Tough Puzzle: Alphacipher
Anthony Steed
In England there is a magazine called Tough Puzzles full of various devious
types of logic problem. There's one type that has me stumped for a simple
and short method to get a handle on the solution, my methods lead to reams
and reams of scribbled notes. Here's a sample, does anyone have a bright idea,
about how it _should_ be done?
******
Alphacipher
The numbers 1 - 26 have been randomly assigned to the letters of the alphabet.
The numbers beside each word are the total of the values assigned to the
letters in the word. e.g for LYRE L,Y,R,E might equal 5,9,20 and
13 respectively or any other combination that add up to 47.
The problem - What is the value of D ?
BALLET 45 POLKA 59
CELLO 43 QUARTET 50
CONCERT 74 SAXOPHONE 134
FLUTE 30 SCALE 51
FUGUE 50 SOLO 37
GLEE 66 SONG 61
JAZZ 58 SOPRANO 82
LYRE 47 THEME 72
OBOE 53 VIOLIN 100
OPERA 65 WALTZ 34
******
Solution soon
Anthony Steed
David Grabiner
> ******
> Alphacipher
> The numbers 1 - 26 have been randomly assigned to the letters of the
> alphabet. The numbers beside each word are the total of the values
> assigned to the letters in the word. e.g for LYRE L,Y,R,E might equal
> 5,9,20 and 13 respectively or any other combination that add up to 47.
> The problem - What is the value of D ?
> BALLET 45 POLKA 59
> CELLO 43 QUARTET 50
> CONCERT 74 SAXOPHONE 134
> FLUTE 30 SCALE 51
> FUGUE 50 SOLO 37
> GLEE 66 SONG 61
> JAZZ 58 SOPRANO 82
> LYRE 47 THEME 72
> OBOE 53 VIOLIN 100
> OPERA 65 WALTZ 34
For anyone who wants to check answers without much of a spoiler, the
value of CHORD follows.
CHORD is 67. A full solution follows another page break.
Spoiler warning...
Subtracting one word from another gives the following equations:
a.NG-LO=24
b.GU-LT=20
c.SA-LO=8
d.SON-E=17
e.CAE-OO=14
f.GE-CLO=23
We can now derive these:
g.CN-E=1 (a-f)
h.SO-C=16 (d-g)
i.LOO-AC=8 (h-c)
j.EL=22 (e+i)
k.G-L=22 (GLEE-2j)
l.T-U=2 (k-b)
m.FUT=8 (FLUTE-j)
The only sets which sum to 8 are 1,2,5 and 1,3,4, so the only way to
satisfy l and m is F=4, T=3, U=1. Since L is at most 4 by k, L=2, and
thus G=24, E=20.
Taking what remains of some words:
BA=18
CO=19
OBO=33
SCA=29
SOO=35
SON=37
Now it is easier to see how to proceed. We have S-B=2, and thus BCA=27.
This gives C=9 since BA=18; thus O=10, B=13, A=5, S=15, N=12, and now we
can get several other letters by substitution.
CONCET=63 so R=11
LRE=33 so Y=14
OERA=46 so P=19
POLA=36 so K=23
UARTET=43 so Q=7
The remaining words become:
JZZ=53
XH=43
HM=29
VII=76
WZ=24
VII=76 must be 26+25+25.
X-M=14, and 22 and 8 are the only remaining pair; thus H=21.
JZZ=53 can now only be 17+18+18, and thus W=6.
1 2 3 4 5 6 7 8 9 10 11 12 13
U L T F A W Q M C O R N B
14 15 16 17 18 19 20 21 22 23 24 25 26
Y S J Z P E H X K G I V
This leaves D=16.
--
David Grabiner, grab...@zariski.harvard.edu
"We are sorry, but the number you have dialed is imaginary."
"Please rotate your phone 90 degrees and try again."
Disclaimer: I speak for no one and no one speaks for me.
Ted Schuerzinger
st...@dcs.qmw.ac.uk (Anthony Steed) writes:
> Alphacipher
> The numbers 1 - 26 have been randomly assigned to the letters of the alphabet.
> The numbers beside each word are the total of the values assigned to the
> letters in the word. e.g for LYRE L,Y,R,E might equal 5,9,20 and
> 13 respectively or any other combination that add up to 47.
>
> The problem - What is the value of D ?
I don't know, but I believe ALBUQUERQUE = 102
:-)
--Ted Schuerzinger
email: J.Theodore....@Dartmouth.EDU
The opinions expressed are not those of Dartmouth President James O.
Freedman. Then again, I wouldn't want to express such opinions.
Larry Baum
so after a bit of study we see:
SOPRANO = 82
SON + OPERA = 82 + E
SON = 17 + E
SONG = 17 + GE
44 = GE
FUU = FUGUE - GE = 6
So F = 4,
U = 1
LE = GLEE - GE = 22
From FLUTE, T = 3
Also LE = 22 & GE = 44 => G = L + 22
So G = 24,
L = 2 (since 1, 3 and 4 are used)
From FUGUE, E = 20
From OBOE, BOO = 33
From SOLO, SOO = 35 => S = B + 2
From BALLET, BA = 18 => SA = 20
From SCALE, SCA = 29
So C = 9
From CELLO, O = 10
From OBOE, B = 13
From SOLO, S = 15
From SCALE, A = 5
From SONG, N = 12
From CONCERT, R = 11
From LYRE, Y = 14
From OPERA, P = 19
From POLKA, K = 23
From, QUARTET, Q = 7
From VIOLIN, VII = 76 => either V = 24 (but G=24) or V = 26
So, V = 26
I = 25
Our remaining letters are: D H J M W X Z
Our remaining numbers are: 6 8 16 17 18 21 22
From THEME, HM = 29 => H,M are 8,21 in some order
From SAXOPHONE, XH = 43 => X,H are 21,22 in some order
So H = 21
M = 8
X = 22
From JAZZ, JZZ = 53 => J is odd. 17 is the only odd number left;
So J = 17
Z = 18
From WALTZ, W = 6
So D = 16 by elimination!
To recap:
A B C D E F G H I J K L M
5 13 9 16 20 4 24 21 25 17 23 2 8
N O P Q R S T U V W X Y Z
12 10 19 7 11 15 3 1 26 6 22 14 18
Ralph 'Hairy' Moonen
> > The problem - What is the value of D ?
>
> > BALLET 45 POLKA 59
> > CELLO 43 QUARTET 50
> > CONCERT 74 SAXOPHONE 134
> > FLUTE 30 SCALE 51
> > FUGUE 50 SOLO 37
> > GLEE 66 SONG 61
> > JAZZ 58 SOPRANO 82
> > LYRE 47 THEME 72
> > OBOE 53 VIOLIN 100
> > OPERA 65 WALTZ 34
SPOLIER!!!
Took me hours, but D=16.
Ian Collier
>In England there is a magazine called Tough Puzzles full of various devious
>types of logic problem. There's one type that has me stumped for a simple
>and short method to get a handle on the solution, my methods lead to reams
>and reams of scribbled notes. Here's a sample, does anyone have a bright idea,
>about how it _should_ be done?
Warning: this is *not* a "nice" solution...
> [what is the value of D given all the following sums of letters?]
>BALLET 45 POLKA 59
>CELLO 43 QUARTET 50
>CONCERT 74 SAXOPHONE 134
>FLUTE 30 SCALE 51
>FUGUE 50 SOLO 37
>GLEE 66 SONG 61
>JAZZ 58 SOPRANO 82
>LYRE 47 THEME 72
>OBOE 53 VIOLIN 100
>OPERA 65 WALTZ 34
Spoiler:
Well, I used a program to transform the above mechanically into 20
simultaneous equations, added "D=const, H=const, I=const, J=const, K=const,
L=const" to make 26 non-singular simultaneous equations in 26 unknowns with
6 unknowns on the right-hand side, and used Matlab to obtain 26 equations of
the form
A = -38 + 3K - 13L
B = 37 - 2K + 11L
...
Z = 48 - .5J - 1.5K + 6.5L
by inverting the matrix of coefficients (the choice of HIJKL as unknowns in
the above was arbitrary, except that my first choice of ABCEF made the
equations singular).
Then, noting that the equation for A guarantees L<4 and K>(38+13L)/3, I
wrote a program to search the solution space in K and L for one in which no
number was duplicated or out of range, ignoring all the letters that depend
on H, I or J. Fortunately, only one solution (K=23, L=2) resulted. I
amended the program to search the resulting space for values of J, and found
two solutions (17 or 21). It was easy enough to finish off by hand.
A=5 B=13 C=9 D=16 E=20 F=4 G=24 H=21 I=25 J=17 K=23 L=2 M=8 N=12 O=10 P=19
Q=7 R=11 S=15 T=3 U=1 V=26 W=6 X=22 Y=14 Z=18
The value of D is 16.
Ian Collier
Ian.C...@prg.ox.ac.uk | i...@ecs.ox.ac.uk