Things I noticed about Z340
Here's a few things I noticed about Z340 when doing different experiments with it. Things that stand out, but didn't lead to any breakthroughs (obviously). I thought I'd share my observations just in case someone else can make some further conclusions, or it might help lead to the solution in some other way.
Splitting Z340 into odd and even symbols (i.e. only symbols at odd or even positions) produces 2 very different sequences:
Odd symbols
Even symbols
Statistically they should be very similar, as English language has absolutely no preference towards odd/even letters that I know of, or odd/even lengths of words, etc. And yet...
Odd symbols:
Very low number of bigram repeats (only 2!).
No trigram repeats, or higher N-grams.
58 unique symbols, so 5 symbols are missing: & 1 @ A t.
Even symbols:
Good number of bigram repeats (9). About the expected for the cipher this length — Z408 had 10 bigram repeats in the first 170 symbols.
2 trigram repeats. Again, about expected — Z408 also had 2 trigram repeats in the first 170 symbols. These 2 trigram repeats are not overlapping with the 2 trigram repeats in the full Z340 in any way, so they were created anew.
54 unique symbols, so 9 symbols are missing: 3 7 ; J M U X b j.
The 14 symbols that are not shared is especially significant, I think. That's 22% our of 63 symbols. I don't think it can happen by chance. For comparison, I've split Z408 the same way, and there is only 1 symbol that's present in odds and not in evens.
Symbol counts (frequencies) do not match at all either. For example, the 3rd most frequent symbol for "odds" ('M') is completely missing in "evens". 2nd most frequent symbol for "evens" ('2') only appears twice in "odds". The only exception is '+', which is the top symbol for both, but most of '+'s went into "evens": 14 vs 10 for "odds".
What could this possibly mean? The obvious conclusion — the odd symbols are just random noise, filler inserted to confuse decryptors, they are nulls that should be dropped. However, the odds sequence does have a fairly expected frequency distribution, so it might be part of the message, just encrypted differently from the evens sequence.
I tried feeding each of the 2 sequences into an auto-solver, but didn't get a coherent solution. The issue might be the very high multiplicity, so it could be that we have the correct extra transposition that Z did to Z340, but the resulting 170 symbol cipher is too short to be cracked. I've tried combining the 2 sequences, by appending one to another, and then the other way around, but didn't get a solve either way. Maybe they should be recombined in some other way?
Another issue with this (and I'm playing devil's advocate now :)), is that Z clearly liked to use a matrix with 17 characters in a row. It's an odd number, so odd/even positions alternate (or flip around) for every other row. I'll demonstrate with a matrix of width 7, and use 0 for odd positions and 1 for even positions:
0101010
1010101
0101010
1010101
So if he did use odds/evens split, it would make more sense, and would be easier to do with a pen and paper, if done per row. I.e. restart odd/even counts for each new row. I've tried decomposing Z340 this way, but ended up with 2 unremarkable sequences, of slightly different lengths, but with very similar stats. On the other hand, maybe Z realized that doing odds/evens across the whole cipher would result in a "checkerboard pattern" I showed above, and maybe he thought that it would be more confusing and harder to crack, and went with it.
Splitting Z340 into odd and even symbols (i.e. only symbols at odd or even positions) produces 2 very different sequences:
Odd symbols
Even symbols
Statistically they should be very similar, as English language has absolutely no preference towards odd/even letters that I know of, or odd/even lengths of words, etc. And yet...
Odd symbols:
Very low number of bigram repeats (only 2!).
No trigram repeats, or higher N-grams.
58 unique symbols, so 5 symbols are missing: & 1 @ A t.
Even symbols:
Good number of bigram repeats (9). About the expected for the cipher this length — Z408 had 10 bigram repeats in the first 170 symbols.
2 trigram repeats. Again, about expected — Z408 also had 2 trigram repeats in the first 170 symbols. These 2 trigram repeats are not overlapping with the 2 trigram repeats in the full Z340 in any way, so they were created anew.
54 unique symbols, so 9 symbols are missing: 3 7 ; J M U X b j.
The 14 symbols that are not shared is especially significant, I think. That's 22% our of 63 symbols. I don't think it can happen by chance. For comparison, I've split Z408 the same way, and there is only 1 symbol that's present in odds and not in evens.
Symbol counts (frequencies) do not match at all either. For example, the 3rd most frequent symbol for "odds" ('M') is completely missing in "evens". 2nd most frequent symbol for "evens" ('2') only appears twice in "odds". The only exception is '+', which is the top symbol for both, but most of '+'s went into "evens": 14 vs 10 for "odds".
What could this possibly mean? The obvious conclusion — the odd symbols are just random noise, filler inserted to confuse decryptors, they are nulls that should be dropped. However, the odds sequence does have a fairly expected frequency distribution, so it might be part of the message, just encrypted differently from the evens sequence.
I tried feeding each of the 2 sequences into an auto-solver, but didn't get a coherent solution. The issue might be the very high multiplicity, so it could be that we have the correct extra transposition that Z did to Z340, but the resulting 170 symbol cipher is too short to be cracked. I've tried combining the 2 sequences, by appending one to another, and then the other way around, but didn't get a solve either way. Maybe they should be recombined in some other way?
Another issue with this (and I'm playing devil's advocate now :)), is that Z clearly liked to use a matrix with 17 characters in a row. It's an odd number, so odd/even positions alternate (or flip around) for every other row. I'll demonstrate with a matrix of width 7, and use 0 for odd positions and 1 for even positions:
0101010
1010101
0101010
1010101
So if he did use odds/evens split, it would make more sense, and would be easier to do with a pen and paper, if done per row. I.e. restart odd/even counts for each new row. I've tried decomposing Z340 this way, but ended up with 2 unremarkable sequences, of slightly different lengths, but with very similar stats. On the other hand, maybe Z realized that doing odds/evens across the whole cipher would result in a "checkerboard pattern" I showed above, and maybe he thought that it would be more confusing and harder to crack, and went with it.