Zodiackillersite.com

I have a possibly naive question. Is is possible to take a handful of the highest scoring cycles, use frequency analysis to determine their probable plaintext, and plug those into your computer programs to see what happens. Can we do that now, or is that premature?

Thanks for your in-depth analysis smokie,

You did catch on to some things here and there but it shows that some randomness can go a long way. I came up with the idea of hiding the 1:1 substitutes by adding some extra randomness to a few cycles that had similar counts. And it seems to have worked, you could see this as a worst case scenario. I guess that 1:1 substitutes are not entirely unplausable for the 340.

Now we have to create an actual wildcard cipher by hand. I will do this saturday or sunday. After that I have something else which I would like you to look at.

The cipher your analysed:

Code: Select all

p44.txt
-----------------
Symbols: 63
Characters: 340
Multiplicity: 0.1852941
Sum of non-repeats: 4175
Index of coincidence: 0.0172479
-----------------
Symbols numbered by order of appearance:
-----------------
1 = k (c=5)
2 = u (c=8)
3 = o (c=5)
4 = , (c=7)
5 = e (c=8)
6 = r (c=5)
7 = 1 (c=6)
8 = F (c=4)
9 = . (c=4)
10 = 7 (c=11)
11 = 3 (c=13)
12 = [ (c=6)
13 = g (c=5)
14 = % (c=3)
15 = t (c=6)
16 = V (c=4)
17 = ^ (c=6)
18 = J (c=12)
19 = R (c=5)
20 = ` (c=5)
21 = v (c=8)
22 = _ (c=6)
23 = > (c=5)
24 = w (c=5)
25 = ! (c=5)
26 = < (c=4)
27 = \ (c=5)
28 = O (c=8)
29 = X (c=8)
30 = p (c=4)
31 = P (c=8)
32 = @ (c=8)
33 = E (c=9)
34 = T (c=10)
35 = ' (c=4)
36 = N (c=12)
37 = 6 (c=2)
38 = G (c=4)
39 = I (c=6)
40 = - (c=5)
41 = ( (c=4)
42 = h (c=9)
43 = + (c=5)
44 = z (c=10)
45 = Q (c=2)
46 = L (c=5)
47 = : (c=4)
48 = i (c=7)
49 = ; (c=7)
50 = Z (c=1)
51 = U (c=3)
52 = l (c=3)
53 = y (c=3)
54 = q (c=2)
55 = d (c=2)
56 = n (c=4)
57 = # (c=4)
58 = C (c=2)
59 = / (c=3)
60 = Y (c=2)
61 = = (c=1)
62 = ? (c=2)
63 = m (c=1)
-----------------
Symbolic cipher:
-----------------
kuuo,er1F.73[g%tV
^J^R`v_>,w!<\OXpP
@E3JTX3[g'N767wvE
utGI-(vh.+JzPX<@Q
TL:OzoNi`-;7Z71`t
TvULh3Ep,kXr\VFQ@
E!RNP3z+>T'Ie[Jl`
.@yquTGhd%L+n#iu(
woIig:Cle^3t\EJz_
OrNTP<7#-Jz^pXk/-
@VY;7R>_1vN[P3:=P
JzwgO\!'Nh;J,eL.E
nh3GN^XNT1@qh;t[;
?3y+J`/1iU(VikoXC
ErRIi;7F!h>,LhevN
_<N67nwOY'@GIe3Ed
T;J%(E1+uPO\JT@go
,73iN7v!e3ruhXv^R
I:tPeOzp-#nTz/uU?
kJzm_#l3,_yFOz>N[
-----------------
Nummeric cipher:
-----------------
1,2,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16
17,18,17,19,20,21,22,23,4,24,25,26,27,28,29,30,31
32,33,11,18,34,29,11,12,13,35,36,10,37,10,24,21,33
2,15,38,39,40,41,21,42,9,43,18,44,31,29,26,32,45
34,46,47,28,44,3,36,48,20,40,49,10,50,10,7,20,15
34,21,51,46,42,11,33,30,4,1,29,6,27,16,8,45,32
33,25,19,36,31,11,44,43,23,34,35,39,5,12,18,52,20
9,32,53,54,2,34,38,42,55,14,46,43,56,57,48,2,41
24,3,39,48,13,47,58,52,5,17,11,15,27,33,18,44,22
28,6,36,34,31,26,10,57,40,18,44,17,30,29,1,59,40
32,16,60,49,10,19,23,22,7,21,36,12,31,11,47,61,31
18,44,24,13,28,27,25,35,36,42,49,18,4,5,46,9,33
56,42,11,38,36,17,29,36,34,7,32,54,42,49,15,12,49
62,11,53,43,18,20,59,7,48,51,41,16,48,1,3,29,58
33,6,19,39,48,49,10,8,25,42,23,4,46,42,5,21,36
22,26,36,37,10,56,24,28,60,35,32,38,39,5,11,33,55
34,49,18,14,41,33,7,43,2,31,28,27,18,34,32,13,3
4,10,11,48,36,10,21,25,5,11,6,2,42,29,21,17,19
39,47,15,31,5,28,44,30,40,57,56,34,44,59,2,51,62
1,18,44,63,22,57,52,11,4,22,53,8,28,44,23,36,12
-----------------
Nummeric cipher for ZKDecrypto:
-----------------
1   2   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16 
17  18  17  19  20  21  22  23  4   24  25  26  27  28  29  30  31 
32  33  11  18  34  29  11  12  13  35  36  10  37  10  24  21  33 
2   15  38  39  40  41  21  42  9   43  18  44  31  29  26  32  45 
34  46  47  28  44  3   36  48  20  40  49  10  50  10  7   20  15 
34  21  51  46  42  11  33  30  4   1   29  6   27  16  8   45  32 
33  25  19  36  31  11  44  43  23  34  35  39  5   12  18  52  20 
9   32  53  54  2   34  38  42  55  14  46  43  56  57  48  2   41 
24  3   39  48  13  47  58  52  5   17  11  15  27  33  18  44  22 
28  6   36  34  31  26  10  57  40  18  44  17  30  29  1   59  40 
32  16  60  49  10  19  23  22  7   21  36  12  31  11  47  61  31 
18  44  24  13  28  27  25  35  36  42  49  18  4   5   46  9   33 
56  42  11  38  36  17  29  36  34  7   32  54  42  49  15  12  49 
62  11  53  43  18  20  59  7   48  51  41  16  48  1   3   29  58 
33  6   19  39  48  49  10  8   25  42  23  4   46  42  5   21  36 
22  26  36  37  10  56  24  28  60  35  32  38  39  5   11  33  55 
34  49  18  14  41  33  7   43  2   31  28  27  18  34  32  13  3   
4   10  11  48  36  10  21  25  5   11  6   2   42  29  21  17  19 
39  47  15  31  5   28  44  30  40  57  56  34  44  59  2   51  62 
1   18  44  63  22  57  52  11  4   22  53  8   28  44  23  36  12 
-----------------
Symbolic cycles:
-----------------
Old symbol: a, homophone(s): kF%+:-U  (kF%-+:-UkF+%+:-k-:+UkF%+:-UkF)
Old symbol: l, homophone(s): uln#  (uuulun#ul#nnuu#nu#l)
Old symbol: t, homophone(s): orR>'G(  (orR>'G(orR>'G(orR>'G(orR>'G(orR>)
Old symbol: h, homophone(s): ,NI  (,,NIN,NIINNN,NNI,NNI,NI,N)
Old symbol: e, homophone(s): Ti7he  (e7T77hTi77ThTeThiieT77hehThiii7hhe7eTT7i7eheT)
Old symbol: r, homophone(s): 1!L;  (1!L;1L!L;1!;L1;;1;!L;1!)
Old symbol: c, homophone(s): .<p  (.<p.<p.<p.<p)
Old symbol: s, homophone(s): 3X  (3X3X3X3X33X33X3X333X3)
Old symbol: o, homophone(s): [w\@  ([w\@[w@\@[@w\@[w\@[w@\@[)
Old symbol: f, homophone(s): gtV  (gtVgttVgtVgtVgt)
Old symbol: i, homophone(s): ^_P  (^^_PPP^_P^_PP^_P^P__)
Old symbol: n, homophone(s): JO  (JOJJOJJOJJOJJOJOJOJO)
Old symbol: y, homophone(s): `  (`````)
Old symbol: w, homophone(s): v  (vvvvvvvv)
Old symbol: u, homophone(s): E  (EEEEEEEEE)
Old symbol: b, homophone(s): 6C  (6CC6)
Old symbol: d, homophone(s): z  (zzzzzzzzzz)
Old symbol: v, homophone(s): QZ  (QZQ)
Old symbol: m, homophone(s): y?  (y?y?y)
Old symbol: p, homophone(s): qd  (qdqd)
Old symbol: g, homophone(s): /Y  (/Y/Y/)
Old symbol: k, homophone(s): =m  (=m)
-----------------
Nummeric cycles:
-----------------
Old symbol: a, homophone(s): 1,8,14,43,47,40,51  (1,8,14,40,43,47,40,51,1,8,43,14,43,47,40,1,40,47,43,51,1,8,14,43,47,40,51,1,8)
Old symbol: l, homophone(s): 2,52,56,57  (2,2,2,52,2,56,57,2,52,57,56,56,2,2,57,56,2,57,52)
Old symbol: t, homophone(s): 3,6,19,23,35,38,41  (3,6,19,23,35,38,41,3,6,19,23,35,38,41,3,6,19,23,35,38,41,3,6,19,23,35,38,41,3,6,19,23)
Old symbol: h, homophone(s): 4,36,39  (4,4,36,39,36,4,36,39,39,36,36,36,4,36,36,39,4,36,36,39,4,36,39,4,36)
Old symbol: e, homophone(s): 34,48,10,42,5  (5,10,34,10,10,42,34,48,10,10,34,42,34,5,34,42,48,48,5,34,10,10,42,5,42,34,42,48,48,48,10,42,42,5,10,5,34,34,10,48,10,5,42,5,34)
Old symbol: r, homophone(s): 7,25,46,49  (7,25,46,49,7,46,25,46,49,7,25,49,46,7,49,49,7,49,25,46,49,7,25)
Old symbol: c, homophone(s): 9,26,30  (9,26,30,9,26,30,9,26,30,9,26,30)
Old symbol: s, homophone(s): 11,29  (11,29,11,29,11,29,11,29,11,11,29,11,11,29,11,29,11,11,11,29,11)
Old symbol: o, homophone(s): 12,24,27,32  (12,24,27,32,12,24,32,27,32,12,32,24,27,32,12,24,27,32,12,24,32,27,32,12)
Old symbol: f, homophone(s): 13,15,16  (13,15,16,13,15,15,16,13,15,16,13,15,16,13,15)
Old symbol: i, homophone(s): 17,22,31  (17,17,22,31,31,31,17,22,31,17,22,31,31,17,22,31,17,31,22,22)
Old symbol: n, homophone(s): 18,28  (18,28,18,18,28,18,18,28,18,18,28,18,18,28,18,28,18,28,18,28)
Old symbol: y, homophone(s): 20  (20,20,20,20,20)
Old symbol: w, homophone(s): 21  (21,21,21,21,21,21,21,21)
Old symbol: u, homophone(s): 33  (33,33,33,33,33,33,33,33,33)
Old symbol: b, homophone(s): 37,58  (37,58,58,37)
Old symbol: d, homophone(s): 44  (44,44,44,44,44,44,44,44,44,44)
Old symbol: v, homophone(s): 45,50  (45,50,45)
Old symbol: m, homophone(s): 53,62  (53,62,53,62,53)
Old symbol: p, homophone(s): 54,55  (54,55,54,55)
Old symbol: g, homophone(s): 59,60  (59,60,59,60,59)
Old symbol: k, homophone(s): 61,63  (61,63)


smokie treats wrote:I have a possibly naive question. Is is possible to take a handful of the highest scoring cycles, use frequency analysis to determine their probable plaintext, and plug those into your computer programs to see what happens. Can we do that now, or is that premature?

It's an idea similar to what I was suggesting. To come down from 63 symbols to 20-26 letters. Scroll back to my previous post to read it. There's a catch, maybe you could help me with it. I'll explain it more thoroughly. A handful of high scoring cycles do not fill up an entire cipher, and they may be wrong. Too many variations.

We need a measurement system that is such that the total score of all cycles in the cipher is the global optimum. The global optimum is the highest score available when you consider all other possible cycle distributions over the letters. Why? Because otherwise the hill climbing algorithm will not get it, it climbs to the highest score and not to some point in the middle.

In the program there are some keys. Key 1 is the number of letters. We don't know how many letters are used in the 340 so we gradually increment key 1 (letters=20,letters=21,letters=22,...) to see what amount of letters gives the best score.

There is also a key 2 which holds a cycle for each letter in key 1. The cycles are at first distributed randomly and then are tried to improve upon by making changes. The next 4 following steps describe the hill climbing algorithm in such a program.

1) Make a random change to key 2.
2) Score all cycles individually (of key 2) with the measurement system, add them together and remember this value as the score.
3a) If the score is higher than the previous score then keep the change.
3b) If the score is lower than the previous score then revert the random change.
4) Go back to step 1.

I believe this variation is called a stochastic hill climber and it's what I use in AZdecrypt with some extras thrown in. So all I need to get this to work is a measurement system as I described. Your ideas are welcome, and if you have an entirely different idea I would certainly like to consider it.

Also, we know that from solvers like ZKDecrypto (which use the hill climbing approach) that a full grid english solution in a left-to-right, top-to-bottom manner is not available for the 340. That's something for which you can take my word on. Something else is going on. Interruptions would certainly do the trick - such as your wildcard theory - given that they are plenty. In this thread we were able to recover the plaintext of the 408 with a wildcard simulation using 24 symbols. pi suggested it was easy but given that it only scored 32000 it was a borderline solve.

If wildcards are actual in the 340 and the wildcard symbols are correct (there are no more or less) I may be able to extract a partial solve from the 340 this weekend with my own solver using the consonant/vowel technique I have been talking about earlier in this thread.

Jarlve:

I understand about making small changes and trying over and over until getting higher scores.

I am surprised about Symbols 32 and 33. I thought for sure that you made that cycle because it was so long. But 33 was a 1:1 and 32 was included in a randomized cycle. The result was an (improbable?) cycle relationship. 32 33 33 32 33 32 33 32 33 32 33 32 33 32 33 33 32.

I will be waiting in the starting blocks for any new coded message.

doranchak: I read your explanation, and grasp the basic concept. I have some thoughts, however. When I manually try to find cycles in the message, I could find ABCDABCD. However, because of the way that I have my ciphertext numbered, the first cycle, ABCD will always appear no matter what. It's the probability of getting another, second ABCD is what I wonder about. In your example, you give some possibilities such as ABDCABDC. But that cannot occur the way that, at least I, have the ciphertext numbered. If you take into account the fact that B could never come before A in the first cycle, does your analysis change?

My comments about your requiring two contiguous cycles for scoring was misplaced; I wasn't thinking straight and had been looking at L-7.

S.T.

smokie treats wrote:doranchak: I read your explanation, and grasp the basic concept. I have some thoughts, however. When I manually try to find cycles in the message, I could find ABCDABCD. However, because of the way that I have my ciphertext numbered, the first cycle, ABCD will always appear no matter what. It's the probability of getting another, second ABCD is what I wonder about. In your example, you give some possibilities such as ABDCABDC. But that cannot occur the way that, at least I, have the ciphertext numbered. If you take into account the fact that B could never come before A in the first cycle, does your analysis change?

That makes sense. So, looking back at my example, if we apply your symbol numbering scheme to the 207,900 possible symbolic cipher texts based on my sample alphabet, then there are only 51,975 possible cipher texts (exactly one forth of the original symbolic cipher texts): http://zodiackillerciphers.com/mini-cip ... erical.txt

This is because each numerical representation can stand for exactly 4 of the original cipher texts, since we are mapping to symbol order instead of a strictly to specific symbols.

I think this means the probabilities do not change, since if you have a numerical cipher such as 0 1 2 3 0 1 2 3 4 5 4 5, it can represent 4 different original symbol ciphers, so we still have to count all 4 of the originals.

Here are all the 144 distinct ways the L=4 cycles can repeat in the numerical version: http://zodiackillerciphers.com/mini-cip ... les-L4.txt
Here are all the 15 distinct ways the L=4 cycles can repeat in the numerical version: http://zodiackillerciphers.com/mini-cip ... les-L2.txt

So you can see the L=4 cycles are still more likely to occur by chance than the L=2 cycles. I did notice, though, that 144 times 4 = 576, but I previously calculated that 720 (not 576) L=4 cycles should appear among all the possible sample ciphertexts. Not sure if I made a small mistake somewhere (maybe I overcounted in the symbol version, or undercounted in the numerical version).

I'm working on pushing a better solve out of the 408 wildcard simulation, the results seem to have improved a bit (judging from example 1 in contrast to the other example I posted a few pages back). When example 1 is the dominant scheme that comes out of the solver I will move back to the 340. The problem currently is that example 2 scores higher and therefore is the dominant scheme.

Example 1:

Code: Select all

ilittallingpeople
auceitionbrunhusa
idrourfunshaillin
gwlerdistahorsete
neusraintherostag
grtudeneralofallt
otilloodthaggiaan
ehersthrillingspe
rndicaenmetssuesg
ettingyoursontohi
shegalthdntparthi
sietherwhateiwill
messmounspareanen
tallthrihadtilywl
lmorerycleasoiwil
osgaasoursaremana
udyouwilltu


Example 2:

Code: Select all

ngessyhgarlgeight
errendermisrhmrly
alsoftwranninther
ligasealotmuchedt
herosenmonesinhil
ltorearesehowitho
usethriedbyllactm
entsnobcnthealoge
thearterstonofall
edonalforcouhsrme
nbelygonemdgecoma
nneobatintheainth
stoosofrlgicayhea
heggobtancesngwit
hsistsfrtecorning
onlycoursliststhe
reforingtof


Jarlve wrote:I have an interesting idea. You guys have been looking at probabilities of individual cycles (well mostly). I propose looking at the probabilities of a full distribution of cycles with the goal to reduce the number of symbols from 63 to somewhere 20-26. A hill climber that sorts out the number of letters and the cycle per letter.

Key1: number of letters.
Key2: symbols (part of cycle) per letter.
Operations: a) increment/decrement letter, b) swap symbols.
Measurement system has to be such that the total score of the actual cycles in the cipher is the global optimum.

There may be some problems with this system, it may or not have multiplicity issues. If some transposition was applied after or during encoding of the 340 then it will fail. If the wildcard hypothesis is true then it may need adjustement. It may have problems when randomisation of the cycles is actual.

Doranchak do you think it is possible to come up with such a measurement system? If you think it's possible and you see some merit in this approach by all means feel free to try it. If you think it's unprobable or don't have the time to try this approach could you point me in the right direction? Thank you.

I think this kind of approach might be really effective. It reminds me of King and Bahler's method - have you seen it? http://www.oranchak.com/king-homophonic-ciphers.pdf I think their method exploits qualities that arise from applying numerical numbering of cipher symbols, and it avoids frequency analysis, statistics, and backtracking. But I think it is sensitive to randomness in the cycles, so it will fail to detect some strong cycles if they are sufficiently imperfect.

I do suspect that the total score of the actual cycles in the cipher will not be the global optimum, though. This would have to be tested.

Some general thoughts. I really think that we are on the right track. I like the idea of working with the cycles as a group, and the symbols as a group to try to identify characteristics about each so that we can break them down into subgroups. Sort of looking at trees individually, and then looking at groups of trees, and then looking at the forest.

I also think that we may be on the right track with the wildcard hypothesis. I am not emotionally attached to it. I think that the high frequency symbols that do not have cyclical relationships could also be 1:1 substitution or filler. But the more I think about it, the more wildcards makes sense. Zodiac knew about diffusion. Why would he defeat the purpose of using multiple symbols to represent the letters with mid-range count, and then make frequency attack easy for the high count letters? Fillers is a possibility as well.

I have been analyzing the 340 a little bit this morning, here is the table of all higher frequency symbols:



Also, here is a simple scatter graph of the analysis, with count on the X axis and total score on the Y axis. Note that if a symbol is in a cycle with other symbols, then their count will be roughly the same. They would cluster together, at least vertically, on the graph. The upper left is a large clustering of mid-range count symbols that have a lot of cycle relationships with each other. On the lower left would be a few low-range count symbols that are probably 1:1 substitution for low frequency letters.



Some of the below is redundant, but I wanted to be more thorough today.

The proposed wildcards are in red. They are high in count and do not have strong cycle relationships with other symbols.

5 has some cycling with 29 (5 5 29 5 5 5 5 29 5 29 5 29 5 29 5 29 5). Score 59%.
51 has some cycling with 23 (23 23 51 51 23 51 23 51 23 51 23 51 51 51 23 51 23 51 23 23). Score 55%.
51 has some cycling with 36 (36 51 51 51 36 51 36 51 36 51 51 36 51 36 36 36 51 36 51 36). Score 50%.

Part of my analysis is a little bit subjective. You could flip a coin all day long and get patterns like these of which there are thousands. But see the Alternative Possibilities for 11, 23, 36, and 51 below.

Note 26 and 50 (purple) have counts of 6 and 7, but have little or no cycle relationships. all of the other symbols of that count are in cycles, but these are not.That's what separates them out as a group. They are either 1:1, wildcards or filler. I am suggesting that they could represent low frequency letters. Zodiac didn't feel the need to put them in cycles, but they appear in higher frequency than Zodiac anticipated because of his choice of words. Thus, they set apart from the other symbols.

Symbols 16 and 40 (blue) are in a strong cycle together (16 40 16 40 16 16 40 16 40 16 40 16 40 16 40 16 16 40 40), however I learned with Experiment 3 J-ST that this can happen quite by chance, and be a false cycle. Or they could simply represent a letter with a 19 count as a two symbol cycle. 16 and 40 do have some weak to medium cycling with other symbols, and I don't know what to make of this. I am hoping that they are not cycled wildcards because of their count, but have considered that possibility.

Symbols 11 and 36 (black) cycle fairly strong together (11 36 11 36 11 36 11 36 11 11 36 36 11 36 36 11 36 11 36 11), and the analysis is the same as above. They both cycle weak to medium with several other symbols.

Symbol 23 (green) cycles with several other symbols, including:

with 31 (23 31 23 31 23 31 23 31 23 31 23 31 23 23 23 23 31). Score 65%, could be random.
and 37 (23 37 23 37 23 23 37 23 37 23 37 23 37 23 23 37 23). Score 65%, could be random.
and 51 ((23 23 51 51 23 51 23 51 23 51 23 51 51 51 23 51 23 51 23 23). Score 55%.

Symbols 3, 6, 7, 21, 30 and 31 (orange) mostly all have cycles with each other.

Alternative Possibilities for 11, 23, 36, and 51

11, 23, 36 and 51 all have the same count of 10, so theoretically there could be combinations of these symbols in cycles where Zodiac made symbol choices at random, which could exclude 51 from the group of proposed wildcards. Here are some possibilities:



Conclusion

Symbols 5, 19 and 20 are still the best candidates for wildcards, because they do not cycle with each other or any other symbol well. However, I now think Symbol 51 could possibly be in a cycle with Symbol 23 and represent the same letter. If that is the case, then Zodiac could have used only a few symbols to represent high frequency letters. I would have to put 51 in a borderline category.

S.T.

Something strange I noticed (as someone who is more English/thespian minded than Math minded):

Stage directions grid's sort of look like the 9 sectioned grid people make to figure out ciphers. If Z was interested in plays and stage directions in theory, could he gave used it in some way for his cipher code? Again, this is probably a dumb crazy outlandish idea bc again I have no clue how this works. Just noticed similar structures. Thank you.

Btw thank goodness for you guys. I have NO clue what you are talking about - you all are way too intelligent for this ole gal.

I have an observation to make.

I have been scanning through the cycles that doranchak found:

https://docs.google.com/spreadsheets/d/ ... sp=sharing

Starting with L=4, but increasingly with L=5 through L=7 as L becomes a higher number of symbols, I see a pattern. I am looking at bracketed cycles, which are perfect. And I am also looking at a lot of random ordering of symbols. The bracketed cycles are shifted to the left in many cases. We get two contiguous cycles, and then randomness.

So I am wondering what you guys think about that. Zodiac did the same thing on the 408 with high frequency letters such as E, I, O and T. See: viewtopic.php?f=81&t=267.

No doubt you guys are aware of that, but maybe that could help us narrow down the search for 20 or so needles in a haystack. Perhaps try and make a brute force search for cycles in the first, say, 170 symbols of the message (picked this number out of the air). See how may cycles we get as compared to a search in the entire 340. Do this for one L value first to see what happens. It seems to me that the probability of finding a highly improbable false cycle will be lower when working with a smaller number of symbols. Probably all of the cycles we are looking for are in the first half of the message. What I am wondering is, what if we made the haystack a lot smaller? Would that help?

On the other hand, we are looking for only 20 or so cycles. But there are way more than 20 cycles in the spreadsheet showing two contiguous perfect cycles and then randomness. What are we looking at here?

S.T.

Guys you have to excuse me for a while. Woke up dead tired and I feel the need to sit on the bench for a while.

@doranchak, I had not seen it. They are indeed exploring the same concept.

@smokie, I really like the ideas in your last post. I think we do need to take in account the positions on which the cycles are found and believe that a cycle with a more uniform spread over the cipher will be more likely to be actual.

@PinkPhantom and morf13, welcome to the thread.