Zodiackillersite.com

Hi,

I’ve searched the forum for threads about the polybius square and found some posts in the huge „Homophonic Substitution“ thread. In those posts the polybius square was used in combination with bifid. Maybe I have overlooked something but I have not found a topic where the polybius cipher was considered in its basic form in combination with homophonic substitution. If someone had posted the following idea before, I may delete my new thread.

Allright then, let’s have a look at a polybius square in combination with a homophonic substitution. To keep things simple I don’t use a keyword to build the square:



As you can see, the polybius square results in a monoalphabetic substitution. Now consider the numbers „1“ to „5“ as the plaintext. You can apply a homophonic substitution and get the results found in the image above.


Now let’s have a look at another variation:



This time two different keys are used. One for the numbers in rows and one for the numbers in columns. The keys consists of the same symbols but in a different order. To encode a polybius message in this way gives you great freedom when assembling the final cipher. You can make parts of the ciphertext look cyclic or like a pattern. You can even put fake bigrams in the cipher to fool everyone. You can imagine how easy it would be to create even such patterns like the pivots.

If Zodiac had used such a system then two things gives me headache:
- The plaintext consists only of 170 letters which is much harder to decipher
- I have no idea how to solve such a cipher

What do you think?

Merry Christmas!

Edit: Fixed a "bug" in my sample images

Merry christmass.

Largo wrote:If Zodiac had used such a system then two things gives me headache:
- The plaintext consists only of 170 letters which is much harder to decipher
- I have no idea how to solve such a cipher

What do you think?

It could be useful to create a couple of these ciphers to look for statistics that stand out. To solve it, convert a language corpus to polybius with suspected key and train ngrams on that.

Largo,

What you propose is a plaintext encoded with polybius:

Code: Select all

34 33 24 45 33 15 43 24 53 44 15 15 33 44 52 34 44
23 34 45 43 11 33 14 11 33 14 21 24 21 44 15 15 33
44 42 45 32 35 11 33 33 34 45 33 13 15 14 23 24 43
13 11 33 14 24 14 11 13 54 21 34 42 35 42 15 43 24
14 15 33 44 34 21 44 23 15 45 33 24 44 15 14 43 44
11 44 15 43 11 44 44 42 45 32 35 44 34 52 15 42 24
33 33 15 52 54 34 42 25 13 24 44 54 24 33 44 23 15
43 35 15 15 13 23 44 42 45 32 35 14 42 15 52 11 44
44 15 33 44 24 34 33 44 34 14 34 32 15 43 44 24 13
24 43 43 45 15 43 43 45 13 23 11 43 24 31 31 15 22
11 31 24 32 32 24 22 42 11 44 24 34 33 34 21 21 43
23 34 42 24 33 22 34 21 11 32 15 42 24 13 11 33 24
34 12 43 44 23 15 45 43 33 11 44 24 34 33 11 31 14
15 12 44 11 33 14 24 43 31 11 32 24 13 44 15 42 42
34 42 24 43 32 52 23 24 13 23 11 31 31 42 15 32 11
24 33 15 14 31 11 42 22 15 44 23 15 32 15 43 14 45
42 24 33 22 44 23 15 13 11 32 35 11 24 22 33 23 15
11 31 43 34 11 33 33 34 45 33 13 15 14 23 24 43 13
11 32 35 11 24 22 33 43 31 34 22 11 33 32 11 25 15
11 32 15 42 24 13 11 22 42 15 11 44 11 22 11 24 33


And then remove the spaces and treat each single digit as a symbol and encode it with homophonic substitution:

Code: Select all

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


So, if this is the case in the 340 then we would indeed have only 170 original plaintext letters:

Code: Select all

491_TY('&><IN6E]I
K;VRX=/"CJ(HZ?N]V
.+#5(*5GO0$43B=O4
10M\3T;:^QJ!H8*_C
L2(,ND?-+'F@#G$41
]VI_&O>0X35%6E]VK
^"=OC+05ZB39^!=?/
Y.[+SU\#HQTW*QG7J
)MPD$4B9+YT)SU7#P
DMD>6E;QJ<HRLJ_L2
P<(5RHJ&=O(N]V5TB
*471I9V_5DA0'B:Z9
&+#13IX./V_5B\TH8
&79YS*UAYK#SU:43D
M1_'[IWX<&,G;QU-N
/$%4HR]VE=*D7O0M;
QJ3'1_]VA5&]>1V5C
B^9?Y$:)I94YSAUP?
:M;+Q#JH<'LG;$2IR
4,1=O(_-N&0>3^I%E


I'm not sure how to solve it, I posted an idea in my previous post but it seems unpractical. Merging by pairs results in 278 unique symbols for a 340 ciphertext so that is out of the question.

What I find interesting is that if such a scheme is actual, then probably only 5 to about 10 symbols would have to be homophonically substituted. So, if Zodiac then cycled like in the 408 we would come to conclusion that there were only 5 to 10 original symbols to begin with and we could strongly suspect such a scheme. Thus, perhaps Zodiac in all his wisdom foresaw this and messed up the cycles.

Could we come up with a test that estimates the number of symbols before encoding?