Zodiackillersite.com

I've independently tried a similar idea. FWIW, I'm cross-correlating somewhat different fragments.

1) The 0L++0 fragment is objectively the most constrained part of the 340 and if you want to limit the keyspace as much as possible, this sequence of 5 characters helps more than any other. Only a few English words or phrases really make sense there, especially with all the other repeated structures around them. The most common ones I get in natural language are "STREET" "LITTLE" "THATTHE". There are a few with double-S too and could fit your hypothesis of + = S. Could you point me to where in the thread you conclude that and why?

2) Have you considered choosing to assume that the beginning starts with an intact word? If you assume "HER>PL^V" isn't some arbitrary fragment in the midsection of a word, that actually decreases the keyspace significantly. There's also good correlation between the first few characters, and most of line 4, and line 10.



3) Expanding on the above, H / P / R / L / ^ / V and + all interact significantly at character 48-77: "8HJSPP7^l32V1PO++RK". So knowing the first 2 words of the cipher would almost fully determines the 3-4 words on line 4 and also 2-3 words on line 10. And not too many fragments in natural language have patterns that can conform to the double P / third P / double ++ combo in line 4. Some examples from real writing include:

Code: Select all

UNLESSTHEREHASBEENA193962
VENINNOCUOUSONESSHO90059
HISSEEMSTOHAVEOCCUR70722
NTBUTTHEYMAYSTILLPM27977
ORTHEENVIRONMENTTHA15341
ADTHEEXACTSAMEISSUE15089
ORNARRATIONFORALLST14967
LAYASSUCHITHASBEENA13811
VETHEEXACTSAMEISSUE13464
TSFORREACHINGRALLTO13382
NGTHEEXACTSAMEISSUE12806
GHTASSUCHITHASBEENA12427
TREALLYMEANSALOTTOM12118
WAYWEENCOURAGEALLOF11184
IMNOTTRYINGTOTELLYO9349
BETWEENDIFFERENTTYP9142
EQUALLYREGARDLESSOF9122

Everything you wrote is correct imo. Also thought about the start of the cipher but did not yet go for it because in the beginning there are sort of no structures at all, making this approach very difficult (but not impossible). Nice find regarding the cross-connection to line 4 & 10.

Bernoulli was posted here:
viewtopic.php?f=81&t=3052&p=47706&hilit=Bernoulli#p47706
(see below 'LDefinition' file attached)

At that time, in 2016, I still believed that Bernoulli leads to the letter 'L'. However, depending on which letter frequency table one might use, 'S' shows up, too - and has a better overall frequency for the + symbol. All other letters are rather a no-go or at least (strong) statistical outliers.

You may search for the term 'Bernoulli' for further reading.
search.php?keywords=Bernoulli&terms=all&author=&sc=1&sf=all&sr=posts&sk=t&sd=d&st=0&ch=300&t=0&submit=Search

My idea was to look into the probability of two symbols occurring twice, one after another, with regard to common double letter frequencies (like pulling colored balls out of a bowl). The Bernoulli approach will remain an estimation as we do not know yet the exact amount of homophones for each alphabetical letter. I relied on how Z did with the 408, adapted the amount of homophones for each letter (e.g. 7-8 for the letter E instead of approx. 5).

I also agree that if we knew the first eight symbols, the cipher could - easily - be solved. Even tried some phrases but there are too many words a sentence can start with. Instead of one long word I'd expect a set of words such as "HOW TO THE PIGS", "IT IS ME AGAIN" or something.

QT

A couple of posts back, Quicktrader stipulates;
"The only rule: Never allow two different alphabetical letters for one and the same symbol."

There is some evidence that this assumption may be flawed, for a specific subset
of symbols and in the case of a specific circumstance. The specific subset of symbols
are those 26 cipher symbols (used in the Z340) that are the letters of the English
alphabet, A to Z, and the specific circumstance is that where the Zodiac may have
performed no (single) homophonic substitution upon a (z340) letter of the English
alphabet but instead occaisionally allowed a letter to simply represent... its usual
self.

To state it plainly, allowance should be made for the 26 symbols (used in the Z340)
that are the letters of the English alphabet, A to Z, such that for that subset of
symbols an individual (Z340) symbol can be either;
1. a single homophonic (Z340) letter that represents a (different) single English
plaintext letter,
or
2. the symbol letter is not actioned as a homophone but instead is read as simply
being its usual self.

Why the Zodiac may have occaisionally actioned a circumstance where
an English (Z340) letter simply acts as its usual self, is currently unknown.
(In an unpublished tract about the Z408 that I circulated to a couple of forum
members in August last year, I suggested a hypothesis as to why the Zodiac might
action such a particular circumstance in the Z340. For completeness, I will quote
the relevant parts of that tract below;)

...
" {Note; I adopt this convention when discussing the Z ciphers - a specific cipher
is referred to as e.g. Z408, and a specific position within that specific
cipher will be referred to by that position within the cipher e.g. Z408#17 is
the 17th position (which in this case is seen to be, a cipher symbol 'B' which
represents the plaintext letter 'L') and e.g. Z340#11 (which in this case is seen
to be the plaintext letter 'I' apparently emplaced within the cipher symbol text
of the Z340 cipher). If the context of the discussion is obviously referring to a
specific cipher, then the cipher name may be dropped and the position only stated,
e.g. discussing the Z408, then #392 is position 392 in that cipher (which in this
case is seen to be, a cipher symbol 'V' which represents the plaintext letter 'B'.} "
...
"A further facet of the Z340 cipher,"..."can be conjectured.
Looking at the beginning of the Z340 cipher, (it is well known that) Z340#1 to
#3 are the letters 'HER' which are thought not to be cipher symbols, but rather
these letters are considered to simply be plaintext letters (for which no
decipherment step is necessary or applicable). Reading further along the cipher
text of the Z340, can be found other instances where plaintext letters are
believed to have been inserted in the cipher text (instead of, for instance,
using an applicable cipher symbol which could then have been deciphered to a
plaintext letter).

If an examination is made of all actual words of the Z340 that have so far been
deciphered (some ~15 words) and take stock of those plaintext letters that are
known to be inserted amongst the cipher text symbols, and also extrapolate
one instance whereat there could be a plaintext letter inserted amongst the
cipher text symbols, it is found that;
Z340#1 to #3 are 'HER", #11 is 'I', #14 is 'T', #24 is 'O', #50 can be extrap-
olated to be 'H', #52 is 'S' and #127 is 'O'. This can be more simply listed as;
'HER I TOHS O', which can be de-anagrammed to 'HER I SHOT O'.

Taking into account that the plaintext 'O' at #127 is some four lines further
away from the other letters, it is reasonable to theorize that there could be
other plaintext letter instances, thus far unknown, that are placed somewhere
before and somewhere after that #127 letter 'O'.

Therefore, there is a reasonable probability that other instances of plaintext
letters inserted amongst the cipher text will occur, and if such do occur, a
possible message text may be embedded therein, a message such as;
'HER I SHOT FOR'...
(assuming any applicable de-anagramming step is also actioned).

This particular facet may be termed as the 'Plaintext Message embedded within
the Cipher Text' speculation. "
_
I hope the above rather wordy explanation will not annoy readers, particularly
Quicktrader, unduly.

Cheers

Absolutely not...Z was dyslectic...thus, YES he did use the same homophone for different letters. Hopefully not more often than 5 times or so. However, I doubt that 15 words have been deciphered, so far. IF so, that was nice.

QT

buyerninety quoted himself;
"If an examination is made of all actual words of the Z340 that have so far been
deciphered (some ~15 words)"...

Quicktrader said;
"I doubt that 15 words have been deciphered"...

I would agree that there is continuing uncertainty as to whether, or how many, words
in the Z340 have been deciphered (by using homophonic substitution and using just
an A-to-Z letter as its plain self). I think it is fair to say that a large proportion of
people regard some parts of the Z340 as having been deciphered, such as the
'SEEANAME' and the 'THESEFOOLSHALL' parts, for instance. Opinions vary -
as an example, my working theory regarding Z340#52 to #68 is that those symbols
may decipher to 'SEEANAMEBEEPLLDSE' , which de-anagramms to read
'SEE A NAME BE SPELLED'.
Cheers
Note; the P, D, and S in 'EPLLDSE above, may not be in the decryption order I have shown,
but may be in a different order (i.e. unsure which of the cipher symbols at Z340#63, #66
and #67 have assigned to them the the decrypted plaintext letters of P, D and S.

Quicktrader wrote:Absolutely not...Z was dyslectic...thus, YES he did use the same homophone for different letters. Hopefully not more often than 5 times or so. However, I doubt that 15 words have been deciphered, so far. IF so, that was nice.

QT

dyslexic yea i agree..
my guy spells his sons name Micheal

Update..

For the first time 'ever', it was possible to identify five words of length >4 simultaneously, with overlapping homophones in different sections, by FCCP method.

Short resume about some advantages of the method:

- different pre-settings can cover wide range of possibilities
- strong cracking due to identical homophones in different sections of the cipher
- method to find all but only one combinations of words
- my favorite: possibility to continue with the cracking process over other parts of the cipher

The picture below shows only one of multiple potential (partial) solutions. One of the results was entered into Doranchaks webtoy. The blue lines show the sections analyzed, the green lines the words from the dictionary that had been found. Z's cleartext could be amongst the list of those partial solutions.



Next step, of course, is to find a sixth, seventh etc. word until the cipher is complete. IF the pre-settings were correct (e.g. + = 'S', dictionary etc.), the cipher will be solved. If not, other pre-settings should and can be tried out by the same way (e.g. + = 'L'). Computation takes a while...each step/word approximately 2-7 days at least. Progressing slowly, we'll see.

QT

Wow, very interesting QT!!

Yes, but to be honest it is only the start. First of all, finding 5 words is not very much. ZDK tool sometimes finds eight or more. Second, the presetting is critical. I compute e.g. with the most 1,000 frequent trigrams..nice but if the cleartext has the 1,254 my computation will not work. Same with the sequences..they are short, eight, nine letters only. If there is only one sequence with multiple short words, the method is lost. Anyway, it's worth a try (as I said before 2,000 years of computation with Tianghe-2 if you really want to get a solution).

Nevertheless..the more you get into that computation stuff, the more possibilities occur to improve the program. Not easy to keep the overview but well.

QT

FOUND:

Seven words of length >4...an example:

Code: Select all

SAGENTBASSBEGGED SGUPPTHATS ABSTAMPE AGENTPGMA UNSPENTNGD PESSNAKE

The words found in this set of strings were

AGENT
BEGGED
THATS
STAMP
AGENT (again...but based on different homophones)
SPENT
SNAKE

Words were found in strings starting from lines 17 (2x), 5, 6, 19, 1, thus from all over the cipher.

Approximately 14,000 of such variations (most of them nearly identical), however some of the strings were quite short (e.g. 8 letters/homophones only). If there is no 5-letter word in such string, the method fails. If the pre-set variables were set up wrongly, the method fails. But we still can use 4-letter words, too. And modify the pre-set. Currently cross-checking the results against an additional, completely different string (line 8). If there is only one result (out of those 14,000), it most likely is Z's CLEARTEXT.

Nevertheless, still a long way to go. Adding 4-letter words. Trying the letter "O" or "L" instead of "S" for the + symbol, adding names to the dictionary, etc..etc...

QT