Help with a Alphacipher
Leigh Parry
I've done this old alphacipher but can't get a definite answer for 'D' as it
appears
there is two possible values.
I can't find the solution as I've just dug the puzzle out of an olf drawer.
Thanks in advance
BALLET 45
CELLO 43
CONCERT 74
FLUTE 30
FUGUE 50
GLEE 66
JAZZ 58
LYRE 47
OBOE 53
OPERA 65
POLKA 59
QUARTET 50
SAXOPHONE 134
SCALE 51
SOLO 37
SONG 61
SOPRANO 82
THEME 72
VIOLIN 100
WALTZ 34
Richard Heathfield
> Hi, this is my first post here, can anybody help.
> I've done this old alphacipher but can't get a definite answer for 'D' as
> it appears
> there is two possible values.
I think there's a bug in the puzzle. Explanation follows.
>
> I can't find the solution as I've just dug the puzzle out of an olf
> drawer.
>
> Thanks in advance
>
>
> BALLET 45
> CELLO 43
> CONCERT 74
> FLUTE 30
> FUGUE 50
> GLEE 66
> JAZZ 58
> LYRE 47
> OBOE 53
> OPERA 65
> POLKA 59
> QUARTET 50
> SAXOPHONE 134
> SCALE 51
> SOLO 37
> SONG 61
> SOPRANO 82
> THEME 72
> VIOLIN 100
> WALTZ 34
SOPRANO = OPERA + 17. Cancelling:
SNO = E + 17
SONG = 61
E + 17 + G = 61
E + G = 44
FUGUE = 50
FUU = 6
UU is even, so is 2, 4, or 6. 6 gives invalid value 0 to F. 2 gives
duplicate value 2 to F. So U is 4 and F is 1.
FLUTE = 30 so LET is 25.
Playing off GLEE = 66 and GE = 44, so LE is 22, giving T = 3, and we get the
following table:
L E G
2 20 24
3 19 25
5 17 27
The last row, and all higher values of L, give G too high a value. So we
have either L = 2 or L = 3. But T = 3, so L must be 2.
F = 1
L = 2
T = 3
U = 4
E = 20
G = 24
Time to substitute those values into the original equations:
BA 18
CO 19
NCR 32
JAZZ 58
YR 25
OBO 33
OPRA 45
POKA 57
QAR 20
SAXOPHON 114
SCA 29
SOO 35
SON 37
HM 29
VIOIN 98
WAZ 29
We note that SOO = 35, SON = 37, so N = O + 2, so CN = CO + 2, so CN = 21,
and NCR is 32, so R = 11:
1 2 3 4 5 6 7 8 9 10
F L T U
11 12 13 14 15 16 17 18 19 20
R E
21 22 23 24 25 26
G
All remaining letters must be at least value 5, so let's look at O, with
reference to SOO, SON, CO, and OBO:
O S N C B Invalid because...
5 25 7 14 23 BA = 18, B > 13
6 23 8 13 21 BA = 18, B > 13
7 21 9 12 19 BA = 18, B > 13
8 19 10 11 17 C conflicts with R
9 17 11 10 15 N conflicts with R
10 15 12 9 13
11 13 13 8 11 O = B, S = N
12 11 14 7 9 S conflicts with R
13 9 15 6 7
14 7 16 5 5 C conflicts with B
That leaves:
O S N C B A
10 15 12 9 13 5
13 9 15 6 7 11 clashes with R
So now our solution matrix reads:
1 2 3 4 5 6 7 8 9 10
F L T U A C O
11 12 13 14 15 16 17 18 19 20
R N B S E
21 22 23 24 25 26
G
Equations:
JZZ 53
Y 14
P 19
K 23
Q 4 Conflicts with U
XH 43
HM 29
VII 72
WZ 24
So either I've made a mistake, which is possible, or there's a bug in the
puzzle.
--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/2005
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
Richard Heathfield
> Leigh Parry said:
>
>> Hi, this is my first post here, can anybody help.
>> I've done this old alphacipher but can't get a definite answer for 'D' as
>> it appears
>> there is two possible values.
>
> I think there's a bug in the puzzle. Explanation follows.
Er, there's a bug in my solution! Hack, and hopefully a real solution,
coming up:
Nonsense.
FUU = 6, so:
U F
1 4
2 2
3 0
So F is 4 and U is 1. Let's see how that plays out, below:
> FLUTE = 30 so LET is 25.
>
> Playing off GLEE = 66 and GE = 44, so LE is 22, giving T = 3, and we get
> the following table:
>
> L E G
> 2 20 24
> 3 19 25
> 5 17 27
>
> The last row, and all higher values of L, give G too high a value. So we
> have either L = 2 or L = 3. But T = 3, so L must be 2.
>
> F = 1
> L = 2
> T = 3
> U = 4
> E = 20
> G = 24
Should be:
U = 1
L = 2
T = 3
F = 4
E = 20
G = 24
>
> Time to substitute those values into the original equations:
>
> BA 18
> CO 19
> NCR 32
> JAZZ 58
> YR 25
> OBO 33
> OPRA 45
> POKA 57
> QAR 20
QAR is actually 23.
> SAXOPHON 114
> SCA 29
> SOO 35
> SON 37
> HM 29
> VIOIN 98
> WAZ 29
> We note that SOO = 35, SON = 37, so N = O + 2, so CN = CO + 2, so CN = 21,
> and NCR is 32, so R = 11:
>
> 1 2 3 4 5 6 7 8 9 10
> F L T U
Swap F and U.
Swap F and U.
> Equations:
>
> JZZ 53
> Y 14
> P 19
> K 23
> Q 4 Conflicts with U
Q is actually 7, which is still free.
> XH 43
> HM 29
> VII 72
> WZ 24
>
> So either I've made a mistake, which is possible, or there's a bug in the
> puzzle.
So I made a mistake. Our solution matrix now looks like this:
1 2 3 4 5 6 7 8 9 10
U L T F A Q C O
11 12 13 14 15 16 17 18 19 20
R N B Y S P E
21 22 23 24 25 26
K G
Equations:
JZZ = 53
XH = 43
HM = 29
VII = 72
WZ = 24
If VII is 72, then both V and I must be at least 20, and only 21, 22, 25, 26
are available:
I II V
21 42 30
22 44 28
25 50 22
26 52 20
Of these, only 25/22 is legal and non-clashing.
1 2 3 4 5 6 7 8 9 10
U L T F A Q C O
11 12 13 14 15 16 17 18 19 20
R N B Y S P E
21 22 23 24 25 26
V K G I
JZZ = 53
XH = 43
HM = 29
WZ = 24
WZ = 24, so we're looking at 6/18 or 8/16. The very highest Z can be, then,
is 18, so ZZ is at most 36, so J must be at least 53-36 = 17. Also, J must
be odd. So it's either 17 or 21.
J Z W
17 18 6
21 16 8
Now, XH is 43 and HM is 29, so XH = HM + 14, so X = M + 14. Our gaps are at
6, 8, 16, 17, 18, 21, 26. Adding 14 to each of these:
20, 22, 30...
So we seem to have hit another bug. Let's check the original values:
SAXOPHONE = 134
SA is 20, so XOPHONE = 114
O is 10, and we have two, so XPHNE = 94
P is 19, so XHNE is 75
So XH does indeed look like it's 43.
THEME = 72, so HM is indeed 29. Seems like I'm stuck again. (sigh)
isr...@math.ubc.ca
Oops. No, that would make FUU=9. You want U=1 and F=4.
I get only one solution with all distinct values from 1 to 26:
S
P
O
I
L
E
R
S
P
A
C
E
A = 5, B = 13, C = 9, 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 only value left for D is 16.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
David J. Grabiner
> Hi, this is my first post here, can anybody help.
> I've done this old alphacipher but can't get a definite answer for 'D' as it
> appears
> there is two possible values.
Clarification: the letters A-Z are 1-26 is some order, and each word is
the sum of its letters.
> I can't find the solution as I've just dug the puzzle out of an olf drawer.
This puzzle was posed in rec.puzzles twelve years ago; the original
credits an English magazine "Tough Puzzles".
There is a unique solution, according to several posts in that thread,
including two fully worked-out solutions.
> Thanks in advance
>
>
> BALLET 45
> CELLO 43
> CONCERT 74
> FLUTE 30
> FUGUE 50
> GLEE 66
> JAZZ 58
> LYRE 47
> OBOE 53
> OPERA 65
> POLKA 59
> QUARTET 50
> SAXOPHONE 134
> SCALE 51
> SOLO 37
> SONG 61
> SOPRANO 82
> THEME 72
> VIOLIN 100
> WALTZ 34
>
>
--
David Grabiner, grab...@alumni.princeton.edu, http://remarque.org/~grabiner
Baseball labor negotiations FAQ: http://remarque.org/~grabiner/laborfaq.html
Shop at the Mobius Strip Mall: Always on the same side of the street!
Klein Glassworks, Torus Coffee and Donuts, Projective Airlines, etc.
donnaash...@gmail.com
soprano - opera : e+17 = son
song - son leaves g + e = 44
fuu = 6
F = 4
U = 1
let = 25
bal = 20
as ge = 44, le = 22
T = 3
yr = 25
sca = 29
l + 22 = g
L must be 2 as g can't be 23, 25 or 26
G = 24
E = 20
ballet - llet = 18 = ba
oboe less e = boo = 33
a + 15 = oo
solo = 37; sal = 22; sa = 20
b + 2 = s
as sca = 29
C = 9
O = 10
B = 13
S = 15
pr = 30
yr = 25
nr = 23
pk = 42
qr = 18
xphn = 79
hm = 29
song = 61, N 0 12
R = 11
P = 19
Y = 14
Q = 7
vii = 76
I = 25
V = 26
K = 23
wz = 24, jzz = 53
W = 6
Z = 18
J = 17
hm = 29, xh = 43
H =21
M = 8
X = 22
D = 16