!]
@book{Bennett76,
author = "Bennett, William Ralph",
title = "Scientific and Engineering Problem Solving with the Computer",
address = "Englewood Cliffs, NJ",
publisher = "Prentice-Hall",
year = 1976}
@book{dImperio78,
author = "D'Imperio, M E",
title = "The Voynich manuscript: An Elegant Enigma",
publisher= "Aegean Park Press",
year = 1978}
@article{Friedman62,
author = "Friedman, Elizebeth Smith",
title = "``The Most Mysterious Manuscript'' Still Mysterious",
booktitle = "Washington Post",
month = "August 5",
notes = "Section E",
pages = "1,5",
year = 1962}
@book{Kahn67,
author = "Kahn, David",
title = "The Codebreakers",
publisher = "Macmillan",
year = "1967"}
@article{Manly31,
author = "Manly, John Matthews",
title = "Roger Bacon and the Voynich MS",
boooktitle = "Speculum VI",
pages = "345--91",
year = 1931}
@article{ONeill44,
author = "O'Neill, Hugh",
title = "Botanical Remarks on the Voynich MS",
journal = "Speculum XIX",
pages = "p.126",
year = 1944}
@book{Poundstone88,
author = "Poundstone, W.",
title = "Labyrinths of Reason",
publisher = "Doubleday",
address = "New York",
month = "November",
year = 1988}
@article{Zimanski70,
author = "Zimanski, C.",
title = "William Friedman and the Voynich Manuscript",
journal = "Philological Quarterly",
year = "1970"}
@article{Guy91b,
author = "Guy, J. B. M.",
title = "Statistical Properties of Two Folios of the Voynich Manuscript",
journal = "Cryptologia",
volume = "XV",
number = "4",
pages = "pp. 207--218",
month = "July",
year = 1991}
@article{Guy91a,
author = "Guy, J. B. M.",
title = "Letter to the Editor Re Voynich Manuscript",
journal = "Cryptologia",
volume = "XV",
number = "3",
pages = "pp. 161--166",
year = 1991}
This is by no means a complete list. It doesn't include Newbold's
(largely discredited) work, nor work by Feely and Stong.
In addition, there is the proposed decryption by Leo Levitov (also
largely discredited):
"Solution of the Voynich Manuscript: A Liturgical Manual for the
Endura Rite of the Cathari Heresy, the Cult of Isis_, available from
Aegean Park Press, P. O. Box 2837, Laguna Hills CA 92654-0837."
According to Earl Boebert, this book is reviewed in
Cryptologia XII, 1 (January 1988). I should add that Brumbaugh's book
above gives a third, also largely discredited, decryption of the Voynich.
According to s...@att.ulysses.com, Aegean Park Press does mail-order
business and can be reached at the above address or at 714-586-8811
(an answering machine).
Micheal Roe has explained how one get microfilms of the whole
manuscript:
"The Beinecke Rare Book Library, Yale University sells a microfilm of the
manuscript. Their catalog number for the original is MS 408, ``The Voynich
`Roger Bacon' Cipher MS''. You should write to them.
The British Library [sic - should be Museum] has a photocopy of the MS
donated to them by John Manly circa 1931. They apparently lost it until
12 March 1947, when it was entered in the catalogue (without
cross-references under Voynich, Manly, Roger Bacon or any other useful
keywords...)
It appears as ``MS Facs 461: Positive rotographs of a Cipher MS (folios 1-56)
acquired in 1912 by Wilfred M. Voynich in Southern Europe.'
Correspondance between Newbold, Manly and various British Museum experts
appears under ``MS Facs 439: Leaves of the Voynich MS, alleged to be in
Roger Bacon's cypher, with correspondence and other pertinent material''
See John Manly's 1931 article in Speculum and Newbold's book for what the
correspondance was about! There are also a number of press cuttings.
Both of these in are in the manuscript collection, for which special
permission is needed in addition to a normal British Library reader's pass."
Also, Jim Gillogly has been extremely kind in making available
part of the manuscript that was transcribed and keyed in by Mary
D'Imperio (see above), using Prescott Currier's notation. It appears to
consist of 166 of the total 232 pages. I hope to do some statistical
studies on this, and I encourage others to do the same and let me know
what they find! As Jim notes, the file is pub/jim/voynich.tar.Z and is
available by anonymous ftp at rand.org. I've had a little trouble with
this file at page 165, where I read "1650voynich 664" etc., with page
166 missing. If anyone else notes this let Jim or I know.
Jim says he has confirmed by correlations between digraph matrices the
discovery by Prescott Crurrier that the manuscript is written in two
visibly distinct hands. These are marked "A" and "B" in the file
voynich.tar.Z.
Because of the possibility that the Voynich is nonsense, it would be
interesting to compare the Voynich to the Codex Seraphinianus, which
Kevin McCarty kindly reminded me of. He writes:
"This is very odd. I know nothing of the Voynich manuscript, but
I know of something which sounds very much like it and was created
by an Italian artist, who it now seems was probably influenced
by this work. It a book titled "Codex Seraphinianus", written in
a very strange script. The title page contains only the book's title
and the publisher's name: Abbeville Press, New York. The only clues
in English (in *any* recognizable language) are some blurbs on the
dust jacket that identify it as a modern work of art, and the copyright
notice, in fine print, which reads
"Library of Congress Cataloging in Publication Data
Serafini, Luigi.
Codex Seraphinianus.
1. Imaginary Languages. 2. Imaginary societies.
3. Encyclopedias and Dictionaries-- Miscellanea.
I. Title.
PN6381.S4 1983 818'.5407 83.-7076
ISBN 0-89659-428-9
First American Edition, 1983.
Copyright (c) 1981 by Franco Maria Ricci. All rights reserved
by Abbeville Press. No part of this book may be reproduced...
without permission in writing from the publisher. Inquiries should
be addressed to Abbeville Press, Inc., 505 Park Avenue, New York
10022. Printed and bound in Italy."
The book is remarkable and bizarre. It *looks* like an encyclopedia
for an imaginary world. Page after page of beautiful pictures
of imaginary flora and fauna, with annotations and captions in
a completely strange script. Machines, architecture, umm, 'situations',
arcane diagrams, implements, an archeologist pointing at a Rosetta stone
(with phony hieroglyphics), an article on penmanship (with unorthodox
pens), and much more, finally ending with a brief index.
The script in this work looks vaguely similar to the Voynich orthography
shown in Poundstone's book (I just compared them); the alphabets
look quite similar, but the Codex script is more cursive and less
bookish than Voynich. It runs to about 200 pages, and probably
ought to provide someone two things:
- a possible explanation of what the Voynich manuscript is
(a highly imaginative work of art)
- a textual work which looks like it was inspired by it and might
provide an interesting comparison for statistical study."
I suppose it would be too much to hope that someone has already
transcribed parts of the Codex, but nonetheless, if anyone has any in
electronic form, I would love to have a copy for comparative statistics.
Jacques Guy kindly summarized his analysis (in Cryptologia, see above)
of the Voynich as follows:
"I transcribed the two folios in Bennett's book and submitted them to
letter-frequency counts, distinguishing word-initial, word-medial,
word-final, isolated, line-initial, and line-final positions. I also
submitted that transcription to Sukhotin's algorithm which, given a text
written in an alphabetical system, identifies which symbols are vowels and
which are consonants. The letter transcribed CT in Bennett's system came
out as a consonant, the one transcribed CC as vowel. Now it so happens
that CT is exactly the shape of the letter "t" in the Beneventan script
(used in medieval Spain and Northern Italy), and CC is exactly the shape
of "a" in that same script. I concluded that the author had a knowledge
of that script, and that the values of CT and CC probably were "t" and
"a". There's a lot more, but more shaky."
By popular demand I've put a machine-readable copy of the Voynich Manuscript
up for anonymous ftp:
Host: rand.org
File: pub/jim/voynich.tar.Z
It uses Prescott Currier's notation, and was transcribed by Mary D'Imperio.
If you use it in any analysis, be sure to give credit to D'Imperio, who put
in a lot of effort to get it right.
--
Jim Gillogly
j...@rand.org
This post is essentially a summary of the fruit of a short research
quest at the local library.
Brief description of the Voynich manuscript:
The Voynich manuscript was bought (in about 1586) by the Holy Roman
Emperor Rudolf II. He believed it to be the work of Roger Bacon
an english 13th century philosopher. The manuscript consisted of about
200 pages with many illustrations. It is believed that the manuscript
contains some secret scientific or magical knowledge since it is entirely
written in secret writing (presumably in cipher).
The Voynich Manuscript is often abbreviated "Voynich MS" in all of the
books I have read on Voynich. This is done without explanation. I
suppose it is just a convention started by the founding analysts of
the manuscript to call it that.
William R. Newbold, one of the original analysts of the Voynich MS after
Voynich, claims to have arrived at a partial decipherment of the entire
manuscript. His book The Cipher of Roger Bacon [2] contains a history
of the unravelment of the cipher *and* keys to the cipher itself. As well
as translations of several pages of the manuscript.
Newbold derives his decipherment rules through a study of the medeival
mind (which he is a leading scholar in) as well as the other writings
of Roger Bacon. Says Newbold, ciphers in Roger Bacon's writings are not
new, as Bacon discusses in other works the need for monks to use
encipherment to protect their knowlege.
Newbold includes many partial decipherments from the Voynich MS but most of
them are presented in Latin only.
Newbolds deciphering rules (from The Cipher of Roger Bacon [1])
---------------------------------------------------------------
1. Syllabification: [double all but the first and last letters of each
word, and divide the product into biliteral groups or symbols.]
2. Translation: [translate these symbols into the alphabetic values]
3. Reversion: [change the alphabetic values to the phonetic values, by use
of the reversion alphabet]
4. Recomposition: [ rearrange the letters in order, and thus recompose the
true text].
The text I copied this from failed to note step 0 which was:
0. Ignore. [ignore the actual shape of every symbol and analyze only the
(random?) properties of the direction of swirl and crosshatch patterns
of the characters when viewed under a microscope. 14 distinct contruction
patterns can be identified among the (much larger) set of symbols]
John M. Manly in The Most Mysterious Manuscript [3], suggests that Newbold's
method of decipherment is totally invalid. Manly goes on to show that it
is not difficult to obtain *ANY DESIRABLE* message from the Voynich MS
using Newbold's rules. He shows that after fifteen minutes deciphering
a short sequence of letters he arrives at the plaintext message
"Paris is lured into loving vestals..."
and quips that he will furnish a continuation of the translation upon
request!
The reason I have spent so much time explaining Newbold's method is that
Newbold presents the most convincing argument for how he arrived at his
conclusions. Notwithstanding the fact that he invented the oija board of
deciphering systems.
Joseph Martin Feely, in his book on the Voynich MS [2] , claims to have
found the key to deciphering at least one page of the Voynich MS. His entire
book on the topic of the Voynich manuscript is devoted to the deciphering of
the single page 78. Feely presents full tables of translation of the page 78
from its written form into latin (and english). It seems that Feely was using
the exhaustive analysis method to determine the key.
Feely suggests the following translation of (the first fiew lines of) page
78 of the Voynich MS:
"the combined stream when well humidified, ramifies; afterward it is broken
down smaller; afterward, at a distance, into the fore-bladder it comes [1].
Then vesselled, it is after-a-while ruminated: well humidified it is
clothed with veinlets [2]. Thence after-a-bit they move down; tiny
teats they provide (or live upon) in the outpimpling of the veinlets.
They are impermiated; are thrown down below; they are ruminated; they are
feminized with the tiny teats. .... "
... and so on for three more pages of "english plaintext".
The descriptions by Feely say that this text is accompanied in the Voynich MS
by an illustration that (he says) is unmistakably the internal female
reproductive organs (I saw the plate myself and they DO look like fallopian
tubes *AFTER* I read the explanation).
The most informative work that I found (I feel) was "The Most Mysterious
Manuscript". Of the five books on Voynich that I found, this was the only
one that didn't claim to have found the key but was, rather, a collection
of essays on the history of the Voynich MS and criticisms of various attempts
by earlier scientists. It was also the *latest* book that I was able to
consult, being published in 1978.
My impression from the black and white plates of the Voynich MS I've seen, are
that the illustrations are very weird when compared to other 'illuminated'
manuscripts of this time. Particularly I would say that there is emphasis
on the female nude that is unusual for the art of this period. I can't say
that I myself believe the images to have ANYTHING to do with the text.
My own conjecture is that the manuscript is a one-way encipherment. A
cipher so clever that the inventor didn't even think of how it could be
deciphered. Sorta like an /etc/passwd file.
Bibliography
------------
1. William R. Newbold. _The Cipher of Roger Bacon_Roland G Kent, ed.
University of Pennsylvania Press, 1928.
2. Joseph Martin Feely. _Roger Bacon's Cipher: The Right Key Found_
Rochester N.Y.:Joseph Martin Feely, pub., 1943.
3. _The Most Mysterious Manuscript_ Robert S. Brumbaugh, ed. Southern Illinois
Press, 1978
Unix filters are so wonderful. Massaging the machine readable file, we find:
4182 "words", of which 1284 are used more than once, 308 used 8+ times,
184 used 15+ times, 23 used 100+ times.
Does this tell us anything about the language (if any) the text is written
in?
For those who may be interested, here are the 23 words used 100+ times:
121 2
115 4OFAE
114 4OFAM
155 4OFAN
195 4OFC89
162 4OFCC89
101 4OFCC9
189 89
111 8AE
492 8AM
134 8AN
156 8AR
248 OE
148 OR
111 S9
251 SC89
142 SC9
238 SOE
150 SOR
244 ZC89
116 ZC9
116 ZOE
Could someone email the Voynich Ms. ref list that appeared here not
very long ago? Thanks in advance...
Also... I came across the following ref that is fun(?):
The Voynich manuscript: an elegant enigma / M. E. D'Imperio
Fort George E. Mead, Md. : National Security Agency(!)
Central Security Service(?), 1978. ix, 140 p. : ill. ; 27 cm.
The (?!) are mine... Sorry if this was already on the list, but the
mention of the NSA (and what's the CSS?) made it jump out at me...
--
Ron Carter | rca...@nyx.cs.du.edu rcarter GEnie 70707.3047 CIS
Director | Center for the Study of Creative Intelligence
Denver, CO | Knowledge is power. Knowledge to the people. Just say know.
Distribution: na
Organization: Wetware Diversions, San Francisco
Keywords:
From sci.archaeology:
>From: ja...@cs.sfu.ca (Jamie Andrews)
>Date: 16 Nov 91 00:49:08 GMT
>
> It seems like the person who would be most likely to solve
>this Voynich manuscript cipher would have
>(a) knowledge of the modern techniques for solving more complex
> ciphers such as Playfairs and Vigineres; and
>(b) knowledge of the possible contemporary and archaic languages
> in which the plaintext could have been written.
An extended discussion of the Voynich Manuscript may be found in the
tape of the same name by Terence McKenna. I'm not sure who is currently
publishing this particular McKenna tape but probably one of:
Dolphin Tapes, POB 71, Big Sur, CA 93920
Sounds True, 1825 Pearl St., Boulder, CO 80302
Sound Photosynthesis, POB 2111, Mill Valley, CA 94942
The Spring 1988 issue of Gnosis magazine contained an article by McKenna
giving some background of the Voynich Manuscipt and attempts to decipher
it, and reviewing Leo Levitov's "Solution of the Voynich Manuscript"
(published in 1987 by Aegean Park Press, POB 2837 Laguna Hills, CA 92654).
Levitov's thesis is that the manuscript is the only surviving primary
document of the Cathar faith (exterminated on the orders of the Pope in
the Albigensian Crusade in the 1230s) and that it is in fact not
encrypted material but rather is a highly polyglot form of Medieval
Flemish with a large number of Old French and Old High German loan
words, written in a special script.
As far as I know Levitov's there has been no challenge to Levitov's
claims so far.
Michael Barlow, who had reviewed Levitov's book in Cryptologia, had sent me
photocopies of the pages where much of the language was described
(pp.21-31). I have just found them, and am looking at them now as I am
typing this. Incidentally, I do not believe this has anything to do with
cryptology proper, but the decipherment of texts in unknown languages. So
if you are into cryptography proper, skip this.
Looking at the "Voynich alphabet" pp.25-27, I made a list of the letters of
the Voynich language as Levitov interprets them, and I added phonetic
descriptions of the sounds I *think* Levitov meant to describe. Here it is:
Letter# Phonetic Phonetic descriptions
(IPA) in linguists' jargon: in plain English:
1 a low open, central unrounded a as in father
e mid close, front, unrounded ay as in May
O mid open, back, rounded aw as in law
or o as in got
(British
pronunciation)
2 s unvoiced dental fricative s as in so
3 d voiced dental stop d
4 E mid, front, unrounded e as in wet
5 f unvoiced labiodental fricative f
6 i short, high open, front, i as in dim
unrounded
7 i: long, high, front, unrounded ea as in weak
8 i:E (?) I can't make head nor tail of Levitov's
explanations. Probably like "ei" in "weird"
dragging along the "e": "weeeird"! (British
pronunciation, with a silent "r")
9 C unvoiced palatal fricative ch in German ich
10 k unvoived velar stop k
11 l lateral, can't be more precise from
description, probably like l in "loony"
12 m voiced bilabial nasal m
13 n voiced dental nasal n
14 r (?) cannot tell precisely from Scottish r?
description Dutch r?
15 t no description; dental stop? t
16 t another form for #15 t
17 T (?) no description th as in this?
th as in thick?
18 TE (?) again, no description
or ET (?)
19 v voiced labiodental fricative v as in rave
20 v ditto, same as #19 ditto
(By now, you will have guessed what my conclusion about Levitov's
decipherment was)
In the column headed "Phonetic (IPA)" I have used capital letters for lack
of the special international phonetic symbols:
E for the Greek letter "epsilon"
O for the letter that looks like a mirror-image of "c"
C for c-cedilla
T for the Greek letter "theta"
The colon (:) means that the sound represented by the preceding letter is
long, e.g. "i:" is a long "i".
The rest, #21 to 25, are not "letters" proper, but represent groups
of two or more letters, just like #18 does. They are:
21 av
22a Ev
22b vE
23 CET
24 kET
25 sET
That gives us a language with 6 vowels: a (#1), e (#1 again), O (#1 again),
E (#4), i (#6), and i: (#7). Letter #8 is not a vowel, but a combination
of two vowels: i: (#7) and probably E (#4). Levitov writes that the
language is derived from Dutch. If so, it has lost the "oo" sound (English
spelling; "oe" in Dutch spelling), and the three front rounded vowels of
Dutch: u as in U ("you", polite), eu as in deur ("door"), u as in vlug
("quick"). Note that out of six vowels, three are confused under the same
letter (#1), even though they sound very different from one another: a, e,
O. Just imagine that you had no way of distinguishing between "last",
"lest" and "lost" when writing in English, and you'll have a fair idea of
the consequences.
Let us look at the consonants now. I will put them in a matrix, with the
points of articulation in one dimension, and the manner of articulation in
the other (it's all standard procedure when analyzing a language). Brackets
around a letter will mean that I could not tell where to place it exactly,
and just took a guess.
labial dental palatal velar
nasal m n
voiced stop d
unvoiced stop t k
voiced fricative v (T)
unvoiced fricative f s C
lateral l
trill (?) (r)
Note that there are only twelve consonant sounds. That is unheard of for a
European language. No European language has so few consonant sounds.
Spanish, which has very few sounds (only five vowels), has seventeen
distinct consonants sounds, plus two semi-consonants. Dutch has from 18 to
20 consonants (depending on speakers, and how you analyze the sounds.
Warning: I just counted them on the back of an envelope; I might have
missed one or two). What is also extraordinary in Levitov's language is
that it lacks a "g", and *BOTH* "b" and "p". I cannot think of one single
language in the world that lacks both "b" and "p". Levitov also says that
"m" occurs only word-finally, never at the beginning, nor in the middle of
a word. That's true: the letter he says is an "m" is always word-final in
the reproductions I have seen of the Voynich MS. But no language I know of
behaves like that. All have an "m" (except one American Indian language,
which is very famous for that, and the name of which escapes me right now),
but, if there is a position where "m" never appears in some languages, that
position is word-finally. Exactly the reverse of Levitov's language.
What does Levitov say about the origin of the language?
"The language was very much standardized. It was an application of a
polyglot oral tongue into a literary language which would be understandable
to people who did not understand Latin and to whom this language could be
read."
At first reading, I would dismiss it all as nonsense: "polyglot oral
tongue" means nothing in linguistics terms. But Levitov is a medical
doctor, so allowances must be made. The best meaning I can read into
"polyglot oral tongue" is "a language that had never been written before
and which had taken words from many different languages". That is perfectly
reasonable: English for one, has done that. Half its vocabulary is Norman
French, and some of the commonest words have non-Anglo-Saxon origins.
"Sky", for instance, is a Danish word. So far, so good.
Levitov continues: "The Voynich is actually a simple language because it
follows set rules and has a very limited vocabulary.... There is a
deliberate duality and plurality of words in the Voynich and much use of
apostrophism".
By "duality and plurality of words" Levitov means that the words are highly
ambiguous, most words having two or more different meanings. I can only
guess at what he means by apostrophism: running words together, leaving
bits out, as we do in English: can not --> cannot --> can't, is not -->
ain't.
Time for a tutorial in the Voynich language as I could piece it together
from Levitov's description. Because, according to Levitov, letter #1
represent 3 vowels sounds, I will represent it by just "a", but remember:
it can be pronounced a, e, or o. But I will distinguish, as does Levitov,
between the two letters which he says were both pronounced "v", using "v"
for letter #20 and "w" for letter #21.
Some vocabulary now. Some verbs first, which Levitov gives in the
infinitive. In the Voynich language the infinitive of verbs ends in -en,
just like in Dutch and in German. I have removed that grammatical ending in
the list which follows, and given probable etymologies in parentheses
(Levitov gives doesn't give any):
ad = to aid, help ("aid")
ak = to ache, pain ("ache")
al = to ail ("ail")
and = to undergo the "Endura" rite ("End[ura]", probably)
d = to die ("d[ie]")
fad = to be for help (from f= for and ad=aid)
fal = to fail ("fail")
fil = to be for illness (from: f=for and il=ill)
il = to be ill ("ill")
k = to understand ("ken", Dutch and German "kennen" meaning "to know")
l = to lie deathly ill, in extremis ("lie", "lay")
s = to see ("see", Dutch "zien")
t = to do, treat (German "tun" = to do)
v = to will ("will" or Latin "volo" perhaps)
vid = to be with death (from vi=with and d=die)
vil = to want, wish, desire (German "willen")
vis = to know ("wit", German "wissen", Dutch "weten")
vit = to know (ditto)
viT = to use (no idea, Latin "uti" perhaps?)
vi = to be the way (Latin "via")
eC = to be each ("each")
ai:a = to eye, look at ("eye", "oog" in Dutch)
en = to do (no idea)
Example given by Levitov: enden "to do to death" made up of "en"
(to do), "d" (to die) and "en" (infinitive ending). Well, to me,
that's doing it the hard way. What's wrong with just "enden" = to
end (German "enden", too!)
More vocabulary:
em = he or they (masculine) ("him")
er = her or they (feminine) ("her")
eT = it or they ("it" or perhaps "they" or Dutch "het")
an = one ("one", Dutch "een")
"There are no declensions of nouns or conjugation of verbs. Only the
present tense is used" says Levitov.
Examples:
den = to die (infinitive) (d = die, -en = infinitive)
deT = it/they die (d = die, eT = it/they)
diteT = it does die (d = die, t = do, eT = it/they, with an "i" added to
make it easier to pronounce, which is quite common and natural
in languages)
But Levitov contradicts himself immediately, giving another tense (known
as present progressive in English grammar):
dieT = it is dying
But I may be unfair there, perhaps it is a compound: d = die, i = is
...-ing, eT = it/they.
Plurals are formed by suffixing "s" in one part of the MS, "eT" in another:
"ans" or "aneT" = ones.
More:
wians = we ones (wi = we, wie in Dutch, an = one, s = plural)
vian = one way (vi = way, an = one)
wia = one who (wi = who, a = one)
va = one will (v = will, a = one)
wa = who
wi = who
wieT = who, it (wi = who, eT = it)
witeT = who does it (wi = who, t = do, eT = it/they)
weT = who it is (wi = who, eT = it, then loss of "i", giving "weT")
ker = she understands (k = understand, er =she)
At this stage I would like to comment that we are here in the presence of a
Germanic language which behaves very, very strangely in the way of the
meanings of its compound words. For instance, "viden" (to be with death) is
made up of the words for "with", "die" and the infinitive suffix. I am sure
that Levitov here was thinking of a construction like German "mitkommen"
which means "to come along" (to "withcome"). I suppose I could say "Bitte,
sterben Sie mit" on the same model as "Bitte, kommen Sie mit" ("Come with
me/us, please), thereby making up a verb "mitsterben", but that would mean
"to die together with someone else", not "to be with death".
Let us see how Levitov translates a whole sentence. Since he does not
explain how he breaks up those compound words I have tried to do it using
the vocabulary and grammar he provides in those pages. My tentative
explanations are in parenthesis.
TanvieT faditeT wan aTviteT anTviteT atwiteT aneT
TanvieT = the one way (T = the (?), an = one, vi =way, eT = it)
faditeT = doing for help (f = for, ad = aid, i = -ing, t = do, eT = it)
wan = person (wi/wa = who, an = one)
aTviteT = one that one knows (a = one, T = that, vit = know, eT = it.
Here, Levitov adds one extra letter which is not in the text,
getting "aTaviteT", which provide the second "one" of his
translation)
anTviteT = one that knows (an =one, T = that, vit = know, eT = it)
atwiteT = one treats one who does it (a = one, t = do, wi = who,
t = do, eT = it. Literally: "one does [one] who does it".
The first "do" is translated as "treat", the second "one" is
added in by Levitov: he added one letter, which gives him
"atawiteT")
aneT = ones (an = one, -eT = the plural ending)
Levitov's translation of the above in better English: "the one way for
helping a person who needs it, is to know one of the ones who do treat
one".
Need I say more? Does anyone still believe that Levitov's translations are
worth anything?
As an exercise, here is the last sentence on p.31, with its word-for-word
translation by Levitov. I leave you to work it out, and to figure out what
it might possibly mean. Good luck!
tvieT nwn anvit fadan van aleC
tvieT = do the ways
nwn = not who does (but Levitov adds a letter to make it "nwen")
anvit = one knows
fadan = one for help
van = one will
aleC = each ail
==> cryptology/swiss.colony.p <==
What are the 1987 Swiss Colony ciphers?
==> cryptology/swiss.colony.s <==
Did anyone solve the 1987 'Crypto-gift' contest that was run by
Swiss Colony? My friend and I worked on it for 4 months, but
didn't get anywhere. My friend solved the 1986 puzzle in
about a week and won $1000. I fear that we missed some clue that
makes it incredibly easy to solve. I'm including the code, clues
and a few notes for those of you so inclined to give it a shot.
197,333,318,511,824,
864,864,457,197,333,
824,769,372,769,864,
865,457,153,824,511,223,845,318,
489,953,234,769,703,489,845,703,
372,216,457,509,333,153,845,333,
511,864,621,611,769,707,153,333,
703,197,845,769,372,621,223,333,
197,845,489,953,223,769,216,223,
769,769,457,153,824,511,372,223,
769,824,824,216,865,845,153,769,
333,704,511,457,153,333,824,333,
953,372,621,234,953,234,865,703,
318,223,333,489,944,153,824,769,
318,457,234,845,318,223,372,769,
216,894,153,333,511,611,
769,704,511,153,372,621,
197,894,894,153,333,953,
234,845,318,223
CHRIS IS BACK WITH GOLD FOR YOU
HIS RHYMES CONTAIN THE SECRET.
YOU SCOUTS WHO'VE EARNED YOUR MERIT BADGE
WILL QUICKLY LEARN TO READ IT.
SO WHEN YOUR CHRISTMAS HAM'S ALL GONE
AND YOU'RE READY FOR THE TUSSLE,
BALL UP YOUR HAND INTO A FIST
AND SHOW OUR MOUSE YOUR MUSCLE.
PLEASE READ THESE CLUES WE LEAVE TO YOU
BOTH FINE ONES AND THE COARSE;
IF CARE IS USED TO HEED THEM ALL
YOU'LL SUFFER NO REMORSE.
Notes:
The puzzle comes as a jigsaw that when assembled has the list of
numbers. They are arranged as indicated on the puzzle, with commas.
The lower right corner has a drawing of 'Secret Agent Chris Mouse'.
He holds a box under his arm which looks like the box
the puzzle comes in. The upper left
corner has the words 'NEW 1987 $50,000 Puzzle'. The lower
left corner is empty. The clues are printed on the
entry form in upper case, with the punctuation as shown.
Ed Rupp
...!ut-sally!oakhill!ed
Motorola, Inc., Austin Tx.
==> cryptology/vcrplus.p <==
What is the code used by VCR+?
==> cryptology/vcrplus.s <==
This program will decode codes 1 through 1000.
/*
* Copyright 1991 Ken Shirriff shir...@sprite.Berkeley.EDU
*/
char *tn[8] = { "6:30", "4:00", "7:30", "4:30", "3:30", "5:30", "6:00", "2:30"};
main(argc,argv)
int argc;
char **argv;
{
int num, month;
int line, day;
int time, chan;
int shift;
int wrap;
int decnum;
int num0;
int table[32][32];
if (argc != 3) {
printf("Usage: decode num month\n");
exit(-1);
}
num = atoi(argv[1]);
num0 = num;
month = atoi(argv[2]);
decnum = decode100(num%100);
if (num==103 || num==387 || num==474 || num==536 || num==658 ||
num==745 || num==929) {
printf("number %d does not fall into the range of the others\n", num);
} else if (num <= 100) {
/*
* Swap 1-9 decoded with 1-9 encoded
*/
if (1<= num && num<=9) {
decnum = num;
} else if (decnum<=9) {
decnum = decode100(decnum);
}
day = 1 + (decnum-1)/32;
line = (decnum+day-1)%32;
} else {
retry:
/*
* We decode the last two digits.
* Then we shift according to the first digit.
* Each shift moves us 3 days over and 7 lines down.
* But since we are using the sheared table, the 3 days over
* results in moving 4 days down.
*/
shift = (11 + num/100 - shift100(decnum))%10;
day = 1 + (decnum-1)/32; /* 1-4 */
line = (decnum-1)%32; /* 0-31 sheared table */
line += shift*4; /* 0-31 + wrap */
/*
* If we've moved down more than 32 lines, we have to wrap back.
*/
wrap = line/32;
day += shift*3 + wrap;
line += day; /* Undo the shear */
decnum = ((line-day-1)%31)+(day-1)*32+1; /* sheared table number*/
/*
* If we decode a number >100 into something in the first 100,
* we have to take the number there and start over.
* This ensures that numbers 1-100 map into codes 1-100.
*/
if (decnum<100) {
/*
* Get the appropriate entry from the first columns, and start over.
*/
num = decode100(num%100);
decnum = decode100(num);
goto retry;
}
}
/*
* Apply the month correction.
*/
line = (line+day*month)&31;
/*
* Decode the line into the time and channel.
*/
time = ((line&16)>>2) | ((line&4)>>1) | (line&1);
chan = (((line&8)>>2) | ((line&2)>>1))+1;
printf("Code %d in month %d = %s, ch %d on day %d\n", num0, month,
tn[time], chan, day);
}
/*
* Decode 0-99 into a sequential number 1-100:
* 1
* .. 33
* .. 65
* .. .. .. 97
* .. .. .. ..
* .. .. .. 100
* 32
* 64
* 96
*/
int
decode100(num)
{
int day;
int row, col, rem, div;
/*
* 4 special cases that make the modulo operations messy
*/
switch (num) {
case 87:
return 97;
case 58:
return 98;
case 29:
return 99;
case 0:
return 100;
}
/*
* Break up into 7 rows of 5 columns on 3 days.
* The numbers are broken mod 29 and then broken in half again.
*/
rem = num%29;
div = num/29;
if (rem<16-div) {
row = 3-div;
} else {
row = 6-div;
rem -= 13;
}
col = 4-(rem-1)/3;
day = (rem-1)%3;
/*
* The numbers are then assigned consecutively down the columns.
*/
return col*7+row + day*31;
}
/*
* Compute the 100's digit shift.
*/
int
shift100(num)
{
int shift;
int i,j;
i = (num+30)%31;
j = (num+30)/31;
shift = ((i+1)/10)*7 + j*4 + i*3;
if ((i==8 || i==28) && (j==2 || j==3)) shift += 7;
if (i==6 && j==4) shift += 8;
if ((i==17 || i==18) && j==3) shift += 7;
shift = shift%10;
return shift;
}
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> geometry/K3,3.p <==
Can three houses be connected to three utilities without the pipes crossing?
_______ _______ _______
| oil | |water| | gas |
|_____| |_____| |_____|
_______ _______ _______
|HOUSE| |HOUSE| |HOUSE|
| one | | two | |three|
==> geometry/K3,3.s <==
The problem you describe is to draw a bipartite graph of 3 nodes
connected in all ways to 3 nodes, all embedded in the plane. The graph
is called K3,3. A famous theorem of Kuratowsky says that all graphs
can be embedded in the plane, EXCEPT those containing a subgraph that
is topologically equivalent to K3,3 or K5 (the complete graph on 5
vertices, i.e., the graph with 5 nodes and 10 edges). So your problem
is a minimal example of a graph that cannot be embedded in the plane.
The proofs that K5 and K3,3 are non-planar are really quite easy, and
only depend on Euler's Theorem that F-E+V=2 for a planar graph. For
K3,3 V is 6 and E is 9, so F would have to be 5. But each face has at
least 4 edges, so E >= (F*4)/2 = 10, contradiction. For K5 V is 5 and
E is 10, so F = 7. In this case each face has at least 3 edges, so E >=
(F*3)/2 = 10.5, contradiction.
The difficult part of Kuratowsky is the proof in the other direction!
A quick, informal proof by contradiction without assuming Euler's Theorem:
Using a map in which the houses are 1, 2, and 3 and the utilities are
A, B, and C, there must be continuous lines that connect the buildings and
divide the area into three sections bounded by the loops A-1-B-2-A,
A-1-B-3-A, and A-2-B-3-A. (One of the areas is the infinite plane *around*
whichever loop is the outer edge of the network.) C must be in one of these
three areas; whichever area it is in, either 1, or 2, or 3, is *not* part of
the loop that rings its area and hence is inaccessible to C.
The usual quibble is to solve the puzzle by running one of the pipes
underneath one of houses on its way to another house; the puzzle's
instructions forbid crossing other *pipes*, but not crossing other *houses*.
==> geometry/bear.p <==
If a hunter goes out his front door, goes 50 miles south, then goes 50
miles west, shoots a bear, goes 50 miles north and ends up in front of
his house. What color was the bear?
==> geometry/bear.s <==
The hunter's door is in one of two locations. One is a foot or so from the
North Pole, facing north, such that his position in front of the door is
precisely upon the North Pole. Since that's a ridiculous place to build a
house and since bears do not roam within fifty miles of the pole, the bear
is either imaginary or imported, and there is no telling what color it is.
There is another place (actually a whole set) on earth from which one
can go fifty miles south, fifty miles west, and fifty miles north and
end up where one started. Consider the parallel of latitude close
enough to the South Pole that its length is 50/n miles, for some
integer n.
Take any point on that parallel of latitude and pick the point fifty miles
north of it. Situate the hunter's front porch there. The hunter goes fifty
miles south from his porch and is at a point we'll call A. He travels fifty
miles west, circling the South Pole n times, and is at A again, where he shoots
the bear. Fifty miles north from A he is back home. Since bears are not
indigenous to the Antarctic, again the bear is either imaginary or imported
and there is no telling what color it might be.
==> geometry/bisector.p <==
Prove if two angle bisectors of a triangle are equal, then the triangle is
isosceles (more specifically, the sides opposite to the two angles
being bisected are equal).
==> geometry/bisector.s <==
PROVE: <ABC = <BCA (i.e. triangle ABC is an isosceles triangle)
A
/ \
/ \
D E XP normal to AB
/ \ / \ XQ normal to AC
P /----X----\ Q
/ / \ \
/ / \ \
/ / \ \
B/_______________\C
PROOF :
Let XP and XQ be normals to AC and AB.
Since the three angle bisectors are concurrent, AX bisects angle A
also and therefore XP = XQ.
Let's assume XD > XE.
Then ang(PDX) < ang(QEX)
Now considering triangles BXD and CXE,
the last condition requires that
ang(DBX) > ang(ECX)
OR ang(XBC) > ang(XCB)
OR XC > XB
Thus our assumption leads to :
XC + XD > XE + XB
OR CD > BE
which is a contradiction.
Similarly, one can show that XD < XE leads to a contradiction too.
Hence XD = XE => CX = BX
From which it is easy to prove that the triangle is isosceles.
-- Manish S Prabhu (mpr...@magnus.acs.ohio-state.edu)
==> geometry/calendar.p <==
Build a calendar from two sets of cubes. On the first set, spell the
months with a letter on each face of three cubes. Use lowercase
three-letter English abbreviations for the names of all twelve months
(e.g., "jan", "feb", "mar"). On the second set, number the days with a
digit on each face of two cubes (e.g., "01", "02", etc.).
==> geometry/calendar.s <==
First note that there are *nineteen* different letters in the
month abbreviations (abcdef gjlmno prstuv y) so to get them all on the
eighteen faces of 3 cubes, you know right away you're going to have to
resort to trickery.
So I wrote them all down and looked at which ones could be
reversed to make another letter in the set. The only pair that jumped
out at me was the d/p pair. Now I knew that it was at least feasible,
as long as it wasn't necessary to duplicate any letters.
Then I scanned the abbreviations to find ones that had a lot of
common letters. The jan-jun-jul series looked like a good place to
start:
j a n
u l
was a good beginning but I realized
right away that I had no room for duplicate letters and the second cube
had both a and u so aug was going to be impossible. In fact I almost
posted that answer. Then I realized that if Martin Gardner wrote about
it, it must have a solution. :-) So I went back to the letter list.
I don't put tails on my u's so it didn't strike me the first
time through that n and u could be combined.
Cube 1 Cube 2 Cube 3
j a n/u
n/u l
would let me get away with putting the g
on the first cube to get aug, so I did.
j a n/u
g n/u l (1)
Now came the fun part. The a was placed so I had to work around
it for the other months that had an a in them (mar, apr, may).
m a r
d/p y (2)
Now the d/p was placed so I had to work around that for sep and dec.
This one was easy since they shared an e as well.
d/p e s
c (3)
Now the e was placed so feb had to be worked in.
f e b (4)
The two months left (oct, nov) were far more complex. Not only
did they have two "set" letters (c, n/u), there were two possible n/u's
to be set with. That's why I left them for last.
o t c
n/u v (5)
So now I had five pieces to fit together, so that no set would
have more than six letters in it. Trial and error provided:
j a n/u a b e
g n/u l or, c d/p g
r s m alphabetically: f l j
y c d/p n/u m o
e v t s n/u r
o f b v t y
Without some gimmick the days cannot be done. Because of the dates 11 and
22, there must be a 1 and a 2 on each cube. Thus there are 8 remaining spaces
for the 8 remaining numbers, and because of 30, we put 3 and 0 on different
cubes. I don't think the way you allocate the others matter. Now 6 numbers on
each cube can produce at most 36 distinct pairs, and we need 31 distinct pairs
to represent all possible dates. But since 3 each of {4,5,6,7,8,9} are on each
cube, there are at least 9 representable numbers which can't be dates.
Therefore there are at most 27 distinct numbers which are dates on the two
cubes, and it can't be done. In particular, not all of {04,05,06,07,08,09} can
be represented.
The gimmick solution would be to represent the numbers in a stylised format
(like say, on a digital clock or on a computer screen) such that the 6 can be
turned upside down to be a 9. Then you can have 012 on both cubes, and three
each of {3,4,5,6,7,8} on the other faces. Done.
Example: 012468 012357
==> geometry/circles.and.triangles.p <==
Find the radius of the inscribed and circumscribed circles for a triangle.
==> geometry/circles.and.triangles.s <==
Let a, b, and c be the sides of the triangle. Let s be the
semiperimeter, i.e. s = (a + b + c) / 2. Let A be the area
of the triangle, and let x be the radius of the incircle.
Divide the triangle into three smaller triangles by drawing
a line segment from each vertex to the incenter. The areas
of the smaller triangles are ax/2, bx/2, and cx/2. Thus,
A = ax/2 + bx/2 + cx/2, or A = sx.
We use Heron's formula, which is A = sqrt(s(s-a)(s-b)(s-c)).
This gives us x = sqrt((s-a)(s-b)(s-c)/s).
The radius of the circumscribed circle is given by R = abc/4A.
==> geometry/coloring/cheese.cube.p <==
A cube of cheese is divided into 27 subcubes. A mouse starts at one
corner and eats each subcube, one at a time. Can it finish in the middle?
==> geometry/coloring/cheese.cube.s <==
Give the subcubes a checkerboard-like coloring so that no two adjacent
subcubes have the same color. If the corner subcubes are black, the
cube will have 14 black subcubes and 13 white ones. The mouse always
alternates colors and so must end in a black subcube. But the center
subcube is white, so the mouse can't end there.
==> geometry/coloring/triominoes.p <==
There is a chess board (of course with 64 squares). You are given 21
"triominoes" of size 3-by-1 (the size of an individual square on a
chess board is 1-by-1). Which square on the chess board can you cut out
so that the 21 triominoes exactly cover the remaining 63 squares? Or is
it impossible?
==> geometry/coloring/triominoes.s <==
||||||||
||||||||
||||||||
---***+*
---...+*
---*+O+*
---*+...
---*+***
There is only one way to remove a square, aside from rotations and
reflections. To see that there is at most one way, do this: Label
all the squares of the chessboard with A, B or C in sequence by rows
starting from the top:
ABCABCAB
CABCABCA
BCABCABC
ABCABCAB
CABCABCA
BCABCABC
ABCABCAB
CABCABCA
Every triomino must cover one A, one B and one C. There is one extra
A square, so an A must be removed. Now label the board again by
rows starting from the bottom:
CABCABCA
ABCABCAB
BCABCABC
CABCABCA
ABCABCAB
BCABCABC
CABCABCA
ABCABCAB
The square removed must still be an A. The only squares that got
marked with A both times are these:
........
........
..A..A..
........
........
..A..A..
........
........
==> geometry/construction/4.triangles.6.lines.p <==
Can you construct 4 equilateral triangles with 6 toothpicks?
==> geometry/construction/4.triangles.6.lines.s <==
Use the toothpicks as the edges of a tetrahedron.
==> geometry/construction/5.lines.with.4.points.p <==
Arrange 10 points so that they form 5 rows of 4 each.
==> geometry/construction/5.lines.with.4.points.s <==
Draw a 5 pointed star, put a point where any two lines meet.
==> geometry/construction/square.with.compass.p <==
Construct a square with only a compass and a straight edge.
==> geometry/construction/square.with.compass.s <==
Draw a circle (C1 at P1). Now draw a diameter D1 (intersects at P2 and
P3). Set the compass larger than before. From each of points P2 and
P3 draw another larger circle (C2 and C3). Where these two circles
cross, draw a line (D2). This line should go through the center of
circle C1 at a right angle to the original diameter line. This line
should cross circle C1 at P4 and P5. Points P2, P4, P3, P5 form a
square.
For extra credit:
Reset the compass to its original size. From P2 and P4 draw a circle
(C4 and C5). These circles intersect at P6 and P1. Connect P6, P2,
P1, P4 for a square of the same size as the original compass setting.
==> geometry/corner.p <==
A hallway of width A turns through 90 degrees into a hallway of width
B. A ladder is to be passed around the corner. If the movement is
within the horizontal plane, what is the maximum length of the ladder?
==> geometry/corner.s <==
------\---------
B / \.......|..B/sin(theta)
theta\ |
---|-----X |
|\ |
| \...|..A/cos(theta)
| \ |
| \ |
| A \|
Theta is the angle off horizontal.
Minimize length = A/cos(theta) + B/sin(theta)
d(length)/d(theta)
= A*sin(theta)/cos(theta)^2 - B*cos(theta)/sin(theta)^2 (?)
= 0
A*sin(theta)/cos(theta)^2 = B*cos(theta)/sin(theta)^2
B/A = sin(theta)^3/cos(theta()^3 = tan(theta)^3
theta = inverse_tan(cube_root(B/A))
If you use the trigonometric formulas cos^2 x = 1/(1 + tan^2 x)
and sin x = tan x cos x, and plug through the algebra, I believe
that the formula for the length reduces to
(A^(2/3) + B^(2/3))^(3/2)
At any rate this is symmetric in A and B as one would expect, and
has the right values at A=B and as either A-->0 or B-->0.
==> geometry/cover.earth.p <==
A thin membrane covers the surface of the (spherical) earth. One
square meter is added to the area of this membrane to form a larger
sphere. How much is added to the radius and volume of this membrane?
==> geometry/cover.earth.s <==
We know that V = (4/3)*pi*r^3 and A = 4*pi*r^2.
We need to find out how much V increases if A increases by 1 m^2.
dV / dr = 4 * pi * r^2
dA / dr = 8 * pi * r
dV / dA = (dV / dr) / (dA / dr)
= (4 * pi * r^2) / (8 * pi * r)
= r/2
= 3,250,000 m
If the area of the cover is increased by 1 square meter,
then the volume it contains is increased by about 3.25 million cubic meters.
We seem to be getting a lot of mileage out of such a small square of cotton.
However, the new cover would not be very high above the surface of the
planet -- about 6 nanometers (calculate dr/dA).
==> geometry/cycle.polynomial.p <==
What are the cycle polynomials for the Platonic solids?
==> geometry/cycle.polynomial.s <==
For future reference, here are the cycle polynomials for the five platonic
solids (and I threw in the tesseract for good measure). Most combinatoric
coloring problems are simple plug-ins to these polynomials. For details,
see any book on combinatorics that presents Polya counting theory.
tetrahedron: (x1^4+3x2^2+8x1*x3)/12
cube: (x1^6+6x2^3+3x1^2*x2^2+8x3^2+6x1^2*x4)/24
octahedron: (x1^8+9x2^4+8x1^2*x3^2+6x4^2)/24
dodecahedron: (x1^12+15x2^6+20x3^4+24x1^2*x5^2)/60
icosahedron: (x1^20+15x2^10+20x1^2*x3^6+24x5^4)/60
tesseract: (32x6^4+x2^12+48x8^3+x1^24+24x1^2*x2^11+12x2^2*x4^5+32x3^8+12x4^6
+18x1^4*x2^10+12x1^4*x4^5)/192
==> geometry/dissections/disk.p <==
Can a disk be cut into similar pieces without point symmetry about the
midpoint? Can it be done with a finite number of pieces?
==> geometry/dissections/disk.s <==
Yes. Draw a circle inside the circumference of the disk, sharing a
common point on the right. Now draw another circle inside the second,
sharing a point at the left. Now draw another inside the third,
sharing a point at the right. Continue in this way, coloring in every
other region thus generated. Now, all the colored regions touch, so
count this as one piece and the uncolored regions as a second piece.
So the circle has been divided into two similar pieces and there is no
point symmetry about the midpoint. Maybe it is cheating to call these
single pieces, though.
==> geometry/dissections/hexagon.p <==
Divide the hexagon into:
1) 3 identical rhombuses.
2) 6 identical kites(?).
3) 4 identical trapezoids (trapeziums in Britain).
4) 8 identical shapes (any shape).
5) 12 identical shapes (any shape).
==> geometry/dissections/hexagon.s <==
What is considered 'identical' for these questions? If mirror-image shapes
are allowed, these are all pretty trivial. If not, the problems are rather
more difficult...
1. Connect the center to every second vertex.
2. Connect the center to the midpoint of each side.
3. This is the hard one. If you allow mirror images, it's trivial:
bisect the hexagon from vertex to vertex, then bisect with a
perpendicular to that, from midpoint of side to midpoint of side.
4. This one's neat. Let the side length of the hexagon be 2 (WLOG).
We can easily partition the hexagon into equilateral triangles
with side 2 (6 of them), which can in turn be quartered into
equilateral triangles with side 1. Thus, our original hexagon
is partitioned into 24 unit equilateral triangles. Take the
trapezoid formed by 3 of these little triangles. Place one such
trapezoid on the inside of each face of the original hexagon, so
that the long side of the trapezoid coincides with the side of the
hexagon. This uses 6 trapezoids, and leaves a unit hexagon in the
center as yet uncovered. Cover this little hexagon with two of
the trapezoids. Voila. An 8-identical-trapezoid partition.
5. Easy. Do the rhombus partition in #1. Quarter each rhombus by
connecting midpoints of opposite sides. This produces 12 small
rhombi, each of which is equivalent to two adjacent small triangles
as in #4.
Except for #3, all of these partitions can be achieved by breaking up the
hexagon into unit equilateral triangles, and then building these into the
shapes desired. For #3, though, this would require (since there are 24 small
triangles) trapezoids formed from 6 triangles each. The only trapezoid that
can be built from 6 identical triangles is a parallelogram; I assume that the
poster wouldn't have asked for a trapezoid if you could do it with a special
case of trapezoid. At any rate, that parallelogram doesn't work.
==> geometry/dissections/largest.circle.p <==
What is the largest circle that can be assembled from two semicircles cut from
a rectangle with edges a and b?
==> geometry/dissections/largest.circle.s <==
There are two methods:
Method M1:
The diameters of the semicircles have to be on the longer sides,
starting at an endpoint of the rectangle. The two semicircles touch
each other in the middle M of the rectangle.
a
D._______________________.C
| |
| |
b | . M |
| |
| |
|_______.___.___________|
A R X B
R should be the center of the semicircle, and because of RA = RM,
it holds that:
r^2 = (a/2 - r)^2 + (b/2)^2
Solving for r gives:
r = min[b,(a^2+b^2)/(4a)], where a >= b.
Method M2:
We'll cut on the line y = c x, where c will turn out to be slightly
less than d, the slope of the diagonal. We describe the semicircle
lying above the line y = c x, having this line as the straight part of
the semi-circle. The center P of the semicircle will be taken on the
line y = d - x, and will be tangent to the left and top of the
rectangle. Clearly the lower down P is on this line, the better. The
naive solution is not optimal because the upper place where the
semicircle meets the diagonal is interior to the rectangle. So we try
to determine c in such a way that this latter point actually lies
slightly down from the top, on the right side of the rectangle. This
involves solving the quartic:
4r^4 - (4a+16b)r^3 + (16b^2+a^2+8ab)r^2 - (6b^3+4ab^2+2ba^2)r + b^4+(ab)^2 = 0,
where r < b, the details of which will be left to the reader.
The other semicircle is the reflection of the first through the origin.
After a few calculations, we find that the value of r given
by M2 is greater than the one given by M1 only if 1 < a/b < 2.472434.
==> geometry/dissections/square.70.p <==
Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 square be dissected into
24 squares of size 1x1, 2x2, 3x3, etc.?
==> geometry/dissections/square.70.s <==
Martin Gardner asked this in his Mathematical Games column in the
September 1966 issue of Scientific American. William Cutler was the first
of 24 readers who reduced the uncovered area to 49, using all but the 7x7
square. All the patterns were the same except for interchanging the
squares of orders 17 and 18 and rearranging the squares of orders 1, ...,
6, 8, 9, and 10. Nobody proved that the solution is minimal.
+----------------+-------------+----------------------+---------------------+
| | | | |
| | | | |
| | 11 | | |
| | | | |
| 16 | | | |
| +-----+--+----+ 22 | 21 |
| | | 2| | | |
| | 5 +--+----+ | |
| | | | | |
+----------------+--+--+ 6 | | |
| | 3| | | |
| ++-+-------+ | |
| || | ++--------------------+
| || 8 +----------------------++ |
| 18 || | | |
| || | | |
| ++---------+ | |
| | | | 20 |
| | 9 | | |
+------------------++ | 23 | |
| || | | |
| ++----------+ | |
| | | +---++---------------+
| | | | || |
| 17 | 10 | | 4 || |
| | +---------------+-------+---++ |
| +-+---------+---------------+ | 15 |
| | | | | |
| | | | 12 | |
+------------------+-+ | +-+-------------+
| | | |1| |
| | +------------+-+ |
| | 24 | | |
| | | | |
| 19 | | 13 | 14 |
| | | | |
| | | | |
| | | | |
+--------------------+-------------------------+--------------+-------------+
==> geometry/dissections/square.five.p <==
Can you dissect a square into 5 parts of equal area with just a straight edge?
==> geometry/dissections/square.five.s <==
1. Prove you can reflect points which lie on the sides of the square
about the diagonals.
2. Construct two different rectangles whose vertices lie on the square
and whose sides are parallel to the diagonals.
3. Construct points A, A', B, B' on one (extended) side of the square
such that A/A' and B/B' are mirror image pairs with respect to another
side of the square.
4. Construct the mirror image of the center of the square in one
of the sides.
5. Divide the original square into 4 equal squares whose sides are
parallel to the sides of the original square.
6. Divide one side of the square into 8 equal segments.
7. Construct a trapezoid in which one base is a square side and one
base is 5/8 of the opposite square side.
8. Divide one side of the square into 5 equal segments.
9. Divide the square into 5 equal rectangles.
==> geometry/dissections/tesseract.p <==
If you suspend a cube by one corner and slice it in half with a
horizontal plane through its centre of gravity, the section face is a
hexagon. Now suspend a tesseract (a four dimensional hypercube) by one
corner and slice it in half with a hyper-horizontal hyperplane through
its centre of hypergravity. What is the shape of the section
hyper-face?
==> geometry/dissections/tesseract.s <==
The 4-cube is the set of all points in [-1,1]^4 .
The hyperplane { (x,y,z,w) : x + y + z + w = 0 } cuts the 4-cube
in the desired manner.
Now, { (.5,.5,-.5,-.5), (.5,-.5,.5,-.5), (.5,-.5,-.5,.5) } is an
orthonormal basis for the hyperplane. Let (a,b,c) be a point on the
hyperplane with respect to this basis. (a,b,c) is in the 4-cube if and
only if |a| + |b| + |c| <= 2. The shape of the intersection is a
regular octahedron.
==> geometry/duck.and.fox.p <==
A duck is swimming about in a circular pond. A ravenous fox (who cannot
swim) is roaming the edges of the pond, waiting for the duck to come close.
The fox can run faster than the duck can swim. In order to escape,
the duck must swim to the edge of the pond before flying away. Assume that
the duck can't fly until it has reached the edge of the pond.
How much faster must the fox run that the duck swims in order to be always
able to catch the duck?
==> geometry/duck.and.fox.s <==
Assume the ratio of the fox's speed to the duck's is a, and the radius
of the pond is r. The duck's best strategy is:
1. Swim around a circle of radius (r/a - delta) concentric with the
pond until you are diametrically opposite the fox (you, the fox, and
the center of the pond are colinear).
2. Swim a distance delta along a radial line toward the bank opposite
the fox.
3. Observe which way the fox has started to run around the circle.
Turn at a RIGHT ANGLE in the opposite direction (i.e. if you started
swimming due south in step 2 and the fox started running to the east,
i.e. clockwise around the pond, then start swimming due west). (Note:
If at the beginning of step 3 the fox is still in the same location as
at the start of step 2, i.e. directly opposite you, repeat step 2
instead of turning.)
4. While on your new course, keep track of the fox. If the fox slows
down or reverses direction, so that you again become diametrically
opposite the fox, go back to step 2. Otherwise continue in a straight
line until you reach the bank.
5. Fly away.
The duck should make delta as small as necessary in order to be able
to escape the fox.
The key to this strategy is that the duck initially follows a
radial path away from the fox until the fox commits to running either
clockwise or counterclockwise around the pond. The duck then turns onto
a new course that intersects the circle at a point MORE than halfway
around the circle from the fox's starting position. In fact, the duck
swims along a tangent of the circle of radius r/a. Let
theta = arc cos (1/a)
then the duck swims a path of length
r sin theta + delta
but the fox has to run a path of length
r*(pi + theta) - a*delta
around the circle. In the limit as delta goes to 0, the duck will
escape as long as
r*(pi + theta) < a*r sin theta
that is,
pi + arc cos (1/a) - a * sqrt(a^2 - 1) < 0
Maximize a in the above: a = 4.6033388487517003525565820291030165130674...
The fox can catch the duck as long as he can run about 4.6 times as fast as
the duck can swim.
"But wait," I hear you cry, "When the duck heads off to that spot
'more than halfway' around the circle, why doesn't the fox just double
back? That way he'll reach that spot much quicker." That is why the
duck's strategy has instructions to repeat step 2 under certain
circumstances. Note that at the end of step 2, if the fox has started
to run to head off the duck, say in a clockwise direction, he and the
duck are now on the same side of some diameter of the circle. This
continues to be true as long as both travel along their chosen paths
at full speed. But if the fox were now to try to reach the duck's
destination in a counterclockwise direction, then at some instant he
and the duck must be on a diameter of the pond. At that instant, they
have exactly returned to the situation that existed at the end of step
1, except that the duck is a little closer to the edge than she was
before. That's why the duck always repeats step 2 if the fox is ever
diametrically opposite her. Then the fox must commit again to go one
way or the other. Every time the fox fails to commit, or reverses his
commitment, the duck gets a distance delta closer to the edge. This
is a losing strategy for the fox.
The limiting ratio of velocities that this strategy works against
cannot be improved by any other strategy, i.e., if the ratio of
the duck's speed to the fox's speed is less than a then the duck
cannot escape given the best fox strategy.
Given a ratio R of speeds less than the above a, the fox is sure to
catch the duck (or keep it in water indefinitely) by pursuing the
following strategy:
Do nothing so long as the duck is in a radius of R around the centre.
As soon as it emerges from this circle, run at top speed around the
circumference. If the duck is foolish enough not to position itself
across from the center when it comes out of this circle, run "the short
way around", otherwise run in either direction.
To see this it is enough to verify that at the circumference of the
circle of radius R, all straight lines connecting the duck to points
on the circumference (in the smaller segment of the circle cut out
by the tangent to the smaller circle) bear a ratio greater than R
with the corresponding arc the fox must follow. That this is enough
follows from the observation that the shortest curve from a point on
a circle to a point on a larger concentric circle (shortest among all
curves that don't intersect the interior of the smaller circle) is
either a straight line or an arc of the smaller circle followed by a
tangential straight line.
==> geometry/earth.band.p <==
How much will a band around the equator rise above the surface if it is
made one meter longer? Assume the equator is a circle.
==> geometry/earth.band.s <==
The formula for the circumference of a circle is 2 * pi * radius. Therefore,
if you increase the circumference by 1 meter, you increase the radius by
1/(2 * pi) meters, or about 0.16 meters.
==> geometry/fence.p <==
A farmer wishes to enclose the maximum possible area with 100 meters of fence.
The pasture is bordered by a straight cliff, which may be used as part of the
fence. What is the maximum area that can be enclosed?
==> geometry/fence.s <==
A circle is the plane figure with highest ratio of area to perimeter.
The cliff can be used to bisect a circle of radius 100/pi meters. By
symmetry, this will form the pen of largest area. The resulting pen
will contain 5000/pi meters squared.
==> geometry/ham.sandwich.p <==
Consider a ham sandwich, consisting of two pieces of bread and one of
ham. Suppose the sandwich was dropped into a machine and spindled,
torn and mutilated. Is it still possible to divide the ham sandwich
with a straight knife cut such that both the ham and each slice of
bread are divided in two parts of equal volume?
==> geometry/ham.sandwich.s <==
Yes. There is a theorem in topology called the Ham Sandwich Theorem,
which says: Given 3 (finite) volumes (each may be of any shape, and in
several pieces), there is a plane that cuts each volume in half. You
would learn about it typically in a first course in algebraic topology,
or maybe in a course on introductory topology (if you studied the
fundamental group).
==> geometry/hike.p <==
You are hiking in a half-planar woods, exactly 1 mile from the edge,
when you suddenly trip and lose your sense of direction. What's the
shortest path that's guaranteed to take you out of the woods? Assume
that you can navigate perfectly relative to your current location and
(unknown) heading.
==> geometry/hike.s <==
Go 2/sqrt(3) away from the starting point, turn 120 degrees and head
1/sqrt(3) along a tangent to the unit circle, then traverse an arc of
length 7*pi/6 along this circle, then head off on a tangent 1 mile.
This gives a minimum of sqrt(3) + 7*pi/6 + 1 = 6.397...
It remains to prove this is the optimal answer.
==> geometry/hole.in.sphere.p <==
Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six inches long.
Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!
==> geometry/hole.in.sphere.s <==
The volume of the leftover material is equal to the volume of a 6" sphere.
First, lets look at the 2 dimensional equivalent of this problem. Two
concentric circles where the chord of the outer circle that is tangent
to the inner circle has length D. What is the annular area between the
circles?
It is pi * (D/2)^2. The same area as a circle with that diameter.
Proof:
big circle radius is R
little circle radius is r
2 2
area of donut = pi * R - pi * r
2 2
= pi * (R - r )
Draw a right triangle and apply the Pythagorean Theorem to see that
2 2 2
R - r = (D/2)
so the area is
2
= pi * (D/2)
Start with a sphere of radius R (where R > 6"), drill out the 6"
high hole. We will now place this large "ring" on a plane. Next to it
place a 6" high sphere. By Archemedes' theorem, it suffices
to show that for any plane parallel to the base plane, the cross-
sectional area of these two solids is the same.
Take a general plane at height h above (or below) the center
of the solids. The radius of the circle of intersection on the sphere is
radius = srqt(3^2 - h^2)
so the area is
pi * ( 3^2 - h^2 )
For the ring, once again we are looking at the area between two concentric
circles. The outer circle has radius sqrt(R^2 - h^2),
The area of the outer circle is therefore
pi (R^2 - h^2)
The inner circle has
radius sqrt(R^2 - 3^2). So the area of the inner circle is
pi * ( R^2 - 3^2 )
the area of the doughnut is therefore
pi(R^2 - h^2) - pi( R^2 - 3^2 )
= pi (R^2 - h^2 - R^2 + 3^2)
= pi (3^2 - h^2)
Therefore the areas are the same for every plane intersecting the solids.
Therefore their volumes are the same.
QED
There also is a meta-theoretic answer to this puzzle. Assume the puzzle
can be solved. Then it must be solvable with a hole of any diameter, even
zero. But if you drill a hole of zero diameter that is six inches long,
you leave behind the volume of a six inch diameter sphere.
==> geometry/hypercube.p <==
How many vertices, edges, faces, etc. does a hypercube have?
==> geometry/hypercube.s <==
Take any vertex of the hypercube, and ask how many k-V's it
participates in. To make a k-V it needs to combine with k adjacent and
orthogonal vertices, and there are (nCk) distinct ways of doing this
[that is, choose k directions out of n possible ones]. Then multiply
by 2^n, the total number of vertices. But this involves multiple
counting, since each k-V is shared by 2^k vertices. So divide by 2^k,
and this yields the answer: (nCk)*2^{n-k}.
For example, 12d hypercube:
0-v: 4,096
1-v: 24,576
2-v: 67,584
3-v: 112,640
4-v: 126,720
5-v: 101,376
6-v: 59,136
7-v: 25,344
8-v: 7,920
9-v: 1,760
10-v: 264
11-v: 24
12-v: 1
==> geometry/kissing.number.p <==
How many n-dimensional unit spheres can be packed around one unit sphere?
==> geometry/kissing.number.s <==
From the Feb. 1992 issue of Scientific American:
Kissing Numbers
Dimension Lower limit Upper limit
1* 2 2
2* 6 6
3* 12 12
4 24 25
5 40 46
6 72 82
7 126 140
8* 240 240
9 306 380
10 500 595
11 582 915
12 840 1,416
13 1,130 2,233
14 1,582 3,492
15 2,564 5,431
16 4,320 8,313
17 5,346 12,215
18 7,398 17,877
19 10,688 25,901
20 17,400 37,974
21 27,720 56,852
22 49,896 86,537
23 93,150 128,096
24* 196,560 196,560
* = dimensions for which the answer is known.
REFERENCES (from the Sci. Am. article)
The Problem of the Thirteen Spheres. John Leech in Mathematical Gazette,
Vol. 40, No. 331, pages 22-23; February 1956
Sphere Packings, Lattics and Groups. John Horton Conway and Neil J. A.
Sloane. Springer-Verlag, 1988.
Sphere Packings and Spherical Geometry--Kepler's Conjecture and Beyond,
preprint. Wu-Yi Hsiang. Center for Pure and Applied Mathematics,
University of California, Berkeley, July 1991.
--
David Radcliffe
radc...@csd4.csd.uwm.edu
==> geometry/konigsberg.p <==
Can you draw a line through each edge on the diagram below without crossing
any edge twice and without lifting your pencil from the paper?
+---+---+---+
| | | |
+---+-+-+---+
| | |
+-----+-----+
==> geometry/konigsberg.s <==
This is solved in the same way as the famous "Seven Bridges of
Konigsberg" problem first solved by Euler. In that problem, the task
was to find a closed path that crossed each of the seven bridges of
Konigsberg (now Kaliningrad, Russia) exactly once. For reasons given
below, no such path existed. In this version, you cannot draw such a
line without cheating by:
(1) drawing a line along one of the edges, or
(2) inscribing the diagram on a torus, or
(3) defining a line segment as the entire length of each straight line, or
(4) adding a vertex on one of the line segemnts, or
(5) defining "crossing" as touching the endpoint of a segment.
The method for determining if paths exist in all similar problems is
given below.
Turn each "room" into a point. Turn each line segment into a line
connecting the two points representing the rooms it abuts. You should
be able to see that drawing one continuous line across all segments in
your drawing is equivalent to traversing all the edges in the resulting
graph. Euler's Theorem states that for a graph to be traversable, the
number of vertices with an odd number of edges proceeding from them
must be either zero or two. For this graph, that number is four, and it
cannot be traversed.
+---+---+---+
| 1 | 2 | 3 |
+---+-+-+---+ 6 (outside)
| 4 | 5 |
+-----+-----+
Number of edges proceeding from each vertex:
1: 4
2: 5 (*odd*)
3: 4
4: 5 (*odd*)
5: 5 (*odd*)
6: 9 (*odd*)
To prove Euler's Theorem, think of walking along the graph from vertex to
vertex. Each vertex must be entered as many times as it is exited, except
for where you start and where you end. So, each vertex must have an
even number of edges, except possibly for two vertices. And if there are
two vertices with an odd number of edges, the path must start at one and
end at the other.
==> geometry/ladders.p <==
Two ladders form a rough X in an alley. The ladders are 11 and 13 meters
long and they cross 4 meters off the ground. How wide is the alley?
==> geometry/ladders.s <==
Ladders 1 and 2, denoted L1 and L2, respectively, will rest along two
walls (taken to be perpendicular to the ground), and they will
intersect at a point O = (a,s), a height s from the ground. Find the
largest s such that this is possible. Then find the width of the
alley, w = a+b, in terms of L1, L2, and s. This diagram is not to
scale.
B D
|\ L1 L2 /|
| \ / | BC = length of L1
| \ / | AD = length of L2
| \ O / | s = height of intersection
x| \ / |y A = (0,0)
| /|\ | AE = a
| m / | \ n | EC = b
| / |s \ | AO = m
| / | \ | CO = n
|/________|________\|
(0,0) = A a E b C
-----------------------------------------------------------------------------
Without loss of generality, let L2 >= L1.
Observe that triangles AOB and DOC are similar. Let r be the ratio of
similitude, so that x=ry. Consider right triangles CAB and ACD. By
the Pythagorean theorem, L1^2 - x^2 = L2^2 - y^2. Substituting x=ry,
this becomes y^2(1-r^2) = L2^2 - L1^2. Letting L= L2^2 -L1^2 (L>=0),
and factoring, this becomes
(*) y^2 (1+r)(1-r) = L
Now, because parallel lines cut L1 (a transversal) in proportion, r =
x/y = (L1-n)/n, and so L1/n = r+1. Now, x/s = L1/n = r+1, so ry = x =
s(r+1). Solving for r, one obtains the formula r = s/(y-s).
Substitute this into (*) to get
(**) y^2 (y) (y-2s) = L (y-s)^2
NOTE: Observe that, since L>=0, it must be true that y-2s>=0.
Now, (**) defines a fourth degree polynomial in y. It can be written in the
form (by simply expanding (**))
(***) y^4 - 2s_y^3 - L_y^2 + 2sL_y - Ls^2 = 0
L1 and L2 are given, and so L is a constant. How large can s be? Given L,
the value s=k is possible if and only if there exists a real solution, y',
to (***), such that 2k <= y' < L2. Now that s has been chosen, L and s are
constants, and (***) gives the desired value of y. (Make sure to choose the
value satisfying 2s <= y' < L2. If the value of s is "admissible" (i.e.,
feasible), then there will exist exactly one such solution.)
Now, w = sqrt(L2^2 - y^2), so this concludes the solution.
L1 = 11, L2 = 13, s = 4. L = 13^2-11^2 = 48, so (***) becomes
y^4 - 8_y^3 - 48_y^2 + 384_y - 768 = 0
Numerically find root y ~= 9.70940555, which yields w ~= 8.644504.
==> geometry/lattice/area.p <==
Prove that the area of a triangle formed by three lattice points is integer/2.
==> geometry/lattice/area.s <==
The formula for the area is
A = | x1*y2 + x2*y3 + x3*y1 - x1*y3 - x2*y1 - x3*y2 | / 2
If the xi and yi are integers, A is of the form (integer/2)
==> geometry/lattice/equilateral.p <==
Can an equlateral triangle have vertices at integer lattice points?
==> geometry/lattice/equilateral.s <==
No.
Suppose 2 of the vertices are (a,b) and (c,d), where a,b,c,d are integers.
Then the 3rd vertex lies on the line defined by
(x,y) = 1/2 (a+c,b+d) + t ((d-b)/(c-a),-1) (t any real number)
and since the triangle is equilateral, we must have
||t ((d-b)/(c-a),-1)|| = sqrt(3)/2 ||(c,d)-(a,b)||
which yields t = +/- sqrt(3)/2 (c-a). Thus the 3rd vertex is
1/2 (a+c,b+d) +/- sqrt(3)/2 (d-b,a-c)
which must be irrational in at least one coordinate.
==> geometry/manhole.cover.p <==
Why is a manhole cover round?
==> geometry/manhole.cover.s <==
It will not fall into the hole, even if rotated, tipped, etc.
It gives maximal area for a given amount of material.
It does not have to be carried, but can be rolled.
Human beings are roughly round in horizontal cross section.
Orientation of the cover with the access hole is not of concern.
Orientation of the access hole with the ladder in the pipe below is not
of concern.
==> geometry/pentomino.p <==
Arrange pentominos in 3x20, 4x15, 5x12, 6x10, 2x3x10, 2x5x6 and 3x4x5 forms.
==> geometry/pentomino.s <==
I've seen several different naming schemes used for pentominoes. This is
the system I'm using (I think only F & N require a bit of imagination):
FF I L N PP TTT U U V V W W W X X Y ZZ
FF I L NN PP T UUU V V W W X YY Z
F I L N P T V X X Y ZZ
I LL N Y
A 3x20 solution (the other solution is easily obtained by a rotation of
the section from the Z to the L inclusive):
UUXPPPZYYYYWTFNNNVVV
UXXXPPZZZYWWTFFFNNLV
UUXIIIIIZWWTTTFLLLLV
A 4x15 solution:
IIIIINNLLLLTVVV
UUXNNNFZZWLTTTV
UXXXYFFFZWWTPPV
UUXYYYYFZZWWPPP
2 5x6 rectangles. Joined side-to-side, end-to-end, or stacked, these
enable construction of the 6x10 & 5x12 rectangles, and the 2x5x6 prism:
NFVVV YYYYI
NFFFV LLYZI
NNFXV LZZZI
PNXXX LZWTI
PPUXU LWWTI
PPUUU WWTTT
The 2x3x10 and 3x4x5 solutions are tricky to show - I hope these diagrams
make sense:
A 2x3x10 solution (shown as 2 layers; Y and L are shared between the
2 layers):
VVVZIIIIIF UUXTTTWWPP
VZZZNNNFFF UXXXTWWPPP
VZYYYYNNFL UUXYTWLLLL
A 3x4x5 solution (3 layers, V F W & L shared between 2 or more layers):
VUUXF VZFFF VNYFW
VUXXX TZZZW NNYPW
VUUXW TTTZW NYYPP
IIIII TLLLL NLYPP
--
+------------------- pe...@bignode.equinox.gen.nz -------------------+
| The effort to understand the universe is one of the very few things |
| that lifts human life above the level of farce, and gives it some |
| of the grace of tragedy - Steven Weinberg |
+---------------------------------------------------------------------+
==> geometry/points.in.sphere.p <==
What is the expected distance between two random points inside a sphere?
Assume the points are uniformly and independently distributed.
==> geometry/points.in.sphere.s <==
Use spherical polar coordinates, and w.l.o.g. choose the polar axis
through one of the points. Now the distance between the two points is
sqrt ( r1^2 + r2^2 - 2 r1 r2 cos(theta))
and cos(theta) is (conveniently) uniformly distributed between -1 and
+1, while r1 and r2 have densities 3 r1^2 d(r1) and 3 r2^2 d(r2). Split
the total integral into two (equal) parts with r1 < r2 and r1 > r2, and
it all comes down to integrating polynomials.
More generally, the expectation of the n'th power of the distance
between the two points is
2^n . 72 / ((n+3)(n+4)(n+6))
So the various means are:
the (arithmetic) mean distance is 36/35 = 1.028571...
the root mean square distance is sqrt(6/5) = 1.095445...
the geometric mean distance is 2exp(-3/4) = 0.944733...
the harmonic mean distance is 5/6 = 0.833333...
the inverse root mean inverse square distance is
2/3 = 0.666666...
==> geometry/points.on.sphere.p <==
What are the odds that n random points on a sphere lie in the same hemisphere?
==> geometry/points.on.sphere.s <==
1 - [1-(1/2)^(n-2)]^n
where n is the # of points on the sphere.
The question will become a lot easier if you restate it as the following:
What is the probability in finding at least one point such that all the other
points on the sphere are on one side of the great circle going through this
point.
When n=2, the probability= 1 ,
when n=infinity, it becomes 0.
In his Scientific American column which was titled "Curious Maps",
Martin Gardner ponders the fact that most of the land mass of the Earth
is in one hemisphere and refers to a paper which models continents
by small circular caps. He gives the above result.
See "The Probability of Covering a Sphere With N Circular Caps" by
E. N. Gilbert in Biometrika 52, 1965, p323.
==> geometry/revolutions.p <==
A circle with radius 1 rolls without slipping once around a circle with radius
3. How many revolutions does the smaller circle make?
==> geometry/revolutions.s <==
4 if the smaller circle rolls on the outside of the larger circle; 2 if
it rolls on the inside.
Imagine you are rolling a wheel by pushing it along the equator of the
earth. Suppose the circumference of the wheel is one third of that of
the earth. By the time you return to your starting point, the wheel
finishes 3 revolutions relative to you. But do not forget you yourself
also finishes 1 revolution in the same direction. As a result, the
number of absolute revolutions is 3+1=4.
But if the small circle is rolling inside the large circle, the answer
is then 3-1=2, because in this case the wheel makes a counter-revolution
as you walk once around.
==> geometry/rotation.p <==
What is the smallest rotation that returns an object to its original state?
==> geometry/rotation.s <==
720 degrees.
Objects are made of bosons (integer-spin particles) and fermions
(half-odd-integer spin particles), and the wave function of a fermion
changes sign upon being rotated by 360 degrees. To get it back to its
original state you must rotate by another 360 degrees, for a total of
720 degrees. This fact is the basis of Fermi-Dirac statistics, the
Pauli Exclusion Principle, electron orbits, chemistry, and life.
Mathematically, this is due to the continuous double cover of SO(2) by
SO(3), where SO(2) is the internal symmetry group of fermions and SO(3)
is the group of rotations in three dimensional space.
A fermion can be modeled by a sphere with strings attached. It is
possible to see that a 360 degree rotation will entangle the strings,
which another 360 degree rotation will disentangle. You can also
demonstrate this with a tray, which you hold in your right hand with
the arm lowered, then rotate twice as you raise your arm and end up
with the tray above your head, rotated twice about its vertical axis,
but without having twisted your arm.
Hospitals have machines which take out your blood, centrifuge it to take out
certain parts, then return it to your veins. Because of AIDS they must never
let your blood touch the inside of the machine which has touched others'
blood. So the inside is lined with a single piece of disposable branched
plastic tubing. This tube must rotate rapidly in the centrifuge where
several branches come out. Thus the tube should twist and tangle up the
branches. But the machine untwists the branches as in the above discussion.
At several hundred rounds per minute!
References
R. Penrose and W. Rindler
Spinors and Space-time, vol. 1, p. 43
Cambridge University Press, 1984
R. Feynman and S. Weinberg
Elementary Particles and the Laws of Physics, p. 29
Cambridge University Press, 1987
M. Gardner
The New Ambidextrous Universe, Revised (Third) Edition, pp. 329-332
W. H. Freeman, 1990
==> geometry/shephard.piano.p <==
What's the maximum area shape that will fit around a right-angle corner?
==> geometry/shephard.piano.s <==
This problem is unsolved. A simple solution called the "Shephard
piano" has area 2.2074+, but this can be improved upon with local
modifications. A solution exists with area 2.215649+. It is known
that a maximum area exists, but not whether the shape achieving it is
symmetric, smooth, or even unique.
See Problem G5 in Croft, Falconer, and Guy, _Unsolved Porblems in
Geometry_, Springer-Verlag, 1991.
==> geometry/smuggler.p <==
Somewhere on the high sees smuggler S is attempting, without much
luck, to outspeed coast guard G, whose boat can go faster than S's. G
is one mile east of S when a heavy fog descends. It's so heavy that
nobody can see or hear anything further than a few feet. Immediately
after the fog descends, S changes course and attempts to escape at
constant speed under a new, fixed course. Meanwhile, G has lost track
of S. But G happens to know S's speed, that it is constant, and that S
is sticking to some fixed heading, unknown to G.
How does G catch S?
G may change course and speed at will. He knows his own speed and
course at all times. There is no wind, G does not have radio or radar,
there is enough space for maneuvering, etc.
==> geometry/smuggler.s <==
One way G can catch S is as follows (it is not the fastest way).
G waits until he knows that S has traveled for one mile. At that time, both
S and G are somewhere on a circle with radius one mile, and with its center
at the original position of S. G then begins to travel with a velocity that
has a radially outward component equal to that of S, and with a tangential
component as large as possible, given G's own limitation of total speed. By
doing so, G and S will always both be on an identical circle having its
center at the original position of S. Because G has a tangential component
whereas S does not, G will always catch S (actually, this is not proven
until you solve the o.d.e. associated with the problem).
If G can go at 40 mph and S goes at 20 mph, you can work out that it will
take G at most 1h 49m 52s to catch S. On average, G will catch S in:
( -2pi + sqrt(3) ( exp(2pi/sqrt(3)) - 1 )) / 40pi hours,
which is, 27 min and 17 sec.
==> geometry/spiral.p <==
How far must one travel to reach the North Pole if one starts from the
equator and always heads northwest?
==> geometry/spiral.s <==
One can't reach the North Pole by going northwest. If one could, then
one could leave the North Pole by going southeast. But from the Pole
all directions are due south.
If one heads northwest continuously, one will spiral closer and closer
to the North Pole, until finally one can't turn that sharply.
==> geometry/table.in.corner.p <==
Put a round table into a (perpendicular) corner so that the table top
touches both walls and the feet are firmly on the ground. If there is
a point on the perimeter of the table, in the quarter circle between
the two points of contact, which is 10 cm from one wall and 5 cm from
the other, what's the diameter of the table?
==> geometry/table.in.corner.s <==
Consider the +X axis and the +Y axis to be the corner. The table has
radius r which puts the center of the circle at (r,r) and makes the
circle tangent to both axis. The equation of the circle (table's
perimeter) is
(x-r)^2 + (y-r)^2 = r^2 .
This leads to
r^2 - 2(x+y) + x^2 + y^2 = 0
Using x = 10, y = 5 we get the solutions 25 and 5. The former is the
radius of the table. It's diameter is 50 cm.
The latter number is the radius of a table that has a point which
satisfies the conditions but is not on the quarter circle nearest
the corner.
==> geometry/tetrahedron.p <==
Suppose you have a sphere of radius R and you have four planes that are
all tangent to the sphere such that they form an arbitrary tetrahedron
(it can be irregular). What is the ratio of the surface area of the
tetrahedron to its volume?
==> geometry/tetrahedron.s <==
For each face of the tetrahedron, construct a new tetrahedron with that
face as the base and the center of the sphere as the fourth vertex.
Now the original tetrahedron is divided into four smaller ones, each of
height R. The volume of a tetrahedron is Ah/3 where A is the area of
the base and h the height; in this case h=R. Combine the four
tetrahedra algebraically to find that the volume of the original
tetrahedron is R/3 times its surface area.
==> geometry/tiling/count.1x2.p <==
Count the ways to tile an MxN rectangle with 1x2 dominos.
==> geometry/tiling/count.1x2.s <==
The number of ways to tile an MxN rectangle with 1x2 dominos is
2^(M*N/2) times the product of
{ cos^2(m*pi/(M+1)) + cos^2(n*pi/(N+1)) } ^ (1/4)
over all m,n in the range 0<m<M+1, 0<n<N+1.
[Exercises:
0) Why does this work for M*N odd?
1) When M<3 the count can be determined directly;
check that it agrees with the above formula.
2) Prove directly this formula gives an integer for all M,N, and
further show that if M=N it is a perfect square when 4|N and
twice a square otherwise.
]
Where does this come from? For starters note that, with the usual checker-
board coloring, each domino must cover one light and one dark square. Assume
that M*N is even (but as it happens our formula will work also when both
M,N are odd --- see exercise 0 above). Form a square matrix of size
M*N/2 whose rows and columns are indexed by the light and dark squares,
and whose (j,k) entry is 1 if the j-th light and k-th dark square are
adjacent and zero otherwise. There are now three key ideas:
First, the number of tilings is the number of ways to match each light
square with an adjacent dark square; thus it is the _permanent_ of our
matrix (recall that the permanent of a rxr matrix is a sum of the same
r! terms that occur in its determinant, except without the usual +1/-1
sign factors).
Second, that by modifying this matrix slightly we can convert the
permanent to a determinant; this is nice because determinants are generally
much easier to evaluate than permanents. One way to do this is to replace
all the 1's that correspond to vertical adjacency to i's, and multiply the
whole thing by a suitable power of i (which will disappear when we raise
it to a fourth power).
[Exercise 3: check that this transformation actually works as advertised!]
Third, that we can diagonalize the resulting matrix A --- or, more
conveniently, the square matrix of A' order M*N whose order-(M*N/2)
blocks are 0,A;A-transpose,0 , whence det(A') = +-(det(A))^2. Then
the rows and columns of A' are indexed by squares of either hue on our
generalized checkerboard, and its entries are 1 for horizontally adjacent
squares, i for vertically adjacent ones, and 0 for nonadjacent (including
coincident) squares. This A' can be diagonalized by using the trigonometric
basis of vectors v_ab (a,b as in the formula above) whose coordinate at
the (m,n)-th square is sin(a*m*pi/(M+1)) * sin(b*n*pi/(N+1)).
Exercise 4: verify that these are in fact orthogonal eigenvectors of A',
determine their eigenvalues, and complete the proof of the above formula.
(None of this is new, but it does not seem to be well-known: indeed
each of the above steps seems to have been discovered independently
several times, and I'm not sure whom to credit with the first discovery
of this particular application of the method. For different approaches
to exactly solvable problems involving the enumeration of domino tilings,
see the two papers of G.Kuperberg, Larsen, Propp and myself on
"Alternating-Sign Matrices and Domino Tilings" in the first volume of
the _Journal of Algebraic Combinatorics_.)
--Noam D. Elkies (elk...@zariski.harvard.edu)
Dept. of Mathematics, Harvard University
==> geometry/tiling/rational.sides.p <==
A rectangular region R is divided into rectangular areas. Show that if
each of the rectangles in the region has at least one side with
rational length then the same can be said of R.
==> geometry/tiling/rational.sides.s <==
"Fourteen proofs of a result about tiling a rectangle" (Stan Wagon)
_The American Mathematical Monthly_, Aug-Sep 1987, Vol 94 #7. There
was also a fifteenth proof published a few issues later, attributed to
a (University of Kentucky?) student.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> induction/handshake.p <==
A married couple organizes a party. They only invite other married
couples. At least one person of an invited couple is acquainted to
at least the host or the hostess (so between sets {host,hostess} and
{male of invited couple, female of invited couple} there exists at
least one relation, but two, three or four relations is also possible).
Upon arrival at the party, each person shakes hands with all other
guests he/she doesn't know yet (it is assumed everybody knows
him/herself and his/her partner).
When all couples have arrived and all the handshaking has been done,
the host mingles between the guests and ask everybody (including his
wife): "How many hands did you shake?". To his surprise, all responses
are different.
With the above information, you must be able to determine how many
hands the host and hostess each shook.
==> induction/handshake.s <==
Assume there were 2n people (including host and hostess)
in the party.
1. When the host asked the question he must have got
2n-1 responses (including from his wife).
2. All of the responses were different.
The responses have to be (0, 1, 2, 3, ..., 2n-2)
to satisfy the above requirements. As 2n-2 is the maximum
possible handshakes any person in this party could have made.
/** Below,
P{x} - means a person who shook x hands.
H - means the host
**/
H: <-------->2n-2 0
2n-3 1
2n-4 2
2n-5 3
. .
. .
. .
n n-1 n-2
(There are 2n-1 on the circle.)
P{2n-2} must have handshaked with H (because in the circle he
can handshake with only 2n-3. He has to exclude himself also.)
P{2n-3} must have handshaked with H (because in the circle he
can handshake with only 2n-4.)
P{2n-4} must have handshaked with H (because in the circle he
can handshake with only 2n-5.)
P{n} must have handshaked with H (because in the circle he
can handshake with only n-1.)
from P{n-1} to P{0} nobody handshakes with H, because, for them
the handshake numbers are satisfied on the circle itself.
This leaves H with (n-1) handshakes.
----------------------------------
P{0} must be the spouse of P{2n-2} (since P{2n-2} handshakes with
everbody else.)
.
.
.
same logic till P{n-2}
leaving the Hostess to be P{n-1}.
----------------------------------------
So,
Host - made (n-1) handshakes.
Hostess - made (n-1) handshakes.
where n is the number of couple including
the host couple.
----------------------------------------
==> induction/hanoi.p <==
Is there an algorithm for solving the Hanoi tower puzzle for any number
of towers? Is there an equation for determining the minimum number of
moves required to solve it, given a variable number of disks and towers?
==> induction/hanoi.s <==
The best way of thinking of the Towers of Hanoi problem is inductively.
To move n disks from post 1 to post 2, first move (n-1) disks
from post 1 to post 3, then move disk n from post 1 to post 2, then move
(n-1) disks from post 3 to post 2 (same procedure as moving (n-1) disks
from post 1 to post 3). In order to figure out how to move (n-1) disks
from post 1 to post 3, first move (n-2) disks . . . .
As far as an algorithm which straightens out any legal position is
concerned, the algorithm would go something like this:
1. Find the smallest disk. Call the post that it's on post 1.
2. Find the smallest disk which is not on post 1. This disk is, say,
size k. (I am calling the smallest disk size 1 here.) Call the post
that disk k is on post 2. Disks 1 through (k-1) are then all stacked up
correctly on post 1 disk k is on top of post 2. This follow from the
fact that the disks are in a legal postition.
3. Move disks 1 through (k-1) from post 1 to post 2, ignoring the other
disks. This is just the standard Tower of Hanoi problem for (k-1)
disks.
4. If the disks are not yet correctly arranged, repeat from step 1.
In fact, this gives the straightening with the fewest number of moves.
With 3 towers, for N number of disks, the formula for the minimum number of
moves to complete the puzzle correctly is:
(2^N) - 1
This bit of ancient folklore was invented by De Parville in 1884.
``In the great temple at Benares, says he, beneath the dome which
marks the centre of the world, rests a brass plate in which are
fixed three diamond needles, each a cubit high and as thick as
the body of a bee. On one of these needles, at the creation,
God placed sixty-four discs of pure gold, the largest disc resting
on the brass plate, and the others getting smaller and smaller
up to the top one. This is the Tower of Bramah. Day and night
unceasingly the priests transfer the discs from one diamond needle
to another according to the fixed and immutable laws of Bramah,
which require that the priest on duty must not move more than one
disc at a time and that he must place this disc on a needle so
that there is no smaller disc below it. When the sixty-four
discs shall have been thus transferred from the needle on which
at the creation God placed them to one of the other needles,
tower, temple, and Brahmins alike will crumble into dust, and
with a thunderclap the world will vanish.'' (W W R Ball,
MATHEMATICAL RECREATIONS AND ESSAYS, p. 304)
This has been discussed by several authors, e.g.
Er, Info Sci 42 (1987) 137-141.
Graham, Knuth and Patashnik, _Concrete_Mathematics_.
There are many papers claiming to solve this, and they are probably
all correct but they rely on the unproven "Frame's conjecture".
In particular, for the 4 peg case the conjecture states that an optimal
solution begins by forming a substack of the k smallest discs, then moving
the rest, and then moving those k again; k to be determined.
Here is a extensible bc program that does the same work. The output
format is not that great. We get 300 numbers as output. The first
hundred represent N, the next 100 represent f(N) and the last hundred
represent i, which is the number of discs to move to tmp1 using f(N).
For convenience, I have here some values for N <= 48. Enjoy.
Sharma
N 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
f(N) 1 3 5 9 13 17 25 33 41 49 65 81 97 113 129 161 193 225 257
i 0 1 1 2 2 3 3 4 5 6 6 7 8 9 10 10 11 12 13
N 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
f(N) 289 321 385 449 513 577 641 705 769 897 1025 1153 1281 1409 1537 1665
i 14 15 15 16 17 18 19 20 21 21 22 23 24 25 26 27
N 36 37 38 39 40 41 42 43 44 45 46 47 48
f(N) 1793 2049 2305 2561 2817 3073 3329 3585 3841 4097 4609 5121 5633
i 28 28 29 30 31 32 33 34 35 36 36 37 38
/* This is the bc program that gives f(N) for 4 peg case */
w = 101; /* This represents the number of disks */
m[0] = 0;
m[1] = 1;
m[2] = 3;
m[3] = 5;
m[4] = 9;
m[5] = 13;
m[6] = 17;
/* f(n) is the function that gives the min # of moves for 4 peg case */
define f(n) {
return (m[n]);
}
/* g(n) is the function that fives the min # of moves for 3 peg case */
define g(n) {
return (2^n - 1);
}
/* x(n) is the Optimization Routine */
define x(n) {
auto j
auto r
auto i
if(n == 1) return (1);
j = f(1) + g(n-1);
for(i = 2; i < n; i++) {
r = f(i) + g(n-i);
if(r < j) { j = r; d = i; }
}
return (j);
}
/* main program */
for(q = 4; q < w; q++) {
t = x(q-1);
m[q] = 2 * t + 1;
d[q] = d;
};
/*This for loop prints the number of discs from 1 <= n <= w*/
for(q = 1; q < w; q++) {
q;
}
/*This for loop prints f(n) for 1 <= n <= w */
for(q = 1; q < w; q++) {
m[q];
}
/*This for loop prints i for 1 <= n <= w
i represents the number of disks to be moved to tmp location using f(n)
N-i-1 will be moved using g(n) */
for(q = 1; q < w; q++) {
d[q];
}
--
sha...@sharma.warren.mentorg.com
There is a solution to the Tower of Hanoi problem which does not require
recursion. On odd-numbered moves, move the smallest sized disk clockwise.
On even-numbered moves, make the single other move which is possible.
==> induction/n-sphere.p <==
With what odds do three random points on an n-sphere form an acute triangle?
==> induction/n-sphere.s <==
Select three points a, b, and c, randomly with respect to the surface of an
n-sphere. These three points determine a fourth, x, which is the intersection
of the sphere with the axis perpendicular to the abc plane. (Choose the pole
nearest the plane.) I could have, just as easily, selected x, a distance d
from x, and three points d units away from x. The distribution of d is not
uniform, but that's ok. For every x and d, the three points abc form an acute
triangle with probability p[n-1]. By induction, p[n] = 1/4.
==> induction/paradox.p <==
Is there a non-trivial property that holds for the first 10,000 positive
integers, then fails?
==> induction/paradox.s <==
Consider the sequences defined by:
s(1) = a; s(2) = b; s(n) = least integer such that s(n)/s(n-1) > s(n-1)/s(n-2).
In other words, s(n) = 1+floor(s(n-1)^2/s(n-2)) for n >= 3. These
sequences are similar in some ways to the classically-studied Pisot
sequences. For example, if a = 1, b = 2, then we get the odd-indexed
Fibonacci numbers.
D. Boyd of UBC, an expert in Pisot sequences, pointed out the following.
If we let a = 8, b = 55 in the definition above, then the resulting
sequence s(n) appears to satisfy the following linear recurrence
of order 4:
s(n) = 6s(n-1) + 7s(n-2) - 5s(n-3) - 6s(n-4)
Indeed, it does satisfy this linear recurrence for the
first 11,056 terms. However, it fails at the 11,057th term!
And s(11057) is a 9270 digit number.
(The reason for this coincidence depends on a remarkable fact
about the absolute values of the roots of the polynomial
x^4 - 6x^3 - 7x^2 + 5x + 6.)
==> induction/party.p <==
You're at a party. Any two (different) people at the party have exactly one
friend in common (the friend is also at the party). Prove that there is at
least one person at the party who is a friend of everyone else. Assume that
the friendship relation is symmetric and not reflexive.
==> induction/party.s <==
Here is an easy solution by induction. Let P be the set of people in the
party, and n the size of P. If n=2 or 3, the result is trivial. Suppose now
that n>3 and that the result is true for n-1.
For any two distinct x,y in P, write x & y to mean that `x is a friend of y',
and x ~& y to mean that `x is not a friend of y'.
Take q in P. The hypothesis on the relation & is still satisfied on P-{q}; by
induction, the result is thus true for P-{q}, and there is some p in P-{q}
such that p & x for any x in P-{p,q}. We have two cases:
a) p & q. Then the result holds for P with p.
b) p ~& q. By hypothesis, there is a unique r in P-{p,q} such that p & r & q.
For any x in P-{p,q}, if x & q, then p & x & q, and so x=r. Thus r is the
unique friend of q. Now for any s in P-{q,r} there exists some x such that s &
x & q, and so x=r. This means that r & s for any s in P-{q,r}, and as r & q,
the results holds in P with r.
The problem can also be solved by applying the spectral theory of graphs
(see for instance Bollobas' excellent book, _Extremal Graph Theory_).
The problem's condition is vacuous if there is only N=1 person at the "party",
impossible if N=2 (If you aren't your own friend, nor I mine, somebody *else*
must be our mutual friend), and trivial if N=3 (everybody must be everyone
else's friend). Henceforth assume N>3.
Let A,B be two friends, and C their mutual friend. Let a be the number
of A's friends other than B and C, and likewise b,c. Each of A's friends
is also friendly with exactly one other of A's friends, and with none of
B and C's other friends (if A1,B1 are friends of A,B resp. and of each other
then A1 and B have more than one mutual friend); likewise for B and C.
Let M=N-(a+b+c+3) be the number of people not friendly with any of A,B,C.
Each of them is friendly with exactly one of A's and one of B's friends;
and each pair of a friend of A and a friend of B must have exactly
one of them as a mutual friend. Thus M=ab; likewise M=ab=ac=bc. Thus
either M and two of a,b,c vanish, or a=b=c=k (say), M=k^2, and N=k^3+3k+3.
In the first case, say b=c=0; necessarily a is even, and A is a friend of
everybody else at the party, each of whom is friendly with exactly one other
person; clearly any such configuration (a graph of k/2+1 triangles with a
common vertex) satisfies the problem's conditions).
It remains to show that the second case is impossible. Since N=k^2+3k+3
does not depend on A,B,C, neither does k, and it quickly follows that the
party's friendship graph is regular with reduced matrix
[ 0 k+2 0 ]
[ 1 1 k ]
[ 0 1 k+1 ]
and eigenvalues k+2 and +-sqrt(k+1) and multiplicities 1,m1,m2 for some
*integers* m1 and m2 such that (m1-m2)*sqrt(k+1)=-(k+2) (because the graph's
matrix has trace zero). Thus sqrt(k+1) divides k+2 and k+1 divides
(k+2)^2=(k+1)(k+3)+1
which is only possible if k=0, Q.E.D.
==> induction/roll.p <==
An ordinary die is thrown until the running total of the throws first
exceeds 12. What is the most likely final total that will be obtained?
==> induction/roll.s <==
Claim: If you throw a die until the running total exceeds n>=5, a final
outcome of n+1 is more likely than any other.
Assume we throw an m for a total n+k>n+1, and assume m-k>=0. Now, it
is just as likely to throw an m as an m-k+1, which means that the sum
n+1 is just as likely as any other. Now consider the series of throws
consisting of n-5 1's followed by a 6 and note that we cannot achieve
more than an n+1 by changing the last die roll. Hence, a total of n+1
is more likely than any other.
==> induction/takeover.p <==
After graduating from college, you have taken an important managing position
in the prestigious financial firm of "Mary and Lee".
You are responsible for all the decisions concerning take-over bids.
Your immediate concern is whether to take over "Financial Data".
There is no doubt that you will be successful if you are the first to
bid and that this will be profitable for the firm and you in the long
run. However, you know that there exist another n financial firms,
similar to "Mary and Lee", that are also considering the possibility.
Although you are likely to be the first one to move, you know that
just after a take-over there is a lot of adjustment that needs to be
done. In fact, for a period of time following any take-over the
successful firm becomes a prime candidate for a take-over which will
cost the job of whoever is responsible for take-overs. Among all
financial firms it is common knowledge that the managers responsible
for take-overs are rational and intelligent. What is your best response?
==> induction/takeover.s <==
Assume the takeover is wise for n. The takeover is then unwise for
n+1, as the other companies now find themselves in the same situation
as you for n. If the decision is unwise for n, by similar reasoning
it is wise to takeover FD for n+1. Now note that for n=1 the takeover
decision is clearly unwise, hence by induction you should takeover
FD iff n is even.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> language/close.antonyms.p <==
What words are similar to their antonyms in other langauges?
==> language/close.antonyms.s <==
Swedish: tryck (push), Norwegian: trekk (pull)
German: kalte (cold), Italian: caldo (hot)
==> language/dutch/dutch.record.p <==
What are some Dutch words with unusual properties?
==> language/dutch/dutch.record.s <==
Spelling
Letter Patterns
Entire Word
longest word levensverzekeringsmaatschappijen projectontwikkelingsmaatschappij (32,2)
longest palindrome meetsysteem (11,1)
longest beginning with a palindrome aha ara (3,2)
longest beginning with b palindrome bob (3,1)
longest beginning with c palindrome ?
longest beginning with d palindrome daad deed dood (4,3)
longest beginning with e palindrome etste (5,1)
longest beginning with f palindrome ?
longest beginning with g palindrome gezeg (5,1)
longest beginning with h palindrome ?
longest beginning with i palindrome ?
longest beginning with j palindrome jij (3,1)
longest beginning with k palindrome kajak kazak (5,2)
longest beginning with l palindrome lepel (5,1)
longest beginning with m palindrome meetsysteem (11,1)
longest beginning with n palindrome nebben neggen nekken nellen neppen nessen netten (6,7)
longest beginning with o palindrome oho (3,1)
longest beginning with p palindrome paap (4,1)
longest beginning with q palindrome ?
longest beginning with r palindrome redder rekker remmer renner (6,4)
longest beginning with s palindrome serres staats stoots (6,3)
longest beginning with t palindrome temet (5,1)
longest beginning with u palindrome ?
longest beginning with v palindrome ?
longest beginning with w palindrome ?
longest beginning with x palindrome ?
longest beginning with y palindrome ?
longest beginning with z palindrome ?
longest with middle a palindrome staats (6,1)
longest with middle b palindrome nebben (6,1)
longest with middle c palindrome ?
longest with middle d palindrome redder (6,1)
longest with middle e palindrome reeer (5,1)
longest with middle f palindrome ?
longest with middle g palindrome neggen (6,1)
longest with middle h palindrome aha oho (3,2)
longest with middle i palindrome gig jij kik lil pip sis (3,6)
longest with middle j palindrome kajak (5,1)
longest with middle k palindrome nekken rekker (6,2)
longest with middle l palindrome nellen (6,1)
longest with middle m palindrome remmer (6,1)
longest with middle n palindrome meeneem (7,1)
longest with middle o palindrome stoots (6,1)
longest with middle p palindrome neppen (6,1)
longest with middle q palindrome ?
longest with middle r palindrome serres (6,1)
longest with middle s palindrome nessen (6,1)
longest with middle t palindrome netten (6,1)
longest with middle u palindrome lul mum sus (3,3)
longest with middle v palindrome neven (5,1)
longest with middle w palindrome ?
longest with middle x palindrome ?
longest with middle y palindrome meetsysteem (11,1)
longest with middle z palindrome gezeg kazak (5,2)
longest tautonym blauwblauw (10,1)
longest beginning with a tautonym ?
longest beginning with b tautonym blauwblauw (10,1)
longest beginning with c tautonym ?
longest beginning with d tautonym dumdum (6,1)
longest beginning with e tautonym enen (4,1)
longest beginning with f tautonym ?
longest beginning with g tautonym genegene (8,1)
longest beginning with h tautonym haha (4,1)
longest beginning with i tautonym inging (6,1)
longest beginning with j tautonym jaja jojo (4,2)
longest beginning with k tautonym kerker (6,1)
longest beginning with l tautonym lala (4,1)
longest beginning with m tautonym mama (4,1)
longest beginning with n tautonym ?
longest beginning with o tautonym ?
longest beginning with p tautonym papa (4,1)
longest beginning with q tautonym ?
longest beginning with r tautonym ?
longest beginning with s tautonym ?
longest beginning with t tautonym taaitaai (8,1)
longest beginning with u tautonym ?
longest beginning with v tautonym verver (6,1)
longest beginning with w tautonym ?
longest beginning with x tautonym ?
longest beginning with y tautonym ?
longest beginning with z tautonym zozo (4,1)
longest head 'n' tail kentekent kindskind (9,2)
longest with middle a head 'n' tail kak lal mam pap sas (3,5)
longest with middle b head 'n' tail uitbuit (7,1)
longest with middle c head 'n' tail ?
longest with middle d head 'n' tail enden (5,1)
longest with middle e head 'n' tail kentekent (9,1)
longest with middle f head 'n' tail ?
longest with middle g head 'n' tail aangaan (7,1)
longest with middle h head 'n' tail echec oehoe (5,2)
longest with middle i head 'n' tail gig jij kik lil pip sis (3,6)
longest with middle j head 'n' tail ?
longest with middle k head 'n' tail enken erker omkom sekse (5,4)
longest with middle l head 'n' tail inklink (7,1)
longest with middle m head 'n' tail aanmaan (7,1)
longest with middle n head 'n' tail ene (3,1)
longest with middle o head 'n' tail verover (7,1)
longest with middle p head 'n' tail ?
longest with middle q head 'n' tail ?
longest with middle r head 'n' tail derde (5,1)
longest with middle s head 'n' tail kindskind (9,1)
longest with middle t head 'n' tail enten (5,1)
longest with middle u head 'n' tail teute (5,1)
longest with middle v head 'n' tail ?
longest with middle w head 'n' tail inwin (5,1)
longest with middle x head 'n' tail ?
longest with middle y head 'n' tail ?
longest with middle z head 'n' tail ?
Subset of Word
longest internal palindrome apothekersrekening apothekersrekeningen kilometervreter kilometervreters marktkramer marktkramers meetsysteem trekkebekken voortstroomde (9,9)
longest internal tautonym noordnoordoost noordnoordwest onvergelijkelijker onvergelijkelijkst tingelingeling (10,5)
longest repeated prefix noordnoordoost noordnoordwest (10,2)
most consecutive doubled letters bevoorraadde bevoorraadden (4,2)
most doubled letters arrondissementscommissarissen (5,1)
longest two cadence beredeneren degenereren eretekenen gedegenereerd gedegenereerde gedegenereerden geregenereerd geregenereerde geregenereerden regenereren ... (5,11)
longest three cadence adembeklemmende adembeklemmender behoeftenbevrediging desbetreffende ettergezwellen inconvenienten keukengedeelte overeengestemde overeengestemden zielverheffende ... (5,11)
longest four cadence afgetelefoneerde afgetelefoneerden beeldendienaren beschermbeelden beschermgeesten beslissingswedstrijd beslissingswedstrijden diepzeeonderzoek eindexamenfeest expresgoederen ... (4,49)
longest five cadence academievergaderingen boedelvereffeningen boekenverzamelingen feestvergaderingen geheugeninterferentie gevangenispersoneel gouvernementsgebouwen identiteitsbewijzen inschikkelijkheid jeugdherinneringen ... (4,18)
Letter Counts
Lipograms
longest letters from first half geleidelijkheid gemakkelijkheid (15,2)
longest letters from second half voortsproot (11,1)
longest without ab schilderijententoonstellingen (29,1)
longest without abcd tentoonstellingsterreinen (25,1)
longest without a to h rijksinstituut stroomlijnvorm (14,2)
longest without a to k vooroploopt voortsproot voortstormt (11,3)
longest without a to n voortsproot (11,1)
longest without a to q rustuur (7,1)
longest without a to s uw (2,1)
longest without e brandwaarborgmaatschappij (25,1)
longest without et landbouwhuishoudschool (22,1)
longest without eta bijscholingscursus (18,1)
longest without etai schooljuffrouw (14,1)
longest without etain schooljuffrouw (14,1)
longest without etains chloorzuur chloroform droogkoord omhoogklom polychroom (10,5)
Letter Choices
Vowels
longest all vowels aai eau oei ooi (3,4)
longest each vowel once verbrandingsprodukt (19,1)
longest each extended vowel once loyauteit (9,1)
shortest each vowel once douarie (7,1)
shortest each extended vowel once loyauteit (9,1)
shortest vowels in order lateihout (9,1)
shortest extended vowels in order ?
longest vowels in order arbeidsonrust (13,1)
longest extended vowels in order ?
shortest vowels in reverse order ?
shortest extended vowels in reverse order ?
longest vowels in reverse order ?
longest extended vowels in reverse order ?
longest one vowel schurftst (9,1)
longest two vowels strijdschrift vlijmscherpst (13,2)
longest containing a univocalic afschraapschaar maarschalksstaf paltsgraafschap straatlantaarns (15,4)
longest containing e univocalic wegenverkeersreglement (22,1)
longest containing i univocalic inschrijvingslijst (18,1)
longest containing o univocalic noordnoordoost (14,1)
longest containing u univocalic turfschuur (10,1)
longest containing y univocalic lyncht (6,1)
longest alternating extended vowel-consonant veramerikaniseren (17,1)
longest alternating vowel-consonant veramerikaniseren (17,1)
Consonants
longest consonant string aandachtsstreep aandachtsstrepen ambachtsscholen ambachtsschool dichtschroef dichtschroefde dichtschroefden dichtschroeft dichtschroeven gezichtsscherm ... (7,18)
longest one consonant aaneennaaien (12,1)
longest two consonant aaneennaaiden (13,1)
Isograms
longest isogram dampkringslucht (15,1)
longest pair isogram steunstukken (12,1)
longest trio isogram ?
longest tetrad isogram ?
longest polygram salarisregelingen (17,1)
longest pyramid aaneengehangene (15,1)
most repeated letters projectontwikkelingsmaatschappij (11,1)
highest containing a repeated aanvaardbaar aanvaardbaarder aanvaardbaarheid aanvaardbaarst aanvaarpaal afbraakmateriaal bastaardnachtegaal brandwaarborgmaatschappij brandwaarborgmaatschappijen onaanvaardbaar ... (6,11)
highest containing b repeated hobbeldebobbel (5,1)
highest containing c repeated accijnsrechten architectonisch architectonische bacchantisch bacchantische bacchantischer bacchantischt beschrijftechnisch beschrijftechnische calciumcarbid ... (3,73)
highest containing d repeated driehonderdduizend driehonderdduizendste (5,2)
highest containing e repeated weledelzeergeleerde (9,1)
highest containing f repeated buffetjuffrouw buffetjuffrouwen diffusiecoefficient diffusiecoefficienten (4,4)
highest containing g repeated beslaglegging beslagleggingen buigingsuitgang buigingsuitgangen drooglegging geaggregeerd geaggregeerde geaggregeerden gegaggeld gegaggelde ... (4,34)
highest containing h repeated huishoudhandschoenen (4,1)
highest containing i repeated liquiditeitssituatie (6,1)
highest containing j repeated jojootje jojootjes jokkernijtje jokkernijtjes kijverijtje kijverijtjes pijjekkertje pijjekkertjes vrijerijtje vrijerijtjes ... (3,13)
highest containing k repeated koekoeksklokken (6,1)
highest containing l repeated beleidsdoelstelling bloembollencultuur bloembollenveld bloembollenvelden knolleschillen krokodillevellen krullebollen leerlingstelsel leliebollen luchtdoelartillerie ... (4,37)
highest containing m repeated maximumthermometer maximumthermometers minimumprogramma minimumthermometer minimumthermometers (5,5)
highest containing n repeated koninginnenkronen (7,1)
highest containing o repeated noordnoordoost oostnoordoost (6,2)
highest containing p repeated aardappelmeelpap appelflappen begrippenapparaat klapperdoppen moppentapper moppentappers oppepper opperheerschappij opproppen papiersnipper ... (4,18)
highest containing q repeated quinquagesima quiproquo (2,2)
highest containing r repeated karrenverhuurder karrenverhuurders meerderjarigheidsverklaring meerderjarigheidsverklaringen meerderjarigverklaring meerderjarigverklaringen prikkeldraadversperring prikkeldraadversperringen programmeertaalvertaler programmeertaalvertalers (5,10)
highest containing s repeated arrondissementscommissarissen (7,1)
highest containing t repeated activiteitstraktement activiteitstraktementen (6,2)
highest containing u repeated cultuurprodukt cultuurprodukten landbouwhuishoudkunde musculatuur rubbercultuur structuurformule structuurformules suikercultuur (4,8)
highest containing v repeated overlevingsvermogen vastenavondvreugde veevervoer verveelvoudig verveelvoudigd verveelvoudigde verveelvoudigden verveelvoudigen verveelvoudigt verversingshaven ... (3,35)
highest containing w repeated gewiewauwd nieuwbouwwijk warmwaterverwarming weduwvrouw weduwvrouwen wiewauw wiewauwde wiewauwden wiewauwen wiewauwt ... (3,13)
highest containing x repeated explosievenexpert (2,1)
highest containing y repeated cyclostyle cyclostyleer cyclostyleerde cyclostyleerden cyclostyleert cyclostyleren cyclostyles gecyclostyleerd gecyclostyleerde gecyclostyleerden ... (2,25)
highest containing z repeated jazzmuziek zigzagsgewijze (3,2)
most different letters brandverzekeringsmaatschappij brandverzekeringsmaatschappijen (18,2)
highest ratio length/letters aaneengehangenen (320,1)
highest ratio length/letters (no tautonyms) aaneengehangenen (320,1)
lowest length 16 ratio length/letters computerafdeling elektrodynamisch honkbalwedstrijd (106,3)
lowest length 17 ratio length/letters achtergrondmuziek ambachtsonderwijs bevruchtingsproef bloedverwantschap elektrodynamische huwelijkscontract veraanschouwelijk zelfoverschatting (113,8)
lowest length 18 ratio length/letters veraanschouwelijkt (112,1)
lowest length 19 ratio length/letters achtergrondmuziekje produktomschrijving veraanschouwelijkte (118,3)
lowest length 20 ratio length/letters bewustzijnsverhogend rijkslandbouwscholen veraanschouwelijking veraanschouwelijkten (125,4)
lowest length 21 ratio length/letters scheepsmontagebedrijf (123,1)
lowest length 22 ratio length/letters bondgenootschappelijke vliegtuigmoederschepen (137,2)
lowest length 23 ratio length/letters onderwijzersgenootschap veiligheidsfunctionaris (143,2)
lowest length 24 ratio length/letters handschriftenverzameling taalverwerkingsproblemen verzekeringsmaatschappij waterleidingmaatschappij (150,4)
lowest length 25 ratio length/letters rijkslandbouwproefstation (147,1)
Letter Appearance
longest short letters zwemmerseczeem (14,1)
longest tall letters bijblijft (9,1)
longest vertical-symmetry letters automaat haamhout taaitaai uithouwt uitwaait vaamhout vaathout (8,7)
longest horizontal-symmetry letters behoedde dooddeed kiekeboe kookboek (8,4)
longest full-symmetry letters hooi (4,1)
highest ratio of dotted letters lijzij (66,1)
Typewriter
longest top row pirouetteert portretteert (12,2)
longest middle row halfslag kaakslag kalklaag (8,3)
longest bottom row ?
longest in order eerroof tuipaal weeraal weeroog weertij (7,5)
longest in reverse order blaatte blootte kapotte naaapte (7,4)
longest left hand verbeteraarsters (16,1)
longest right hand minimumloon (11,1)
longest alternating hands roemruchtigheid wispelturigheid (15,2)
longest one finger deed (4,1)
longest adjacent keys redresseerde (12,1)
Puzzle
longest palindrome in Morse code weerkreeg (9,1)
longest formed with chemical symbols binnenhuisarchitecten (21,1)
longest formed with US postal codes innemend landpaal landwind moordend paasmaal waarmede (8,6)
longest formed with amino acid abbreviations metser valser (6,2)
longest formed with piano notes bedaagde begaafde gedaagde (8,3)
Letter Order
Alphabetical
longest letters in order afklopt afkorst beklopt beknopt beknort bekorst bemorst (7,7)
longest letters in order with repeats afknoopt bijkoopt bijloopt (8,3)
longest letters in reverse order wolkige (7,1)
longest letters in reverse order with repeats vuurrood wrokkige (8,2)
longest roller-coaster veramerikanisering (18,1)
longest no letters in place levensverzekeringsmaatschappijen (32,1)
most letters in place aandoenlijk aandoenlijke aandoenlijker aandoenlijkheid aandoenlijkst aaneenspijker aaneenspijkerde aaneenspijkerden aaneenspijkeren aaneenspijkert ... (5,46)
most letters in place shifted beweeglijk beweeglijke beweeglijker beweeglijkheid beweeglijkst boodschappenlopers franciscanenkloosters kijkvensters koperstuk nobelprijswinnaars ... (6,20)
most consecutive letters in order consecutively achterstuk achterstukken altaarstuk altaarstukken bijklank bijklanken dijklasten dijkleger doorsturen doorstuur ... (4,106)
most consecutive letters in order alarmknop ambachtsgilde ambachtsgilden ambachtsonderwijs bedrijfsklimaat bondgenootschappelijk bondgenootschappelijke deftigheid droefgeestigheid eerstehulpverlening ... (5,56)
most consecutive letters financieringsmogelijkheden (13,1)
highest ratio of consecutive letters to length koolmijn (87,1)
==> language/english/equations.p <==
1 = E. on a C.
1 = G. L. for M.
1 = K. K. on the E. S. B.
1 = S. C. in D. P.
1 = S. S.
1 = S. S. for a M.
1 = T. that O. K.
1 = W. on a U.
1 = if by L.
1 = the L. N.
2 = B. G.
2 = C. B. as B. as O.
2 = C. in the T. by C. D.
2 = E. of E.
2 = H. in a W.
2 = H. of C.
2 = L. in the H. B.
2 = N. in a D.
2 = O. in a D. P.
2 = P. J. P.
2 = P. in a B.
2 = P. in a P.
2 = Q. in a H.-D.
2 = S. of a L.'s T.
2 = T. in the W. T. C.
2 = T. the P. A. R.
2 = W. S. in "C."
2 = W. that D.'t M. a R.
2 = if by S.
3 = B. M. (S. H. T. R.!)
3 = B. in "G."
3 = B. in "L. of the R."
3 = B. in the H. E.
3 = B. of the U. S. G.
3 = D. for J. in the B. of the W.
3 = D. of the C.
3 = F. in a Y.
3 = L. P.
3 = L. on a T.
3 = M. and a B.
3 = O. K. C.'s F.
3 = P. in a B. S.
3 = P. in a H. G.
3 = P. of an I.
3 = R. in a C.
3 = S. and Y.'re O.
3 = S. on C.' F. V.
3 = S. in a T.
3 = S. to the W.
3 = W. M. at B.
3 = W. in "M."
3 = W. on a T.
3 = on a M.
4 = B. in a W.
4 = C. in the H. H.
4 = G. M. in C.
4 = H. of the A.
4 = I's in M.
4 = L. in a D. W.
4 = M. in a S.
4 = M. of the B.
4 = Q. in a F. G.
4 = Q. in a G.
4 = S's in M.
4 = S. in a D. of C.
4 = S. in a R.
4 = S. in a S.
4 = S. in a Y.
4 = T. FDR E. P.
4 = W. on a C.
4 = Y. in an O.
5 = B. in the T.
5 = D. in a Z. C.
5 = F. on a H.
5 = G. L.
5 = M. B. in "J. and the B."
5 = M. in a C. R.
5 = P. in a N.
5 = S. in a P.
5 = S. in the S. C.
5 = T. on a F.
5 = V. O. in the H. B.
6 = D. of the C.
6 = "B. C."
6 = A. of the A.
6 = D. of C.
6 = H. D. of O. for R. C.
6 = P. in a P.
6 = P. in a P. T.
6 = S. in a H.
6 = T. Z. in the U. S.
6 = W. of H. VIII
6 = of O. and a H. D. of the O.
7 = "P. of W."
7 = A. of M.
7 = C. in a R.
7 = D. in "S. W."
7 = H. of R.
7 = K. of F. in H. P.
7 = P. of E.
7 = S. of S. the S.
7 = W. of the A. W.
8 = C. in a M.
8 = D. in H.
8 = K. H. of E.
8 = L. on a S.
8 = L. on an O.
8 = S. in a O.
8 = S. on a S. S.
8 = S.'s R.
8 = G. T. in a L. B. C.
9 = C. of H. in the "D. C."
9 = D. in a Z. C., with the S. C.
9 = G. M.
9 = I. in a B. G.
9 = J. on the U. S. S. C.
9 = L. of a C.
9 = M. in a H. B.'s G. P.
9 = P. in the S. S.
9 = S. by B.
10 = A. in the B. of R.
10 = C. G. to M.
10 = F. in B.
10 = L. I. B.
10 = P. in C.
10 = Y. in a D.
11 = F. in a C.
11 = P. on a C. T.
11 = P. on a F. T.
11 = S. in the C.
12 = A. at the L. S.
12 = C. of A. R.
12 = D. of C.
12 = D. of J.
12 = L. of H.
12 = S. of the Z.
12 = T. of I.
13 = a B.'s D.
13 = O. C.
13 = S. on the A. F.
14 = D. in a F.
14 = L. in a S.
14 = P. in W. W.'s P.
15 = M. of F. per P., A. to A. W.
15 = M. on a D. M.'s C.
16 = O. in a P.
16 = O. of M.
17 = S. in a H.
17 = P. that D. N. and S. V.
18 = H. on a G. C.
18 = I. in a C.
19 = G. S. for the V. P.
20 = C. in a P.
20 = Y. S. by R. V. W.
21 = G. S. for the P.
22 = C.
23 = P. on a G. J.
24 = B. B. in a P.
24 = B. B. to a C.
24 = D. P. to the I. for M. I.
24 = H. in a D.
24 = S. Z. of I. T.
25 = Y. of M. for a S. A.
25 = P. in a Q.
26 = L. of the A.
27 = D. W. W. by G. G.
27 = A. to the U. S. C.
28 = D. in F.
29 = D. in F. in a L. Y.
30 = D. H. S.
30 = D. in J.
30 = P. of S. G. to J.
31 = D. in O.
32 = D. F. at which W. F.
36 = I. on a Y. S.
36 = R. M.
39 = L. in M.' L.
40 = D. and N. of the G. F.
40 = D. in L.
40 = Y. in the D. for I.
43 = B. in E. C. of N.
45 = D. in a O.
46 = C. in a H. C.
47 = P. P. in E.'s E.
48 = C. in a P. D.
49 = R. A. in a R.
50 = S. in the U.
50 = W. to L. Y. L.
52 = W. in a Y.
52 = C. in a D.
54 = C. in a D. (with the J.)
57 = H. V.
61 = H. R. H. by R. M. in O. S.
64 = S. on a C.
64 = Y. O. W. I H. Y. W. S. N. M.
66 = B. in the K. J. B.
70 = D. of C.
75 = Y. of M. for the D. A.
76 = T. L. the B. P.
78 = C. in a T. D.
80 = C. in a M.
80 = D. to G. A. the W.
87 = F. S. and S. Y.
88 = C. in the S.
88 = P. K.
90 = D. in a R. A.
95 = M. L.'s T.
96 = T., by ?
99 = B. of B. on a W.
100 = C. in a D.
100 = Y. in a F. F.
101 = D.
101 = a S. M. L.
102 = F. in the E. S. B.
116 = Y. in the H. Y.'s W.
144 = U. in a G.
150 = P. in the B.
180 = D. to R. D.
200 = D. for P. G. in M.
200 = M. in a C.
206 = B. in the H. B.
235 = I. of U. in the A. B.
300 = B. in the T. of C.
365 = D. in a Y.
366 = D. in a L. Y.
432 = P. in a H.
435 = S. in the H. of R.
451 = D. F. at which B. B.
500 = F. C.
500 = M. in the I. F. H.
500 = S. in a R.
666 = N. of the B. in the B. of R.
707 = Y. of C.
755 = H. runs H. by H. A.
800 = W. in V.
880 = N. of M. S. of O. F.
1000 = I. in N. Y.
1000 = P. of L.
1000 = S. L. by the F. of H. of T.
1000 = W. that a P. is W.
1001 = A. N.
1440 = M. in a D.
1760 = Y. in a M.
5280 = F. in a M.
20000 = L. U. the S.
==> language/english/equations.s <==
This puzzle originally consisted of 24 "equations" by Will Shortz, printed
in the May-June 1981 issue of Games magazine, with an acknowledgement to
Morgan Worthy. Games ran several followups in subsequent issues, and
reported that people kept resubmitting the puzzle to them, sometimes as
original work! Many people have now added to the list of equations.
The 24 original ones are starred (*) below.
1 = E. on a C. (Eye on a Cyclops)
1 = G. L. for M. (Giant Leap for Mankind)
1 = K. K. on the E. S. B. (King Kong on the Empire State Building)
1 = S. C. in D. P. (Single Calorie in Diet Pepsi)
1 = S. S. (Singular Sensation)
1 = S. S. for a M. (Small Step for a Man)
1 = T. that O. K. (Time that Opportunity Knocks)
1 = W. on a U. (Wheel on a Unicycle *)
1 = if by L. (if by Land)
1 = the L. N. (the Loneliest Number)
2 = B. G. (Brothers Grimm)
2 = C. B. as B. as O. (Can Be as Bad as One)
2 = C. in the T. by C. D. (Cities in the Tale by Charles Dickens)
2 = E. of E. (Elizabeths of England)
2 = H. in a W. (Halves in a Whole)
2 = H. of C. (Houses of Congress)
2 = L. in the H. B. (Lungs in the Human Body)
2 = N. in a D. (Nickels in a Dime)
2 = O. in a D. P. (Outs in a Double Play)
2 = P. J. P. (Pope John Pauls)
2 = P. in a B. (Pieces in a Bikini)
2 = P. in a P. (Peas in a Pod)
2 = Q. in a H.-D. (Quarters in a Half-Dollar)
2 = S. of a L.'s T. (Shakes of a Lamb's Tail)
2 = T. in the W. T. C. (Towers in the World Trade Center)
2 = T. the P. A. R. (Times the Postman Always Rings)
2 = W. S. in "C." (Wicked Stepsisters in "Cinderella")
2 = W. that D.'t M. a R. (Wrongs that Don't Make a Right)
2 = if by S. (if by Sea)
3 = B. M. (S. H. T. R.!) (Blind Mice, See How They Run! *)
3 = B. in "G." (Bears in "Goldilocks")
3 = B. in "L. of the R." (Books in "Lord of the Rings")
3 = B. in the H. E. (Bones in the Human Ear)
3 = B. of the U. S. G. (Branches of the United States Government)
3 = D. for J. in the B. of the W. (Days for Jonah in the Belly of the Whale)
3 = D. of the C. (Days of the Condor -- movie)
3 = F. in a Y. (Feet in a Yard)
3 = L. P. (Little Pigs)
3 = L. on a T. (Legs on a Tripod)
3 = M. and a B. (Men and a Baby)
3 = O. K. C.'s F. (Old King Cole's Fiddlers)
3 = P. in a B. S. (Pieces in a Business Suit)
3 = P. in a H. G. (Periods in a Hockey Game)
3 = P. of an I. (Parts of an Insect)
3 = R. in a C. (Rings in a Circus)
3 = S. and Y.'re O. (Strikes and You're Out)
3 = S. on C.' F. V. (Ships on Columbus' First Voyage)
3 = S. in a T. (Sides in a Triangle)
3 = S. to the W. (Sheets to the Wind)
3 = W. M. at B. (Wise Men at Bethlehem)
3 = W. in "M." (Witches in "Macbeth")
3 = W. on a T. (Wheels on a Tricycle)
3 = on a M. (on a March)
4 = B. in a W. (Balls in a Walk)
4 = C. in the H. H. (Chambers in the Human Heart)
4 = G. M. in C. (Gang Members in China)
4 = H. of the A. (Horsemen of the Apocalypse)
4 = I's in M. (I's in Mississippi)
4 = L. in a D. W. (Letters in a Dirty Word)
4 = M. in a S. (Movements in a Symphony)
4 = M. of the B. (Members of the Beatles)
4 = Q. in a F. G. (Quarters in a Football Game)
4 = Q. in a G. (Quarts in a Gallon *)
4 = S's in M. (S's in Mississippi)
4 = S. in a D. of C. (Suits in a Deck of Cards)
4 = S. in a R. (Sides in a Rectangle)
4 = S. in a S. (Sides in a Square)
4 = S. in a Y. (Seasons in a Year)
4 = T. FDR E. P. (Terms FDR Elected President)
4 = W. on a C. (Wheels on a Car)
4 = Y. in an O. (Years in an Olympiad)
5 = B. in the T. (Books in the Torah)
5 = D. in a Z. C. (Digits in a Zip Code *)
5 = F. on a H. (Fingers on a Hand)
5 = G. L. (Great Lakes)
5 = M. B. in "J. and the B." (Magic Beans in "Jack and the Beanstalk")
5 = M. in a C. R. (Miles in a Camptown Racetrack)
5 = P. in a N. (Pennies in a Nickel)
5 = S. in a P. (Sides in a Pentagon)
5 = S. in the S. C. (Stars in the Southern Cross)
5 = T. on a F. (Toes on a Foot)
5 = V. O. in the H. B. (Vital Organs in the Human Body)
6 = D. of the C. (Days of the Condor -- book)
6 = "B. C." ("Brandenburg Concertos")
6 = A. of the A. (Avenue of the Americas)
6 = D. of C. (Days of Creation)
6 = H. D. of O. for R. C. (Holy Days of Obligation for Roman Catholics)
6 = P. in a P. (Pigs in a Poke)
6 = P. in a P. T. (Pockets in a Pool Table)
6 = S. in a H. (Sides in a Hexagon)
6 = T. Z. in the U. S. (Time Zones in the United States)
6 = W. of H. VIII (Wives of Henry VIII)
6 = of O. and a H. D. of the O. (of One and a Half Dozen of the Other)
7 = "P. of W." ("Pillars of Wisdom")
7 = A. of M. (Ages of Man)
7 = C. in a R. (Colors in a Rainbow : ROYGBIV)
7 = D. in "S. W." (Dwarfs in "Snow White")
7 = H. of R. (Hills of Rome)
7 = K. of F. in H. P. (Kinds of Fruit in Hawaiian Punch)
7 = P. of E. (Plagues of Egypt)
7 = S. of S. the S. (Seas of Sinbad the Sailor)
7 = W. of the A. W. (Wonders of the Ancient World *)
8 = C. in a M. (Candles in a Menorah)
8 = D. in H. (Days in Hanukkah)
8 = K. H. of E. (King Henrys of England)
8 = L. on a S. (Legs on a Spider)
8 = L. on an O. (Legs on an Octopus)
8 = S. in a O. (Sides in a Octagon)
8 = S. on a S. S. (Sides on a Stop Sign *)
8 = S.'s R. (Santa's Reindeer)
8 = G. T. in a L. B. C. (Great Tomatoes in a Little Bitty Can)
9 = C. of H. in the "D. C." (Circle of Hell in the "Divine Comedy")
9 = D. in a Z. C., with the S. C. (Digits in a Zip Code, with the Street Code)
9 = G. M. (Greek Muses)
9 = I. in a B. G. (Innings in a Baseball Game)
9 = J. on the U. S. S. C. (Justices on the United States Supreme Court)
9 = L. of a C. (Lives of a Cat)
9 = M. in a H. B.'s G. P. (Months in a Human Baby's Gestation Period)
9 = P. in the S. S. (Planets in the Solar System *)
9 = S. by B. (Symphonies by Beethoven)
10 = A. in the B. of R. (Amendments in the Bill of Rights)
10 = C. G. to M. (Commandments Given to Moses)
10 = F. in B. (Frames in Bowling)
10 = L. I. B. (Little Indian Boys)
10 = P. in C. (Provinces in Canada)
10 = Y. in a D. (Years in a Decade)
11 = F. in a C. (Fathoms in a Chain)
11 = P. on a C. T. (Players on a Cricket Team)
11 = P. on a F. T. (Players on a Football Team *)
11 = S. in the C. (States in the Confederacy)
12 = A. at the L. S. (Apostles at the Last Supper)
12 = C. of A. R. (Caesars of Ancient Rome)
12 = D. of C. (Days of Christmas)
12 = D. of J. (Disciples of Jesus)
12 = L. of H. (Labors of Hercules)
12 = S. of the Z. (Signs of the Zodiac *)
12 = T. of I. (Tribes of Israel)
13 = a B.'s D. (a Baker's Dozen)
13 = O. C. (Original Colonies)
13 = S. on the A. F. (Stripes on the American flag *)
14 = D. in a F. (Days in a Fortnight)
14 = L. in a S. (Lines in a Sonnet)
14 = P. in W. W.'s P. (Points in Woodrow Wilson's Plan)
15 = M. of F. per P., A. to A. W. (Minutes of Fame per Person, According to Andy Warhol)
15 = M. on a D. M.'s C. (Men on a Dead Man's Chest)
16 = O. in a P. (Ounces in a Pound)
16 = O. of M. (Orders of Mammals)
17 = S. in a H. (Syllables in a Haiku)
17 = P. that D. N. and S. V. (Parallel that Divided North and South Vietnam)
18 = H. on a G. C. (Holes on a Golf Course *)
18 = I. in a C. (Inches in a Cubit)
19 = G. S. for the V. P. (Gun Salute for the Vice President)
20 = C. in a P. (Cigarettes in a Pack)
20 = Y. S. by R. V. W. (Years Slept by Rip Van Winkle)
21 = G. S. for the P. (Gun Salute for the President)
22 = C. (Catch)
23 = P. on a G. J. (People on a Grand Jury)
24 = B. B. in a P. (Black Birds Baked in a Pie)
24 = B. B. to a C. (Beer Bottles to a Case)
24 = D. P. to the I. for M. I. (Dollars Paid to the Indians for Manhattan Island)
24 = H. in a D. (Hours in a Day *)
24 = S. Z. of I. T. (Standard Zones of International Time)
25 = Y. of M. for a S. A. (Years of Marriage for a Silver Anniversary)
25 = P. in a Q. (Pennies in a Quarter)
26 = L. of the A. (Letters of the Alphabet *)
27 = D. W. W. by G. G. (Different Wigs Worn by Gregory Giggs)
27 = A. to the U. S. C. (Amendments to the United States Constitution)
28 = D. in F. (Days in February)
29 = D. in F. in a L. Y. (Days in February in a Leap Year *)
30 = D. H. S. (Days Hath September)
30 = D. in J. (Days in June)
30 = P. of S. G. to J. (Pieces of Silver Given to Judas)
31 = D. in O. (Days in October)
32 = D. F. at which W. F. (Degrees Fahrenheit at which Water Freezes *)
36 = I. on a Y. S. (Inches on a Yard Stick)
36 = R. M. (Righteous Men)
39 = L. in M.' L. (Lashes in Moses' Law)
40 = D. and N. of the G. F. (Days and Nights of the Great Flood *)
40 = D. in L. (Days in Lent)
40 = Y. in the D. for I. (Years in the Desert for Israelites)
43 = B. in E. C. of N. (Beans in Each Cup of Nescafe)
45 = D. in a O. (Degrees in an Octant)
46 = C. in a H. C. (Chromosomes in a Human Cell)
47 = P. P. in E.'s E. (Pythagorean Proposition in Euclid's Elements)
48 = C. in a P. D. (Cards in a Pinochle Deck)
49 = R. A. in a R. (Reasons All in a Row)
50 = S. in the U. (States in the Union)
50 = W. to L. Y. L. (Ways to Leave Your Lover)
52 = W. in a Y. (Weeks in a Year)
52 = C. in a D. (Cards in a Deck)
54 = C. in a D. (with the J.) (Cards in a Deck with the Jokers *)
57 = H. V. (Heinz Varieties *)
61 = H. R. H. by R. M. in O. S. (Home Runs Hit by Roger Maris in One Season)
64 = S. on a C. (Squares on a Chessboard *)
64 = Y. O. W. I H. Y. W. S. N. M. (Years Old When I Hope You Will Still Need Me)
66 = B. in the K. J. B. (Books in the King James Bible)
70 = D. of C. (Disciples of Christ)
75 = Y. of M. for the D. A. (Years of Marriage for the Diamond Anniversary)
76 = T. L. the B. P. (Trombones Led the Big Parade)
78 = C. in a T. D. (Cards in a Tarot Deck)
80 = C. in a M. (Chains in a Mile)
80 = D. to G. A. the W. (Days to Go Around the World)
87 = F. S. and S. Y. (Four Score and Seven Years)
88 = C. in the S. (Constellations in the Sky)
88 = P. K. (Piano Keys *)
90 = D. in a R. A. (Degrees in a Right Angle *)
95 = M. L.'s T. (Martin Luther's Theses)
96 = T., by ? (Tears, by ?)
99 = B. of B. on a W. (Bottles of Beer on a Wall)
100 = C. in a D. (Cents in a Dollar)
100 = Y. in a F. F. (Yards in a Football Field)
101 = D. (Dalmatians)
101 = a S. M. L. (a Silly Millimeter Longer)
102 = F. in the E. S. B. (Floors in the Empire State Building)
116 = Y. in the H. Y.'s W. (Years in the Hundred Year's War)
144 = U. in a G. (Units in a Gross)
150 = P. in the B. (Psalms in the Bible)
180 = D. to R. D. (Degrees to Reverse Direction)
200 = D. for P. G. in M. (Dollars for Passing Go in Monopoly *)
200 = M. in a C. (Milligrams in a Carat)
206 = B. in the H. B. (Bones in the Human Body)
235 = I. of U. in the A. B. (Isotope of Uranium in the Atomic Bomb)
300 = B. in the T. of C. (Bees in the Tomb of Childeric)
365 = D. in a Y. (Days in a Year)
366 = D. in a L. Y. (Days in a Leap Year)
432 = P. in a H. (Pints in a Hogshead)
435 = S. in the H. of R. (Seats in the House of Representatives)
451 = D. F. at which B. B. (Degrees Fahrenheit at which Books Burn)
500 = F. C. (Fortune Companies)
500 = M. in the I. F. H. (Miles in the Indianapolis Five Hundred)
500 = S. in a R. (Sheets in a Ream)
666 = N. of the B. in the B. of R. (Number of the Beast in the Book of Revelation)
707 = Y. of C. (Year of Confusion)
755 = H. runs H. by H. A. (Home runs Hit by Hank Aaron)
800 = W. in V. (Warriors in Valhalla)
880 = N. of M. S. of O. F. (Number of Magic Squares of Order Four)
1000 = I. in N. Y. (Islands in New York)
1000 = P. of L. (Points of Light)
1000 = S. L. by the F. of H. of T. (Ships Launched by the Face of Helen of Troy)
1000 = W. that a P. is W. (Words that a Picture is Worth *)
1001 = A. N. (Arabian Nights *)
1440 = M. in a D. (Minutes in a Day)
1760 = Y. in a M. (Yards in a Mile)
5280 = F. in a M. (Feet in a Mile)
20000 = L. U. the S. (Leagues Under the Sea)
==> language/english/etymology/acronym.p <==
What acronyms have become common words or are otherwise interesting?
==> language/english/etymology/acronym.s <==
The following is the list of acronyms which have become common nouns.
An acronym is "a word formed from the initial letter or letters of each
of the successive parts or major parts of a compound term" (Webster's Tenth).
A common noun will occur uncapitalized in Webster's Tenth.
Entries in the following table include the year in which they first
entered the language (according to the Tenth), and the Merriam-Webster
dictionary that first contains them. The following symbols are used:
NI1 New International (1909)
NI1+ New Words section of the New International (1931)
NI2 New International Second Edition (1934)
NI2+ Addendum section of the Second (1959, same as 1954)
NI3 Third New International (1961)
9C Ninth New Collegiate (1983)
10C Collegiate - Tenth Edition (1993)
12W 12,000 Words (separately published addendum to the Third, 1986)
asdic Anti-Submarine Detection Investigation Committee (1940, NI2+)
dopa DihydrOxyPhenylAlanine (1917, NI3)
fido Freaks + Irregulars + Defects + Oddities (1966, 9C)
jato Jet-Assisted TakeOff (1947, NI2+)
laser Light Amplification by Stimulated Emission of Radiation (1957, NI3)
lidar LIght Detection And Ranging (1963, 9C)
maser Microwave Amplification by Stimulated Emission of Radiation (1955, NI3)
nitinol NIckel + TIn + Naval Ordinance Laboratory (1968, 9C)
rad Radiation Absorbed Dose (1918, NI3)
radar RAdio Detection And Ranging (ca. 1941, NI2+)
rem Roentgen Equivalent Man (1947, NI3)
rep Roentgen Equivalent Physical (1947, NI3)
scuba Self-Contained Underwater Breathing Apparatus (1952, NI3)
snafu Situation Normal -- All Fucked (Fouled) Up (ca. 1940, NI2+)
sofar SOund Fixing And Ranging (1946, NI2+)
sonar SOund NAvigation Ranging (1945, NI2+)
tepa Tri-Ethylene Phosphor-Amide (1953, 9C)
zip Zone Improvement Plan (1963, 9C)
Below are blends that technically are also acronyms:
alnico ALuminum + NIckel + CObalt (1935, NI2+)
avgas AViation GASoline (1943, NI3)
biopic BIOgraphical PICture (12W)
boff Box OFFice (1946, NI3)
canola CANada Oil + Low Acid (1979, 10C)
ceramal CERAMic ALloy (ca. 1948, NI2+)
cermet CERamic METal (1948, NI2+)
comsymp COMmunist SYMPathizer (ca. 1961, 9C)
cyborg CYBernetic ORGanism (ca. 1962, 9C)
dorper DORset horn + blackhead PERsian (1949, NI3)
elhi ELementary school + HIgh school (1948, 9C)
gox Gaseous OXygen (1959, 9C)
hela HEnrietta LAcks (1953, 9C)
kip KIlo- + Pound (1914, NI2)
linac LINear ACcelerator (1950, 9C)
loran LOng-RAnge Navigation (ca. 1932, NI2+)
lox Liquid OXygen (1923, 9C)
mascon MASs CONcentration (1968, 9C)
maximin MAXImum + MINimum (1951, 9C)
minimax MINImum + MAXimum (1918, 9C)
modem MOdulator + DEModulator (ca. 1952, 9C)
motocross MOTOr + CROSS-country (1951, 9C)
napalm NAphthenic and PALMitic acids (1942, NI2+)
parsec PARallax SECond (ca. 1913, NI1+)
redox REDuction + OXidation (1828, NI2)
selsyn SELf-SYNchronizing (1936, NI2+)
shoran SHOrt-RAnge Navigation (ca. 1932, NI2+)
silvex SILVa + EXterminator (1961, 9C)
sitcom SITuation COMedy (1965, 9C)
teleran TELEvision-RAdar Navigation (1946, NI2+)
telex TELeprinter EXchange (ca. 1943, 9C)
vidicon VIDeo + ICONoscope (1950, NI3)
wilco WILl COmply (ca. 1938, NI3)
Acronyms from other languages:
agitprop AGITatsiya + PROPaganda (Russian, ca. 1926, NI2+)
flak FLiegerAbwehrKanonen (German, 1938, NI2+)
gestapo GEheime STAatsPOlizei (German, 1934, NI2+)
gulag Glavnoe Upravlenie ispravitel'notrudovykh LAGerei (Russian, 1974, 9C)
kolkhoz KOLlektivnoe KHOZyaistvo (Russian, 1921, NI2)
moped MOtor + PEDal (Swedish, ca. 1955, 9C)
sambo SAMozashchita Bez Oruzhiya (Russian, 1972, 9C)
tokamak TOroidal'naya KAMera s AKsial'nym magnitnym polem (Russian, 1965, 9C)
Selected near misses:
athodyd Aero-THermODYnamic Duct (1945, NI2+) -- blend
awol Absent WithOut Leave (1919, NI2+) -- usually capitalized
benday BENjamin DAY (1903, NI1+) -- blend
deet Di-Ethyl Tolumide (1962, 9C) -- pronunciation of D. E. T.
dew Distant Early Warning (1953, 9C) -- only in phrase "dew line"
echovirus Enteric Cytopathogenic Human Orphan VIRUS (1955, 9C) -- blend
hi-fi HIgh FIdelity (1948, NI2+) -- hyphenated
ibuprofen Iso-BUtyl PROpionic PHENyl (1969, 12W) -- PH pronounced f
jaygee Junior Grade (1943, NI3) -- pronunciation of J. G.
jayvee Junior Varsity (1937, NI3) -- pronunciation of J. V.
jeep General Purpose (1940, NI2+) -- pronunciation of G. P.
nazi NAtionalsoZIalist (German, 1930, NI2) -- shorten & alter
nystatin New York STATe + -IN (1952, NI3) -- extraneous suffix
op-ed OPposite EDitorial (1970, 9C) -- hyphenated
pixel PIX + ELement (1969, 10C) -- blend
pj's PaJamas (1951, NI3) -- punctuated
reovirus Respiratory Enteric Orphan VIRUS (1959, 9C) -- blend
sci-fi SCIence FIction (1955, 9C) -- hyphenated
siloxane SILicon + OXygen + methANE (1922, NI3) -- blend
tradevman TRAining DEVices MAN (ca. 1947, NI3) -- blend
updo UPswept hairDO (1946, NI2+) -- blend
veep Vice President (1940, NI2+) -- pronunciation of V. P.
warfarin Wisconsin Alumni Research Foundation + coumARIN (ca. 1950, NI3) - blend
yuppie Young Urban Professional + -PIE (1983, 9C) -- extraneous suffix
Acronyms that should be in Webster's Tenth:
fifo First In, First Out (NI2+)
lifo Last In, First Out (NI2+)
nomic NO Metal In Composition (NI3) (John Bulten)
quango QUAsi-Non Governmental Organization (12W)
shazam Solomon Hercules Atlas Zeus Achilles Mercury (12W)
tacan TACtical Air Navigation (12W)
Self-referential acronyms:
gnu Gnu's Not Unix
nbi Nothing But Initials
swan System Without A Name
Supposed (but not real) acronyms:
fuck For Unlawful Carnal Knowledge
posh Port Out, Starboard Home
spiff Sales Productivity Incentive Fund
tip To Insure (should be Ensure) Politeness (or Promptness)
wog Worthy (Wealthy) Oriental Gentleman
==> language/english/etymology/fossil.p <==
What are some examples of idioms that include obsolete words?
==> language/english/etymology/fossil.s <==
These are called fossil expresions -- words that have dropped out of
common use but hang around in idioms. Not all of them are separate
words, some are part of other words or have prefixes or suffixes
attached. There are also words which have current meaning, but the
meaning in the idiom is unrelated to it.
idiom fossil meaning of fossil
--------------------------------------------------
swashbuckler buckler small shield
newfangled fangled siezed
rank and file file column
to and fro fro from
gormless gorm attention
hem and haw haw make the sound "haw"
hem and haw hem make the sound "hem"
hue and cry hue outcry
kit and kaboodle kaboodle collection
out of kilter kilter order
kith and kin kith friends
let or hindrance let hindrance
footpad pad highwayman
pratfall prat buttocks
rank and file rank row
raring to go raring enthusiastic
ruthless ruth compassion
short shrift shrift confession
spick-and-span span chunk of wood
spick-and-span spick nail (spike)
swashbuckler swash bluster or stagger
bank teller tell to count
==> language/english/etymology/portmanteau.p <==
What are some words formed by combining together parts of other words?
==> language/english/etymology/portmanteau.s <==
Such words are called "Portmanteau" words. Here is a very incomplete list:
beefalo beef, buffalo
brunch breakfast, lunch
chortle chuckle, snort
fantabulous fantastic, fabulous
flare flame, glare
flounder flounce, founder
glimmer gleam, shimmer
glitz glamour, ritz
liger lion, tiger
motel motor, hotel
smash smack, mash
smog smoke, fog
squiggle squirm, wiggle
tangelo tangerine, pomelo
tigon tiger, lion
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/frequency.p <==
In the English language, what are the most frequently appearing:
1) letters overall?
2) letters BEGINNING words?
3) final letters?
4) digrams (ordered pairs of letters)?
==> language/english/frequency.s <==
web2 = word list from Webster's Second Unabridged
web2a = hyphenated words and phrases from Webster's Second Unabridged
both = web2 + web2a
net = several gigabytes of Usenet traffic
1) Most frequently appearing letters overall:
web2: eiaorn tslcup mdhygb fvkwzx qj
both: eairon tslcud pmhgyb fwvkzx qj
net: etaoin srhldc umpfgy wbvkxj qz
2) Most frequently appearing letters BEGINNING words:
web: spcaut mbdrhi eofgnl wvkjqz yx
both: spcatb umdrhf eigowl nvkqjz yx
net: taisow cmbphd frnelu gyjvkx qz
3) Most frequent final letters:
web: eysndr ltacmg hkopif xwubzv jq
both: eydsnr tlagcm hkpoiw fxbuzv jq
net: estndr yolafg mhipuk cwxbvz jq
4) Most frequent digrams (ordered pairs of letters)
web: er in ti on te al an at ic en is re ra le ri ro st ne ar ...
both: er in te ti on an re al at le en ra ic ar st ri ro ed ne ...
net: th he in er re an on at te es or en ar ha is ou it to st nd ...
Program to compute this from word list in standard input:
#include <stdio.h>
#include <ctype.h>
typedef struct {
int count;
char name[3];
} FREQ;
FREQ all[256],initial[256],terminal[256],digram[65536];
int compare(p,q)
FREQ *p,*q;
{ return q->count - p->count;
}
void sort_and_print(freq,count,description)
FREQ *freq;
int count;
char *description;
{ register FREQ *p;
(void)qsort(freq,count,sizeof(*freq),compare);
puts(description);
for (p=freq;p<freq+count;p++)
if (p->count) printf("%s %d\n",p->name,p->count);
}
main()
{ char s[BUFSIZ];
register char *p;
register int i;
while (gets(s)!=NULL) {
if (islower(*s)) {
initial[*s].count++;
sprintf(initial[*s].name,"%c",*s);
for (p=s;*p;p++) {
if (isalpha(*p)) {
all[*p].count++;
sprintf(all[*p].name,"%c",*p);
if (isalpha(p[1])) {
i = p[0]*256 + p[1];
digram[i].count++;
sprintf(digram[i].name,"%c%c",p[0],p[1]);
}
}
}
terminal[*--p].count++;
sprintf(terminal[*p].name,"%c",*p);
}
}
sort_and_print(all,256,"overall character distribution: ");
sort_and_print(initial,256,"initial character distribution: ");
sort_and_print(terminal,256,"terminal character distribution: ");
sort_and_print(digram,65536,"digram distribution: ");
}
==> language/english/idioms.p <==
List some idioms that say the opposite of what they mean.
==> language/english/idioms.s <==
I fell head over heels.
I could care less.
I turned my life around 360 degrees.
==> language/english/less.ness.p <==
Find a word that forms two other words, unrelated in meaning, when "less"
and "ness" are added.
==> language/english/less.ness.s <==
base -> baseless, baseness
light -> lightless, lightness
sound -> soundless, soundness
wit -> witless, witness
==> language/english/letter.rebus.p <==
Define the letters of the alphabet using self-referential common phrases (e.g.,
"first of all" defines "a").
==> language/english/letter.rebus.s <==
A first of all, midday
B fifth of bourbon, starting block
C fifth of scotch
D end of the world, back of my hand
E end of the line, beginning of the end
F starting friction, front
G middle of the night, starting gate
H end of the earth, top of the heap, middle of nowhere
I next of kin
J center of project
K bottom of the deck, two of a kind
L bottom of the barrel, starting line
M top of my head
N center of attention, final countdown, end run
O second in command
P bottom of the heap, the first of painters, starting point
Q at the front of the queue, top quality
R middle of the road, center of inertia
S _Last of the Mohicans_, start of something big
T top o' the morning, one's wit's end, bottom of my heart, last, central
U second guess
V center of gravity
W end of the rainbow, top of the world
X wax finish, climax
Y top of your head, center of the cyclone, early years, final extremity
Z led zeppelin
==> language/english/malaprop.p <==
List some phrases with the same meaning that differ by one sound.
==> language/english/malaprop.s <==
The car caree(n/r)s down the road.
Can you (e/i)nsure that all is well?
We must inst(a/i)ll some discipline.
It's coming down the pi(p/k)e.
The sun pe(a/c)ked out from behind clouds.
-Patricia D. Cornwell, _All That Remains_, Avon, New York, 1993, p. 36
I've been ra(c)king my brain the whole time.
-Patricia D. Cornwell, _All That Remains_, Avon, New York, 1993, p. 231
Listen to the blabbing/babbling brook.
-Norm Crosby
==> language/english/piglatin.p <==
What words in pig latin also are words?
==> language/english/piglatin.s <==
cess -> essay
coke -> okay
lawn -> onlay
lout -> outlay
plover -> overplay
plunder -> underplay
sass -> assay
stout -> outstay
trash -> ashtray
wear -> airway
wonder -> underway
==> language/english/pleonasm.p <==
What are some redundant terms that occur frequently (like "ABM missile")?
==> language/english/pleonasm.s <==
11.5% APR (although "APR" means that interest is charged on the balance only)
ABM missile
ABS system
AC current
ACT test
AMOCO Oil Co.
APL programming language
ATM machine
BASIC Code
BBS System
CAD design
CNN news network
DC current
DMZ zone
DOS operating system
GMT time
Geirangerfjorden (Fjord Fjord Fjord)
HIV virus
ISBN number
ISDN network
LCD display
LED diode
MIDI Interface
Mount Fujiyama (Mount Mountain)
NATO organization
NFS File System
PIN number
RAM (or ROM) memory
Rio Grande River (Big River River)
Ruidoso River (Noisy River River)
SALT talks or SALT treaty
SAT test
SCSI Interface
SEATO organization
START talks or START treaty
The La Brea Tar Pits (The The Tar Tar Pits)
The Los Altos Hills (The The Hills Hills)
VIN number
floccinoccinihlipilification (from 4 latin words meaning "nothing")
the hoi polloi (the the many)
hot water heater
==> language/english/plurals/collision.p <==
Two words, spelled and pronounced differently, have plurals spelled
the same but pronounced differently.
==> language/english/plurals/collision.s <==
axe and axis -> axes
base and basis -> bases
ellipse and ellipsis -> ellipses
==> language/english/plurals/doubtful.number.p <==
A little word of doubtful number,
a foe to rest and peaceful slumber.
If you add an "s" to this,
great is the metamorphosis.
Plural is plural now no more,
and sweet what bitter was before.
What am I?
==> language/english/plurals/doubtful.number.s <==
cares -> caress
==> language/english/plurals/drop.terminal.p <==
What words have their plurals formed by dropping the final letter?
==> language/english/plurals/drop.terminal.s <==
There are a number of words (e.g., "jinni") where there is a variant
spelling sans the terminal letter (e.g., "jinn") and where the plural
is the same as the singular. I don't think this is in keeping with the
spirit of the question.
There are a number of Italian words where the final "o" is dropped to
form a plural:
arpeggio
capriccio
carroccio
fascio
impresario
intarsio
latticinio
pasticcio
preludio
scenario
seraglio
solfeggio
Other words are:
hemiepes n pl hemiepe
:a dactylic tripody having a spondaic third foot or lacking the two
short syllables of the third foot
koruna n pl korun or koruny also korunas
:the basic monetary unit of Czechoslovakia
microstomia also microstomus n pl microstomias also microstomi
:an abnormally small mouth
necropolis n pl necropolises or necropoles also necropoleis or necropoli
:a large elaborate cemetery of an ancient city
samanid n pl samanids or samani
:a member of a 9th and 10th century Persian dynasty ruling from Bokhara
and encouraging literature and art
volksdeutscher n pl volksdeutsche
:a person of German ethnic origin long settled in a central or east
European country, repatriated for political reasons by the Nazi regime,
and expelled into West Germany after World II
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/plurals/endings.p <==
List a plural ending with each letter of the alphabet.
==> language/english/plurals/endings.s <==
Legend
0 = plural formed (basically) by adding letter
1 = plural spelled differently from singular
2 = ditto, plural contains punctuation
3 = plural spelled the same as singular
All entries are from Merriam-Webster's Ninth Collegiate Dictionary,
except those marked "(NI3)", which are from the Third International.
Entries in brackets are probable dictionary artifacts.
A 0 VAS VASA
B 1 SLUBBI SLEYB (NI3)
C 0 CALPULLI CALPULLEC (NI3)
D 2 GRANT-IN-AID GRANTS-IN-AID
E 0 ALA ALAE
F 1 SHARIF ASHRAF (NI3)
G 0 AIRE AIRIG (NI3)
H 0 LIRA LIROTH
I 0 BAN BANI
J 1 KHARIJITE KHAWARIJ (NI3)
K 0 PULI PULIK
L 1 ARMFUL ARMSFUL
M 0 GOY GOYIM
N 0 KRONE KRONEN
O 2 DERRING-DO DERRINGS-DO (NI3) [1 MEO MIAO/MIXTECA MIXTECO/PAPIOPIO PAPIO/SUMU SUMO (NI3)]
P 2 AIDE-DE-CAMP AIDES-DE-CAMP
Q 3 QARAQALPAQ QARAQALPAQ (NI3)
R 0 KRONE KRONER
S 0 A AS
T 0 MATZO MATZOT
U 0 HALER HALERU
V 3 TIV TIV (NI3)
W 2 SON-IN-LAW SONS-IN-LAW [1 KWAPA QUAPAW (NI3)]
X 0 EAU EAUX
Y 0 GROSZ GROSZY
Z 3 HERTZ HERTZ
==> language/english/plurals/man.p <==
Words ending with "man" make their plurals by adding "s".
==> language/english/plurals/man.s <==
caiman
doberman
German
human
leman
ottoman
pitman
Pullman
Roman
shaman
talisman
==> language/english/plurals/switch.first.p <==
What plural is formed by switching the first two letters?
==> language/english/plurals/switch.first.s <==
falaj -> aflaj (Chambers English Dictionary)
==> language/english/potable.color.p <==
Find words that are both beverages and colors.
==> language/english/potable.color.s <==
burgundy
champagne
chartreuse
chocolate
claret
cocoa
coffee
cream
midori (Japanese for green. Does Japanese count?)
rose
wine
==> language/english/pronunciation/autonym.p <==
What is the longest word whose phonetic and normal spellings are the same?
==> language/english/pronunciation/autonym.s <==
spendthrift (11 letters)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/pronunciation/homograph/different.pronunciation.p <==
What sequence of letters has the most different pronunciations?
==> language/english/pronunciation/homograph/different.pronunciation.s <==
OUGH (28)
common words:
<sw> BOROUGH, THOROUGH
<sw>f TOUGH
<sw>p HICCOUGH
<a"> NOUGHT
<au> BOUGH
<a">f COUGH
<o_> THOUGH
<o.> OUGHT
<o.>f COUGH
<u"> THROUGH
uncommon words:
<sw>k TURLOUGH (dialectical, obsolete)
<sw><k_> BROUGH (dialectical)
<sw>w BOROUGH, THOROUGH
<a">ft TROUGH (dialectical)
<a">k LOUGH (dialectical)
<a"><k_> DOUGHT, LOUGH (dialectical)
<a">th TROUGH (dialectical)
<o_>g SKEOUGH (dialectical, obsolete)
<o_>k WOUGH (dialectical, obsolete)
<o_><k_> BOUGHT, HOUGH (dialectical)
<o.>ft TROUGH (dialectical)
<o.>th TROUGH (dialectical)
<u">f SOUGH (dialectical, obsolete)
<u">g SLOUGHI
<u">k OUGH (obsolete)
<u"><k_> OUGH
<u.>f SOUGH (dialectical, obsolete)
<u.><k_> OUGH
Pronunciation Key:
<sw> the, humdrum
<a"> bother, cot
<a.> father
<au> now, loud
<o_> bone, snow
<o.> saw, all
<u"> rule, fool
<u.> pull, wood
<k_> German ch in ich-laut
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/pronunciation/homograph/homographs.p <==
List some homographs (words spelled the same but pronounced differently)
==> language/english/pronunciation/homograph/homographs.s <==
This list organized by Mark Brader <m...@sq.com>
Classes:
A - All of the following "defects" absent
B - Basic meanings are related
C - Capitalization differs ("capitonyms")
D - Different spellings also exist (US vs UK, hyphenation, etc.)
E - Equal pronunciations also exist (US vs UK, regional, etc.)
F - Foreign word, or may be distinguished with accent marks
G - Gcontrived :-), coined, jargon, or other uncommon word
3 - 3-way homograph
4 - 4-way homograph
B abstract {corresponding noun and verb; henceforth abbreviated NV}
B abuse {NV}
B addict {NV}
B advocate {NV}
BG affect {alter; emotion}
B affiliate {NV}
B affix {NV}
G agape {wide open; form of love}
B aggregate {NV}
G ai {sloth; ouch!}
BE ally {NV}
B alternate {NV}
BD analyses {plural noun; singular verb (UK)}
B animate {verb; adjective}
A appropriate {take posession of; suitable}
B approximate {verb; adjective}
E are {form of to be; unit of area}
B arithmetic {noun; adjective}
B articulate {verb; adjective}
4DFG as {like; Roman coin; Persian card game; pl. of a}
B aspirate {NV}
B associate {NV}
A ate {past of eat; recklessness}
B attribute {NV}
C august
A axes {plural of ax; plural of axis}
A bases {plural of base; plural of basis}
A bass {~ fiddle; fishing for ~}
BE blessed {prosperous; he ~ her}
A bow(ed) {~ and arrow; ~ to the king}
E buffet {jostle; ~ lunch}
B bustier {undergarment; more busty}
B close {~ call; ~ the door}
B closer {door ~; more close}
B coagulate {NV}
G coax {urge; coaxial cable}
3FG colon {":"; colonial farmer; Costa Rican monetary unit}
B combat {NV}
B combine {NV}
A commune {take Communion; administrative district}
A compact {closely arranged; treaty}
B compound {NV}
B compress {NV}
B conduct {NV}
B confect {NV}
B confines {NV}
B conflict {NV}
B conglomerate {NV}
B conjugate {NV}
BE conserve {preserve; jam}
A console {soothe; keyboard desk}
B consort {NV}
B construct {NV}
B consummate {verb; adjective}
BE contact {NV}
E content {what is contained; satisfied}
B contest {NV}
B contract {NV}
B contrast {NV}
BG convent {nunnery; convene}
A converse {logic term; to talk}
B convert {NV}
B convict {NV}
BE coordinate {NV}
FG dame {woman; term in the game of Go}
DE decameter {poetic line with 10 feet; 10 meters (US)}
B defect {flaw; turn traitor}
E defense {sports term; fortification}
BE delegate {NV}
B deliberate {adjective; verb}
A desert {leave alone; Sahara ~}
B desolate {adjective; verb}
D dingy {dull; small boat}
BE discharge {NV}
B discriminate {distinguish; opposite of indiscriminate}
E divers {plural diver; various}
F do {perform; tonic note of scale}
A does {~ the buck see the ~?}
A dove {dived; pigeon}
F dozen {12; stun (Scottish)}
B drawer {one who draws; chest of ~s}
B duplicate {NV}
B elaborate {verb; adjective}
A entrance {door; delight}
BDE envelop(e) {NV}
G ergotism {logical reasoning; ergot poisoning}
B escort {NV}
B essay {piece of writing; try}
B estimate {NV}
CFG ewe {female sheep; African language}
B excuse {NV}
B exploit {NV}
BF expose {NV}
B ferment {NV}
E fiasco {failure; bottle}
BDE fillet {cut of meat/fish; band of ribbon/wood}
A flower {one who flows; bloom}
G formal {ceremonious; methylal}
DEG genet {civetlike animal; horselike animal}
A gill {volume unit; organ in fish}
A glower {sullen look; one that glows}
B graduate {NV}
F he {pronoun; Hebrew letter}
CE herb {name; plant}
A hinder {hamper; posterior}
B house {NV}
B import {NV}
A incense {infuriate; perfume for burning}
B increase {NV}
B initiate {NV}
B insert {NV}
B insult {NV}
B intern {NV}
A intimate {~ relations; to suggest}
A invalid {cripple; erroneous}
B invite {NV}
G is {form of to be; plural of i}
B jagged {slashed or cut; having a zigzag edge}
C Job
BCF jubilate {rejoice; joyous song}
CF junker/Junker
3A lather {suds; lath worker; lathe worker}
A lead {~ pipe; ~ astray}
BE legged {past tense verb; adjective}
CF Lima
B live {~ in peace; ~ audience}
B lives {~ in peace; for all of our ~}
D lower {to let down; frown}
F manes {plural of mane; Roman gods}
F mate {friend; type of tea}
D micrometer {measuring device; .000001 meter (US)}
A minute {60 seconds; tiny}
B misconduct {NV}
BE mobile {movable; wind-blown sculpture}
B moderate {NV}
EG molar {back tooth; chemical term}
A moped {brooded; fun vehicle}
BE mouse {rodent; to hunt them}
B mouth {NV}
A mow {pile of hay; to cut down}
B multiply {verb; adverb}
A number {decimal ~; more numb}
B object {thing; complain}
E offense {sports term; attack}
3DG os {bone; esker; pl. of o}
A overage {too old; surplus}
BD paralyses {plural noun; singular verb (UK)}
A pasty {pastelike; British meat pie}
3FG pate {head; food paste; porcelain paste for ceramics}
A peaked {sharply pointed; unhealthy looking}
A peer {equal; one who pees}
B perfect {verb; adjective}
G periodic {regularly occurring; ~ acids, HIO4 and related substances}
B permit {NV}
C Placer
C polish
A poll {head; group of students}
B predicate {NV}
BE premise {NV}
B present {NV}
E primer {intro book/material (US); device for priming}
B proceeds {goes; income}
B produce {give rise to; fruits and vegetables}
B progress {to move forward; work in ~}
A project {planned undertaking; to throw forward}
BE prospect {NV}
B protest {NV}
A pussy {cat; infected}
B putter/putting {golf club; one that puts}
DG rabat {clerical garment; pottery piece used for polishing}
DG rabbi {clerical garment; Jewish religious official}
B ragged {teased; tattered}
F re {pertaining to; 2nd note of scale}
B read {present tense; past tense}
C Reading
F real {actual; former Spanish coin}
B rebel {NV}
B recess {NV}
B recoil {NV}
B record {NV}
D recreate {relax; create again}
3BD redress {compensate; compensation; dress again}
B refill {NV}
B refund {NV}
B refuse {NV}
B regress {NV}
B reject {NV}
E repent {regret; creeping}
B replay {NV}
D represent {stand for; present again}
B rerun {NV}
D research {investigate; search again}
A resent {be indignant; sent again}
D reserve {hold back; serve again}
D resign {quit; sign again}
D resolve {settle dispute; solve again}
D resort {vacation spot; sort again}
F resume {work summary; restart}
A river {watercourse; one who rives}
F rose {flower; wine}
DE routing {making a route for (US spelling); woodworking term}
A row {a fight; ~,~,~ your boat}
DF sake {purpose; Japanese drink}
3AF salve {ointment; salvage; hail!}
3BE second {2nd; 1/60 minute; transfer to temporary job}
B segment {NV}
B separate {NV}
A severer {cutter; more severe}
A sewer {storm ~; one who sews}
A shower {one who shows; ~ stall}
A singer {one who singes; one who sings}
A skied {past tense of ski; past tense of sky}
A slaver {slave taker; drool}
A slough {swamp; cast-off}
A sow {~ seeds; female pig}
A stingy {meager; able to sting}
EG stipulate {request explicitly; having stipules}
EG stupe {hot compress as counterirritant; stupid person}
B subject {NV}
A supply {in a supple way; ~ and demand}
B survey {NV}
B suspect {NV}
G swinger {whopper; one that swings}
G swingle {flax beating tool; single person who "swings"}
B syndicate {NV}
CF tang {flavor; Chinese dynasty}
A tarry {covered in tar; dawdle}
A tear {~ down; shed a ~}
A thou {you; slang for thousand}
A thymic {of thyme; of thymus}
A tier {one who ties; row or rank}
B torment {NV}
A tower {one who tows; leaning ~}
B transfer {NV}
B transplant {NV}
B transport {NV}
DG unionized {~ labor; ~ hydrogen}
B upset {NV}
G us {we; plural of u}
B use {NV}
A violist {viol player; viola player}
A wind {~ the clock; north ~}
CF worms
A wound {injury; wrapped around}
E yak {ox; laugh}
==> language/english/pronunciation/homophone/homophones.alphabet.p <==
Homophones can be confusing when used to exemplify a letter. For example,
"g as in gnu" or "k as in knot." Give one for each letter.
==> language/english/pronunciation/homophone/homophones.alphabet.s <==
A as in isle (aisle)
B as in dell (bdell-) (combining form)
C as in teen (ctene)
D as in gin (djin)
E as in air (ere)
F as in phrase (frays) ('f' sound present)
G as in new (gnu)
H as in air (heir)
I as in use (ius)
J as in yuca (juca)
K as in need (knead)
L as in have (halve) (not first letter)
M as in nemic (mnemic) (contains m)
N as in dam (damn) (not first letter)
O as in wed (oued)
P as in salmon (psammon)
Q as in key (quay)
R as in foxhole (forecastle) (not first letter; pronunciation off)
S as in eyelet (islet) (not first letter)
T as in sign (tsine)
U as in wang (uang)
V as in fro (vrow)
W as in rest (wrest)
X as in jurel (xurel)
Y as in the (ye)
Z as in sabra (zabra)
***
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/pronunciation/homophone/homophones.letter.p <==
For each letter, list homophones that differ by that letter.
==> language/english/pronunciation/homophone/homophones.letter.s <==
o(a)r
lam(b)
s(c)ent
le(d)ger
do(e)
waf(f)
rei(g)n
(h)our
wa(i)ve
haj(j)i
(k)not
ha(l)ve
prim(m)er
dam(n)
j(o)ust
(p)salter
?
car(r)ies
(s)cent
taro(t)
b(u)y
civ(v)ies
t(w)o
beau(x)
re(y)
biz(z)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/pronunciation/homophone/homophones.most.p <==
What words have four or more spellings that sound alike?
==> language/english/pronunciation/homophone/homophones.most.s <==
a: ba, baa, bah (3)
<a_>: nae, nay, nee, neigh (4)
<a.>:
<a">: pa, pah, pas (3)
a<u.>: dhow, dow, tao (3)
b: cob, cobb, kob (3)
ch:
d: road, rode, rowed (3)
e:
<e_>: cee, sea, see, si (4)
f: leaf, leif (2)
g:
h:
hw:
i: boy, buoy (2)
<i_>: ai, ay, aye, eye, I (5)
j: gage, gauge (2)
k: peak, peek, peke, pique (4)
<k_>: dreich, dreigh, driech, driegh (4)
l: weal, weel, we'll, wheal, wheel (5)
m: combe, coom, coomb, cwm (4)
n: coign, coin, quoin (3)
<o.>: tau, taw (2)
<o_>: eau, eaux, O, oh, owe (5)
p: nap, knap, nappe (3)
r: air, are, ayr, ayre, e'er, ere, err, eyre, heir (9)
s: cense, cents, scents, sense (4)
sh:
t: right, rite, wright, write (4)
th:
<th>:
u:
<u">: tew, to, too, two (4)
v:
w:
y:
z: raise, rase, rays, raze, res (5)
zh: greige, griege (2)
***
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> decision/allais.p <==
The Allais Paradox involves the choice between two alternatives:
A. 89% chance of an unknown amount
10% chance of $1 million
1% chance of $1 million
B. 89% chance of an unknown amount (the same amount as in A)
10% chance of $2.5 million
1% chance of nothing
What is the rational choice? Does this choice remain the same if the
unknown amount is $1 million? If it is nothing?
==> decision/allais.s <==
This is "Allais' Paradox".
Which choice is rational depends upon the subjective value of money.
Many people are risk averse, and prefer the better chance of $1
million of option A. This choice is firm when the unknown amount is
$1 million, but seems to waver as the amount falls to nothing. In the
latter case, the risk averse person favors B because there is not much
difference between 10% and 11%, but there is a big difference between
$1 million and $2.5 million.
Thus the choice between A and B depends upon the unknown amount, even
though it is the same unknown amount independent of the choice. This
violates the "independence axiom" that rational choice between two
alternatives should depend only upon how those two alternatives
differ.
However, if the amounts involved in the problem are reduced to tens of
dollars instead of millions of dollars, people's behavior tends to
fall back in line with the axioms of rational choice. People tend to
choose option B regardless of the unknown amount. Perhaps when
presented with such huge numbers, people begin to calculate
qualitatively. For example, if the unknown amount is $1 million the
options are:
A. a fortune, guaranteed
B. a fortune, almost guaranteed
a tiny chance of nothing
Then the choice of A is rational. However, if the unknown amount is
nothing, the options are:
A. small chance of a fortune ($1 million)
large chance of nothing
B. small chance of a larger fortune ($2.5 million)
large chance of nothing
In this case, the choice of B is rational. The Allais Paradox then
results from the limited ability to rationally calculate with such
unusual quantities. The brain is not a calculator and rational
calculations may rely on things like training, experience, and
analogy, none of which would be help in this case. This hypothesis
could be tested by studying the correlation between paradoxical
behavior and "unusualness" of the amounts involved.
If this explanation is correct, then the Paradox amounts to little
more than the observation that the brain is an imperfect rational
engine.
==> decision/division.p <==
N-Person Fair Division
If two people want to divide a pie but do not trust each other, they can
still ensure that each gets a fair share by using the technique that one
person cuts and the other person chooses. Generalize this technique
to more than two people. Take care to ensure that no one can be cheated
by a coalition of the others.
==> decision/division.s <==
N-Person Fair Division
Number the people from 1 to N. Person 1 cuts off a piece of the pie.
Person 2 can either diminish the size of the cut off piece or pass.
The same for persons 3 through N. The last person to touch the piece
must take it and is removed from the process. Repeat this procedure
with the remaining N - 1 people, until everyone has a piece.
References:
Luce and Raiffa, "Games and Decisions", Wiley, 1957, p. 366
Kenneth Rebman, "How To Get (At Least) A Fair Share of the Cake",
in Mathematical Plums, Ross Honsberger, ed, Dolciani Mathematical
Expostions Number 4, published by the MAA.
There is a cute result in combinatorics called the Marriage Theorem.
A village has n men and n women, such that for all 0 < k <= n and for any
set of k men there are at least k women, each of whom is in love with at least
one of the k men. All of the men are in love with all of the women :-}.
The theorem asserts that there is a way to arrange the village into n
monogamous couplings.
The Marriage Theorem can be applied to the Fair Pie-Cutting Problem.
One player cuts the pie into n pieces. Each of the players labels
some non-null subset of the pieces as acceptable to him. For reasons
given below he should "accept" each piece of size > 1/n, not just the
best piece(s). The pie-cutter is required to "accept" all of the pieces.
Given a set S of players let S' denote the set of pie-pieces
acceptable to at least one player in S. Let t be the size of the largest
set (T) of players satisfying |T| > |T'|. If there is no such set, the
Marriage Theorem can be applied directly. Since the pie-cutter accepts
every piece we know that t < n.
Choose |T| - |T'| pieces at random from outside T', glue them
together with the pieces in T' and let the players in T repeat the game
with this smaller (t/n)-size pie. This is fair since they all rejected
the other n-t pieces, so they believe this pie is larger than t/n.
The remaining n-t players can each be assigned one of the remaining
n-t pie-pieces without further ado due to the Marriage Theorem. (Otherwise
the set T above was not maximal.)
The problem of getting not just a fair solution, but an envy-free solution,
is not solved. A reference to this problem:
David Gale, "Dividing a Cake," in Mathematical Entertainments,
Mathematical Intelligencer, Vol. 15, No. 1, Winter 1993, p. 50,
contains references to work by Steven Breams and Alan Taylor.
==> decision/dowry.p <==
Sultan's Dowry
A sultan has granted a commoner a chance to marry one of his hundred
daughters. The commoner will be presented the daughters one at a time.
When a daughter is presented, the commoner will be told the daughter's
dowry. The commoner has only one chance to accept or reject each
daughter; he cannot return to a previously rejected daughter.
The sultan's catch is that the commoner may only marry the daughter with
the highest dowry. What is the commoner's best strategy assuming
he knows nothing about the distribution of dowries?
==> decision/dowry.s <==
Solution
Since the commoner knows nothing about the distribution of the dowries,
the best strategy is to wait until a certain number of daughters have
been presented then pick the highest dowry thereafter. The exact number to
skip is determined by the condition that the odds that the highest dowry
has already been seen is just greater than the odds that it remains to be
seen AND THAT IF IT IS SEEN IT WILL BE PICKED. This amounts to finding the
smallest x such that:
x/n > x/n * (1/(x+1) + ... + 1/(n-1)).
Working out the math for n=100 and calculating the probability gives:
The commoner should wait until he has seen 37 of the daughters,
then pick the first daughter with a dowry that is bigger than any
preceding dowry. With this strategy, his odds of choosing the daughter
with the highest dowry are surprisingly high: about 37%.
(cf. F. Mosteller, "Fifty Challenging Problems in Probability with Solutions",
Addison-Wesley, 1965, #47; "Mathematical Plums", edited by Ross Honsberger,
pp. 104-110)
Here's a twist on the sultan's dowry problem I hope hasn't been posted yet.
I became interested in an iterated version of this problem, which goes as
follows:
There's a long line of suitors outside the sultan's palace, and one by one
they come in. If a suitor gets the right girl, he marries her, as before.
Unfortunately (for the suitor, at least), if he doesn't, he gets his head
lopped off, and the next suitor comes in.
Anyway, the first few dozen guys all know their probability theory, so they
know that the best strategy is to skip the first 37 girls, and then pick
the first girl who is the best up to that point. Alas, each one assumes
that he's the only one who knows that strategy, so one by one, these few
dozen guys get their heads lopped off.
After the 49th head has just rolled down the hill, and the sultan's vizier
has just cried out, "Next!" the next guy in line says, "This isn't working
out. We might all be doing the same thing. It doesn't hurt any of us to
tell the rest what strategy we'll be using, so that none of us sets out to
pick the same girl over and over again. I might as well just tell you,
though, that I'm going to get her! I know this great strategy where you
skip the first 37 blah blah blah..." Naturally, a few moments later, head
number 50 comes rolling down.
"Next!" cries the vizier.
Well, suitor number 51 is in a quandary. He's all set to skip 37, etc, etc,
except now he knows, that's not the right strategy. But he doesn't know if
the last guy skipped the right girl because she was in the first 37, or if
he didn't meet her yet because he stopped too early.
QUESTION 1: What is his best strategy?
ANSWER 1: His best strategy is:
"Skip the first 14. Take the first girl in [15,37] who is better
than the first 14. If there isn't one, take the SECOND girl in [38,100]
who is the best up to that point."
Unfortunately, head number 51 rolls down the hill. "Next!" cries the vizier,
who is getting a little hoarse, and wishes he didn't have this job.
QUESTION 2: What is suitor number 52's best strategy?
ANSWER 2: His best strategy is:
"Skip the first 5. Take the first girl in [6,14] who is better than
the first 5. If there isn't one, take the SECOND girl in [15,37]
who is the best up to that point. If there isn't one, take the THIRD
girl in [38,100] who is the best up to that point."
By the end of the day, of course, a wedding will be set, won't it?
MORE QUESTIONS: If each suitor uses the best strategy at that point, how
many suitors will it take before the right girl is certain to be found?
Does each succeeding suitor always have a better chance of winning
than the preceding one?
SPECULATION: The last strategy is "Pick the last girl." Its probability
of success is 1. And it is strategy number 100. (The corresponding
conditions hold for 3, 4, and 5 daughters.)
Does anyone have any observations on this one?
byron elbows
(mail to br...@cs.ucla.edu)
==> decision/envelope.p <==
Someone has prepared two envelopes containing money. One contains twice as
much money as the other. You have decided to pick one envelope, but then the
following argument occurs to you: Suppose my chosen envelope contains $X,
then the other envelope either contains $X/2 or $2X. Both cases are
equally likely, so my expectation if I take the other envelope is
.5 * $X/2 + .5 * $2X = $1.25X, which is higher than my current $X, so I
should change my mind and take the other envelope. But then I can apply the
argument all over again. Something is wrong here! Where did I go wrong?
In a variant of this problem, you are allowed to peek into the envelope
you chose before finally settling on it. Suppose that when you peek you
see $100. Should you switch now?
==> decision/envelope.s <==
Let's follow the argument carefully, substituting real numbers for
variables, to see where we went wrong. In the following, we will assume
the envelopes contain $100 and $200. We will consider the two equally
likely cases separately, then average the results.
First, take the case that X=$100.
"I have $100 in my hand. If I exchange I get $200. The value of the exchange
is $200. The value from not exchanging is $100. Therefore, I gain $100
by exchanging."
Second, take the case that X=$200.
"I have $200 in my hand. If I exchange I get $100. The value of the exchange
is $100. The value from not exchanging is $200. Therefore, I lose $100
by exchanging."
Now, averaging the two cases, I see that the expected gain is zero.
So where is the slip up? In one case, switching gets X/2 ($100), in the
other case, switching gets 2X ($200), but X is different in the two
cases, and I can't simply average the two different X's to get 1.25X.
I can average the two numbers ($100 and $200) to get $150, the expected
value of switching, which is also the expected value of not switching,
but I cannot under any circumstances average X/2 and 2X.
This is a classic case of confusing variables with constants.
OK, so let's consider the case in which I looked into the envelope and
found that it contained $100. This pins down what X is: a constant.
Now the argument is that the odds of $50 is .5 and the odds of $200
is .5, so the expected value of switching is $125, so we should switch.
However, the only way the odds of $50 could be .5 and the odds of $200
could be .5 is if all integer values are equally likely. But any
probability distribution that is finite and equal for all integers
would sum to infinity, not one as it must to be a probability distribution.
Thus, the assumption of equal likelihood for all integer values is
self-contradictory, and leads to the invalid proof that you should
always switch. This is reminiscent of the plethora of proofs that 0=1;
they always involve some illegitimate assumption, such as the validity
of division by zero.
Limiting the maximum value in the envelopes removes the self-contradiction
and the argument for switching. Let's see how this works.
Suppose all amounts up to $1 trillion were equally likely to be
found in the first envelope, and all amounts beyond that would never
appear. Then for small amounts one should indeed switch, but not for
amounts above $500 billion. The strategy of always switching would pay
off for most reasonable amounts but would lead to disastrous losses for
large amounts, and the two would balance each other out.
For those who would prefer to see this worked out in detail:
Assume the smaller envelope is uniform on [$0,$M], for some value
of $M. What is the expectation value of always switching? A quarter of
the time $100 >= $M (i.e. 50% chance $X is in [$M/2,$M] and 50% chance
the larger envelope is chosen). In this case the expected switching
gain is -$50 (a loss). Thus overall the always switch policy has an
expected (relative to $100) gain of (3/4)*$50 + (1/4)*(-$50) = $25.
However the expected absolute gain (in terms of M) is:
/ M
| g f(g) dg, [ where f(g) = (1/2)*Uniform[0,M)(g) +
/-M (1/2)*Uniform(-M,0](g). ]
= 0. QED.
OK, so always switching is not the optimal switching strategy. Surely
there must be some strategy that takes advantage of the fact that we
looked into the envelope and we know something we did not know before
we looked.
Well, if we know the maximum value $M that can be in the smaller envelope,
then the optimal decision criterion is to switch if $100 < $M, otherwise stick.
The reason for the stick case is straightforward. The reason for the
switch case is due to the pdf of the smaller envelope being twice as
high as that of the larger envelope over the range [0,$M). That is, the
expected gain in switching is (2/3)*$100 + (1/3)*(-$50) = $50.
What if we do not know the maximum value of the pdf? You can exploit
the "test value" technique to improve your chances. The trick here is
to pick a test value T. If the amount in the envelope is less than the
test value, switch; if it is more, do not. This works in that if T happens
to be in the range [M,2M] you will make the correct decision. Therefore,
assuming the unknown pdf is uniform on [0,M], you are slightly better off
with this technique.
Of course, the pdf may not even be uniform, so the "test value" technique
may not offer much of an advantage. If you are allowed to play the game
repeatedly, you can estimate the pdf, but that is another story...
==> decision/exchange.p <==
At one time, the Canadian and US dollars were discounted by 10 cents on
each side of the border (i.e., a Canadian dollar was worth 90 US cents
in the US, and a US dollar was worth 90 Canadian cents in Canada). A
man walks into a bar on the US side of the border, orders 10 US cents
worth of beer, pays with a US dollar and receives a Canadian dollar in
change. He then walks across the border to Canada, orders 10 Canadian
cents worth of beer, pays with a Canadian dollar and receives a US
dollar in change. He continues this throughout the day, and ends up
dead drunk with the original dollar in his pocket.
Who pays for the drinks?
==> decision/exchange.s <==
The man paid for all the drinks. But, you say, he ended up with the
same amount of money that he started with! However, as he transported
Canadian dollars into Canada and US dollars into the US, he performed
"economic work" by moving the currency to a location where it was in
greater demand (and thus valued higher). The earnings from this work
were spent on the drinks.
Note that he can only continue to do this until the Canadian bar runs
out of US dollars, or the US bar runs out of Canadian dollars, i.e.,
until he runs out of "work" to do.
==> decision/high.or.low.p <==
I pick two numbers, randomly, and tell you one of them. You are supposed
to guess whether this is the lower or higher one of the two numbers I
picked. Can you come up with a method of guessing that does better than
picking the response "low" or "high" randomly (i.e. probability to guess
right > .5) ?
==> decision/high.or.low.s <==
Pick any cumulative probability function P(x) such that a > b ==> P(a)
> P(b). Now if the number shown is y, guess "low" with probability
P(y) and "high" with probability 1-P(y). This strategy yields a
probability of > 1/2 of winning since the probability of being correct
is 1/2*( (1-P(a)) + P(b) ) = 1/2 + (P(b)-P(a)), which is > 1/2 by
assumption.
==> decision/monty.hall.p <==
You are a participant on "Let's Make a Deal." Monty Hall shows you
three closed doors. He tells you that two of the closed doors have a
goat behind them and that one of the doors has a new car behind it.
You pick one door, but before you open it, Monty opens one of the two
remaining doors and shows that it hides a goat. He then offers you a
chance to switch doors with the remaining closed door. Is it to your
advantage to do so?
==> decision/monty.hall.s <==
Under reasonable assumptions about Monty Hall's motivation, your chance
of picking the car doubles when you switch.
The problem is confusing for two reasons: first, there are hidden
assumptions about Monty's motivation that cloud the issue; and second,
novice probability students do not see that the opening of the door
gave them any new information.
Monty can have one of three basic motives:
1. He randomly opens doors.
2. He always opens the door he knows contains nothing.
3. He only opens a door when the contestant has picked the grand prize.
These result in very different strategies:
1. No improvement when switching.
2. Double your odds by switching.
3. Don't switch!
Most people, myself included, think that (2) is the intended
interpretation of Monty's motive. Interviews with Monty Hall
indicate that he did indeed try to lure the contestant who had picked
the car with cash incentives to switch. However, if Monty always
adopted this strategy, contestants would soon learn never to switch,
so one presumes that occasionally Monty offered another door even when
the contestant had picked a goat. At any rate, analyzing the problem
with this strategy is difficult, since it requires knowing something
about Monty's probability of bluffing.
A good way to see that Monty is giving you information by opening doors
that he knows are valueless is to increase the number of doors from
three to 100. If there are 100 doors, and Monty shows that 98 of them
are valueless, isn't it pretty clear that the chance the prize is behind
the remaining door is 99/100?
The original Monty Hall problem (and solution) appears to be due to
Steve Selvin, and appears in American Statistician, Feb 1975, V. 29,
No. 1, p. 67 under the title ``A Problem in Probability.'' It should
be of no surprise to readers of this group that he received several
letters contesting the accuracy of his solution, so he responded two
issues later (American Statistician, Aug 1975, V. 29, No. 3, p. 134).
However, the principles that underlie the problem date back at least to
the fifties, and probably are timeless. See the references below for
details.
Reference (too numerous to mention, but these contain bibliographies):
Leonard Gillman, "The Car and the Goats", AMM 99:1 (Jan 1992), p. 3
Ed Barbeau, "The Problem of the Car and Goats", CMJ 24:2 (Mar 1993), p. 149
The second reference contains a list of equivalent or related problems.
==> decision/newcomb.p <==
Newcomb's Problem
A being put one thousand dollars in box A and either zero or one million
dollars in box B and presents you with two choices:
(1) Open box B only.
(2) Open both box A and B.
The being put money in box B only if it predicted you will choose option (1).
The being put nothing in box B if it predicted you will do anything other than
choose option (1) (including choosing option (2), flipping a coin, etc.).
Assuming that you have never known the being to be wrong in predicting your
actions, which option should you choose to maximize the amount of money you
get?
==> decision/newcomb.s <==
This is "Newcomb's Paradox".
You are presented with two boxes: one certainly contains $1000 and the
other might contain $1 million. You can either take one box or both.
You cannot change what is in the boxes. Therefore, to maximize your
gain you should take both boxes.
However, it might be argued that you can change the probability that
the $1 million is there. Since there is no way to change whether the
million is in the box or not, what does it mean that you can change
the probability that the million is in the box? It means that your
choice is correlated with the state of the box.
Events which proceed from a common cause are correlated. My mental
states lead to my choice and, very probably, to the state of the box.
Therefore my choice and the state of the box are highly correlated.
In this sense, my choice changes the "probability" that the money is
in the box. However, since your choice cannot change the state of the
box, this correlation is irrelevant.
The following argument might be made: your expected gain if you take
both boxes is (nearly) $1000, whereas your expected gain if you take
one box is (nearly) $1 million, therefore you should take one box.
However, this argument is fallacious. In order to compute the
expected gain, one would use the formulas:
E(take one) = $0 * P(predict take both | take one) +
$1,000,000 * P(predict take one | take one)
E(take both) = $1,000 * P(predict take both | take both) +
$1,001,000 * P(predict take one | take both)
While you are given that P(do X | predict X) is high, it is not given
that P(predict X | do X) is high. Indeed, specifying that P(predict X
| do X) is high would be equivalent to specifying that the being could
use magic (or reverse causality) to fill the boxes. Therefore, the
expected gain from either action cannot be determined from the
information given.
==> decision/prisoners.p <==
Three prisoners on death row are told that one of them has been chosen
at random for execution the next day, but the other two are to be
freed. One privately begs the warden to at least tell him the name of
one other prisoner who will be freed. The warden relents: 'Susie will
go free.' Horrified, the first prisoner says that because he is now
one of only two remaining prisoners at risk, his chances of execution
have risen from one-third to one-half! Should the warden have kept his
mouth shut?
==> decision/prisoners.s <==
Each prisoner had an equal chance of being the one chosen to be
executed. So we have three cases:
Prisoner executed: A B C
Probability of this case: 1/3 1/3 1/3
Now, if A is to be executed, the warden will randomly choose either B or C,
and tell A that name. When B or C is the one to be executed, there is only
one prisoner other than A who will not be executed, and the warden will always
give that name. So now we have:
Prisoner executed: A A B C
Name given to A: B C C B
Probability: 1/6 1/6 1/3 1/3
We can calculate all this without knowing the warden's answer.
When he tells us B will not be executed, we eliminate the middle two
choices above. Now, among the two remaining cases, C is twice
as likely as A to be the one executed. Thus, the probability that
A will be executed is still 1/3, and C's chances are 2/3.
==> decision/red.p <==
I show you a shuffled deck of standard playing cards, one card at a
time. At any point before I run out of cards, you must say "RED!".
If the next card I show is red (i.e. diamonds or hearts), you win. We
assume I the "dealer" don't have any control over what the order of
cards is.
The question is, what's the best strategy, and what is your
probability of winning ?
==> decision/red.s <==
If a deck has n cards, r red and b black, the best strategy wins
with a probability of r/n. Thus, you can say "red" on the first card,
the last card, or any other card you wish.
Proof by induction on n. The statement is clearly true for one-card decks.
Suppose it is true for n-card decks, and add a red card.
I will even allow a nondeterministic strategy, meaning you say "red"
on the first card with probability p. With probability 1-p,
you watch the first card go by, and then apply the "optimal" strategy
to the remaining n-card deck, since you now know its composition.
The odds of winning are therefore: p * (r+1)/(n+1) +
(1-p) * ((r+1)/(n+1) * r/n + b/(n+1) * (r+1)/n).
After some algebra, this becomes (r+1)/(n+1) as expected.
Adding a black card yields: p * r/(n+1) +
(1-p) * (r/(n+1) * (r-1)/n + (b+1)/(n+1) * r/n).
This becomes r/(n+1) as expected.
==> decision/rotating.table.p <==
Four glasses are placed upside down in the four corners of a square
rotating table. You wish to turn them all in the same direction,
either all up or all down. You may do so by grasping any two glasses
and, optionally, turning either over. There are two catches: you are
blindfolded and the table is spun after each time you touch the
glasses. Assuming that a bell rings when you have all the glasses up,
how do you do it?
==> decision/rotating.table.s <==
1. Turn two adjacent glasses up.
2. Turn two diagonal glasses up.
3. Pull out two diagonal glasses. If one is down, turn it up and you're done.
If not, turn one down and replace.
4. Take two adjacent glasses. Invert them both.
5. Take two diagonal glasses. Invert them both.
References
"Probing the Rotating Table"
W. T. Laaser and L. Ramshaw
_The Mathematical Gardner_,
Wadsworth International, Belmont CA 1981.
... we will see that such a procedure exists if and
only if the parameters k and n satisfy the inequality
k >= (1-1/p)n, where p is the largest prime factor
of n.
The paper mentions (without discussing) two other generalizations:
more than two orientations of the glasses (Graham and Diaconis)
and more symmetries in the table, e.g. those of a cube (Kim).
==> decision/stpetersburg.p <==
What should you be willing to pay to play a game in which the payoff is
calculated as follows: a coin is flipped until it comes up heads on the
nth toss and the payoff is set at 2^n dollars?
==> decision/stpetersburg.s <==
Classical decision theory says that you should be willing to pay any
amount up to the expected value of the wager. Let's calculate the
expected value: The probability of winning at step n is 2^-n, and the
payoff at step n is 2^n, so the sum of the products of the
probabilities and the payoffs is:
E = sum over n (2^-n * 2^n) = sum over n (1) = infinity
So you should be willing to pay any amount to play this game. This is
called the "St. Petersburg Paradox."
The classical solution to this problem was given by Bernoulli. He
noted that people's desire for money is not linear in the amount of
money involved. In other words, people do not desire $2 million twice
as much as they desire $1 million. Suppose, for example, that people's
desire for money is a logarithmic function of the amount of money.
Then the expected VALUE of the game is:
E = sum over n (2^-n * C * log(2^n)) = sum over n (2^-n * C' * n) = C''
Here the C's are constants that depend upon the risk aversion of the
player, but at least the expected value is finite. However, it turns
out that these constants are usually much higher than people are really
willing to pay to play, and in fact it can be shown that any
non-bounded utility function (map from amount of money to value of
money) is prey to a generalization of the St. Petersburg paradox. So
the classical solution of Bernoulli is only part of the story.
The rest of the story lies in the observation that bankrolls are always
finite, and this dramatically reduces the amount you are willing to bet
in the St. Petersburg game.
To figure out what would be a fair value to charge for playing the game
we must know the bank's resources. Assume that the bank has 1 million
dollars (1*K*K = 2^20). I cannot possibly win more than $1 million
whether I toss 20 tails in a row or 2000.
Therefore my expected amount of winning is
E = sum n up to 20 (2^-n * 2^n) = sum n up to 20 (1) = $20
and my expected value of winning is
E = sum n up to 20 (2^-n * C * log(2^n)) = some small number
This is much more in keeping with what people would really pay to
play the game.
Incidentally, T.C. Fry suggested this change to the problem in 1928
(see W.W.R. Ball, Mathematical Recreations and Essays, N.Y.: Macmillan,
1960, pp. 44-45).
The problem remains interesting when modified in this way,
for the following reason. For a particular value of the bank's
resources, let
e denote the expected value of the player's winnings; and let
p denote the probability that the player profits from the game, assuming
the price of getting into the game is 0.8e (20% discount).
Note that the expected value of the player's profit is 0.2e. Now
let's vary the bank's resources and observe how e and p change. It
will be seen that as e (and hence the expected value of the profit)
increases, p diminishes. The more the game is to the player's
advantage in terms of expected value of profit, the less likely it is
that the player will come away with any profit at all. This
is mildly counterintuitive.
==> decision/truel.p <==
A, B, and C are to fight a three-cornered pistol duel. All know that
A's chance of hitting his target is 0.3, C's is 0.5, and B never misses.
They are to fire at their choice of target in succession in the order
A, B, C, cyclically (but a hit man loses further turns and is no longer
shot at) until only one man is left. What should A's strategy be?
==> decision/truel.s <==
This is problem 20 in Mosteller _Fifty Challenging Problems in Probability_
and it also appears (with an almost identical solution) on page 82 in
Larsen & Marx _An Introduction to Probability and Its Applications_.
Here's Mosteller's solution:
A is naturally not feeling cheery about this enterprise. Having the
first shot he sees that, if he hits C, B will then surely hit him, and
so he is not going to shoot at C. If he shoots at B and misses him,
then B clearly {I disagree; this is not at all clear!} shoots the more
dangerous C first, and A gets one shot at B with probability 0.3 of
succeeding. If he misses this time, the less said the better. On the
other hand, suppose A hits B. Then C and A shoot alternately until one
hits. A's chance of winning is (.5)(.3) + (.5)^2(.7)(.3) +
(.5)^3(.7)^2(.3) + ... . Each term corresponds to a sequence of misses
by both C and A ending with a final hit by A. Summing the geometric
series we get ... 3/13 < 3/10. Thus hitting B and finishing off with
C has less probability of winning for A than just missing the first shot.
So A fires his first shot into the ground and then tries to hit B with
his next shot. C is out of luck.
As much as I respect Mosteller, I have some serious problems with this
solution. If we allow the option of firing into the ground, then if
all fire into the ground with every shot, each will survive with
probability 1. Now, the argument could be made that a certain
strategy for X that both allows them to survive with probability 1
*and* gives less than a probability of survival of less than 1 for
at least one of their foes would be preferred by X. However, if
X pulls the trigger and actually hits someone what would the remaining
person, say Y, do? If P(X hits)=1, clearly Y must try to hit X, since
X firing at Y with intent to hit dominates any other strategy for X.
If P(X hits)<1 and X fires at Y with intent to hit, then
P(Y survives)<1 (since X could have hit Y). Thus, Y must insure that
X can not follow this strategy by shooting back at X (thus insuring
that P(X survives)<1). Therefore, I would conclude that the ideal
strategy for all three players, assuming that they are rational and
value survival above killing their enemies, would be to keep firing
into the ground. If they don't value survival above killing their
enemies (which is the only a priori assumption that I feel can be
safely made in the absence of more information), then the problem
can't be solved unless the function each player is trying to maximize
is explicitly given.
--
-- cl...@remus.rutgers.edu (Chris Long)
OK - I'll have a go at this.
How about the payoff function being 1 if you win the "duel" (i.e. if at some
point you are still standing and both the others have been shot) and 0
otherwise? This should ensure that an infinite sequence of deliberate misses
is not to anyone's advantage. Furthermore, I don't think simple survival
makes a realistic payoff function, since people with such a payoff function
would not get involved in the fight in the first place!
[ I.e. I am presupposing a form of irrationality on the part of the
fighters: they're only interested in survival if they win the duel. Come
to think of it, this may be quite rational - spending the rest of my life
firing a gun into the ground would be a very unattractive proposition to
me :-)
]
Now, denote each position in the game by the list of people left standing,
in the order in which they get their turns (so the initial position is
(A,B,C), and the position after A misses the first shot (B,C,A)). We need to
know the value of each possible position for each person.
By definition:
valA(A) = 1 valB(A) = 0 valC(A) = 0
valA(B) = 0 valB(B) = 1 valC(B) = 0
valA(C) = 0 valB(C) = 0 valC(C) = 1
Consider the two player position (X,Y). An infinite sequence of misses has
value zero to both players, and each player can ensure a positive payoff by
trying to shoot the other player. So both players deliberately missing is a
sub-optimal result for both players. The question is then whether both
players should try to shoot the other first, or whether one should let the
other take the first shot. Since having the first shot is always an
advantage, given that some real shots are going to be fired, both players
should try to shoot the other first. It is then easy to establish that:
valA(A,B) = 3/10 valB(A,B) = 7/10 valC(A,B) = 0
valA(B,A) = 0 valB(B,A) = 1 valC(B,A) = 0
valA(B,C) = 0 valB(B,C) = 1 valC(B,C) = 0
valA(C,B) = 0 valB(C,B) = 5/10 valC(C,B) = 5/10
valA(C,A) = 3/13 valB(C,A) = 0 valC(C,A) = 10/13
valA(A,C) = 6/13 valB(A,C) = 0 valC(A,C) = 7/13
Now for the three player positions (A,B,C), (B,C,A) and (C,A,B). Again, the
fact that an infinite sequence of misses is sub-optimal for all three
players means that at least one player is going to decide to fire. However,
it is less clear than in the 2 player case that any particular player is
going to fire. In the 2 player case, each player knew that *if* it was
sub-optimal for him to fire, then it was optimal for the other player to
fire *at him* and that he would be at a disadvantage in the ensuing duel
because of not having got the first shot. This is not necessarily true in
the 3 player case.
Consider the payoff to A in the position (A,B,C). If he shoots at B, his
expected payoff is:
0.3*valA(C,A) + 0.7*valA(B,C,A) = 9/130 + 0.7*valA(B,C,A)
If he shoots at C, his expected payoff is:
0.3*valA(B,A) + 0.7*valA(B,C,A) = 0.7*valA(B,C,A)
And if he deliberately misses, his expected payoff is:
valA(B,C,A)
Since he tries to maximise his payoff, we can immediately eliminate shooting
at C as a strategy - it is strictly dominated by shooting at B. So A's
expected payoff is:
valA(A,B,C) = MAX(valA(B,C,A), 9/130 + 0.7*valA(B,C,A))
A similar argument shows that C's expected payoffs in the (C,A,B) position are:
For shooting at A: 0.5*valC(A,B,C)
For shooting at B: 35/130 + 0.5*valC(A,B,C)
For missing: valC(A,B,C)
So C either shoots at B or deliberately misses, and:
valC(C,A,B) = MAX(valC(A,B,C), 35/130 + 0.5*valC(A,B,C))
Each player can obtain a positive expected payoff by shooting at one of the
other players, and it is known that an infinite sequence of misses will
result in a zero payoff for all players. So it is known that some player's
strategy must involve shooting at another player rather than deliberately
missing.
Now look at this from the point of view of player B. He knows that *if* it
is sub-optimal for him to shoot at another player, then it is optimal for at
least one of the other players to shoot. He also knows that if the other
players choose to shoot, they will shoot *at him*. If he deliberately
misses, therefore, the best that he can hope for is that they miss him and
he is presented with the same situation again. This is clearly less good for
him than getting his shot in first. So in position (B,C,A), he must shoot at
another player rather than deliberately miss.
B's expected payoffs are:
For shooting at A: valB(C,B) = 5/10
For shooting at C: valB(A,B) = 7/10
So in position (B,C,A), B shoots at C for an expected payoff of 7/10. This
gives us:
valA(B,C,A) = 3/10 valB(B,C,A) = 7/10 valC(B,C,A) = 0
So valA(A,B,C) = MAX(3/10, 9/130 + 21/100) = 3/10, and A's best strategy is
position (A,B,C) is to deliberately miss, giving us:
valA(A,B,C) = 3/10 valB(A,B,C) = 7/10 valC(A,B,C) = 0
And finally, valC(C,A,B) = MAX(0, 35/130 + 0) = 7/26, and C's best strategy
in position (C,A,B) is to shoot at B, giving us:
valA(C,A,B) = 57/260 valB(C,A,B) = 133/260 valC(C,A,B) = 7/26
I suspect that, with this payoff function, all positions with 3 players can
be resolved. For each player, we can establish that if their correct
strategy is to fire at another player, then it is to fire at whichever of
the other players is more dangerous. The most dangerous of the three players
then finds that he has nothing to lose by firing at the second most
dangerous.
Questions:
(a) In the general case, what are the optimal strategies for the other two
players, possibly as functions of the hit probabilities and the cyclic
order of the three players?
(b) What happens in the 4 or more player case?
-- David Seal <ds...@armltd.co.uk>
In article <1993Mar25.0...@cs.cornell.edu>, ka...@cs.cornell.edu (David Karr) writes:
> The Good, the Bad, and the Ugly are standing at three equidistant
"P" "Q" "R" -- allow me these alternate names.
> points around a very large circle, about to fight a three-way duel to
> see who gets the treasure. They all know that the Good hits with
> probability p=.9, the Bad hits with probability q=.7, and the Ugly
> hits with probability r=.5.
>
> Yes, I know this sounds like decision/truel from the rec.puzzles
> archive. But here's the difference:
>
> At some instant, all three will fire simultaneously, each at a target
> of his choice. Then any who survive that round fire simultaneously
> again, until at most one remains. Note that there are then four
> possible outcomes: the Good wins, the Bad wins, the Ugly wins, or all
> are killed.
A multi-round multi-person game can get complicated if implicit
alliances are formed or the players deduce each other's strategies.
For simplicity let's disallow communication and assume the players
forget who shot at whom after each round.
>
> Now the questions:
These are not easy questions, even with the simplifying
assumptions I've made.
>
> 1. What is each shooter's strategy?
Each player has two possible strategies so there are eight cases
to consider; unfortunately none of the players has a strictly
dominant strategy:
P aims at Q aims at R aims at P survival Q survival R survival Noone lives
--------- --------- --------- ---------- ---------- ---------- -----------
Q P P 0.0649 0.0355 0.7991 0.1005
Q P Q 0.1371 0.0146 0.6966 0.1517
* Q R P 0.3946 0.0444 0.1470 0.4140
Q R Q 0.8221 0.0026 0.0152 0.1601
R P P 0.0381 0.8221 0.0152 0.1246
* R P Q 0.1824 0.3443 0.0426 0.4307
R R P 0.1371 0.5342 0.0027 0.3260
R R Q 0.6367 0.0355 0.0008 0.3270
(The similarity of, say, the 4th and 5th lines here looks wrong:
the intermediate expressions are quite different. I can't
explain *why* P_survival(q,r,q) = Q_survival(r,p,p) = 0.8221
but I *have* double-checked this result.)
If I *know* who my opponents are going to aim at, I should shoot
at the better shooter if they're both aiming at me or neither is
aiming at me. Otherwise I should aim at whoever is *not* aiming
at me. There are two equilibrium points (marked "*" above):
Good aims at Bad; Bad aims at Ugly; Ugly aims at Good.
and
Good aims at Ugly; Bad aims at Good; Ugly aims at Bad.
Here, unlike for zero-sum two-person games, the equilibria
are *not* equivalent and "solution", if any, may lie elsewhere.
Perhaps a game-theorist lurking in r.p can offer a better comment.
Note that the probability all three shooters die is highest at
the equilibria! This seems rather paradoxical and rather sad :-(
>
> 2. Who is most likely to survive?
Good, Bad, or Ugly, depending on the strategies.
>
> 3. Who is least likely to survive?
>
Bad or Ugly, depending on the strategies.
> 4. Can you change p, q, and r under the constraint p > q > r so that
> the answers to questions 2 and 3 are reversed? Which of the six
> possible permutations of the three shooters is a possible ordering
> of probability of survival under the constraint p > q > r?
Yes. Of the six possible survival-probability orderings,
five can be obtained readily:
p q r P_surv Q_surv R_surv Order
--- --- --- ------ ------ ------ -------
0.255 0.25 0.01 0.408 0.413 0.172 Q P R
0.26 0.25 0.01 0.412 0.406 0.173 P Q R
0.75 0.25 0.01 0.675 0.076 0.242 P R Q
0.505 0.50 0.01 0.325 0.324 0.344 R P Q
0.505 0.50 0.02 0.314 0.320 0.353 R Q P
Unlike the p=.9, q=.7, r=.5 case we are given, the five cases
in this table *do* have simple pure solutions: in each case
p shoots at q, while q and r each shoot at p. (I've found no
case with a "simple pure" solution other than this "obvious"
p aims at q, q aims at p, r aims at p choice.)
>
> 5. Are there any value of p, q, and r for which it is ever in the
> interest of one of the shooters to fire into the ground?
No. It can't hurt to shoot at one's stronger opponent.
This is the easiest of the questions ... but it's still
not easy enough for me to construct an elegant proof
in English.
>
> -- David Karr (ka...@cs.cornell.edu)
>
Speaking of decision/truel, I recall a *very* interesting
analysis (I *might* have seen it here in rec.puzzles) suggesting
that the N-person "truel" (N-uel?) has a Cooperative Solution
(ceasefire) if and only if N = 3. But I don't see this in the
FAQL; anyone care to repost it?
-- James Allen
In article <1993Apr1.1...@vax5.cit.cornell.edu> m...@vax5.cit.cornell.edu writes:
>In article <1993Mar25.0...@cs.cornell.edu>, ka...@cs.cornell.edu (David Karr) writes:
[...]
>> 5. Are there any value of p, q, and r for which it is ever in the
>> interest of one of the shooters to fire into the ground?
>>
> Yes, p=1, q=1, r=1. The only way for one to survive is to have the other
> two shoot at eachother. Shooting at anyone has no effect on ones personal
> survival.
I assume by "has no effect on" you mean "does not improve."
> If all follow the same logic, they will keep shooting into the
> ground and thus all live.
I was very pleased by this answer but I had to think about it. First
of all, it assumes that continuing the fight forever has a positive
value for each shooter. My preferred assumption is that it doesn't.
But even if each shooter is simply trying to maximize his probability
of never being shot, I wonder about the "has no effect" statement.
Suppose that in round 1 the Good fires into the ground and the Bad
shoots at the Good. Then the Ugly lives if he shoots the Bad and dies
if he does anything else. (The Bad will surely shoot at the Ugly if
he can in round 2, since this dominates any other strategy.) So it
definitely makes a difference to the Ugly in this case to shoot at the
Bad.
But all this is under the assumption that no shooter can tell what
the others are about to do until after all have shot. This isn't
entirely unreasonable--we can certainly set up a game that plays
this way--but suppose we assume instead:
All three start out with guns initially holstered.
Each one is a blindingly fast shot: he can grab his holstered gun,
aim, and fire in 0.6 second.
A shooter can redirect his unholstered gun at a different target and
fire in just 0.4 second.
The reaction time of each shooter is just 0.2 second. That is, any
decision he makes to act can be based only on the actions of the
other two up to 0.2 second before he initiates his own action.
The bullets travel between shooters in less than 0.1 second and
stop any further action when they hit.
Then I *think* the conclusion holds for p=q=r=1: The best strategy is
to wait for someone else to grab for their gun, then shoot that
person, therefore nobody will shoot at anyone. At least I haven't yet
thought of a case in which you improve your survival by shooting at
anyone. Of course this is only good if you don't mind waiting around
the circle forever.
-- David Karr (ka...@cs.cornell.edu)
In article <1993Apr5.2...@cs.cornell.edu>,
ka...@cs.cornell.edu (David Karr) writes:
> In article <1993Apr1.1...@vax5.cit.cornell.edu> m...@vax5.cit.cornell.edu writes:
>>In article <1993Mar25.0...@cs.cornell.edu>, ka...@cs.cornell.edu (David Karr) writes:
> [...]
>>> 5. Are there any value of p, q, and r for which it is ever in the
>>> interest of one of the shooters to fire into the ground?
>>>
>> Yes, p=1, q=1, r=1. The only way for one to survive is to have the other
>> two shoot at eachother. Shooting at anyone has no effect on ones personal
>> survival.
>
> I assume by "has no effect on" you mean "does not improve."
>
>> If all follow the same logic, they will keep shooting into the
>> ground and thus all live.
>
> I was very pleased by this answer but I had to think about it. First
> of all, it assumes that continuing the fight forever has a positive
> value for each shooter. My preferred assumption is that it doesn't.
> But even if each shooter is simply trying to maximize his probability
> of never being shot, I wonder about the "has no effect" statement.
>
> Suppose that in round 1 the Good fires into the ground and the Bad
> shoots at the Good. Then the Ugly lives if he shoots the Bad and dies
> if he does anything else. (The Bad will surely shoot at the Ugly if
> he can in round 2, since this dominates any other strategy.) So it
> definitely makes a difference to the Ugly in this case to shoot at the
> Bad.
>
Here's where the clincher comes in! If we "assume" the object of the game
is to survive, and that there exists _one_ unique method for survival, then
all the shooters will behave in the same fashion. Obviously the above case
will not hold. How do we distinguish between the good, the bad and the ugly?
If the command is "Shoot" then all will shoot and somebody is going to wind up
lucky (Prob that it is you is 1/3). If the command is "No Shoot", then all
will fire into the ground (or just give up and go home--no sense waitin' around
wastin' time, ya know)...
But wait, you cry! What if there exists _more than one_ solution for optimal
survival. Then what? Will the Good the Bad and the Ugly each randomly decide
between "Shoot" and "No Shoot" with .5 probability? If this is true, then is
it in your best interest to shoot someone? If it is, then we arrive back at
square one: since we assume all shooters are geniouses, then all will shoot--
arriving at an optimal solution of "Shooting". If the answer is "No Shooting",
we arrive at an optimal solution of "No Shooting". If there is no effect on
your personal survival, then do we analyze this with another .5 probability
between the chances of soemone shooting or not shooting? If the answer to this
is "Shoot" then we arrive at square one: all will Shoot; if no, then all will
withold. If there is no effect, then we arrive at another .5 probability...
Obviously you can see the recursion of this process.
Perhaps this would be easier to discuss if we let p=1, q=1, r=0. Obviously, in
terms of survival, shooting at the ugly would be wasting a shot. Thus we have
made a complex problem more simple but retaining the essence of the paradox:
If there are two gunmen who shoot and think with perfect logic and are kept
inside a room and are allowed to shoot at discrete time intervals without
being able to "see" what your opponent will do, what will happen?
Let's say that you are one of the gunmen (the Good). You reason "My probability
to survive the next round is independent on whether or not I fire at him." So
you say to yourself, "Fire at the opponent! I'll get to stop playing this
blasted game." But then you realize that your opponent will also think the same
way...so you might think that you might as well not shoot. But if your
opponent thinks that way, then you know that: 1. You can survive the next
round. 2. You can shoot him if you wish on this round (if you like). So you
say to yourself, "Fire at the opponent!". But you know the opponent thinks the
same way so... you're dead. But really, you say. Is there a way of "knowing"
what the opponent thinks? Of course not. You reason that you can know your
probability of shooting your opponent (either 1 or 0). You reason that your
opponent has a variable probability of shooting you. Thus from your
perspective, p=1 and r<1. We already discussed this case and said "Shoot!".
But wait you cry! What if the opponent figures this out too: p<1, r=1? Sorry,
you're both dead. 'nuff said! This applies to the p=r=q=1 case as well.
> But all this is under the assumption that no shooter can tell what
> the others are about to do until after all have shot.
Ay, there's the rub!
>This isn't entirely unreasonable--we can certainly set up a game that plays
> this way--but suppose we assume instead:
>
> All three start out with guns initially holstered.
> Each one is a blindingly fast shot: he can grab his holstered gun,
> aim, and fire in 0.6 second.
> A shooter can redirect his unholstered gun at a different target and
> fire in just 0.4 second.
> The reaction time of each shooter is just 0.2 second. That is, any
> decision he makes to act can be based only on the actions of the
> other two up to 0.2 second before he initiates his own action.
> The bullets travel between shooters in less than 0.1 second and
> stop any further action when they hit.
>
> Then I *think* the conclusion holds for p=q=r=1: The best strategy is
> to wait for someone else to grab for their gun, then shoot that
> person, therefore nobody will shoot at anyone. At least I haven't yet
> thought of a case in which you improve your survival by shooting at
> anyone. Of course this is only good if you don't mind waiting around
> the circle forever.
Hmmn...alternate ploy:
0.0 You begin to unholster your gun
0.2 Opponents begin unholstering guns. You aim into the ground for .2 sec.
0.4 Opponents are unholstered you are unholstered. They note you aren't
aiming at them. They haven't aimed at anyone yet.
What happens now? I'll have to think about it, but I haven't seen anything
fundamentally different between this and the above case yet.
More ideas to consider:
You begin unholstering your gun but only for .1 sec (you place it by .2 )
You begin unholstering your gun but only for .09 sec (you place it by .19)
You start to aim for .1 sec and then stop aiming.
You start to aim for .1 sec and then turn and aim at another.
You start to aim for .09 sec and then stop aiming (or aim at another)
-Greg
Looking at the answer for decision/truel, I came across the following:
>Each player can obtain a positive expected payoff by shooting at one of the
>other players, and it is known that an infinite sequence of misses will
>result in a zero payoff for all players. So it is known that some player's
>strategy must involve shooting at another player rather than deliberately
>missing.
This may be true but it's not obvious to me. For example, suppose A, B,
and C are passengers in a lifeboat in a storm. If they all stay aboard,
the lifeboat is certain to sink eventually, taking all three to the
bottom with it. If anyone jumps overboard, the two remaining in the
boat are guaranteed to survive, while the person who jumped has a 1%
chance of survival.
It seems to me the lifeboat satisfies the quoted conditions, in the
sense that if nobody jumps then the payoff for all is zero, and the
payoff for jumping is 0.01 which is positive. But it is not clear to
me that the three shouldn't just all sit still until someone goes nuts
and jumps overboard despite everything, for this strategy gives a 67%
chance of survival (assuming everyone is equally likely to "crack"
first) vs. only 1% for jumping by choice. Even if there is a wave
about to swamp the boat, I'd wonder if the situation wouldn't just
reduce to a game of "chicken," with each person waiting until the last
minute and jumping only if it seems the other two have decided to sink
with the boat if you don't jump.
On the other hand, this situation is set up so it is always worse to
be the first person to jump. In the truel I don't think this is true,
but only because of the asymmetry of the odds, and to determine
whether *anyone* shoots, it is easiest to proceed directly to
considering B's point of view.
Whenever it is B's turn to shoot, B can divide the possible courses of
action into four possibilities (actually there are seven, but three
are ruled out a priori by obvious optimizations of each individual's
strategy):
Nobody ever shoots (expected value 0)
A shoots first (at B, expected value <= .7)
C shoots first (at B, expected value <= .5)
B shoots first (at C, expected value .7)
In fact the value of "A shoots first" is strictly less than .7 because
in case A misses, the same four possibilities recur, and all have
expected payoff < 1.
So the value of "B shoots first" uniquely maximizes B's value function,
ergo B will always shoot as soon as possible.
The rest of the analysis then follows as in the archive.
-- David Karr (ka...@cs.cornell.edu)
> Looking at the answer for decision/truel, I came across the following:
>
> >Each player can obtain a positive expected payoff by shooting at one of the
> >other players, and it is known that an infinite sequence of misses will
> >result in a zero payoff for all players. So it is known that some player's
> >strategy must involve shooting at another player rather than deliberately
> >missing.
>
> This may be true but it's not obvious to me. For example, suppose A, B,
> and C are passengers in a lifeboat in a storm. If they all stay aboard,
> the lifeboat is certain to sink eventually, taking all three to the
> bottom with it. If anyone jumps overboard, the two remaining in the
> boat are guaranteed to survive, while the person who jumped has a 1%
> chance of survival.
>
> It seems to me the lifeboat satisfies the quoted conditions, in the
> sense that if nobody jumps then the payoff for all is zero, and the
> payoff for jumping is 0.01 which is positive. But it is not clear to
> me that the three shouldn't just all sit still until someone goes nuts
> and jumps overboard despite everything, for this strategy gives a 67%
> chance of survival (assuming everyone is equally likely to "crack"
> first) vs. only 1% for jumping by choice. ...
Yes and no. Yes in the sense that if you treat the game as a psychological
one, the best strategy is as you say. But treating it as a mathematical
game, you've got to adhere to your strategy and you've got to assume that
others will adhere to theirs.
I.e. as a mathematical game, "Don't jump at all" and "Don't jump unless I
crack" are different strategies, and the first one is often (not always)
superior - e.g. if I take "Don't jump at all" and the others take "Don't
jump unless I crack", I'm certain to survive and the others each have a
50.5% chance, which is better from my point of view than a 67% chance of
survival for all of us. As a psychological game, some of the mathematical
strategies may simply not be available - i.e. you cannot control what you
will do if you crack, and so we commonly use "Don't jump" to mean "Don't
jump unless I crack", since "Don't jump at all" is not an available strategy
for most real humans. But for mathematical analysis, the problem has to tell
you what strategies you are not allowed to take.
What the argument above shows is that "Don't jump at all" is not a stable
strategy, in the sense that if everyone takes it, it is in everyone's
interest to change strategy. I.e. it shows that someone will jump
eventually, even if it's only the result of someone actually having taken
"Don't jump unless I crack".
Applied to the truel, the argument above *does* show that someone's strategy
will involve shooting at another player: the strategy "Don't shoot at all"
is unstable in exactly the same way as "Don't jump at all" was. But I agree
it allows for a lot of leeway about how and when the deadlock gets broken,
and your argument showing that it is always in B's interest to shoot is more
satisfactory.
David Seal
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> language/english/spelling/operations.on.words/transposition.p <==
What exceptional words turn into other words by transposition of letters?
==> language/english/spelling/operations.on.words/transposition.s <==
longest reversal DESSERTS,STRESSED (8)
longest well-mixed transposal
CINEMATOGRAPHER, MEGACHIROPTERAN (15)
longest transposition list
APERS, APRES, ASPER, PARES, PARSE, PEARS, PRASE, PRESA, RAPES, REAPS, SPARE,
SPEAR (12)
ANGRIEST, ANGRITES, ASTRINGE, GAIRTENS, GANISTER, GANTRIES, GRANITES,
INGRATES, RANGIEST, TEARINGS (10) [SATING(ER), SIGNATE(R), TANGIER(S) (3)]
ANORETICS, ATROSCINE, CANOTIERS, CERTOSINA, CONARITES, CREATIONS, REACTIONS,
TRICOSANE (8)
transposition with deletion, insertion, or substitution
longest well-mixed transdeletion
SONOLUMINESCENCES, UNECONOMICALNESSES (17/18)
longest word transdeletable to a single letter
CONCENTRATIONS-CONSTERNATION-CONTORNIATES-TRANSECTION-
STENTORIAN-TRANSIENT-ENTRAINS-NASTIER-ASTERN-TEARS-SATE-TEA-AT-A (14)
longest Baltimore transdeletion (word transdeletable on every letter)
IDOLATERS: DELATORS, SOTERIAL, DILATERS, ASTEROID,
STOLIDER, SOREDIAL, DILATORS, DIASTOLE, TAILORED (9)
shortest word that cannot be transadded to another word SYZYGY (6)
longest well-mixed transubstitution
MICROELECTROPHORESIS, SPECTROCOLORIMETRIES (20)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/spelling/operations.on.words/words.within.words.p <==
What exceptional words contain other words?
==> language/english/spelling/operations.on.words/words.within.words.s <==
longest non-trivial charade IN-DISC-RIM-IN-A-TI-ON (16)
longest forward and reverse charade
MAT-HE-MA-TI-CAL, LAC-IT-AM-EH-TAM
longest snowball or rhopalic T-EM-PER-AMEN-TALLY (15)
longest reverse rhopalic HETERO-TRANS-PLAN-TAT-IO-N (21)
highest ratio of subwords/length (logogram)
FIRESTONE: RE, TO, ON, NO, IF, FIR, IRE, RES, TON, ONE, NOT, RIF, FIRE,
IRES, REST, TONE, FIRES, STONE, SERIF (20/9)
longest charlinkade FORESTALL: FOREST, ALL; FORE, REST, TALL (9)
longest alternade TRIENNIALLY: TINILY, RENAL (11)
shortest three-letter-minimum word deletions
PILGRIMAGE: RIM, GAG, PILE; GRIM, LAG, PIE
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/spelling/palindromes.p <==
What are some long palindromes?
==> language/english/spelling/palindromes.s <==
The first words spoken were a palindrome:
Madam, I'm Adam.
or perhaps:
Madam in Eden, I'm Adam.
The response, of course, must have been:
Eve
Napolean's lament:
Able was I ere I saw Elba.
Has been improved with:
Unremarkable was I ere I saw Elba, Kramer, nu?
A fish is a:
laminar animal
The timeless classic:
A man, a plan, a canal; Panama?
Has been improved by:
A dog, a plan, a canal: pagoda!
-- Roger Angell
A man, a plan, a cat, a canal; Panama?
-- Jim Saxe, plan file @ CMU, 9 October 1983
A man, a plan, a cat, a ham, a yak, a yam, a hat, a canal--Panama!
-- Guy Jacobson, plan file @ CMU late 1983
A man, a plan, a caret, a ban, a myriad, a sum, a lac, a liar, a hoop, a
pint, a catalpa, a gas, an oil, a bird, a yell, a vat, a caw, a pax, a wag,
a tax, a nay, a ram, a cap, a yam, a gay, a tsar, a wall, a car, a luger, a
ward, a bin, a woman, a vassal, a wolf, a tuna, a nit, a pall, a fret, a
watt, a bay, a daub, a tan, a cab, a datum, a gall, a hat, a fag, a zap, a
say, a jaw, a lay, a wet, a gallop, a tug, a trot, a trap, a tram, a torr, a
caper, a top, a tonk, a toll, a ball, a fair, a sax, a minim, a tenor, a
bass, a passer, a capital, a rut, an amen, a ted, a cabal, a tang, a sun, an
ass, a maw, a sag, a jam, a dam, a sub, a salt, an axon, a sail, an ad, a
wadi, a radian, a room, a rood, a rip, a tad, a pariah, a revel, a reel, a
reed, a pool, a plug, a pin, a peek, a parabola, a dog, a pat, a cud, a nu,
a fan, a pal, a rum, a nod, an eta, a lag, an eel, a batik, a mug, a mot, a
nap, a maxim, a mood, a leek, a grub, a gob, a gel, a drab, a citadel, a
total, a cedar, a tap, a gag, a rat, a manor, a bar, a gal, a cola, a pap, a
yaw, a tab, a raj, a gab, a nag, a pagan, a bag, a jar, a bat, a way, a
papa, a local, a gar, a baron, a mat, a rag, a gap, a tar, a decal, a tot, a
led, a tic, a bard, a leg, a bog, a burg, a keel, a doom, a mix, a map, an
atom, a gum, a kit, a baleen, a gala, a ten, a don, a mural, a pan, a faun,
a ducat, a pagoda, a lob, a rap, a keep, a nip, a gulp, a loop, a deer, a
leer, a lever, a hair, a pad, a tapir, a door, a moor, an aid, a raid, a
wad, an alias, an ox, an atlas, a bus, a madam, a jag, a saw, a mass, an
anus, a gnat, a lab, a cadet, an em, a natural, a tip, a caress, a pass, a
baronet, a minimax, a sari, a fall, a ballot, a knot, a pot, a rep, a
carrot, a mart, a part, a tort, a gut, a poll, a gateway, a law, a jay, a
sap, a zag, a fat, a hall, a gamut, a dab, a can, a tabu, a day, a batt, a
waterfall, a patina, a nut, a flow, a lass, a van, a mow, a nib, a draw, a
regular, a call, a war, a stay, a gam, a yap, a cam, a ray, an ax, a tag, a
wax, a paw, a cat, a valley, a drib, a lion, a saga, a plat, a catnip, a
pooh, a rail, a calamus, a dairyman, a bater, a canal--Panama.
--Dan Hoey, 'discovered' in 1984.
Dan goes on to say "...a little work on the search algorithm could make
it several times as long."
Those wonderful proper names:
Dennis, Nell, Edna, Leon, Nedra, Anita, Rolf, Nora, Alice, Carol, Leo,
Jane, Reed, Dena, Dale, Basil, Rae, Penny, Lana, Dave, Denny, Lena,
Ida, Bernadette, Ben, Ray, Lila, Nina, Jo, Ira, Mara, Sara, Mario, Jan,
Ina, Lily, Arne, Bette, Dan, Reba, Diane, Lynn, Ed, Eva, Dana, Lynne,
Pearl, Isabel, Ada, Ned, Dee, Rena, Joel, Lora, Cecil, Aaron, Flora,
Tina, Arden, Noel, and Ellen sinned.
A poem:
Mood's mode!
Pallas, I won!
(Diaper pane, sold entire.)
Melt till ever sere, hide it.
Drown a more vile note;
(Tar of rennet.)
Ah, trowel, baton, eras ago.
The reward? A "nisi." Two nag.
Otary tastes putrid, yam was green.
Odes up and on; stare we.
Rats nod. Nap used one-erg saw.
(May dirt upset satyr?)
A toga now; 'tis in a drawer, eh?
Togas are notable.
(Worth a tenner for Ate`.)
Tone liver. O Man, word-tied I.
Here's revel!
Little merit, Ned? Lose, Nap?
Repaid now is all apedom's doom.
-- Hubert Phillips:
Headmaster's Palindromic List on his Memo Pad:
Test on Erasmus Dr of Law
Deliver slap Stop dynamo (OTC)
Royal: phone no.? Tel: Law re Kate Race
Ref. Football. Caps on for prep
Is sofa sitable on? Pots- no tops
XI--Staff over Knit up ties ('U')
Sub-edit Nurse's order Ned (re paper)
Caning is on test (snub slip-up) Eve's simple hot dish (crib)
Birch (Sid) to help Miss Eve Pupil's buns
Repaper den T-set: no sign in a/c
Use it Red roses
Put inkspot on stopper Run Tide Bus?
Prof.--no space Rev off at six
Caretaker (wall, etc.) Noel Bat is a fossil
Too many d*** pots Lab to offer one 'Noh' play--
Wal for duo? (I'd name Dr. O) or 'Pals Reviled'?
See few owe fees (or demand IOU?) Sums are not set.
-- Joyce Johnson
(_New_Statesman_ competition in 1967. 126 words, 467 letters)
The entire book _Satire: Veritas_ is a palindrome, it starts
"Sir, I stra..." and ends "... Art, sir, is Satire: Veritas."
Other palindromes in ascending length (drum roll please):
O, no!
Do God!
We few.
Go, dog!
Old? Lo!
Top pot.
A Toyota.
Redivider
Top spot.
Live evil.
Semi limes
Stun nuts.
Taft: fat!
Go do, dog.
Never even.
Stop spots.
Regal lager.
Dot sees Tod.
Lepers repel.
Ma has a ham.
Dennis sinned.
Sir, I'm Iris.
Dee saw a seed.
Del saw a sled.
I met System I.
Lee had a heel.
Lonely Tylenol.
Not so, Boston.
Senile felines.
Sup not on pus.
"Naomi," I Moan.
He won snow, eh?
Laid on no dial.
Name no one man.
Peel Sam asleep.
Redraw a warder.
Step on no pets.
Do geese see God?
Ed is on no side.
Emil saw a slime.
Dot saw I was Tod.
Eros' sis is sore.
Gateman's nametag.
God saw I was dog!
Lem saw I was Mel.
Naomi, did I moan?
Never odd or even.
Rise to vote, sir.
Sleep on no peels.
Enid and Edna dine.
He stops spots, eh?
Revenge Meg? Never!
Sleep not on peels.
Was it a rat I saw?
Wendel's sled: new?
A Toyota's a Toyota.
No lemons, no melon.
O, stone, be not so.
God, a saw was a dog!
Ma is a nun, as I am.
No, I met System Ion.
No, it is opposition.
Now I see bees I won.
Delia saw I was ailed.
Dr. Lime, 121 Emil Rd.
He goddam mad dog, eh?
He harasses Sarah, eh?
Niagara, O roar again!
No evil shah lives on.
Dennis and Edna sinned.
Rats drown in WordStar.
Star comedy: Democrats!
Drat Saddam mad dastard.
He lived as a devil, eh?
He won a Toyota now, eh?
Red rum, sir, is murder.
Sit on a potato pan Otis!
A slut nixes sex in Tulsa.
A slut nixes sex in Tulsa.
Ma is as selfless as I am.
Nurse, I spy gupsies, run!
Rats live on no evil star.
Slap a ham on Omaha, pals!
Top step's pup's pet spot.
A dog! A panic in a pagoda!
Go deliver a dare, vile dog.
I won, Karen, an era know I.
May a moody baby doom a yam?
Murder for a jar of red rum.
No "x" in "Mr. R. M. Nixon"?
Swen, on gnus, sung no news.
Ten animals I slam in a net.
Was it Eliot's toilet I saw?
Was it a car or a cat I saw?
Al lets Della call Ed Stella.
No, she stops spots, eh, son?
Eros? Sidney, my end is sore!
Devil Natasha, ah, Satan lived.
Draw, O Caeser, erase a coward.
Emil's Nip eels sleep in slime!
Did Eve salt an atlas? Eve did.
I, man, am regal; a German am I.
Golf? No sir, prefer prison flog.
No pinot noir on Orion to nip on.
O, memsahib Bart, rabbi has memo.
Trap all afoot; I too fall apart.
Evil I did dwell; lewd did I live.
Norma is as selfless as I am, Ron.
"Naomi, sex at noon taxes," I moan.
Yo, bad anaconda had no Canada boy.
"Asi, Buenos Dias!" said Son Eubisa.
Egad! A base tone denotes a bad age.
Egad, a base life defiles a bad age.
Kay, a red nude, peeped under a yak.
Satan, oscillate my metallic sonatas.
Sums are not set as a test on Erasmus.
Marge lets Norah see Sharon's telegram.
Red dude kill lion. No ill-liked udder.
Anna: "Did Otto peep?" Otto: "Did Anna?"
Address of Emil Otto Lime: Fosse Rd., DA.
Aim a Toyota tatami mat at a Toyota, Mia.
Did I draw Della too tall, Edward? I did?
I roamed under it as a tired, nude Maori.
He won't, ah, wander, Edna. What now, eh?
To Peru, named llama mall 'De Manure Pot'.
Anne, I vote more cars race Rome to Vienna.
Stop, Syrian! I start at rats in airy spots.
I roamed under rats as a starred, nude Maori.
Paget saw an Irish tooth, Sir, in a waste gap.
We few erase cares, Al; laser aces are we few.
Level? No, I tan, I'm at no contamination level.
Straw? No, too stupid a fad; I put soot on warts.
Live dirt up a side track carted is a putrid evil.
No gnu have never after fret far, even Eva hung on.
Now, Ned, I am a maiden nun; Ned, I am a maiden won.
We repaid a no name Pacific ape man on a diaper, ew!
Here we no got conical ill lilac in octogon ewer, eh?
Salamander a ton now. Raw war won not, a Red Nam, alas.
I prevent U-boats, tar, evil live rats, Tao but never pi.
Fool! A dog lives sad a boxer, Rex. O bad ass evil god aloof!
'Tenor Octopus Night' netted a cadet tenth ginsu pot, coronet.
Won total, I am a pro. Bali radar I labor. Pa, mail a tot now!
Yo, boy! Trap gnus, nude. 'Kangaroo Rag' naked unsung party, O boy!
Did I strap red nude, red rump, also slap murdered underparts? I did!
Doc, note: I dissent. A fast never prevents a fatness. I diet on cod.
So regards Rat's Lib: regrets no more hero monster gerbil stars' drag Eros.
Degas, are we not drawn onward, we freer few, drawn onward to new eras aged?
Garret, I ogle. Enemy democrats party; trap star comedy men, eel goiter rag.
Named undenominationally rebel, I rile Beryl? La, no! I tan. I'm, O Ned, nude, man!
Sagas emit taxes, rat snot, or pastrami. I'm Arts, a proton star - sex at times a gas.
Dr. Ana, Cataracts. Uranium enema smarts if fist rams, Amen! Emu in a rust car at a canard.
T. Eliot, top bard, notes putrid tang emanating, is sad; I'd assign it a name: gnat dirt upset on drab pot toilet.
Some word (not letter) palindromes:
So patient a doctor to try to doctor a patient so.
You can cage a swallow, can't you, but you can't swallow a cage, can you?
Girl, bathing on Bikini, eyeing boy, finds boy eyeing
bikini on bathing girl.
Amusing is that company of fond people bores people fond of
company that is amusing.
Line palindrome poems:
As I was passing near the jail
I met a man, but hurried by.
His face was ghastly, grimly pale.
He had a gun. I wondered why
He had. A gun? I wondered... why,
His face was *ghastly*! Grimly pale,
I met a man, but hurried by,
As I was passing near the jail.
DOPPELGANGER (umlaut on the A)
Entering the lonely house with my wife,
I saw him for the first time
Peering furtively from behind a bush--
Blackness that moved,
A shape amid the shadows,
A momentary glimpse of gleaming eyes
Revealed in the ragged moon.
A closer look (he seemed to turn) might have
Put him to flight forever--
I dared not
(For reasons that I failed to understand),
Though I knew I should act at once.
I puzzled over it, hiding alone,
Watching the woman as she neared the gate.
He came, and I saw him crouching
Night after night.
Night after night
He came, and I saw him crouching,
Watching the woman as she neared the gate.
I puzzled over it, hiding alone--
Though I knew I should act at once,
For reasons that I failed to understand
I dared not
Put him to flight forever.
A closer look (he seemed to turn) might have
Revealed in the ragged moon
A momentary glimpse of gleaming eyes,
A shape amid the shadows,
Blackness that moved.
Peering furtively from behind a bush,
I saw him, for the first time,
Entering the lonely house with my wife.
Some palindromic names:
Lon Nol
(former premier of Cambodia)
U Nu
(former premier of Burma)
Revilo P. Oliver
(professor at U. of Illinois; his father and grandfather have/had
the same name, which was deliberately contrived)
Wassamassaw
(a swamp north of Charleston SC)
Yreka Bakery
(a former business in Yreka CA)
Yrella Gallery
(a business which later occupied the same premises)
References:
Palindromes and Anagrams
Howard W. Bergerson
Dover Publications
New York, 1973
ISBN 0-486-20664-5.
The Oxford Guide to Word Games, chapter 11, titled "Palindromes"
Tony Augarde
Lid Off a Daffodil
John Pool
Holt, Rinehart and Wilson
1982
The Palindromicon
Jeff Grant
Word Ways
Spring Valley Road
Morristown, NJ 07960
From A to Zotamorf: The Dictionary of Palindromes
Stephen Chism
Word Ways
Spring Valley Road
Morristown, NJ 07960
==> language/english/spelling/sets.of.words/ladder.p <==
Find the shortest word ladders stretching between the following pairs:
hit - ace
pig - sty
four - five
play - game
green - grass
wheat - bread
order - chaos
order - impel
sixth - hubby
speedy - comedy
chasing - robbers
effaces - cabaret
griming - goblets
vainest - injects
vainest - infulae
==> language/english/spelling/sets.of.words/ladder.s <==
Using every unabridged dictionary available, the best yet found are:
hit ait act ace
pig peg seg sey sty
four foud fond find fine five
play blay bray bras baas bams gams game
green grees greys grays grass
wheat theat treat tread bread
order older elder eider cider cides codes coles colls coals chals chaos
order ormer armer ammer amper imper impel
sixth sixty silty silly sally sably sabby nabby nubby hubby
speedy speeds steeds steers sheers shyers sayers payers papers papery popery
popely pomely comely comedy
griming priming prising poising toising toiling coiling colling collins collies
dollies doilies dailies bailies bailees bailers failers fablers gablers gabbers
gibbers gibbets gobbets goblets
chasing ceasing cessing messing massing masting marting martins martens martels
cartels carpels carpers campers cambers combers cobbers combers robbers
vainest fainest fairest sairest saidest saddest maddest middest mildest wildest
wiliest winiest waniest caniest cantest contest confest confess confers conners
canners fanners fawners pawners pawnees pawnces paunces jaunces jaunced jaunted
saunted stunted stented stenned steined stained spained splined splines salines
savines savings pavings parings earings enrings endings ondings ondines undines
unlines unlives unwives unwires unwares unbares unbared unpared unpaged uncaged
incaged incased incised incises incites indites indices indicts inducts indults
insults insulas insulae infulae
This is not another travelling salesman - it is merely finding the diameter of
connected components of that graph. The simple algorithm for this is to do
one depth first search from each word, resulting in an O(n*m) worst case
algorithm (where n is the number of words, and m is the number of arcs). In
practice, it is actually somewhat better, since the graph breaks down into
many connected components. However, the diameters (and solutions) depend on
what dictionary is used. Here are the results from various dictionaries:
From /usr/dict/words (restricted to words all lower case alphabetical) (19,694
words): sixth - hubby (46 steps)
From the official scrabble players dictionary (94,276 words): effaces -
cabaret (57 steps)
From the british official scrabble words (134,051 words): vainest - infulae
(73 steps)
From webster's ninth new collegiate dictionary (abridged) (78, 167 words):
griming - goblets (56 steps)
From all of the above, merged (180,676 words): vainest - injects (58 steps)
To see the effect the dictionary has on paths, here are the lengths of the
shortest paths these pairs, and for the ones mentioned in previous posts, for
each dictionary (a - means that there is no path using only words from that
dictionary):
UDW OSPD OSW W9 ALL
hit - ace 5 3 3 5 3
pig - sty - 5 4 5 4
four - five 6 6 5 7 5
play - game 8 7 7 8 7
green - grass 13 4 4 7 4
wheat - bread 6 6 6 6 6
sixth - hubby 46 9 9 - 9
effaces - cabaret - 57 - - 33
vainest - infulae - - 73 - 52
griming - goblets - 22 19 56 15
vainest - injects - - 72 - 58
==> language/english/spelling/sets.of.words/nots.and.crosses.p <==
What is the most number of letters that can be fit into a three by three grid
of words, such that no letter is repeated in any row, column or diagonal?
==> language/english/spelling/sets.of.words/nots.and.crosses.s <==
Games magazine ran a contest on this. The winner had 62:
proxying| buckwash | veldt
------------------------------------
stumbled| j | zincography
------------------------------------
whack | providently | bumfs
Here are some good tries:
backsword |thumpingly | fez
----------------------------------
vexingly | q | throwbacks = 61
----------------------------------
thump | beadworks | jingly
backsword | thumpingly| vex
----------------------------------
vexingly | q | throwbacks = 60
----------------------------------
thump | bedrocks | flying
subjack |downrightly| fez
----------------------------------
novelwright| q | backups = 59
----------------------------------
pyx | subface | downright
krafts | exhuming | blowzy
----------+-----------+-----------
phylum | j | transfixed = 56
----------+-----------+-----------
vexing | folkways | chump
klutz | cymograph | fend
----------+-----------+-----------
exscind | j | kymograph = 54
----------+-----------+-----------
myograph |flunked | vibs
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/spelling/sets.of.words/perfect.ladder.p <==
A "perfect" ladder comprises five-letter words where every letter is
the new letter once, no position is changed twice in a row, and fifth
letter changes are better than fourth letter changes are better than
third letter changes, etc. (as defined by "Games" magazine, July 1993)
==> language/english/spelling/sets.of.words/perfect.ladder.s <==
The winning entry:
Score: 9, 4, 8, 1, 4
SOAPS
SOUPS U 3
COUPS C 1
COMPS M 3
COMBS B 4
COMBO O 5
COMPO P 4
COMPT T 5
COMET E 4
COMES S 5
COXES X 3
COXEY Y 5
COZEY Z 3
COZEN N 5
COVEN V 3
COVED D 5
CAVED A 2
CAVEL L 5
JAVEL J 1
JAVER R 5
JAGER G 3
WAGER W 1
WAKER K 3
FAKER F 1
FAKIR I 4
FAQIR Q 3
FAQIH H 5
==> language/english/spelling/sets.of.words/squares.p <==
What are some exceptional word squares (square crosswords with no blanks)?
==> language/english/spelling/sets.of.words/squares.s <==
Word squares are a particular example of a type of crossword known
as "forms". They were more popular early in the late 19th and early
20th century than they are now, but people still like to compose and
solve them. Forms appear every month in the _Enigma_ (as well as many
other puzzle types), which is the monthly publication of the National
Puzzlers' League. The membership fee is $13 for the first year, $11
a year thereafter. Information (or a free sample) may be obtained from:
Joseph J. Adamski
2507 Almar
Jenison, MI 49428
All members have the option of choosing a nom de plume ("nom"); for
example, I go by the nom "Cubist". Another good place to find information
on forms is in _Word Ways_, which is a quarterly journal of recreational
linguistics:
Word Ways
Faith and Ross Eckler, editors
Spring Valley Road
Morristown, NJ 07960
I had a paper appear in the February 1993 issue (Vol. 26 Num. 1) on the
mathematics of word squares, and the ideas extend to more general forms.
Word squares come in two traditional types, regular and double. In
regular word squares the words are the same across and down; in double
word squares all words are different. The largest "legitimate" word
square has order 9 (although Jeff Grant has come close to the 10), and
what is considered to be one of the finest examples was discovered by
Eric Albert via computer search:
necessism
existence
circumfer
escarping
sturnidae
sempitern
infidelic
scenarize
mergences
All words appear in from Webster's New International Dictionary, Second
Edition. It's the *only* single-source 9-square known, and the only
(minor) flaw is that "Sturnidae" is a proper (capitalized) word. All
words are also solid-form (no phrases, spaces, punctuation marks, etc.).
Eric was using about 63,000 words when he discovered his square. Using
about 78,500 9-letter words, I found the square on the left; adding
another 4,000 I found the square on the right:
bortsches karatekas
overtrust apocopate
reparence rosecolor
trabeatae acetoxime
strestell tokokinin
creatural epoxidize
hunterite kalinites
escalates atomizers
steelless serenesse
For the left square, all are in the OED, except for "trabeatae", which
is in NI2. This makes this square arguably the second-best ever
discovered. All words are uncapitalized and solid-form, and it is the
only known 9-square that uses only uncapitalized, solid-form dictionary
words. I consider the square on the right to be one of the most
interesting ever found, as it has two rare letters ("x" and "z") not on
the main diagonal. Since then I've found four additional squares, which
will be appearing in a _Word Ways_ article sometime in the near future.
There are about 1000 9-squares known, all of which were constructed
by hand except for the seven noted above. Almost all of these use
very obscure sources of words. As a general rule of thumb, if you
discover a new square via computer search, it is probably going to
be of high quality, since it is hard to obtain computer-readable word
lists that contain very obscure words.
The largest known double word-squares are of order 8. They are
considered to be about as hard to construct as a regular word
square of order-9, and this is substantiated by the work I've
done on the mathematics of square construction. The following
fine example was constructed by Jeff Grant (see his article in
_Word Ways_, Vol. 25 Num. 1, pp. 9-12):
trattled
hemerine
apotomes
metapore
nailings
aloisias
tentmate
assessed
All are dictionary terms, but there are some weak entries, e.g.
Aloisias: individuals named Aloisia, a feminine form of Aloysius
occurring in the 16th and 17th century in parish registers of
Hinton Charterhouse, England (The Oxford Dictionary of English
Christian Names, 3rd Edition, E.G. Withycombe, 1977)
Such words are, however, dear to the heart of logologists! For
other examples of double squares see the article mentioned above.
One addition to this article is that I've discovered a new double-7
square which may be the best found to date:
smashes
pontine
ingrate
relater
asinine
lingots
sagenes
A new type of form which is in a certain sense as natural as the
regular and double square is the inversion square (so named by
Frank Rubin). So far I've discovered the only known proper
inversion 10-square:
detasseled
exercitate
tectonical
arthrolite
scorpionis
sinoiprocs
etilorhtra
lacinotcet
etaticrexe
delessated
Based on analysis I've done inversion 10-squares are about as rare as
regular 9-squares.
Some interesting foreign language squares I've discovered include:
Dutch German Italian Norwegian Swedish
zaklamp waelzte accosto kaskade apropaa
acribie abhauen ciascun apparat primaer
krijsen ehrtest campato spinett riddare
lijmers latente ospiter kantate omdomen
absente zuenden scatola arealer paamind
miertje testest tuteler dattera aerende
penseel entente onorare etterat arenden
And an 8-square:
French
marbrier
amarante
rabattes
brasiers
rationna
intenses
eternels
ressassa
These aren't the largest known. For example, a French 10-square has
been constructed.
Polyglot 9-square that uses 6 different languages:
absorbera Swedish
betoertem German
storpiavo Italian
oorhanger Dutch
repandent French
brinderai Italian
etagerons French
revenants English
amortisse French
Polyglot double 5-square that uses 10 different languages:
a a g j e Dutch
f a l o t French
f l i r t English
y t t r a Swedish
r o t o l Italian
D F G S N
a i e p o
n n r a r
i n m n w
s i a i e
h s n s g
h h i
a
n
There are also many other types of forms. Some of the most common
are pyramids, stars, and diamonds, and some come in regular and double
varieties, and some are inherently double (e.g. rectangles).
How hard is it to discover a square, anyway, and how many are there?
As a data point, my program using the main (Air Force) entries in
NI2 (26,332 words), found only seven 8x8 squares. This took an hour
to run. They are:
outtease appetite unabated acetated interact repeated repeated
unweaned prenaris nopinene cadinene neomenia evenmete evenmete
twigsome perscent apostate edentate toxicant pectosic pectosic
teguexin ensconce bistered tindered emittent entresol entresol
easement taconite antehall antehall rectoral amoebula amoebula
anoxemia irenicum tearable tearable anaerobe tessular tessular
seminist tincture entellus entellus cinnabar etiolate etiolate
edentate esteemer deedless deedless tattlery declarer declared
If the heuristic mathematics are worked out, the number of different
words in your word-list before you'd expect to find a regular word
square of order-n (the "support") is about e^{(n-1)/2}, where e ~ 15.8.
For a double word square of order-n the support is about e^{n/2}.
There is a simple algorithm which is more precise, and this gives a
support of 75,641 for a regular 9-square, and a support of 272,976
for a double 9-square (using my 9-letter word list).
Bibliography:
Albert, E. "The Best 9x9 Square Yet" _Word Ways_ Vol. 24 Num. 4
Borgmann, D. "More Quality Word Squares" _Word Ways_ Vol. 21 Num. 1
Brooke, M. "A Word Square Update" _Word Ways_ Vol. 16 Num. 4
Grant, J. "9x9 Word Squares" _Word Ways_ Vol. 13 Num. 4
Grant, J. "Ars Magna: The Ten-Square" _Word Ways_ Vol. 18 Num. 4
Grant, J. "Double Word Squares"_Word Ways_, Vol. 25 Num. 1
Grant, J. "In Search of the Ten-Square" _Word Ways_ Vol. 23 Num. 4
Long, C. "Mathematics of Square Construction" _Word Ways_ Vol. 26 Num. 1
Ropes, G. H. "Further Struggles with a 10-Square" _Word Ways_ Vol. 23 Num. 1
Rubin, F. "Inversion Squares" _Word Ways_ Vol. 23 Num. 3
Chris Long
265 Old York Road
Bridgewater, NJ 08807
cl...@remus.rutgers.edu
==> language/english/spelling/sets.of.words/variogram.p <==
What is the largest known variogram (word square where repeated letters count
as one square)?
==> language/english/spelling/sets.of.words/variogram.s <==
o c c u p a t e s
c h i n a m i n e
c i g a r e t t e
u n a ll o t t e d
p a r o c h i n e
a m e t h y s t a
t i t t i s h e r
e n t e n t e s t
s e e d e a r t h
--
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
==> language/english/spelling/sets.of.words/word.torture.p <==
What is the longest word all of whose contiguous subsequences are words?
==> language/english/spelling/sets.of.words/word.torture.s <==
This problem was discussed in _Word Ways_ in 1974-5. In August 1974,
Ralph Beaman, in an article titled "Word Torture", offered the word
SHADES, from which one obtains HADES, SHADE; ADES, HADE, SHAD; DES, ADE,
HAD, SHA; ES, DE, AD, HA, SH; S, E, D, A, H. All of these are words
given in Webster's Third.
Since that time, a serious search has been launched for a seven-letter
word. The near misses so far are:
Date Person Word Missing
Aug 74 Ralph Beaman GAMINES INES, GAMI, NES, INE
Nov 74 Dmitri Borgmann ABASHED INE, NES, ABASHE, BASHE, ASHE (all in OED)
May 75 David Robinson GUNITES GU, GUNIT (using Webster's Second)
May 75 David Robinson ETAMINE ETAMI, TAMI (using Webster's Second)
May 75 Ralph Beaman MORALES RAL (using Webster's Second)
Aug 75 Tom Pulliam SHEAVES EAV (using Webster's Second)
Webster's Second has been used for most of the attempts since it
contains so many more words than Webster's Third. The seven-letter
plateau remains to be achieved.
==> language/english/spelling/single.words.p <==
What words have exceptional lengths, patterns, etc.?
==> language/english/spelling/single.words.s <==
Spelling
Letter Patterns
Entire Word
longest word trinitrophenylmethylnitramine (29,1)
longest palindrome kinnikinnik (11,1)
longest beginning with a palindrome adinida (7,1)
longest beginning with b palindrome boob (4,1)
longest beginning with c palindrome carac civic (5,2)
longest beginning with d palindrome deified devoved (7,2)
longest beginning with e palindrome ecce esse (4,2)
longest beginning with f palindrome FF (2,1) !
longest beginning with g palindrome goog (4,1)
longest beginning with h palindrome hagigah halalah (7,2)
longest beginning with i palindrome igigi imami (5,2)
longest beginning with j palindrome j (1,1)
longest beginning with k palindrome kinnikinnik (11,1)
longest beginning with l palindrome lemel level lysyl (5,3)
longest beginning with m palindrome malayalam (9,1)
longest beginning with n palindrome nauruan (7,1)
longest beginning with o palindrome oppo otto (4,2)
longest beginning with p palindrome peeweep (7,1)
longest beginning with q palindrome qazaq (5,1)
longest beginning with r palindrome reviver rotator (7,2)
longest beginning with s palindrome sawbwas seesees seities sememes (7,4)
longest beginning with t palindrome terret tibbit tippit (6,3)
longest beginning with u palindrome uku ulu utu (3,3)
longest beginning with v palindrome vav (3,1)
longest beginning with w palindrome waw wow (3,2)
longest beginning with x palindrome x (1,1)
longest beginning with y palindrome yaray (5,1)
longest beginning with z palindrome z (1,1)
longest with middle a palindrome halalah rotator (7,2)
longest with middle b palindrome sawbwas (7,1)
longest with middle c palindrome soccos succus (6,2)
longest with middle d palindrome murdrum (7,1)
longest with middle e palindrome sememes (7,1)
longest with middle f palindrome deified (7,1)
longest with middle g palindrome degged (6,1)
longest with middle h palindrome aha ihi oho (3,3)
longest with middle i palindrome hagigah reviver (7,2)
longest with middle j palindrome kajak (5,1)
longest with middle k palindrome kinnikinnik (11,1)
longest with middle l palindrome hallah selles (6,2)
longest with middle m palindrome sammas (6,1)
longest with middle n palindrome adinida (7,1)
longest with middle o palindrome devoved (7,1)
longest with middle p palindrome tippit (6,1)
longest with middle q palindrome q (1,1)
longest with middle r palindrome nauruan (7,1)
longest with middle s palindrome seesees (7,1)
longest with middle t palindrome seities (7,1)
longest with middle u palindrome alula arura (5,2)
longest with middle v palindrome civic level rever tevet (5,4)
longest with middle w palindrome peeweep (7,1)
longest with middle x palindrome sexes (5,1)
longest with middle y palindrome malayalam (9,1)
longest with middle z palindrome kazak qazaq (5,2)
longest tautonym tangantangan (12,1)
longest beginning with a tautonym akeake atlatl (6,2)
longest beginning with b tautonym bellabella (10,1)
longest beginning with c tautonym caracara chowchow couscous (8,3)
longest beginning with d tautonym dugdug dumdum (6,2)
longest beginning with e tautonym EE ee (2,2)
longest beginning with f tautonym froufrou (8,1)
longest beginning with g tautonym ganggang greegree guitguit (8,3)
longest beginning with h tautonym hotshots (8,0) ?
longest beginning with i tautonym ipilipil (8,1)
longest beginning with j tautonym juju (4,1)
longest beginning with k tautonym kavakava kawakawa khuskhus kohekohe kouskous kukukuku (8,6)
longest beginning with l tautonym lapulapu lavalava lomilomi (8,3)
longest beginning with m tautonym mahimahi makomako matamata murumuru (8,4)
longest beginning with n tautonym nagnag (6,1)
longest beginning with o tautonym oo (2,1)
longest beginning with p tautonym palapala pioupiou piripiri poroporo (8,4)
longest beginning with q tautonym quiaquia (8,1)
longest beginning with r tautonym riroriro (8,1)
longest beginning with s tautonym sweeswee (8,1)
longest beginning with t tautonym tangantangan (12,1)
longest beginning with u tautonym ulaula (6,1)
longest beginning with v tautonym valval verver (6,2)
longest beginning with w tautonym wallawalla (10,1)
longest beginning with x tautonym ? (0,0) ?
longest beginning with y tautonym yariyari (8,1)
longest beginning with z tautonym zoozoo (6,1)
longest head 'n' tail einsteins muckamuck okeydokey overcover pungapung tarantara trinitrin (9,7)
longest with middle a head 'n' tail muckamuck pungapung (9,2)
longest with middle b head 'n' tail SBs aba (3,2)
longest with middle c head 'n' tail overcover (9,1)
longest with middle d head 'n' tail okeydokey (9,1)
longest with middle e head 'n' tail arear caeca (5,2)
longest with middle f head 'n' tail efe ofo (3,2)
longest with middle g head 'n' tail aggag algal edged magma (5,4)
longest with middle h head 'n' tail outshouts (9,0) ?
longest with middle i head 'n' tail trinitrin (9,1)
longest with middle j head 'n' tail anjan (5,1)
longest with middle k head 'n' tail arkar kokko (5,2)
longest with middle l head 'n' tail ingling khalkha (7,2)
longest with middle m head 'n' tail bamba bombo mamma pampa (5,4)
longest with middle n head 'n' tail tarantara (9,1)
longest with middle o head 'n' tail ingoing mesomes (7,2)
longest with middle p head 'n' tail SPs apa (3,2)
longest with middle q head 'n' tail q (1,1)
longest with middle r head 'n' tail adrad kurku ugrug verve (5,4)
longest with middle s head 'n' tail hotshot (7,1)
longest with middle t head 'n' tail einsteins (9,1)
longest with middle u head 'n' tail mauma shush siusi veuve (5,4)
longest with middle v head 'n' tail ava eve (3,2)
longest with middle w head 'n' tail abwab (5,1)
longest with middle x head 'n' tail manxman (7,1)
longest with middle y head 'n' tail calycal (7,1)
longest with middle z head 'n' tail z (1,1)
Subset of Word
longest internal palindrome kinnikinniks sensuousness sensuousnesses (11,3)
longest internal tautonym anhydrohydroxyprogesterone anhydrohydroxyprogesterones kinnikinnick kinnikinnicks kinnikinnics kinnikinniks magnetophotophoresis methylethylpyridine micromicrofarad neuroneuronal ... (10,11)
longest repeated prefix kinnikinnick kinnikinnicks kinnikinnics kinnikinniks micromicrofarad neuroneuronal (10,6)
most consecutive doubled letters bookkeeper bookkeeping (3,2)
most doubled letters possessionlessness possessionlessnesses successlessness successlessnesses (4,4)
longest two cadence humuhumunukunukuapuaa humuhumunukunukuapuaas (8,2)
longest three cadence effervescence effervescences extendednesses neglectednesses pervertednesses redheadednesses reflectednesses unexpectednesses vallabhacharya vallabhacharyas (5,10)
longest four cadence alveolopalatal coproporphyrinuria coproporphyrinurias distributivities gastroschisises humuhumunukunukuapuaa humuhumunukunukuapuaas inevitabilities roentgenometries somesthesises ... (4,12)
longest five cadence indecipherablenesses recollectivenesses (4,2)
Letter Counts
Lipograms
longest letters from first half hamamelidaceae (14,1)
longest letters from second half nonsupports (11,0) ?
longest without ab hydroxydesoxycorticosterone (27,1)
longest without abcd philoprogenitivenesses (22,1)
longest without a to h supposititiously (16,1)
longest without a to k monotonously synonymously tumultuously voluptuously (12,4)
longest without a to n prototropy zoosporous (10,2)
longest without a to q susurrus (8,1)
longest without a to s tutty (5,1)
longest without e humuhumunukunukuapuaas macracanthorhynchiasis phonocardiographically prorhipidoglossomorpha supradiaphragmatically (22,5)
longest without et humuhumunukunukuapuaas phonocardiographically prorhipidoglossomorpha (22,3)
longest without eta coccidioidomycosis (18,1)
longest without etai phyllospondylous (16,1)
longest without etain chlorophyllous chromosomology chrysochlorous phyllomorphous polymorphously scolopophorous (14,6)
longest without etains promorphology (13,1)
Letter Choices
Vowels
longest all vowels aiee ieie (4,2)
longest each vowel once entwicklungsroman (17,1)
longest each extended vowel once cylindrocellular phosphuranylites ventriculography (16,3)
shortest each vowel once eulogia eutocia eutopia isourea sequoia (7,5)
shortest each extended vowel once oxyuridae (9,1)
shortest vowels in order caesious (8,1)
shortest extended vowels in order facetiously (11,1)
longest vowels in order abstentious (11,1)
longest extended vowels in order abstemiously (12,1)
shortest vowels in reverse order muroidea (8,1)
shortest extended vowels in reverse order ? (80,0) ?
longest vowels in reverse order subcontinental (14,1)
longest extended vowels in reverse order ? (0,0) ?
longest one vowel strengths (9,1)
longest two vowels schwartzbrots (13,1)
longest containing a univocalic tathagatagarbhas (16,1)
longest containing e univocalic strengthlessnesses (18,1)
longest containing i univocalic instinctivistic (15,1)
longest containing o univocalic loxolophodonts (14,1)
longest containing u univocalic struldbrugs (11,1)
longest containing y univocalic glycyls gypsyfy khlysts khlysty phytyls pyrryls qyrghyz rhythms styryls thymyls ... (7,11)
longest alternating extended vowel-consonant hypovitaminosises (17,1)
longest alternating vowel-consonant aluminosilicates diketopiperazine epicoracohumeral (16,3)
Consonants
longest consonant string bergschrund bergschrunds catchphrase eschscholtzia eschscholtzias festschrift festschriften festschrifts goldschmidtine goldschmidtines ... (6,24)
longest one consonant assessees coccaceae (9,2)
longest two consonant nauseousnesses sensuousnesses (14,2)
Isograms
longest isogram dermatoglyphics (15,1)
longest pair isogram scintillescent (14,1)
longest trio isogram deeded (6,1)
longest tetrad isogram kukukuku (8,1)
longest polygram unprosperousnesses (18,0) ?
longest pyramid chachalaca deadheaded disseisees evennesses keennesses kinnikinic rememberer sassanians sereneness sleeveless ... (10,11)
most repeated letters dihydroxycholecalciferol hydroxydesoxycorticosterone hysterosalpingographies methyldihydromorphinone microspectrophotometrically octamethylpyrophosphoramide phosphatidylethanolamine pseudohermaphroditism tetrabromophenolphthalein tetraiodophenolphthalein ... (9,11)
highest containing a repeated palaeacanthocephala tathagatagarbha tathagatagarbhas (6,3)
highest containing b repeated bubbybush flibbertigibbet flibbertigibbets flibbertigibbety (4,4)
highest containing c repeated chroococcaceae chroococcaceous circumcrescence circumcrescences echinococcic micrococcaceae (5,6)
highest containing d repeated condiddled dadded deadheaded dendrodendritic diddered diddled diddledees didodecahedron disbudded dodded ... (4,32)
highest containing e repeated ethylenediaminetetraacetate (7,1)
highest containing f repeated chiffchaff chiffchaffs giffgaff giffgaffed giffgaffing giffgaffs riffraff (4,7)
highest containing g repeated aggregating aggreging chugalugging gagging gaggling ganggang ganggangs gigging giggling gigglingly ... (4,18)
highest containing h repeated ichthyophthiriasis ichthyophthirius ichthyophthiriuses rhamphorhynchid rhamphorhynchids rhamphorhynchoid rhamphorhynchus (4,7)
highest containing i repeated dirigibilities discriminabilities distinguishabilities divisibilities ignitibilities indiscernibilities indiscerptibilities indistinguishability indivisibility infinitesimalities ... (6,12)
highest containing j repeated ajonjoli ajonjolis avijja avijjas djokjakarta gastrojejunal gastrojejunostomy hajj hajjes hajji ... (2,53)
highest containing k repeated kakkak kakkaks knickknack knickknackatories knickknackatory knickknackeries knickknackery knickknacky kukukuku kukukukus (4,10)
highest containing l repeated allochlorophyll allochlorophylls alloplastically intellectualistically lillypillies lillypilly polysyllabically (5,7)
highest containing m repeated dynamometamorphism hamamelidanthemum immunocompromised mammatocumulus mammectomies mammectomy mammiform mammilliform mammogram mammonism ... (4,23)
highest containing n repeated inconvenientness inconvenientnesses nannoplankton nannoplanktonic nondenominational nondenominationalism nonentanglement nonintervention noninterventionist syngenesiotransplantation ... (5,12)
highest containing o repeated monogonoporous pseudomonocotyledonous (6,2)
highest containing p repeated aplopappus haplopappus hyperleptoprosopic hyperleptoprosopy snippersnapper whippersnapper (4,6)
highest containing q repeated qaraqalpaq qaraqalpaqs (3,2)
highest containing r repeated ferriprotoporphyrin ferroprotoporphyrin (5,2)
highest containing s repeated possessionlessnesses (9,1)
highest containing t repeated ethylenediaminetetraacetate tetrasubstituted throttlebottom totipotentiality yttrotantalite (5,5)
highest containing u repeated humuhumunukunukuapuaa humuhumunukunukuapuaas (9,2)
highest containing v repeated overconservative ovoviviparity ovoviviparous ovoviviparously ovoviviparousness vulvovaginitis (3,6)
highest containing w repeated bowwow bowwows powwow powwowed powwowing powwows swallowwort whillywhaw whillywhaws whitlowwort ... (3,17)
highest containing x repeated dextropropoxyphene executrix executrixes exlex exlexes exonarthex exotoxic exotoxin hexachlorocyclohexane hexahydroxy ... (2,37)
highest containing y repeated acetylphenylhydrazine acetylphenylhydrazines anhydrohydroxyprogesterone anhydrohydroxyprogesterones brachydactyly chylophylly cryptozygy cystopyelography cytophysiologically cytophysiology ... (3,91)
highest containing z repeated pizzazz pizzazzes razzmatazz razzmatazzes (4,4)
most different letters blepharoconjunctivitis pseudolamellibranchiata pseudolamellibranchiate psychogalvanometric (16,4)
highest ratio length/letters kukukuku (400,1)
highest ratio length/letters (no tautonyms) senselessnesses (375,1)
lowest length 16 ratio length/letters ventriculography (106,1)
lowest length 17 ratio length/letters entwicklungsroman hydrobasaluminite pterygomandibular (113,3)
lowest length 18 ratio length/letters carboxyhemoglobins entwicklungsromane hyperglobulinemias psychogalvanometer ventriculographies (120,5)
lowest length 19 ratio length/letters psychogalvanometric (118,1)
lowest length 20 ratio length/letters brachycephalizations dimethyltubocurarine encephalomyocarditis magnetofluiddynamics moschellandsbergites (133,5)
lowest length 21 ratio length/letters diphenylthiocarbazone pseudolamellibranchia sphygmomanometrically (140,3)
lowest length 22 ratio length/letters blepharoconjunctivitis (137,1)
lowest length 23 ratio length/letters pseudolamellibranchiata pseudolamellibranchiate (143,2)
lowest length 24 ratio length/letters diphenylaminechlorarsine laryngotracheobronchitis meningoencephalomyelitis (171,3)
lowest length 25 ratio length/letters spectroheliokinematograph (166,1)
Letter Appearance
longest short letters erroneousnesses verrucosenesses (15,2)
longest tall letters lighttight lillypilly (10,2)
longest vertical-symmetry letters homotaxia thymomata (9,2)
longest horizontal-symmetry letters checkbook checkhook chookchie (9,3)
longest full-symmetry letters ohio (4,1)
highest ratio of dotted letters jinjili (71,1)
Typewriter
longest top row proprietory proterotype rupturewort (11,3)
longest middle row shakalshas (10,1)
longest bottom row MNC (3,1) !
longest in order wettish (7,1)
longest in reverse order bourree chapote chappie chappow gouttee (7,5)
longest left hand tesseradecades (14,0) ?
longest right hand hypolimnion kinnikinnik (11,2)
longest alternating hands leucocytozoans (14,1)
longest one finger deeded humhum hummum muhuhu muumuu (6,5)
longest adjacent keys assessees redresser redresses seeresses sweeswees (9,5)
Puzzle
longest palindrome in Morse code footstool (9,1)
longest formed with chemical symbols nonrepresentationalism (22,1)
longest formed with US postal codes convallarias (12,1)
longest formed with amino acid abbreviations hisser proser serval valval (6,4)
longest formed with piano notes cabbaged fabaceae fagaceae (8,3)
Letter Order
Alphabetical
longest letters in order aegilops (8,1)
longest letters in order with repeats aegilops (8,1)
longest letters in reverse order sponged wronged (7,2)
longest letters in reverse order with repeats trollied (8,1)
longest roller-coaster decriminalizations provincializations (18,2)
longest no letters in place trinitrophenylmethylnitramine (29,1)
most letters in place abudefduf agammaglobulinemias archencephalon archetypical archetypically syngenesiotransplantation (5,6)
most letters in place shifted cooperatively daughterlinesses definitivenesses gymnoplast gymnoplasts inoperative inoperativeness inopportunely intraoperatively neighborlinesses ... (6,16)
most consecutive letters in order consecutively bierstube bierstuben bierstubes gymnopaedia gymnopaedias gymnopaedic gymnopedia gymnopedias gymnophiona gymnoplast ... (4,27)
most consecutive letters in order aborticide aborticides abscinded absconded abscondence abscondences alimentotherapy aluminographies aluminography aluminotype ... (5,38)
most consecutive letters appropinquates appropinquations appropinquities equiponderates equiponderations perquisition perquisitions preconquest propinquities quadruplications ... (8,12)
highest ratio of consecutive letters to length klompen (85,1)
==> language/english/spoonerisms.p <==
List some exceptional spoonerisms.
==> language/english/spoonerisms.s <==
Original by Spooner himself:
I am afraid you have tasted the whole worm, and must
therefore take the next town drain.
Some years ago in the Parliament, a certain member known for his quick and
rapier wit, cut across a certain other member who was trying to make some
bad joke. He called him a "Shining Wit" then apologized for making a
Spoonerism.
Another famous broadcast fluff was on the Canadian Broadcasting
Corporation, which an announcer identified as the "Canadian
Broadcorping Castration."
Oh yes, another radio announcer one that has sort of crept into
common English usage is "one swell foop".
A friend of mine had just eaten dinner in the school
cafeteria, and he didn't look very happy. Another of
my friends said, "John, what's wrong?" Knowing exactly
what he was saying, he said, "It's the bound grief I
had for dinner!"
A radio announcer, talking about a royal visit (or some such) said the
visitor would be greeted with a "twenty one sun galoot".
There are several fractured fables based on spoonerisms, such as:
A king on a desert island was so beloved by his people, they decided to
give him a very special gift for the anniversary of his coronation. So
after much thought, they decided to make him a throne out of seashells,
which were plentiful on the island. And when it was finished, they
presented it to the king, who loved it. But he soon discovered it was
very uncomfortable to sit on. So he told his subjects it was too
special to use everyday (so as not to hurt their feelings) and put it in
the attic of his palace (which was, of course, a hut like all the other
dwellings on the island), planning to use it just for special occasions.
But that night, it fell through the ceiling of his bedroom and landed
on top of him, killing him instantly. And the moral of the story is:
Those who live in grass houses shouldn't stow thrones!
==> language/english/synonyms/ambiguous.p <==
What word in the English language is the most ambiguous?
What is the greatest number of parts of speech that a single word
can be used for?
==> language/english/synonyms/ambiguous.s <==
In Webster's Ninth, "set" occupies 1.2 columns, has 25 vb entries, 11 vi
entries, 23 noun entries, 7 adjective entries; "take" occupies 1.3 columns,
has 19 vb entries, 8 vi entries, 4 noun entries.
The word "like" occupies eight parts of speech:
verb "Fruit flies like a banana."
noun "We may never see its like again."
adjective "People of like tastes agree."
adverb "The rate is more like 12 percent."
preposition "Time flies like an arrow."
conjunction "They acted like they were scared."
interjection "Like, man, that was far out."
verbal auxiliary "So loud I like to fell out of bed."
==> language/english/synonyms/antonym.p <==
What words, when a single letter is added, reverse their meanings?
Exclude words that are obtained by adding an "a-" to the beginning.
==> language/english/synonyms/antonym.s <==
e: fast -> feast, fiancee -> fiance
h: treat -> threat
r: fiend -> friend
s: he -> she
t: here -> there
==> language/english/synonyms/contradictory.proverbs.p <==
What are some proverbs that contradict one another?
==> language/english/synonyms/contradictory.proverbs.s <==
Beware of Greeks bearing gifts.
Never look a gift horse in the mouth.
Look before you leap.
He who hesitates is lost.
Nothing venture, nothing gain.
Fools rush in where angels fear to tread.
Seek and ye shall find.
Curiosity killed the cat.
Save for a rainy day.
Tomorrow will take care of itself.
Life is what we make it.
What is to be will be.
Too many cooks spoil the broth.
Many hands make light work.
One man's meat is another man's poison.
Sauce for the goose is sauce for the gander.
With age comes wisdom.
Out of the mouths of babes and sucklings come all wise sayings.
Bear ye one another's burdens. (Gal. 6:2)
For every man shall bear his own burden. (Gal. 6:5)
Great minds run in the same channel.
Fools think alike.
A rolling stone gathers no moss.
A setting hen never lays.
A hollow pot makes the most noise.
The squeaky wheel gets the grease.
Faint heart never won fair lady.
The meek shall inherit the Earth.
To thine own self be true.
The nail that stands out gets hammered down.
==> language/english/synonyms/contranym.p <==
What words are their own antonym?
==> language/english/synonyms/contranym.s <==
In his 1989 book _Crazy English_, Richard Lederer calls such words
contranyms and lists more than 35, although some are phrases instead of
words. These can be divided into homographs (same spelling) and
homophones (same pronunciation).
A partial list of homographs:
aught = all, nothing
bill = invoice, money
cleave = to separate, to join
clip = cut apart, fasten together
comprise = contain, compose
custom = usual, special
dust = to remove, add fine particles
fast = rapid, unmoving
literally = actually, figuratively
model = archetype, copy
moot = debatable, academic
note = promise to pay, money
oversight = care, error
peep = look quietly, beep
peer = noble, person of equal rank
put = lay, throw
puzzle = pose problem, solve problem
quantum = very small, very large (quantum leap)
ravel = entangle, disentangle
resign = to quit, to sign up again
sanction = to approve of, to punish
sanguine = murderous, optimistic
scan = to examine closely, to glance at quickly
set = fix, flow
skin = to cover with, remove outer covering
speak = express verbally, express nonverbally
stipulate = request explicitly, agree to
strike = miss (baseball), hit
table = propose [British], set aside
temper = calmness, passion
trim = cut things off, put things on
A very short list of homophones:
aural, oral = heard, spoken
fiance, fiancee = female betrothed, male betrothed
raise, raze = erect, tear down
A pair of French words which can be very confusing:
La symetrie (symmetry) and L'asymetrie (asymmetry).
Latin:
immo = yes, no
Possibilities:
draw (curtains, open or close) (money, withdraw, accumulate interest)
eke
==> language/english/synonyms/double.synonyms.p <==
What words have two different synonymous meanings?
==> language/english/synonyms/double.synonyms.s <==
list, roll: set of names (noun), tilt (verb)
==> language/finnish/finnish.plural.p <==
What Finnish word is the anagram of its plural?
==> language/finnish/finnish.plural.s <==
matto/matot (Finnish for "carpet"/"carpets").
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> language/english/pronunciation/homophone/trivial.p <==
Consider the free non-abelian group on the twenty-six letters of the
alphabet with all relations of the form <word1> = <word2>, where <word1>
and <word2> are homophones (i.e. they sound alike but are spelled
differently). Show that every letter is trivial.
For example, be = bee, so e is trivial.
==> language/english/pronunciation/homophone/trivial.s <==
be = bee ==> e is trivial;
ail = ale ==> i is trivial;
week = weak ==> a is trivial;
lie = lye ==> y is trivial;
to = too ==> o is trivial;
two = to ==> w is trivial;
hour = our ==> h is trivial;
faggot = fagot ==> g is trivial;
bowl = boll ==> l is trivial;
gell = jel ==> j is trivial;
you = ewe ==> u is trivial;
damn = dam ==> n is trivial;
limb = limn ==> b is trivial;
bass = base ==> s is trivial;
cede = seed ==> c is trivial;
knead = need ==> k is trivial;
add = ad ==> d is trivial;
awful = offal ==> f is trivial;
gram = gramme ==> m is trivial;
grip = grippe ==> p is trivial;
cue = queue ==> q is trivial;
carrel = carol ==> r is trivial;
butt = but ==> t is trivial;
lox = locks ==> x is trivial;
tsar = czar ==> z is trivial;
vlei = flay ==> v is trivial.
For a related problem, see _The Jimmy's Book_ (_The American Mathematical
Monthly_, Vol. 93, Num. 8 (Oct. 1986), p. 637):
Consider the free group on twenty-six letters A, ..., Z. Mod out by
the relation that defines two words to be equivalent if (a) one is a
permutation of the other and (b) each appears as a legitimate English
word in the dictionary. Identify the center of this group.
-- cl...@remus.rutgers.edu (Chris Long)
==> language/english/pronunciation/oronym.p <==
List some oronyms (phrases or sentences that can be read in two ways
with the same sound).
==> language/english/pronunciation/oronym.s <==
Phrases:
a name an aim
a nice man an ice man
a notion an ocean
append up end
bang cat bank at
be quiet Beek Wyatt
bean ice be nice
bee feeder beef eater
beer drips beard rips
buys ink buy zinc
catch it cat shit
catch ooze cat chews
Cato Kay toe
damn pegs damp eggs
field red feel dread
forced air four stair
fork reeps four creeps
form ate four mate
freed Annie free Danny
grade A gray day
grasp rice grass price
great ape grey tape
her butter herb utter
hiatus Hy ate us
homemaker hoe-maker
I scream ice cream
I stink iced ink
it sprays it's praise
it swings its wings
keep sticking keeps ticking
known ocean no notion
lawn chair launch air
may cough make off
new Deal nude eel
new direction nude erection
night rate nitrate
pawn shop paunch op
peace talks pea stalks
pinch air pin chair
play taught plate ought
plum pie plump eye
scar face scarf ace
seal eyeing see lying
see Mabel seem able
see the meat see them eat
seize ooze see zoos
sick squid six quid
slide rule sly drool
standards-based standard-spaced
stay dill stayed ill
that's tough that stuff
the suns rays meet the sons raise meat
thing call think all
tour an two ran
tulips two lips
twenty six ones twenty sick swans
we'll own we loan
well done other weld another
white shoes why choose
yelp at yell Pat
your crimes York rhymes
youth read you thread
Sentences:
A politician's fate often hangs in a [delicate / delegate] balance.
Any [grey day/grade A] would be bad news for one professor I know.
Are you aware of the words you have [just uttered / just stuttered]?
He would kill Hamlet for [that reason / that treason].
How did you do in the [contest / Kant test]?
I don't know how [mature/much your] people enjoy such a show.
I have [known oceans/no notions] that you yourself couldn't imagine.
I like [sadder day/ Saturday].
If you listen you can hear the [night rain / night train].
I'm taking [a nice / an ice] cold shower.
Oh, no! [This guy/The sky] is falling!
Reading in the library is sometimes [allowed / aloud].
[Some others / Some mothers] I've seen...
That reflects the [secretariat's sphere / secretariat's fear] of competence.
That's the [biggest hurdle / biggest turtle] I've ever seen!
The [stuffy nose / stuff he knows] can lead to problems.
Where is the [spice center / spy center]?
[White shoes: / Why choose] the trademark of Pat Boone[./?]
You'd be surprised to see a [mint spy / mince pie] in your bank.
See also:
the archive entry "telegrams"
References:
Barry, W J, 1981, Internal Juncture and Speech Communication,
Arbeitsberichte. Institut fu"r Phonetik. University of Kiel. Vol 16.
Brandreth, Giles, _The Joy of Lex_, 1980, New York: William Morrow and Co.,
pp. 58-59, who coined the word "oronym"
Cutler & Butterfield, Rhythmic Cues to Speech Segmentation. Evidence
from juncture misperception, 1992, Journal of Memory and Language, Vol
31(2) 218-236 Provides materials in context frames where the
alternative segmentations lead to one Vs two word parsings: in furs Vs
infers.
Grice, Martine and Hazan, Valerie, 1989, The assessment of synthetic
speech intelligibility using semantically unpredictable sentences,
Speech, Hearing and Language, Work in Progress, University College
London. The inferior quality of the synthetic speech often caused more
than one type of error at once (bright eye -> dry tie).
Hoard, James E., `Juncture and syllable structure', Phonetica 15,
1966, 96-109
Hockett, Charles F. Hockett, _A Course in Modern Linguistics_, New York:
Macmillan, 1958, 54-61
Lehiste, Ilse. 1960. An acoustic-phonetic study of internal open
juncture. Phonetica 5 (supplement). pp. 5-54.
Nakatani L. H. & Dukes, K.D., 1977, Locus of Segmental Cues to Word
Juncture, Journal of the Acoustical Society of America, Vol 62, pp
714-719 and
Price P, Ostendorf M, Shattuck-Huffnagel S & Fong C (1991) "The use of
prosody in syntactic disambibuation" JASA 90 (6) pp2956-2970. It has
some good examples,
==> language/english/pronunciation/phonetic.letters.p <==
What does "FUNEX" mean?
==> language/english/pronunciation/phonetic.letters.s <==
FUNEX? (Have you any eggs?)
SVFX. (Yes, we have eggs.)
FUNEM? (Have you any ham?)
SVFM. (Yes, we have ham.)
FUMNX? (Have you ham and eggs?)
S,S:VFM,VFX,VFMNX! (Yes, yes: we have ham, we have eggs, we have ham and eggs!)
CD ED BD DUCKS? (See the itty bitty ducks?)
MR NOT DUCKS! (Them are not ducks!)
OSAR, CDEDBD WINGS? (Oh yes they are, see the itty bitty wings?)
LILB MR DUCKS! (Well I'll be, them are ducks!)
In Spanish:
SOCKS. (Eso si que es.)
==> language/english/pronunciation/rhyme.p <==
What English words are hard to rhyme?
"Rhyme is the identity in sound of an accented vowel in a word...and
of all consonantal and vowel sounds following it; with a difference in
the sound of the consonant immediately preceding the accented vowel."
(From The Complete Rhyming Dictionary by Clement Wood). Appropriately
Wood says a couple of pages later, "If a poet commences, 'October is
the wildest month' he has estopped himself from any rhyme; since
"month" has no rhyme in English."
==> language/english/pronunciation/rhyme.s <==
Unless noted otherwise, all words occur in Webster's Ninth Collegiate
Dictionary, Merriam-Webster, Springfield, MA, 1986.
NI3 = Merriam-Webster's Third New International Dictionary
NI2 = Merriam-Webster's New International Dictionary, Second Edition
RHD = Random House Unabridged Dictionary
+ means slang, foreign, obsolete, dialectical, etc.
Word Rhyme Assonance
--------------- --------------------------------------- --------------------
aitch brache (NI2+), taich (NI2+) naish
angry unangry (NI2+) aggry
angst lanx
beards weirds
breadth death
bulb pulp
carpet charpit
chimney timne, polymny (NI2+)
cusp wusp (NI2) bust
depth stepped
eighth faith
else belts
exit direxit (RHD+) sexist
fiends teinds, piends
filched hilched (NI3+), milched (NI2) zilch
filth spilth, tilth
fifth drift
film pilm (NI3+) kiln
fluxed luxed (NI3+), muxed (NI3+) ducked
glimpsed rinsed
gospel hostile
gulf pulse
jinxed outminxed (?) blinked
leashed niched, tweesht (NI2+)
liquid wicked
mollusk smallest
mouthed southed
month grumph
mulcts bulks
mulched gulched (NI3+) bulged
ninth pint
oblige bides
oomph sumph (NI3+)
orange sporange
pint jint (NI2+) bind
poem phloem, proem
pregnant regnant
purple curple (NI3+), hirple (NI3+)
puss schuss
rhythm smitham
scalds balds, caulds (NI3+), faulds (NI3+)
scarce clairce (NI2), hairse (NI2+) cares
sculpts gulps
silver chilver (NI3+)
sixth kicks
spirit squiret (NI2+)
tenth nth bent
tsetse baronetcy, intermezzi, theetsee
tuft yuft
twelfth health
widow kiddo
width bridge
window indo, lindo
wolf bulls
==> language/english/pronunciation/silent.letter.p <==
For each letter, what word contains that letter silent?
==> language/english/pronunciation/silent.letter.s <==
aisle
comb
indict
handsome
twitched
halfpenny
gnome
myrrh
business
marijuana
knock
talk
mnemonic
autumn
people
psyche
cinqcents
forecastle
viscount
hautboy
plaque
fivepence
write
tableaux
prayer
rendezvous
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/pronunciation/silent.most.p <==
What word has the most silent letters in a row?
==> language/english/pronunciation/silent.most.s <==
BROUGHAM (4, UGHA)
for each letter AISLE, COMB, INDICT,
HANDSOME, TWITCHED, HALFPENNY, GNOME, MYRRH, BUSINESS, MARIJUANA, KNOCK,
TALK, MNEMONIC, AUTUMN, PEOPLE, PSYCHE, CINQCENTS, FORECASTLE, VISCOUNT,
HAUTBOY, PLAQUE, FIVEPENCE, WRITE, TABLEAUX, PRAYER, RENDEZVOUS
homophones, for each letter O(A)R, LAM(B), S(C)ENT,
LE(D)GER, DO(E), WAF(F), REI(G)N, (H)OUR, WA(I)VE, HAJ(J)I, (K)NOT, HA(L)VE,
PRIM(M)ER, DAM(N), J(O)UST, (P)SALTER, ?, CAR(R)IES, (S)CENT, TARO(T),
B(U)Y, ?, T(W)O, ?, RE(Y), BIZ(Z)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/pronunciation/syllable.p <==
What words have an exceptional number of letters per syllable?
==> language/english/pronunciation/syllable.s <==
longest for each number of syllables
one SCRAUNCHED [SQUIRRELLED (11)] two SCRATCHBRUSHED (14)
one, for each letter ARCHED, BROUGHAMS, CRAUNCHED, DRAUGHTS,
EARTHED, FLINCHED, GROUCHED, HAUNCHED, ITCHED, JOUNCED, KNIGHTS, LAUNCHED,
MOOCHED, NAUGHTS, OINKED, PREACHED, QUETCHED, REACHED, SCRAUNCHED,
THOUGHTS, UMPHS, VOUCHED, WREATHED, XYSTS, YEARNED, ZOUAVES
two, for each letter ARCHFIENDS, BREAKTHROUGHS, CLOTHESHORSE,
DRAUGHTBOARDS, EARTHTONGUES, FLAMEPROOFED, GREATHEART, HAIRSBREADTHS,
INTHRALLED, JUNETEENTHS, KNICKKNACKS, LIGHTWEIGHTS, MOOSETONGUES,
NIGHTCLOTHES, OUTSTRETCHED, PLOUGHWRIGHTS, QUICKTHORNS, ROUGHSTRINGS,
SCRATCHBRUSHED, THROATSTRAPS, UNSTRETCHED, VERSESMITHS, WHERETHROUGH,
XANTHINES, YOURSELVES, ZEITGEISTS
shortest for each number of syllables
two AA (2) three AREA (4) [O'IO (3)] four IEIE (4) five OXYOPIA (7)
six ONIOMANIA (9) [AMIOIDEI (8)] seven EPIDEMIOLOGY (12) [OMOHYOIDEI (10)]
eight EPIZOOTIOLOGY (13) nine EPIZOOTIOLOGICAL (16)
ten EPIZOOTIOLOGICALLY (18) twelve HUMUHUMUNUKUNUKUAPUAA (21)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/pronunciation/telegrams.p <==
Since telegrams cost by the word, phonetically similar messages can be cheaper.
See if you can decipher these extreme cases:
UTICA CHANSON MIGRATE INVENTION ANNUAL KNOBBY SORRY IN FACTUAL BEEN CLOVER.
WEED LICHEN ICE CHEST FOREARM OTHER DISGUISE DELIMIT.
CANCEL MYOCARDIA ITS INFORMAL FUNCTION.
YEARN AFFIX, LOST UKASE, UGANDA JAIL, CONSERVE TENURES YACHT APPEAL.
EYELET SHEILA INDIA HOUSE SHEILAS TURKEY.
BOB STILT SEA, CANTANKEROUS BOAT, HUMUS GOAD IMMORTAL DECOS GUARD.
MARY SINBAD SHEER TOURNEY AUGUSTA WIND NOCTURNE TOOTHBRUSH.
WHINE YOSEMITE NAMES SOY CAN PHILATELIST.
ALBEIT DETRACT UNIVERSE EDIFY MUSTAFA TICKET TICKET IN.
==> language/english/pronunciation/telegrams.s <==
These are from an old "Games" magazine:
UTICA CHANSON MIGRATE INVENTION ANNUAL KNOBBY SORRY IN FACTUAL BEEN CLOVER.
You take a chance on my great invention and you'll not be sorry.
In fact, you'll be in clover.
WEED LICHEN ICE CHEST FOREARM OTHER DISGUISE DELIMIT.
We'd like a nice chest for our mother; the sky's the limit.
CANCEL MYOCARDIA ITS INFORMAL FUNCTION.
Can't sell my ol' car dear; it's in for malfunction.
YEARN AFFIX, LOST UKASE, UGANDA JAIL, CONSERVE TENURES YACHT APPEAL.
You're in a fix. Lost your case. You goin' to jail.
Can serve ten years. You ought to appeal.
EYELET SHEILA INDIA HOUSE SHEILAS TURKEY.
I let Sheila in their house; she lost her key.
BOB STILT SEA, CANTANKEROUS BOAT, HUMUS GOAD IMMORTAL DECOS GUARD.
Bob's still at sea; can't anchor his boat. You must go to him
or tell the coast guard.
MARY SINBAD SHEER TOURNEY AUGUSTA WIND NOCTURNE TOOTHBRUSH.
Mary's in bed; she hurt her knee. A gust of wind
knocked her into the brush.
WHINE YOSEMITE NAMES SOY CAN PHILATELIST.
Why don't you (why'n'ya) send me the names, so I can
fill out a list.
ALBEIT DETRACT UNIVERSE EDIFY MUSTAFA TICKET TICKET IN.
I'll be at the track and I have a receipt if I must have a ticket to
get in.
Here are some others not from "Games":
BRUISES HURT ERASED AFFORD ERECTED TOO
Bruce is hurt. He raced a Ford. He wrecked it, too.
==> language/english/puns.p <==
Where can I find a collection of puns?
==> language/english/puns.s <==
garbo.uwasa.fi:/pc/ts/tspun12.zip
==> language/english/rare.trigraphs.p <==
What trigraphs (three-letter combinations) occur in only one word?
==> language/english/rare.trigraphs.s <==
Here is a list of all the trigraphs which occur exactly once in the union of
_Official Scrabble Words_ (First Edition), the _Official Scrabble Players
Dictionary_ and _Webster's Unabridged Dictionary (Second Edition)_,
together with the words in which they occur.
The definition of "word" is a problematic. For example, lots of words
starting deoxy- contain the trigraph `eox', but no others do. Should
`eox' be on the list?
Common words are marked with a *.
aae baaed
adq*headquarter headquarters
ajs svarajs
aqs talaqs
bks nabks
bze subzero
cda ducdame
dph*headphone headphones
dsf*handsful
dts veldts
dzu kudzu kudzus
ekd*weekday weekdays
evh evhoe
evz evzone evzones
exv sexvalent
ezv*rendezvous
fhu cliffhung
fjo fjord fjords
fsp*offspring offsprings
gds smaragds
ggp*eggplant eggplants
gnb signboard signboards
gnp*signpost signposted signposting signposts
gnt sovereignty
gty hogtying
gza*zigzag zigzagged zigzagging zigzaggy zigzags
hds camanachds
hky droshky
hlr kohlrabi kohlrabies kohlrabis
hrj lehrjahre
hyx asphyxia asphyxias asphyxiate asphyxies asphyxy
itv mitvoth
iwy skiwy
ixg sixgun
jds slojds
jje hajjes
jki pirojki pirojki
jym jymold
kky yukky
ksg*thanksgiving
kuz yakuza
kvo mikvoth
kyj*skyjack skyjacked skyjacker skyjackers skyjacking skyjackings skyjacks
llj killjoy killjoys
lmd filmdom filmdoms
ltd*meltdown meltdowns
lxe calxes
lzy schmalzy
mds fremds
mfy comfy
mhs ollamhs
mky dumky
mmm dwammming
mpg*campground
mss bremsstrahlung
muo muon muonic muonium muoniums muons
nhs sinhs
njy benjy
nuu continuum
obg hobgoblin hobgoblins
ojk pirojki
okc*bookcase bookcases
ovk sovkhoz sovkhozes sovkhozy
pev*grapevine grapevines
pfs dummkopfs
php ephphatha
pss topssmelt
pyj pyjama pyjamaed pyjamas
siq physique physiques
slt juslted
smk besmkes
spb*raspberries raspberry
spt claspt
swy swythe
syg*easygoing
szy groszy
tux*tux tuxedo tuxedoes tuxedos tuxes
tvy outvying
tzu tzuris
ucd ducdame
vho evhoe
vkh sovkhoz sovkhozes sovkhozy sovkhos
vly vly
vns eevns
voh evohe
vun avuncular
wcy gawcy
wdu*sawdust sawdusted sawdusting sawdusts sawdusty
wfr bowfront
wft ewftes
xeu exeunt
xgl foxglove foxgloves
xiw taxiway taxiways
xls cacomixls
xtd nextdoor
xva sexvalent
yks bashlyks
yrf gyrfalcon gyrfalcons
ysd paysd
yxy asphyxy
zhk pirozhki
zow zowie
zwo*buzzword buzzwords
zzs*buzzsaw
==> language/english/self.ref/self.ref.letters.p <==
Construct a true sentence of the form: "This sentence contains _ a's, _ b's,
_ c's, ...," where the numbers filling in the blanks are spelled out.
==> language/english/self.ref/self.ref.letters.s <==
A little history of the problem, culled from the pages of _Metamagical
Themas_, Hofstadter's collection of his _Scientific American_ columns.
First mention of it is in the Jan. '82 column. Lee Sallows opened the
field with a sentence that began "Only the fool would take trouble to
verify that his sentence was composed of ten a's ...." etc.
Then in the addendum to the Jan.'83 column on viral sentences, Hofstadter
quotes Sallows describing his Pangram Machine, "a clock-driven cascade of
sixteen Johnson-counters," to tackle the problem. An early success was:
"This pangram tallies five a's, one b, one c, two d's, twenty-
eight e's, eight f's, six g's, eight h's, thirteen i's, one j,
one k, three l's, two m's, eighteen n's, fifteen o's, two p's,
one q, seven r's, twenty-five s's, twenty-two t's, four u's, four
v's, nine w's, two x's, four y's, and one z."
Sallows wagered ten guilders that no-one could create a perfect self-
documenting sentence beginning, "This computer-generated pangram contains
...." within ten years.
It was solved very quickly, after Sallows' challenge appeared in Dewdney's
Oct. '84 SA column. Larry Tesler solved it by a method Hofstadter calls
"Robinsonizing," which involves starting with an arbitrary set of values
for each letter, getting the true values when the sentence is made, and
plugging the new values back in, making a feedback loop. Eventually, you
can zero in on a set of values that work. Tesler's sentence:
This computer-generated pangram contains six a's, one b, three
c's, three d's, thirty-seven e's, six f's, three g's, nine h's,
twelve i's, one j, one k, two l's, three m's, twenty-two n's,
thirteen o's, three p's, one q, fourteen r's, twenty-nine s's,
twenty-four t's, five u's, six v's, seven w's, four x's, five
y's, and one z.
The method of solution (called "Robinsonizing," after the logician Raphael
Robinson) is as follows:
1) Fix the count of a's.
2) Fix the count of b's.
3) Fix the count of c's.
...
26) Fix the count of z's.
Then, if the sentence is still wrong, go back to step 1.
Most attempts will fall into long loops (what Hofstadter calls attractive
orbits), but with a good computer program, it's not too hard to find a
Robinsonizing sequence that zeros in on a fixed set of values.
The February and May 1992 _Word Ways_ have articles on this subject,
titled "In Quest of a Pangram, (Part 1)" by Lee Sallows. It tells of his
search for a self-referential pangram of the form, "This pangram
contains _ a's, ..., and one z." (He built special hardware to search
for them.) Two such pangrams given in the article are:
This pangram lists four a's, one b, one c, two d's,
twenty-nine e's, eight f's, three g's, five h's, eleven i's,
one j, one k, three l's, two m's twenty-two n's, fifteen o's,
two p's, one q, seven r's, twenty-six s's, nineteen t's, four
u's, five v's, nine w's, two x's, four y's, and one z.
This pangram contains four a's, one b, two c's, one d, thirty
e's, six f's, five g's, seven h's, eleven i's, one j, one k,
two l's, two m's eighteen n's, fifteen o's, two p's, one q,
five r's, twenty-seven s's, eighteen t's, two u's, seven v's,
eight w's, two x's, three y's, & one z.
It also contains one in Dutch by Rudy Kousbroek:
Dit pangram bevat vijf a's, twee b's, twee c's, drie d's,
zesenveertig e's, vijf f's, vier g's, twee h's, vijftien i's,
vier j's, een k, twee l's, twee m's, zeventien n's, een o,
twee p's, een q, zeven r's, vierentwintig s's, zestien t's,
een u, elf v's, acht w's, een x, een y, and zes z's.
References:
Dewdney, A.K. Scientific American, Oct. 1984, pp 18-22.
Sallows, L.C.F. Abacus, Vol.2, No.3, Spring 1985, pp 22-40.
Sallows, L.C.F. Word Ways, Feb. & May 1992
Hofstadter, D. Scientific American, Jan. 1982, pp 12-17.
==> language/english/self.ref/self.ref.numbers.p <==
What true sentence has the form: "There are _ 0's, _ 1's, _ 2's, ...,
in this sentence"? The "'s" can be dropped if there is only one of any
number.
==> language/english/self.ref/self.ref.numbers.s <==
There are 1 0, 7 1's, 3 2's, 2 3's, 1 4, 1 5, 1 6, 2 7's, 1 8, and 1 9
in this sentence.
There are 1 0, 11 1's, 2 2's, 1 3, 1 4, 1 5, 1 6, 1 7, 1 8, and 1 9
in this sentence.
==> language/english/self.ref/self.ref.words.p <==
What sentence describes its own word, syllable and letter count?
==> language/english/self.ref/self.ref.words.s <==
This sentence contains ten words, eighteen syllables, and sixty-four letters.
==> language/english/sentences/behead.p <==
Is there a sentence that remains a sentence when all its words are beheaded?
==> language/english/sentences/behead.s <==
Show this bold Prussian that praises slaughter, slaughter brings rout.
==> language/english/sentences/charades.p <==
A ....... surgeon was ....... to operate because he had .......
==> language/english/sentences/charades.s <==
A notable surgeon was not able to operate because he had no table.
==> language/english/sentences/emphasis.p <==
List some sentences that change meaning when the emphasis is moved.
==> language/english/sentences/emphasis.s <==
Give what you WANT to her and keep the rest for yourself.
Give what YOU want to HER and keep the rest for yourself.
==> language/english/sentences/pangram.p <==
A "pangram" is a sentence containing all 26 letters.
What is the shortest pangram (measured by number of letters or words)?
What is the shortest word list using all 26 letters in alphabetical order?
In reverse alphabetical order?
==> language/english/sentences/pangram.s <==
The single-letter words that have meanings unrelated to their letter shapes
or sounds, position in the alphabet, etc. are:
a - indefinite article; on; in; at; to; he; him; she; her; they; them; it; I;
have; of; all
c - 100; cocaine; programming language
d - 500
e - base of natural logs; eccentricity; enlarging
g - acceleration of gravity; general ability; $1000; general audience
i - one; unit vector in x direction; personal pronoun; in; aye
j - one; unit vector in y direction
k - 1000; 1024; strikeout; unit vector in z direction
l - 50; ell; elevated railroad
m - 1000; em; pica; an antigen of human blood
n - an indefinite number; en; an antigen of human blood
o - oh
q - quality of oscillatory circuit
R - one of the three Rs; restricted audience
t - t-shirt
u - upper class
v - five
w - w particle
x - unknown quantity; atmospherics; adults only
y - unknown quantity; YMCA
z - unknown quantity; buzzing sound; z particle
It is therefore advisable to exclude single-letter words, with the
possible exception of 'a'.
As always, word acceptability varies with the dictionaries used. We use these:
9C - Merriam-Webster's Ninth New Collegiate Dictionary, 1986
NI3 - Merriam-Webster's Third New International Dictionary, 1961
NI2 - Merriam-Webster's New International Dictionary, Second Edition, 1935
OED - Oxford English Dictionary with Supplements, 1933 - 85
'+' indicates obsolete, dialectical, slang, or otherwise substandard word.
Some exceptional pangrams:
Using only words in 9C:
Sympathizing would fix Quaker objectives. (5 words, 36 letters)
Quick brown fox, jump over the lazy dogs. (8 words, 32 letters)
Pack my box with five dozen liquor jugs. (8 words, 32 letters)
Jackdaws love my big sphinx of quartz. (7 words, 31 letters)
The five boxing wizards jump quickly. (6 words, 31 letters)
How quickly daft jumping zebras vex. (6 words, 30 letters)
Dub waltz, nymph, for quick jigs vex. (7 words, 28 letters,
Stephen Smith, from a palindrome by Gyles Brandreth)
Quartz glyph job vex'd cwm finks. (6 words?, 26 letters)
Cwm, fjord-bank glyphs vext quiz. (6 words, 26 letters, Dmitri Borgmann)
Using words in 9C and NI3:
Veldt jynx grimps waqf zho buck. (6 words, 26 letters, Michael Jones)
Using words in 9C, NI3 and NI3+:
Squdgy fez, blank jimp crwth vox. (6 words, 26 letters, Claude Shannon)
Using words in 9C, NI3, NI2 and NI2+:
Phlegms fyrd wuz qvint jackbox. (5 words, 26 letters, Dmitri Borgmann)
Some exceptional panalphabetic word lists:
jackbox viewfinder phlegmy quartz (4 words, 31 letters, Mary Hazard)
benzoxycamphors quick-flowing juventude (3 words, 36 letters, Darryl Francis)
Some exceptional nearly panalphabetic isogrammatic word lists:
blacksmith gunpowdery (2 words, 20 letters)
humpbacks frowzy tingled (3 words, 22 letters)
Some exceptional panalphabetic word lists with letters in alphabetical order:
Using only words in 9C:
a BCD ef ghi jack limn op querist ulva wax oyez (11 words, 37 letters)
ABC defog hijack limn op querist ulva wax oyez (9 words, 38 letters)
Using words in 9C and NI3:
a BCD ef ghi jak limn op qres to uva wax oyez (12 words, 34 letters)
ABC defy ghi jak limn op qres to uva wax oyez (11 words, 35 letters)
ABC defy ghi jak limn opaquers turves wax oyez (9 words, 38 letters)
scabicide afghani jokul manrope querist purview oxygenize (7 words, 51 letters)
Using words in 9C, NI3 and NI3+:
a BCD ef ghi jak limn op QRS to uva wax yez (12 words, 32 letters)
ABC defy ghi jak limn op QRS to uva wax yez (11 words, 33 letters)
ab cad ef ghi jak limn op qre stun vow ox yez (12 words, 34 letters)
ABC defy ghi jak limn op querist uva wax yez (10 words, 35 letters)
Using words in 9C, NI3, NI3+, NI2 and NI2+:
ABC def ghi jak limn op qre struv wax yez (10 words, 32 letters)
ABC def ghi jak limn opaquer struv wax yez (9 words, 34 letters)
Using words in 9C, NI3, NI3+, NI2, NI2+ and the OED:
ABC defog hij klam nop QRS tu vow XYZ (9 words, 29 letters, Jeff Grant)
ABC def ghi jak limn op qres tu vow XYZ (10 words, 30 letters)
ABC defog hij klam nop querist uvrow XYZ (8 words, 33 letters, Jeff Grant)
ABC defyghe bij sklim nop querist uvrow XYZ (8 words, 36 letters)
ABC defog hijack limnophil querist uvrow XYZ (7 words, 38 letters, Jeff Grant)
Some exceptional panalphabetic word lists with letters in reverse alpha order:
Using words from 9C:
lazy ox ow vug tsar quip on milk jib hag fed cab a (13 words, 38 letters)
lazy ox wave uts reequip on milk jihad gifted cabal (10 words, 42 letters)
Using words from 9C and NI3:
lazy ox ow vug tsar quip on milk jib hag fed caba (12 words, 38 letters)
lazy ox wave uts roque pon milk jihad gifted caba (10 words, 40 letters)
Using words from 9C, NI3 and NI2:
zo yex wu vug tsar quip on milk jib hag fed caba (12 words, 37 letters)
zo yex wave uts roque pon milk jihad gifted caba (10 words, 39 letters)
All words are main entries in 9C except the following:
9C: ghi (at 'ghee')
NI3: caba, fyrd, jak, opaquers, pon, qre(s), squdgy, uva
NI3+: jimp, QRS (at 'QRS complex'), sklim, vox (at 'vox populi'), yez
NI2: benzoxycamphors, jackbox, limnophil, quick-flowing, yex, zo
NI2+: def, juventude, klam, quar, qvint, struv, tu, wuz
OED: defyghe (at 'defy'), bij (at 'buy'), hij, nop, uvrow (at 'yuffrouw'),
XYZ (at 'X')
The first time I saw this pangram was in Gyles Brandeth's _The Joy of Lex_.
It appeared there as:
Waltz, nymph, for quick jigs vex Bud. (7 words, 28 letters, proper noun.)
I always wondered why they didn't try modifying it as:
Waltz, nymph, for quick jigs vex buds. (7 words, 29 letters, no proper noun.)
However, why fast dances would irritate incipient flowers is beyond me,
so I tried again:
Waltz, dumb nymph, for quick jigs vex. (7 words, 29 letters, no proper noun,
makes more sense.)
However, sounds kind of sexist, and we can maybe chop off a letter and
eliminate the sexism, although suffering some loss of sense:
Waltz, bud nymph, for quick jigs vex. (7 words, 28 letters, no proper noun,
makes less sense.)
There are river nymphs and tree nymphs and mountain nymphs, so there can
be nymphs of the aforementioned incipent flowers, right? Sense is a matter
of opinion, so you can move the bud around or turn it into another imperative
verb rather than a noun-as-adjective:
Waltz, nymph, bud, for quick jigs vex. (7 words, 28 letters, no proper noun,
sense is dubious.)
[We've all heard of budding youth, right?]
Waltz, nymph, for quick bud jigs vex. (7 words, 28 letters, no proper noun,
sense is dubious.)
[Yeah, we've all learned to dance a merry jig that looks like one of those
infamous incipient flowers.]
Dub waltz, nymph, for quick jigs vex. (7 words, 28 letters, no proper noun,
came up with this on the spot and
actually it looks pretty good!)
[The idea being that a nymph, being in control of the soundtrack for a TV
sitcom, has to change the music to which a grandmother is listening, from
something from Ireland to something from Strauss.]
-- Stephen Joseph Smith <sjs...@cs.umd.edu>
It is fairly straightforward, if time-consuming, to search for minimal
pangrams given a suitable lexicon, and the enclosed program does this.
The run time is of the order of 20 MIPS-days if fed `Official Scrabble
Words', a document nominally listing all sufficiently short words
playable in tournament Scrabble in Britain.
I also enclose a lexicon which will reproduce the OSW results much more
quickly.
The results are dominated by onomatopoeic interjections (`pst', `sh',
etc.), and words borrowed from Welsh (`cwm', `crwth') and Arabic (`qat',
`suq'). Other lexicons will contain a very different leavening of such
words, and yield a very different set of pangrams.
Readers are invited to form sentences (or, less challenging, newspaper
headlines) from these pangrams. Few are amenable to this sort of thing.
-- Steve Thomas
-----cut-----here-----
#include <stdio.h>
#include <ctype.h>
extern void *malloc ();
extern void *realloc ();
long getword ();
#define MAXWORD 26
int list[MAXWORD];
int lp;
struct list {
struct list *next;
char *word;
};
struct word {
long mask;
struct list *list;
} *w;
int wp;
int wsize;
char wordbuf[BUFSIZ];
char *letters = "qxjzvwfkbyhpmgcdultnoriase";
int cmp (ap, bp)
struct word *ap, *bp;
{
char *p;
long a = ap->mask, b = bp->mask;
for (p = letters; *p; p++)
{
long m = 1L << (*p - 'a');
if ((a & m) != (b & m))
if ((a & m) != 0)
return -1;
else
return 1;
}
return 0;
}
void *
newmem (p, size)
void *p;
int size;
{
if (p)
p = realloc (p, size);
else
p = malloc (size);
if (p == NULL) {
fprintf (stderr, "Out of memory\n");
exit (1);
}
return p;
}
char *
dupstr (s)
char *s;
{
char *p = newmem ((void *)NULL, strlen (s) + 1);
strcpy (p, s);
return p;
}
main (argc, argv)
int argc;
char **argv;
{
long m;
int i, j;
while ((m = getword (stdin)) != 0)
{
if (wp >= wsize)
{
wsize += 1000;
w = newmem (w, wsize * sizeof (struct word));
}
w[wp].mask = m;
w[wp].list = newmem ((void *)NULL, sizeof (struct list));
w[wp].list->word = dupstr (wordbuf);
w[wp].list->next = (struct list *)NULL;
wp++;
}
qsort (w, wp, sizeof (struct word), cmp);
for (i = 1, j = 1; j < wp; j++)
{
if (w[j].mask == w[i - 1].mask) {
w[j].list->next = w[i - 1].list;
w[i - 1].list = w[j].list;
} else
w[i++] = w[j];
}
wp = i;
pangram (0L, 0, letters);
exit (0);
}
pangram (sofar, min, lets)
long sofar;
int min;
char *lets;
{
register int i;
register long must;
if (sofar == 0x3ffffff) {
print ();
return;
}
for (; *lets; lets++)
if ((sofar & (1 << (*lets - 'a'))) == 0)
break;
must = 1 << (*lets - 'a');
for (i = min; i < wp; i++)
{
if (w[i].mask & sofar)
continue;
if ((w[i].mask & must) == 0)
continue;
list[lp++] = i;
pangram (w[i].mask | sofar, i + 1, lets);
--lp;
}
}
long
getword (fp)
FILE *fp;
{
long mask, m;
char *p;
char c;
while (fgets (wordbuf, sizeof (wordbuf), fp) != NULL) {
p = wordbuf;
mask = 0L;
while ((c = *p++) != '\0') {
if (!islower (c))
break;
m = 1L << (c - 'a');
if ((mask & m) != 0)
break;
mask |= m;
}
if (c == '\n')
p[-1] = c = '\0';
if (c == '\0' && mask)
return mask;
}
return 0;
}
print ()
{
int i;
for (i = 0; i < lp; i++)
{
struct word *p = &w[list[i]];
struct list *l;
if (p->list->next == NULL)
printf ("%s", p->list->word);
else {
printf ("(");
for (l = p->list; l; l = l->next) {
printf ("%s", l->word);
if (l->next)
printf (" ");
}
printf (")");
}
if (i != lp - 1)
printf (" ");
}
printf ("\n");
fflush (stdout);
}
-----and-----here-----
ankh
bad
bag
bald
balk
balks
band
bandh
bang
bank
bap
bard
barf
bark
bed
beds
beg
bend
benj
berk
berks
bez
bhang
bid
big
bight
bilk
bink
bird
birds
birk
bisk
biz
blad
blag
bland
blank
bled
blend
blight
blimp
blin
blind
blink
blintz
blip
blitz
block
blond
blunk
blunks
bod
bods
bog
bok
boks
bold
bond
bong
bonk
bonks
bop
bord
bords
bosk
box
brad
brank
bred
brink
brinks
brod
brods
brog
brogh
broghs
bud
bug
bugs
bulk
bulks
bump
bumps
bund
bunds
bung
bungs
bunk
bunks
burd
burds
burg
burgh
burghs
burk
burks
burp
busk
by
ch
crwth
cwm
cwms
dab
dag
dak
damp
dap
deb
debs
debt
deft
delf
delfs
delft
delph
delphs
depth
derv
dervs
dhak
dib
dig
dight
dink
dinks
dirk
disk
div
divs
dob
dobs
dog
dop
dorp
dowf
drab
draft
drib
dribs
drip
drop
drub
drubs
drunk
drunks
dub
dug
dugs
dung
dunk
dunks
dup
dusk
dwarf
dzo
dzos
fad
fag
falx
fank
fard
fax
fed
fend
fends
fenks
fez
fib
fid
fig
fight
fink
firk
fisk
fix
fiz
fjord
fjords
flab
flag
flak
flank
flap
flax
fled
fleg
flex
flight
flimp
flip
flisk
flix
flog
flogs
flong
flongs
flop
flops
flub
flump
flumps
flung
flunk
flux
fob
fobs
fog
fogs
fold
folds
folk
folks
fond
fonds
fop
fops
ford
fords
fork
forks
fox
frab
frank
frap
fremd
fright
friz
frog
frond
frump
frumps
fub
fud
fug
fugs
fund
funds
funk
funks
fy
fyrd
fyrds
gab
gad
gamb
gamp
gap
gawk
gawp
ged
geld
gib
gid
gif
gild
gink
gip
gju
gjus
gled
glib
glid
glift
glitz
glob
globs
glyph
glyphs
gob
god
gold
golf
golfs
golp
golps
gonk
gov
govs
gowd
gowf
gowfs
gowk
gowks
graft
graph
grub
grypt
gub
gubs
gulf
gulfs
gulp
gulph
gulphs
gunk
gup
gym
gymp
gyp
gyps
hadj
hank
hyp
hyps
jab
jag
jak
jamb
jap
jark
jerk
jerks
jib
jibs
jig
jimp
jink
jinks
jinx
jird
jirds
jiz
job
jobs
jog
jogs
jud
juds
jug
jugs
junk
junks
jynx
kang
kant
kaw
keb
kebs
ked
kef
kefs
keg
kelp
kemb
kemp
kep
kerb
kerbs
kerf
kerfs
kex
khan
khud
khuds
kid
kids
kif
kifs
kight
kild
kiln
kilp
kind
kinds
king
kip
klepht
knag
knight
knob
knobs
knub
knubs
kob
kobs
kond
kop
kops
kraft
krantz
kranz
kvetch
ky
kynd
kynds
lav
lev
lez
link
luz
lynx
mawk
nabk
nth
pad
park
pawk
pax
ped
peg
pegh
peghs
pelf
pelfs
penk
perk
perv
pervs
phang
phiz
phlox
pig
pight
pix
pleb
plebs
pled
plight
plink
plod
plongd
plonk
pluck
plug
plumb
plumbs
plunk
ply
pod
polk
polks
pong
pork
pox
poz
prex
prod
prof
prog
pst
pub
pud
pug
pugh
pulk
pulks
punk
pyx
qat
qats
qibla
qiblas
quark
quiz
sh
skelf
skid
skrump
skug
sphinx
spiv
squawk
st
sunk
suq
swiz
sylph
tank
thilk
tyg
vamp
van
vang
vant
veg
veld
velds
veldt
vend
vends
verb
verbs
vet
vex
vibs
vild
vint
vly
vox
vug
vugs
vuln
waltz
wank
welkt
whack
zag
zap
zarf
zax
zed
zek
zeks
zel
zig
zigs
zimb
zimbs
zing
zings
zip
zips
zit
zurf
zurfs
==> language/english/sentences/repeated.words.p <==
What is a sentence with the same word several times repeated? Do not use
quotation marks, proper names, a language other than English, or anything
else distasteful.
==> language/english/sentences/repeated.words.s <==
Five "had"s in a row:
The parents were unable to conceive, so they hired someone else to
be a surrogate.
The parents had had a surrogate have their child.
The parents had had had their child.
The child had had no breakfast.
The child whose parents had had had had had no breakfast.
It is arguably possible to construct sentences with arbitrarily many
repetitions of a word. For example:
1. Bulldogs fight
2. Bulldogs bulldogs fight fight.
(i.e., bulldogs (that) bulldogs fight, (themselves) fight)
3. Bulldogs bulldogs bulldogs fight fight fight.
(i.e., bulldogs (that) bulldogs (that) bulldogs fight, (themselves) fight,
(themselves) fight)
...
n. etc.
==> language/english/sentences/sentence.p <==
Find a sentence with words beginning with the letters of the alphabet, in order.
==> language/english/sentences/sentence.s <==
After boxes containing dynamite exploded furiously, generating hellish
inferno jet killing laboring miners, novice operator, paralyzed,
quickly refuses surgical treatment until veteran workers x-ray youth
zealously.
A big cuddly dog emitted fierce growls, happily ignoring joyful kids
licking minute nuts on pretty queer rotten smelly toadstools underneath
vampires who x-rayed young zombies.
==> language/english/sentences/snowball.p <==
Construct the longest coherent sentence you can such that the nth word
is n letters long.
==> language/english/sentences/snowball.s <==
I
do
not
know
where
family
doctors
acquired
illegibly
perplexing
handwriting;
nevertheless,
extraordinary
pharmaceutical
intellectuality,
counterbalancing
indecipherability,
transcendentalizes
intercommunications'
incomprehensibleness.
If you add the condition that the letters in each word must be in
inverse alphabetic order, you have:
A
zo
fed
upon
solid
toffee
zyxomma.
==> language/english/sentences/weird.p <==
Make a sentence containing only words that violate the "i before e" rule.
==> language/english/sentences/weird.s <==
From the May, 1990 _Word Ways_:
That is IE - Or, Is That EI?
by Paul Leopold
Stockholm, Sweden
"Seeing wherein neither weirdly-veiled sovereign deigned
agreeing, their feisty heirs, leisurely eyeing eight heinous
deity-freightened reindeer sleighs, counterfeited spontaneity,
freeing rein (reveille, neighing!); forfeited obeisance,
fleeing neighborhood. Kaleidoscopically-veined foreign
heights being seized, either reigned, sleight surfeited,
therein; reinvented skein-dyeing; reiteratedly inveighed,
feigning weighty seismological reinforcement."
The above passage appears in a book on the ecological conservation
measures of the enlightened plutocracies of antiquity, Ancient
Financier Aristocracies' Conscientious Scientific Species Policies,
by Creighton Leigh Peirce and Keith Leiceister Reid. . . .
Any beings decreeing such ogreish, albeit nonpareil,
homogeneity must be nucleic protein-deficient from sauteing
pharmacopoeial caffeine and codeine!
From an 'fgrep cie /usr/dict/words', with similiar words removed.
ancient coefficient concierge conscience conscientious deficient efficient
financier glacier hacienda Muncie omniscient proficient science
Societe(?) society species sufficient
A search through Webster's on-line dictionary produced the following exceptions:
Word: *cie*
Possible matches are:
1. -facient 2. abortifacient 3. ancien regime
4. ancient 5. ancientry 6. boccie
7. cenospecies 8. christian science 9. coefficient
10. concierge 11. conscience 12. conscience money
13. conscientious 14. conscientious objector15. deficiency
16. deficiency disease 17. deficient 18. domestic science
19. earth science 20. ecospecies 21. efficiency
22. efficiency engineer 23. efficient 24. facies
25. fancier 26. financier 27. genospecies
28. geoscience 29. glacier 30. glacier theory
31. habeas corpus ad subjiciendum32. hacienda 33. inconscient
34. inefficiency 35. inefficient 36. insufficience
37. insufficiency 38. insufficient 39. international scientific vocabulary
40. library science 41. liquefacient 42. mental deficiency
43. mutafacient 44. natural science 45. nescience
46. omniscience 47. omniscient 48. physical science
49. political science 50. precieux 51. prescience
52. prescientific 53. prima facie 54. proficiency
55. proficient 56. pseudoscience 57. rubefacient
58. science 59. science fiction 60. scient
61. sciential 62. scientific 63. scientific method
64. scientism 65. scientist 66. scientistic
67. secret society 68. self-sufficiency 69. self-sufficient
70. social science 71. social scientist 72. societal
73. society 74. society verse 75. somnifacient
76. specie 77. species 78. stupefacient
79. sub specie aeternitatis80. subspecies 81. sufficiency
82. sufficient 83. sufficient condition 84. superficies
85. type species 86. unscientific 87. valenciennes
88. vers de societe
==> language/english/sentences/word.boundaries.p <==
List some sentences that can be radically altered by changing word boundaries
and punctuation.
==> language/english/sentences/word.boundaries.s <==
Issues topping our mail: manslaughter.
Is Sue stopping our mailman's laughter?
The real ways I saw it.
There always is a wit.
You read evil tomes, Tim, at Ed's issue.
"You're a devil, Tom!" estimated sis Sue.
==> language/english/spelling/gry.p <==
Find three completely different words ending in "gry."
==> language/english/spelling/gry.s <==
Aside from "angry" and "hungry" and words derived therefrom, there is
only one word ending with "-gry" in Webster's Third Unabridged: "aggry."
However, this word is defective in that it is part of a phrase "aggry beads."
The OED's usage examples all talk about "aggry beads."
Moving to older dictionaries, we find that "gry" itself is a word in Webster's
Second Unabridged (and the OED):
gry, n. [L. gry, a trifle; Gr. gry, a grunt]
1. a measure equal to one-tenth of a line. [Obs.] (Obs. = obsolete)
2. anything very small. [Rare.]
This is a list of 100 words, phrases and names ending in "gry":
[Explanation of references is given at the end of the list.]
aggry [OED:1:182; W2; W3]
Agry Dagh (Mount Agry) [EB11]
ahungry [OED:1:194; FW; W2]
angry [OED; FW; W2; W3]
anhungry [OED:1:332; W2]
Badagry [Johnston; EB11]
Ballingry [Bartholomew:40; CLG:151; RD:164, pl.49]
begry [OED:1:770,767]
bewgry [OED:1:1160]
bowgry [OED:1:1160]
braggry [OED:1:1047]
Bugry [TIG]
Chockpugry [Worcester]
Cogry [BBC]
cony-gry [OED:2:956]
conyngry [OED:2:956]
Croftangry [DFC, as "Chrystal Croftangry"]
dog-hungry [W2]
Dshagry [Stieler]
Dzagry [Andree]
eard-hungry [CED (see "yird"); CSD]
Echanuggry [Century:103-104, on inset map, Key 104 M 2]
Egry [France; TIG]
ever-angry [W2]
fire-angry [W2]
Gagry [EB11]
gry (from Latin _gry_) [OED:4/2:475; W2]
gry (from Romany _grai_) [W2]
haegry [EDD (see "hagery")]
half-angry [W2]
hangry [OED:1:329]
heart-angry [W2]
heart-hungry [W2]
higry pigry [OED:5/1:285]
hogry [EDD (see "huggerie"); CSD]
hogrymogry [EDD (see "huggerie"); CSD (as "hogry-mogry")]
hongry [OED:5/1:459; EDD:3:282]
huggrymuggry [EDD (see "huggerie"); CSD (as "huggry-muggry")]
hungry [OED; FW; W2; W3]
Hungry Bungry [Daily Illini, in ad for The Giraffe, Spring 1976]
iggry [OED]
Jagry [EB11]
kaingry [EDD (see "caingy")]
land-hungry [OED; W2]
leather-hungry [OED]
Langry [TIG; Times]
Lisnagry [Bartholomew:489]
MacLoingry [Phillips (as "Flaithbhertach MacLoingry")]
mad-angry [OED:6/2:14]
mad-hungry [OED:6/2:14]
magry [OED:6/2:36, 6/2:247-48]
malgry [OED:6/2:247]
man-hungry [OED]
Margry [Indians (see "Pierre Margry" in bibliog., v.2, p.1204)]
maugry [OED:6/2:247-48]
mawgry [OED:6/2:247]
meagry [OED:6/2:267]
meat-hungry [W2]
menagry [OED (see "managery")]
messagry [OED]
nangry [OED]
overangry [RH1; RH2]
Pelegry [CE (in main index as "Raymond de Pelegry")]
Pingry [Bio-Base; HPS:293-94, 120-21]
podagry [OED; W2 (below the line)]
Pongry [Andree (Supplement, p.572)]
pottingry [OED:7/2:1195; Jamieson:3:532]
puggry [OED:8/1:1573; FW; W2]
pugry [OED:8/1:1574]
rungry [EDD:5:188]
scavengry [OED (in 1715 quote under "scavengery")]
Schtschigry [LG/1:2045; OSN:97]
Seagry [TIG; EB11]
Segry [Johnston; Andree]
self-angry [W2]
self-hungry ?
Shchigry [CLG:1747; Johnson:594; OSN:97,206; Times:185,pl.45]
shiggry [EDD]
Shtchigry [LG/1:2045; LG/2:1701]
Shtshigry [Lipp]
skugry [OED:9/2:156, 9/1:297; Jamieson:4:266]
Sygry [Andree]
Tangry [France]
Tchangry [Johnson:594; LG/1:435,1117]
Tchigry [Johnson:594]
tear-angry [W2]
tike-hungry [CSD]
Tingry [France; EB11 (under "Princesse de Tingry")]
toggry [Simmonds (as "Toggry", but all entries are capitalized)]
ulgry [Partridge; Smith:24-25]
unangry [OED; W2]
vergry [OED:12/1:123]
Virgy [CLG:2090]
Wirgy [CLG:2090; NAP:xxxix; Times:220, pl.62; WA:948]
wind-angry.
wind-hungry [W2]
yeard-hungry [CED (see "yird")]
yerd-hungry [CED (see "yird"); OED]
yird-hungry [CED (see "yird")]
Ymagry [OED:1:1009 (col. 3, 1st "boss" verb), (variant of "imagery")]
This list was gathered from the following articles:
George H. Scheetz, In Goodly Gree: With Goodwill, Word Ways 22:195 (Nov. 1989)
Murray R. Pearce, Who's Flaithbhertach MacLoingry?, Word Ways 23:6 (Feb. 1990)
Harry B. Partridge, Gypsy Hobby Gry, Word Ways 23:9 (Feb. 1990)
A. Ross Eckler, -Gry Words in the OED, Word Ways 25:4 (Nov. 1992)
References:
(Many references are of the form [Source:volume:page] or [Source:page].)
Andree, Richard. Andrees Handatlas (index volume). 1925.
Bartholomew, John. Gazetteer of the British Isles: Statistical and
Topographical. 1887.
BBC = BBC Pronouncing Dictionary of English Names.
Bio-Base. (Microfiche) Detroit: Gale Research Company. 1980.
CE = Catholic Encyclopedia. 1907.
CED = Chambers English Dictionary. 1988.
Century = "India, Northern Part." The Century Atlas of the World. 1897, 1898.
CLG = The Colombia Lippincott Gazetteer of the World. L.E.Seltzer, ed. 1952.
CSD = Chambers Scots Dictionary. 1971 reprint of 1911 edition.
Daily Illini (University of Illinois at Urbana-Champaign).
DFC = Dictionary of Fictional Characters. 1963.
EB11 = Encyclopedia Britannica, 11th ed.
EDD = The English Dialect Dictionary. Joseph Wright, ed. 1898.
France = Map Index of France. G.H.Q. American Expeditionary Forces. 1918.
FW = Funk & Wagnalls New Standard Dictionary of the English Language. 1943.
HPS = The Handbook of Private Schools: An Annual Descriptive Survey of
Independent Education, 66th ed. 1985.
Indians = Handbook of American Indians North of Mexico. F. W. Hodge. 1912.
Jamieson, John. An Etymological Dictionary of the Scottish Language. 1879-87.
Johnston, Keith. Index Geographicus... 1864.
LG/1 = Lippincott's Gazetteer of the World: A Complete Pronouncing Gazetteer
or Geographical Dictionary of the World. 1888.
LG/2 = Lippincott's New Gazetteer: ... 1906.
Lipp = Lippincott's Pronouncing Gazetteer of the World. 1861, undated
edition from late 1800's; 1902.
NAP = Narodowy Atlas Polski. 1973-1978 [Polish language]
OED = The Oxford English Dictionary. 1933. [Form: OED:volume/part number if
applicable:page]
OSN: U.S.S.R. Volume 6, S-T. Official Standard Names Approved by the United
States Board on Geographic Names. Gazetteer #42, 2nd ed. June 1970.
Partridge, Harry B. "Ad Memoriam Demetrii." Word Ways, 19 (Aug. 1986): 131.
Phillips, Lawrence. Dictionary of Biographical Reference. 1889.
RD = The Reader's Digest Complete Atlas of the British Isles, 1st ed. 1965.
RH1 = Random House Dictionary of the English Language, Unabridged. 1966.
RH2 = Random House Dictionary of the English Language, Second Edition
Unabridged. 1987.
Simmonds, P.L. Commercial Dictionary of Trade Products. 1883.
Smith, John. The True Travels, Adventvres and Observations: London 1630.
Stieler, Adolph. Stieler's Handatlas (index volume). 1925.
TIG = The Times Index-Gazetteer of the World. 1965.
Times = The Times Atlas of the World, 7th ed. 1985.
W2 = Webster's New International Dictionary of the English Language,
Second Edition, Unabridged. 1934.
W3 = Webster's Third New International Dictionary of the English Language,
Unabridged. 1961.
WA = The World Atlas: Index-Gazetteer. Council of Ministires of the USSR, 1968.
Worcester, J.E. Universal Gazetteer, Second Edition. 1823.
Some words containing "gry" that do not end with "gry": agrypnia,
agrypnotic, Gryllidae, gryllid, gryllus, Gryllus, grylloblattid,
Gryllotalpa, gryllos, grypanian, Gryphaea, Gryll, Gryphaea, gryposis,
grysbok, gryphon, Gryphosaurus, Grypotherium, grysbuck. Most of these
are in Webster's Second also with one from Webster's Third Edition and
one from the Random House Dictionary, Second Edition Unabridged.
==> language/english/spelling/j.ending.p <==
What words and names end in j?
==> language/english/spelling/j.ending.s <==
Following is a compilation of words ending in j from various
dictionaries. Capitalized words and words marked as foreign
are included, but to keep the list to a managable size,
personal and place names are excluded.
aflaj plural of falaj (Cham)
benj variant of bhang - hemp plant (NI2)
bhimraj the rachet-tailed drongo (F&W)
Bhumij branch of Munda tribes in India (NI3)
Chuj a people of Northwestern Guatemala (NI3)
esraj an Indian musical instrument with 3 or 4 strings (OED2)
falaj a water channel as part of the ancient irrigation
system of Oman (Cham)
Funj variant of Fung - a people dominant in Sennar (NI3)
gaj Omanese coin (NI2)
genj a common type of cotton cloth in Sudan (F&W)
gunj a grannery in India (NI2)
hadj variant of hajj (NI3)
haj variant of hajj (NI3)
hajilij the bito - a small scrubby tree that grows in dry
parts of Africa and Asia (NI2)
hajj pilgimage to Mecca (NI3)
hij obsolete form of hie or high (OED2)
Jubaraj variant of Yuvaraja - the male heir to an Indian
pricipality (OED2)
kaleej variant of kalij (NI3)
kalij any of several crested Indian pheasants (NI3)
kankrej guzerat - a breed of Indian cattle (NI3)
kharaj a tax on unbelievers (NI2)
Khawarij plural of Kharijite - a member of the oldest
religious sect of Islam (NI3)
khiraj variant of kharaj (NI2)
kilij a Turkish saber with a crescent shaped blade (RHD)
kurunj variant of kurung - the Indian beech (NI2)
Maharaj variant of Maharaja - East Indian prince (OED2)
munj a tough Asiatic grass (NI3)
naranj Maldive Island name for mancala - an Arabian board
game (CD)
pakhawaj a doubleheaded drum used in Indian music (OED2)
raj rule (NI3)
saj the Indian laurel (NI2)
samaj Hindu religious society (NI3)
sohmaj variant of samaj (NI2)
somaj variant of samaj (NI2)
svaraj variant of swaraj (F&W)
swaraj local self-government in India (NI3)
taj a tall conical cap worn by Moslems (NI3)
tedj variant of tej (OED2)
tej Ethiopian mead (OED2)
Viraj in Hindu Mythology, the mysterious primeval being
when differentiating itself into male and female (F&W)
Yuvaraj same as Jubaraj (OED2)
Yuveraj same as Jubaraj (OED2)
Yuvraj same as Jubaraj (OED2)
zij Persian astronomical tables (F&W)
This list is almost certainly not complete. For example, on
page 187 of Beyond Language, Dmitri Borgmann has "Udruj" in a
word list. What reference he dug this word out of is unknown;
the combined efforts of the NPL electronic mailing list could
not produce the source of this word. So additions to this list
will be welcomed by the author.
REFERENCES
CD - The Century Dictionary and Cyclopedia, 1911
Cham - Chambers English Dictionary, 1988
F&W - Funk & Wagnall's New Standard Dictionary of the English
Language, 1941
NI2 - Webster's New International Dictionary, Second Edition,
1942
NI3 - Webster's Third New International Dictionary, 1981
OED2 - Oxford English Dictionary, Second Edition, 1989
RHD - Random House Dictionary of the English Language, 1966
---
Dan Tilque -- da...@logos.WR.TEK.COM
==> language/english/spelling/lipograms.p <==
What books have been written without specific letters, vowels, etc.?
==> language/english/spelling/lipograms.s <==
Such a book is called a lipogram.
A novel-length example in English (omitting e) exists, titled _Gadsby_.
Georges Perec wrote a French novel titled _La Disparition_ which does
not contain the letter 'e', except in a few bits of text that the
publisher had to include in or on the book somewhere -- such as the
author's name :-). But these were all printed in red, making them
somehow ``not count''.
Perec also wrote another novel in which `e' was the only vowel.
In _La Disparition_, unlike _Gadsby_, the lipogrammatic
technique is reflected in the story. Objects disappear or become
invisible. We know, however, more or less why the characters can't
find things like eggs or even remember their names -- because the
words for them can't be used.
Amazingly, it's been ``translated'' into English (by Harry Mathews, I
think).
Another work which manages to [almost] adhere to restrictive
alphabetic rules while also remaining readable as well as providing
amusement and literary satisfaction (though you have to like
disjointed fiction) is _Alphabetical Africa_ by Walter Abish. The
rules (which of course he doesn't explain, you can't help noticing
most of them) have to do with initial letters of words. There are 52
chapters. In the first, all words begin with `a'; in the second, all
words begin with either `a' or `b'; etc, until all words are allowed
in chapter 26. Then in the second half, the letters are taken away
one by one. It's remarkable when, for instance, you finally get `the'
and realize how much or little you missed it; earlier, when `I' comes
in, you feel something like the difference between third- and
first-person narration. As one of the blurbs more or less says (I
don't have it here to quote), reading this is like slowly taking a
deep breath and letting it out again.
----
Mitch Marks mitc...@cs.uchicago.edu
==> language/english/spelling/longest.p <==
What is the longest word in the English language?
==> language/english/spelling/longest.s <==
The longest word to occur in both English and American "authoritative"
unabridged dictionaries is "pneumonoultramicroscopicsilicovolcanoconiosis."
The following is a brief citation history of this "word."
New York Herald Tribune, February 23, 1935, p. 3
"Pneumonoultramicroscopicsilicovolcanokoniosis succeeded
electrophotomicrographically as the longest word in the English
language recognized by the National Puzzlers' League at the opening
session of the organization's 103d semi-annual meeting held yesterday
at the Hotel New Yorker.
The puzzlers explained that the forty-five-letter word is the name of a
special form of silicosis caused by ultra-microscopic particles of
siliceous volcanic dust."
Everett M. Smith (b. 1/1/1894), President of NPL and Radio News Editor
of the Christian Science Monitor, cited the word at the convention.
Smith was also President of the Yankee Puzzlers of Boston.
It is not known whether Smith coined the word.
"Bedside Manna. The Third Fun in Bed Book.", edited by Frank Scully,
Simon and Schuster, New York, 1936, p. 87
"There's been a revival in interest in spelling, but Greg Hartswick,
the cross word king and world's champion speller, is still in control
of the situation. He'd never get any competition from us, that's
sure, though pronouncing, let alone spelling, a 44 letter word like:
Pneumonoultramicrosopicsilicovolkanakoniosis,
a disease caused by ultra-microscopic particles of sandy volcanic dust
might give even him laryngitis."
It is likely that Scully, who resided in New York in February 1935,
read the Herald Tribune article and slightly misremembered the word.
Supplement to the Oxford English Dictionary, 1936
Both "-coniosis" and "-koniosis" are cited.
"a factitious word alleged to mean 'a lung disease caused by the inhalation
of very fine silica dust' but occurring chiefly as an instance of a very long
word."
Webster's first cite is "-koniosis" in the addendum to the Second Edition.
The Third Edition changes the "-koniosis" to "-coniosis."
I conjecture that this "word" was coined by word puzzlers, who then
worked assiduously to get it into the major unabridged dictionaries
(perhaps with a wink from the editors?) to put an end to the endless
squabbling about what is the longest word.
==> language/english/spelling/most.p <==
What word has the most variant spellings?
==> language/english/spelling/most.s <==
catercorner
There's eight spellings in Webster's Third.
catercorner
cater-cornered
catacorner
cata-cornered
catty-corner
catty-cornered
kitty-corner
kitty-cornered
If you look in Random House, you will find one more which doesn't appear
in Web3, but it only differs by a hyphen:
cater-corner
---
Dan Tilque -- da...@techbook.com
==> language/english/spelling/near.palindrome.p <==
What are some long near palindromes, i.e., words that except for one
letter would be palindromes?
==> language/english/spelling/near.palindrome.s <==
Here are the longest near palindromes in Webster's Ninth Collegiate:
catalatic footstool red pepper
detonated locofocos red spider
dew-clawed nabataean retreater
eisegesis possessor stargrass
foolproof ratemeter webmember
==> language/english/spelling/operations.on.words/deletion.p <==
What exceptional words turn into other words by deletion of letters?
==> language/english/spelling/operations.on.words/deletion.s <==
longest beheadable word P(REDETERMINATION) (16/15)
longest for each letter (6-88,181,198,213,13-159,14-219,15-155,16-96,220,
17-85) APATHETICALLY, BLITHESOME, CHASTENING, DEMULSIFICATION,
EMOTIONLESSNESS, FUTILITARIANISM, GASTRONOMICALLY, HEDRIOPHTHALMA,
IDENTIFICATION, JUNCTIONAL, KINAESTHETIC, LIMITABLENESS, METHYLACETYLENE,
NEOPALEOZOIC, OENANTHALDEHYDE, PREDETERMINATION, QUINTA, REVOLUTIONARILY,
SELECTIVENESS, TREASONABLENESS, UPRAISER, VINDICATION, WHENCEFORWARD,
XANTHOPHYLLITE, YOURSELVES, ZOOSPORIFEROUS
longest beheadable down to a single letter PRESTATE (8)
longest curtailable word (not a plural) (COUNTERMAN)D, (UNDERCOVER)T (11)
longest curtailable down to a single letter LAMBASTES
longest alternately beheadable and curtailable word ASHAMED (7)
longest arbitrarily beheadable and curtailable (all subsequences words)
SHADES (6)
longest terminal ellision word D(EPILATION)S (11)
longest letter subtraction down to a single letter STRANGLING,
STRANGING, STANGING, STAGING, SAGING, AGING, GING, GIN, IN, I
longest charitable word (subtract letter anywhere)
PLEATS: LEATS,PEATS,PLATS,PLEAS,PLEAT
shortest stingy word (no deletion possible) PRY (3)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/spelling/operations.on.words/insertion.and.deletion.p <==
What exceptional words turn into other words by both insertion and
deletion of letters?
==> language/english/spelling/operations.on.words/insertion.and.deletion.s <==
longest word both charitable and hospitable
AMY: AM,AY,MY;GAMY,ARMY,AMOY,AMYL
shortest word both stingy and hostile IMPETUOUS (9)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/spelling/operations.on.words/insertion.p <==
What exceptional words turn into other words by insertion of letters?
==> language/english/spelling/operations.on.words/insertion.s <==
longest hydration (double reheadment) (D,R)EVOLUTIONIST (12/13)
longest hospitable word (insert letter anywhere)
CARES: SCARES, CHARES, CADRES, CARIES, CARETS, CARESS
shortest hostile word (no deletion possible) SYZYGY (6)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/spelling/operations.on.words/movement.p <==
What exceptional words turn into other words by movement of letters?
==> language/english/spelling/operations.on.words/movement.s <==
longest word allowing exchange of letters (metallege)
CONSERVATIONAL, CONVERSATIONAL
longest head-to-tail shift
SPECULATION, PECULATIONS
longest double head-to-tail shift
STABLE-TABLES-ABLEST
longest complete cyclic transposal ATE-TEA-EAT (3)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> language/english/spelling/operations.on.words/substitution.p <==
What exceptional words turn into other words by substitution of letters?
==> language/english/spelling/operations.on.words/substitution.s <==
longest onalosi (substitution in every position possible)
PASTERS: MASTERS,POSTERS,PALTERS,PASSERS,PASTORS,PASTELS,PASTERN
shortest isolano (no substitution possible)
ECRU
longest word, all letters changed to other letters in minimum number of
steps, yielding another word THUMBING-THUMPING-TRUMPING-TRAMPING-
TRAPPING-CRAPPING-CRAPPIES-CRAPPOES
longest word girders BADGER/SUNLIT, BUDLET/SANGIR (6)
longest word with full vowel substitution
CL(A,E,I,O,U)CKING (8) also Y D(A,E,I,O,U,Y)NE (4)
longest words with vowel substitutions
DESTRUCTIBILITIES, DISTRACTIBILITIES (17)
longest word constant-letter-shifted to another PRIMERO-SULPHUR (7)
arithmetical-letter-shifted DREAM-ETHER (5)
constant-shift-with-transposal (shiftgrams) AEROPHANE-SILVERITE (9)
longest word pair shifted one position on typewriter keyboard WAXIER-ESCORT (6)
longest word pair confusable on a telephone keypad AMOUNTS-CONTOUR (7)
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> geometry/tiling/rectangles.with.squares.p <==
Given two sorts of squares, (axa) and (bxb), what rectangles can be tiled?
==> geometry/tiling/rectangles.with.squares.s <==
A rectangle can be tiled with (axa) and (bxb) squares, iff
(i) gcd(a,b)=1 , and any of the following hold:
either: both sides of the rectangle are multiples of a;
or: both sides of the rectangle are multiples of b;
or: one side is a multiple of (ab), and the other is any length EXCEPT
one of a finite number of "bad" lengths: those numbers which are
NOT positive integer combinations of a & b. { By Sylvester's theorem
there are (a-1)(b-1)/2 of these, the largest being (a-1)(b-1)-1. }
(ii) gcd(a,b) = d .
Then merely apply (i) to the problem with a,b replaced
by a/d, b/d and the rectangle lengths also divided by d.
i.e. all cells must appear in (dxd) subsquares.
------
PROOF
It is clear that (ii) follows from (i), and that simple constructions give
the "if" part of (i). For the "only if" part, we prove that...
(S) If one side of the rectangle is not divisible by a, and the other is
not divisible by b, then the tiling is impossible.
The results in (i) follow immediately from (S).
To prove (S): ( Chakraborty-Hoey style ).
~~~~~~~~~~~~~~~~
Let the width of the rectangle be a NON-(a-multiple). Then the number of
bxb squares starting (i.e. top edge) at row 1 must be a NON-a-multiple.
Thus the number of bxb starting at row 2 must BE an a-multiple. Similarly
for the number starting at rows 3,4,...,b . Then the number starting at
row (b+1) must be a NON-a-multiple again.
Similarly the number starting at rows (2b+1), (3b+1), (4b+1),... must all be
non-a-multiples. So if the number of rows is NOT a multiple of b, (call it
bx+r), then row (bx+1) must have a NON-a-multiple of bxb squares starting
there, i.e. at least one, and there is no room left to squeeze it in. [QED]
----
A Rickard-style proof of (S) is ..BBB....BBWWW...WBBB....BBWWW...W(..etc)
~~~~~~~ also possible, by ..BBB....BBWWW...WBBB....BBWWW...W
coloring the rectangle in ..BBB....BBWWW...WBBB....BBWWW...W
vertical strips as shown here: <- a ->< b-a ><- a ->< b-a >
Every square tile covers an a-multiple of black squares. But if the width
is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there
are a NON-a-multiple of black squares in total. [QED]
(Note: the coloring must have 1 column of blacks on the right, and any
==== spare columns of whites on the left.)
===================
Bill Taylor. w...@math.canterbury.ac.nz
>A Rickard-style proof of (S) is ..BBB....BBWWW...WBBB....BBWWW...W(..etc)
> ~~~~~~~ also possible, by ..BBB....BBWWW...WBBB....BBWWW...W
>coloring the rectangle in ..BBB....BBWWW...WBBB....BBWWW...W
>vertical strips as shown here: <- a ->< b-a ><- a ->< b-a >
>
>Every square tile covers an a-multiple of black squares. But if the width
>is a NON-b-multiple, and the number of rows is a NON-a-multiple, then there
>are a NON-a-multiple of black squares in total. [QED]
>
>(Note: the coloring must have 1 column of blacks on the right, and any
> ==== spare columns of whites on the left.)
This statement of how to position the colouring isn't good enough, I'm
afraid. Take a=4, b=7 and consider e.g. a 19x10 rectangle. Coloured your
way, you get:
BWWWBBBBWWWBBBBWWWB
BWWWBBBBWWWBBBBWWWB
:::::::::::::::::::
BWWWBBBBWWWBBBBWWWB
BWWWBBBBWWWBBBBWWWB
The result has 10*10=100 black squares in it, which *is* a multiple of a=4,
despite the fact that 19 is not a multiple of 7 and 10 is not a multiple of
4.
Of course, there is an alternative offset for the pattern that does give you
the result you want:
WWBBBBWWWBBBBWWWBBB
WWBBBBWWWBBBBWWWBBB
:::::::::::::::::::
WWBBBBWWWBBBBWWWBBB
WWBBBBWWWBBBBWWWBBB
To show this happens in general: because the width of the rectangle is a
non-multiple of b, it is possible to position it on the pattern so that the
leftmost column in the rectangle is white and the column just right of the
right edge of the rectangle is black. Suppose N columns are black with this
positioning. Then the rectangle contains N*H black cells, where H is the
height of the rectangle.
If we then shift the rectangle right by one, the number of black columns
increases by 1 and it contains (N+1)*H black cells. The difference between
these two numbers of black cells is H, which is not a multiple of a.
Therefore N*H and (N+1)*H cannot both be multiples of a, and so one of these
two positionings of the pattern will suit your purposes.
David Seal
ds...@armltd.co.uk
==> geometry/tiling/scaling.p <==
A given rectangle can be entirely covered (i.e. concealed) by an
appropriate arrangement of 25 disks of unit radius.
Can the same rectangle be covered by 100 disks of 1/2 unit radius?
==> geometry/tiling/scaling.s <==
Yes. The same configuration of circles, when every distance is reduced
by half (including the diameters), will cover a similar rectangle whose
sides are one half of the original one. The original rectangle is the
union of four such rectangles.
==> geometry/tiling/seven.cubes.p <==
Consider 7 cubes of equal size arranged as follows. Place 5 cubes so
that they form a Swiss cross or a + (plus) (4 cubes on the sides and
1 in the middle). Now place one cube on top of the middle cube and the
seventh below the middle cube, to effectively form a 3-dimensional
Swiss cross.
Can a number of such blocks (of 7 cubes each) be arranged so that they
are able to completely fill up a big cube (say 10 times the size of
the small cubes)? It is all right if these blocks project out of the
big cube, but there should be no holes or gaps.
==> geometry/tiling/seven.cubes.s <==
Let n be a positive integer. Define the function f from Z^n to Z by
f(x) = x_1+2x_2+3x_3+...+nx_n. For x in Z^n, say y is a neighbor of x
if y and x differ by one in exactly one coordinate. Let S(x) be the
set consisting of x and its 2n neighbors. It is easy to check that
the values of f(y) for y in S(x) are congruent to 0,1,2,...,2n+1 (mod
2n+1) in some order. Using this, it is easy to check that every y in
Z^n is a neighbor of one and only one x in Z^n such that f(x) is
congruent to 0 (mod 2n+1). So Z^n can be tiled by clusters of the
form S(x), where f(x) is congruent to 0 mod 2n+1.
==> geometry/topology/fixed.point.p <==
A man hikes up a mountain, and starts hiking at 2:00 in the afternoon
on a Friday. He does not hike at the same speed (a constant rate), and
stops every once in a while to look at the view. He reaches the top in
4 hours. After spending the night at the top, he leaves the next day
on the same trail at 2:00 in the afternoon. Coming down, he doesn't
hike at a constant rate, and stops every once in a while to look at the
view. It takes him 3 hours to get down the mountain.
Q: What is the probability that there exists a point along the trail
that the hiker was at on the same time Friday as Saturday?
You can assume that the hiker never backtracked.
==> geometry/topology/fixed.point.s <==
100%. Superimpose the days: Friday starts walking up at 2:00,
Saturday starts walking down at 2:00. Since they are on the same
path, they must meet.
==> geometry/touching.blocks.p <==
Can six 1x2x4 blocks be arranged so that each block touches n others, for all n?
==> geometry/touching.blocks.s <==
n=0: 6 separate blocks
n=1: 3 pairs
n=2: 2 threesomes
n=3: a 3x3 grid
n=4: a box (each sides touches the four adjoining sides, but not the opposite)
n=5:
Crude ascii:
Front view: Side view:
/\ /\ -----
/ \/ \ | | |
/ /\ \ | | |
/ / \ \ | | |
\ /----\ / ---|.|.|---
\/| |\/ | | | | |
----------- -----------
| | | |
----------- -----
--
stein....@tf.tele.no [X.400] stein....@nta.no [internet]
Place block A onto the x-y plane so that four of its corners are at
(0,0), (0,1), (4,0), (4,1) (I give x and y coordinates only because
the z coordinate will always be obvious). Place block B so four of
its corners are at (2,1), (2,2), (6,1), (6,2). Now place block C with
one 4x1 face on the x-y plane with one corner at (0,1) (tangent to
block A) and tangent to block B at (2,1). Note that the angle between
block A and block C is arctan(1/2), and a corner of block C will be at
a point with approximate coordinates (3.5777, 2.7888). Call this
point P.
Now place an identical configuration of blocks on top of the first
three as follows: block D with corners at (3.4,0.4), (4.4,0.4),
(3.4,4.4), (4.4,4.4), block E with corners at (2.4,2.4), (3.4,2.4),
(2.4,-1.6), (3.4,-1.6), and block F with one corner tangent to D at
(3.4,4.4) and one side tangent to E at (2.4,2.4).
If you have been plotting this on graph paper, then the following
will be clear:
Every block touches every other in its own layer. And A and B each
touch D and E, and block C touches F. Point P falls under block D, so
blocks C and D touch, and by symmetry so do blocks F and A. And the
edge of block C intersects the edge of block E at (2.4,2.2) so blocks
C and E touch, and by symmetry so do blocks F and B. Done!
-- David Karr (ka...@cs.cornell.edu)
All the blocks are placed with their 2x4 face UP, although any face up
would have worked, as it turns out. The blocks are called AAAA BBBB CCCC,
etc.
AAAA
AAAA /_______
BBCC \
BBCC
BBCC
BBCC
/\
||
The two arrows point to the intersection of AC and BC.
Now take block "D" and place the top edge along the diagonal (between the
two arrows) so that the block extends SOUTH EAST of the line. Likely now
the block does not touch either A or B. So slide the block towards the
NORTH WEST so that it just touches A and B. You can now easily place
blocks E and F perpendicular to block "D" so that they both touch all of
ABC.....QED
--
Guy Cousineau
==> geometry/trigonometry/euclidean.numbers.p <==
For what numbers x is sin(x) expressible using only integers, +, -, *, / and
square root?
==> geometry/trigonometry/euclidean.numbers.s <==
Numbers generated by +, -, *, /, and sqrt from the integers are the
Euclidean numbers, so called because they are those for which line
segments can be constructed by use of straightedge and compass the
ratio of whose lengths has that value.
Using degrees, sin (360*M/N) (where (M,N)=1) is Euclidean if and only
if the regular polygon with N sides can be constructed by straightedge
and compass. This is true if (Gauss) and only if (easier) N is a power
of 2 times the product of different Fermat primes (3, 5, 17, 257, 65537
and probably no more). So sin (3/17) = sin (360/(2^3*3*5*17)) is
Euclidean, for example.
Some particular values:
sin(54) = (1 + sqrt(5))/4
sin(3) = sqrt(8 - sqrt(3) - sqrt(15) - sqrt(10 - 2*sqrt(5)))/4
==> geometry/trigonometry/inequality.p <==
Show that (sin x)^(sin x) < (cos x)^(cos x) when 0 < x < pi/4.
==> geometry/trigonometry/inequality.s <==
The function f(x) = x^(1/sqrt(1-x^2)) is monotonically increasing for
0 < x < 1, easily verified by taking the derivative.
Since 0 < sin x < cos x < 1 for 0 < x < pi/4, f(sin x) < f(cos x).
But f(sin x) = (sin x)^(1/cos x) and f(cos x) = (cos x)^(1/sin x).
Raising both sides to the power (cos x.sin x), we get the desired
result.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> competition/tests/math/putnam/putnam.1987.p <==
WILLIAM LOWELL PUTNAM MATHEMATICAL COMPETITION
FORTY EIGHTH ANNUAL Saturday, December 5, 1987
Examination A;
Problem A-1
------- ---
Curves A, B, C, and D, are defined in the plane as follows:
A = { (x,y) : x^2 - y^2 = x/(x^2 + y^2) },
B = { (x,y) : 2xy + y/(x^2 + y^2) = 3 },
C = { (x,y) : x^3 - 3xy^2 + 3y = 1 },
D = { (x,y) : 3yx^2 - 3x - y^3 = 0 }.
Prove that the intersection of A and B is equal to the intersection of
C and D.
Problem A-2
------- ---
The sequence of digits
1 2 3 4 5 6 7 8 9 1 0 1 1 1 3 1 4 1 5 1 6 1 7 1 8 1 9 2 0 2 1 ...
is obtained by writing the positive integers in order. If the 10^n th
digit in this sequence occurs in the part of the sequence in which the
m-digit numbers are placed, define f(n) to be m. For example f(2) = 2
because the 100th digit enters the sequence in the placement of the
two-digit integer 55. Find, with proof, f(1987).
Problem A-3
------- ---
For all real x, the real valued function y = f(x) satisfies
y'' - 2y' + y = 2e^x.
(a) If f(x) > 0 for all real x, must f'(x) > 0 for all real x? Explain.
(b) If f'(x) > 0 for all real x, must f(x) > 0 for all real x? Explain.
Problem A-4
------- ---
Let P be a polynomial, with real coefficients, in three variables and F
be a function of two variables such that
P(ux,uy,uz) = u^2*F(y-x,z-x) for all real x,y,z,u,
and such that P(1,0,0) = 4, P(0,1,0) = 5, and P(0,0,1) = 6. Also let
A,B,C be complex numbers with P(A,B,C) = 0 and |B-A| = 10. Find
|C-A|.
Problem A-5
------- ---
^
Let G(x,y) = ( -y/(x^2 + 4y^2) , x/(x^2 + 4y^2), 0 ). Prove or disprove
that there is a vector-valued function
^
F(x,y,z) = ( M(x,y,z) , N(x,y,z) , P(x,y,z) )
with the following properties:
(1) M,N,P have continuous partial derivatives for all
(x,y,z) <> (0,0,0) ;
^ ^
(2) Curl F = 0 for all (x,y,z) <> (0,0,0) ;
^ ^
(3) F(x,y,0) = G(x,y).
Problem A-6
------- ---
For each positive integer n, let a(n) be the number of zeros in the
base 3 representation of n. For which positive real numbers x does
the series
inf
-----
\ x^a(n)
> ------
/ n^3
-----
n = 1
converge?
--
Examination B;
Problem B-1
------- ---
4/ (ln(9-x))^(1/2) dx
Evaluate | --------------------------------- .
2/ (ln(9-x))^(1/2) + (ln(x+3))^(1/2)
Problem B-2
------- ---
Let r, s, and t be integers with 0 =< r, 0 =< s, and r+s =< t.
Prove that
( s ) ( s ) ( s ) ( s )
( 0 ) ( 1 ) ( 2 ) ( s ) t+1
----- + ------- + ------- + ... + ------- = ---------------- .
( t ) ( t ) ( t ) ( t ) ( t+1-s )( t-s )
( r ) ( r+1 ) ( r+2 ) ( r+s ) ( r )
( n ) n(n-1)...(n+1-k)
( Note: ( k ) denotes the binomial coefficient ---------------- .)
k(k-1)...3*2*1
Problem B-3
------- ---
Let F be a field in which 1+1 <> 0. Show that the set of solutions to
the equation x^2 + y^2 = 1 with x and y in F is given by
(x,y) = (1,0)
r^2 - 1 2r
and (x,y) = ( ------- , ------- ),
r^2 + 1 r^2 + 1
where r runs through the elements of F such that r^2 <> -1.
Problem B-4
------- ---
Let ( x(1), y(1) ) = (0.8,0.6) and let
x(n+1) = x(n)*cos(y(n)) - y(n)*sin(y(n))
and y(n+1) = x(n)*sin(y(n)) + y(n)*cos(y(n))
for n = 1,2,3,... . For each of the limits as n --> infinity of
x(n) and y(n), prove that the limit exists and find it or prove that
the limit does not exist.
Problem B-5
------- ---
Let O(n) be the n-dimensional zero vector (0,0,...,0). Let M be a
2n x n matrix of complex numbers such that whenever
( z(1), z(2), ..., z(2n)*M = O(n), with complex z(i), not all zero,
then at least one of the z(i) is not real. Prove that for arbitrary
real number r(1), r(2), ..., r(2n), there are complex numbers w(1),
w(2), ..., w(n) such that
( ( w(1) ) ) ( r(1) )
( ( . ) ) ( . )
Re ( M*( . ) ) = ( . ) .
( ( . ) ) ( . )
( ( w(n) ) ) ( r(2n) )
(Note: If C is a matrix of complex numbers, Re(C) is the matrix whose
entries are the real parts of entries of C.)
Problem B-6
------- ---
Let F be the field of p^2 elements where p is an odd prime. Suppose S
is a set of (p^2-1)/2 distinct nonzero elements of F with the property
that for each a <> 0 in F, exactly one of a and -a is in S. Let N be
the number of elements in the intersection of S with { 2a : a e S }.
Prove that N is even.
--
==> competition/tests/math/putnam/putnam.1987.s <==
Problem A-1
------- ---
Let z=x+i*y. Then A and B are the real and imaginary parts of
z^2=3i+1/z, and C, D are likewise Re and Im of z^3-3iz=1, and
the two equations are plainly equivalent. Alternatively, having
seen this, we can formulate a solution that avoids explicitly
invoking the complex numbers, starting with C=xA-yB, D=yA+xB.
Problem A-2
------- ---
Counting, we see that the numbers from 1 to n digits take
up (10^n*(9n-1)+1)/9 spaces in the above sequence. Hence we need
to find the least n for which 10^n*(9n-1)+1 > 9*10^1987, but it
is easy to see that n = 1984 is the minimum such. Therefore
f(1987) = 1984.
In general, I believe, f(n) = n + 1 - g(n), where g(n) equals
the largest value of m such that (10^m-1)/9 + 1 =< n if n>1,
and g(0) = g(1) is defined to be 0.
Hence, of course, g(n) = [log(9n-8)] if n>0. Therefore
f(0) = 1,
f(n) = n + 1 - [log(9n-8)] for n>0.
Q.E.D.
Problem A-3
------- ---
We have a differential equation, solve it. The general solution is
y = f(x) = e^x*(x^2 + a*x + b),
where a and b are arbitrary real constants. Now use completing the
square and the fact that e^x > 0 for all real x to deduce that
(1) f(x) > 0 for all real x iff 4b > a^2.
(2) f'(x) > 0 for all real x iff 4b > a^2 + 4.
It is now obvious that (2) ==> (1) but (1) /==> (2).
Q.E.D.
Problem A-4
------- ---
Setting x=0, u=1 we find F(y,z)=P(0,y,z) so F is a polynomial; keeping
x=0 but varying u we find F(uy,uz)=u^2*F(y,z), so F is homogeneous of
degree 2, i.e. of the form Ay^2+Byz+Cz^2, so
P(x,y,z)=R(y-x)^2+S(y-x)(z-x)+T(z-x)^2
for some real R,S,T. The three given values of P now specify three
linear equations on R,S,T, easily solved to give (A,B,C)=(5,-7,6).
If now P(A,B,C)=0 then (C-A)=r(B-A), r one of the two roots of
5-7r+6r^2. This quadratic has negative discrminant (=-71) so its
roots are complex conjugates; since their product is 5/6, each
one has absolute value sqrt(5/6). (Yes, you can also use the
Quadratic Equation.) So if B-A has absolute value 10, C-A must
have absolute value 10*sqrt(5/6)=5*sqrt(30)/3.
Problem A-5
------- ---
There is no such F. Proof: assume on the contrary that G extends
to a curl-free vector field on R^3-{0}. Then the integral of G
around any closed path in R^3-{0} vanishes because such a path
bounds some surface [algebraic topologists, read: because
H^2(R^3-{0},Z)=0 :-) ]. But we easily see that the integral
of F around the closed path z=0, x^2+4y^2=1 (any closed path
in the xy-plane that goes around the origin will do) is nonzero---
contradiction.
Problem A-6
------- ---
For n>0 let
T(n) = x^a(n)/n^3 and U(n) = T(3n) + T(3n+1) + T(3n+2)
and for k>=0 let
Z(k) = sum {n=3^k to 3^(k+1)-1} T(n)
We have
Z(k+1) = sum {n=3^(k+1) to 3^(k+2)-1} T(n)
= sum {n=3^k to 3^(k+1)-1} [T(3n) + T(3n+1) + T(3n+2)]
= sum {n=3^k to 3^(k+1)-1} U(n)
Let us compare U(n) to T(n). We have a(3n)=a(n)+1 and a(3n+1)=a(3n+2)=a(n).
Thus
U(n) = x^[a(n)+1]/(3n)^3 + x^a(n)/(3n+1)^3 + x^a(n)/(3n+2)^3
and so U(n) has as upper bound
x^a(n) * (x+2)/(3n)^3 = T(n) * (x+2)/27
and as lower bound
x^a(n) * (x+2)/(3n+2)^3 = T(n) * (x+2)/(3+2/n)^3
in other words U(n) = T(n) * (x+2)/(27+e(n)), where e(n)<(3+2/n)^3-27 tends to
0 when n tends to infinity. It follows then that
Z(k+1)= Z(k)*(x+2)/(27+f(k))
where f(k)<(3+2/3^k)^3-27 tends to 0 for n tending to infinity.
Now the series is the sum of all Z(k). Thus for x>25 we have Z(k+1)>Z(k) for k
large enough, and the series diverges; for x<25 we have Z(k+1)< r * Z(k) (with
r=(x+2)/27<1) for every k, and the series converges. For x=25 the series
diverges too (I think so), because Z(k+1)/Z(k) tends to 1 for k tending to
infinity.
Another proof:
I would say,for x<25. Let S(m) be the sum above taken over 3^m <= n < 3^(m+1).
Then for the n's in S(m), the base 3 representation of n has m+1 digits.
Hence we can count the number of n's with a(n)=k as being the number
of ways to choose a leading nonzero digit, times the number of ways
to choose k positions out of the m other digits for the k zeroes, times
the number of ways to choose nonzero digits for the m-k remaining positions,
namely
( m ) m-k
2 ( ) 2 .
( k )
Hence we have
3^(m+1)-1 m
----- -----
\ a(n) \ ( m ) m-k k
> x = > 2 ( ) 2 x
/ / ( k )
----- -----
n=3^m k=0
m
= 2 (x+2) .
m -3m m -3(m+1)
Hence we can bound S(m) between 2 (x+2) 3 and 2 (x+2) 3 .
It is then clear that the original sum converges just when
inf
-----
\ m -3m
> (x+2) 3 converges, or when x<25.
/
-----
m=0
Problem B-1
------- ---
Write the integrand as
(ln(x+3))^(1/2)
1 - --------------------------------- .
(ln(9-x))^(1/2) + (ln(x+3))^(1/2)
Use the change of variables x = 6-t on the above and the fact that
the two are equal to deduce that the original is equal to 1.
QED.
Problem B-3
------- ---
First note that the above values for x and y imply that
x^2 + y^2 = 1. On the other foot note that if x<>1 ,x^2 + y^2 = 1,
and 2 <> 0, then (x,y) is of the required form, with r = y/(1-x).
Also note that r^2 <> -1, since this would imply x = 1.
Derivation of r: We want x = (r^2-1)/(r^2+1) ==> 1-x = 2/(r^2+1),
and also y = 2r/(r^2+1) ==> 1-x = (2y)/(2r) if 2<>0. Hence if
2<>0, r = y/(1-x).
The above statement is false in some cases if 1+1 = 0 in F. For
example, in Z(2) the solution (0,1) is not represented.
QED.
Problem B-4
------- ---
Observe that the vector (x(n+1), y(n+1)) is obtained from (x(n), y(n))
by a rotation through an angle of y(n). So if Theta(n) is the inclination
of (x(n), y(n)), then for all n,
Theta(n+1) = Theta(n) + y(n)
Furthermore, all vectors have the same length, namely that of (x1, y1),
which is 1. Therefore y(n) = sin (Theta(n)) and x(n) = cos (Theta(n)).
Thus the recursion formula becomes
(*) Theta(n+1) = Theta(n) + sin (Theta(n))
Now 0 < Theta(1) < pi. By induction 0 < sin(Theta(n)) = sin(pi - Theta(n))
< pi - Theta(n), so 0 < Theta(n+1) < Theta(n) + (pi - Theta(n)) = pi.
Consequently, {Theta(n)} is an increasing sequence bounded above by pi, so
it has a limit, Theta. From (*) we get Theta = Theta + sin(Theta),
so with Theta in the interval (0,pi], the solution is Theta = pi.
Thus lim (x(n),y(n)) = (cos(Theta), sin(Theta)) = (-1, 0).
Problem B-5
------- ---
First note that M has rank n, else its left nullspace N has C-dimension >n
and so R-dimension >2n, and thus nontrivially intersects the R-codimension
2n subspace of vectors all of whose coordinates are real. Thus the
subspace V of C^(2n) spanned by M's columns has C-dimsension n and so
R-dimension 2n, and to prove the R-linear map Re: V-->R^(2n) surjective,
we need only prove it injective. So assume on the contrary that v is
a nonzero vector in V all of whose coordinates are purely imaginary,
and let W be the orthogonal complement of <v>; this is a subspace of
C^(2n) of C-dim. 2n-1 and R-dim. 4n-2 . W contains N,
which we've seen has R-dimension 2n; it also contains the
orthogonal complement of <i*v> in R^(2n), which has R-dimension 2n-1.
Since (2n)+(2n-1) > (4n-2), these two real subspaces of W intersect
nontrivially, producing a nonzero real vector in N---contradiction.
So Re: V-->R^(2n) indeed has zero kernel and cokernel, Q.E.D. .
Problem B-6
------- ---
Let P be the product of elements of S; then P'=2^|S|*P, the product of
the elements of 2S, is either P or -P according to whether |2S-S| is
even or odd. (each element of 2S is either in S or in -S, so match
the factors in the products for P and P'.) But by Fermat's little
theorem, 2^(p-1)=1, and since |S|=(p^2-1)/2=(p-1)*(p+1)/2 is a multiple
of p-1, also 2^|S|=1 and P=P', Q.E.D. .
This solution--analogous to one of Gauss' proof of Quadratic
Reciprocity--is undoubtedly what the proposers had in mind, and had
it been the only solution, B-6 would be a difficult problem on a par
with B-6 problems of previous years. Unfortunately, just knowing
that F* is a cyclic group of order |F|-1 for any finite field F,
one can split F* into cosets of the subgroup generated by 2 and -1
and produce a straightforward, albeit plodding and uninspired, proof.
I wonder how many of the contestants' answers to B-6 went this way
and how many found the intended solution.
Another proof:
Given such a set S, it is immediate to verify that for any a in S, if
one deletes a and adjoins -a to obtain a new set S' then the number
of elements in the intersection of S' and 2S' is congruent (modulo 2)
to the number of elements in the intersection of S and 2S. If S and
S' are any two sets meeting the condition of this problem, then S can
be changed to S' by repeating this operation several times. So, it
suffices to prove the result for any one set S meeting the condition of
the problem. A simple candidate for such an S is obtained by letting
(u, v) be a basis for F over the field of p elements and letting S
be the unions of the sets {au + bv: 1 <= u <= (p-1)/2, 0 <= b < p} and
{bv: 0 <= b < (p-1)/2}. An elementary counting argument completes the
proof.
==> competition/tests/math/putnam/putnam.1988.p <==
Problem A-1: Let R be the region consisting of the points (x,y) of the
cartesian plane satisfying both |x| - |y| <= 1 and |y| <= 1. Sketch
the region R and find its area.
Problem A-2: A not uncommon calculus mistake is to believe that the
product rule for derivatives says that (fg)' = f'g'. If f(x) =
exp(x^2), determine, with proof, whether there exists an open interval
(a,b) and a non-zero function g defined on (a,b) such that the wrong
product rule is true for x in (a,b).
Problem A-3: Determine, with proof, the set of real numbers x for
which sum from n=1 to infinity of ((1/n)csc(1/n) - 1)^x converges.
Problem A-4:
(a) If every point on the plane is painted one of three colors, do
there necessarily exist two points of the same color exactly one inch
apart?
(b) What if "three" is replaced by "nine"?
Justify your answers.
Problem A-5: Prove that there exists a *unique* function f from the
set R+ of positive real numbers to R+ such that
f(f(x)) = 6x - f(x) and f(x) > 0 for all x > 0.
Problem A-6: If a linear transformation A on an n-dimensional vector
space has n + 1 eigenvectors such that any n of them are linearly
independent, does it follow that A is a scalar multiple of the
identity? Prove your answer.
---------------------------------------------------------------------------
Problem B-1: A *composite* (positive integer) is a product ab with a
and b not necessarily distinct integers in {2,3,4,...}. Show that
every composite is expressible as xy + xz + yz + 1, with x, y, and z
positive integers.
Problem B-2: Prove or disprove: If x and y are real numbers with y >= 0
and y(y+1) <= (x+1)^2, then y(y-1) <= x^2.
Problem B-3: For every n in the set Z+ = {1,2,...} of positive
integers, let r(n) be the minimum value of |c-d sqrt(3)| for all
nonnegative integers c and d with c + d = n. Find, with proof, the
smallest positive real number g with r(n) <= g for all n in Z+.
Problem B-4: Prove that if sum from n=1 to infinity a(n) is a
convergent series of positive real numbers, then so is
sum from n=1 to infinity (a(n))^(n/(n+1)).
Problem B-5: For positive integers n, let M(n) be the 2n + 1 by 2n + 1
skew-symmetric matrix for which each entry in the first n subdiagonals
below the main diagonal is 1 and each of the remaining entries below
the main diagonal is -1. Find, with proof, the rank of M(n).
(According to the definition the rank of a matrix is the largest k
such that there is a k x k submatrix with non-zero determinant.)
One may note that M(1) = ( 0 -1 1 ) and M(2) = ( 0 -1 -1 1 1 )
( 1 0 -1 ) ( 1 0 -1 -1 1 )
( -1 1 0 ) ( 1 1 0 -1 -1 )
( -1 1 1 0 -1 )
( -1 -1 1 1 0 )
Problem B-6: Prove that there exist an infinite number of ordered
pairs (a,b) of integers such that for every positive integer t the
number at + b is a triangular number if and only if t is a triangular
number. (The triangular numbers are the t(n) = n(n + 1)/2 with n in
{0,1,2,...}.)
==> competition/tests/math/putnam/putnam.1988.s <==
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Chris Long, Rutgers University \hfill January 22, 1989}
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\problem {A-1:}
Let $R$ be the region consisting of the points $(x,y)$ of the cartesian
plane satisfying both $|x| - |y| \le 1$ and $|y| \le 1$. Sketch the
region $R$ and find its area.
\solution {Solution:}
The area is $6$; the graph I leave to the reader.
\problem {A-2:}
A not uncommon calculus mistake is to believe that the product rule for
derivatives says that $(fg)' = f'g'$. If $f(x) = e^{x^{2}}$, determine,
with proof, whether there exists an open interval $(a,b)$ and a non-zero
function $g$ defined on $(a,b)$ such that the wrong product rule
is true for $x$ in $(a,b)$.
\solution {Solution:}
We find all such functions. Note that $(fg)' = f'g' \Rightarrow f'g'
= f'g + fg'$ hence if $g(x), f'(x) - f(x) \neq 0$ we get that $g'(x)/g(x)
= f'(x)/(f'(x) - f(x))$. For the particular $f$ given, we then get that
$g'(x)/g(x) = (2x)e^{x^2}/( (2x-1) (e^{x^2}) ) \Rightarrow g'(x)/g(x) =
2x/(2x-1)$ (since $e^{x^2} > 0$). Integrating, we deduce that $\ln{|g(x)|}
= x + (1/2)\ln{|2x-1|} + c$ (an arbitrary constant) $\Rightarrow |g(x)|
= e^{c} \sqrt{|2x-1|} e^{x} \Rightarrow g(x) = C \sqrt{|2x-1|} e^{x}, C$
arbitrary $\neq 0$. We finish by noting that any $g(x)$ so defined is
differentiable on any open interval that does not contain $1/2$.
Q.E.D.
\problem {A-3:}
Determine, with proof, the set of real numbers $x$ for which
$\sum_{n=1}^{\infty} {( {1\over n} \csc ({1\over n}) - 1)}^{x}$
converges.
\solution {Solution:}
The answer is $x > {1\over 2}$. To see this, note that by Taylor's
theorem with remainder $\sin( {1\over{n}} ) = \sum_{i=1}^{k-1}
{ {(-1)}^{i-1} {n}^{-(2i+1)} } + c { {(-1)}^{k-1} {n}^{-(2k+1)} }$,
where $0 \leq c \leq {1\over n}$. Hence for $n\geq 1 ( 1/n )/( 1/n -
1/(3! n^{3}) + 1/(5! n^{5}) - 1 < (1/n) \csc(1/n) - 1 < ( 1/n )/( 1/n
- 1/(3! n^{3}) ) - 1 \Rightarrow$ for $n$ large enough, $ (1/2) 1/(3! n^{2})
< (1/n) \csc(1/n) - 1 < 2\cdot 1/(3! n^{2})$. Applying the p-test and the
comparison test, we see that $\sum_{n=1}^{\infty} {( {1\over n}
\csc({1\over n}) - 1)}^{x}$ converges iff $x > {1\over 2}$.
Q.E.D.
\problem {A-4:}
Justify your answers.
\subproblem {(a)}
If every point on the plane is painted one of three colors, do there
necessarily exist two points of the same color exactly one inch apart?
\solution {Solution:}
The answer is yes. Assume not and consider two equilateral
triangles with side one that have exactly one common face $\Rightarrow$
all points a distance of $\sqrt{3}$ apart are the same color; now
considering a triangle with sides $\sqrt{3}, \sqrt{3}, 1$ we reach the
desired contradiction.
Here is a pretty good list of references for the chromatic number of
the plane (i.e., how many colors do you need so that no two points 1
away are the same color) up to around 1982 (though the publication
dates are up to 1985). This asks for the chromatic number of the graph
where two points in R^2 are connected if they are distance 1 apart.
Let this chromatic number be chi(2) and in general let chi(n) be the
chromatic number of R^n. By a theorem in [2] this is equivalent to
finding what the maximum chromatic number of a finite subgraph of this
infinite graph is.
[1] H. Hadwiger, ``Ein Ueberdeckungssatz f\"ur den
Euklidischen Raum,'' Portugal. Math. #4 (1944), p.140-144
This seems to be the original reference for the problem
[2] N.G. de Bruijn and P. Erd\"os, ``A Color Problem for Infinite
Graphs and a Problem in the Theory of Relations,'' Nederl. Akad.
Wetensch. (Indag Math) #13 (1951), p. 371-373.
[3] H. Hadwiger, ``Ungel\"oste Probleme No. 40,'' Elemente der Math.
#16 (1961), p. 103-104.
Gives the upper bound of 7 with the hexagonal tiling and also
a reference to a Portugese journal where it appeared.
[4] L. Moser and W. Moser, ``Solution to Problem 10,'' Canad. Math.
Bull. #4 (1961), p. 187-189.
Shows that any 6 points in the plane only need 3 colors but
gives 7 points that require 4 (``the Moser Graph'' see [7]).
[5] Paul Erd\"os, Frank Harary, and William T. Tutte, ``On the
Dimension of a Graph,'' Mathematika #12 (1965), p. 118-122.
States that 3<chi(2)<8. Proves that chi(n) is finite for all n.
[6] P. Erd\"os, ``Problems and Results in Combinatorial Geometry,''
in ``Discrete Geometry and Convexity,'' Edited by Jacob E. Goodman,
Erwin Lutwak, Joseph Malkevitch, and Richard Pollack, Annals of
the New York Academy of Sciences Vol. 440, New York Academy of
Sciences 1985, Pages 1-11.
States that 3<chi(n)<8 and ``I am almost sure that chi(2)>4.''
States a question of L. Moser: Let R be large and S a measurable
set in the circle of radius R so that no two points of S have
distance 1. Denote by m(S) the measure of S. Determine
Lim_{R => infty} max m(S)/R^2.
Erd\"os conjectures that this limit is less than 1/4.
Erd\"os asks the following: ``Let S be a subset of the plane. Join
two points of S if their distances is 1. This gives a graph G(S).
Assume that the girth (shortest circuit) of G(S) is k. Can its
chromatic number be greater than 3? Wormald proved that such
a graph exists for k<6. The problem is open for k>5. Wormald
suggested that this method may work for k=6, but probably a
new idea is needed for k>6. A related (perhaps identical)
question is: `Does G(S) have a subgraph that has girth k and
chromatic number 4?' ''
[7] N. Wormald, ``A 4-chromatic graph with a special plane drawing,''
J. Austr. Math. Soc. Ser. A #28 (1970), p. 1-8.
The reference for the above question.
[8] R.L. Graham, ``Old and New Euclidean Ramsey Theorems,''
in ``Discrete Geometry and Convexity,'' Edited by Jacob E. Goodman,
Erwin Lutwak, Joseph Malkevitch, and Richard Pollack, Annals of
the New York Academy of Sciences Vol. 440, New York Academy of
Sciences 1985, Pages 20-30.
States that the best current bounds are 3<chi(2)<8. Calls the
graph in [3] the Moser graph. Quotes the result of Frankl and
Wilson [8] that chi(n) grows exponentially in n settling an
earlier conjecture of Erd\"os (I don't know the reference for
this). The best available bounds for this are
(1+ o(1)) (1.2)^n \le chi(n) \le (3+ o(1))^n.
[9] P. Frankl and R.M. Wilson, ``Intersection Theorems with Geometric
Consequences,'' Combinatorica #1 (1981), p. 357-368.
[10] H. Hadwiger, H. Debrunner, and V.L. Klee, ``Combinatorial
Geometry in the Plane,'' Holt, Rinehart & Winston, New York
(English edition, 1964).
[11] D.R. Woodall, ``Distances Realized by Sets Covering the Plane,''
Journal of Combinatorial Theory (A) #14 (1973), p. 187-200.
Among other things, shows that rational points in the plane can
be two colored.
[12] L. A. Sz\'ekely, ``Measurable Chromatic Number of Geometric
Graphs and Sets without some Distances in Euclidean Space,''
Combinatorica #4 (1984), p.213-218.
Considers \chi_m(R^2), the measurable chromatic number,
where sets of one color must be Lebesgue measurable.
He conjectures that \chi_m(R^2) is not equal to
\chi(R^2) (if the Axiom of Choice is false).
[13] Martin Gardner, ``Scientific American,'' October 1960, p. 160.
[14] Martin Gardner, ``Wheels, Life and other Mathematical Amusements,''
W.H. Freeman and Co., New York 1983, pages 195-196.
This occurs in a chapter on mathematical problems including
the 3x+1 problem. I think that his references are wrong, including
attributing the problem to Erd\"os and claiming that Charles Trigg
had original solutions in ``Problem 133,'' Crux Mathematicorum,
Vol. 2, 1976, pages 144-150.
Q.E.D.
\subproblem {(b)}
What if "three" is replaced by "nine"?
In this case, there does not necessarily exist two points of the same
color exactly one inch apart; this can be demonstrated by considering
a tessellation of the plane by a $3\times 3$ checkboard with side $2$,
with each component square a different color (color of boundary points
chosen in an obvious manner).
Q.E.D.
The length of the side of the checkerboard is not critical (the reader
my enjoy showing that $3/2 <$ side $< 3\sqrt{2}/2$ works).
\problem {A-5:}
Prove that there exists a {\it unique} function $f$ from the set
${\R}^{+}$ of positive real numbers to ${\R}^{+}$ such that $f(f(x))
= 6x - f(x)$ and $f(x) > 0$ for all $x > 0$.
\solution {Solution 1:}
Clearly $f(x) = 2x$ is one such solution; we need to show that it is the
{\it only} solution. Let $f^{1}(x) = f(x), f^{n}(x) = f(f^{n-1}(x))$ and
notice that $f^{n}(x)$ is defined for all $x>0$. An easy induction
establishes that for $n>0 f^{n}(x) = a_{n} x + b_{n} f(x)$, where $a_{0}
= 0, b_{0} = 1$ and $a_{n+1} = 6 b_{n}, b_{n+1} = a_{n} - b_{n}
\Rightarrow b_{n+1} = 6 b_{n-1} - b_{n}$. Solving this latter equation
in the standard manner, we deduce that $\lim_{n\to \infty} a_{n}/b_{n}
= -2$, and since we have that $f^{n}(x) > 0$ and since $b_{n}$ is
alternately negative and positive; we conclude that $2x \leq f(x) \leq 2x$
by letting $n \rightarrow \infty$.
Q.E.D.
\solution {Solution 2:}
(Dan Bernstein, Princeton)
As before, $f(x) = 2x$ works. We must show that if $f(x) = 2x + g(x)$
and $f$ satisfies the conditions then $g(x) = 0$ on ${\R}^{+}$.
Now $f(f(x)) = 6x - f(x)$ means that $2f(x) + g(f(x)) = 6x - 2x - g(x)$,
i.e., $4x + 2g(x) + g(f(x)) = 4x - g(x)$, i.e., $3g(x) + g(f(x)) = 0$.
This then implies $g(f(f(x))) = 9g(x)$. Also note that $f(x) > 0$
implies $g(x) > -2x$. Suppose $g(x)$ is not $0$ everywhere. Pick $y$
at which $g(y) \neq 0$. If $g(y) > 0$, observe $g(f(y)) = -3g(y) < 0$,
so in any case there is a $y_{0}$ with $g(y_{0}) < 0$. Now define $y_{1}
= f(f(y_{0})), y_{2} = f(f(y_{1}))$, etc. We know $g(y_{n+1})$ equals
$g(f(f(y_{n}))) = 9g(y_{n})$. But $y(n+1) = f(f(y_{n})) = 6y_{n} -
f(y_{n}) < 6y_{n}$ since $f>0$. Hence for each $n$ there exists
$y_{n} < 6^{n} y_{0}$ such that $g(y_{n}) = 9^{n} g(y_{0})$.
The rest is obvious: $0 > g(y_{0}) = 9^{-n} g(y_{n}) > -2\cdot 9^{-n}
y_{n} > -2 (6/9)^{n} y_{0}$, and we observe that as $n$ goes to infinity
we have a contradiction.
Q.E.D.
\problem {A-6:}
If a linear transformation $A$ on an $n$-dimensional vector space has
$n+1$ eigenvectors such that any $n$ of them are linearly independent,
does it follow that $A$ is a scalar multiple of the identity? Prove your
answer.
\solution {Solution:}
The answer is yes. First note that if $x_{1}, \ldots, x_{n+1}$ are the
eigenvectors, then we must have that $a_{n+1} x_{n+1} = a_{1} x_{1}
+ \cdots + a_{n} x_{n}$ for some non-zero scalars $a_{1}, \ldots, a_{n+1}$.
Multiplying by $A$ on the left we see that $\lambda_{n+1} a_{n+1} x_{n+1}
= \lambda_{1} a_{1} x_{1} + \cdots + \lambda_{n} a_{n} x_{n}$, where
$\lambda_{i}$ is the eigenvalue corresponding to the eigenvectors $x_{i}$.
But since we also have that $\lambda_{n+1} a_{n+1} x_{n+1} = \lambda_{n+1}
a_{1} x_{1} + \cdots + \lambda_{n+1} a_{n} x_{n}$ we conclude that
$\lambda_{1} a_{1} x_{1} + \cdots + \lambda_{n} a_{n} x_{n} = \lambda_{n+1}
a_{1} x_{1} + \cdots + \lambda_{n+1} a_{n} x_{n} \Rightarrow a_{1}
(\lambda_{1} - \lambda_{n+1}) x_{1} + \cdots + a_{n} (\lambda_{n} -
\lambda_{n+1}) x_{1} = 0 \Rightarrow \lambda_{1} = \cdots = \lambda_{n+1}
= \lambda$ since $x_{1}, \ldots, x_{n}$ are linearly independent. To
finish, note that the dimension of the eigenspace of $\lambda$ is equal
to $n$, and since this equals the dimension of the nullspace of $A -
\lambda I$ we conclude that the rank of $A - \lambda I$ equals $n - n =
0 \Rightarrow A - \lambda I = 0$.
Q.E.D.
\problem {B-1:}
A {\it composite} (positive integer) is a product $ab$ with $a$ and $b$
not necessarily distinct integers in $\{ 2,3,4,\ldots \}$. Show that
every composite is expressible as $xy + xz + yz + 1$, with $x, y$, and
$z$ positive integers.
\solution {Solution:}
Let $x = a-1, y = b-1, z = 1$; we then get that $xy + xz + yz + 1
= (a-1)(b-1) + a-1 + b-1 + 1 = ab$.
Q.E.D.
\problem {B-2:}
Prove or disprove: If $x$ and $y$ are real numbers with $y \geq 0$
and $y(y+1) \leq {(x+1)}^2$, then $y(y-1) \leq x^2$.
\solution {Solution:}
The statement is true. If $x+1 \geq 0$ we have that $\sqrt{y(y+1)}
- 1 \leq x \Rightarrow x^{2} \geq y^{2} + y + 1 - 2 \sqrt{y^{2}+y} \geq
y^{2} - y$ since $2y + 1 \geq 2 \sqrt{y^{2}+y}$ since ${(2y + 1)}^{2}
\geq 4 (y^{2}+y)$ if $y\geq 0$. If $x+1 < 0$, we see that $\sqrt{y(y+1)}
\leq -x - 1 \Rightarrow x^{2} \geq y^{2} + y + 1 + 2 \sqrt{y^{2}+y}
\geq y^{2} - y$.
Q.E.D.
\problem {B-3:}
For every $n$ in the set ${\Z}^{+} = \{ 1,2,\ldots \}$ of positive
integers, let $r(n)$ be the minimum value of $|c-d \sqrt{3}|$ for all
nonnegative integers $c$ and $d$ with $c + d = n$. Find, with proof,
the smallest positive real number $g$ with $r(n) \leq g$ for all $n$
in ${\Z}^{+}$.
\solution {Solution:}
The answer is $(1 + \sqrt{3})/2$. We write $|c-d\sqrt{3}|$ as
$|(n-d) - d\sqrt{3}|$; I claim that the minimum over all $d, 0 \leq d
\leq n$, occurs when $d = e = \floor{n/(1+\sqrt{3})}$ or when $d = f =
e+1 = \floor{n/(1+\sqrt{3})} + 1$. To see this, note that $(n-e) - e
\sqrt{3} > 0$ and if $e'<e$, then $(n-e') - e' \sqrt{3} > (n-e) - e
\sqrt{3}$, and similarly for $f'>f$. Now let $r = n/(1+\sqrt{3}) -
\floor{n/(1+\sqrt{3})}$ and note that $|(n-e) - e \sqrt{3}| = r
(1+\sqrt{3})$ and $|(n-f) - f \sqrt{3}| = (1-r) (1+\sqrt{3})$. Clearly
one of these will be $\leq (1+\sqrt{3})/2$. To see that $(1+\leq{3})/2$
cannot be lowered, note that since $1+\sqrt{3}$ is irrational, $r$ is
uniformly distributed $\mod{1} $.
Q.E.D.
\notes {Notes:}
We do not really need the result that $x$ irrational $\Rightarrow x n
- \floor{x n}$ u. d. $\mod{1} $, it would suffice to show that $x$
irrational $\Rightarrow x n - \floor{x n}$ is dense in $(0,1)$. But
this is obvious, since if $x$ is irrational there exists arbitrarily
large $q$ such that there exists $p$ with $(p,q) = 1$ such that $p/q <
x < (p+1)/q$. The nifty thing about the u. d. result is that it answers
the question: what number $x$ should we choose such that the density
of $\{ n : r(n) < x \}$ equals $t, 0 < t < 1$? The u. d. result implies
that the answer is $t (1+\sqrt{3})/2$. The u. d. result also provides
the key to the question: what is the average value of $r(n)$? The
answer is $(1+\sqrt{3})/4$.
\problem {B-4:}
Prove that if $\sum_{n=1}^{\infty} a(n)$ is a convergent series of
positive real numbers, then so is $\sum_{n=1}^{\infty} {(a(n))}^{n/(n+1)}$.
\solution {Solution:}
Note that the subseries of terms ${a(n)}^{n\over{n+1}}$ with
${a(n)}^{1\over{n+1}} \leq {1\over 2}$ converges since then
${a(n)}^{n\over{n+1}}$ is dominated by $1/2^{n}$, the subseries of
terms ${a(n)}^{n\over{n+1}}$ with ${a(n)}^{1\over{n+1}} > {1\over 2}$
converges since then ${a(n)}^{n\over{n+1}}$ is dominated by $2 a(n)$,
hence $\sum_{n=1}^{\infty} {a(n)}^{n\over{n+1}}$ converges.
Q.E.D.
\problem {B-5:}
For positive integers $n$, let $M(n)$ be the $2n + 1$ by $2n + 1$
skew-symmetric matrix for which each entry in the first $n$ subdiagonals
below the main diagonal is $1$ and each of the remaining entries below
the main diagonal is $-1$. Find, with proof, the rank of $M(n)$.
(According to the definition the rank of a matrix is the largest $k$
such that there is a $k \times k$ submatrix with non-zero determinant.)
One may note that \break\hfill
$M(1) = \left( \matrix{0&-1&1 \cr 1&0&-1
\cr -1&1&0} \right)$ and $M(2) = \left( \matrix{0&-1&-1&1&1
\cr 1&0&-1&-1&1 \cr 1&1&0&-1&-1 \cr -1&1&1&0&-1 \cr -1&-1&1&1&0}
\right)$.
\solution {Solution 1:}
Since $M(n)$ is skew-symmetric, $M(n)$ is singular for all $n$, hence
the rank can be at most $2n$. To see that this is indeed the answer,
consider the submatrix $M_{i}(n)$ obtained by deleting row $i$ and column
$i$ from $M(n)$. From the definition of the determinant we have that
$\det(M_{i}(n)) = \sum {(-1)}^{\delta(k)} a_{1 k(1)} \cdots a_{(2n) k(2n)}$,
where $k$ is member of $S_{2n}$ (the group of permutations on
$\{1,\ldots,2n\}$) and $\delta(k)$ is $0$ if $k$ is an even permutation or
$1$ if $k$ is an odd permutation. Now note that ${(-1)}^{\delta(k)}
a_{1 k(1)} \cdots a_{(2n) k(2n)}$ equals either $0$ or $\pm 1$, and is
non-zero iff $k(i) \neq i$ for all $i$, i.e. iff $k$ has no fixed points.
If we can now show that the set of all elements $k$ of $S_{2n}$, with
$k(i) \neq i$ for all $i$, has odd order, we win since this would imply that
$\det(M_{i}(n))$ is odd $\Rightarrow \det(M_{i}) \neq 0$. To show this,
let $f(n)$ equal the set of all elements $k$ of $S_n$ with $k(i) \neq
i$ for all $i$. We have that $f(1) = 0, f(2) = 1$ and we see that
$f(n) = (n-1) ( f(n-1) + f(n-2) )$ by considering the possible values of
$f(1)$ and whether or not $f(f(1)) = 1$; an easy induction now establishes
that $f(2n)$ is odd for all $n$.
Q.E.D.
\notes {Notes:}
In fact, it is a well-known result that $f(n) = n! ( 1/2! - 1/3! + \cdots
+ {(-1)}^{n}/n! )$.
\solution {Solution 2:}
As before, since $M(n)$ is skew-symmetric $M(n)$ is singular for all $n$
and hence can have rank at most $2n$. To see that this is the rank,
let $M_{i}(n)$ be the submatrix obtained by deleting row $i$ and column
$i$ from $M(n)$. We finish by noting that ${M_{i}(n)}^{2} \equiv
I_{2n} \Mod{2} $, hence $M_{i}(n)$ is nonsingular.
Q.E.D.
\problem {B-6:}
Prove that there exist an infinite number of ordered pairs $(a,b)$ of
integers such that for every positive integer $t$ the number $at + b$
is a triangular number if and only if $t$ is a triangular number.
(The triangular numbers are the $t(n) = n(n + 1)/2$ with $n$ in
$ \{ 0,1,2,\ldots \}$ ).
\solution {Solution:}
Call a pair of integers $(a,b)$ a {\it triangular pair} if $at + b$
is a triangular number iff $t$ is a triangular number. I claim that
$(9,1)$ is a triangular pair. Note that $9 (n(n+1)/2) + 1 =
(3n+1)(3n+2)/2$ hence $9t+1$ is triangular if $t$ is. For the other
direction, note that if $ 9t+1 = n(n+1)/2 \Rightarrow n = 3k+1$
hence $ 9t+1 = n(n+1)/2 = 9(k(k+1)/2)+1 \Rightarrow t = k(k+1)/2$,
therefore $t$ is triangular. Now note that if $(a,b)$ is a triangular
pair then so is $(a^{2},(a+1)b)$, hence we can generate an infinite
number of triangular pairs starting with $(9,1)$.
Q.E.D.
\notes {Notes:}
The following is a proof of necessary and sufficient conditions for $(a,b)$
to be a triangular pair.
I claim that $(a,b)$ is a triangular pair iff for some odd integer $o$
we have that $a = o^{2}, b = (o^{2}-1)/8$. I will first prove the
direction $\Leftarrow$. Assume we have $a = o^{2}, b = (o^{2}-1)/8$.
If $t = n(n+1)/2$ is any triangular number, then the identity $o^{2}
n(n+1)/2 + (o^{2}-1)/8 = (on + (o-1)/2) (on + (o+1)/2)/2$ shows that
$at + b$ is also a triangular number. On the other hand if $o^{2} t +
(o^{2}-1)/8 = n(n+1)/2$, the above identity implies we win if we can show
that $( n - (o-1)/2 )/o$ is an integer, but this is true since $o^{2} t +
(o^{2}-1)/8 \equiv n(n+1)/2 \Mod{o^{2}} \Rightarrow 4n^{2} + 4n \equiv -1
\Mod{o^{2}} \Rightarrow {(2n + 1)}^{2} \equiv 0 \Mod{o^{2}} \Rightarrow 2n + 1
\equiv 0 \Mod{o} \Rightarrow n \equiv (o-1)/2 \Mod{o} $. For the direction
$\Rightarrow$ assume that $(a,b)$ and $(a,c), c\ge b$, are both triangular
pairs; to see that $b = c$ notice that if $at + b$ is triangular for all
triangular numbers $t$, then we can choose $t$ so large that if $c>b$ then
$at + c$ falls between two consecutive triangular numbers; contradiction hence
$b = c$. Now assume that $(a,c)$ and $(b,c)$ are both triangular pairs;
I claim that $a = b$. But this is clear since if $(a,c)$ and $(b,c)$
are triangular pairs $\Rightarrow (ab,bc+c)$ and $(ab,ac+c)$ are
triangular pairs $\Rightarrow bc+c = ac+c$ by the above reasoning
$\Rightarrow bc=ac \Rightarrow$ either $a=b$ or $c=0 \Rightarrow a=b$
since $c=0 \Rightarrow a=b=1$. For a proof of this last assertion,
assume $(a,0), a>1$, is a triangular pair; to see that this gives a
contradiction note that if $(a,0)$ is a triangular pair $\Rightarrow
(a^{2},0)$ is also triangular pair, but this is impossible since then
we must have that $a(a^{3}+1)/2$ is triangular (since $a^{2} a (a^{3}+1)/2$
is triangular) but $(a^{2}-1)a^{2}/2 < a(a^{3}+1)/2 < a^{2}(a^{2}+1)/2$
(if $a>1$). We are now done, since if $(a,b)$ is a triangular pair
$\Rightarrow a 0 + b = n(n+1)/2$ for some $n\geq 0 \Rightarrow b =
({(2n+1)}^{2} - 1)/8$.
Q.E.D.
\bye
==> competition/tests/math/putnam/putnam.1990.p <==
Problem A-1
How many primes among the positive integers, written as usual in base
10, are such that their digits are alternating 1's and 0's, beginning
and ending with 1?
Problem A-2
Evaluate \int_0^a \int_0^b \exp(\max{b^2 x^2, a^2 y^2}) dy dx where a and
b are positive.
Problem A-3
Prove that if 11z^10 + 10iz^9 + 10iz - 11 = 0, then |z| = 1. (Here z is
a complex number and i^2 = -1.)
Problem A-4
If \alpha is an irrational number, 0 < \alpha < 1, is there a finite
game with an honest coin such that the probability of one player winning
the game is \alpha? (An honest coin is one for which the probability of
heads and the probability of tails are both 1/2. A game is finite if
with probability 1 it must end in a finite number of moves.)
Problem A-5
Let m be a positive integer and let G be a regular (2m + 1)-gon
inscribed in the unit circle. Show that there is a positive constant A,
independent of m, with the following property. For any point p inside G
there are two distinct vertices v_1 and v_2 of G such that
1 A
| |p - v_1| - |p - v_2| | < --- - ---.
m m^3
Here |s - t| denotes the distance between the points s and t.
Problem A-6
Let \alpha = 1 + a_1 x + a_2 x^2 + ... be a formal power series with
coefficients in the field of two elements. Let
/ 1 if every block of zeros in the binary expansion of n
/ has an even number of zeros in the block,
a_n = {
\ 0 otherwise.
(For example, a_36 = 1 because 36 = 100100_2 and a_20 = 0 because 20 =
10100_2.) Prove that \alpha^3 + x \alpha + 1 = 0.
Problem B-1
A dart, thrown at random, hits a square target. Assuming that any two
points of the target of equal area are equally likely to be hit, find
the probability that the point hit is nearer to the center than to any
edge. Express your answer in the form (a \sqrt(b) + c) / d, where a, b,
c, d are integers.
Problem B-2
Let S be a non-empty set with an associative operation that is left and
right cancellative (xy = xz implies y = z, and yx = zx implies y = z).
Assume that for every a in S the set { a^n : n = 1, 2, 3, ... } is
finite. Must S be a group?
Problem B-3
Let f be a function on [0,\infty), differentiable and satisfying
f'(x) = -3 f(x) + 6 f(2x)
for x > 0. Assume that |f(x)| <= \exp(-\sqrt(x)) for x >= 0 (so that
f(x) tends rapidly to 0 as x increases). For n a non-negative integer,
define
\mu_n = \int_0^\infty x^n f(x) dx
(sometimes called the nth moment of f).
a. Express \mu_n in terms of \mu_0.
b. Prove that the sequence { \mu_n 3^n / n! } always converges, and that
the limit is 0 only if \mu_0 = 0.
Problem B-4
Can a countably infinite set have an uncountable collection of non-empty
subsets such that the intersection of any two of them is finite?
Problem B-5
Label the vertices of a trapezoid T (quadrilateral with two parallel
sides) inscribed in the unit circle as A, B, C, D so that AB is parallel
to CD and A, B, C, D are in counterclockwise order. Let s_1, s_2, and d
denote the lengths of the line segments AB, CD, and OE, where E is the
point of intersection of the diagonals of T, and O is the center of the
circle. Determine the least upper bound of (s_1 - s_2) / d over all such
T for which d \ne 0, and describe all cases, if any, in which it is
attained.
Problem B-6
Let (x_1, x_2, ..., x_n) be a point chosen at random from the
n-dimensional region defined by 0 < x_1 < x_2 < ... < x_n < 1.
Let f be a continuous function on [0, 1] with f(1) = 0. Set x_0 = 0 and
x_{n+1} = 1. Show that the expected value of the Riemann sum
\sum_{i = 0}^n (x_{i+1} - x_i) f(x_{i+1})
is \int_0^1 f(t)P(t) dt, where P is a polynomial of degree n,
independent of f, with 0 \le P(t) \le 1 for 0 \le t \le 1.
==> competition/tests/math/putnam/putnam.1990.s <==
Problem A-1
How many primes among the positive integers, written as usual in base
10, are such that their digits are alternating 1's and 0's, beginning
and ending with 1?
Solution:
Exactly one, namely 101. 1 is not prime; 101 is prime. The sum
100^n + 100^{n - 1} + ... + 1 is divisible by 101 if n is odd,
10^n + 10^{n - 1} + ... + 1 if n is even. (To see the second part,
think about 101010101 = 102030201 - 1020100 = 10101^2 - 1010^2.)
Problem A-2
Evaluate \int_0^a \int_0^b \exp(\max{b^2 x^2, a^2 y^2}) dy dx where a and
b are positive.
Solution:
Split the inner integral according to the max{}. The easy term becomes
an integral of t e^{t^2}. The other term becomes an easy term after you
switch the order of integration. Your answer should have an e^{(ab)^2}.
Problem A-3
Prove that if 11z^10 + 10iz^9 + 10iz - 11 = 0, then |z| = 1. (Here z is
a complex number and i^2 = -1.)
Solution:
z is not zero, so divide by z^5 to make things a bit more symmetric.
Now write z = e^{i \theta} and watch the formula dissolve into a simple
trigonometric sum. The 11 sin 5 \theta term dominates the sum when that
sine is at its maximum; by this and similar considerations, just *write
down* enough maxima and minima of the function that it must have ten
real roots for \theta. (This cute solution is due to Melvin Hausner,
an NYU math professor.)
Problem A-4
If \alpha is an irrational number, 0 < \alpha < 1, is there a finite
game with an honest coin such that the probability of one player winning
the game is \alpha? (An honest coin is one for which the probability of
heads and the probability of tails are both 1/2. A game is finite if
with probability 1 it must end in a finite number of moves.)
Solution:
Yes. Write \alpha in binary---there's no ambiguity since it's irrational.
At the nth step (n >= 0), flip the coin. If it comes up heads, go to the
next step. If it comes up tails, you win if the nth bit of \alpha is 1.
Otherwise you lose. The probability of continuing forever is zero. The
probability of winning is \alpha.
This problem could have been better stated. Repeated flips of the coin
must produce independent results. The note that ``finite'' means only
``finite with probability 1'' is hidden inside parentheses, even though
it is crucial to the result. In any case, this problem is not very
original: I know I've seen similar problems many times, and no serious
student of probability can take more than ten minutes on the question.
Problem A-5
Let m be a positive integer and let G be a regular (2m + 1)-gon
inscribed in the unit circle. Show that there is a positive constant A,
independent of m, with the following property. For any point p inside G
there are two distinct vertices v_1 and v_2 of G such that
1 A
| |p - v_1| - |p - v_2| | < --- - ---.
m m^3
Here |s - t| denotes the distance between the points s and t.
Solution:
Place G at the usual roots of unity. Without loss of generality assume
that p = re^{i\theta} is as close to 1 as to any other vertex; in other
words, assume |\theta| <= 2\pi / (4m + 2) = \pi / (2m + 1). Now take the
distance between p and the two farthest (not closest!) vertices. Make
sure to write | |X| - |Y| | as the ratio of | |X|^2 - |Y|^2 | to |X| + |Y|.
I may have miscalculated, but I get a final result inversely proportional
to (4m + 2)^2, from which the given inequality follows easily with, say,
A = 0.01.
Alternate solution:
The maximum distance between p and a point of G is achieved between two
almost-opposite corners, with a distance squared of double 1 + \cos\theta
for an appropriately small \theta, or something smaller than 2 - A/m^2
for an appropriate A. Now consider the set of distances between p and
the vertices; this set is 2m + 1 values >= 0 and < 2 - A/m^2, so that
there are two values at distance less than 1/m - A/m^3 as desired.
Problem A-6
Let \alpha = 1 + a_1 x + a_2 x^2 + ... be a formal power series with
coefficients in the field of two elements. Let
/ 1 if every block of zeros in the binary expansion of n
/ has an even number of zeros in the block,
a_n = {
\ 0 otherwise.
(For example, a_36 = 1 because 36 = 100100_2 and a_20 = 0 because 20 =
10100_2.) Prove that \alpha^3 + x \alpha + 1 = 0.
Solution:
(Put a_0 = 1, of course.) Observe that a_{4n} = a_n since adding two zeros
on the right end does not affect the defining property; a_{4n + 2} = 0
since the rightmost zero is isolated; and a_{2n + 1} = a_n since adding
a one on the right does not affect the defining property. Now work in the
formal power series ring Z_2[[x]]. For any z in that ring that is a
multiple of x, define f(z) as a_0 + a_1 z + a_2 z^2 + ... . Clearly
f(z) = f(z^4) + z f(z^2) by the relations between a's. Now over Z_2,
(a + b)^2 = a^2 + b^2, so f(z) = f(z)^4 + z f(z)^2. Plug in x for z and
cancel the f(x) to get 1 = \alpha^3 + x \alpha as desired.
Problem B-1
A dart, thrown at random, hits a square target. Assuming that any two
points of the target of equal area are equally likely to be hit, find
the probability that the point hit is nearer to the center than to any
edge. Express your answer in the form (a \sqrt(b) + c) / d, where a, b,
c, d are integers.
Solution:
This is straightforward. The closer-to-the-center region is centered on
a square of side length \sqrt 2 - 1; surrounding the square and meeting
it at its corners are parabolic sections extending out halfway to the
edge. b is 2 and d is 6; have fun.
Problem B-2
Let S be a non-empty set with an associative operation that is left and
right cancellative (xy = xz implies y = z, and yx = zx implies y = z).
Assume that for every a in S the set { a^n : n = 1, 2, 3, ... } is
finite. Must S be a group?
Solution:
Yes. There is a minimal m >= 1 for which a^m = a^n for some n with n > m;
by cancellation, m must be 1. We claim that a^{n-1} is an identity in S.
For ba = ba^n = ba^{n-1}a, so by cancellation b = ba^{n-1}, and similarly
on the other side. Now a has an inverse, a^{n-2}. This problem is not new.
Problem B-3
Let f be a function on [0,\infty), differentiable and satisfying
f'(x) = -3 f(x) + 6 f(2x)
for x > 0. Assume that |f(x)| <= \exp(-\sqrt(x)) for x >= 0 (so that
f(x) tends rapidly to 0 as x increases). For n a non-negative integer,
define
\mu_n = \int_0^\infty x^n f(x) dx
(sometimes called the nth moment of f).
a. Express \mu_n in terms of \mu_0.
b. Prove that the sequence { \mu_n 3^n / n! } always converges, and that
the limit is 0 only if \mu_0 = 0.
Solution:
The only trick here is to integrate \mu_n by parts the ``wrong way,''
towards a higher power of x. A bit of manipulation gives the formula for
\mu_n as \mu_0 times n! / 3^n times the product of 2^k / (2^k - 1) for
1 <= k <= n. Part b is straightforward; the product converges since the
sum of 1 / (2^k - 1) converges (absolutely---it's positive).
Problem B-4
Can a countably infinite set have an uncountable collection of non-empty
subsets such that the intersection of any two of them is finite?
Solution:
Yes. A common example for this very well-known problem is the set of
rationals embedded in the set of reals. For each real take a Cauchy
sequence converging to that real; those sequences form the subsets of
the countably infinite rationals, and the intersection of any two of
them had better be finite since the reals are Archimedian. Another
example, from p-adics: Consider all binary sequences. With sequence
a_0 a_1 a_2 ... associate the set a_0, a_0 + 2a_1, a_0 + 2a_1 + 4a_2,
etc.; or stick 1 bits in all the odd positions to simplify housekeeping
(most importantly, to make the set infinite). Certainly different
sequences give different sets, and the intersection of two such sets
is finite.
Alternative solution:
Let C be a countable collection of non-empty subsets of A with the property
that any two subsets have finite intersection (from now
on we call this property, countable intersection property). Clearly
such a collection exists. We will show that C is not maximal, that is,
there exists a set which does not belong to C and it intersects finitely
with any set in C. Hence by Zorn's lemma, C can be extended to an
uncountable collection.
Let A1, A2, .... be an enumeration of sets in C. Then by axiom of choice,
pick an element b sub i from each of A sub i - Union {from j=1 to i-1} of
A sub j. It is easy to see that each such set is non-empty. Let B be the
set of all b sub i's. Then clearly B is different from each of the A sub i's
and its intersection with each A sub i is finite.
Yet another alternative solution:
Let the countable set be the lattice points of the plane. For each t in
[0,pi) let s(t) be the lattice points in a strip with angle of inclination
t and width greater than 1. Then the set of these strips is uncountable.
The intersection of any two is bounded, hence finite.
More solutions:
The problem (in effect) asks for an uncountable collection of
sets of natural numbers that are "almost disjoint," i.e., any two
have a finite intersection. Here are two elementary ways to
get such a collection.
1. For any set A={a, b, c, ...} of primes, let A'={a, ab, abc, ...}.
If A differs from B then A' has only a finite intersection with B'.
2. For each real number, e.g. x=0.3488012... form the set
S_x={3, 34, 348, 3488, ...}. Different reals give almost disjoint sets.
Problem B-5
Label the vertices of a trapezoid T (quadrilateral with two parallel
sides) inscribed in the unit circle as A, B, C, D so that AB is parallel
to CD and A, B, C, D are in counterclockwise order. Let s_1, s_2, and d
denote the lengths of the line segments AB, CD, and OE, where E is the
point of intersection of the diagonals of T, and O is the center of the
circle. Determine the least upper bound of (s_1 - s_2) / d over all such
T for which d \ne 0, and describe all cases, if any, in which it is
attained.
Solution:
Center the circle at the origin and rotate the trapezoid so that AB and
CD are horizontal. Assign coordinates to A and D, polar or rectangular
depending on your taste. Now play with s_1 - s_2 / d for a while;
eventually you'll find the simple form, after which maximization is
easy. The answer, if I've calculated right, is 2, achieved when rotating
the trapezoid by 90 degrees around the circle would take one vertex into
another. (A right triangle, with the hypoteneuse the length-two diamater
and d = 1, is a degenerate example.)
Alternative solution:
Let a be the distance from O (the center of the circle) to AB (that is
the side with length s1), and b the distance from O to CD. Clearly,
a = sqrt(1-s1*s1/4) and b = sqrt(1-s2*s2/4). Then with some mathematical
jugglery, one can show that (s1-s2)/d = (s1*s1-s2*s2)/(b*s1-a*s2).
Then differentiating this with respect to s1 and s2 and equating to
0 yields s1*s1+s2*s2=4, and hence s1=2*b and s2=2*a. The value of (s1-s2)/d
for these values is then 2. Hence (s1-s1)/d achieves its extremeum when
s1*s1+s2*s2=4 (that this value is actually a maximum is then easily seen),
and the lub is 2.
Problem B-6
Let (x_1, x_2, ..., x_n) be a point chosen at random from the
n-dimensional region defined by 0 < x_1 < x_2 < ... < x_n < 1.
Let f be a continuous function on [0, 1] with f(1) = 0. Set x_0 = 0 and
x_{n+1} = 1. Show that the expected value of the Riemann sum
\sum_{i = 0}^n (x_{i+1} - x_i) f(x_{i+1})
is \int_0^1 f(t)P(t) dt, where P is a polynomial of degree n,
independent of f, with 0 \le P(t) \le 1 for 0 \le t \le 1.
Solution:
Induct right to left. Show that for each k, given x_{k-1}, the
expected value at a point chosen with x_{k-1} < x_k < ... < x_n < 1
is a polynomial of the right type with the right degree. It's pretty
easy once you find the right direction. 0 \le P(t) \le 1 comes for
free: if P(t) is out of range at a point, it is out of range on an
open interval, and setting f to the characteristic function of that
interval produces a contradiction.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> language/italian/italian.record.p <==
What are some Italian words with unusual properties?
==> language/italian/italian.record.s <==
Spelling
Letter Patterns
Entire Word
longest word contemporaneamente contraddistinguere insufficientemente reinizializzazione (18,4)
longest palindrome onorarono (9,1)
longest beginning with a palindrome avallava (8,1)
longest beginning with b palindrome ?
longest beginning with c palindrome ?
longest beginning with d palindrome ?
longest beginning with e palindrome esse (4,1)
longest beginning with f palindrome ?
longest beginning with g palindrome ?
longest beginning with h palindrome ?
longest beginning with i palindrome inni (4,1)
longest beginning with j palindrome ?
longest beginning with k palindrome ?
longest beginning with l palindrome ?
longest beginning with m palindrome ?
longest beginning with n palindrome non (3,1)
longest beginning with o palindrome onorarono (9,1)
longest beginning with p palindrome ?
longest beginning with q palindrome ?
longest beginning with r palindrome ?
longest beginning with s palindrome ?
longest beginning with t palindrome ?
longest beginning with u palindrome ?
longest beginning with v palindrome ?
longest beginning with w palindrome ?
longest beginning with x palindrome ?
longest beginning with y palindrome ?
longest beginning with z palindrome ?
longest with middle a palindrome onorarono (9,1)
longest with middle b palindrome ?
longest with middle c palindrome ?
longest with middle d palindrome ?
longest with middle e palindrome arera aveva (5,2)
longest with middle f palindrome afa (3,1)
longest with middle g palindrome ?
longest with middle h palindrome ?
longest with middle i palindrome i (1,1)
longest with middle j palindrome ?
longest with middle k palindrome ?
longest with middle l palindrome avallava (8,1)
longest with middle m palindrome ama (3,1)
longest with middle n palindrome inni (4,1)
longest with middle o palindrome non (3,1)
longest with middle p palindrome ?
longest with middle q palindrome ?
longest with middle r palindrome aerea (5,1)
longest with middle s palindrome esse osso (4,2)
longest with middle t palindrome otto (4,1)
longest with middle u palindrome ?
longest with middle v palindrome ivi (3,1)
longest with middle w palindrome ?
longest with middle x palindrome ?
longest with middle y palindrome ?
longest with middle z palindrome ?
longest tautonym restereste (10,1)
longest beginning with a tautonym ?
longest beginning with b tautonym ?
longest beginning with c tautonym ?
longest beginning with d tautonym ?
longest beginning with e tautonym ?
longest beginning with f tautonym ?
longest beginning with g tautonym ?
longest beginning with h tautonym ?
longest beginning with i tautonym ?
longest beginning with j tautonym ?
longest beginning with k tautonym ?
longest beginning with l tautonym ?
longest beginning with m tautonym ?
longest beginning with n tautonym nana (4,1)
longest beginning with o tautonym ?
longest beginning with p tautonym ?
longest beginning with q tautonym ?
longest beginning with r tautonym restereste (10,1)
longest beginning with s tautonym ?
longest beginning with t tautonym ?
longest beginning with u tautonym ?
longest beginning with v tautonym vivi (4,1)
longest beginning with w tautonym ?
longest beginning with x tautonym ?
longest beginning with y tautonym ?
longest beginning with z tautonym ?
longest head 'n' tail conoscono (9,1)
longest with middle a head 'n' tail leale (5,1)
longest with middle b head 'n' tail ?
longest with middle c head 'n' tail ?
longest with middle d head 'n' tail iodio (5,1)
longest with middle e head 'n' tail diedi (5,1)
longest with middle f head 'n' tail afa (3,1)
longest with middle g head 'n' tail ?
longest with middle h head 'n' tail ?
longest with middle i head 'n' tail i (1,1)
longest with middle j head 'n' tail ?
longest with middle k head 'n' tail ?
longest with middle l head 'n' tail melme palpa salsa tolto (5,4)
longest with middle m head 'n' tail bimbi mamma pompo pompo` (5,4)
longest with middle n head 'n' tail nonno tanta tinti tonto (5,4)
longest with middle o head 'n' tail non (3,1)
longest with middle p head 'n' tail pappa (5,1)
longest with middle q head 'n' tail ?
longest with middle r head 'n' tail barba gorgo sorso torto (5,4)
longest with middle s head 'n' tail conoscono (9,1)
longest with middle t head 'n' tail tette (5,1)
longest with middle u head 'n' tail ?
longest with middle v head 'n' tail ivi (3,1)
longest with middle w head 'n' tail ?
longest with middle x head 'n' tail ?
longest with middle y head 'n' tail ?
longest with middle z head 'n' tail ?
Subset of Word
longest internal palindrome altererete ammacammo ammalammo cessassero fissassimo ignorarono interpreta interpretare interpretassi interpretata ... (7,28)
longest internal tautonym assassina assassinai assassini assassino battettero chiamiamolo contentera contentero gargarismi gargarismo ... (6,86)
longest repeated prefix assassina assassinai assassini assassino gargarismi gargarismo mormorii mormorio tartaruga tartarughe ... (6,52)
most consecutive doubled letters abbellii addii avvii inghiottii lineetta lineette muggii rifuggii rullii sbigottii ... (2,14)
most doubled letters abbassammo abbassasse abbassassi abbattemmo abbattesse abbattessi abbattette abbattetti abbellii abbellimmo ... (3,86)
longest two cadence eleverete genererete inimicizia inimicizie recederete venererete (5,6)
longest three cadence abbastanza consentendone decrescente eccellente espediente esperienze immissioni incipriati incipriavi inconveniente ... (4,19)
longest four cadence abbraccia abbracciai acclamata acclamava acclamera` acclamerai affiatata affiatava affiatera affiaterai ... (3,211)
longest five cadence identifichi inerentemente inserimenti permanentemente posizionandosi visualizzabili visualizzati visualizzazione (3,8)
Letter Counts
Lipograms
longest letters from first half illeggibili (11,1)
longest letters from second half sottoposto (10,1)
longest without ab insufficientemente (18,1)
longest without abcd sottoespressione sottoespressioni (16,2)
longest without a to h intromissioni (13,1)
longest without a to k prorompono proromputo sottoposto sottosuolo (10,4)
longest without a to n sottoposto (10,1)
longest without a to q su tu (2,2)
longest without a to s tu (2,1)
longest without e incontrollabili (15,1)
longest without et configurazioni posizionandosi visualizzabili (14,3)
longest without eta corrispondono riconoscibili (13,2)
longest without etai compongono concludono concorrono corrompono prorompono scrupoloso soccombono suppongono (10,8)
longest without etain scrupoloso (10,1)
longest without etains cocuzzolo (9,1)
Letter Choices
Vowels
longest all vowels ai io (2,2)
longest each vowel once aggiungendo computative concettuali documentati progettuali (11,5)
longest each extended vowel once ?
shortest each vowel once aiuole (6,1)
shortest each extended vowel once ?
shortest vowels in order ?
shortest extended vowels in order ?
longest vowels in order ?
longest extended vowels in order ?
shortest vowels in reverse order ?
shortest extended vowels in reverse order ?
longest vowels in reverse order ?
longest extended vowels in reverse order ?
longest one vowel gran nord (4,2)
longest two vowels sblocchi scndendo sgranchi` spranghe spranghi standard stringhe stronchi (8,8)
longest containing a univocalic abbastanza affrancata affrancava aggrappata aggrappava amalgamata amalgamava fracassata fracassava raccattata ... (10,26)
longest containing e univocalic precedentemente (15,1)
longest containing i univocalic intimiditi intimidivi invisibili irrigiditi ripristini significhi stillicidi (10,7)
longest containing o univocalic compongono concorrono corrompono prorompono protocollo soccombono sottoposto (10,7)
longest containing u univocalic blu gru pur sud sul (3,5)
longest containing y univocalic ?
longest alternating extended vowel-consonant denominatore disabilitare manipolatori recuperabile ridefinibile ridefinibili (12,6)
longest alternating vowel-consonant denominatore disabilitare manipolatori recuperabile ridefinibile ridefinibili (12,6)
Consonants
longest consonant string scndendo substrati substrato (4,3)
longest one consonant acciacco aggiogai aiutiate annoiano attuiate tatuiate (8,6)
longest two consonant abbassasse abbassassi abbattette abbattetti abbattiate accettiate acciaieria acciaierie accorcerai accorcerei ... (10,173)
Isograms
longest isogram acquistero` adulterino asciughero calpestino censuriamo combinaste compariste comperasti compilaste completavi ... (10,166)
longest pair isogram abbattesse abboccammo abbondando abbozzammo accennasse accettasse addobbammo addossammo afferrasse affettasse ... (10,39)
longest trio isogram ?
longest tetrad isogram ?
longest polygram abbattesse abbattette abboccammo abbondando abbozzammo accennasse accettasse addobbammo addossammo afferrasse ... (10,53)
longest pyramid abbassasse abbassassi ammassammo ammassasse ammassassi ammazzammo assassinai istituissi istituisti obererebbe ... (10,21)
most repeated letters caratteristiche commercializzato contemporaneamente contraddistinguere dicotomicamente duecentocinquanta interfacciamento rappresentativi riorganizzazione (6,9)
highest containing a repeated amalgamata amalgamava (5,2)
highest containing b repeated abolirebbe abuserebbe babbi babbo babbuini babbuino bacerebbe baderebbe bagnerebbe ballerebbe ... (3,36)
highest containing c repeated acciacchi acciacco (4,2)
highest containing d repeated addendi addensando addentando additando addobbando addossando condividendo dedicando degradando denudando ... (3,21)
highest containing e repeated precedentemente (6,1)
highest containing f repeated affabile affabili affabilita affaccera affaccerai affaccerei affaccero affacci affaccia affacciai ... (2,1260)
highest containing g repeated aggeggi gorgheggi gorgheggia gorgheggio (4,4)
highest containing h repeated chiaccherare chimiche chioschi chirurghi piluchhero` rachitiche schiocchi (2,7)
highest containing i repeated distinguibili inimicizia inimicizie inizializzati inizializzazione insiemistici intimidii intimiditi intimidivi invisibili ... (5,15)
highest containing j repeated gjiaccerei (1,1)
highest containing k repeated monokini (1,1)
highest containing l repeated allocarlo allodola allodole ballabile ballabili cancellarlo cellula cellule collezionarli colonnelli ... (3,39)
highest containing m repeated ammacammo ammalammo ammaliammo ammassammo ammazzammo ammirammo ammontammo camminammo commutammo immigrammo ... (4,12)
highest containing n repeated annientano annientino annunziano annunzino condannano condannino consentendone danneranno inconveniente indipendentemente ... (4,17)
highest containing o repeated coinvolgono compongono concorrono conoscono coronarono corrispondono corrompono doloroso logorarono monopolio ... (4,39)
highest containing p repeated apppare appropri appropria appropriai appropriato approprino approprio pappa pappagalli pappagallo ... (3,20)
highest containing q repeated qualunque soqquadro (2,2)
highest containing r repeated percorrere precorrere ricorrere rincorrere (4,4)
highest containing s repeated assestasse assestassi salassasse salassassi scassasse scassassi sgrassasse sgrassassi squassasse squassassi ... (5,12)
highest containing t repeated abbattette abbattetti altrettante attentaste attentasti attentata attentate attentati attentato attentiate ... (4,67)
highest containing u repeated qualunque tumultuera` tumultuero` usufruendo usufruiate usufruimmo usufruira usufruirai usufruire usufruirei ... (3,26)
highest containing v repeated ravvivava ravvivavi ravvivavo (4,3)
highest containing w repeated software (1,1)
highest containing x repeated uxorcidi uxorcidio xenofobia xilofono (1,4)
highest containing y repeated ?
highest containing z repeated inizializzazione reinizializzazione (4,2)
most different letters contraddistinguere (12,1)
highest ratio length/letters allattata attaccata errerebbe tintinnii (300,4)
highest ratio length/letters (no tautonyms) allattata attaccata errerebbe tintinnii (300,4)
lowest length 16 ratio length/letters contrapposizione schematizzazione tradizionalmente (145,3)
lowest length 17 ratio length/letters multidimensionali proporzionalmente (154,2)
lowest length 18 ratio length/letters contraddistinguere (150,1)
lowest length 19 ratio length/letters ?
lowest length 20 ratio length/letters ?
lowest length 21 ratio length/letters ?
lowest length 22 ratio length/letters ?
lowest length 23 ratio length/letters ?
lowest length 24 ratio length/letters ?
lowest length 25 ratio length/letters ?
Letter Appearance
longest short letters accasarono accasavamo accasavano accaseremo accecarono accecavamo accecavano accennammo accennasse accusarono ... (10,226)
longest tall letters diffidi litighi (7,2)
longest vertical-symmetry letters avvitavamo (10,1)
longest horizontal-symmetry letters chiocce ciocche dediche dedichi eccedei obbedii (7,6)
longest full-symmetry letters ho io (2,2)
highest ratio of dotted letters iniziai rifinii (57,2)
Typewriter
longest top row irrequiete irrequieti irrequieto irriterete otturerete perequerei pitturerei proiettero` proiettore proiettori ... (10,18)
longest middle row salassa (7,1)
longest bottom row ?
longest in order toppa (5,1)
longest in reverse order capire lauree sapore vapore (6,4)
longest left hand abbassasse abbassaste abbattesse abbatteste abbattette abbeverare abbeverata abbeverate abbeverava abbeverera` ... (10,199)
longest right hand illuminino (10,1)
longest alternating hands riformando rifornendo ritornando (10,3)
longest one finger eccede (6,1)
longest adjacent keys eresse essere fredde popolo popolo` sedere (6,6)
Puzzle
longest palindrome in Morse code infilai (7,1)
longest formed with chemical symbols rappresentativi (15,1)
longest formed with US postal codes allagavamo allattiamo arrivavamo caricavamo comandiamo cominciamo cominciava cominciavi conciavamo conciliamo ... (10,32)
longest formed with amino acid abbreviations pro val (3,2)
longest formed with piano notes accadda accadde (7,2)
Letter Order
Alphabetical
longest letters in order chino degno (5,2)
longest letters in order with repeats accenno (7,1)
longest letters in reverse order solida sponda tromba tronca (6,4)
longest letters in reverse order with repeats solida spicca sponda trecca tromba tronca (6,6)
longest roller-coaster compilatore compilatori preliminare preliminari (11,4)
longest no letters in place reinizializzazione (18,1)
most letters in place abbreviai acciacchi accrediti dannegghi predefinite predefinito rendergli (4,7)
most letters in place shifted definibili ghiaccino operativo ridefinibile ridefinibili (5,5)
most consecutive letters in order consecutively abroghiamo abroghiate abroghino affoghiamo affoghiate affoghino agghiacci agghiaccia agghiaccio aggioghino ... (3,622)
most consecutive letters in order restaurava restauravi restauravo restituiva restituivi restituivo (5,6)
most consecutive letters perquisito (7,1)
highest ratio of consecutive letters to length fughe fughi stura sturi sturo sturo` (80,6)
==> language/multi.palindromes.p <==
List some multi-lingual palindromes.
==> language/multi.palindromes.s <==
Hebrew: aba or abba, English: dad
German: tat, English: deed
==> language/norwegian/norwegian.record.p <==
What are some Norwegian words with unusual properties?
==> language/norwegian/norwegian.record.s <==
Spelling
Letter Patterns
Entire Word
longest word arbeidstakerorganisasjonene korrekturlesingsprogrammene statstjenestemannskartellet undervisningsorganisasjoner (27,4)
longest palindrome stillits (8,1)
longest beginning with a palindrome amma anna (4,2)
longest beginning with b palindrome ?
longest beginning with c palindrome ?
longest beginning with d palindrome da@d do"d dyd (3,3)
longest beginning with e palindrome ergre (5,1)
longest beginning with f palindrome ?
longest beginning with g palindrome gniding (7,1)
longest beginning with h palindrome ?
longest beginning with i palindrome i (1,1)
longest beginning with j palindrome ?
longest beginning with k palindrome kik kok (3,2)
longest beginning with l palindrome ?
longest beginning with m palindrome madam (5,1)
longest beginning with n palindrome nebben (6,1)
longest beginning with o palindrome obo (3,1)
longest beginning with p palindrome pep pip pop prp (3,4)
longest beginning with q palindrome ?
longest beginning with r palindrome rallar rekker renner retter (6,4)
longest beginning with s palindrome stillits (8,1)
longest beginning with t palindrome tekket tellet teppet tillit (6,4)
longest beginning with u palindrome ?
longest beginning with v palindrome vev viv vov (3,3)
longest beginning with w palindrome ?
longest beginning with x palindrome ?
longest beginning with y palindrome ?
longest beginning with z palindrome ?
longest with middle a palindrome stats (5,1)
longest with middle b palindrome nebben (6,1)
longest with middle c palindrome ?
longest with middle d palindrome gniding stedets (7,2)
longest with middle e palindrome sees (4,1)
longest with middle f palindrome ?
longest with middle g palindrome ergre (5,1)
longest with middle h palindrome aha (3,1)
longest with middle i palindrome eie kik pip rir tit viv (3,6)
longest with middle j palindrome ?
longest with middle k palindrome rekker tekket (6,2)
longest with middle l palindrome stillits (8,1)
longest with middle m palindrome amma (4,1)
longest with middle n palindrome renner (6,1)
longest with middle o palindrome kok non pop ror sos vov (3,6)
longest with middle p palindrome teppet (6,1)
longest with middle q palindrome ?
longest with middle r palindrome prp (3,1)
longest with middle s palindrome nesen reser (5,2)
longest with middle t palindrome retter settes (6,2)
longest with middle u palindrome sus tut (3,2)
longest with middle v palindrome neven (5,1)
longest with middle w palindrome ?
longest with middle x palindrome ?
longest with middle y palindrome dyd ryr (3,2)
longest with middle z palindrome ?
longest tautonym barbar berber bonbon cancan erverv farfar mormor purpur tartar (6,9)
longest beginning with a tautonym ?
longest beginning with b tautonym barbar berber bonbon (6,3)
longest beginning with c tautonym cancan (6,1)
longest beginning with d tautonym ?
longest beginning with e tautonym erverv (6,1)
longest beginning with f tautonym farfar (6,1)
longest beginning with g tautonym ?
longest beginning with h tautonym ?
longest beginning with i tautonym ?
longest beginning with j tautonym ?
longest beginning with k tautonym kaka (4,1)
longest beginning with l tautonym ?
longest beginning with m tautonym mormor (6,1)
longest beginning with n tautonym nn (2,1)
longest beginning with o tautonym ?
longest beginning with p tautonym purpur (6,1)
longest beginning with q tautonym ?
longest beginning with r tautonym ?
longest beginning with s tautonym ?
longest beginning with t tautonym tartar (6,1)
longest beginning with u tautonym ?
longest beginning with v tautonym veve (4,1)
longest beginning with w tautonym ?
longest beginning with x tautonym ?
longest beginning with y tautonym ?
longest beginning with z tautonym ?
longest head 'n' tail barnebarn endevende (9,2)
longest with middle a head 'n' tail miami (5,1)
longest with middle b head 'n' tail obo (3,1)
longest with middle c head 'n' tail ?
longest with middle d head 'n' tail enden (5,1)
longest with middle e head 'n' tail barnebarn (9,1)
longest with middle f head 'n' tail lefle (5,1)
longest with middle g head 'n' tail avgav engen magma (5,3)
longest with middle h head 'n' tail aha (3,1)
longest with middle i head 'n' tail seise veive (5,2)
longest with middle j head 'n' tail ?
longest with middle k head 'n' tail enken omkom saksa takta tekte (5,5)
longest with middle l head 'n' tail erler etlet oslos telte volvo (5,5)
longest with middle m head 'n' tail ermer mamma (5,2)
longest with middle n head 'n' tail ganga nenne tanta tente (5,4)
longest with middle o head 'n' tail kok non pop ror sos vov (3,6)
longest with middle p head 'n' tail pappa (5,1)
longest with middle q head 'n' tail ?
longest with middle r head 'n' tail terte verve (5,2)
longest with middle s head 'n' tail etset sesse tasta teste (5,4)
longest with middle t head 'n' tail enten erter satsa tette (5,4)
longest with middle u head 'n' tail sausa stust (5,2)
longest with middle v head 'n' tail endevende (9,1)
longest with middle w head 'n' tail ?
longest with middle x head 'n' tail ?
longest with middle y head 'n' tail dyd ryr (3,2)
longest with middle z head 'n' tail ?
Subset of Word
longest internal palindrome pennevennen (9,1)
longest internal tautonym allmennmenneskelig allmennmenneskelige (8,2)
longest repeated prefix barbaren barbariet barbarisere barbarisk barbariske berberis berberisen berberiss bonbonen cancanen ... (6,27)
most consecutive doubled letters irreell kassaapparat kassaapparatet (3,3)
most doubled letters tilleggsstraff tilleggsstraffa tilleggsstraffen (4,3)
longest two cadence degenerere degenererer regenerere (5,3)
longest three cadence medbestemmelsesrett (6,1)
longest four cadence fengselsvesenet rekordforbedring (4,2)
longest five cadence arbeidsdepartementet arbeidsdepartementets vindusinnfatningen (4,3)
Letter Counts
Lipograms
longest letters from first half allmenndannende (15,1)
longest letters from second half oppsta@tt oppstuss prototyp stuttorv torvstro" (8,5)
longest without ab retrospeksjonretroversjon (25,1)
longest without abcd retrospeksjonretroversjon (25,1)
longest without a to h sta@lkonstruksjon (16,1)
longest without a to k motorsport motorstopp (10,2)
longest without a to n oppsta@tt oppstuss prototyp stuttorv torvstro" (8,5)
longest without a to q to"rrva"r utsto"tt utstyrt (7,3)
longest without a to s ta@tt tutt tytt uto"y va"tt va@tt (4,6)
longest without e byggforskningsinstitutt (23,1)
longest without et ungdomsorganisasjon (19,1)
longest without eta forskningsopphold (17,1)
longest without etai ungdomsforbunds (15,1)
longest without etain bruksforhold (12,1)
longest without etains koldjomfru va"rforhold (10,2)
Letter Choices
Vowels
longest all vowels aue eau eie (3,3)
longest each vowel once ?
longest each extended vowel once ?
shortest each vowel once ?
shortest each extended vowel once ?
shortest vowels in order ?
shortest extended vowels in order ?
longest vowels in order ?
longest extended vowels in order ?
shortest vowels in reverse order ?
shortest extended vowels in reverse order ?
longest vowels in reverse order ?
longest extended vowels in reverse order ?
longest one vowel skjelmsk (8,1)
longest two vowels skjo"nnskrift stridsskrift (12,2)
longest containing a univocalic farskapssaka kassaapparat salgsapparat (12,3)
longest containing e univocalic medbestemmelsesrett (19,1)
longest containing i univocalic prisstigning spiritistisk stridsskrift (12,3)
longest containing o univocalic oppholdsrom (11,1)
longest containing u univocalic brukskunst (10,1)
longest containing y univocalic lydtrykk (8,1)
longest alternating extended vowel-consonant demilitarisere originaliteten vitamininisere (14,3)
longest alternating vowel-consonant demilitarisere originaliteten vitamininisere (14,3)
Consonants
longest consonant string punktskrift tilleggsstraff tilleggsstraffa tilleggsstraffen (6,4)
longest one consonant unionene (8,1)
longest two consonant appenninene autoriteter statssto"tte territoriet (11,4)
Isograms
longest isogram godtkjo"psvaren (14,1)
longest pair isogram musikkinstrumenter (18,1)
longest trio isogram ?
longest tetrad isogram ?
longest polygram musikkinstrumenter (18,1)
longest pyramid ekserserer statskassa statsskatt (10,3)
most repeated letters differanseapproksimasjon kommunikasjonshjelpemidler musikkinstrumenter tverrkontraksjonstallet (9,4)
highest containing a repeated abrakadabra abrakadabraet administrasjonsapparatet administrasjonsapparatets kassaapparat kassaapparatet (5,6)
highest containing b repeated babbel babbelet ballklubben barneklubb bedriftsklubb bedriftsklubben bensinbombe bensinbomber bergnebbe breiflabb ... (3,13)
highest containing c repeated baccardi backfisch boccia cabincruiser cabincruiseren cancan cancanen cancer canceren ccm ... (2,49)
highest containing d repeated do"dfo"dd driftsunderskudd middelaldrende skreddersydd (4,4)
highest containing e repeated aldersgrensebestemmelsene (8,1)
highest containing f repeated diffusjonseffekt (4,1)
highest containing g repeated beregningsgrunnlaget boligbyggelag boligbygging boligbyggingen tilleggsbevilgning tilleggsbevilgningen (4,6)
highest containing h repeated avhengighetsforhold avhengighetsforholdet (3,2)
highest containing i repeated akklimatiseringsperiode aluminiumsindustri aluminiumsindustrien artilleristilling binomialutvikling differensialligning differensialligningen differensialligningene differensiallikning differensiallikningen ... (4,78)
highest containing j repeated aksjeemisjonen aksjeemisjoner aksjekjo"pet aksjemajoriteten aksjonslinje aksjonslinjen artisjokkhjerte bataljonssjef bekjentgjo"relse bekjentgjo"relsen ... (2,86)
highest containing k repeated alkoholikerklinikk konstruksjonsteknikken vekkeklokke vekkerklokke (5,4)
highest containing l repeated fellingstillatelse (5,1)
highest containing m repeated familiemedlemmer komitemedlemmenes mellomrommet (4,3)
highest containing n repeated annonseomkostningene grunnutdanningen tangensialspenningen vindusinnfatningen (6,4)
highest containing o repeated onomatopoetikon onomatopoetikonet onomatopoietikon (5,3)
highest containing p repeated oppstopper oppstopperen opptrapping opptrappinga opptrappingen (4,5)
highest containing q repeated aqua diplomatique enquete enqueteen enqueten iq mannequin mannequinen mannequiner mocambique ... (1,25)
highest containing r repeated arbeiderpartiregjering korrekturlesingsprogrammene korrekturlesingsprogrammet retrospeksjonretroversjon terroriserer (5,5)
highest containing s repeated administrasjonsmessig assosiasjonsrikdom bero"ringsassosiasjon beslutningsprosess beslutningsprosessen bessemerprosessen diffusjonsprosess eksistensialistisk livsforsikringsselskaper mikroprosessorsystem ... (5,13)
highest containing t repeated sentralinstituttet statstjenestemannskartellet (6,2)
highest containing u repeated akupunktur aluminiumfluorid aluminiumforbruket aluminiumsindustri aluminiumsindustrien aluminiumsprodukter aluminiumsvinduer dumdumkule futurum futurumet ... (3,25)
highest containing v repeated veloverveid virvelvind virvelvinden (3,3)
highest containing w repeated knowhow newsweek warszawa wienerwurst wigwam (2,5)
highest containing x repeated xerxes (2,1)
highest containing y repeated arbeidslo"ysetrygd belysningsstyrke belysningsstyrken belysningsutstyr bygdo"y byggelo"yve byggryn byggrynet bystyre bystyret ... (2,86)
highest containing z repeated jazz jazzen jibbenzzsjo"uttr mezzoforte mezzosopran mezzosopranen razzia razziaen zanzibar zanzibarisk (2,10)
most different letters publikasjonsvirksomhet (17,1)
highest ratio length/letters attentatmennene etterrett refererer (300,3)
highest ratio length/letters (no tautonyms) attentatmennene etterrett refererer (300,3)
lowest length 16 ratio length/letters nyhetsformidling redaksjonsutvalg utviklingsarbeid (114,3)
lowest length 17 ratio length/letters riksma@lsforbundet (113,1)
lowest length 18 ratio length/letters utvekslingsprogram (120,1)
lowest length 19 ratio length/letters avgiftsforho"yelsene (126,1)
lowest length 20 ratio length/letters avskjedsforestilling popula"rvitenskapelig (133,2)
lowest length 21 ratio length/letters administrasjonsflo"yen opplysningsvirksomhet popula"rvitenskapelige (140,3)
lowest length 22 ratio length/letters publikasjonsvirksomhet (129,1)
lowest length 23 ratio length/letters databehandlingsfunksjon tekstbehandlingsprogram (143,2)
lowest length 24 ratio length/letters folkeopplysningsarbeidet (150,1)
lowest length 25 ratio length/letters koordinattransformasjoner (192,1)
Letter Appearance
longest short letters reassuransen sommervarmen (12,2)
longest tall letters flittig flyktig pliktig tilbydd tilbygg tilgift tilgitt (7,7)
longest vertical-symmetry letters automat ho"ya"tta mahatma mottatt (7,4)
longest horizontal-symmetry letters bebodde (7,1)
longest full-symmetry letters ohio (4,1)
highest ratio of dotted letters bikini isjias jiujitsu (50,3)
Typewriter
longest top row opprioritere uprioriterte (12,2)
longest middle row allslags do"dsfall salgslag (8,3)
longest bottom row bbc ccm (3,2)
longest in order etikk etiop etisk ettal oppad topas toppa topps tuppa yppal (5,10)
longest in reverse order blaute blautt blotte kaputt kjappe spotte spytte (6,7)
longest left hand breddegrader (12,1)
longest right hand poliklinikk (11,1)
longest alternating hands subjektivisme (13,1)
longest one finger hunn kikk munn (4,3)
longest adjacent keys dressere dressert sedertre (8,3)
Puzzle
longest palindrome in Morse code rekker retter settes tekket teppet (6,5)
longest formed with chemical symbols alkoholikerklinikk vindusinnfatningen (18,2)
longest formed with US postal codes mottakinga utmattinga utnyttinga (10,3)
longest formed with amino acid abbreviations valser (6,1)
longest formed with piano notes bededag (7,1)
Letter Order
Alphabetical
longest letters in order forsta@ (6,1)
longest letters in order with repeats addert begikk forsta@ oppsta@ (6,4)
longest letters in reverse order utrolige (8,1)
longest letters in reverse order with repeats utrolige (8,1)
longest roller-coaster operativsystemet (16,1)
longest no letters in place statstjenestemannskartellet undervisningsorganisasjoner (27,2)
most letters in place aksjekapitalen (5,1)
most letters in place shifted aksjekapitalen avdelingskontorer databehandlingsutstyr deflasjonen differensialoperator forstuva forstuve forstuvet frekvensforskyver gjenoppretta ... (5,29)
most consecutive letters in order consecutively forstuva forstuve forstuvet (5,3)
most consecutive letters in order akklimatiseringsperiode allmenneuropeiske fo"rsteutgaven forstuva forstuve forstuvet kulminasjon kulminasjonen vekselstro"msgenerator vekselstro"msgeneratoren (5,10)
most consecutive letters databehandlingsfunksjon kommunikasjonshjelpemidler publikasjonsvirksomhet (9,3)
highest ratio of consecutive letters to length heftig (83,1)
==> language/repeated.word.p <==
In any language, construct a sentence by repeating one word four times.
==> language/repeated.word.s <==
In Annamite:
Ba ba ba ba.
(Three ladies gave a box on the ear to the favorite of the Prince.)
==> language/swedish/swedish.record.p <==
What are some Swedish words with unusual properties?
==> language/swedish/swedish.record.s <==
Spelling
Letter Patterns
Entire Word
longest word programmeringsfo"rutsa"ttningar (29,1)
longest palindrome rattar (6,1)
longest beginning with a palindrome Anna alla amma (4,3)
longest beginning with b palindrome ?
longest beginning with c palindrome ?
longest beginning with d palindrome da@d do"d (3,2)
longest beginning with e palindrome esse (4,1)
longest beginning with f palindrome ?
longest beginning with g palindrome ?
longest beginning with h palindrome ?
longest beginning with i palindrome ini (3,1)
longest beginning with j palindrome ?
longest beginning with k palindrome ka"k ka@k kik ko"k (3,4)
longest beginning with l palindrome ?
longest beginning with m palindrome mim (3,1)
longest beginning with n palindrome ?
longest beginning with o palindrome oro (3,1)
longest beginning with p palindrome pdp pip pop (3,3)
longest beginning with q palindrome ?
longest beginning with r palindrome rattar (6,1)
longest beginning with s palindrome sa@s ses sis sos sus (3,5)
longest beginning with t palindrome ta"t ta@t tut (3,3)
longest beginning with u palindrome ?
longest beginning with v palindrome varav (5,1)
longest beginning with w palindrome ?
longest beginning with x palindrome ?
longest beginning with y palindrome ?
longest beginning with z palindrome ?
longest with middle a palindrome rar (3,1)
longest with middle b palindrome ?
longest with middle c palindrome ?
longest with middle d palindrome radar (5,1)
longest with middle e palindrome ses vev (3,2)
longest with middle f palindrome ?
longest with middle g palindrome ?
longest with middle h palindrome aha (3,1)
longest with middle i palindrome kik mim pip sis viv (3,5)
longest with middle j palindrome aja (3,1)
longest with middle k palindrome eke (3,1)
longest with middle l palindrome alla (4,1)
longest with middle m palindrome ramar (5,1)
longest with middle n palindrome Anna (4,1)
longest with middle o palindrome pop sos vov (3,3)
longest with middle p palindrome apa (3,1)
longest with middle q palindrome ?
longest with middle r palindrome varav (5,1)
longest with middle s palindrome esse (4,1)
longest with middle t palindrome rattar (6,1)
longest with middle u palindrome sus tut (3,2)
longest with middle v palindrome ?
longest with middle w palindrome ?
longest with middle x palindrome ?
longest with middle y palindrome ?
longest with middle z palindrome ?
longest tautonym berber mormor (6,2)
longest beginning with a tautonym ?
longest beginning with b tautonym berber (6,1)
longest beginning with c tautonym ?
longest beginning with d tautonym ?
longest beginning with e tautonym ?
longest beginning with f tautonym ?
longest beginning with g tautonym ?
longest beginning with h tautonym ?
longest beginning with i tautonym ?
longest beginning with j tautonym jojo (4,1)
longest beginning with k tautonym ?
longest beginning with l tautonym ?
longest beginning with m tautonym mormor (6,1)
longest beginning with n tautonym ?
longest beginning with o tautonym ?
longest beginning with p tautonym ?
longest beginning with q tautonym ?
longest beginning with r tautonym ?
longest beginning with s tautonym ?
longest beginning with t tautonym toto (4,1)
longest beginning with u tautonym ?
longest beginning with v tautonym ?
longest beginning with w tautonym ?
longest beginning with x tautonym ?
longest beginning with y tautonym ?
longest beginning with z tautonym ?
longest head 'n' tail Volvo annan armar avgav dadda elfel mamma pappa samsa sansa ... (5,11)
longest with middle a head 'n' tail rar (3,1)
longest with middle b head 'n' tail ?
longest with middle c head 'n' tail ?
longest with middle d head 'n' tail dadda (5,1)
longest with middle e head 'n' tail ses vev (3,2)
longest with middle f head 'n' tail elfel (5,1)
longest with middle g head 'n' tail avgav (5,1)
longest with middle h head 'n' tail aha (3,1)
longest with middle i head 'n' tail kik mim pip sis viv (3,5)
longest with middle j head 'n' tail aja (3,1)
longest with middle k head 'n' tail eke (3,1)
longest with middle l head 'n' tail Volvo (5,1)
longest with middle m head 'n' tail armar mamma samsa (5,3)
longest with middle n head 'n' tail annan sansa (5,2)
longest with middle o head 'n' tail pop sos vov (3,3)
longest with middle p head 'n' tail pappa (5,1)
longest with middle q head 'n' tail ?
longest with middle r head 'n' tail oro (3,1)
longest with middle s head 'n' tail asa (3,1)
longest with middle t head 'n' tail satsa (5,1)
longest with middle u head 'n' tail sus tut (3,2)
longest with middle v head 'n' tail ?
longest with middle w head 'n' tail ?
longest with middle x head 'n' tail ?
longest with middle y head 'n' tail ?
longest with middle z head 'n' tail ?
Subset of Word
longest internal palindrome dragningar fo"redragningar fo"rsvagningar hoptagningar interpretator interpretatorer interpretatorn personalplaneringsdelar promemoria promemorian ... (7,12)
longest internal tautonym a@rsperspektiv (8,1)
longest repeated prefix mormorsmors (6,1)
most consecutive doubled letters ideell ideella ideellt kommitteer reell reella reellt skadeeffekt (2,8)
most doubled letters kommitteer labbtillfa"llen labbtillfa"llet omsta"llningstillfa"llet otillfredssta"llande rapporteringstillfa"lle tillverkningssta"lle (3,7)
longest two cadence levererera leverereras (4,2)
longest three cadence sammanfalla sammanfatta sammankallar sammanlagda sammansatta skalbaggarna (4,6)
longest four cadence a"ndringarna atlantpakten attrahera beslutsma"ssighet civilisation distribution fo"ra"ldraga@rd ha@rdvaruservice ha@rdvaruverktyg handlingarna ... (3,33)
longest five cadence a"rendehantering anknytningen anla"ggningen anpassningen anva"ndningen formuleringarna granbarkborre ho"gerparenteser inloggningen inriktningen ... (3,21)
Letter Counts
Lipograms
longest letters from first half kemikaliebehandla (17,1)
longest letters from second half a@tra@va"rt prototyp storspov torvstro" (8,4)
longest without ab produktionsstyrningssystem (26,1)
longest without abcd synkroniseringssekvenser (24,1)
longest without a to h ma"nniskokropp tullkontorist (13,2)
longest without a to k mormorsmors (11,1)
longest without a to n a@tra@va"rt prototyp storspov torvstro" (8,4)
longest without a to q a@tra@va"rt (8,1)
longest without a to s tva"tt (5,1)
longest without e anva"ndarsammanslutningar cirkulationstja"nstgo"ring (24,2)
longest without et bokfo"ringsgranskning (20,1)
longest without eta uppkopplingsfo"rso"k (18,1)
longest without etai la"nsskolna"mnds (14,1)
longest without etain mormorsmors (11,1)
longest without etains blomkrukor dva"rgbjo"rk jordloppor (10,3)
Letter Choices
Vowels
longest all vowels a@a"o" (3,1)
longest each vowel once ?
longest each extended vowel once ?
shortest each vowel once ?
shortest each extended vowel once ?
shortest vowels in order ?
shortest extended vowels in order ?
longest vowels in order ?
longest extended vowels in order ?
shortest vowels in reverse order ?
shortest extended vowels in reverse order ?
longest vowels in reverse order ?
longest extended vowels in reverse order ?
longest one vowel dtlkrets (8,1)
longest two vowels blixtsnabbt gransknings (11,2)
longest containing a univocalic sammankallar skalbaggarna (12,2)
longest containing e univocalic reglementet televerkets (11,2)
longest containing i univocalic tidningsklipp (13,1)
longest containing o univocalic plockkontroll (13,1)
longest containing u univocalic slutspurt (9,1)
longest containing y univocalic nybyggd (7,1)
longest alternating extended vowel-consonant digitaliserade (14,1)
longest alternating vowel-consonant digitaliserade (14,1)
Consonants
longest consonant string a"ngsstrit programmeringsspra@k regeringsskrivelse (6,3)
longest one consonant Aurora (6,1)
longest two consonant barria"rer intention intuition outtestat preparera (9,5)
Isograms
longest isogram Ha"gerstalund o"verska@dligt oo"verska@dlig ta"vlingsform tidsluckorna tva@hjulingar (12,6)
longest pair isogram gireringen (10,1)
longest trio isogram ?
longest tetrad isogram ?
longest polygram mormorsmors (11,1)
longest pyramid anamma fiffig minimi mottot pannan papaya samsas svassa tafatt (6,9)
most repeated letters korrigeringstransaktion materialstyrningssystem (9,2)
highest containing a repeated anva"ndarsammanslutningar (5,1)
highest containing b repeated bearbetningsbeskrivning blixtsnabbt bubbelminne bubbelminnet bubbla bubblorna (3,6)
highest containing c repeated Chicago acceleration accelerator acceleratorer accelererad accent accentuera acceptabel acceptabelt acceptabla ... (2,35)
highest containing d repeated skra"ddarsydd (4,1)
highest containing e repeated beteendevetenskapliga (6,1)
highest containing f repeated Friluftsfra"mjandet Friluftsfra"mjandets fiffig fiffigt fiffla fo"ra"ldratra"ff fo"rbluffa fo"rfattarfond fo"rtra"fflighet friluftsfra"mjandets ... (3,12)
highest containing g repeated smugglingslagstiftning (5,1)
highest containing h repeated hemho"righet (3,1)
highest containing i repeated initialisering (5,1)
highest containing j repeated jockej jojk jojo projektuppfo"ljningen tjej tjejer (2,6)
highest containing k repeated kyrkvakterskan (4,1)
highest containing l repeated omsta"llningstillfa"llet (6,1)
highest containing m repeated gemensammma (4,1)
highest containing n repeated anknytningen anpassningsutrustning anva"ndarfo"reningen anva"ndarsammanslutningar anva"ndningen anva"ndningsomra@den anva"ndningsteknik betalningsbevakningen ledningsdragningen minnesanteckningsar ... (5,16)
highest containing o repeated socialantropolog (4,1)
highest containing p repeated papperstyp trappuppga@ngar uppkoppling uppkopplingsbega"ran uppkopplingsfo"rso"k uppsluppen vippomkopplare vippomkopplaren vippomkopplarens vippomkopplarna (4,10)
highest containing q repeated Nyqvist Osquar Osquarulda Osqulda lindqvist squaw toque (1,7)
highest containing r repeated programmerarkretsen programmeringsfo"rutsa"ttningar ra@varuresurser systemprogrammerarkursen (5,4)
highest containing s repeated produktionsstyrningssystem programmeringsassistans synkroniseringssekvenser sysselsa"tts (5,4)
highest containing t repeated utnyttjandestatistik (6,1)
highest containing u repeated maskinvarustrukturer (3,1)
highest containing v repeated Ulvsundava"gen Volvo a"rtvivel anva"ndarbeskrivning anva"ndarkravspecifikation anva"ndarutveckling anva"ndarva"nliga anva"ndarva"nlighet anva"ndarva"nligt anvvisning ... (2,130)
highest containing w repeated knowhow (2,1)
highest containing x repeated xenix (2,1)
highest containing y repeated materialstyrningssystem nybyggare nybyggaren nybyggd nybyggnad nyrekrytering produktionsstyrningssystem royalty synonym synonyma ... (2,12)
highest containing z repeated intermezzo intermezzot jazz (2,3)
most different letters fasfo"rskjutningsproblem (18,1)
highest ratio length/letters mormorsmors (275,1)
highest ratio length/letters (no tautonyms) mormorsmors (275,1)
lowest length 16 ratio length/letters funktionsfo"rma"ga (114,1)
lowest length 17 ratio length/letters skolmyndigheterna (113,1)
lowest length 18 ratio length/letters Friluftsfra"mjandet a@klagarmyndigheten anva"ndarutveckling befolkningstrycket datorbestyckningen uppdateringsfa"lten uppfo"ljningssystem utbildningslokaler (128,8)
lowest length 19 ratio length/letters upphandlingsbehovet va"xlingsfo"rfarandet (126,2)
lowest length 20 ratio length/letters a@lderdomsskro"plighet fo"rbrukningsmaterial fo"rbrukningsmateriel kompileringsa@tga"rder ordbehandlingssystem (133,5)
lowest length 21 ratio length/letters ha@rdvarubesta"llningar upplysningsverksamhet (131,2)
lowest length 22 ratio length/letters fo"rfattningshandbo"cker tra"dplanteringskampanj tra"dplanteringsprojekt (146,3)
lowest length 23 ratio length/letters fasfo"rskjutningsproblem (127,1)
lowest length 24 ratio length/letters cirkulationstja"nstgo"ring (160,1)
lowest length 25 ratio length/letters teckenomvandlingstabeller (156,1)
Letter Appearance
longest short letters mormorsmors (11,1)
longest tall letters liktydigt (9,1)
longest vertical-symmetry letters automat maximum oha"mmat (7,3)
longest horizontal-symmetry letters dioxid ebcdic ekbock (6,3)
longest full-symmetry letters ?
highest ratio of dotted letters idioti minimi (50,2)
Typewriter
longest top row prioritet (9,1)
longest middle row flaskhals (9,1)
longest bottom row mm (2,1)
longest in order uppha"v (6,1)
longest in reverse order ba"ttre kaputt va"ggar (6,3)
longest left hand Westerberg adressater vedertaget (10,3)
longest right hand homonym million minimum monopol oja"mlik opinion polynom (7,7)
longest alternating hands provisoriskt (12,1)
longest one finger hunn hymn (4,2)
longest adjacent keys dessa dress edert freds yttre (5,5)
Puzzle
longest palindrome in Morse code esse no"ja sans tant (4,4)
longest formed with chemical symbols banbeskrivningen replikerbarheten representationer (16,3)
longest formed with US postal codes Gunvor Pascal indela inlade inland inviga memoar utdela utlade utland ... (6,12)
longest formed with amino acid abbreviations asp pro (3,2)
longest formed with piano notes bagage (6,1)
Letter Order
Alphabetical
longest letters in order abort ansta@ begin behov bersa@ besta@ bista@ borst deist dekor ... (5,13)
longest letters in order with repeats accent access belopp (6,3)
longest letters in reverse order o"vriga poliga trolig utreda (6,4)
longest letters in reverse order with repeats utsedda (7,1)
longest roller-coaster parameterfa"lten (15,1)
longest no letters in place programmeringsfo"rutsa"ttningar (29,1)
most letters in place abstraktion akademi akademiker akademin akademisk alldeles anma"lningsblankett anskaffning anskaffningar ansta"llningsnummer ... (3,42)
most letters in place shifted kooperativ masterslavetyp operativ operativa operativt ordbehandlingssystem textbehandlingssystem torparstugan upphandlingsbeslut (5,9)
most consecutive letters in order consecutively gra"ddstuvad gra"ddstuvade gra"stuva gra"stuvan stuva timmerstuga torparstugan (4,7)
most consecutive letters in order gra"ddstuvad gra"ddstuvade gra"stuva gra"stuvan (5,4)
most consecutive letters fo"rfattningshandbo"cker (9,1)
highest ratio of consecutive letters to length vurst (100,1)
==> language/synonymous.reversals.p <==
What words are synonymous with their reversals in other langauges?
==> language/synonymous.reversals.s <==
Dutch: nier (kidney), French: rein
Almost:
French: etats (plural), English: state (singular)
==> language/vowels.repeated.p <==
In any language, what word contains the same vowel repeated four times in a row?
==> language/vowels.repeated.s <==
Article 23235 of rec.puzzles:
Newsgroups: rec.puzzles
Path: questrel!news.cerf.net!usc!cs.utexas.edu!uwm.edu!linac!att!princeton!phoenix.Princeton.EDU!cgseife
From: cgs...@phoenix.Princeton.EDU (Charles Geoffrey Seife)
Subject: Re: possibly original puzzle...
Message-ID: <1993May1.0...@Princeton.EDU>
Originator: news@nimaster
Sender: ne...@Princeton.EDU (USENET News System)
Nntp-Posting-Host: phoenix.princeton.edu
Organization: Princeton University
References: <93118.1933...@vma.cc.nd.edu> <1993Apr29...@vaxc.stevens-tech.edu> <C6B4M...@dartvax.dartmouth.edu>
Date: Sat, 1 May 1993 03:49:10 GMT
Lines: 28
In article <C6B4M...@dartvax.dartmouth.edu> z...@Dartmouth.EDU (Ted Schuerzinger) writes:
>In article <1993Apr29...@vaxc.stevens-tech.edu>
>u96_am...@vaxc.stevens-tech.edu writes:
>
>
>Unfortunately, that's not what the question asked -- it states that the
>first two vowels must be the same.
>
>For three vowels in a row, I came up with words like "booing" and
>"cooing". A better one might be "agreeable". I haven't come up with
>one for four yet, but I read in an old edition of Guinness that there's
>an Estonian word that contains has the same vowel four times in a row.
>Since the question didn't specify English words, I guess this would be
>acceptable, as would words in other agglutinative languages (eg.
>Finnish) :-)
>
>
>--Ted Schuerzinger
The Estonian word 'Ja"a"a"a"rne', which means "standing by the edge
of the ice."
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> logic/29.p <==
Three people check into a hotel. They pay $30 to the manager and go
to their room. The manager finds out that the room rate is $25 and
gives $5 to the bellboy to return. On the way to the room the bellboy
reasons that $5 would be difficult to share among three people so
he pockets $2 and gives $1 to each person.
Now each person paid $10 and got back $1. So they paid $9 each,
totalling $27. The bellboy has $2, totalling $29.
Where is the remaining dollar?
==> logic/29.s <==
Each person paid $9, totalling $27. The manager has $25 and the bellboy $2.
The bellboy's $2 should be added to the manager's $25 or subtracted from
the tenants' $27, not added to the tenants' $27.
==> logic/ages.p <==
1) Ten years from now Tim will be twice as old as Jane was when Mary was
nine times as old as Tim.
2) Eight years ago, Mary was half as old as Jane will be when Jane is one year
older than Tim will be at the time when Mary will be five times as old as
Tim will be two years from now.
3) When Tim was one year old, Mary was three years older than Tim will be when
Jane is three times as old as Mary was six years before the time when Jane
was half as old as Tim will be when Mary will be ten years older than Mary
was when Jane was one-third as old as Tim will be when Mary will be three
times as old as she was when Jane was born.
HOW OLD ARE THEY NOW?
==> logic/ages.s <==
The solution: Tim is 3, Jane is 8, and Mary is 15. A little grumbling
is in order here, as clue number 1 leads to the situation a year and a
half ago, when Tim was 1 1/2, Jane was 6 1/2, and Mary was 13 1/2.
This sort of problem is easy if you write down a set of equations. Let
t be the year that Tim was born, j be the year that Jane was born, m be
the year that Mary was born, and y be the current year. As indefinite
years come up, let y1, y2, ... be the indefinite years. Then the
equations are
y + 10 - t = 2 (y1 - j)
y1 - m = 9 (y1 - t)
y - 8 - m = 1/2 (y2 - j)
y2 - j = 1 + y3 - t
y3 - m = 5 (y + 2 - t)
t + 1 - m = 3 + y4 - t
y4 - j = 3 (y5 - 6 - m)
y5 - j = 1/2 (y6 - t)
y6 - m = 10 + y7 - m
y7 - j = 1/3 (y8 - t)
y8 - m = 3 (j - m)
t = y - 3
j = y - 8
m = y - 15
==> logic/attribute.p <==
All the items in the first list share a particular attribute. The second
list is of some items lacking the attribute.
Set#1
with: battery, key, yeast, bookmark
w/out: stapler, match, Rubik's cube, pill bottle
Set#2
with: Rubik's cube, chess set, electrical wiring, compass needle
w/out: clock, rope, tic-tac-toe, pencil sharpener
Set#3:
with: koosh, small intestine, Yorkshire Terrier, Christmas Tree
w/out: toothbrush, oak chair, soccer ball, icicle
Points to realize:
1.
There may be exceptions to any item on the list, for instance a particular
clock may share the properties of the 'with' list of problem two, BUT MOST
ORDINARY clocks do not. All the properties apply the vast majority of the
the items mentioned. Extraordinary exceptions should be ignored.
2.
Pay the most attention to the 'with' list. The 'without' list is only
present to eliminate various 'stupid' answers.
==> logic/attribute.s <==
The attribute puzzle format is a traditional format in math education.
It occurs in logic materials developed in the sixties by EDC in Boston,
with visual objects. Example:
these are gloops: A B C D E
these are not gloops: F G J L N
which of these are gloops? O P Q R S
Set#1
with: battery, key, yeast, bookmark
w/out: stapler, match, Rubik's cube, pill bottle
Needs to be placed inside something else when used for its usual purpose.
Set#2
with: Rubik's cube, chess set, electrical wiring, compass needle
w/out: clock, rope, tic-tac-toe, pencil sharpener
Uses color to distinguish between otherwise identical parts.
Set#3:
with: koosh, small intestine, Yorkshire Terrier, Christmas Tree
w/out: toothbrush, oak chair, soccer ball, icicle
Villiform.
==> logic/bookworm.p <==
A bookworm eats from the first page of an encyclopedia to the last
page. The bookworm eats in a straight line. The encyclopedia consists
of ten 1000-page volumes and is sitting on a bookshelf in the usual
order. Not counting covers, title pages, etc., how many pages does the
bookworm eat through?
==> logic/bookworm.s <==
On a book shelf the first page of the first volume is on the "inside"
__ __
B| | | |F
A|1 |...........................|10|R
C| | | |O
K| | | |N
| | | |T
----------------------------------
so the bookworm eats only through the cover of the first volume, then 8 times
1000 pages of Volumes 2 - 9, then through the cover to the 1st page of Vol 10.
He eats 8,000 pages.
If the bookworm ate the first page and the last page, it ate 8,004 pages.
==> logic/boxes.p <==
Which Box Contains the Gold?
Two boxes are labeled "A" and "B". A sign on box A says "The sign
on box B is true and the gold is in box A". A sign on box B says
"The sign on box A is false and the gold is in box A". Assuming there
is gold in one of the boxes, which box contains the gold?
==> logic/boxes.s <==
The problem cannot be solved with the information given.
The sign on box A says "The sign on box B is true and the gold is in box A".
The sign on box B says "The sign on box A is false and the gold is in box A".
The following argument can be made: If the statement on box A is true, then
the statement on box B is true, since that is what the statement on box A
says. But the statement on box B states that the statement on box A is false,
which contradicts the original assumption. Therefore, the statement on box A
must be false. This implies that either the statement on box B is false or
that the gold is in box B. If the statement on box B is false, then either
the statement on box A is true (which it cannot be) or the gold is in box B.
Either way, the gold is in box B.
However, there is a hidden assumption in this argument: namely, that
each statement must be either true or false. This assumption leads to
paradoxes, for example, consider the statement: "This statement is
false." If it is true, it is false; if it is false, it is true. The
only way out of the paradox is to deny that the statement is either true
or false and label it meaningless instead. Both of the statements on the
boxes are therefore meaningless and nothing can be concluded from them.
In general, statements about the truth of other statements lead to
contradictions. Tarski invented metalanguages to avoid this problem.
To avoid paradox, a statement about the truth of a statement in a language
must be made in the metalanguage of the language.
Common sense dictates that this problem cannot be solved with the information
given. After all, how can we deduce which box contains the gold simply by
reading statements written on the outside of the box? Suppose we deduce that
the gold is in box B by whatever line of reasoning we choose. What is to stop
us from simply putting the gold in box A, regardless of what we deduced?
(cf. Smullyan, "What Is the Name of This Book?", Prentice-Hall, 1978, #70)
==> logic/camel.p <==
An Arab sheikh tells his two sons to race their camels to a distant
city to see who will inherit his fortune. The one whose camel is
slower will win. The brothers, after wandering aimlessly for days, ask
a wise man for advise. After hearing the advice they jump on the
camels and race as fast as they can to the city. What does the wise
man say?
==> logic/camel.s <==
The wiseman tells them to switch camels.
==> logic/centrifuge.p <==
You are a biochemist, working with a 12-slot centrifuge. This is a gadget
that has 12 equally spaced slots around a central axis, in which you can
place chemical samples you want centrifuged. When the machine is turned on,
the samples whirl around the central axis and do their thing.
To ensure that the samples are evenly mixed, they must be distributed in the
12 slots such that the centrifuge is balanced evenly. For example, if you
wanted to mix 4 samples, you could place them in slots 12, 3, 6 and 9
(assuming the slots are numbered from 1 to 12 like a clock).
Problem: Can you use the centrifuge to mix 5 samples?
==> logic/centrifuge.s <==
The superposition of any two solutions is yet another solution, so given
that the factors > 1 of 12 (2, 3, 4, 6, 12) are all solutions, the
only thing to check about, for example, the proposed solution 2+3 is
that not all ways of combining 2 & 3 would have centrifuge tubes
from one subsolution occupying the slot for one of the tubes in
another solution. For the case 2+3, there is no problem: Place 3
tubes, one in every 4th position, then place the 4th and 5th
diametrically opposed (each will end up in a slot adjacent to one of
the first 3 tubes). The obvious generalization is, what are the
numbers of tubes that cannot be balanced? Observing that there are
solutions for 2,3,4,5,6 tubes and that if X has a solution, 12-X has
also one (obtained by swapping tubes and holes), it is obvious that
1 and 11 are the only cases without solutions.
Here is how this problem is often solved in practice: A dummy tube
is added to produce a total number of tubes that is easy to balance.
For example, if you had to centrifuge just one sample, you'd add a
second tube opposite it for balance.
==> logic/chain.p <==
What is the least number of links you can cut in a chain of 21 links to be able
to give someone all possible number of links up to 21?
==> logic/chain.s <==
Two.
OOO C OOOOO C OOOOOOOOOOO
(where Os are chained unbroken links, and the Cs are the unchained broken links)
And equivalently:
OOO C OOOOOO C OOOOOOOOOO
==> logic/children.p <==
A man walks into a bar, orders a drink, and starts chatting with the
bartender. After a while, he learns that the bartender has three
children. "How old are your children?" he asks. "Well," replies the
bartender, "the product of their ages is 72." The man thinks for a
moment and then says, "that's not enough information." "All right,"
continues the bartender, "if you go outside and look at the building
number posted over the door to the bar, you'll see the sum of the
ages." The man steps outside, and after a few moments he reenters and
declares, "Still not enough!" The bartender smiles and says, "My
youngest just loves strawberry ice cream."
How old are the children?
A variant of the problem is for the sum of the ages to be 13 and the
product of the ages to be the number posted over the door. In this
case, it is the oldest that loves ice cream.
Then how old are they?
==> logic/children.s <==
First, determine all the ways that three ages can multiply together to get 72:
72 1 1 (quite a feat for the bartender)
36 2 1
24 3 1
18 4 1
18 2 2
12 6 1
12 3 2
9 4 2
9 8 1
8 3 3
6 6 2
6 4 3
As the man says, that's not enough information; there are many possibilities.
So the bartender tells him where to find the sum of the ages--the man now knows
the sum even though we don't. Yet he still insists that there isn't enough
info. This must mean that there are two permutations with the same sum;
otherwise the man could have easily deduced the ages.
The only pair of permutations with the same sum are 8 3 3 and 6 6 2, which both
add up to 14 (the bar's address). Now the bartender mentions his
"youngest"--telling us that there is one child who is younger than the other
two. This is impossible with 8 3 3--there are two 3 year olds. Therefore the
ages of the children are 6, 6, and 2.
Pedants have objected that the problem is insoluble because there could be
a youngest between two three year olds (even twins are not born exactly at
the same time). However, the word "age" is frequently used to denote the
number of years since birth. For example, I am the same age as my wife,
even though technically she is a few months older than I am. And using the
word "youngest" to mean "of lesser age" is also in keeping with common parlance.
So I think the solution is fine as stated.
In the sum-13 variant, the possibilities are:
11 1 1
10 2 1
9 3 1
9 2 2
8 4 1
8 3 2
7 5 1
7 4 2
7 3 3
6 6 1
6 5 2
6 4 3
The two that remain are 9 2 2 and 6 6 1 (both products equal 36). The
final bit of info (oldest child) indicates that there is only one
child with the highest age. This cancels out the 6 6 1 combination, leaving
the childern with ages of 9, 2, and 2.
==> logic/condoms.p <==
How can a man have mutually safe sex with three women with only two condoms?
How can three men have mutually safe sex with one woman with two condoms?
==> logic/condoms.s <==
Use both condoms on the first woman. Take off the outer condom (turning it
inside-out in the process) and set it aside. Use the inner condom alone on
the second woman. Put the outer condom back on. Use it on the third woman.
First man uses both condoms. Take off the outer condom (do NOT reverse
it) and have second man use it. First man takes off the inner condom
(turning it inside-out). Third man puts on this condom, followed by
second man's condom.
==> logic/dell.p <==
How can I solve logic puzzles (e.g., as published by Dell) automatically?
==> logic/dell.s <==
#include <setjmp.h>
#define EITHER if (S[1] = S[0], ! setjmp((S++)->jb)) {
#define OR } else EITHER
#define REJECT longjmp((--S)->jb, 1)
#define END_EITHER } else REJECT;
/* values in tmat: */
#define T_UNK 0
#define T_YES 1
#define T_NO 2
#define Val(t1,t2) (S->tmat[t1][t2])
#define CLASS(x) \
(((x) / NUM_ITEM) * NUM_ITEM)
#define EVERY_TOKEN(x) \
(x = 0; x < TOT_TOKEN; x++)
#define EVERY_ITEM(x, class) \
(x = CLASS(class); x < CLASS(class) + NUM_ITEM; x++)
#define BEGIN \
struct state { \
char tmat[TOT_TOKEN][TOT_TOKEN]; \
jmp_buf jb; \
} States[100], *S = States; \
\
main() \
{ \
int token; \
\
for EVERY_TOKEN(token) \
yes(token, token); \
EITHER
/* Here is the problem-specific data */
#define NUM_ITEM 5
#define NUM_CLASS 6
#define TOT_TOKEN (NUM_ITEM * NUM_CLASS)
#define HOUSE_0 0
#define HOUSE_1 1
#define HOUSE_2 2
#define HOUSE_3 3
#define HOUSE_4 4
#define ENGLISH 5
#define SPANISH 6
#define NORWEG 7
#define UKRAIN 8
#define JAPAN 9
#define GREEN 10
#define RED 11
#define IVORY 12
#define YELLOW 13
#define BLUE 14
#define COFFEE 15
#define TEA 16
#define MILK 17
#define OJUICE 18
#define WATER 19
#define DOG 20
#define SNAIL 21
#define FOX 22
#define HORSE 23
#define ZEBRA 24
#define OGOLD 25
#define PLAYER 26
#define CHESTER 27
#define LSTRIKE 28
#define PARLIA 29
char *names[] = {
"HOUSE_0", "HOUSE_1", "HOUSE_2", "HOUSE_3", "HOUSE_4",
"ENGLISH", "SPANISH", "NORWEG", "UKRAIN", "JAPAN",
"GREEN", "RED", "IVORY", "YELLOW", "BLUE",
"COFFEE", "TEA", "MILK", "OJUICE", "WATER",
"DOG", "SNAIL", "FOX", "HORSE", "ZEBRA",
"OGOLD", "PLAYER", "CHESTER", "LSTRIKE", "PARLIA",
};
BEGIN
yes(ENGLISH, RED); /* Clue 1 */
yes(SPANISH, DOG); /* Clue 2 */
yes(COFFEE, GREEN); /* Clue 3 */
yes(UKRAIN, TEA); /* Clue 4 */
EITHER /* Clue 5 */
yes(IVORY, HOUSE_0);
yes(GREEN, HOUSE_1);
OR
yes(IVORY, HOUSE_1);
yes(GREEN, HOUSE_2);
OR
yes(IVORY, HOUSE_2);
yes(GREEN, HOUSE_3);
OR
yes(IVORY, HOUSE_3);
yes(GREEN, HOUSE_4);
END_EITHER
yes(OGOLD, SNAIL); /* Clue 6 */
yes(PLAYER, YELLOW); /* Clue 7 */
yes(MILK, HOUSE_2); /* Clue 8 */
yes(NORWEG, HOUSE_0); /* Clue 9 */
EITHER /* Clue 10 */
yes(CHESTER, HOUSE_0);
yes(FOX, HOUSE_1);
OR
yes(CHESTER, HOUSE_4);
yes(FOX, HOUSE_3);
OR
yes(CHESTER, HOUSE_1);
EITHER yes(FOX, HOUSE_0);
OR yes(FOX, HOUSE_2);
END_EITHER
OR
yes(CHESTER, HOUSE_2);
EITHER yes(FOX, HOUSE_1);
OR yes(FOX, HOUSE_3);
END_EITHER
OR
yes(CHESTER, HOUSE_3);
EITHER yes(FOX, HOUSE_2);
OR yes(FOX, HOUSE_4);
END_EITHER
END_EITHER
EITHER /* Clue 11 */
yes(PLAYER, HOUSE_0);
yes(HORSE, HOUSE_1);
OR
yes(PLAYER, HOUSE_4);
yes(HORSE, HOUSE_3);
OR
yes(PLAYER, HOUSE_1);
EITHER yes(HORSE, HOUSE_0);
OR yes(HORSE, HOUSE_2);
END_EITHER
OR
yes(PLAYER, HOUSE_2);
EITHER yes(HORSE, HOUSE_1);
OR yes(HORSE, HOUSE_3);
END_EITHER
OR
yes(PLAYER, HOUSE_3);
EITHER yes(HORSE, HOUSE_2);
OR yes(HORSE, HOUSE_4);
END_EITHER
END_EITHER
yes(LSTRIKE, OJUICE); /* Clue 12 */
yes(JAPAN, PARLIA); /* Clue 13 */
EITHER /* Clue 14 */
yes(NORWEG, HOUSE_0);
yes(BLUE, HOUSE_1);
OR
yes(NORWEG, HOUSE_4);
yes(BLUE, HOUSE_3);
OR
yes(NORWEG, HOUSE_1);
EITHER yes(BLUE, HOUSE_0);
OR yes(BLUE, HOUSE_2);
END_EITHER
OR
yes(NORWEG, HOUSE_2);
EITHER yes(BLUE, HOUSE_1);
OR yes(BLUE, HOUSE_3);
END_EITHER
OR
yes(NORWEG, HOUSE_3);
EITHER yes(BLUE, HOUSE_2);
OR yes(BLUE, HOUSE_4);
END_EITHER
END_EITHER
/* End of problem-specific data */
solveit();
OR
printf("All solutions found\n");
exit(0);
END_EITHER
}
no(a1, a2)
{
int non1, non2, token;
if (Val(a1, a2) == T_YES)
REJECT;
else if (Val(a1, a2) == T_UNK) {
Val(a1, a2) = T_NO;
no(a2, a1);
non1 = non2 = -1;
for EVERY_ITEM(token, a1)
if (Val(token, a2) != T_NO)
if (non1 == -1)
non1 = token;
else
break;
if (non1 == -1)
REJECT;
else if (token == CLASS(a1) + NUM_ITEM)
yes(non1, a2);
for EVERY_TOKEN(token)
if (Val(token, a1) == T_YES)
no(a2, token);
}
}
yes(a1, a2)
{
int token;
if (Val(a1, a2) == T_NO)
REJECT;
else if (Val(a1, a2) == T_UNK) {
Val(a1, a2) = T_YES;
yes(a2, a1);
for EVERY_ITEM(token, a1)
if (token != a1)
no(token, a2);
for EVERY_TOKEN(token)
if (Val(token, a1) == T_YES)
yes(a2, token);
else if (Val(token, a1) == T_NO)
no(a2, token);
}
}
solveit()
{
int token, tok2;
for EVERY_TOKEN(token)
for (tok2 = token; tok2 < TOT_TOKEN; tok2++)
if (Val(token, tok2) == T_UNK) {
EITHER
yes(token, tok2);
OR
no(token, tok2);
END_EITHER;
return solveit();
}
printf("Solution:\n");
for EVERY_ITEM(token, 0) {
for (tok2 = NUM_ITEM; tok2 < TOT_TOKEN; tok2++)
if (Val(token, tok2) == T_YES)
printf("\t%s %s\n",names[token],names[tok2]);
printf("\n");
}
REJECT;
}
---
ja...@crc.ricoh.com (James Allen)
==> logic/elimination.p <==
97 baseball teams participate in an annual state tournament.
The way the champion is chosen for this tournament is by the same old
elimination schedule. That is, the 97 teams are to be divided into
pairs, and the two teams of each pair play against each other.
After a team is eliminated from each pair, the winners would
be again divided into pairs, etc. How many games must be played
to determine a champion?
==> logic/elimination.s <==
In order to determine a winner all but one team must lose.
Therefore there must be at least 96 games.
==> logic/flip.p <==
How can a toss be called over the phone (without requiring trust)?
==> logic/flip.s <==
A flips a coin. If the result is heads, A multiplies 2 prime numbers
containing about 90 digits; if the result is tails, A multiplies 3
prime numbers containing about 60 digits. A tells B the result of the
multiplication. B now calls either heads or tails and tells A. A then
supplies B with the original numbers to verify the flip.
Consider what is involved in multiplying 90 digit numbers. Using the method
of long multiplication that we all learned in grade school, you have maybe
90 or so strings of 90 characters (or "digits") each. That's no problem for
a computer to deal with. The magnitude of the number represented isn't
much of a factor; we're only manipulating the string of digits.
Consider what is involved in factoring 90 digit numbers. There are of course,
little tricks for determining factorability by small integers which we all
learned in grade school. (Is the last digit even? Is the sum of all the
digits divisible by 9? And so on.) But these are of little use in factoring
large numbers with large factors. In fact, there is no essentially better
method than checking every number smaller that the number to be factored and
seeing if it one divides the other evenly. We means we could be checking some
100,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
nummbers. This is very hard to do, even for a supercomputer. Here, the
number of digits this number has is of little consequence. It is the
magnitude of the number that we have to worry about, and there just isn't
enough time in the world to do this properly.
Where does A find a list of 60- and 90- digit prime numbers? Well, that's not
to hard to come by. The simplest method to find a few prime numbers is simply
to choose a 90-digit number (or 60-digit number, as the case warrants)
randomly, and check to see if it is prime. You won't have to wait too long
before you stumble across a prime number.
"But wait!" you cry. "I thought you just said it was incredibly difficult
to factor large numbers. If that's the case, then how are you going to know
if the number you randomly choose is prime?" A good question. Here we enter
into the strange an wacky world of number theory. It turns out (and I won't
explain how unless someone asks) there are "probabalistic" algorithms,
which depend on us choosing numbers at random. There are probablistic
algorithms that when given a prime number, will quickly tell us that it is
a prime number, and when given a composite number, will either tell us that
it is a composite number (with very, very high probability) or will tell us
that it is a prime number (with very, very low probability.) What's the use
of an algorithm that only returns the right answer "with very, very high
probability?" Well, we can make this probability as high as we want, just by
running the algorithm longer. In fact, it is within our technological
abilities to quickly run this algorithm for 90-digit numbers so that the
probability of it giving a wrong answer is less than the probability of a
cosmic ray striking a vital part of the computer at some vital time and causing
the computer to spit out the wrong answer anyway. That's what I mean by "very,
very high."
==> logic/flowers.p <==
How many flowers do I have if all of them are roses except two, all of
them are tulips except two, and all of them are daisies except two?
==> logic/flowers.s <==
There are two solutions:
Three flowers: rose, tulip, daisy
Two flowers: carnation, geranium
==> logic/friends.p <==
Any group of 6 or more contains either 3 mutual friends or 3 mutual strangers.
Prove it.
==> logic/friends.s <==
Take a person X. Of the five other people, there must either be at least three
acquaintances of X or at least three strangers of X. Assume wlog that X has
three strangers A,B,C. Unless A,B,C is the required triad of acquaintances,
they must include a pair of strangers, wlog A,B. Then X,A,B is the required
triad of strangers, QED.
==> logic/hofstadter.p <==
In first-order logic, find a predicate P(x) which means "x is a power of 10."
==> logic/hofstadter.s <==
Well, one summer, I decided to tackle the problem. Not having any great
knowledge of number theory, I used a more brute force approach. Note
that for greater comprehensibility, I have broken the resulting formula
up into several stages, but it would not be difficult to put it
back together into one vast formula:
{e is prime:}
PRIME(e) :=
~Eb:Ec:((b+2)*(c+2) = e)
{if e is a prime, true iff a is a power of e:}
PPOWER(a,e) :=
Ab:Ac:((b*c = a) -> ((b = 1) or (Ed: (e*d) = b)))
{if e is a prime, and a is a power of e, true iff d is the
(log_e a)th digit (counting from the right, starting with 0)
in the base-e expansion of n:}
DIG(a,e,d,n) :=
(d < e) & Eb:Ec:((c < a) & (n = (b*e*a) + (d*a) + c))
{if e is prime, and a is a power of e, true iff n has exactly
(log_e a) digits in its base-e expansion (0 is counted as having 0
digits:}
LENGTH(e,a,n):=
(n < a) & Ab:((PPOWER(b,e) & (b < a)) -> (b <= n))
POWER_OF_TEN(x):=
Ee:(PRIME(e) & (e > x) &
En:Ea:(LENGTH(e,a,n) &
DIG(1,e,1,n) &
Ai:Aj:Au:( (PPOWER(u,e) & ((e*u) < a)
& DIG(u,e,i,n) & DIG(e*u,e,j,n))
-> j = (10 * i) ) &
Eu:(PPOWER(u,e) & (e*u = a) & DIG(u,e,x,n))
) )
The basic idea is that you are asserting that for some prime e greater
than x, we can find a number n which, when expressed in base-e notation,
will have 1 in its units place, 10 in its e's place, 100 in its (e^2)'s
place, and in general have the "digit" in each place be 10 times
greater than the one to its right, such that the leftmost digit is our x.
To translate into Hofstadter's notation, of course, we must use horse-shoes
instead of ->'s, big carets instead of &'s, letters a through e (followed
by however many ''s) exclusively, and so forth. (We must also replace <'s
and <= with appropriate expansions, and substitute in for our capitalized
formula abbreviations.) This is left as an exercise to the reader.
Kevin Wald
wa...@husc.harvard.edu
==> logic/hundred.p <==
A sheet of paper has statements numbered from 1 to 100. Statement n says
"exactly n of the statements on this sheet are false." Which statements are
true and which are false? What if we replace "exactly" by "at least"?
==> logic/hundred.s <==
From _Litton's Problematical Recreations_
It is tempting to argue as follows:
At most one statement can be true (they are mutually contradictory).
If they are all false, statement 100 would be true, which is no good.
So 99 are false, and statement 99 is true.
If replaced by "at least", and the "real" number of false statements is
x, then statements x+1 to 100 will be false (since they falsely claim
that there are more false statements than there actually are). So, 100-x are
false, ie. x=100-x, so x=50. The first 50 statements are true, and statements
51 to 100 are false.
However, there is a hidden and incorrect assumption in this argument.
To see this, suppose that there is one statement on the sheet and it
says "One statement is false" or "At least one statement is false,"
either way it implies "this statement is false," which is a familiar
paradoxical statement. We have learned that this paradox arises because
of the false assumption that all statements are either true or false.
This is the hidden assumption in the above reasoning.
If it is acknowledged that some of the statements on the page may be
neither true nor false (i.e., meaningless), then nothing whatsoever can
be concluded about which statements are true or false.
This problem has been carefully contrived to appear to be solvable (like
the vacuous statement "this statement is true"). By changing the
numbers in some statements and changing "true" to "false," various
circular forms of the liar's paradox can be constructed. A much
more complicated version of the same problem is:
1) At least one of the last two statements in this list is true.
2) This is either the first true or the first false statement in the
list.
3) There exist three consecutive false statements.
4) The difference beween the numbers of the last true statement and
the first true statement is a factor of the unknown number.
5) The sum of the numbers of the true statements is the unknown
number.
6) This is not the last true statement.
7) Each true statement's number is a factor of the unknown number.
8) The unknown number equals the percentage of these statements which
are true.
9) The number of different factors which the unknown number has
(excluding 1 and itself) is more than the sum of the numbers of the
true statements.
10) There are no three cosecutive true statements.
What is the number?
The incorrect but plausible solution is:
By 2, either way 1 must be false, and then so must both 9 and 10.
6, if false, says "This is the last true statement", which gives
a paradox, thus 6 must be true, and so must 7 and/or 8.
7 and 8 cannot both be true, as the number had to be a multiple
of 6,7,8 , that is a multiple of 168 (by 7), and less than 100 (by 8)
If 8 is false, then 3 is true (8,9,10 is false), if 8 is true, then
3 is false (3 cannot be true when both 6 and 8 are true, then there
are no three consecutive statements left).
So we have either
A) F X F X X T F T F F
or
B) F X T X X T T F F F
In A), 4 and 5 must be true (by false 10), and 2 may be true or false.
So by 5 the number shall be either 27 (2+3+4+5+6+7) or 25 (3+4+5+6+7).
None of these can fullfill 8, though, so A) is out leaving us with B)
Now, (by 7) 3,6 and 7 are factors, the number must be a multiple of
42, as 2+3+4+5+6+7=27, 5 must be false.
By false 10, 2 and 4 must be true, that is 5 shall be a multiple
of the number. Now the number must be a multiple of 2,3,4,5,6 and 7,
that is a multiple of 3*4*5*7=420. 420 has 22 different factors
(2,3,4,5,6,7,10,12,14,15,20,21,28,30,35,42,60,70,84,105,140,210)
and the sum 2+3+4+6+7 = 22, so the only multiple of 420 that
fulfills false 9 is 420.
==> logic/inverter.p <==
Can a digital logic circuit with two inverters invert N independent inputs?
The circuit may contain any number of AND or OR gates.
==> logic/inverter.s <==
It can be shown that N inverters can invert 2^N-1 independent inputs,
given an unlimited supply of AND and OR gates. The classic version of
this puzzle is to invert 3 independent inputs using AND gates, OR
gates, and only 2 inverters.
This is solved by:
n1 = not(i1 and i2 or i1 and i3 or i2 and i3);
n2 = not((i1 or i2 or i3) and n1 or i1 and i2 and i3);
o1 = (i2 or i3 or n2) and n1 or i2 and i3 and n2;
o2 = (i1 or i3 or n2) and n1 or i1 and i3 and n2;
o3 = (i1 or i2 or n2) and n1 or i1 and i2 and n2;
i1, i2, and i3 are the inputs, n1 and n2 are the inverted signals, and
o1, o2, and o3 are the outputs. "and" has higher precedence than "or".
So, start with N inverters. Replace 3 of them with 2.
Keep doing that until you're down to 2 inverters.
I was skeptical at first, because such a design requires so much feedback
that I was sure the system would oscillate when switching between two
particular states. But after writing a program to test every possible state
change (32^2), it appears that this system settles after a maximum of
3 feedback logic iterations. I did not include gate delays in the simulation,
however, which could increase the number of iterations before the system
settles.
In any case, it appears that the world needs only 2 inverters! :-)
==> logic/josephine.p <==
The recent expedition to the lost city of Atlantis discovered scrolls
attributted to the great poet, scholar, philosopher Josephine. They
number eight in all, and here is the first.
THE KINGDOM OF MAMAJORCA, WAS RULED BY QUEEN HENRIETTA I. IN MAMAJORCA
WOMEN HAVE TO PASS AN EXTENSIVE LOGIC EXAM BEFORE THEY ARE ALLOWED TO
GET MARRIED. QUEENS DO NOT HAVE TO TAKE THIS EXAM. ALL THE WOMEN IN
MAMAJORCA ARE LOYAL TO THEIR QUEEN AND DO WHATEVER SHE TELLS THEM TO.
THE QUEENS OF MAMAJORCA ARE TRUTHFUL. ALL SHOTS FIRED IN MAMAJORCA CAN
BE HEARD IN EVERY HOUSE. ALL ABOVE FACTS ARE KNOWN TO BE COMMON
KNOWLEDGE.
HENRIETTA WAS WORRIED ABOUT THE INFIDELITY OF THE MARRIED MEN IN
MAMAJORCA. SHE SUMMONED ALL THE WIVES TO THE TOWN SQUARE, AND MADE
THE FOLLOWING ANNOUNCEMENT. "THERE IS AT LEAST ONE UNFAITHFUL HUSBAND
IN MAMAJORCA. ALL WIVES KNOW WHICH HUSBANDS ARE UNFAITHFUL, BUT HAVE
NO KNOWLEDGE ABOUT THE FIDELITY OF THEIR OWN HUSBAND. YOU ARE
FORBIDDEN TO DISCUSS YOUR HUSBAND'S FAITHFULNESS WITH ANY OTHER WOMAN.
IF YOU DISCOVER THAT YOUR HUSBAND IS UNFAITHFUL, YOU MUST SHOOT HIM AT
PRECISELY MIDNIGHT OF THE DAY YOU FIND THAT OUT."
THIRTY-NINE SILENT NIGHTS FOLLOWED THE QUEEN'S ANNOUNCEMENT. ON THE
FORTIETH NIGHT, SHOTS WERE HEARD. QUEEN HENRIETTA I IS REVERED IN
MAMAJORCAN HISTORY.
As with all philosophers Josephine doesn't provide the question, but leaves
it implicit in his document. So figure out the questions - there are two -
and answer them.
Here is Josephine's second scroll.
QUEEN HENRIETTA I WAS SUCCEEDED BY DAUGHTER QUEEN HENRIETTA II. AFTER
A WHILE HENRIETTA LIKE HER FAMOUS MOTHER BECAME WORRIED ABOUT THE
INFIDELITY PROBLEM. SHE DECIDED TO ACT, AND SENT A LETTER TO HER
SUBJECTS (WIVES) THAT CONTAINED THE EXACT WORDS OF HENRIETTA I'S
FAMOUS SPEECH. SHE ADDED THAT THE LETTERS WERE GUARENTEED TO REACH
ALL WIVES EVENTUALLY.
QUEEN HENRIETTA II IS REMEMBERED AS A FOOLISH AND UNJUST QUEEN.
What is the question and answer implied by this scroll?
==> logic/josephine.s <==
The two questions for scroll #1 were:
1. How many husbands were shot on that fateful night?
2. Why is Queen Henrietta I revered in Mamajorca?
The answers are:
If there are n unfaithful husbands (UHs), every wife of an UH knows of
n-1 UH's while every wife of a faithful husband knows of n UHs. [this
because everyone has perfect information about everything except the
fidelity of their own husband]. Now we do a simple induction: Assume
that there is only one UH. Then all the wives but one know that there
is just one UH, but the wife of the UH thinks that everyone is
faithful. Upon hearing that "there is at least one UH", the wife
realizes that the only husband it can be is her own, and so shoots
him. Now, imagine that there are just two UH's. Each wife of an UH
assumes that the situation is "only one UH in town" and so waits to
hear the other wife (she knows who it is, of course) shoot her husband
on the first night. When no one is shot, that can only be because her
OWN husband was a second UH. The wife of the second UH makes the same
deduction when no shot is fired the first night (she was waiting, and
expecting the other to shoot, too). So they both figure it out after
the first night, and shoot their husbands the second night. It is
easy to tidy up the induction to show that the n UHs will all be shot
just on the n'th midnight.
The question for scroll #2 is:
3. Why is Queen Henrietta II not?
The answer is:
The problem now is that QHII didn't realize that it is *critical* that
all of the wives, of faithful and UH's alike, to
*BEGIN*AT*THE*SAME*MOMENT*. The uncertainty of having a particular
wife's notice come a day or two late makes the whole logic path fall
apart. That's why she's foolish. She is unjust, because some wives,
honed and crack logicians all, remember, will *incorrectly* shoot
faithful husbands. Let us imagine the situation with just a SINGLE UH
in the whole country. And, wouldn't you know it, the notice to the
wife of the UH just happens to be held up a day, whereas everyone
else's arrived the first day. Now, all of the wives that got the
notice the first day know that there is just one UH in the country.
And they know that the wife of that UH will think that everyone is
faithful, and so they'll expect her to figure it out and shoot her
husband the first night. BUT SHE DIDN"T GET THE NOTICE THE FIRST
NIGHT.... BUT THE OTHER WIVES HAVE NO WAY OF KNOWING THAT. So, the
wife of the UH doesn't know that anything is going on and so (of
course) doesn't do anything the first night. The next day she gets
the notice, figures it all out, and her husband will be history come
that midnight. BUT... *every* other wife thought that there should
have been a shooting the first night, and since there wasn't there
must have been an additional UH, and it can only have been _her_
husband. So on the second night **ALL** of the husbands are shot.
Things are much more complicated if the mix of who gets the notice
when is less simple than the one I mentioned above, but it is always
wrong and/or tragic.
NOTE: if the wives *know* that the country courier service (or however
these things get delivered) is flaky, then they can avoid the
massacre, but unless the wives exchange notes no one will ever be shot
(since there is always a chance that rather than _your_ husband being
an UH, you could reason that it might be that the wife of one of the
UH's that you know about just hasn't gotten her copy of the scroll
yet). I guess you could call this case "unjust", too, since the UH's
evade punishment, despite the perfect logic of the wives.
==> logic/locks.and.boxes.p <==
You want to send a valuable object to a friend. You have a box which
is more than large enough to contain the object. You have several
locks with keys. The box has a locking ring which is more than large enough
to have a lock attached. But your friend does not have the key to any
lock that you have. How do you do it? Note that you cannot send a key
in an unlocked box, since it might be copied.
==> logic/locks.and.boxes.s <==
Attach a lock to the ring. Send it to her. She attaches her own lock
and sends it back. You remove your lock and send it back to her. She
removes her lock.
==> logic/min.max.p <==
In a rectangular array of people, which will be taller, the tallest of the
shortest people in each column, or the shortest of the tallest people in each row?
==> logic/min.max.s <==
Let T denote shortest of tall
Let S denote tallest of short
-------------------------------
| |
| |
| S |
| |
| |
| T X |
| |
| |
| |
| |
-------------------------------
So T >= X >= S.
==> logic/mixing.p <==
Start with a half cup of tea and a half cup of coffee. Take one tablespoon
of the tea and mix it in with the coffee. Take one tablespoon of this mixture
and mix it back in with the tea. Which of the two cups contains more of its
original contents?
==> logic/mixing.s <==
Mixing Liquids
The two cups end up with the same volume of liquid they started with. The same
amount of tea was moved to the coffee cup as coffee to the teacup. Therefore
each cup contains the same amount of its original contents.
==> logic/monty.52.p <==
Monty and Waldo play a game with N closed boxes. Monty hides a
dollar in one box; the others are empty. Monty opens the empty boxes
one by one. When there are only two boxes left Waldo opens either box;
he wins if it contains the dollar. Prior to each of the N-2 box
openings Waldo chooses one box and locks it, preventing Monty from opening
it next. That box is then unlocked and cannot be so locked twice in a row.
What are the optimal strategies for Monty and Waldo and what is the
fair price for Waldo to pay to play the game?
==> logic/monty.52.s <==
The fair price for large N is $0.6321205588285576784; I will offer
a proof along with optimal strategies.
Denote the game as G_N(). After (N-M) rounds of play, the game will have
the same form as G_M(). Depending on the strategies each of the M boxes
will have a probability p_i of containing the dollar. Let Waldo lock
the M'th box (renumbering the boxes if necessary). Denote the game and
Waldo's expected winnings in the game by
G_M(p_1, p_2, ..., p_M)
where
p_1 + p_2 + ... + p_M = 1
When
p_2 = p_3 = p_4 = ... = p_M
we adopt the abbreviation
G_M(b) = G_M(1 + b - Mb, b, b, b, b, ..., b)
and note that since probabilities are never negative:
1 + b - Mb >= 0, or
0 <= b <= 1 / (M-1)
Various G_M(p_1, p_2, ..., p_M) have difficult solutions but we are asked
only to solve G_M(1/M) and it turns out we can accomplish this by
considering only the games
G_M(b) where 1/M <= b <= 1/(M-1) [1]
Games of this form will be said to satisfy constraint [1].
Induction is used for one of the theorems, so we'd better solve G_2 and G_3
for our basis.
G_2(p_1, p_2) = max (p_1, p_2)
G_3(p_1, p_2, p_3) = max (p_1 + p_2, p_3)
since after Monty opens box #1, box #2 will have probability (p_1 + p_2)
and vice versa. When the probabilities satisfy constraint [1]:
G_2(b) = G_2(1-b, b) = b
G_3(b) = G_3(1-2b, b, b) = 1 - b
The proof of Theorem 1 is based on the probability p_i that box #i
contains the dollar. (Of course this is Waldo's perceived probability:
Monty's probability would be 0 or 1.) It may seem wrong for Waldo to
"forget" the game history and remember only the computed p_i. For
example he may have previously locked some but not all of the boxes
and this fact is ignored except in the calculation of p_i. Or Monty may
have some higher level "plan" which mightn't seem to translate directly
into probabilities. But probability algebra obeys some simplifying
linearity rules and, if Monty keeps a "poker face", the probability model
is the only thing Waldo has to act on.
Especially paradoxical is the derivation of Waldo's p_i in his trivial
strategy below: he can adopt inferior but "correct" p_i to simplify the
analysis.
Theorem 1)
If b >= 1/M then
G_M(b) = G_[M-1]( (1-b) / (M-2) )
Proof)
We will show that Monty and Waldo each have a strategy in G_M(b) to
reduce the game to G_[M-1](b, q, q, ..., q) where q = (1-b) / (M-2)
and where the boxes have been renumbered so that box #1 was box #M
(the one Waldo locked) from the prior round and the new box #(M-1)
is the one Waldo locks next. Note that if Monty indeed arranges
the probability mixture G_[M-1](b, q, q, q, q, ..., q) it won't
matter which box Waldo locks (Box #1 has the only non-equal
probability but Waldo cannot lock the same box twice in a row);
this is a typical property of "saddlepoint" strategy.
We will derive the necessary and sufficient condition for Monty to
reduce any game G_M(p_1, p_2, p_3, ..., p_M) to a single game with
the form G_[M-1](b, q, q, ..., q). Using the numbering of G_M()
let R(i,j) be the joint probability that Box #i contains the dollar
and Box #j is opened by Monty in G_M(). We need consider only
M >= 3
Clearly,
R(i, j) >= 0
R(i, i) = 0
R(i, M) = 0, i < M
sum_over_j R(i,j) = p_i
and to achieve q_2 = q_3 = ... = q_[M-1] in G_[M-1],
R(1, j) = R(k, j)
for 1 < j,k < M and j != k
R(2, 1) = R(k, 1)
for 2 < k < M
and to make G_[M-1] be independent of Monty's play
R(M, j)/R(1, j) = R(M, 2)/R(1, 2)
for 2 < j < M
R(M, 2)/R(1, 2) = R(M, 1)/R(2, 1)
The above have a simple unique solution:
R(i, j) = (1 - p_M) / (M - 2) - p_j [2]
for i,j < M and i != j
R(M, j) = p_M - p_j * p_M / (1 - p_M) [3]
for j < M
p_j * (M-2) + p_M <= 1 [4]
For the theorem we are given that G_M(b) satisfies constraint [1]
1 / M <= b <= 1 / (M - 1)
which implies the weaker inequality
(M - 3) / (M^2 - 3M + 1) <= b <= 1 / (M - 1)
and since for the constraint-[1] compliant G_M()
p_j = b or p_j = (1+b-Mb) for all j
the inequality [4] follows directly.
Hence Monty can transpose G_M(b) into G_[M-1]( (1-b) / (M-2) )
whenever b >= 1/M and M >= 3. (This G_[M-1] also happens to
satisfy constraint [1] as needed for the next theorem.)
It should be easy to argue that this strategy is optimal for Monty,
but we want to derive Waldo's best strategy anyway and if it
guarantees the same value we know we're at the "saddlepoint".
If Waldo knows Monty has a non-optimal strategy he can take
advantage of it, but we will just derive a strategy good enough
to achieve the saddle-point value.
Monty must transform G_M(b) into some
G_[M-1](b, q_2, q_3, ..., q_[M-1])
where Waldo has the choice of locking any of boxes #2 through #(M-1).
If Waldo locks each of the (M-2) available boxes with probability
1/(M-2) it is easily seen that the average probability that he
locks the box with the dollar is (1-b) / (M-2) and the probabilities
q_2, q_3, ..., q_[M-2] will also have the average value (1-b)/(M-2).
If Waldo pretends to "forget" which box he picked before, he can
take q_2 = q_3 = ... = (1-b)/(M-2) thereby constructing the same
game Monty constructed with his saddlepoint strategy!
In the above Waldo in effect "degraded" the accuracy of his
probability estimates with the substitutions
q_2' = (q_2 + q_3 + ... + q_[M-1]) / (M - 2)
q_3' = (q_2 + q_3 + ... + q_[M-1]) / (M - 2)
et cetera
If Waldo "knows" more than this, he can pretend he doesn't!
For example he can ask Monty to secretly shuffle the boxes.
Thus either player can reduce
G_M(b), b >= 1/M
to
G_[M-1]((1-b)/(M-2))
so this must be the saddlepoint.
Q.E.D.
Theorem 2)
If b >= 1/M then
G_M(b) = 1 - 1/2! + 1/3! - ... - (1-b)(-1)^M/(M-2)!
= - sum (-1)^i/i! - (1-b)(-1)^M/(M-2)!
where the sum is over i = 1, 2, 3, ..., M-3
Proof)
The proof is by induction. We know the theorem holds for M = 3
and we will assume it holds for (M-1). Set
c = (1-b) / (M-2)
We noted earlier that b <= 1/(M-1): otherwise p_1 = (1 + b - Mb)
is negative; hence we obtain
c = (1-b)/(M-2) >= (1 - 1/(M-1)) / (M-2)
or simply
c >= 1/(M-1)
Thus the condition of the inductive hypothesis is satisfied and
G_[M-1](c) = 1 - 1/2! + 1/3! - ... + (1-c)(-1)^M/(M-3)!
But from Theorem 1
G_M(b) = G_[M-1](c)
and from the definition of c,
c/(M-3)! = (1-b)/(M-2)!
which establishes the theorem.
Theorem 3)
G_M(1/M) = G_M(1/M, ..., 1/M) = 1 - 1/2! + 1/3! - ... -(-1)^M/M!
Proof)
This follows directly from Theorem 2 and the observation that
(1/M)/(M-2)! = 1/(M-1)! - 1/M!
For large M, G_M(1/M) approaches (1 - 1/e). It will be a little bigger
when M is odd and a little smaller when M is even. I've appended the
numeric values below.
% dc
[[Solution for M =]Plb1+pdsb]sy
65k1sa1sblyx2sc[la1lc/-dsaplclyx*scla1lc/+dsaplclyx*sclzx]dszx
Solution for M =2
0.50000000000000000000000000000000000000000000000000000000000000000
Solution for M =3
0.66666666666666666666666666666666666666666666666666666666666666666
Solution for M =4
0.62500000000000000000000000000000000000000000000000000000000000000
Solution for M =5
0.63333333333333333333333333333333333333333333333333333333333333333
Solution for M =6
0.63194444444444444444444444444444444444444444444444444444444444445
Solution for M =7
0.63214285714285714285714285714285714285714285714285714285714285714
Solution for M =8
0.63211805555555555555555555555555555555555555555555555555555555556
Solution for M =9
0.63212081128747795414462081128747795414462081128747795414462081129
Solution for M =10
0.63212053571428571428571428571428571428571428571428571428571428572
. . .
Solution for M =52
0.63212055882855767840447622983853913255418886896823216549216319831
P. S. There are related unsolved problems:
(a) what about G_M(p_1, p_2, ..., p_M) that do not fit the pattern used
in the above solution?
(b) what if two boxes contain dollars? (first, what should the rules be?)
-- ja...@crc.ricoh.com (James Allen)
==> logic/number.p <==
Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce
any truth from any set of axioms. Two integers (not necessarily unique) are
somehow chosen such that each is within some specified range. Mr. S.
is given the sum of these two integers; Mr. P. is given the product of these
two integers. After receiving these numbers, the two logicians do not
have any communication at all except the following dialogue:
<<1>> Mr. P.: I do not know the two numbers.
<<2>> Mr. S.: I knew that you didn't know the two numbers.
<<3>> Mr. P.: Now I know the two numbers.
<<4>> Mr. S.: Now I know the two numbers.
Given that the above statements are absolutely truthful, what are the two
numbers?
==> logic/number.s <==
The answer depends upon the ranges from which the numbers are chosen.
The unique solution for the ranges [2,62] through [2,500+] is:
SUM PRODUCT X Y
17 52 4 13
The unique solution for the ranges [3,94] through [3,500+] is:
SUM PRODUCT X Y
29 208 13 16
There are no unique solutions for the ranges starting with 1,
and there are no solutions for ranges starting with numbers above 3.
A program to compute the possible pairs is included below.
#include <stdio.h>
/*
BEGINNING OF PROBLEM STATEMENT:
Mr. S. and Mr. P. are both perfect logicians, being able to correctly deduce
any truth from any set of axioms. Two integers (not necessarily unique) are
somehow chosen such that each is within some specified range. Mr. S.
is given the sum of these two integers; Mr. P. is given the product of these
two integers. After receiving these numbers, the two logicians do not
have any communication at all except the following dialogue:
<<1>> Mr. P.: I do not know the two numbers.
<<2>> Mr. S.: I knew that you didn't know the two numbers.
<<3>> Mr. P.: Now I know the two numbers.
<<4>> Mr. S.: Now I know the two numbers.
Given that the above statements are absolutely truthful, what are the two
numbers?
END OF PROBLEM STATEMENT
*/
#define SMALLEST_MIN 1
#define LARGEST_MIN 10
#define SMALLEST_MAX 50
#define LARGEST_MAX 500
long P[(LARGEST_MAX + 1) * (LARGEST_MAX + 1)]; /* products */
long S[(LARGEST_MAX + 1) + (LARGEST_MAX + 1)]; /* sums */
find(long min, long max)
{
long i, j;
/*
* count factorizations in P[]
* all P[n] > 1 satisfy <<1>>.
*/
for(i = 0; i <= max * max; ++i)
P[i] = 0;
for(i = min; i <= max; ++i)
for(j = i; j <= max; ++j)
++P[i * j];
/*
* decompose possible SUMs and check factorizations
* all S[n] == min - 1 satisfy <<2>>.
*/
for(i = min + min; i <= max + max; ++i) {
for(j = i / 2; j >= min; --j)
if(P[j * (i - j)] < 2)
break;
S[i] = j;
}
/*
* decompose SUMs which satisfy <<2>> and see which products
* they produce. All (P[n] / 1000 == 1) satisfy <<3>>.
*/
for(i = min + min; i <= max + max; ++i)
if(S[i] == min - 1)
for(j = i / 2; j >= min; --j)
if(P[j * (i - j)] > 1)
P[j * (i - j)] += 1000;
/*
* decompose SUMs which satisfy <<2>> again and see which products
* satisfy <<3>>. Any (S[n] == 999 + min) satisfies <<4>>
*/
for(i = min + min; i <= max + max; ++i)
if(S[i] == min - 1)
for(j = i / 2; j >= min; --j)
if(P[j * (i - j)] / 1000 == 1)
S[i] += 1000;
/*
* find the answer(s) and print them
*/
printf("[%d,%d]\n",min,max);
for(i = min + min; i <= max + max; ++i)
if(S[i] == 999 + min)
for(j = i / 2; j >= min; --j)
if(P[j * (i - j)] / 1000 == 1)
printf("{ %d %d }: S = %d, P = %d\n",
i - j, j, i, (i - j) * j);
}
main()
{
long min, max;
for (min = SMALLEST_MIN; min <= LARGEST_MIN; min ++)
for (max = SMALLEST_MAX; max <= LARGEST_MAX; max++)
find(min,max);
}
-------------------------------------------------------------------------
= Jeff Kenton (617) 894-4508 =
= jke...@world.std.com =
-------------------------------------------------------------------------
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> logic/situation.puzzles.p <==
Jed's List of Situation Puzzles
"A man lies dead in a room with fifty-three bicycles in front of him.
What happened?"
This is a list of what I refer to (for lack of a better name) as situation
puzzles. In the game of situation puzzles, a situation like the one above is
presented to a group of players, who must then try to find out more about the
situation by asking further questions. The person who initially presented
the situation can only answer "yes" or "no" to questions (or occasionally
"irrelevant" or "doesn't matter").
My list has been divided into two sections. Section 1 consists of
situation puzzles which are set in a realistic world; the situations could
all actually occur. Section 2 consists of puzzles which involve double
meanings for one or more words and those which could not possibly take place
in reality as we know it, plus a few miscellaneous others.
See the end of the list for more notes and comments.
The answers to these puzzles are available in a separate file.
Section 1: "Realistic" situation puzzles.
1.1. In the middle of the ocean is a yacht. Several corpses are floating in
the water nearby. (SJ)
1.2. A man is lying dead in a room. There is a large pile of gold and
jewels on the floor, a chandelier attached to the ceiling, and a large open
window. (DVS; partial JM wording)
1.3. A woman came home with a bag of groceries, got the mail, and walked
into the house. On the way to the kitchen, she went through the living room
and looked at her husband, who had blown his brains out. She then continued
to the kitchen, put away the groceries, and made dinner. (partial JM
wording)
1.4. A body is discovered in a park in Chicago in the middle of summer. It
has a fractured skull and many other broken bones, but the cause of death was
hypothermia. (MI, from _Hill Street Blues_)
1.5. A man lives on the twelfth floor of an apartment building. Every
morning he takes the elevator down to the lobby and leaves the building. In
the evening, he gets into the elevator, and, if there is someone else in the
elevator -- or if it was raining that day -- he goes back to his floor
directly. However, if there is nobody else in the elevator and it hasn't
rained, he goes to the 10th floor and walks up two flights of stairs to his
room. (MH)
1.6. A woman has incontrovertible proof in court that her husband was
murdered by her sister. The judge declares, "This is the strangest case I've
ever seen. Though it's a cut-and-dried case, this woman cannot be punished."
(This is different from #1.43.) (MH)
1.7. A man walks into a bar and asks for a drink. The bartender pulls out a
gun and points it at him. The man says, "Thank you," and walks out. (DVS)
1.8. A man is returning from Switzerland by train. If he had been in a
non-smoking car he would have died. (DVS; MC wording)
1.9. A man goes into a restaurant, orders abalone, eats one bite, and kills
himself. (TM and JM wording)
1.10. A man is found hanging in a locked room with a puddle of water under
his feet. (This is different from #1.11.)
1.11. A man is dead in a puddle of blood and water on the floor of a locked
room. (This is different from #1.10.)
1.12. A man is lying, dead, face down in the desert wearing a backpack.
(This is different from #1.13, #2.11, and #2.12.)
1.13. A man is lying face down, dead, in the desert, with a match near his
outstretched hand. (This is different from #1.12, #2.11, and #2.12.) (JH;
partial JM wording)
1.14. A man is driving his car. He turns on the radio, listens for five
minutes, turns around, goes home, and shoots his wife. (This is different
from #1.15.)
1.15. A man driving his car turns on the radio. He then pulls over to the
side of the road and shoots himself. (This is different from #1.14.)
1.16. Music stops and a woman dies. (DVS)
1.17. A man is dead in a room with a small pile of pieces of wood and
sawdust in one corner. (from "Coroner's Inquest," by Marc Connelly)
1.18. A flash of light, a man dies. (ST original)
1.19. A rope breaks. A bell rings. A man dies. (KH)
1.20. A woman buys a new pair of shoes, goes to work, and dies. (DM)
1.21. A man is riding a subway. He meets a one-armed man, who pulls out a
gun and shoots him. (SJ)
1.22. Two women are talking. One goes into the bathroom, comes out five
minutes later, and kills the other.
1.23. A man is sitting in bed. He makes a phone call, saying nothing, and
then goes to sleep. (SJ)
1.24. A man kills his wife, then goes inside his house and kills himself.
(DH original, from "Nightmare in Yellow," by Fredric Brown)
1.25. Abel walks out of the ocean. Cain asks him who he is, and Abel
answers. Cain kills Abel. (MWD original)
1.26. Two men enter a bar. They both order identical drinks. One lives;
the other dies. (CR; partial JM wording)
1.27. Joe leaves his house, wearing a mask and carrying an empty sack. An
hour later he returns. The sack is now full. He goes into a room and turns
out the lights. (AL)
1.28. A man takes a two-week cruise to Mexico from the U.S. Shortly after
he gets back, he takes a three-day cruise which doesn't stop at any other
ports. He stays in his cabin all the time on both cruises. As a result, he
makes $250,000. (MI, from "The Wager")
1.29. Hans and Fritz are German spies during World War II. They try to
enter America, posing as returning tourists. Hans is immediately arrested.
(JM)
1.30. Tim and Greg were talking. Tim said "The terror of flight." Greg
said "The gloom of the grave." Greg was arrested. (MPW original, from "No
Refuge Could Save," by Isaac Asimov)
1.31. A man is found dead in his parked car. Tire tracks lead up to the car
and away. (SD)
1.32. A man dies in his own home. (ME original)
1.33. A woman in France in 1959 is waiting in her room, with all the doors
locked from the inside, for her husband to come home. When he arrives, the
house has burned to the ground and she's dead. (JM, from _How Come --
Again?_)
1.34. A man gets onto an elevator. When the elevator stops, he knows his
wife is dead. (LA; partial KH wording)
1.35. Three men die. On the pavement are pieces of ice and broken glass.
(JJ)
1.36. She lost her job when she invited them to dinner. (DS original)
1.37. A man is running along a corridor with a piece of paper in his hand.
The lights flicker and the man drops to his knees and cries out, "Oh no!"
(MP)
1.38. A car without a driver moves; a man dies. (EMS)
1.39. As I drive to work on my motorcycle, there is one corner which I go
around at a certain speed whether it's rainy or sunny. If it's cloudy but
not raining, however, I usually go faster. (SW original)
1.40. A woman throws something out a window and dies. (JM)
1.41. An avid birdwatcher sees an unexpected bird. Soon he's dead. (RSB
original)
1.42. There are a carrot, a pile of pebbles, and a pipe lying together in
the middle of a field. (PRO; partial JM wording)
1.43. Two brothers are involved in a murder. Though it's clear that one of
them actually committed the crime, neither can be punished. (This is
different from #1.6.) (from "Unreasonable Doubt," by Stanley Ellin)
1.44. An ordinary American citizen, with no passport, visits over thirty
foreign countries in one day. He is welcomed in each country, and leaves
each one of his own accord. (PRO)
1.45. If he'd turned on the light, he'd have lived. (JM)
1.46. A man is found dead on the floor in the living room. (ME original)
1.47. A man is found dead outside a large building with a hole in him. (JM,
modified from PRO)
1.48. A man is found dead in an alley lying in a red pool with two sticks
crossed near his head. (PRO)
1.49. A man lies dead next to a feather. (PRO)
1.50. There is blood on the ceiling of my bedroom. (MI original)
1.51. A man wakes up one night to get some water. He turns off the light
and goes back to bed. The next morning he looks out the window, screams, and
kills himself. (CR; KK wording)
1.52. She grabbed his ring, pulled on it, and dropped it. (JM, from _Math
for Girls_)
1.53. A man sitting on a park bench reads a newspaper article headlined
"Death at Sea" and knows a murder has been committed.
1.54. A man tries the new cologne his wife gave him for his birthday. He
goes out to get some food, and is killed. (RW original)
1.55. A man in uniform stands on the beach of a tropical island. He takes
out a cigarette, lights it, and begins smoking. He takes out a letter and
begins reading it. The cigarette burns down between his fingers, but he
doesn't throw it away. He cries. (RW)
1.56. A man went into a restaurant, had a large meal, and paid nothing for
it. (JM original)
1.57. A married couple goes to a movie. During the movie the husband
strangles the wife. He is able to get her body home without attracting
attention. (from _Beyond the Easy Answer_)
1.58. A man ran into a fire, and lived. A man stayed where there was no
fire, and died. (Eric Wang original)
1.59. A writer with an audience of millions insisted that he was never to be
interrupted while writing. After the day when he actually was interrupted,
he never wrote again. (JM, from _How Come?_)
1.60. Beulah died in the Appalachians, while Craig died at sea. Everyone
was much happier with Craig's death. (JM, from _How Come?_)
1.61. Mr. Browning is glad the car ran out of gas. (JM, from _Home Come?_)
1.62. A man is sitting suspended over two pressurized containers. Suddenly,
he dies. (NK original)
1.63. A man leaves a motel room, goes to his car, and honks the horn. (AS
original)
1.64. Two dead people sit in their cars on a street. (AG)
1.65. A woman lies dead in the street near a car. (AG)
1.66. A riverboat filled with passengers suddenly capsized, drowning most of
those aboard. (from _How Come -- Again?_)
Section 2: Double meanings, fictional settings, and miscellaneous others.
2.1. A man shoots himself, and dies. (HL) (This is different from #2.2.)
2.2. A man walks into a room, shoots, and kills himself. (HL) (This is
different from #2.1.)
2.3. Adults are holding children, waiting their turn. The children are
handed (one at a time, usually) to a man, who holds them while a woman shoots
them. If the child is crying, the man tries to stop the crying before the
child is shot. (ML)
2.4. Hiking in the mountains, you walk past a large field and camp a few
miles farther on, at a stream. It snows in the night, and the next day you
find a cabin in the field with two dead bodies inside. (KL; KD and partial
JM wording)
2.5. A man marries twenty women in his village but isn't charged with
polygamy.
2.6. A man is alone on an island with no food and no water, yet he does not
fear for his life. (MN)
2.7. Joe wants to go home, but he can't go home because the man in the mask
is waiting for him. (AL wording)
2.8. A man is doing his job when his suit tears. Fifteen minutes later,
he's dead. (RM)
2.9. A dead man lies near a pile of bricks and a beetle on top of a book.
(MN)
2.10. At the bottom of the sea there lies a ship worth millions of dollars
that will never be recovered. (TF original)
2.11. A man is found dead in the arctic with a pack on his back. (This is
different from #1.12, #1.13, and #2.12.) (PRO)
2.12. There is a dead man lying in the desert next to a rock. (This is
different from #1.12, #1.13, and #2.11.) (GH)
2.13. As a man jumps out of a window, he hears the telephone ring and
regrets having jumped. (from "Some Days are Like That," by Bruce J.
Balfour; partial JM wording)
2.14. Two people are playing cards. One looks around and realizes he's
going to die. (JM original)
2.15. A man lies dead in a room with fifty-three bicycles in front of him.
2.16. A horse jumps over a tower and lands on a man, who disappears. (ES
original)
2.17. A train pulls into a station, but none of the waiting passengers move.
(MN)
2.18. A man pushes a car up to a hotel and tells the owner he's bankrupt.
(DVS; partial AL and JM wording)
2.19. Three large people try to crowd under one small umbrella, but nobody
gets wet. (CC)
2.20. A black man dressed all in black, wearing a black mask, stands at a
crossroads in a totally black-painted town. All of the streetlights in town
are broken. There is no moon. A black-painted car without headlights drives
straight toward him, but turns in time and doesn't hit him. (AL and RM
wording)
2.21. Bob and Carol and Ted and Alice all live in the same house. Bob and
Carol go out to a movie, and when they return, Alice is lying dead on the
floor in a puddle of water and glass. It is obvious that Ted killed her but
Ted is not prosecuted or severely punished.
2.22. A man rides into town on Friday. He stays one night and leaves on
Friday. (KK)
2.23. Bruce wins the race, but he gets no trophy. (EMS)
2.24. A woman opens an envelope and dyes. (AL)
2.25. A man was brought before a tribal chief, who asked him a question. If
he had known the answer, he probably would have died. He didn't, and lived.
(MWD original)
2.26. Two men are found dead outside of an igloo. (SK original)
2.27. A man is born in 1972 and dies in 1952 at the age of 25. (DM)
Attributions key:
When I know who first told me the current version of a puzzle, I've put
initials in parentheses after the puzzle statement; this is the key to those
acknowledgments. The word "original" following an attribution means that, to
the best of my knowledge, the cited person invented that puzzle. If a given
puzzle isn't marked "original" but is attributed, that just means that's the
first person I heard it from. I would appreciate it if attributions for
originals were not removed; however, this list is hereby entered into the
public domain, so do with it what you wish.
LA == Laura Almasy RSB == Ranjit S. Bhatnagar
CC == Chris Cole MC == Matt Crawford
MWD == Matthew William Daly KD == Ken Duisenberg
SD == Sylvia Dutcher ME == Marguerite Eisenstein
TF == Thomas Freeman AG == Andreas Gammel
JH == Joaquin Hartman MH == Marcy Hartman
KH == Karl Heuer GH == Geoff Hopcraft
DH == David Huddleston MI == Mark Isaak
SJ == Steve Jacquot JJ == J|rgen Jensen
KK == Karen Karp NK == Nev King
SK == Shelby Kilmer KL == Ken Largman
AL == Andy Latto HL == Howard Lazoff
ML == Merlyn LeRoy DM == Dan Murray
RM == "Reaper Man" (real name unknown)
TM == Ted McCabe JM == Jim Moskowitz
DM == Damian Mulvena MN == Jan Mark Noworolski
PRO == Peter R. Olpe (from his list)
MP == Martin Pitwood CR == Charles Renert
EMS == Ellen M. Sentovich (from her list)
AS == Annie Senghas ES == Eric Stephan
DS == Diana Stiefbold ST == Simon Travaglia
DVS == David Van Stone RW == Randy Whitaker
MPW == Matthew P Wiener SW == Steve Wilson (not sure of name)
Special thanks to Jim Moskowitz, Karl Heuer, and Mark Brader, for a lot of
discussion of small but important details and wording.
Notes and comments:
My outtakes list (items removed from this list for various reasons, most
of which came down to the fact that I didn't like them) is now available from
the rec.puzzles archive server.
There are many possible wordings for most of the puzzles in this list.
Most of them have what I consider the best wording of the variants I've
heard; if you think there's a better way of putting one or more of them, or
if you don't like my categorization of any of them, or if you have any other
comments or suggestions, please drop me a note. If you know others not on
this list, please send them to me.
Of course, in telling a group of players one of these situations, you can
add or remove details, either to make getting the answer harder or easier, or
simply to throw in red herrings. I've made a few specific suggestions along
these lines in the answer list, available in a separate file. Also in the
answer list are variant problem statements and variant answers.
Bibliography:
The game of situation puzzles is also known by a variety of other names:
mystery questions, story riddles, lateral thinking puzzles, mini-mysteries,
minute mysteries, missing links, how come?, situational puzzles, law school
puzzles, quistels (in the Netherlands and other parts of Europe), mystery
puzzles, and so on. I prefer the term 'situation puzzles,' but I change my
mind every few years when a new term that I like more comes along. At any
rate, here are some sources for these puzzles, under a variety of names.
Unfortunately, almost all of these books are out of print and extremely
difficult to find. Try inter-library loan, and be prepared to wait. I don't
know of any such books outside of the US (though at least the Sloane book is
also printed in Canada, Europe, and Australia), but I'd be happy to include
references to such in future editions if anyone sends me bibliographical
info.
On this edition of my list, I have included a few puzzles from these books
which I didn't previously have. I've paraphrased them and cited the sources,
which I hope should be good enough to avoid copyright infringement; however,
I hope to contact the various copyright holders soon and get explicit
permission to include more of their puzzles. If I fail to get that
permission, a few of the items on this list may go away in the next edition.
_Games_ magazine (bibliographical data currently unavailable). They ran a
situation-puzzle contest recently, but I have yet to see any of the results.
_Math for Girls_ (bibliographical data unavailable).
Rogers, Agnes, _How Come?_ (1953: Doubleday & Company, Inc., New York).
Library of Congress catalog number 53-5756. OCLC #1612919. The author may
also be listed as Agnes Rogers Allen. With its sequel (see below), the
classic volume on the subject; is probably the original source for quite a
few standard situation puzzles, though Rogers says she does not know who
invented the form. Nor does she know the source of most of those she
includes -- like all good folklore, situation puzzles are difficult to trace
to their origins. Unfortunately, both these books are long out of print.
Besides their historical value, these two come furnished with delightful
illustrations of various wrong approaches to some of the puzzles. These
versions were definitely intended to be read from the book, though; the
puzzle statements are much more long-winded than the versions in my list.
Rogers, Agnes, and Sheehan, Richard G., _How Come -- Again?_ (1960:
Doubleday & Company, Inc., New York). Library of Congress catalog number
60-13745. OCLC #2580602.
Sloane, Paul, _Lateral Thinking Puzzlers_ (1992: Sterling Publishing Co.,
Inc., 387 Park Avenue South, New York, 10016). ISBN 0-8069-8227-6. There's
a lot of overlap here with the rec.puzzles archives, including a lot of
puzzles that I wouldn't even consider doing as situation puzzles (such as the
infamous "12 balls" problem). Still, it does have one or two nice situation
puzzles in it. Warning: these are not lateral thinking puzzles in the sense
in which I like to use that phrase -- each puzzle has a definite correct
answer, and creativity and sideways leaps of logic aren't rewarded unless
they result in that answer. Cover price $US 4.95; should be available (or
orderable) in most chain bookstores in the US.
_Stories With Holes_ (bibliographical data unavailable).
Weintraub, Richard, and Krieger, Richard, _Beyond the Easy Answer:
exploring new perspectives through creative problem-solving games_ (1979:
Zenger Publications, Inc., Gateway Station 802, Culver City, CA 90230). ISBN
0-934508-00-3. Contains a variety of puzzles and games, most of which aren't
really situation puzzles (and many of which are in the rec.puzzles archives),
plus some creativity games. Out of print.
History of List:
original compilation 11/28/87
major revision 08/09/89
further additions 08/23/89 - 10/21/90
variants added to answer list 07/04/90
editing and renumbering 07/25/90 - 11/11/90
items removed; title changed 09/20/90 - 11/11/90
editing and additions 02/26/92 - 09/17/92
more additions (incl. biblio.) 03/31/93 - 05/03/93
--Jed Hartman
lo...@random.esd.sgi.com (as of 5/93)
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> competition/tests/math/putnam/putnam.1992.p <==
Problem A1
Prove that f(n) = 1 - n is the only integer-valued function defined on
the integers that satisfies the following conditions.
(i) f(f(n)) = n, for all integers n;
(ii) f(f(n + 2) + 2) = n for all integers n;
(iii) f(0) = 1.
Problem A2
Define C(alpha) to be the coefficient of x^1992 in the power series
expansion about x = 0 of (1 + x)^alpha. Evaluate
\int_0^1 C(-y-1) ( 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/(y+1992) ) dy.
Problem A3
For a given positive integer m, find all triples (n,x,y) of positive
integers, with n relatively prime to m, which satisfy (x^2 + y^2)^m = (xy)^n.
Problem A4
Let f be an infinitely differentiable real-valued function defined on
the real numbers. If
f(1/n) = n^2/(n^2 + 1), n = 1,2,3,...,
compute the values of the derivatives f^(k)(0), k = 1,2,3,... .
Problem A5
For each positive integer n, let
a_n = { 0 if the number of 1's in the binary representation of n is even,
1 if the number of 1's in the binary representation of n is odd.
Show that there do not exist positive integers k and m such that
a_{k+j} = a_{k+m+j} = a_{k+2m+j}, for 0 <= j <= m - 1.
Problem A6
Four points are chosen at random on the surface of a sphere. What is the
probability that the center of the sphere lies inside the tetrahedron
whose vertices are at the four points? (It is understood that each point
is independently chosen relative to a uniform distribution on the sphere.)
Problem B1
Let S be a set of n distinct real numbers. Let A_S be the set of numbers
that occur as averages of two distinct elements of S. For a given n >= 2,
what is the smallest possible number of distinct elements in A_S?
Problem B2
For nonnegative integers n and k, define Q(n,k) to be the coefficient of
x^k in the expansion of (1+x+x^2+x^3)^n. Prove that
Q(n,k) = \sum_{j=0}^n {n \choose j} {n \choose k - 2j},
where {a \choose b} is the standard binomial coefficient. (Reminder: For
integers a and b with a >= 0, {a \choose b} = a!/b!(a-b)! for 0 <= b <= a,
and {a \choose b} = 0 otherwise.)
Problem B3
For any pair (x,y) of real numbers, a sequence (a_n(x,y))_{n>=0} is
defined as follows:
a_0(x,y) = x
a_{n+1}(x,y) = ( (a_n(x,y))^2 + y^2 ) / 2, for all n >= 0.
Find the area of the region { (x,y) | (a_n(x,y))_{n>=0} converges }.
Problem B4
Let p(x) be a nonzero polynomial of degree less than 1992 having no
nonconstant factor in common with x^3 - x. Let
( d^1992 / dx^1992 ) ( p(x) / x^3 - x ) = f(x) / g(x)
for polynomials f(x) and g(x). Find the smallest possible degree of f(x).
Problem B5
Let D_n denote the value of the (n-1) by (n-1) determinant
| 3 1 1 1 ... 1 |
| 1 4 1 1 ... 1 |
| 1 1 5 1 ... 1 |
| 1 1 1 6 ... 1 |
| . . . . ... . |
| 1 1 1 1 ... n+1 |
Is the set {D_n/n!}_{n >= 2} bounded?
Problem B6
Let M be a set of real n by n matrices such that
(i) I \in M, where I is the n by n identity matrix;
(ii) if A \in M and B \in M, then either AB \in M or -AB \in M, but not both;
(iii) if A \in M and B \in M, then either AB = BA or AB = -BA;
(iv) if A \in M and A \noteq I, there is at least one B \in M such that
AB = -BA.
Prove that M contains at most n^2 matrices.
==> competition/tests/math/putnam/putnam.1992.s <==
Now the unofficial solutions. Thanks to Noam Elkies for the B6 solution;
thanks also to Peter Akemann, Greg John, and Peter Montgomery.
The Putnam exam deserves criticism this year for an exceptional number
of typos and poorly worded problems. How can someone taking the Putnam
be sure that his solutions will be graded carefully, if the exam itself
shows sloppy typography, grammar, and style?
Problem A1
Prove that f(n) = 1 - n is the only integer-valued function defined on
the integers that satisfies the following conditions.
(i) f(f(n)) = n, for all integers n;
(ii) f(f(n + 2) + 2) = n for all integers n;
(iii) f(0) = 1.
(The comma in (i) is wrong. Either ``conditions.'' should be
``conditions:'' or the semicolons should be periods. Little things...)
Solution: Certainly if f(n) = 1 - n then (i), (ii), and (iii) hold.
Conversely, say (i), (ii), and (iii) hold. We show that f(k) = 1 - k
and f(1 - k) = k for every k >= 0. For k = 0 and 1 we have f(0) = 1 and
f(1) = f(f(0)) = 0. For k >= 2, by induction we have f(k - 2) = 3 - k
and f(3 - k) = k - 2. Note that f(n + 2) + 2 = f(f(f(n + 2) + 2)) = f(n).
So f(k) = f(k - 2) - 2 = 1 - k and f(1 - k) = f(3 - k) + 2 = k as desired.
As k and 1 - k for k >= 1 cover the integers, we are done.
Problem A2
Define C(alpha) to be the coefficient of x^1992 in the power series
expansion about x = 0 of (1 + x)^alpha. Evaluate
\int_0^1 C(-y-1) ( 1/(y+1) + 1/(y+2) + 1/(y+3) + ... + 1/(y+1992) ) dy.
Solution: C(alpha) is given by the usual binomial coefficient formula
{alpha \choose 1992} = \alpha (\alpha - 1) ... (\alpha - 1991) / 1992!.
Hence C(-y-1) = (-y-1)(-y-2)...(-y-1992)/1992! = (y+1)...(y+1992)/1992!.
Set f(y) = (y+1)(y+2)...(y+1992). Now f has logarithmic derivative
f'/f = ( 1/(y+1) + 1/(y+2) + ... + 1/(y+1992) ). Hence our integral
equals \int_0^1 f(y)/1992! f'(y)/f(y) dy = \int_0^1 f'(y) dy/1992! =
(f(1) - f(0))/1992! = (1993! - 1992!)/1992! = 1993 - 1 = 1992.
Problem A3
For a given positive integer m, find all triples (n,x,y) of positive
integers, with n relatively prime to m, which satisfy (x^2 + y^2)^m = (xy)^n.
Solution: Certainly xy < x^2 + y^2 so m < n. Put n = m + k, k > 0.
Let d be the greatest common divisor of x and y. Say x = ds, y = dt.
Now d^{2m}(s^2 + t^2)^m = d^{2n}(st)^n, so (s^2 + t^2)^m = d^{2k}(st)^n.
If a prime p divides s then p divides d^{2k}(st)^n = (s^2 + t^2)^m,
so p must divide t. But s and t are relatively prime. Hence s = 1.
Similarly t = 1. So d^{2k} = 2^m. Hence d is a power of 2, say d = 2^j,
and m = 2jk. As n = m + k = 2jk + k we see that k is a common factor of
m and n. Thus k = 1. So m = 2j, n = 2j + 1, x = y = d = 2^j. If m is odd
then there are no solutions; if m is even then the only possible solution
is (m + 1,2^{m/2},2^{m/2}). This does satisfy the given conditions.
Problem A4
Let f be an infinitely differentiable real-valued function defined on
the real numbers. If
f(1/n) = n^2/(n^2 + 1), n = 1,2,3,...,
compute the values of the derivatives f^(k)(0), k = 1,2,3,... .
(``Let f be a foo. If bar, compute blah.'' Does this mean that one need
only answer the question if ``bar'' happens to be true? May one choose
his favorite f, such as f(x) = x, observe that it does not satisfy ``bar'',
and [successfully] answer the question without computing anything?
``If ..., compute'' should be ``Assume ... . Compute''.)
Solution: We first observe that the function g(x) = 1/(1 + x^2) satisfies
all the known properties of f: it is infinitely differentiable, and
g(1/n) = n^2/(n^2 + 1). From the Taylor expansion of g around 0,
g(x) = 1 - x^2 + x^4 - x^6 + ..., we see that g^(k)(0) is 0 for k odd,
(-1)^{k/2}k! for k even.
Can f differ from g in its derivatives at 0? The difference h = f - g
is an infinitely differentiable function with roots at 1/n for every
positive n. We show that any such function has all derivatives 0 at 0.
Suppose not. Take the least k >= 0 for which h^(k)(0) is nonzero.
Without loss of generality say it is positive. By continuity h^(k) is
positive in an open interval U around 0. Let S be the set of positive
elements of U. Now h^(k-1) is increasing on S, and by minimality of k
we have h^(k-1)(0) = 0, so h^(k-1) is positive on S. Then h^(k-2) is
increasing on S, and h^(k-2)(0) = 0, so h^(k-2) is positive on S. In
general h^j is positive on S for j down from k to 0 by induction. In
particular h is positive on S, but this is impossible, as 1/n is in S
for n sufficiently large.
Thus f has all the same derivatives as g: f^(k)(0) is 0 for k odd,
(-1)^{k/2}k! for k even.
Alternative solution:
Let g(x) = 1/(1 + x^2). We may compute the derivatives as before, and again
the problem reduces to showing that all derivatives of f(x)-g(x) = h(x)
vanish at 0.
By continuity, h^(0) vanishes at 0. We now proceed by induction using the
nth mean value theorem, which states that h(x) = h(0) + h'(0) + ... +
h^(n-1)(0)/(n-1)! + h^(n)(\theta)/n!, where 0 <= \theta <= x. By induction,
the derivatives up to the (n-1)-th vanish at 0, and if we take x = 1, 1/2,
... we get h^(n)(\theta_n) = 0, where 0 <= \theta_n <= 1/n. Hence \{\theta_n\}
tends to 0. But h is infinitely differentiable, so h^n is infinitely
differentiable and hence continuous. It follows that h^(n)(0) = 0.
Yet another solution:
Consider only n divided by l.c.m/(1,2,\ldots,k).
f^(k)(0) = \lim_{n\rightarrow\infty}
( \sum_{i=0}^k {k\choose i}(-1)^{i+1} f(i/n) ) / (1/n)^k
= \lim_{n\rightarrow\infty}
( \sum_{i=0}^k {k\choose i}(-1)^{i+1} g(i/n) ) / (1/n)^k
=g^{k}(0)= \cos(\pi k/2) k!
Or replace n by n*l.c.m.(1,2,\ldots,k) in the above and allow n to be any
positive integer.
Problem A5
For each positive integer n, let
a_n = { 0 if the number of 1's in the binary representation of n is even,
1 if the number of 1's in the binary representation of n is odd.
Show that there do not exist positive integers k and m such that
a_{k+j} = a_{k+m+j} = a_{k+2m+j}, for 0 <= j <= m - 1.
(Is every student supposed to know what the writer means by ``the binary
representation of n''? Anyway, this problem is well known in some circles.
I don't think Putnam writers should copy problems.)
Solution: Let us suppose the opposite. Pick m minimal such that, for
some k, a_{k+j} = a_{k+m+j} for every 0 <= j <= 2m - 1. The minimality
guarantees that m is odd. (Indeed, say m = 2m'. If k = 2k' is even,
then put j = 2j' for 0 <= j' <= m - 1 = 2m' - 1. Now a_n = a_{2n} so
a_{k'+j'} = a_{k+j} = a_{k+m+j} = a_{k'+m'+j'} as desired. If on the
other hand k = 2k' - 1 is odd, then put j = 2j' + 1 for 0 <= j' <= 2m' - 1.
Now a_{k'+j'} = a_{k+j} = a_{k+m+j} = a_{k'+m'+j'} as desired.)
We cannot have m = 1. Indeed, if a_k = a_{k+1} = a_{k+2}, then we have
a_{2n} = a_{2n+1} for n = k/2 if k is even or n = (k+1)/2 if k is odd.
But a_{2n} + a_{2n+1} = 1 always. Hence m is an odd number greater than 1.
Define b_n = (a_n + a_{n-1}) mod 2. For 1 <= j <= 2m - 1 we have
b_{k+j} = b_{k+m+j}. This range of j covers at least 5 values; we can
certainly choose j so as to make k+j equal to 2 mod 4. Then k+m+j is
odd. But b_n is 0 when n is 2 mod 4, and b_n is 1 when n is odd, so we
have a contradiction.
Problem A6
Four points are chosen at random on the surface of a sphere. What is the
probability that the center of the sphere lies inside the tetrahedron
whose vertices are at the four points? (It is understood that each point
is independently chosen relative to a uniform distribution on the sphere.)
Solution: Pick three points A, B, C, and consider the spherical triangle
T defined by those points. The center of the sphere lies in the convex
hull of A, B, C, and another point P if and only if it lies in the convex
hull of T and P. This happens if and only if P is antipodal to T. So the
desired probability is the expected fraction of the sphere's surface area
which is covered by T.
Denote the antipode to a point P by P'. We consider the eight spherical
triangles ABC, A'BC, AB'C, A'B'C, ABC', A'BC', AB'C', A'B'C'. Denote
these by T_0, T_1, T_2, T_3, T_4, T_5, T_6, T_7; we regard each T_i
as a function of the random variables A, B, C. There is an automorphism
of our probability space defined by (A,B,C) -> (A,B,C'). Hence T_0 and
T_4 have the same distribution, and similarly T_1 and T_5, T_2 and T_6,
and T_3 and T_7. Of course the same applies to B, so that T_0 and T_2,
T_1 and T_3, T_4 and T_6, and T_5 and T_7 all have the same distribution.
Finally T_0 and T_1, T_2 and T_3, T_4 and T_5, and T_6 and T_7 all have
the same distribution. We conclude that all the T_i have exactly the
same distribution. In particular the fractional area A_i of T_i has the
same distribution for all i.
On the other hand the total fractional area of all the T_i is exactly
1: the eight triangles cover the sphere exactly once. Hence each T_i
has expected fractional area 1/8. In particular, T_0, the probability we
wanted, has expected value 1/8.
Note that this proof does not require the full strength of uniform
distribution in the usual measure; nor does it require independence
between all the variables. It requires only certain automorphisms of
the probability space.
Problem B1
Let S be a set of n distinct real numbers. Let A_S be the set of numbers
that occur as averages of two distinct elements of S. For a given n >= 2,
what is the smallest possible number of distinct elements in A_S?
(``Smallest possible number of distinct elements in A_S''? Who talks
about ``the number of _distinct_ elements'' of a set? How about just
``the number of elements''? Or ``the size''? Aargh. The quantifiers
here are completely out of whack: n has to be fixed [not ``given'']
before anything else, and it's not immediately clear that ``smallest
possible'' refers to ``the minimum over all S''. Proposed rewrite:
``Fix n >= 2. For any set S of n real numbers, let A_S be the set of
averages of two distinct elements of S. What is the minimum, over all
S, of the size of A_S?'')
Solution: We put the elements of S in order: s_1 < s_2 < ... < s_n.
We have s_1 + s_2 < s_1 + s_3 < ... < s_1 + s_{n-1} < s_1 + s_n <
s_2 + s_n < s_3 + s_n < ... < s_{n-1} + s_n. Hence the 2n - 3 averages
(s_i + s_j)/2, i < j, with i = 1 or j = n, are all distinct. So A_S
has size at least 2n - 3. This is achieved if, for instance,
S = {1,2,...,n}.
Problem B2
For nonnegative integers n and k, define Q(n,k) to be the coefficient of
x^k in the expansion of (1+x+x^2+x^3)^n. Prove that
Q(n,k) = \sum_{j=0}^n {n \choose j} {n \choose k - 2j},
where {a \choose b} is the standard binomial coefficient. (Reminder: For
integers a and b with a >= 0, {a \choose b} = a!/b!(a-b)! for 0 <= b <= a,
and {a \choose b} = 0 otherwise.)
Solution: (1+x^2)(1+x) = 1+x+x^2+x^3, so (1+x^2)^n(1+x)^n = (1+x+x^2+x^3)^n,
so (\sum {n\choose j} x^{2j}) (\sum {n\choose m} x^m) = \sum Q(n,k)x^k.
The coefficient of x^k on the left is the sum of {n\choose j}{n\choose m}
over all j,m with m + 2j = k, i.e., \sum_j {n\choose j}{n\choose k-2j}.
Problem B3
For any pair (x,y) of real numbers, a sequence (a_n(x,y))_{n>=0} is
defined as follows:
a_0(x,y) = x
a_{n+1}(x,y) = ( (a_n(x,y))^2 + y^2 ) / 2, for all n >= 0.
Find the area of the region { (x,y) | (a_n(x,y))_{n>=0} converges }.
(The parentheses in (a_n(x,y))^2 are confusing, as the writer also
uses parentheses to denote the entire sequence of a_n.)
Solution: Note that (x,y) and (x,-y) produce the same sequence, and
(-x,y) and (x,y) produce the same sequence after the first step. So
we will restrict attention to nonnegative x and y and quadruple our
final answer.
Fix x and y. Set f(z) = ( z^2 + y^2 ) / 2, so that a_n(x,y) =
f(a_{n-1}(x,y)). Now f'(z) = z, so f is increasing on the positive reals.
So (a_n(x,y))_n is monotone---either increasing, decreasing, or constant.
We consider several (non-exclusive) possibilities.
Case 1. Say y > 1. Then f(z) > (1 + z^2)/2 + (y - 1) >= z + (y - 1), so
a_n(x,y) increases by at least y - 1 at each step.
Case 2. Say f(x) < x. Then we have 0 < a_n(x,y) < a_{n-1}(x,y) <= x for
every n. (Indeed, for n = 1 we have 0 < f(x) < x. For n >= 2 we have
a_{n-1}(x,y) < a_{n-2}(x,y) by induction. So a_n(x,y) < a_{n-1}(x,y),
as f is increasing.) As (a_n(x,y))_n is decreasing and bounded below,
it converges.
Case 3. Say f(x) > x > 1. Define g(z) = f(z) - z, so that g(x) > 0.
We have g'(z) = z - 1, so g is increasing past 1. Now a_n(x,y) >=
x + ng(x). (Indeed, for n = 1 we have a_1(x,y) = f(x) = x + g(x).
For n >= 2 set a = a_{n-1}(x,y). We have a >= x + (n-1)g(x) > x by
induction. So g(a) > g(x), and a_n(x,y) = f(a) = a + g(a) > a + g(x) >=
x + ng(x) as desired.) So a_n increases without bound.
Case 4. Say x < 1, y < 1. Then f(x) < f(1) < (1 + 1)/2 = 1. Similarly
a_n(x,y) < 1 for every n. As (a_n(x,y))_n is bounded and monotone, it
converges.
Let's put this all together. For y > 1 the sequence diverges. For y < 1
and x < 1 the sequence does converge. For y < 1 and x > 1, the sequence
converges if f(x) < x, and diverges if f(x) > x. The points we miss in
this tally---y = 1, x = 1, f(x) = x---have zero total area.
The condition f(x) < x is equivalent to (x-1)^2 + y^2 < 1, which
describes a quarter-circle of radius 1 in the region y > 0, x > 1. Thus
the total area for positive x and y is 1 (for the square y < 1, x < 1)
plus pi/4 (for the quarter-circle). The final answer is quadruple this,
or 4 + pi.
Problem B4
Let p(x) be a nonzero polynomial of degree less than 1992 having no
nonconstant factor in common with x^3 - x. Let
( d^1992 / dx^1992 ) ( p(x) / (x^3 - x) ) = f(x) / g(x)
for polynomials f(x) and g(x). Find the smallest possible degree of f(x).
(The second sentence is backwards---``let'' should be followed
immediately by the variable being introduced. Would you say ``Let
2 equal x + y for integers x and y''?)
Solution: First divide p(x) by x^3 - x: p(x) = (x^3 - x)q(x) + r(x),
with r of degree at most 2. Now f(x)/g(x) = D^1992 (q(x) + r(x)/(x^3-x))
= D^1992 (r(x)/(x^3-x)), as q has degree less than 1992; here we write
D for d/dx. We expand r(x)/(x^3-x) in partial fractions as -r(0)/x +
r(1)/2(x-1) + r(-1)/2(x+1). Now the 1992nd derivative of this is
Cr(0)/x^1993 + Cr(1)/(x-1)^1993 + Cr(-1)/(x+1)^1993 for a certain
nonzero constant C which we don't care about. This then equals
(Cr(0)(x^2-1)^1993 + Cr(1)(x^2+x)^1993 + Cr(-1)(x^2-x)^1993)/(x^3-x)^1993.
The numerator and denominator here are coprime, for none of x, x-1, x+1
divide the numerator. (If, for instance, x divided the numerator, then
r(0) would have to be 0, but then p(x) would have a factor of x in
common with x^3-x, contrary to hypothesis.) So f(x) is a multiple of
the numerator and g(x) is a multiple of the denominator. Our question
is thus ``What is the smallest possible degree of the polynomial P =
U(x^2-1)^1993 + V(x^2+x)^1993 + W(x^2-x)^1993, over all U, V, W which
can arise as U=Cr(0), V=Cr(1), W=Cr(-1)?''
P has degree at most 2.1993. Its 2.1993 coefficient is U + V + W. Its
2.1993-1 coefficient is 1993V - 1993W. Its 2.1993-2 coefficient is
-1993U + 1993.(1992/2)V + 1993.(1992/2)W. If all three of these are
zero then by linear algebra all of U, V, W are zero, which is not
possible. Hence P, and thus also f, has degree at least 2.1993-2, or
double 1992. This is achieved if, for instance, p(x) = r(x) = 3x^2 - 2,
so that r(0)+r(1)+r(-1)=-2+1+1=0 and r(1)=r(-1).
(The degree restriction on p in this problem seems somewhat strange,
though it simplifies the solution slightly. Noam Elkies notes that
the ``nonzero constant C'' above will be zero---so that f will be 0---
if we're working over a field with characteristic dividing 1992!.
Should the problem have explicitly identified the ground field as
the reals?)
Problem B5
Let D_n denote the value of the (n-1) by (n-1) determinant
| 3 1 1 1 ... 1 |
| 1 4 1 1 ... 1 |
| 1 1 5 1 ... 1 |
| 1 1 1 6 ... 1 |
| . . . . ... . |
| 1 1 1 1 ... n+1 |
Is the set {D_n/n!}_{n >= 2} bounded?
(``The value of the determinant''? Why not just ``the determinant''?
Why talk about ``the set'' when it's much more common to talk about
``the sequence''? And where's the period on the first sentence?)
Solution: No, it is the harmonic series.
We subtract the first row from each of the other rows, to get a matrix
3 1 1 1 ... 1, -2 3 0 0 ... 0, -2 0 4 0 ... 0, ..., -2 0 0 0 ... n.
Then we subtract 1/3 of the new second row from the first, 1/4 of the
new third row from the first, and so on, to kill all the 1's along the
top. We are left with a triangular matrix, with diagonal X 3 4 ... n,
where X equals 3 - (-2)/3 - (-2)/4 - ... - (-2)/n =
3 + 2/3 + 2/4 + ... + 2/n = 2(1 + 1/2 + 1/3 + 1/4 + ... + 1/n). Thus
the determinant is n! times 1 + 1/2 + 1/3 + ... + 1/n. Q. E. D.
Problem B6
Let M be a set of real n by n matrices such that
(i) I \in M, where I is the n by n identity matrix;
(ii) if A \in M and B \in M, then either AB \in M or -AB \in M, but not both;
(iii) if A \in M and B \in M, then either AB = BA or AB = -BA;
(iv) if A \in M and A \noteq I, there is at least one B \in M such that
AB = -BA.
Prove that M contains at most n^2 matrices.
Solution (courtesy Noam Elkies): Fix A in M. By (iii) AB = eBA, where e
is either +1 or -1, for any B in M. Then AAB = AeBA = eABA = e^2BAA = BAA.
So A^2 commutes with any B in M. Of course the same is true of -A^2. On
the other hand by (ii) A^2 or -A^2 is in M. Pick C = A^2 or C = -A^2 so
that C is in M.
If C is not I, then by (iv) we can find a B in M such that CB = -BC. But
we know CB = BC for any B in M. Thus CB = 0, which is impossible, as by
(ii) no two elements of M can multiply to 0.
We conclude that C must be I. In other words, for any A in M, either A^2
or -A^2 must be I.
Now suppose M has more than n^2 matrices. The space of real n by n
matrices has dimension n^2, so we can find a nontrivial linear relation
\sum_{D in M} x_D D = 0. Pick such a relation with the smallest possible
number of nonzero x_D. We will construct a smaller relation, obtaining a
contradiction and finishing the proof.
Pick an A with x_A nonzero, and multiply by it: \sum_{D in M} x_D DA = 0.
In light of (ii) the matrices DA run over M modulo sign. Hence we have a
new relation \sum_{E in M} y_E E = 0. The point of this transformation is
that now the coefficient y_I of I is +- x_A, which is nonzero.
Pick any C other than I such that y_C is nonzero. By (iv) pick B in M
such that CB = -BC. Multiply \sum_{E in M} y_E E = 0 by B on both the left
and the right, and add: \sum_{E in M} y_E (BE + EB) = 0. Now by (iii) we
have BE + EB = (1 + e_{BE})BE, where e_{BE} is either +1 or -1. In
particular e_{BI} = 1 (clear) and e_{BC} = -1 (by construction of B).
So we get \sum_{E in M} y_E (1 + e_{BE}) BE = 0, where at least one term
does not disappear and at least one term does disappear. As before the
matrices BE run over M modulo sign. So we have a relation with fewer
terms as desired.
---Dan Bernstein, brn...@ocf.berkeley.edu, 7 December 1992
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> language/finnish/finnish.record.p <==
What are some Finnish words with unusual properties?
==> language/finnish/finnish.record.s <==
Spelling
Letter Patterns
Entire Word
longest word pitka"nta"hta"yksenkokonaistaloudelliseensuhdannekehitykseen (57,1)
longest palindrome naamioimaan (11,1)
longest beginning with a palindrome avattava (8,1)
longest beginning with b palindrome bob bsb (3,2)
longest beginning with c palindrome cfc cic (3,2)
longest beginning with d palindrome dad did do"d (3,3)
longest beginning with e palindrome ettette (7,1)
longest beginning with f palindrome f (1,1)
longest beginning with g palindrome gorog (5,1)
longest beginning with h palindrome hannah (6,1)
longest beginning with i palindrome illalli iskuksi (7,2)
longest beginning with j palindrome jj (2,1)
longest beginning with k palindrome kayak (5,1)
longest beginning with l palindrome lehel level (5,2)
longest beginning with m palindrome mkm mmm mqm mtm mum (3,5)
longest beginning with n palindrome naamioimaan (11,1)
longest beginning with o palindrome ookoo (5,1)
longest beginning with p palindrome piip (4,1)
longest beginning with q palindrome q (1,1)
longest beginning with r palindrome radar reger reser reyer rodor rosor rovor (5,7)
longest beginning with s palindrome syo"ppo"ys (8,1)
longest beginning with t palindrome tavattavat (10,1)
longest beginning with u palindrome uupuu (5,1)
longest beginning with v palindrome vv (2,1)
longest beginning with w palindrome w (1,1)
longest beginning with x palindrome x (1,1)
longest beginning with y palindrome yly yry yty (3,3)
longest beginning with z palindrome z (1,1)
longest with middle a palindrome assassa illalli (7,2)
longest with middle b palindrome abba ebbe (4,2)
longest with middle c palindrome ecce (4,1)
longest with middle d palindrome radar rodor sadas (5,3)
longest with middle e palindrome ettette (7,1)
longest with middle f palindrome afa cfc (3,2)
longest with middle g palindrome reger (5,1)
longest with middle h palindrome lehel nahan nuhun (5,3)
longest with middle i palindrome neien taiat (5,2)
longest with middle j palindrome aja eje rjr (3,3)
longest with middle k palindrome ookoo sokos tikit (5,3)
longest with middle l palindrome tallat tollot tullut (6,3)
longest with middle m palindrome naamaan (7,1)
longest with middle n palindrome hannah (6,1)
longest with middle o palindrome naamioimaan (11,1)
longest with middle p palindrome syo"ppo"ys (8,1)
longest with middle q palindrome mqm (3,1)
longest with middle r palindrome suuruus (7,1)
longest with middle s palindrome nissin tissit (6,2)
longest with middle t palindrome tavattavat (10,1)
longest with middle u palindrome iskuksi (7,1)
longest with middle v palindrome level rovor tavat (5,3)
longest with middle w palindrome w (1,1)
longest with middle x palindrome exe ixi (3,2)
longest with middle y palindrome kayak reyer (5,2)
longest with middle z palindrome z (1,1)
longest tautonym tytta"rentytta"ren (16,1)
longest beginning with a tautonym asiasi (6,1)
longest beginning with b tautonym baba bibi (4,2)
longest beginning with c tautonym cancan (6,1)
longest beginning with d tautonym dandan (6,1)
longest beginning with e tautonym ee (2,1)
longest beginning with f tautonym ?
longest beginning with g tautonym giogio (6,1)
longest beginning with h tautonym ha"nha"n heihei (6,2)
longest beginning with i tautonym isa"nisa"n (8,1)
longest beginning with j tautonym jnejne joojoo (6,2)
longest beginning with k tautonym kookoo (6,1)
longest beginning with l tautonym lili lyly (4,2)
longest beginning with m tautonym mormor (6,1)
longest beginning with n tautonym nana nono (4,2)
longest beginning with o tautonym ohoh (4,1)
longest beginning with p tautonym peffapeffa pojanpojan (10,2)
longest beginning with q tautonym ?
longest beginning with r tautonym roro (4,1)
longest beginning with s tautonym sitsit (6,1)
longest beginning with t tautonym tytta"rentytta"ren (16,1)
longest beginning with u tautonym uu (2,1)
longest beginning with v tautonym voivoivoivoi (12,1)
longest beginning with w tautonym ?
longest beginning with x tautonym ?
longest beginning with y tautonym ?
longest beginning with z tautonym zouzou (6,1)
longest head 'n' tail koiraskoira sellaisella vastaavasta (11,3)
longest with middle a head 'n' tail vastaavasta (11,1)
longest with middle b head 'n' tail b (1,1)
longest with middle c head 'n' tail c (1,1)
longest with middle d head 'n' tail ada ede idi ldl (3,4)
longest with middle e head 'n' tail eeeee (5,1)
longest with middle f head 'n' tail kafka (5,1)
longest with middle g head 'n' tail edged magma (5,2)
longest with middle h head 'n' tail jahja ohhoh vahva yhhyh (5,4)
longest with middle i head 'n' tail sellaisella (11,1)
longest with middle j head 'n' tail aijai (5,1)
longest with middle k head 'n' tail kuukkuu (7,1)
longest with middle l head 'n' tail kulku lalla lilli oslos salsa ta"lta" tultu volvo (5,8)
longest with middle m head 'n' tail chamcha (7,1)
longest with middle n head 'n' tail tarantara (9,1)
longest with middle o head 'n' tail nenonen (7,1)
longest with middle p head 'n' tail pappa peppe uupuu (5,3)
longest with middle q head 'n' tail mqm (3,1)
longest with middle r head 'n' tail giorgio (7,1)
longest with middle s head 'n' tail koiraskoira (11,1)
longest with middle t head 'n' tail A"llita"lli tavattava (9,2)
longest with middle u head 'n' tail laula vauva (5,2)
longest with middle v head 'n' tail ava eve (3,2)
longest with middle w head 'n' tail w (1,1)
longest with middle x head 'n' tail exe ixi (3,2)
longest with middle y head 'n' tail ta"yta" va"yva" (5,2)
longest with middle z head 'n' tail z (1,1)
Subset of Word
longest internal palindrome bisneskeskukseksi rahoituskeskukseksi terveyskeskukseksi ulkomaalaiskeskukseksi (11,4)
longest internal tautonym A"teritsiputeritsipuolilautatsija"nka" (18,1)
longest repeated prefix kansankansanvaalilla kiittikiittijaka"yma"a"ntaas lapsenlapsenlapsia lapsenlapsensa victorvictoria (12,5)
most consecutive doubled letters aatteessaan liikkeella"a"n pa"a"tteelleen pa"a"tteessa"a"n periaatteellaan tietokonepa"a"tteella"a"n (5,6)
most doubled letters lennonsuunnittelupa"a"llikko" suhteellisuusperiaatteella tuotannonsuunnittelupa"a"llikko" (7,3)
longest two cadence alabamalaista pajavasara sahatavara sahatavaraa sahatavaran sahatavarassa sahatavarasta salakavala satamamakasiinissa tasasakarainen ... (5,11)
longest three cadence asianharrastajia (6,1)
longest four cadence rajakauppaorganisaatio rajakauppaorganisaationsa unkarilaispakolaisia (5,3)
longest five cadence artikkelistafinancial asiakasrekisterista" ateriakorvauksia autonasentajalta azerbaidzhanissa azerbaidzhanista dieettisuosituksista diplomiarkkitehti ekspressionistisesti epa"inhimillisissa"kin ... (4,61)
Letter Counts
Lipograms
longest letters from first half nikkeliniemimaalla (18,1)
longest letters from second half tytto"ysta"va"sta" (14,1)
longest without ab yhdeksa"sta"kymmenesta"yhdeksa"sta" ympa"risto"nsuojeluinvestointien (30,2)
longest without abcd ympa"risto"nsuojeluinvestointien (30,1)
longest without a to h ka"a"nnynna"ispiiripa"a"lliko"ita" (27,1)
longest without a to k ymma"rta"ma"tto"myytta"a"n (20,1)
longest without a to n tytto"ysta"va"sta" (14,1)
longest without a to q tytto"ysta"va"sta" (14,1)
longest without a to s to"o"tta"a"va"t (10,1)
longest without e apulaisosastotarkastajapa"a"lliko"n (32,1)
longest without et kunnallisvaalikampanjassaan (27,1)
longest without eta kirkkomusiikkipa"ivilla" (22,1)
longest without etai lukuporoma"a"ra"nsa" (16,1)
longest without etain hyo"kka"yska"skya" va"ha"ssa"kyro"ssa" (14,2)
longest without etains hyo"kka"a"va"mpa"a" polkupyo"ra"lla" (13,2)
Letter Choices
Vowels
longest all vowels eeeee uuuii (5,2)
longest each vowel once perusla"hto"kohtia va"hemmisto"suojan va"hemmisto"suojat (16,3)
longest each extended vowel once lennonvarmistustyo"ssa" (21,1)
shortest each vowel once kaukona"ko"inen (13,1)
shortest each extended vowel once syo"pa"kuolemia (13,1)
shortest vowels in order ?
shortest extended vowels in order ?
longest vowels in order ?
longest extended vowels in order ?
shortest vowels in reverse order ?
shortest extended vowels in reverse order ?
longest vowels in reverse order ?
longest extended vowels in reverse order ?
longest one vowel kfsdistr schlicht schmaltz schmolck schwa"rmt schwartz (8,6)
longest two vowels mA@ngbranschf schwartzkopf (12,2)
longest containing a univocalic harrastajakalastajat (20,1)
longest containing e univocalic schellenberger (14,1)
longest containing i univocalic inhimillisiin nihilistisiin siviilitripin (13,3)
longest containing o univocalic octophoros stockholms volkogonov (10,3)
longest containing u univocalic kukkuluuruu punkjuttuun tunnusluvut (11,3)
longest containing y univocalic kynnyskysymys (13,1)
longest alternating extended vowel-consonant tavarajunaveturina (18,1)
longest alternating vowel-consonant tavarajunaveturina (18,1)
Consonants
longest consonant string entschluss franskspra@kiga gcdkssa" goldschmieds hdtvssa" hdtvsta" mkrtchian skdllle skdllta" skdlssa" ... (6,16)
longest one consonant voivoivoivoi (12,1)
longest two consonant muuttumattomuutta (17,1)
Isograms
longest isogram va"esto"pohjakin ympa"risto"hanke (14,2)
longest pair isogram itsena"isyytena" kokemattomalle teknisestika"a"n tunnetuimmille (14,4)
longest trio isogram pojanpojanpojan (15,1)
longest tetrad isogram voivoivoivoi (12,1)
longest polygram kiireellisyysja"rjestyksesta" (27,1)
longest pyramid kaikenkaikkiaan keskikokoisiksi sisilialaisessa sunniittisissit (15,4)
most repeated letters pitka"nta"hta"yksenkokonaistaloudelliseensuhdannekehitykseen (14,1)
highest containing a repeated oppilaastaanrakastajattarestaan (11,1)
highest containing b repeated babbino babbitt babib barabbakseen belsebub bensinbomben bobbi bobbie bobby bobbyn ... (3,13)
highest containing c repeated acapriccio accademica acceptance acconci bocaccio caccini capecchi capecchin cappuccilli capricciata ... (3,41)
highest containing d repeated jodlalaadididuaadi (5,1)
highest containing e repeated pitka"nta"hta"yksenkokonaistaloudelliseensuhdannekehitykseen (8,1)
highest containing f repeated fofanoffit peffapeffa (4,2)
highest containing g repeated vergangenheitsbewa"ltigung (4,1)
highest containing h repeated harhahiihdosta hrushtshevhan rahghhh tphhhh vaihtoehtohoitoihin (4,5)
highest containing i repeated luterilaispedagogismannerheimila"istannerheimila"isia" (10,1)
highest containing j repeated harjujensuojeluohjelman ja"rjestelyasiakirjoja ja"rjesto"johtajaa ja"rjesto"johtajat ja"rjestyssijoja ja"senja"rjesto"jen ja"senja"rjesto"jensa ja"senja"rjesto"jensa" johtajanlahjojaan ka"sikirjoittajaohjaaja ... (4,18)
highest containing k repeated hiukkastutkimuskeskuksen hiukkastutkimuskeskuksessa joukkokulutushyo"dykkeeksi kaupunkikaksikko keskuspankkipolitiikkaa kirkkorakennuksissakin kirsikkakukkapuutarha korkeakoulupaikkakunnilla pikkupikkukesakoihin pitka"nta"hta"yksenkokonaistaloudelliseensuhdannekehitykseen ... (6,11)
highest containing l repeated nelja"lla"kymmenella"kuudella paikallisellala"a"nitasolla yritystenhintakilpailukykyynkansainva"lisilla"markkinoilla (7,3)
highest containing m repeated mammuttimaisemmat (6,1)
highest containing n repeated nelja"a"nmiljoonaankolmeensataantuhanteen (8,1)
highest containing o repeated kokooomusjohtoisen (6,1)
highest containing p repeated piipputupakkapakkaukseen poppanareppu (5,2)
highest containing q repeated acoustiques acqua acquaintance acquanetta ahlqvist ahlqvistia ahlqvistillekin ahlqvistin almquist almqvist ... (1,388)
highest containing r repeated baskiterroristija"rjesto" cairnterrierinsa" carrerras eurooppalaisterroristeihin eurooppalaisterroristeja harrivirran herrareservi itsemurhaterroristi kamariorkerteri karria"a"rivirkamiehia" ... (4,33)
highest containing s repeated itsena"isyysjulistuksessa kansainva"lisyyskasvatuksessa kansallissosialistisesta kustannuslaskentaperusteisissa sadassaviidessa"kymmenessa" saksalaissuuntauksessa sisustusarkkitehtitoimistossa sisustusratkaisuissa sisustussuunnittelussa sotasyyllisyysasiassa ... (7,15)
highest containing t repeated kertaka"ytto"tuotteita neuvostoliittoinstituutista tavoittamattomuutta tekstinka"sittelylaitteesta toteuttamistoimenpiteet toteuttamistoimenpiteita" tuotantotavoitteet tytta"rentytta"resta" tytto"tertsetti (7,9)
highest containing u repeated suuronnettomuustutkimuslautakuntaa (8,1)
highest containing v repeated arvonlisa"veroliikevaihtovero mandariininviipaleenvirtaviivainen vahingonkorvausvelvollisuuden voivoivoivoi (4,4)
highest containing w repeated kwakwakwak (3,1)
highest containing x repeated exxon exxonilla maxixe mexx mexxeineen xerxeen xerxes xxe xxii xxiiin ... (2,11)
highest containing y repeated sotasyyllisyyskysymyksista" syyllisyyskysymykseen syyllisyyskysymyksesta" syyllisyyskysymysten (7,4)
highest containing z repeated dzhordzhadze (3,1)
most different letters ympa"risto"nsuojeluneuvottelukunnan ympa"risto"nsuojeluvaatimukset (18,2)
highest ratio length/letters eeeee (500,1)
highest ratio length/letters (no tautonyms) eeeee (500,1)
lowest length 16 ratio length/letters perusla"hto"kohdan ympa"risto"ongelma (106,2)
lowest length 17 ratio length/letters lyhenta"miskorvaus ympa"risto"ohjelman (106,2)
lowest length 18 ratio length/letters vormundschaftliche ympa"risto"nsuojelua (112,2)
lowest length 19 ratio length/letters ympa"risto"vahinkojen (111,1)
lowest length 20 ratio length/letters ympa"risto"nvaihdoksen (117,1)
lowest length 21 ratio length/letters kymmenvuotisjuhlarevy purjehduskyvytto"ma"ksi valtameripurjehduksen ympa"risto"nsuojelijain ympa"risto"puoluettakin (131,5)
lowest length 22 ratio length/letters ra"ja"hdysvoimakkuudesta ympa"risto"nsuojelijoita (137,2)
lowest length 23 ratio length/letters radiopuhelinja"rjestelma" viimeistelyharjoituksen ympa"risto"nsuojelijoiden ympa"risto"nsuojelullisia (143,4)
lowest length 24 ratio length/letters va"hemmisto"suojapyka"lilla" ympa"risto"nsuojeluohjelma (141,2)
lowest length 25 ratio length/letters ympa"risto"nsuojeluohjelman (147,1)
Letter Appearance
longest short letters messusaarnassaan (16,1)
longest tall letters kiihdytti pitkittyi ylikiltti (9,3)
longest vertical-symmetry letters tavoittamattomuutta (19,1)
longest horizontal-symmetry letters heikohko (8,1)
longest full-symmetry letters ihhii ohhoh (5,2)
highest ratio of dotted letters hiipii (66,1)
Typewriter
longest top row prototyypit prototyyppi reportterit terroriteot (11,4)
longest middle row lahjakkaalla (12,1)
longest bottom row bbcn cnnn nbcn nvcn (4,4)
longest in order erilla"a"n ryyppa"a"n (8,2)
longest in reverse order bourree jaoitte na"ittte (7,3)
longest left hand verrattaessa (12,1)
longest right hand huippuhyppa"a"ja"na" myyma"la"pa"a"lliko"n (16,2)
longest alternating hands turhanta"rkeys (13,1)
longest one finger juhuuuuh (8,1)
longest adjacent keys juhuuuuh (8,1)
Puzzle
longest palindrome in Morse code ettette naisina waiting (7,3)
longest formed with chemical symbols kasviyhdyskuntasukkessioita (27,1)
longest formed with US postal codes victorvictoria (14,1)
longest formed with amino acid abbreviations valmet valser (6,2)
longest formed with piano notes baggage (7,1)
Letter Order
Alphabetical
longest letters in order almost delors (6,2)
longest letters in order with repeats innostuu (8,1)
longest letters in reverse order polkea tonkia vuomia wolfea (6,4)
longest letters in reverse order with repeats A"speeaa polkkia uuuuiih (7,3)
longest roller-coaster matkustajasataman metsa"osakeyhtio"ta" (17,2)
longest no letters in place pitka"nta"hta"yksenkokonaistaloudelliseensuhdannekehitykseen (57,1)
most letters in place arkkitehtuuritoimisto avomerihylkeina" kiireellisyysja"rjestyksesta" laajenemismahdollisuudet soidensuojelun ta"ydellisyydenhimosta titaanidioksidipa"a"sto"ja" (5,7)
most letters in place shifted elinkeinoverotuksesta elinkeinoverotus havainnollistuvat (7,3)
most consecutive letters in order consecutively aikuistuvan aikuistuvana alistuvaisuuteen alistuvan alistuvia altistuvat astuva astuvalla astuvan astuvansa ... (4,227)
most consecutive letters in order kulminaatiopiste kulminaatiopisteeseen kulminaatiopisteissa" kulminaatiopisteita" pyha"inja"a"nno"skokoelma (6,5)
most consecutive letters limapohjakengilla" (10,1)
highest ratio of consecutive letters to length heftig ja"lkim kolmen virsut (83,4)
==> language/french/french.palindromes.p <==
List some French palindromes.
==> language/french/french.palindromes.s <==
Elu par cette crapule.
L'a^me su^re ruse mal.
A l'e'tape, e'pate-la!
Sexe ve^tu, tu te vexes.
Esope reste et se repose.
L'ame des uns n'use de mal.
Esope reste ici et se repose.
Leon a trop par rapport a Noel.
Se've`re mal a` l'a^me, re^ves.
Tu l'as trop e'crase', Ce'sar, ce Port-Salut !
9691
EDNA D'NILU
O. MU. ACERE. PSEG ROEG
Trace l'inegal palindrome. Neige. Bagatelle, dira Hercule. Le brut
repentir, cet ecrit ne Perec. L'arc lu pese trop, lis a vice-versa.
Perte . Cerise d'une verite banale, le Malstrom, Alep, mort
edulcore, crepe porte de ce desir brise d'un iota. Livre si aboli, tes
sacres ont ereinte, cor cruel, nos albatros. Etre las, autel bati,
miette vice-versa du jeu que fit, nacre, medical, le selenite relaps,
ellipsoidal.
Ivre il bat, la turbine bat, l'isole me ravale : le verre si obei du
Pernod - eh, port su ! - obsedante sonate teintee d'ivresse.
Ce reve se mit - peste ! - a blaguer. Beh ! L'art sec n'a si peu
qu'algebre, s'elabore de l'or evalue. Idiome etire, hesite, batard
replie, l'os nu. Si, a la gene secrete - verbe nul a l'instar de cinq
occis --, rets amincis, drailles inegales, il, avatar espace, caresse
ce noir Belzebuth, oeil offense, tire !
L'echo fi (a desert) : Salut, sang, robe et ete.
Fievres.
Adam, rauque ; il ecrit : Abrupt ogre, eh, cerceuil, l'avenir tu,
effile, genial a la rue (murmure sud eu ne tire vaseline separee ;
l'epeire gelee rode : Hep, mortel ?) lia ta balafre native.
Litige. Regagner (et ne m'...).
Ressac. Il fremit, se sape, na ! Eh, cavale ! Timide, il nia ce
sursaut.
Hasard repu, tel, le magicien a morte me lit. Un ignare le rapsode,
lacs emu, mixa, mela :
Hep, Oceano Nox, o, bechamel azur ! Ejaculer ! Topaze !
Le cedre, malabar faible, Arsinoe le macule, mante ivre, glauque,
pis, l'air atone (sic). Art sournois : si, medicinale, l'autre glace
(Melba ?) l'un ? N'alertai ni pollen (reteter : gerce, repu, dente...)
ni tobacco.
Tu, desir, brio rime, eh, prolixe necrophore, tu ferres l'avenir
velu, ocre, cromant-ne ?
Rage, l'ara. Veuglaire. Sedan, tes elzevirs t'obsedent. Romain ?
Exact. Et Nemrod selle ses Samson !
Et nier teocalli ?
Cave canem (car ce nu trop minois - rembuscade d'eruptives a babil -
admonestra, fil accru, Tetebleu ! qu'Ariane evitat net.
Attention, ebenier factice, ressorti du reel. Ci-git. Alpaga, gnome,
le heros se lamente, trompe, chocolat : ce laid totem, ord, nil
aplati, rituel biscornu ; ce sacre bedeau (quel bat ce Jesus !).
Palace piege, Torpedo drue si a fellah tot ne peut ni le Big a ruer
bezef.
L'eugeniste en rut consuma d'art son epi d'eolinne ici rot (eh...
rut ?). Toi, d'idem gin, elevera, elu, bifocal, lithos et notre pathos
a la hauteur de sec salamlec ?
Elucider. Ion eclate : Elle ? Tenu. Etna but (item mal fame), degre
vide, julep : macedoine d'axiomes, sac seme d'Ecole, veniel, ah, le
verbe enivre (ne sucer ni arreter, eh ca jamais !) lu n'abolira le
hasard ?
Nu, ottoman a l'echo, l'art su, oh, tara zero, belle Deborah, o,
sacre ! Pute, vertubleu, qualite si vertu a la part tarife (decalitres
?) et nul n'a lu trop s'il seria de ce basilic Iseut.
Il a prie bonzes, Samaritain, Tora, vilains monstres (idolatres DNA
en sus), reves, evapores :
Arbalete (betes) en noce du Tell ivre-mort, emeri tu : O, trapu a
elfe, il lie l'os, il lia jeremiade lucide. Petard ! Rate ta reinette,
bigleur cruel, non a ce lot ! Si, farcis-toi dito le coeur !
Lied a monstre velu, ange ni bete, sec a pseudo delire : Tsarine
(sellee, la), Cid, Aretin, abruti de Ninive, Dejanire...
Le Phenix, eve de sables, ecarte, ne peut egarer racines radiales en
mana : l'Oubli, fetiche en argile.
Foudre.
Prix : Ile de la Gorgone en roc, et, o, Licorne ecartelee,
Sirene, rumb a bannir a ma (Red n'osa) niere de mimosa :
Paysage d'Ourcq ocre sous ive d'ecale ;
Volcan. Roc : tarot cele du Pere.
Livres.
Silene bavard, replie sur sa nullite (nu a je) belge : ipseite
banale. L' (eh, ca!) hydromel a ri, psalterion. Erree Lorelei...
Fi ! Marmelade derive d'Aladine. D'or, Noel : creche (l'an ici
taverne gelee des bol...) a santon givre, fi !, cule de l'ane vairon.
Lapalisse elu, gnoses sans orgeuil (ecru, sale, sec).
Saluts : angiome. T'es si craneur !
[ici, repartir vers le debut, pour arriver a la signature: ]
Georges Perec,
Au Moulin d'Ande, 1969
==> language/french/french.record.p <==
What are some French words with unusual properties?
==> language/french/french.record.s <==
Spelling
Letter Patterns
Entire Word
longest word anticonstitutionnellement (25,1)
longest palindrome ressasser (9,1)
longest beginning with a palindrome aviva (5,1)
longest beginning with b palindrome bob (3,1)
longest beginning with c palindrome cc (2,1)
longest beginning with d palindrome d' (1,1)
longest beginning with e palindrome elle erre erre' esse (4,4)
longest beginning with f palindrome ?
longest beginning with g palindrome gag (3,1)
longest beginning with h palindrome h (1,1)
longest beginning with i palindrome ici (3,1)
longest beginning with j palindrome j' (1,1)
longest beginning with k palindrome kayak (5,1)
longest beginning with l palindrome l l' (1,2)
longest beginning with m palindrome mm (2,1)
longest beginning with n palindrome nanan (5,1)
longest beginning with o palindrome o^ (1,1)
longest beginning with p palindrome ?
longest beginning with q palindrome ?
longest beginning with r palindrome ressasser (9,1)
longest beginning with s palindrome selle's selles serre's serres (6,4)
longest beginning with t palindrome to^t tut (3,2)
longest beginning with u palindrome ?
longest beginning with v palindrome ?
longest beginning with w palindrome ?
longest beginning with x palindrome x (1,1)
longest beginning with y palindrome y (1,1)
longest beginning with z palindrome ?
longest with middle a palindrome ressasser (9,1)
longest with middle b palindrome ?
longest with middle c palindrome ici (3,1)
longest with middle d palindrome radar (5,1)
longest with middle e palindrome re'er (4,1)
longest with middle f palindrome re'ifier (7,1)
longest with middle g palindrome sagas (5,1)
longest with middle h palindrome h (1,1)
longest with middle i palindrome aviva (5,1)
longest with middle j palindrome j' (1,1)
longest with middle k palindrome ?
longest with middle l palindrome selle's selles (6,2)
longest with middle m palindrome se`mes seme's (5,2)
longest with middle n palindrome nanan sanas sonos (5,3)
longest with middle o palindrome bob non to^t (3,3)
longest with middle p palindrome ?
longest with middle q palindrome ?
longest with middle r palindrome serre's serres (6,2)
longest with middle s palindrome esse (4,1)
longest with middle t palindrome rotor (5,1)
longest with middle u palindrome eue sus tut (3,3)
longest with middle v palindrome re^ver se`ves (5,2)
longest with middle w palindrome ?
longest with middle x palindrome sexes (5,1)
longest with middle y palindrome kayak (5,1)
longest with middle z palindrome ?
longest tautonym traintrain (10,1)
longest beginning with a tautonym ?
longest beginning with b tautonym be'ribe'ri (8,1)
longest beginning with c tautonym chercher couscous (8,2)
longest beginning with d tautonym dada (4,1)
longest beginning with e tautonym ?
longest beginning with f tautonym flonflon froufrou (8,2)
longest beginning with g tautonym glouglou (8,1)
longest beginning with h tautonym ?
longest beginning with i tautonym ?
longest beginning with j tautonym joujou (6,1)
longest beginning with k tautonym ?
longest beginning with l tautonym ?
longest beginning with m tautonym me^me (4,1)
longest beginning with n tautonym ?
longest beginning with o tautonym ?
longest beginning with p tautonym papa (4,1)
longest beginning with q tautonym ?
longest beginning with r tautonym rentrent (8,1)
longest beginning with s tautonym ?
longest beginning with t tautonym traintrain (10,1)
longest beginning with u tautonym ?
longest beginning with v tautonym ?
longest beginning with w tautonym ?
longest beginning with x tautonym ?
longest beginning with y tautonym youyou (6,1)
longest beginning with z tautonym ?
longest head 'n' tail cherche cherche' de'rader de'rider de'roder entrent hercher quelque raierai raseras (7,10)
longest with middle a head 'n' tail de'rader (7,1)
longest with middle b head 'n' tail ?
longest with middle c head 'n' tail hercher (7,1)
longest with middle d head 'n' tail d' (1,1)
longest with middle e head 'n' tail raierai raseras (7,2)
longest with middle f head 'n' tail ?
longest with middle g head 'n' tail magma (5,1)
longest with middle h head 'n' tail e'chec (5,1)
longest with middle i head 'n' tail de'rider (7,1)
longest with middle j head 'n' tail j' (1,1)
longest with middle k head 'n' tail ?
longest with middle l head 'n' tail quelque (7,1)
longest with middle m head 'n' tail pampa (5,1)
longest with middle n head 'n' tail sense' tente tente' (5,3)
longest with middle o head 'n' tail de'roder (7,1)
longest with middle p head 'n' tail ?
longest with middle q head 'n' tail ?
longest with middle r head 'n' tail cherche cherche' entrent (7,3)
longest with middle s head 'n' tail esses teste teste' (5,3)
longest with middle t head 'n' tail antan (5,1)
longest with middle u head 'n' tail veuve (5,1)
longest with middle v head 'n' tail ?
longest with middle w head 'n' tail ?
longest with middle x head 'n' tail texte (5,1)
longest with middle y head 'n' tail y (1,1)
longest with middle z head 'n' tail ?
Subset of Word
longest internal palindrome ressasse`rent ressassera ressasserai ressasseraient ressasserais ressasserait ressasseras ressasserez ressasseriez ressasserions ... (9,12)
longest internal tautonym cherche`rent chercherai chercheraient chercherais chercherait chercheras chercherez chercheriez chercherions chercherons ... (8,40)
longest repeated prefix cherche`rent chercherai chercheraient chercherais chercherait chercheras chercherez chercheriez chercherions chercherons ... (8,22)
most consecutive doubled letters irre'elle irre'elles (3,2)
most doubled letters commissionne'e commissionne'es (4,2)
longest two cadence de'ce'le're`rent de'ge'ne're`rent divisibilite' re'ge'ne're`rent re'ge'ne'rerez (5,5)
longest three cadence effervescence effervescente effervescentes (5,3)
longest four cadence enclencheraient entrelaceraient entreme^leraient entreposeraient insatisfaisants manutentionnions satisfaisants satisfaisions satisfassions (4,9)
longest five cadence individualiserai individualiseraient individualiserais individualiserait (4,4)
Letter Counts
Lipograms
longest letters from first half de'calcifie'e (11,1)
longest letters from second half pourvoyons provoquons supportons (10,3)
longest without ab constitutionnellement interprofessionnelles socioprofessionnelles (21,3)
longest without abcd interprofessionnelles (21,1)
longest without a to h soumissionnions (15,1)
longest without a to k pourvoyons provoquons supportons surmontons (10,4)
longest without a to n pourtours (9,1)
longest without a to q trusts (6,1)
longest without a to s tut (3,1)
longest without e institutionnalisation (21,1)
longest without et approfondissions approvisionnions individualisions (16,3)
longest without eta circonscririons circonscrivions soumissionnions (15,3)
longest without etai subordonnons (12,1)
longest without etain corrompus froufrous glouglous surplombs (9,4)
longest without etains corrompu courroux froufrou glouglou (8,4)
Letter Choices
Vowels
longest all vowels oui"e (4,1)
longest each vowel once assujettissons chiropracteurs contradicteurs croustillantes grandiloquents insupportables insurmontables obscurantistes raccourcissent reconstituants ... (14,16)
longest each extended vowel once hydrographiques prophylactiques (15,2)
shortest each vowel once oiseau (6,1)
shortest each extended vowel once noyautiez (9,1)
shortest vowels in order ?
shortest extended vowels in order ?
longest vowels in order ?
longest extended vowels in order ?
shortest vowels in reverse order rudoiera tutoiera (8,2)
shortest extended vowels in reverse order ?
longest vowels in reverse order rudoieras tutoieras (9,2)
longest extended vowels in reverse order ?
longest one vowel prompts putschs scripts sprints stricts (7,5)
longest two vowels frustrants patchworks sphincters stressants tranchants transcrits transferts transports tremblants (10,9)
longest containing a univocalic abracadabrants (14,1)
longest containing e univocalic encheve^trements (15,1)
longest containing i univocalic distinctifs indistincts instinctifs (11,3)
longest containing o univocalic confondrons confondront confrontons coordonnons corroborons corromprons corrompront morfondrons morfondront (11,9)
longest containing u univocalic cumulus putschs summums surplus tumulus (7,5)
longest containing y univocalic lynx thym (4,2)
longest alternating extended vowel-consonant de'solidariseras de'solidariserez re'capitulatives (15,3)
longest alternating vowel-consonant de'solidariseras de'solidariserez re'capitulatives (15,3)
Consonants
longest consonant string abstraction abstractions abstraie abstraient abstraies abstraira abstrairai abstrairaient abstrairais abstrairait ... (4,565)
longest one consonant saisissais suissesses (10,2)
longest two consonant assainissions dessaisissais ressaisissais (13,3)
Isograms
longest isogram surplombaient (13,1)
longest pair isogram concerteront e'parpillerai entrai^nerait occasionnais renai^traient tenaillaient trai^neraient (12,7)
longest trio isogram resserre's resserres (9,2)
longest tetrad isogram ?
longest polygram concentrationnaire (18,1)
longest pyramid desserre'es ente^te`rent ente^tement rassasiais re'gresse'es redresse'es resserrera saisissait (10,8)
most repeated letters concentrationnaire concentrationnaires contractualiserions internationaliserions internationaliserons photographieraient re'approvisionnerais re'approvisionneras standardiseraient (8,9)
highest containing a repeated abracadabrant abracadabrante abracadabrantes abracadabrants (5,4)
highest containing b repeated baobab baobabs bibliobus (3,3)
highest containing c repeated concupiscence (4,1)
highest containing d repeated de'bandade de'bandades dividende dividendes rhododendron rhododendrons superdividende superdividendes (3,8)
highest containing e repeated de'ge'ne'rescence e've'nementielle e've'nementielles (6,3)
highest containing f repeated affectif affectifs affirmatif affirmatifs effectif effectifs fieffe' fieffe'e fieffe'es fieffe's ... (3,14)
highest containing g repeated gigogne gigognes grignotage (3,3)
highest containing h repeated hachisch haschich haschisch (3,3)
highest containing i repeated indivisibilite' (6,1)
highest containing j repeated joujou joujoux (2,2)
highest containing k repeated bookmaker bookmakers drakkar drakkars enkikiner kaki kakis kamikaze kamikazes kapok ... (2,26)
highest containing l repeated intellectuelle intellectuellement intellectuelles libellule libellules (4,5)
highest containing m repeated accommodement accommodements anticommunisme commandement commandements comme'mora comme'moraient comme'morais comme'morait comme'morant ... (3,112)
highest containing n repeated anticonstitutionnellement conventionnaient conventionnant conventionne`rent conventionnement conventionnements conventionnent conventionneraient conventionnerions conventionnerons ... (5,25)
highest containing o repeated coordonnerions coordonnerons coordonneront coordonnions coordonnons corroboration corroborerions corroborerons corroboreront corroborions ... (4,40)
highest containing p repeated appropria appropriaient appropriais appropriait appropriant appropriation approprie approprie' approprie'e approprie'es ... (3,70)
highest containing q repeated aquatique aquatiques e'quivoque e'quivoquer e'quivoques enquiquiner qqn qqu'un qqun quelconque ... (2,32)
highest containing r repeated arre'rager carbonitrurer contrecarre`rent contrecarrer contrecarrera contrecarrerai contrecarreraient contrecarrerais contrecarrerait contrecarreras ... (4,177)
highest containing s repeated dessaisissais dessaisissements dessaisisses dessaisissions dessaisissons ressaisissais ressaisisses ressaisissions ressaisissons suissesses (6,10)
highest containing t repeated anticonstitutionnellement trottinette trottinettes (5,3)
highest containing u repeated tumultueuse tumultueuses tumultueux (4,3)
highest containing v repeated aviva avivaient avivais avivait avivant avive avive' avive'e avive'es avive's ... (2,349)
highest containing w repeated bowling bowlings bungalow bungalows clown clownerie clowneries clowns crawl crawler ... (1,97)
highest containing x repeated anxieux expe'rimentaux hexagonaux lexicaux luxueux luxurieux maximaux paradoxaux (2,8)
highest containing y repeated hydrolyse hydrolyser hydrolyses hypophyse polyptyque polyptyques polysyllabique polysyllabiques psychanalysant psychanalyse ... (2,28)
highest containing z repeated zigzaguerez zigzagueriez zigzaguez zigzaguiez (3,4)
most different letters dactylographiques (16,1)
highest ratio length/letters resserre'es saisissais (333,2)
highest ratio length/letters (no tautonyms) resserre'es saisissais (333,2)
lowest length 16 ratio length/letters dactylographique (106,1)
lowest length 17 ratio length/letters dactylographiques (106,1)
lowest length 18 ratio length/letters cine'matographiques dactylographierons (120,2)
lowest length 19 ratio length/letters dactylographierions ste'nodactylographie (126,2)
lowest length 20 ratio length/letters dactylographieraient (142,1)
lowest length 21 ratio length/letters anticonstitutionnelle constitutionnellement institutionnalise`rent institutionnaliserait institutionnaliseront interprofessionnelles re'approvisionneraient socioprofessionnelles (210,8)
lowest length 22 ratio length/letters e'lectroence'phalogramme (183,1)
lowest length 23 ratio length/letters e'lectroence'phalogrammes (176,1)
lowest length 24 ratio length/letters ?
lowest length 25 ratio length/letters anticonstitutionnellement (227,1)
Letter Appearance
longest short letters concurrencerons (15,1)
longest tall letters digit hippy (5,2)
longest vertical-symmetry letters vouvoyait (9,1)
longest horizontal-symmetry letters de'coche'e (8,1)
longest full-symmetry letters oh (2,1)
highest ratio of dotted letters infini inimitie' initia initiais initiait initie initie' initiiez jailli simili (50,10)
Typewriter
longest top row re'pertorie'e re'pertorier (11,2)
longest middle row alfas dadas dalla flash galas halls lassa sagas shahs (5,9)
longest bottom row cc cm mm (2,3)
longest in order e'tuis qqu'un (5,2)
longest in reverse order chapitre cloi^tre'e (8,2)
longest left hand extraterrestres (15,1)
longest right hand moinillon (9,1)
longest alternating hands authentifiaient authentiquement (15,2)
longest one finger de'ce'de'e (7,1)
longest adjacent keys desserre'es redresse'es ressasse'es resserre'es (10,4)
Puzzle
longest palindrome in Morse code refiler tintait (7,2)
longest formed with chemical symbols internationaliserions (21,1)
longest formed with US postal codes calamine cocorico concilia concorde concorde' de'coinc/a lavandin mandarin me'morial moutarde (8,10)
longest formed with amino acid abbreviations argile hisser promet serval valser (6,5)
longest formed with piano notes de'ce'de'e de'gage'e de'gagea efface'e (7,4)
Letter Order
Alphabetical
longest letters in order bijoux dehors (6,2)
longest letters in order with repeats accent afflux bijoux billot dehors effort (6,6)
longest letters in reverse order rongea songea spolia spolie spolie' (6,5)
longest letters in reverse order with repeats spolie'e (7,1)
longest roller-coaster de'budge'tiserions remilitarisation syste'matisations syste'matiserions (16,4)
longest no letters in place constitutionnellement interprofessionnelles (21,2)
most letters in place abasourdiraient acade'micien acade'micienne acade'miciennes acade'miciens accre'diteraient accre'ditiez accre'ditif accre'ditifs accre'ditions ... (4,40)
most letters in place shifted anticolonialiste anticolonialistes be'ne'ficierions congratulerez coope'rative coope'ratives de'christianise'es de'christianiser de'finiraient de'finissions ... (5,27)
most consecutive letters in order consecutively abce'der abce`s anophe`le anophe`les astuce astuces astucieuse astucieusement astucieuses astucieux ... (3,975)
most consecutive letters in order abracadabrante abracadabrantes compliquerais compliqueras compliquerions compliquerons me'canographique me'canographiques misanthropique misanthropiques (5,10)
most consecutive letters anthropome'triques cine'matographiques empaquetterions empaquetterons misanthropique misanthropiques sporadiquement (9,7)
highest ratio of consecutive letters to length quarts survit vertus (83,3)
==> language/german/german.palindromes.p <==
List some German palindromes.
==> language/german/german.palindromes.s <==
Ein Neger mit Gazelle zagt im Regen nie.
==> language/german/german.record.p <==
What are some German words with unusual properties?
==> language/german/german.record.s <==
Spelling
Letter Patterns
Entire Word
longest word Arbeitslosenversicherungsbeitra"ge Bundesausbildungsfo"rderungsgesetz Krankenhaustagegeldversicherungen Wirtschaftspru"fungsgesellschaften (33,4)
longest palindrome Rentner (7,1)
longest beginning with a palindrome A"ra (3,1)
longest beginning with b palindrome Bub (3,1)
longest beginning with c palindrome ?
longest beginning with d palindrome ?
longest beginning with e palindrome Egge esse (4,2)
longest beginning with f palindrome ?
longest beginning with g palindrome ?
longest beginning with h palindrome ?
longest beginning with i palindrome ibi (3,1)
longest beginning with j palindrome ?
longest beginning with k palindrome ?
longest beginning with l palindrome ?
longest beginning with m palindrome minim (5,1)
longest beginning with n palindrome nennen netten (6,2)
longest beginning with o palindrome Otto (4,1)
longest beginning with p palindrome ?
longest beginning with q palindrome ?
longest beginning with r palindrome Rentner (7,1)
longest beginning with s palindrome Staats (6,1)
longest beginning with t palindrome Tat tot (3,2)
longest beginning with u palindrome Uhu (3,1)
longest beginning with v palindrome ?
longest beginning with w palindrome ?
longest beginning with x palindrome ?
longest beginning with y palindrome ?
longest beginning with z palindrome ?
longest with middle a palindrome Staats (6,1)
longest with middle b palindrome Neben neben (5,2)
longest with middle c palindrome ?
longest with middle d palindrome ?
longest with middle e palindrome Stets stets (5,2)
longest with middle f palindrome ?
longest with middle g palindrome Reger (5,1)
longest with middle h palindrome Ehe Uhu ehe oho (3,4)
longest with middle i palindrome ?
longest with middle j palindrome ?
longest with middle k palindrome ?
longest with middle l palindrome ?
longest with middle m palindrome ?
longest with middle n palindrome Renner nennen (6,2)
longest with middle o palindrome tot (3,1)
longest with middle p palindrome ?
longest with middle q palindrome ?
longest with middle r palindrome A"ra (3,1)
longest with middle s palindrome esse (4,1)
longest with middle t palindrome Rentner (7,1)
longest with middle u palindrome Neuen neuen (5,2)
longest with middle v palindrome ?
longest with middle w palindrome ?
longest with middle x palindrome Sexes (5,1)
longest with middle y palindrome ?
longest with middle z palindrome ?
longest tautonym Zinseszinses (12,1)
longest beginning with a tautonym ?
longest beginning with b tautonym Barbar Bonbon blabla (6,3)
longest beginning with c tautonym ?
longest beginning with d tautonym ?
longest beginning with e tautonym ?
longest beginning with f tautonym ?
longest beginning with g tautonym ?
longest beginning with h tautonym ?
longest beginning with i tautonym ?
longest beginning with j tautonym ?
longest beginning with k tautonym ?
longest beginning with l tautonym ?
longest beginning with m tautonym Mama Momo (4,2)
longest beginning with n tautonym nennen (6,1)
longest beginning with o tautonym ?
longest beginning with p tautonym Purpur (6,1)
longest beginning with q tautonym ?
longest beginning with r tautonym ?
longest beginning with s tautonym ?
longest beginning with t tautonym Tamtam (6,1)
longest beginning with u tautonym ?
longest beginning with v tautonym ?
longest beginning with w tautonym Wowo (4,1)
longest beginning with x tautonym ?
longest beginning with y tautonym ?
longest beginning with z tautonym Zinseszinses (12,1)
longest head 'n' tail Helfershelfer (13,1)
longest with middle a head 'n' tail Tat rar sas" (3,3)
longest with middle b head 'n' tail Ausbaus (7,1)
longest with middle c head 'n' tail ?
longest with middle d head 'n' tail enden (5,1)
longest with middle e head 'n' tail testetest (9,1)
longest with middle f head 'n' tail ?
longest with middle g head 'n' tail abgab engen (5,2)
longest with middle h head 'n' tail Lehle (5,1)
longest with middle i head 'n' tail Karteikarte (11,1)
longest with middle j head 'n' tail ?
longest with middle k head 'n' tail Aufkauf (7,1)
longest with middle l head 'n' tail Auflauf (7,1)
longest with middle m head 'n' tail ?
longest with middle n head 'n' tail nenne (5,1)
longest with middle o head 'n' tail sto"s"t (5,1)
longest with middle p head 'n' tail Pippi (5,1)
longest with middle q head 'n' tail ?
longest with middle r head 'n' tail A"ra (3,1)
longest with middle s head 'n' tail Helfershelfer (13,1)
longest with middle t head 'n' tail ?
longest with middle u head 'n' tail genu"gen (7,1)
longest with middle v head 'n' tail ?
longest with middle w head 'n' tail ?
longest with middle x head 'n' tail Texte texte (5,2)
longest with middle y head 'n' tail ?
longest with middle z head 'n' tail ?
Subset of Word
longest internal palindrome Achselho"hlen Afrikareferat Arbeitstier Assoziativita"t Badegelegenheit Begegnungen Betoniereinrichtung Bierpreiserho"hung Bu"hnenhumorist Darlehensnehmer ... (7,134)
longest internal tautonym Frischfischfa"nger Menschenschicksale Menschenschla"ge Menschenschlag Menschenschlages Menschenschlangen menschenscheu menschenscheuem menschenscheuen menschenscheuer ... (10,17)
longest repeated prefix Viervierteltakt wortwo"rtlich (8,2)
most consecutive doubled letters Allee Alleebaum Alleen Alliierten Angriffsseite Angriffsspiel Auslo"seeffekt Aussaat Barentssee Beifallsstu"rme ... (2,127)
most doubled letters Kommunikationsschnittstelle Mes"stellennummer Schnittstellennummer Spannungsschnittstelle (4,4)
longest two cadence Panamakanal Panamakanalzone (5,2)
longest three cadence Ferienreisenden Ferienreisendenrn Getreidepreisen Vermessenheiten elfenbeinerne herbestellende herbestellendem herbestellender herbestellendes hineindenkenden ... (5,26)
longest four cadence Aufnahmeantra"gen Beistellnetzteil Betriebsmeister Diskussionsvorschlag Europarekordlerin Feingliedrigkeit Haushaltsausgaben Investmentberater Investmentberatern Investmentvehikel ... (4,39)
longest five cadence A"nderungswu"nschen Beilagena"nderungen Benutzerinterface Diplomingenieurtitel Geburtenregelungen Gedankenleserinnen Gesetzesa"nderungen Kreditwesengesetze Neigungssensoren Textsteuerbefehle ... (4,11)
Letter Counts
Lipograms
longest letters from first half klammheimliche (14,1)
longest letters from second half Prototyps Stos"trupp (9,2)
longest without ab Feuchtigkeitsunempfindlichkeit (30,1)
longest without abcd Illustrierungskomponenten Insektenvertilgungsmittel Unternehmersteuervorteile (25,3)
longest without a to h Kompositionsprinzip (19,1)
longest without a to k Motorsports (11,1)
longest without a to n Prototyps Stos"trupp (9,2)
longest without a to q stu"rzt stu"tzt stutzt surrst wu"rzst (6,5)
longest without a to s Tutu (4,1)
longest without e Fus"ballnationalmannschaft Konfigurationsinformation (25,2)
longest without et Diskussionsvorschlag (20,1)
longest without eta Bildungskommission Zwillingsforschung (18,2)
longest without etai Fu"hrungszuordnung (17,1)
longest without etain Bohrvorschubs (13,1)
longest without etains Chlorophyll (11,1)
Letter Choices
Vowels
longest all vowels ?
longest each vowel once Schlichtungsvorschla"gen Schlichtungsvorschlages (23,2)
longest each extended vowel once Hydraulikkolbens (16,1)
shortest each vowel once Autopsie (8,1)
shortest each extended vowel once Olympiafu"nfte (13,1)
shortest vowels in order Arbeitsordnung (14,1)
shortest extended vowels in order ?
longest vowels in order Arbeitsordnung (14,1)
longest extended vowels in order ?
shortest vowels in reverse order ?
shortest extended vowels in reverse order ?
longest vowels in reverse order ?
longest extended vowels in reverse order ?
longest one vowel schluchzst schnarchst schrumpfst (10,3)
longest two vowels Schlachtschiff (14,1)
longest containing a univocalic Landtagswahlkampf (17,1)
longest containing e univocalic Geschlechtsverkehrs erstrebenswertesten erstrebenswertester erstrebenswertestes schreckenerregendem schreckenerregenden schreckenerregendes zweckentfremdeteren zweckentfremdeterer zweckentfremdeteres ... (19,16)
longest containing i univocalic pflichtwidrig spiritistisch zinspflichtig (13,3)
longest containing o univocalic Mordkomplott Vollkornbrot (12,2)
longest containing u univocalic Buchdruckkunst Flugstu"tzpunkt Luftstu"tzpunkt (14,3)
longest containing y univocalic Lynch (5,1)
longest alternating extended vowel-consonant Analysegera"te Pilotenexamen Relativita"ten Solitarita"ten Totalisatoren dazugeho"rigen dazugeho"riger dazugeho"riges hexadezimalem hexadezimalen ... (13,20)
longest alternating vowel-consonant Pilotenexamen Relativita"ten Solitarita"ten Totalisatoren dazugeho"rigen dazugeho"riger dazugeho"riges hexadezimalem hexadezimalen karikativerem ... (13,19)
Consonants
longest consonant string Angstschweis" Angstschweis"es Angstschwelle Gerichtsschreiber Gerichtsschreibern Geschichtsschreiber Geschichtsschreibers Geschichtsschreibung Glu"ckwunschschreiben (8,9)
longest one consonant Unionen rarerer tutetet (7,3)
longest two consonant eingegangenen (13,1)
Isograms
longest isogram Bildungsprojekt Machtverfilzung (15,2)
longest pair isogram Parita"tspreise (14,1)
longest trio isogram ?
longest tetrad isogram ?
longest polygram proportionierteren (18,1)
longest pyramid beendenden besessenen ero"rterter erretteten errettetes gegessenes geretteter gesessenen (10,8)
most repeated letters Wirtschaftspru"fungsgesellschaften (12,1)
highest containing a repeated A"quatorialafrika Antiquariatskatalog Antiquariatskataloge Apparateglasbla"ser Bausparkassenkatastrophe Fra"sparametervariation Fus"ballnationalmannschaft Generalstaatsanwalt Haushaltungsapparate Panamakanal ... (5,16)
highest containing b repeated Betriebsbeschreibung Bibelbearbeitungen (4,2)
highest containing c repeated Hochdruckschlauch Schnickschnacklose Schreckensnachrichten gleichgeschlechtlicher (4,4)
highest containing d repeated Mindestdividende (4,1)
highest containing e repeated Gewerbesteuerbescheid Herzensangelegenheiten Unternehmersteuervorteile aufsehenerregendere aufsehenerregenderem aufsehenerregenderen aufsehenerregenderes bemerkenswerterweise ekelerregenderem ekelerregenderen ... (7,25)
highest containing f repeated Angriffswaffen Briefaufschrifft Elfmeterpfiff Frachtschiffsflotte Kaffeelo"ffel Kaffeelo"ffeln Koffergriff Luftwaffenchef Offensivangriff Schiffsfrachtbrief ... (4,12)
highest containing g repeated entgegengegangene entgegengegangenen entgegengegangener (5,3)
highest containing h repeated Ho"chstwahrscheinlich ho"chstwahrscheinlich (5,2)
highest containing i repeated Initialisierungsbit ptimistischeroptimistic (6,2)
highest containing j repeated Jesaja Jubila"umsjahr Jungja"ger (2,3)
highest containing k repeated Wolkenkuckucksla"nder Wolkenkuckucksland Wolkenkuckuckslands (4,3)
highest containing l repeated Halbintellektueller Parallelfall Ultraschallwellen Vollwellengleichrichtung (5,4)
highest containing m repeated zusammenbekommenem zusammengekommenem zusammengenommenem (5,3)
highest containing n repeated Entspannungstendenzen (7,1)
highest containing o repeated Koalitionsprotokoll Monopolorganisation (5,2)
highest containing p repeated Papierpuppen (5,1)
highest containing q repeated Liquidita"tsquellen (2,1)
highest containing r repeated Frequenzprogrammierbarer (6,1)
highest containing s repeated Bundestagsausschusses Mes"werterfassungssystems Schlichtungsausschusses Wissenschaftsausschus" (7,4)
highest containing t repeated Bestattungsinstitut Bestattungsinstitute Bestattungsinstituten Frontplattentastatur Schnittstellenattribute (6,5)
highest containing u repeated Grundstu"cksausnutzung Lusuxausfu"hrung U"berbru"ckungsru"ckzahlung Untersuchungsausschus" (5,4)
highest containing v repeated Alternativvorschlag Altkonservativer Bundesvermo"gensverwaltung Devalvation Devalvierung Devisenvergehen Direktversicherungsvertra"ge Dividendenverzicht Effektivverzinsung Exklusivvertrag ... (2,338)
highest containing w repeated Baumwollweis"waren (3,1)
highest containing x repeated Artaxerxes Exportpraxis Pixelmatrix (2,3)
highest containing y repeated Hydrauliksystemen Hydraulikzylinder Hydraulikzylindern Hydraulikzylinders Lymphozyten Mylady Polystyrol Psychoanalyse Psychoanalytiker Psychoanalytikern ... (2,24)
highest containing z repeated Einzelhandelsumsatzzahlen Einzelzimmerzuschlag Jazzszene Lizenzzahlungen hinzuzuziehen zuru"ckzuzahlen zuru"ckzuziehen (3,7)
most different letters Zylinderkopfschraube (18,1)
highest ratio length/letters entgegengegangenen entgegengetretenen (360,2)
highest ratio length/letters (no tautonyms) entgegengegangenen entgegengetretenen (360,2)
lowest length 16 ratio length/letters Abschottzylinder Computerhandling Dschungelka"mpfer Maiglo"ckchenduft Mikroschaltungen Schwarzgeldkonto Vierpolschaltung Vorschubzylinder Zwischenprodukte aufdringlichstem ... (106,12)
lowest length 17 ratio length/letters Bandstahlwerkzeug Denkmalschutzjahr Dschungelka"mpfern Entwicklungsfirma Gewerkschaftsbund Grundsa"tzlichkeit Kaufverpflichtung Mietwohnungsblock Parkverbotsschild Pflichtversa"umnis ... (113,23)
lowest length 18 ratio length/letters Bevo"lkerungsdichte Druckkopfbelastung Flu"chtlingsproblem Publikumsnachfrage Vo"lkergemeinschaft vormundschaftliche zwangsverpflichtet (112,7)
lowest length 19 ratio length/letters Bevo"lkerungszuwachs Bundessozialgericht Bundestagswahlkampf Datenblocksicherung Flu"chtlingsproblems Softwarebezeichnung vormundschaftlichen vormundschaftlicher vormundschaftliches (118,9)
lowest length 20 ratio length/letters Zylinderkopfschraube (111,1)
lowest length 21 ratio length/letters Bevo"lkerungswachstums (123,1)
lowest length 22 ratio length/letters Profilschneidautomatik Stempeldruckabha"ngiges (129,2)
lowest length 23 ratio length/letters Bekleidungsvorschriften Gemeinschaftsproduktion (135,2)
lowest length 24 ratio length/letters Verwendungsmo"glichkeiten Wirtschaftspublikationen (141,2)
lowest length 25 ratio length/letters Gemeinschaftsproduktionen Handlungsbevollma"chtigter U"berwachungsmo"glichkeiten Vera"nderungsmo"glichkeiten (147,4)
Letter Appearance
longest short letters Wasserressourcen (16,1)
longest tall letters billigt (7,1)
longest vertical-symmetry letters Atomium Automat (7,2)
longest horizontal-symmetry letters bedecke (7,1)
longest full-symmetry letters hoi oho (3,2)
highest ratio of dotted letters Bikini (50,1)
Typewriter
longest top row erweiterter erzitterter erzittertet irritierter irritiertet quittierter repetierter repetiertet rezitiertet zerru"tteter (11,10)
longest middle row Allahs (6,1)
longest bottom row ?
longest in order Wertzoll (8,1)
longest in reverse order Blatte Kapuze kaputt (6,3)
longest left hand Staatsvertrages Vertragsstaates wetterfestester (15,3)
longest right hand Poliklinik (10,1)
longest alternating hands vorzeichenrichtig (17,1)
longest one finger Mu"nz Mumm (4,2)
longest adjacent keys Fresser wassert (7,2)
Puzzle
longest palindrome in Morse code Retter (6,1)
longest formed with chemical symbols Knotenrechnerapplikation Kontingentbeschra"nkungen Pressevero"ffentlichungen (24,3)
longest formed with US postal codes Maschine Matthias (8,2)
longest formed with amino acid abbreviations Pro arg pro (3,3)
longest formed with piano notes Abgabe Bagdad (6,2)
Letter Order
Alphabetical
longest letters in order Chintz Chlors Delors ablo"st dehnst filmst (6,6)
longest letters in order with repeats Beginnt beehrst beeilst beginnt beirrst (7,5)
longest letters in reverse order wolkige (7,1)
longest letters in reverse order with repeats sonnige trommle vo"llige wolkige wollige wonnige (7,6)
longest roller-coaster paramilita"rischen verfu"hrerischeren verfu"hrerischerer verfu"hrerischeres zivilisatorischem zivilisatorischen zivilisatorischer (17,7)
longest no letters in place Geschwindigkeitsu"berschreitungen Selbstverwirklichungsmo"glichkeit (32,2)
most letters in place Geldentwicklung Goldentwicklung (6,2)
most letters in place shifted Entwicklungskosten Entwicklungskostenbeteiligung Entwicklungssystem Entwicklungssysteme Softwareentwicklungssystem (7,5)
most consecutive letters in order consecutively Anwendungsunterstu"tzung Applikationsunterstu"tzung Arbeitslosenunterstu"tzung Arbeitslosenunterstu"tzungen Architekturstudenten Bierstuben Bildersturm Bravourstu"ck Bravourstu"ckchen Bu"cherstube ... (4,170)
most consecutive letters in order Abschirmdienst Abschlus"widerstand Abschottzylinder Alarmknopf Berufsgeheimnis Blumentopf Bundesverfassungsgericht Christusvisionen Datenerfassungsrechners Elefantengeda"chtnis ... (5,41)
most consecutive letters Fallschirmja"gerbrigade (10,1)
highest ratio of consecutive letters to length fa"hige (83,1)
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> logic/smullyan/priest.p <==
In a small town there are N married couples in which one of the pair
has committed adultery. Each adulterer has succeeded in keeping their
dalliance a secret from their spouse. Since it is a small town,
everyone knows about everyone else's infidelity. In other words, each
spouse of an adulterer thinks there are N - 1 adulterers, but everyone
else thinks there are N adulterers.
People of this town have a tradition of denouncing their spouse in
church if they are guilty of adultery. So far, of course, no one has
been denounced. In addition, people of this town are all amateur
logicians of sorts, and can be expected to figure out the implications
of anything they know.
A priest has heard the confession of all the people in the town, and is
troubled by the state of moral turpitude. He cannot break the
confessional, but knowing of his flock's logical turn of mind, he hits
upon a plan to do God's work. He announces in Mass one Sunday that
there is adultery in the town.
Is the priest correct? Will this result in every adulterer being denounced?
==> logic/smullyan/priest.s <==
Yes. Let's start with the simple case that N = 1. The offended spouse
reasons as follows: the priest knows there is at least one adulterer,
but I don't know who this person is, and I would if it were anyone
other than me, so it must be me. What happens if N = 2? On the first
Sunday, the two offended spouses each calmly wait for the other to get
up and condemn their spouses. When the other doesn't stand, they
think: They do not think that they are a victim. But if they do not
think they are victims, then they must think there are no adulterers,
contrary to what the priest said. But everyone knows the priest speaks
with the authority of God, so it is unthinkable that he is mistaken.
The only remaining possibility is that they think there WAS another
adulterer, and the only possibility is: MY SPOUSE! So, they know that
they too must be victims. So on the next Sunday, they will get up.
What if N = 3? On the first Sunday, each victim believes that the other
two will not get up because they did not know about the other person
(in other words, they believe that each of the two other victims thought
there was only one adulterer). However, each victim reasons, the two
will now realize that there must be two victims, for the reasons given
under the N = 2 case above. So they will get up next Sunday. This
excuse lasts until the next Sunday, when still no one gets up, and now
each victim realizes that either the priest was mistaken (unthinkable!)
or there are really three victims, and I am ONE! So, on the third
Sunday, all three get up. This reasoning can be repeated inductively
to show that no one will do anything (except use up N - 1 excuses as to
why no one got up) until the Nth Sunday, when all N victims will arise
in unison.
The induction can also be run "top down" on the priest's statement. What
must everyone believe about what everyone else believes?
N victims of adultery believe there are N - 1 victims, and that all of these
N - 1 victims believe that there are N - 2 victims, and that all of these
N - 2 victims believe that there are N - 3 victims, and that all of these
...
2 victims believe that there is 1 victim, and that this
1 victim believes there are no victims.
Suppose the priest says, "There are N adulterers in this town." Then
all the adulterers will know immediately that their spouses have been
unfaithful, and will denounce them the next Sunday. Now, suppose the
priest says, "There are at least N - 1 adulterers in this town." On
the first Sunday, the offended spouses all wait for each other to stand
up. When none do, they all know that they have all been horribly
mistaken, and they stand up on the following Sunday. But what if the
priest says, "There are at least N - 2 adulterers in this town." On
the first Sunday, the victims reason, those N - 1 victims are going to
be surprised when no one stands up, and they'll stand up next Sunday.
On the second Sunday, the victims reason, wait a minute, since they
didn't stand up, I must be one of the deluded victims. And everyone
stands up on the third Sunday. This resoning applies inductively,
adding one Sunday at each step, until the priest's original statement
is reached, which takes N Sundays for everyone to unravel.
By the way, the rest of the town, which thinks there are N adulterers,
is about to conclude that their perfectly innocent spouses have been
unfaithful too. This includes the adulterous spouses, who are about to
conclude that the door swings both ways. So the priest is playing a
dangerous game. A movie plot in there somewhere?
This problem is an analogue of the "dirty children" problem discussed in
the seminal paper on common knowledge by Halpern and Moses (JACM 1990).
If the information of each victim is less than perfect, the problem is
related to the "frame" problem of AI, cf. J. M. McCarthy & P. J. Hayes,
"Some philosophical problems from the standpoint of artificial intelligence"
(1968) in _Readings in Artificial Intelligence_ (pp. 431-450),
Tioga Publishing Co., Palo Alto, CA (1981).
==> logic/smullyan/stamps.p <==
The moderator takes a set of 8 stamps, 4 red and 4 green, known to the
logicians, and loosely affixes two to the forehead of each logician so that
each logician can see all the other stamps except those 2 in the moderator's
pocket and the two on her own head. He asks them in turn
if they know the colors of their own stamps:
A: "No"
B: "No"
C: "No"
A: "No
B: "Yes"
What are the colors of her stamps, and what is the situation?
==> logic/smullyan/stamps.s <==
B says: "Suppose I have red-red. A would have said on her
second turn: 'I see that B has red-red. If I also have red-red, then all
four reds would be used, and C would have realized that she had green-green.
But C didn't, so I don't have red-red. Suppose I have green-green. In that
case, C would have realized that if she had red-red, I would have seen
four reds and I would have answered that I had green-green on my first
turn. On the other hand, if she also has green-green [we assume that
A can see C; this line is only for completeness], then B would have seen
four greens and she would have answered that she had two reds. So C would
have realized that, if I have green-green and B has red-red, and if
neither of us answered on our first turn, then she must have green-red.
"'But she didn't. So I can't have green-green either, and if I can't have
green-green or red-red, then I must have green-red.'
So B continues: "But she (A) didn't say that she had green-red, so
the supposition that I have red-red must be wrong. And as my logic applies
to green-green as well, then I must have green-red."
So B had green-red, and we don't know the distribution of the others
certainly.
(Actually, it is possible to take the last step first, and deduce
that the person who answered YES must have a solution which would work
if the greens and reds were switched -- red-green.)
==> logic/supertasks.p <==
You have an empty urn, and an infinite number of labeled balls. Each
has a number written on it corresponding to when it will go in. At a
minute to the hour, you take the first ten balls and put them in the
urn, and remove the last ball. At the next half interval, you put in
the next ten balls, and remove ball number 20. At the next half
interval, you put in ten more balls and remove ball 30. This continues
for the whole minute.... how many balls are in the urn at this point?
(infinite)
You have the same urn, and the same set of balls. This time, you put
in 10 balls and remove ball number 1. Then you put in another ten
balls and remove ball number 2. Then you put in another ten balls and
remove ball number 3. After the minute is over, how many balls are
left in the urn now? (zero)
Are the above answers correct, and why or why not?
==> logic/supertasks.s <==
Almost all people will intuitively feel that the first experiment
(where only balls labeled with multiples of 10 are removed) results in
an urn with an infinite number of balls.
The real excitement starts with the experiment where balls are removed
in increasing order, but 10 times slower than they are added. Some
feel that the urn will not get empty, due to the slowness of removing.
Some others feel that the urn does get empty, since each ball is
removed at some time during the experiment. The remaining people claim
that the experiment is not well defined, that it is not possible to do
something an infinite number of times, or something similar,
effectively dismissing the experiment.
Just to put a bit of doubt in some peoples mind, I will add a third experment:
Let us suppose that at 1 minute to 12 p.m. balls nummbered 1 through 9
are placed in the urn, and instead of withdrawing a ball we add a zero
to the label of ball number 1 so that its label becomes 10. At 1/2 minute
to 12 p.m., balls numbered 11 through 19 are placed in the urn, and we
add a zero to the label of ball number 2 so that it becomes ball number 20.
At 1/4 minute to 12 p.m., balls numbered 21 through 29 are placed in the
urn and ball number 3 becomes ball number 30, and so on.
At each instant, instead of withdrawing the ball with the smallest label
we add a zero to its label so that its number is multiplied by 10.
How many balls are in the urn at 12 p.m. and what are their labels?
If we look at this experiment, at any point in time the inside of the
urn looks exactly like the inside during the execution of the original
paradoxical experiment. However, since no balls leave the urn, it is
now impossible to conclude that the urn will be empty at 12 p.m.
Still, there is no natural number that is the label of any ball in the
urn. Instead, each ball in the urn will have as its lable a natural
number followed by an infinite number of zero's.
A possible question is now: does this support that the outcome of the
original experiment where balls are removed in increasing order is that
there are an infinite number of balls in the urn? Possibly also with
'infinite natural numbers' as their labels, or are these experiments so
different that the answer is still a clear 'zero'?
I now come to the main points.
1. Our normal mathematical models do not cater for the COMPLETION of infinite
tasks (called super tasks by Thomson in 1954).
2. Since we intuitively feel that for many of these experiments there
are obvious outcomes, we would like to enhance our model to describe the
outcomes of these experiments.
3. In the enhancement of the model continuity should play an important role.
We include statement 3, since a model in which the conclusion of all
these experiments is that, at 12 p.m. the urn contains "exactly 7
balls, all red" is not desirable, nor useful.
It can be easily shown that general continuity is unattainable. For
instance the sentence "it is before midnight" is true during the
experiment, but is suddenly false after the experiment.
The people claiming that in the second experiment the urn will contain
an infinite number of balls, base this on the fact that the number of
balls in the urn during the experiment, is 9n at (1/2)^(n-1) minute
before 12. They thus assume that this statement is continuous. This
remains to be seen, however.
We have not come to a clear set of criteria which decide whether a
given statement is continuous with respect to performing supertasks. We
did define a "kinematical principle of continuity", which is roughly
formalised as:
If at some moment before 12 p.m. a ball comes to rest at a particular
position, which it does not leave till 12 p.m., then it is still at
that position at 12 p.m.
If we look at the three experiments mentioned, then we can see that in
each case we can come to a conclusion on the contents of the urn.
1. In the first experiment, with the 10-folds being removed, each ball
which number is a multiple of 10 comes to rest outside the urn (just
after being removed) and thus is outside the urn at 12 p.m. All other
balls come to rest inside the urn (just after being placed there), and
thus are inside the urn at 12 p.m. Therefore the urn contains an infinite
number of balls at 12 p.m.
2. In the second experiment, with the balls being removed in increasing order,
each balls comes to rest outside the urn. Thus all balls involved are not
in the urn. Thus the urn is empty.
3. In the third experiment, all balls come to rest inside the urn and thus the
urn contains an infinite number of balls. The labels of these balls are
naturall number followed by an infinite number of zero's (since each of the
numbers is not changed, and zero's once added remain at the label, we can
draw this conclusion).
The first and third experiment are rather straightforward, while the
second is paradoxical, but not inconsistent. Please note that is just
one way of extending our model to include super tasks. We have only
shown that for these experiments, in our model, we come to consistent
conclusions. It does not mean that there are no other models which lead
to different, but also, within that model, consistent solutions.
A final remark: while thinking about these matters, we have wondered
whether we could create a model in which the second experiment would
lead to an urn containing an infinite number of balls. A possibility is
assuming that if a position is continuously occupied by a ball,
although the occupant ball may be swapped every now and again for
another ball, that at 12 p.m. the position is occupied by a so-called
LIMIT BALL. For the second experiment we could than place balls 1, 10,
100 .. 2, 20, 200, .. each at its own spot in the urn. Each spot in
the urn, once occupied is than continuously occupied with a ball,
leading to limit balls.
This idea of continuity is stronger than the kinematic principle
suggested above, and we have not followed these ideas up enough to
decide whether this extended principle can be made consistent. If any
of the readers have feelings whether this can or cannot be done, I
would be interested to hear their arguments.
I conclude by stating that the result of the super task depends on how
our standard models are enlarged to include the execution of
supertasks. We have given one extension which leads to consistent
results for the supertasks suggested by Ross. Other models may lead to
different, but also consistent, conclusions.
Reference:
Victor Allis and Teunis Koetsier (1991).
On Some Paradoxes of the Infinite.
Brit. J. Phil. Sci. 42 pp. 187-194.
-- al...@cs.rulimburg.nl (Victor Allis)
I am interested in the origin of the puzzle. As far as I know in this
form the puzzle occurs for the first time in Littlewood's "Mathematical
Miscellanea", which is an amusing little booklet from the 1950s (it may
be even older). Littlewood does not discuss the puzzle. DOES ANYONE
KNOW OF EARLIER REFERENCES TO THIS PUZZLE? The puzzle also occurs in
S. Ross's "A first course in probability", New York and London, 1988,
without critical comment.
-- te...@cs.vu.nl (Teun Koetsier)
==> logic/timezone.p <==
Two people are talking long distance on the phone; one is in an East-
Coast state of the US, the other is in a West-Coast state of the US.
The first asks the other "What time is it?", hears the answer, and
says, "That's funny. It's the same time here!"
==> logic/timezone.s <==
One is in Eastern Oregon (in Mountain time), the other in
Western Florida (in Central time), and it's daylight-savings
changeover day at 1:30 AM.
==> logic/unexpected.p <==
Swedish civil defense authorities announced that a civil defense drill would
be held one day the following week, but the actual day would be a surprise.
However, we can prove by induction that the drill cannot be held. Clearly,
they cannot wait until Friday, since everyone will know it will be held that
day. But if it cannot be held on Friday, then by induction it cannot be held
on Thursday, Wednesday, or indeed on any day.
What is wrong with this proof?
==> logic/unexpected.s <==
This problem has generated a vast literature (see below). Several
solutions of the paradox have been proposed, but as with most paradoxes
there is no consensus on which solution is the "right" one.
The earliest writers (O'Connor, Cohen, Alexander) see the announcement as
simply a statement whose utterance refutes itself. If I tell you that I
will have a surprise birthday party for you and then tell you all the
details, including the exact time and place, then I destroy the surprise,
refuting my statement that the birthday will be a surprise.
Soon, however, it was noticed that the drill could occur (say on Wednesday),
and still be a surprise. Thus the announcement is vindicated instead of
being refuted. So a puzzle remains.
One school of thought (Scriven, Shaw, Medlin, Fitch, Windt) interprets
the announcement that the drill is unexpected as saying that the date
of the drill cannot be deduced in advanced. This begs the question,
deduced from which premises? Examination of the inductive argument
shows that one of the premises used is the announcement itself, and in
particular the fact that the drill is unexpected. Thus the word
"unexpected" is defined circularly. Shaw and Medlin claim that this
circularity is illegitimate and is the source of the paradox. Fitch
uses Godelian techniques to produce a fully rigorous self-referential
announcement, and shows that the resulting proposition is
self-contradictory. However, none of these authors explain how it can
be that this illegitimate or self-contradictory announcement
nevertheless appears to be vindicated when the drill occurs. In other
words, what they have shown is that under one interpretation of "surprise"
the announcement is faulty, but their interpretation does not capture the
intuition that the drill really is a surprise when it occurs and thus
they are open to the charge that they have not captured the essence of
the paradox.
Another school of thought (Quine, Kaplan and Montague, Binkley,
Harrison, Wright and Sudbury, McClelland, Chihara, Sorenson) interprets
"surprise" in terms of "knowing" instead of "deducing." Quine claims
that the victims of the drill cannot assert that on the eve of the last
day they will "know" that the drill will occur on the next day. This
blocks the inductive argument from the start, but Quine is not very
explicit in showing what exactly is wrong with our strong intuition
that everybody will "know" on the eve of the last day that the drill
will occur on the following day. Later writers formalize the paradox
using modal logic (a logic that attempts to represent propositions
about knowing and believing) and suggest that various axioms about
knowing are at fault, e.g., the axiom that if one knows something, then
one knows that one knows it (the "KK axiom"). Sorenson, however,
formulates three ingenious variations of the paradox that are
independent of these doubtful axioms, and suggests instead that the
problem is that the announcement involves a "blindspot": a statement
that is true but which cannot be known by certain individuals even if
they are presented with the statement. This idea was foreshadowed by
O'Beirne and Binkley. Unfortunately, a full discussion of how this
blocks the paradox is beyond the scope of this summary.
Finally, there are two other approaches that deserve mention. Cargile
interprets the paradox as a game between ideally rational agents and finds
fault with the notion that ideally rational agents will arrive at the same
conclusion independently of the situation they find themselves in. Olin
interprets the paradox as an issue about justified belief: on the eve of
the last day one cannot be justified in believing BOTH that the drill will
occur on the next day AND that the drill will be a surprise even if both
statements turn out to be true; hence the argument cannot proceed and the
drill can be a surprise even on the last day.
For those who wish to read some of the literature, good papers to start with
are Bennett-Cargile and both papers of Sorenson. All of these provide
overviews of previous work and point out some errors, and so it's helpful to
read them before reading the original papers. For further reading on the
"deducibility" side, Shaw, Medlin and Fitch are good representatives. Other
papers that are definitely worth reading are Quine, Binkley, and Olin.
D. O'Connor, "Pragmatic Paradoxes," Mind 57:358-9, 1948.
L. Cohen, "Mr. O'Connor's 'Pragmatic Paradoxes,'" Mind 59:85-7, 1950.
P. Alexander, "Pragmatic Paradoxes," Mind 59:536-8, 1950.
M. Scriven, "Paradoxical Announcements," Mind 60:403-7, 1951.
D. O'Connor, "Pragmatic Paradoxes and Fugitive Propositions," Mind 60:536-8,
1951
P. Weiss, "The Prediction Paradox," Mind 61:265ff, 1952.
W. Quine, "On A So-Called Paradox," Mind 62:65-7, 1953.
R. Shaw, "The Paradox of the Unexpected Examination," Mind 67:382-4, 1958.
A. Lyon, "The Prediction Paradox," Mind 68:510-7, 1959.
D. Kaplan and R. Montague, "A Paradox Regained," Notre Dame J Formal Logic
1:79-90, 1960.
G. Nerlich, "Unexpected Examinations and Unprovable Statements," Mind
70:503-13, 1961.
M. Gardner, "A New Prediction Paradox," Brit J Phil Sci 13:51, 1962.
K. Popper, "A Comment on the New Prediction Paradox," Brit J Phil Sci 13:51,
1962.
B. Medlin, "The Unexpected Examination," Am Phil Q 1:66-72, 1964.
F. Fitch, "A Goedelized Formulation of the Prediction Paradox," Am Phil Q
1:161-4, 1964.
R. Sharpe, "The Unexpected Examination," Mind 74:255, 1965.
J. Chapman & R. Butler, "On Quine's So-Called 'Paradox,'" Mind 74:424-5, 1965.
J. Bennett and J. Cargile, Reviews, J Symb Logic 30:101-3, 1965.
J. Schoenberg, "A Note on the Logical Fallacy in the Paradox of the
Unexpected Examination," Mind 75:125-7, 1966.
J. Wright, "The Surprise Exam: Prediction on the Last Day Uncertain," Mind
76:115-7, 1967.
J. Cargile, "The Surprise Test Paradox," J Phil 64:550-63, 1967.
R. Binkley, "The Surprise Examination in Modal Logic," J Phil 65:127-36,
1968.
C. Harrison, "The Unanticipated Examination in View of Kripke's Semantics
for Modal Logic," in Philosophical Logic, J. Davis et al (ed.), Dordrecht,
1969.
P. Windt, "The Liar in the Prediction Paradox," Am Phil Q 10:65-8, 1973.
A. Ayer, "On a Supposed Antinomy," Mind 82:125-6, 1973.
M. Edman, "The Prediction Paradox," Theoria 40:166-75, 1974.
J. McClelland & C. Chihara, "The Surprise Examination Paradox," J Phil Logic
4:71-89, 1975.
C. Wright and A. Sudbury, "The Paradox of the Unexpected Examination,"
Aust J Phil 55:41-58, 1977.
I. Kvart, "The Paradox of the Surprise Examination," Logique et Analyse
337-344, 1978.
R. Sorenson, "Recalcitrant Versions of the Prediction Paradox," Aust J Phil
69:355-62, 1982.
D. Olin, "The Prediction Paradox Resolved," Phil Stud 44:225-33, 1983.
R. Sorenson, "Conditional Blindspots and the Knowledge Squeeze: A Solution to
the Prediction Paradox," Aust J Phil 62:126-35, 1984.
C. Chihara, "Olin, Quine and the Surprise Examination," Phil Stud 47:191-9,
1985.
R. Kirkham, "The Two Paradoxes of the Unexpected Hanging," Phil Stud
49:19-26, 1986.
D. Olin, "The Prediction Paradox: Resolving Recalcitrant Variations," Aust J
Phil 64:181-9, 1986.
C. Janaway, "Knowing About Surprises: A Supposed Antinomy Revisited," Mind
98:391-410, 1989.
==> logic/verger.p <==
A very bright and sunny Day
The Priest did to the Verger say:
"Last Monday met I strangers three
None of which were known to Thee.
I ask'd Them of Their Age combin'd
which amounted twice to Thine!
A Riddle now will I give Thee:
Tell Me what Their Ages be!"
So the Verger ask'd the Priest:
"Give to Me a Clue at least!"
"Keep Thy Mind and Ears awake,
And see what Thou of this can make.
Their Ages multiplied make plenty,
Fifty and Ten Dozens Twenty."
The Verger had a sleepless Night
To try to get Their Ages right.
"I almost found the Answer right.
Please shed on it a little Light."
"A little Clue I give to Thee,
I'm older than all Strangers three."
After but a little While
The Verger answered with a Smile:
"Inside my Head has rung a Bell.
Now I know the answer well!"
Now, the question is:
How old is the PRIEST??
======
==> logic/verger.s <==
The puzzler tried to take the test;
Intriguing rhymes he wished to best.
But "Fifty and ten dozens twenty"
made his headache pound aplenty.
When he finally found some leisure,
He took to task this witty treasure.
"The product of the age must be
Twenty-Four Hundred Fifty!"
Knowing that, he took its primes,
permuted them as many times
as needed, til he found amounts
equal to, by all accounts,
twice the Verger's age, so that
He would have that next day's spat.
The reason for the lad's confusion
was due to multiple solution!
Hence he needed one more clue
to give the answer back to you!
Since only one could fit the bill,
and then confirm the priest's age still,
the eldest age of each solution
by one could differ, with no coercion. <=(Sorry)
Else, that last clue's revelation
would not have brought information!
With two, two, five, seven, and seven,
construct three ages, another set of seven.
Two sets of three yield sixty-four,
Examine them, yet one time more.
The eldest age of each would be
forty-nine, and then, fifty!
With lack of proper rhyme and meter,
I've tried to be the first completor
of this poem and a puzzle;
my poetry, you'd try to muzzle!
And lest you think my wit is thrifty,
The answer, of course, must be fifty!
If dispute, you wish to tender,
note my addresss, as the sender!
--
Kevin Nechodom <kne...@stacc.med.utah.edu>
==> logic/weighing/balance.p <==
You are given 12 identical-looking coins, one of which is counterfeit
and weighs slightly more or less (you don't know which) than the
others. You are given a balance scale which lets you put the same
number of coins on each side and observe which side (if either) is
heavier. How can you identify the counterfeit and tell whether it
is heavy or light, in 3 weighings?
More generally, you are given N coins, one of which is heavy or light.
==> logic/weighing/balance.s <==
Martin Gardner gave a neat solution to this problem.
Assume that you are allowed W weighings. Write down the 3^W possible
length W strings of the symbols '0', '1', and '2'. Eliminate the three
such strings that consist of only one symbol repeated W times.
For each string, find the first symbol that is different from the symbol
preceeding it. Consider that pair of symbols. If that pair is not 01,
12, or 20, cross out that string. In other words, we only allow strings
of the forms 0*01.*, 1*12.*, or 2*20.* ( using ed(1) regular expressions ).
You will have (3^W-3)/2 strings left. This is how many coins you can
handle in W weighings.
Perform W weighings as follows:
For weighing I, take all the coins that have a 0 in string
position I, and weigh them against all the coins that have
a 2 in string position I.
If the side with the 0's in position I goes down, write down
a 0. If the other side goes down, write down a 2. Otherwise,
write down a 1.
After the W weighings, you have written down an W symbol string. If
your string matches the string on one of the coins, then that is the
odd coin, and it is heavy. If none of them match, than change every
2 to a 0 in your string, and every 0 to a 2. You will then have a
string that matches one of the coins, and that coin is lighter than
the others.
Note that if you only have to identify the odd coin, but don't have to
determine if it is heavy or light, you can handle (3^W-3)/2+1 coins.
Label the extra coin with a string of all 1's, and use the above
method.
Note also that you can handle (3^W-3)/2+1 coins if you *do* have to
determine whether it is heavy or light, provided you have a single reference
coin available, which you know has the correct weight. You do this by
labelling the extra coin with a string of all 2s. This results in it being
placed on the same side of the scales each time, and in that side of the
scales having one more coin than the other each time. So you put the
reference coin on the other side of the scales to the "all 2s" coin on each
weighing.
Proving that this works is straightforward, once you notice that the
method of string construction makes sure that in each position, 1/3
of the strings have 0, 1/3 have 1, and 1/3 have 2, and that if a
string occurs, then the string obtained by replacing each 0 with a
2 and each 2 with a 0 does not occur.
If you already know the odd coin is heavy (or light), you can handle
3^W coins. Given W weighings, there can only be 3^W possible
combinations of balances, left pan heavy, and right pan heavy.
The algorithm in this case:
Divide the coins into three equal groups... A, B, and C. Weigh A
against B. If a pan sinks, it contains the heavy coin, otherwise, the
heavy coin is in group C. If your group size is 1, you've found the coin,
otherwise recurse on the group containing the heavy coin.
==> logic/weighing/box.p <==
You have ten boxes; each contains nine balls. The balls in one box
weigh 0.9 kg; the rest weigh 1.0 kg. You have one weighing on an
accurate scale to find the box containing the light balls. How do you
do it?
==> logic/weighing/box.s <==
Number the boxes 0-9. Take 0 balls from box 0, 1 ball from box 1, 2
balls from box 2, etc. Now weigh all those balls and follow this
table:
If odd box is Weight is
0 45 kg
1 44.9 kg
2 44.8 kg
3 44.7 kg
4 44.6 kg
5 44.5 kg
6 44.4 kg
7 44.3 kg
8 44.2 kg
9 44.1 kg
==> logic/weighing/find.median.p <==
What is the least number of pairwise comparisons needed to find the median of
2n+1 distinct real numbers?
==> logic/weighing/find.median.s <==
Consider median of three numbers {a,b,c}.
Let G[a,b]=max(a,b) and L[a,b]=min(a,b)
If we assume that median of {a,b,c} can be computed by two
comparisons, then the solution must be one of the following:
G[c,G[a,b]], G[c,L[a,b]], L[c,G[a,b]], L[c,L[a,b]].
However, it is easily seen that none of these provides
a solution. Therefore, we need more than two comparisons to
get median of three numbers.
Now, consider median of 5 numbers {a,b,c,d,e}.
There are two possible ways to start the computation.
Let Ci[a,b] denote the ith comparison between {a} and {b}.
First, we could start with C1[a,b] and C2[c,d].
Second, we could start with C2[a,C1[b,c]].
We ignore other trivialities such as C1[a,a] or C2[a,C1[a,b]].
In the first case, the next operation is to find
median of S={e,C1[a,b],C2[c,d]} which requires at least three
comparisons in addition to the two comparisons already performed.
So the total cost of the first approach is at least 5 comparisons.
However, if the median is not equal to {e} then we can always
choose C1 and C2 such that the median is not in S.
In the second case, the next step is to find median of S={C2[a,C1[b,c]],d,e}.
Again, the total cost of this approach is at least five comparisons.
If median is not equal to {d} or {e}, we can again always
choose C1 and C2 such that the median is not in S.
Other starting sets, such as {e,d,c,b,a}, can always be ordered
as {a,b,c,d,e}. This shows that the argument covers all possible cases.
Navid,
had...@sipi.usc.edu
==> logic/weighing/gummy.bears.p <==
Real gummy drop bears have a mass of 10 grams, while imitation gummy
drop bears have a mass of 9 grams. Spike has 7 cartons of gummy drop bears,
4 of which contain real gummy drop bears, the others imitation.
Using a scale only once and the minimum number of gummy drop bears, how
can Spike determine which cartons contain real gummy drop bears?
==> logic/weighing/gummy.bears.s <==
Spike uses 51 gummy drop bears: from the 7 boxes he takes respectively
0, 1, 2, 4, 7, 13, and 24 bears.
The notion is that each box of imitation bears will subtract its
number of bears from the total "ideal" weight of 510 grams (1 gram of
missing weight per bear), so Spike weighs the bears, subtracts the
result from 510 to obtain a number N, and finds the unique combination
of 3 numbers from the above list (since there are 3 "imitation" boxes)
that sum to N.
The trick is for the sums of all triples selected from the set S of
numbers of bears to be unique. To accomplish this, I put numbers into
S one at a time in ascending order, starting with the obvious choice,
0. (Why is this obvious? If I'd started with k > 0, then I could
have improved on the resulting solution by subtracting k from each
number) Each new number obviously had to be greater than any previous,
because otherwise sums are not unique, but also the sums it made when
paired with any previous number had to be distinct from all previous
pairs (otherwise when this pair is combined with a third number you
can't distinguish it from the other pair)--except for the last box,
where we can ignore this point. And most obviously all the new
triples had to be distinct from any old triples; it was easy to find
what the new triples were by adding the newest number to each old sum
of pairs.
Now, in case you're curious, the possible weight deficits and their
unique decompositions are:
3 = 0 + 1 + 2
5 = 0 + 1 + 4
6 = 0 + 2 + 4
7 = 1 + 2 + 4
8 = 0 + 1 + 7
9 = 0 + 2 + 7
10 = 1 + 2 + 7
11 = 0 + 4 + 7
12 = 1 + 4 + 7
13 = 2 + 4 + 7
14 = 0 + 1 + 13
15 = 0 + 2 + 13
16 = 1 + 2 + 13
17 = 0 + 4 + 13
18 = 1 + 4 + 13
19 = 2 + 4 + 13
20 = 0 + 7 + 13
21 = 1 + 7 + 13
22 = 2 + 7 + 13
24 = 4 + 7 + 13
25 = 0 + 1 + 24
26 = 0 + 2 + 24
27 = 1 + 2 + 24
28 = 0 + 4 + 24
29 = 1 + 4 + 24
30 = 2 + 4 + 24
31 = 0 + 7 + 24
32 = 1 + 7 + 24
33 = 2 + 7 + 24
35 = 4 + 7 + 24
37 = 0 + 13 + 24
38 = 1 + 13 + 24
39 = 2 + 13 + 24
41 = 4 + 13 + 24
44 = 7 + 13 + 24
Note that there had to be (7 choose 3) distinct values; they end up
ranging from 3 to 44 inclusive with 7 numbers missing: 4, 23, 34, 36,
40, 42, and 43.
-- David Karr (ka...@cs.cornell.edu)
==> logic/weighing/optimal.weights.p <==
What is the smallest set of weights that allow you to weigh on a
balance scale every integer number of kilograms up to some number N?
==> logic/weighing/optimal.weights.s <==
a) EQUATION
-----------
Obviously (I hate this word! :-) any weight Y that can be weighed
by X1, X2, ... Xn can be written as:
Y = A1*X1 + A2*X2 + .... + An*Xn
where Ai is equal to -1, or 0, or 1.
b) UPPER BOUND FOR Y(n)
-----------------------
Each Ai can have one of three values, so the total number of
combinations of values for A1,A2, ... An is 3^n. At least one of these
combinations gives Y=0 (A1=A2=...=An=0). Out of the remaining 3^n-1
combinations, some give a negative Y (for example A1=A2=...=An=-1).
and some a positive Y (and some might also give zero values, e.g. if
X1=X2, and A1=1, A2=-1).
Because of symmetry it's easy to see that the combinations that give Y>0
are at most half i.e. (3^n-1)/2. It is also possible that different
combinations might give the same value of Y, and it is also possible
that these values of Y are not successive.
However, to obtain an upper bound, lets assume the best case i.e.
all (3^n-1)/2 combinations give different, positive, and successive
values, so:
Y(n) <= (3^n-1)/2
c) AN OPTIMAL ALGORITHM FOR CHOOSING Xn
---------------------------------------
I will present an algorithm for choosing the weights X1,X2,...Xn.
Then we will prove that it is optimal.
For n=1, we choose X1=1, and of course Y(1) = 1.
Let's assume that we have already chosen n weights X1, X2 ... Xn,
and that we can weigh all Y where 1<=Y<=Y(n).
We are allowed to get an extra new weight Xn+1.
If we choose:
Xn+1 = 2*Y(n)+1
then we get
Y(n+1) = Y(n) + Xn+1 = 3*Y(n)+1
Proof:
for 1<= Y <= Y(n):
use the weights X1...Xn (ignoring the new one)
for Y(n)+1 <= Y < Xn+1 = 2*Y(n)+1
Put Xn+1 on one side of the scale, and on the other side put
the unknown weight, and a combination of X1...Xn so that
this combination weighs "Xn+1 - Y" (which is a number
in the range 0...Y(n), so *there exists* such a combination)
for 2*Y(n)+1 <= Y <= 3*Y(n)+1:
put the unknown weight on one side, and on the other side
put Xn+1, and combination of X1...Xn with a weight equal to
"Y - Xn+1" (which again is a number in the range 0...Y(n),
so there exists such a combination)
So, to summarize, if we use such an algorithm, we have:
X1 = 1;
Y(1) =1;
Xn+1 = 2*Y(n)+1
Y(n+1) = Y(n) + Xn+1 = 3*Y(n) + 1
It's easy to prove (e.g. by induction) that:
Y(n) = (3^n-1)/2
X(n) = 3^n
So, Y(n) is equal to the upper bound we found before, so this algorithm is
optimal, and the weights you must choose are powers of 3.
Spyros Potamianos
pota...@hpl.hp.com
==> logic/weighing/weighings.p <==
Some of the supervisors of Scandalvania's n mints are producing bogus coins.
It would be easy to determine which mints are producing bogus coins but,
alas, the only scale in the known world is located in Nastyville,
which isn't on very friendly terms with Scandalville. In fact, Nastyville's
king will only let you use the scale twice. Your job? You must determine
which of the n mints are producing the bogus coins using only two weighings
and the minimum number of coins (your king is rather parsimonious, to put it
nicely). This is a true scale, i.e. it will tell you the weight of whatever
you put on it. Good coins are known to have a weight of 1 ounce and it is
also known that all the bogus mints (if any) produce coins that are
light or heavy by the same amount.
Some examples: if n=1 then we only need 1 coin, if n=2 then clearly 2
coins suffice, one from each mint.
What are the solutions for n=3,4,5? What can be said for general n?
==> logic/weighing/weighings.s <==
Oh gracious and wise king, I have solved this problem by first
simplifying and then expanding. That is, consider the problem of
being allowed only a single weighing. Stop reading right now if you
want to think about it further.
There are three possible outcomes for each mint (light, OK, heavy)
which may be represented as (-1, 0, +1). Now, let each mint represent
one place in base 3. Thus, the first mint is the ones place, the
second the threes place, the third is the nines place and so on. The
number of coins from each mint must equal the place. That is, we'll
have 1 coin from mint 1, 3 from mint 2, 9 from mint 3, and, in
general, 3^(n-1) from mint n.
By weighing all coins at once, we will get a value between 1 + 3 + 9 +
... and -1 + -3 + -9 + ... In fact, we notice that that value will
be unique for any mint outcomes. Thus, for the one weighing problem,
we need
sum for i=1 to n (3^(i-1))
which evaluates to (3^n - 1)/2
I'm fairly satisfied that this is a minimum for a single weighing.
What does a second weighing give us? Well, we can divide the coins
into two groups and use the same method. That is, if we have 5 mints,
one weighing will be:
1 coin from mint 1 + 3 coins from mint 2 + 9 coins from mint 3
while the other weighing will be:
1 coin from mint 4 + 3 coins from mint 5
It's pretty plain that this gives us a total coinage of:
3^(n/2) - 1 for even n and, after some arithmetic agitation:
2 * 3^((n-1)/2) - 1 for odd n
I think the flaw in this solution is that we don't know ahead of time
the amount by which the coins are off weight. So if you weigh 1 coin
from mint 1 together with 3 coins from mint 2 and the result is heavy
by 3x units, you still don't know whether the bogus coins are from
mint 3 (heavy by x units) or from mint 1 (heavy by 3x units). Note
that we're not given the error amount, only the fact that is is equal
for all bogus coins.
Here is my partial solution:
After considering the above, it would seem that on each of the two
weighings we must include coins from all of the mints (except for the
special cases of small n). So let ai (a sub i) be the number of coins
from mint i on weighing 1 and bi be the number of coins from mint i on
weighing 2. Let the error in the bogus coins have a value x, and let
ci be a the counterfeit function: ci is 0 if mint i is good, 1
otherwise.
Then
Sum ai ci x = delta1 error on weighing 1
Sum bi ci x = delta2 error on weighing 2
Now the ratio of delta1 to delta2 will be rational regardless of the
value of x, since x will factor out; let's call this ratio p over q (p
and q relatively prime). We would like to choose { ai } and { bi }
such that for any set of mints J, which will be a subset of { 1 , 2 ,
... , n }, that
Sum aj ( = Sum ai ci ) is relatively prime to Sum bj.
If this is true then we can determine the error x; it will simply be
delta1/p, which is equal to delta2/q.
If the { ai } have been carefully chosen, we should be able to figure
out the bogus mints from one of the weighings, provided that
all subsets ( { { aj } over all J } ) have unique sums.
This was the strategy proposed above, where is was suggested
that ai = 3 ** (i-1) ; note that you can use base 2 instead
of base 3 since all the errors have the same sign.
Well, for the time being I'm stumped.
This agrees with the analysis I've been fighting with. I actually
came up with a pair of functions that "almost" works. So that the
rest of you can save some time (in case you think the way I did):
Weighing 1: 1 coin from each mint
Weighing 2: 2^(k-1) coins from mint k, for 1...k...n
(total 2^n - 1 coins)
Consider the n mints to be one-bit-each -- bit set -> mint makes bogus
coins. Then we can just state that we're trying to discover "K",
where K is a number whose bit pattern _just_ describes the bogosity of
each mint. OK - now, assuming we know 'x', and we only consider the
*difference* of the weighing from what it should be, for weighing 1,
the devaiation is just the Hamming weight of K -- that is the number
of 1-bits in it -- that is, the number of bogosifying mints. For
weighing 2, the deviation is just K! When the nth bit of K is set,
then that mint contributes just 2^n to the deviation, and so the total
deviation will just be K.
So that set me in search of a lemma: given H(x) is the hamming weight
of x, is f(x) = x / H(x) a 1-1 map integers into rationals? That is,
if x/H(x) = y/H(y) can we conclude that x = y?
The answer (weep) is NO. The lowest pair I could find are 402/603
(both give the ratio 100.5). Boy it sure looked like a good
conjecture for a while! Sigh.
There are two parts to the problem. First let us try to come up with a
solution to finding the answer in 2 weighings - then worry about using the
min. number of coins.
Solutions are for GENERAL n.
Let N = set of all mints, 1 to n. Card(N) = n.
Let P = set of all bogus mints. Let Card(P) = p.
Weighing I: Weigh n coins, 1 from each mint.
Since each "good" coins weighs one ounce, let delta1 be the error in weighing.
Since all bogus coins are identical, let delta1 be abs(error).
If x is the weight by which one bogus coin differs from a good coin,
delta1 = p * x.
Weighing II: The coins to be weighed are composed thusly.
Let a1 be the number of coins from mint 1, a2 # from mint2 .. and an from
mint n. All ai's are distinct integers.
Let A = Set of all ai's.
Let delta2 = (abs.) error in weighing 2 = x * k
where k is the number of coins that are bogus in weighing two.
Or more formally
k = sigma(ai)
(over all i in P)
Assuming p is not zero (from Weighing I - in that case go back and get beheaded
for giving the king BAAAAAD advice),
Let ratio = delta1/delta2 = p/k.
Let IR = delta2/delta1 = k/p = inverse-ratio (for later proof).
Let S(i) be the bag of all numbers generated by summing i distinct elements
from A. Clearly there will be nCi (that n comb. i) elements in S(i).
[A bag is a set that can have the same element occur more than once.]
So S(1) = A
and S(n) will have one element that is the sum of all the elements of A.
Let R(i) = {x : For-all y in S(i), x = i/y} expressed as p/q (no common
factors).
(R is a bag too).
Let R-A = Bag-Union(R(i) for 1>= i >=n). (can include same element twice)
Choose A, such that all elements of R-A are DISTINCT, i.e. Bag(R-A) = Set(R-A).
Let the sequence a1, a2, .. an, be an L-sequence if the above property is
true. Or more simply, A is in L.
**********************************************************************
CONJECTURE: The bogus mint problem is solved in two weighings if A is in L.
Sketchy proof: R(1) = all possible ratios (= delta1/delta2) when p=1.
R(i) = all possible ratio's when p=i.
Since all possible combinations of bogus mints are reflected in R, just match
the actual ratio with the generated table for n.
************************************************************************
A brief example. Say n=3. Skip to next line if you want.
Let A=(2,3,7).
p=1 possible ratios = 1/2 1/3 1/7
p=2 possible ratios = 2/5 2/9 1/5(2/10)
p=3 possible ratios = 1/4(3/12) (lots of blood in Scandalvania).
As all outcomes are distinct, and the actual ratio MUST be one of these,
match it to the answer, and start sharpening the axe.
Note that the minimum for n=3 is A=(0,1,3)
possible ratios are
p=1 infinity (delta2=0),1,1/3
p=2 2/1,2/3,1/2
p=3 3/4
************************************************************************
All those with the determination to get this far are saying OK, OK how do we
get A.
I propose a solution that will generate A, to give you the answer in two
weighings, but will not give you the optimal number of coins.
Let a1=0
For i>=2 >=n
ai = i*(a1 + a2 + ... + ai-1) + 1
*****************************************
* i-1 *
* ai = i* [Sigma(aj)] + 1 * ****Generator function G*****
* j=1 *
*****************************************
If A is L, all RATIO's are unique. Also all inverse-ratio's (1/ratio) are
unique. I will prove that all inverse-ratio's (or IR's) are unique.
Let A(k), be the series generated by the first k elements from eqn. G.(above)
************************************************************************
PROOF BY INDUCTION.
A(1) = {0} is in L.
A(2) = {0,1} is in L.
ASSUME A(k) = {0,1, ..., ak} is in L.
T.P.T. A(k+1) = {0,1, ..., ak, D) is in L where D is generated from G.
We know that all IR's(inverse ratio's) from A(k) are distinct.
Let K = set of all IR's of A(k).
Since A(k+1) contains A(k), all IR's of A(k) will also be IR's of A(K+1).
So for all P, such that (k+1) is not in P, we get a distinct IR.
So consider cases when (k+1) is in P.
p=1 (i.e. (k+1) = only bogus mint), IR = D
______________________________________________________________________
CONJECTURE: Highest IR for A(k) = max(K) = ak
Proof: Since max[A(k)] = ak,
for p'= 1, max IR = ak/1 = ak
for p'= 2, max IR (max sum of 2 ai's)/2
= (ak + ak-1)/2 < ak (as ak>ak-1).
for p'= i max IR sum of largest i elements of A(k)
--------------------------------
i
< i * ak/i = ak.
So max. IR for A(k) is ak.
______________________________________________________________________
D > ak
So for p=1 IR is distinct.
Let Xim be the IR formed by choosing i elements from A(k+1).
Note: We are choosing D and (i-1) elements from A(k).
m is just an index to denote each distinct combination of
(i-1) elemnts of A(i).
______________________________________________________________________
CONJECTURE : For p=j, all new IR's Xjm are limited to the range
D/(j-1) > Xjm > D/j.
Proof:
Xjm = (D + {j-1 elements of A(k)})/j
Clearly Xjm > D/j.
To show: max[Xjm] < D/(j-1)
Note: a1 + a2 .. + ak < D/(k+1)
max[Xjm] = (D + ak + ak-1 + ... + a(k-j+1))/j
< (D + D/(k+1))/j
= D (k+2)/(k+1)j
= [D/(j-1)] * alpha.
alpha = (j-1)/(j) * (k+2)/(k+1)
Since j <= k, (j-1)/j <= (k-1)/k < (k+1)/(k+2)
IMPLIES alpha < 1.
Conjecture proved.
______________________________________________________________________
CONJECTURE : For a given p, all newly generated IR's are distinct.
Proof by contradiction:
Assume this is not so.
Implies
(D + (p-1) elements of A(k))/p
= (D + some other (p-1) elements of A(k))/p
Implies SUM[(p-1) elements of A(k)] = SUM[ some other (p-1) elements of A(k)]
Implies SUM[(p-1) elements of A(k)]/(p-1)
= SUM[some other (p-1) elements]/(p-1)
Implies A(k) is NOT in L.
Contra.
Hence conjecture.
______________________________________________________________________
CONJECTURE: A(k+1) is in L.
Since all newly generated IR's are distinct from each other, and all newly generated IR's are greater than previous IR's, A(k+1) is in L.
==> logic/zoo.p <==
I took some nephews and nieces to the Zoo, and we halted at a cage marked
Tovus Slithius, male and female.
Beregovus Mimsius, male and female.
Rathus Momus, male and female.
Jabberwockius Vulgaris, male and female.
The eight animals were asleep in a row, and the children began to guess
which was which. "That one at the end is Mr Tove." "No, no! It's Mrs
Jabberwock," and so on. I suggested that they should each write down
the names in order from left to right, and offered a prize to the one
who got most names right.
As the four species were easily distinguished, no mistake would arise in
pairing the animals; naturally a child who identified one animal as Mr
Tove identified the other animal of the same species as Mrs Tove.
The keeper, who consented to judge the lists, scrutinised them carefully.
"Here's a queer thing. I take two of the lists, say, John's and Mary's.
The animal which John supposes to be the animal which Mary supposes to be
Mr Tove is the animal which Mary supposes to be the animal which John
supposes to be Mrs Tove. It is just the same for every pair of lists,
and for all four species.
"Curiouser and curiouser! Each boy supposes Mr Tove to be the animal
which he supposes to be Mr Tove; but each girl supposes Mr Tove to be
the animal which she supposes to be Mrs Tove. And similarly for the oth-
er animals. I mean, for instance, that the animal Mary calls Mr Tove
is really Mrs Rathe, but the animal she calls Mrs Rathe is really Mrs
Tove."
"It seems a little involved," I said, "but I suppose it is a remarkable
coincidence."
"Very remarkable," replied Mr Dodgson (whom I had supposed to be the
keeper) "and it could not have happened if you had brought any more
children."
How many nephews and nieces were there? Was the winner a boy or a girl?
And how many names did the winner get right? [by Sir Arthur Eddington]
==> logic/zoo.s <==
Given that there is at least one boy and one girl (John and Mary are
mentioned) then the answer is that there were 3 nephews and 2 nieces,
the winner was a boy who got 4 right.
Number the animals 1 through 8, such that the females are even and the
males are odd, with members of the same species consecutive; i.e. 1 is
Mr. Tove, 2 Mrs. Tove, etc.
Then each childs guesses can be represented by a permutation. I use
the standard notation of a permutation as a set of orbits. For
example: (1 3 5)(6 8) means 1 -> 3, 3 -> 5, 5 -> 1, 6 -> 8, 8 -> 6 and
2,4,7 are unchanged.
[1] Let P be any childs guesses. Then P(mate(i)) = mate(P(i)).
[2] If Q is another childs guesses, then [P,Q] = T, where [P,Q] is the
commutator of P and Q (P composed with Q composed with P inverse
composed with Q inverse) and T is the special permutation (1 2) (3 4)
(5 6) (7 8) that just swaps each animal with its spouse.
[3] If P represents a boy, then P*P = I (I use * for composition, and
I for the identity permutation: (1)(2)(3)(4)(5)(6)(7)(8)
[4] If P represents a girl, then P*P = T.
[1] and [4] together mean that all girl's guesses must be of the form:
(A B C D) (E F G H) where A and C are mates, as are B & D,
E & F G & H.
So without loss of generality let Mary = (1 3 2 4) (5 7 6 8)
Without to much effort we see that the only possibilities for other
girls "compatible" with Mary (I use compatible to mean the relation
expressed in [2]) are:
g1: (1 5 2 6) (3 8 4 7)
g2: (1 6 2 5) (3 7 4 8)
g3: (1 7 2 8) (3 5 4 6)
g4: (1 8 2 7) (3 6 4 5)
Note that g1 is incompatible with g2 and g3 is incompatible with g4.
Thus no 4 of Mary and g1-4 are mutually compatible. Thus there are at
most three girls: Mary, g1 and g3 (without loss of generality)
By [1] and [3], each boy must be represented as a product of
transpostions and/or singletons: e.g. (1 3) (2 4) (5) (6) (7) (8) or
(1) (2) (3 4) (5 8) (6 7).
Let J represent John's guesses and consider J(1).
If J(1) = 1, then J(2) = 2 (by [1]) using [2] and Mary J(3) = 4, J(4) =
3, and g1 & J => J(5) = 6, J(6) = 5, & g3 & J => J(8) = 7 J(7) = 8
i.e. J = (1)(2)(3 4)(5 6)(7 8). But the [J,Mary] <> T. In fact, we
can see that J must have no fixed points, J(i) <> i for all i, since
there is nothing special about i = 1.
If J(1) = 2, then we get from Mary that J(3) = 3. contradiction.
If J(1) = 3, then J(2) = 4, J(3) = 1, J(4) = 2 (from Mary) =>
J(5) = 7, J(6) = 8, J(7) = 5, J(8) = 6 => J = (1 3)(2 4)(5 7)(6 8)
(from g1)
But then J is incompatible with g3.
A similar analysis shows that J(1) cannot be 4,5,6,7 or 8; i.e. no J
can be compatible with all three girls. So without loss of generality,
throw away g3.
We have Mary = (1 3 2 4) (5 7 6 8)
g1 = (1 5 2 6) (3 8 4 7)
The following are the only possible boy guesses which are compatible
with
both of these:
B1: (1)(2)(3 4)(5 6)(7)(8)
B2: (1 2)(3)(4)(5)(6)(7 8)
B3: (1 3)(2 4)(5 7)(6 8)
B4: (1 4)(2 3)(5 8)(6 7)
B5: (1 5)(2 6)(3 8)(4 7)
B6: (1 6)(2 5)(3 7)(4 8)
Note that B1 & B2 are incombatible, as are B3 & B4, B5 & B6, so at most
three of them are mutually compatible. In fact, Mary, g1, B1, B3 and
B5 are all mutually compatible (as are all the other possibilities you
can get by choosing either B1 or B2, B3 or B4, B5 or B6. So if there
are 2 girls there can be 3 boys, but no more, and we have already
eliminated the case of 3 girls and 1 boy.
The only other possibility to consider is whether there can be 4 or
more boys and 1 girl. Suppose there are Mary and 4 boys. Each boy
must map 1 to a different digit or they would not be mutually
compatible. For example if b1 and b2 both map 1 to 3, then they both
map 3 to 1 (since a boy's map consists of transpositions), so both
b1*b2 and b2*b1 map 1 to 1. Furthermore, b1 and b2 cannot map 1 onto
spouses. For example, if b1(1) = a and b is the spouse of a, then
b1(2) = b. If b2(1) = b, then b2(2) = a. Then b1*b2(1) = b1(b) = 2
and b2*b1(1) = b2(a) = 2 (again using the fact that boys are all
transpostions). Thus the four boys must be:
B1: (1)(2)... or (1 2)....
B2: (1 3)... or (1 4) ...
B3: (1 5) ... or (1 6) ...
B4: (1 7) ... or (1 8) ...
Consider B4. The only permutation of the form (1 7)... which is
compatible
with Mary ( (1 3 2 4) (5 7 6 8) ) is:
(1 7)(2 8)(3 5)(4 6)
The only (1 8)... possibility is:
(1 8)(2 7)(3 6)(4 5)
Suppose B4 = (1 7)(2 8)(3 5)(4 6)
If B3 starts (1 5), it must be (1 5)(2 6)(3 8)(4 7) to be compatible
with B4. This is compatible with Mary also.
Assuming this and B2 starts with (1 3) we get B2 = (1 3)(2 4)(5 8)(6 7)
in order to be compatible with B4. But then B2*B3 and B3*B2 moth map 1
to 8. I.e. no B2 is mutually compatible with B3 & B4.
Similarly if B2 starts with (1 4) it must be (1 4)(2 3)(5 7)(6 8) to
work with B4, but this doesn't work with B3.
Likewise B3 starting with (1 6) leads to no possible B2 and the
identical reasoning eliminates B4 = (1 8)...
So no B4 is possible!
I.e at most 3 boys are mutually compatiblw with Mary, so 2 girls & 3
boys is optimal.
Thus:
Mary = (1 3 2 4) (5 7 6 8)
Sue = (1 5 2 6) (3 8 4 7)
John = (1)(2)(3 4)(5 6)(7)(8)
Bob = (1 3)(2 4)(5 7)(6 8)
Jim = (1 5)(2 6)(3 8)(4 7)
is one optimal solution, with the winner being John (4 right: 1 2 7 & 8)
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> pickover/pickover.01.p <==
Title: Cliff Puzzle 1: Can you beat the numbers game?
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please include your name,
address, affiliation, e-mail address. If you like, tell me a little bit
about yourself. You might also directly mail me a copy of your response
in addition to any responding you do in the newsgroup. I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover
* * *
At a recent trip to the Ontario Science Center in Toronto, Canada I came
across an interesting puzzle. The center is located minutes from
downtown Toronto and it's a vast playground of science with hundreds of
exhibits inviting you to touch, try, test, and titillate your curiosity.
The puzzle I saw there can be stated as follows. In the 10 boxes below,
write a 10-digit number. The digit in the first box indicates the total
number of zeros in the entire number. The box marked "1" indicates the
total number of 1's in the number. The box marked "2" indicates the
total number of 2's in the number, and so on. For example, the "3" in
the box labeled "0" would indicate that there must be exactly three 0's
in the 10-digit number.
-------------------------------
| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
| 3| | | | | | | | | |
-------------------------------
Stop And Think
1. Is there a solution to this problem? Are there many solutions to this
problem?
2. A more advanced an interesting problem is to continue to
generate a sequence in a recursive fashion such that each row becomes
the sequence for the previous. For example, start with the usual
0 through 9 digits in row 1:
Row 1: 0 1 2 3 4 5 6 7 8 9
Assume Row 2 is your solution to the puzzle. I've just inserted random
digits below so as not to give away the solution:
Row 1: 0 1 2 3 4 5 6 7 8 9 S(1)
Row 2: 9 3 2 3 3 1 6 7 8 9 S(2)
Row 3: S(3)
Row 2 is now the starting point, and your next job is to form row 3, row 4,
etc. using the same rules. In the previous example, a digit in the
first box would indicate how many 9's there are in the next 10-digit number,
and so forth.
Contest: I am looking for the longest sequence of numbers users can come
up with using these rules. Can you find a Row 2 or Row 3?
Is it even possible to generate a "row 2" or "row 3"?
==> pickover/pickover.01.s <==
1) 0 1 2 3 4 5 6 7 8 9
2) 6 2 1 0 0 0 1 0 0 0
3) 0 0 0 4 4 4 0 4 4 4
4) 6 6 6 0 0 0 6 0 0 0
5) 0 0 0 4 4 4 0 4 4 4
.
.
.
and so on, repeating rows 3 and 4.
I don't know yet whether there are multiple solutions, but
I'll keep looking.
Mark Hayes
Goddard Space Flight Center (GSFC) / Interferometrics, Inc.
m...@gemini.gsfc.nasa.gov
GSFC Code 926.9
Greenbelt, MD 20771
-------------------------
In article <1992Sep14.1...@watson.ibm.com>, you write:
|> The puzzle I saw there can be stated as follows. In the 10 boxes below,
|> write a 10-digit number. The digit in the first box indicates the total
|> number of zeros in the entire number. The box marked "1" indicates the
|> total number of 1's in the number. The box marked "2" indicates the
|> total number of 2's in the number, and so on. For example, the "3" in
|> the box labeled "0" would indicate that there must be exactly three 0's
|> in the 10-digit number.
|>
|> -------------------------------
|> | 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
|> | 3| | | | | | | | | |
|> -------------------------------
|>
|>
|> Stop And Think
|>
|> 1. Is there a solution to this problem? Are there many solutions to this
|> problem?
This is an old puzzle, but I'll solve it as if it was new because I
find your extension below to be interesting.
Since all possible digits must be "counted" once, the ten digits must
add up to 10. Consider the first digit (= the amount of zeroes used):
9: Impossible, since all the other digits would have to be zero.
8: Also impossible, since we must mark a 1 under the 8, and the other
digits then must be zeroes.
7: We must mark a 1 under the 7, and we have one more non-zero digit
to assign. We've used a 1, so we must put a non-zero digit under the 1.
However, if we put a 1 there, it's wrong because we've used two ones,
and if we put a two that's also wrong. So 7 zeroes doesn't work.
6: Begin as before, putting a 1 under the 6. Now we must mark under the 1,
but putting a 1 is wrong, so put a 2. Now we have one non-zero digit
left to assign, and marking a 1 under the two works. 6210001000 works.
5: Now we take a different approach to analyze this. If there are only
five zeroes, then there are five non-zeroes, one of which is the 5 we
marked under the zero. Obviously a 1 must be marked under the 5 and
zeroes under 6-9, so we have 5----10000, where the dashes contain one
zero and three other numbers, which must add up to four (since all
ten digits must add up to ten) - so we have two ones and a two. But then
the digits we have are described by 5310010000, which is not the set of
digits we have, so there is no solution. Similar proofs show that there
cannot be 4,3,2, or 1 zero.
0: Impossible, since you would have to use a zero to indicate you didn't have
a zero.
|> 2. A more advanced an interesting problem is to continue to
|> generate a sequence in a recursive fashion such that each row becomes
|> the sequence for the previous. For example, start with the usual
|> 0 through 9 digits in row 1:
|>
|> Row 1: 0 1 2 3 4 5 6 7 8 9
|>
|> Assume Row 2 is your solution to the puzzle. I've just inserted random
|> digits below so as not to give away the solution:
|>
|>
|> Row 1: 0 1 2 3 4 5 6 7 8 9 S(1)
|> Row 2: 9 3 2 3 3 1 6 7 8 9 S(2)
|> Row 3: S(3)
|>
|> Row 2 is now the starting point, and your next job is to form row 3, row 4,
|> etc. using the same rules. In the previous example, a digit in the
|> first box would indicate how many 9's there are in the next 10-digit number,
|> and so forth.
|>
|> Contest: I am looking for the longest sequence of numbers users can come
|> up with using these rules. Can you find a Row 2 or Row 3?
|> Is it even possible to generate a "row 2" or "row 3"?
Well, first off, our handy rule about all the digits adding up to ten no
longer applies. Let's see if we can find an answer:
Row 1: 0 1 2 3 4 5 6 7 8 9
Row 2: 6 2 1 0 0 0 1 0 0 0
Row 3: ?
All the same digits must be placed under all the zeroes in row 2, or some
of them would be wrong, and this digit cannot be larger than 4 since six
non-zeroes are used under the zeroes in row 2. So, consider the cases:
4: If we put 4's under all the zeroes, we must put zeroes everywhere else.
0004440444 works.
3: Now we must place one non-zero digit under either the 6 or the 2, since
there are two 1's that must stay alike. Putting any non-zero digit under
the 6 is wrong since there aren't any sixes, unless you put a 6 under
the 6, which is still wrong. Similarly no digit works under the two.
2: Now we must put a non-zero digit under the 2, since we already used 6 of
them. We must also have two zeroes, which can only go under the ones.
This gives us --02220222. However, we must put a non-zero under the 6,
and we can't put a one, since we must have zeroes under the ones. Any
number greater than one is wrong, because we don't have that many 6's.
1: OK, we start with ---111-111, and one of the -'s must be a zero. This
zero must go under the 2 or the 6, because the ones must be alike (and
we've already used some ones). Suppose we put 6's under the ones, and
don't use any more ones. Then we need a 2 under the 6, and we need
a one under the 2, which breaks what we did before. So, instead put
7's under the ones. Now we must put a 1 and a 0 in the other two spots,
but either arrangement is wrong. We can't put a higher number under the
ones because there aren't enough spaces left, so there is no solution
with 1 zero.
0: Self-contradiction, as in the original problem.
So now we have a unique third row. Can we make a fourth?
Row 1: 0 1 2 3 4 5 6 7 8 9
Row 2: 6 2 1 0 0 0 1 0 0 0
Row 3: 0 0 0 4 4 4 0 4 4 4
Now there can only be two different digits used in the next number. Consider
the possibilities:
No zero is used: We need to mark this by putting zeroes under the zeroes
Self-contradiction.
Some zeroes are used: They can't go under the zeroes, so put zeroes under
the fours. Now six zeroes are used, so put 6's under the zeroes.
6660006000 works.
The same logic used to find row four shows that row five must be 0004440444
again, and we get into an infinite cycle alternating between these two.
--
----w-w--------------Joseph De Vincent...@owlnet.rice.edu----------------
( ^ ) Disclaimer: My opinions do not represent those of Owlnet.
(O O) Owlnet: George R. Brown School of Engineering Educational Network.
v-v (Unauthorized use is prohibited.) (Being uwop-ap!sdn is allowed.)
Snail mail: Rice U., 6100 S. Main, Houston TX 77005.
-------------------------
In rec.puzzles you write:
>Title: Cliff Puzzle 1: Can you beat the numbers game?
>From: cl...@watson.ibm.com
[...]
>1. Is there a solution to this problem? Are there many solutions to this
>problem?
Yes. No.
>2. A more advanced an interesting problem is to continue to
>generate a sequence in a recursive fashion such that each row becomes
>the sequence for the previous. For example, start with the usual
>0 through 9 digits in row 1:
[...]
>Contest: I am looking for the longest sequence of numbers users can come
>up with using these rules. Can you find a Row 2 or Row 3?
>Is it even possible to generate a "row 2" or "row 3"?
My program produces the following output:
0123456789
6210001000
no solutions found
So I believe that the result for row 2 is unique and that there is no
result for row 3.
[ I am including the program at the end of this message just for your interest ]
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address. If you like, tell me a little bit
>about yourself. You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup. I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
The name, address etc should appear in my signature. As for myself,
I'm a PhD student due to finish much too shortly who likes solving
puzzles.
Pauli
Paul Dale | gr...@cs.uq.oz.au
Department of Computer Science | +61 7 365 2445
University of Queensland |
Australia, 4072 | Did you know that there are 41 two letter
| words containing the letter 'a'?
The program I used follows:
--------------------------------------8<------------------------------
#include <stdio.h>
#include <stdlib.h>
#define START(in) for(in=0;in<9;in++) { \
if(sum+in > 10) \
break; \
else \
sum = sum+in; \
counts[digits[in]]++;
#define STOP(in) counts[digits[in]]--; \
sum -= in; \
}
main() {
short counts[10];
short i, sum;
short i0,i1,i2,i3,i4,i5,i6,i7,i8,i9;
static short digits[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
short solns[10][100];
short solcnt=0;
printf("0123456789\n");
again:
for(i=0;i<10;i++) counts[i]=0;
sum = 0;
START(i0)
START(i1)
START(i2)
START(i3)
START(i4)
START(i5)
START(i6)
START(i7)
START(i8)
START(i9)
if(counts[0]==digits[i0] && counts[1]==digits[i1] && counts[2]==digits[i2] &&
counts[3]==digits[i3] && counts[4]==digits[i4] &&
counts[5]==digits[i5] && counts[6]==digits[i6] &&
counts[7]==digits[i7] && counts[8]==digits[i8] &&
counts[9]==digits[i9]) {
printf("%d%d%d%d%d%d%d%d%d%d\n", i0,i1,i2,i3,i4,i5,
i6,i7,i8,i9);
for(i=0;i<10;i++)
solns[0][solcnt] = i0;
solns[1][solcnt] = i1;
solns[2][solcnt] = i2;
solns[3][solcnt] = i3;
solns[4][solcnt] = i4;
solns[5][solcnt] = i5;
solns[6][solcnt] = i6;
solns[7][solcnt] = i7;
solns[8][solcnt] = i8;
solns[9][solcnt] = i9;
solcnt++;
}
STOP(i9)
STOP(i8)
STOP(i7)
STOP(i6)
STOP(i5)
STOP(i4)
STOP(i3)
STOP(i2)
STOP(i1)
STOP(i0)
if(solcnt == 0) {
printf("no solutions found\n");
} else if(solcnt == 1) {
for(i=0;i<10;i++)
digits[i] = solns[i][0];
solcnt = 0;
goto again;
} else
printf("multiple solutions found\n");
}
--------------------------------------8<------------------------------
-------------------------
In article <1992Sep14.1...@watson.ibm.com> you write:
>Title: Cliff Puzzle 1: Can you beat the numbers game?
>From: cl...@watson.ibm.com
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address. If you like, tell me a little bit
>about yourself. You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup. I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
>
>* * *
>At a recent trip to the Ontario Science Center in Toronto, Canada I came
>across an interesting puzzle. The center is located minutes from
>downtown Toronto and it's a vast playground of science with hundreds of
>exhibits inviting you to touch, try, test, and titillate your curiosity.
>The puzzle I saw there can be stated as follows. In the 10 boxes below,
>write a 10-digit number. The digit in the first box indicates the total
>number of zeros in the entire number. The box marked "1" indicates the
>total number of 1's in the number. The box marked "2" indicates the
>total number of 2's in the number, and so on. For example, the "3" in
>the box labeled "0" would indicate that there must be exactly three 0's
>in the 10-digit number.
>
>-------------------------------
>| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
>| 3| | | | | | | | | |
>-------------------------------
>
>
>Stop And Think
>
>1. Is there a solution to this problem? Are there many solutions to this
>problem?
A. Since there are ten digits in the number, the sum of the digits in the bottom
row must be 10.
B. If x appears under y there must be x appearences of y, hence x*y<10
So, the MAXIMUM that can appear under each number is:
---------------------
|0|1|2|3|4|5|6|7|8|9|
|9|9|4|3|2|1|1|1|1|1| max
---------------------
C. In fact, under the numbers 5..9 there can be AT MOST one non-zero (1) answer
since otherwise two numbers of the 5..9 veriaty would appear and violate rule A.
D. So there must be at least 4 zeros. If there were exactly 4 zeros, then the
numbers 1..4 will all have under them non-zeros (as the zeros are used up for
the 5..9 group). There is also at least one number that is 5 or greater. Well,
there is a 5 (or more), a 4 (under zero), a 1 (under the 5..9 category) and
something above zero under the other 1..4 digits for a total above 10. This
violates rule A.
E. So there must be at least 5 zeros. So a (exactly one) number that is at
least 5 has a 1 under it. (since under zero would appear a >=5 number).
F. Under 1 there must be at least 1 since the solution has at least one 1 (the
one under a 5..9 number). However it could not be exactly 1 as then there would
be 2 (or more) 1's in the solution.
G. If there were 3 or more ones, then they must be under 2..9 . But then there
would be a 5 (or more) under zero + a 3 (or more) under one + a 1 under three
(or more) other places for a total above 10.
H. So there must be at exactly 2 ones in the solution. And hence, at least 1
under two.
We can summerize:
---------------------
|0|1|2|3|4|5|6|7|8|9|
|5|2|1|0|0|----1----| min
|6|2|2|1|1|----1----| max
---------------------
where the maximum under each digit is 10 - SUM(minimum of all others)
I. Since no 3 or 4 is now possible, those two numbers must have a zero under
them.
J. So there are six zeros. Hence:
---------------------
|0|1|2|3|4|5|6|7|8|9|
|6|2|1|0|0|0|1|0|0|0| min
|6|2|2|0|0|0|1|0|0|0| max
---------------------
K. Notice that "min" is a solution, while "max" is not. Hence, "min is the
*ONLY* solution!
My name is Dan Shoham. This is the only fact about me I care to make public.
You are free to attribute it, but provide me a note when you do so.
sho...@ll.mit.edu
-------------------------
>From cl...@romulus.rutgers.edu (Chris Long) Tue Sep 15 06:08:45 1992
Path: igor.rutgers.edu!romulus.rutgers.edu!clong
From: cl...@romulus.rutgers.edu (Chris Long)
Newsgroups: rec.puzzles
Subject: Re: Puzzle 1 (SPOILER)
Message-ID: <Sep.15.06.08...@romulus.rutgers.edu>
Date: 15 Sep 92 10:08:45 GMT
References: <1992Sep14.1...@watson.ibm.com> <1992Sep15.0...@questrel.com>
Organization: Rutgers Univ., New Brunswick, N.J.
Lines: 62
In article <1992Sep15.0...@questrel.com>, Chris Cole writes:
Chris, don't forget to include my name on my solutions in the FAQ,
please. My old article should be replaced with the following in the
FAQ, anyway:
--Cut here--
Solution prepared by Chris Long.
Unfortunately, this isn't completely new, since I believe a similar
puzzle I posted and answered are in the FAQ. However, it *is* different
enough to be interesting.
In article <1992Mar3....@hls.com>, ra...@hls.com writes:
> Here's a small number puzzle :
> Generate numbers such that the each digit in the number specifies
> the number of the occurences of the position of the digit ( postions starting
> with 0 from the left ). Example
> The number 1210
...
My guess is only:
1210
21200
3211000
42101000
521001000
6210001000
No 1, 2, or 3 digit numbers are possible. Letting x_i be the ith
digit, starting with 0, we see that (1) x_0 + ... + x_n = n+1 and
(2) 0*x_0 + ... + n*x_n = n+1, where n+1 is the number of digits.
I'll first prove that x_0 > n-3 if n>4. Assume not, then this
implies that at least four of the x_i with i>0 are non-zero. But
then we would have \sum_i i*x_i >= 10 by (2), impossible unless n=9,
but it isn't possible in this case (51111100000 isn't valid).
Now I'll prove that x_0 < n-1. x_0 clearly can't equal n; assume
x_0 = n-1 ==> x_{n-1} = 1 by (2) if n>3. Now only one of the
remaining x_i may be non-zero, and we must have that x_0 + ... + x_n
= n+1, but since x_0 + x_{n-1} = n ==> the remaining x_i = 1 ==> by
(2) that x_2 = 1. But this can't be, since x_{n-1} = 1 ==> x_1>0.
Now assuming x_0 = n-2 we conclude that x_{n-2} = 1 by (2) if n>5
==> x_1 + ... + x_{n-3} + x_{n-1} + x_n = 2 and 1*x_1 + ... +
(n-3)*x_{n-3} + (n-1)*x_{n-1} + n*x_n = 3 ==> x_1=1 and x_2=1,
contradiction.
Case n>5:
We have that x_0 = n-3 and if n>=7 ==> x_{n-3}=1 ==> x_1=2 and
x_2=1 by (1) and (2). For the case n=6 we see that x_{n-3}=2
leads to an easy contradiction, and we get the same result. The
cases n=4,5 are easy enough to handle, and lead to the two solutions
above.
--
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
--
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
-------------------------
The number "2020" was left off my list by mistake ... sorry.
-Chris
-------------------------
> * * *
> At a recent trip to the Ontario Science Center in Toronto, Canada I came
> across an interesting puzzle. The center is located minutes from
> downtown Toronto and it's a vast playground of science with hundreds of
> exhibits inviting you to touch, try, test, and titillate your curiosity.
> The puzzle I saw there can be stated as follows. In the 10 boxes below,
> write a 10-digit number. The digit in the first box indicates the total
> number of zeros in the entire number. The box marked "1" indicates the
> total number of 1's in the number. The box marked "2" indicates the
> total number of 2's in the number, and so on. For example, the "3" in
> the box labeled "0" would indicate that there must be exactly three 0's
> in the 10-digit number.
>
> -------------------------------
> | 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
> | 3| | | | | | | | | |
> -------------------------------
>
>
> Stop And Think
>
> 1. Is there a solution to this problem? Are there many solutions to this
> problem?
>
[Second question and contest problem omitted]
Good puzzle! I am wondering though whether the second question (which
I have not tried to solve yet) is moe amenable to computer search.
It seems to me that there should not be so many cases to consider, so
that even exhaustive search should work.
So, here is my ten minutes work on the first question.
I think there is a unique solution which is: 6210001000.
Here is the reasoning.
Let the number be (in Tex notation)
d_0 d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9.
By definition
d_0 + d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7 + d_8 + d_9 = 10. (1)
Moreover, d_0 > 0, since d_0 = 0 contradicts itself.
Let d_0 = c for some integer 9 >= c >= 1.
If c = 9, then d_9 = 1, contradiction since d_1 should both be 0 and 1 then.
If 9 > c >= 1, we rewrite (1) removing all d_i s that are zeros
c + d_(i_1) + d_(i_2) + ... + d_(i_(9-c)) = 10
<=> d_(i_1) + d_(i_2) + ... + d_(i_(9-c)) = 10 -c (2)
where all the d_(i_j) >= 1, j=1,...,9-c (3)
(2) & (3) imply that the d_(i_j)s are 8-c 1s and one 2.
Since there exists ONE 2, then there exists at least one 1.
So the only digits in the number are 0, 1, 2, and c (if different than 1 and 2).
If c is either 1 or 2, we have 3 different digits in the number, which
implies d_1 <= 3, impossible since d_1 = 8 - c >= 6.
If c> 2, we have four different digits in the number, and in fact
d_0 = c, d_1 = 8-c, d_2 = 1, d_c = 1, which leaves us with 6 0s. QED
I hope I did not miss any other cases.
Do you plan to post answers or comments later?
Leonidas
--------------------------------------------------------------------------------
Leonidas Palios The Geometry Center
1300 South Second Str
pal...@geom.umn.edu Minneapolis, Minnesota 55454
-------------------------
-------------------------------
| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
-------------------------------
| 6| 2| 1| 0| 0| 0| 1| 0| 0| 0|
| 0| 0| 0| 4| 4| 4| 0| 4| 4| 4| <-
| 6| 6| 6| 0| 0| 0| 6| 0| 0| 0| |
| 0| 0| 0| 4| 4| 4| 0| 4| 4| 4| <-
.
.
.
I must be missing something in my understanding of your rules.
I found the second row by imagining that I'd need lots of zeros
and putting nine in the 0 column, then skipping back and forth
adjusting things. I had to put a tic in the 9 column, then
I had to put one in the 1 column, then I realized that had to
change that to a two since now there were two ones, and at the
same time another required tic in the 2 column balanced the
change of one to two in the 1 column, and then of course there
weren't nine zeros anymore, but there were still six and so by
changing the nine in the 1 column to a six, the one in the 9
column sould just migrate down to the 6 column. But it almost
seems like cheating to use fours in the second row when there
were none in the second row to necessitate this kind of adjusting.
*shrug* If this is right, the series is infinite, obviously.
Please let me know if I'm interpreting something wrong.
Thanks, and nice puzzle. :)
Grant Culbertson
gr...@minos.nmt.edu
dg...@sirius.nmt.edu
==> pickover/pickover.02.p <==
Title: Cliff Puzzle 2: Grid of the Gods
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please include your name,
address, affiliation, e-mail address. If you like, tell me a little bit
about yourself. You might also directly mail me a copy of your response
in addition to any responding you do in the newsgroup. I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover
* * *
Consider a grid of infinitesimal dots spaced 1 inch apart in a cube with
an edge equal in length to the diameter of the sun (4.5x10**9 feet).
For conceptual purposes, you can think of the dots as having unit
spacing, being precisely placed at 1.00000...., 2.00000....,
3.00000...., etc. Next choose one of the dots and draw a line through it
which extends from that dot to the edge of the huge cube in both
directions.
Stop And Think
1. What is the probability that your line will intersect another dot
in the fine grid of dots within the cube the size of the sun?
Would your answer be different if the cube were the size of the
solar system?
2. Could a computer program be written to simulate this process?
3. Answer the two questions above, but this time assume the line
to have some finite thickness, T.
==> pickover/pickover.02.s <==
-------------------------
In article <1992Sep14.1...@watson.ibm.com> you write:
>Title: Cliff Puzzle 2: Grid of the Gods
>From: cl...@watson.ibm.com
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address. If you like, tell me a little bit
>about yourself. You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup. I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
>
>* * *
>
>Consider a grid of infinitesimal dots spaced 1 inch apart in a cube with
>an edge equal in length to the diameter of the sun (4.5x10**9 feet).
>For conceptual purposes, you can think of the dots as having unit
>spacing, being precisely placed at 1.00000...., 2.00000....,
>3.00000...., etc. Next choose one of the dots and draw a line through it
>which extends from that dot to the edge of the huge cube in both
>directions.
>
>Stop And Think
>
>1. What is the probability that your line will intersect another dot
>in the fine grid of dots within the cube the size of the sun?
>Would your answer be different if the cube were the size of the
>solar system?
That depends on the manner the dot and the direction of the line were choosen.
If both process used uniform (or any other continous) distribution, then - of
course - the probability would be zero. If, for instance, the direction of
the line is always choosen to be parallel to one of the cube's edges, then the
probability is one.
>
>2. Could a computer program be written to simulate this process?
Not a meaningfull question. Simple minded programs could never simulate
infinitesimal points, but well thought out algorithm could express anything
that can be shown analytically.
>
>3. Answer the two questions above, but this time assume the line
>to have some finite thickness, T.
>
This is equivelent to making each dot of diameter T, and keeping the line thin.
For T> (1 inch / 4.5*10^9 ft) inches, the probability -> 1.
A simple minded computer program could simulate this.
Dan Shoham
sho...@ll.mit.edu
-------------------------
In article <1992Sep14.1...@watson.ibm.com> you write:
>1. What is the probability that your line will intersect another dot
>in the fine grid of dots within the cube the size of the sun?
About 50%, because I usually draw horizontal lines.
I.e., YOU DIDN'T GIVE THE DISTRIBUTION OF "lines".
cf the puzzle of "what is the probability that a randomly selected
chord of a circle is longer than the radius of that circle." The
answer depends on how you "randomly select."
_________________________________________________________
Matt Crawford cra...@fncent.fnal.gov Fermilab
==> pickover/pickover.03.p <==
Title: Cliff Puzzle 3: Too many 3's
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please include your name,
address, affiliation, e-mail address. If you like, tell me a little bit
about yourself. You might also directly mail me a copy of your response
in addition to any responding you do in the newsgroup. I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover
* * *
How many numbers have at least one digit -- a three?
In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
which contain the digit 3. This means that 1/10 or 10% of the numbers
have the number 1 in the first 10 numbers. In the first 100 numbers the
occurrence of numbers with at least one three seems to be growing. In
fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93,
30,31,32,34,35,36,37,38,39. This means that about 19% of the digits
contain the number 3 in the first 100 numbers.
We can make a table showing the percentage of numbers with
at least one 3-digit for the first N numbers.
N %
10 1
100 19
1000 27
10000 34
The percentages rapidly increase to 100% indicating that almost all of
the numbers have a 3 in them! In fact, a formula describing the
proportion of 3's can be written: 1-(9/10)**N. The proportion gets
very close to 1 as N increases.
Stop And Think
1. How can it be that almost all of the numbers have a 3 in them?
==> pickover/pickover.03.s <==
-------------------------
You wrote (in article <1992Sep14.1...@watson.ibm.com>):
>Title: Cliff Puzzle 3: Too many 3's
>1. How can it be that almost all of the numbers have a 3 in them?
Because as the numbers get larger, they contain more digits,
increasing the probability that one of the digits in them might be a
3. In fact, the probability that a 3 will _not_ appear in a very long
number is very low.
I like this puzzle. Simple, but it made me think for a moment.
A three in every number? Preposterous! ;)
As for the other information you requested from responders: You have
my name and email address now, I don't give out my home address unless
it's necessary, and what sort of 'affiliation' are you seeking --
religious, business, or what?
<< Brian >>
--
_/_/_/ Brian Kendig Macintosh Jedi Live never to be ashamed
_/_/ Starfleet Captain Oracle Employee if anything you do or say
_/ Intrepid Adventurer Saturn SL2 Owner is published around the world
bske...@netcom.com Wizard of Frobozz -- even if what is published
Princeton '92! BSE/CS Writer/Actor/Singer is not true.
-------------------------
In article <1992Sep14.1...@watson.ibm.com> you write:
>Title: Cliff Puzzle 3: Too many 3's
>From: cl...@watson.ibm.com
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address. If you like, tell me a little bit
>about yourself. You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup. I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
>
>* * *
>
>How many numbers have at least one digit -- a three?
>
>In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
>which contains the digit 3. This means that 1/10 or 10% of the numbers
>have the number 1 in the first 10 numbers. In the first 100 numbers the
>occurrence of numbers with at least one three seems to be growing. In
>fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93,
>30,31,32,34,35,36,37,38,39. This means that about 19% of the digits
>contain the number 3 in the first 100 numbers.
>
>We can make a table showing the percentage of numbers with
>at least one 3-digit for the first N numbers.
>N %
>10 1
>100 19
>1000 27
>10000 34
>
>The percentages rapidly increase to 100% indicating that almost all of
>the numbers have a 3 in them! In fact, a formula describing the
>proportion of 3's can be written: 1-(9/10)**N. The proportion gets
>very close to 1 as N increases.
>
>Stop And Think
>
>1. How can it be that almost all of the numbers have a 3 in them?
>
Thaddeus Crews, 509 Windsor Green Blvd., Goodlettsville, TN, 37072
Graduate Student (Ph.D.) @ Vanderbilt University, Computer Science
thad...@vuse.vanderbilt.edu
This problem seems a little bit simple to me, but I was never that
great at math problems so I am not betting the farm on this answer.
The percentages you show for # of the first N numbers with at least
one 3-digit is also true (about) for the # of the first N numbers
with at least one 4-digit, at least one 5-digit, etc...
Basically, as N increases, so does the number of digits in N, and
therefore so does the number of chances for the digit 3 to appear
(as well as all other digits). Given a number N with enough (?)
digits, there is a 100% chance of all digits 0-9 appearing in that
number (of course, 1.0E10000000000) does not have a 3 in it, but
if you take the next 1.0E10000000000 numbers the percent that has
a 3 will be (I suspect) 100%.
My proof is clearly weak, but the claim is this: as N increases,
the number of digits in N also increases. As N approaches
infinity, the number of digits in N approaches infinity (at a
slower rate, however). As the number of digits approaches infinity,
the likelyhood of any specific digit appearing at least once
approaches 100%.
I think the real question (to be answered by someone with a better
math training) would be "At what number N does the statistical
likelyhood become 100% of at least one 3-digit appearing in the
first N numbers."
Hope this helps....
--
-- Thad Crews (email thad...@vuse.vanderbilt.edu)
--------------------------------------------------------------------------
"Some people have a way with words, and some people, ... oh ... *not* have
a way, I suppose..." -- Steve Martin
-------------------------
Heh. As the numbers get larger, they have more digits. Assuming a random occu
various digits in the larger numbers (not unreasonable when n-> infinity) the pr
number NOT having a 3 is very low.
-john 'I know it's not a proof...' karakash-
-------------------------
>Title: Cliff Puzzle 3: Too many 3's
Seth Breidbart
PO Box 5157
New York, NY 10185
Morgan Stanley & Co.
>1. How can it be that almost all of the numbers have a 3 in them?
The probability that a random sequence of n digits does not contain a
3 is .9^n; as n->infinity, this probability -> 0. Since almost all
numbers have a lot of digits (there are only a finite number of
integers with <n digits, and infinitely many with >n), the limiting
probability is 0.
-------------------------
In article <1992Sep14.1...@watson.ibm.com> you write:
>Title: Cliff Puzzle 3: Too many 3's
>From: cl...@watson.ibm.com
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address. If you like, tell me a little bit
>about yourself. You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup. I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
>
>* * *
>
>How many numbers have at least one digit -- a three?
>
>In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
>which contains the digit 3. This means that 1/10 or 10% of the numbers
>have the number 1 in the first 10 numbers. In the first 100 numbers the
>occurrence of numbers with at least one three seems to be growing. In
>fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93,
>30,31,32,34,35,36,37,38,39. This means that about 19% of the digits
>contain the number 3 in the first 100 numbers.
>
>We can make a table showing the percentage of numbers with
>at least one 3-digit for the first N numbers.
>N %
>10 1
>100 19
>1000 27
>10000 34
>
>The percentages rapidly increase to 100% indicating that almost all of
>the numbers have a 3 in them! In fact, a formula describing the
>proportion of 3's can be written: 1-(9/10)**N. The proportion gets
>very close to 1 as N increases.
>
>Stop And Think
>
>1. How can it be that almost all of the numbers have a 3 in them?
>
No problem. In fact almost all very large numbers have all digits in them.
It is rather hard for a number with zillions of digits to avoid "3"s (or any
other digit).
It fact, the sequences "15", "172", and "666" (and any other finite sequence)
are also contained (in order) within almost all numbers.
Dan Shoham
sho...@ll.mit.edu
-------------------------
Before I forget:
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
--
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
-------------------------
>Title: Cliff Puzzle 3: Too many 3's
>From: cl...@watson.ibm.com
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address. If you like, tell me a little bit
>about yourself. You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup. I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
>* * *
>How many numbers have at least one digit -- a three?
>In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
>which contains the digit 3. This means that 1/10 or 10% of the numbers
>have the number 1 in the first 10 numbers. In the first 100 numbers the
>occurrence of numbers with at least one three seems to be growing. In
>fact there are 19 numbers: 3,13,23,33,43,53,63,73,83,93,
>30,31,32,34,35,36,37,38,39. This means that about 19% of the digits
>contain the number 3 in the first 100 numbers.
>
>We can make a table showing the percentage of numbers with
>at least one 3-digit for the first N numbers.
>N %
>10 1
>100 19
>1000 27
>10000 34
>The percentages rapidly increase to 100% indicating that almost all of
>the numbers have a 3 in them! In fact, a formula describing the
>proportion of 3's can be written: 1-(9/10)**N. The proportion gets
>very close to 1 as N increases.
>Stop And Think
>1. How can it be that almost all of the numbers have a 3 in them?
I'm not sure this is the answer you are looking for, but:
9 = 9
9*9 = 81
9*9*9 = 729
9*9*9*9 = 6561
etc.
The probability of having 3 as the digit in a one-digit number is 1/10.
" of not having 3 " is 9/10.
In a two-digit number, the prob. of NOT having 3 as the first digit or
the second digit, ie. not having 3 in the two-digit number, is simply
the product of (NOT having 3 in first digit) times (NOT having 3 in second):
(9/10)*(9/10) = 81/100
= 0.81
For a three-digit number: (9/10)*(9/10)*(9/10) = 729/1000
= 0.729
For an n-digit number: (9/10)**n = probability.
We can see that as "n" becomes larger and larger, the
probability of NOT having a three at all in the number
becomes smaller and smaller. Indeed, as "n" approaches
infinity, this probability approaches zero. In other
words, it is very rare for a large number NOT to have 3
as one of its digits. In fact, it is very rare for a
large number NOT to have any of the ten possible integers
represented at least once.
[Aside, N %
10 1 = (10 - 9)/1 times 100
100 19 = (100 - 81)/100 times 100
1000 27 = (1000 - 729)/1000 times 100
10000 34 = (10000 - 6561)/10000 times 100
etc. ]
Kumar
ku...@ug.cs.dal.ca
ps: I'll leave it as a small exercise to tie up the loose ends.
==> pickover/pickover.04.p <==
Title: Cliff Puzzle 4: Time in a Bottle
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please include your name,
address, affiliation, e-mail address. If you like, tell me a little bit
about yourself. PLEASE ALSO directly mail me a copy of your response
in addition to any responding you do in the newsgroup. I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover
* * *
Consider a chain of bottles (B) each connected to one another by a thin
tube. A marble is placed in bottle 1.
Each tube contains a one-way valve so marbles can only
go from left to right in the tubes which are symbolized with "-" marks:
1 2 3 4
B - B - B - B -
The tubes are thin so it takes
1 hour of constant random shaking to get the marble from B1 to B2.
Likewise for each bottle.
I have not fully described the bottle collection. Each bottle
has a backward 1-way tube to bottle 1. I've tried to diagram these
with "*" symbols. Each time the marble enters bottle B(N) it has
a 50% probability of going back to bottle 1 via these tubes.
****<********
* *
***<***** *
* * *
* * * * *
1 2 3 4
B - B - B - B -
Stop And Think
1. In how many hours will you expect to get the marble out of bottle 10
after placing the marble in bottle 1?
2. Is there a general formula for the amount of time
required to get the ball out of bottle N into bottle N+1 given
a probability P of backwards motion (given as 50% in this problem)?
3. In how many hours will you expect to get the marble out of bottle 10
after placing the marble in bottle 1 given two backward tubes for each
bottle instead of one backward tube?
==> pickover/pickover.04.s <==
-------------------------
Subject: Re: Cliff Puzzle 4 (SPOILER)
Newsgroups: rec.puzzles
References: <1992Sep15....@watson.ibm.com>
In article <1992Sep15....@watson.ibm.com>, Cliff writes:
> 1. In how many hours will you expect to get the marble out of bottle 10
> after placing the marble in bottle 1?
The expected amount of time to go from state n-1 to n (state 11 is an
absorbing state) is
E(n-1,n) = 1 + E(1,n)/2 for 1<n<11;
also
E(n-1,n+1) = E(n-1,n) + E(n,n+1) for 1<n<11.
If n=2 then E(1,2) = 1 + E(1,2)/2 ==> E(1,2) = 2 (it should be clear
that no E is infinite for this problem).
E(2,3) = 1 + E(1,3)/2 = 1 + E(1,2)/2 + E(2,3)/2 ==> E(2,3)/2 = 2
==> E(1,3) = 6.
I claim that in general E(1,n) = 2^n - 2 and E(n-1,n) = 2^(n-1).
Assume true for n, then E(n,n+1) = 1 + E(1,n+1)/2 = 1 + E(1,n)/2 +
E(n,n+1)/2 ==> E(n,n+1)/2 = 1 + E(1,n)/2 ==> E(n,n+1) = 2 + E(1,n)
==> E(n,n+1) = 2 + 2^n - 2 = 2^n. Furthermore E(1,n+1) = E(1,n) +
E(n,n+1) = 2^n - 2 + 2^n = 2^(n+1) - 2. Therefore by induction the
result is established.
Now E(1,11) = E(1,10) + 1 (ball can't go back to bottle 1 after
leaving bottle 10) = 2^10 - 1.
> 2. Is there a general formula for the amount of time
> required to get the ball out of bottle N into bottle N+1 given
> a probability P of backwards motion (given as 50% in this problem)?
I'd have to check my work, but I get E(n,n+1) = q^n, where q = 1/(1-p).
> 3. In how many hours will you expect to get the marble out of bottle 10
> after placing the marble in bottle 1 given two backward tubes for each
> bottle instead of one backward tube?
I get E(1,n) = (q^n - q)/(q-1), so E(1,11) = E(1,10) + 1 =
(3^10 - 3)/2 + 1.
-------------------------
In regards to your fourth problem, the following comments (marked
with a ">") should be added. I thought the answer was quite surprising!
---
The expected amount of time to go from state n-1 to n (state 11 is an
absorbing state) is
E(n-1,n) = 1 + E(1,n)/2 for 1<n<11
> since we expect it'll take an hour for the ball to leave bottle n-1,
> and it then has a probability of 1/2 of returning to the first bottle;
also
E(n-1,n+1) = E(n-1,n) + E(n,n+1) for 1<n<11
> since the only way of getting to state n+1 from n-1 is to first
> go from state n-1 to n, and then from n to n+1; the total expected
> time is the sum of the two.
==> pickover/pickover.05.p <==
Title: Cliff Puzzle 5: Mystery Sequence
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address. If you like, tell me a little bit
about yourself so I can cite you appropriately if you provide unique
information. PLEASE ALSO directly mail me a copy of your response in
addition to any responding you do in the newsgroup. I will assume it is
OK to describe your answer in any article or publication I may write in
the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover
* * *
What is the next term in the Mystery Sequence:
22.45906, 17600.22, 0.34714E+12,
==> pickover/pickover.05.s <==
-------------------------
Some serious roundoff error going on here, but...
The sequence 22.45906, 17600.22, 0.34714E+22 is clearly meant to be:
Pi^e, (Pi^e)^Pi, ((Pi^e)^Pi)^e,
so the next term should be (((Pi^e)^pi)^e)^pi = 1.80169E36.
Actually, it looks like you were using "pi" = 3.14159 and "e" = 2.71828, possibl
with other intermediate rounding off, so you may have been looking for something
more like 1.8011E36.
James
jlay...@grissom.jpl.nasa.gov
-------------------------
In article <+M_UUYZ8!@linac.fnal.gov> you write:
>cl...@watson.ibm.com (cliff) writes:
>>What is the next term in the Mystery Sequence:
>>
>>22.45906, 17600.22, 0.34714E+12,
>
>I disagree about the last couple of significant digits, but otherwise
>the series is pi^e, (pi^e)^pi, ((pi^e)^pi)^e, ... and the next term
>is about 1.8017E+36.
>_________________________________________________________
>Matt Crawford cra...@fncent.fnal.gov Fermilab
>
>
Background:
I recognized the approximate value of the first term from figuring
out (during high school, about 20 years ago) the old question of
which is larger, e^pi or pi^e. After that it was a mater of taking
ratios of logs of the terms.
_________________________________________________________
Matt Crawford cra...@fncent.fnal.gov Fermilab
BS 1978 Applied Math & Physics; PhD 1985 Physics
-------------------------
Before I go barking up a wrong tree, I notice that
e
Pi = 22.45916
>22.45906, 17600.22, 0.34714E+12,
which seems suspiciously close to your first constant. Which should I assume?
"Typo. It should read 22.45916",
"Coincidence.",
or "No Comment -- no clues."
???
/Alan Paeth
-------------------------
In article <1992Sep17.1...@watson.ibm.com> you write:
>What is the next term in the Mystery Sequence:
>22.45906, 17600.22, 0.34714E+12,
As a one-time math major, I thought I recognized that telltale 22.45906 ...
The sequence continues with 1.8016851E+36
Steve
--
-- mon...@diablo.amd.com -- (512) 462-4013
__ | signature designed by | One thing about kneading that pizza dough by
(_ | (and ripped off from) | hand -- it sure gets your fingernails clean!
__)teve | Stephen Wayne Miller | Pizzaria Friend of Danny
==> pickover/pickover.06.p <==
Title: Cliff Puzzle 6: Star Chambers
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address. If you like, tell me a little bit
about yourself so I can cite you appropriately if you provide unique
information. PLEASE ALSO directly mail me a copy of your response in
addition to any responding you do in the newsgroup. I will assume it is
OK to describe your answer in any article or publication I may write in
the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover
* * *
As many of you probably know, 5-sided stars produced by drawing a
continuous line with your pencil can nest inside each other. (One star
can sit inside the pentagon produced by the larger star. Each of the
5 points of the small star coincide with the 5 points of the
internal pentagon of the large star.)
Start with a five sided star formed with 5 line segments, each 1 inch
long. Continually nest stars so that the assembly of stars gets bigger
and bigger.
Questions:
1. How many nestings N are required to make star N
have an edge-length equal to the diameter of the sun (4.5E9 feet)?
2. How many nestings N are required to make the cumulative length
of lines of all the nested stars equal to the diameter of the sun?
==> pickover/pickover.06.s <==
-------------------------
Cliff Pickover,
So here I am, waiting to see if one of my long Grobner basis
calculations is going to finish before the machine goes down.
This is a good time to read news, and I came across this trivial
problem in rec.games.puzzles. I'm not sure why I'm responding,
perhaps the hour, or perhaps curiousity to see what will come
of this, but I could have done this the day in high school
when I learned how to compute cos(pi/5). The ratio between
side lengths of successive pentagrams is r = (3+sqrt(5))/2
= 1 + golden ratio = 2.618... . The smallest N for which
r^N > 5.48e10 (slightly more accurate value for sun's diameter
in inches) is 26, with r^26 = 7.37e10. The smallest N for which
5[r^(N+1)-1]/(r-1) > 5.48e10 is 24, with 5(r^25 - 1)/(r-1) = 8.70e10.
This seems too trivial to post, but do with this response as you like.
Bob Holt
-------------------------
I just started reading 'rec.puzzles', so have just seen this one and
the one before (#5)... and to be honest I'm not sure why you put this one
out, since it is pretty straightforward.
>Start with a five sided star formed with 5 line segments, each 1 inch
>long. Continually nest stars so that the assembly of stars gets bigger
>and bigger.
The analytical (and general) answer to this problem comes from the
basic relationship of a "chord" of a regular pentagon, which is defined
as follows:
_=*=_
_=/ / \=_
_=/ | \=_
_=/ | \=_
* / *
| | <-- "chord" |
\ | /
| / |
\ | /
| / |
*-------------*
compared to the length of one of the sides is the golden ratio, i.e.
_
1 + \/5
--------- .
2
It can then be derived that the length of the "chord" (i.e. segment
length) of the next bigger Star compared to the length of the "chord"
of its incribed Star is the square of the golden ratio, or the golden
ratio plus one, same thing.
>Questions:
>1. How many nestings N are required to make star N
>have an edge-length equal to the diameter of the sun (4.5E9 feet)?
Back-of-envelope calculations as follows:
4.5E9 * 12 = total of 5.22E10 inches.
ratio of Star sizes approx. 2.618.
The best answer is 27 nested Stars, although it produces a Star with
a "chord" length of 7.366E10 inches, i.e. a bit bigger. The first, and
smallest Star, is assumed to be the one with "chord" length of 1 inch.
>2. How many nestings N are required to make the cumulative length
>of lines of all the nested stars equal to the diameter of the sun?
This is just the sum of a geometric sequence with the ratio being
the golden ratio squared (or the golden ratio plus one).
_
/ 1 + \/5 \ 2
So, S = 1 inch, and S = S | --------- |
0 n n-1 \ 2 /
The sum is just the standard geometric summation, which I can't remember
offhand, but the contributing terms in the sum (other than the last term),
are less than one 1.6th of the total (by conincidence the inverse of the
golden ratio ;-). This means that the 25th Star (term) is the determining
factor, and in this case is the answer with a total length of 8.694E10
inches amoung all of them, and 5.373E10 inches for just the sum of the
segments of the 25th Star (again, counting the first one as side length
of 1 inch, or sum of 5 inches).
Well, that's it, I guess. Sorry to be so exhaustive, but I like to
use analytical methods to be sure I have the right answer.
My .signature explains most of what you need to know. What I mean
by "Honorary Grad Student" is that I have been taking Grad math classes
since my sophomore year, and for all intensive purposes might as well
be one. My Snail-mail address is 1521 S.W. 66th Ave., Portland, OR
97225. As to info about myself... I love learning about things, and
mathematics and consequently computers tend to be a great focus.
Anyway, if you have any more puzzles, pass them along... I am still
pondering on that sequence (puzzle #5) that you posted.
Thanks for your time.
Erich
--
"I haven't lost my mind; I know exactly where it is."
/ -- Erich Stefan Boleyn -- \ --=> *Mad Genius wanna-be* <=--
{ Honorary Grad. Student (Math) } Internet E-mail: <er...@gemini.mth.pdx.edu>
\ Portland State University / WARNING: INTERESTED AND EXCITABLE
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> competition/games/rubiks/rubiks.cube.p <==
What is known about bounds on solving Rubik's cube?
==> competition/games/rubiks/rubiks.cube.s <==
The "official" world record was set by Minh Thai at the 1982 World
Championships in Budapest Hungary, with a time of 22.95 seconds.
Keep in mind mathematicians provided standardized dislocation patterns
for the cubes to be randomized as much as possible.
The fastest cube solvers from 19 different countries had 3 attempts each
to solve the cube as quickly as possible. Minh and several others have
unofficially solved the cube in times between 16 and 19 seconds.
However, Minh averages around 25 to 26 seconds after 10 trials, and my
best average of ten trials is about 27 seconds (although it is usually
higher).
Consider that in the World Championships 19 of the world's fastest cube
solvers each solved the cube 3 times and no one solved the cube in less
than 20 seconds...
God's algorithm is the name given to an as yet (as far as I know)
undiscovered method to solve the rubik's cube in the least number of
moves; as opposed to using 'canned' moves.
The known lower bound is 18 moves. This is established by looking at
things backwards: suppose we can solve a position in N moves. Then by
running the solution backwards, we can also go from the solved position
to the position we started with in N moves. Now we count how many
sequences of N moves there are from the starting position, making
certain that we don't turn the same face twice in a row:
N=0: 1 (empty) sequence;
N=1: 18 sequences (6 faces can be turned, each in 3 different ways)
N=2: 18*15 sequences (take any sequence of length 1, then turn any of the
five faces which is not the last face turned, in any of 3
different ways); N=3: 18*15*15 sequences (take any sequence of
length 2, then turn any of the five faces which is not the last
face turned, in any of 3 different ways); : : N=i: 18*15^(i-1)
sequences.
So there are only 1 + 18 + 18*15 + 18*15^2 + ... + 18*15^(n-1)
sequences of moves of length n or less. This sequence sums to
(18/14)*(15^n - 1) + 1. Trying particular values of n, we find that
there are about 8.4 * 10^18 sequences of length 16 or less, and about
1.3 times 10^20 sequences of length 17 or less.
Since there are 2^10 * 3^7 * 8! * 12!, or about 4.3 * 10^19, possible
positions of the cube, we see that there simply aren't enough sequences
of length 16 or less to reach every position from the starting
position. So not every position can be solved in 16 or less moves -
i.e. some positions require at least 17 moves.
This can be improved to 18 moves by being a bit more careful about
counting sequences which produce the same position. To do this, note
that if you turn one face and then turn the opposite face, you get
exactly the same result as if you'd done the two moves in the opposite
order. When counting the number of essentially different sequences of N
moves, therefore, we can split into two cases:
(a) Last two moves were not of opposite faces. All such sequences can be
obtained by taking a sequence of length N-1, choosing one of the 4 faces
which is neither the face which was last turned nor the face opposite
it, and choosing one of 3 possible ways to turn it. (If N=1, so that the
sequence of length N-1 is empty and doesn't have a last move, we
can choose any of the 6 faces.)
(b) Last two moves were of opposite faces. All such sequences can be
obtained by taking a sequence of length N-2, choosing one of the 2
opposite face pairs that doesn't include the last face turned, and
turning each of the two faces in this pair (3*3 possibilities for how it
was turned). (If N=2, so that the sequence of length N-2 is empty and
doesn't have a last move, we can choose any of the 3 opposite face
pairs.)
This gives us a recurrence relation for the number X_N of sequences of
length N:
N=0: X_0 = 1 (the empty sequence)
N=1: X_1 = 18 * X_0 = 18
N=2: X_2 = 12 * X_1 + 27 * X_0 = 243
N=3: X_3 = 12 * X_2 + 18 * X_1 = 3240
:
:
N=i: X_i = 12 * X_(i-1) + 18 * X_(i-2)
If you do the calculations, you find that X_0 + X_1 + X_2 + ... + X_17 is
about 2.0 * 10^19. So there are fewer essentially different sequences of
moves of length 17 or less than there are positions of the cube, and so some
positions require at least 18 moves.
The upper bound of 50 moves is I believe due to Morwen Thistlethwaite, who
developed a technique to solve the cube in a maximum of 50 moves. It
involved a descent through a chain of subgroups of the full cube group,
starting with the full cube group and ending with the trivial subgroup
(i.e. the one containing the solved position only). Each stage involves a
careful examination of the cube, essentially to work out which coset of the
target subgroup it is in, followed by a table look-up to find a sequence to put
it into that subgroup. Needless to say, it was not a fast technique!
But it was fascinating to watch, because for the first three quarters or
so of the solution, you couldn't really see anything happening - i.e. the
position continued to appear random! If I remember correctly, one of the
final subgroups in the chain was the subgroup generated by all the double
twists of the faces - so near the end of the solution, you would suddenly
notice that each face only had two colours on it. A few moves more and the
solution was complete. Completely different from most cube solutions, in
which you gradually see order return to chaos: with Morwen's solution, the
order only re-appeared in the last 10-15 moves.
* Mark's Update/Comments ----------------------------------------------
* Despite some accounts of Thistlethwaite's method, it is possible to
* follow the progression of his algorithm. Clearly after Stage 2 is
* completed the L and R faces will have L and R colours on them only.
* After Stage 3 is complete the characteristics of the square's group
* is clearly visible on all 6 sides. It is harder to understand what
* is accomplished in Stage 1.
*
* ---------------------------------------------------------------------
With God's algorithm, of course, I would expect this effect to be even more
pronounced: someone solving the cube with God's algorithm would probably
look very much like a film of someone scrambling the cube, run in
reverse!
Finally, something I'd be curious to know in this context: consider the
position in which every cubelet is in the right position, all the corner
cubelets are in the correct orientation, and all the edge cubelets are
"flipped" (i.e. the only change from the solved position is that every edge
is flipped). What is the shortest sequence of moves known to get the cube
into this position, or equivalently to solve it from this position? (I know
of several sequences of 24 moves that do the trick.)
* Mark's Update/Comments ----------------------------------------------
*
* This is from my file cmoves.txt which includes the best known
* sequences for interesting patterns:
*
* p3 12 flip R1 L1 D2 B3 L2 F2 R2 U3 D1 R3 D2 F3 B3 D3 F2 D3 (20)
* R2 U3 F2 D3
*
* ---------------------------------------------------------------------
The reason I'm interested in this particular position: it is the unique
element of the centre of the cube group. As a consequence, I vaguely suspect
(I'd hardly like to call it a conjecture :-) it may lie "opposite" the
solved position in the cube graph - i.e. the graph with a vertex for each
position of the cube and edges connecting positions that can be transformed
into each other with a single move. If this is the case, then it is a good
candidate to require the maximum possible number of moves in God's
algorithm.
-- David Seal ds...@armltd.co.uk
To my knowledge, no one has ever demonstrated a specific cube position
that takes 15 moves to solve. Furthermore, the lower bound is known to
be greater than 15, due to a simple proof.
* ---------------------------------------------------------------------
* Mark> Definitely sequences exist in the square's group of length 15
* > e.g. Antipode 2 R2 B2 D2 F2 D2 F2 T2 L2 T2 D2 F2 T2 L2 T2 B2
* ---------------------------------------------------------------------
The way we know the lower bound is by working backwards counting how
many positions we can reach in a small number of moves from the solved
position. If this is less than 43,252,003,274,489,856,000 (the total
number of positions of Rubik's cube) then you need more than that
number of moves to reach the other positions of the cube. Therefore,
those positions will require more moves to solve.
The answer depends on what we consider a move. There are three common
definitions. The most restrictive is the QF metric, in which only a
quarter-turn of a face is allowed as a single move. More common is
the HF metric, in which a half-turn of a face is also counted as a
single move. The most generous is the HS metric, in which a quarter-
turn or half-turn of a central slice is also counted as a single move.
These metrics are sometimes called the 12-generator, 18-generator, and
27-generator metrics, respectively, for the number of primitive moves.
The definition does not affect which positions you can get to, or even
how you get there, only how many moves we count for it.
The answer is that even in the HS metric, the lower bound is 16,
because at most 17,508,850,688,971,332,784 positions can be reached
within 15 HS moves. In the HF metric, the lower bound is 18, because
at most 19,973,266,111,335,481,264 positions can be reached within 17
HF moves. And in the QT metric, the lower bound is 21, because at
most 39,812,499,178,877,773,072 positions can be reached within 20 QT
moves.
-- jjf...@skcla.monsanto.com writes:
Lately in this conference I've noted several messages related to Rubik's
Cube and Square 1. I've been an avid cube fanatic since 1981 and I've
been gathering cube information since.
Around Feb. 1990 I started to produce the Domain of the Cube Newsletter,
which focuses on Rubik's Cube and all the cube variants produced to
date. I include notes on unproduced prototype cubes which don't even
exist, patent information, cube history (and prehistory), computer
simulations of puzzles, etc. I'm up to the 4th issue.
Anyways, if you're interested in other puzzles of the scramble by
rotation type you may be interested in DOTC. It's available free to
anyone interested. I am especially interested in contributing articles
for the newsletter, e.g. ideas for new variants, God's Algorithm.
Anyone ever write a Magic Dodecahedron simulation for a computer?
Drop me a SASE (say empire size) if you're interested in DOTC or if you
would like to exchange notes on Rubik's Cube, Square 1 etc.
I'm also interested in exchanging puzzle simulations, e.g. Rubik's Cube,
Twisty Torus, NxNxN Simulations, etc, for Amiga and IBM computers. I've
written several Rubik's Cube solving programs, and I'm trying to make
the definitive puzzle solving engine. I'm also interested in AI programs
for Rubik's Cube and the like.
Ideal Toy put out the Rubik's Cube Newsletter, starting with
issue #1 on May 1982. There were 4 issues in all.
They are: #1, May 1982
#2, Aug 1982
#3, Winter 1983
#4, Aug 1983
There was another sort of magazine, published in several languages
called Rubik's Logic and Fantasy in space. I believe there were 8
issues in all. Unfortunately I don't have any of these! I'm willing
to buy these off anyone interesting in selling. I would like to get the
originals if at all possible...
I'm also interested in buying any books on the cube or related puzzles.
In particular I am _very_ interested in obtaining the following:
Cube Games Don Taylor, Leanne Rylands
Official Solution to Alexander's Star Adam Alexander
The Amazing Pyraminx Dr. Ronald Turner-Smith
The Winning Solution Minh Thai
The Winning Solution to Rubik's Revenge Minh Thai
Simple Solutions to Cubic Puzzles James G. Nourse
I'm also interested in buying puzzles of the mechanical type.
I'm still missing the Pyraminx Star (basically a Pyraminx with more tips
on it), Pyraminx Ball, and the Puck.
If anyone out here is a fellow collector I'd like to hear from you.
If you have a cube variant which you think is rare, or an idea for a
cube variant we could swap notes.
I'm in the middle of compiling an exhaustive library for computer
simulations of puzzles. This includes simulations of all Uwe Meffert's
puzzles which he prototyped but _never_ produced. In fact, I'm in the
middle of working on a Pyraminx Hexagon solver. What? Never heard of it?
Meffert did a lot of other puzzles which never were made.
I invented some new "scramble by rotation" puzzles myself. My favourite
creation is the Twisty Torus. It is a torus puzzle with segments (which
slide around 360 degrees) with multiple rings around the circumference.
The computer puzzle simulation library I'm forming will be described
in depth in DOTC #4 (The Domain of the Cube Newsletter). So if you
have any interesting computer puzzle programs please email me and
tell me all about them!
Also to the people interested in obtaining a subscription to DOTC,
who are outside of Canada (which it seems is just about all of you!)
please don't send U.S. or non-Canadian stamps (yeah, I know I said
Self-Addressed Stamped Envelope before). Instead send me an
international money order in Canadian funds for $6. I'll send you
the first 4 issues (issue #4 is almost finished).
Mark Longridge
Address: 259 Thornton Rd N, Oshawa Ontario Canada, L1J 6T2
Email: mark.lo...@canrem.com
One other thing, the six bucks is not for me to make any money. This
is only to cover the cost of producing it and mailing it. I'm
just trying to spread the word about DOTC and to encourage other
mechanical puzzle lovers to share their ideas, books, programs and
puzzles. Most of the programs I've written and/or collected are
shareware for C64, Amiga and IBM. I have source for all my programs
(all in C or Basic) and I am thinking of providing a disk with the
4th issue of DOTC. If the response is favourable I will continue
to provide disks with DOTC.
-- Mark Longridge <mark.lo...@canrem.com> writes:
It may interest people to know that in the latest issue of "Cubism For Fun" %
(# 28 that I just received yesterday) there is an article by Herbert Kociemba
from Darmstadt. He describes a program that solves the cube. He states that
until now he has found no configuration that required more than 21 turns to
solve.
He gives a 20 move manoeuvre to get at the "all edges flipped/
all corners twisted" position:
DF^2U'B^2R^2B^2R^2LB'D'FD^2FB^2UF'RLU^2F'
or in Varga's parlance:
dofitabiribirilobadafodifobitofarolotifa
Other things #28 contains are an analysis of Square 1, an article about
triangular tilings by Martin Gardner, and a number of articles about other
puzzles.
--
% CFF is a newsletter published by the Dutch Cubusts Club NKC.
Secretary:
Anneke Treep
Postbus 8295
6710 AG Ede
The Netherlands
Membership fee for 1992 is DFL 20 (about$ 11).
--
-- dik t. winter <d...@cwi.nl>
References:
Mathematical Papers
-------------------
Rubik's Revenge: The Group Theoretical Solution
Mogens Esrom Larsen A.M.M. Vol. 92 No. 6, June-July 1985, pages
381-390
Rubik's Groups
E. C. Turner & K. F. Gold, American Mathematical Monthly,
vol. 92 No. 9 November 1985, pp. 617-629.
Cubelike Puzzles - What Are They and How Do You Solve Them?
J.A. Eidswick A.M.M. Vol. 93 No. 3 March 1986, pp 157-176
The Slice Group in Rubik's Cube, David Hecker, Ranan Banerji
Mathematics Magazine, Vol. 58 No. 4 Sept 1985
The Group of the Hungarian Magic Cube
Chris Rowley Proceedings of the First Western Austrialian
Conference on Algebra, 1982
Books
-----
Rubik's Cubic Compendium
Erno Rubik, Tamas Varga, et al, Edited by David Singmaster
Oxford University Press, 1987
Enslow Publishers Inc
==> competition/games/rubiks/rubiks.magic.p <==
How do you solve Rubik's Magic?
==> competition/games/rubiks/rubiks.magic.s <==
The solution is in a 3x3 grid with a corner missing.
+---+---+---+ +---+---+---+---+
| 3 | 5 | 7 | | 1 | 3 | 5 | 7 |
+---+---+---+ +---+---+---+---+
| 1 | 6 | 8 | | 2 | 4 | 6 | 8 |
+---+---+---+ +---+---+---+---+
| 2 | 4 | Original Shape
+---+---+
To get the 2x4 "standard" shape into this shape, follow this:
1. Lie it flat in front of you (4 going across).
2. Flip the pair (1,2) up and over on top of (3,4).
3. Flip the ONE square (2) up and over (1).
[Note: if step 3 won't go, start over, but flip the entire original shape
over (exposing the back).]
4. Flip the pair (2,4) up and over on top of (5,6).
5. Flip the pair (1,2) up and toward you on top of (blank,4).
6. Flip the ONE square (2) up and left on top of (1).
7. Flip the pair (2,4) up and toward you.
Your puzzle won't be completely solved, but this is how to get the shape.
Notice that 3,5,6,7,8 don't move.
==> competition/games/scrabble.p <==
What are some exceptional Scrabble Brand Crossword Game (TM) games?
==> competition/games/scrabble.s <==
The shortest Scrabble game:
The Scrabble Players News, Vol. XI No. 49, June 1983, contributed by
Kyle Corbin of Raleigh, NC:
[J]
J U S
S O X
[X]U
which can be done in 4 moves, JUS, SOX, [J]US, and [X]U.
In SPN Vol. XI, No. 52, December 1983, Alan Frank presented what
he claimed is the shortest game where no blanks are used, also
four moves:
C
WUD
CUKES
DEY
S
This was followed in SPN, Vol. XII No. 54, April 1984, by Terry Davis
of Glasgow, KY:
V
V O[X]
[X]U,
which is three moves. He noted that the use of two blanks prevents
such plays as VOLVOX. Unfortunately, it doesn't prevent SONOVOX.
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
Record for the highest Scrabble score in a single turn (in a legal position):
According to the Scrabble Players Newspaper (since renamed to
Scrabble Players News) issue 44, p13, the highest score for one
turn yet discovered, using the Official Scrabble Players
Dictionary, 1st ed. (the 2nd edition is now in use in club and
tournament play) and the Websters 9th New Collegiate Dictionary,
was the following:
d i s e q u i l i b r a t e D
. . . . . . . e . . . . . . e
. . . . . . . e . . . . . o m
r a d i o a u t o g r a p(h)Y
. . . . . . . . . . . w a s T
. . . . . . . . . . b e . . h
. . . . . . . . . . a . . g o
. . . c o n j u n c t i v a L
. . . . . . . . . . . . . n o
. . . . . . . f i n i k i n G
. . . . . . . a . . . (l) e i
. . . . . . . d . s p e l t Z
. . . . . . w e . . . . . . e
. . . . . . r . . . . . . o r
m e t h o x y f l u r a n e S
for 1682 points.
According to the May 1986 issue of GAMES, the highest known score achievable
in one turn is 1,962 points. The word is BENZOXYCAMPHORS formed across the
three triple-word scores on the bottom of the board. Apparently it was
discovered by Darryl Francis, Ron Jerome, and Jeff Grant.
As for other Scrabble trivia, the highest-scoring first move based on the
Official Scrabble Players Dictionary is 120 points, with the words JUKEBOX,
QUIZZED, SQUEEZE, or ZYMURGY. If Funk & Wagnall's New Standard Dictionary
is used then ZYXOMMA, worth 130 points, can be formed.
The highest-scoring game, based on Webster's Second and Third and on the
Oxford English Dictionary, was devised by Ron Jerome and Ralph Beaman and
totalled 4,142 points for the two players. The highest-scoring words in
the game were BENZOXYCAMPHORS, VELVETEEN, and JACKPUDDINGHOOD.
The following example of a SCRABBLE game produced a score of 2448 for one
player and 1175 for the final word. It is taken from _Beyond Language_ (1967)
by Dmitri Borgman (pp. 217-218). He credits this solution to Mrs. Josefa H.
Byrne of San Francisco and implies that all words can be found in _Webster's
Second Edition_. The two large words (multiplied by 27 as they span 3 triple
word scores) are ZOOPSYCHOLOGIST (a psychologist who treats animals rather
than humans) and PREJUDICATENESS (the condition or state of being decided
beforehand). The asterisks (*) represent the blank tiles. (Please excuse
any typo's).
Board Player1 Player2
Z O O P S Y C H O L O G I S T ABILITY 76 ERI, YE 9
O N H A U R O W MAN, MI 10 EN 2
* R I B R O V E I FEN, FUN 14 MANIA 7
L T I K E G TABU 12 RIB 6
O L NEXT 11 AM 4
G I AX 9 END 6
I T IT, TIKE 10 LURE 6
* Y E LEND, LOGIC*AL 79 OO*LOGICAL 8
A R FUND, JUD 27 ATE, MA 7
L E N D M I ROVE 14 LO 2
E A Q DARE, DE 13 ES, ES, RE 6
W A X F E N U RE, ROW 14 IRE, IS, SO 7
E T A B U I A DARED, QUAD 22 ON 4
E N A M D A R E D WAX, WEE 27 WIG 9
P R E J U D I C A T E N E S S CHIT, HA 14 ON 2
PREJUDICATENESS,
AN, MANIAC,
QUADS, WEEP 911 OOP 8
ZOOPSYCHOLOGIST,
HABILITY, TWIG,
ZOOLOGICAL 1175
--------------------------------------
Total: 2438 93
F, N, V, T in
loser's hand: +10 -10
--------------------------------------
Final Score: 2448 83
---------------------------------------------------------------------------
It is possible to form the following 14 7-letter OSPD words from the
non-blank tiles:
HUMANLY
FATUOUS
AMAZING
EERIEST
ROOFING
TOILERS
QUIXOTE
JEWELRY
CAPABLE
PREVIEW
BIDDERS
HACKING
OVATION
DONATED
==> competition/games/set.p <==
What is the size of the largest collection of cards from which NO "set"
can be selected ?
==> competition/games/set.s <==
I can get 20:
1ROL
1GDL
1GDM
1GOM
1GSL
1PDH
1PDM
1POL
1POM
2RSL
2PDM
3ROL
3ROM
3RSL
3RSH
3GDM
3GOL
3GSL
3GSM
3POM
This collection of 20 is a local maximum.
The small C progam shown below was used to check for all possible
extensions to a collection of 21.
Of course this leaves open the question whether there exists a completely
different collection of 21 from which no "set" can be selected.
-- Gene Miller
------- C Program enclosed -------
#define N 21
static int data[N][4]= {
1, 1, 2, 1, /* 00 */
1, 2, 1, 1, /* 01 */
1, 2, 1, 2, /* 02 */
1, 2, 2, 2, /* 03 */
1, 2, 3, 1, /* 04 */
1, 3, 1, 3, /* 05 */
1, 3, 1, 2, /* 06 */
1, 3, 2, 1, /* 07 */
1, 3, 2, 2, /* 08 */
2, 1, 3, 1, /* 09 */
2, 3, 1, 2, /* 10 */
3, 1, 2, 1, /* 11 */
3, 1, 2, 2, /* 12 */
3, 1, 3, 1, /* 13 */
3, 1, 3, 3, /* 14 */
3, 2, 1, 2, /* 15 */
3, 2, 2, 1, /* 16 */
3, 2, 3, 1, /* 17 */
3, 2, 3, 2, /* 18 */
3, 3, 2, 2, /* 19 */
0, 0, 0, 0 /* 20 */ /* leave space for Nth combo */
};
main()
{
int x, y, z, w;
for (x=1; x<=3; x++) /* check all combos */
for (y=1; y<=3; y++)
for (z=1; z<=3; z++)
for (w=1; w<=3; w++)
printf ("%d %d %d %d -> sets=%d\n", x, y, z, w,
check (x, y, z, w));
}
int check(x,y,z,w)
int x, y, z, w;
{
int i,j,k,m;
int errors, sets;
for (i=0; i<N; i++) /* check for pre-existing combos */
if (x==data[i][0] &&
y==data[i][1] &&
z==data[i][2] &&
w==data[i][3] ) {
return -1; /* discard pre-existing*/
}
data[N-1][0] = x; /* make this the Nth combo */
data[N-1][1] = y;
data[N-1][2] = z;
data[N-1][3] = w;
sets = 0; /* start counting sets */
for (i=0; i<N; i++) /* look for sets */
for (j=i+1; j<N; j++)
for (k=j+1; k<N; k++) {
errors = 0;
for (m=0; m<4; m++) {
if (data[i][m] == data[j][m]) {
if (data[k][m] != data[i][m]) errors++;
if (data[k][m] != data[j][m]) errors++;
}
else {
if (data[k][m] == data[i][m]) errors++;
if (data[k][m] == data[j][m]) errors++;
}
}
if (errors == 0) /* no errors means is a set */
sets++; /* increment number of sets */
}
return sets;
}
--
I did some more experimenting. With the enclosed C program, I looked at many
randomly generated collections. In an earlier version of this program I
got one collection of 20 from a series of 100 trials. The rest were collections
ranging in size from 16 to 19. Unfortunately, in an attempt to make this
program more readable and more general, I changed the algorithm slightly and
I haven't been able to achieve 20 since then. I can't remember enough about
my changes to be able to get back to the previous version. In the most recent
1000 trials all of the maximaml collections range in size from 16 to 18.
I think that this experiment has shed very little light on what is the
global maximum, since the search space is many orders of magnitude larger
than what can be tried in a reasonable amount of time through random
searching.
I assume that Mr. Ring found his collection of 20 by hand. This indicates
that an intelligent search may be more fruitful than a purely random one.
------------------ Program enclosed -------------
int n;
int data[81][4];
main()
{
int i;
for (i=0; i<1000; i++) { /* Do 1000 independent trials */
printf ("Trial %d:\n", i);
try();
}
}
try()
{
int i;
int x, y, z, w;
n = 0; /* set collection size to zero */
for (i=0; i<100; i++) { /* try 100 random combos */
x = 1 + rand()%3;
y = 1 + rand()%3;
z = 1 + rand()%3;
w = 1 + rand()%3;
check (x, y, z, w);
}
for (x=1; x<=3; x++) /* check all combos */
for (y=1; y<=3; y++)
for (z=1; z<=3; z++)
for (w=1; w<=3; w++)
check (x, y, z, w);
printf (" collection size=%d\n", n);
}
int check(x, y, z, w) /* check whether a new combo can be added */
int x, y, z, w;
{
int i,j,k,m;
int errors, sets;
for (i=0; i<n; i++) /* check for pre-existing combos */
if (x==data[i][0] &&
y==data[i][1] &&
z==data[i][2] &&
w==data[i][3] ) {
return -1; /* discard pre-existing*/
}
data[n][0] = x; /* make this the nth combo */
data[n][1] = y;
data[n][2] = z;
data[n][3] = w;
sets = 0; /* start counting sets */
for (i=0; i<=n; i++) /* look for sets */
for (j=i+1; j<=n; j++)
for (k=j+1; k<=n; k++) {
errors = 0;
for (m=0; m<4; m++) {
if (data[i][m] == data[j][m]) {
if (data[k][m] != data[i][m]) errors++;
if (data[k][m] != data[j][m]) errors++;
}
else {
if (data[k][m] == data[i][m]) errors++;
if (data[k][m] == data[j][m]) errors++;
}
}
if (errors == 0) /* no errors means is a set */
sets++; /* increment number of sets */
}
if (sets == 0) {
n++; /* increment collection size */
printf ("%d %d %d %d\n", x, y, z, w);
}
return sets;
}
------------------ end of enclosed program -------------
-- Gene
--
Gene Miller Multimedia Communications
NYNEX Science & Technology Phone: 914 644 2834
500 Westchester Avenue Fax: 914 997 2997, 914 644 2260
White Plains, NY 10604 Email: ge...@nynexst.com
==> competition/games/soma.p <==
What is the solution to Soma Cubes?
==> competition/games/soma.s <==
The soma cube is dissected in excruciating detail in volume 2 of
"Winning Ways" by Conway, Berlekamp and Guy, in the same chapter as the
excruciatingly detailed dissection of Rubik's Cube.
==> competition/games/square-1.p <==
Does anyone have any hints on how to solve the Square-1 puzzle?
==> competition/games/square-1.s <==
SHAPES
1. There are 29 different shapes for a side, counting reflections:
1 with 6 corners, 0 edges
3 with 5 corners, 2 edges
10 with 4 corners, 4 edges
10 with 3 corners, 6 edges
5 with 2 corners, 8 edges
2. Naturally, a surplus of corners on one side must be compensated
by a deficit of corners on the other side. Thus there are 1*5 +
3*10 + C(10,2) = 5 + 30 + 55 = 90 distinct combinations of shapes,
not counting the middle layer.
3. You can reach two squares from any other shape in at most 7 transforms,
where a transform consists of (1) optionally twisting the top, (2)
optionally twisting the bottom, and (3) flipping.
4. Each transform toggles the middle layer between Square and Kite,
so you may need 8 transforms to reach a perfect cube.
5. The shapes with 4 corners and 4 edges on each side fall into four
mutually separated classes. Side shapes can be assigned values:
0: Square, Mushroom, and Shield; 1: Left Fist and Left Paw; 2:
Scallop, Kite, and Barrel; 3. Right Fist and Right Paw. The top
and bottom's sum or difference, depending on how you look at them,
is a constant. Notice that the side shapes with bilateral symmetry
are those with even values.
6. To change this constant, and in particular to make it zero, you must
attain a position that does not have 4 corners and 4 edges on each
side. Almost any such position will do, but returning to 4 corners
and 4 edges with the right constant is left to your ingenuity.
7. If the top and bottom are Squares but the middle is a Kite, just flip
with the top and bottom 30deg out of phase and you will get a cube.
COLORS
1. I do not know the most efficient way to restore the colors. What
follows is my own suboptimal method. All flips keep the yellow
stripe steady and flip the blue stripe.
2. You can permute the corners without changing the edges, so first
get the edges right, then the corners.
3. This transformation sends the right top edge to the bottom
and the left bottom edge to the top, leaving the other edges
on the same side as they started: Twist top 30deg cl, flip,
twist top 30deg ccl, twist bottom 150deg cl, flip, twist bottom
30deg cl, twist top 120deg cl, flip, twist top 30deg ccl, twist
bottom 150deg cl, flip, twist bottom 30deg cl. Cl and ccl are
defined looking directly at the face. With this transformation
you can eventually get all the white edges on top.
4. Check the parity of the edge sequence on each side. If either is
wrong, you need to fix it. Sorry -- I don't know how! (See any
standard reference on combinatorics for an explanation of parity.)
5. The following transformation cyclically permutes ccl all the top edges
but the right one and cl all the bottom edges but the left one. Apply
the transformation in 3., and turn the whole cube 180deg. Repeat.
This is a useful transformation, though not a cure-all.
6. Varying the transformation in 3. with other twists will produce other
results.
7. The following transformation changes a cube into a Comet and Star:
Flip to get Kite and Kite. Twist top and bottom cl 90deg and flip to get
Barrel and Barrel. Twist top cl 30 and bottom cl 60 and flip to get
Scallop and Scallop. Twist top cl 60 and bottom cl 120 and flip to
get Comet and Star. The virtue of the Star is that it contains only
corners, so that you can permute the corners without altering the edges.
8. To reach a Lemon and Star instead, replace the final bottom cl 120 with
a bottom cl 60. In both these transformation the Star is on the bottom.
9. The following transformation cyclically permutes all but the bottom
left rear. It sends the top left front to the bottom, and the bottom
left front to the top. Go to Comet and Star. Twist star cl 60.
Go to Lemon and Star -- you need not return all the way to the cube, but
do it if you're unsure of yourself by following 7 backwards. Twist star
cl 60. Return to cube by following 8 backwards. With this transformation
you should be able to get all the white corners on top.
10. Check the parity of the corner sequences on both sides. If the bottom
parity is wrong, here's how to fix it: Go to Lemon and Star. The
colors on the Star will run WWGWWG. Twist it 180 and return to cube.
11. If the top parity is wrong, do the same thing, except that when you
go from Scallop and Scallop to Lemon and Star, twist the top and bottom
ccl instead of cl. The colors on the Star should now run GGWGGW.
12. Once the parity is right on both sides, the basic method is to
go to Comet and Star, twist the star 120 cl (it will be WGWGWG),
return to cube, twist one or both sides, go to Comet and Star,
undo the star twist, return to cube, undo the side twists.
With no side twists, this does nothing. If you twist the top,
you will permute the top corners. If you twist the bottom,
you will permute the bottom corners. Eventually you will get
both the top and the bottom right. Don't forget to undo the
side twists -- you need to have the edges in the right places.
Happy twisting....
--
Col. G. L. Sicherman
g...@windmill.att.COM
==> competition/games/think.and.jump.p <==
THINK & JUMP: FIRST THINK, THEN JUMP UNTIL YOU
ARE LEFT WITH ONE PEG! O - O O - O
/ \ / \ / \ / \
O---O---O---O---O
BOARD DESCRIPTION: To the right is a model of \ / \ / \ / \ /
the Think & Jump board. The O---O---O---O---O---O
O's represent holes which / \ / \ / \ / \ / \ / \
contain pegs. O---O---O---O---O---O---O
\ / \ / \ / \ / \ / \ /
O---O---O---O---O---O
DIRECTIONS: To play this brain teaser, you begin / \ / \ / \ / \
by removing the center peg. Then, O---O---O---O---O
moving any direction in the grid, \ / \ / \ / \ /
jump over one peg at a time, O - O O - O
removing the jumped peg - until only
one peg is left. It's harder then it looks.
But it's more fun than you can imagine.
SKILL CHART:
10 pegs left - getting better
5 pegs left - true talent
1 peg left - you're a genius
Manufactured by Pressman Toy Corporation, NY, NY.
==> competition/games/think.and.jump.s <==
Three-color the board in the obvious way. The initial configuration has 12
of each color, and each jump changes the parity of all three colors. Thus,
it is impossible to achieve any position where the colors do not have the
same parity; in particular, (1,0,0).
If you remove the requirement that the initially-empty cell must be at the
center, the game becomes solvable. The demonstration is left as an exercise.
Karl Heuer rutgers!harvard!ima!haddock!karl ka...@haddock.ima.isc.com
Here is one way of reducing Think & Jump to two pegs.
Long simplifies Balsley's scintillating snowflake solution:
1 U-S A - B C - D
2 H-U / \ / \ / \ / \
3 V-T E---F---G---H---I
4 S-H \ / \ / \ / \ /
5 D-M J---K---L---M---N---O
6 F-S / \ / \ / \ / \ / \ / \
7 Q-F P---Q---R---S---T---U---V
8 A-L \ / \ / \ / \ / \ / \ /
9 S-Q W---X---Y---Z---a---b
10 P-R / \ / \ / \ / \
11 Z-N c---d---e---f---g
12 Y-K \ / \ / \ / \ /
13 h-Y h - i j - k
14 k-Z
The board should now be in the snowflake pattern, i.e. look like
o - * * - o
/ \ / \ / \ / \
*---o---*---o---*
\ / \ / \ / \ /
*---*---*---*---*---*
/ \ / \ / \ / \ / \ / \
o---o---o---o---o---o---o
\ / \ / \ / \ / \ / \ /
*---*---*---*---*---*
/ \ / \ / \ / \
*---o---*---o---*
\ / \ / \ / \ /
o - * * - o
where o is empty and * is a peg. The top and bottom can now be reduced
to single pegs individually. For example, we could continue
15 g-T
16 Y-a
17 i-Z
18 T-e
19 j-Y
20 b-Z
21 c-R
22 Z-X
23 W-Y
24 R-e
which finishes the bottom. The top can be done in a similar manner.
--
Chris Long
==> competition/games/tictactoe.p <==
In random tic-tac-toe, what is the probability that the first mover wins?
==> competition/games/tictactoe.s <==
Count cases.
First assume that the game goes on even after a win. (Later figure
out who won if each player gets a row of three.) Then there are
9!/5!4! possible final boards, of which
8*6!/2!4! - 2*6*4!/0!4! - 3*3*4!/0!4! - 1 = 98
have a row of three Xs. The first term is 8 rows times (6 choose 2)
ways to put down the remaining 2 Xs. The second term is the number
of ways X can have a diagonal row plus a horizontal or vertical row.
The third term is the number of ways X can have a vertical and a
horizontal row, and the 4th term is the number of ways X can have two
diagonal rows. All the two-row configurations must be subtracted to
avoid double-counting.
There are 8*6!/1!5! = 48 ways O can get a row. There is no double-
counting problem since only 4 Os are on the final board.
There are 6*2*3!/2!1! = 36 ways that both players can have a
row. (6 possible rows for X, each leaving 2 possible rows for O
and (3 choose 2) ways to arrange the remaining row.) These
cases need further consideration.
There are 98 - 36 = 62 ways X can have a row but not O.
There are 48 - 36 = 12 ways O can have a row but not X.
There are 126 - 36 - 62 - 12 = 16 ways the game can be a tie.
Now consider the 36 configurations in which each player has a row.
Each such can be achieved in 5!4! = 2880 orders. There are 3*4!4!
= 1728 ways that X's last move completes his row. In these cases O
wins. There are 2*3*3!3! = 216 ways that Xs fourth move completes
his row and Os row is already done in three moves. In these cases O
also wins. Altogether, O wins 1728 + 216 = 1944 out of 2880 times
in each of these 36 configurations. X wins the other 936 out of
2880.
Altogether, the probability of X winning is ( 62 + 36*(936/2880) ) / 126.
win: 737 / 1260 ( 0.5849206... )
lose: 121 / 420 ( 0.2880952... )
draw: 8 / 63 ( 0.1269841... )
The computer output below agress with this analysis.
1000000 games: won 584865, lost 288240, tied 126895
Instead, how about just methodically having the program play every
possible game, tallying up who wins?
Wonderful idea, especially since there are only 9! ~ 1/3 million
possible games. Of course some are identical because they end in
fewer than 8 moves. It is clear that these should be counted
multiple times since they are more probable than games that go
longer.
The result:
362880 games: won 212256, lost 104544, tied 46080
#include <stdio.h>
int board[9];
int N, move, won, lost, tied;
int perm[9] = { 0, 1, 2, 3, 4, 5, 6, 7, 8 };
int rows[8][3] = {
{ 0, 1, 2 }, { 3, 4, 5 }, { 6, 7, 8 }, { 0, 3, 6 },
{ 1, 4, 7 }, { 2, 5, 8 }, { 0, 4, 8 }, { 2, 4, 6 }
};
main()
{
do {
bzero((char *)board, sizeof board);
for ( move=0; move<9; move++ ) {
board[perm[move]] = (move&1) ? 4 : 1;
if ( move >= 4 && over() )
break;
}
if ( move == 9 )
tied++;
#ifdef DEBUG
printf("%1d%1d%1d\n%1d%1d%1d w %d, l %d, t %d\n%1d%1d%1d\n\n",
board[0], board[1], board[2],
board[3], board[4], board[5], won, lost, tied,
board[6], board[7], board[8]);
#endif
N++;
} while ( nextperm(perm, 9) );
printf("%d games: won %d, lost %d, tied %d\n", N, won, lost, tied);
exit(0);
}
int s;
int *row;
over()
{
for ( row=rows[0]; row<rows[8]; row+=3 ) {
s = board[row[0]] + board[row[1]] + board[row[2]];
if ( s == 3 )
return ++won;
if ( s == 12 )
return ++lost;
}
return 0;
}
nextperm(c, n)
int c[], n;
{
int i = n-2, j=n-1, t;
while ( i >= 0 && c[i] >= c[i+1] )
i--;
if ( i < 0 )
return 0;
while ( c[j] <= c[i] )
j--;
t = c[i]; c[i] = c[j]; c[j] = t;
i++; j = n-1;
while ( i < j ) {
t = c[i]; c[i] = c[j]; c[j] = t;
i++; j--;
}
return 1;
}
==> competition/tests/analogies/long.p <==
1. Host : Guest :: Cynophobia : ?
2. Mountain : Plain :: Acrocephalic : ?
3. Lover : Believer :: Philofelist : ?
4. 4 : 6 :: Bumblebee : ?
5. 2 : 1 :: Major : ?
6. 1 : 24 :: Group : ?
7. 4 : 64 :: Crotchet : ?
8. Last : First :: Grave : ?
9. 7 : 9 :: Throne : ?
10. Pride : Hatred :: Beelzebub : ?
11. Dollar : Bond :: Grant : ?
12. Ek : Sankh :: 1 : ?
==> competition/tests/analogies/long.s <==
1. Lyssophobia
Cynophobia is the fear of dogs; lyssophobia is the fear of rabies. As
Rodney Adams pointed out, a word meaning the fear of fleas would be
even better, but I've been unable to locate such a word (although
surely one must exists).
2. Homalocephalic
Acrocephalic is having a pointed head; homalocephalic is having a flat
head. Rodney Adamas suggested "planoccipital", but on checking this
apparently refers to having a flat back of the skull, so he only gets
partial credit.
3. Galeanthropist
A philofelist is a cat-lover (also commonly called an ailurophile);
a galeanthropist is a person who believes they are a cat.
4. Blue Bird
A Camp Fire Girl who is 4 is in the Bumblebee Club; a Campfire Girl
who is 6 in the Blue Bird Club. I should have had "4 or 5" instead
of "4" to remove ambiguity (e.g. Mark Brader suggested "triplane").
5. Brigadier
A 2-star general in the army is a major general; a 1-star general
in the army is a brigadier general.
6. Field Army
Army groupings; there are 24 "groups" in a "field army".
7. Hemidemisemiquaver
A crotchet is a quarter-note; a hemidemisemiquaver is a sixty-fourth
note. Rodney Adams and Mark Brader both got this.
8. Prestissimo
In music prestissimo means extremely fast; grave means very slow to
the point of solemnity. This question was poorly worded (I received
both "womb" and "cradle" as answers).
9. Seraph
In the ninefold hierarchy of angels, a throne ranks 7th and a seraph
9th (9th being most powerful). Rodney Adams got this one.
10. Sonneillon
In Father Sebastien Machaelis's (one of the more famous exorcists)
hierarchy of devils, Beelzebub is responsible for pride, Sonneillon
for hatred.
11. Roosevelt
Grant is on the $50 bill; Roosevelt on the $50 savings bond.
12. 10^14
Ek is 1 in Hindi; sankh is 10^14 (one hundred quadrillion).
--
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
==> competition/tests/analogies/pomfrit.p <==
1. NATURAL: ARTIFICIAL :: ANKYLOSIS: ?
2. RESCUE FROM CHOKING ON FOOD, etc.: HEIMLICH :: ADJUSTING MIDDLE EAR PRESSURE: ?
3. LYING ON OATH: PERJURY :: INFLUENCING A JURY: ?
4. RECTANGLE: ELLIPSE :: MERCATOR: ?
5. CLOSED: CLEISTOGAMY :: OPEN: ?
6. FO'C'SLE: SYNCOPE :: TH'ARMY: ?
7. FILMS: OSCAR :: MYSTERY NOVELS: ?
8. QUANTITATIVE DATA: STATISTICS :: HUMAN SETTLEMENTS: ?
9. 7: 19 : : SEPTIMAL: ?
10. 3 TO 2: SESQUILATERAL :: 7 TO 5: ?
11. SICILY: JAPAN :: MAFIA: ?
12. CELEBRITIES: SYCOPHANTIC :: ANCESTORS: ?
13. 95: 98 :: VENITE: ?
14. MINCES: EYES :: PORKIES: ?
15. POSTAGE STAMPS: PHILATELIST: MATCHBOX LABELS: ?
16. MALE: FEMALE :: ARRENOTOKY: ?
17 TAILOR: DYER :: SARTORIAL: ?
18. HERMES: BACCHUS :: CADUCEUS: ?
19. 2823: 5331 :: ELEPHANT: ?
20. CENTRE OF GRAVITY: BARYCENTRIC :: ROTARY MOTION: ?
21. CALIFORNIA: EUREKA :: NEW YOKK: ?
22. MARRIAGE: DIGAMY :: COMING OF CHRIST: ?
23. 6: 5 :: PARR: ?
24. GROUP: INDIVIDUAL :: PHYLOGENESIS: ?
25. 12: 11 :: EPSOM: ?
==> competition/tests/analogies/pomfrit.s <==
1. ARTHRODESIS
2. VALSALVA
3. EMBRACERY
4. MOLLWEIDE
5. CHASMOGAMY
6. SYNAL(O)EPHA
7. EDGAR
8. EKISTICS
9. DECENNOVAL
10. SUPERBIQUINTAL
11. YAKUZA
12. FILIOPETISTIC
13. CANTATE
14. LIES
15. PHILLUMENIST
16. THELYTOKY
17. TINCTORIAL
18. THYRSUS
19. ANTIQUARIAN
20. TROCHILIC
21. EXCELSIOR (mottos)
22. PAROUSIA
23. HOWARD (wives of Henry VIII)
24. ONTOGENESIS
25. GLAUBER (salts)
==> competition/tests/analogies/quest.p <==
1. Mother: Maternal :: Stepmother: ?
2. Club: Axe :: Claviform: ?
3. Cook Food: Pressure Cooker :: Kill Germs: ?
4. Water: Air :: Hydraulic: ?
5. Prediction: Dirac :: Proof: ?
6. Raised: Sunken :: Cameo: ?
7. 1: 14 :: Pound: ?
8. Malay: Amok :: Eskimo Women: ?
9. Sexual Intercourse: A Virgin :: Bearing Children: ?
10. Jaundice, Vomiting, Hemorrhages: Syndrome :: Jaundice: ?
11. Guitar: Cello :: Segovia: ?
12. Bars: Leaves :: Eagle: ?
13. Roll: Aileron :: Yaw: ?
14. 100: Century :: 10,000: ?
15. Surface: Figure :: Mobius: ?
16. Logic: Philosophy :: To Know Without Conscious Reasoning: ?
17. Alive: Parasite :: Dead: ?
18. Sea: Land :: Strait: ?
19. Moses: Fluvial :: Noah: ?
20. Remnant: Whole :: Meteorite: ?
21. Opossum, Kangaroo, Wombat: Marsupial :: Salmon, Sturgeon, Shad: ?
22. Twain/Clemens: Allonym :: White House/President: ?
23. Sculptor: Judoka :: Fine: ?
24. Dependent: Independent :: Plankton: ?
25. Matthew, Mark, Luke, John: Gospels :: Joshua-Malachi: ?
26. Luminous Flux: Lumen :: Sound Absorption: ?
27. 2: 3 :: He: ?
28. Growth: Temperature :: Pituitary Gland: ?
29. Spider: Arachnoidism :: Snake: ?
30. Epigram: Anthology :: Foreign Passages: ?
31. Pathogen: Thermometer :: Lethal Wave: ?
32. Russia: Balalaika :: India: ?
33. Involuntary: Sternutatory :: Voluntary: ?
34. Unusual Hunger: Bulimia :: Hunger for the Unusual: ?
35. Blind: Stag :: Tiresias: ?
36. River: Fluvial :: Rain: ?
37. Country: City :: Tariff: ?
38. $/Dollar: Logogram :: 3, 5, 14, 20/Cent: ?
39. Lung Capacity: Spirometer :: Arterial Pressure: ?
40. Gold: Ductile :: Ceramic: ?
41. 7: 8 :: Uranium: ?
42. Judaism: Messiah :: Islam: ?
43. Sight: Amaurosis :: Smell: ?
44. Oceans: Cousteau :: Close Encounters of the Third Kind: ?
45. Diamond/Kimberlite: Perimorph :: Fungus/Oak: ?
46. Compulsion to Pull One's Hair: Trichotillomania ::
Imagine Oneself As a Beast: ?
47. Cross: Neutralism :: Hexagram: ?
48. Wing: Tail :: Fuselage: ?
49. Bell: Loud :: Speak: ?
50. Benevolence: Beg :: Philanthropist: ?
51. 10: Decimal :: 20: ?
52. Five-sided Polyhedron: Pentahedron :: ?
Faces of Parallelepiped Bounded by a Square: ?
53. Motor: Helicopter :: Airflow: ?
54. Man: Ant :: Barter: ?
55. United States: Soviet Union :: Cubism: ?
56. State: Stipend :: Church: ?
57. Motorcycle: Bicycle :: Motordrome: ?
58. Transparent: Porous :: Obsidian: ?
59. pi*r^2*h: 1/3*pi*r^2*h :: Cylinder: ?
==> competition/tests/analogies/quest.s <==
Annotated solutions.
If there is more than one word that fits the analogy, we list the best
word first. Goodness of fit considers many factors, such as parallel
spelling, pronunciation or etymology. In general, a word that occurs
in Merriam-Webster's Third New International Dictionary is superior to
one that does not. If we are unsure of the answer, we mark it with
a question mark.
Most of these answers are drawn from Herbert M. Baus, _The Master
Crossword Puzzle Dictionary_, Doubleday, New York, 1981. The notation
in parentheses refers to the heading and subheading, if any, in Baus.
1. Mother: Maternal :: Stepmother: Novercal (STEPMOTHER, pert.)
2. Club: Axe :: Claviform: Dolabriform, Securiform (AXE, -shaped)
"Claviform" is from Latin "clava" for "club"; "securiform" is from
Latin "secura" for "axe"; "dolabriform" is from Latin "dolabra" for "to
hit with an axe." Thus "securiform" has the more parallel etymology.
However, only "dolabriform" occurs in Merriam-Webster's Third New
International Dictionary.
3. Cook Food: Pressure Cooker :: Kill Germs: Autoclave (PRESSURE, cooker)
4. Water: Air :: Hydraulic: Pneumatic (AIR, pert.)
5. Prediction: Dirac :: Proof: Anderson (POSITRON, discoverer)
6. Raised: Sunken :: Cameo: Intaglio (GEM, carved)
7. 1: 14 :: Pound: Stone (ENGLAND, weight)
8. Malay: Amok :: Eskimo Women: Piblokto (ESKIMO, hysteria)
9. Sexual Intercourse: A Virgin :: Bearing Children: A Nullipara
10. Jaundice, Vomiting, Hemorrhages: Syndrome :: Jaundice: Symptom (EVIDENCE)
11. Guitar: Cello :: Segovia: Casals (SPAIN, cellist)
12. Bars: Leaves :: Eagle: Stars (INSIGNIA)
13. Roll: Aileron :: Yaw: Rudder (AIRCRAFT, part)
14. 100: Century :: 10,000: Myriad, Banzai? (NUMBER)
"Century" usually refers to one hundred years, while "myriad" refers
to 10,000 things, but "century" can also mean 100 things. "Banzai"
is Japanese for 10,000 years.
15. Surface: Figure :: Mobius: Klein
16. Logic: Philosophy ::
To Know Without Conscious Reasoning: Theosophy (MYSTICISM)
There are many schools of philosophy that tout the possibility of
knowledge without conscious reasoning (e.g., intuitionism).
"Theosophy" is closest in form to the word "philosophy."
17. Alive: Parasite :: Dead: Saprophyte (SCAVENGER)
18. Sea: Land :: Strait: Isthmus (CONNECTION)
19. Moses: Fluvial :: Noah: Diluvial (FLOOD, pert.)
20. Remnant: Whole :: Meteorite: Meteoroid? (METEOR)
A meteorite is the remains of a meteoroid after it has
partially burned up in the atmosphere. The original meteoroid
may have come from an asteroid, comet, dust cloud, dark matter,
supernova, interstellar collision or other sources as yet unknown.
21. Opossum, Kangaroo, Wombat: Marsupial ::
Salmon, Sturgeon, Shad: Andromous (SALMON)
22. Twain/Clemens: Allonym :: White House/President: Metonym (FIGURE, of speech)
23. Sculptor: Judoka :: Fine: Martial (SELF, -defense)
24. Dependent: Independent :: Plankton: Nekton (ANIMAL, free-swimming)
25. Matthew, Mark, Luke, John: Gospels ::
Joshua-Malachi: Nebiim (HEBREW, bible books)
26. Luminous Flux: Lumen :: Sound Absorption: Sabin (SOUND, absorption unit)
27. 2: 3 :: He: Li (ELEMENT)
28. Growth: Temperature :: Pituitary Gland: Hypothalamus (BRAIN, part)
29. Spider: Arachnoidism :: Snake: Ophidism, Ophidiasis, Ophiotoxemia
None of these words is in Webster's Third.
30. Epigram: Anthology :: Foreign Passages: Chrestomathy, Delectus (COLLECTION)
These words are equally good answers.
31. Pathogen: Thermometer :: Lethal Wave: Dosimeter? (X-RAY, measurement)
What does "lethal wave" refer to? If it is radiation, then
a dosimeter measures the dose, not the effect, as does a thermometer.
32. Russia: Balalaika :: India: Sitar, Sarod (INDIA, musical instrument)
Both are guitar-like instruments (lutes) native to India.
33. Involuntary: Sternutatory :: Voluntary: Expectorant? (SPIT)
A better word would be an agent that tends to cause snorting or
exsufflation, which is the voluntary, rapid expulsion of air from
the lungs.
34. Unusual Hunger: Bulimia ::
Hunger for the Unusual: Allotriophagy, Pica (HUNGER, unusual)
These words are synonyms.
35. Blind: Stag :: Tiresias: Actaeon (STAG, changed to)
36. River: Fluvial :: Rain: Pluvial (RAIN, part.)
37. Country: City :: Tariff: Octroi (TAX, kind)
38. $/Dollar: Logogram :: 3, 5, 14, 20/Cent: Cryptogram (CODE)
39. Lung Capacity: Spirometer ::
Arterial Pressure: Sphygmomanometer (PULSE, measurer)
40. Gold: Ductile :: Ceramic: Fictile (CLAY, made of)
41. 7: 8 :: Uranium: Neptunium (ELEMENT, chemical)
42. Judaism: Messiah :: Islam: Mahdi (MOHAMMEDAN, messiah)
43. Sight: Amaurosis :: Smell: Anosmia, Anosphresia (SMELL, loss)
These words are synonyms.
44. Oceans: Cousteau :: Close Encounters of the Third Kind: Spielburg, Truffaut
Steven Spielburg was the person most responsible for the movie;
Francois Truffaut was a French person appearing in the movie.
45. Diamond/Kimberlite: Perimorph ::
Fungus/Oak: Endophyte, Endoparasite (PARASITE, plant)
An endoparasite is parasitic, while an endophyte may not be. Which
answer is best depends upon the kind of fungus.
46. Compulsion to Pull One's Hair: Trichotillomania ::
Imagine Oneself As a Beast: Zoanthropy, Lycanthropy
Neither word is exactly right: "zoanthropy" means imagining oneself
to be an animal, while "lycanthropy" means imagining oneself to be
a wolf.
47. Cross: Neutralism :: Hexagram: Zionism (ISRAEL, doctrine)
48. Wing: Tail :: Fuselage: Empennage, Engines, Waist? (TAIL, kind)
"Empennage" means the tail assemply of an aircraft, which is more a
synonym for "tail" than "wing" is for "fuselage." The four primary
forces on an airplane are: lift from the wings, negative lift from
the tail, drag from the fuselage, and thrust from the engines. The
narrow part at the end of the fuselage is called the "waist."
49. Bell: Loud :: Speak: Hear, Stentorian?
The Sanskrit root of "bell" means "he talks" or "he speaks"; the
Sanskrit root of "loud" means "he hears". A bell that makes a lot
of noise is loud; a speaker who makes a lot of noise is stentorian.
50. Benevolence: Beg :: Philanthropist: Mendicant, Mendicate?
If the analogy is attribute: attribute :: noun: noun, the answer
is "mendicant"; if the analogy is noun: verb :: noun: verb the
answer is "mendicate."
51. 10: Decimal :: 20: Vigesimal (TWENTY, pert.)
52. Five-sided Polyhedron: Pentahedron ::
Faces of Parallelepiped Bounded by a Square: ?
Does this mean a parallelepiped all of whose faces are bounded by
a square (and what does "bounded" mean), or does it mean all six
parallelograms that form the faces of a parallelepiped drawn in a
plane inside of a square?
53. Motor: Helicopter :: Airflow: Autogiro (HELICOPTER)
54. Man: Ant :: Barter: Trophallaxis
55. United States: Soviet Union :: Cubism: ? (ART, style)
If the emphasis is on opposition and collapse, there were several
movements that opposed Cubism and that died out (e.g., Purism,
Suprematism, Constructivism). If the emphasis is on freedom of
perspective versus constraint, there were several movements that
emphasized exact conformance with nature (e.g., Naturalism, Realism,
Photo-Realism). If the emphasis is on dominating the art
scene, the only movement that was contemporary with Cubism and
of the same popularity as Cubism was Surrealism. A better answer
would be an art movement named "Turkey-ism", since the Soviet Union
offered to exchange missiles in Cuba for missiles in Turkey during
the Cuban Missile Crisis.
56. State: Stipend :: Church: Prebend (STIPEND)
57. Motorcycle: Bicycle :: Motordrome: Velodrome (CYCLE, track)
58. Transparent: Porous :: Obsidian: Pumice (GLASS, volcanic)
59. pi*r^2*h: 1/3*pi*r^2*h :: Cylinder: Cone
==> competition/tests/math/putnam/putnam.1967.p <==
In article <584...@hpesoc1.HP.COM>, nich...@hpesoc1.HP.COM (Ron Nicholson) writes:
> Say that we have a hallway with n lockers, numbered from sequentialy
> from 1 to n. The lockers have two possible states, open and closed.
> Initially all the lockers are closed. The first kid who walks down the
> hallway flips every locker to the opposite state, that is, opens them
> all. The second kid flips the first locker door and every other locker
> door to the opposite state, that is, closes them. The third kid flips
> every third door, opening some, closing others. The forth kid does every
> fourth door, etc.
>
> After n kid have passed down the hallway, which lockers are open, and
> which are closed?
B4. (a) Kids run down a row of lockers, flipping doors (closed doors
are opened and opened doors are closed). The nth boy flips every nth
lockers' door. If all the doors start out closed, which lockers will
remain closed after infinitely many kids?
==> competition/tests/math/putnam/putnam.1967.s <==
B4. (a) Only lockers whose numbers have an odd number of factors will remain
closed, which are the squares.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> group/group.01.p <==
AEFHIKLMNTVWXYZ BCDGJOPQRSU
==> group/group.01.s <==
AEFHIKLMNTVWXYZ drawn with straight lines
BCDGJOPQRSU not drawn with straight lines
==> group/group.01a.p <==
147 0235689
==> group/group.01a.s <==
147 drawn with straight lines
0235689 not drawn with straight lines
==> group/group.02.p <==
ABEHIKMNOTXZ CDFGJLPQRSUVWY
==> group/group.02.s <==
ABEHIKMNOTXZ resembles corresponding Greek letter
CDFGJLPQRSUVWY does not resemble corresponding Greek letter
==> group/group.03.p <==
BEJQXYZ DFGHLPRU KSTV CO AIW MN
==> group/group.03.s <==
BEJQXYZ no US state starting with this letter
DFGHLPRU one US state starting with this letter
KSTV two US states starting with this letter
CO three US states starting with this letter
AIW four US states starting with this letter
five US states starting with this letter
six US states starting with this letter
seven US states starting with this letter
MN eight US states starting with this letter
==> group/group.04.p <==
BDO P ACGIJLMNQRSUVWZ EFTY HKX
==> group/group.04.s <==
All answers assume a sans serif type font:
BDO no endpoint
P one endpoint
ACGIJLMNQRSUVWZ two endpoints
EFTY three endpoints
HKX four endpoints
==> group/group.05.p <==
CEFGHIJKLMNSTUVWXYZ ADOPQR B
==> group/group.05.s <==
CEFGHIJKLMNSTUVWXYZ no enclosed area
ADOPQR one enclosed area (in some fonts, Q encloses two)
B two enclosed areas
==> group/group.06.p <==
BCEGKMQSW DFHIJLNOPRTUVXYZ
==> group/group.06.s <==
BCEGKMQSW prime numbers
DFHIJLNOPRTUVXYZ composites
==> group/group.07.p <==
CDEFLOPTZ ABGHIJKMNQRSUVWXY
==> group/group.07.s <==
Letters found on the eye chart
==> group/group.08.p <==
COS ABDEFGHIJKLMNPQRTUVWXYZ
==> group/group.08.s <==
Letters that contain only curves
==> group/group.09.p <==
CDILMVX ABEFGHJKNOPQRSTUWYZ
==> group/group.09.s <==
Letters that are Roman numerals
==> group/group.10.p <==
AHIMOTUVWXY BCDEFGJKLNPQRSZ
==> group/group.10.s <==
Letters with horizontal symmetry
==> group/group.11.p <==
BCDIJLMNOPQRSUVWZ AEFGHKTXY
==> group/group.11.s <==
Letters that can be drawn without taking a pen off of the paper
==> group/group.12.p <==
COPSUVWXZ ABDEFGHIJKLMNQRTY
==> group/group.12.s <==
Capital letters that look like lower case
==> group/group.13.p <==
BCDEHIKOX AFGJLMNPQRSTUVWYZ
==> group/group.13.s <==
Letters with vertical symmetry
==> group/group.14.p <==
EHIS MOT ABCDFGJKLNPQRUVWXYZ
==> group/group.14.s <==
Letters which in Morse code consist of only dots, only dashes, or both.
==> group/group.15.p <==
HIOX ABCDEFGJKLMNPQRSTUVWYZ
==> group/group.15.s <==
Letters with vertical and horizontal symmetry
==> group/group.16.p <==
IJ ABCDEFGHKLMNOPQRSTUVWXYZ
==> group/group.16.s <==
Letters that are dotted in lower case.
==> group/group.17.p <==
J BDFHIKLT GPQY ACEMNORSUVWXZ
==> group/group.17.s <==
Letters that have ascenders or dots or descenders in lower case.
Letters that have ascenders or dots but not descenders in lower case.
Letters that have descenders but not ascenders or dots in lower case.
Letters that do not have ascenders or dots or descenders in lower case.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> physics/balloon.p <==
A helium-filled balloon is tied to the floor of a car that makes a
sharp right turn. Does the balloon tilt while the turn is made?
If so, which way? The windows are closed so there is no connection
with the outside air.
==> physics/balloon.s <==
Because of buoyancy, the helium balloon on the string will want to move
in the direction opposite the effective gravitational field existing
in the car. Thus, when the car turns the corner, the balloon will
deflect towards the inside of the turn.
==> physics/brick.p <==
What is the maximum overhang you can create with an infinite supply of bricks?
==> physics/brick.s <==
You can create an infinite overhang.
Let us reverse the problem: how far can brick 1 be from brick 0?
Let us assume that the brick is of length 1.
To determine the place of the center of mass a(n):
a(1)=1/2
a(n)=1/n[(n-1)*a(n-1)+[a(n-1)+1/2]]=a(n-1)+1/(2n)
Thus
n 1 n 1
a(n)=Sum -- = 1/2 Sum - = 1/2 H(n)
m=1 2m m=1 m
Needless to say the limit for n->oo of half the Harmonic series is oo.
==> physics/bubbles.p <==
In a universe with the same physical laws, but which is mostly water
with little bubbles in it, do the bubbles attract, repel, or what?
==> physics/bubbles.s <==
A bubble should produce a gravitational field that is the negative of
that produced by an equal volume of water in an empty universe. This is
because a point in space would be affected only by the water in the
symmetric image of the bubble with respect to that point. The effect
on another bubble in that field would be to attract that bubble since
it would be pushing the water around it away. Therefore, the bubbles
should attract.
==> physics/cannonball.p <==
A person in a boat drops a cannonball overboard; does the water level change?
==> physics/cannonball.s <==
The cannonball in the boat displaces an amount of water equal to the MASS
of the cannonball. The cannonball in the water displaces an amount of water
equal to the VOLUME of the cannonball. Water is unable to support the
level of salinity it would take to make it as dense as a cannonball, so the
first amount is definitely more than the second amount, and the water level
drops.
==> physics/magnets.p <==
You have two bars of iron. One is magnetized along its length, the
other is not. Without using any other instrument (thread, filings,
other magnets, etc.), find out which is which.
==> physics/magnets.s <==
Take the two bars, and put them together like a T, so that one bisects the
other.
___________________
bar A ---> |___________________|
| |
| |
| |
| |
bar B ------------> | |
| |
| |
|_|
If they stick together, then bar B is the magnet. If they don't, bar A is
the magnet. (reasoning follows)
Bar magnets are "dead" in their centers (i.e., there is no magnetic force,
since the two poles cancel out). So, if bar A is the magnet, then bar B
won't stick to its center.
However, bar magnets are quite "alive" at their edges (i.e., the magnetic
force is concentrated). So, if bar B is the magnet, then bar A will stick
nicely to its end.
==> physics/milk.and.coffee.p <==
You are just served a hot cup of coffee and want it to be as hot as
possible later. If you like milk in your coffee, should you add it
when you get the cup or just before you drink it?
==> physics/milk.and.coffee.s <==
Normalize your temperature scale so that 0 degrees = room temperature.
Assume that the coffee cools at a rate proportional to the difference
in temperature, and that the amount of milk is sufficiently small that
the constant of proportinality is not changed when you add the milk.
An early calculus homework problem is to compute that the temperature
of the coffee decays exponentially with time,
T(t) = exp(-ct) T0, where T0 = temperature at t=0.
Let l = exp(-ct), where t is the duration of the experiment.
Assume that the difference in specific heats of coffee and milk are
negligible, so that if you add milk at temperature M to coffee at
temperature C, you get a mix of temperature aM+bC, where a and b
are constants between 0 and 1, with a+b=1. (Namely, a = the fraction
of final volume that is milk, and b = fraction that is coffee.)
If we let C denote the original coffee temperature and M the milk
temperature, we see that
Add milk later: aM + blC
Add milk now: l(aM+bC) = laM+blC
The difference is d=(1-l)aM. Since l<1 and a>0, we need to worry about
whether M is positive or not.
M>0: Warm milk. So d>0, and adding milk later is better.
M=0: Room temp. So d=0, and it doesn't matter.
M<0: Cold milk. So d<0, and adding milk now is better.
Of course, if you wanted to be intuitive, the answer is obvious if you
assume the coffee is already at room temperature and the milk is
either scalding hot or subfreezing cold.
Moral of the story: Always think of extreme cases when doing these puzzles.
They are usually the key.
Oh, by the way, if we are allowed to let the milk stand at room
temperature, then let r = the corresponding exponential decay constant
for your milk container.
Add acclimated milk later: arM + blC
We now have lots of cases, depending on whether
r<l: The milk pot is larger than your coffee cup.
(E.g, it really is a pot.)
r>l: The milk pot is smaller than your coffee cup.
(E.g., it's one of those tiny single-serving things.)
M>0: The milk is warm.
M<0: The milk is cold.
Leaving out the analysis, I compute that you should...
Add warm milk in large pots LATER.
Add warm milk in small pots NOW.
Add cold milk in large pots NOW.
Add cold milk in small pots LATER.
Of course, observe that the above summary holds for the case where the
milk pot is allowed to acclimate; just treat the pot as of infinite
size.
==> physics/mirror.p <==
Why does a mirror appear to invert the left-right directions, but not up-down?
==> physics/mirror.s <==
Mirrors invert front to back, not left to right.
The popular misconception of the inversion is caused by the fact that
a person when looking at another person expects him/her to face her/him,
so with the left-hand side to the right. When facing oneself (in the
mirror) one sees an 'uninverted' person.
See Martin Gardner, ``Hexaflexagons and other mathematical
diversions,'' University of Chicago Press 1988, Chapter 16. A letter
by R.D. Tschigi and J.L. Taylor published in this book states that the
fundamental reason is: ``Human beings are superficially and grossly
bilaterally symmetrical, but subjectively and behaviorally they are
relatively asymmetrical. The very fact that we can distinguish our
right from our left side implies an asymettry of the perceiving
system, as noted by Ernst Mach in 1900. We are thus, to a certain
extent, an asymmetrical mind dwelling in a bilaterally symmetrical
body, at least with respect to a casual visual inspection of our
external form.''
Martin Gardner has also written the book ``The Ambidextrous Universe.''
==> physics/monkey.p <==
Hanging over a pulley there is a rope, with a weight at one end.
At the other end hangs a monkey of equal weight. What happens if
the monkey starts to ascend the rope? Assume that the mass of the
rope and pulley are negligible, and the pulley is frictionless.
==> physics/monkey.s <==
The monkey is pulling down on the rope hard enough to pull itself up. This
increases the tension in the rope just enough to cause the weight to rise at
the same rate as the monkey, since they are of equal mass.
==> physics/pole.in.barn.p <==
Accelerate a pole of length l to a constant speed of 90% of the speed of
light (.9c). Move this pole towards an open barn of length .9l (90%
the length of the pole). Then, as soon as the pole is fully inside the
barn, close the door. What do you see and what actually happens?
==> physics/pole.in.barn.s <==
What the observer sees depends upon where the observer is, due to
the finite speed of light.
For definiteness, assume the forward end of the pole is marked "A" and
the after end is marked "B". Let's also assume there is a light source
inside the barn, and that the pole stops moving as soon as end "B" is
inside the barn.
An observer inside the barn next to the door will see the following
sequence of events:
1. End "A" enters the barn and continues toward the back.
2. End "B" enters the barn and stops in front of the observer.
3. The door closes.
4. End "A" continues moving and penetrates the barn at the far end.
5. End "A" stops outside the barn.
An observer at the other end of the barn will see:
1. End "A" enters the barn.
2. End "A" passes the observer and penetrates the back of the barn.
3. If the pole has markings on it, the observer will notice the part
nearest him has stopped moving. However, both ends are still
moving.
4. End "A" stops moving outside the barn.
5. End "B" continues moving until it enters the barn and then stops.
6. The door closes.
After the observers have subtracted out the effects of the finite speed
of light on what they see, both observers will agree on what happened:
The pole entered the barn; the door closed so that the pole was
completely contained within the barn; as the pole was being stopped it
elongated and penetrated the back wall of the barn.
Things are different if you are riding along with the pole. The pole
is never inside the barn since it won't fit. End A of the pole penetrates
the rear wall of the barn before the door is closed.
If the wall of the barn is impenetrable, in all the above scenarios insert
the wording "End A of the pole explodes" for "End A penetrates the barn."
==> physics/resistors.p <==
What is the resistance between various pairs of vertices on a lattice
of unit resistors in the shape of a
1. Cube,
2. Platonic solid,
3. N dimensional Hypercube,
4. Infinite square lattice,
and
5. between two small terminals on a continuous sheet?
==> physics/resistors.s <==
1. Cube
The key idea is to observe that if you can show that two
points in a circuit must be at the same potential, then you can
connect them, and no current will flow through the connection and the
overall properties of the circuit remain unchanged. In particular, for
the cube, there are three resistors leaving the two "connection
corners". Since the cube is completely symmetrical with respect to the
three resistors, the far sides of the resistors may be connected
together. And so we end up with:
|---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
| | |---WWWWWW---| | |
*--+---WWWWWW---+-+---WWWWWW---+-+---WWWWWW---+---*
| | |---WWWWWW---| | |
|---WWWWWW---| |---WWWWWW---| |---WWWWWW---|
|---WWWWWW---|
This circuit has resistance 5/6 times the resistance of one resistor.
2. Platonic Solids
Same idea for 8, 12 and 20, since you use the symmetry to identify
equi-potential points. The tetrahedron is a hair more subtle:
*---|---WWWWWW---|---*
|\ /|
W W W W
W W W W
W W W W
| \ / |
\ || |
\ | /
\ W /
\ W / <-------
\ W /
\|/
+
By symmetry, the endpoints of the marked resistor are equi-potential. Hence
they can be connected together, and so it becomes a simple:
*---+---WWWWW---+----*
| |
+-WWW WWW-+
| |-| |
|-WWW WWW-|
3. Hypercube
Think of injecting a constant current I into the start vertex.
It splits (by symmetry) into n equal currents in the n arms; the current of
I/n then splits into I/n(n-1), which then splits into I/[n(n-1)(n-1)] and so
on till the halfway point, when these currents start adding up. What is the
voltage difference between the antipodal points? V = I x R; add up the voltages
along any of the paths:
n even: (n-2)/2
V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1) )}
n odd: (n-3)/2
V = 2{I/n + I/(n(n-1)) + I/(n(n-1)(n-1)) + ... + I/(n(n-1) )} (n-1)/2
+ I/(n(n-1) )
And R = V/I i.e. replace the Is in the above expression by 1s.
For the 3-cube: R = 2{1/3} + 1/(3x2) = 5/6 ohm
For the 4-cube: R = 2{1/4 + 1/(4x3)} = 2/3 ohm
This formula yields the resistance from root to root of
two (n-1)-ary trees of height n/2 with their end nodes identified
(-when n is even; something similar when n is odd).
Coincidentally, the 4-cube is such an animal and thus the answer
2/3 ohms is correct in that case.
However, it does not provide the solution for n >= 5, as the hypercube
does not have quite as many edges as were counted in the formula above.
4. The Infinite Plane
For an infinite lattice: First inject a constant current I at a point; figure
out the current flows (with heavy use of symmetry). Remove that current. Draw
out a current I from the other point of interest (or inject a negative current)
and figure out the flows (identical to earlier case, but displaced and in the
other direction). By the principle of superposition, if you inject a current I
into point a and take out a current I at point b at the same time, the currents
in the paths are simply the sum of the currents obtained in the earlier two
simpler cases. As in the n-cube, find the voltage between the points of
interest, divide by I and voila`!
As an illustration, in the adjacent points case: we have a current of I/4 in
each of the four resistors:
^ |
| v
<--o--> -->o<--
| ^
v |
(inject) (take out)
And adding the currents, we have I/2 in the resistor connecting the two points.
Therefore V=(1 ohm) x I/2 and effective resistance between the points = 1/2 ohm.
We do not derive it, but the equivalent resistance between two nodes k
diagonal units apart is (2/pi)(1+1/3+1/5+...+1/(2k-1)); that, plus
symmetry and the known equivalent resistance between two adjacent
nodes, is sufficient to derive all equivalent resistances in the
lattice.
5. Continuous sheet
I think the answer is (rho/dz)log(L/r)/pi where rho is the resistivity,
dz is the sheet thickness, L is the separation, r is the terminal radius.
cf. "Random Walks and Electric Networks", by Doyle and Snell, published by the
Mathematical Association of America.
==> physics/sail.p <==
A sailor is in a sailboat on a river. The current is 3 knots with respect
to the land. The wind (air velocity) is zero, with respect to the
land. The sailor wants to proceed downriver as quickly as possible,
maximizing his downstream speed with respect to the land.
Should he raise the sail, or not?
==> physics/sail.s <==
Depends on the sail. If the boat is square-rigged, then not, since
raising the sail will simply increase the air resistance.
If the sailor has a fore-and-aft rig, then he should, since he can then
tack into the wind. (Imagine the boat in still water with a 3-knot head
wind).
==> physics/shoot.sun.p <==
If you are standing at the equator at sunrise, where must you point a laser
cannon to hit the Sun dead center? Assume that the Sun is stationary and
that the Earth's orbit around it is circular.
==> physics/shoot.sun.s <==
You aim it at the horizon. The sun is exactly in the place where it
appears to be. It is true that the sun wasn't on the horizon 8 minutes
ago (the specific number is 2 degrees), when it emitted the light you
are now seeing. However, "the sun wasn't on the horizon" doesn't mean
the sun moved; it means the horizon moved.
==> physics/skid.p <==
What is the fastest way to make a 90 degree turn on a slippery road?
==> physics/skid.s <==
For higher speeds (measured at a small distance from the point of initiation
of a sharp turn) the fastest way round is to "outside loop" - that is, steer
away from the curve, and do a skidding 270.
This technique is taught in advanced driving schools.
References:
M. Freeman and P. Palffy, American Journal of Physics, vol 50, p. 1098, 1982.
P. Palffy and Unruh, American Journal of Physics, vol 49, p. 685, 1981.
==> physics/spheres.p <==
Two spheres are the same size and weight, but one is hollow. They are
each made of uniform material, though of course not the same material.
With a minimum of apparatus, how can I tell which is hollow?
==> physics/spheres.s <==
Since the balls have equal diameter and equal mass, their volume and
density are also equal. However, the mass distribution is not equal,
so they will have different moments of inertia - the hollow sphere has
its mass concentrated at the outer edge, so its moment of inertia will
be greater than the solid sphere. Applying a known torque and observing
which sphere has the largest angular acceleration will determine which
is which. An easy way to do this is to "race" the spheres down an
inclined plane with enough friction to prevent the spheres from sliding.
Then, by conservation of energy:
mgh = 1/2 mv^2 + 1/2 Iw^2
Since the spheres are rolling without sliding, there is a relationship
between velocity and angular velocity:
w = v / r
so
mgh = 1/2 mv^2 + 1/2 I (v^2 / r^2) = 1/2 (m + I/r^2) v^2
and
v^2 = 2mgh / (m + I / r^2)
From this we can see that the sphere with larger moment of inertia (I) will
have a smaller velocity when rolled from the same height, if mass and radius
are equal with the other sphere. Thus the solid sphere will roll faster.
==> physics/wind.p <==
Is a round-trip by airplane longer or shorter if there is wind blowing?
==> physics/wind.s <==
It will take longer, by the ratio (s^2)/(s^2 - w^2) where s is the
plane's speed, and w is the wind speed. The stronger the wind the
longer it will take, up until the wind speed equals the plane's speed,
at which point the plane will run out of fuel before too long.
Math:
s = plane's speed
w = wind speed
d = distance in one direction
d / (s + w) = time to complete leg flying with the wind
d / (s - w) = time to complete leg flying against the wind
d / (s + w) + d / (s - w) = round trip time
d / (s + w) + d / (s - w) = ratio of flying with wind to
------------------------- flying with no wind (bottom of
d / s + d / s equation is top with w = 0)
this simplifies to s^2 / (s^2 - w^2).
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> real-life/icecubes.p <==
You have an old-fashioned refrigerator with a small freezer compartment
which could hold seven ice cube trays stacked vertically, but there are
no shelves to separate the trays. You have an unlimited supply of trays,
each of which can make a dozen cubes, but if you stand one on top of
another before it's frozen, it will nest part way into it and you won't
get full cubes from the bottom tray. So, what is the fastest way to
make ice cubes?
==> real-life/icecubes.s <==
By using frozen cubes as spacers to hold the trays apart, you can make
84 cubes in the time it takes to freeze two trays. Fill one tray,
freeze it and remove the cubes. Place two cubes in the opposite
corners of six trays, and fill the rest with water. Freeze all six,
plus a seventh you put on top, at the same time.
==> real-life/microwave.p <==
Every morning when I warm my milk for breakfast, I put one cup of milk
in the microwave (which is in working order) for exactly 84 seconds. Why?
==> real-life/microwave.s <==
When you put your cup in the microwave, the handle on the cup is
pointing towards you. Your microwave has a revolving plate on which the
cup is placed. After 84 seconds, the handle is again in it's initial
position, so you can take the cup out without burning your hands.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> references/books/bloopers.p <==
What are some errors made in puzzle books?
==> references/books/bloopers.s <==
geometry/fence.s
Charles W. Trigg, Mathematical Quickies, Dover, 1985, #93
logic/hundred.s
Angela Dunn, ed., Mathematical Bafflers, Dover, 1980, p. 112
David Wells, The Penguin Book of Curious and Interesting Puzzles, Penguin,
1992, #373 & #554
==> references/books/masquerade.p <==
What is the solution to _Masquerade_ by Kit Williams?
==> references/books/masquerade.s <==
The original book:
_Masquerade_ by Kit Williams, Jonathan Cape, London, 1979
The answer book:
_Masquerade The Complete Book with the Answer Explained_ by Kit Williams,
Jonathan Cape, London, 1982
The whole story:
_The Quest for the Golden Hare_ by Bamber Gascoigne, Jonathan Cape,
London, 1983
_Masquerade_ contains fifteen very detailed one- or two-page paintings
rendered in the fantastic style typical of a high quality children's
book, together with a dreamy story containing characters such as Jack
Hare, Tara Tree-tops and the Lady Moon. Most of the very lifelike
people in the paintings are actual friends of Mr. Williams. This book
set off a frenzy of solving activity unequalled by any subsequent book,
even though its imitators offered much higher prizes, culminating in
the $500,000 of the book _Treasure_ with puzzle by Paul Hoffman (a.k.a.
Dr. Crypton).
The solution to Masquerade is simplicity itself, and is fully in
keeping with the nature of the book: namely, a picture book. First of
all, the text has nothing to do with it; the pictures alone contain the
answer. Secondly, the answer is literally pointed to by the pictures.
Each picture is bordered by letters, which is a dead giveaway since the
letters have no reason for being there if they are not part of the
puzzle. By drawing a line from the eyes of the various creatures in
the pictures, through their longest fingers, biggest toes, etc., and
extending to the bordering letters, this message is found:
CATHERINES LONG FINGER OVER SHADOWS EARTH BURIED YELLOW AMULET MIDDAY
POINTS THE HOUR IN LIGHT OF EQUINOX LOOK YOU.
The first letter from each page spells:
CLOSE BY AMPTHILL
This method of solution is hinted to on the title page with the rhyme:
To solve the hidden riddle, you must use your eyes,
And find the hare in every picture that may point you to the prize.
Armed with this information, it is a simple matter to discover that
there is a statue of Catherine of Aragon in a public park near the
village of Ampthill. By doing a little amateur astronomy, the exact
spot pointed to by the statue's long finger can be determined without
waiting for the equinox. Beneath this spot was the treasure, a golden
hare. The book also contains a number of confirming clues.
_Quest_ chronicles some of the amazingly far-fetched approaches taken
by Masqueraders. Mr. Gascoigne, a respected author on the arts,
accompanied Mr. Williams the night he buried the treasure. He also
read the tens of thousands of letters received by Mr. Williams. The
hare was found three years after the book was published by a shadowy
figure with pseudonym Ken Thomas. Mr. "Thomas" found the hare by
researching Mr. Williams' life, going to places that he had lived, and
doing a lot of digging with the occasional help of some of the
confirming clues. Two British physicists did finally solve the puzzle
with the help of a hint published by Mr. Williams in the Sunday Times,
but they were a little too late.
After the announcement that the hare was unearthed, many fanatical
Masqueraders tried to prove that their approaches could lead to the
correct solution. For example, someone discovered that the word
"thill" means a fleck of paint (according to some obscure dictionary),
and he thought he saw an inexplicable fleck of paint in each painting.
He also thought he saw the word "amp" hidden in each painting. For
example, in one picture a girl is floating in the air above houses.
And a volt (vault) over an ohm (home) is an amp. Mr. Gascoigne
summarizes his observations thus:
Tens of thousands of letters from Masqueraders have convinced me that
the human mind has an equal capacity for pattern-matching and
self-deception. While some addicts were busy cooking the riddle,
others were more single-mindedly continuing their own pursuit of the
hare quite regardless of the news that it had been found. Their own
theories had come to seem so convincing that no exterior evidence could
refute them. These most determined of Masqueraders may grudgingly have
accepted that a hare of some sort was dug up at Ampthill, but they
believed there would be another hare, or a better solution, awaiting
them at their favourite spot. Kit would expect them to continue
undismayed by the much publicised diversion at Ampthill and would be
looking forward to the day when he would greet them as the real
discoverers of the real puzzle of Masquerade. Optimistic expeditions
were still setting out, with shovels and maps, throughout the summer of
1982.
==> references/books/maze.p <==
What is the solution to _Maze_ by Christopher Manson?
==> references/books/maze.s <==
In room 29, a door to room 17 is hidden to look like a table. Using this door
this 16-step tour exists: 1 26 30 42 4 29 17 45 23 8 12 39 4 15 37 20 1.
The riddle of room 45 remains to be solved.
==> references/books/treasure.p <==
What is the solution to _Treasure_ by Dr. Crypton?
==> references/books/treasure.s <==
"Treasure" was a puzzle by Dr. Crypton (Paul Hoffman) released
simultaneously in 1984 as a book, a videotape and a laserdisk. The book
and video versions include a number of mysterious pictures and images
connected by a loose plot involving the theft of a golden horse. The
1-kilo golden horse itself was buried, and the mysterious images were
supposed to give instructions on how to find it. The lucky winners would
get the golden horse and $500,000. The clues were interesting and
obscure; it was impossible to tell which of the puzzles were relevant to
the solution and which weren't. Enough of them were sort of solvable to
give people hope that they were on the right track. For example, some
clues written on an umbrella gave the birth and death years of Mary, Queen
of Scots; and a chess game turned out to be identifiable as Anderssen vs.
Kieseritzky, the "Shower of Gold" game. Evidently neither of these
observations was relevant to the solution in the end.
It was alleged that during the production of the video enough people
were let in on the secret that the location had to be changed... but
that very little of the puzzle was changed to reflect the new location.
Nobody solved the puzzle in time -- i.e. by midnight of 26 May 1989.
The horse was dug up by the promoters and the prize donated to a charity:
Big Brothers and Sisters of America. However, the promoters and Dr.
Crypton refused to make the solution public. Seven months later two
men, Nick Boone and Anthony Castaneda, went to Tennessee Pass in Colorado
and dug up a vial with congratulations inside. They wrote a description
of their thought processes that left other frustrated treasure-seekers
suspicious and annoyed: their "solution" appeared to be motivated very
little by anything in the puzzle itself, so that it seemed apparent to
many that they were virtually guided to that location by the promoters.
This suspicion has not been confirmed or denied.
--Jim Gillogly <uunet!rand.org!James_Gillogly>
==> references/books/unnamed.p <==
What is the solution to the unnamed book by Kit Williams?
==> references/books/unnamed.s <==
The title is "The Bee on the Comb."
In the first picture, there are two "hybrid" animals, one half-mouse,
half-horse, the other half-cat, half-toad. If you've read
"Masquerade", the drawings remind you of the circle of animals in one
of the pictures in that book, and there's even a footnote there
explaining the names of the animals in that picture. Using the same
reasoning, the two animals in "The Bee on the Comb" ought to be called
a "morse" and a "coad". So the obvious conclusion is that this is a
clue indicating that Morse code is involved. The Morse code is around
the frame of the gardening picture, and spells out "All animals are
equal in a tale of tail to tail, end to end to end." This is the same
message that is around the picture in "Masquerade."
Each picture in "The Bee on the Comb" contains a hidden animal. Ignore
all the naturalistic animals: you're looking just for one animal hidden
in some visually punning way. For example, in the first picture,
there's a parrot hidden in the young man's vest--turn the page upside
down and the leaves pictured on his vest become the parrot's feathers.
If you write down all fifteen hidden animals and take their last
letters, "end to end to end", it spells out "The Bee on the Comb". I
recall that we found the hidden animal in the picture on the kitchen
(the one with the box of Oxo cubes on the mantel) particularly
difficult to find, though I expect that'll vary from person to person.
The hidden animals are wonderfully cleverly hidden. Oh, and the animal
ending in C is rather obscure; I think we had to figure out its name
only after we'd figured out the title of the book and knew it ended in
C.
If you count the number of bees in each picture and convert it to
letters, using A = 1, B = 2, etc., you get "Bees Only Sting". By
looking at the honeycomb that obscures the title on the cover, you can
see how many letters the words in the title contain, and "Bees Only
Sting" does not work.
There's at least one other indication that the bees are a red herring. The
fourth line from the end of the text reads "the bees they are of little
consequence". I'm not positive that this isn't a coincidence, but it sure
looks like it might be a message to ignore the bees.
Scott Marley
hu...@well.sf.ca.us
==> references/faq.p <==
Where should I look if I can't find the answer here?
==> references/faq.s <==
FAQs are available via ftp from rtfm.mit.edu.
*******
physics
*******
Sci.Physics is an unmoderated newsgroup dedicated to the discussion
of physics, news from the physics community, and physics-related social
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the following:
This Frequently Asked Questions List is posted monthly, at or near
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material which is worth preserving. If you have corrections or answers to
other frequently asked questions that you would like included in this
posting, send E-mail to sic...@csa2.lbl.gov (Scott I. Chase).
Index of Subjects
-----------------
1. An Introduction to Sci.Physics
2. Gravitational Radiation
3. Energy Conservation in Cosmology and Red Shift
4. Effects Due to the Finite Speed of Light
5. The Top Quark
6. Tachyons
7. Special Relativistic Paradoxes
(a) The Barn and the Pole
(b) The Twin Paradox
8. The Particle Zoo
9. Olbers' Paradox
10. What is Dark Matter?
11. Hot Water Freezes Faster than Cold!
12. Which Way Will my Bathtub Drain?
13. Why are Golf Balls Dimpled?
14. Why do Mirrors Reverse Left and Right?
15. What is the Mass of a Photon?
16. How to Change Nuclear Decay Rates
17. Baryogenesis - Why Are There More Protons Than Antiprotons?
18.*Time Travel - Fact or Fiction?
19.*The Nobel Prize for Physics
20. Open Questions
21. Accessing and Using Online Physics Resources
****
math
****
This is a list of frequently asked questions for sci.math (version 3.5).
Any contributions/suggestions/corrections are most welcome. Please use
* e-mail * on any comment concerning the FAQ list.
Changes of version will be important enough to deserve reading the FAQ
list again. Additions are marked with a # on the table of contents.
Still you may kill all versions of FAQ using the * wildcard. (Ask your
local unix guru for ways to do so). The FAQ is available via ftp in
rtfm.mit.edu (18.172.1.27).
The list of contributors to this FAQ list is to large to include here;
but thanks are due to all of them (you know who you are folks).
Table of Contents
-----------------
1Q.- Fermat's Last Theorem, status of ..
2Q.- Four Colour Theorem, proof of ..
3Q.- Values of Record Numbers
4Q.- General Netiquette
5Q.- Computer Algebra Systems, application of ..
6Q.- Computer Algebra Systems, references to ..
7Q.- Fields Medal, general info ..
8Q.- 0^0=1. A comprehensive approach ..
9Q.- 0.999... = 1. Properties of the real numbers ..
10Q.- Digits of Pi, computation and references ..
11Q.- There are three doors, The Monty Hall problem ..
12Q.- Surface and Volume of the n-ball
13Q.- f(x)^f(x)=x, name of the function ..
14Q.- Projective plane of order 10 ..
15Q.- How to compute day of week of a given date ....
16Q.- Axiom of Choice and/or Continuum Hypothesis?
17Q.- Cutting a sphere into pieces of larger volume
18Q.- Pointers to Quaternions
***********
weird stuff
***********
This is the sci.skeptic FAQ. It is intended to provide a factual base
for most of the commonly discussed topics on sci.skeptic.
Unfortunately I don't have much time to do this in, and anyway a FAQ
should be the Distilled Wisdom of the Net rather than just My Arrogant
Opinion, so I invite submissions and let all the net experts out there
fill in the details. Submissions from any point of view and on any
sci.skeptic topic are welcomed, but please keep them short and to the
point. The ideal submission is a short summary with one or two
references to other literature. I have added comments in square
brackets where I think more information is particularly needed, but
don't let that stop you sending something else.
Background
----------
0.1: What is sci.skeptic for?
0.2: What is sci.skeptic not for?
0.3: What is CSICOP? Whats their address? +
0.4: What is "Prometheus"?
0.5: Who are some prominent skeptics? +
0.6: Aren't all skeptics just closed-minded bigots?
0.7: Aren't all paranormalists just woolly-minded fools?
0.8: What is a "conspiracy theory"?
The Scientific Method
---------------------
1.1: What is the scientific method?
1.2: What is the difference between a fact, a theory and a hypothesis?
1.3: Can science ever really prove anything?
1.4: If scientific theories keep changing, where is the Truth?
1.5: What evidence is needed for an extraordinary claim?
1.6: What is Occam's Razor?
1.7: Galileo was persecuted, just like researchers into <X> today.
1.8: What is the "Experimenter effect".
1.9: How much fraud is there in science? *
1.9.1: Did Mendel fudge his results? *
Psychic Powers
--------------
2.1: Is Uri Geller for real? *
2.2: I have had a psychic experience. +
2.3: What is "sensory leakage"?
2.4: Who are the main psi researchers? +
2.5: Does dowsing work? +
2.6: Could psi be inhibited by the presence of skeptics?
UFOs/Flying Saucers
-------------------
3.1 What are UFOs?
3.1.1: Are UFOs alien spacecraft?
3.1.2: Are UFOs natural phenomena?
3.1.3: But isn't it possible that aliens are visiting Earth?
3.2: Is it true that the US government has a crashed flying saucer?
(MJ-12)? +
3.3: What is "channeling"?
3.4: How can we test a channeller?
3.5: I am in telepathic contact with the aliens.
3.6: Some bozo has just posted a load of "teachings" from a UFO. What
should I do?
3.7: Are crop circles made by flying saucers?
3.7.1: Are crop circles made by "vortices"?
3.7.2: Are crop circles made by hoaxers?
3.7.3: Are crop circles radioactive?
3.7.4: What about cellular changes in plants within crop circles?
3.8: Have people been abducted by UFOs?
3.9: What is causing the strange cattle deaths? *
3.10: What is the face on Mars?
3.11: Did Ezekiel See a Flying Saucer?
Faith Healing and Alternative Therapies
---------------------------------------
4.1: Isn't western medicine reductionistic and alternatives holistic? +
4.2: What is a double-blind trial? What is a placebo?
4.3: Why should scientific criteria apply to alternative therapies?
4.4: What is homeopathy? +
4.5: What is aroma therapy?
4.6: What is reflexology? +
4.7: Does acupuncture work?
4.8: What about psychic surgery?
4.9: What is Crystal Healing?
4.10: Does religious healing work? +
4.11: What harm does it do anyway?
Creation versus Evolution
-------------------------
5.1: Is the Bible evidence of anything? +
5.2: Could the Universe have been created old?
5.3: What about Carbon-14 dating?
5.4: What is "dendrochronology"?
5.5: What is evolution? Where do I find out more?
5.6: The second law of thermodynamics says....
5.7: How could living organisms arise "by chance"?
5.8: But doesn't the human body seem to be well designed?
5.9: What about the thousands of scientists who have become Creationists?
Fire-walking
-----------
6.1: Is fire-walking possible?
6.2: Can science explain fire-walking?
New Age
-------
7.1: What do New Agers believe?
7.2: What is the Gaia hypothesis?
7.3: Was Nostradamus a prophet?
7.4: Does astrology work? *
7.4.1: Could astrology work by gravity? *
7.4.2: What is the `Mars Effect'? *
Strange Machines: Free Energy and Anti-Gravity
----------------------------------------------
8.1: Why don't electrical perpetul motion machines work?
8.2: Why don't magnetic perpetual motion machines work?
8.3: Why don't mechanical perpetual motion machines work?
8.4: Magnets can levitate. Where is the energy from?
8.5: But its been patented!
8.6: The oil companies are conspiring to suppress my invention
8.7: My machine gets its free energy from <X>
8.8: Can gyroscopes neutralise gravity?
8.9: My prototype gets lighter when I turn it on.
AIDS
----
9.1: What about these theories on AIDS?
9.1.1: The Mainstream Theory
9.1.2: Strecker's CIA Theory
9.1.3: Duesberg's Risk-Group Theory
==> references/magazines.p <==
What magazines and journals contain puzzles?
==> references/magazines.s <==
AMAYC Review, The
Algorithm
A. K. Dewdney's magazine devoted to recreational computing.
Quarterly
$19.95 per year US, $24.95 Canada, $23.95 elsewhere (all prices US)
Louis Magguilli
Algorithm
P.O. Box 29237
Westmount Postal Outlet
785 Wonderland Road S.
London, Ontario N6K 1M6
Canada
American Mathematical Monthly
Monthly
$32US/year for MAA members
Mathematical Association of America
1529 Eighteenth Street, N.W.
Washington, DC 20036-1385
Arbelos (full of problems)
Bent, The
BIT
Bulletin of the Institute of Mathematics and Its Applications
College Mathematics Journal
Five times per year
$20US/year for MAA members
Mathematical Association of America
1529 Eighteenth Street, N.W.
Washington, DC 20036-1385
Consortium
Crux Mathematicorum (formerly: EUREKA -- all problems)
$25C/year
Dr. Kenneth S. Williams
Canadian Mathematical Society
577 King Edward Avenue
Ottawa
Ontario
Canada K1N 6N5
Cubism For Fun
CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
Cubists Club). It appears a bit irregular, but a few times a year.
Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
approximately $ 15.-. Institutional membership is also possible.
Information is available from the editor:
Anton Hanegraaf
Heemskerkstraat 9
6662 AL Elst
The Netherlands
Delta (Waukesha)
Discrete Mathematics
EATCS Bulletin
Fibonacci Quarterly, The
Games
The best-known puzzle and game publication. A wide variety of puzzles
and articles in every issue.
Bimonthly
$17.97 per year US, $22.97 Canada, $27.97 elsewhere (all prices US)
Games
P.O. Box 605
Mt. Morris, IL 61054-0605
1-(800)-827-1256
James Cook Mathematical Notes
Journal of Algorithms
Journal of Automated Reasoning
Journal of Recreational Mathematics
A must for anyone interested in recreational mathematics.
Quarterly
$23.45 per year for US and Canada, $28.30 elsewhere
Baywood Publishing Company, Inc.
26 Austin Avenue
P.O. Box 337
Amityville, NY 11701
Mathematical Digest
Mathematical Gazette, The
Mathematical Intelligencer
Mathematical Spectrum
Mathematics and Computer Education (formerly: The AMATYC Journal)
Mathematics Magazine
Bimonthly
$16US/year for MAA members
Mathematical Association of America
1529 Eighteenth Street, N.W.
Washington, DC 20036-1385
Mathematics Teacher, The
Ontario Secondary School Mathematics Bulletin
Parabola
Pentagon, The
Pi Mu Epsilon
Problem Solver, The
PS News
PuzzleSIGns
Publication of the Mensa "Puzzle" SIG. This fledgling newsletter
contains a variety of puzzles in every issue. Sample issue $1.
Quarterly
$7 per year for Mensa members, $8 non-members, $10 foreign
Chuck Murphy
Puzzle SIGns Coordinator
11430 East Palomino Road
Scottsdale, AZ 85259
Real Analysis Exchange (only "queries")
REC (Recreational & Educational Computing)
Devoted to recreational computing.
8 issues per year
$27 per year US, $28 Canada, $36 elsewhere
Michael Ecker
909 Violet Terrace
Clarks Summit, PA 18411
Science of Computer Programming
School Science and Mathematics
SIAM Review
Technology Review
Word Fun
Publication of the Mensa "Fun with Word" SIG, but anyone may
subscribe. A variety of wordplay and puzzles; fantastic bargain.
Sample issue $.50 stamps per coin (no checks) + business-size SASE.
Bimonthly
$5 per year US and Canada, $10 elsewhere
Jill Conway
Rte. 6
3001 Johnson Lane
Columbia, MO 65202
Word Ways
An absolutely fantastic journal devoted to recreational linguistics;
a must for anyone who loves words or word puzzles.
Quarterly
$17 per year
Faith W. Eckler
Spring Valley Road
Morristown, NJ 07960
German language:
Berita Matematik
Elemente der Mathematik
Kvant
Matematicko - Fizicki Lijt
Mathematik in der Schule
Mathematika Tanitasa, A
Menemui Mathematik
Nieuw Archief voor Wiskunde
French language:
Nouvel Archimede, le
Revue des Mathematiques Speciales (mostly problems from entrance exams)
Dutch language:
Euclides
Italian language:
Matematiche, Le
Russian language:
Matematika v Shkole
Scandinavian language:
Normat (formerly Nordisk Matematisk Tidskrift)
Hungarian language:
Kozepiskolai Matematikai Lapok (koMaL)
Matematikai Lapok (but the problems are stated in English)
Ceased publication:
Graham Dial, The
Mathematics Student Journal, The
Nabla
==> references/organizations.p <==
What organizations exist for puzzle lovers?
==> references/organizations.s <==
American Cryptogram Association
Publication:
The Cryptogram
Bimonthly
Dues:
See below
Treasurer:
ACA Treasurer
18789 West Hickory St.
Mundelein, IL 60060
Comments:
Devoted to cryptography. Every issue of the journal contains
several thoughtful articles and a large number of puzzles, including
aristocrats, patristocrats, xenocrypts, cipher exchanges and
cryptarithms. Members have the option of picking a "nom" (nom de
plume), e.g. the president is Gizmo. As it is a specialized
organization, you should request a sample issue first (I don't
know the procedure for this, but $1 and a SASE should do it).
The National Puzzlers' League
Publication:
The Enigma
Monthly
Dues:
See below
Editor:
Judith E. Bagai
Box 82289
Portland, OR 97282
Comments:
Simply the best organization devoted to word puzzles. The _Enigma_
contains over 80 word puzzles per issue, ranging in difficulty from
easy to extremely difficult and in type from the familiar anagrams
and riddles to such obscure forms as spoonergrams and acrostical
enigmas. Each issue also includes a member-written cryptic. Members
get to pick a "nom" (nom de plume), e.g. I'm Cubist and Chris Cole
is Canon. The NPL is a somewhat specialized organization, so you
should send a SASE with a request for a mini-sample to the editor
to see if it's for you.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> trivia/area.codes.p <==
When looking at a map of the distribution of telephone area codes for
North America, it appears that they are randomly distributed. I am
doubtful that this is the case, however. Does anyone know how the area
codes were/are chosen?
==> trivia/area.codes.s <==
Originally, back in the middle 1950's when direct dialing of long
distance calls first became possible, the idea was to assign area codes
with the 'shortest' dialing time required to the larger cities.
Touch tone dialing was very rare. Most dialed calls were with 'rotary'
dials. Area codes like 212, 213, 312 and 313 took very little time to
dial (while waiting for the dial to return to normal) as opposed, for
example, to 809, 908, 709, etc ...
So the 'quickest to dial' area codes were assigned to the places which
would probably receive the most direct dialed calls, i.e. New York City
got 212, Chicago got 312, Los Angeles got 213, etc ... Washington, DC got
202, which is a little longer to dial than 212, but much shorter than
others.
In order of size and estimated amount of telephone traffic, the numbers
got larger: San Francisco got 415, which is sort of in the middle, and
Miami got 305, etc. At the other end of the spectrum came places like
Hawaii (it only got statehood as of 1959) with 808, Puerto Rico
with 809, Newfoundland with 709, etc.
The original (and still in use until about 1993) plan is that area codes
have a certain construction to the numbers:
The first digit will be 2 through 9.
The second digit will always be 0 or 1.
The third digit will be 1 through 9.
Three digit numbers with two zeros will be special codes, ie. 700, 800 or
900. Three digit numbers with two ones are for special local codes,
i.e. 411 for local directory assistance, 611 for repairs, etc.
Three digit codes ending in '10', i.e. 410, 510, 610, 710, 810, 910 were
'area codes' for the AT&T (and later on Western Union) TWX network. This
rule has been mostly abolished, however 610 is still Canadian TWX, and
910 is still used by Western Union TWX. Gradually the '10' codes are
being converted to regular area codes.
We are running out of possible combinations of numbers using the above
rules, and it is estimated that beginning in 1993-94, area codes will
begin looking like regular telephone prefix codes, with numbers other than
0 or 1 as the second digit.
I hope this gives you a basic idea. There were other rules at one time
such as not having an area code with zero in the second digit in the same
state as a code with one in the second digit, etc .. but after the initial
assignment of numbers back almost forty years ago, some of those rules
were dropped when it became apparent they were not flexible enough.
Patrick Townson
TELECOM Digest Moderator
--
Patrick Townson
pat...@chinet.chi.il.us / ptow...@eecs.nwu.edu / US Mail: 60690-1570
FIDO: 115/743 / AT&T Mail: 529-6378 (!ptownson) / MCI Mail: 222-4956
==> trivia/body.parts.p <==
Name ten body parts that are spelled with three letters. No slang words.
==> trivia/body.parts.s <==
arm, ear, eye, gum, hip, jaw, leg, lip, rib, toe
Not strictly body parts or slang: ass, box, bud, bum, fat, fin, gam, gut, lap,
lid, mug, ora, orb, ova, paw, pin, pit, pup, pus, tit, wax, yap
With two letters: os
==> trivia/coincidence.p <==
Name some amazing coincidences.
==> trivia/coincidence.s <==
The answer to the question, "Who wrote the Bible," is, of
course, Shakespeare. The King James Version was published in
1611. Shakespeare was 46 years old then (he turned 47 later in
the year). Look up Psalm 46. Count 46 words from the beginning of
the Psalm. You will find the word "Shake." Count 46 words from
the end of the Psalm. You will find the word "Spear." An obvious
coded message. QED.
How many inches in the pole-to-pole diameter of the Earth? The
answer is almost exactly 500,000,000 inches. Proof that the inch
was defined by spacemen.
The speed of light is within 0.1% of 300,000,000 meters/second. The
meter and second were defined with respect to the size and rotation rate
of the Earth. Proof that the Earth was built by spacemen.
==> trivia/eskimo.snow.p <==
How many words do the Eskimo have for snow?
==> trivia/eskimo.snow.s <==
Couple of weeks ago, someone named D.K. Holm in the Boston Phoenix came up
with the list, drawn from the Inupiat Eskimo Dictionary by Webster and
Zibell, and from Thibert's English-Eskimo Eskimo-English Dictionary.
The words may remind you of generated passwords.
Eskimo English Eskimo English
---------------------------------+----------------------------
apun snow | pukak sugar snow
apingaut first snowfall | pokaktok salt-like snow
aput spread-out snow | miulik sleet
kanik frost | massak snow mixed with water
kanigruak frost on a | auksalak melting snow
living surface | aniuk snow for melting
ayak snow on clothes | into water
kannik snowflake | akillukkak soft snow
nutagak powder snow | milik very soft snow
aniu packed snow | mitailak soft snow covering an
aniuvak snowbank | opening in an ice floe
natigvik snowdrift | sillik hard, crusty snow
kimaugruk snowdrift that | kiksrukak glazed snow in a thaw
blocks something | mauya snow that can be
perksertok drifting snow | broken through
akelrorak newly drifting snow | katiksunik light snow
mavsa snowdrift overhead | katiksugnik light snow deep enough
and about to fall | for walking
kaiyuglak rippled surface | apuuak snow patch
of snow | sisuuk avalanche
=*=
==> trivia/federal.reserve.p <==
What is the pattern to this list:
Boston, MA
New York, NY
Philadelphia, PA
Cleveland, OH
Richmond, VA
Atlanta, GA
Chicago, IL
St. Louis, MO
Minneapolis, MN
Kansas City, MO
Dallas, TX
San Francisco, CA
==> trivia/federal.reserve.s <==
Each of the cities is a location for a Federal Reserve. The cities
are listed in alphabetical order based on the letter that represents each
city on a dollar bill.
==> trivia/jokes.self-referential.p <==
What are some self-referential jokes?
==> trivia/jokes.self-referential.s <==
Q: What is alive, green, lives all over the world, and has seventeen legs?
A: Grass. I lied about the legs.
The two rules for success are:
1. Never tell them everything you know.
There are three kinds of people in the world: those who can count,
and those who cannot.
Q: Why did Douglas Hofstadter cross the road?
A: To make this riddle possible.
Song from the Sheri Lewis Lambchop hour:
This is the song that doesn't end
Yes it goes on and on my friend
Some people starting singing it not knowing what it was
Now they'll continue singing it forever just because
(repeat)
How long is the answer to this question?
Ten letters.
(There are endless variations on this theme)
==> trivia/memory.tricks.p <==
When asked to name a color, many people answer "red." What are some other
examples of this phenomenon?
==> trivia/memory.tricks.s <==
What's 3 + 7? What's 4 + 6? What's 8 + 2? Name a vegetable.
Carrot.
Pick a number from 1 to 10.
Multiply by 9.
Subtract 5.
Sum the digits, repeat this step until you have a one digit number.
For whatever number you have pick that letter of the alphabet.
Think of a country that begins with that letter.
Now think of an animal that begins with the second letter of the country.
Think of a color usually associated with the animal.
So are you a grey elephant from Denmark?
==> trivia/quotations.p <==
Where can I find the source for a quotation?
==> trivia/quotations.s <==
The Quotations Archive
All the quotations that fit the guidelines are stored at a publicly
available ftp site: wilma.cs.brown.edu:pub/alt.quotations/Archive.
In the future there will be an organized index system. Right now,
just the raw postings are available.
The quotes are grouped primarily by subject, but there are indexes
by author, keyword, type of source (movie, play, book), and
meta-subject (humor is a meta-subject, humor-about-cars is a
subject).
Movie and television quotes have a tendency to mean nothing to
people who haven't seen the show, and bring back fond memories to
people who have. That doesn't make them real quotations, but since
they are so popular, a part of the archive will be set aside for
these media related quotes.
The index is labeled either ``exact'', or ``incomplete''. If you
can give the exact wording to a quote marking ``incomplete'',
please write j...@cs.brown.edu. We are trying to keep paraphrasing
to a minimum.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> logic/riddle.p <==
Who makes it, has no need of it. Who buys it, has no use for it. Who
uses it can neither see nor feel it.
Tell me what a dozen rubber trees with thirty boughs on each might be?
As I went over London Bridge
I met my sister Jenny
I broke her neck and drank her blood
And left her standing empty
It is said among my people that some things are improved by death.
Tell me, what stinks while living, but in death, smells good?
All right. Riddle me this: what goes through the door without
pinching itself? What sits on the stove without burning itself? What
sits on the table and is not ashamed?
What work is it that the faster you work, the longer it is before
you're done, and the slower you work, the sooner you're finished?
Whilst I was engaged in sitting I spied the dead carrying the living.
I know a word of letters three. Add two, and fewer there will be.
I give you a group of three. One is sitting down, and will never get
up. The second eats as much as is given to him, yet is always hungry.
The third goes away and never returns.
Whoever makes it, tells it not. Whoever takes it, knows it not. And
whoever knows it wants it not.
Two words, my answer is only two words.
To keep me, you must give me.
Sir, I bear a rhyme excelling
In mystic force and magic spelling
Celestial sprites elucidate
All my own striving can't relate
There is not wind enough to twirl
That one red leaf, nearest of its clan,
Which dances as often as dance it can.
Half-way up the hill, I see thee at last
Lying beneath me with thy sounds and sights --
A city in the twilight, dim and vast,
With smoking roofs, soft bells, and gleaming lights.
I am, in truth, a yellow fork
From tables in the sky
By inadvertent fingers dropped
The awful cutlery.
Of mansions never quite disclosed
And never quite concealed
The apparatus of the dark
To ignorance revealed.
Many-maned scud-thumper,
Maker of worn wood,
Shrub-ruster,
Sky-mocker,
Rave!
Make me thy lyre, even as the forests are.
What if my leaves fell like its own --
The tumult of thy mighty harmonies
Will take from both a deep autumnal tone.
This darksome burn, horseback brown,
His rollock highroad roaring down,
In coop and in comb the fleece of his foam
Flutes and low to the body falls home.
I've measured it from side to side,
'Tis three feet long and two feet wide.
It is of compass small, and bare
To thirsty suns and parching air.
My love, when I gaze on thy beautiful face,
Careering along, yet always in place --
The thought has often come into my mind
If I ever shall see thy glorious behind.
Then all thy feculent majesty recalls
The nauseous mustiness of forsaken bowers,
The leprous nudity of deserted halls --
The positive nastiness of sullied flowers.
And I mark the colours, yellow and black,
That fresco thy lithe, dictatorial thighs.
When young, I am sweet in the sun.
When middle-aged, I make you gay.
When old, I am valued more than ever.
I am always hungry,
I must always be fed,
The finger I lick
Will soon turn red.
All about, but cannot be seen,
Can be captured, cannot be held,
No throat, but can be heard.
I am only useful
When I am full,
Yet I am always
Full of holes.
If you break me
I do not stop working,
If you touch me
I may be snared,
If you lose me
Nothing will matter.
If a man carried my burden
He would break his back.
I am not rich,
But leave silver in my track.
Until I am measured
I am not known,
Yet how you miss me
When I have flown.
I drive men mad
For love of me,
Easily beaten,
Never free.
When set loose
I fly away,
Never so cursed
As when I go astray.
I go around in circles
But always straight ahead,
Never complain
No matter where I am led.
Lighter than what
I am made of,
More of me is hidden
Than is seen.
I turn around once,
What is out will not get in.
I turn around again,
What is in will not get out.
Each morning I appear
To lie at your feet,
All day I will follow
No matter how fast you run,
Yet I nearly perish
In the midday sun.
Weight in my belly,
Trees on my back,
Nails in my ribs,
Feet I do lack.
Bright as diamonds,
Loud as thunder,
Never still,
A thing of wonder.
My life can be measured in hours,
I serve by being devoured.
Thin, I am quick
Fat, I am slow
Wind is my foe.
To unravel me
You need a simple key,
No key that was made
By locksmith's hand,
But a key that only I
Will understand.
I am seen in the water
If seen in the sky,
I am in the rainbow,
A jay's feather,
And lapis lazuli.
Glittering points
That downward thrust,
Sparkling spears
That never rust.
You heard me before,
Yet you hear me again,
Then I die,
'Till you call me again.
Three lives have I.
Gentle enough to soothe the skin,
Light enough to caress the sky,
Hard enough to crack rocks.
You can see nothing else
When you look in my face,
I will look you in the eye
And I will never lie.
Lovely and round,
I shine with pale light,
grown in the darkness,
A lady's delight.
At the sound of me, men may dream
Or stamp their feet
At the sound of me, women may laugh
Or sometimes weep
When I am filled
I can point the way,
When I am empty
Nothing moves me,
I have two skins
One without and one within.
My tines be long,
My tines be short
My tines end ere
My first report.
What am I?
With thieves I consort,
With the vilest, in short,
I'm quite at ease in depravity;
Yet all divines use me,
And savants can't lose me,
For I am the center of gravity.
As a whole, I am both safe and secure.
Behead me, and I become a place of meeting.
Behead me again, and I am the partner of ready.
Restore me, and I become the domain of beasts.
What am I?
I sought my first in starry skies
Where shines the April sun;
My second came before my eyes,
And warned me to be done.
'Tis very hard to lose one's sight;
I'm blind as bat or mole;
Once hills and fields were my delight,
Now I'm no more my whole.
My first is high,
My second damp,
My whole a tie,
A writer's cramp.
A hundred and one
by fifty divide,
And if a cipher
is rightly applied,
The answer is one from nine.
What does man love more than life
Fear more than death or mortal strife
What the poor have, the rich require,
and what contented men desire,
What the miser spends and the spendthrift saves
And all men carry to their graves?
I build up castles.
I tear down mountains.
I make some men blind,
I help others to see.
What am I?
Ripped from my mother's womb,
Beaten and burned,
I become a blood-thirsty slayer
What am I?
Five hundred begins it, five hundred ends it,
Five in the middle is seen;
First of all figures, the first of all letters,
Take up their stations between.
Join all together, and then you will bring
Before you the name of an eminent king.
==> logic/riddle.s <==
Who makes it, has no need of it. Who buys it, has no use for it. Who
uses it can neither see nor feel it.
coffin
Tell me what a dozen rubber trees with thirty boughs on each might be?
months of the year
As I went over London Bridge
I met my sister Jenny
I broke her neck and drank her blood
And left her standing empty
gin
It is said among my people that some things are improved by death.
Tell me, what stinks while living, but in death, smells good?
pig
All right. Riddle me this: what goes through the door without
pinching itself? What sits on the stove without burning itself? What
sits on the table and is not ashamed?
the sun
What work is it that the faster you work, the longer it is before
you're done, and the slower you work, the sooner you're finished?
roasting meat on a spit
Whilst I was engaged in sitting I spied the dead carrying the living.
a ship
I know a word of letters three. Add two, and fewer there will be.
'few'
I give you a group of three. One is sitting down, and will never get
up. The second eats as much as is given to him, yet is always hungry.
The third goes away and never returns.
stove, fire, and smoke
Whoever makes it, tells it not. Whoever takes it, knows it not. And
whoever knows it wants it not.
counterfeit money
Two words, my answer is only two words.
To keep me, you must give me.
your word
Sir, I bear a rhyme excelling
In mystic force and magic spelling
Celestial sprites elucidate
All my own striving can't relate
Pi (digits given by length of words)
There is not wind enough to twirl
That one red leaf, nearest of its clan,
Which dances as often as dance it can.
the sun, Samuel Taylor Coleridge
Half-way up the hill, I see thee at last
Lying beneath me with thy sounds and sights --
A city in the twilight, dim and vast,
With smoking roofs, soft bells, and gleaming lights.
the past, Longfellow
I am, in truth, a yellow fork
From tables in the sky
By inadvertent fingers dropped
The awful cutlery.
Of mansions never quite disclosed
And never quite concealed
The apparatus of the dark
To ignorance revealed.
lightning, Emily Dickinson
Many-maned scud-thumper,
Maker of worn wood,
Shrub-ruster,
Sky-mocker,
Rave!
Portly pusher,
Wind-slave.
the ocean, John Updike
Make me thy lyre, even as the forests are.
What if my leaves fell like its own --
The tumult of thy mighty harmonies
Will take from both a deep autumnal tone.
the west wind, Percy Bysshe Shelley
This darksome burn, horseback brown,
His rollock highroad roaring down,
In coop and in comb the fleece of his foam
Flutes and low to the body falls home.
river, Gerard Manley Hopkins
I've measured it from side to side,
'Tis three feet long and two feet wide.
It is of compass small, and bare
To thirsty suns and parching air.
the grave of a child, Wordsworth
My love, when I gaze on thy beautiful face,
Careering along, yet always in place --
The thought has often come into my mind
If I ever shall see thy glorious behind.
the moon, Sir Edmund Gosse
Then all thy feculent majesty recalls
The nauseous mustiness of forsaken bowers,
The leprous nudity of deserted halls --
The positive nastiness of sullied flowers.
And I mark the colours, yellow and black,
That fresco thy lithe, dictatorial thighs.
spider, Francis Saltus Saltus
When young, I am sweet in the sun.
When middle-aged, I make you gay.
When old, I am valued more than ever.
wine
I am always hungry,
I must always be fed,
The finger I lick
Will soon turn red.
fire
All about, but cannot be seen,
Can be captured, cannot be held,
No throat, but can be heard.
wind
I am only useful
When I am full,
Yet I am always
Full of holes.
sieve (or sponge)
If you break me
I do not stop working,
If you touch me
I may be snared,
If you lose me
Nothing will matter.
heart
If a man carried my burden
He would break his back.
I am not rich,
But leave silver in my track.
snail
Until I am measured
I am not known,
Yet how you miss me
When I have flown.
time
I drive men mad
For love of me,
Easily beaten,
Never free.
gold
When set loose
I fly away,
Never so cursed
As when I go astray.
a fart
I go around in circles
But always straight ahead,
Never complain
No matter where I am led.
wagon wheel
Lighter than what
I am made of,
More of me is hidden
Than is seen.
iceberg
I turn around once,
What is out will not get in.
I turn around again,
What is in will not get out.
stopcock
Each morning I appear
To lie at your feet,
All day I will follow
No matter how fast you run,
Yet I nearly perish
In the midday sun.
shadow
Weight in my belly,
Trees on my back,
Nails in my ribs,
Feet I do lack.
ship
Bright as diamonds,
Loud as thunder,
Never still,
A thing of wonder.
waterfall? (fireworks?)
My life can be measured in hours,
I serve by being devoured.
Thin, I am quick
Fat, I am slow
Wind is my foe.
candle
To unravel me
You need a simple key,
No key that was made
By locksmith's hand,
But a key that only I
Will understand.
cipher
I am seen in the water
If seen in the sky,
I am in the rainbow,
A jay's feather,
And lapis lazuli.
blue
Glittering points
That downward thrust,
Sparkling spears
That never rust.
icicle
You heard me before,
Yet you hear me again,
Then I die,
'Till you call me again.
echo
Three lives have I.
Gentle enough to soothe the skin,
Light enough to caress the sky,
Hard enough to crack rocks.
water
You can see nothing else
When you look in my face,
I will look you in the eye
And I will never lie.
your reflection
Lovely and round,
I shine with pale light,
grown in the darkness,
A lady's delight.
pearl
At the sound of me, men may dream
Or stamp their feet
At the sound of me, women may laugh
Or sometimes weep
music
When I am filled
I can point the way,
When I am empty
Nothing moves me,
I have two skins
One without and one within.
glove
My tines be long,
My tines be short
My tines end ere
My first report.
What am I?
lightning
With thieves I consort,
With the vilest, in short,
I'm quite at ease in depravity;
Yet all divines use me,
And savants can't lose me,
For I am the center of gravity.
The letter 'v'.
As a whole, I am both safe and secure.
Behead me, and I become a place of meeting.
Behead me again, and I am the partner of ready.
Restore me, and I become the domain of beasts.
What am I?
stable
I sought my first in starry skies
Where shines the April sun;
My second came before my eyes,
And warned me to be done.
'Tis very hard to lose one's sight;
I'm blind as bat or mole;
Once hills and fields were my delight,
Now I'm no more my whole.
?
My first is high,
My second damp,
My whole a tie,
A writer's cramp.
?
A hundred and one
by fifty divide,
And if a cipher
is rightly applied,
The answer is one from nine.
?
What does man love more than life
Fear more than death or mortal strife
What the poor have, the rich require,
and what contented men desire,
What the miser spends and the spendthrift saves
And all men carry to their graves?
nothing
I build up castles.
I tear down mountains.
I make some men blind,
I help others to see.
What am I?
sand
Ripped from my mother's womb,
Beaten and burned,
I become a blood-thirsty slayer
What am I?
?
Five hundred begins it, five hundred ends it,
Five in the middle is seen;
First of all figures, the first of all letters,
Take up their stations between.
Join all together, and then you will bring
Before you the name of an eminent king.
DAVID (Roman numerals)
==> logic/river.crossing.p <==
Three humans, one big monkey and two small monkeys are to cross a river:
a) Only humans and the big monkey can row the boat.
b) At all times, the number of human on either side of the
river must be GREATER OR EQUAL to the number of monkeys
on THAT side. ( Or else the humans will be eaten by the monkeys!)
==> logic/river.crossing.s <==
The three columns represent the left bank, the boat, and the right bank
respectively. The < or > indicates the direction of motion of the boat.
HHHMmm . .
HHHm Mm> .
HHHm <M m
HHH Mm> m
HHH <M mm
HM HH> mm
HM <Hm Hm
Hm HM> Hm
Hm <Hm HM
mm HH> HM
mm <M HHH
m Mm> HHH
m <M HHHm
. Mm> HHHm
. . HHHMmm
==> logic/ropes.p <==
Two fifty foot ropes are suspended from a forty foot ceiling, about
twenty feet apart. Armed with only a knife, how much of the rope can
you steal?
==> logic/ropes.s <==
Almost all of it. Tie the ropes together. Climb up one of them. Tie
a loop in it as close as possible to the ceiling. Cut it below the
loop. Run the rope through the loop and tie it to your waist. Climb
the other rope (this may involve some swinging action). Pull the rope
going through the loop tight and cut the other rope as close as
possible to the ceiling. You will swing down on the rope through the
loop. Lower yourself to the ground by letting out rope. Pull the
rope through the loop. You will have nearly all the rope.
==> logic/same.street.p <==
Sally and Sue have a strong desire to date Sam. They all live on the
same street yet neither Sally or Sue know where Sam lives. The houses
on this street are numbered 1 to 99.
Sally asks Sam "Is your house number a perfect square?". He answers.
Then Sally asks "Is is greater than 50?". He answers again.
Sally thinks she now knows the address of Sam's house and decides to
visit.
When she gets there, she finds out she is wrong. This is not
surprising, considering Sam answered only the second question
truthfully.
Sue, unaware of Sally's conversation, asks Sam two questions.
Sue asks "Is your house number a perfect cube?". He answers.
She then asks "Is it greater than 25?". He answers again.
Sue thinks she knows where Sam lives and decides to pay him a visit.
She too is mistaken as Sam once again answered only the second
question truthfully.
If I tell you that Sam's number is less than Sue's or Sally's,
and that the sum of their numbers is a perfect square multiplied
by two, you should be able to figure out where all three of them
live.
==> logic/same.street.s <==
Sally asks Sam "Is your house number a perfect square?". He answers.
Then Sally asks "Is is greater than 50?". He answers again.
Sally thinks she now knows the address of Sam's house and decides to
visit.
Since Sally thinks that she has enough information, I deduce
that Sam answered that his house number was a perfect square
greater than 50. There are two of these {64,81} and Sally must
live in one of them in order to have decided she knew where Sam
lives.
When she gets there, she finds out she is wrong. This is not
surprising, considering Sam answered only the second question
truthfully.
So Sam's house number is greater than 50, but not a perfect
square.
Sue, unaware of Sally's conversation, asks Sam two questions.
Sue asks "Is your house number a perfect cube?". He answers.
She then asks "Is it greater than 25?". He answers again.
Observation: perfect cubes greater than 25 are {27, 64}, less
than 25 are {1,8}.
Sue thinks she knows where Sam lives and decides to pay him a visit.
She too is mistaken as Sam once again answered only the second
question truthfully.
Since Sam's house number is greater than 50, he told Sue that
it was greater than 25 as well. Since Sue thought she knew
which house was his, she must live in either of {27,64}.
If I tell you that Sam's number is less than Sue's or Sally's,
Since Sam's number is greater than 50, and Sue's is even
bigger, she must live in 64. Assuming Sue and Sally are not
roommates (although awkward social situations of this kind are
not without precedent), Sally lives in 81.
and that the sum of their numbers is a perfect square multiplied
by two, you should be able to figure out where all three of them
live.
Sue + Sally + Sam = 2 p^2 for p an integer
64 + 81 + Sam = 2 p^2
Applying the constraint 50 < Sam < 64, looks like Sam = 55 (p = 10).
In summary,
Sam = 55
Sue = 64
Sally = 81
-- Tom Smith <t...@ulysses.att.com>
==> logic/self.ref.p <==
Find a number ABCDEFGHIJ such that A is the count of how many 0's are in the
number, B is the number of 1's, and so on.
==> logic/self.ref.s <==
6210001000
For other numbers of digits:
n=1: no sequence possible
n=2: no sequence possible
n=3: no sequence possible
n=4: 1210, 2020
n=5: 21200
n=6: no sequence possible
n=7: 3211000
n=8: 42101000
n=9: 521001000
n=10: 6210001000
n>10: (n-4), 2, 1, 0 * (n-7), 1, 0, 0, 0
No 1, 2, or 3 digit numbers are possible. Letting x_i be the ith
digit, starting with 0, we see that (1) x_0 + ... + x_n = n+1 and
(2) 0*x_0 + ... + n*x_n = n+1, where n+1 is the number of digits.
I'll first prove that x_0 > n-3 if n>4. Assume not, then this
implies that at least four of the x_i with i>0 are non-zero. But
then we would have \sum_i i*x_i >= 10 by (2), impossible unless n=9,
but it isn't possible in this case (51111100000 isn't valid).
Now I'll prove that x_0 < n-1. x_0 clearly can't equal n; assume
x_0 = n-1 ==> x_{n-1} = 1 by (2) if n>3. Now only one of the
remaining x_i may be non-zero, and we must have that x_0 + ... + x_n
= n+1, but since x_0 + x_{n-1} = n ==> the remaining x_i = 1 ==> by
(2) that x_2 = 1. But this can't be, since x_{n-1} = 1 ==> x_1>0.
Now assuming x_0 = n-2 we conclude that x_{n-2} = 1 by (2) if n>5
==> x_1 + ... + x_{n-3} + x_{n-1} + x_n = 2 and 1*x_1 + ... +
(n-3)*x_{n-3} + (n-1)*x_{n-1} + n*x_n = 3 ==> x_1=1 and x_2=1,
contradiction.
Case n>5:
We have that x_0 = n-3 and if n>=7 ==> x_{n-3}=1 ==> x_1=2 and
x_2=1 by (1) and (2). For the case n=6 we see that x_{n-3}=2
leads to an easy contradiction, and we get the same result. The
cases n=4,5 are easy enough to handle, and lead to the two solutions
above.
--
-- cl...@romulus.rutgers.edu (Chris Long)
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> pickover/pickover.12.p <==
Title: Cliff Puzzle 12: Slides in Hell
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
Consider a metallic slide with 10 large holes in it equally spaced from
top to bottom. If you attempt to slide down the slide you have a 50%
probability of sliding through each hole in the slide into an oleaginous
substance beneath the slide during each encounter with a hole.
1. If you were a gambling person, which hole would you bet a person
would fall through?
2. If you were a gambling person, how many attempts would it require
for a person to slide from the top of the slide to the bottom without
falling through a single hole.
3. If all the people on earth lined up to go down the slide, and they
slid down a more horrifying slide with 100 holes at a rate of 1 person
per second, when would you expect the first person to arrive at the
bottom of the slide without falling through.
An hour? A day? A decade? ...
Received: from uoft02.utoledo.edu by watson.ibm.com (IBM VM SMTP V2R2) with TCP;
Title: Cliff Puzzle 12: Slides in Hell
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom. If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an
>oleaginous substance beneath the slide during each encounter with a
>hole.
>
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
None. The best chance is the first hole but I got a 50-50 chance. Why
bother? (2nd hole is 1/4, 3rd 2**-3, ...)
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
No gurantee. Each slide is an independent event. Now, if you are
talking mere probability, on the average, one in 1024 slides may make
it through all 10 holes.
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through. An hour? A day? A decade?
Again, can't tell. It could be the first one, it could be none. Probablity
can not foretell actual events. But if you have infinite number of people
sliding down till eternity, on the average, you may see 1 person slide over
all holes every (2**100)/(365*24*69*6) years. This number is many times
bigger than the world population for now.
==> pickover/pickover.12.s <==
-------------------------
In article <1992Oct23.1...@watson.ibm.com> you write:
: Consider a metallic slide with 10 large holes in it equally spaced from
: top to bottom. If you attempt to slide down the slide you have a 50%
: probability of sliding through each hole in the slide into an oleaginous
: substance beneath the slide during each encounter with a hole.
:
: 1. If you were a gambling person, which hole would you bet a person
: would fall through?
The chance of falling thru the first hole is 50%. For the second hole, it
is (.5)(.5) = 25%, the thrid is (.5)^3 = .125. The chance by the tenth
hole is about .0097 %. Obviously, since I am limited to one hole, I would
place my money on hole #1 (best chance).
: 2. If you were a gambling person, how many attempts would it require
: for a person to slide from the top of the slide to the bottom without
: falling through a single hole.
The sum of the prob for falling thru a hole is .5 + .5^2 + .5^3 +...+.5^10.
This is about 99.902% = .99902. So about 98 times out of 100000, someone
will make it through without falling. This is about 1 time out of 1020.
So give or take about 1020 tries....
:
: 3. If all the people on earth lined up to go down the slide, and they
: slid down a more horrifying slide with 100 holes at a rate of 1 person
: per second, when would you expect the first person to arrive at the
: bottom of the slide without falling through.
: An hour? A day? A decade? ...
The prob for falling thru the last hole is .5^100 = 7.88x10^-31. There must
be some chance less than this that one WILL make it thru the slide. The MIN
number of tries that it must take is 1/.5^100 = 1.26x10^30. At the given rate
this is about 9.647 x 10^23 years, much older than the universe if I remeber
correctly.
Also, the chance of making it must be GREATER than .5^101. or with
all the math, the MAX amount of time is 1.929x10^24 years. So give or
take about 1.5x10^24 years....
--
Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD!
// | Senior, Chemical Engineering, Univ. of Toledo
\\ // Only the | Summer Intern, NASA Lewis Research Center
\ \X/ AMIGA! | mne...@jupiter.cse.utoledo.edu /
--------+ How do YOU spell 'potato'? How 'bout 'lousy'? +----------
"Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
-------------------------
In rec.puzzles you write:
>Title: Cliff Puzzle 12: Slides in Hell
>From: cl...@watson.ibm.com
>
>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
Jeff Rogers
Rensselaer Polytechnic institute
rog...@rpi.edu
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom. If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an oleaginous
>substance beneath the slide during each encounter with a hole.
>
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
The first one. There's only a 50% chance of them getting past it, and a
small chance of them falling into each succeeding hole.
hole # percent chance of reaching and falling into
1 50
2 25
3 12.5
4 6.25
5 3.125
6 1.5625
7 0.78125
8 0.390625
9 0.1953125
10 0.09765625
>
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
The chances for reaching each succeeding hole are the same as reaching and
falling into the previous one. Therefore, the chances of passing all the
holes are the same as reaching and falling into the last hole (see previous
answer for stats), which makes the probability .0009765625, so
statistically, 1024 slides would be required to guarantee reaching the
bottom. If I was a gambling person, I'd probably bet about half this,
because the actual events can happen in any order, and on average, I'd guess
that he'd get down in about 512 slides.
>
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...
This is solved similarly; it is represented by powers of 2. To successfully
get past the last hole, it would require (statistically, at least) 2^100
or (by my trusty pocket calculator) 1.2676506 *10^30 slides.
More significant figures? dc! Which gives 1267650600228229401496703205376.
In similar logic as the last problem, I'd expect about half that, or
633825300114114700748351602688 slides. How much time would this be? Excluding
leap years, I calculate 20098468420665737593491 years. That's 20 sextillion
years, significantly more than the age of the universe, by about 11 orders
of magnitude. So I'd guess that no one will ever reach the bottom, they'll
all try and fail (assuming everyone only gets to go once), or die waiting in
line.
Diversion
--
"I can see 'em | "Want me to create a diversion?"
I can see 'em | Diversion
Someone wake me when it's over" | rog...@rpi.edu
-------------------------
In article <1992Oct23.1...@watson.ibm.com> you write:
Title: Cliff Puzzle 12: Slides in Hell
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom. If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an
>oleaginous substance beneath the slide during each encounter with a
>hole.
>
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
None. The best chance is the first hole but I got a 50-50 chance. Why
bother? (2nd hole is 1/4, 3rd 2**-3, ...)
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
No gurantee. Each slide is an independent event. Now, if you are
talking mere probability, on the average, one in 1024 slides may make
it through all 10 holes.
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through. An hour? A day? A decade?
Again, can't tell. It could be the first one, it could be none. Probablity
can not foretell actual events. But if you have infinite number of people
sliding down till eternity, on the average, you may see 1 person slide over
all holes every (2**100)/(365*24*69*6) years. This number is many times
bigger than the world population for now.
-------------------------
Some answers to your questions:
1. As the puzzle states there is a 50% chance of falling into each
hole, I would bet a person would fall into the first hole -- in a large
enough sample, 1/2 of the people will fall through the first hole, 1/4
through the second, 1/8 through the third, etc.
2. In a large sample, 1/(2^10) people would make it all the way down
the slide without falling through any of the holes (1/1024). This means
that 1023 out of 1024 people would fall through a hole. Using the
formula (1023/1024)^x=1/2, we can determine out of the first x people
to go down the slide, there is a 50% chance that one person will make
it down without falling through a hole. The answer to this equation is
x=709.4 Thus I would bet that a person would make it all the way down
on one of the first 710 attempts.
3. As 2^100=1.2676*10^30 (roughly), and (including leaps years under
the Gregorian calendar) there are 31556952 seconds in the average year,
then statistically one person should make it down the slide every
4.017*10^22 YEARS. However, and this is a very rough estimate, I figure
the log of (1-1/(1.2676*10^30)) to be about -5.5*10^(-29). [I'm doing
the calculations on a scientific calculator which only has 10 places.]
Thus, using the formula xlog(1-1/2^100)=log(1/2), I get x=5.5*10^27.
Thus, there's about a 50% chance that after 5.5*10^27 seconds, someone
will have made it down the slide. To be on the safe side, I'd bet only
if I were given at least 6*10^27 seconds, a value which equals
1.901*10^20 YEARS.
I hope this answers the questions.
Ted Schuerzinger
email: J.Theodore....@Dartmouth.EDU
snailmail: HB 3819
Dartmouth College
Hanover, NH 03755
USA
In case you're wondering, I'm just a junior at Dartmouth who's
interested in puzzles like these. I'm not even a math major -- I'm a
double major in government and Russian.
-------------------------
In article <1992Oct23.1...@watson.ibm.com> you write:
>Title: Cliff Puzzle 12: Slides in Hell
>From: cl...@watson.ibm.com
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom. If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an oleaginous
>substance beneath the slide during each encounter with a hole.
>
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
There's a 50% chance of falling through the first hole, 25% the
second, 2^-n the n'th. If the odds offered were the same, I'd go for
the first hole.
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
You expect to make it 1 out of 1024 times; after 710 tries, the chance
of someone succeeding exceeds 1/2. (Log base (1023/1024) of 1/2 is
709.4).
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...
Never. OK, 1/2^100 will make it. There being under 2^33 people on
the planet, ...
After 4.2e22 years, the expected number of people who succeeded is 1;
after about 2.9e22 years, the chance of someone having succeeded is
about 1/2.
Like I said, never.
Seth se...@fid.morgan.com
-------------------------
In rec.puzzles you write:
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
If the pay-back odds were the same regardless of the hole, then obviously,
I'd bet on the first hole! There's a 1:2 chance the person falls through
the first hole, a 1:4 combined chance of the person falling though the
second hole, etc...
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
1024 is the median value for this case... There's a 1:2**n chance of
a person falling through the nth hole, having missed all of the holes
before n. Since the probability of falling through = the probability
passing over the hole safely (vs not ever getting there), the
probability that a person makes it to the end is also 1:1024.
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...
There is a 1:2**(100-Log2(5 billion people)) chance that somebody makes
it through... Given a finite # of people on the planet (approx 5 bil.)
I think we'll run out first...
--Joseph Zbiciak im1...@camelot.bradley.edu
-------------------------
Subject: Re: Cliff Puzzle 12: Slides in Hell (SPOILER)
Newsgroups: rec.puzzles
References: <1992Oct23.1...@watson.ibm.com>
In article <1992Oct23.1...@watson.ibm.com>, Cliff Pickover writes:
> Consider a metallic slide with 10 large holes in it equally spaced from
> top to bottom. If you attempt to slide down the slide you have a 50%
> probability of sliding through each hole in the slide into an oleaginous
> substance beneath the slide during each encounter with a hole.
> 1. If you were a gambling person, which hole would you bet a person
> would fall through?
The probability of falling into hole i is (1/2)^i, so your best bet
would be hole 1.
> 2. If you were a gambling person, how many attempts would it require
> for a person to slide from the top of the slide to the bottom without
> falling through a single hole.
The probability of success is p = (1/2)^10, and as each trial is
independant the expected number of trials before success is 1/p or
2^10.
> 3. If all the people on earth lined up to go down the slide, and they
> slid down a more horrifying slide with 100 holes at a rate of 1 person
> per second, when would you expect the first person to arrive at the
> bottom of the slide without falling through.
In this case the number of expected trials is 2^100, which is much
larger than the total number of people.
> An hour? A day? A decade? ...
Try about 10^24 years. As another problem, assuming a large enough
supply of sliders estimate when the slide will wear through from
friction.
-------------------------
In article <1992Oct23.1...@watson.ibm.com> you write:
>Title: Cliff Puzzle 12: Slides in Hell
>From: cl...@watson.ibm.com
>
>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
>you provide unique information. PLEASE ALSO directly mail me a copy of
>your response in addition to any responding you do in the newsgroup. I
>will assume it is OK to describe your answer in any article or
>publication I may write in the future, with attribution to you, unless
>you state otherwise. Thanks, Cliff Pickover
>
> * * *
>
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom. If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an oleaginous
>substance beneath the slide during each encounter with a hole.
>
>1. If you were a gambling person, which hole would you bet a person
>would fall through?
I'd bet that they fell through the first hole. The probability of that
happening is 50%. The probability of them falling through the second
hole is:
P(didn't fall through the first)*P(fell through the second) = 50%*50% = 25%
In general, P(falls through hole n)=
P(no fall through 1)*P(no fall through 2)*...*P(no fall through n-1)
*P(fell through hole n).
For this problem, P(falls through hole n) is (50%)^n, where n is the hole #
from the top.
>2. If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.
(Hey, after the first failed attempt, they're screwed, no?)
P(success)=P(no fail)=P(no fall 1)P(no fall 2)...P(no fall 10)
=50%^10
=1/1024
They should make it at least one time in 1024.
>3. If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...
Oh, one in about 4.02*10^22 years... I wouldn't hold my breath.
-Richard
-------------------------
1. I would bet on the first hole, as there is a 0.5 probability of a person's
falling into it, which is the highest such probability.
2. The probability of reaching the end of the slide on a particular try is
1/2^10 = 1/1024. In 709 tries, there is an approximately 0.5 probability of
3. Beats me - the even money bet is for a number of tries (approximately) equal
((2^100 - 1)/(2^100))
calculate it.
--
_______________________________________________________________________
Dan Blum Institute for the Learning Sciences Room 327
bl...@ils.nwu.edu 1890 Maple Ave., Evanston, IL 60201 708-467-2306
"Let it be granted that a controversy may be raised about any question,
and at any distance from that question."
Lewis Carroll
_______________________________________________________________________
==> pickover/pickover.13.p <==
Title: Cliff Puzzle 13: Ladders to Heaven
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
Consider the following scenario. A standard ladder stretches from each
country on the earth upward a distance equal to the distance from the
earth to the moon.
Assume:
1. the ladder is made out of a strong metal such as
titanium, which will not break.
2. the ladder is inclined at a very steep angle, 70 degrees, for
each country.
3. there is a breathable atmosphere.
4. the people (or teams of people) are allowed to use standard
mountain climbing and camping gear, e.g. ropes, backpacks, etc. but not
sophisticated electrical mechanisms, engines, etc.
5. a reward is given to whomever reaches the top of the ladder
first: 1 million dollars to that person. In addition the country's
national debt is wiped out.
Questions:
1. Approximate how long it would take a person (or team of people) to
reach the top of the ladder. Days? Weeks? Years?
2. Which country would be the first?
3. Is there any novel method you would suggest to achieve this goal?
4. Is this task impossible to carry out.
==> pickover/pickover.13.s <==
-------------------------
Interesting puzzle... Just one question though: Is there a moon,
i.e. is it possible to use the gravitational field of the moon to your
advantage by "falling upwards" once you have reached the point where
the moon's gravity is bigger than the erath's (and do we also assume that
the the climber(s) must survive the fall?? :-) or shall we assume that the
earth is alone in the universe?
Spyros Potamianos
pota...@hpl.hp.com
-------------------------
Newsgroups: rec.puzzles
Subject: Re: Cliff Puzzle 13: Ladders to Heaven
References: <1992Oct23.1...@watson.ibm.com>
Organization: The Chrome Plated Megaphone of Destiny
>1. Approximate how long it would take a person (or team of people) to
>reach the top of the ladder. Days? Weeks? Years?
Note that after you're 22,300 miles from the earth's axis, you get to
"fall" the rest of the way, as long as you don't lose contact with
the ladder.
>2. Which country would be the first?
It has already been pointed out that countries on the equator have an
advantage. I suppose you could consider that countries with a large
national debt have extra motivation. :-)
>3. Is there any novel method you would suggest to achieve this goal?
I would suggest a bicycle-like vehicle clamped to the ladder. By
pulling a light but strong rope on a pulley (perhaps obtained form
the same source as this fantastic ladder material), riders could be
changed fairly quickly, thanks to a crew of brawny pulley-pullers
with a variable-geared linkage to the rope.
For the rider to pull this ever-longer rope seems impossible, but I
think shorter segments could be lifted and linked. Or the ground
crew could help the rider by pulling down rope from a hub of lesser
diameter than the wheels of the vehicle.
>4. Is this task impossible to carry out.
No. I thought it might be impossible to halt at the far end of the
ladder and return, due to centrifugal acceleration, but that
acceleration turns out to be only about 5 cm/s^2.
__________________________________________________________
Matt Crawford ma...@severian.chi.il.us Java Man
-------------------------
> How do we get food to the people?
I would have the riders change so often that they'd only need some
high-carbohydrate snacks and a couple quarts of fluid. I think the
brawny ground crew could pull up the next rider, with his supplies
and another pulley and segment of rope, at an acceleration of about
0.5 g or better. That would be under 90 minutes for each shift-
change up to the synchronous orbit level.
I haven't figured out yet how to link each new piece of rope that's
pulled up with a rider to the pulley that's at the high point reached
by the previous rider. Linking is easy, but it would be nice to find
a way that lets the next pulled-up rider go from one segment to the
other without interruption. Well, since the sky-buckets at
Disneyland do this trick at each end, I know it can be done.
I didn't know you'd written any books, but it was clear you're
working on one now. Sure, send a list, but I have access to some
on-line catalogs, so maybe I can find them anyway.
Matt Crawford
-------------------------
> Consider the following scenario. A standard ladder stretches from each
> country on the earth upward a distance equal to the distance from the
> earth to the moon.
>
> Assume:
> 1. the ladder is made out of a strong metal such as
> titanium, which will not break.
> 2. the ladder is inclined at a very steep angle, 70 degrees, for
> each country.
> 3. there is a breathable atmosphere.
> 4. the people (or teams of people) are allowed to use standard
> mountain climbing and camping gear, e.g. ropes, backpacks, etc. but not
> sophisticated electrical mechanisms, engines, etc.
> 5. a reward is given to whomever reaches the top of the ladder
> first: 1 million dollars to that person. In addition the country's
> national debt is wiped out.
I would imagine that one would be able to fashion a hot air balloon given
condition 4. Also, given condition 3, the hot air balloon would be able
to cover the entire distance. One would then only need to attach a sliding
hookup between the ladder and the balloon and wait.
===M.Graf==gr...@island.com==================================================
==> pickover/pickover.14.p <==
Title: Cliff Puzzle 14: Geography Genuflection
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
1. How would the world be different today, geopolitically speaking, if
the ancient land masses had never drifted apart and, therefore,
today's world consisted of a single supercontintent?
2. What would today's world be like if the land mass which formed the
Greek peninsula never existed?
3. What would today's world be like if the land bridge which joined
Alaska to Asia never existed?
4. Why do all the major peninsulas on earth point south? See for
example: Italy, Greece, Florida, and Baja, and the tips of Africa,
South America, India, Norway, Sweden, Greenland, and many other
landmasses.
==> pickover/pickover.14.s <==
-------------------------
In rec.puzzles you write:
>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
>you provide unique information.
>
Mike Neergaard
University of Wisconsin
neer...@math.wisc.edu
I'm not a professional at this sort of thing, so I just summarized my
conclusions. I'm sure they would be ripped to shreds by any competent
whatsit-type-individual-who-knows-all-about-this-kind-of-stuff.
>1. How would the world be different today, geopolitically speaking, if
>the ancient land masses had never drifted apart and, therefore,
>today's world consisted of a single supercontintent?
We would all speak German.
>2. What would today's world be like if the land mass which formed the
>Greek peninsula never existed?
>
We would know a low more about fluid dynamics.
>3. What would today's world be like if the land bridge which joined
>Alaska to Asia never existed?
Christopher Columbus would be a national hero, instead of being vulnerable
to counter-claims of genocide. America would have been settled several
decades later, due to a dearth of demonstrable natural resources.
>4. Why do all the major peninsulas on earth point south? See for
>example: Italy, Greece, Florida, and Baja, and the tips of Africa,
>South America, India, Norway, Sweden, Greenland, and many other
>landmasses.
I just work here . . .
--
I really don't make any claim at all to know what I'm talking about.
Actually, I make no claim to know what YOU'RE talking about, either.
In fact, now I've forgotten what we were talking about . . .
-------------------------
In article <1992Oct26.1...@watson.ibm.com> you write:
>Title: Cliff Puzzle 14: Geography Genuflection
>From: cl...@watson.ibm.com
>
>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
>you provide unique information. PLEASE ALSO directly mail me a copy of
>your response in addition to any responding you do in the newsgroup. I
>will assume it is OK to describe your answer in any article or
>publication I may write in the future, with attribution to you, unless
>you state otherwise. Thanks, Cliff Pickover
>
> * * *
>
Okay, administrative trivia first. My name is Martin Eiger, you don't
need my address (home or business?), I don't want you citing my
affiliation if you quote me, and my e-mail address is
m...@thumper.bellcore.com.
>1. How would the world be different today, geopolitically speaking, if
>the ancient land masses had never drifted apart and, therefore,
>today's world consisted of a single supercontintent?
My theory is that mankind would never have evolved. The dominant
species would still be some sort of mammal, but not us. This renders
a large number of geopolitical questions irrelevant. For example,
elephant-like creatures are unlikely to care whether there is one or
two Germanys.
>2. What would today's world be like if the land mass which formed the
>Greek peninsula never existed?
A tough one, since I'm not up on my Greek influences in the evolution
of civilization. My guess is that civilization would have evolved
anyway, probably not too differently than it did. It might not have
evolved as fast, i.e., we might now be where we were a thousand years
ago or so, but over the long haul, human history would follow a
similar course.
>3. What would today's world be like if the land bridge which joined
>Alaska to Asia never existed?
Pretty much the same, I bet. People would have colonized North
America anyway. After all, they got to Hawaii, so somebody could
probably have gotten to North America. And whether or not people
colonized North America from across the Pacific, people from Europe
would have paved the place over just the same.
>4. Why do all the major peninsulas on earth point south? See for
>example: Italy, Greece, Florida, and Baja, and the tips of Africa,
>South America, India, Norway, Sweden, Greenland, and many other
>landmasses.
First of all, you have to define what's a major peninsula. Secondly,
I don't like your list. Norway and Sweden are on the same peninsula,
and Greenland is an island, not a peninsula. And third, there are
plenty of perfectly fine peninsulas that don't point south: Alaska,
Siberia, Michigan (two peninsulas for the price of one), Yucatan,
Arabia (points kind of southeast), and Iberia, for instance. And
fourth, you missed a few good southern-pointing ones, such as Korea,
Crimea, the Sinai, and the one that kind of points from eastern
Siberia toward Japan that I'm sure has a name but I don't know it. So
while there are lots of peninsulas pointing lots of directions, a
majority of them do seem to point south, and I have no idea why.
==> pickover/pickover.15.p <==
Title: Cliff Puzzle 15: Cherries in Wine Glasses
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
Consider a 9x9 grid of beautiful crystal wineglasses. Throw 32 cherries
at the grid. A glass is considered occupied if it contains at least one
cherry. (With each throw a cherry goes into one of the glasses.) How
many different patterns of occupied glasses can you make? (A glass with
more than one cherry is considered the same as a glass with one cherry
in the pattern).
2. Same as above except that you place 8 cherries in glasses (x,y) and
then determine the other positions by placing cherries at (x,-y),
(-x,y), (-x,-y) leading to 32 cherries in the grid. Consider the array
of glasses centered at the origin. How many different patterns of
occupied glasses can you make? (A glass with more than one cherry is
considered the same as a glass with one cherry in the pattern).
3. Can your results be extrapolated to an NxN grid with M cherries
thrown at it for both problems?
==> pickover/pickover.15.s <==
In article <1992Oct30.1...@watson.ibm.com> you write:
: Consider a 9x9 grid of beautiful crystal wineglasses. Throw 32 cherries
: at the grid. A glass is considered occupied if it contains at least one
: cherry. (With each throw a cherry goes into one of the glasses.) How
: many different patterns of occupied glasses can you make? (A glass with
: more than one cherry is considered the same as a glass with one cherry
: in the pattern).
Assuming that rotated patterns are allowed, then it is (simply)
sum( 81!/(81-n)! , n=1->32) . Since, if a total of n different classes are
filled, then the number of combinations is 81!/(81-n)!. Since there can
be from 1 to 32 glasses filled, the total # is just the sum of these...
:
: 2. Same as above except that you place 8 cherries in glasses (x,y) and
: then determine the other positions by placing cherries at (x,-y),
: (-x,y), (-x,-y) leading to 32 cherries in the grid. Consider the array
: of glasses centered at the origin. How many different patterns of
: occupied glasses can you make? (A glass with more than one cherry is
: considered the same as a glass with one cherry in the pattern).
This limitation basically reduces the number of available spots, from 9x9
to 5x5. Also, I only have to worry about 8 occupied spaces. Soo...
#of comb. = sum( (25!/(25-n)!, n=1->8)
:
: 3. Can your results be extrapolated to an NxN grid with M cherries
: thrown at it for both problems?
With a odd N, and M = 4k (evenly divs by 4), then
for 1....
#of comb = sum( (N^2)!/(N^2-n)! , n=1->M)
for 2....
#of comb = sum( (((N+1)/2)^2)!/(((N+1)/2)^2-n)! , n=1->M/4)
--
Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD!
// | Senior, Chemical Engineering, Univ. of Toledo
\\ // Only the | Summer Intern, NASA Lewis Research Center
\ \X/ AMIGA! | mne...@jupiter.cse.utoledo.edu /
--------+ How do YOU spell 'potato'? How 'bout 'lousy'? +----------
"Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
==> pickover/pickover.16.p <==
Title: Cliff Puzzle 16: Undulating Squares
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
A square number is of the form y=x**2. For example, 25 is a square
number.
Undulating numbers are of the form: ababababab... For example, the
following are undulating numbers: 1717171, 282828, etc.
1. Are there any undulating square numbers?
2. Are there any undulating cube numbers?
==> pickover/pickover.16.s <==
-------------------------
In article <1992Oct30.1...@watson.ibm.com> you write:
: 1. Are there any undulating square numbers?
11^2 = 121
: 2. Are there any undulating cube numbers?
7^3 = 343
(yes, I know they're short, but they qualify!)
--
Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD!
// | Senior, Chemical Engineering, Univ. of Toledo
\\ // Only the | Summer Intern, NASA Lewis Research Center
\ \X/ AMIGA! | mne...@jupiter.cse.utoledo.edu /
--------+ How do YOU spell 'potato'? How 'bout 'lousy'? +----------
"Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
-------------------------
In article <1992Oct30.2...@watson.ibm.com> you write:
>Hi, I was interested in non-trivial cases. Those with greater
>than 3 digits. Award goes to the person who finds the largest
>undulating square or cube number. Thanks, Cliff
343 and 676 aren't trivial (unlike 121 and 484 it doesn't come from
obvious algebraic identities). The chance that a "random"
number around x should be a perfect square is about 1/sqrt(x);
more generally, x^(-1+1/d) for a perfect d-th power. Since
there are for each k only 90 k-digit undulants you expect
to find only finitely many of these that are perfect powers,
and none that are very large. But provably listing all cases
is probably only barely, if at all, possible by present-day
methods for treating exponential Diophantine equations, unless
(as was shown in a rec.puzzles posting re your puzzles on
arith. prog. of squares with common difference 10^k) there is
some ad-hoc trick available. At any rate the largest undulating
power is probably 69696=264^2, though 211^3=9393931 comes
remarkably close.
--Noam D. Elkies
-------------------------
In article <1992Oct30.1...@watson.ibm.com>, you write...
>1. Are there any undulating square numbers?
>
Other than the obvious 11**2, 22**2, and 26**2, there is 264**2
which equals 69696.
>2. Are there any undulating cube numbers?
>
Just 7**3 as far as I can tell, though I'm limited to IEEE computational
reals.
PauL M SchwartZ (-Z-) | Follow men's eyes as they look to the skies
v206...@ubvms.BitNet | the shifting shafts of shining
p...@geog.buffalo.edu | weave the fabric of their dreams
p...@acsu.buffalo.edu | - RUSH -
==> pickover/pickover.17.p <==
Title: Cliff Puzzle 17: Weird Recursive Sequence
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
Consider the simple yet weird recursive formula
a(n) = a(a(n-1)) + a(n-a(n-1))
The sequences starts with a(1) = 1, and a(2) = 1. The "future" values
at higher values of n depend on past values in intricate recursive ways.
Can you determined the third member of the sequence? At first, this may
seem a little complicated to evaluate, but you can being slowly, by
inserting values for n, as in the following:
a(3) = a(a(2)) + a(3-a(2))
a(3) = a(1) + a(3-1) =
a(3) = 1+1 = 2
Therefore, the 3rd value of the sequence a(3) is 2.
The sequence a(n) seems simple enough: 1, 1, 2, 2, 3, 4, 4, 4, 5, ...
Try computing a few additional numbers. Can you find any interesting
patterns? The prolific mathematician John H Conway presented this
recursive sequence at a recent talk entitled "Some Crazy Sequences." He
noticed that the value a(n)/n approaches 1/2 as the sequence grows and n
becomes larger. Can you find a value, N, above which the sequence the
value of a(n)/n is always within 0.05 of the value 1/2? (In other
words,
.eq vbar a(n)/n -1/2 vbar lt 0.05.
The bars indicate the absolute value.)
A difficult problem? you ask.
John Conway offered $10,000 to the person to find the s-m-a-l-l-e-s-t
such N. A month after Conway made the offer, Colin Mallows of AT&T
solved the $10,000 question: N = 3,173,375,556. Manfred Shroeder has
noted that the sequence is "replete with appealing self-similarities
that contain the clue to the problem's solution." Can you find any
self-similarities? As I write this, no-one on the planet has found a
value for the smallest N such that a(n)/n is always within 0.01 of the
value 1/2.
.eq (vbar a(n)/n -1/2 vbar lt 0.01. )
==> pickover/pickover.17.s <==
-------------------------
In article <1992Nov06.1...@watson.ibm.com> you write:
: Title: Cliff Puzzle 17: Weird Recursive Sequence
: Consider the simple yet weird recursive formula
: a(n) = a(a(n-1)) + a(n-a(n-1))
The first 32 terms, and the ratio a(n)/n for each is as follows...
n a(n) a(n)/n
1 1 1.0
2 1 1.0
3 2 .666
4 2 .5
5 3 .6
6 4 .666
7 4 .5714
8 4 .5
9 5 .5555
10 6 .6
11 7 .6363
12 7 .5833
13 8 .6153
14 8 .5714
15 8 .5333
16 8 .5
17 9 .5294
18 10 .5555
19 11 .5789
20 12 .6
21 12 .5714
22 13 .5909
23 14 .6086
24 14 .5833
25 15 .6
26 15 .5769
27 15 .5555
28 16 .5714
29 16 .5517
30 16 .5333
31 16 .5161
32 16 .5
33 17 .... and so and....
off the top, we can see that on the 2^k (k a pos. int) terms, the
ratio goes to .5
between each of these, the ratio goes up and then drops back to .5
(ignoring the variances due to integer arithmatic)
the value of n at the maximum in each jump is halfway between the two
2^k points. The value of a(n) at those points seems to be
2^(k-1) - f(k), where f(k) is some function that I cannot determine
without more computing power.... *sniff*
Therefore, we must find a value of x such that...
(2^(x-1)-f(x))/2^x - .5 <.05 (or whatever)
or
f(x)/2^x < .05
and then N would be .5*(2^x-2^(x-1))
if I could see the next terms up to 128, I might be able to calculate it...
--
Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD!
// | Senior, Chemical Engineering, Univ. of Toledo
\\ // Only the | Summer Intern, NASA Lewis Research Center
\ \X/ AMIGA! | mne...@jupiter.cse.utoledo.edu /
--------+ How do YOU spell 'potato'? How 'bout 'lousy'? +----------
"Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
-------------------------
In article <1992Nov06.1...@watson.ibm.com> you write:
>John Conway offered $10,000 to the person to find the s-m-a-l-l-e-s-t
>such N. A month after Conway made the offer, Colin Mallows of AT&T
>solved the $10,000 question: N = 3,173,375,556.
As I pointed out in my posting, this is incorrect, and differs from
Mallows' correct answer published in his article. But a bit of
investigation shows that the above N is hardly a random guess, either.
Conway's sequence is best understood by analyzing it on "levels",
where the k'th level is the set of integers between 2^k and 2^(k+1).
It turns out that Mallows' correct answer, 6083008742, lies on level 32,
and the largest candidate answer on level 31 is N=3173375556, the
number quoted above.
Where did you see the above value of N given as the answer to Conway's
question?
p.s. As I found out when I edited my posted response to your message,
you either use lines longer than 80 characters in your postings,
or else your editor appends extra linefeeds to each line. Since
both conditions could be problematic for a lot of people who read
your messages on rec.puzzles, you might want to correct this
condition.
==> pickover/pickover.18.p <==
Title: Cliff Puzzle 18: Difficult Nested Roots
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
Consider the following nested set of square roots.
.eq ? = sqrt <1 + G sqrt <1+(G+1) sqrt < 1 + ... >>>
Here, G indicates "Googol" or 10**100.
The "<" and ">" symbols indicate where the beginning and ends of the
the nested roots.
1. What is the value for in this infinite set of nested roots.
2. What is the next term under the root?
Hint:
In 1911, the famous mathematical prodigy Srinivasa Ramanujan posed the
following question (#298) in a new mathematical journal called the
:Journal of the Indian Mathematical Society.
.eq ? = sqrt <1 + 2 sqrt <1+3 sqrt <1 + ... >>>
==> pickover/pickover.18.s <==
-------------------------
In article <1992Nov11.2...@watson.ibm.com> you write:
: Title: Cliff Puzzle 18: Difficult Nested Roots
: From: cl...@watson.ibm.com
: Consider the following nested set of square roots.
:
: ? = sqrt <1 + G sqrt <1+(G+1) sqrt < 1 + ... >>>
:
: Here, G indicates "Googol" or 10**100.
: The "<" and ">" symbols indicate where the beginning and ends of the
: the nested roots.
:
: 1. What is the value for in this infinite set of nested roots.
: 2. What is the next term under the root?
: Hint:
: In 1911, a twenty-three-year-old Indian clerk named Srinivasa Ramanujan
: posed the following question (#298) in a new mathematical journal called
: the Journal of the Indian Mathematical Society.
:
: ? = sqrt <1 + 2 sqrt <1+3 sqrt <1 + ... >>>
:
Doing a n-depth thing-a-ding on this.....
n=1 v=1
2 1.732
3 2.236
4 2.5598
5 2.7551
6 2.867
....
20 2.99999376
....
so I expect that the sum is actually 3. Or in the general case when the
2 (or the G from above) is replaced by m, then the evaluation of the series
is m+1. This CAN be shown as follows....
m+1 = sqrt(1+m sqrt(1+(m+1)*sqrt(....))
m^2 + 2m +1 = 1 + m *sqrt(1 + (m+1)*sqrt(...))
m^2 + 2m = m*sqrt(1+(m+1)*sqrt(...))
m+2 = sqrt(1+(m+1)*sqrt(1+(m+2)*sqrt(...))
Thus if m+1 is then sum when the series is based off m, then m+2 is then
sum when the series is based off m+1. Since this works for m=2 (as shown
above), then it must work for all whole numbers (mathematical induction is
such a wonderful thing...)
Therefore, the sum with m=G is G+1.
The next term, as show above, is (1+(m+2)*sqrt(1+....))
--
Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD!
// | Senior, Chemical Engineering, Univ. of Toledo
\\ // Only the | Summer Intern, NASA Lewis Research Center
\ \X/ AMIGA! | mne...@jupiter.cse.utoledo.edu /
--------+ How do YOU spell 'potato'? How 'bout 'lousy'? +----------
"Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> probability/amoeba.p <==
A jar begins with one amoeba. Every minute, every amoeba
turns into 0, 1, 2, or 3 amoebae with probability 25%
for each case ( dies, does nothing, splits into 2, or splits
into 3). What is the probability that the amoeba population
eventually dies out?
==> probability/amoeba.s <==
If p is the probability that a single amoeba's descendants will die
out eventually, the probability that N amoebas' descendents will all
die out eventually must be p^N, since each amoeba is independent of
every other amoeba. Also, the probability that a single amoeba's
descendants will die out must be independent of time when averaged
over all the possibilities. At t=0, the probability is p, at t=1 the
probability is 0.25(p^0+p^1+p^2+p^3), and these probabilities must be
equal. Extinction probability p is a root of f(p)=p. In this case,
p = sqrt(2)-1.
The generating function for the sequence P(n,i), which gives the
probability of i amoebas after n minutes, is f^n(x), where f^n(x) ==
f^(n-1) ( f(x) ), f^0(x) == x . That is, f^n is the nth composition
of f with itself.
Then f^n(0) gives the probability of 0 amoebas after n minutes, since
f^n(0) = P(n,0). We then note that:
f^(n+1)(x) = ( 1 + f^n(x) + (f^n(x))^2 + (f^n(x))^3 )/4
so that if f^(n+1)(0) -> f^n(0) we can solve the equation.
The generating function also gives an expression for the expectation
value of the number of amoebas after n minutes. This is d/dx(f^n(x))
evaluated at x=1. Using the chain rule we get f'(f^(n-1)(x))*d/dx(f^(n-1)(x))
and since f'(1) = 1.5 and f(1) = 1, we see that the result is just
1.5^n, as might be expected.
==> probability/apriori.p <==
An urn contains one hundred white and black balls. You sample one hundred
balls with replacement and they are all white. What is the probability
that all the balls are white?
==> probability/apriori.s <==
This question cannot be answered with the information given.
In general, the following formula gives the conditional probability
that all the balls are white given you have sampled one hundred balls
and they are all white:
P(100 white | 100 white samples) =
P(100 white samples | 100 white) * P(100 white)
-----------------------------------------------------------
sum(i=0 to 100) P(100 white samples | i white) * P(i white)
The probabilities P(i white) are needed to compute this formula. This
does not seem helpful, since one of these (P(100 white)) is just what we
are trying to compute. However, the following argument can be made:
Before the experiment, all possible numbers of white balls from zero to
one hundred are equally likely, so P(i white) = 1/101. Therefore, the
odds that all 100 balls are white given 100 white samples is:
P(100 white | 100 white samples) =
1 / ( sum(i=0 to 100) (i/100)^100 ) =
63.6%
This argument is fallacious, however, since we cannot assume that the urn
was prepared so that all possible numbers of white balls from zero to one
hundred are equally likely. In general, we need to know the P(i white)
in order to calculate the P(100 white | 100 white samples). Without this
information, we cannot determine the answer.
This leads to a general "problem": our judgments about the relative
likelihood of things is based on past experience. Each experience allows
us to adjust our likelihood judgment, based on prior probabilities. This
is called Bayesian inference. However, if the prior probabilities are not
known, then neither are the derived probabilities. But how are the prior
probabilities determined? For example, if we are brains in the vat of a
diabolical scientist, all of our prior experiences are illusions, and
therefore all of our prior probabilities are wrong.
All of our probability judgments indeed depend upon the assumption that
we are not brains in a vat. If this assumption is wrong, all bets are
off.
==> probability/bayes.p <==
One urn contains black marbles, and the other contains white or black
marbles with even odds. You pick a marble from an urn; it is black;
you put it back; what are the odds that you will draw a black marble on
the next draw? What are the odds after n black draws?
==> probability/bayes.s <==
Every time you draw a black marble, you throw out (from your
probability space) half of those possible urns that contain both
colors. So you have 1/2^n times as many ways to have a white marble in
the urn after n draws, all black, as at the start. But you have
exactly the same number of ways to have both marbles black. The
numbers (mixed cases vs. all-black cases) go as 1:1, 1:2, 1:4, 1:8,...
and the chance of having a white marble in the urn goes as 1/2, 1/3,
1/5, 1/9, ..., 1/(1+2^(n-1)), hence the odds of drawing a white marble
on the nth try after n-1 consecutive drawings of black are
1/4 the first time
1/6 the second time
1/10 the third time
...
1/(2+2^n) the nth time
==> probability/birthday/line.p <==
At a movie theater, the manager announces that they will give a free ticket
to the first person in line whose birthday is the same as someone who has
already bought a ticket. You have the option of getting in line at any
time. Assuming that you don't know anyone else's birthday, that birthdays
are distributed randomly throughtout the year, etc., what position in line
gives you the greatest chance of being the first duplicate birthday?
==> probability/birthday/line.s <==
Suppose you are the Kth person in line. Then you win if and only if the
K-1 people ahead all have distinct birtdays AND your birthday matches
one of theirs. Let
A = event that your birthday matches one of the K-1 people ahead
B = event that those K-1 people all have different birthdays
Then
Prob(you win) = Prob(B) * Prob(A | B)
(Prob(A | B) is the conditional probability of A given that B occurred.)
Now let P(K) be the probability that the K-th person in line wins,
Q(K) the probability that the first K people all have distinct
birthdays (which occurs exactly when none of them wins). Then
P(1) + P(2) + ... + P(K-1) + P(K) = 1 - Q(K)
P(1) + P(2) + ... + P(K-1) = 1 - Q(K-1)
P(K) = Q(K-1) - Q(K) <--- this is what we want to maximize.
Now if the first K-1 all have distinct birthdays, then assuming
uniform distribution of birthdays among D days of the year,
the K-th person has K-1 chances out of D to match, and D-K+1 chances
not to match (which would produce K distinct birthdays). So
Q(K) = Q(K-1)*(D-K+1)/D = Q(K-1) - Q(K-1)*(K-1)/D
Q(K-1) - Q(K) = Q(K-1)*(K-1)/D = Q(K)*(K-1)/(D-K+1)
Now we want to maximize P(K), which means we need the greatest K such
that P(K) - P(K-1) > 0. (Actually, as just given, this only
guarantees a local maximum, but in fact if we investigate a bit
farther we'll find that P(K) has only one maximum.)
For convenience in calculation let's set K = I + 1. Then
Q(I-1) - Q(I) = Q(I)*(I-1)/(D-I+1)
Q(I) - Q(I+1) = Q(I)*I/D
P(K) - P(K-1) = P(I+1) - P(I)
= (Q(I) - Q(I+1)) - (Q(K-2) - Q(K-1))
= Q(I)*(I/D - (I-1)/(D-I+1))
To find out where this is last positive (and next goes negative), solve
x/D - (x-1)/(D-x+1) = 0
Multiply by D*(D+1-x) both sides:
(D+1-x)*x - D*(x-1) = 0
Dx + x - x^2 - Dx + D = 0
x^2 - x - D = 0
x = (1 +/- sqrt(1 - 4*(-D)))/2 ... take the positive square root
= 0.5 + sqrt(D + 0.25)
Setting D=365 (finally deciding how many days in a year!),
desired I = x = 0.5 + sqrt(365.25) = 19.612 (approx).
The last integer I for which the new probability is greater then the old
is therefore I=19, and so K = I+1 = 20. You should try to be the 20th
person in line.
Computing your chances of actually winning is slightly harder, unless
you do it numerically by computer. The recursions you need have already
been given.
-- David Karr (ka...@cs.cornell.edu)
==> probability/birthday/same.day.p <==
How many people must be at a party before you have even odds or better
of two having the same bithday (not necessarily the same year, of course)?
==> probability/birthday/same.day.s <==
23.
See also:
archive entry "coupon"
==> probability/cab.p <==
A cab was involved in a hit and run accident at night. Two cab companies,
the Green and the Blue, operate in the city. Here is some data:
a) Although the two companies are equal in size, 85% of cab
accidents in the city involve Green cabs and 15% involve Blue cabs.
b) A witness identified the cab in this particular accident as Blue.
The court tested the reliability of the witness under the same circumstances
that existed on the night of the accident and concluded that the witness
correctly identified each one of the two colors 80% of the time and failed
20% of the time.
What is the probability that the cab involved in the accident was
Blue rather than Green?
If it looks like an obvious problem in statistics, then consider the
following argument:
The probability that the color of the cab was Blue is 80%! After all,
the witness is correct 80% of the time, and this time he said it was Blue!
What else need be considered? Nothing, right?
If we look at Bayes theorem (pretty basic statistical theorem) we
should get a much lower probability. But why should we consider statistical
theorems when the problem appears so clear cut? Should we just accept the
80% figure as correct?
==> probability/cab.s <==
The police tests don't apply directly, because according to the
wording, the witness, given any mix of cabs, would get the right
answer 80% of the time. Thus given a mix of 85% green and 15% blue
cabs, he will say 20% of the green cabs and 80% of the blue cabs are
blue. That's 20% of 85% plus 80% of 15%, or 17%+12% = 29% of all the
cabs that the witness will say are blue. Of those, only 12/29 are
actually blue. Thus P(cab is blue | witness claims blue) = 12/29.
That's just a little over 40%.
Think of it this way... suppose you had a robot watching parts on a
conveyor belt to spot defective parts, and suppose the robot made a
correct determination only 50% of the time (I know, you should
probably get rid of the robot...). If one out of a billion parts are
defective, then to a very good approximation you'd expect half your
parts to be rejected by the robot. That's 500 million per billion.
But you wouldn't expect more than one of those to be genuinely
defective. So given the mix of parts, a lot more than 50% of the
REJECTED parts will be rejected by mistake (even though 50% of ALL the
parts are correctly identified, and in particular, 50% of the
defective parts are rejected).
When the biases get so enormous, things starts getting quite a bit
more in line with intuition.
For a related real-life example of probability in the courtroom see
People v. Collins, 68 Cal 2d319 (1968).
==> probability/coupon.p <==
There is a free gift in my breakfast cereal. The manufacturers say that
the gift comes in four different colors, and encourage one to collect
all four (& so eat lots of their cereal). Assuming there is an equal
chance of getting any one of the colors, what is the expected number
of boxes I must consume to get all four? Can you generalise to n
colors and/or unequal probabilities?
==> probability/coupon.s <==
The problem is well known under the name of "COUPON COLLECTOR PROBLEM".
The answer for n equally likely coupons is
(1) C(n)=n*H(n) with H(n)=1+1/2+1/3+...+1/n.
In the unequal probabilities case, with p_i the probability of coupon i,
it becomes
(2) C(n)=int_0^infty [1-prod_{i=1}^n (1-exp(-p_i*t))] dt
which reduces to (1) when p_i=1/n (An easy exercise).
In the equal probabilities case C(n)~n log(n). For a Zipf law,
from (2), we have C(n)~n log^2(n).
A related problem is the "BIRTHDAY PARADOX" familiar to people
interested in hashing algorithms: With a party of 23 persons,
you are likely (i.e. with probability >50%) to find two with
the same birthday. The non equiprobable case was solved by:
M. Klamkin and D. Newman, Extensions of the birthday
surprise, J. Comb. Th. 3 (1967), 279-282.
==> probability/darts.p <==
Peter throws two darts at a dartboard, aiming for the center. The
second dart lands farther from the center than the first. If Peter now
throws another dart at the board, aiming for the center, what is the
probability that this third throw is also worse (i.e., farther from
the center) than his first? Assume Peter's skilfulness is constant.
==> probability/darts.s <==
Since the three darts are thrown independently,
they each have a 1/3 chance of being the best throw. As long as the
third dart is not the best throw, it will be worse than the first dart.
Therefore the answer is 2/3.
Ranking the three darts' results from A (best) to C
(worst), there are, a priori, six equiprobable outcomes.
possibility # 1 2 3 4 5 6
1st throw A A B B C C
2nd throw B C A C A B
3rd throw C B C A B A
The information from the first two throws shows us that the first
throw will not be the worst, nor the second throw the best. Thus
possibilities 3, 5 and 6 are eliminated, leaving three equiprobable
cases, 1, 2 and 4. Of these, 1 and 2 have the third throw worse than
the first; 4 does not. Again the answer is 2/3.
==> probability/derangement.p <==
12 men leave their hats with the hat check. If the hats are randomly
returned, what is the probability that nobody gets the correct hat?
==> probability/derangement.s <==
Suppose we are handing out hats to n people. First we start with all
the possible outcomes. Then we subtract off those that assign the
right hat to a given person, for each of the n people. But this
double-counts each outcome that assigned 2 hats correctly, so we have
to add those outcomes back in. But now we've counted one net copy of
each outcome with 3 correct hats in our set, so we have to subtract
those off again. But now we've taken away each 4-correct-hat outcome
once too often, and have to put it back in, and so forth ... until we
add or subtract the outcome that involves all n people getting the
correct hats.
Putting it all in probabilities, the measure of the original set is 1.
For a given subset of k people, the probability that they all get
their correct hats is (n-k)!/n!, while there are (n choose k) such
subsets of k people altogether. But then
(n choose k)*(n-k)!/n! = (n!/((n-k)!*k!))*(n-k)!/n! = 1/k!
is the total probability measure we get by counting each subset of k
people once each. So we end up generating the finite series
1 - 1/1! + 1/2! - 1/3! +- ... +/- 1/n!
which is of course just the first n+1 terms of the Taylor series
expansion for f(x) = e^x centered at 0 and evaluated at -1, which
converges to 1/e quite fast. You can compute the exact probability for
any n >= 1 simply by rounding n!/e to the nearest whole number, then
dividing again by n!. Moreover I think you will find you are always
rounding down for odd n and rounding up for even n.
For example,
12! / e = 176214840.95798...
which is within 0.05 (absolute error, not relative) of the correct
intermediate result, 176214841.
Another fact is that if you set the probability of no matching hats
equal to m/n!, then m is an integer divisible by (n-1). That's
because the number of ways you can give hat #2 to person #1 is exactly
the same as the number of ways you can give that person hat #3,
likewise for hat #4, and so forth.
-- David Karr (ka...@cs.cornell.edu)
==> probability/family.p <==
Suppose that it is equally likely for a pregnancy to deliver
a baby boy as it is to deliver a baby girl. Suppose that for a
large society of people, every family continues to have children
until they have a boy, then they stop having children.
After 1,000 generations of families, what is the ratio of males
to females?
==> probability/family.s <==
The ratio will be 50-50 in both cases. We are not killing off any
fetuses or babies, and half of all conceptions will be male, half
female. When a family decides to stop does not affect this fact.
==> probability/flips/once.in.run.p <==
What are the odds that a run of one H or T (i.e., THT or HTH) will occur
in n flips of a fair coin?
==> probability/flips/once.in.run.s <==
References:
John P. Robinson, Transition Count and Syndrome are Uncorrelated, IEEE
Transactions on Information Theory, Jan 1988.
First we define a function or enumerator P(n,k) as the number of length
"n" sequences that generate "k" successes. For example,
P(4,1)= 4 (HHTH, HTHH, TTHT, and THTT are 4 possible length 4 sequences).
I derived two generating functions g(x) and h(x) in order to enumerate
P(n,k), they are compactly represented by the following matrix
polynomial.
_ _ _ _ _ _
| g(x) | | 1 1 | (n-3) | 4 |
| | = | | | |
| h(x) | | 1 x | |2+2x |
|_ _| |_ _| |_ _|
The above is expressed as matrix generating function. It can be shown
that P(n,k) is the coefficient of the x^k in the polynomial
(g(x)+h(x)).
For example, if n=4 we get (g(x)+h(x)) from the matrix generating
function as (10+4x+2x^2). Clearly, P(4,1) (coefficient of x) is 4 and
P(4,2)=2 ( There are two such sequences THTH, and HTHT).
We can show that
mean(k) = (n-2)/4 and sd= square_root(5n-12)/4
We need to generate "n" samples. This can be done by using sequences of length
(n+2). Then our new statistics would be
mean = n/4
sd = square_root(5n-2)/4
Similar approach can be followed for higher dimensional cases.
==> probability/flips/twice.in.run.p <==
What is the probability in n flips of a fair coin that there will be two
heads in a row?
==> probability/flips/twice.in.run.s <==
Well, the question is then how many strings of n h's and t's contain
hh? I would guess right off hand that its going to be easier to
calculate the number of strings that _don't_ contain hh and then
subtract that from the total number of strings.
So we want to count the strings of n h's and t's with no hh in them.
How many h's and t's can there be? It is fairly clear that there must
be from 0 to n/2 h's, inclusive. (If there were (n/2+1) then there
would have to be two touching.)
How many strings are there with 0 h's? 1
How many strings are there with 1 h? Well, there are (n-1) t's, so
there are a total of n places to put the one h. So the are nC1 such
strings. How many strings are there with 2 h's? Well, there are (n-1)
places to put the two h's, so there are (n-1)C2 such strings.
Finally, with n/2 h's there are (n/2+1) places to put them, so there
are (n/2+1)C(n/2) such strings.
Therefore the total number of strings is
Sum (from i=0 to n/2) of (n-i+1)C(i)
Now, here's where it get's interesting. If we play around with Pascal's
triangle for a while, we see that this sum equals none other than the
(n+2)th Fibonacci number.
So the probability that n coin tosses will give a hh is:
2^n-f(n+2)
----------
2^n
(where f(x) is the xth Fibanocci number (so that f(1) is and f(2) are both 1))
==> probability/flips/unfair.p <==
Generate even odds from an unfair coin. For example, if you
thought a coin was biased toward heads, how could you get the
equivalent of a fair coin with several tosses of the unfair coin?
==> probability/flips/unfair.s <==
Toss twice. If both tosses give the same result, repeat this process
(throw out the two tosses and start again). Otherwise, take the first
of the two results.
==> probability/flips/waiting.time.p <==
Compute the expected waiting time for a sequence of coin flips, or the
probabilty that one sequence of coin flips will occur before another.
==> probability/flips/waiting.time.s <==
Here's a C program I had lying around that is relevant to the
current discussion of coin-flipping. The algorithm is N^3 (for N flips)
but it could certainly be improved. Compile with
cc -o flip flip.c -lm
-- Guy
_________________ Cut here ___________________
#include <stdio.h>
#include <math.h>
char *progname; /* Program name */
#define NOT(c) (('H' + 'T') - (c))
/* flip.c -- a program to compute the expected waiting time for a sequence
of coin flips, or the probabilty that one sequence
of coin flips will occur before another.
Guy Jacobson, 11/1/90
*/
main (ac, av) int ac; char **av;
{
char *f1, *f2, *parseflips ();
double compute ();
progname = av[0];
if (ac == 2) {
f1 = parseflips (av[1]);
printf ("Expected number of flips until %s = %.1f\n",
f1, compute (f1, NULL));
}
else if (ac == 3) {
f1 = parseflips (av[1]);
f2 = parseflips (av[2]);
if (strcmp (f1, f2) == 0) {
printf ("Can't use the same flip sequence.\n");
exit (1);
}
printf ("Probability of flipping %s before %s = %.1f%%\n",
av[1], av[2], compute (f1, f2) * 100.0);
}
else
usage ();
}
char *parseflips (s) char *s;
{
char *f = s;
while (*s)
if (*s == 'H' || *s == 'h')
*s++ = 'H';
else if (*s == 'T' || *s == 't')
*s++ = 'T';
else
usage ();
return f;
}
usage ()
{
printf ("usage: %s {HT}^n\n", progname);
printf ("\tto get the expected waiting time, or\n");
printf ("usage: %s s1 s2\n\t(where s1, s2 in {HT}^n for some fixed n)\n",
progname);
printf ("\tto get the probability that s1 will occur before s2\n");
exit (1);
}
/*
compute -- if f2 is non-null, compute the probability that flip
sequence f1 will occur before f2. With null f2, compute
the expected waiting time until f1 is flipped
technique:
Build a DFA to recognize (H+T)*f1 [or (H+T)*(f1+f2) when f2
is non-null]. Randomly flipping coins is a Markov process on the
graph of this DFA. We can solve for the probability that f1 precedes
f2 or the expected waiting time for f1 by setting up a linear system
of equations relating the values of these unknowns starting from each
state of the DFA. Solve this linear system by Gaussian Elimination.
*/
typedef struct state {
char *s; /* pointer to substring string matched */
int len; /* length of substring matched */
int backup; /* number of one of the two next states */
} state;
double compute (f1, f2) char *f1, *f2;
{
double solvex0 ();
int i, j, n1, n;
state *dfa;
int nstates;
char *malloc ();
n = n1 = strlen (f1);
if (f2)
n += strlen (f2); /* n + 1 states in the DFA */
dfa = (state *) malloc ((unsigned) ((n + 1) * sizeof (state)));
if (!dfa) {
printf ("Ouch, out of memory!\n");
exit (1);
}
/* set up the backbone of the DFA */
for (i = 0; i <= n; i++) {
dfa[i].s = (i <= n1) ? f1 : f2;
dfa[i].len = (i <= n1) ? i : i - n1;
}
/* for i not a final state, one next state of i is simply
i+1 (this corresponds to another matching character of dfs[i].s
The other next state (the backup state) is now computed.
It is the state whose substring matches the longest suffix
with the last character changed */
for (i = 0; i <= n; i++) {
dfa[i].backup = 0;
for (j = 1; j <= n; j++)
if ((dfa[j].len > dfa[dfa[i].backup].len)
&& dfa[i].s[dfa[i].len] == NOT (dfa[j].s[dfa[j].len - 1])
&& strncmp (dfa[j].s, dfa[i].s + dfa[i].len - dfa[j].len + 1,
dfa[j].len - 1) == 0)
dfa[i].backup = j;
}
/* our dfa has n + 1 states, so build a system n + 1 equations
in n + 1 unknowns */
eqsystem (n + 1);
for (i = 0; i < n; i++)
if (i == n1)
equation (1.0, n1, 0.0, 0, 0.0, 0, -1.0);
else
equation (1.0, i, -0.5, i + 1, -0.5, dfa[i].backup, f2 ? 0.0 : -1.0);
equation (1.0, n, 0.0, 0, 0.0, 0, 0.0);
free (dfa);
return solvex0 ();
}
/* a simple gaussian elimination equation solver */
double *m, **M;
int rank;
int neq = 0;
/* create an n by n system of linear equations. allocate space
for the matrix m, filled with zeroes and the dope vector M */
eqsystem (n) int n;
{
char *calloc ();
int i;
m = (double *) calloc (n * (n + 1), sizeof (double));
M = (double **) calloc (n, sizeof (double *));
if (!m || !M) {
printf ("Ouch, out of memory!\n");
exit (1);
}
for (i = 0; i < n; i++)
M[i] = &m[i * (n + 1)];
rank = n;
neq = 0;
}
/* add a new equation a * x_na + b * x_nb + c * x_nc + d = 0.0
(note that na, nb, and nc are not necessarily all distinct.) */
equation (a, na, b, nb, c, nc, d) double a, b, c, d; int na, nb, nc;
{
double *eq = M[neq++]; /* each row is an equation */
eq[na + 1] += a;
eq[nb + 1] += b;
eq[nc + 1] += c;
eq[0] = d; /* column zero holds the constant term */
}
/* solve for the value of variable x_0. This will go nuts if
therer are errors (for example, if m is singular) */
double solvex0 ()
{
register i, j, jmax, k;
register double max, val;
register double *maxrow, *row;
for (i = rank; i > 0; --i) { /* for each variable */
/* find pivot element--largest value in ith column*/
max = 0.0;
for (j = 0; j < i; j++)
if (fabs (M[j][i]) > fabs (max)) {
max = M[j][i];
jmax = j;
}
/* swap pivot row with last row using dope vectors */
maxrow = M[jmax];
M[jmax] = M[i - 1];
M[i - 1] = maxrow;
/* normalize pivot row */
max = 1.0 / max;
for (k = 0; k <= i; k++)
maxrow[k] *= max;
/* now eliminate variable i by subtracting multiples of pivot row */
for (j = 0; j < i - 1; j++) {
row = M[j];
if (val = row[i]) /* if variable i is in this eq */
for (k = 0; k <= i; k++)
row[k] -= maxrow[k] * val;
}
}
/* the value of x0 is now in constant column of first row
we only need x0, so no need to back-substitute */
val = -M[0][0];
free (M);
free (m);
return val;
}
_________________________________________________________________
Guy Jacobson (201) 582-6558 AT&T Bell Laboratories
uucp: {att,ucbvax}!ulysses!guy 600 Mountain Avenue
internet: g...@ulysses.att.com Murray Hill NJ, 07974
==> probability/flush.p <==
Which set contains proportionately more flushes than the set of all
possible poker hands?
(1) Hands whose first card is an ace
(2) Hands whose first card is the ace of spades
(3) Hands with at least one ace
(4) Hands with the ace of spades
==> probability/flush.s <==
An arbitrary hand can have two aces but a flush hand can't. The
average number of aces that appear in flush hands is the same as the
average number of aces in arbitrary hands, but the aces are spread out
more evenly for the flush hands, so set #3 contains a higher fraction
of flushes.
Aces of spades, on the other hand, are spread out the same way over
possible hands as they are over flush hands, since there is only one of
them in the deck. Whether or not a hand is flush is based solely on a
comparison between different cards in the hand, so looking at just one
card is necessarily uninformative. So the other sets contain the same
fraction of flushes as the set of all possible hands.
==> probability/hospital.p <==
A town has two hospitals, one big and one small. Every day the big
hospital delivers 1000 babies and the small hospital delivers 100
babies. There's a 50/50 chance of male or female on each birth.
Which hospital has a better chance of having the same number of boys
as girls?
==> probability/hospital.s <==
The small one. If there are 2N babies born, then the probability of an
even split is
(2N choose N) / (2 ** 2N) ,
where (2N choose N) = (2N)! / (N! * N!) .
This is a DECREASING function.
If there are two babies born, then the probability of a split is 1/2
(just have the second baby be different from the first). With 2N
babies, If there is a N,N-1 split in the first 2N-1, then there is a
1/2 chance of the last baby making it an even split. Otherwise there
can be no even split. Therefore the probability is less than 1/2
overall for an even split.
As N goes to infinity the probability of an even split approaches zero
(although it is still the most likely event).
==> probability/icos.p <==
The "house" rolls two 20-sided dice and the "player" rolls one
20-sided die. If the player rolls a number on his die between the
two numbers the house rolled, then the player wins. Otherwise, the
house wins (including ties). What are the probabilities of the player
winning?
==> probability/icos.s <==
It is easily seen that if any two of the three dice agree that the
house wins. The probability that this does not happen is 19*18/(20*20).
If the three numbers are different, the probability of winning is 1/3.
So the chance of winning is 19*18/(20*20*3) = 3*19/200 = 57/200.
==> probability/intervals.p <==
Given two random points x and y on the interval 0..1, what is the average
size of the smallest of the three resulting intervals?
==> probability/intervals.s <==
In between these positions the surface forms a series of planes.
Thus the volume under it consists of 2 pyramids each with an
altitude of 1/3 and an (isosceles triangular) base of area 1/2,
yielding a total volume of 1/9.
==> probability/killers.and.pacifists.p <==
You enter a town that has K killers and P pacifists. When a
pacifist meets a pacifist, nothing happens. When a pacifist meets a
killer, the pacifist is killed. When two killers meet, both die.
Assume meetings always occur between exactly two persons and the pairs
involved are completely random. What are your odds of survival?
==> probability/killers.and.pacifists.s <==
Regardless of whether you are a pacifist or a killer, you may disregard
all events in which a pacifist other than yourself is involved and
consider only events in which you are killed or a pair of killers other
than yourself is killed.
Thus we may assume that there are N killers and yourself. If N is odd,
your odds of surviving are zero. If N is even, it doesn't matter to
you whether you are a killer or not. So WLOG assume you are. Then your
probability of survival is 1/(N+1).
==> probability/leading.digit.p <==
What is the probability that the ratio of two random reals starts with a 1?
What about 9?
==> probability/leading.digit.s <==
What is the distribution of y/x for (x,y) chosen with uniform distribution
from the unit square?
First you want y/x in one of the intervals ... [0.01,0.02), [0.1,0.2),
[1,2), [10/20), ... . This corresponds to (x,y) lying in one of
several triangles with height 1 and bases on either the right or top
edges of the square. The bases along the right edge have lengths 0.1
(for 0.1 <= y/x < 0.2), 0.01, 0.001, ... ; the sum of this series is
1/9. The bases along the top edge have lengths 0.5 (for 0.5 < x/y <=
1), 0.05, 0.005, ... ; the sum of this series is 5/9. So you have a
total base length of 6/9 = 2/3, height 1, so the area is 1/3. The
total area of the square is 1/3, so the probability that y/x starts
with a 1 is 1/3 ~= 0.333333.
In the second case you have the same lengths (but in different places)
on the right edge, total 1/9. But on the top edge, 9 <= y/x < 10 gives
you 1/10 < x/y <= 1/9 gives you a base of length 1/9 - 1/10 = 1/90,
and the series proceeds 1/900, 1/9000, ... ; the sum is 1/81. Total
base length then is 9/81 + 1/81 = 10/81, height 1, total area (and
hence probability of a leading 9) is 5/81 ~= 0.061728.
==> probability/lights.p <==
Waldo and Basil are exactly m blocks west and n blocks north from
Central Park, and always go with the green light until they run out of
options. Assuming that the probability of the light being green is 1/2
in each direction, that if the light is green in one direction it is
red in the other, and that the lights are not synchronized, find the
expected number of red lights that Waldo and Basil will encounter.
==> probability/lights.s <==
Let E(m,n) be this number, and let (x)C(y) = x!/(y! (x-y)!). A model
for this problem is the following nxm grid:
^ B---+---+---+ ... +---+---+---+ (m,0)
| | | | | | | | |
N +---+---+---+ ... +---+---+---+ (m,1)
<--W + E--> : : : : : : : :
S +---+---+---+ ... +---+---+---+ (m,n-1)
| | | | | | | | |
v +---+---+---+ ... +---+---+---E (m,n)
where each + represents a traffic light. We can consider each
traffic light to be a direction pointer, with an equal chance of
pointing either east or south.
IMHO, the best way to approach this problem is to ask: what is the
probability that edge-light (x,y) will be the first red edge-light
that the pedestrian encounters? This is easy to answer; since the
only way to reach (x,y) is by going south x times and east y times,
in any order, we see that there are (x+y)C(x) possible paths from
(0,0) to (x,y). Since each of these has probability (1/2)^(x+y+1)
of occuring, we see that the the probability we are looking for is
(1/2)^(x+y+1)*(x+y)C(x). Multiplying this by the expected number
of red lights that will be encountered from that point, (n-k+1)/2,
we see that
m-1
-----
\
E(m,n) = > ( 1/2 )^(n+k+1) * (n+k)C(n) * (m-k+1)/2
/
-----
k=0
n-1
-----
\
+ > ( 1/2 )^(m+k+1) * (m+k)C(m) * (n-k+1)/2 .
/
-----
k=0
Are we done? No! Putting on our Captain Clever Cap, we define
n-1
-----
\
f(m,n) = > ( 1/2 )^k * (m+k)C(m) * k
/
-----
k=0
and
n-1
-----
\
g(m,n) = > ( 1/2 )^k * (m+k)C(m) .
/
-----
k=0
Now, we know that
n
-----
\
f(m,n)/2 = > ( 1/2 )^k * (m+k-1)C(m) * (k-1)
/
-----
k=1
and since f(m,n)/2 = f(m,n) - f(m,n)/2, we get that
n-1
-----
\
f(m,n)/2 = > ( 1/2 )^k * ( (m+k)C(m) * k - (m+k-1)C(m) * (k-1) )
/
-----
k=1
- (1/2)^n * (m+n-1)C(m) * (n-1)
n-2
-----
\
= > ( 1/2 )^(k+1) * (m+k)C(m) * (m+1)
/
-----
k=0
- (1/2)^n * (m+n-1)C(m) * (n-1)
= (m+1)/2 * (g(m,n) - (1/2)^(n-1)*(m+n-1)C(m)) - (1/2)^n*(m+n-1)C(m)*(n-1)
therefore
f(m,n) = (m+1) * g(m,n) - (n+m) * (1/2)^(n-1) * (m+n-1)C(m) .
Now, E(m,n) = (n+1) * (1/2)^(m+2) * g(m,n) - (1/2)^(m+2) * f(m,n)
+ (m+1) * (1/2)^(n+2) * g(n,m) - (1/2)^(n+2) * f(n,m)
= (m+n) * (1/2)^(n+m+1) * (m+n)C(m) + (m-n) * (1/2)^(n+2) * g(n,m)
+ (n-m) * (1/2)^(m+2) * g(m,n) .
Setting m=n in this formula, we see that
E(n,n) = n * (1/2)^(2n) * (2n)C(n),
and applying Stirling's theorem we get the beautiful asymptotic formula
E(n,n) ~ sqrt(n/pi).
==> probability/lottery.p <==
There n tickets in the lottery, k winners and m allowing you to pick another
ticket. The problem is to determine the probability of winning the lottery
when you start by picking 1 (one) ticket.
A lottery has N balls in all, and you as a player can choose m numbers
on each card, and the lottery authorities then choose n balls, define
L(N,n,m,k) as the minimum number of cards you must purchase to ensure that
at least one of your cards will have at least k numbers in common with the
balls chosen in the lottery.
==> probability/lottery.s <==
This relates to the problem of rolling a random number
from 1 to 17 given a 20 sided die. You simply mark the numbers 18,
19, and 20 as "roll again".
The probability of winning is, of course, k/(n-m) as for every k cases
in which you get x "draw again"'s before winning, you get n-m-k similar
cases where you get x "draw again"'s before losing. The example using
dice makes it obvious, however.
L(N,k,n,k) >= Ceiling((N-choose-k)/(n-choose-k)*
(n-1-choose-k-1)/(N-1-choose-k-1)*L(N-1,k-1,n-1,k-1))
= Ceiling(N/n*L(N-1,k-1,n-1,k-1))
The correct way to see this is as follows: Suppose you have an
(N,k,n,k) system of cards. Look at all of the cards that contain the
number 1. They cover all ball sets that contain 1, and therefore these
cards become an (N-1,k-1,n-1,k-1) covering upon deletion of the number
1. Therefore the number 1 must appear at least L(N-1,k-1,n-1,k-1).
The same is true of all of the other numbers. There are N of them but
n appear on each card. Thus we obtain the bound.
==> probability/oldest.girl.p <==
You meet a stranger on the street, and ask how many children he has. He
truthfully says two. You ask "Is the older one a girl?" He truthfully
says yes. What is the probability that both children are girls? What
would the probability be if your second question had been "Is at least
one of them a girl?", with the other conditions unchanged?
==> probability/oldest.girl.s <==
There are four possibilities:
Oldest child Youngest child
1. Girl Girl
2. Girl Boy
3. Boy Girl
4. Boy Boy
If your friend says "My oldest child is a girl," he has eliminated cases
3 and 4, and in the remaining cases both are girls 1/2 of the time. If
your friend says "At least one of my children is a girl," he has
eliminated case 4 only, and in the remaining cases both are girls 1/3
of the time.
==> probability/particle.in.box.p <==
A particle is bouncing randomly in a two-dimensional box. How far does it
travel between bounces, on average?
Suppose the particle is initially at some random position in the box and is
traveling in a straight line in a random direction and rebounds normally
at the edges.
==> probability/particle.in.box.s <==
Let theta be the angle of the point's initial vector. After traveling a
distance r, the point has moved r*cos(theta) horizontally and r*sin(theta)
vertically, and thus has struck r*(sin(theta)+cos(theta))+O(1) walls. Hence
the average distance between walls will be 1/(sin(theta)+cos(theta)). We now
average this over all angles theta:
2/pi * intg from theta=0 to pi/2 (1/(sin(theta)+cos(theta))) dtheta
which (in a computation which is left as an exercise) reduces to
2*sqrt(2)*ln(1+sqrt(2))/pi = 0.793515.
==> probability/pi.p <==
Are the digits of pi random (i.e., can you make money betting on them)?
==> probability/pi.s <==
No, the digits of pi are not truly random, therefore you can win money
playing against a supercomputer that can calculate the digits of pi far
beyond what we are currently capable of doing. This computer selects a
position in the decimal expansion of pi -- say, at 10^100. Your job is
to guess what the next digit or digit sequence is. Specifically, you
have one dollar to bet. A bet on the next digit, if correct, returns
10 times the amount bet; a bet on the next two digits returns 100 times
the amount bet, and so on. (The dollar may be divided in any fashion,
so we could bet 1/3 or 1/10000 of a dollar.) You may place bets in any
combination. The computer will tell you the position number, let you
examine the digits up to that point, and calculate statistics for you.
It is easy to set up strategies that might win, if the supercomputer
doesn't know your strategy. For example, "Always bet on 7" might win,
if you are lucky. Also, it is easy to set up bets that will always
return a dollar. For example, if you bet a penny on every two-digit
sequence, you are sure to win back your dollar. Also, there are
strategies that might be winning, but we can't prove it. For example,
it may be that a certain sequence of digits never occurs in pi, but we
have no way of proving this.
The problem is to find a strategy that you can prove will always win
back more than a dollar.
The assumption that the position is beyond the reach of calculation
means that we must rely on general facts we know about the sequence of
digits of pi, which is practically nil. But it is not completely nil,
and the challenge is to find a strategy that will always win money.
A theorem of Mahler (1953) states that for all integers p, q > 1,
-42
|pi - p/q| > q
This says that pi cannot have a rational approximation that is
extremely tight.
Now suppose that the computer picks position N. I know that the next
41 * N digits cannot be all zero. For if they were, then the first N
digits, treated as a fraction with denominator 10^N, satisfies:
|pi - p / 10^N| < 10^(-42 N)
which contradicts Mahler's theorem.
So, I split my dollar into 10^(41N) - 1 equal parts, and bet on each of
the sequences of 41N digits, except the one with all zeroes. One of
the bets is sure to win, so my total profit is about 10(^-41N) of a
dollar!
This strategy can be improved a number of ways, such as looking for
other repeating patterns, or improvements to the bound of 42 -- but the
earnings are so pathetic, it hardly seems worth the effort.
Are there other winning strategies, not based on Mahler's theorem? I
believe there are algorithms that generate 2N binary digits of pi,
where the computations are separate for each block of N digits. Maybe
from something like this, we can find a simple subsequence of the
binary digits of pi which is always zero, or which has some simple
pattern.
==> probability/random.walk.p <==
Waldo has lost his car keys! He's not using a very efficient search;
in fact, he's doing a random walk. He starts at 0, and moves 1 unit
to the left or right, with equal probability. On the next step, he
moves 2 units to the left or right, again with equal probability. For
subsequent turns he follows the pattern 1, 2, 1, etc.
His keys, in truth, were right under his nose at point 0. Assuming
that he'll spot them the next time he sees them, what is the
probability that poor Waldo will eventually return to 0?
==> probability/random.walk.s <==
I can show the probability that Waldo returns to 0 is 1. Waldo's
wanderings map to an integer grid in the plane as follows. Let
(X_t,Y_t) be the cumulative sums of the length 1 and length 2 steps
respectively taken by Waldo through time t. By looking only at even t,
we get the ordinary random walk in the plane, which returns to the
origin (0,0) with probability 1. In fact, landing at (2n, n) for any n
will land Waldo on top of his keys too. There's no need to look at odd
t.
Similar considerations apply for step sizes of arbitrary (fixed) size.
==> probability/reactor.p <==
There is a reactor in which a reaction is to take place. This reaction
stops if an electron is present in the reactor. The reaction is started
with 18 positrons; the idea being that one of these positrons would
combine with any incoming electron (thus destroying both). Every second,
exactly one particle enters the reactor. The probablity that this particle
is an electron is 0.49 and that it is a positron is 0.51.
What is the probability that the reaction would go on for ever?
Note: Once the reaction stops, it cannot restart.
==> probability/reactor.s <==
Let P(n) be the probability that, starting with n positrons, the
reaction goes on forever. Clearly P'(n+1)=P'(0)*P'(n), where the
' indicates probabilistic complementation; also note that
P'(n) = .51*P'(n+1) + .49*P'(n-1). Hence we have that P(1)=(P'(0))^2
and that P'(0) = .51*P'(1) ==> P'(0) equals 1 or 49/51. We thus get
that either P'(18) = 1 or (49/51)^19 ==> P(18) = 0 or 1 - (49/51)^19.
The answer is indeed the latter. A standard result in random walks
(which can be easily derived using Markov chains) yields that if p>1/2
then the probability of reaching the absorbing state at +infinity
as opposed to the absorbing state at -1 is 1-r^(-i), where r=p/(1-p)
(p is the probability of moving from state n to state n-1, in our
case .49) and i equals the starting location + 1. Therefore we have
that P(18) = 1-(.49/.51)^19.
==> probability/roulette.p <==
You are in a game of Russian roulette, but this time the gun (a 6
shooter revolver) has three bullets _in_a_row_ in three of the
chambers. The barrel is spun only once. Each player then points the
gun at his (her) head and pulls the trigger. If he (she) is still
alive, the gun is passed to the other player who then points it at his
(her) own head and pulls the trigger. The game stops when one player
dies.
Now to the point: would you rather be first or second to shoot?
==> probability/roulette.s <==
All you need to consider are the six possible bullet configurations
B B B E E E -> player 1 dies
E B B B E E -> player 2 dies
E E B B B E -> player 1 dies
E E E B B B -> player 2 dies
B E E E B B -> player 1 dies
B B E E E B -> player 1 dies
One therefore has a 2/3 probability of winning (and a 1/3 probability of
dying) by shooting second. I for one would prefer this option.
==> probability/transitivity.p <==
Can you number dice so that die A beats die B beats die C beats die A?
What is the largest probability p with which each event can occur?
==> probability/transitivity.s <==
Yes. The actual values on the dice faces don't matter, only their
ordering. WLOG we may assume that no two faces of the same or
different dice are equal. We can assume "generalised dice", where the
faces need not be equally probable. These can be approximated by dice
with equi-probable faces by having enough faces and marking some of
them the same.
Take the case of three dice, called A, B, and C. Picture the different
values on the faces of the A die. Suppose there are three:
A A A
The values on the B die must lie in between those of the A die:
B A B A B A B
With three different A values, we need only four different B values.
Similarly, the C values must lie in between these:
C B C A C B C A C B C A C B C
Assume we want A to beat B, B to beat C, and C to beat A. Then the above
scheme for the ordering of values can be simplified to:
B C A B C A B C A B C
since for example, the first C in the previous arrangement can be moved
to the second with the effect that the probability that B beats C is
increased, and the probabilities that C beats A or A beats B are
unchanged. Similarly for the other omitted faces.
In general we obtain for n dice A...Z the arrangement
B ... Z A B ... Z ...... A B ... Z
where there are k complete cycles of B..ZA followed by B...Z. k must be
at least 1.
CONJECTURE: The optimum can be obtained for k=1.
So the arrangement of face values is B ... Z A B ... Z. For three dice
it is BCABC. Thus one die has just one face, all the other dice have two
(with in general different probabilities).
CONJECTURE: At the optimum, the probabilities that each die beats the
next can be equal.
Now put probabilities into the BCABC arrangement:
B C A B C
x y 1 x' y'
Clearly x+x' = y+y' = 1.
Prob. that A beats B = x'
B beats C = x + x'y'
C beats A = y
Therefore x' = y = x + x'y'
Solving for these gives x = y' = 1-y, x' = y = (-1 + sqrt(5))/2 = prob.
of each die beating the next = 0.618...
For four dice one obtains the probabilities:
B C D A B C D
x y z 1 x' y' z'
A beats B: x'
B beats C: x + x'y'
C beats D: y + y'z'
D beats A: z
CONJECTURE: for any number of dice, at the optimum, the sequence of
probabilities abc...z1a'b'c...z' is palindromic.
We thus have the equalities:
x+x' = 1
y+y' = 1
z+z' = 1
x' = z = x + x'y' = x + x'y'
y = y' (hence both = 1/2)
Solving this gives x = 1/3, z = 2/3 = prob. of each die beating the next.
Since all the numbers are rational, the limit is attainable with
finitely many equiprobable faces. E.g. A has one face, marked 0. C has
two faces, marked 2 and -2. B has three faces, marked 3, -1, -1. D has
three faces, marked 1, 1, -3. Or all four dice can be given six faces,
marked with numbers in the range 0 to 6.
Finding the solution for 5, 6, or n dice is left as an exercise.
-- ____
Richard Kennaway __\_ / School of Information Systems
Internet: j...@sys.uea.ac.uk \ X/ University of East Anglia
uucp: ...mcsun!ukc!uea-sys!jrk \/ Norwich NR4 7TJ, U.K.
Martin Gardner (of course!) wrote about notransitive dice, see the Oct '74
issue of Scientific American, or his book "Wheels, Life and Other Mathematical
Amusements", ISBN 0-7167-1588-0 or ISBN 0-7167-1589-9 (paperback).
In the book, Gardner cites Bradley Efron of Stanford U. as stating that
the maximum number for three dice is approx .618, requiring dice with more
than six sides. He also mentions that .75 is the limit approached as the
number of dice increases. The book shows three sets of 6-sided dice, where
each set has 2/3 as the advantage probability.
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> series/series.00.p <==
Are "complete this series" problems well defined?
==> series/series.00.s <==
Since there are infinitely many formulas that will fit any finite
series, many people object that such problems have no good answer.
But isn't this a special case of the general observation that theory
is underdetermined by experience (in other words, that there are a
lot of world views that are consistent with all the facts that we
know)? And if so, doesn't this objection really apply to all puzzles?
Isn't it just more obvious in the case of series puzzles?
As a long-time observer of rec.puzzles nit-picking, I have never seen
a puzzle answer that could not be challenged. The list of assumptions
made in solving any puzzle is neverending. Luckily, most of us share
all or nearly all of these assumptions, so that we can agree on an
answer when we see it.
All of this has a lot to do with topics such as computational
complexity, algorithmic compressibility, Church's thesis, intelligence
and life.
==> series/series.01.p <==
M, N, B, D, P ?
==> series/series.01.s <==
T. If you say the sounds these letters make out loud, you will see
that the next letter is T. The letters are in the order of where the
sounds are formed in the mouth, from back to front.
==> series/series.02.p <==
H, H, L, B, B, C, N, O, F ?
==> series/series.02.s <==
Answer 1: N, N, M, A The first letters of the symbols for the elements.
Answer 2: N, S, M, A The first letters of the names of the elements.
==> series/series.03.p <==
W, A, J, M, M, A, J?
==> series/series.03.s <==
J, V, H, T, P, T, F, P, B, L. Presidents of the US.
==> series/series.03a.p <==
G, J, T, J, J, J, A, M, W, J, J, Z, M, F, J, ?
==> series/series.03a.s <==
G, J, T, J, J, J, A, M, W, J, J, Z, M, F, J, A. US Presidents' first names.
==> series/series.03b.p <==
A, J, B, C, G, T, C, V, J, T, D, F, K, B, H, ?
==> series/series.03b.s <==
A, J, B, C, G, T, C, V, J, T, D, F, K, B, H, J. US Vice Presidents.
==> series/series.03c.p <==
M, A, M, D, E, L, R, H, ?
==> series/series.03c.s <==
M, A, M, D, E, L, R, H, A. US Presidents' wives' first names.
==> series/series.04.p <==
A, E, H, I, K, L, ?
==> series/series.04.s <==
M, N, O, P, U, W. Letters in the Hawaiian alphabet.
==> series/series.05.p <==
A B C D E F G H?
==> series/series.05.s <==
M. The names of the cross-streets travelling west on (say)
Commonwealth Avenue from Boston's Public Garden: Arlington, Berkeley,
Clarendon, Dartmouth, Exeter, Fairfield, Gloucester, Hereford,
Massachusetts Ave. The alphabet does continue in the Fenway; after
Massachuetts Ave., there is Charlesgate, and then Ipswich, Jersey,
Kilmarnock.
==> series/series.06.p <==
Z, O, T, T, F, F, S, S, E, N?
==> series/series.06.s <==
T. The name of the integers starting with zero.
==> series/series.06a.p <==
F, S, T, F, F, S, ?
==> series/series.06a.s <==
The words "first", "second", "third", etc. The same as the previous from this
point on.
==> series/series.07.p <==
1, 1 1, 2 1, 1 2 1 1, ...
What is the pattern and asymptotics of this series?
==> series/series.07.s <==
Each line is derived from the last by the transformation (for example)
... z z z x x y y y ... ->
... 3 z 2 x 3 y ...
John Horton Conway analyzed this in "The Weird and Wonderful Chemistry
of Audioactive Decay" (T M Cover & B Gopinath (eds) OPEN PROBLEMS IN
COMMUNICATION AND COMPUTATION, Springer-Verlag (1987)). You can also
find his most complete FRACTRAN paper in this collection.
First, he points out that under this sequence, you frequently get
adjacent subsequences XY which cannot influence each other in any
future derivation of the sequence rule. The smallest such are
called "atoms" or "elements". As Conway claims to have proved,
there are 92 atoms which show up eventually in every sequence, no
matter what the starting value (besides <> and <22>), and always in
the same non-zero limiting proportions.
Conway named them after some other list of 92 atoms. As a puzzle,
see if you can recreate the list from the following, in decreasing
atomic number:
U Pa Th Ac Ra Fr Rn Ho.AT Po Bi Pm.PB Tl Hg Au Pt Ir Os Re Ge.Ca.W Ta
HF.Pa.H.Ca.W Lu Yb Tm ER.Ca.Co HO.Pm Dy Tb Ho.GD EU.Ca.Co Sm PM.Ca.Zn
Nd Pr Ce LA.H.Ca.Co Ba Cs Xe I Ho.TE Eu.Ca.SB Pm.SN In Cd Ag Pd Rh
Ho.RU Eu.Ca.TC Mo Nb Er.ZR Y.H.Ca.Tc SR.U Rb Kr Br Se As GE.Na Ho.GA
Eu.Ca.Ac.H.Ca.ZN Cu Ni Zn.CO Fe Mn CR.Si V Ti Sc Ho.Pa.H.CA.Co K Ar
Cl S P Ho.SI Al Mg Pm.NA Ne F O N C B Be Ge.Ca.LI He Hf.Pa.H.Ca.Li
Uranium is 3, Protactinium is 13, etc. Rn => Ho.AT means the following:
Radon forms a string that consists of two atoms, Holmium on the left,
and Astatine on the right. I capitalize the symbol for At to remind
you that Astatine, and not Holmium, is one less than Radon in atomic
number. As a check, against you or me making a mistake, Hf is 111xx,
Nd is 111xxx, In and Ni are 111xxxxx, K is 111x, and H is 22.
Next see if you can at least prove that any atom other than Hydrogen,
eventually (and always thereafter) forms strings containing all 92 atoms.
The grand Conway theorem here is that every string eventually forms (within
a universal time limit) strings containing all the 92 atoms in certain
specific non-zero limiting proportions, and that digits N greater than 3
are eventually restricted to one of two atomic patterns (ie, abc...N and
def...N for some {1,2,3} sequences abc... and def...), which Conway calls
isotopes of Np and Pu. (For N=2, these are He and Li), and that these
transuranic atoms have a zero limiting proportion.
The longest lived exotic element is Methuselum (2233322211N) which takes
about 25 applications to reduce to the periodic table.
-Matthew P Wiener (wee...@libra.wistar.upenn.edu)
Conway gives many results on the ultimate behavior of strings under
this transformation: for example, taking the sequence derived from 1
(or any other string except 2 2), the limit of the ratio of length of
the (n+1)th term to the length of the nth term as n->infinity is a
fixed constant, namely
1.30357726903429639125709911215255189073070250465940...
This number is from Ilan Vardi, "Computational Recreations in Mathematica",
Addison Wesley 1991, page 13.
Another sequence that is related but not nearly as interesting is:
1, 11, 21, 1112, 3112, 211213, 312213, 212223, 114213, 31121314, 41122314,
31221324, 21322314,
and 21322314 generates itself, so we have a cycle.
==> series/series.08a.p <==
G, L, M, B, C, L, M, C, F, S, ?
==> series/series.08a.s <==
US Army officer ranks, descending.
==> series/series.08b.p <==
A, V, R, R, C, C, L, L, L, E, ?
==> series/series.08b.s <==
US Navy officer ranks, descending.
==> series/series.09a.p <==
S, M, S, S, S, C, P, P, P, ?
==> series/series.09a.s <==
Army non-comm ranks, descending.
==> series/series.09b.p <==
M, S, C, P, P, P, S, S, S, ?
==> series/series.09b.s <==
Navy non-comm ranks, descending.
==> series/series.10.p <==
D, P, N, G, C, M, M, S, ?
==> series/series.10.s <==
N, V, N, N, R. US States in Constitution ratification order.
==> series/series.11.p <==
R O Y G B ?
==> series/series.11.s <==
V or I V. Colors.
==> series/series.12.p <==
A, T, G, C, L, ?
==> series/series.12.s <==
V, L, S, S, C, A, P. Zodiacal signs.
==> series/series.13.p <==
M, V, E, M, J, S, ?
==> series/series.13.s <==
U, N, P or U, P, N. Names of the Planets.
==> series/series.14.p <==
A, B, D, O, P, ?
==> series/series.14.s <==
Q, R. Only letters with an inside as printed.
==> series/series.14a.p <==
A, B, D, E, G, O, P, ?
==> series/series.14a.s <==
Q. Letters with insides in cursive form.
==> series/series.15.p <==
A, E, F, H, I, ?
==> series/series.15.s <==
L, M, N, O, R, S, U, X. Letters whose English names start with vowels.
==> series/series.16.p <==
A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, X, Y?
==> series/series.16.s <==
Z. Letters whose English names have one syllable.
==> series/series.17.p <==
T, P, O, F, O, F, N, T, S, F, T, F, E, N, S, N?
==> series/series.17.s <==
T, T, T, E, F, S. Digits of Pi.
==> series/series.18.p <==
10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, ___ , 100, 121, 10000
==> series/series.18.s <==
10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, 31 , 100, 121, 10000
Sixteen in base n for n=16, 15, ..., 2.
==> series/series.19.p <==
0 01 01011 0101101011011 0101101011011010110101101101011011 etc.
Each string is formed from the previous string by substituting '01' for '0'
and '011' for '1' simultaneously at each occurance.
Notice that each string is an initial substring of the previous string so
that we may consider them all as initial substrings of an infinite string.
The puzzle then is, given n, determine if the nth digit is 0 or 1 without
having to construct all the previous digits. That is, give a non-recursive
formula for the nth digit.
==> series/series.19.s <==
Let G equal the limit string generated by the above process and define
the string F by
F[0] = "0",
F[n] = "1" if n = floor(phi*m) for some positive integer m,
F[n] = "0" if n = floor(phi^2*m) for some positive integer m,
where floor(x) is the greatest integer =< x and phi = (1 + \/5)/2;
I claim that F = G.
I will try to motivate my solution. Let g[0]="0" and define g[n+1]
to be the string that results from replacing "0" in g[n] with "01"
and "1" with "011"; furthermore, let s(n) and t(n) be the number of
"0"'s and "1"'s in g[n], respectively. Note that we have the
following recursive formulas : s(n+1) = s(n) + t(n) and t(n+1) =
s(n) + 2t(n). I claim that s(n) = Fib(2n-1) and t(n) = Fib(2n),
where Fib(m) is the mth Fibonacci number (defined by Fib(-1) = 1,
Fib(0) = 0, Fib(n+1) = Fib(n) + Fib(n-1) for n>=0); this is easily
established by induction. Now noting that Fib(2n)/Fib(2n-1) -> phi
as n -> infinity, we see that if the density of the "0"'s and "1"'s
exists, they must be be 1/phi^2 and 1/phi, respectively. What is
the simplest generating sequence which has this property? Answer:
the one given above.
Proof: We start with
Beatty's Theorem: if a and b are positive irrational numbers such
that 1/a + 1/b = 1, then every positive integer has a representation
of the form floor(am) or floor(bm) (m a positive integer), and this
representation is unique.
This shows that F is well-defined. I now claim that
Lemma: If S(n) and T(n) (yes, two more functions; apparently today's
the day that functions have their picnic) represent the number of
"0"'s and "1"'s in the initial string of F of length n, then S(n)
= ceil(n/phi^2) and T(n) = floor(n/phi) (ceil(x) is the smallest
integer >= x).
Proof of lemma: using the identity phi^2 = phi + 1 we see that S(n)
+ T(n) = n, hence for a given n either S(n) = S(n-1) + 1 or T(n) =
T(n-1) + 1. Now note that if F[n-1]="1" ==> n-1 = floor(phi*m) for
some positive integer m and since phi*m-1 < floor(phi*m) < phi*m ==>
m-1/phi < (n-1)/phi < m ==> T(n) = T(n-1) + 1. To finish, note that
if F[n-1]="0" ==> n-1 = floor(phi^2*m) for some positive integer m
and since phi^2*m-1 < floor(phi^2*m) < phi^2*m ==> m-1/phi^2 <
(n-1)/phi^2 < m ==> S(n) = S(n-1) + 1. Q.E.D.
I will now show that F is invariant under the operation of replacing
"0" with "01" and "1" with "011"; it will then follow that F=G.
Note that this is equivalent to showing that F[2S(n) + 3T(n)]
= "0", F[2S(n) + 3T(n) + 1] = "1", and that if n = [phi*m] for some
positive integer m, then F[2S(n) + 3T(n) + 2] = "1". One could
waste hours trying to prove some fiendish identities; watch how
I sidestep this trap. For the first part, note that by the above
lemma F[2S(n) + 3T(n)] = F[2*ceil(n/phi^2) + 3*floor(n/phi)] =
F[2n + floor(n/phi)] = F[2n + floor(n*phi-n)] = F[floor(phi*n+n)]
= F[floor(phi^2*n)] ==> F[2S(n) + 3T(n)] = "0". For the second, it
is easy to see that since phi^2>2, if F[m]="0" ==> F[m]="1" hence
the first part implies the second part. Finally, note that if n =
[phi*m] for some positive integer m, then F[2S(n) + 3T(n) + 3] =
F[2S(n+1) + 3T(n+1)] = "0", hence by the same reasoning as above
F[2S(n) + 3T(n) + 2] = "1".
Q.E.D.
-- cl...@remus.rutgers.edu (Chris Long)
==> series/series.20.p <==
1 2 5 16 64 312 1812 12288
==> series/series.20.s <==
ANSWER: 95616
The sum of factorial(k)*factorial(n-k) for k=0,...,n.
==> series/series.21.p <==
5, 6, 5, 6, 5, 5, 7, 5, ?
==> series/series.21.s <==
The number of letters in the ordinal numbers.
First 5
Second 6
Third 5
Fourth 6
Fifth 5
Sixth 5
Seventh 7
Eighth 6
Ninth 5
etc.
==> series/series.22.p <==
3 1 1 0 3 7 5 5 2 ?
==> series/series.22.s <==
ANSWER: 4
The digits of pi expressed in base eight.
==> series/series.23.p <==
22 22 30 13 13 16 16 28 28 11 ?
==> series/series.23.s <==
ANSWER: 15
The birthdays of the Presidents of the United States.
==> series/series.24.p <==
What is the next letter in the sequence: W, I, T, N, L, I, T?
==> series/series.24.s <==
S. First letters of words in question.
==> series/series.25.p <==
1 3 4 9 10 12 13 27 28 30 31 36 37 39 40 ?
==> series/series.25.s <==
1 3 4 9 10 12 13 27 28 30 31 36 37 39 40 ...
Positive integers in binary, treated as a base 3 number and converted
to decimal.
==> series/series.26.p <==
1 3 2 6 7 5 4 12 13 15 14 10 11 9 8 24 25 27 26 ?
==> series/series.26.s <==
1 3 2 6 7 5 4 12 13 15 14 10 11 9 8 24 25 27 26 ...
Positive integers in binary, for each 1 bit (not changed) flip the next bit.
This can also be phrased in reversing sequences of numbers.
More simply, just the integers in reflective-Gray-code order.
==> series/series.27.p <==
0 1 1 2 1 2 1 3 2 2 1 3 1 2 2 4 1 3 1 3 2 2 1 4 2 ?
==> series/series.27.s <==
2 3 3 1 3 1 5 2 2 2 4 1 2 2 4 1 3 1 3 3 2 1 5 2 3 2 3 1 4 2 4 2 2 1 ...
Number of factors in prime factorization of positive integers.
==> series/series.28.p <==
0 2 3 4 5 5 7 6 6 7 11 7 13 9 8 8 17 8 19 9 10 13 23 9 10 ?
==> series/series.28.s <==
15 9 11 29 10 31 10 14 19 12 10 37 21 16 11 41 12 43 15 11 25 47 11 14 12 ...
Sum of factors in prime factorization of positive integers.
==> series/series.29.p <==
1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 ?
==> series/series.29.s <==
1 1 2 1 2 2 3 1 2 2 3 2 3 3 4 1 2 2 3 2 3 3 4 2 3 3 4 3 4 ...
The number of 1s in the binary expansion of the positive integers.
==> series/series.30.p <==
I I T Y W I M W Y B M A D
==> series/series.30.s <==
? (first letters of "If I tell you what it means will you buy me a drink?")
==> series/series.31.p <==
6 2 5 5 4 5 6 3 7
==> series/series.31.s <==
6. The number of segments on a standard calculator display it takes
to represent the digits starting with 0.
_ _ _ _ _ _ _ _
| | | _| _| |_| |_ |_ | |_| |_|
|_| | |_ _| | _| |_| | |_| _|
==> series/series.32.p <==
0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1
==> series/series.32.s <==
0 -> 1 01 -> 10 0110 -> 1001 01101001 -> 10010110
Recursively append the inverse.
This sequence is known as the Morse-Thue sequence. It can be defined
non-recursively as the nth term is the mod 2 count of 1s in n written
in binary:
0->0 1->1 10->1 11->0 100->1 101->0 110->0 111->1 etc.
Reference:
Dekking, et. al., "Folds! I,II,III"
The Mathematical Intelligencer, v4,#3,#4,#4.
==> series/series.33.p <==
2 12 360 75600
==> series/series.33.s <==
2 = 2^1
12 = 2^2 * 3^1
360 = 2^3 * 3^2 * 5^1
75600 = 2^4 * 3^3 * 5^2 * 7^1
174636000 = 2^5 * 3^4 * 5^3 * 7^2 * 11^1
==> series/series.34.p <==
3 5 4 4 3 5 5 4 3
==> series/series.34.s <==
The number of letters in the English words for the counting numbers.
==> series/series.35.p <==
1 2 3 2 1 2 3 4 2 1 2 3 4 2 2 3
==> series/series.35.s <==
The number of letters in the Roman numeral representation of the numbers.
==> series/series.36.p <==
ETIANMSURWDKGO
==> series/series.36.s <==
HVF and U with an umlaut.
The letters are sorted by their international Morse code representations:
E .
T -
I ..
A .-
N -.
M --
S ...
U ..-
etc.
==> series/series.37.p <==
10^3 10^9 10^27 10^2 0 4 8 3
==> series/series.37.s <==
10^3 = one thousAnd
10^9 = one Billion
10^27 = one oCtillion
10^2 = one hunDreD
0 = zEro
4 = Four
8 = eiGht
3 = tHree
5 = fIve
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> pickover/pickover.07.p <==
Title: Cliff Puzzle 7: 3x3 Recursion
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
Consider the 3x3 array below. All nine digits are used exactly once.
1 9 2
3 8 4
5 7 6
Notice that "384" is twice the number in the first row, and that
"576" is three times the number in the first row.
Questions:
1. Are there other ways of arranging the number to produce the same
result using each digit only once and the same rules?
Remember, the second row must be twice the first. The third row
must be 3 times the first row.
2. Start with the number in the last row (e.g "576" or any other
solution you may find) and continue to form another 3x3 matrix using the
same rules with the new starting number. In other words, the number in
the second row must be twice the first. The third row must be three
times the first. (For this problem you may truncate any digits in the
beginning. For example, 1384 would become 384.)
Keep going. How many matrices can you create before it is impossible
to continue. Again, each digit must be used only once
in each matrix.
==> pickover/pickover.07.s <==
-------------------------
> Title: Cliff Puzzle 7: 3x3 Recursion
> Consider the 3x3 array below. All nine digits are used exactly once.
> 1 9 2
> 3 8 4
> 5 7 6
> Questions:
> 1. Are there other ways of arranging the numbers to produce the same
> result using each digit only once and the same rules?
YES .
2 1 9 2 7 3 3 2 7
4 3 8 5 4 6 6 5 4
6 5 7 8 1 9 9 8 1
> same rules with the new starting number. In other words,
> the last number becomes the first, and the number in
> the new second row must be twice the first. The third row must be three
> times the first. (For this problem you may truncate any digits in the
> beginning. For example, 1384 would become 384.)
NONE. There is no solution to the continuation problem.
Bye.
Amit Agarwal
> to continue? Again, each digit must be used only once
> in each matrix.
>
>
-------------------------
By exhaustive search I found that there are only four such arrays.
Here they are:
1 9 2 2 1 9 2 7 3 3 2 7
3 8 4 4 3 8 5 4 6 6 5 4
5 7 6 6 5 7 8 1 9 9 8 1
Since these are the only four it is clear from inspection that
none of the last numbers ever begin another array with the desired
properties.
Bob Murphy (rmu...@aludra.usc.edu)
-------------------------
Well, I think I have an answer to both parts. I did what should have been
a complete analysis of all possible column combinations, but it is
certainly possible that I missed a point somewhere. If you don't get any
answers contradicting this one, I'd be happy to send you my analysis.
Anyway - for part 1, I found the following three matrices (in additionn
to the one you gave):
2 1 9 2 7 3 3 2 7
4 3 8 5 4 6 6 5 4
6 5 7 8 1 9 9 8 1
Note that the first one of these can be generated from yours by moving
the third column to the first position, and the third one can be generated
from the second similarly. In both cases, one column does not receive
or produce any carryovers, so it can be placed at either end.
For part 2, there is obviously no solution, assuming that these are indeed
the only four matrices satisfying the requirements. In my analysis, I
included columns with carryovers in all positions, so if there were any
matrices that would satisfy the relaxed condition of part 2 I should
have found them.
Dan Blum
Institute for the
Learning Sciences
Northwestern U.
bl...@ils.nwu.edu
-------------------------
Cliff,
In article <1blrk9...@aludra.usc.edu> (Bob Murphy) writes:
>By exhaustive search I found that there are only 4 starting numbers
>which produce a 3x3 array with the desired property. Here they are:
>
> 1 9 2 2 1 9 2 7 3 3 2 7
> 3 8 4 4 3 8 5 4 6 6 5 4
> 5 7 6 6 5 7 8 1 9 9 8 1
For each of these solutions I happened to notice that the sum of each row
is a constant:
sum(row1) = 12
sum(row2) = 15
sum(row3) = 18
(necessary but not sufficient condition)
And the sums all differ by the same constant (3). I wonder if this
property may somehow be generalized to matrices of higher degree?
Regards,
-- Greg Schmidt schm...@iccgcc.decnet.ab.com
-------------------------
> If you respond to this puzzle, if possible please send me your name, address,
> affiliation, and e-mail address so I can credit you in the future if needed.
> If you like, tell me a little bit about yourself so I can cite you
> appropriately if you provide unique information. PLEASE ALSO directly mail
> me a copy of your response in addition to any responding you do in the
> newsgroup. I will assume it is OK to describe your answer in any article or
> publication I may write in the future, with attribution to you, unless you
> state otherwise.
> Thanks, Cliff Pickover
>
> Consider the 3x3 array below. All nine digits are used exactly once.
>
> 1 9 2
> 3 8 4
> 5 7 6
>
> Notice that "384" is twice the number in the first row, and that
> "576" is three times the number in the first row.
>
> Questions:
> 1. Are there other ways of arranging the numbers to produce the same
> result using each digit only once and the same rules?
> Remember, the second row must be twice the first. The third row
> must be 3 times the first row.
>
> 2. Start with the number in the last row (e.g "576" or any other
> solution you may find) and continue to form another 3x3 matrix using the
> same rules with the new starting number. In other words,
> the last number becomes the first, and the number in
> the new second row must be twice the first. The third row must be three
> times the first. (For this problem you may truncate any digits in the
> beginning. For example, 1384 would become 384.)
>
> Keep going. How many matrices can you create before it is impossible
> to continue? Again, each digit must be used only once
> in each matrix.
Well, this is probably not news to you by now, but I only get four solutions
to the original problem:
1 9 2 2 1 9 2 7 3 3 2 7
3 8 4 4 3 8 5 4 6 6 5 4
5 7 6 6 5 7 8 1 9 9 8 1
If we relax the rules slightly and allow zeroes, and just specify that the nine
numbers only have to be different, then we get two more solutions:
0 7 8 2 6 7
1 5 6 5 3 4
2 3 4 8 0 1
both of which use the digits 0-8, which may be of interest.
The second problem (in either form) has only the above solutions, with only one
matrix in each solution.
If we switch to base 9 (where we must use a zero), there is no solution to the
first, and only one solution to the second (with only one matrix):
4 8 1
0 7 2
5 6 3
In fact, I considered 3 versions of problem 2. In all cases zeroes were
allowed, but the 9 numbers had to be different. For each of them the first 3x3
matrix has to meet the original specifications; where they differ is in how the
succeeding matrices are constructed. In the ensuing discussion, the original
number is called 'n'. So in the example given with the problem, n is 192.
Version A: The second matrix consists of rows with n*2, n*3 and n*4 in them.
(The last three digits of those, anyway.) The next would have n*3,
n*4, and n*5, then n*4, n*5, n*6, etc.
Version B: The second matrix consists of n*3, n*6, n*9. (This is essentially
the second problem as given.)
Version C: The second matrix consists of n*4, n*5, n*6. The next would have
n*7, n*8, n*9 etc.
Results for various bases:
Base 9:
A, B, C: 4 8 1
0 7 2
5 6 3
Base 10:
A, B, C: 0 7 8 1 9 2 2 1 9 2 6 7 2 7 3 3 2 7
1 5 6 3 8 4 4 3 8 5 3 4 5 4 6 6 5 4
2 3 4 5 7 6 6 5 7 8 0 1 8 1 9 9 8 1
In addition, with version C, we get a second matrix for 219.
2 1 9 8 7 6
4 3 8 ==> 0 9 5
6 5 7 3 1 4
Base 11: (A, B, etc. represent 10, 11, etc..)
A, B, C: 18 solutions. From now on, I'll only show the multiple matrix ones.
A: 5 A 1 0 9 2 6 7 4 2 3 8
0 9 2 ==> 6 8 3 2 3 8 ==> 9 0 1
6 8 3 1 7 4 9 0 1 4 7 5
B: 9 3 4 5 A 1
7 6 8 ==> 0 9 2
5 A 1 6 8 3
C: 8 9 1 2 3 4
6 7 2 ==> 0 1 5
4 5 3 8 A 6
(Note that the B solution ends in an A solution matrix!)
Base 12: 19 solutions
A: 7 3 4 2 6 8
2 6 8 ==> 9 A 0
9 A 0 5 1 4
B: None
C: 3 5 7 1 A 4
6 B 2 ==> 5 3 B
A 4 9 8 9 6
Base 13: 71 solutions...and it rapidly increases from here....
The number of solutions rises rapidly, as we might expect, as the more possible
values for digits there are in the base, the more likely the set of 9 will be
distinct. If we look at solutions which only involve the digits 1-9, then the
following is a list of all solutions (for all bases):
Base 10: 1 9 2 2 1 9 2 7 3 3 2 7
3 8 4 4 3 8 5 4 6 6 5 4
5 7 6 6 5 7 8 1 9 9 8 1
Base 11: 7 8 3 8 4 6 8 9 1
4 5 6 5 9 1 6 7 2
1 2 9 3 2 7 4 5 3
(Tested all cases until base 17. After that, no solution can involve a carry.
But there are no solutions without carries. So, no more solutions.)
I hope this is of some interest.
Cheers,
Geoff.
-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ft...@cs.su.oz.au | Gameplayer by vocation.
-------------------------------------------------------------------------------
-------------------------
> Ref: Your note of Mon, 19 Oct 92 22:24:47 EST
>
> Where are you from?
Whoops, knew I forgot to put something in. I'm currently a student at the
University of Sydney (Australia), doing Computer Science (Honours).
Cheers,
Geoff.
-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ft...@cs.su.oz.au | Gameplayer by vocation.
-------------------------------------------------------------------------------
-------------------------
By the way, I tried searching for analogous solutions for other sizes and other
bases. So the new problems become:
Consider an n by n matrix containing the 'digits' from 1 to n^2 in a base b,
where b > n^2. The i'th row of the matrix consists of the last n 'digits' of
i*(first row). The versions differ in how succeeding matrices may be
constructed. Let f be the first row.
Version A: The next matrix has rows with 2*f, 3*f, ... , (n+1)*f
The j'th matrix has rows j*f, (j+1)*f, ... , (n+1-j)*f
Version B: The next matrix has rows with n*f, 2*n*f, ... , n*n*f
The j'th matrix has rows (n^(j-1))*f, 2*(n^(j-1))*f, .... , (n^j)*f
Version C: The next matrix has rows with (n+1)*f, (n+2)*f, ... , 2*n*f
The j'th matrix has rows (1+n*(j-1))*f, (2+n*(j-1))*f, ..., j*n*f
Basically these are analogies of the three versions I wrote to you about
before. The results I get are:
n: 1 base: any above 1
A, B, C: 1
n: 2 base: 5
A, B, C: 3 2 4 1
1 4 3 2
In case B, the second one extends:
4 1 ==> 3 2
3 2 1 4
In case C, the second one also extends:
4 1 ==> 2 3
3 2 1 4
base: 6
A, B, C: 1 4 3 4
3 2 1 2
Note that the only solution to the first problem (no overflow allowed) is
1 4 (in base 6)
3 2
n: 3 base: 10
A, B, C: 1 9 2 2 1 9 2 7 3 3 2 7
3 8 4 4 3 8 5 4 6 6 5 4
5 7 6 6 5 7 8 1 9 9 8 1
base: 11
A, B, C: 7 8 3 8 4 6 8 9 1
4 5 6 5 9 1 6 7 2
1 2 9 3 2 7 4 5 3
Note the base 10 solutions all solve the first problem, while none of the
base 11 solutions do, and there is no second matrix for any of them.
n: 4 base: 18
A, B, C: 1 15 14 4 1 15 16 2 2 1 15 16 2 3 13 16
3 13 10 8 3 13 14 4 4 3 13 14 4 7 9 14
5 11 6 12 5 11 12 6 6 5 11 12 6 11 5 12
7 9 2 16 7 9 10 8 8 7 9 10 8 15 1 10
3 13 14 4 3 13 16 2 4 1 15 14 4 3 13 14
7 9 10 8 7 9 14 4 8 3 13 10 8 7 9 10
11 5 6 12 11 5 12 6 12 5 11 6 12 11 5 6
15 1 2 16 15 1 10 8 16 7 9 2 16 15 1 2
In case C, two of them extend:
1 15 16 2 9 7 8 10 2 1 15 16 10 9 7 8
3 13 14 4 ==> 11 5 6 12 4 3 13 14 ==> 12 11 5 6
5 11 12 6 13 3 4 14 6 5 11 12 14 13 3 4
7 9 10 8 15 1 2 16 8 7 9 10 16 15 1 2
Note all of these solutions solve the first problem (no overflow).
Unfortunately, my algorithm is O((n!)^2), so any results for n = 5 are not
going to be forthcoming soon.
Cheers,
Geoff.
-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ft...@cs.su.oz.au | Gameplayer by vocation.
-------------------------------------------------------------------------------
==> pickover/pickover.08.p <==
Title: Cliff Puzzle 8: Squares and Squares and Squares ....
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
1. What is the smallest square with leading digit 1 which remains a
square when the leading 1 is replaced by a 2?
In other words, if x**2 = 1.........., is there a y**2 = 2......... ?
2. What is the smallest square with leading digit 1 which remains a
square when the leading 1 is replaced by a 2 and also remains a square
when the leading digit is replaced by a 3?
3. What is the smallest square with leading digit 1 which remains a
square when the leading 1 is replaced by a 2, and also remains a square
when the leading digit is replaced by a 3, and also remains a square
when the leading digit is replaced by a 4?
==> pickover/pickover.08.s <==
-------------------------
> 1. What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2?
>
11025 ( 105 * 105 ) ---- 21025 ( 145 * 145 )
>
> 2. What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2 and also remains a square
> when the leading digit is replaced by a 3?
>
No solution till 1,000,000,000.
> 3. What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2, and also remains a square
> when the leading digit is replaced by a 3, and also remains a square
> when the leading digit is replaced by a 4?
>
>
No solution till 1,000,000,000.
The property that you are looking for ( however with different leading
digits ) is owned by the following numbers.
2025 3025
-------------
11025 21025
57600 67600
---------------
202500 302500
342225 442225
------------------
1102500 2102500
3515625 4515625
5760000 6760000
-------------------
11390625 21390625
20250000 30250000
34222500 44222500
----------------------
110250000 210250000
196700625 296700625
351562500 451562500
576000000 676000000
-------------------------
This is probably of no use to you, but, anyway.
-------------------------
In article <1992Oct20.1...@watson.ibm.com> you write:
>Title: Cliff Puzzle 8: Squares and Squares and Squares ....
>1. What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2?
>In other words, if x**2 = 1.........., is there a y**2 = 2......... ?
(Isn't this first part an old puzzle?)
105^2=11025; 145^2=21025. In general we want 10^k=(y-x)(y+x) and
1.5 < (y/x)^2 < 2. Thus y+x and y-x must be factors of 10^k of
the same parity whose ratio is between 5.828... and 9.899...
(these are (t+1)/(t-1) for t^2=2 and 1.5 respectively). The
smallest solution (x,y)=(105,145) corresponds to the factorization
10^4=40*250; gp/pari's "fordiv" function allows one to easily list
all primitive solutions [i.e. not obtained from a smaller solution
by multiplying x,y by the same power of 10] with x^2 and y^2 each
having at most (say) 50 digits:
[x,y]=
[145, 105]
[17225, 14025]
[454625, 326625]
[53948125, 43708125]
[1425503125, 1015903125]
[168971890625, 136203890625]
[529265958203125, 424408358203125]
[1657888279384765625, 1322343959384765625]
[5193483785077392578125, 4119741961077392578125]
In fact it can be seen that the primitive solutions correspond to
integer linear combinations of log(2) and log(5) lying in a certain
fixed interval (determined by the bounds 5.828... and 9.899...),
which probably explains the regular growth of this list.
>2. What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2 and also remains a square
>when the leading digit is replaced by a 3?
There is no such beast, since the three squares would constitute an
arithmetic progression of integer squares of common difference 10^k,
and so give an A.P. of 3 rational squares of common difference 1 or 10 ---
which is known to be impossible by a "2-descent" argument (the case of
common difference 1 is already due to Fermat). [We were lucky here:
in a different number system this argument might fail; for instance the
squares of 7/2, 17/2, 23/2 are an A.P. of common difference 60, the
sexagesimal base. (Some numerology: 7,17,23 are the first three primes
of which 2 is a quadratic residue.) Still, given the base b, the general
theory of elliptic curves indicates that the rational solutions of
Y^2-X^2=Z^2-X^2=b are rather sparsely distributed (the number of d-digit
solutions growing as some power of d), and the extra condition that they
arise by changing only the initial digits of three integer squares is
strong enough to ensure that there are at most finitely many solutions;
with yet more powerful methods one can even provably list them all.]
>3. What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2, and also remains a square
>when the leading digit is replaced by a 3, and also remains a square
>when the leading digit is replaced by a 4?
Of course the above solution to part 2 also disposes of this part;
alternatively I could apeal to another classic result of Fermat:
there is no 4-term A.P. of rational squares.
My question: why all the blank spaces at the end of every line?
--Noam D. Elkies (elk...@zariski.harvard.edu)
Dept. of Math., Harvard Univ., Cambridge, MA 02138
-------------------------
I dunno the direct answer to your squares problem. I do know,
however, that phi (from the Golden Ratio--approx 0.61), which is
defined as the number x such that x + 1 = x^2. It just so happens
that phi+1 and (phi+1)^2 differ by only 1. (1.61 and 2.61) The
rest of the digits are the SAME! Phi = (Sqrt(5)-1)/2.
Phi+1 = (Sqrt(5)+1)/2 phi+2 = (Sqrt(5)+3)/2
(Phi+1)^2= (5+2*Sqrt(5)*1+1)/4 = (2*Sqrt(x)+6)/4 = (Sqrt(x) + 3)/2
Notice how that all works out? Perhaps this property will bring you
closer to an answer. I just sent you all my personal data in
a previous letter concerning your 123 problem. Let me know
what you think of this approach, ok? Thanks in advance!
--Joseph Zbiciak im1...@camelot.bradley.edu
-------------------------
In article <1992Oct20.1...@watson.ibm.com> you write:
: 2. What is the smallest square with leading digit 1 which remains a
: square when the leading 1 is replaced by a 2 and also remains a square
: when the leading digit is replaced by a 3?
This is not possible. One of these numbers would leave a remainder
of 2 when divided by 3, and no square is congruent to 2 modulo 3.
--
David Radcliffe
radc...@csd4.csd.uwm.edu
-------------------------
In article <1992Oct20.1...@watson.ibm.com> you write:
: 1. What is the smallest square with leading digit 1 which remains a
: square when the leading 1 is replaced by a 2?
11025. I found, by hand, all integral solutions to
(x+y)(x-y) = 10000. The solution (145,105) is the only
one with 10000 < y^2 < 20000.
You have permission to use my solution, but not my name.
--
David Radcliffe
radc...@csd4.csd.uwm.edu
-------------------------
Well, as a previous poster already mentioned on Rec.puzzles, there are only 4
solutions to the initial problem. They are 192, 219, 293, and 327. None of
these solutions can be connected to others as in part 2 of your problem.
I first extended the problem to allow any multipliers. So the second row must
be some multiple of the first and the third some other multiple of the first.
I found 19 solutions to this problem. However, there is still no way to chain
a second solution to the first.
Then I allowed 0s. Now there are 134 solutions. There are also 17 2-chains.
There are two 3-chains which I will list here:
192 394
*2= 384 *3=1182
*3= 576 *4=1576
*7=4032 now the same as the other solution.
*9=5184
*4= 736
*5= 920
I will be more than happy to send you all 134 solutions if you really want
them! I also have Pascal source code.
Comments on some of your other problems will follow.
Dan Cory
Senior, Stanford
perm. address:
55 Cedar St.
Chapel Hill, NC 27514
school address:
PO Box 13113
Stanford, CA 94309
Should you use any of my results, please send a copy of the work to the
permanent address above.
-------------------------
In article <1992Oct20.1...@watson.ibm.com>, you write:
|> Title: Cliff Puzzle 8: Squares and Squares and Squares ....
|> 1. What is the smallest square with leading digit 1 which remains a
|> square when the leading 1 is replaced by a 2?
11025 = 105^2, 21025 = 145^2.
|> 2. What is the smallest square with leading digit 1 which remains a
|> square when the leading 1 is replaced by a 2 and also remains a square
|> when the leading digit is replaced by a 3?
|>
|> 3. What is the smallest square with leading digit 1 which remains a
|> square when the leading 1 is replaced by a 2, and also remains a square
|> when the leading digit is replaced by a 3, and also remains a square
|> when the leading digit is replaced by a 4?
These two cases never occur.
Proof: (This was a LOT harder than I thought it would be when I started!)
The original problem can be reduced to:
"Find positive integers x,y,n such that
y^2-x^2 = 10^n and 10^n < x^2 < 2*10^n." [1]
The second problem amounts to finding x,y,z,n which meet the above
conditions, plus z^2-y^2=10^n.
For the second problem, look at the set of solutions to
z^2-y^2 = 10^n, 2*10^n < y^2 < 3*10^n. [2]
A solution to the second problem consists of x,y,z,n, where x,y,n solve
the original problem and y,z,n solve the above system.
The first equation in [1] can be factored into (y-x)(y+x) = 10^n = 2^n * 5^n.
Similarly (z-y)(z+y) = 10^n. Since x,y,z are integers, we must have
y+x = 2^a * 5^b, y-x = 2^(n-a) * 5^(n-b)
z+y = 2^c * 5^d, z-y = 2^(n-c) * 5^(n-d)
where a,b,c,d are integers. When a=c and b=d, y+x = z+y and y-x = z-y,
which leads to a contradiction.
Then 2y = 2^a * 5^b + 2^(n-a) * 5^(n-b) = 2^c * 5^d + 2^(n-c) * 5^(n-d)
However, in the last equality above, divide both sides by 2^f, where f is
the smallest of a, c, n-a, and n-c. The result is:
2^(a-f) * 5^b + 2^(n-a-f) * 5^(n-b) = 2^(c-f) * 5^d + 2^(n-c-f) * 5^(n-d) [3]
Now, at least one of the four products above is a product of only 5's, and
is odd. Only one is odd unless a=c, 2a=n, or 2c=n.
If a=c, then either b=d (contradiction) or z+y is at least
a factor of 5 larger than y+x. However, considering
sqrt(3)*sqrt(10^n) < z < 2*sqrt(10^n)
sqrt(2)*sqrt(10^n) < y < sqrt(3)*sqrt(10^n)
sqrt(10^n) < x < sqrt(2)*sqrt(10^n)
we have:
(sqrt(3)+sqrt(2))*sqrt(10^n) < z+y < (2+sqrt(3))*sqrt(10^n)
(1+sqrt(2))*sqrt(10^n) < y+x < (sqrt(3)+sqrt(2))*sqrt(10^n)
and then (z+y)/(y+x) < (2+sqrt(3))/(1+sqrt(2)) < 5.
If a exactly equals n/2:
In the case that b=a=n/2, y+x = y-x, so x=0 (not possible).
If b<n/2, y-x>y+x, but we want x to be positive, so b>n/2. Since b and
n/2 are integers (remember n/2=a), b-(n-b) >= 2, and (y+x)/(y-x) >= 25.
This gives (y+x) >= 25(y-x),
(y+x+y-x) = 2y >= 26(y-x),
y >= 13y-13x,
13x >= 12y,
x/y >= 12/13
x^2/y^2 >= 144/169
However, we know 10^n < x^2 < 2*10^n, and y^2 = x^2 + 10^n, so x^2/y^2
varies between 1/2 and 2/3, and cannot be greater than 144/169.
Similarly, when c=n/2, the same argument applies, and in the final step
we know y^2/z^2 varies between 2/3 and 3/4.
Finally, we've eliminated all cases where more than one of the terms in [3]
is odd. With exactly one term odd, we have odd=even, a contradiction,
so there is no solution.
--
----w-w--------------Joseph De Vincent...@owlnet.rice.edu----------------
( ^ ) Disclaimer: My opinions do not represent those of Owlnet.
(O O) Owlnet: George R. Brown School of Engineering Educational Network.
v-v (Unauthorized use is prohibited.) (Being uwop-ap!sdn is allowed.)
-------------------------
G'day Cliff!
> * * *
>
> 1. What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2?
>
> In other words, if x**2 = 1.........., is there a y**2 = 2......... ?
The smallest I could find was 105**2 = 11025
145**2 = 21025
Indeed, an exhaustive search shows that this is the smallest.
The other pairs I found (after a few minutes playing with pen and paper - I
could probably write a program to generate them ad nauseum, but I've got a
draft thesis to write...) were:
3375**2 = 11390625, 4625**2 = 21390625
14025**2 = 196700625, 17225**2 = 296700625
326625**2 = 106683890625, 454625**2 = 206683890625
I don't know what pattern there is in them. Of course, if x is a solution,
then so is 10*x. So these give solutions for 1050*1050 = 1102500, etc.
> 2. What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2 and also remains a square
> when the leading digit is replaced by a 3?
>
> 3. What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2, and also remains a square
> when the leading digit is replaced by a 3, and also remains a square
> when the leading digit is replaced by a 4?
I'll answer part 3 first. If such a square exists, then observe that we have
4 squares in arithmetic progression (common difference a power of 10). There
is a well known theorem that there is no set of four squares in arithmetic
progression, so there is no solution to part 3.
Now, for part 2. We have 3 squares in arithmetic progression. Another well
known (and not too hard to derive) theorem states that for three squares in
arithmetic progression, their common difference is of the form:
D = 4 * K^2 * m * n * (m^2 - n^2) = 4 * K^2 * m * n * (m + n) * (m - n)
Now, this value is a power of 10. So the only primes in its factorisation are
2 and 5. Hence neither m nor n is divisible by 3. So (m^2 - n^2) is
divisible by 3. Hence a power of 10 is divisible by 3. Contradiction. So
now such set of three squares exist (which also proves part 3).
Cheers,
Geoff.
PS: I assume you still have whatever details of mine you care about.
-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm) | Programmer by trade --
ft...@cs.su.oz.au | Gameplayer by vocation.
-------------------------------------------------------------------------------
-------------------------
Here is the solution I just posted to rec.puzzles. Note that I changed my mind
on this puzzle!
Dan Cory
Senior, Stanford
PO Box 13113
Stanford, CA 94309
yp...@leland.stanford.edu
Newsgroups: rec.puzzles
Subject: Re: Cliff Puzzle 8: Squares and Squares ... (SPOILER)
Approved: news-answe...@MIT.Edu
Summary: solutions to part 1, no solutions to parts 2 or 3
Expires:
References: <1992Oct20.1...@watson.ibm.com>
Sender:
Followup-To:
Distribution:
Organization: DSG, Stanford University, CA 94305, USA
Keywords: squares, cliff, 8, gcd
In article <1992Oct20.1...@watson.ibm.com> cl...@watson.ibm.com (cliff)
>1. What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2?
>In other words, if x**2 = 1.........., is there a y**2 = 2......... ?
We write this condition as the following equations with x,y,a integers:
y^2-x^2=10^a
1*10^a<=x^2<=2*10^a
2*10^a<=y^2<=3*10^a
We factor the first equation:
(y-x)(y+x)=10^a.
Let u=x+y. Then 10^a/u=x-y. Since x+y>x-y, u>10^a/u so u>10^(a/2)
Then x=(u-10^a/u)/2 and y=(u+10^a/u)/2.
Subsitute these equations into the inequalities above.
For x we get:
10^a<=((u-10^a/u)/2)^2<=2*10^a
Take the square root of both (all three?) sides:
10^(a/2)<=(u-10^a/u)/2<=sqrt(2)*10^(a/2)
Multiply through by 2 and divide through by 10^(a/2).
2<=u/10^(a/2)-10^(a/2)/u<=2*sqrt(2)
Let v=u/10^(a/2). So v>1. Then:
2<=v-1/v<=2*sqrt(2).
We solve these two inequalities. First the left:
v-1/v>=2
v^2-2v-1>=0
v>=(1+sqrt(2)) or v<=(1-sqrt(2)).
v-1/v<=2*sqrt(2)
v>=(sqrt(2)+sqrt(3)) or v<=(sqrt(2)-sqrt(3)).
Since v>1, we drop the negative solutions and find:
1+sqrt(2) <= v <= sqrt(2)+sqrt(3).
or
1+sqrt(2) <= u/10^(a/2) <= sqrt(2)+sqrt(3).
We can do the same for y but we will find the same restriction on u.
Now we remember that u|10^a (u divides 10^a). Therefore u must be a power of
2 times a power of 5. Let u=5^b*2^c with b,c integers less than or equal to a.
Since we are going to divide it by 2, we must have c>=1.
Then we need to find a,b,c such that:
1+sqrt(2) <= 5^b*2^c/10^(a/2) <= sqrt(2)+sqrt(3)
These will give us u which will in turn determine x and y.
So take the log base 10 of all three sides. Since log is increasing, we do not
change the direction of inequality. Thus:
log(1+sqrt(2)) <= b*log(5)+c*log(2)-a/2 <= log(sqrt(2)+sqrt(3))
Multiply through by 2:
2*log(1+sqrt(2)) <= 2*b*log(5)+2*c*log(2)-a <= 2*log(sqrt(2)+sqrt(3))
If we approximate log(5) and log(2), this is sort of a Diophantine equation.
Since log(5) is very very close to 0.7 and log(2) is very very close to 0.3,
our approximations will be okay to find low solutions. If we want big solutions
then we need to use better convergents. We can calculate the boundary logs
as accurately as necessary. So:
0.77 <= 7/5*b+3/5*c-a <= 0.99
Multiply through by 5:
3.8 <= 7*b+3*c-5*a <= 4.9
So we must find a,b,c such that 7*b+3*c-5*a = 4, with a>b>=0 and a>=c>0.
There are many good ways to solve this but we will just pick a small solution.
b=3, c=1, a=4 (7*3+3-5*4=21+3-20=4)
Then u=5^3*2^1=250.
So y+x=250 and y-x=10^a/u=10^4/250=40.
Then y=145 and x=105.
y^2=21025 and x^2=11025.
This is, in fact, the smallest solution (it is easy to show that there is no
solution to the 7*b+3*c-5*a with a<4 and a>b>=0,a>=c>0).
>2. What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2 and also remains a square
>when the leading digit is replaced by a 3?
We note from above that y=(5^b*2^c+10^a/(5^b*2^c)/2 or
2y=5^b*2^c+5^(a-b)*2^(a-c).
Should we now repeat the problem for a square with leading digit 2 that is
replaced by a 3, everything is the same except that y is now the smaller of the
pair. Thus:
2y=5^B*2^C-5^(a-B)*2^(a-C)
where B and C are different from b and c above but a is necessarily the same
(since we want the difference to be the same power of 10 for each transition).
Combining the two we get:
5^b*5^c+5^(a-b)*2^(a-b)=5^B*2^C-5^(a-B)*2^(a-C).
The proof that this has no solutions is too small to fit in the margin of this
posting.
>3. What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2, and also remains a square
>when the leading digit is replaced by a 3, and also remains a square
>when the leading digit is replaced by a 4?
There is no solution since there is no solution to part 2.
Dan Cory
==> pickover/pickover.09.p <==
Title: Cliff Puzzle 9: 3-Atoms and Growth
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
Start with 3 digits: 1, 2, and 3.
Each succeding row repeats the previous three rows, in order,
as you can see from the following diagram.
1
2
3
123
23123
312323123
12323123312323123
2312331232312312323123312323123
1. What is the sum of digits in the 100th row?
2. Get rid of all the twos. Here I've replaced each of them with a "."
1
.
3
1.3
.31.3
31.3.31.3
1.3.31.331.3.31.3
.31.331.3.31.31.3.31.331.3.31.3
In the last row of this diagram, there are three different species: 31,
331 and 3. How many different species are there in row 30?
3. When the sequence first hits a three, it now undergoes an enzymatic
cleavage, and the digits on the right of the 3 are swapped with the
digits on the left.
1
2
3
123
23123 now becomes 12323
312312323 now becomes 123123233
Now answer the question posed in question 2.
==> pickover/pickover.09.s <==
-------------------------
Subject: Re: Cliff Puzzle 9: 3-Atoms and Growth (PARTIAL SPOILER)
Newsgroups: rec.puzzles
References: <1992Oct20.1...@watson.ibm.com>
In article <1992Oct20.1...@watson.ibm.com>, Cliff Pickover writes:
> Start with 3 digits: 1, 2, and 3.
> Each succeding row repeats the previous three rows, in order
> as you can see from the following diagram.
> 1
> 2
> 3
> 123
> 1. What is the sum of digits in the 100th row?
This one's easy. You basically have a Tribonacci sequence with
the initial conditions S_1 = 1, S_2 = 2, S_3 = 3 and S_n = S_{n-1} +
S_{n-2} + S_{n-3} for n>3. Thus, it's possible to find a closed
form of the type c_1*r_1^n + c_2*r_2^n + c_3*r_3^n. Indeed, letting
T_i be the standard Tribonnaci sequence which has initial values
T_1 = 1, T_2 = 1, T_3 = 1 we can play a little game by noting the
T's go 1 1 1 3 5, and so by linearity S_i = ( T_i + T_{i+2} )/2, hence
S_100 = ( T_100 + T_102 )/2.
-------------------------
Dear Mr. Pickover,
I found your "123" problem interesting. Here's the answers that I
came up with. (Note: my personal info that you requested that I
include is at the end of the document.)
> * * *
>Start with 3 digits: 1, 2, and 3.
>Each succeding row repeats the previous three rows, in order,
>as you can see from the following diagram.
>1
>2
>3
>123
>23123
>312323123
>12323123312323123
>2312331232312312323123312323123
>1. What is the sum of digits in the 100th row?
Define an arithmetic series as follows:
(Note: Read a_1 as "a sub 1" and a_(n-1) as "a sub n-1". I have
to do this because I can't use subscripts here.)
a_1 = 1
a_2 = 2
a_3 = 3
a_n = a_(n-3) + a_(n-2) + a_(n-1); n>=4
The sum of each line is the sum of it's parts, so therefore, the
sum of each row is the sum of the previous three rows' sums.
a_30 = 45152016 (I wrote a simple basic program to calculate it.)
>2. Get rid of all the twos. Here I've replaced each of them with a "."
>.31.3
>31.3.31.3
>1.3.31.331.3.31.3
>.31.331.3.31.31.3.31.331.3.31.3
>In the last row of this diagram, there are three different species: 31,
>331 and 3. How many different species are there in row 30?
First, let me show that no "new" species will develop, other than those
seen in the sample few lines above:
First, notice that there are four unique species above:
"1","3","31","331". Next, notice that the first species
on a line goes in cycles of 3. (Remember how we're building
successive rows. The first row repeated on a line is the
row three back. Hence the repeating pattern.) Also notice
that the ends of the rows do not change, this time because
the last row represented on the current row is the row
directly previous (and hence, it ends the same.)
Because we are building successive rows via concatination,
then only locations within new rows where "new" species may
be found ("new" meaning not seen in any previous rows) is
where the ends of two rows meet in the new row. Since we
know that the "end" of each row is limited to ".3" and
the "beginnings" of each row cycle through "31", "1", ".",
the only possible combinations we can make are "331", "31",
and "3". Since we alreadly have seen these, it is now
obvious that we will create no more new species.
Next, let me show what species we WILL see:
The species "3" is on the end of every line. Therefore
it will be in row 30.
The species "31" and the species "331" are both imbedded
in a row previous to row 30. Therefore they will be in
row 30, because the "middle parts" of each row are
duplicated down the list, not modified.
The species "1" only shows up every third row. It happens
to occur on rows such that (Row #) mod 3 = 1. Because
30 mod 3=0, the species "1" will NOT occur in row 30.
Hence, we have the three species "3","31","331" occuring
in row 30.
>3. When the sequence first hits a three, it now undergoes an enzymatic
>cleavage, and the digits on the right of the 3 are swapped with the
>digits on the left.
>1
>2
>3
>123
>23123 now becomes 12323
>312312323 now becomes 123123233
>Now answer the question posed in question 2.
I'm not taking the time to work this one out entirely. It appears that
this algorithm forces 1's out in front all of the time, and keeps
appending 3's on the end of the row. Hence, you'll see a proliferation
of species such as "3331","33331","333331", etc. It also appears that
in row 30, you will have all the species from "3" , "31", "331","3331",
"33331", etc up to "33333333333333333333333331". Now, I haven't
doublechecked my work here... I've been up all night, and am too
tired to double check my conjecture here. But, I believe that I am
right, or at least on the right track.
I hope these answers help you. I have two questions in return:
"Are you the 'pickover' responsible for many of the Fractint
fractal types?" and "Were my answers above even close?" I apologize
if my answers seemed a little rough & non-formal at points. I
hope you understand my explanation above.
Thanks for the mental workout. I hope that this helps you, once again.
Hope to hear from you soon!
-- Joseph Zbiciak im1...@camelot.bradley.edu
Here's that personal data to requested that I include:
I am Joseph Zbiciak, an Electrical/Computer Engineering Major at
Bradley University, Peoria, IL. My current address is as follows:
Room 121, Heitz Hall
912 N Elmwood,
Peoria, IL 61606
My e-mail address is im1...@camelot.bradley.edu.
Other info: Year in school: Freshman, DOB: 08/29/75
Academic standing: good Favorite toy: his computer
Favorite hobby: spelunking through the internet looking
for tidbits like this question here.
If you need any more information, let me know.
Note: I did not post this on the nn yet. Feel free to for me, however.
Thanks!
--
-------------------------
|> 3.When the sequence first hits a three, it now undergoes an enzymatic
|> cleavage, and the digits on the right of the 3 are swapped with the
|> digits on the left.
|>
|> 1
|> 2
|> 3
|> 123
|> 23123 now becomes 12323
|> 312312323 now becomes 123123233
>From how I understand the descriptive rule I get:
1
2
3
123 becomes 312
23123 becomes 12332
331223123 becomes 312231233
>From your example it seems that the trailing 3 is not regarded as a
'first' 3 (123 is not changed), nor is it regarded as a digit to be
swapped (as in the two other examples).
Is this how the rule should be interpreted?
And ... Keep up the good work, these are really good puzzles!!
--
stein....@nta.no (Norwegian Telecom Research)
'When murders are committed by mathematics, they can be solved by
mathematics. Most of them aren't, and this one wasn't'
- Nick Charles (Dashiell Hammett's "The Thin Man")
-------------------------
Dear Dr. Pickover,
I found your "123" problem interesting. Here's the answers that I
came up with. (Note: my personal info that you requested that I
include is at the end of the document.)
> * * *
>Start with 3 digits: 1, 2, and 3.
>Each succeding row repeats the previous three rows, in order,
>as you can see from the following diagram.
>1
>2
>3
>123
>23123
>312323123
>12323123312323123
>2312331232312312323123312323123
>1. What is the sum of digits in the 100th row?
Define an arithmetic series as follows:
(Note: Read a_1 as "a sub 1" and a_(n-1) as "a sub n-1". I have
to do this because I can't use subscripts here.)
a_1 = 1
a_2 = 2
a_3 = 3
a_n = a_(n-3) + a_(n-2) + a_(n-1); n>=4
The sum of each line is the sum of it's parts, so therefore, the
sum of each row is the sum of the previous three rows' sums.
a_30 = 45152016 (I wrote a simple basic program to calculate it.)
>2. Get rid of all the twos. Here I've replaced each of them with a "."
>.31.3
>31.3.31.3
>1.3.31.331.3.31.3
>.31.331.3.31.31.3.31.331.3.31.3
>In the last row of this diagram, there are three different species: 31,
>331 and 3. How many different species are there in row 30?
First, let me show that no "new" species will develop, other than those
seen in the sample few lines above:
First, notice that there are four unique species above:
"1","3","31","331". Next, notice that the first species
on a line goes in cycles of 3. (Remember how we're building
successive rows. The first row repeated on a line is the
row three back. Hence the repeating pattern.) Also notice
that the ends of the rows do not change, this time because
the last row represented on the current row is the row
directly previous (and hence, it ends the same.)
Because we are building successive rows via concatination,
then only locations within new rows where "new" species may
be found ("new" meaning not seen in any previous rows) is
where the ends of two rows meet in the new row. Since we
know that the "end" of each row is limited to ".3" and
the "beginnings" of each row cycle through "31", "1", ".",
the only possible combinations we can make are "331", "31",
and "3". Since we alreadly have seen these, it is now
obvious that we will create no more new species.
Next, let me show what species we WILL see:
The species "3" is on the end of every line. Therefore
it will be in row 30.
The species "31" and the species "331" are both imbedded
in a row previous to row 30. Therefore they will be in
row 30, because the "middle parts" of each row are
duplicated down the list, not modified.
The species "1" only shows up every third row. It happens
to occur on rows such that (Row #) mod 3 = 1. Because
30 mod 3=0, the species "1" will NOT occur in row 30.
Hence, we have the three species "3","31","331" occuring
in row 30.
>3. When the sequence first hits a three, it now undergoes an enzymatic
>cleavage, and the digits on the right of the 3 are swapped with the
>digits on the left.
>1
>2
>3
>123
>23123 now becomes 12323
>312312323 now becomes 123123233
>Now answer the question posed in question 2.
I'm not taking the time to work this one out entirely. It appears that
this algorithm forces 1's out in front all of the time, and keeps
appending 3's on the end of the row. Hence, you'll see a proliferation
of species such as "3331","33331","333331", etc. It also appears that
in row 30, you will have all the species from "3" , "31", "331","3331",
"33331", etc up to "33333333333333333333333331". Now, I haven't
doublechecked my work here... I've been up all night, and am too
tired to double check my conjecture here. But, I believe that I am
right, or at least on the right track.
Thanks for the mental workout. I anxiously await more such puzzles!
Hope to hear from you soon!
-- Joseph Zbiciak im1...@camelot.bradley.edu
Here's that personal data to requested that I include:
I am Joseph Zbiciak, an Electrical/Computer Engineering Major at
Bradley University, Peoria, IL. My current address is as follows:
Room 121, Heitz Hall
B
912 N Elmwood,
Peoria, IL 61606
My e-mail address is im1...@camelot.bradley.edu.
Other info: Year in school: Freshman, DOB: 08/29/75
Academic standing: good Favorite toy: his computer
Favorite hobby: spelunking through the internet looking
for tidbits like this question here.
==> pickover/pickover.10.p <==
Title: Cliff Puzzle 10: The Ark Series
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
1. Given a large ark containing 2 individuals of every animal species
in the world, what would be the approximate total weight of all the
organisms? How would your answer differ if you included every plant,
bacterial, and fungal organism?
2. Assume that all other organisms on earth were dead except for those
on the ark in question 1, and that the animals were released 1000 years
ago. What would you expect to be surviving today? (Assume that, where
applicable, a male and female were used for each species.)
3. Assume that the year is 1992 and that it rained for 40 days, and the
rain covered all the land on the earth. Further assume that the flood
waters receded to pre-flood days within several months.
What would be the geopolitical changes as a result of the
temporary flood?
What would be the ecological changes as a result of the
temporary flood?
==> pickover/pickover.10.s <==
-------------------------
In article <1992Oct20.1...@watson.ibm.com> you write:
|> Title: Cliff Puzzle 10: The Ark Series
|> From: cl...@watson.ibm.com
|>
[ lotsa lines deleted ]
|>
|> 2. Assume that all other organisms on earth were dead except for those
|> on the ark in question 1, and that the animals were released 1000 years
|> ago. What would you expect to be surviving today? (Assume that, where
^^^^^^^
|> applicable, a male and female were used for each species.)
Were you thinking of parthenogenesis or something ???
|>
|> 3. Assume that the year is 1992 and that it rained for 40 days, and the
|> rain covered all the land on the earth. Further assume that the flood
|> waters receded to pre-flood days within several months.
|>
|> What would be the geopolitical changes as a result of the
|> temporary flood?
Dunno about this but it's a safe bet that the Netherlands _wouldn't_ get flooded
We've been blocking the sea out for hundreds of years, so we've more experience
at it than anyone else.
|>
|> What would be the ecological changes as a result of the
|> temporary flood?
Andy.
Just my opinions, nobody else's, especially not Oracle's
-------------------------
> 1. Given a large ark containing 2 individuals of every animal species
> in the world, what would be the approximate total weight of all the
> organisms? How would your answer differ if you included every plant,
> bacterial, and fungal organism?
1000 tons (guessed 10 million species with an average weight of 100 grams,
insects push this number down with their huge number of species).
No increase through bacteriae or fungi, but maybe with plants.
(You were unspecific: All living species?)
> 2. Assume that all other organisms on earth were dead except for those
> on the ark in question 1, and that the animals were released 1000 years
> ago. What would you expect to be surviving today? (Assume that, where
> applicable, a male and female were used for each species.)
None. I think it's common knowledge with biologists that you need at least
~50 individuals of a species to keep genetic health --- aside from the
problem of both a male and female baby surviving.
> 3. Assume that the year is 1992 and that it rained for 40 days, and the
> rain covered all the land on the earth. Further assume that the flood
> waters receded to pre-flood days within several months.
"Covers the land." How deep? To cover *all* land (Himalaya) evenly, you
need a depth of 9000 m in most regions, so the question is, how fast will
it rise? Do we just have time to put some tins in the boat? Most people
don't have one. Most airplanes cannot land but maybe some of them swim.
One has to calculate the distribution of swimming things in usual locations.
For if people have to swim 500-1000 m in cold water to a beam, most will
drown.
> What would be the geopolitical changes as a result of the
> temporary flood?
With the survival of at most 1 percent of the population there will be a
completely new beginning. Don't know if they would make the same mistakes,
though. Technology will be thrown back, and science more than that.
Niven/Pournelle's "Lucifer's Hammer" is an accurate description.
> What would be the ecological changes as a result of the
> temporary flood?
Lack of most animals, especially those dependent of plants (many of them
can't live without a day of food). Most plants will grow again after some
time.
--ralf
************************************************************************
After some tests, I decided to put 4 lines of sig here, because I really
like the optical effect. Now there's the problem what to write in it...
************************************************************************
==> pickover/pickover.11.p <==
Title: Cliff Puzzle 11: The Leviathan Number
From: cl...@watson.ibm.com
If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information. PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup. I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise. Thanks, Cliff Pickover
* * *
Many interesting observations have recently been published
concerning various number theory properties of the "number of the
beast", 666. In this new
puzzle here I ask you to consider the monstrous
"leviathan number",
a number so large as to make the number of electrons,
protons, and neutrons in the universe (10**79) pale in comparison. (It
also makes a googol (10**100) look kind of small).
The leviathan number is defined as (10**666)!, where the "!" indicates
factorial.
1. What are the first 6 digits of the leviathan number? Hint: you
need not actually compute the leviathan to determine this. If you can
determine the first 6 digits, please carefully spell out your method.
2. Could modern supercomputers compute the leviathan, or will this
beyond the realm of humankind for the next century?
3. Even if we cannot compute the leviathan, how many other
characteristics of this number can we write down.
==> pickover/pickover.11.s <==
-------------------------
Subject: Re: Cliff Puzzle 11: The Leviathan Number (PARTIAL SPOILER)
Newsgroups: rec.puzzles
References: <1992Oct21.1...@watson.ibm.com>
In article <1992Oct21.1...@watson.ibm.com>, Cliff Pickover writes:
> The leviathan number is defined as (10**666)!, where the "!" indicates
> factorial.
> 1. What are the first 6 digits of the leviathan number?
The simplest technique would be to use Stirling's formula to compute
the mantissa, i.e. frac( log(n) ) = frac( log(2*pi)/2 + log(n)/2
n*(log(n)-log(e)) ). In our case n = 10^666, so this equals
frac( log(2*pi)/2 + 333 + 10^666*(666-log(e)) ) =
frac( log(2*pi)/2 + 10^666*(1-log(e)) ), so we'd basically need
to know something like 10 digits to the right of the decimal point
for log(2*pi)/2, and something like 700 digits for log(e) (which is
easily doable). We then compute (1-log(e)), shift the digits 666
spaces to the left, and we're all set.
> 2. Could modern supercomputers compute the leviathan, or will this
> beyond the realm of humankind for the next century?
The number of digits is more than 10^668, and this compares
unfavorably to the number of particles in the universe. Furthermore,
even if a googol digits could be output per second, you'd never
make it before the end of the universe. So, I'd say it's beyond
the realm of humanity, period.
> 3. Even if we cannot compute the leviathan, how many other
> characteristics of this number can we write down.
As another puzzle, how many zeroes does it end with, and what are
the last two non-zero digits?
.qq
&EXIT
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Date: Thu, 22 Oct 1992 07:12 EDT
From: <FRAMEM@UNION>
Subject: RE: googol!
To: CLIFF@YKTVMV
Original_To: Jnet%"CLIFF@YKTVMV"
Hi, Cliff.
The log10(e) comes from applying Stirling's approximation
for the factorial: for large n, n! is approximately
sqrt(2*pi*n)*((n/e)^n). Substitute googol for n, take
log10 of both sides, and recall the mantissa of the log10
gives the digits of the original number.
In these days of fast symbolic packages allowing exact
computation of large factorials (though presumably not
so large as a googol), people forget Stirling's formula.
Until a few years ago, this was the only way to find
factorials (albeit, only approximately) for large numbers.
Mike
Chris Cole
Last-modified: 17 Aug 1993
Version: 4
==> logic/situation.puzzles.s <==
Jed's List of Situation Puzzles
(with answers)
"A man lies dead in a room with fifty-three bicycles in front of him.
What happened?"
This is a list of what I refer to (for lack of a better name) as situation
puzzles. In the game of situation puzzles, a situation like the one above is
presented to a group of players, who must then try to find out more about the
situation by asking further questions. The person who initially presented
the situation can only answer "yes" or "no" to questions (or occasionally
"irrelevant" or "doesn't matter").
My list has been divided into two sections. Section 1 consists of
situation puzzles which are set in a realistic world; the situations could
all actually occur. Section 2 consists of puzzles which involve double
meanings for one or more words and those which could not possibly take place
in reality as we know it, plus a few miscellaneous others.
See the end of the list for more notes and comments.
This version of the list contains answers to the puzzles, as well as
variants.
Section 1: "Realistic" situation puzzles.
1.1. In the middle of the ocean is a yacht. Several corpses are floating in
the water nearby. (SJ)
1.1. A bunch of people are on an ocean voyage in a yacht. One afternoon,
they all decide to go swimming, so they put on swimsuits and dive off the
side into the water. Unfortunately, they forget to set up a ladder on the
side of the boat, so there's no way for them to climb back in, and they
drown.
1.1a. Variant answer: The same situation, except that they set out a ladder
which is just barely long enough. When they all dive into the water, the
boat, without their weight, rises in the water until the ladder is just
barely out of reach. (also from Steve Jacquot)
1.2. A man is lying dead in a room. There is a large pile of gold and
jewels on the floor, a chandelier attached to the ceiling, and a large open
window. (DVS; partial JM wording)
1.2. The room is the ballroom of an ocean liner which sank some time ago.
The man ran out of air while diving in the wreck.
1.2a. Variant which puts this in section 2: same statement, ending with "a
large window through which rays are coming." Answer: the rays are manta rays
(this version tends to make people assume vampires are involved, unless they
notice the awkwardness of the phrase involving rays).
1.3. A woman came home with a bag of groceries, got the mail, and walked
into the house. On the way to the kitchen, she went through the living room
and looked at her husband, who had blown his brains out. She then continued
to the kitchen, put away the groceries, and made dinner. (partial JM
wording)
1.3. The husband killed himself a while ago; it's his ashes in an urn on the
mantelpiece that the wife looks at. It's debatable whether this belongs in
section 2 for double meanings.
1.4. A body is discovered in a park in Chicago in the middle of summer. It
has a fractured skull and many other broken bones, but the cause of death was
hypothermia. (MI, from _Hill Street Blues_)
1.4. A poor peasant from somewhere in Europe wants desperately to get to the
U.S. Not having money for airfare, he stows away in the landing gear
compartment of a jet. He dies of hypothermia in mid-flight, and falls out
when the landing gear compartment opens as the plane makes its final
approach.
1.4a. Variant: A man is lying drowned in a dead forest. Answer: He's scuba
diving when a firefighting plane lands nearby and fills its tanks with water,
sucking him in with the water. He runs out of air while the plane is in
flight; the plane then dumps its load of water, with him in it, onto a
burning forest. (from Jim Moskowitz)
1.5. A man lives on the twelfth floor of an apartment building. Every
morning he takes the elevator down to the lobby and leaves the building. In
the evening, he gets into the elevator, and, if there is someone else in the
elevator -- or if it was raining that day -- he goes back to his floor
directly. However, if there is nobody else in the elevator and it hasn't
rained, he goes to the 10th floor and walks up two flights of stairs to his
room. (MH)
1.5. The man is a midget. He can't reach the upper elevator buttons, but he
can ask people to push them for him. He can also push them with his
umbrella. I've usually heard this stated with more details: "Every morning
he wakes up, gets dressed, eats, goes to the elevator..." Ron Carter
suggests a nice red herring: the man lives on the 13th floor of the building.
1.6. A woman has incontrovertible proof in court that her husband was
murdered by her sister. The judge declares, "This is the strangest case I've
ever seen. Though it's a cut-and-dried case, this woman cannot be punished."
(This is different from #1.43.) (MH)
1.6. The sisters are Siamese twins.
1.6a. Variant: A man and his brother are in a bar drinking. They begin to
argue (as always) and the brother won't get out of the man's face, shouting
and cursing. The man, finally fed up, pulls out a pistol and blows his
brother's brains out. He sits down to die. Answer: They are Siamese twins.
In the original story, the argument started when one complained about the
other's bad hygiene and bad breath. The shooter bled to death (from his
brother's wounds) by the time the police arrived. (from Randy Whitaker,
based on a 1987 _Weekly World News_ story)
1.7. A man walks into a bar and asks for a drink. The bartender pulls out a
gun and points it at him. The man says, "Thank you," and walks out. (DVS)
1.7. The man has hiccups; the bartender scares them away by pulling a gun.
1.8. A man is returning from Switzerland by train. If he had been in a
non-smoking car he would have died. (DVS; MC wording)
1.8. The man used to be blind; he's now returning from an eye operation
which restored his sight. He's spent all his money on the operation, so when
the train (which has no internal lighting) goes through a tunnel he at first
thinks he's gone blind again and almost decides to kill himself.
Fortunately, the light of the cigarettes people are smoking convinces him
that he can still see.
1.8a. Variant: A man dies on a train he does not ordinarily catch. Answer:
The man (a successful artist) has had an accident in which he injured his
eyes. His head is bandaged and he has been warned not to remove the bandages
under any circumstances lest the condition be irreversibly aggravated. He
catches the train home from the hospital and cannot resist peeking. Seeing
nothing at all (the same train-in-tunnel situation as above obtains, but
without the glowing cigarettes this time), he assumes he is blinded and kills
himself in grief. I like this version a lot, except that it makes much less
sense that he'd be traveling alone. (from Bernd Wechner)
1.9. A man goes into a restaurant, orders abalone, eats one bite, and kills
himself. (TM and JM wording)
1.9. The man was in a ship that was wrecked on a desert island. When there
was no food left, another passenger brought what he said was abalone but was
really part of the man's wife (who had died in the wreck). The man suspects
something fishy, so when they finally return to civilization, he orders
abalone, realizes that what he ate before was his wife, and kills himself.
1.9a. Variant: same problem statement but with albatross instead of abalone.
Answer: In this version, the man was in a lifeboat, with his wife, who died.
He hallucinated an albatross landing in the boat which he caught and killed
and ate; he thought that his wife had been washed overboard. When he
actually eats albatross, he discovers that he had actually eaten his wife.
1.9b. Variant answer to 1.9a, with a slightly different problem statement:
the man already knew that he had been eating human flesh. He asks the waiter
in the restaurant what kind of soup is available, and the waiter responds,
"Albatross soup." Thinking that "albatross soup" means "human soup," and
sickened by the thought of such a society (place in a foreign country if
necessary), he kills himself. (from Mike Neergaard)
1.10. A man is found hanging in a locked room with a puddle of water under
his feet. (This is different from #1.11.)
1.10. He stood on a block of ice to hang himself. The fact that there's no
furniture in the room can be added to the statement, but if it's mentioned in
conjunction with the puddle of water the answer tends to be guessed more
easily.
1.11. A man is dead in a puddle of blood and water on the floor of a locked
room. (This is different from #1.10.)
1.11. He stabbed himself with an icicle.
1.12. A man is lying, dead, face down in the desert wearing a backpack.
(This is different from #1.13, #2.11, and #2.12.)
1.12. He jumped out of an airplane, but his parachute failed to open. Minor
variant wording (from Joe Kincaid): he's on a mountain trail instead of in a
desert. Minor variant wording (from Mike Reymond): he's got a ring in his
hand (it came off of the ripcord).
1.12a. Silly variant: same problem statement, with the addition that one of
the man's shoelaces is untied. Answer: He pulled his shoelace instead of the
ripcord.
1.12b. Variant answer: The man was let loose in the desert with a pack full
of poisoned food. He knows it's poisoned, and doesn't eat it -- he dies of
hunger. (from Mike Neergaard)
1.13. A man is lying face down, dead, in the desert, with a match near his
outstretched hand. (This is different from #1.12, #2.11, and #2.12.) (JH;
partial JM wording)
1.13. He was with several others in a hot air balloon crossing the desert.
The balloon was punctured and they began to lose altitude. They tossed all
their non-essentials overboard, then their clothing and food, but were still
going to crash in the middle of the desert. Finally, they drew matches to
see who would jump over the side and save the others; this man lost. Minor
variant wording: add that the man is nude.
1.14. A man is driving his car. He turns on the radio, listens for five
minutes, turns around, goes home, and shoots his wife. (This is different
from #1.15.)
1.14. The radio program is one of the call-up-somebody-and-ask-them-a-
question contest shows; the announcer gives the phone number of the man's
bedroom phone as the number he's calling, and a male voice answers. It's
been suggested that such shows don't usually give the phone number being
called; so instead the wife's name could be given as who's being called, and
there could be appropriate background sounds when the other man answers the
phone.
1.15. A man driving his car turns on the radio. He then pulls over to the
side of the road and shoots himself. (This is different from #1.14.)
1.15. He worked as a DJ at a radio station. He decided to kill his wife,
and so he put on a long record and quickly drove home and killed her,
figuring he had a perfect alibi: he'd been at work. On the way back he turns
on his show, only to discover that the record is skipping.
1.15a. Variant: The music stops and the man dies. Answer: The same, except
it's a tape breaking instead of a record skipping. (from Michael Killianey)
(See also #1.16, #1.19e, and #1.34a.)
1.16. Music stops and a woman dies. (DVS)
1.16. The woman is a tightrope walker in a circus. Her act consists of
walking the rope blindfolded, accompanied by music, without a net. The
musician (organist, or calliopist, or pianist, or whatever) is supposed to
stop playing when she reaches the end of the rope, telling her that it's safe
to step off onto the platform. For unknown reasons (but with murderous
intent), he stops the music early, and she steps off the rope to her death.
1.16a. Variant answer: The woman is a character in an opera, who "dies" at
the end of her song.
1.16b. Variant answer: The "woman" is the dancing figure atop a music box,
who "dies" when the box runs down. (Both of the above variants would
probably require placing this puzzle in section 2 of the list.)
1.16c. Variant: Charlie died when the music stopped. Answer: Charlie was an
insect sitting on a chair; the music playing was for the game Musical Chairs.
(from Bob Philhower)
(See also #1.15a, #1.19e, and #1.34a.)
1.17. A man is dead in a room with a small pile of pieces of wood and
sawdust in one corner. (from "Coroner's Inquest," by Marc Connelly)
1.17. The man is a blind midget, the shortest one in the circus. Another
midget, jealous because he's not as short, has been sawing small pieces off
of the first one's cane every night, so that every day he thinks he's taller.
Since his only income is from being a circus midget, he decides to kill
himself when he gets too tall.
1.17a. Slightly variant answer: Instead of sawing pieces off of the midget's
cane, someone has sawed the legs off of his bed. He wakes up, stands up, and
thinks he's grown during the night.
1.17b. Variant: A pile of sawdust, no net, a man dies. Answer: A midget is
jealous of the clown who walks on stilts. He saws partway through the
stilts; the clown walks along and falls and dies when they break. (from
Peter R. Olpe)
1.18. A flash of light, a man dies. (ST original)
1.18. The man is a lion-tamer, posing for a photo with his lions. The lions
react badly to the flash of the camera, and the man can't see properly, so he
gets mauled.
1.18a. Variant: He couldn't find a chair, so he died. Answer: He was a
lion-tamer. This one is kind of silly, but I like it, and it sounds possible
to me (though I'm told a whip is more important than a chair to a
lion-tamer). (from "Reaper Man," with Karl Heuer wording)
1.19. A rope breaks. A bell rings. A man dies. (KH)
1.19. A blind man enjoys walking near a cliff, and uses the sound of a buoy
to gauge his distance from the edge. One day the buoy's anchor rope breaks,
allowing the buoy to drift away from the shore, and the man walks over the
edge of the cliff.
1.19a. Variant: A bell rings. A man dies. A bell rings. Answer: A blind
swimmer sets an alarm clock to tell him when and what direction to go to
shore. The first bell is a buoy, which he mistakenly swims to, getting tired
and drowning. Then the alarm clock goes off. In other variations, the first
bell is a ship's bell, and/or the second bell is a hand-bell rung by a friend
on shore at a pre-arranged time.
1.19b. Variant answer to 1.19a: The man falls off a belltower, pulling the
bell-cord (perhaps he was climbing a steeple while hanging onto the rope),
and dies. The second bell is one rung at his funeral. Could also be a
variant on 1.19 (as suggested by Mike Neergaard): the bell-cord breaks when
he falls (and there's no second bell involved).
1.19c. Variant answer to 1.19a: The man is a boxer. The first bell signals
the start of a round; the second is either the end of the round or a funeral
bell after he dies during the match. Could also be a variant on 1.19 (as
suggested by Mike Neergaard): a boxing match in which the top rope breaks,
tumbling a boxer to the floor (and he dies of a concussion).
1.19d. Variant: The wind stopped blowing and the man died. Answer: The sole
survivor of a shipwreck reached a desert isle. Unfortunately, he was blind.
Luckily, there was a freshwater spring on the island, and he rigged the
ship's bell (which had drifted to the island also) at the spring's location.
The bell rang in the wind, directing him to water. When he was becalmed for
a week, he could not find water again, and so he died of thirst. (from Peter
R. Olpe)
1.19e. Variant: The music stopped and the man died. Answer: Same as 1.19a,
but the blind swimmer kept a portable transistor radio on the beach instead
of a bell. When the batteries gave out, he got lost and drowned. (from Joe
Kincaid) (See also #1.15a, #1.16, and #1.34a.)
1.20. A woman buys a new pair of shoes, goes to work, and dies. (DM)
1.20. The woman is the assistant to a (circus or sideshow) knife-thrower.
The new shoes have higher heels than she normally wears, so that the thrower
misjudges his aim and one of his knives kills her during the show.
1.20a. Variant: A woman sees her husband entering a certain place of
business and insists on dissolving their partnership. Answer: The husband is
a knife-thrower; the woman is his assistant as well as his wife. She sees
him going into an optometrist's office and decides that if he's having
trouble with his eyes she doesn't want him throwing knives at her. (from
_How Come -- Again?_)
1.21. A man is riding a subway. He meets a one-armed man, who pulls out a
gun and shoots him. (SJ)
1.21. Several men were shipwrecked together. They agreed to survive by
eating each other a piece at a time. Each of them in turn gave up an arm,
but before they got to the last man, they were rescued. They all demanded
that the last man live up to his end of the deal. Instead, he killed a bum
and sent the bum's arm to the others in a box to "prove" that he had
fulfilled the bargain. Later, one of them sees him on the subway, holding
onto an overhead ring with the arm he supposedly cut off; the other realizes
that the last man cheated, and kills him.
1.21a. Variant wording: A man sends a package to someone in Europe and gets
a note back saying "Thank you. I received it." Answer: This is just a
simpler version; the shipwreck situation is the same, and the man actually
did send his own arm.
1.21b. Variant wording: Two men throw a box off of a cliff. Answer: Exactly
the same situation as in 1.21a (one slight variation has a hand in the box
instead of a whole arm), with the two men being two of the fellow passengers
who had already lost their arms.
1.21c. Variant wording: A man in a Sherlock Holmes-style cape walks into a
room, places a box on the table and leaves. Answer: In this one he's wearing
the cape either to disguise the fact that he hasn't really cut off his
arm/hand as required, or else simply in order to hide his now-missing limb.
(from Joe Kincaid)
1.22. Two women are talking. One goes into the bathroom, comes out five
minutes later, and kills the other.
1.22. Both women are white; the one whose house this takes place in is
single. A black friend of the other woman, the one who goes into the
bathroom, was recently killed, reportedly by the KKK. The woman who goes
into the bathroom discovers a bloodstained KKK robe in the other's laundry
hamper, picks up a nail file from the medicine cabinet (or some other
impromptu weapon), and goes out and kills the other.
1.22a. Variant: A man goes to hang his coat and realizes he will die that
day. Answer: The man (who is black) has car trouble and is in need of a
telephone. He asks at the nearest house and on being invited in goes to hang
his coat, whereupon he notices the white robes of the Ku Klux Klan in the
closet. (from Bernd Wechner)
1.23. A man is sitting in bed. He makes a phone call, saying nothing, and
then goes to sleep. (SJ)
1.23. He is in a hotel, and is unable to sleep because the man in the
adjacent room is snoring. He calls the room next door (from his own room
number he can easily figure out his neighbor's, and from the room number, the
telephone number). The snorer wakes up, answers the phone. The first man
hangs up without saying anything and goes to sleep before the snorer gets
back to sleep and starts snoring again.
1.23a. Slightly variant answer: It's a next-door neighbor in an apartment
building who's snoring, rather than in a hotel. The caller thus knows his
neighbor and the phone number.
1.24. A man kills his wife, then goes inside his house and kills himself.
(DH original, from "Nightmare in Yellow," by Fredric Brown)
1.24. It's the man's fiftieth birthday, and in celebration of this he plans
to kill his wife, then take the money he's embezzled and move on to a new
life in another state. His wife takes him out to dinner; afterward, on their
front step, he kills her. He opens the door, dragging her body in with him,
and all the lights suddenly turn on and a group of his friends shout
"Surprise!" He kills himself. (Note that the whole first part, including
the motive, isn't really necessary; it was just part of the original story.)
1.25. Abel walks out of the ocean. Cain asks him who he is, and Abel
answers. Cain kills Abel. (MWD original)
1.25. Abel is a prince of the island nation that he landed on. A cruel and
warlike prince, he waged many land and naval battles along with his father
the king. In one naval encounter, their ship sank, the king died, and the
prince swam to a deserted island where he spent several months building a
raft or small boat. In the meantime, a regent was appointed to the island
nation, and he brought peace and prosperity. When Prince Abel returned to
his kingdom, Cain (a native fisherman) realized that the peace of the land
would only be maintained if Abel did not reascend to his throne, and killed
the prince (with a piece of driftwood or some other impromptu weapon).
1.26. Two men enter a bar. They both order identical drinks. One lives;
the other dies. (CR; partial JM wording)
1.26. The drinks contain poisoned ice cubes; one man drinks slowly, giving
them time to melt, while the other drinks quickly and thus doesn't get much
of the poison. The fact that they drink at different speeds could be added
to the statement, possibly along with red herrings such as saying that one of
the men is big and burly and the other short and thin.
1.27. Joe leaves his house, wearing a mask and carrying an empty sack. An
hour later he returns. The sack is now full. He goes into a room and turns
out the lights. (AL)
1.27. Joe is a kid who goes trick-or-treating for Halloween.
1.28. A man takes a two-week cruise to Mexico from the U.S. Shortly after
he gets back, he takes a three-day cruise which doesn't stop at any other
ports. He stays in his cabin all the time on both cruises. As a result, he
makes $250,000. (MI, from "The Wager")
1.28. He's a smuggler. On the first cruise, someone brings the contraband
to his cabin, and he hides it in an air conditioning duct. Returning to the
U.S., he leaves without the contraband, and so passes through customs with no
trouble. On the second trip, he has the same cabin on the same ship.
Because it doesn't stop anywhere, he doesn't have to go through customs when
he returns, so he gets the contraband off safely.
1.29. Hans and Fritz are German spies during World War II. They try to
enter America, posing as returning tourists. Hans is immediately arrested.
(JM)
1.29. Hans and Fritz do everything right up until they're filling out a
personal-information form and have to write down their birthdays. Fritz'
birthday is, say, July 7, so he writes down 7/7/15. Hans, however, was born
on, say, June 20, so he writes down 20/6/18 instead of what an American would
write, 6/20/18. Note that this is only a problem because they *claim* to be
returning Americans; as has been pointed out to me, there are lots of other
nations which use the same date ordering.
1.30. Tim and Greg were talking. Tim said "The terror of flight." Greg
said "The gloom of the grave." Greg was arrested. (MPW original, from "No
Refuge Could Save," by Isaac Asimov)
1.30. Another WWII story. Greg is a German spy. His "friend" Tim is
suspicious, so he plays a word-association game with him. When Tim says "The
land of the free," Greg responds with "The home of the brave." Then Tim says
"The terror of flight," and Greg says "The gloom of the grave." Any U.S.
citizen knows the first verse of the national anthem, but only a spy would
have memorized the third verse. (Why Tim knew the third verse is left as an
exercise to the reader.)
1.31. A man is found dead in his parked car. Tire tracks lead up to the car
and away. (SD)
1.31. The dead man was the driver in a hit-and-run accident which paralyzed
its victim. The victim did manage to get the license plate number of the
car; now in a wheelchair, he eventually tracked down the driver and shot and
killed him.
1.32. A man dies in his own home. (ME original)
1.32. His home is a houseboat and he has run out of water while on an
extended cruise.
1.32a. Variant wording: A man dies of thirst in his own home. This version
goes more quickly because it gives more information; but it may be less
likely to annoy people who think the original statement is too vague.
1.33. A woman in France in 1959 is waiting in her room, with all the doors
locked from the inside, for her husband to come home. When he arrives, the
house has burned to the ground and she's dead. (JM, from _How Come --
Again?_)
1.33. This is apparently a true story. The hot sun reflected from the
woman's large mirror (which I speculate may have been imperfectly flat and
therefore focused the sunlight, but I don't know for sure) and heated the
lingerie she was wearing to the burning point. She was absorbed in a book at
the time and didn't notice the heat until her clothing was afire. Nobody
could get to her to help because her doors were locked from the inside.
Please disregard the version of this answer from previous editions of this
list; it's not true.
1.34. A man gets onto an elevator. When the elevator stops, he knows his
wife is dead. (LA; partial KH wording)
1.34. He's leaving a hospital after visiting his wife, who's on heavy
life-support. When the power goes out, he knows she can't live without the
life-support systems (he assumes that if the emergency backup generator were
working, the elevator wouldn't lose power; this aspect isn't entirely
satisfactory, so in a variant, the scene is at home rather than in a
hospital).
1.34a. Variant: The music stops and a woman dies. Answer: The woman is
confined in an iron lung, and the music is playing on her radio or stereo.
The power goes out. (from Randy Whitaker) (See also #1.15a, #1.16, and
#1.19e.)
1.35. Three men die. On the pavement are pieces of ice and broken glass.
(JJ)
1.35. A large man comes home to the penthouse apartment he shares with his
beautiful young wife, taking the elevator up from the ground floor. He sees
signs of lovemaking in the bedroom, and assumes that his wife is having an
affair; her beau has presumably escaped down the stairs. The husband looks
out the French windows and sees a good-looking man just leaving the main
entrance of the building. The husband pushes the refrigerator out through
the window onto the young man below. The husband dies of a heart attack from
overexertion; the young man below dies from having a refrigerator fall on
him; and the wife's boyfriend, who was hiding inside the refrigerator, also
dies from the fall.
1.36. She lost her job when she invited them to dinner. (DS original)
1.36. Let's say "she" is named Suzy, and "they" are named Harry and Jane.
Harry is an elderly archaeologist who has found a very old skeleton, which
he's dubbed "Jane" (a la "Lucy"). Suzy is a buyer for a museum; she's
supposed to make some sort of purchase from Harry, so she invites him to have
a business dinner with her (at a restaurant). When she calls to invite him,
he keeps talking about "Jane," so Suzy assumes that Jane is his wife and says
to bring her along. Harry, offended, calls Suzy's boss and complains; since
Suzy should've known who Jane was, she gets fired.
1.37. A man is running along a corridor with a piece of paper in his hand.
The lights flicker and the man drops to his knees and cries out, "Oh no!"
(MP)
1.37. The man is delivering a pardon, and the flicker of the lights
indicates that the person to be pardoned has just been electrocuted.
1.38. A car without a driver moves; a man dies. (EMS)
1.38. The murderer sets the car on a slope above the hot dog stand where the
victim works. He then wedges an ice block in the car to keep the brake pedal
down, and puts the car in neutral, after which he flies to another city to
avoid suspicion. It's a warm day; when the ice melts, the car rolls down the
hill and strikes the hot dog man at his roadside stand, killing him.
1.39. As I drive to work on my motorcycle, there is one corner which I go
around at a certain speed whether it's rainy or sunny. If it's cloudy but
not raining, however, I usually go faster. (SW original)
1.39. There's a car wash on that corner. On rainy days, the rain reduces
traction. On sunny days, water from the car wash has the same effect. If
rain is threatening, though, the car wash gets little business and thus
doesn't make the road wet, so I can take the corner faster.
1.40. A woman throws something out a window and dies. (JM)
1.40. The object she throws is a boomerang. It flies out, loops around, and
comes back and hits her in the head, killing her. Boomerangs do not often
return so close to the point from which they were thrown, but I believe it's
possible for this to happen.
1.40a. Silly variant answer: She's in a submarine or spacecraft and throws a
heavy object at the window, which breaks.
1.41. An avid birdwatcher sees an unexpected bird. Soon he's dead. (RSB
original)
1.41. He is a passenger in an airplane and sees the bird get sucked into an
engine at 20,000 feet.
1.42. There are a carrot, a pile of pebbles, and a pipe lying together in
the middle of a field. (PRO; partial JM wording)
1.42. They're the remains of a melted snowman.
1.43. Two brothers are involved in a murder. Though it's clear that one of
them actually committed the crime, neither can be punished. (This is
different from #1.6.) (from "Unreasonable Doubt," by Stanley Ellin)
1.43. One of the brothers (A) confesses to the murder. At his trial, his
brother (B) is called as the only defense witness; B immediately confesses,
in graphic detail, to having committed the crime. The defense lawyer refuses
to have the trial stopped, and A is acquitted under the "reasonable doubt"
clause. Immediately afterward, B goes on trial for the murder; A is called
as the only defense witness and HE confesses. B is declared innocent; and
though everyone knows that ONE of them did it, how can they tell who?
Further, neither can be convicted of perjury until it's decided which of them
did it... I don't know if that would actually work under the US legal
system, but someone else who heard the story said that his father was on the
jury for a VERY similar case in New York some years ago. Mark Brader points
out that the brothers might be convicted of conspiracy to commit perjury or
to obstruct justice, or something of that kind.
1.44. An ordinary American citizen, with no passport, visits over thirty
foreign countries in one day. He is welcomed in each country, and leaves
each one of his own accord. (PRO)
1.44. He is a mail courier who delivers packages to the different foreign
embassies in the United States. The land of an embassy belongs to the
country of the embassy, not to the United States.
1.45. If he'd turned on the light, he'd have lived. (JM)
1.45. A man was shot during a robbery in his store one night. He staggered
into the back room, where the telephone was, and called home, dialing by feel
since he hadn't turned on the light. Once the call went through he gasped,
"I'm at the store. I've been shot. Help!" or words to that effect. He set
the phone down to await help, but none came; he'd treated the telephone
pushbuttons like cash register numbers, when the arrangements of the numbers
are upside down reflections of each other. The stranger he'd dialed had no
way to know where "the store" was.
1.46. A man is found dead on the floor in the living room. (ME original)
1.46. The dead man was playing Santa Claus, for whatever reason; he slipped
while coming down the chimney and broke his neck.
1.46a. Variant answer: The dead man WAS Santa Claus. This moves the puzzle
to section 2 as far as I'm concerned.
1.47. A man is found dead outside a large building with a hole in him. (JM,
modified from PRO)
1.47. The man was struck by an object thrown from the roof of the Empire
State Building. Originally I had the object being a penny, but several
people suggested that a penny probably wouldn't be enough to penetrate
someone's skull. Something aerodynamic and heavier, like a dart, was
suggested, but I don't know how much mass would be required.
1.47a. Variant: A man is found dead outside a large marble building with
three holes in him. Answer: The man was a paleontologist working with the
Archaeological Research Institute. He was reviving a triceratops frozen in
the ice age when it came to life and killed him. This couldn't possibly
happen because triceratops didn't exist during the ice age. (from Peter R.
Olpe)
1.48. A man is found dead in an alley lying in a red pool with two sticks
crossed near his head. (PRO)
1.48. The man died from eating a poisoned popsicle.
1.49. A man lies dead next to a feather. (PRO)
1.49. The man was a sword swallower in a carnival side-show. While he was
practicing, someone tickled his throat with the feather, causing him to gag.
1.50. There is blood on the ceiling of my bedroom. (MI original)
1.50. A mosquito bit me, and I swatted it when it later landed on my ceiling
(so the blood is my own as well as the mosquito's).
1.51. A man wakes up one night to get some water. He turns off the light
and goes back to bed. The next morning he looks out the window, screams, and
kills himself. (CR; KK wording)
1.51. The man is a lighthouse keeper. He turns off the light in the
lighthouse and during the night a ship crashes on the rocks. Seeing this the
next morning, the man realizes what he's done and commits suicide.
1.51a. Variant, similar to #1.15: The light goes out and a man dies.
Answer: The lighthouse keeper uses his job as an alibi while he's elsewhere
committing a crime, but the light goes out and a ship crashes, thereby
disproving the alibi. The lighthouse keeper kills himself when he realizes
his alibi is no good. (From Eric Wang)
1.51b. Variant answer to 1.51a: Someone else's alibi is disproven. (A man
commits a heinous crime, claiming as his alibi that he was onboard a certain
ship. When he learns that it was wrecked without reaching port safely, he
realizes that his alibi is disproven and commits suicide to avoid being sent
to prison.) (From Eric Wang)
1.52. She grabbed his ring, pulled on it, and dropped it. (JM, from _Math
for Girls_)
1.52. They were skydiving. He broke his arm as he jumped from the plane by
hitting it on the plane door; he couldn't reach his ripcord with his other
arm. She pulled the ripcord for him.
1.52a. Sketch of variant answer: The ring was attached to the pin of a
grenade that he was holding. Develop a situation from there.
1.53. A man sitting on a park bench reads a newspaper article headlined
"Death at Sea" and knows a murder has been committed.
1.53. The man is a travel agent. He had sold someone two tickets for an
ocean voyage, one round-trip and one one-way. The last name of the man who
bought the tickets is the same as the last name of the woman who "fell"
overboard and drowned on the same voyage, which is the subject of the article
he's reading.
1.54. A man tries the new cologne his wife gave him for his birthday. He
goes out to get some food, and is killed. (RW original)
1.54. The man is a beekeeper, and the bees attack en masse because they
don't recognize his fragrance. Randy adds that this is based on something
that actually happened to his grandfather, a beekeeper who was severely
attacked by his bees when he used a new aftershave for the first time in 10
or 20 years.
1.55. A man in uniform stands on the beach of a tropical island. He takes
out a cigarette, lights it, and begins smoking. He takes out a letter and
begins reading it. The cigarette burns down between his fingers, but he
doesn't throw it away. He cries. (RW)
1.55. He is a guard / attendant in a leper colony. The letter (to him)
tells him that he has contracted the disease. The key is the cigarette
burning down between his fingers -- leprosy is fairly unique in killing off
sensory nerves without destroying motor ability.
1.56. A man went into a restaurant, had a large meal, and paid nothing for
it. (JM original)
1.56. The man was a famous artist. A woman who collected autographs saw him
dining; after he left the restaurant, she purchased the check that he used to
pay for the meal from the restaurant manager. The check was therefore never
cashed, so the artist never paid for the meal.
1.57. A married couple goes to a movie. During the movie the husband
strangles the wife. He is able to get her body home without attracting
attention. (from _Beyond the Easy Answer_)
1.57. The movie is at a drive-in theatre.
1.58. A man ran into a fire, and lived. A man stayed where there was no
fire, and died. (Eric Wang original)
1.58. The two men were working in a small room protected by a carbon dioxide
gas fire extinguisher system, when a fire broke out in an adjoining room.
One of the men ran through the fire and escaped with only minor burns. The
other one stayed in the room until the fire extinguishers kicked in, and died
of oxygen starvation. (This originally involved a halon gas extinguisher,
but those don't work that way; fortunately, Gisle Hannemyr pointed out that
CO2 extinguishers do work that way. Gisle says a CO2 extinguisher on a
Norwegian ship a few years ago did go off accidentally when there was no
fire, killing everyone in the engine room.)
1.59. A writer with an audience of millions insisted that he was never to be
interrupted while writing. After the day when he actually was interrupted,
he never wrote again. (JM, from _How Come?_)
1.59. He was a skywriter whose plane crashed into another plane.
1.60. Beulah died in the Appalachians, while Craig died at sea. Everyone
was much happier with Craig's death. (JM, from _How Come?_)
1.60. Beulah and Craig were hurricanes.
1.61. Mr. Browning is glad the car ran out of gas. (JM, from _Home Come?_)
1.61. Mr. and Mrs. Browning had just gotten married. Mrs. Browing was
subject to fits of depression. They had their first fight soon after they
were married; Mr. Browning stormed out of the house, and Mrs. Browning went
into the garage and started up the car, intending to kill herself by filling
the garage with car exhaust. But the car ran out of gas quickly, and Mr.
Browning, returning home to apologize, found Mrs. Browning in time to summon
help and restore her to health.
1.62. A man is sitting suspended over two pressurized containers. Suddenly,
he dies. (NK original)
1.62. He's riding a bicycle or motorcycle, and he crashes and dies.
1.63. A man leaves a motel room, goes to his car, and honks the horn. (AS
original)
1.63. It's the middle of the night. The man goes outside to get something
from his car, but as the parking lot is set apart from the building, he
forgets which room he was in. His wife is deaf, so he honks the car horn
loudly, waking up everyone else in the motel. The other residents all get up
and turn on their room lights; the man then returns to the one dark room.
1.64. Two dead people sit in their cars on a street. (AG)
1.64. Because there was a heavy fog, two people driving in opposite
directions on the same road both stuck their heads out of their windows to
better see the road's center line. Their heads hit each other at high speed,
killing them both. Andreas says this is based on an actual accident.
1.65. A woman lies dead in the street near a car. (AG)
1.65. She was on a motorcycle, and her long hair got caught on the car's
antenna. It ripped out part of her scalp and she bled to death. Andreas
says this is also based on an actual accident.
1.66. A riverboat filled with passengers suddenly capsized, drowning most of
those aboard. (from _How Come -- Again?_)
1.66. The boat was moving along a river in India when a large snake dropped
onto the deck. The passengers all rushed to the other side of the boat,
thereby overturning it. This is apparently based on a true incident reported
in the _World Almanac_.
Section 2: Double meanings, fictional settings, and miscellaneous others.
2.1. A man shoots himself, and dies. (HL) (This is different from #2.2.)
2.1. The man is a heroin addict, and has contracted AIDS by using an
infected needle. In despair, he shoots himself up with an overdose, thereby
committing suicide.
2.2. A man walks into a room, shoots, and kills himself. (HL) (This is
different from #2.1.)
2.2. The man walks into a casino and goes to the craps table. He bets all
the money he owns, and shoots craps. Since he is now broke, he becomes
despondent and commits suicide.
2.3. Adults are holding children, waiting their turn. The children are
handed (one at a time, usually) to a man, who holds them while a woman shoots
them. If the child is crying, the man tries to stop the crying before the
child is shot. (ML)
2.3. Kids getting their pictures taken with Santa. I see #2.1, #2.2, and
#2.3 as different enough from each other to merit separate numbers, although
they all rely on the same basic gimmick of alternate meanings of the word
"shoot."
2.4. Hiking in the mountains, you walk past a large field and camp a few
miles farther on, at a stream. It snows in the night, and the next day you
find a cabin in the field with two dead bodies inside. (KL; KD and partial
JM wording)
2.4. It's the cabin of an airplane that crashed there because of the
snowstorm.
2.4a. Variant wording: A cabin, on the side of a mountain, locked from the
inside, is opened, and 30 people are found dead inside. They had plenty of
food and water. (from Ron Carter)
2.5. A man marries twenty women in his village but isn't charged with
polygamy.
2.5. He's a priest; he is marrying them to other people, not to himself.
2.6. A man is alone on an island with no food and no water, yet he does not
fear for his life. (MN)
2.6. The "island" is a traffic island.
2.7. Joe wants to go home, but he can't go home because the man in the mask
is waiting for him. (AL wording)
2.7. A baseball game is going on. The base-runner sees the catcher waiting
at home plate with the ball, and so decides to stay at third base to avoid
being tagged out.
2.7a. Variant: Two men are in a field. One is wearing a mask. The other
man is running towards him to avoid him. Answer: the same, but the catcher
isn't right at home plate; the runner is trying to get home before the
catcher can. (from Hal Lowery, by way of Chris Riley) This phrasing would
allow the puzzle to migrate to section 1, but I don't like it as much.
2.8. A man is doing his job when his suit tears. Fifteen minutes later,
he's dead. (RM)
2.8. The man is an astronaut out on a space walk.
2.9. A dead man lies near a pile of bricks and a beetle on top of a book.
(MN)
2.9. The man was an amateur mechanic, the book is a Volkswagen service
manual, the beetle is a car, and the pile of bricks is what the car fell off
of.
2.10. At the bottom of the sea there lies a ship worth millions of dollars
that will never be recovered. (TF original)
2.10. The Eagle landed in the Sea of Tranquility and will likely remain
there for the foreseeable future.
2.11. A man is found dead in the arctic with a pack on his back. (This is
different from #1.12, #1.13, and #2.12.) (PRO)
2.11. It's a wolf pack; they've killed and eaten (most of) the man.
2.12. There is a dead man lying in the desert next to a rock. (This is
different from #1.12, #1.13, and #2.11.) (GH)
2.12. The dead man is Superman; the rock is Green Kryptonite. Invent a
reasonable scenario from there.
2.13. As a man jumps out of a window, he hears the telephone ring and
regrets having jumped. (from "Some Days are Like That," by Bruce J.
Balfour; partial JM wording)
2.13. This is a post-holocaust scenario of some kind; for whatever reason,
the man believes himself to be the last human on earth. He doesn't want to
live by himself, so he jumps, just before the telephone rings... (of course,
it could be a computer calling, but he has no way of knowing).
2.14. Two people are playing cards. One looks around and realizes he's
going to die. (JM original)
2.14. The one who looks around sees his own reflection in the window (it's
dark outside), but not his companion's. Thus, he realizes the other is a
vampire, and that he's going to be killed by him.
2.15. A man lies dead in a room with fifty-three bicycles in front of him.
2.15. The "bicycles" are Bicycle playing cards; the man was cheating at
cards, and when the extra card was found, he was killed by the other players.
2.15a. Variant: There are 53 bees instead of 53 bicycles. Answer: The same
(Bee is another brand of playing cards).
2.15b. Variant: There are 51 instead of 53. Answer: Someone saw the guy
conceal a card, and proved the deck was defective by turning it up and
pointing out the missing ace. Or, the game was bridge, and the others
noticed the cheating when the deal didn't come out even. The man had palmed
an ace during the shuffle and meant to put it in his own hand during the
deal, but muffed it. (both answers from Mark Brader)
2.16. A horse jumps over a tower and lands on a man, who disappears. (ES
original)
2.16. A chess game; knight takes pawn.
2.16a. Variant: It's the year 860 A.D., at Camelot. Two priests are sitting
in the castle's chapel. The queen attacks the king. The two priests rise,
shake hands, and leave the room. Answer: The two priests are playing chess;
one of them just mated by moving his queen. (from Ellen M. Sentovich)
2.16b. Variant: A black leader dies in Africa. Answer: The black leader is
a chess king, and the game was played in Africa. (from Erick Brethenoux)
2.17. A train pulls into a station, but none of the waiting passengers move.
(MN)
2.17. It's a model train set.
2.17a. Variant: The Orient Express is derailed and a kitten plays nearby.
Answer: The Orient Express is a model train which has been left running
unattended. The kitten has playfully derailed it. (from Bernd Wechner)
2.18. A man pushes a car up to a hotel and tells the owner he's bankrupt.
(DVS; partial AL and JM wording)
2.18. It's a game of Monopoly.
2.18a. Variant: The car came out of the blue and the man came into some
money. Answer: The same; in this case the car token passes Go and the player
collects $200. (from "Mo," whose full name I missed)
2.19. Three large people try to crowd under one small umbrella, but nobody
gets wet. (CC)
2.19. The sun is shining; there's no rain.
2.20. A black man dressed all in black, wearing a black mask, stands at a
crossroads in a totally black-painted town. All of the streetlights in town
are broken. There is no moon. A black-painted car without headlights drives
straight toward him, but turns in time and doesn't hit him. (AL and RM
wording)
2.20. It's daytime; the sun is out.
2.21. Bob and Carol and Ted and Alice all live in the same house. Bob and
Carol go out to a movie, and when they return, Alice is lying dead on the
floor in a puddle of water and glass. It is obvious that Ted killed her but
Ted is not prosecuted or severely punished.
2.21. Alice is a goldfish; Ted is a cat.
2.21a. A very common variant uses the names Romeo and Juliet instead, to
further mislead audiences. For example: Romeo is looking down on Juliet's
dead body, which is on the floor surrounded by water and broken glass. (from
Adam Carlson)
2.21b. Minor variant: Tom and Jean lay dead in a puddle of water with broken
pieces of glass and a baseball nearby. Answer: Tom and Jean are both fish;
it was a baseball, rather than a cat, that broke their tank. (from Mike
Reymond)
2.22. A man rides into town on Friday. He stays one night and leaves on
Friday. (KK)
2.22. Friday is a horse.
2.22a. Variant with the same basic gimmick: A woman comes home, sees
Spaghetti on the wall and kills her husband. Answer: Spaghetti was the name
of her pet dog. Her husband had it stuffed and mounted after it made a mess
on his rug. (Simon Travaglia original)
2.23. Bruce wins the race, but he gets no trophy. (EMS)
2.23. Bruce is a horse.
2.24. A woman opens an envelope and dyes. (AL)
2.24. Should be done orally; the envelope is an envelope of dye, and she's
dying some cloth, but it sounds like "opens an envelope and dies" if said out
loud.
2.25. A man was brought before a tribal chief, who asked him a question. If
he had known the answer, he probably would have died. He didn't, and lived.
(MWD original)
2.25. The native chief asked him, "What is the third baseman's name in the
Abbot and Costello routine 'Who's on First'?" The man, who had no idea, said
"I don't know," the correct answer. However, he was a major smartass, so if
he had known the answer he would have pointed out that What was the SECOND
baseman's name. The chief, being quite humorless, would have executed him on
the spot. This is fairly silly, but I like it too much to remove it from the
list.
2.26. Two men are found dead outside of an igloo. (SK original)
2.26. The men have gone spelunking and have taken an Igloo cooler with them
so they can have a picnic down in the caves. They cleverly used dry ice to
keep their beer cold, not realizing that as the dry ice sublimed (went from
solid state to vapor state) it would push the lighter oxygen out of the cave
and they would suffocate.
2.27. A man is born in 1972 and dies in 1952 at the age of 25. (DM)
2.27. He's born in room number 1972 of a hospital and dies in room number
1952. The numbers can of course vary; it was originally set up with those
numbers reversed (born in 1952, died in 1972), but I like it better this way.
Attributions key:
Notes and comments:
Bibliography:
History of List:
==> logic/smullyan/black.hat.p <==
Three logicians, A, B, and C, are wearing hats, which they know are either
black or white but not all white. A can see the hats of B and C; B can see
the hats of A and C; C is blind. Each is asked in turn if they know the color
of their own hat. The answers are:
A: "No."
B: "No."
C: "Yes."
What color is C's hat and how does she know?
==> logic/smullyan/black.hat.s <==
A must see at least one black hat, or she would know that her hat is black
since they are not all white. B also must see at least one black hat, and
further, that hat had to be on C, otherwise she would know that her
hat was black (since she knows A saw at least one black hat). So C knows
that her hat is black, without even seeing the others' hats.
==> logic/smullyan/fork.three.men.p <==
Three men stand at a fork in the road. One fork leads to Someplaceorother;
the other fork leads to Nowheresville. One of these people always answers
the truth to any yes/no question which is asked of him. The other always
lies when asked any yes/no question. The third person randomly lies and
tells the truth. Each man is known to the others, but not to you.
What is the least number of yes/no questions you can ask of these men and
pick the road to Someplaceorother? Does the answer change if the third
man randomly answers?
==> logic/smullyan/fork.three.men.s <==
One question, and you only need one man of any type:
"If I were to ask you whether the left fork leads to Someplaceorother,
and you chose to answer that question with the same degree of truth as
you answer this question, would you then answer 'yes'?"
The truthteller will say "yes" if the left fork leads to Someplaceorother,
and "no" otherwise. The liar will answer the same, since he will lie about
where the left fork leads, and he will lie about lying. The randomizer
may either lie or tell the truth about this one question, but either way
he is behaving like either the truthteller or the liar and thus must
correctly report the road to Someplaceorother.
If however the third person randomly answers yes or no it is clear that
you must ask at least two questions, since you might be asking the
first one of the randomizer and there is nothing you can tell from his
answers.
Start by asking A "Is B more likely to tell the truth than C?"
If he answers "yes", then:
If A is truthteller, B is randomizer, C is liar.
If A is liar, B is randomizer, C is truthteller.
If A is randomizer, C is truthteller or liar.
If he answers "no", then:
If A is truthteller, B is liar, C is randomizer.
If A is liar, B is truthteller, C is randomizer.
If A is randomizer, B is truthteller or liar.
In either case, we now know somebody (C or B, respectively) who is
either a truthteller or liar. Now, use the technique for finding
information from a truthteller/liar, viz., you ask him the following
question: "If I were to ask you if the left fork leads to
Someplaceorother, would you say 'yes'?"
If the answer is "yes", take the left fork, if "no" take the right fork.
==> logic/smullyan/fork.two.men.p <==
Two men stand at a fork in the road. One fork leads to Someplaceorother; the
other fork leads to Nowheresville. One of these people always answers the
truth to any yes/no question which is asked of him. The other always lies
when asked any yes/no question. By asking one yes/no question, can you
determine the road to Someplaceorother?
==> logic/smullyan/fork.two.men.s <==
The fact that there are two is a red herring - you only need one of
either type. You ask him the following question: "If I were to ask
you if the left fork leads to Someplaceorother, would you say 'yes'?"
If the person asked is a truthteller, he will answer "yes" if the left
fork leads to Someplaceorother, and "no" otherwise. But so will the
liar. So, either way, go left is the answer is "yes", and right otherwise.
It is possible, of course, that the liars are malicious, and they will tell
the truth if they figure out that you are trying to trick them.
==> logic/smullyan/integers.p <==
Two logicians place cards on their foreheads so that what is written on the
card is visible only to the other logician. Consecutive positive integers
have been written on the cards. The following conversation ensues:
A: "I don't know my number."
B: "I don't know my number."
A: "I don't know my number."
B: "I don't know my number."
... n statements of ignorance later ...
A or B: "I know my number."
What is on the card and how does the logician know it?
==> logic/smullyan/integers.s <==
If A saw 1, she would know that she had 2, and would say so. Therefore,
A did not see 1. A says "I don't know my number."
If B saw 2, she would know that she had 3, since she knows that A did not see
1, so B did not see 1 or 2. B says "I don't know my number."
If A saw 3, she would know that she had 4, since she knows that B did not
see 1 or 2, so A did not see 1, 2 or 3. A says "I don't know my number."
If B saw 4, she would know that she had 5, since she knows that A did not
see 1, 2 or 3, so B did not see 1, 2, 3 or 4. B says "I don't know my number."
... n statements of ignorance later ...
If X saw n, she would know that she had n + 1, since she knows that ~X did not
see 1 ... n - 1, so X did see n. X says "I know my number."
And the number in n + 1.
==> logic/smullyan/painted.heads.p <==
While three logicians were sleeping under a tree, a malicious child painted
their heads red. Upon waking, each logician spies the child's handiwork as
it applied to the heads of the other two. Naturally they start laughing.
Suddenly one falls silent. Why?
==> logic/smullyan/painted.heads.s <==
The one who fell silent, presumably the quickest of the three, reasoned
that his head must be painted also. The argument goes as follows.
Let's call the quick one Q, and the other two D and S. Let's assume
Q's head is untouched. Then D is laughing because S's head is painted,
and vice versa. But eventually, D and S will realize that their head
must be painted, because the other is laughing. So they will quit
laughing as soon as they realize this. So, Q waits what he thinks is
a reasonable amount of time for them to figure this out, and when they
don't stop laughing, his worst fears are confirmed. He concludes that
his assumption is invalid and he must be crowned in crimson too.
smith22...@gmail.com
smith22...@gmail.com
> Archive-name: puzzles/archive/Instructions
> Version: 4
>
>
> Instructions for Accessing rec.puzzles Archive
>
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>
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>
>
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>
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>
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>
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>
> The archive is NOT the original work of the editor (just in case you were
> wondering :^).
>
> In keeping with the general net practice on FAQ's, I do not as a rule assign
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> 1. The archive is about the answers to the questions, not about assigning
> credit.
> 2. Many people, in providing free answers to the net, do not have the time
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> to come up with as complete an answer as possible.
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>
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>
> Address corrections or comments to archive...@questrel.com.
>
> INDEX
>
> ==> bicycle (analysis) <==
> A boy, a girl and a dog go for a 10 mile walk. The boy and girl can
>
> ==> boy.girl.dog (analysis) <==
> A boy, a girl and a dog are standing together on a long, straight road.
>
> ==> bugs (analysis) <==
> Four bugs are placed at the corners of a square. Each bug walks always
>
> ==> c.infinity (analysis) <==
> What function is zero at zero, strictly positive elsewhere, infinitely
>
> ==> cache (analysis) <==
> Cache and Ferry (How far can a truck go in a desert?)
>
> ==> calculate.pi (analysis) <==
> How can I calculate many digits of pi?
>
> ==> cats.and.rats (analysis) <==
> If 6 cats can kill 6 rats in 6 minutes, how many cats does it take to
>
> ==> dog (analysis) <==
> A body of soldiers form a 50m-by-50m square ABCD on the parade ground.
>
> ==> e.and.pi (analysis) <==
> Without finding their numerical values, which is greater, e^(pi) or (pi)^e?
>
> ==> functional/distributed (analysis) <==
> Find all f: R -> R, f not identically zero, such that
>
> ==> functional/linear (analysis) <==
> Suppose f is non-decreasing with
>
> ==> integral (analysis) <==
> If f is integrable on (0,inf) and differentiable at 0, and a > 0, and:
>
> ==> irrational.stamp (analysis) <==
> You have an ink stamp which is so amazingly precise that, when inked
>
> ==> minimum.time (analysis) <==
> N people can walk or drive in a two-seater to go from city A to city B. What
>
> ==> particle (analysis) <==
> What is the longest time that a particle can take in travelling between two
>
> ==> period (analysis) <==
> What is the least possible integral period of the sum of functions
>
> ==> rubberband (analysis) <==
> A bug walks down a rubber band which is attached to a wall at one end and a car
>
> ==> sequence (analysis) <==
> Show that in the sequence: x, 2x, 3x, .... (n-1)x (x can be any real number)
>
> ==> snow (analysis) <==
> Snow starts falling before noon on a cold December day. At noon a
>
> ==> tower (analysis) <==
> R = N ^ (N ^ (N ^ ...)). What is the maximum N>0 that will yield a finite R?
>
> ==> 7-11 (arithmetic/part1) <==
> A customer at a 7-11 store selected four items to buy, and was told
>
> ==> arithmetic.progression (arithmetic/part1) <==
> Is there an arithmetic progression of 20 or more primes?
>
> ==> clock/day.of.week (arithmetic/part1) <==
> It's restful sitting in Tom's cosy den, talking quietly and sipping
>
> ==> clock/palindromic (arithmetic/part1) <==
> How many times per day does a digital clock display a palindromic number?
>
> ==> clock/reversible (arithmetic/part1) <==
> How many times per day can the hour and minute hands on an analog clock switch
>
> ==> clock/right.angle (arithmetic/part1) <==
> How many times per day do the hour and minute hands of a clock form a
>
> ==> clock/thirds (arithmetic/part1) <==
> Do the 3 hands on a clock ever divide the face of the clock into 3
>
> ==> consecutive.composites (arithmetic/part1) <==
> Are there 10,000 consecutive non-prime numbers?
>
> ==> consecutive.product (arithmetic/part1) <==
> Prove that the product of three or more consecutive positive integers cannot
>
> ==> consecutive.sums (arithmetic/part1) <==
> Find all series of consecutive positive integers whose sum is exactly 10,000.
>
> ==> conway (arithmetic/part1) <==
> Describe the sequence a(1)=a(2)=1, a(n) = a(a(n-1)) + a(n-a(n-1)) for n>2.
>
> ==> digits/6.and.7 (arithmetic/part1) <==
> Does every number which is not divisible by 5 have a multiple whose
>
> ==> digits/all.ones (arithmetic/part1) <==
> Prove that some multiple of any integer ending in 3 contains all 1s.
>
> ==> digits/arabian (arithmetic/part1) <==
> What is the Arabian Nights factorial, the number x such that x! has 1001
>
> ==> digits/circular (arithmetic/part1) <==
> What 6 digit number, with 6 different digits, when multiplied by all integers
>
> ==> digits/divisible (arithmetic/part1) <==
> Find the least number using 0-9 exactly once that is evenly divisible by each
>
> ==> digits/equations/123456789 (arithmetic/part1) <==
> In how many ways can "." be replaced with "+", "-", or "" (concatenate) in
>
> ==> digits/equations/1992 (arithmetic/part1) <==
> 1 = -1+9-9+2. Extend this list to 2 through 100 on the left side of
>
> ==> digits/equations/24 (arithmetic/part1) <==
> Form an expression that evaluates to 24 that contains two 3's, two 7's,
>
> ==> digits/equations/383 (arithmetic/part1) <==
> Make 383 out of 1,2,25,50,75,100 using +,-,*,/.
>
> ==> digits/equations/find (arithmetic/part1) <==
> Write a program for finding expressions built out of given numbers and using
>
> ==> digits/extreme.products (arithmetic/part1) <==
> What are the extremal products of three three-digit numbers using digits 1-9?
>
> ==> digits/labels (arithmetic/part1) <==
> You have an arbitrary number of model kits (which you assemble for
>
> ==> digits/least.significant/factorial (arithmetic/part1) <==
> What is the least significant non-zero digit in the decimal expansion of n!?
>
> ==> digits/least.significant/tower.of.power (arithmetic/part1) <==
> What are the least significant digits of 9^(8^(7^(6^(5^(4^(3^(2^1))))))) ?
>
> ==> digits/most.significant/googol (arithmetic/part1) <==
> What digits does googol! start with?
>
> ==> digits/most.significant/powers (arithmetic/part1) <==
> What is the probability that 2^N begins with the digits 603245?
>
> ==> digits/nine.digits (arithmetic/part1) <==
> Form a number using 0-9 once with its first n digits divisible by n.
>
> ==> digits/palindrome (arithmetic/part1) <==
> Does the series formed by adding a number to its reversal always end in
>
> ==> digits/palintiples (arithmetic/part1) <==
> Find all numbers that are multiples of their reversals.
>
> ==> digits/power.two (arithmetic/part1) <==
> Prove that for any 9-digit number (base 10) there is an integral power
>
> ==> digits/prime/101 (arithmetic/part1) <==
> How many primes are in the sequence 101, 10101, 1010101, ...?
>
> ==> digits/prime/all.prefix (arithmetic/part1) <==
> What is the longest prime whose every proper prefix is a prime?
>
> ==> digits/prime/change.one (arithmetic/part1) <==
> What is the smallest number that cannot be made prime by changing a single
>
> ==> digits/prime/prefix.one (arithmetic/part1) <==
> 2 is prime, but 12, 22, ..., 92 are not. Similarly, 5 is prime
>
> ==> digits/reverse (arithmetic/part1) <==
> Is there an integer that has its digits reversed after dividing it by 2?
>
> ==> digits/rotate (arithmetic/part1) <==
> Find integers where multiplying them by single digits rotates their digits
>
> ==> digits/sesqui (arithmetic/part1) <==
> Find the least number where moving the first digit to the end multiplies by 1.5.
>
> ==> digits/squares/change.leading (arithmetic/part1) <==
> What squares remain squares when their leading digits are incremented?
>
> ==> digits/squares/length.22 (arithmetic/part1) <==
> Is it possible to form two numbers A and B from 22 digits such that
>
> ==> digits/squares/length.9 (arithmetic/part1) <==
> Is it possible to make a number and its square, using the digits from 1
>
> ==> digits/squares/three.digits (arithmetic/part2) <==
> What squares consist entirely of three digits (e.g., 1, 4, and 9)?
>
> ==> digits/squares/twin (arithmetic/part2) <==
> Let a twin be a number formed by writing the same number twice,
>
> ==> digits/sum.of.digits (arithmetic/part2) <==
> Find sod ( sod ( sod (4444 ^ 4444 ) ) ).
>
> ==> digits/zeros/million (arithmetic/part2) <==
> How many zeros occur in the numbers from 1 to 1,000,000?
>
> ==> digits/zeros/trailing (arithmetic/part2) <==
> How many trailing zeros are in the decimal expansion of n!?
>
> ==> magic.squares (arithmetic/part2) <==
> Are there large squares, containing only consecutive integers, all of whose
>
> ==> pell (arithmetic/part2) <==
> Find integer solutions to x^2 - 92y^2 = 1.
>
> ==> subset (arithmetic/part2) <==
> Prove that all sets of n integers contain a subset whose sum is divisible by n.
>
> ==> sum.of.cubes (arithmetic/part2) <==
> Find two fractions whose cubes total 6.
>
> ==> sums.of.powers (arithmetic/part2) <==
> Partition 1,2,3,...,16 into two equal sets, such that the sums of the
>
> ==> tests.for.divisibility/eleven (arithmetic/part2) <==
> What is the test to see if a number is divisible by eleven?
>
> ==> tests.for.divisibility/nine (arithmetic/part2) <==
> What is the test to see if a number is divisible by nine?
>
> ==> tests.for.divisibility/seven (arithmetic/part2) <==
> What is the test to see if a number is divisible by seven?
>
> ==> tests.for.divisibility/three (arithmetic/part2) <==
> What is the test to see if a number is divisible by three?
>
> ==> alphabet.blocks (combinatorics) <==
> What is the minimum number of dice painted with one letter on all six sides
>
> ==> coinage/combinations (combinatorics) <==
> Assuming you have enough coins of 1, 5, 10, 25 and 50 cents, how many
>
> ==> coinage/dimes (combinatorics) <==
> "Dad wants one-cent, two-cent, three-cent, five-cent, and ten-cent
>
> ==> coinage/impossible (combinatorics) <==
> What is the smallest number of coins that you can't make a dollar with?
>
> ==> color (combinatorics) <==
> An urn contains n balls of different colors. Randomly select a pair, repaint
>
> ==> full (combinatorics) <==
> Consider a string that contains all substrings of length n. For example,
>
> ==> gossip (combinatorics) <==
> n people each know a different piece of gossip. They can telephone each other
>
> ==> grid.dissection (combinatorics) <==
> How many (possibly overlapping) squares are in an mxn grid? Assume that all
>
> ==> permutation (combinatorics) <==
> Compute the nth permutation of k numbers (or objects).
>
> ==> subsets (combinatorics) <==
> Out of the set of integers 1,...,100 you are given ten different
>
> ==> transitions (combinatorics) <==
> How many n-bit binary strings (0/1) have exactly k transitions
>
> ==> contests/games.magazine (competition/part1) <==
> What are the best answers to various contests run by _Games_ magazine?
>
> ==> contests/national.puzzle/npc.1993 (competition/part1) <==
> What are the solutions to the Games magazine 1993 National Puzzle Contest?
>
> ==> games/bridge (competition/part1) <==
> Are there any programs for solving double-dummy Bridge?
>
> ==> games/chess/knight.control (competition/part1) <==
> How many knights does it take to attack or control the board?
>
> ==> games/chess/knight.most (competition/part1) <==
> What is the maximum number of knights that can be put on n x n chessboard
>
> ==> games/chess/knight.tour (competition/part1) <==
> For what size boards are knight tours possible?
>
> ==> games/chess/mutual.stalemate (competition/part1) <==
> What's the minimal number of pieces in a legal mutual stalemate?
>
> ==> games/chess/queen.control (competition/part1) <==
> How many queens does it take to attack or control the board?
>
> ==> games/chess/queen.most (competition/part1) <==
> How many non-mutually-attacking queens can be placed on various sized boards?
>
> ==> games/chess/queens (competition/part1) <==
> How many ways can eight queens be placed so that they control the board?
>
> ==> games/chess/rook.paths (competition/part1) <==
> How many non-overlapping paths can a rook take from one corner to the opposite
>
> ==> games/chess/size.of.game.tree (competition/part1) <==
> How many different positions are there in the game tree of chess?
>
> ==> games/cigarettes (competition/part1) <==
> The game of cigarettes is played as follows:
>
> ==> games/connect.four (competition/part1) <==
> Is there a winning strategy for Connect Four?
>
> ==> games/craps (competition/part1) <==
> What are the odds in craps?
>
> ==> games/crosswords (competition/part1) <==
> Are there programs to make crosswords? What are the rules for cluing cryptic
>
> ==> games/cube (competition/part2) <==
> What are some games involving cubes?
>
> ==> games/go-moku (competition/part2) <==
> For a game of k in a row on an n x n board, for what values of k and n is
>
> ==> games/hi-q (competition/part2) <==
> What is the quickest solution of the game Hi-Q (also called Solitaire)?
>
> ==> games/jeopardy (competition/part2) <==
> What are the highest, lowest, and most different scores contestants
>
> ==> games/nim (competition/part2) <==
> Place 10 piles of 10 $1 bills in a row. A valid move is to reduce
>
> ==> games/online/online.scrabble (competition/part2) <==
> How can I play Scrabble online on the Internet?
>
> ==> games/online/unlimited.adventures (competition/part2) <==
> Where can I find information about unlimited adventures?
>
> ==> games/othello (competition/part2) <==
> How good are computers at Othello?
>
> ==> games/pc/best (competition/part2) <==
> What are the best PC games?
>
> ==> games/pc/reviews (competition/part2) <==
> Are reviews of PC games available online?
>
> ==> games/pc/solutions (competition/part2) <==
> What are the solutions to various popular PC games?
>
> ==> games/poker.face.up (competition/part2) <==
> In Face-Up Poker, two players each select five cards from a face-up deck,
>
> ==> games/risk (competition/part2) <==
> What are the odds when tossing dice in Risk?
>
> ==> games/rubiks/rubiks.clock (competition/part2) <==
> How do you quickly solve Rubik's clock?
>
> ==> games/rubiks/rubiks.cube (competition/part3) <==
>
>
>
>
> What is the solution to Soma Cubes?
>
> ==> games/square-1 (competition/part3) <==
>
>
>
>
>
>
>
>
> ==> tests/math/putnam/putnam.1987 (competition/part4) <==
>
>
> Problem A-1
>
> ==> tests/math/putnam/putnam.1992 (competition/part5) <==
> Problem A1
>
> ==> Beale (cryptology) <==
> What are the Beale ciphers?
>
> ==> Feynman (cryptology) <==
> What are the Feynman ciphers?
>
> ==> Voynich (cryptology) <==
> What are the Voynich ciphers?
>
> ==> swiss.colony (cryptology) <==
> What are the 1987 Swiss Colony ciphers?
>
> ==> vcrplus (cryptology) <==
> What is the code used by VCR+?
>
> ==> allais (decision) <==
>
> N-Person Fair Division
>
> ==> dowry (decision) <==
> Sultan's Dowry
>
> ==> envelope (decision) <==
>
>
>
>
> Newcomb's Problem
>
> ==> prisoners (decision) <==
>
>
> Four glasses are placed upside down in the four corners of a square
>
> ==> stpetersburg (decision) <==
>
>
> Can three houses be connected to three utilities without the pipes crossing?
>
> ==> bear (geometry/part1) <==
> If a hunter goes out his front door, goes 50 miles south, then goes 50
>
> ==> bisector (geometry/part1) <==
> Prove if two angle bisectors of a triangle are equal, then the triangle is
>
> ==> calendar (geometry/part1) <==
> Build a calendar from two sets of cubes. On the first set, spell the
>
> ==> circles.and.triangles (geometry/part1) <==
> Find the radius of the inscribed and circumscribed circles for a triangle.
>
> ==> coloring/cheese.cube (geometry/part1) <==
> A cube of cheese is divided into 27 subcubes. A mouse starts at one
>
> ==> coloring/triominoes (geometry/part1) <==
> There is a chess board (of course with 64 squares). You are given 21
>
> ==> construction/4.triangles.6.lines (geometry/part1) <==
> Can you construct 4 equilateral triangles with 6 toothpicks?
>
> ==> construction/5.lines.with.4.points (geometry/part1) <==
> Arrange 10 points so that they form 5 rows of 4 each.
>
> ==> construction/square.with.compass (geometry/part1) <==
> Construct a square with only a compass and a straight edge.
>
> ==> corner (geometry/part1) <==
> A hallway of width A turns through 90 degrees into a hallway of width
>
> ==> cover.earth (geometry/part1) <==
> A thin membrane covers the surface of the (spherical) earth. One
>
> ==> cycle.polynomial (geometry/part1) <==
> What are the cycle polynomials for the Platonic solids?
>
> ==> dissections/disk (geometry/part1) <==
> Can a disk be cut into similar pieces without point symmetry about the
>
> ==> dissections/hexagon (geometry/part1) <==
> Divide the hexagon into:
>
> ==> dissections/largest.circle (geometry/part1) <==
> What is the largest circle that can be assembled from two semicircles cut from
>
> ==> dissections/square.70 (geometry/part1) <==
> Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 square be dissected into
>
> ==> dissections/square.five (geometry/part1) <==
> Can you dissect a square into 5 parts of equal area with just a straight edge?
>
> ==> dissections/tesseract (geometry/part1) <==
> If you suspend a cube by one corner and slice it in half with a
>
> ==> duck.and.fox (geometry/part1) <==
> A duck is swimming about in a circular pond. A ravenous fox (who cannot
>
> ==> earth.band (geometry/part1) <==
> How much will a band around the equator rise above the surface if it is
>
> ==> fence (geometry/part1) <==
> A farmer wishes to enclose the maximum possible area with 100 meters of fence.
>
> ==> ham.sandwich (geometry/part1) <==
> Consider a ham sandwich, consisting of two pieces of bread and one of
>
> ==> hike (geometry/part1) <==
> You are hiking in a half-planar woods, exactly 1 mile from the edge,
>
> ==> hole.in.sphere (geometry/part1) <==
> Old Boniface he took his cheer,
>
> ==> hypercube (geometry/part1) <==
> How many vertices, edges, faces, etc. does a hypercube have?
>
> ==> kissing.number (geometry/part1) <==
> How many n-dimensional unit spheres can be packed around one unit sphere?
>
> ==> konigsberg (geometry/part1) <==
> Can you draw a line through each edge on the diagram below without crossing
>
> ==> ladders (geometry/part1) <==
> Two ladders form a rough X in an alley. The ladders are 11 and 13 meters
>
> ==> lattice/area (geometry/part1) <==
> Prove that the area of a triangle formed by three lattice points is integer/2.
>
> ==> lattice/equilateral (geometry/part1) <==
> Can an equlateral triangle have vertices at integer lattice points?
>
> ==> manhole.cover (geometry/part1) <==
> Why is a manhole cover round?
>
> ==> pentomino (geometry/part1) <==
> Arrange pentominos in 3x20, 4x15, 5x12, 6x10, 2x3x10, 2x5x6 and 3x4x5 forms.
>
> ==> points.in.sphere (geometry/part1) <==
> What is the expected distance between two random points inside a sphere?
>
> ==> points.on.sphere (geometry/part1) <==
> What are the odds that n random points on a sphere lie in the same hemisphere?
>
> ==> revolutions (geometry/part1) <==
> A circle with radius 1 rolls without slipping once around a circle with radius
>
> ==> rotation (geometry/part1) <==
> What is the smallest rotation that returns an object to its original state?
>
> ==> shephard.piano (geometry/part1) <==
> What's the maximum area shape that will fit around a right-angle corner?
>
> ==> smuggler (geometry/part1) <==
> Somewhere on the high sees smuggler S is attempting, without much
>
> ==> spiral (geometry/part1) <==
> How far must one travel to reach the North Pole if one starts from the
>
> ==> table.in.corner (geometry/part1) <==
> Put a round table into a (perpendicular) corner so that the table top
>
> ==> tetrahedron (geometry/part1) <==
> Suppose you have a sphere of radius R and you have four planes that are
>
> ==> tiling/count.1x2 (geometry/part1) <==
> Count the ways to tile an MxN rectangle with 1x2 dominos.
>
> ==> tiling/rational.sides (geometry/part1) <==
> A rectangular region R is divided into rectangular areas. Show that if
>
> ==> tiling/rectangles.with.squares (geometry/part2) <==
>
>
>
>
> Can six 1x2x4 blocks be arranged so that each block touches n others, for all n?
>
> ==> trigonometry/euclidean.numbers (geometry/part2) <==
>
>
> AEFHIKLMNTVWXYZ BCDGJOPQRSU
>
> ==> group.01a (group) <==
> 147 0235689
>
> ==> group.02 (group) <==
> ABEHIKMNOTXZ CDFGJLPQRSUVWY
>
> ==> group.03 (group) <==
>
>
> CEFGHIJKLMNSTUVWXYZ ADOPQR B
>
> ==> group.06 (group) <==
> BCEGKMQSW DFHIJLNOPRTUVXYZ
>
> ==> group.07 (group) <==
> CDEFLOPTZ ABGHIJKMNQRSUVWXY
>
> ==> group.08 (group) <==
> COS ABDEFGHIJKLMNPQRTUVWXYZ
>
> ==> group.09 (group) <==
> CDILMVX ABEFGHJKNOPQRSTUWYZ
>
> ==> group.10 (group) <==
> AHIMOTUVWXY BCDEFGJKLNPQRSZ
>
> ==> group.11 (group) <==
> BCDIJLMNOPQRSUVWZ AEFGHKTXY
>
> ==> group.12 (group) <==
> COPSUVWXZ ABDEFGHIJKLMNQRTY
>
> ==> group.13 (group) <==
> BCDEHIKOX AFGJLMNPQRSTUVWYZ
>
> ==> group.14 (group) <==
> EHIS MOT ABCDFGJKLNPQRUVWXYZ
>
> ==> group.15 (group) <==
> HIOX ABCDEFGJKLMNPQRSTUVWYZ
>
> ==> group.16 (group) <==
> IJ ABCDEFGHKLMNOPQRSTUVWXYZ
>
> ==> group.17 (group) <==
> J BDFHIKLT GPQY ACEMNORSUVWXZ
>
> ==> handshake (induction) <==
> A married couple organizes a party. They only invite other married
>
> ==> hanoi (induction) <==
> Is there an algorithm for solving the Hanoi tower puzzle for any number
>
> ==> n-sphere (induction) <==
> With what odds do three random points on an n-sphere form an acute triangle?
>
> ==> paradox (induction) <==
> Is there a non-trivial property that holds for the first 10,000 positive
>
> ==> party (induction) <==
> You're at a party. Any two (different) people at the party have exactly one
>
> ==> roll (induction) <==
> An ordinary die is thrown until the running total of the throws first
>
> ==> takeover (induction) <==
> After graduating from college, you have taken an important managing position
>
> ==> close.antonyms (language/part1) <==
> What words are similar to their antonyms in other langauges?
>
> ==> dutch/dutch.record (language/part1) <==
> What are some Dutch words with unusual properties?
>
> ==> english/equations (language/part1) <==
> 1 = E. on a C.
>
> ==> english/etymology/acronym (language/part1) <==
> What acronyms have become common words or are otherwise interesting?
>
> ==> english/etymology/fossil (language/part1) <==
> What are some examples of idioms that include obsolete words?
>
> ==> english/etymology/portmanteau (language/part1) <==
> What are some words formed by combining together parts of other words?
>
> ==> english/frequency (language/part1) <==
> In the English language, what are the most frequently appearing:
>
> ==> english/idioms (language/part1) <==
> List some idioms that say the opposite of what they mean.
>
> ==> english/less.ness (language/part1) <==
> Find a word that forms two other words, unrelated in meaning, when "less"
>
> ==> english/letter.rebus (language/part1) <==
> Define the letters of the alphabet using self-referential common phrases (e.g.,
>
> ==> english/malaprop (language/part1) <==
> List some phrases with the same meaning that differ by one sound.
>
> ==> english/piglatin (language/part1) <==
> What words in pig latin also are words?
>
> ==> english/pleonasm (language/part1) <==
> What are some redundant terms that occur frequently (like "ABM missile")?
>
> ==> english/plurals/collision (language/part1) <==
> Two words, spelled and pronounced differently, have plurals spelled
>
> ==> english/plurals/doubtful.number (language/part1) <==
> A little word of doubtful number,
>
> ==> english/plurals/drop.terminal (language/part1) <==
> What words have their plurals formed by dropping the final letter?
>
> ==> english/plurals/endings (language/part1) <==
> List a plural ending with each letter of the alphabet.
>
> ==> english/plurals/man (language/part1) <==
> Words ending with "man" make their plurals by adding "s".
>
> ==> english/plurals/switch.first (language/part1) <==
> What plural is formed by switching the first two letters?
>
> ==> english/potable.color (language/part1) <==
> Find words that are both beverages and colors.
>
> ==> english/pronunciation/autonym (language/part1) <==
> What is the longest word whose phonetic and normal spellings are the same?
>
> ==> english/pronunciation/homograph/different.pronunciation (language/part1) <==
> What sequence of letters has the most different pronunciations?
>
> ==> english/pronunciation/homograph/homographs (language/part1) <==
> List some homographs (words spelled the same but pronounced differently)
>
> ==> english/pronunciation/homophone/homophones.alphabet (language/part1) <==
> Homophones can be confusing when used to exemplify a letter. For example,
>
> ==> english/pronunciation/homophone/homophones.letter (language/part1) <==
> For each letter, list homophones that differ by that letter.
>
> ==> english/pronunciation/homophone/homophones.most (language/part1) <==
> What words have four or more spellings that sound alike?
>
> ==> english/pronunciation/homophone/trivial (language/part2) <==
>
> List some oronyms (phrases or sentences that can be read in two ways
>
> ==> english/pronunciation/phonetic.letters (language/part2) <==
> What does "FUNEX" mean?
>
> ==> english/pronunciation/rhyme (language/part2) <==
>
>
>
>
>
>
>
>
>
>
> Is there a sentence that remains a sentence when all its words are beheaded?
>
> ==> english/sentences/charades (language/part2) <==
>
>
>
>
>
>
>
>
>
>
>
>
>
> What are some long near palindromes, i.e., words that except for one
>
> ==> english/spelling/operations.on.words/deletion (language/part2) <==
>
>
>
>
>
> What exceptional words turn into other words by transposition of letters?
>
> ==> english/spelling/operations.on.words/words.within.words (language/part3) <==
>
> What are some long palindromes?
>
> ==> english/spelling/sets.of.words/ladder (language/part3) <==
>
>
>
>
>
>
>
> List some exceptional spoonerisms.
>
> ==> english/synonyms/ambiguous (language/part3) <==
> What word in the English language is the most ambiguous?
>
> ==> english/synonyms/antonym (language/part3) <==
>
>
>
> What words have two different synonymous meanings?
>
> ==> finnish/finnish.plural (language/part3) <==
>
>
> List some French palindromes.
>
> ==> french/french.record (language/part4) <==
>
> List some German palindromes.
>
> ==> german/german.record (language/part4) <==
>
> What are some Italian words with unusual properties?
>
> ==> multi.palindromes (language/part5) <==
> List some multi-lingual palindromes.
>
> ==> norwegian/norwegian.record (language/part5) <==
> What are some Norwegian words with unusual properties?
>
> ==> repeated.word (language/part5) <==
>
> What are some Swedish words with unusual properties?
>
> ==> synonymous.reversals (language/part5) <==
>
> In any language, what word contains the same vowel repeated four times in a row?
>
> ==> 29 (logic/part1) <==
>
>
>
>
>
> An Arab sheikh tells his two sons to race their camels to a distant
>
> ==> centrifuge (logic/part1) <==
>
>
>
>
>
>
>
>
>
> In first-order logic, find a predicate P(x) which means "x is a power of 10."
>
> ==> hundred (logic/part1) <==
>
>
>
>
>
>
>
>
> Who makes it, has no need of it. Who buys it, has no use for it. Who
>
> ==> river.crossing (logic/part2) <==
>
>
>
>
>
>
>
>
>
>
>
On Wednesday, August 18, 1993 at 8:04:18 PM UTC+14, Chris Cole wrote:
> Archive-name: puzzles/archive/Instructions
> Version: 4
>
>
> Instructions for Accessing rec.puzzles Archive
>
> INTRODUCTION
>
> Below is a list of puzzles, categorized by subject area. Each puzzle
> includes a solution, compiled from various sources, which is supposed
> to be definitive.
>
>
> To request a puzzle, send a message to archive...@questrel.com like:
>
> return_address your_name@your_site.your_domain
> send requested_puzzle_name
>
> For example, if your net address is "mic...@disneyland.com", to request
> "decision/allais.p", send the message:
>
> return_address mic...@disneyland.com
> send allais
>
> To request multiple puzzles, use several "send" lines in a message.
> Please refrain from requesting the entire archive via email. Use FTP.
>
> FTP
>
> The archive has been posted to news.answers and rec.answers, which are
> archived in the periodic posting archive on rtfm.mit.edu in the
> anonymous ftp directory /pub/usenet.
>
> Other archives are:
>
> ftp.cs.ruu.nl [131.211.80.17] in the anonymous ftp
> directory /pub/NEWS.ANSWERS (also accessible via mail
> server requests to mail-...@cs.ruu.nl)
> cnam.cnam.fr [192.33.159.6] in the anonymous ftp directory /pub/FAQ
> ftp.uu.net [137.39.1.9 or 192.48.96.9] in the anonymous ftp
> directory /usenet
> ftp.win.tue.nl [131.155.70.100] in the anonymous ftp directory
> /pub/usenet/news.answers
> grasp1.univ-lyon1.fr [134.214.100.25] in the anonymous ftp
> directory /pub/faq (also accessible via mail server
> requests to list...@grasp1.univ-lyon1.fr), which is
> best used by EASInet sites and sites in France that do
> not have better connectivity to cnam.cnam.fr (e.g.
> Lyon, Grenoble)
>
> Note that the periodic posting archives on rtfm.mit.edu are also
> accessible via Prospero and WAIS (the database name is "usenet" on port
> 210).
>
> CREDIT
>
> The archive is NOT the original work of the editor (just in case you were
> wondering :^).
>
> In keeping with the general net practice on FAQ's, I do not as a rule assign
> credit for solutions. There are many reasons for this:
> 1. The archive is about the answers to the questions, not about assigning
> credit.
> 2. Many people, in providing free answers to the net, do not have the time
> to cite their sources.
> 3. I cut and paste freely from several people's solutions in most cases
> to come up with as complete an answer as possible.
> 4. I use sources other than postings.
> 5. I am neither qualified nor motivated to assign credit.
>
> However, I do whenever possible put bibliographies in archive entries, and
> I see the inclusion of the net addresses of interested parties as a
> logical extension of this practice. In particular, if you wrote a
> program to solve a problem and posted the source code of the program,
> you are presumed to be interested in corresponding with others about
> the problem. So, please let me know the entries you would like to be
> listed in and I will be happy to oblige.
>
> Address corrections or comments to archive...@questrel.com.
>
> INDEX
>
> ==> bicycle (analysis) <==
> A boy, a girl and a dog go for a 10 mile walk. The boy and girl can
>
> ==> boy.girl.dog (analysis) <==
> A boy, a girl and a dog are standing together on a long, straight road.
>
> ==> bugs (analysis) <==
> Four bugs are placed at the corners of a square. Each bug walks always
>
> ==> c.infinity (analysis) <==
> What function is zero at zero, strictly positive elsewhere, infinitely
>
> ==> cache (analysis) <==
> Cache and Ferry (How far can a truck go in a desert?)
>
> ==> calculate.pi (analysis) <==
> How can I calculate many digits of pi?
>
> ==> cats.and.rats (analysis) <==
> If 6 cats can kill 6 rats in 6 minutes, how many cats does it take to
>
> ==> dog (analysis) <==
> A body of soldiers form a 50m-by-50m square ABCD on the parade ground.
>
> ==> e.and.pi (analysis) <==
> Without finding their numerical values, which is greater, e^(pi) or (pi)^e?
>
> ==> functional/distributed (analysis) <==
> Find all f: R -> R, f not identically zero, such that
>
> ==> functional/linear (analysis) <==
> Suppose f is non-decreasing with
>
> ==> integral (analysis) <==
> If f is integrable on (0,inf) and differentiable at 0, and a > 0, and:
>
> ==> irrational.stamp (analysis) <==
> You have an ink stamp which is so amazingly precise that, when inked
>
> ==> minimum.time (analysis) <==
> N people can walk or drive in a two-seater to go from city A to city B. What
>
> ==> particle (analysis) <==
> What is the longest time that a particle can take in travelling between two
>
> ==> period (analysis) <==
> What is the least possible integral period of the sum of functions
>
> ==> rubberband (analysis) <==
> A bug walks down a rubber band which is attached to a wall at one end and a car
>
> ==> sequence (analysis) <==
> Show that in the sequence: x, 2x, 3x, .... (n-1)x (x can be any real number)
>
> ==> snow (analysis) <==
> Snow starts falling before noon on a cold December day. At noon a
>
> ==> tower (analysis) <==
> R = N ^ (N ^ (N ^ ...)). What is the maximum N>0 that will yield a finite R?
>
> ==> 7-11 (arithmetic/part1) <==
> A customer at a 7-11 store selected four items to buy, and was told
>
> ==> arithmetic.progression (arithmetic/part1) <==
> Is there an arithmetic progression of 20 or more primes?
>
> ==> clock/day.of.week (arithmetic/part1) <==
> It's restful sitting in Tom's cosy den, talking quietly and sipping
>
> ==> clock/palindromic (arithmetic/part1) <==
> How many times per day does a digital clock display a palindromic number?
>
> ==> clock/reversible (arithmetic/part1) <==
> How many times per day can the hour and minute hands on an analog clock switch
>
> ==> clock/right.angle (arithmetic/part1) <==
> How many times per day do the hour and minute hands of a clock form a
>
> ==> clock/thirds (arithmetic/part1) <==
> Do the 3 hands on a clock ever divide the face of the clock into 3
>
> ==> consecutive.composites (arithmetic/part1) <==
> Are there 10,000 consecutive non-prime numbers?
>
> ==> consecutive.product (arithmetic/part1) <==
> Prove that the product of three or more consecutive positive integers cannot
>
> ==> consecutive.sums (arithmetic/part1) <==
> Find all series of consecutive positive integers whose sum is exactly 10,000.
>
> ==> conway (arithmetic/part1) <==
> Describe the sequence a(1)=a(2)=1, a(n) = a(a(n-1)) + a(n-a(n-1)) for n>2.
>
> ==> digits/6.and.7 (arithmetic/part1) <==
> Does every number which is not divisible by 5 have a multiple whose
>
> ==> digits/all.ones (arithmetic/part1) <==
> Prove that some multiple of any integer ending in 3 contains all 1s.
>
> ==> digits/arabian (arithmetic/part1) <==
> What is the Arabian Nights factorial, the number x such that x! has 1001
>
> ==> digits/circular (arithmetic/part1) <==
> What 6 digit number, with 6 different digits, when multiplied by all integers
>
> ==> digits/divisible (arithmetic/part1) <==
> Find the least number using 0-9 exactly once that is evenly divisible by each
>
> ==> digits/equations/123456789 (arithmetic/part1) <==
> In how many ways can "." be replaced with "+", "-", or "" (concatenate) in
>
> ==> digits/equations/1992 (arithmetic/part1) <==
> 1 = -1+9-9+2. Extend this list to 2 through 100 on the left side of
>
> ==> digits/equations/24 (arithmetic/part1) <==
> Form an expression that evaluates to 24 that contains two 3's, two 7's,
>
> ==> digits/equations/383 (arithmetic/part1) <==
> Make 383 out of 1,2,25,50,75,100 using +,-,*,/.
>
> ==> digits/equations/find (arithmetic/part1) <==
> Write a program for finding expressions built out of given numbers and using
>
> ==> digits/extreme.products (arithmetic/part1) <==
> What are the extremal products of three three-digit numbers using digits 1-9?
>
> ==> digits/labels (arithmetic/part1) <==
> You have an arbitrary number of model kits (which you assemble for
>
> ==> digits/least.significant/factorial (arithmetic/part1) <==
> What is the least significant non-zero digit in the decimal expansion of n!?
>
> ==> digits/least.significant/tower.of.power (arithmetic/part1) <==
> What are the least significant digits of 9^(8^(7^(6^(5^(4^(3^(2^1))))))) ?
>
> ==> digits/most.significant/googol (arithmetic/part1) <==
> What digits does googol! start with?
>
> ==> digits/most.significant/powers (arithmetic/part1) <==
> What is the probability that 2^N begins with the digits 603245?
>
> ==> digits/nine.digits (arithmetic/part1) <==
> Form a number using 0-9 once with its first n digits divisible by n.
>
> ==> digits/palindrome (arithmetic/part1) <==
> Does the series formed by adding a number to its reversal always end in
>
> ==> digits/palintiples (arithmetic/part1) <==
> Find all numbers that are multiples of their reversals.
>
> ==> digits/power.two (arithmetic/part1) <==
> Prove that for any 9-digit number (base 10) there is an integral power
>
> ==> digits/prime/101 (arithmetic/part1) <==
> How many primes are in the sequence 101, 10101, 1010101, ...?
>
> ==> digits/prime/all.prefix (arithmetic/part1) <==
> What is the longest prime whose every proper prefix is a prime?
>
> ==> digits/prime/change.one (arithmetic/part1) <==
> What is the smallest number that cannot be made prime by changing a single
>
> ==> digits/prime/prefix.one (arithmetic/part1) <==
> 2 is prime, but 12, 22, ..., 92 are not. Similarly, 5 is prime
>
> ==> digits/reverse (arithmetic/part1) <==
> Is there an integer that has its digits reversed after dividing it by 2?
>
> ==> digits/rotate (arithmetic/part1) <==
> Find integers where multiplying them by single digits rotates their digits
>
> ==> digits/sesqui (arithmetic/part1) <==
> Find the least number where moving the first digit to the end multiplies by 1.5.
>
> ==> digits/squares/change.leading (arithmetic/part1) <==
> What squares remain squares when their leading digits are incremented?
>
> ==> digits/squares/length.22 (arithmetic/part1) <==
> Is it possible to form two numbers A and B from 22 digits such that
>
> ==> digits/squares/length.9 (arithmetic/part1) <==
> Is it possible to make a number and its square, using the digits from 1
>
> ==> digits/squares/three.digits (arithmetic/part2) <==
> What squares consist entirely of three digits (e.g., 1, 4, and 9)?
>
> ==> digits/squares/twin (arithmetic/part2) <==
> Let a twin be a number formed by writing the same number twice,
>
> ==> digits/sum.of.digits (arithmetic/part2) <==
> Find sod ( sod ( sod (4444 ^ 4444 ) ) ).
>
> ==> digits/zeros/million (arithmetic/part2) <==
> How many zeros occur in the numbers from 1 to 1,000,000?
>
> ==> digits/zeros/trailing (arithmetic/part2) <==
> How many trailing zeros are in the decimal expansion of n!?
>
> ==> magic.squares (arithmetic/part2) <==
> Are there large squares, containing only consecutive integers, all of whose
>
> ==> pell (arithmetic/part2) <==
> Find integer solutions to x^2 - 92y^2 = 1.
>
> ==> subset (arithmetic/part2) <==
> Prove that all sets of n integers contain a subset whose sum is divisible by n.
>
> ==> sum.of.cubes (arithmetic/part2) <==
> Find two fractions whose cubes total 6.
>
> ==> sums.of.powers (arithmetic/part2) <==
> Partition 1,2,3,...,16 into two equal sets, such that the sums of the
>
> ==> tests.for.divisibility/eleven (arithmetic/part2) <==
> What is the test to see if a number is divisible by eleven?
>
> ==> tests.for.divisibility/nine (arithmetic/part2) <==
> What is the test to see if a number is divisible by nine?
>
> ==> tests.for.divisibility/seven (arithmetic/part2) <==
> What is the test to see if a number is divisible by seven?
>
> ==> tests.for.divisibility/three (arithmetic/part2) <==
> What is the test to see if a number is divisible by three?
>
> ==> alphabet.blocks (combinatorics) <==
> What is the minimum number of dice painted with one letter on all six sides
>
> ==> coinage/combinations (combinatorics) <==
> Assuming you have enough coins of 1, 5, 10, 25 and 50 cents, how many
>
> ==> coinage/dimes (combinatorics) <==
> "Dad wants one-cent, two-cent, three-cent, five-cent, and ten-cent
>
> ==> coinage/impossible (combinatorics) <==
> What is the smallest number of coins that you can't make a dollar with?
>
> ==> color (combinatorics) <==
> An urn contains n balls of different colors. Randomly select a pair, repaint
>
> ==> full (combinatorics) <==
> Consider a string that contains all substrings of length n. For example,
>
> ==> gossip (combinatorics) <==
> n people each know a different piece of gossip. They can telephone each other
>
> ==> grid.dissection (combinatorics) <==
> How many (possibly overlapping) squares are in an mxn grid? Assume that all
>
> ==> permutation (combinatorics) <==
> Compute the nth permutation of k numbers (or objects).
>
> ==> subsets (combinatorics) <==
> Out of the set of integers 1,...,100 you are given ten different
>
> ==> transitions (combinatorics) <==
> How many n-bit binary strings (0/1) have exactly k transitions
>
> ==> contests/games.magazine (competition/part1) <==
> What are the best answers to various contests run by _Games_ magazine?
>
> ==> contests/national.puzzle/npc.1993 (competition/part1) <==
> What are the solutions to the Games magazine 1993 National Puzzle Contest?
>
> ==> games/bridge (competition/part1) <==
> Are there any programs for solving double-dummy Bridge?
>
> ==> games/chess/knight.control (competition/part1) <==
> How many knights does it take to attack or control the board?
>
> ==> games/chess/knight.most (competition/part1) <==
> What is the maximum number of knights that can be put on n x n chessboard
>
> ==> games/chess/knight.tour (competition/part1) <==
> For what size boards are knight tours possible?
>
> ==> games/chess/mutual.stalemate (competition/part1) <==
> What's the minimal number of pieces in a legal mutual stalemate?
>
> ==> games/chess/queen.control (competition/part1) <==
> How many queens does it take to attack or control the board?
>
> ==> games/chess/queen.most (competition/part1) <==
> How many non-mutually-attacking queens can be placed on various sized boards?
>
> ==> games/chess/queens (competition/part1) <==
> How many ways can eight queens be placed so that they control the board?
>
> ==> games/chess/rook.paths (competition/part1) <==
> How many non-overlapping paths can a rook take from one corner to the opposite
>
> ==> games/chess/size.of.game.tree (competition/part1) <==
> How many different positions are there in the game tree of chess?
>
> ==> games/cigarettes (competition/part1) <==
> The game of cigarettes is played as follows:
>
> ==> games/connect.four (competition/part1) <==
> Is there a winning strategy for Connect Four?
>
> ==> games/craps (competition/part1) <==
> What are the odds in craps?
>
> ==> games/crosswords (competition/part1) <==
> Are there programs to make crosswords? What are the rules for cluing cryptic
>
> ==> games/cube (competition/part2) <==
> What are some games involving cubes?
>
> ==> games/go-moku (competition/part2) <==
> For a game of k in a row on an n x n board, for what values of k and n is
>
> ==> games/hi-q (competition/part2) <==
> What is the quickest solution of the game Hi-Q (also called Solitaire)?
>
> ==> games/jeopardy (competition/part2) <==
> What are the highest, lowest, and most different scores contestants
>
> ==> games/nim (competition/part2) <==
> Place 10 piles of 10 $1 bills in a row. A valid move is to reduce
>
> ==> games/online/online.scrabble (competition/part2) <==
> How can I play Scrabble online on the Internet?
>
> ==> games/online/unlimited.adventures (competition/part2) <==
> Where can I find information about unlimited adventures?
>
> ==> games/othello (competition/part2) <==
> How good are computers at Othello?
>
> ==> games/pc/best (competition/part2) <==
> What are the best PC games?
>
> ==> games/pc/reviews (competition/part2) <==
> Are reviews of PC games available online?
>
> ==> games/pc/solutions (competition/part2) <==
> What are the solutions to various popular PC games?
>
> ==> games/poker.face.up (competition/part2) <==
> In Face-Up Poker, two players each select five cards from a face-up deck,
>
> ==> games/risk (competition/part2) <==
> What are the odds when tossing dice in Risk?
>
> ==> games/rubiks/rubiks.clock (competition/part2) <==
> How do you quickly solve Rubik's clock?
>
> ==> games/rubiks/rubiks.cube (competition/part3) <==
>
>
>
>
> What is the solution to Soma Cubes?
>
> ==> games/square-1 (competition/part3) <==
>
>
>
>
>
>
>
>
> ==> tests/math/putnam/putnam.1987 (competition/part4) <==
>
>
> Problem A-1
>
> ==> tests/math/putnam/putnam.1992 (competition/part5) <==
> Problem A1
>
> ==> Beale (cryptology) <==
> What are the Beale ciphers?
>
> ==> Feynman (cryptology) <==
> What are the Feynman ciphers?
>
> ==> Voynich (cryptology) <==
> What are the Voynich ciphers?
>
> ==> swiss.colony (cryptology) <==
> What are the 1987 Swiss Colony ciphers?
>
> ==> vcrplus (cryptology) <==
> What is the code used by VCR+?
>
> ==> allais (decision) <==
>
> N-Person Fair Division
>
> ==> dowry (decision) <==
> Sultan's Dowry
>
> ==> envelope (decision) <==
>
>
>
>
> Newcomb's Problem
>
> ==> prisoners (decision) <==
>
>
> Four glasses are placed upside down in the four corners of a square
>
> ==> stpetersburg (decision) <==
>
>
> Can three houses be connected to three utilities without the pipes crossing?
>
> ==> bear (geometry/part1) <==
> If a hunter goes out his front door, goes 50 miles south, then goes 50
>
> ==> bisector (geometry/part1) <==
> Prove if two angle bisectors of a triangle are equal, then the triangle is
>
> ==> calendar (geometry/part1) <==
> Build a calendar from two sets of cubes. On the first set, spell the
>
> ==> circles.and.triangles (geometry/part1) <==
> Find the radius of the inscribed and circumscribed circles for a triangle.
>
> ==> coloring/cheese.cube (geometry/part1) <==
> A cube of cheese is divided into 27 subcubes. A mouse starts at one
>
> ==> coloring/triominoes (geometry/part1) <==
> There is a chess board (of course with 64 squares). You are given 21
>
> ==> construction/4.triangles.6.lines (geometry/part1) <==
> Can you construct 4 equilateral triangles with 6 toothpicks?
>
> ==> construction/5.lines.with.4.points (geometry/part1) <==
> Arrange 10 points so that they form 5 rows of 4 each.
>
> ==> construction/square.with.compass (geometry/part1) <==
> Construct a square with only a compass and a straight edge.
>
> ==> corner (geometry/part1) <==
> A hallway of width A turns through 90 degrees into a hallway of width
>
> ==> cover.earth (geometry/part1) <==
> A thin membrane covers the surface of the (spherical) earth. One
>
> ==> cycle.polynomial (geometry/part1) <==
> What are the cycle polynomials for the Platonic solids?
>
> ==> dissections/disk (geometry/part1) <==
> Can a disk be cut into similar pieces without point symmetry about the
>
> ==> dissections/hexagon (geometry/part1) <==
> Divide the hexagon into:
>
> ==> dissections/largest.circle (geometry/part1) <==
> What is the largest circle that can be assembled from two semicircles cut from
>
> ==> dissections/square.70 (geometry/part1) <==
> Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 square be dissected into
>
> ==> dissections/square.five (geometry/part1) <==
> Can you dissect a square into 5 parts of equal area with just a straight edge?
>
> ==> dissections/tesseract (geometry/part1) <==
> If you suspend a cube by one corner and slice it in half with a
>
> ==> duck.and.fox (geometry/part1) <==
> A duck is swimming about in a circular pond. A ravenous fox (who cannot
>
> ==> earth.band (geometry/part1) <==
> How much will a band around the equator rise above the surface if it is
>
> ==> fence (geometry/part1) <==
> A farmer wishes to enclose the maximum possible area with 100 meters of fence.
>
> ==> ham.sandwich (geometry/part1) <==
> Consider a ham sandwich, consisting of two pieces of bread and one of
>
> ==> hike (geometry/part1) <==
> You are hiking in a half-planar woods, exactly 1 mile from the edge,
>
> ==> hole.in.sphere (geometry/part1) <==
> Old Boniface he took his cheer,
>
> ==> hypercube (geometry/part1) <==
> How many vertices, edges, faces, etc. does a hypercube have?
>
> ==> kissing.number (geometry/part1) <==
> How many n-dimensional unit spheres can be packed around one unit sphere?
>
> ==> konigsberg (geometry/part1) <==
> Can you draw a line through each edge on the diagram below without crossing
>
> ==> ladders (geometry/part1) <==
> Two ladders form a rough X in an alley. The ladders are 11 and 13 meters
>
> ==> lattice/area (geometry/part1) <==
> Prove that the area of a triangle formed by three lattice points is integer/2.
>
> ==> lattice/equilateral (geometry/part1) <==
> Can an equlateral triangle have vertices at integer lattice points?
>
> ==> manhole.cover (geometry/part1) <==
> Why is a manhole cover round?
>
> ==> pentomino (geometry/part1) <==
> Arrange pentominos in 3x20, 4x15, 5x12, 6x10, 2x3x10, 2x5x6 and 3x4x5 forms.
>
> ==> points.in.sphere (geometry/part1) <==
> What is the expected distance between two random points inside a sphere?
>
> ==> points.on.sphere (geometry/part1) <==
> What are the odds that n random points on a sphere lie in the same hemisphere?
>
> ==> revolutions (geometry/part1) <==
> A circle with radius 1 rolls without slipping once around a circle with radius
>
> ==> rotation (geometry/part1) <==
> What is the smallest rotation that returns an object to its original state?
>
> ==> shephard.piano (geometry/part1) <==
> What's the maximum area shape that will fit around a right-angle corner?
>
> ==> smuggler (geometry/part1) <==
> Somewhere on the high sees smuggler S is attempting, without much
>
> ==> spiral (geometry/part1) <==
> How far must one travel to reach the North Pole if one starts from the
>
> ==> table.in.corner (geometry/part1) <==
> Put a round table into a (perpendicular) corner so that the table top
>
> ==> tetrahedron (geometry/part1) <==
> Suppose you have a sphere of radius R and you have four planes that are
>
> ==> tiling/count.1x2 (geometry/part1) <==
> Count the ways to tile an MxN rectangle with 1x2 dominos.
>
> ==> tiling/rational.sides (geometry/part1) <==
> A rectangular region R is divided into rectangular areas. Show that if
>
> ==> tiling/rectangles.with.squares (geometry/part2) <==
>
>
>
>
> Can six 1x2x4 blocks be arranged so that each block touches n others, for all n?
>
> ==> trigonometry/euclidean.numbers (geometry/part2) <==
>
>
> AEFHIKLMNTVWXYZ BCDGJOPQRSU
>
> ==> group.01a (group) <==
> 147 0235689
>
> ==> group.02 (group) <==
> ABEHIKMNOTXZ CDFGJLPQRSUVWY
>
> ==> group.03 (group) <==
>
>
> CEFGHIJKLMNSTUVWXYZ ADOPQR B
>
> ==> group.06 (group) <==
> BCEGKMQSW DFHIJLNOPRTUVXYZ
>
> ==> group.07 (group) <==
> CDEFLOPTZ ABGHIJKMNQRSUVWXY
>
> ==> group.08 (group) <==
> COS ABDEFGHIJKLMNPQRTUVWXYZ
>
> ==> group.09 (group) <==
> CDILMVX ABEFGHJKNOPQRSTUWYZ
>
> ==> group.10 (group) <==
> AHIMOTUVWXY BCDEFGJKLNPQRSZ
>
> ==> group.11 (group) <==
> BCDIJLMNOPQRSUVWZ AEFGHKTXY
>
> ==> group.12 (group) <==
> COPSUVWXZ ABDEFGHIJKLMNQRTY
>
> ==> group.13 (group) <==
> BCDEHIKOX AFGJLMNPQRSTUVWYZ
>
> ==> group.14 (group) <==
> EHIS MOT ABCDFGJKLNPQRUVWXYZ
>
> ==> group.15 (group) <==
> HIOX ABCDEFGJKLMNPQRSTUVWYZ
>
> ==> group.16 (group) <==
> IJ ABCDEFGHKLMNOPQRSTUVWXYZ
>
> ==> group.17 (group) <==
> J BDFHIKLT GPQY ACEMNORSUVWXZ
>
> ==> handshake (induction) <==
> A married couple organizes a party. They only invite other married
>
> ==> hanoi (induction) <==
> Is there an algorithm for solving the Hanoi tower puzzle for any number
>
> ==> n-sphere (induction) <==
> With what odds do three random points on an n-sphere form an acute triangle?
>
> ==> paradox (induction) <==
> Is there a non-trivial property that holds for the first 10,000 positive
>
> ==> party (induction) <==
> You're at a party. Any two (different) people at the party have exactly one
>
> ==> roll (induction) <==
> An ordinary die is thrown until the running total of the throws first
>
> ==> takeover (induction) <==
> After graduating from college, you have taken an important managing position
>
> ==> close.antonyms (language/part1) <==
> What words are similar to their antonyms in other langauges?
>
> ==> dutch/dutch.record (language/part1) <==
> What are some Dutch words with unusual properties?
>
> ==> english/equations (language/part1) <==
> 1 = E. on a C.
>
> ==> english/etymology/acronym (language/part1) <==
> What acronyms have become common words or are otherwise interesting?
>
> ==> english/etymology/fossil (language/part1) <==
> What are some examples of idioms that include obsolete words?
>
> ==> english/etymology/portmanteau (language/part1) <==
> What are some words formed by combining together parts of other words?
>
> ==> english/frequency (language/part1) <==
> In the English language, what are the most frequently appearing:
>
> ==> english/idioms (language/part1) <==
> List some idioms that say the opposite of what they mean.
>
> ==> english/less.ness (language/part1) <==
> Find a word that forms two other words, unrelated in meaning, when "less"
>
> ==> english/letter.rebus (language/part1) <==
> Define the letters of the alphabet using self-referential common phrases (e.g.,
>
> ==> english/malaprop (language/part1) <==
> List some phrases with the same meaning that differ by one sound.
>
> ==> english/piglatin (language/part1) <==
> What words in pig latin also are words?
>
> ==> english/pleonasm (language/part1) <==
> What are some redundant terms that occur frequently (like "ABM missile")?
>
> ==> english/plurals/collision (language/part1) <==
> Two words, spelled and pronounced differently, have plurals spelled
>
> ==> english/plurals/doubtful.number (language/part1) <==
> A little word of doubtful number,
>
> ==> english/plurals/drop.terminal (language/part1) <==
> What words have their plurals formed by dropping the final letter?
>
> ==> english/plurals/endings (language/part1) <==
> List a plural ending with each letter of the alphabet.
>
> ==> english/plurals/man (language/part1) <==
> Words ending with "man" make their plurals by adding "s".
>
> ==> english/plurals/switch.first (language/part1) <==
> What plural is formed by switching the first two letters?
>
> ==> english/potable.color (language/part1) <==
> Find words that are both beverages and colors.
>
> ==> english/pronunciation/autonym (language/part1) <==
> What is the longest word whose phonetic and normal spellings are the same?
>
> ==> english/pronunciation/homograph/different.pronunciation (language/part1) <==
> What sequence of letters has the most different pronunciations?
>
> ==> english/pronunciation/homograph/homographs (language/part1) <==
> List some homographs (words spelled the same but pronounced differently)
>
> ==> english/pronunciation/homophone/homophones.alphabet (language/part1) <==
> Homophones can be confusing when used to exemplify a letter. For example,
>
> ==> english/pronunciation/homophone/homophones.letter (language/part1) <==
> For each letter, list homophones that differ by that letter.
>
> ==> english/pronunciation/homophone/homophones.most (language/part1) <==
> What words have four or more spellings that sound alike?
>
> ==> english/pronunciation/homophone/trivial (language/part2) <==
>
> List some oronyms (phrases or sentences that can be read in two ways
>
> ==> english/pronunciation/phonetic.letters (language/part2) <==
> What does "FUNEX" mean?
>
> ==> english/pronunciation/rhyme (language/part2) <==
>
>
>
>
>
>
>
>
>
>
> Is there a sentence that remains a sentence when all its words are beheaded?
>
> ==> english/sentences/charades (language/part2) <==
>
>
>
>
>
>
>
>
>
>
>
>
>
> What are some long near palindromes, i.e., words that except for one
>
> ==> english/spelling/operations.on.words/deletion (language/part2) <==
>
>
>
>
>
> What exceptional words turn into other words by transposition of letters?
>
> ==> english/spelling/operations.on.words/words.within.words (language/part3) <==
>
> What are some long palindromes?
>
> ==> english/spelling/sets.of.words/ladder (language/part3) <==
>
>
>
>
>
>
>
> List some exceptional spoonerisms.
>
> ==> english/synonyms/ambiguous (language/part3) <==
> What word in the English language is the most ambiguous?
>
> ==> english/synonyms/antonym (language/part3) <==
>
>
>
> What words have two different synonymous meanings?
>
> ==> finnish/finnish.plural (language/part3) <==
>
>
> List some French palindromes.
>
> ==> french/french.record (language/part4) <==
>
> List some German palindromes.
>
> ==> german/german.record (language/part4) <==
>
> What are some Italian words with unusual properties?
>
> ==> multi.palindromes (language/part5) <==
> List some multi-lingual palindromes.
>
> ==> norwegian/norwegian.record (language/part5) <==
> What are some Norwegian words with unusual properties?
>
> ==> repeated.word (language/part5) <==
>
> What are some Swedish words with unusual properties?
>
> ==> synonymous.reversals (language/part5) <==
>
> In any language, what word contains the same vowel repeated four times in a row?
>
> ==> 29 (logic/part1) <==
>
>
>
>
>
> An Arab sheikh tells his two sons to race their camels to a distant
>
> ==> centrifuge (logic/part1) <==
>
>
>
>
>
>
>
>
>
> In first-order logic, find a predicate P(x) which means "x is a power of 10."
>
> ==> hundred (logic/part1) <==
>
>
>
>
>
>
>
>
> Who makes it, has no need of it. Who buys it, has no use for it. Who
>
> ==> river.crossing (logic/part2) <==
>
>
>
>
>
>
>
>
>
>
>
please mail this to a Ex President, 2000 of them, first to Idenitfy if Senator Lewis on the tape of redo of goveremnt that narrowly changed in 1993 so that Superior couts could corrupt the generations of people I respect and want to help , this man said he was no relation to my father lewis, explain if he went to my dad colleges and schools in a classroom under a different name or his real, and if he was confused looks to be one of steve lewis cousins but when he say or notice that man refused to have knowledge of being family , even so if the innocence is true than another pool of aka alis using dad name to subliment world history of dad real elections to senate but people used the perfect people from the upper room government the crazys and the none to organize special needs and sucided to be the peace tready of parents never allowing anyone back inside the home after someone wife dies, so that they can trow away family phones and not ever have true family take the property or witness when hhe died, they allow ghost spirit or vitural reality internet game people to push a hallucination and program, this heighths the bullshit around kknowing how is Christ jesus and how to begain a life if you sufferec from oxegun out recessatation you don't know if your dead or alive but one thing stands out the government powers are not allows the one in power or if they are they love to push program to ramsack wisdom the puppy sitcom defect of laughing and lugon run embarrassment is evidence enough that a haughty weird acult reality rule over the universe, seen in laguna beach aa meeting when they started to fuck make fun o f my dog of the eightys always using voices to say my do g is kidnap in itally put up propagranda my ca is a kitten with marking in check point Charlie germanym is true many secrets of the world and I witness animal life that was out of hell with dark mark of direty and gross reality of dead peopke back in Newport, not hallucination because the enet is used by god angels and they since rising of the dead presentation dolphien reader demo, prior that 92 I read my bible and riped it up because it read different, it never did that before for 23 years, it said many rose from grave when jesus got out of thomb, pissed I was because it made it seem Jeuss was not the first to do it orspecail, but it means that in the future when you hear of this, now it is close at hand his arrival but he has to hide his custom be cause of the jackles fox and wolf and satan I run into, can decide if a bullshit government plan the pulp pope john was to meet me, they say laeave brian out of the data of Karen and steve an d off net becacuse he gets too mean not adaption to family life for women he off and on try to rescure seem anti clear about his netreality of the golfers and the delutions and if he is talking to obomba what about, why the laughting . mail the transcript of communication off this cpu and all the archieve comment from qestrel.com my friend rellaford got on to this enet work by buying me a lap top he owen me coming feed and he should be allowed to get on the net until eh pays me the fee of 600000 10 list 19 of 888888888 salivary not 375 an hour or 8 bseine circle Irvine cal four copyt one to linda one to stev