rec.puzzles FAQ, part 1 of 15
Chris Cole
Last-modified: 1992/09/20
Version: 3
Instructions for Accessing rec.puzzles Frequently Asked Questions List
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 letter to uunet!questrel!faql-request
containing one or more lines of the form:
send <puzzle_name>
For example, to request decision/allais.p, send the line:
send decision/allais.p
or just:
send allais
The puzzle will be mailed via return email to the address in your
request's "From:" line. If you are unsure of this address, and cannot
edit this line, then include in your message BEFORE the first "send" line
the line:
return_address <your_return_email_address>
FTP
The FAQL has been posted to news.answers. News.answers is archived in
the periodic posting archive on pit-manager.mit.edu [18.172.1.27].
Postings are located in the anonymous ftp directory
/pub/usenet/news.answers, and are archived by "Archive-name". Other
subdirectories of /pub/usenet contain periodic postings that may not
appear in news.answers.
Other news.answers/FAQ archives (which carry some or all of the FAQs
in the pit-manager archive) are:
archive.cs.ruu.nl [131.211.80.5] 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 pit-manager.mit.edu are
also accessible via Prospero and WAIS (the database name is "usenet"
on port 210).
CREDIT
The FAQL is NOT the original work of the editor (just in case you were
wondering :^).
In keeping with the general net practice on FAQL's, I do not as a rule assign
credit for FAQL solutions. There are many reasons for this:
1. The FAQL 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 FAQL 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 uunet!questrel!faql-comment.
INDEX
==> analysis/bugs.p <==
Four bugs are placed at the corners of a square. Each bug walks directly
toward the next bug in the clockwise direction. The bugs walk with
constant speed always directly toward their clockwise neighbor. Assuming
the bugs make at least one full circuit around the center of the square
==> analysis/c.infinity.p <==
What function is zero at zero, strictly positive elsewhere, infinitely
differentiable at zero and has all zero derivitives at zero?
==> 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.
==> 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/e.and.pi.p <==
Which is greater, e^(pi) or (pi)^e ?
==> 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/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/integral.p <==
If f is integrable on (0,inf), and differentiable at 0, and a > 0, show:
inf ( f(x) - f(ax) )
==> analysis/period.p <==
What is the least possible integral period of the sum of functions
of periods 3 and 6?
==> analysis/rubberband.p <==
A bug walks down a rubberband 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 rubberband is uniformly and
infinitely elastic, will the bug ever reach the car?
==> analysis/series.p <==
Show that in the series: 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/snow.p <==
Snow starts falling before noon on a cold December day.
At noon a snowplow 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??
==> analysis/tower.p <==
A number is raised to its own power. The same number is then raised to
the power of this result. The same number is then raised to the power
of this second result. This process is continued forever. What is the
maximum number which will yield a finite result from this process?
==> 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
==> 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.
==> 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/consecutive.product.p <==
Prove that the product of three or more consecutive natural numbers cannot be a
perfect square.
==> arithmetic/consecutive.sums.p <==
Find all series of consecutive positive integers whose sum is exactly 10,000.
==> arithmetic/digits/all.ones.p <==
Prove that some multiple of any integer ending in 3 contains all 1s.
==> 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/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
==> arithmetic/digits/divisible.p <==
Find the least number using 0-9 exactly once that is evenly divisible by each
of these digits?
==> 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/1992.p <==
1 = -1+9-9+2. Extend this list to 2 - 100 on the left side of the equals sign.
==> arithmetic/digits/equations/383.p <==
Make 383 out of 1,2,25,50,75,100 using +,-,*,/.
==> arithmetic/digits/extreme.products.p <==
What are the extremal products of three three-digit numbers using digits 1-9?
==> arithmetic/digits/googol.p <==
What digits does googol! start with?
==> 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
==> arithmetic/digits/nine.digits.p <==
Form a number using 0-9 once with its first n digits divisible by n.
==> arithmetic/digits/palindrome.p <==
Does the series formed by adding a number to its reversal always end in
a palindrome?
==> arithmetic/digits/palintiples.p <==
Find all numbers that are multiples of their reversals.
==> 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/prime/101.p <==
How many primes are in the sequence 101, 10101, 1010101, ...?
==> arithmetic/digits/prime/all.prefix.p <==
What is the longest prime whose every proper prefix is a prime?
==> 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/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/reverse.p <==
Is there an integer that has its digits reversed after dividing it by 2?
==> arithmetic/digits/rotate.p <==
Find integers where multiplying them by single digits rotates their digits.
==> arithmetic/digits/sesqui.p <==
Find the least number where moving the first digit to the end multiplies by 1.5.
==> arithmetic/digits/squares/leading.7.to.8.p <==
What is the smallest square with leading digit 7 which remains a square
when leading 7 is replaced by an 8?
==> 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.9.p <==
Is it possible to make a number and its square, using the digits from 1 through
9 exactly once?
==> arithmetic/digits/squares/three.digits.p <==
What squares consist entirely of three digits (e.g., 1, 4, and 9)?
==> 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/sum.of.digits.p <==
Find sod ( sod ( sod (4444 ^ 4444 ) ) ).
==> arithmetic/digits/zeros/factorial.p <==
How many zeros are in the decimal expansion of n!?
==> arithmetic/digits/zeros/lsd.factorial.p <==
What is the least significant non-zero digit in the decimal expansion of n!?
==> arithmetic/digits/zeros/million.p <==
How many zeros occur in the numbers from 1 to 1,000,000?
==> 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/pell.p <==
Find integer solutions to x^2 - 92y^2 = 1.
==> arithmetic/prime/arithmetic.progression.p <==
Is there an arithmetic progression of 20 or more primes?
==> arithmetic/prime/consecutive.composites.p <==
Are there 10,000 consecutive non-prime numbers?
==> arithmetic/sequence.p <==
Prove that all sets of n integers contain a subset whose sum is divisible by n.
==> arithmetic/sum.of.cubes.p <==
Find two fractions whose cubes total 6.
==> arithmetic/tests.for.divisibility/eleven.p <==
What is the test to see if a number is divisible by eleven?
==> arithmetic/tests.for.divisibility/nine.p <==
What is the test to see if a number is divisible by nine?
==> arithmetic/tests.for.divisibility/seven.p <==
What is the test to see if a number is divisible by 7?
==> arithmetic/tests.for.divisibility/three.p <==
Prove that if a number is divisible by 3, the sum of its digits is likewise.
==> combinatorics/coinage/combinations.p <==
How many ways are there to make change for a dollar? Count
combinations of coins, not permuations.
==> 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?
==> 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),
==> 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/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/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/grid.dissection.p <==
How many (possibly overlapping) squares are in an mxn grid?
==> 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 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
==> cryptology/Beale.p <==
What are the Beale ciphers?
==> cryptology/Feynman.p <==
What are the Feynman ciphers?
==> cryptology/Voynich.p <==
What are the Voynich ciphers?
==> cryptology/swiss.colony.p <==
What are the 1987 Swiss Colony ciphers?
==> decision/allais.p <==
The Allais Paradox involves the choice between two alternatives:
A. 89% chance of an unknown amount
10% chance of $1 million
==> 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
==> 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.
==> 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
==> decision/exchange.p <==
At one time, the Mexican and American dollars were devalued by 10 cents on each
side of the border (i.e. a Mexican dollar was 90 cents in the US, and a US
dollar was worth 90 cents in Mexico). A man walks into a bar on the American
side of the border, orders 10 cents worth of beer, and tenders a Mexican dollar
==> 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:
==> 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
==> 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
==> 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
==> 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 in comes up heads on the
nth toss and the payoff is set at 2^n dollars?
==> decision/switch.p <==
Switch? (The Monty Hall Problem)
Two black marbles and a red marble are in a bag. You choose one marble from the
bag without looking at it. Another person chooses a marble from the bag and it
==> 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
==> english/acronym.p <==
What acronyms have become common words?
==> english/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?
==> english/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.
==> english/behead.p <==
Is there a sentence that remains a sentence when all its words are beheaded?
==> english/capital.p <==
What words change pronunciation when capitalized (e.g., polish -> Polish)?
==> english/charades.p <==
A ....... surgeon was ....... to operate because he had .......
==> english/contradictory.proverbs.p <==
What are some proverbs that contradict one another?
==> english/contranym.p <==
What words are their own antonym?
==> english/element.p <==
The name of what element ends in "h"?
==> english/equations.p <==
Each equation below contains the initials of words that will make the phrase
correct. Figure out the missing words. Lower case is used only to help the
initials stand out better.
==> english/fossil.p <==
What are some examples of idioms that include obsolete words?
==> english/frequency.p <==
In the English language, what are the most frequently appearing:
1) letters overall?
2) letters BEGINNING words?
3) final letters?
==> english/gry.p <==
Find three completely different words ending in "gry."
==> english/homographs.p <==
List all homographs (words that are spelled the same but pronounced differently)
==> english/homophones.p <==
What words have four or more spellings that sound alike?
==> english/j.ending.p <==
What words and names end in j?
==> english/ladder.p <==
Find the shortest word ladders stretching between the following pairs:
hit - ace
pig - sty
four - five
==> english/less.ness.p <==
Find a word that forms two other words, unrelated in meaning, when "less"
and "ness" are added.
==> english/letter.rebus.p <==
Define the letters of the alphabet using self-referential common phrases (e.g.,
"first of all" defines "a").
==> english/lipograms.p <==
What books have been written without specific letters, vowels, etc.?
==> english/multi.lingual.p <==
What words in multiple languages are related in interesting ways?
==> english/near.palindrome.p <==
What are some long near palindromes, i.e., words that except for one
letter would be palindromes?
==> english/palindromes.p <==
What are some long palindromes?
==> english/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?
==> english/phonetic.letters.p <==
What does "FUNEX" mean?
==> english/piglatin.p <==
What words in pig latin also are words?
==> english/pleonasm.p <==
What are some redundant terms that occur frequently (like "ABM missile")?
==> english/plurals/collision.p <==
Two words, spelled and pronounced differently, have plurals spelled
the same but pronounced differently.
==> 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.
==> english/plurals/drop.s.p <==
What plural is formed by DROPPING the terminal "s" in a word?
==> english/plurals/endings.p <==
List a plural ending with each letter of the alphabet.
==> english/plurals/french.p <==
What English word, when spelled backwards, is its French plural?
==> english/plurals/man.p <==
Words ending with "man" make their plurals by adding "s".
==> english/plurals/switch.first.p <==
What plural is formed by switching the first two letters?
==> english/portmanteau.p <==
What are some words formed by combining together parts of other words?
==> english/potable.color.p <==
Find words that are both beverages and colors.
==> english/rare.trigraphs.p <==
What trigraphs (three-letter combinations) occur in only one word?
==> english/records/pronunciation/silent.p <==
What words have an exceptional number of silent letters?
==> english/records/pronunciation/spelling.p <==
What words have exceptional ways to spell sounds?
==> english/records/pronunciation/syllable.p <==
What words have an exceptional number of letters per syllable?
==> english/records/spelling/longest.p <==
What is the longest word in the English language?
==> english/records/spelling/most.p <==
What word has the most variant spellings?
==> english/records/spelling/operations.on.words/deletion.p <==
What exceptional words turn into other words by deletion of letters?
==> english/records/spelling/operations.on.words/insertion.and.deletion.p <==
What exceptional words turn into other words by both insertion and
deletion of letters?
==> english/records/spelling/operations.on.words/insertion.p <==
What exceptional words turn into other words by insertion of letters?
==> english/records/spelling/operations.on.words/movement.p <==
What exceptional words turn into other words by movement of letters?
==> english/records/spelling/operations.on.words/substitution.p <==
What exceptional words turn into other words by substitution of letters?
==> english/records/spelling/operations.on.words/transposition.p <==
What exceptional words turn into other words by transposition of letters?
==> english/records/spelling/operations.on.words/words.within.words.p <==
What exceptional words contain other words?
==> english/records/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?
==> english/records/spelling/sets.of.words/squares.p <==
What are some exceptional word squares (square crosswords with no blanks)?
==> english/records/spelling/single.words.p <==
What words have exceptional lengths, patterns, etc.?
==> english/repeat.p <==
What is a sentence containing the most repeated words, without:
using quotation marks,
using proper names,
using a language other than English,
==> english/repeated.words.p <==
What is a sentence with the same word several times repeated?
==> english/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
==> english/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.
==> english/self.ref.numbers.p <==
What true sentence has the form: "There are _ 0's, _ 1's, _ 2's, ...,
in this sentence"?
==> english/self.ref.words.p <==
What sentence describes its own word, syllable and letter count?
==> english/sentence.p <==
Find a sentence with words beginning with the letters of the alphabet, in order.
==> english/snowball.p <==
Construct the longest coherent sentence you can such that the nth
word is n letters long.
==> english/spoonerisms.p <==
List some exceptional spoonerisms.
==> english/states.p <==
What long words have all bigrams either a postal state code or its reverse?
==> english/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.
==> english/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.
==> english/weird.p <==
Make a sentence containing only words that violate the "i before e" rule.
==> english/word.boundaries.p <==
List some sentences that can be radically altered by changing word boundaries
and punctuation.
==> english/word.torture.p <==
What is the longest word all of whose contiguous subsequences are words?
==> games/chess/knight.control.p <==
How many knights does it take to attack or control the board?
==> games/chess/mutual.check.p <==
What position is a stalemate for both sides and is reachable in a legal game
(including the requirement to prevent check)?
==> games/chess/mutual.stalemate.p <==
What's the minimal number of pieces in a legal mutual stalemate?
==> games/chess/queens.p <==
How many ways can eight queens be placed so that they control the board?
==> games/chess/size.of.game.tree.p <==
How many different positions are there in the game tree of chess?
==> 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 looses by not
==> games/connect.four.p <==
Is there a winning strategy for Connect Four?
==> games/craps.p <==
What are the odds in craps?
==> games/crosswords/cryptic/clues.p <==
What are some clues (indicators) used in cryptics?
==> games/crosswords/cryptic/double.p <==
Each clue has two solutions, one for each diagram; one of the answers
to 1ac. determines which solutions are for which diagram.
All solutions are in Chamber's and Webster's Third except for one solution
==> games/crosswords/cryptic/intro.p <==
What are the rules for cluing cryptic crosswords?
==> 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?
==> games/hi-q.p <==
What is the quickest solution of the game Hi-Q (also called Solitair)?
For those of you who aren't sure what the game looks like:
==> games/jeopardy.p <==
What are the highest, lowest, and most different scores contestants
can achieve during a single game of Jeopardy?
==> games/knight.tour.p <==
For what board sizes is a knight's tour possible?
==> 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
==> games/othello.p <==
How good are computers at Othello?
==> games/risk.p <==
What are the odds when tossing dice in Risk?
==> games/rubiks.clock.p <==
How do you quickly solve Rubik's clock?
==> games/rubiks.cube.p <==
What is known about bounds on solving Rubik's cube?
==> games/rubiks.magic.p <==
How do you solve Rubik's Magic?
==> games/scrabble.p <==
What are some exceptional scrabble games?
==> games/square-1.p <==
Does anyone have any hints on how to solve the Square-1 puzzle?
==> 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
==> games/tictactoe.p <==
In random tic-tac-toe, what is the probability that the first mover wins?
==> geometry/K3,3.p <==
Can three houses be connected to three utilities without the pipes crossing?
_______ _______ _______
| oil | |water| | gas |
==> 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/bisector.p <==
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/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 abbreviations for the names of all
twelve months (e.g., "jan", "feb", "mar"). On the second set,
==> geometry/circles.and.triangles.p <==
Find the radius of the inscribed and circumscribed circles for a triangle.
==> 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/dominoes.p <==
There is a chess board (of course with 64 squares). You are given
21 dominoes 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 dominoes exactly cover the remaining
==> geometry/construction/4.triangles.6.lines.p <==
Can you construct 4 equilateral triangles with 6 toothpicks?
==> geometry/construction/5.lines.with.4.points.p <==
Arrange 10 points so that they form 5 rows of 4 each.
==> geometry/construction/square.with.compass.p <==
Construct a square with only a compass and a straight edge.
==> geometry/cover.earth.p <==
A thin membrane covers the surface of the earth. One square meter is
added to the area of this membrane. How much is added to the radius and
volume of this membrane?
==> geometry/dissections/circle.p <==
Can a circle be cut into similar pieces without point symmetry
about the midpoint? Can it be done with a finite number of pieces?
==> geometry/dissections/hexagon.p <==
Divide the hexagon into:
1) 3 indentical rhombuses.
2) 6 indentical kites(?).
3) 4 indentical trapezoids.
==> geometry/dissections/square.70.p <==
Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 sqaure be dissected into
24 squares of size 1x1, 2x2, 3x3, etc.?
==> geometry/dissections/square.five.p <==
Can you dissect a square into 5 parts of equal area with just a straight edge?
==> 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
==> geometry/earth.band.p <==
How much will a band around the equator rise above the surface if it
is made one meter longer?
==> 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 mutiliated. Is it still possible to divide the ham sandwich
with a straight knife cut such that both the ham and the bread are
==> 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
==> 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.
==> 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/lattice/area.p <==
Prove that the area of a triangle formed by three lattice points is integer/2.
==> geometry/lattice/equilateral.p <==
Can an equlateral triangle have vertices at integer lattice points?
==> geometry/rotation.p <==
What is the smallest rotation that returns an object to its original state?
==> 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
==> 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
==> geometry/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
==> 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/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/rectangles.with.squares.p <==
Given two sorts of squares, (axa) and (bxb), what rectangles can be tiled?
==> 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/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
==> group/group.01.p <==
AEFHIKLMNTVWXYZ BCDGJOPQRSU
==> group/group.01a.p <==
147 0235689
==> group/group.02.p <==
ABEHIKMNOPTXZ CDFGJLQRSUVWY
==> group/group.03.p <==
BEJQXYZ DFGHLPRU KSTV CO AIW MN
==> group/group.04.p <==
BDO P ACGJLMNQRSUVWZ EFTY HIKX
==> group/group.05.p <==
CEFGHIJKLMNSTUVWXYZ ADOPQR B
==> group/group.06.p <==
BCEGKMQSW DFHIJLNOPRTUVXYZ
==> induction/hanoi.p <==
Is there an algorithom 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/n-sphere.p <==
With what odds do three random points on an n-sphere form an acute triangle?
==> induction/paradox.p <==
What simple property holds for the first 10,000 integers, then fails?
==> 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/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/takeover.p <==
After graduating from college, you have taken an important managing position
in the prestigious financial firm of "Mary and Lee".
You are responsable for all the decisions concerning take-over bids.
Your immediate concern is whether to take over "Financial Data".
==> 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
==> 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
==> 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. Not counting covers, title pages, etc., how many pages
does the bookworm eat through?
==> 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
==> logic/calibans.will.p <==
----------------------------------------------
| Caliban's Will by M.H. Newman |
----------------------------------------------
==> logic/camel.p <==
An Arab sheikh tells his two sons that are 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 wiseman for advise. After hearing the advice they jump on the
==> 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.
==> 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
==> logic/condoms.p <==
How can you have mutually safe sex with three women with only two condoms?
==> logic/dell.p <==
How can I solve logic puzzles (e.g., as published by Dell) automatically?
==> 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.
==> logic/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.
==> logic/flip.p <==
How can a toss be called over the phone (without requiring trust)?
==> logic/friends.p <==
Any group of 6 or more contains either 3 mutual friends or 3 mutual strangers.
Prove it.
==> 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/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/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.
==> 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
==> 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/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
==> 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?
==> 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
==> 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/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.
==> 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/situation.puzzles.outtakes.p <==
The following puzzles have been removed from my situation puzzles list,
or never made it onto the list in the first place. There are a wide
variety of reasons for the non-inclusion: some I think are obvious,
some don't have enough of a story, some involve gimmicks that annoy me,
==> logic/situation.puzzles.p <==
Jed's List of Situation Puzzles
History:
original compilation 11/28/87
==> 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:
==> 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
==> 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
==> 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."
==> logic/smullyan/liars.et.al.p <==
Of a group of n men, some always lie, some never lie, and the rest sometimes
lie. They each know which is which. You must determine the identity of each
man by asking the least number of yes-or-no questions.
==> 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/priest.p <==
A priest takes confession of all the inhabitants in a small town. He
discovers that in N married pairs in the town, one of the pair has
committed adultery. Assume that the spouse of each adulterer does not
know about the infidelity of his or her spouse, but that, since it is
==> 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
==> logic/timezone.p <==
Two people are talking long distance on the phone; one is in an East-
Coast state, the other is in a West-Coast state. The first asks the other
"What time is it?", hears the answer, and says, "That's funny. It's the
same time here!"
==> 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
==> logic/verger.p <==
A very bright and sunny Day
The Priest didst to the Verger say:
"Last Monday met I strangers three
None of which were known to Thee.
==> logic/weighing/balance.p <==
You are given N balls and a balance scale and told that
one ball is slightly heavier or lighter than the other identical
ones. The scale lets you put the same number of balls on each side
and observe which side (if either) is heavier.
==> 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 a
scale to find the box containing the light balls. How do you do it?
==> 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
==> 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
==> 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.
==> 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/bicycle.p <==
A boy, a girl and a dog go for a 10 mile walk. The boy and girl can
walk 2 mph and the dog can trot at 4 mph. They also have bicycle
which only one of them can use at a time. When riding, the boy and
girl can travel at 12 mph while the dog can peddle at 16 mph.
==> physics/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.
==> physics/brick.p <==
What is the maximum overhang you can create with an infinite supply of bricks?
==> physics/cannonball.p <==
A person in a boat drops a cannonball overboard; does the water level change?
==> physics/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
==> physics/magnets.p <==
You have two bars of iron. One is magnetic, the other is not. Without
using any other instrument (thread, filings, other magnets, etc.), find
out which is which.
==> physics/milk.and.coffee.p <==
You are just served a hot cup of coffee and want it to be as hot as possible
when you drink it some number of minutes later. Do you add milk when you get
the cup or just before you drink it?
==> physics/mirror.p <==
Why does a mirror appear to invert the left-right directions, but not up-down?
==> 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. The rope weighs
4 ounces per foot. The combined ages of the monkey and it's mother
is 4 years. The weight of the monkey is as many pounds as the mother
==> physics/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?
==> 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/resistors.p <==
What are the resistances between lattices of resistors in the shape of a:
1. Cube
==> physics/sail.p <==
A sailor is in a sailboat on a river. The water (current) is flowing
downriver at a velocity of 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
==> physics/skid.p <==
What is the fastest way to make a 90 degree turn on a slippery road?
==> physics/spheres.p <==
Two spheres are the same size and weight, but one is hollow. They are
made of uniform material, though of course not the same material. Without
a minimum of apparatus, how can I tell which is hollow?
==> physics/wind.p <==
Is a round-trip by airplane longer or shorter if there is wind blowing?
==> 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
==> 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/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
==> probability/coincidence.p <==
Name some amazing coincidences.
==> probability/coupon.p <==
There is a free gift in my breakfast cereal. The manufacturers say
that the gift comes in four different colours, 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 colours, what is the
==> 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
==> probability/flips.p <==
Consider a run of coin tosses: HHTHTTHTTTHTTTTHHHTHHHHHTHTTHT
Define a success as a run of one H or T (as in THT or HTH). Use two
different methods of sampling. The first method would consist of
==> probability/flush.p <==
Which set contains more flushes than the set of all possible 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
==> 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
==> 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
==> 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/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 and
that if the light is green in one direction it is red in the other, find the
==> 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.
==> probability/particle.in.box.p <==
A particle is bouncing randomly in a two-dimensional box. How far does it
travel between bounces, on avergae?
Suppose the particle is initially at some random position in the box and is
==> probability/pi.p <==
Are the digits of pi random (i.e., can you make money betting on them)?
==> 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
==> 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,
==> 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
==> probability/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?
==> series/series.01.p <==
M, N, B, D, P ?
==> series/series.02.p <==
H, H, L, B, B, C, N, O, F ?
==> series/series.03.p <==
W, A, J, M, M, A, J?
==> series/series.03a.p <==
G, J, T, J, J, J, A, M, W, J, J, Z, M, F, J, ?
==> series/series.03b.p <==
A, J, B, C, G, T, C, V, J, T, D, F, K, B, H, ?
==> series/series.03c.p <==
M, A, M, D, E, L, R, H, ?
==> series/series.04.p <==
A, E, H, I, K, L, ?
==> series/series.05.p <==
A B C D E F G H?
==> series/series.06.p <==
Z, O, T, T, F, F, S, S, E, N?
==> series/series.06a.p <==
F, S, T, F, F, S, ?
==> series/series.07.p <==
1, 1 1, 2 1, 1 2 1 1, ...
What is the pattern and asymptotics of this series?
==> series/series.08a.p <==
G, L, M, B, C, L, M, C, F, S, ?
==> series/series.08b.p <==
A, V, R, R, C, C, L, L, L, E, ?
==> series/series.09a.p <==
S, M, S, S, S, C, P, P, P, ?
==> series/series.09b.p <==
M, S, C, P, P, P, S, S, S, ?
==> series/series.10.p <==
D, P, N, G, C, M, M, S, ?
==> series/series.11.p <==
R O Y G B ?
==> series/series.12.p <==
A, T, G, C, L, ?
==> series/series.13.p <==
M, V, E, M, J, S, ?
==> series/series.14.p <==
A, B, D, O, P, ?
==> series/series.14a.p <==
A, B, D, E, G, O, P, ?
==> series/series.15.p <==
A, E, F, H, I, ?
==> 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.17.p <==
T, P, O, F, O, F, N, T, S, F, T, F, E, N, S, N?
==> series/series.18.p <==
10, 11, 12, 13, 14, 15, 16, 17, 20, 22, 24, ___ , 100, 121, 10000
==> series/series.19.p <==
1 01 01011 0101101011011 0101101011011010110101101101011011 etc.
Each string is formed from the previous string by substituting '01' for '1'
and '011' for '0' simultaneously at each occurance.
==> series/series.20.p <==
1 2 5 16 64 312 1812 12288
==> series/series.21.p <==
5, 6, 5, 6, 5, 5, 7, 5, ?
==> series/series.22.p <==
3 1 1 0 3 7 5 5 2 ?
==> series/series.23.p <==
22 22 30 13 13 16 16 28 28 11 ?
==> series/series.24.p <==
What is the next letter in the sequence: W, I, T, N, L, I, T?
==> series/series.25.p <==
1 3 4 9 10 12 13 27 28 30 31 36 37 39 40 ?
==> 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.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.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.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.30.p <==
I I T Y W I M W Y B M A D
==> series/series.31.p <==
6 2 5 5 4 5 6 3 7
==> series/series.32.p <==
0 1 1 0 1 0 0 1 1 0 0 1 0 1 1 0 1
==> series/series.33.p <==
2 12 360 75600
==> series/series.34.p <==
3 5 4 4 3 5 5 4 3
==> series/series.35.p <==
1 2 3 2 1 2 3 4 2 1 2 3 4 2 2 3
==> 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/eskimo.snow.p <==
How many words do the Eskimo have for snow?
==> trivia/federal.reserve.p <==
What is the pattern to this list:
Boston, MA
New York, NY
Philadelphia, PA
==> trivia/jokes.self-referential.p <==
What are some self-referential jokes?
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> analysis/bugs.p <==
Four bugs are placed at the corners of a square. Each bug walks directly
toward the next bug in the clockwise direction. The bugs walk with
constant speed always directly toward their clockwise neighbor. Assuming
the bugs make at least one full circuit around the center of the square
before meeting, how much closer to the center will a bug be at the end
of its first full circuit?
==> analysis/bugs.s <==
Amorous Bugs
ANSWER: 1 - e^(-2*pi)
Let O(t) be the angle at time t of bug 1 relative to its starting
point and r(O(t)) be its distance from the center of the square.
Bug 1's vector trajectory is (using a Cartesian coordinate system with
the origin at the center of the square):
(1) X1 = [r(O) * cos(O), r(O) * sin(O)]
By symmetry, bug 2's trajectory is the same only rotated by pi/2, viz.:
(2) X2 = [-r(O) * sin(O), r(O) * cos(O)]
Since bug 1 walks directly toward bug 2, the velocity of bug 1 must be
proportional to the vector from bug 1 to bug 2:
(3) d(X1)/d(t) = k * (X2 - X1)
Equating each component of the vector equation (3) yields:
(4) (d(r)/d(O) * cos(O) - r * sin(O)) * d(O)/d(t) =
k * (-r * cos(O) - r * sin(O))
(5) (d(r)/d(O) * sin(O) + r * cos(O)) * d(O)/d(t) =
k * (-r * sin(O) + r * cos(O))
These equations are solved by:
(6) k = d(O)/d(t)
and:
(7) d(r)/d(O) = -r(O)
(7) is solved by:
(8) r(O) = e^-O
Constant speed gives:
(9) v^2 = constant = ((d(r)/d(O))^2+r^2)*(d(O)/d(t))^2
Substituting (8) into (9) yields (let V = v/sqrt(2)):
(10) d(O)/d(t) = V * e^O
Which is solved (using the boundary condition O(0) = 0) by:
(11) O(t) = -ln(1 - V * t)
Substituting (11) into (8) yields:
(12) r(t) = r(0) - V * t
The bug has made a full circle when O(T) = 2*pi; using (11):
(13) T = 1/V * (1 - e^(-2*pi))
Substituting T into (12) yields the answer:
(14) r(T) - r(0) = 1 - e^(-2*pi)
==> analysis/c.infinity.p <==
What function is zero at zero, strictly positive elsewhere, infinitely
differentiable at zero and has all zero derivitives at zero?
==> analysis/c.infinity.s <==
exp(-1/x^2)
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.
Otherwise, the "Cache and Ferry" 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 1955.5509 (I'm sure that this N = 139 distance is not 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 | "If you have anything to tell me of importance,
McDonnell Douglas | for God's sake begin at the end."
sou...@mdcbbs.com | Sara Jeanette Duncan
@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@
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/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/e.and.pi.p <==
Which is greater, e^(pi) or (pi)^e ?
==> analysis/e.and.pi.s <==
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, 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/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.
==> analysis/rubberband.p <==
A bug walks down a rubberband 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 rubberband 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/series.p <==
Show that in the series: 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/series.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 snowplow 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 proportioned
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 <==
A number is raised to its own power. The same number is then raised to
the power of this result. The same number is then raised to the power
of this second result. This process is continued forever. What is the
maximum number which will yield a finite result from this process?
==> analysis/tower.s <==
Tower of Exponentials
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)
==> 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/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.
==> 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.
==> arithmetic/consecutive.product.p <==
Prove that the product of three or more consecutive natural numbers 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/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. By previous reasoning 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
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 - 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/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/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/googol.p <==
What digits does googol! start with?
==> arithmetic/digits/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/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/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.
==> 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.
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> 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/leading.7.to.8.p <==
What is the smallest square with leading digit 7 which remains a square
when leading 7 is replaced by an 8?
==> arithmetic/digits/squares/leading.7.to.8.s <==
x=2996282391593370361328125
y=2824483699753370361328125
x^2=8977708170172487211329625006796419620513916015625
y^2=7977708170172487211329625006796419620513916015625
==> 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 <==
567 and 854.
==> 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
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/factorial.p <==
How many zeros are in the decimal expansion of n!?
==> arithmetic/digits/zeros/factorial.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/digits/zeros/lsd.factorial.p <==
What is the least significant non-zero digit in the decimal expansion of n!?
==> arithmetic/digits/zeros/lsd.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/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/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 <==
Here is an 8x8 example:
01 56 48 25 33 24 16 57
63 10 18 39 31 42 50 07
62 11 19 38 30 43 51 06
04 53 45 28 36 21 13 60
05 52 44 29 37 20 12 61
59 14 22 35 27 46 54 03
58 15 23 34 26 47 55 02
08 49 41 32 40 17 09 64
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/prime/arithmetic.progression.p <==
Is there an arithmetic progression of 20 or more primes?
==> arithmetic/prime/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/prime/consecutive.composites.p <==
Are there 10,000 consecutive non-prime numbers?
==> arithmetic/prime/consecutive.composites.s <==
9973!+2 through 9973!+10006 are composite.
==> arithmetic/sequence.p <==
Prove that all sets of n integers contain a subset whose sum is divisible by n.
==> arithmetic/sequence.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/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 7?
==> 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 2009.
Subtract (2009 mod 10) * 2 from (2009 div 10)
- 9 * 2 + 200
= 182
Subtract (182 mod 10) * 2 from (182 div 10)
- 2 * 2 + 18
= 14
so 2009 is divisible by 7.
==> arithmetic/tests.for.divisibility/three.p <==
Prove that if a number is divisible by 3, the sum of its digits is likewise.
==> arithmetic/tests.for.divisibility/three.s <==
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
==> combinatorics/coinage/combinations.p <==
How many ways are there to make change for a dollar? Count
combinations of coins, not permuations.
==> combinatorics/coinage/combinations.s <==
Assuming that you had coins of one cent, five cents, ten cents, 25 cents,
50 cents, and 100 cents, there are 293 ways to make change for a dollar.
This can be calculated by determining the number of ways to make change
using only a penny and then a penny and nickel, then penny, nickel, and
dime, etc.
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
1.0 1 2 4 6 9 13 18 24 31 39 49 62 77 93 112 134 159 187 218 253 293
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?
[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 nickle 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 professors (A, B, C, and D) as
the "core group". Each professor 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?
==> 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/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 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 <==
First, a couple of points:
(1) All empty subsets of the 10 integers are disjoint and have the same sum.
This doesn't make for a very interesting problem. Thus, we impose the
additional restriction that S and T be non-empty.
(2) The 10 integers must be pairwise distinct. Consider, e.g., the 10
integers 1, 1, 1, 1, 1, 1, 1, 1, 1, and 1. There are no non-empty
disjoint subsets with equal sums.
Proof of puzzle:
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) - sub(C) = sum(B) - sum(C) = sum(B-C) = sum(T).
QED
==> 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 either 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
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> 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.
==> 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.
(cf. Luce and Raiffa, "Games and Decisions", Wiley, 1957, p. 366)
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.)
==> 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)
==> 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 Mexican and American dollars were devalued by 10 cents on each
side of the border (i.e. a Mexican dollar was 90 cents in the US, and a US
dollar was worth 90 cents in Mexico). A man walks into a bar on the American
side of the border, orders 10 cents worth of beer, and tenders a Mexican dollar
in change. He then walks across the border to Mexico, orders 10 cents worth of
beer and tenders 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 Mexican
dollars into Mexico 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 Mexican bar runs out
of US dollars, or the US bar runs out of Mexican dollars, i.e., until
he runs out of "work" to do.
==> 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.
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> english/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
==> english/less.ness.p <==
Find a word that forms two other words, unrelated in meaning, when "less"
and "ness" are added.
==> english/less.ness.s <==
base -> baseless, baseness
light -> lightless, lightness
sound -> soundless, soundness
wit -> witless, witness
==> english/letter.rebus.p <==
Define the letters of the alphabet using self-referential common phrases (e.g.,
"first of all" defines "a").
==> 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
==> english/lipograms.p <==
What books have been written without specific letters, vowels, etc.?
==> english/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
==> english/multi.lingual.p <==
What words in multiple languages are related in interesting ways?
==> english/multi.lingual.s <==
Synonymous reversals:
Dutch: nier (kidney), French: rein
French: etats, English: state
==> english/near.palindrome.p <==
What are some long near palindromes, i.e., words that except for one
letter would be palindromes?
==> english/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
==> english/palindromes.p <==
What are some long palindromes?
==> english/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
Other palindromes in ascending length (drum roll please):
Dennis sinned.
Sir, I'm Iris.
Sup not on pus.
Name no one man.
Naomi, did I moan?
Enid and Edna dine.
Revenge Meg? Never!
No lemons, no melon.
A Toyota's a Toyota.
Ma is a nun, as I am.
He harasses Sarah, eh?
Niagara, O roar again!
He lived as a devil, eh?
Nurse, I spy gupsies, run!
Sit on a potato pan, Otis!
Slap a ham on Omaha, pals!
A slut nixes sex in Tulsa.
Rats live on no evil star.
Ten animals I slam in a net.
Go deliver a dare, vile dog.
Was it a car or a cat I saw?
Was it Eliot's toilet I saw?
Al lets Della call Ed Stella.
Draw, O Caeser, erase a coward.
Did Eve salt an atlas? Eve did.
No pinot noir on Orion to nip on.
Naomi, sex at noon taxes! I moan.
Evil I did dwell; lewd did I live.
Yo, bad anaconda had no Canada boy .
Egad! A base tone denotes a bad age.
Satan, oscillate my metallic sonatas.
Red dude kill lion. No ill-liked udder.
I roamed under it as a tired, nude Maori.
To Peru, named llama mall 'De Manure Pot'.
Straw? No, too stupid a fad. I put soot on warts.
Now, Ned, I am a maiden nun; Ned, I am a maiden won.
Here we no got conical ill lilac in octogon ewer, eh?
Salamander a ton now. Raw war won not, a Red Nam, alas.
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.
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.
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 soap 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)
Canning is on test (snub slip-up) Eve's simple hot dish (crib)
Birch (Sid) to help Miss Eve Pupil's buns
Reaper 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 mand 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)
Some word (not letter) palindromes:
So patient a doctor to doctor a patient so.
Girl, bathing on Bikini, eyeing boy, finds boy eyeing bikini on bathing girl.
In German:
Ein Neger mit Gazelle zagt im Regen nie.
In Serbo-Croat:
Ana voli Milovana.
Ana nabra par banana.
Imena Amen nema, a me mi.
U pero soli i los o repu.
Ako jad moli silom daj oka.
Odano mati pita: a ti pitam, o nado?
Evo sam iza padam mada pazim asove.
v v v v
A krt u razu mi laze no one zalim u zaru trka.
Palindromes in other languages that are palindromes in English:
Hebrew: aba or abba, English: dad
German: tat, English: deed
The timeless classic:
A man, a plan, a canal; Panama?
Has been improved by:
A dog, a plan, a canal: pagoda!
-- anonymous
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."
The entire book _Satire: Veritas_ is a palindrome, it starts
"Sir, I stra..." and ends "... Art, sir, is Satire: Veritas."
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
==> english/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?
==> english/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)
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
==> english/phonetic.letters.p <==
What does "FUNEX" mean?
==> english/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.)
==> english/piglatin.p <==
What words in pig latin also are words?
==> english/piglatin.s <==
cess -> essay
coke -> okay
lawn -> onlay
lout -> outlay
lover -> overlay
plover -> overplay
plunder -> underplay
sass -> assay
stout -> outstay
trash -> ashtray
wear -> airway
wonder -> underway
==> english/pleonasm.p <==
What are some redundant terms that occur frequently (like "ABM missile")?
==> english/pleonasm.s <==
11.5% APR
ABM missile
ABS system
AC current
ACT tests
AMOCO Oil Co.
APL programming language
ATM macine
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
La Brea Tar Pits
Los Altos Hills (The Hills Hills)
MIDI Interface
Mount Fujiyama (Mount Mountain)
NATO organization
NFS File System
PCV valve
PIN number
RAM (or ROM) memory
Ruidoso River (Noisy River River)
SALT talks
SAT test
SCSI Interface
SEATO organization
VIN number
floccinoccinihlipilification (from 4 latin words meaning "nothing")
hoi polloi (a genuine bilingual redundancy)
hot water heater
==> english/plurals/collision.p <==
Two words, spelled and pronounced differently, have plurals spelled
the same but pronounced differently.
==> english/plurals/collision.s <==
axe and axis -> axes
base and basis -> bases
ellipse and ellipsis -> ellipses
==> 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?
==> english/plurals/doubtful.number.s <==
cares -> caress
==> english/plurals/drop.s.p <==
What plural is formed by DROPPING the terminal "s" in a word?
==> english/plurals/drop.s.s <==
necropolis -> necropoli
==> english/plurals/endings.p <==
List a plural ending with each letter of the alphabet.
==> 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
==> english/plurals/french.p <==
What English word, when spelled backwards, is its French plural?
==> english/plurals/french.s <==
state/etats
==> english/plurals/man.p <==
Words ending with "man" make their plurals by adding "s".
==> english/plurals/man.s <==
caiman
doberman
German
human
leman
ottoman
pitman
Pullman
Roman
shaman
talisman
==> english/plurals/switch.first.p <==
What plural is formed by switching the first two letters?
==> english/plurals/switch.first.s <==
falaj -> aflaj (Chambers English Dictionary)
==> english/portmanteau.p <==
What are some words formed by combining together parts of other words?
==> english/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.
==> english/potable.color.p <==
Find words that are both beverages and colors.
==> english/potable.color.s <==
burgundy
champagne
chartreuse
chocolate
claret
cocoa
coffee
cream
midori (Japanese for green. Does Japanese count?)
rose
wine
==> english/rare.trigraphs.p <==
What trigraphs (three-letter combinations) occur in only one word?
==> 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
==> english/records/pronunciation/silent.p <==
What words have an exceptional number of silent letters?
==> english/records/pronunciation/silent.s <==
longest sequence 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.
==> english/records/pronunciation/spelling.p <==
What words have exceptional ways to spell sounds?
==> english/records/pronunciation/spelling.s <==
same spelling, different sound -OUGH (7)
BOUGH, COUGH, DOUGH, HICCOUGH, LOUGH, ROUGH, THROUGH
different spelling, same sound AIR (9)
AIR, AIRE, ARE, AYR, AYER, E'ER, ERE, ERR, HEIR
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> english/records/pronunciation/syllable.p <==
What words have an exceptional number of letters per syllable?
==> english/records/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 three AREA (4) [O'IO (3)] four IEIE (4) five OXYOPIA (7)
six ONIOMANIA [AMIOIDEI (8)] seven EPIDEMIOLOGY (12) [OMOHYOIDEI (10)]
eight EPIZOOTIOLOGY nine EPIZOOTIOLOGICAL (16) ten EPIZOOTIOLOGICALLY
twelve HUMUHUMUNUKUNUKUAPUAA (21)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> english/records/spelling/longest.p <==
What is the longest word in the English language?
==> english/records/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.
==> english/records/spelling/most.p <==
What word has the most variant spellings?
==> english/records/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
==> english/records/spelling/operations.on.words/deletion.p <==
What exceptional words turn into other words by deletion of letters?
==> english/records/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) (BULLETIN)G (9)
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.
==> english/records/spelling/operations.on.words/insertion.and.deletion.p <==
What exceptional words turn into other words by both insertion and
deletion of letters?
==> english/records/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.
==> english/records/spelling/operations.on.words/insertion.p <==
What exceptional words turn into other words by insertion of letters?
==> english/records/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.
==> english/records/spelling/operations.on.words/movement.p <==
What exceptional words turn into other words by movement of letters?
==> english/records/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.
==> english/records/spelling/operations.on.words/substitution.p <==
What exceptional words turn into other words by substitution of letters?
==> english/records/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)
****
Unless noted otherwise, all words occur in Webster's Third New International
Dictionary, Merriam-Webster, Springfield, MA, 1961.
==> english/records/spelling/operations.on.words/transposition.p <==
What exceptional words turn into other words by transposition of letters?
==> english/records/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.
==> english/records/spelling/operations.on.words/words.within.words.p <==
What exceptional words contain other words?
==> english/records/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.
==> english/records/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?
==> english/records/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.
==> english/records/spelling/sets.of.words/squares.p <==
What are some exceptional word squares (square crosswords with no blanks)?
==> english/records/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 20th century than
they are now, but people still like to compose and solve them. Forms
appear every month in the _Enigma_, which is the monthly publication
of the National Puzzlers' League. The membership fee is $13 for
the first year, and information may be obtained from:
David A. Rosen
207 East 27th St. #3K
New York, NY 10016
All members have the option of choosing a nom de plume; 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_
Spring Valley Road
Morristown, NJ 07960
I'll have a paper appearing here at some point on the "support" of a
form (which I'll discuss below).
Word squares come in two flavors, 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 the finest example 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 its only
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 an additional square:
bortsches
overtrust
reparence
trabeatae
strestell
creatural
hunterite
escalates
steelless
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, but it has the flaw of using
more than one source. It is, however, the *only* known 9-square that
uses only uncapitalized, solid-form dictionary words.
There are about 2000 9-squares known, all of which were constructed
by hand except for the two noted above. Almost all of these use
very obscure sources of words. As a general rule of thumb, if you
discover a new form via computer search, it is probably going to be
of high quality, since it is hard to obtain computer-readable word
lists that contain *really* 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 form 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.
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 ~ 15.7.
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), which agrees
well with reality.
--
Chris Long, 265 Old York Rd., Bridgewater, NJ 08807-2618
cl...@remus.rutgers.edu
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> english/records/spelling/single.words.s <==
Word Records from Webster's Third
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 f (1,1)
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 (2,1)
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 aba (3,1)
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 apa (3,1)
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 trimethylethylene (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 stresslessness stresslessnesses (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 vowel & y once cylindrocellular phosphuranylites ventriculography (16,3)
shortest each vowel once eulogia eutocia eutopia isourea sequoia (7,5)
shortest each vowel & y once oxyuridae (9,1)
shortest vowels in order caesious (8,1)
shortest vowels & y in order facetiously (11,1)
longest vowels in order abstentious (11,1)
longest vowels & y in order abstemiously (12,1)
shortest vowels in reverse order muroidea (8,1)
shortest vowels & y in reverse order ? (80,0) ?
longest vowels in reverse order subcontinental (14,1)
longest vowels & y 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 tyddyns (7,11)
longest alternating vowel-consonant hypovitaminosises (17,1)
longest alternating vowel-consonant excluding y aluminosilicates diketopiperazine epicoracohumeral (16,3)
Consonants
longest consonant string bergschrund bergschrunds catchphrase eschscholtzia eschscholtzias festschrift festschriften festschrifts goldschmidtine goldschmidtines goldschmidtite goldschmidtites lachsschinken lachsschinkens latchstring mischsprache mischsprachen nachschlag nachschlage nachschlags promptscript veldtschoen weltschmerz weltschmerzes (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 susurruses (10,11)
most repeated letters dihydroxycholecalciferol hydroxydesoxycorticosterone hysterosalpingographies methyldihydromorphinone microspectrophotometrically octamethylpyrophosphoramide phosphatidylethanolamine pseudohermaphroditism tetrabromophenolphthalein tetraiodophenolphthalein trinitrophenylmethylnitramine (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 doddered doddled driddled dunderheaded dunderheadedness dunderheadednesses dyakisdodecahedral dyakisdodecahedron dyakisdodecahedrons fiddledeedee fiddleheaded granddaddy lepidodendrid lepidodendrids lepidodendroid muddleheaded muddleheadedness muddleheadednesses muddyheaded puddingheaded skedaddled woodshedded (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 glugging goggling grigging grogging guggling lallygagging lollygagging zigzagging (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 intelligibilities invincibilities (6,12)
highest containing j repeated ajonjoli ajonjolis avijja avijjas djokjakarta gastrojejunal gastrojejunostomy hajj hajjes hajji hajjis jajman jajmani jajmans jajoba jejuna jejunal jejune jejunely jejuneness jejunenesses jejunities jejunity jejunostomies jejunostomy jejunum jeremejevite jeremejevites jimberjawed jimjams jinglejangle jinglejangles jinjili jinjilis jipijapa jipijapas jirajara jirajaras jiujitsu jiujitsus jiujutsu jiujutsus jogjakarta jojoba jujitsu jujitsus juju jujube jujubes jujus jujutsu juj
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 mammonisms mesembryanthemum mesembryanthemums meshummadim mohammedanism mohammedanisms muhammadanism muhammadanisms mummiform tetramethylammonium thermometamorphism zamzummim zamzummims (4,23)
highest containing n repeated inconvenientness inconvenientnesses nannoplankton nannoplanktonic nondenominational nondenominationalism nonentanglement nonintervention noninterventionist syngenesiotransplantation unconvincingness unconvincingnesses (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 williwaw williwaws willowware willowweed willowworm willywaw willywaws (3,17)
highest containing x repeated dextropropoxyphene executrix executrixes exlex exlexes exonarthex exotoxic exotoxin hexachlorocyclohexane hexahydroxy hexaxon hexoxide hydroxydeoxycorticosterone hydroxydesoxycorticosterone maxixe maxixes myxoxanthin oxyhexactine oxyhexaster paxwax paxwaxes paxywaxies paxywaxy saxifrax saxifraxes saxitoxin sextuplex xanthotoxin xanthoxenite xanthoxenites xanthoxylaceae xanthoxyletin xanthoxyletins xanthoxylin xanthoxylins xanthoxylum xanthoxylums (2,37)
highest containing y repeated acetylphenylhydrazine acetylphenylhydrazines anhydrohydroxyprogesterone anhydrohydroxyprogesterones brachydactyly chylophylly cryptozygy cystopyelography cytophysiologically cytophysiology dacryocystorhinostomy dactylosymphysis dihydroxyphenylalanine dyssynergy glycolytically gypsyfy gypsyfying hydrodynamically hydronymy hydroxydeoxycorticosterone hydroxydesoxycorticosterone hydroxyethyl hydroxyethylation hydroxyethylations hydroxylysine hydroxymethyl hydroxymethylation hydrox
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 narrow letters (ACEMNORSUVWXZ) erroneousnesses verrucosenesses (15,2)
longest tall letters (BDFGHIJKLPQTY) lighttight lillypilly (10,2)
longest vertical-symmetry letters (AHIMOTUVWXY) homotaxia thymomata (9,2)
longest horizontal-symmetry letters (BCDEHIKOX) checkbook checkhook chookchie (9,3)
highest ratio of dotted letters (IJ) jinjili (71,1)
Typewriter
longest top row proprietory proterotype rupturewort (11,3)
longest middle row shakalshas (10,1)
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 formed with chemical symbols nonrepresentationalism (22,1)
longest formed with US postal codes convallarias (12,1)
longest formed with compass points newnesses sweeswees (9,2)
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 operatively postoperatively preoperatively undefendablenesses unoperative unspiritually (6,16)
most consecutive letters in order consecutively bierstube bierstuben bierstubes gymnopaedia gymnopaedias gymnopaedic gymnopedia gymnopedias gymnophiona gymnoplast gymnoplasts klavierstuck limnopithecus limnoplankton limnoplanktonic overstudy overstuff overstuffed semnopithecus semnopitheque semnopitheques thamnophile thamnophiles thamnophiline thamnophilus thamnophis understudy (4,27)
most consecutive letters in order aborticide aborticides abscinded absconded abscondence abscondences alimentotherapy aluminographies aluminography aluminotype aluminotypes ambuscade ambuscaded ambuscades helminthosporia helminthosporin helminthosporins helminthosporium helminthosporiums helminthosporoid laminograph laminographic laminographies laminography laminosioptes limnograph limnopithecus limnoplankton limnoplanktonic luminophor luminophors luminoscope opaquers reconstructive reconstructively redist
most consecutive letters appropinquates appropinquations appropinquities equiponderates equiponderations perquisition perquisitions preconquest propinquities quadruplications sesquiterpenoid sesquiterpenoids (8,12)
highest ratio of consecutive letters to length klompen (85,1)
==> english/repeat.p <==
What is a sentence containing the most repeated words, without:
using quotation marks,
using proper names,
using a language other than English,
anything else distasteful.
==> english/repeat.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.
==> english/repeated.words.p <==
What is a sentence with the same word several times repeated?
==> english/repeated.words.s <==
It is true for all that, that that that that that that signifies, is not
the one to which I refer.
Here are some steps to understanding the entire sentence:
That is not the one to which I refer.
That (that that that signifies) is not the one to which I refer.
That that that that that that signifies, is not the one to which I refer.
In Annamite:
Ba ba ba ba.
(Three ladies gave a box on the ear to the favorite of the Prince.)
==> english/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."
==> english/rhyme.s <==
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
==> english/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.
==> english/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, a followup to one on self-
referential sentences. 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 Dewdny'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.
==> english/self.ref.numbers.p <==
What true sentence has the form: "There are _ 0's, _ 1's, _ 2's, ...,
in this sentence"?
==> english/self.ref.numbers.s <==
There are 1 0's, 7 1's, 3 2's, 2 3's, 1 4's, 1 5's, 1 6's, 2 7's, 1 8's,
and 1 9's in this sentence.
There are 1 0's, 11 1's, 2 2's, 1 3's, 1 4's, 1 5's, 1 6's, 1 7's, 1 8's
and 1 9's in this sentence.
==> english/self.ref.words.p <==
What sentence describes its own word, syllable and letter count?
==> english/self.ref.words.s <==
This sentence contains ten words, eighteen syllables, and sixty-four letters.
==> english/sentence.p <==
Find a sentence with words beginning with the letters of the alphabet, in order.
==> english/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 toadstalls underneath vampires
who x-rayed young zombies.
==> english/snowball.p <==
Construct the longest coherent sentence you can such that the nth
word is n letters long.
==> english/snowball.s <==
I
do
not
know
where
family
doctors
acquired
illegibly
perplexing
handwriting;
nevertheless,
extraordinary
pharmaceutical
intellectuality,
counterbalancing
indecipherability,
transcendentalizes
intercommunications'
incomprehensibleness.
==> english/spoonerisms.p <==
List some exceptional spoonerisms.
==> 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!
==> english/states.p <==
What long words have all bigrams either a postal state code or its reverse?
==> english/states.s <==
10 paramarine
10 indentment
10 cacocnemia
9 amendment
9 paramimia
9 paramenia
9 paralinin
9 paralalia
9 palilalia
9 palapalai
8 scalawag
8 memorial
Disallowing reversals of state codes the longest common ones are:
8 malarial
7 malaria
6 scalar
6 marine
5 flaky
Terry Donahue
==> english/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.
==> english/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.
==> english/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.
==> english/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)
==> english/weird.p <==
Make a sentence containing only words that violate the "i before e" rule.
==> english/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
==> english/word.boundaries.p <==
List some sentences that can be radically altered by changing word boundaries
and punctuation.
==> english/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.
==> english/word.torture.p <==
What is the longest word all of whose contiguous subsequences are words?
==> english/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.
==> games/chess/knight.control.p <==
How many knights does it take to attack or control the board?
==> 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_.
==> games/chess/mutual.check.p <==
What position is a stalemate for both sides and is reachable in a legal game
(including the requirement to prevent check)?
==> games/chess/mutual.check.s <==
Put the following configuration in one corner:
|
| x
| P x
|B P P
|K R B
+---------
("x" is a Black pawn), and the same with colors reversed in the h8
corner.
--Noam D. Elkies (elk...@zariski.harvard.edu)
Dept. of Mathematics, Harvard University
==> games/chess/mutual.stalemate.p <==
What's the minimal number of pieces in a legal mutual stalemate?
==> games/chess/mutual.stalemate.s <==
6.
W Kh8 e6 f7 h7 B Kf8 e7
W Kb1 B Ka3 b2 b3 b4 a4
W Kf1 B Kh1 Bg1 f2 f3 h2
==> games/chess/queens.p <==
How many ways can eight queens be placed so that they control the board?
==> 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.
==> games/chess/size.of.game.tree.p <==
How many different positions are there in the game tree of chess?
==> 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 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 demonstarted 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
==> 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 looses 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?
==> 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.
==> games/connect.four.p <==
Is there a winning strategy for Connect Four?
==> 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
==> games/craps.p <==
What are the odds in craps?
==> 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
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> games/crosswords/cryptic/clues.s <==
The following list is derived from indicators used in a variety of
crosswords: the letters in the left column are the letters being
indicated; the right hand column is how these letters might be
indicated in a clue.
Caveat emptor: many of the entries in this list would be considered
unsound in some puzzles (some of these unsound indicators are marked
with a +). Entries marked * are used mostly in advanced cryptics.
I would welcome corrections and additions to the list.
----------------------------------------------------------------------
a Austria
a I
a academician
a accepted
a ace
a acre
a active
a adult
a advanced
a afternoon
a aleph
a alpha
a amateur
a ampere
a an/ane
a angstrom
a answer
a ante
a are (metric)
a articles - English
a associate
a atomic
a ay
a aye
a before
a blood group
a bomb
a effect
a examination
a fifty
a film
a first character
a first class
a first letter
a five hundred
a five thousand
a good
a high class
a it
a key +
a level
a midday
a note +
a one *
a paper
a road
a strings (violin)
a un
a unit
a violin string
a vitamin
a year
aa motoring organisation
ab able seaman
ab hand
ab rating
ab sailor/salt/seaman
ab tar
abbe priest (Fr.)
abe Lincoln
abel first victim
abel murder victim
abel second child
abel third man
able can
able expert
abo native
ac account
ac accountant
ac aircraftsman
ac alternating current
ac before Christ
ac bill
ac current
aca accountant
acas peacemakers
acc account
acc bill
ace card
ace champion
ace expert
ace one
ace pilot
ace service
ace winner
act decree
act performance
actor tree
ad Christian era
ad advertisement
ad after date
ad before the day
ad contemporary
ad in the modern age
ad in the year of our Lord
ad modern times
ad notice
ad now
ad nowadays
ad our time/era
ad period
ad present day
ad promotion
ad puff
ad this era
ad today
adam first character
adam first person
adam number one
add sum
add tot
aden port
admin management
admin running
ado business
ado difficulty
ado fuss
ado row
ado trouble
ae aged
ae poet
aet aged
ag silver
aga Muslim leader
age (long) time
age mature
age period
agent spy
agm annual meeting
agm meeting
agm yearly meeting
ai capital
ai first class
ai good
ai high class
ai main road
ai sloth
ail trouble
ain own (Scot.)
air appearance
air display
air song
aire river
ait island
al Alabama
al Alan
al Albania
al Albert
al Capone
al aluminium
al gangster
al one pound
ala Alabama
ala after the style of
ala in the style of
ala to the (Fr.)
ala wings
alas Alaska
alb one pound
ale beer
aleph Hebrew letter
all completely
all everybody
all everything
alp mountain
alp peak
alph river
alpha Greek letter
alpha beginning
alpha first character
alpha first letter
am America
am American
am I am
am admitting
am boasting
am half day
am hymns
am in the morning
am morning
am self-confessed
amen final word
amen last word
amer American
ammo missiles
amos bookmaker
amp one member
an I
an articles - English
an before
an if (old word)
an one *
an un
ana tales
ane I
ane one
ane one
ankh life symbol
anne princess
anon now
ans answer
ans brief reply
ans collection
ans short answer
ant if it (old word)
ant six-footer
ant social worker
ant soldier
ant worker
ape copy
ape primate
aq water
ar arrive/arrival
ar year of reign
ara academician
ara artist
ara painter
arab horse
arc curve
argo old ship
aria song
arm gun
arm limb
arm member
arr arrive/arrival
art contrivance
art craft
art cunning
art painting
art skill
as Anglo-Saxon
as ayes
as ays
as like
as one's
as specifically
as when
ash remains
ash tree
asia continent
aside one fifteen
asis existing state
asp snake
ass donkey
asti wine
ate goddess
ate mischief
athena goddess
ation at one on
atoll bikini
au gold
au to the (Fr.)
aus Australia
aux to the (Fr.)
av bible
av lived so long
ave average
ave greeting
ave hail
ave road
ave way
aver average
avon county
ay I
ay agreement
ay always
ay ever
ay yes
aye I
aye I say
aye agreement
aye always
aye ever
aye yes
az Azed
az scope, plenty of
b Bach
b Beethoven
b Belgium
b Brahms
b Britain
b British
b a follower
b bachelor
b baron
b bedbug
b bee
b bel
b beta
b beth
b bishop
b black
b blood group
b bloody
b book *
b born
b boron
b bowled
b boy
b breadth
b inferior
b key +
b magnetic flux
b note +
b paper
b second
b second class
b second letter
b three hundred
b three thousand
b vitamin
ba Bachelor of Arts
ba airline
ba bachelor
ba barium
ba degree
ba graduate
ba scholar
bac airline
bacon philosopher
ban curse
ban outlaw
ban prohibition
bar Inn
bar lawyers
bar prevent
bar save
barb horse
bat fly-by-night
bb bees
bb books
bb very black
bc ancient times
bc before Christ
bc period
bd beady
bd bound
bd cleric
bd theologian
be exist
be live
bea airline
bear speculator
bed in bed
bee buzzer
bee group of workers
bee six-footer
bee social worker
bee worker
bef Gort's men
bess queen
beta Greek letter
beth Hebrew letter
bi double
bi two (double)
bird prison
bis two (twice)
bit chewed
bit piece
biz business
bk book
bl British company
bl lawman
bl lawyer
blue Conservative
bm British Museum
bm doctor
bo American man
boa snake
board directors
bob old shilling
bp bishop
br Britain
br British
br British Rail
br bank rate
br branch
br bridge
br brig
br brother
br brown *
br lines/landline
br railway(s)
br trains
br transport
bra female support(er)
bra support
bra undergarment
brass money
brat child
bren gun
brer rabbit
bridal old wedding
bro brother
bs bees
bst summer time
bull American policeman
bull gold
bus transport
c Celsius/centigrade
c Charles
c Conservative
c Cuba
c about (approx.)
c approx(imately)
c around
c cape
c caput
c carbon
c caught
c cedi
c cent/centime
c centi-
c century
c chapter
c circa
c club
c cold
c complex number
c copyright
c coulomb
c electrical capacity
c hundred
c hundred thousand
c key +
c lot
c many
c note +
c roughly
c sea
c see
c speed of light
c spring
c tap
c vitamin
ca about (approx.)
ca accountant
ca approx(imately)
ca calcium
ca roughly
cab transport
cade conspirator
cain first murderer
cain killer
cain murderer
cal California
cam river
can able to +
can is able to
can prison
can vessel
cantuar archbishop
cap chapter
cap international
car carat
car transport
carnation motor race
cart transport
cat jazz fan
cato censor
cattle neat
cave warning
cb Seabee (Amer.)
cc county council
cc seas
cc small measure
cc small quantity
cc two hundred
cd diplomat
cd seedy
ce Church of England
ce church
ce engineers
ce this (Fr.)
cent money
cet this (Fr.)
ch China
ch Companion of Honour
ch Switzerland
ch award
ch central heating
ch champion
ch chapter
ch chief
ch child
ch church
ch companion
ch honour
ch order
cha tea
chai gypsy woman
chair president
chal gypsy
char daily
che guerrilla
che revolutionary
cher dear (Fr.)
chere dear (Fr.)
chi Greek letter
ci Channel Islands
ci hundred and one
cia secret service
cia spies
cid captain
cid chief
cid detectives
cid police
cid spanish hero
cinc commander
cl chlorine
cl class
cl clause
cl gas - chlorine
cl hundred and fifty
co Colombia
co business
co care of
co cobalt
co commander
co commanding officer
co company
co county
co firm
co gas - carbon monoxide
co house
co objector
co officer
cod fish
cod swimmer
col neck
col pass
cole old king
colon stop
com commander
comb hairdresser
composer scorer
con Conservative
con against
con party
con politician
con study
con swindle
con trick
cooler prison
core decentralise
corn naval commander
cot bed
cot house
cow lower
cow neat
cr credit
cr crown
cr king
cs Civil Service
cs Czechoslovakia
cs hundreds
cs seas
ct Connecticut
ct carat
ct caught
ct cent/centime
ct court
ct small weight
ct weight
cu copper
cu see you
cue queue
cure priest (Fr.)
cutie pretty girl
cv autobiography
cy see why
d (old) penny
d Dee
d Democrat
d Deutsch
d Germany
d Schubert's works
d copper
d damn
d date
d daughter
d day
d dead/died
d deci
d degree
d delete
d delta
d deserted
d deuterium
d diameter
d diamond
d differential operator
d electrical flux
d five hundred
d four
d four thousand
d hundreds
d key +
d lot
d many +
d mark
d note +
d notice
d number
d strings (violin)
d violin string
d vitamin
da American lawyer
da District Attorney
da agreement - foreign (Russ.)
da dagger *
da lawman
da lawyer
da yes (Russ.)
dab expert
dad father
dad old man
dail Irish house
dam barrier
dam mother
dam restrain
dame lady
dan tribe
das articles - German
das the (Ger.)
dc Washington
dc Washington
dc current
dd cleric
dd days
dd divine
dd doctor
dd doctor of divinity
dd theologian
de from (Fr.)
de of (Fr.)
dean good man
dec Christmas period
dec last month
decanter Tantalus' prisoner
dee river
deed indeed
deed legal document
deep in the main
deep main
deep sea
deg degree
del of the (Ital.)
dela from the (Fr.)
dela of the (Fr.)
demi half
den retreat
den study
der articles - German
der the (Ger.)
derby horse race
des of the (Fr.)
det detective
di double
di five hundred and one
di princess
di two (double)
die articles - German
die the (Ger.)
dime 12.5 cents
dior designer
dis Hell
dis Pluto
dis underworld
disc circle
disc record
disc ring
dish pretty girl
dit named
dit reported
dit said (Fr.)
dit say (Fr.)
diy amateur's department
dk Denmark
dm mark
do (the) same
do act
do cheat
do cook
do ditto
do note
do party
do work
dodo double act
doh note
don fellow
don nobleman
don put on
don university teacher
down county
dr dead reckoning
dr doctor
dr dram
dr drawer
dr healer
drake bowler
dt alcoholic state
dt psychotic state
du from the (Fr.)
du of the (Fr.)
dutch wife
e Asian
e Edward
e Elizabeth
e England/English
e Spain
e boat
e bridge players
e direction
e east/eastern
e eight
e eight thousand
e energy +
e epsilon
e eta
e five
e five thousand
e key +
e layer
e logarithm base
e low grade
e note +
e orient
e oriental
e point
e quarter
e strings (violin)
e two hundred and fifty
e two hundred and fifty thousand
e universal set
e violin string
e vitamin
ea East Africa
ea each
ea river *
ea running water
ea water
ear listener
ear organ
ear spike *
earp lawman
eat Tanzania
ebor archbishop
ec London district
ec city
eccles Cakesville
ed Edward
ed editor
ed journalist
eden garden
eden old Prime Minister
eden paradise
edison inventor
edit censor
ee ease
eel fish
eel swimmer
eer always
eer ever
eer invariably
eg for example
eg for instance
egg bomb
egg cocktail
egg encourage
eight rowing boat
ein number one (Ger.)
el American railway
el American railway
el articles - Spanish
el measure
el printer's measure
el small measure
el the (Span.)
eld old age
eli priest
eli prophet
elia writer
ell four feet
ell length
ell measure
ely city
ely city
ely see
em measure
em printer's measure
em small measure
em small square
em them
en measure
en printer's measure
en small measure
eng England/English
ent otorhinolaryngology
entry record
eon age
eon time
ep record
er Cockney girl
er QE
er difficulty
er ever
er hesitation
er king
er monarch
er queen
er royal badge
era generation
erasmus old scholar
ere always
ere before
ergo so
eric gradually
erie lake
err blunder
err sin
err wander
erse Gaelic
es French art
es ease
esp sixth sense
esp telepathy
est is (Fr.)
et Egypt
et alien
et and (Fr.)
et exotic
et extraterrestrial
et film
eta Greek letter
eta estimated time of arrival
eta illegal army
eta terrorists
eton college
eton educational establishment
eton school
etty artist
eur continent
eve first lady
eve first mate
eve lady
eve woman
ew bridge partners
ew partnership
ex former
ex from
ex late
ex one time
exe river
eye I
eye I say
eye seer
eyot island
ezra pound
f Fahrenheit
f France
f Friday
f clef
f farad
f farthing
f fathom
f fellow
f female
f feminine
f filly
f fine
f fluorine
f folio
f following
f foot
f force
f forte
f forty
f frequency
f gas - fluorine
f hole
f key +
f loud
f noisy
f note +
f strong
f vitamin
f woman
fa football
fa note
fah note
fal river
fare China area (Far East)
fast firm
fe iron
fed American detective
ff folios
ff followings
ff fortissimo
ff very loud
ff very loud
ff very strong
ff very strong
fig small illustration
fir tree
firm business
firm company
fist duke
fl flourished
fla Florida
flower bloomer
flu illness
fo Foreign Office
fo folio
fob free on board
foc free of charge
fol folio
fool dessert
foot infantry
for free on rail
force police
fore warning
four rowing boat
fr French
fr father
fr fragment
fr franc
fr frequently
ft feet
ft foot
ft measure
fz forced
g Gauss
g George
g Germany
g agent
g clef
g four hundred
g gamma
g gamut
g gee
g girl
g gram(me)
g grand (Amer.)
g gravity
g guinea
g gulf
g key +
g man
g midnight
g note +
g strings (violin)
g suit
g thousand
g thousand
g violin string
g weight
g-man American detective
ga Georgia
gab gift
gad tribe
gael Gaelic
gal girl
gar fish
gat gun
gate old goat
gb Great Britain
gb our islands
gee horse
gee little horse
geiger counter
gel jelly
gen Genesis
gen general
gen information
gen low down
gent fellow
george pilot
gg Gee-gee
gg horse
gg little horse
gi American soldier
gi doughboy
gi fighter
gi government issue
gi private
gi serving man
gi soldier
gladiator old fighter
glc capital authority
gm counter
go bargain
go energy
go in good condition
go ready
go success
go traffic signal
go work
gotham New York
gp doctor
gr Greece
gr Greek
gr King George
gr grain
gr grammar
gr gramme
gr grouse
gr king
gr small weight
gr weight
grant general
grass informer
grist miller's corn
grs gunmen
gs general service
gs general staff
gt fast car
gt sports car
gu guinea
gu old fiddle
gue old fiddle
h Dirac's constant
h Hungary
h Planck's constant
h bomb
h gas - hydrogen
h hand
h hard
h heart
h height
h henry
h horse
h hospital
h hot
h hour
h house
h husband
h hydrant
h hydrogen
h tap
h two hundred
h two hundred thousand
h vitamin
ha half ditch
ha laugh
ha this year
haha ditch
haha laugh
hair net lock keeper
ham (poor) actor
han Chinese dynasty
hand worker
has bears
haw hedge
he (high) explosive
he His Excellency
he ambassador
he excellency
he gas - helium
he governor
he helium
he legate
he male
he our man
he the man
head point
hearth hard ground
hebe goddess
hehe laugh
hen female
hen layer
her female
her the woman's
her woman
hf half
hg Dad's army
hg mercury
hh very hard
hi greeting
hi hello
hic here in Rome
hic this (Lat.)
him male
hm Her Majesty
hm His Majesty
hm king
hm queen
ho house
hobo tramp
hock prison
hol short break
hp hire purchase
hp never-never
hr hour
hs here is
ht high tension
hun German
hun barbarian
i Italy
i a
i an/ane
i ay
i aye
i che
i electrical current
i eye
i first person
i imaginary number
i iodine
i iota
i island
i line
i lunchtime
i number one
i one
i one
i single
i square root of -1
i straight line
i un
i unit
i upright
i yours truly
ia Iowa
iam I am
iam admitting
iam afternoon
iam boasting
iam early morning
iam self-confessed
ian Scot/Scotsman
ians one answer
ib in the same place
ib same place
ibid in the same place
ic I see
ic hundred
ic in charge
ic ninety nine
ice diamond(s)
ice hard water
iceni old people
id I had
id I would
id fish
id genius
id identification
id instinct
id same
id that (Lat.)
ida princess
ide fish
idem said
ie that is (that's)
if condition
if provided/providing
ign gin cocktail/sling
ii eleven
ii eyes
ii two
il Israel
il articles - Italian
il one pound
il the (Ital.)
ilb one pound
ill I shall/will
ill Illinois
ill badly
ill unwell
im I am
im admitting
im boasting
im self-confessed
imp little devil
imp mischevious child
imp one member
impi soldiers
in at home
in batting
in elected
in fashionable
in favoured
in in fashion
in not out
in playing
in trendy
in wearing
ina princess
inch island
ind India
ine oriental
ing gin cocktail/sling
inn local
insect six-footer
intens decimally
io cry of triumph
io joyful cry
io maiden
io ten
io triumphant cry
ioc dime (Amer.)
iom Isle of Man
iom island
iom man
ion number one returning
iota Greek letter
ious credit notes
ious promises to pay
ip trivial sum
ir Iran
ira illegal army
ira terrorists
ire anger
ire rage
irl Ireland
is Iceland
is ayes
is ays
is eyes
is island
is one's
isis goddess
isis river
isle man
ism doctrine
ism theory
iss exists
ist first
it Italian
it sex appeal
it the thing
iv four
iv ivy
iv tea-time
ive I have
ix nine
j Jack
j Japan
j heat
j jay
j joule
j judge
j justice
j knave
j one
j pen
j square root of -1
ja Jamaica
ja agreement - foreign (Ger.)
ja yes (Ger.)
jack sailor/salt/seaman
je I, being French
je In Paris, I
jo little woman
jo sweetheart
job bookmaker
jock Scot/Scotsman
joe American soldier
jolly Marine
jug prison
k Kay
k Khmer Republic
k Kirkpatrick
k Koechel
k Mozart's works
k Scarlatti's works
k constant
k kappa
k kelvin
k kilo
k king
k knight
k monarch
k potassium
k thousand
k twenty
k twenty thousand
k two hundred and fifty
k vitamin
ka double *
ka genius *
ka individuality
key opener
kg cagey
kine cattle
kine neat
kine ox
kish graphite
kl kale
km kilometre
kn cayenne
knee bender
ko decisive blow
ko kick off
ko knock out
ko stunner
kr krypton
kt knight
kv cave (beware)
kv cavy
ky Kentucky
l Labour
l Liberal
l Luxembourg
l angle
l apprentice
l corner
l el
l elevated railway
l ell
l fifty
l fifty thousand
l half century
l hand
l inductance
l inexperienced driver
l la(m)bda
l lake
l lambert
l latin
l latitude
l league
l learner
l learning
l left
l length
l licentiate
l line
l lira/lire
l litre
l long
l lumen
l luminance
l many +
l money
l new driver
l novice
l number plate
l one pound
l overhead railway
l port
l pound
l pupil
l railway
l side
l sovereign
l student
l trainee
l tyro
l vitamin
la Los Angeles
la Louisiana
la articles - French
la articles - Italian
la articles - Spanish
la look *
la note
la the (Fr.)
la the (Ital.)
la the (Span.)
lab Labour
lab laboratory
lab party
lab politician
lab science centre
lac aircraftsman
lac hundred thousand
lah note
lakh hundred thousand
lam beat
lam pound
lambda Greek letter
lamed Hebrew letter
lar Libya
laud archbishop
lay song
lb one pound
lb pound
le articles - French
le the (Fr.)
lear king
lee general
leg limb
leg member
leg on
leg support
lei flowers
lei wreath *
lely artist
les articles - French
les the French
let allow(ed)
let hindrance
let permit(ted)
let service
let with a tenant
lewis gun
lh left hand
li fifty one
lib Liberal
lib book
lib party
lib politician
limn old paint
line shipping company
ling fish
ling swimmer
lips mouthpiece
lips speakers
lit drunk
lit loaded
lit settled
ll ells
ll els
ll fifty pounds
ll fifty-fifty
lner old railway
lo look
lo see
loch lake
log maths function
log record
los articles - Spanish
loser fabulous hare
lot large amount
loti first item in sale
lower cow
lowing neat sound
lp long playing
lp record
ls ells
ls els
lso orchestra
lt Lieutenant
lt officer
lud old king
lum chimney (Scot.)
lv meal ticket
m Bond's boss
m French man
m Malta
m Monday
m em
m lot
m maiden
m maiden over
m male
m man
m many +
m mare
m mark
m married
m masculine
m mass
m master
m measure
m member
m meso-
m meta-
m meter (metre)
m midday
m mile
m modulus
m monsieur
m month
m moon
m motorway
m mu
m noon
m roof
m small square
m spymaster
m thousand
m vitamin
ma Master of Arts
ma academic
ma degree
ma educated man
ma graduate
ma master
ma mother
ma old woman
ma scholar
mab fairy queen
mab queen
mac Scot/Scotsman
main deep
main sea
mal bad French
mam mother
man Friday
man fellow
man fighter
man hand
man husband
man island
man piece (chess)
man soldier
man worker
manifesto show-ring
mass Massachusetts
mass service
maxim gun
may can
mayo tree ring
mb doctor
mc master of ceremonies
mc medal
md doctor
md one thousand five hundred
me first person
me note
me number one
men people
men soldiers
ment intended
ment meant
ment on purpose
ment understood
mer sea (Fr.)
met New York opera
met police
mg magnesium
mi main road
mi motorway
mi note
mig aeroplane
mill economist
min half minute
min thirty seconds
ming Chinese dynasty
ming china
mini car
miss Mississippi
mm French men
mm ems
mm medal
mn Merchant Navy
mo Missouri
mo doctor
mo half minute
mo month
mo second
mo short time
mo time
mo way of working
mog cat
mon Monday
mon Scot/Scotsman
moo low
moo neat sound
mos months
moses lawgiver
mot car test
mot test
moth fly-by-night
mp fairly quiet
mp member
mp member of parliament
mp military police
mp mounted police
mp mountie(s)
mp policeman
mp politician
mp representative
mph rate
mph speed
mps chemist
mr mister
ms ems
ms handwriting
ms manuscript
ms paper
ms text
ms writing
mss manuscripts
mss papers
mt hill
mt mountain
mtb torpedo boat
mu Greek character
mu Greek letter
mum mother
mum quiet
mum silence
mur wall (Fr.)
mutt dog
my gracious me
n Avogadro's number
n Norway
n born
n bridge players
n direction
n en
n gas - nitrogen
n half an em
n indefinite number
n knight
n midday
n name
n neper
n neuter
n new
n newton
n ninety
n ninety thousand
n nitrogen
n noon
n north(ern)
n note
n noun
n nu
n number
n point
n pole +
n quarter
n unfavourable aspect
n unknown number
n unlimited number
na North America
na no (Scot.)
na not (Scot.)
na sodium
nae no (Scot.)
nae not (Scot.)
nag horse
nag little horse
nat born
nation people
nation race
national horse race
nb nota bene
nb note
nco non-commissioned officer
nco sergeant
nd North Dakota
nd no date
ne Durham area
ne Humberside
ne Tyneside
ne born
ne bridge opponents
ne gas - neon
ne neon
ne north-east
ne not (old word)
ne quarter
neat cattle
neat ox
ned donkey
ned little horse
nee born
nemo submariner
ness head
ness loch
ness point
net capture
net fabric
ney Marshal
ng no good
ni Northern Ireland
ni Ulster
ni nickel
nib writer
nick prison
nie close (old word)
nie near (old word)
nil love
nil nothing
nitre chemical
nitre fertiliser
nj New Jersey
nl not clear
nl not far
nl not permitted
nn ens
no New Orleans
no indefinite number
no not out
no number
no play (Jap.)
no refusal
noh play (Jap.)
noi leading
noi no one
noi number one
non no (Fr.)
nose informer
np North Pole
np pole
ns bridge partners
ns ens
ns not specific
ns partnership
nt Holy Writ
nt New Testament
nt Northern Territories
nt book(s)
nt books
nt good book
nt part of Bible
nt preservationists
nt scriptures
nu Greek character
nu Greek letter
nu name unknown
nu unidentified
num miners
number anaesthetic
nun bluetit
nurse shark
nus dawn
nus students' (union)
nut teachers
nv envy
nw Merseyside
nw bridge opponents
nw north-west
nw quarter
ny Gotham
ny New York
o Ohio
o around
o aught
o bald patch
o ball
o blob
o blood group
o cavity
o cipher
o circle
o circuit
o circular letter
o dial
o disc
o duck
o egg
o eleven
o eleven thousand
o empty
o examination
o full moon
o gas - oxygen
o globe
o gulf
o hole
o hollow
o hoop
o loop
o love
o naught
o nil
o no
o nothing
o nought
o oh
o omicron
o opening
o orb
o ortho-
o ought
o owe
o oxygen
o pellet
o ring
o round
o spangle
oaks horse race
oates conspirator
ob old boy
obe award
obe honour
obe order
obit dead/died
obit final message
obit last word
oc commander
oc officer commanding
odo bishop
oe old English
oe old Etonian
offa old king
og ogee
og own goal
og soccer blunder
oic officer in charge
ok acceptable
ok all right
ok approval
ok authorisation
ok correct
ok fine
ok okay
old man captain
ole cry of delight
om award
om honour
om order
omega Greek letter
omega final letter
omega final word
omega last letter
omitting skipping
on about
on acting
on being broadcast
on leg
on on the menu
on operating
on performing
on playing
one lunchtime
one undivided
oo duck's eggs
oo ohs
oo owes
oo spectacles
oom Dutch uncle
op operation
op opposite
op opus
op out of print
op work
optic seer
opus work
or alternative
or alternatively
or before *
or gold
or yellow
oral examination
oral test
ord no way
ord ring road
orion hunter
orpheus classical musician
os Ordinary Seaman
os big
os large letters
os ohs
os old style
os outsize
os owes
os sailor
os very large
ost east (Ger.)
ot Holy Writ
ot Old Testament
ot book(s)
ot good book
ot occupational therapy
ot part of Bible
ot scriptures
ou Open University
oui agreement - foreign (Fr.)
oui yes (Fr.)
ouija Franco-German agreement
ous nothing to America
ouse river
out away
out not at home
out unfashionable
over maiden
ox bull
ox neat
oxonian dark blue
oy oh, why
oz ounce
oz small weight
oz weight
oz wizard place
p Celt
p Kelt
p Portugal
p copper
p four hundred
p four hundred thousand
p page
p park(ing)
p participle
p pawn
p pea
p pedal
p pee
p peg
p penny
p phosphorous
p pi
p piano
p pint
p poise
p power
p president
p prince
p quiet
p small change
p soft
p softly
p vitamin
pa Panama
pa father
pa old man
pan god
par standard
para Brazilian
para airborne soldier
parent source
pas dance
pas step
pate head
pawn piece (chess)
pawnbroker uncle
pb lead
pc copper
pc policeman
pe Peru
pe gym
pe physical education
pe training
peg tee
pen author
pen enclosure
pen prison
pen writer
per by
per for each
per through
pet cherished
pet favourite
pg paying guest
ph local
phi Greek letter
pi Greek character
pi Greek letter
pi confusion *
pi good
pi religious
pi upright
pier mole
pit Hell
pitt old Prime Minister
pl Poland
pla mountain retreat
pla port authority
plato philosopher
plo illegal army
plo terrorists
plot garden
pm Prime Minister
pm afternoon
pm half day
pm in the afternoon
po Italian flower
po Post Office
po airman
po palladium
po pole
po river
polo Merchant of Venice
poly college
pony twenty five pounds
pony twenty-five pounds
pop father
pop old man
port left
port wine
pp pages
pp peas
pp pees
pp pianissimo
pp very quiet/soft(ly)
pr Puerto Rico
pr Romans/Roman people
pr electoral system
pr image building
pr president
pr price
pr prince
pr public relations
pr two
pra academician
pra artist
pra painter
pres president
prison bird
pro expert
pro for
pro public relations officer
prof academic
provos terrorists
ps footnote
ps peas
ps pees
ps postscript
ps second thoughts
pt part
pt physical training
pt pint
pt platinum
pt point
pt post town
pt stop
pt training
pub local
pv peavy
q Celt
q Kelt
q Quebec
q Queensland
q boat
q cue
q electrical charge
q farthing
q five hundred
q five hundred thousand
q koppa
q ninety
q ninety thousand
q quality
q queen
q query
q question
q queue
q quintal
q rational numbers
qp kewpie (doll)
qt cutie
qt quart
qt quiet
qu quart
qu queen
quad prison
que what (Fr.)
qui who (Fr.)
r Reaumur
r Regina
r Republican
r Rex
r Romania
r are
r arithmetic
r castle
r eighty
r eighty thousand
r hand
r king
r monarch
r month
r queen
r radius
r rain
r rand
r reading
r real numbers
r recipe
r resistance
r rho
r right
r river
r road
r rontgen unit
r rook
r royal
r run
r side
r take *
r writing
ra Royal Academician
ra Royal Academy
ra Royal Artillery
ra academician
ra academy
ra artillery
ra artist
ra big guns
ra gunmen
ra gunner(s)
ra painter
ra soldiers
ra sun (god)
rab Butler
rac motoring organisation
race people
rada academy
raf fliers
raf service
rage fashion
rain waterfall
ram butter
ram music school
ram sheep
ran managed
ran smuggled
ras head
ras prince
rat art nouveau
rat desert fighter
rat desert(er)
rat scab
rat strikebreaker
rate speed
rc Roman Catholic
rc church
rd little way
rd road
rd way
re Royal Engineer(s)
re about
re again
re concerning
re engineer(s)
re note
re over
re religious education
re sapper(s)
re soldiers
re touching
rebecca Welsh riots
rec recipe
rec take
red anarchist
red bloody
red cent *
red communist
red leftist
red revolutionary
red socialist
regan princess
regan wicked sister
reine French queen
reme engineers
rene French man
rep agent
rep salesman
rep traveller
res old thing
resh Hebrew letter
ret soak
rev priest
rev vicar
rex cat
rg argy(-bargy)
rh right hand
ri Rhode Island
ri king emperor
rib wife
rid clear
rid free
ring circle
rio port
rip final message
rip last word
river banker
river flower
river runner
rly lines/landline
rly railway
rly transport
rly way
rm Marine
rm Royal Marine(s)
rm jolly
rm old Irish magistrate
rm resident magistrate
rma Sandhurst
rms mailboat
rn Navy
rn Royal Navy
rn fleet
rn sailors
rn service
ro right hand
roc fabulous bird
rock diamond
rod fast car
rod pole
rod sports car
roi French king
roi king (Fr.)
rom gypsy
rose flower
rosinante poor horse
rot corruption
rot decay
rot rubbish
rr Right Reverend
rr Rolls Royce
rr ars
rr bishop
rr car
rsm non-commissioned officer
rt arty
rt right
ru football
ru rugby
ruc Irish police
rue street (Fr.)
run manage
run smuggle
rur Capek's play
rv bible
ry landline
ry line(s)
ry little way
ry rail
ry railway
ry transport
ry way
s Bach's works
s God's
s Sabbath
s Saturday
s Schmieder
s Sweden
s as
s bend
s bob
s bridge players
s direction
s dollar
s es
s ess
s has
s his
s is
s largesse
s old Bob
s old shilling
s paragon
s part of collar
s point
s pole +
s quarter
s saint
s second
s seven
s seven thousand
s shilling
s side
s siemens
s sister
s snow
s society
s son
s south(ern)
s space
s spade
s square
s stokes
s sulphur
s sun
s us
sa South Africa
sa South America
sa essay
sa it
sa sex appeal
sa without date
sad blue
sailor able seaman
saint paragon
salt able seaman
sam uncle
sas soldiers
sat Saturday
satin dressmaker
sc namely
sc self-contained
sc small capitals
sc specifically
sc that is
sc viz
scot fine
scot tax
scr scruple
sculptor blockbuster
sd South Dakota
sd without a day fixed
sdp nationalists
se Home Counties
se London Area
se bridge opponents
se quarter
se south-east
sea deep
sea main
sec dry
sec second
sec short time
sec time
see look
seed children
semi half
semi house
sent ecstatic
set group
set put
seth fourth man
sgt non-commissioned officer
sgt sergeant
sh hush
sh quiet
sh silence
she (the) woman
she female
she lady
she novel
sherpa mountaineer
si South Island
si agreement - foreign (Span.)
si note
si silicon
si yes (Ital.)
si yes (Span.)
sib relation
sic thus
side team
silk dressmaker
sin err
sin evil
sin without
sin wrong
sine maths function
sine without
sir knight
sis sister
sly tinker
sm French king
sm French queen
sm king (Fr.)
sm non-commissioned officer
smith economist
sn Essen
sn partnership
sn tin
so ergo
so note
so therefore
so thus
so well
soh note
sol note
sol sun-god
som county
some approx(imately)
son issue
sop soprano
sos appeal
sp South Pole
sp childless
sp odds
sp pole
sp species
sp starting price
sp without children
spa spring
spire Oxford dreamer
spy agent
spy mole
square unfashionable
sr old railway
sr senior
srn nurse
ss German soldier
s
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> 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
==> 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.
==> 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
==> games/tictactoe.p <==
In random tic-tac-toe, what is the probability that the first mover wins?
==> 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... )
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;
}
==> 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 K3,3 or K5 (the complete graph
on 5 vertices, i.e., the graph with 5 nodes and 10 edges) as a
subgraph. 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 the circumference of the earth at that latitude 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, going n times around the earth, 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 <==
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 <==
The following proof is probably from Altshiller-Court's College
Geometry, since that's where I first saw the problem.
Let the triangle be ABC, with angle bisectors BE and CD.
Let F be such that BEFD is a parallelagram.
Let x = measure of angle CBE = angle DBE,
y = measure of angle BCD = angle DCE,
x' = measure of angle EFC,
y' = measure of angle ECF.
(You will probably want to draw a picture.)
Suppose x > y. Consider the triangles EBC and DCB. Since BC = BC and
BE = CD, we must have CE > BD. Now, since BD = EF, we have that CE >
EF, so that x' > y'. Thus x+x' > y+y'. But, triangle FDC is
isosceles, since DF = BE = DC, so x+x' = y+y', a contradiction.
Similarly, we cannot have x < y. Therefore the base angles of ABC are
equal, making ABC an isosceles triangle. QED
==> 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 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 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.
==> geometry/coloring/dominoes.p <==
There is a chess board (of course with 64 squares). You are given
21 dominoes 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 dominoes exactly cover the remaining
63 squares? Or is it impossible?
==> geometry/coloring/dominoes.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 trimino 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 points P2
and P3 draw another larger circle (C2 and C3). Where these two
circles cross, draw a line (D2). This line should go the center of
circle C1 at a rt angle to the original diameter line. This line
should cross circle C1 at P4 and P5
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.
==> geometry/cover.earth.p <==
A thin membrane covers the surface of the earth. One square meter is
added to the area of this membrane. 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/dissections/circle.p <==
Can a circle be cut into similar pieces without point symmetry
about the midpoint? Can it be done with a finite number of pieces?
==> geometry/dissections/circle.s <==
Yes. Draw a circle inside the original circle, 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 indentical rhombuses.
2) 6 indentical kites(?).
3) 4 indentical trapezoids.
4) 8 indentcal 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/square.70.p <==
Since 1^2 + 2^2 + 3^2 + ... + 24^2 = 70^2, can a 70x70 sqaure 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/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?
==> 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/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 mutiliated. Is it still possible to divide the ham sandwich
with a straight knife cut such that both the ham and the 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. One
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 area of the "doughnut"
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
==> 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/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.
You can 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.
Also, by attaching strings to a sphere, it is possible to see that a
360 degree rotation will entangle the strings, which another 360 degree
rotation will disentangle.
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
P. A. M. Dirac's "scissors demonstration"
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
==> 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/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 on the outside edge of the table.
==> geometry/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/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/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/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.
==> 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.
==> 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 <==
ABEHIKMNOPTXZ CDFGJLQRSUVWY
==> group/group.02.s <==
ABEHIKMNOPTXZ resembles Greek letter
CDFGJLQRSUVWY does not resemble Greek letter
==> group/group.03.p <==
BEJQXYZ DFGHLPRU KSTV CO AIW MN
==> group/group.03.s <==
BEJQXYZ no state starting with this letter
DFGHLPRU one state starting with this letter
KSTV two states starting with this letter
CO three states starting with this letter
AIW four states starting with this letter
five states starting with this letter
six states starting with this letter
seven states starting with this letter
MN eight states starting with this letter
==> group/group.04.p <==
BDO P ACGJLMNQRSUVWZ EFTY HIKX
==> group/group.04.s <==
BDO no endpoint
P one endpoint
ACGJLMNQRSUVWZ two endpoints
EFTY three endpoints
HIKX four endpoints
==> group/group.05.p <==
CEFGHIJKLMNSTUVWXYZ ADOPQR B
==> group/group.05.s <==
CEFGHIJKLMNSTUVWXYZ no enclosed area
ADOPQR one enclosed area
B two enclosed areas
==> group/group.06.p <==
BCEGKMQSW DFHIJLNOPRTUVXYZ
==> group/group.06.s <==
BCEGKMQSW prime numbers
DFHIJLNOPRTUVXYZ composites
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> induction/hanoi.p <==
Is there an algorithom 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
==> 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 <==
What simple property holds for the first 10,000 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 responsable 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 responsable for take-overs. Among all
financial firms it is common knowledge that the managers responsable
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.
==> 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/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. 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.
==> 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/calibans.will.p <==
----------------------------------------------
| Caliban's Will by M.H. Newman |
----------------------------------------------
When Caliban's will was opened it was found to contain the following
clause:
"I leave ten of my books to each of Low, Y.Y., and 'Critic,' who are
to choose in a certain order.
No person who has seen me in a green tie is to choose before Low.
If Y.Y. was not in Oxford in 1920 the first chooser never lent me
an umbrella.
If Y.Y. or 'Critic' has second choice, 'Critic' comes before the one
who first fell in love."
Unfortunately Low, Y.Y., and 'Critic' could not remember any of the
relevant facts; but the family solicitor pointed out that, assuming the
problem to be properly constructed (i.e. assuming it to contain no
statement superfluous to its solution) the relevant data and order
could be inferred.
What was the prescribed order of choosing; and who lent Caliban an
umbrella?
==> logic/calibans.will.s <==
Let T be "person who saw Caliban in a green tie."
Let U be "person who lent Caliban an umbrella."
Then the data are:
(1) No T chooses before Low.
(2) Either Y.Y. was in Oxford in March 1920 or the first chooser is not
a U.
(3) Either Low is second or Critic is not last.
Consider first (3)
If it could be shown that Low is first, then from (3), Critic is not
last and therefore is second; i.e. the order is Low, Critic, Y.Y.
Next (1)
If both Critic and Y.Y. were T's would require Low first and (3) then
gives the order Low-Critic-Y.Y., ie. (2) would be superfluous. Hence
Critic and Y.Y. are not both T's.
If neither Critic nor Y.Y. were a T, (1) would be trivially true for
any ordering and therefore would give no information, i.e. would be
superfluous. Hence just one of Y.Y. and Critic is a T. It follows
that the only possible order in which Low is not first is:
Not T, Low, T
Now (2)
First if Y.Y was in Oxford in March 1920, nothing follows from (2)
about the order and (2) is superfluous. Hence Y.Y. was not in Oxford.
If Low were a U he would not, by (2) come first, and so by (1) the
order would be:
Not T, Low, T
i.e. (1) and (2) alone would fix an order, and (3) would be superfluous.
Hence Low is not a U.
It now follows, by the arguments just given for T's under (1) that just
one of Y.Y. and Critic is a U. If the same one is the T and the U (2)
follows from (1) (since Low is not a U); i.e (2) is superfluous. The
situation is therefore:
T's: just one of Y.Y. and Critic; not Low
U's: the other one of Y.Y; not Low
It now follows that "not T, Low, T" is impossible, for the "not T" is
the "U" and therefore, by (2), is not first. Hence Low is first, and
(3) gives the order:
Low, Critic, Y.Y.
Finally, Y.Y. is a T, and Critic is a U. For if Critic is a T, then
by (1) Low precedes Critic and hence (3) allows only "Low, Critic, Y.Y";
(2) is superfluous. I.e. Critic (only) lent Caliban an umbrella.
The problem is from _Problems Omnibus_ by Hubert Phillips,
Arco Publications, London, 1960. Hubert Phillips was a noted puzzelist
who contributed under his own name and the pseudonyms of "Caliban",
"T.O. Hare", and "The Doc".
==> logic/camel.p <==
An Arab sheikh tells his two sons that are 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 wiseman for advise. After hearing the advice they jump on the
camels and race as fast as they can to the city. What did the wiseman
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/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 you have mutually safe sex with three women with only 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.
==> 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/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?
==> logic/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.
==> 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 90-digit prime
numbers; if the result is tails, A multiplies 3 60-digit prime
numbers. 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.
==> 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/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 <==
It is tempting to argue as follows:
Since only one statement can be true (they are mutually contradictory),
therefore 99 are false. So, all are false except for statement 99.
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.
From _Litton's Problematical Recreations_
==> 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 2N-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.
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?
==> 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/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/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 =
-------------------------------------------------------------------------
==> 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?
==> 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
???
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.
?
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.
sails?
My tines be long,
My tines be short
My tines end ere
My first report.
What am I?
lightning
==> 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.
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> logic/situation.puzzles.s <==
Answers to Jed's List of Situation Puzzles
This is the list of answers to the puzzles in my situation puzzles
list. See that list for more details. This document also contains
variant setups and answers for some of the puzzles.
Section 1: "Realistic" situation puzzles.
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. 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. 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 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. 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. 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. The man has hiccups; the bartender scares them away by pulling a
gun.
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. 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. 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. He stabbed himself with an icicle.
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. 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. 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. 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. 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. 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)
1.17c. Rough sketch of variant: There were a mirror and a bottle on the
table, and sawdust on the floor. He came in and dropped dead. Answer: He
was a midget, but he wasn't aware of it, because the table used to be too
high for him to see his reflection in the mirror, until someone shortened
its legs. He was horrified by the discovery, and the shock killed him.
(vaguely remembered by Ivan A Derzhanski, who adds that this would be best
used as raw material for some elaboration. I agree; it's pretty
implausible as is)
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 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. 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.
1.21. 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. 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 realises 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. 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. 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 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. 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 is a kid who goes trick-or-treating for Halloween.
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 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. 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. The dead man was the driver in a hit-and-run acccident 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. 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. I'm told this is a true story. Windows in Paris at that time were
apparently imperfectly flat; they could act as lenses. One particularly
hot day, the sun shining in through such a window caused a woman's
lingerie (which she was wearing at the time, awaiting her husband's
return) to catch fire, and eventually the entire house caught and burned.
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. 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. 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. 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. 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. 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. 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. He is a passenger in an airplane and sees the bird get sucked into
an engine at 20,000 feet.
1.42. They're the remains of a melted snowman.
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 our 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. 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. 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. 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.
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. The man died from eating a poisoned popsicle.
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. 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. 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. 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. 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. 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. 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. (Randy was told this by
Gary Haas and Chris Englehard)
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. The movie is at a drive-in theatre.
Section 2: Double meanings, fictional settings, and miscellaneous others.
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. 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. 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. 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. He's a priest; he is marrying them to other people, not to himself.
2.6. The "island" is a traffic island.
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. The man is an astronaut out on a space walk.
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. The Eagle landed in the Sea of Tranquility and will likely remain
there for the foreseeable future.
2.11. It's a wolf pack; they've killed and eaten (most of) the man.
2.12. The dead man is Superman; the rock is Green Kryptonite. Invent a
reasonable scenario from there.
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. 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. 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 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. 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. It's a game of Monopoly.
2.19. The sun is shining; there's no rain.
2.20. It's daytime; the sun is out.
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. 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 is a horse.
2.24. Should be done orally; the envelope is an evelope of dye, and she's
dying some cloth, but it sounds like "opens an envelope and dies" if said
out loud.
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. 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.
==> 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?
==> logic/smullyan/fork.three.men.s <==
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 a person of the opposite
type to yourself if the left fork leads to Someplacerother, would he say yes?"
If the person asked is a truthteller, he will tell you what a liar would
say, which would be the wrong information. If the person asked is a liar,
he will either tell you what a liar would say, or he will lie about what a
truthteller would say. In either case, he will report the wrong information.
If the answer is yes, take the right fork, if no take the left 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 question to ask is: "Will the other person say the right fork leads to
Someplaceorother?" If the person asked says yes, then take the left fork,
else take the right fork.
If the person asked is the truthteller, then he correctly reports that the
liar will misinform you about the right fork. If he is the liar, then he
lies about what the truthteller will say. Either way, you should go the
opposite direction from the way that the person asked says the other person
will answer.
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 a
person of the opposite type to yourself if the left fork leads to
Someplacerother, would he say yes?"
If the person asked is a truthteller, he will tell you what a liar would
say, which would be the wrong information. If the person asked is a liar,
he will either tell you what a liar would say, or he will lie about what a
truthteller would say. In either case, he will report the wrong information.
If the answer is yes, take the right fork, if no take the left fork.
This solution also removes the problem that the men may not know the
other's identity.
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/liars.et.al.p <==
Of a group of n men, some always lie, some never lie, and the rest sometimes
lie. They each know which is which. You must determine the identity of each
man by asking the least number of yes-or-no questions.
==> logic/smullyan/liars.et.al.s <==
The real problem is to isolate the sometimes liars.
Consider the case of three men:
Ask man 1: "Does man 2 lie more than 3?"
If the answer is yes, then man 2 cannot be the sometimes liar.
Proof by analyzing the cases:
Case 1: Man 2 is not the sometimes liar.
Case 2: Man 2 is the sometimes liar, man 1 is the truth teller, and man 3 is
the liar. Then man 1 would not say that man 2 lies more than man 3.
Case 3: Man 2 is the sometimes liar, man 3 is the truth teller, and man 1 is
the liar. Then man 1 would not say that man 2 lies more than man 3.
QED.
Similarly, if the answer is no, then man 3 cannot be the sometimes liar.
Now ask the symmetric question of whichever man has been eliminated as the
sometimes liar. The answer will now allow you to determine the identity
of the sometimes liar. To determine the identity of the two remaining men, ask
some question like "Does 1=1?" which is always true.
This is not the only way to solve this problem. You could have asked the
question which is always true (or false) second, which would now establish
the identity of either the liar or the truth teller. Then ask the third
question of this man to find out which of the other two is the sometimes
liar.
This problem requires three questions, whether or not they are yes-or-no
questions. In order to identify all three men, you must identify the
sometimes liar. You cannot identify the sometimes liar in one question
since you may be asking it of the sometimes liar, and any answer from him
conveys no information at all. Therefore at least two questions are
necessary to identify the sometimes liar. Once the sometimes liar is
identified, you still need one more question at least to identify the
remaining men. Therefore, three questions are required.
Suppose we have two truth-tellers, two liars, and two randomizers.
The answer is 8. A proof follows.
For brevity, "T" means truth-teller, "L" liar, "R" randomizer, "P" predictable
(either T or L). Define a _pattern_ to be one of the C(6,2)=15 permutations
of RRPPPP (each of which has C(4,2)=6 interpretations of the Ps as 2 Ts and 2
Ls). For any question Q, let !Q denote the question "If I were to ask you Q,
would you answer Yes?". Note that question !Q directed toward any P will
yield a truthful answer to question Q; in other words, a "Yes" answer to !Q
means that either Q is true or the respondent is an R, whereas "No" means that
either Q is false or the respondent is an R.
Ask #1, !"Are both Rs in the set {#2, #3, #4}?". "No" implies that at most
one of {#2, #3, #4} is an R. "Yes" implies that at most one of {#2, #5, #6}
is an R. Without loss of generality, assume the former.
Ask #2, !"Is #3 an R?". "No" implies that #3 is a P. "Yes" implies that #4
is a P.
Having identified someone as a P, there are at most C(5,2)=10 possible
patterns, and hence at most 10*6=60 possible results. We can determine which
one reflects reality with at most 6 more questions with a binary search. (At
each step, bisect the set of possible answers, and ask the question !"Is the
correct pattern in the first subset?".)
Now, let's show that it can't be done in 7.
After asking your first two questions, renumber if necessary so that the first
question was directed to #1 and the second to #2. (If you asked the same
person twice, you're even worse off than in the analysis below.) You have no
way to rule out the possibility that both are Rs, so pattern RRPPPP yields 6
possibilities. Of the four patterns RPRPPP RPPRPP RPPPRP RPPPPR, your first
question gave no information and the second had one bit; so at best you can
eliminate half of these 4*6 possibilities, leaving 12. Similarly for the four
patterns PRRPPP PRPRPP PRPPRP PRPPPR there remain at least 12 possibilities.
Of the remaining 6 patterns PPRRPP PPRPRP PPRPPR PPPRRP PPPRPR PPPPRR, your
two bits of information can eliminate 3/4 of the 6*6, leaving 9. Thus, after
two questions there are at least 6+12+12+9=39 arrangements that could have
given the answers you heard; your five remaining questions have only 32
possible replies, so you can't distinguish them.
==> 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.
==> logic/smullyan/priest.p <==
A priest takes confession of all the inhabitants in a small town. He
discovers that in N married pairs in the town, one of the pair has
committed adultery. Assume that the spouse of each adulterer does not
know about the infidelity of his or her spouse, but that, 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. The
priest, who is an Old Testament type, decides that he should do
something about the situation. He cannot break the confessional, but
being an amateur logician of sorts, he hits upon a plan to do God's
work. He announces in Mass one Sunday that the spouse of each
adulterer has the moral obligation to punish his or her adulterous
spouse by publicly denouncing them in church, and that he will make
time during his next Sunday service for this, and continue to do so
until all adulterers have been denounced. 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 waits for the other
two to get up. When they do not, they assume that they did not get up
because they did not know about the other person (in other words, they
hypothesize that each of the two other victims thought there was only
one adulterer). However, each victim reasons, the two will now realize
that they 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.
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?
==> 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/timezone.p <==
Two people are talking long distance on the phone; one is in an East-
Coast state, the other is in a West-Coast state. 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 didst 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??
======
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> 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 24 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/flips.p <==
Consider a run of coin tosses: HHTHTTHTTTHTTTTHHHTHHHHHTHTTHT
Define a success as a run of one H or T (as in THT or HTH). Use two
different methods of sampling. The first method would consist of
sampling the above data by taking 7 groups of three. This translates
into the following sequences HHT, HTT, HTT, THT, TTT, HHH, and THH.
In this sample there was one success, THT. The second method of
sampling could be gotten by taking the groups in a serial sequence.
Another way of explaining the method would be to take the tosses 1, 2,
and 3 as the first set then tosses 2, 3, and 4 as the second set and
so on to produce seven samples. The samples would be HHT, HTH, THT,
HTT TTH, THT, and HTT, thus giving two success, HTH and THT. With
these two methods what would be the expected value and the standard
deviation for both methods?
==> probability/flips.s <==
Case 1:
Let us start with a simple case: Define success as T and consider
sequences of length 1. In this case, the two sampling techniques are
equivalent. Assuming that we are examining a truly random sequence of
T and H. Thus if n groups of single sequences are considered or a
sequence of length n is considered we will have the following
statistics on the number of successes:
Mean = n/2, and standard deviation (sd) = square_root(n)/2
Case 2:
Define success as HT or TH: Here the two sampling techniques need to
be distinguished:
Sampling 1: Take "n" groups of two: Here probability of getting success in
any group is 1/2 (TH and HT out of 4 possible cases). So the mean and the
standard deviation is same as described in case 1.
Sampling 2: Generate a sequence of length "n". It is easy to show
that (n-1) samples are generated. The number of successes in this
sequence is same as the number of T to H and H to T transitions. This
problem is solved in John P. Robinson, Transition Count and Syndrome
are Uncorrelated, IEEE Transactions on Information Theory, Jan 1988.
I will just state the results:
mean = (n-1)/2 and standard deviation = square_root(n-1)/2.
In fact, if you want "n" samples in this case you need to generate
length (n+1) sequence. Then the new mean and standard deviation are
the same as described in Sampling 1. (replace n by n+1). The
advantage in this sampling (i.e., sampling 2) is that you need only
generate a sequence length of (n+1) to obtain n sample points as
opposed to 2n (n groups of 2) in Sampling 1.
OBSERVATION: The statistics has not changed.
Case 3:
Define success as THT and HTH.
Sampling 1: This is a simple situation. Let us assume "n" samples
need to be generated. Therefore, "n" groups of three are generated.
The probability of a group being successful is 1/4 (THT and HTH out of
8 cases). Therefore mean and standard deviation are:
mean= n/4 and sd= square_root(7n)/4
Sampling 2: This is not a simple situation. Let us generate a random
sequence of length "n". There will be (n-2) samples in this case.
Use an approach similar to that followed in the Jan 88 IEEE paper. I
will just state the results. 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
If we want to compare Sampling 1 with Sampling 2 then in Sampling 2 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 (same as that in sampling 1)
sd = square_root(5n-2)/4 (not the same as in sampling 1)
sd in sampling 2 < sd in sampling 1.
This difference can be explained by the fact that in serial sampling
there is dependence on the past state. For example, if the past
sample was TTT then we know that the next sample can't be a success.
This was not the case in Case 1 or Case 2 (transition count). For
example, in case 2 success was independent of previous sample. That is
if my past sample was TT then my next sample can be TT or TH. The
probability of success in Case 1 and Case 2 is not influenced by past
history).
Similar approach can be followed for higher dimensional cases.
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 more flushes than the set of all possible 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 flushs.
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 <==
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.
Think about it. 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 <==
You could make a graph of the size of the
smallest interval versus x and y. We know that the height of the
graph is 0 along all the edges of the unit square where it is defined
and on the line x=y. It achieves its maximum of 1/3 at (1/3,2/3) and
(2/3,1/3). Assuming the graph has no curves or bizzare jags, the
volume under the graph must be 1/9 and so the average height must also
be 1/9.
==> 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 and
that if the light is green in one direction it is red in the other, 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/particle.in.box.p <==
A particle is bouncing randomly in a two-dimensional box. How far does it
travel between bounces, on avergae?
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 probablity 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/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/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.
==> 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.
==> series/series.02.p <==
H, H, L, B, B, C, N, O, F ?
==> series/series.02.s <==
Answer 1: N, N, M, A The symbols for the elements.
Answer 2: N, S, M, A 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.
==> 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. 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. 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. 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 Garden: Arlington, Berkeley, Clarendon, Dartmouth,
Exeter, Fairfield, Gloucester, Hereford, Massachusetts Ave.
==> 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 <==
Army officer ranks, descending.
==> series/series.08b.p <==
A, V, R, R, C, C, L, L, L, E, ?
==> series/series.08b.s <==
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. States in Constitution ratification order.
==> series/series.11.p <==
R O Y G B ?
==> series/series.11.s <==
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. 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 cursive insides.
==> series/series.15.p <==
A, E, F, H, I, ?
==> series/series.15.s <==
L, M, N, O, S, U. 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, T, E. 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 <==
1 01 01011 0101101011011 0101101011011010110101101101011011 etc.
Each string is formed from the previous string by substituting '01' for '1'
and '011' for '0' 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 ...
i 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 ...
Take i in binary, for each 1 bit (in i, 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 <==
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 ...
Number of factors in prime factorization of i.
==> 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 <==
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 ...
Sum of factors in prime factorization of i.
==> 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 n.
==> 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.
==> 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 Fransisco 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 about 1958) with 808, Puerto Rico
with 809, Newfounland 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/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.
Chris Cole
Last-modified: 1992/09/20
Version: 3
s Sunday School
ss liner
ss saints
ss ship
ss steamship
st good man
st hush
st little way
st paragon
st road
st saint
st silence
st stone
st street
st stumped
st thoroughfare
st way
st weight
stag speculator
sten gun
stet don't change it
stir prison
stop traffic signal
sts saints
sty filthy place
stye eyesore
su Soviet Union
sub U-boat
sub stand-in
sub substitute
sub warship
supra over
sure certain
sw Cornwall
sw Devon
sw bridge opponents
sw quarter
sw south-west
swift screecher
swiss roll jammed cylinder
sx Essex
t Thailand
t Tuesday
t bandage
t bar
t bone
t cart
t cloth
t cross
t crossed
t half dry
t hundred and sixty
t hundred and sixty thousand
t junction
t model +
t peg
t perfect letter
t plate
t rail
t shirt
t short time
t square
t tau
t te
t tea
t tee
t tesla
t the
t time
t ton(ne)
t tritium
ta Territorial Army
ta army
ta cheers
ta reserves
ta soldiers
ta terriers
ta territorials
ta thank you
ta thanks
ta volunteers
tab label
tace silence
tag label
tan beat
tan brown
tan maths function
tar able seaman
tar art nouveau
tar sailor/salt/seaman
tata Tosti's song
tata goodbye
tate gallery
tau cross
tay river
tb torpedo boat
td medal
te Lawrence
te note
tea leaves
tec detective
ted Edward
ted Heath
tee peg
teen old injury
tees river
tell archer
temp secretary
ten PM's address
tene old injury
tent wine
ter three (triple)
ter thrice
test educational journal
test examination
test match
teth Hebrew letter
the article
the articles - English
ti note
tic note
tic spasm
tic twitching
tier row
time father
times daily
timon misanthrope
tin can
tin cash
tin money
tin vessel
tiny small
tion empty container
tit bird
tit inferior horse
tit poor horse
tnt big banger
tnt explosive
tod fox
todo commotion
toe extremity
toe member
tom big bell
tom cat
tome book
ton fashion
ton hundred
ton large amount
ton weight
tonne weight
tor hell
tor hill
tor mountain
tor point
tor prominence
tory Conservative
tory party
tory politician
tp teepee
tr Turkey
tr transaction
tr translation
tram transport
tree actor
tres very (Fr.)
tri three (triple)
tri thrice
troy ancient city
troy old city
try attempt
try essay
ts teas
ts tees
tt abstaining
tt dry
tt on the wagon
tt race
tt teas
tt tees
tt teetotal
tt teetotaller
tt thank you
tu tradesmen
tuck friar
twelve eec
two company
u Conservative
u Uruguay
u Utah
u about turn
u acceptable
u bend
u boat
u educational establishment
u ewe
u film
u for all to see
u high class
u on view to all
u posh
u socially acceptable
u suitable for children
u superior
u trap
u tube
u turn
u union/Unionist
u universal
u university
u upper class
u uppish
u upsilon
u uranium
u yew
u you
uc you see
uk United Kingdom
uk this country
uk this island
ule rubber
ult last month
um doubt
um hesitation
un United Nations
un international
un number one (Fr.)
un one
un one (dialect)
un peacekeepers
una number one (Ital.)
unco very (Scot.)
une number one (Fr.)
uno international organisation
uno number one (Ital.)
up at university
up excited
up in court
up mounted
up riding
up superior
uq you queue
ur ancient city
ur hesitation
ur old city
ur primitive
ur you are
ure river
uru Uruguay
us America
us American
us as above
us ewes
us no good
us transatlantic
us undersecretary
us use
us useless
us yews
us you and me
usa America
use application
use custom
use employ(ment)
use practice
use practise
ussr Soviet Union
ut note
ute half minute
uu ewes
uu use
uu yews
ux wife
v Vatican
v against
v agent
v bomb
v day
v five
v look
v neck
v neckline
v notch
v opposing
v see
v sign
v vanadium
v vee
v velocity
v verb
v verse
v versus
v very
v victory
v vide
v volt
v volume
v win
va Virginia
vad nurse
vale farewell
vale goodbye
vat tax
vau Hebrew letter
vb verb
ve victory
ver rev up
very light
vet surgeon
vg for example
vi half dozen
vi six
via old way
vid see
vid tanner/sixpence
vide look
vide see
vin French wine
vip big noise
vip tanner/sixpence
vir man/Roman
vis viscount
vj victory
vo left hand
vol book
vol volume
vy various years
w Wednesday
w Welsh
w William
w bridge players
w direction
w point
w quarter
w tungsten
w watt
w weak
w west(ern)
w whole numbers
w wicket
w width
w wife
w woman
ward disadvantage (drawback)
washington young feller
we partnership
we you and I
wee little
wee minor
wee small
who doctor
wi Mayfair
wi West Indies
wi Westminster
winner fabulous tortoise
wise youth leaders
wist knew (old word)
women monstrous regiment
woof bark
wt small weight
wt weight
x Christ
x PM's address
x Xmas
x across
x body
x chi
x chromosome
x cross
x draw
x ex,Exe
x film
x illiterate's signature
x kiss
x particle
x ray
x sign of love
x sign of the times
x spot marked
x ten
x ten thousand
x thousand
x times
x unknown
x vitamin
x vote
x wrong sign
x xi
xc ninety
xi eleven
xi side
xi team
xl excel
xv side
xv team
y alloy
y chromosome
y level
y measure
y moth
y one hundred and fifty
y one hundred and fifty thousand
y track
y unknown
y why
y yard
y year
y yen
y young
y yttrium
yard detectives
yd measure
ye the (old word)
ye you (old word)
yea agreement
yew tree
yr year
yr your
ys wise
ys youth leaders
yt that (old word)
yu jade
yule you will, say
yy wise
z Zambia
z bar
z bend
z cedilla
z final letter
z integers
z izzard
z last character
z last letter
z omega
z seven
z seven thousand
z sound of sleep
z zed
z zee
z zero
z zeta
zo cross *
zr Zaire
zz (sound of) snoring
----------------------------------------------------------------------
--
Ross Beresford, | Email (trusted): rber...@cix.compulink.co.uk
10 Wagtail Close, | (work): ro...@siesoft.co.uk
Twyford, Reading, | (under test): ro...@dickens.demon.co.uk
RG10 9ED, UK |
==> games/crosswords/cryptic/double.p <==
Each clue has two solutions, one for each diagram; one of the answers
to 1ac. determines which solutions are for which diagram.
All solutions are in Chamber's and Webster's Third except for one solution
of each of 1dn, 3dn and 4dn, which can be found in Webster's 2nd. edition.
#######################################################################
#1 |2 | | |3 |4 |5 #1 |2 | | |3 |4 |5 #
# | | | | | | # | | | | | | #
#----+----###########----#----#----#----+----###########----#----#----#
#6 | |7 | | # # #6 | |7 | | # # #
# | | | | # # # | | | | # # #
#----#----#----######----#----#----#----#----#----######----#----#----#
# # # #8 | | | # # # #8 | | | #
# # # # | | | # # # # | | | #
#----#----#----######----#----#----#----#----#----######----#----#----#
#9 | | | # # # #9 | | | # # # #
# | | | # # # # | | | # # # #
#----#----#----######----#----#----#----#----#----######----#----#----#
# # #10 | | | | # # #10 | | | | #
# # # | | | | # # # | | | | #
#----#----#----###########----+----#----#----#----###########----+----#
#11 | | | | | | #11 | | | | | | #
# | | | | | | # | | | | | | #
#######################################################################
Ac.
1. What can have distinctive looking heads spaced about more prominently
right. (7)
6. Vermin that can overrun fish and t'English tor perhaps. (5)
8. Old testament reversal - Adam's conclusion, start of sin.
Felines initially with everything there. (4)
9. Black initiated cut, oozed out naturally. (4)
10. For instance, 11 with spleen dropping I count? (5)
11. Provoked explosion of grenade. (7)
Dn.
1. Some of club taking part in theatrical function, for the equivalent
of a fraction of a pound. (6)
2. Close-in light meter in one formation originally treated as limestone. (6)
3. Xingu River hombres having symmetrical shape. (5)
4. About sex-appeal measure - what waitresses should be? (6)
5. Old penny, least damaged, was preserved. (6)
7. IRA to harm ruling Englishman; extremes could be belonging to group. (5)
==> games/crosswords/cryptic/double.s <==
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|r e d c a p s|d e x t r a l|
+ + +-+-+ + + + + +-+-+ + + +
|o t t e r|o|a|r o a c h|s|a|
+ + + +-+ + + + + + +-+ + + +
|u|a|h|f a l l|a|z|m|t o m s|
+ + + +-+ + + + + + +-+ + + +
|b l e d|r|i|t|c o o n|m|i|t|
+ + + +-+ + + + + + +-+ + + +
|l|o|i r a t e|m|o|n o b l e|
+ + + +-+-+ + + + + +-+-+ + +
|e n r a g e d|a n g e r e d|
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Notes.
Left grid: Ac. 1. R + spaced (anag). 6. T'E tor (anag). 8. F-all. 9. B-led.
10. I-rate. Dn. 1. Ro-ub-le. 2. T.A.L. in one (anag). 4. it in pole.
5. anag of D+least. 7. anag of initial letters.
Right grid: Ac. 1. D-extra-L. 6. 3 mngs. 8. OT (rev) + m-s.
9. initial letters. 10. No.-b(i)le. Dn. Dra-c-ma. 2. Zoo(m) in one (anag).
3. hidden. 4. SA (rev) + mile. 5. anag of D+least. 7. anag of final letters.
--------------------------------------------------------------------
How I built it: it was hard!
Basically, I started with a couple of word pairs which were easy to clue
(e.g. enraged/angered - same meaning and anagrams of each other)
and built a grid around them, trying to ensure corresponding words
had something in common, either in meaning (their, among) or structure,
(EtalON, EOzooN) and making sure that there was at least one word
which could be used to distinguish the two grids (dextral).
The clues were built in one of two ways:
either the words had a common definition, and so a subsidiary indication
which could refer to either was needed; or it was necessary to define each
word in such a way that it was a subsidiary definition for all or part
of the corresponding word, and deal with any remaining parts as before.
I think the single hardest part was finding a definition of "interferometer"
which could also be interpreted as "zoo" or "ozo".
Roy
r...@ukc.ac.uk
==> games/crosswords/cryptic/intro.p <==
What are the rules for cluing cryptic crosswords?
==> games/crosswords/cryptic/intro.s <==
This is a brief set of instructions for solving cryptic crossword puzzles
for those of you who are intrigued by these puzzles, but haven't known how
to begin solving them. For a more complete introduction, send a
self-addressed, stamped envelope to The Atlantic Puzzler, 745 Boylston
Street, Boston, Mass. 02116.
The characteristic common to all cryptic crossword puzzles is the format of
the clues. Each clue is a miniature word puzzle consisting of a straight
definition of the answer and a cryptic definition of the answer. For
example,
Axle is poorly splined (7)
yields SPINDLE. Axle is the straight definition. The cryptic definition
(poorly splined) indicates an anagram of "splined". The number in
parentheses is the number of letters in the answer. Punctuation and
capitalization may be ignored in interpreting the clues.
There are only eight categories of clues, as follows:
1. Anagram
An anagram is a word formed by mixing up the letters of another word. An
anagram clue is indicated by some word that means "mixed up", for
example, out, crazy, bizarre, insane, etc. One or more words may
contribute to the anagram. For example:
Tim goes insane from selfishness (7)
for EGOTISM (anagram of "Tim goes")
2. Double Definition
A double definition is simply two definitions of the word. Most two-word
clues are double definitions. For example:
Release without charge (4)
for FREE
3. Container
A container clue indicates that something is to be put in (or wrapped
around) something else. A container is indicated by phrases such as
eaten by, contains, in, gobbles, etc. For example:
In Missouri, consumed by fear (7)
for AMONGST (MO = Missouri in ANGST = fear)
4. Hidden Word
A hidden word is a word embedded in another word or words. It is
indicated by phrases such as spot in, hides, at the heart of, covers,
etc. For example:
Worn spot in paper at typo (5)
for RATTY (find ratty in "paper at typo")
5. Reversal
A reversal is a definition of a word with the letters reversed. It is
indicated by words such as back, reversed, up (for down clues), leftward
(for across clues), etc. For example:
Egad! Ray entirely reversed the lot of cloth (7)
for YARDAGE ("Egad! Ray" reversed)
6. Homophone
A homophone definition is a definition of a word that sounds the same as
the answer, but is spelled differently. A homophone is indicated by
words such as in audience, I hear, mouthed, verbally, etc. For example:
Regrets prank, I hear (4)
for RUES (the homophone is RUSE = prank)
7. Charade
In a charade, the pieces of the word are "spelled" out in order. There
are no auxiliary words that indicate a charade. For example:
Excite a jerk extremist (7)
for FANATIC (FAN = excite, A, TIC = jerk)
8. Deletion
A deletion is a clue where you are instructed to remove a part of some
word to make another word. For example,
Times with poor wages (4)
for AGES (with-poor WAGES, where with is abbreviated by W)
Often the clue types are combined. Some common examples are 1) hidden word
reversals where the answer is found backwards embedded in other words, and
2) containers or charades where the parts are anagrams. For example:
Car shops have broken gear immersed in gasoline. (7)
for GARAGES (RAGE = gear anagram in GAS = gasoline)
All manner of common abbreviations, acronyms, and other symbology such as
roman numerals are allowed. For example:
c one hundred, cup, or centigrade
vi six
h hot
s small
ca california
Two punctuation marks at the end of the clue have been reserved for special
meaning. A question mark (?) indicates that the straight clue is not
entirely straight (usually a pun). For example:
I tie down mascara holder soundly? (7)
for EYELASH (homophone of "I lash", mascara holder is a punning
definition of EYELASH)
An exclamation point (!) indicates that some part (usually all) of the clue
overlaps. For example, the straight definition may also be the anagram
indicator. Here is an example that entirely overlaps:
A moped also has these! (6)
for PEDALS (hidden word)
Here, the entire clue indicates the hidden word, but the entire clue is
also a straight definition of the answer.
Give it a try! Cryptic crossword puzzles are a lot of fun.
-- Steve Koehler
ucsd.edu!telesoft!koehler
telesoft!koe...@ucsd.edu
koe...@telesoft.com
==> 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?
==> 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.
==> games/hi-q.p <==
What is the quickest solution of the game Hi-Q (also called Solitair)?
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
neighboor (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).
==> 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
==> games/jeopardy.p <==
What are the highest, lowest, and most different scores contestants
can achieve during a single game of Jeopardy?
==> games/jeopardy.s <==
highest: $283,200.00, lowest: -$29,000.00, biggest difference: $309,700.00
(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 luckier 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.
==> games/knight.tour.p <==
For what board sizes is a knight's tour possible?
==> games/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.
==> 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?
==> 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
==> games/othello.p <==
How good are computers at Othello?
==> games/othello.s <==
The interesting game in which computers are undoubted masters of all they
survey is Othello, where Kai-Fu Lee's (CMU) program "Bill" is so good it can
only play itself to learn to get better. Bill has a fantastically
correct and efficient evaluation function, that recently has been further
improved by learning coefficients for additional terms made up of the
pair-wise combination of the four old terms. This improved the quality
of the play approximately as much as searching an extra two ply.
Bill is so good it can beat lots of players with no search at all. Its
6 or 7 ply search sweeps aside all opposition (though Kai-Fu says that some
very good players are now coming along in Japan, and he is not sure whether
Bill would beat them). One interesting question remaining in Othello is
the game theoretic value of the starting position. Bill's results seem
to indicate that the first player has an advantage. It appears that,
since Kai-Fu has published all his evaluation material, someone could
build an Othello machine, and produce a constructive proof (as was done
for Cubic) that it is a win for the first player.
==> games/risk.p <==
What are the odds when tossing dice in Risk?
==> games/risk.s <==
Attacker using 3 dice, Defender using 2:
Probability that Attacker wins 2 = 2323 / 7776
Probability that Attacker wins 1 = 3724 / 7776
Probability that Attacker wins 0 = 1729 / 7776
Attacker using 3 dice, Defender using 1:
Probability that Attacker wins 1 = 855 / 1296
Probability that Attacker wins 0 = 441 / 1296
Attacker using 2 dice, Defender using 2:
Probability that Attacker wins 2 = 225 / 1296
Probability that Attacker wins 1 = 630 / 1296
Probability that Attacker wins 0 = 441 / 1296
Attacker using 2 dice, Defender using 1:
Probability that Attacker wins 1 = 125 / 216
Probability that Attacker wins 0 = 91 / 216
Attacker using 1 dice, Defender using 2:
Probability that Attacker wins 1 = 90 / 216
Probability that Attacker wins 0 = 126 / 216
Attacker using 1 dice, Defender using 1:
Probability that Attacker wins 1 = 15 / 36
Probability that Attacker wins 0 = 21 / 36
==> games/rubiks.clock.p <==
How do you quickly solve Rubik's clock?
==> games/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.
==> games/rubiks.cube.p <==
What is known about bounds on solving Rubik's cube?
==> games/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 by
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 apposed 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.
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.)
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.
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? Anyone
understand Morwen Thistlethwaite's 50 move solution to Rubik's Cube? I'd
love to hear from you.
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, and I'm missing
#3.
I have: #1, May 1982
#2, Aug 1982
#3, Aug 1983
I am willing to trade photocopies with anyone to obtain #3.
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), the Puck, and Hungarian Rings.
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:
E. C. Turner & K. F. Gold, "Rubik's Groups", American Mathematical Monthly,
vol. 92 (1985), pp. 617-629.
Cubelike Puzzles - What Are They and How Do You Solve Them?
J.A. Eidswick A.M.M. March, 1986
Rubik's Revenge: The Group Theoretical Solution
Mogens Esrom Larsen A.M.M. June-July, 1985
The Group of the Hungarian Magic Cube
Chris Rowley Proceedings of the First Western Austrialian
Conference on Algebra, 1982
Rubik's Cubic Compendium
Erno Rubik, Tamas Varga, et al
(Ed by David Singmaster)
Oxford University Press, 1987
(Some chapters on mathematics of the cube.)
David Singmaster, _Notes on Rubik's `Magic Cube'_
"Winning Ways"
by
Berlekamp, Elwyn R.
Conway, John H.
Guy, Richard K.
Volume two, pages 760-768, 808, 809
==> games/rubiks.magic.p <==
How do you solve Rubik's Magic?
==> games/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.
==> games/scrabble.p <==
What are some exceptional scrabble games?
==> 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 tiles:
HUMANLY
FATUOUS
AMAZING
EERIEST
ROOFING
TOILERS
QUIXOTE
JEWELRY
CAPABLE
PREVIEW
BIDDERS
HACKING
OVATION
DONATED
Chris Cole
Last-modified: 1992/09/20
Version: 3
==> 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 N balls and a balance scale and told that
one ball is slightly heavier or lighter than the other identical
ones. The scale lets you put the same number of balls on each side
and observe which side (if either) is heavier.
1. What's the minimum # of weighings X (and way of doing them)
that will always find the unique ball and whether it's heavy or light?
2. If you are told the unique ball is, in fact, heavier than the
others, what's the minimum # of weighings Y to find it?
==> logic/weighing/balance.s <==
Martin Gardner gave a neat solution to this problem.
Assume that you are allowed N weighings. Write down the 3^N possible
length N strings of the symbols '0', '1', and '2'. Eliminate the three
such strings that consist of only one symbol repeated N 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^N-3)/2 strings left. This is how many balls you can
handle in N weighings.
Perform N weighings as follows:
For weighing I, take all the balls that have a 0 in string
position I, and weigh them against all the balls 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 N weighings, you have written down an N symbol string. If
your string matches the string on one of the balls, then that is the
odd ball, 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 balls, and that ball is lighter than
the others.
Note that if you only have to identify the odd ball, but don't have to
determine if it is heavy or light, you can handle (3^N-3)/2+1 balls.
Label the extra ball with a string of all 1's, and use the above
method.
Note also that you can handle (3^N-3)/2+1 balls if you *do* have to
determine whether it is heavy or light, provided you have a single reference
ball available, which you know has the correct weight. You do this by
labelling the extra ball 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 ball than the other each time. So you put the
reference ball on the other side of the scales to the "all 2s" ball 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.
==> 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 a
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 weight 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/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 used 51 gummy drop bears: from the 7 boxes he took 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/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)
==> 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/bicycle.p <==
A boy, a girl and a dog go for a 10 mile walk. The boy and girl can
walk 2 mph and the dog can trot at 4 mph. They also have bicycle
which only one of them can use at a time. When riding, the boy and
girl can travel at 12 mph while the dog can peddle at 16 mph.
What is the shortest time in which all three can complete the trip?
==> physics/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.
==> physics/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?
==> physics/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.
==> 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/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/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?
==> physics/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.
==> physics/magnets.p <==
You have two bars of iron. One is magnetic, 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 (ie there is no magnetic force,
since the two poles cance 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 (ie 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
when you drink it some number of minutes later. Do you add milk 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 Amidextrous 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. The rope weighs
4 ounces per foot. The combined ages of the monkey and it's mother
is 4 years. The weight of the monkey is as many pounds as the mother
is years old. The mother is twice as old as the monkey was when the
mother was half as old as the monkey will be when the monkey is 3 times
as old as the mother was when she was 3 times as old as the monkey.
The weight of the rope and the weight is one-half as much again as the
difference between the weight of the weight and the weight of the weight
plus the weight of the monkey.
How long is the rope?
==> physics/monkey.s <==
The most difficult thing about this puzzle, as you probably expected,
is translating all the convoluted problem statements into equations...
the solution is pretty trivial after that. So...
Let:
m represent monkey
M represent mother of monkey
w represent weight
r represent rope
W[x] = present weight of x (x is m, M, w, or r)
A[x,t] = age of x at time t (x is M or M, t is one of T1 thru T4)
T1 = time at which mother is 3 times as old as monkey
T2 = time at which monkey is 3 times as old as mother at T1
T3 = time at which mother is half as old as monkey at T2
T4 = present time
For the ages, we have:
A[M,T1] = 3*A[m,T1]
A[m,T2] = 3*A[M,T1] = 9*A[m,T1]
A[M,T3] = A[m,T2]/2 = 9*A[m,T1]/2
A[m,T3] = A[m,T1] + (T3-T1)
= A[m,T1] + (A[M,T3]-A[M,T1])
= A[m,T1] + (9*A[M,T1]/2 - 3*A[m,T1])
= 5*A[m,T1]/2
A[M,T4] = 2*A[m,T3] = 5*A[m,T1]
A[m,T4] = A[m,T1] + (T4-T1)
= A[m,T1] + (A[M,T4]-A[M,T1])
= A[m,T1] + (5*A[m,T1] - 3*A[m,T1])
= 3*A[m,T1]
The present ages of monkey and mother sum to 4, so we have
A[m,T4] + A[M,T4] = 4
3*A[m,T1] + 5*A[m,T1] = 4
8*A[m,T1] = 4
A[m,T1] = 1/2
Thus:
A[M,T4] = 5/2
A[m,T4] = 3/2
Now for the weights, translating everything to ounces:
Monkey's weight in lbs = mother's age in years, so:
W[m] = 16*5/2 = 40
Weight and monkey are same weight, so:
W[w] = W[m] = 40
The last paragraph in the problem translates into:
W[r]+W[w] = (3/2)*((W[w]+W[m])-W[w])
W[r]+ 40 = (3/2)*(( 40 + 40 )- 40 )
W[r]+ 40 = 60
W[r] = 20
The rope weighs 4 ounces per foot, so its length is 5 feet.
==> physics/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?
==> physics/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.
==> 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 are the resistances between lattices of resistors in the shape of a:
1. Cube
2. Platonic solid
3. Hypercube
4. Plane sheet
5. 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---|
2. Platonic Solids
Same idea for 8 12 and 20, since you use the symmetry to identify
equi-potential points. The tetrahedron is 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.
You can (and showed how to) use symmetry to obtain the equivalent resistance
of 1/2 between two adjacent nodes; but I doubt that symmetry alone will give you
even the equivalent resistance of 2/pi between two diagonally adjacent nodes.
[More generally, 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
Doesn't the resistance diverge in that case? The problem is that you can't
inject current at a point.
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 water (current) is flowing
downriver at a velocity of 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/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 kidding 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
made of uniform material, though of course not the same material. Without
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 planes speed, at
which point the plane will never finish the trip.
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).
==> 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/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/coincidence.p <==
Name some amazing coincidences.
==> probability/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.
==> probability/coupon.p <==
There is a free gift in my breakfast cereal. The manufacturers say
that the gift comes in four different colours, 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 colours, what is the
expected number of packets I must plough through to get all four?
Can you generalise to n colours and/or unequal probabilities?
Ken Arromdee
>==> 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.
While this answer is certainly correct, the solution omits the "expected"
answer.
The answer you're supposed to give is this: "We are not told what the original
sphere size is, or how wide the hole is. However, we are told that the
problem has a solution. Therefore, the solution does not depend on the size
of the sphere/hole, so we can choose any value we want. Therefore, we can
choose a hole width of 0. This automatically produces the answer that the
volume is equivalent to that of a 6" sphere."
--
"the bogosity in a field equals the bogosity imported from related areas, plus
the bogosity generated internally, minus the bogosity expelled or otherwise
disposed of." -- K. Eric Drexler
Ken Arromdee (UUCP: ....!jhunix!arromdee; BITNET: arromdee@jhuvm;
INTERNET: arro...@jyusenkyou.cs.jhu.edu)
Ken Arromdee
>Is there an algorithom 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.
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.
Ken Arromdee
>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.)
This one is from an Isaac Asimov story.