Article: 2407 of rec.puzzles
From: steed@dcs.qmw.ac.uk (Anthony Steed)
Subject: Tough Puzzle: Alphacipher
Organization: Computer Science Dept, QMW, University of London
Date: Wed, 24 Mar 1993 12:23:21 GMT

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

Anthony Steed

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  S P O I L E R
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Article: 2416 of rec.puzzles
From: grabiner@math.harvard.edu (David Grabiner) 
Date: 24 Mar 93 21:34:59 GMT

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 ?

> 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, grabiner@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.

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Article 2460 of rec.puzzles:
From: lbaum@atc.boeing.com (Larry Baum)
Date: Fri, 26 Mar 1993 00:58:48 GMT

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 ?
>
>
>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

The trick is to analyze the letter patterns and look for close matches;
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

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Article: 2736 of rec.puzzles
From: imc@comlab.ox.ac.uk (Ian Collier)
Date: 28 Mar 93 14:42:22 GMT

Warning: this is *not* a "nice" solution...

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.Collier@prg.ox.ac.uk | imc@ecs.ox.ac.uk

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