My question for both of you…: What kind of message/kind of form of info do you expect to find?

I think and hope that solving the 340 will close the case. In that it may open the way to solve the My name is cipher and/or the Mount Diablo code or will contain other important information. But I won’t be disappointed if it does not.

Form of info? Expecting an English message close to 340 characters.

Hello Mr Doranchak and Mr Jarlve.
I have a question, it seems silly (and should be), saw his lecture on the codes with the Mr Doranchak material (slides) , the time when you talk about the symbols not perterncentes to both codes.

I was thinking of three things.
1-Clerical errors in the words.
2-The use of a single symbol (in z408) for two distinct letters.
3-Filling in the end with the famous EBEORIETEMETHHPITI.
It has been crafted hypothesis of a "code derived"
I say, keep the position of symbols not used in either a new code
I can not imagine the complexity of setting up such a code, if this is possible

Forgive me if this is BS
Marcelo

Guys the way you tackle this decades old mystery code is amazing. I have a lot of respect for you.

Jarvle wrote:

Form of info? Expecting an English message close to 340 characters.

Info can also be "written" without words. The sequenties you guys are finding are increadible but could be part of something else than a text message…

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I am working on smokie14, a variation of smokie9-13 but nothing quite as exotic as the transposition schemes that you guys have been working on. I have been updating my spreadsheets and they are really cool now. I can look at three different period x bigrams side by side for lots of fun hours staring at my computer screen, including being able to show where period x bigrams have symbols that match period y bigrams from one view of the message compared with another. After smokie14, which is optional for you guys to solve, I am going to dig into the 340 and find where diagonal rows of bigram repeat symbols are offset so that I can make a short list of position sets where plaintext s may have been skipped or added during transposition.
.

I ran across my old quadrant analysis code and decided to run it again with a focus on counting repeating bigrams.

Here is a reference to the approach I used to split the cipher text into quadrants: http://zodiackillerciphers.com/wiki/ind … t_analysis

Basically, I test all possible ways to split the cipher into 4 quadrants. I ran it for all possible (i,j) values that define the "split point" (the intersection of the four quadrants). For each possible (i,j), I reorder the 4 quadrants in all possible ways, and apply all possible combinations of flips and rotations to each quadrant before combining them back into a single cipher text. Then I count the bigram repeats, and track the highest count across all the permutations for a given (i,j) split point.

Here is a heat map showing the results (x-axis is j value, y-axis is i value, and the numbers in squares are bigram repeat counts):

The peak of 40 bigram repeats occurs at (i,j) = (14,4), which roughly corresponds to row 15 column 5. The number 15 stood out to be because that is the same as the period that gives us the highest repeated bigrams!

When I get some time, I will try to post the corresponding transformed ciphers and the steps that were used to produce them.

Basically, I test all possible ways to split the cipher into 4 quadrants. I ran it for all possible (i,j) values that define the "split point" (the intersection of the four quadrants). For each possible (i,j), I reorder the 4 quadrants in all possible ways, and apply all possible combinations of flips and rotations to each quadrant before combining them back into a single cipher text. Then I count the bigram repeats, and track the highest count across all the permutations for a given (i,j) split point.

Doranchak, your work answers the question I posed here. It does look like the cipher could possibly be split into quadrants on column 8 and row 9 (the "center" being the most logical place), though there are many other possibilities as your "heat map" demonstrates. Did you try the "knight’s tour" as one of the transposition schemes for the quadrants?

Sounds good smokie and doranchak!

Info can also be "written" without words. The sequenties you guys are finding are increadible but could be part of something else than a text message…

Such as? I think for now it’s best to assume an English message and the FBI suggested the same.

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Another idea for you guys. Score the untransposed period 1 bigram repeats by probability of occurrence and total the score for your different transposition schemes. Maybe you could compare the scores for that in addition to comparing solve scores.

If there are six A’s and four B’s in a message and they appear AB AB AB, then that should get a higher score than twenty four C’s and ten D’s that appear CD CD.

The one on the left just ain’t gonna happen by chance, whereas the one on the right will.
.

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Smokie, that is a great idea. I have scored repeated patterns that way in the past. I think I will use a probability measurement when I finish coding up my transposition explorer.

Here is a spreadsheet showing the best results from the quadrant transposition experiments (Jarlve, the resulting cipher texts are in here if you want to run them through your solver):

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

The best one was at split point (14,4) which divided the cipher into rows 1-14 and 15-20, and into columns 1-4 and 5-17. The resulting cipher text contains 40 bigram repeats. Here is an illustration of the transpositions performed:

First, start with the original cipher:

Apply split point (14,4):

So here are the quadrants. They will be referred to as 0, 1, 2 and 3:

This permutation is selected: {1, 3, 0, 2}. So, reorder the quadrants based on that permutation:

Rotate quadrants 0, 1 and 3 by 180 degrees (quadrant 2 is left alone):

Flip quadrants 2 and 3 horizontally:

We have to stick the quadrants back together, so a concatenation operation is selected: ((A c1 B) c1 C) c1 D, where c1 means "align the blocks by their bottoms". All this means is to write them all out with their bottoms lined up like this:

Finally, write out the new cipher text, reading left to right and top to bottom, skipping over the whitespace between the quadrants:

And here is a visualization of the resulting 40 repeated bigrams and 3 repeated trigrams:

I have a feeling that we’ll be able to find many spurious transformations that produce 40+ bigram repeats, so the real trick will be to find the correct one (hopefully it exists!). I’m going to resume working on a more generalized transposition explorer using multiple measurements to explore the space of possibilities.

Here are the results: https://drive.google.com/open?id=0B5r0r … 3BldjNHLUE

Thanks for doing this doranchak and I don’t mind running them through so send as much you like.

Thanks Jarlve. Here are the words found among those plaintexts: http://pastebin.com/raw.php?i=h0dMv5Th

Did you try the "knight’s tour" as one of the transposition schemes for the quadrants?

I did not try that. There are too many possibilities, unless you limit the kinds of knight’s tours (for instance, the period 19 pattern follows one specific movement, rather than allowing for multiple directions). Even just an 8×8 board has 26,534,728,821,064 possible complete tours. So the search space is high.

Still, I doubt Zodiac would have implemented a random knight’s tour – since the bigrams are peaking at period 19 (and period 15 for the flipped version), maybe simple variations of basic knight’s tours could be explored exhaustively. Still, in those cases, the periodic bigrams reduce to a normal bigram counting problem once the cipher is untransposed with a simple procedure.

Side note: I thought this was interesting. It’s the site of a guy who spent a lot of time trying to count how many possible knight’s tours exist for various grid sizes.

http://magictour.free.fr/

I’m no longer interested in magic knight’s tours
and don’t want to encourage others to waste time on it.

His frustration is amusing, but we all feel it to some degree with this damned Z340!

<OFFTOPIC>
There was an old computer game with puzzles ‘ "The 7th Guest" that required completing a Knight’s Tour as part of the game. HARD!!
</OFFTOPIC>
-glurk

Jarlve wrote:

Such as? I think for now it’s best to assume an English message and the FBI suggested the same.

And assuming this hasn’t helped anyone to solve the cipher in 4 decades at this point.

Z claimed his identity was in his first cipher. Harden has solved the cipher but Z wrote in the solution of it " I will not give you my name ". Did Z really lie about his identitiy in the code? I believe he didn’t…

There are more things you can see as your indentity besides your name. What if there was a second layer of some sort of info (besides the textbased solution) about his identity? He was talking about his identity being in the code, right?

The work you guys do on the second cipher is amazing. But will it lead to a solution (text/message)?

Because Z had a solution/text in his first cipher (which got solved quite quickly) doesn’t mean he did so in the others as well (unsolved for decades). Maybe Zodiac is crackproof (and not lieing about it) because nobody can get a solid solution out of the ciphers because there isn’t one in them?