CIPHER STRUCTURE

[smokie treats]

Hey Jarlve, I have another idea and was wondering what you think.

What about changing the dictionary.txt file to trick Decrypto by adding wildcards into the words. For example, the word "slave" could have six different versions:

slave
*lave
s*ave
sl*ve
sla*e
slav*

Then take out a lot of the words that Zodiac probably didn't use, like the word "ABATTOIR," to mitigate the increased file length.

In Decrypto, lock the alleged polyalphabetic symbols to * (or perhaps don't lock them) and see what happens. If we get a readable solution, substitute the appropriate letters for the *s.

Could something like that work?

[glurk]

With ZKD, the "dictionary.txt" files is used ONLY to list found words in the cipher. It is not used in the actual solving process AT ALL.

So changing it has no effect in any way on the actual solution that is found. The program will work fine without it, in fact.

-glurk

[smokie treats]

Glurk,

Thanks for answering that question. And yes I see that the program will in fact work without the dictionary.text file. I tried it.

Isn't that your program?

We have established that Zodiac did use cycles, and that those symbols and two others do not cycle well with any other symbols.

Do you have any thoughts about whether the "+" symbol, the "q" symbol, or others could be polyalphabetic?

S.T.

[Jarlve]

Hey Jarlve, I have another idea and was wondering what you think.

What about changing the dictionary.txt file to trick Decrypto by adding wildcards into the words. For example, the word "slave" could have six different versions:

slave
*lave
s*ave
sl*ve
sla*e
slav*

Then take out a lot of the words that Zodiac probably didn't use, like the word "ABATTOIR," to mitigate the increased file length.

In Decrypto, lock the alleged polyalphabetic symbols to * (or perhaps don't lock them) and see what happens. If we get a readable solution, substitute the appropriate letters for the *s.

Could something like that work?

Hey smokie,

The files bigraphs, trigraphs, tetragraphs and pentagraphs under the language/eng directory are used to score the ciphers. I changed them to your specifications and used the "Z" letter as wildcard because that letter is unused by ZKDecrypto in the ciphers we usually work with (you can check this by looking at the Init Key distribution and it has to do with its low frequency in English). Then in ZKDecrypto lock every wildcard symbol to the letter "Z" and you are good to go. ZKD ngrams with "Z" added for every position and duplicates removed.

I'm not sure how well this would work and you may need to change IoC weight a bit. Problem with suspected wildcards is that they create a lot of double symbols.

[smokie treats]

Wow thanks Jarlve!

I constructed a message (first 340 letters of the 408) with perfect cycles, no 1:1 or wildcards and the new n-graph files with the Z's worked.

Symbol
Count
4 A 1 2 3 4
1 B 5
2 C 6 7
2 D 8 9
8 E 10 11 12 13 14 15 16 17
1 F 18
2 G 19 20
4 H 21 22 23 24
5 I 25 26 27 28 29
0 J
1 K 30
3 L 31 32 33
1 M 34
4 N 35 36 37 38
5 O 39 40 41 42 43
2 P 44 45
0 Q
3 R 46 47 48
3 S 49 50 51
5 T 52 53 54 55 56
2 U 58 59
1 V 60
2 W 61 62
1 X 57
1 Y 63
0 Z
63

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

Solves in about a minute with both original and wildcard n-graph files

2. Took 20 (the Symbol "B" in the Z340) from G cycle and moved it to Z cycle. Put 20 in the same places where it appears in the 340.

a. Using the original n-graph files, and without locking 20 to Z, solved in less than 2 minutes but thought that 20 was E (maybe because of the expected frequency?).

b. Using the wildcard n-graph files, and without locking 20 to Z, solved in about 3 minutes and found the letter Z where 20 was. Decrypto seemed to have corresponded the n-grams with the Z's in them and used them in the solution.

ILIKEKILLINGPEOPL
EBEZAUSEITISSOMUC
ZFUNITISMOREFUNTH
ANKILLINGWILDGAME
INTHEFORRESTBECAU
SEMANISTHEMOSTDAN
GEROUSANIMALOFALL
TOKILLSOMETHINGGI
VESMETHEMZSTTHRIL
LINGEXPERENCEIZEV
ENBETTERTZANGETTI
NGYOURROCKSOFFWZT
HAGIRLTHEBEZTPART
OFITISTHATWHENIWI
LZBEREBORNINPARAD
ICEAZDALLTHEIHAVE
KILLEDWILLBECZMEM
YSLAVESIWILLNOTGI
VEYOUMYNZMEBZCAUS
EYOUWILLTRYTOSLOW

I also tried experiments with Symbol 19, but they were messed up partly by me and partly by the double symbols I think. I have more experiments to do.

S.T.

[Jarlve]

It's nice to see that it works to some degree. But yes, I also think the problem are doubles (and even a triple).

Do you want me to create ngrams also with double ZZ's for 3, 4 and 5-grams? Would be amazing if that worked!

[smokie treats]

If this works, then we will be able to use Decrypto to find words that have polyalphabetic symbols in them!

But hang on for a bit and let me do some confirmation testing. I want to make sure that I am thinking and doing this right.

[smokie treats]

Wildcard n-gram Experiment 2

Same message with perfect cycles, except that I changed the symbol L values a little bit.

Symbol Count
5 A 1 2 3 4 5
1 B 6
2 C 7 8
3 D 9 10 11
7 E 12 13 14 15 16 17 18
1 F 19
1 G 20
4 H 21 22 23 24
4 I 25 26 27 28
0 J
1 K 29
3 L 30 31 32
2 M 33 34
4 N 35 36 37 38
5 O 39 40 41 42 43
1 P 44
0 Q
4 R 45 46 47 48
4 S 49 50 51 52
5 T 53 54 55 56 57
1 U 58
1 V 59
1 W 60
1 X 61
1 Y 62
0 Z 63

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

Perfect solve in 15 seconds with the wildcard n-gram files.

To make things easy, I made 63 the wildcard and letter Z. I substituted 63 where the q appears in the 340 (not that it really matters exactly where). L=1 for 63.

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

Solved in 4:45. Quite a bit longer, but it did solve with the Z's where they should be. For a while, the program thought that the Z's were S's, maybe because of expected frequency. I don't know.

Note that the two Z's on line 4 where the two q's are on the 340 are at the end of the word "than" and the beginning of the word "killing." So no two-wildcard n-grams were needed to solve.

ILIKZKILLINGPEOPL
EZECAUSEITISSOMUC
HFUNITISMOREFUNTH
AZZILLINGWZLDGAME
INTHEFORRESTBECAU
ZEMANISTHEMOSTDAN
GEROUSANIMALOFALL
TOKILLSOMETHINGGI
VESMETHEMOSTTHRIL
LINGEXPEREZCEISEV
ENBETTERTHAZGETTI
NGYOURROCKSOFFWIT
HAGIRLTHEBESTPART
OFITISTHATWHENIWI
LLBEREBORNINPARAD
ICEANDALLZHEIHAVE
KILLEDWILLBECOMEM
YSLAVESIWILLNOTGZ
VEYOUMYNAMEBECAUS
EYOUWZLLTRYTOSLOW

Symbol 63 (or Z) is E, B, N, K, I, S, N, N, T, I, and I in that order

So yeah, if we want to test whether one of the symbols is polyalphabetic, and there are not two of those symbols in the same n-graph(?), we can use the one-wildcard n-grams to solve. I think that is what I am finding. Assuming that we know the L counts, and I am not sure how differences between set L counts and actual L counts affect things.

I will conduct another experiment before we should make two-wildcard n-grams.

Smokie

[Jarlve]

Glad to see it's working a bit, though wonder what the limits are for this approach.

Made them anyway. And haven't tried them.

By the way, I didn't include the bigraphs for obvious reasons but come to think of it you should probably add a line at the bottom "ZZ : 100 0". The numeric value being the score of the particular ngram, maybe you'll need to put it a bit higher even.

And your not locking the suspected wildcards to the "Z" letter?

zzgrams

[Jarlve]

@glurk

Is it a problem if the ngram files for ZKDecrypto are a bit unsorted?

Thanks