A more complex but not impossible scenario is that z408 was sent by a copy-cat and that z340 was Z’s response.
Copying whom? The z408 was the first Z cipher.
It’s complex and involves time travel.
Ah, time travel. Hadn’t thought of that. Finally, a reasonable Z theory. Maybe Z is Marty McFly.
It’s complex and involves time travel.
Ah, time travel. Hadn’t thought of that. Finally, a reasonable Z theory. Maybe Z is Marty McFly.
No wonder he’s never been caught, we’ve all been wrongly assuming that he lived during the time of his crimes!
Here is his evidence:
http://tsjuzek.com/blog/z340.html-glurk
Somewhat demotivating. 
But there may be one particular weakness to what he is doing here. He says "I have encrypted the forty texts, using a similar encryption to z408". He does not give details about exactly what that entails. However, I think we should not be surprised that other encryptions in the style of z408 have a similar signature to z408. His other point of comparison is a random sequence, i.e., "fake ciphers", and says Z340 looks like these, not like Z408. Ok.
But, wait. What about looking for a way to encode real text with a real cipher that gives the signature of a "fake cipher"? Is that particularly hard to do, or might that even arise naturally with a particular method of hand encoding? Maybe the signature becomes indistinguishable from "fake", if we design the cipher deliberately to obscure the bigrams and trigrams that he’s looking for. Given the larger alphabet of z340 compared to z408, this might be easier to accomplish.
His article is based on the missing bigrams in the horizontal direction of the 340: it should have somewhere around 45 bigrams yet there are only 25.
His other point of comparison is a random sequence, i.e., "fake ciphers", and says Z340 looks like these, not like Z408. Ok.
Transposition of the plaintext would make the cipher look fake in such a test.
Given the larger alphabet of z340 compared to z408, this might be easier to accomplish.
While the 340 has a larger alphabet it is actually more repetitive than the 408 if compared at similar lengths. A good way to measure the "repetitive potential" of a cipher is to look at its index of coincidence, if it is higher then it will be generally more repetitive. To illustrate my point consider the following tests. The average number of bigrams for 1.000.000 shuffles of the 340 is 19.7. The average number of bigrams for 1.000.000 shuffles of a 340 character length 408 is 15.3.
Transposition of the plaintext would make the cipher look fake in such a test.
Agreed. Do you think it is also plausible to simply hand tweak a cipher to remove the bigrams? Or would this run into trouble as removing one tends to creates others, and it would take some kind of advanced procedure to obtain the low occurrences we see in Z340?
Transposition of the plaintext would make the cipher look fake in such a test.
Agreed. Do you think it is also plausible to simply hand tweak a cipher to remove the bigrams? Or would this run into trouble as removing one tends to creates others, and it would take some kind of advanced procedure to obtain the low occurrences we see in Z340?
We looked at this hypothesis back in 2015.
Zodiac could have maintained a list of bigrams on a separate piece of paper as he was writing down the symbols of the cipher. And do "something" whenever a bigram repeats to often. Pick a different homophone, use a polyphone, leave out the symbol, add a symbol, etc. This "something" then should have happened around 20 to 30 times, to get down the probable number of bigrams the cipher should have had to 25. Even with 30 "somethings" the cipher would probably still solve normally and if not any of the hill climbers in AZdecrypt should have made work of it. It has solved many such ciphers that often were much harder.
Period 2 bigrams are also missing. A 340 character 408 has 38 period 2 bigrams, though these are less likely to occur. Period 2 bigrams are 2 positions apart. Normal bigram "AA", period 2 bigram "A A", anything could be in between.
Period 2 bigrams are also missing. A 340 character 408 has 38 period 2 bigrams, though these are less likely to occur. Period 2 bigrams are 2 positions apart. Normal bigram "AA", period 2 bigram "A A", anything could be in between.
Oh wow. To me that obliterates the idea that he could have removed them by inspection. They are not there for some other reason.
Transposition of the plaintext would make the cipher look fake in such a test.
Agreed. Do you think it is also plausible to simply hand tweak a cipher to remove the bigrams? Or would this run into trouble as removing one tends to creates others, and it would take some kind of advanced procedure to obtain the low occurrences we see in Z340?
We looked at this hypothesis back in 2015.
Zodiac could have maintained a list of bigrams on a separate piece of paper as he was writing down the symbols of the cipher. And do "something" whenever a bigram repeats to often. Pick a different homophone, use a polyphone, leave out the symbol, add a symbol, etc. This "something" then should have happened around 20 to 30 times, to get down the probable number of bigrams the cipher should have had to 25. Even with 30 "somethings" the cipher would probably still solve normally and if not any of the hill climbers in AZdecrypt should have made work of it. It has solved many such ciphers that often were much harder.
Period 2 bigrams are also missing. A 340 character 408 has 38 period 2 bigrams, though these are less likely to occur. Period 2 bigrams are 2 positions apart. Normal bigram "AA", period 2 bigram "A A", anything could be in between.
I count 38 period 2 bigrams in the full 408, and 27 in 340. These give ratios of 0.093 and 0.079 (# bigrams/length).
Compared to the ratios of 0.115 and 0.067 for bigrams, it doesn’t seem as conclusive that they are missing in any anomalous way from the 340.
There are 49 period 2 bigrams in the full 408 and 21 in the 340. And there are 38 period 2 bigrams in a 340 character 408.
FreeBASIC period transpose/untranspose code:
for i=1 to period
for j=i to cipher_length step period
k+=1
if transpose=1 then
new_cipher(j)=old_cipher(k)
end if
if untranspose=1 then
new_cipher(k)=old_cipher(j)
end if
next j
next i
Though the missing period 2 bigrams are not as conclusive as it may seem because the 408 has more period 2 bigrams than expected. I did 2 tests that created 10000 test ciphers each of the following: english plaintext + sequential homophonic substitution targetting the symbols and ioc of the compared cipher + period 2 untransposition. These are made with 100 unique plaintext found on the web which may not represent the Zodiac his use of language very well.
Versus 10000 test ciphers a period 2 bigram count of 21 such as in the 340 are -1.34 standard deviations away from the mean (28.21). The 340 has less period 2 bigrams than expected but not by so much.
Combinations processed: 10000/10000 Measurements: - Mean: 28.2187 - Variance: 28.61607030999968 - Standard deviation: 5.349399060642202 - Sigma of 21: -1.349441295773022 - Sigma of 11: -3.21881015134677 - Sigma of 49: 3.884791499833475 - Lowest: 11 (Plaintext(22), Encode: homophonic substitution(2), Period(UTP,2)) - Highest: 49 (Plaintext(60), Encode: homophonic substitution(83), Period(UTP,2))
And again versus 10000 test ciphers a period 2 bigram count of 38 such as in a 340 character 408 are 1.79 standard deviations away from the mean (27.87). The 408 has more period 2 bigrams than expected.
Combinations processed: 10000/10000 Measurements: - Mean: 27.8765 - Variance: 31.88524775000047 - Standard deviation: 5.646702378379834 - Sigma of 38: 1.792816288451998 - Sigma of 10: -3.165830037093113 - Sigma of 53: 4.449233962851164 - Lowest: 10 (Plaintext(33), Encode: homophonic substitution(67), Period(UTP,2)) - Highest: 53 (Plaintext(1), Encode: homophonic substitution(14), Period(UTP,2))
