Zodiackillersite.com

doranchak wrote:Maybe the strong patterns in the original 340 can be ranked by the relative probability, weighted against whatever the shuffle tests show as the distribution of naturally-occurring patterns. Then you'll at least have a sense of which patterns escape the reach of pure randomness.

But the patterns I've introduced by my multiobjective search are arrived at using artificial means (i.e., creating candidate alternative ciphertexts). So the only real ranking we can do is to look at other patterns that might be "unlocked" by our guesses of the wildcard assignments, or by applying a strong solver to each candidate until the correct one is identified.

I don't even know if the approach I'm using is effective, since I haven't applied any test ciphers to it yet.

I'm scoring your ciphers now with up to 6-grams and a very high amount of keys to make sure they are maxed out.

Edit: your ciphers will not be done before I have to go to work. I will make another reply this evening with your results.

doranchak wrote:My view is that we should start simple, and only assume the + is a wildcard. If your solver can deal with those tests, then expand the wildcards to more than one symbol, until the solver can handle all 4 wildcard symbols effectively. Then you'll be fairly confident that you've excluded the hypothesis for the real 340 (or you'll actually crack the damned thing. PLEASE DO THAT! )

Well in this thread we already explored a 340 character part of the 408 with the "+" symbols from the 340 superimposed but that is only 53 symbols. Both ZKDecrypto and AZdecrypt can solve that but the score is quite low because the patterns of text are interrupted and the higher level n-grams cannot lock in as much. I'll make a few ciphers, ranging from 1 to 4 wildcards or so.

By the way I tried something else. I superimposed the 4 suspected wildcards on a 340 part of the 408 and then removed them. Scored this both in ZKDecrypto and AZdecrypt and it didn't catch on. Here is the cipher and the plaintext for which the wildcards were removed also. This is not the plaintext that came out of a solver, it's the actual one.

408 wildcard simulation:

Code: Select all

ilikkillingpeople
auseitissomucfuni
ismorefunthaillin
gwlameinteforrete
cauemaisthemotdan
geousanimalofallt
okillsoethiggives
ehemtthrillingepe
eceiseenbetterang
ettingyourrocksfi
thagilthestpartfi
tisthatwhedeiwill
berebornnparaicen
dalltheihvekildwl
lbomemyslavesiwil
otgiveoumnamebeca
ueyouwilltr

Code: Select all

¼ºP//uBºËOR¥ÐX¥BW
GyF°¼HP¹K‚ÑyÅJy^u
˽ÑTÔNQyDµ£S¼·BPO
RAºÌÑEË^LZJÄÒ\ÐHW
žy+ÑG¼KI£°ÑXµ¤S¢
RNIyEÌO¾ÑGBTQS·BL
Ä/P·B¹XEHMuRRËÃZK
УWѵLMÒ¼·BPDR+¥°
NÅEËFZÐOVWIµ+ÔÌ^R
°HI¼DR¸TyÒ\ÄÅ/¹JP
µM¾RuºL£EKH¥GÒIJË
µ¼¾LMÌNA£Z¤ÐËA¼·B
VW\+VTÔO^¥SÒÌuÅ°D
¤GººIMNË£ÃE/¼ºÆA·
BVXÑWѸF·¾Ã+¹¼A¼º
TµRuÃ+ÄyÑ^SÑWVZÅG
yE¸TyA¼º·LÔ

Then I made a simple change to my solver to remove characters at random in the corpus! This is after I remove all other garbage characters that are not "abcdefg...". I wonder if this idea can be improved, any thoughts?

Code: Select all

if rnd>0.15 then
'process letter
end if


Then scored the 408 example above and amazingly one of the results came out rather decently, given the circumstances.

Code: Select all

ilaffallindpeople
auseinatherumhusa
idrourfunshaillan
dwleroistthorvene
meusraihtherostan
drtuoeneralofallt
ofalltoonhaddinth
ehersthrillandspe
rmoistencetssuesd
entindyoursomftha
shedalthohnparthi
sietherwhtteiwill
cesscounspareamen
tallthrihnofilewl
lcoreryslenstiwil
osdansoursarectma
uoyouwilltu

Some catch though, this is one of 300 files, it's not a high scoring one and if you don't know the plaintext... I did this for the 340 and here are the results, I checked them but I really don't know if there is anything. 1. hypothesis has to be correct. 2. wildcards have to be correct.

Results for the 408 and 340 for various percentages of removed characters from the corpus.

Observations for the results of the 408, many returns seemed to have at least the right consonants/vowels in place. I will add something to my solver that after each return records the consonant/vowel of each position to a map by simply adding or subtracting 1. And then see if I can get at least a lock on the consonants/vowels.

smokie treats wrote:Yes, I agree that symbols 5, 19, 20 and 51 are the four hot candidates for wildcards, fillers or homophones. They are high in count but, unlike the other symbols (excepting those that are very low in count), absolutely do not cycle well with any other symbols.

Made a cipher with 63 symbols, uses perfect cycles (computer generated) but has 4 symbols which have a 1 to 1 substitution, so no homophones (counts are rather similar to the suspected wildcards of the 340). They should also not cycle well, you should be able to identify them by frequency. I wonder what you could say about it. It does solve but that's not the issue which I'd like to explore here.

p70chs63.txt

Code: Select all

[VK]EXk;,nHRix(^m
\B^"wpvS}sH/‚YFDq
$bo|=ƒgblb&_-uwRG
x9Ma^'b1*:Q~.;4,\
p2|X([^}\"TunbUMb
FVQi]@vqDYƒ[-K2Eo
YbU]^g.t$l:&b/Y,n
_Vk'm=p9b*v(;RYF\
4+|XuGb\M}ql-Q.\,
2po_ƒSi|Y:‚b'xYKg
s\u1*(bFVM"/bt9wQ
~DY.[q-,@=EypkBm$
Xnlb}|Sob_:u;Rg^9
MXsx}yHi"U‚\&^/4D
'$v*VQYK\\b1(-.ƒ:
E]koFba=RgB+m\S9Y
,qblVx[Y-‚bnKY_"X
yEv2bwp;k'biƒmUb}
B|S/*:&D=ut‚(M;Ko
sE$i[Vk]mgn^2Ub9F

smokie treats wrote:P.S. I have a question about your computer programs. I see that you have been using your own symbol set. I was just wondering if your programs work with the actual symbols, or if the programs work with numbers and convert to symbols. The reason that I ask is because I noticed that you have started referring to the symbols by number instead of by symbol.

I'm not using the webtoy transcription but I should switch to it because it is easier to work with. In computer programs these symbols are usually converted to numbers (integers) because the CPU can process integers faster.

Jarlve wrote:

smokie treats wrote:Yes, I agree that symbols 5, 19, 20 and 51 are the four hot candidates for wildcards, fillers or homophones. They are high in count but, unlike the other symbols (excepting those that are very low in count), absolutely do not cycle well with any other symbols.

Made a cipher with 63 symbols, uses perfect cycles (computer generated) but has 4 symbols which have a 1 to 1 substitution, so no homophones (counts are rather similar to the suspected wildcards of the 340). They should also not cycle well, you should be able to identify them by frequency. I wonder what you could say about it. It does solve but that's not the issue which I'd like to explore here.

EDIT: 1 to 1 substitutes.

It sounds to me like you are experimenting with your computer programs that solve coded messages, and using models that are similar to what I am suggesting about the sequences, wildcards, and other possibilities for the four candidate symbols.

Your Question

I think that you are wondering if the results of an analysis of this new model would be similar to the results of my analysis of the 340. Using my rudimentary scoring system of two symbol cycles. I think that you are wondering if my method would flush out the same four symbols. I could do that, it would take a few hours at least. But it would be nice if you could convert the symbols to numbers for me, starting with [=1, V=2, etc., just like how my version of the 340 is numbered. I have a simple spreadsheet that compares each symbol to every other symbol, and has 63*62=3906 different two symbol sets to work with and look for patterns. I could convert your symbols to numbers by myself, but it would take a while. I do a lot of this stuff manually. But I would definitely like to try flushing out your four symbols if that is what you are asking for. If I can, it would mean that I could have flushed out four symbols in the 340 that merely have a 1 to 1 substitution.

Is that what you are thinking?

Your model uses perfect cycles, which raises an interesting question for me. Almost all if not all of the long and/ or multi repeated cycles in the 340 are missing symbols. If the wildcard hypothesis is correct, then the total number of missing symbols in the cycles would be roughly the same as the wildcard count. I think that if your model had roughly the same number of missing cycle symbols, I am curious to find what my analysis would show.

*** I would be happy to try with the perfect cycles or with the cycles with missing symbols. ***

Thanks for answering the question about the conversions of symbols to numbers and vice versa. You must a a subroutine that does that as an aside to the work that your computer is doing. Numbers do make programming easier.

Distribution Comparison Method Revised

In the interest of being thorough, I will finish my expressed thoughts on the distribution comparison method for isolating Zodiac's symbol cycles:

In a post above, I was talking about making 30 randomizations, and comparing the mean count of each score to the count for each 340 score. I was using a score system of 0% to 99%, or one hundred possible score slots. However, I mentioned about possible including the standard deviation for each mean count as well. To finish and document this thought for future reference:



To concisely revise, a standard deviation value could be added to the mean. Any 340 cycle that scored higher than the mean plus the standard deviation would be more likely to be Zodiac made rather than random. And the nice thing about the method is that the standard deviation value could be adjusted to 1.2 standard deviations, 1.4 standard deviations, or whatever to eliminate more and more cycles from the huge list of cycles that we are working with.

It's just one possible tool. I really liked the shuffle idea. And there are more methods that we can apply as we learn about the nature of the cycles.

Thanks, S.T.

Here are your ciphers doranchak. I should be able to process them faster next time. The scores are normalised over the amount of characters "score * 340 / c", I could possible also include normalisation over the amount of symbols "score * 340 * 63 / c / s" or anyway you want it. I copied the ciphers of your document as they appear and numbered them from top to bottom.

smokie treats wrote:I think that you are wondering if the results of an analysis of this new model would be similar to the results of my analysis of the 340. Using my rudimentary scoring system of two symbol cycles. I think that you are wondering if my method would flush out the same four symbols. I could do that, it would take a few hours at least. But it would be nice if you could convert the symbols to numbers for me, starting with [=1, V=2, etc., just like how my version of the 340 is numbered. I have a simple spreadsheet that compares each symbol to every other symbol, and has 63*62=3906 different two symbol sets to work with and look for patterns. I could convert your symbols to numbers by myself, but it would take a while. I do a lot of this stuff manually. But I would definitely like to try flushing out your four symbols if that is what you are asking for. If I can, it would mean that I could have flushed out four symbols in the 340 that merely have a 1 to 1 substitution.

Is that what you are thinking?

That's part of what I'm thinking. I would also like to know if you find possible broken cycles in relation to these 1:1 symbols. There should be none but I'm thinking that some may show up out of chance. And how does this compare to the 340? Do you see more or less? Take your time, when you're done I'd like to give you the same cipher but with some randomisation applied to the cycles.

Code: Select all

p70chs63.txt
-----------------
Symbols: 63
Characters: 340
Multiplicity: 0.1852941
Index of Coincidence: 0.0174735
-----------------
Numbered by appearance:
1:            [: 6
2:            V: 7
3:            K: 6
4:            ]: 5
5:            E: 6
6:            X: 6
7:            k: 6
8:            ;: 6
9:            ,: 6
10:           n: 6
11:           H: 3
12:           R: 5
13:           i: 6
14:           x: 5
15:           (: 6
16:           ^: 8
17:           m: 6
18:           \: 11
19:           B: 4
20:           ": 5
21:           w: 4
22:           p: 6
23:           v: 5
24:           S: 5
25:           }: 6
26:           s: 4
27:           /: 5
28:           ‚: 5
29:           Y: 12
30:           F: 6
31:           D: 5
32:           q: 5
33:           $: 5
34:           b: 23
35:           o: 6
36:           |: 6
37:           =: 5
38:           ƒ: 5
39:           g: 6
40:           l: 5
41:           &: 4
42:           _: 5
43:           -: 6
44:           u: 6
45:           G: 2
46:           9: 6
47:           M: 6
48:           a: 2
49:           ': 5
50:           1: 3
51:           *: 5
52:           :: 6
53:           Q: 5
54:           ~: 2
55:           .: 5
56:           4: 3
57:           2: 5
58:           T: 1
59:           U: 5
60:           @: 2
61:           t: 3
62:           +: 2
63:           y: 3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18 19 16 20 21 22 23 24 25 26 11 27 28 29 30 31 32
33 34 35 36 37 38 39 34 40 34 41 42 43 44 21 12 45
14 46 47 48 16 49 34 50 51 52 53 54 55 8 56 9 18
22 57 36 6 15 1 16 25 18 20 58 44 10 34 59 47 34
30 2 53 13 4 60 23 32 31 29 38 1 43 3 57 5 35
29 34 59 4 16 39 55 61 33 40 52 41 34 27 29 9 10
42 2 7 49 17 37 22 46 34 51 23 15 8 12 29 30 18
56 62 36 6 44 45 34 18 47 25 32 40 43 53 55 18 9
57 22 35 42 38 24 13 36 29 52 28 34 49 14 29 3 39
26 18 44 50 51 15 34 30 2 47 20 27 34 61 46 21 53
54 31 29 55 1 32 43 9 60 37 5 63 22 7 19 17 33
6 10 40 34 25 36 24 35 34 42 52 44 8 12 39 16 46
47 6 26 14 25 63 11 13 20 59 28 18 41 16 27 56 31
49 33 23 51 2 53 29 3 18 18 34 50 15 43 55 38 52
5 4 7 35 30 34 48 37 12 39 19 62 17 18 24 46 29
9 32 34 40 2 14 1 29 43 28 34 10 3 29 42 20 6
63 5 23 57 34 21 22 8 7 49 34 13 38 17 59 34 25
19 36 24 27 51 52 41 31 37 44 61 28 15 47 8 3 35
26 5 33 13 1 2 7 4 17 39 10 16 57 59 34 46 30

[VK]EXk;,nHRix(^m
\B^"wpvS}sH/‚YFDq
$bo|=ƒgblb&_-uwRG
x9Ma^'b1*:Q~.;4,\
p2|X([^}\"TunbUMb
FVQi]@vqDYƒ[-K2Eo
YbU]^g.t$l:&b/Y,n
_Vk'm=p9b*v(;RYF\
4+|XuGb\M}ql-Q.\,
2po_ƒSi|Y:‚b'xYKg
s\u1*(bFVM"/bt9wQ
~DY.[q-,@=EypkBm$
Xnlb}|Sob_:u;Rg^9
MXsx}yHi"U‚\&^/4D
'$v*VQYK\\b1(-.ƒ:
E]koFba=RgB+m\S9Y
,qblVx[Y-‚bnKY_"X
yEv2bwp;k'biƒmUb}
B|S/*:&D=ut‚(M;Ko
sE$i[Vk]mgn^2Ub9F


Edit: please disregard this sample I will upload a much better version tomorrow with the actual cycles included.
Is this okay Smokie? Let me know if you still want it different. Could seperate numbers by commas next time?

smokie treats wrote:Your model uses perfect cycles, which raises an interesting question for me. Almost all if not all of the long and/ or multi repeated cycles in the 340 are missing symbols. If the wildcard hypothesis is correct, then the total number of missing symbols in the cycles would be roughly the same as the wildcard count. I think that if your model had roughly the same number of missing cycle symbols, I am curious to find what my analysis would show.

Excellent idea. I'll see if I can make such a cipher for you tomorrow.

Jarlve,

What you have provided is perfect the way it is. I want to find the cycles myself. After I input the numbers, my spreadsheet will show them to me and I can sort and filter them any way that I want. Then I copy and paste the scores to another spreadsheet and total them up. The symbols with the lowest overall scores have fewer cycle relationships with all other symbols.

I will perform the exact same analysis that I performed on the 340 two symbol cycles, and give you my answer as to what symbols do not cycle well with the others. I understand that some of the non-cycling symbols may actually cycle randomly with other symbols. That is a possibility. But I will be able to show any relationships that are in question. It may take a little time; there is going to be a lot of copying and pasting going on over here.

Thanks.

doranchak, how are your results? I would like to see how your mutli-objective hill climber performs on a wildcard cipher of our own.

smokie,

First one was too easy because all 4 of them had the highest frequencies of all symbols. I'm not sure but there may also have been a problem with the previous encoder, which I wrote rather quickly a few weeks ago. Anyway here's a better one, for which 4 symbols have a 1:1 substitution and only 1 is identifiable by frequency. Besides those everything else cycles perfectly, I'm really anxious now if you can identify them!

The cycles are also in the text file both in symbolic and nummeric version, if you don't want to see them just don't scroll down but it's pretty neat! I don't think you'll remember them if you just glance over it.

Here's the new one: (please use this one, sorry if I wasted your time!)

Code: Select all

p70chs63.txt
-----------------
Symbols: 63
Characters: 340
Multiplicity: 0.1852941
Sum of non-repeats: 4387
Index of coincidence: 0.0217074
-----------------
Symbols numbered by order of appearance:
-----------------
1 = Y (c=5)
2 = a (c=4)
3 = c (c=12)
4 = U (c=4)
5 = y (c=11)
6 = F (c=10)
7 = ' (c=11)
8 = p (c=4)
9 = S (c=14)
10 = w (c=5)
11 = C (c=2)
12 = E (c=5)
13 = - (c=4)
14 = k (c=5)
15 = V (c=7)
16 = " (c=8)
17 = m (c=11)
18 = o (c=4)
19 = 1 (c=5)
20 = Z (c=3)
21 = 8 (c=13)
22 = I (c=4)
23 = J (c=10)
24 = N (c=3)
25 = b (c=1)
26 = = (c=4)
27 = n (c=12)
28 = M (c=6)
29 = s (c=5)
30 = _ (c=6)
31 = 4 (c=5)
32 = e (c=23)
33 = u (c=10)
34 = * (c=13)
35 = L (c=4)
36 = q (c=4)
37 = z (c=6)
38 = \ (c=4)
39 = ! (c=6)
40 = : (c=3)
41 = 6 (c=2)
42 = 0 (c=2)
43 = & (c=2)
44 = 3 (c=6)
45 = . (c=3)
46 = > (c=3)
47 = A (c=2)
48 = ; (c=3)
49 = 2 (c=3)
50 = , (c=4)
51 = v (c=4)
52 = / (c=1)
53 = W (c=4)
54 = + (c=3)
55 = Q (c=3)
56 = 5 (c=3)
57 = r (c=2)
58 = @ (c=3)
59 = ) (c=2)
60 = O (c=3)
61 = ^ (c=2)
62 = R (c=1)
63 = B (c=3)
-----------------
Symbolic cipher:
-----------------
YacUyF'pSwCE-kV"c
mo"1Z8IyJNb='nMs_
4eu*LqFeze\!:S6E0
kJ8&"3e.V>*AS;28m
*,SuMv"Fm1/8YeW*e
_+SQ5rwzsnIq@cUyJ
ne,W"u8)4!O\epn*v
3a'Vc-SFeMY_=Enzm
2^8J*0emSu!3:8*mS
58FVwyL*n>'eMkncJ
ZmS._ze!+81;eRuN*
AsnSI3@8rQyB*'oc4
FqVeJSyueMO8pEF"J
*u6kFBC-1U'm\"=2s
_4vzaSncmme.!:8Y>
y,'J3e&LEuo^cmyFn
*VeM+kwn@'eIcn_1J
ByqWeZS;'zeQvc5eu
o8yp!O\s-*)'3S=cF
Ny4LYa'UcJw",WeuV
-----------------
Nummeric cipher:
-----------------
1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,3
17,18,16,19,20,21,22,5,23,24,25,26,7,27,28,29,30
31,32,33,34,35,36,6,32,37,32,38,39,40,9,41,12,42
14,23,21,43,16,44,32,45,15,46,34,47,9,48,49,21,17
34,50,9,33,28,51,16,6,17,19,52,21,1,32,53,34,32
30,54,9,55,56,57,10,37,29,27,22,36,58,3,4,5,23
27,32,50,53,16,33,21,59,31,39,60,38,32,8,27,34,51
44,2,7,15,3,13,9,6,32,28,1,30,26,12,27,37,17
49,61,21,23,34,42,32,17,9,33,39,44,40,21,34,17,9
56,21,6,15,10,5,35,34,27,46,7,32,28,14,27,3,23
20,17,9,45,30,37,32,39,54,21,19,48,32,62,33,24,34
47,29,27,9,22,44,58,21,57,55,5,63,34,7,18,3,31
6,36,15,32,23,9,5,33,32,28,60,21,8,12,6,16,23
34,33,41,14,6,63,11,13,19,4,7,17,38,16,26,49,29
30,31,51,37,2,9,27,3,17,17,32,45,39,40,21,1,46
5,50,7,23,44,32,43,35,12,33,18,61,3,17,5,6,27
34,15,32,28,54,14,10,27,58,7,32,22,3,27,30,19,23
63,5,36,53,32,20,9,48,7,37,32,55,51,3,56,32,33
18,21,5,8,39,60,38,29,13,34,59,7,44,9,26,3,6
24,5,31,35,1,2,7,4,3,23,10,16,50,53,32,33,15
-----------------
Nummeric cipher for ZKDecrypto:
-----------------
1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  3   
17  18  16  19  20  21  22  5   23  24  25  26  7   27  28  29  30 
31  32  33  34  35  36  6   32  37  32  38  39  40  9   41  12  42 
14  23  21  43  16  44  32  45  15  46  34  47  9   48  49  21  17 
34  50  9   33  28  51  16  6   17  19  52  21  1   32  53  34  32 
30  54  9   55  56  57  10  37  29  27  22  36  58  3   4   5   23 
27  32  50  53  16  33  21  59  31  39  60  38  32  8   27  34  51 
44  2   7   15  3   13  9   6   32  28  1   30  26  12  27  37  17 
49  61  21  23  34  42  32  17  9   33  39  44  40  21  34  17  9   
56  21  6   15  10  5   35  34  27  46  7   32  28  14  27  3   23 
20  17  9   45  30  37  32  39  54  21  19  48  32  62  33  24  34 
47  29  27  9   22  44  58  21  57  55  5   63  34  7   18  3   31 
6   36  15  32  23  9   5   33  32  28  60  21  8   12  6   16  23 
34  33  41  14  6   63  11  13  19  4   7   17  38  16  26  49  29 
30  31  51  37  2   9   27  3   17  17  32  45  39  40  21  1   46 
5   50  7   23  44  32  43  35  12  33  18  61  3   17  5   6   27 
34  15  32  28  54  14  10  27  58  7   32  22  3   27  30  19  23 
63  5   36  53  32  20  9   48  7   37  32  55  51  3   56  32  33 
18  21  5   8   39  60  38  29  13  34  59  7   44  9   26  3   6   
24  5   31  35  1   2   7   4   3   23  10  16  50  53  32  33  15 
-----------------
Symbolic cycles:
-----------------
Old symbol: s, homophone(s): YwIqv  (YwIqvYwIqvYwIqvYwIqvYw)
Old symbol: h, homophone(s): a:>+@O  (a:>+@Oa:>+@Oa:>+@Oa)
Old symbol: a, homophone(s): cy'  (cy'cy'cy'cy'cy'cy'cy'cy'cy'cy'cy'c)
Old symbol: m, homophone(s): U,W5  (U,W5U,W5U,W5U,W)
Old symbol: n, homophone(s): FJu  (FJuFJuFJuFJuFJuFJuFJuFJuFJuFJu)
Old symbol: r, homophone(s): p-=L;Q  (p-=L;Qp-=L;Qp-=L;Qp-=L)
Old symbol: e, homophone(s): S8*  (S8*S8*S8*S8*S8*S8*S8*S8*S8*S8*S8*S8*S8*S)
Old symbol: p, homophone(s): Cb  (CbC)
Old symbol: i, homophone(s): Ek1s4  (Ek1s4Ek1s4Ek1s4Ek1s4Ek1s4)
Old symbol: t, homophone(s): VM_z!3  (VM_z!3VM_z!3VM_z!3VM_z!3VM_z!3VM_z!3V)
Old symbol: u, homophone(s): "  ("""""""")
Old symbol: l, homophone(s): m  (mmmmmmmmmmm)
Old symbol: g, homophone(s): oB  (oBoBoBo)
Old symbol: d, homophone(s): ZN6  (ZN6ZN6ZN)
Old symbol: c, homophone(s): n  (nnnnnnnnnnnn)
Old symbol: o, homophone(s): e  (eeeeeeeeeeeeeeeeeeeeeee)
Old symbol: f, homophone(s): \.  (\.\.\.\)
Old symbol: v, homophone(s): 0A  (0A0A)
Old symbol: b, homophone(s): &^  (&^&^)
Old symbol: y, homophone(s): 2r  (2r2r2)
Old symbol: k, homophone(s): /  (/)
Old symbol: w, homophone(s): )R  ()R))
-----------------
Nummeric cycles:
-----------------
Old symbol: s, homophone(s): 1,10,22,36,51  (1,10,22,36,51,1,10,22,36,51,1,10,22,36,51,1,10,22,36,51,1,10)
Old symbol: h, homophone(s): 2,40,46,54,58,60  (2,40,46,54,58,60,2,40,46,54,58,60,2,40,46,54,58,60,2)
Old symbol: a, homophone(s): 3,5,7  (3,5,7,3,5,7,3,5,7,3,5,7,3,5,7,3,5,7,3,5,7,3,5,7,3,5,7,3,5,7,3,5,7,3)
Old symbol: m, homophone(s): 4,50,53,56  (4,50,53,56,4,50,53,56,4,50,53,56,4,50,53)
Old symbol: n, homophone(s): 6,23,33  (6,23,33,6,23,33,6,23,33,6,23,33,6,23,33,6,23,33,6,23,33,6,23,33,6,23,33,6,23,33)
Old symbol: r, homophone(s): 8,13,26,35,48,55  (8,13,26,35,48,55,8,13,26,35,48,55,8,13,26,35,48,55,8,13,26,35)
Old symbol: e, homophone(s): 9,21,34  (9,21,34,9,21,34,9,21,34,9,21,34,9,21,34,9,21,34,9,21,34,9,21,34,9,21,34,9,21,34,9,21,34,9,21,34,9,21,34,9)
Old symbol: p, homophone(s): 11,25  (11,25,11)
Old symbol: i, homophone(s): 12,14,19,29,31  (12,14,19,29,31,12,14,19,29,31,12,14,19,29,31,12,14,19,29,31,12,14,19,29,31)
Old symbol: t, homophone(s): 15,28,30,37,39,44  (15,28,30,37,39,44,15,28,30,37,39,44,15,28,30,37,39,44,15,28,30,37,39,44,15,28,30,37,39,44,15,28,30,37,39,44,15)
Old symbol: u, homophone(s): 16  (16,16,16,16,16,16,16,16)
Old symbol: l, homophone(s): 17  (17,17,17,17,17,17,17,17,17,17,17)
Old symbol: g, homophone(s): 18,63  (18,63,18,63,18,63,18)
Old symbol: d, homophone(s): 20,24,41  (20,24,41,20,24,41,20,24)
Old symbol: c, homophone(s): 27  (27,27,27,27,27,27,27,27,27,27,27,27)
Old symbol: o, homophone(s): 32  (32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32)
Old symbol: f, homophone(s): 38,45  (38,45,38,45,38,45,38)
Old symbol: v, homophone(s): 42,47  (42,47,42,47)
Old symbol: b, homophone(s): 43,61  (43,61,43,61)
Old symbol: y, homophone(s): 49,57  (49,57,49,57,49)
Old symbol: k, homophone(s): 52  (52)
Old symbol: w, homophone(s): 59,62  (59,62,59)


I'm slowly getting caught up. Right now I'm adjusting my measurements and trying to optimize them for speed.

I'm still lost in the Sauce Me coming into these cipher threads is like me walking into a Rolls Royce dealership. It's nice to look, but I know when I leave, I'll be going home in the same car I came in.
It's cool to see you Guys all working together as a team, and hope you Guys find something big!

Jarlve,

Yes, the first one was easy. Here are the results and hopefully no mistakes. I followed the same procedure as with the 340, as if I didn't know the counts.

Experiment 1 J-ST



It seems that you intended me to find 16, 18, 29, and 34.

Symbol 16 does not have any perfect cycles with any other symbol, but does cycle with 20, 41 and 63 in part of the message.

For example, 16 and 20 (the others are similar):



I wrote another spreadsheet, and now all I have to do is enter the numbers and I can examine the one on one relationship scores very quickly. Next time don't give me anything but a table of numbers, the ZKDecrypto version is easiest to work with. On the one you most recently posted, I won't look at anything else.

S.T.