Guys,

I am not saying to give up on trying to get a message out of the cipher. If there is a solid solution in it and someone can decipher it, it it is you guys!

I think it is important to note that Zodiac could have said "In this cipher is my name."
He could easily have said that, if that is what he meant. But he did not. He said "In this cipher is my identity."

Two different words, with two different meanings, aren’t they. Something to consider.

-glurk

Curious…. How does null encryption play or not play in your work here? How do you know you haven’t already deciphered the solution?

Soze

I’m assuming you’re talking about something like https://en.wikipedia.org/wiki/Null_cipher

I guess it depends on the steps the cipher author took.

If the null cipher’s message is hidden within normal readable plaintext, then the current methods will at least (in theory) find the normal readable plaintext, and not necessarily the hidden null cipher message. Finding the normal readable plaintext (the nulls) in the 340 would still be huge progress even if there’s another message hidden in the "distraction" plaintext.

But I think current attacks will not work if the null cipher’s message is hidden within randomized plaintext.

Also, there’s still a chance a null cipher message exists within the 408’s plaintext!

When I asked my question I was thinking about decoys and the steps the Zodiac could have made to manipulate the cryptanalyst in the initial deciphering stage. If he knows people, knows how they think and behave, and knows about ciphers, then how could he psychologically be leading the cryptanalyst astray? Nulls is apparently not where I wanted to go in that thought because it seems to reference the aftertaste of ciphers.

With regards to nulls, however, are there other examples than selecting the first, second or third letter of each word to make a new message? I really do not see that example as being an example of what’s occurring in the 408.

Soze

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If I was Zodiac, I would make one symbol polyalphabetic and use it to disrupt the homophonic cycles. But then I would also make that one polyalphabetic symbol map to plaintext that spells a separate message. That would be cool. I am not Zodiac, though, and think that we may be a ways off from finding out. We are having a very difficult time solving the message as it is.
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If I was Zodiac, I would make one symbol polyalphabetic and use it to disrupt the homophonic cycles.

I’ve ran a solver that picked 3 out of the top 10 symbols (basically all with the count of 8 or more), and expanded them to a unique symbol for each occurence. It took several weeks to iterate through all permutations with a good number of iterations for each combination, but nothing came out. So it’s almost guaranteed that none of the top 10 symbols are polyalphabetic, including any set of 3. I don’t think there is any point in trying the expand the least frequently occurring symbols, as it would’ve partially solved even without expanding.

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I am going to take a little break for a while. Maybe read a classic novel to do something different with my mind.

Until I return, I am going to ask a question:

Is there any possible way to have a 17 x 20 plaintext message, without any transposition of plaintext or encoding of symbols, where there are as many plaintext period 19 bigram repeats as the 340 does with its 63 symbols?

Can anybody do that.
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Until I return, I am going to ask a question:

Is there any possible way to have a 17 x 20 plaintext message, without any transposition of plaintext or encoding of symbols, where there are as many plaintext period 19 bigram repeats as the 340 does with its 63 symbols?

Can anybody do that.
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Yes, it is quite easy and common, simply because there are far fewer symbols in the unencrypted plaintext, resulting in many more opportunities for the bigram repeats to occur.

THECONFESSIONBYSH
EWASYOUNGANDBEAUT
IFULBUTNOWSHEISBA
TTEREDANDDEADSHEI
SNOTTHEFIRSTANDSH
EWILLNOTBETHELAST
ILAYAWAKENIGHTSTH
INKINGABOUTMYNEXT
VICTIMMAYBESHEWIL
LBETHEBEAUTIFULBL
ONDTHATBABYSITSNE
ARTHELITTLESTOREA
NDWALKSDOWNTHEDAR
KALLEYEACHEVENING
ABOUTSEVENORMAYBE
SHEWILLBETHESHAPE
LYBRUNETTTHATSAID
NOWHENIASKEDHERFO
RADATEINHIGHSCHOO
LBUTMAYBEITWILLNO

Transposed to period 19:

TABAITHELNHKANESF
LHSUNRHTXLDESCOSA
OBEYTDSESTBTLDHRH
IRUCONDTLTVEHIOEM
ADATOUOEAAHITATWV
APNDMNNWANSICHTTN
EYEOAAFGSDDTNTEBL
TNBLWTYEAHSSIKIBA
EHIEYHEBSNEHHLIME
BSENSBEIESDIEEANM
AYTDGHRNNIIBSIWYG
AUSOAAEUIHTOEBSIA
AYTIRRBWNAIWNAANL
WBBITEKOIESGIBUTO
LAOEFSAAULTKHLYTT
TNKUSUNNLTLTESLSI
ETOETHLEDLSBTDCNH
FRHTNMEBAWEEEHHHO
EUEEBIYWLRAYVTAEO
WLDFEGNIOTLEEHTRO

Number of bigram repeats: 144

I assume you meant "at least as many period 19 bigram repeats as the 340", rather than "just as many period 19 bigram repeats as the 340".

Yeah but what about without transposition.

Not sure what you mean. The only reason I transposed the plaintext was to convert the period 19 bigrams into period 1 bigrams for counting.

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I think that you might be double counting or I have a spreadsheet issue, which would not be the first time. Here is my list for "the confession" plaintext period 19 bigrams with count >1, which totals 73.

E:B 6
A:A 6
O:E 5
S:I 5
I:E 5
E:H 5
E:E 5
T:N 5
H:T 5
E:S 5
L:T 5
Y:T 4
A:N 4
T:L 4
H:I 4
T:A 3
E:Y 3
N:N 3
E:A 3
N:E 3
B:S 3
T:T 3
L:D 3
B:A 3
T:D 3
R:H 3
H:H 3
I:T 3
H:L 3
A:Y 3
T:E 3
A:E 3
H:S 2
C:O 2
S:D 2
I:B 2
N:A 2
E:U 2
W:L 2
S:U 2
A:H 2
U:N 2
N:D 2
T:O 2
E:T 2
A:I 2
D:T 2
I:W 2
N:L 2
S:E 2
N:S 2
L:W 2
B:I 2
T:H 2
A:O 2
N:M 2
H:E 2
U:S 2
M:E 2
N:I 2
M:A 2
A:U 2
I:O 2
B:T 2
L:E 2
A:T 2
B:L 2
I:R 2
S:A 2
O:A 2
H:R 2
S:B 2
E:O 2

The question is: Did Zodiac transpose the plaintext before encoding? To re-frame the question: Is it possible to make a message that is not transposed which has +/- 70 period 19 bigram repeats? If you have all 1:1 substitutes, then the answer is yes. But can it be done with 63 symbols?
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I verify doranchak’s count of 144.

The question is: Did Zodiac transpose the plaintext before encoding?

It’s one of my original questions for this thread: Is it possible to attribute the 340 not cycling as well as the 408 (despite its higher symbol count) due to some transposition after encoding?

It’s still unknown. I’ve done a great deal of work in this direction in trying to provide an answer. What I’ve learned is that any transposition which has a strong vertical component (when looking at the cipher in a grid) has the tendency to completely randomize the cycles. This is from a post of mine a few pages back, the numbers are my 2-symbol cycle measurement where higher is more cyclic:

—————–
408, normal: 249
408, each row randomized (position of the symbols are randomized locally over the row): 230

408, rows per pair of 2 randomized: 188
340, normal: 180
—————–

So what I did here was check how much information you need to move around (vertical transposition element) in order to get the 408 to be as cyclic as the 340. It seems that randomizing the 408 per 2 rows is about right. So my routine randomizes the symbol positions of row 1 and 2, then 3 and 4, etc. When that is done for the whole 408 it is about as cyclic as the 340.

Update: this means that a period 15 or 19 transposition scheme (if actual) very probably was applied before encoding because it "randomizes" the entire string and in effect almost all of the cycle information is completely randomized. So then we would see the 340 as mostly random rather than cyclic, and after undoing the transposition scheme the cipher would become increasingly more cyclic. The opposite is observed.

So, your wildcard hypothesis may still be actual!

But can it be done with 63 symbols?

From a few pages back:

—————–
I also concluded yesterdays test. In 100.000.000 randomizations of the 340 only 55 had a bigram repeat count of 41 or more @ period 15. Out of these 55, none had a secondary peak of 34 or more @ period 29. As for the significance of the 15, 29 discrepancy, I think it’s quite big because I haven’t been able to reproduce it will a clean transposition scheme. In a clean transposition scheme it should only happen if the period by itself is an outlier.
—————–

Yes I actually had a user issue with the spreadsheet not a spreadsheet issue. I came up with 206 if you count AB AB and AB as three, or 133 if you count them as two. No wraparound at the bottom.

So that is very interesting. I wonder if there is any way to reproduce the period 19 phenomenon without any type of transposition. With 63 symbols.

EDIT: I went to the bookstore and purchased a book. For mental health reasons I am going to try to read the book and take a break from the 340 for a while.

So that is very interesting. I wonder if there is any way to reproduce the period 19 phenomenon without any type of transposition. With 63 symbols.

I believe that my cipher generator can produce them (simple 1:1 substitution ciphers using 63 symbols that have bigram peaks at period 19). It is designed to force those kinds of (seemingly rare) features to appear by using a multiobjective search. Here are some earlier examples of ciphers that are simple substitution but have many of the unusual features we see in the 340: viewtopic.php?f=81&t=2760

Unfortunately, I did not yet add the period 19 anomaly. I will circle back to that as soon as I get a chance. Been making steady progress on my transposition explorer, which will try different manipulations of the 340 in a search for low-probability patterns such as high repeated bigram counts. I am strongly focusing on that project for now because I’m eager to see what it turns up.

EDIT: I went to the bookstore and purchased a book. For mental health reasons I am going to try to read the book and take a break from the 340 for a while.

Probably a good idea. Enjoy the book.

PLOT TWIST: The book is called "Cracking the Zodiac 340 Cipher"