This small study shows that it is highly suggestive that the 340 just like the 408 has increasingly random cycles. By taking that in account in future analyses we should be able to make more sense of the cycles in the 340.
By comparing the 340 and 408 against randomizations of itself with a 3-symbol cycle measurement it can be seen that both ciphers appear to have a sequential homophonic substitution layer. With the 408 being more sequential and the 340 being more random. The farther these sigma values are away from 0 the better the correlation is:
340, 3-symbol cycles: 6.48 sigma
408, 3-symbol cycles: 13.26 sigma
As observed before, the 340 does not scale going from 3 to 4-symbol cycles. Higher n-symbol cycles do not seem to appear in the 340:
340, 4-symbol cycles: 6.75 sigma
408, 4-symbol cycles: 18.74 sigma
If we look at increasingly random cycles then both ciphers appear to fall in that category. With the 340 scaling a bit better now going from 3 to 4-symbol cycles:
340, increasingly random 3-symbol cycles: 4.81 sigma
340, increasingly random 4-symbol cycles: 5.98 sigma
408, increasingly random 3-symbol cycles: 10.38 sigma
408, Increasingly random 4-symbol cycles: 16.80 sigma
Which is not the case for decreasingly random cycles:
340, decreasingly random 3-symbol cycles: -0.04 sigma
408, decreasingly random 3-symbol cycles: -0.56 sigma
A homophonic cipher with about 25% sequential randomness has both increasingly and decreasingly random cycles since the randomness is generally spread out through the cipher:
25% sequential randomness cipher, increasingly random 3-symbol cycles: 3.10 sigma
25% sequential randomness cipher, decreasingly random 3-symbol cycles: 3.15 sigma
Other cycle type hypotheses seem not to be a good match for the 340:
Shortened cycles cipher, 3-symbol cycles: 12.43 sigma
Shortened cycles cipher, increasingly random 3-symbol cycles: 2.29 sigma
Shortened cycles cipher, decreasingly random 3-symbol cycles: 4.30 sigma
Palindromic cycles cipher, 3-symbol cycles: 10.13 sigma
Palindromic cycles cipher, increasingly random 3-symbol cycles: 3.29 sigma
Palindromic cycles cipher, decreasingly random 3-symbol cycles: 1.95 sigma
'cl=cycle length 'cs=cycle size (3-symbol cycles for example) 'the z() array contains a single cycle in numbers (1,2,3,1,2,3,1,2,3) 'loop through all n-symbol cycles and sum the score each time case 1 'normal cycles al=cl-(cs-1) for i=1 to cl-(cs-1) e=1 for j=i to i+(cs-2) for k=j+1 to i+(cs-1) if z(j)=z(k) then e=0 exit for,for end if next k next j z2(i)=e next i a=0 for i=1 to cl-(cs-1) a+=z2(i) next i score=a*(a/al) case 2 'increasingly random cycles al=cl-(cs-1) for i=1 to cl-(cs-1) e=1 for j=i to i+(cs-2) for k=j+1 to i+(cs-1) if z(j)=z(k) then e=0 exit for,for end if next k next j z2(i)=e next i al=cl-(cs-1) if al mod 2=1 then al1=1:al2=(al-1)/2 else al1=0:al2=al/2 end if p1=0:p2=0 for i=1 to al2 p1+=z2(i) p2+=z2(i+al1+al2) next i if p1-p2>0 then score=((p1+p2)/al2)*(p1-p2) case 3 'decreasingly random cycles al=cl-(cs-1) for i=1 to cl-(cs-1) e=1 for j=i to i+(cs-2) for k=j+1 to i+(cs-1) if z(j)=z(k) then e=0 exit for,for end if next k next j z2(i)=e next i al=cl-(cs-1) if al mod 2=1 then al1=1:al2=(al-1)/2 else al1=0:al2=al/2 end if p1=0:p2=0 for i=1 to al2 p1+=z2(i) p2+=z2(i+al1+al2) next i if p2-p1>0 then score=((p1+p2)/al2)*(p2-p1)
I agree that it could be increasingly random cycles.
Something that I have been wondering about recently, could research but will just ask instead.
Your 26 of L=17, measured like this: A*B*****B or this A*****B*B. And these are not consistent with 25% randomization throughout the entire message, correct?
I am pretty sure you also measured like this: B**********B. What did you get for these, and are they consistent with 25% randomization throughout?
Just need a refresher on the B*********B type of measurement.
Just need a refresher on the B************B type of measurement.
Count the lengths of every continuous sequence in the cipher that has no repeating symbols. The first such sequence in the 340 is "HER>pl^VPk|1LTG2dN" with a length of 18. The first sequence stopped at the "N" symbol because the "p" symbol comes after it and that would cause a repeat. The second sequence is "ER>pl^VPk|1LTG2dN" etc. Just increment the position by one and slide through the entire cipher without causing a repeat in each sequence.
A 26 L17 or higher measured as such happens about 1 in a 75 times with 25% encoding randomization. A problem here is that I just match the ioc of the test ciphers to that of the 340 while the exact frequencies may also play a role.
A 25% encoding randomization as we generally have considered it has it randomness spread out through the cipher and that shows:
A homophonic cipher with about 25% sequential randomness has both increasingly and decreasingly random cycles since the randomness is generally spread out through the cipher:
25% sequential randomness cipher, increasingly random 3-symbol cycles: 3.10 sigma
25% sequential randomness cipher, decreasingly random 3-symbol cycles: 3.15 sigma
And is in that regard not a good match for the 340.
You can also see doranchak’s video from about 36:15 to 39:15 for an explanation of how the unique sequences are measured: https://www.youtube.com/watch?v=3Hk__Hk5c9M
Direct link to that time in the video:
https://www.youtube.com/watch?v=3Hk__Hk5c9M&t=36m15s
Slight misunderstanding. I totally get your 26 L17 and have a spreadsheet that detects those, and have tried to make messages with 25% randomization that does it, etc.
I mean this one: pl^VPk|1LTG2dNp without the HER>. Have you tried to duplicate those stats, whatever they are, with 25%?
I mean this one: pl^VPk|1LTG2dNp without the HER>. Have you tried to duplicate those stats, whatever they are, with 25%?
Nope.
The cycles give clues as to which line is to be read in which order imho.
Its not that its cycling randomly its that you are reading the wrong line or have started the cycle at the wrong point.
It was made with pen and paper it will be solved with pen and paper.
Yes, dyslexia is probably my first undiagnosed language.
The cycles give clues as to which line is to be read in which order imho.
Its not that its cycling randomly its that you are reading the wrong line or have started the cycle at the wrong point.
There are many reasons for the imperfect cycles. The reading direction is only one of them. Just as well fillers/blenders can be in the game or whatever. But yes: A route transposition or two interlocked routes can be a reason.
It was made with pen and paper it will be solved with pen and paper.
If you lock an ordinary shackle lock with your hand, this does not mean that you can open it again with your hand. It will be similar with z340. Yes, it was made with pen and paper. And I’m also one of those people who think that a relatively simple method was used, but we just haven’t gotten behind it yet. Nevertheless: Since there are (340 faculty) possibilities with pen & paper, such analyses have to be carried out so that we can better estimate what we are dealing with.
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But yes: A route transposition or two interlocked routes can be a reason.
After sequential homophonic substitution? I’d say it is very unlikely. Can you show me an example?
But yes: A route transposition or two interlocked routes can be a reason.
After sequential homophonic substitution? I’d say it is very unlikely. Can you show me an example?
I have been thinking about hoax lately and how it could be done. Something that would cause the various patterns that we are looking at. Vague idea so far, but a grid, maybe 26 wide x 13 high, with each row having some of the symbols arranged in a pattern LR. Then reading TBLR and transcribing LRTB to create what looks like encoding cycles, P39 and P78, and somehow also P19. It is a fear of mine. I tried it one day already, but I need to try again. So far I don’t see any way to rearrange the symbols in such a grid that shows the patterns though. Something to consider.
Also, we have touched on the idea of masking P1 repeats by subtly rearranging symbols in a pattern that could disrupt the cycles and cause P19.
Hi,
all right, I’ll give it a try:
First let’s assume that Zodiac used the following procedure to substitute a plaintext:
viewtopic.php?f=81&t=3337
I refer to the variant described in the first .PDF. To simplify things, we assume that Zodiac made no mistake and got a perfect cyclic substitution. Here is an example:
Plaintext:
IGOTMYFIRSTREALSI XSTRINGBOUGHTITAT THEFIVEANDDIMEPLA YEDITTILMYFINGERS BLEDWASTHESUMMERO FSIXTYNINEMEANDSO MEGUYSFROMSCHOOLH ADABANDANDWETRIED REALHARDJIMMYQUIT JODYGOTMARRIEDISH OULDVEKNOWNWEDNEV ERGETFAROHWHENILO OKBACKNOWTHATSUMM ERSEEMEDTOLASTFOR EVERANDIFIHADTHEC HOICEYEAHIDALWAYS WANNABETHERETHOSE WERETHEBESTDAYSOF MYLIFEAINTNOUSEIN COMPLAININWHENYOU
Substitution:
hEXL;TIcnowGad:1C -KsV5AJ9=rEFNhLDw s2xIcQ3iqkBC;yYuZ T+H5NLhU;TIc7J0tM 4:Pkldowpg1v;;anX IKC-sTR5Ax;3DqBM= ;yESToIGX;1bF=Xu2 iHZ9d7kDRBl+NVh0H tPiUpZnkmc;;TOrCL m=BTJXw;dGV5gHhKF =v:kQajAXlqlxB73Q ytE+sIDn=2lp0RcuX =j4iejAXlNFZLMS;; PGoga;xHw=Ud1sIXV 3QytDqkCI52iBNp+b F=he0TPZ2cHd:lDTK li7RZ9gLpanxwFXM3 lyG+s204PoNkdT1=I ;TuCIgD5ALqXrKah7 b=;YUicRCAlpxqTXv
I assume that he read in the newspaper how the Hardens cracked z408. If he is now looking for a very simple way to avoid such a vulnerability, he can simply interrupt the text flow. The easiest way to do this is to transfer his draft to the final cipher with regular interruptions. He does not skip symbols from the draft, but leaves gaps in the final cipher. Let’s take the substitution from the previous example:
Now he fills the gaps with the rest of the draft. I’ve chosen a diagonal direction. (Doranchak once posted something similar. It was a link to Klaus Schmeh’s blog. Unfortunately I can’t find it anymore).
I did it like this:
This is the final result:
hEXFL;T0Icn:owGLa d:=1C-TKsVl5AJp9= rhEFNPhLDDws2axIc eQ3iZqkBTC;ynYuZ0 T+H25NLKhU;xTIc47 J0ctM4l:PkwldoPwp gH1v;i;anFXIKoC-s dTR57Ax;X3DqNBM=D ;yERSToMIGXk;1b5F =XZu2i3HZ9dd7kADR B9l+NlVh0THtPLiUp gZnkymc;1;TOqrCL; m=BGTJX=w;dXGV5Yg Hh+KF=Iv:krQajUAX lsqlx;B73KQytiE+s 2IDnT=2lap0RccuXp =j4uiejhAXlRNFZxL MSC;;P7GogCa;xqHw =IUd1bsIXAV3QTytD gqkC=I52liBNXp+bv
Why did I choose diagonal in the example? Quite simple: Now there are always 3 characters which are interrupted by one character each, so about 25% randomness. The result is a rather, but not perfectly cyclic cipher. Autosolvers can’t solve it and depending on the chosen pattern for the gaps, the funniest statistics will result.
PS: If you resize the cipher to 20×17, the gaps form regular columns.
If you do not take a diagonal for the gaps, but columns, you can theoretically explain the pivots. Then they simply consist of repeated trigrams.
I haven’t done many experiments of this kind yet, but I can imagine that such a procedure was used in z340.
I have been thinking about hoax lately and how it could be done. Something that would cause the various patterns that we are looking at. Vague idea so far, but a grid, maybe 26 wide x 13 high, with each row having some of the symbols arranged in a pattern LR. Then reading TBLR and transcribing LRTB to create what looks like encoding cycles, P39 and P78, and somehow also P19. It is a fear of mine. I tried it one day already, but I need to try again. So far I don’t see any way to rearrange the symbols in such a grid that shows the patterns though. Something to consider.
Your hoax theory is interesting. We could try to develop a kind of hoax generator that will scramble different patterns in the way you described. Then we could compare the results with z340.
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Edit:
There are no perfect 4/5 cycles in the sample cipher. If find some time I’ll try to generate a better one.
I have been thinking about hoax lately and how it could be done. Something that would cause the various patterns that we are looking at. Vague idea so far, but a grid, maybe 26 wide x 13 high, with each row having some of the symbols arranged in a pattern LR. Then reading TBLR and transcribing LRTB to create what looks like encoding cycles, P39 and P78, and somehow also P19. It is a fear of mine. I tried it one day already, but I need to try again. So far I don’t see any way to rearrange the symbols in such a grid that shows the patterns though. Something to consider.
Your hoax theory is interesting. We could try to develop a kind of hoax generator that will scramble different patterns in the way you described. Then we could compare the results with z340.
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So far I have not been able to pour the symbols into any shape that results in distinct sets of symbols arranged in columns or rows, set out in any pattern. Thanks for your work.
Hey Largo,
The interruption can’t be periodic since my encoding randomization tests finds such stuff easily:
Another issue with periodic interruption is that it will generate both increasingly and decreasingly random cycles while the 340 has only increasingly random cycles above average.
The thing with transposition after encoding is that it almost always inflates unigram repeats over short distances. Your cipher has 11 unigram repeats before transposition (summed per row) and after it jumps to 30 unigram repeats! The 340 has only 18 unigram repeats while having a higher ioc to boot.
