Zodiackillersite.com

Largo wrote:Okay...check this out:

Rows 1-6, followed by rows 15-20:

I had the same idea, but these cycles may be an artifact of smaller pieces of cipher, hard to say. The first half of the 340 also has many of these. The scoring is based on the ioc formula: cycle_alternations*(cycle_alternations-1).

Largo wrote:If different keys were used alternately, this could be the cause of incomplete but not completely absent cycling. Has anyone ever done some tests before? For example, rows 1-6 and 15-20 with key A, then rows 7-14 encoded with key B and check the cycling? If not, I'd like to put that on my todo list.

Good idea. Should be easy to check since there are only 2^20 row combinations for 2 keys.

I'm bored right now, but I don't feel like programming or analyzing any complicated statistics. So I look at the connections of the different observations again. I noticed the following:

What is characteristic for homophonic encryption? Right! As a rule, the first rows of a the encryption usually contain the fewest repetitions of characters from a statistical point of view. This is especially true if a cyclical encryption was used. z408 was an exception, since the plaintext letter "L" appeared relatively often and the key was not chosen well. This is also an indication that Zodiac chose his key based on the letter frequency of general English texts, not those of the plaintext of z408.

Let's say the following three assumptions are correct:

- Zodiac was able to learn and he improved his key(s)
- Zodiac has used several keys in z340.
- The observation that the upper 6, the middle 8 and the lower 6 rows consist of at least two keys is correct.

Let's take a look at the following picture (source: http://zodiackillerciphers.com/images/z ... olumns.jpg)



It is noticeable that this actually fits together. If several keys have been used, the observations of the character repetitions within the individual rows could be an indication of this.

What do you think?

Hi Largo.

What I tried to do here:

viewtopic.php?f=81&t=3196&start=750

is basically the same as this:



I tried to make a message that would have a P20 spike with transposition at period 20, about 39 symbol positions unique to top and bottom 6 rows, and only four symbol positions unique to the middle 8 rows. "Cake" 6-8-6.

It was very difficult to transpose the plaintext at P20 and use two different keys to get a spike at P20. Assuming, of course, that Zodiac transposed the plaintext before encoding. I had to take an entire book and find the chunk of 340 plaintext that would do it the most. It may be that more than one key was used and the P19 spike is a phantom.

The P20 repeats for smokie 64 were mostly within their 6-8-6 regions, meaning that there could be a P20 bigram in the top 6, but the repeat would occur in the top 6 or bottom 6. Or there would be P20 bigram in the middle 8, but the repeat would also be in the middle 8. There were few that crossed over from one region into the other. And with the 340, the P19 bigram repeats occur in all regions ( right? someone check me on this ).

It was easy to make a message that had a lot of symbols unique to the top and bottom 6 rows.

But I was unable to make a message that only had a few symbols unique to the middle 8 rows. The reason is because with homophonic encoding, a lot of symbols appeared in the middle 8 rows and did not appear in the top and bottom 6 rows. As different from the 340, where only two symbols, or a total of four symbol positions are unique to the middle 8 rows.

That is why I think that one possibility is not two different keys, but that Zodiac selected his symbols from the homophonic groups of symbols that map to the individual plaintext by some pattern that would cause the regional bias.

That's just my two cents, and I could be wrong.

What if the reason the symbols are sparse in the middle but are lush at the top and come back in the bottom is the following:

Instead of a key that looks like this:

E = ABCDEABCDEABCDE (cyclic)

it looks like this:

E = ABCDEFGHGFEDCBA

i.e. palindromic

?

We really need a webtoy that has both Ciphertext and Key, shown side by side.

Fisherman'sFriend wrote:What if the reason the symbols are sparse in the middle but are lush at the top and come back in the bottom is the following:

Instead of a key that looks like this:

E = ABCDEABCDEABCDE (cyclic)

it looks like this:

E = ABCDEFGHGFEDCBA

i.e. palindromic

?

I agree that something like that could be possible.

smokie treats wrote:

Fisherman'sFriend wrote:What if the reason the symbols are sparse in the middle but are lush at the top and come back in the bottom is the following:

Instead of a key that looks like this:

E = ABCDEABCDEABCDE (cyclic)

it looks like this:

E = ABCDEFGHGFEDCBA

i.e. palindromic

?

I agree that something like that could be possible.

It will be very hard to have a say about it as with so many symbols many many millions of cycles are to be considered and such things are bound to happen.

Here are a few of the best palindromic cycles in the 340:

Code: Select all

5-symbol:
H>OYHOO>XOYOYOX>OOHYO>HO: 11
H>OYHOO>;OYOYO;>OOHYO>HO;: 10.08
H>H35>5;555;5>53H>H;: 9

6-symbol:
H>YH3>;Y@Y;>3HY>H;: 8
H>YH3>XY@YX>3HY>H: 8
H>YH3>XYAYX>3HY>HA: 8
>Yf3>XfY@YfX>3fY>: 8
>Yf3>XfYAYfX>3fY>A: 8


Largo wrote:It is noticeable that this actually fits together. If several keys have been used, the observations of the character repetitions within the individual rows could be an indication of this.

What do you think?

I do not recommend to use repeats per row to gauge multiple keys. Your vision is then limited to the length of these rows. Rows 11, 12 and 13 are unlikely to be the start of a new key:

- Rows 1, 2 and 3 of the 340 have 41 unique symbols and ioc of 0.0078. Row 2 has 1 symbol that repeats from row 1.

- Rows 11, 12 and 13 of the 340 have 37 unique symbols and ioc of 0.0149. Row 12 has 7 symbols that repeat from row 11! These are: "+R5|BTM".

smokie treats wrote:It was very difficult to transpose the plaintext at P20 and use two different keys to get a spike at P20. Assuming, of course, that Zodiac transposed the plaintext before encoding. I had to take an entire book and find the chunk of 340 plaintext that would do it the most.

My results speak against that. I suspect that your ciphertexts have a low ioc and that this diminishes the bigram repeats a whole lot.

Largo wrote:

Jarlve wrote:I would be happy if someone could make a few more test ciphers that are in line with borkky's hypothesis

Here you are:

Largo_Sing_Along

Code: Select all

8z0Oap:sI7n;WGkRN
2XvArxKhPlzMS8QVO
nL9:sJ4t3iZNabfBo
uVUx;AF38isL14qyN
lwuMk2erx8o07sVLd
:=r2vpy8e+aCGxTY0
a5K-p7:qPaRdw0PHM
Vit1o3ZGyTFmSIsji
RmbX=uYvZwBp+T;Lj
Ohn:cNS3;yPFQWfA-
0EkZJWUePFxF9T34J
b;z+Q:VA0LFwCyNBP
0UhtDUePFOMonXlaa
zq2U0pmvaeuuVKDx9
awSVgLiucj:37QyCP
L0rd5pmGw8iVHFtp=
Fox3G1jvMWq9SLPY4
FbI+QwCh57OZVpR0:
VRveZTxkuV;3rIXz5
EB=SoF4WfYb1uuOmw


My workstation has been running this for 2 days now without any interesting result. Largo, I am a bit worried that your cipher is harder than the 340, it has 66 symbols and a low ioc, which diminishes internal structure. Please tell me how the keys are distributed so that I can find out if AZdecrypt can crack this cipher when the distribution is known.