There’s a lot more to Zodiac than meets the eye.
There will be doubters of course, but you can’t deny logic.
And it isn’t just the one rabbit hole- it’s the whole damn warren.
Like I said a few months ago- I found an interesting little hole, and I’ve got a big spade tearing the warren down. Why do you think I’ve stopped asking questions?
P.S. This isn’t sending me batty, but that was one interesting (or lack of) diversion which, like all rabbit holes, ended up eliminating and confirming certain things.

And that ends my Zodiac role-playing for now- still lots to do and lots to uncover, check, double check and triple check. About 3 months I’m predicting.
We’ll see though, eh?

Another attack on the 340 (as homophone substitution):

Assume to be both, frequent as well as a frequent double letter (e.g. ‘S’ or ‘L’). Imagine and not being exotic either (as both trigrams actually appear twice despite multiple homophones).

Now take the following fragments of the Z340 out of lines 3, 17, 9, 13 all of those with partially corresponding homophones:

You may now want to try to put in four (!) different words of minimum length 5 (onto those strings).

IMO this is NOT easy at all..so far, I have computed a total of 1,409,582,685,888,000 different letter variations, in about one hour, and have not even found one single combination of four words fitting those patterns simultaneously (dictionary size approx. 4,500).

Please be aware that only three symbols are unrelated to any repeating bigrams/trigrams and/or other strings…let’s see:

Thus, an attack with common bigrams/trigrams has a fairly good chance because only three homophones have to be guessed to be anything from A to Z…

QT

More than seven years ago, this thread was started to approach an attack based on homophone substitution on the Z340. Most of this time, there was the ‘bad feeling’ that even by computing the Z340 it would take ages if not thousands of years to crack the cipher. If at all.

However, with the FCCP theory, selecting different areas from the cipher, many things got improved. So it was additionally by using algorithms for word search or comparable functions, mostly programmed individually according to the cipher text (by using Python). Since two or three years there – still – was nothing but staggering in the dark. Endless lists or no results at all, both hindering any (visible) progress.

But again, the program has now been improved:

1) Reduced focus: Less strings but longer ones, allowing some kind of better ‘certainty’ that any word from the dictionary is found
2) Taking advantage of a total of three bigrams which are repeating in the cipher text
3) Thus, less ‘input’ required to start the program running, e.g. using only the FBc section instead of the IoFBc

It now is possible to proceed as the following:

a.) Guess the + symbol, e.g. ‘L’
b.) Guess the (repeating) FBc trigram, e.g. ‘TER’
c.) RUN the program

Currently the program runs with 3 bigrams out of a (pre-selected) group of bigrams. It then adds another seven (up to nine or more possible) homophones, all of each carefully selected according to overlapping structures amongst those three different strings. Additionally, by usinga a step-by-step computing approach, approx. 98% of the computation has not to be performed at all (e.g. when no word is found in string #1).

Based on this approach, a total of 13 homophones are computed in addition to the four homophones deriving from the input a.) and b.). Thus, one computation runs a total of 26^17 letter variations (1,133 827,315,385,150,725,554,176 – searching for approx. 4,500 words in each of the three strings, on all possible positions…).

After input, e.g. ‘S’ and ‘TER’, the program runs a total of 2,481,152,873,203,736,576 letter variations. Having started one example run at about half an hour ago or so, the program has already progressed to the 2nd bigram, therefore already having covered approximately 3,670,344,486,987,776 different configurations. This being about 0.15% leads to the assumption, that after input / starting a run, it will take about one week to compute one single input setting (but it does..).

Out of those alreadz computed 3.67 quintrillion variations, the program has found two (!) [groups of] results. Also, the program continuously shows the exact status at which it is currently computing (last line):

As you can see, a potential cleartext such as "…PRETTYSS…" is unlikely to be correct. However, either in this input section or another, some better cleartext may (or may not) occur in the near future. Still, if unlucky, the computation of the cipher to find the cleartext could last a very long time..

UPDATE:

About 48 hours later, almost 30 bigrams (for the repeating ones) have been computed. Keep in mind, only four letters had to be ‘guessed’ in addition to that: The + symbol as well as the ‘FBc’ trigram.

Thus, a total of 202,401,616,435,200 variations was not only ‘targeted’ but has even been CHECKED with the dictionary. This number is quite huge. So far, out of those only five (partial) solutions were identified. Therefore, the setting S/TER most likely is NOT the correct guess. The ‘cracking ratio’ of this procedure, however, is approximately 2.5 out of 100,000,000,000,000 different letter variations..most likely the strongest attack run on the Z340 ever.

Assuming an estimated 200 different runs until the correct solution is found, about one year of computing will be necessary (resulting in a list of about 1,000 potential results). One of those potential results could then be the right one..let’s see but my guess is until the end of the year 2021, the Z340 will not be a mystery anymore.

QT

Small update:

So far, a total of 1,301,153,248,512,000 or 1.3 quadrillion different letter variations have been tested: All of those variations have been checked (each single one) against a dictionary of ~4,500 words. Each word on each possible position. Out of those 1,3 quadrillion variations, only six (!) potential results have occurred ("At least one word of length 5 or longer found in each of the three strings").

Still aware of the total size of Z340?

26^63 or 1.39e+89 or 139,098,011,710,742,195,590,974,259,094,800,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 theoretically existing variations (63 homophones). Chess has approximately 10e+120 variations (‘Shannon number‘), which is significantly greater. And, obviously, the Hardens were able to solve a 2.56e+76 cipher. Thus, 1.39e+89 should – still – be in some computational spheres..

Thus, the complexity of the Z340 cipher can be placed somewhere between the Z408 and chess; closer to the Harden’s, I’d say.

Considering the cipher structure itself:

– identical homophones in different strings
– repeating bigrams etc

is ‘key’: The English language dramatically reduces the huge number above. How much? We do not know because this mostly depends on the linguistic patterns of English/American language (dictionary). For example, the word QQQQQ does not exist; thus all of it’s related combinations are excluded, not even computed.

The TASK is to compute as many variations as possible by simultaneously considering the English language with the cipher structure. In fact, this currently happens in some kind of ‘short version’ of 26^17 different variations.

An almost philosophical (cryptoanalytical) question arises:

Will the total potential of the cipher’s encryption method be reduced strongly enough by the English language with regard to its cipher structure to make it computable in a specific time period?

In our case, the latter (‘computable’) is limited to a setting of 26^17, thus computing 17 letters. I simply do not have any faster computer available than the one in front of me. The unexpected answer, however, is YES: This can be seen during the concurrent computing progress as the program ‘skips’ to the next bigram section, practically.

At least, when focussing on common bigrams.

If the pre-set is good, the cipher ‘should’ be cracked over the next 12 months. HOWEVER: It makes a huge difference if the computation runs with the correct set-up, e.g. 30 frequent bigrams (30^3=27,000) or all potentially existing bigram combinations (676^3=308,915,776 – three repeating bigrams are considered). Using all bigrams slows down the computation progress by the factor 1:11,441. Eleven thousand years of computation instead of 12 months. And this just because using all bigrams instead of only the common ones. This somehow shows the ‘sensitivity’ of this cracking process.

At least, there is now a way to ‘cover’ multiple quadrillions of the most likely letter variations. From now on, the question is not ‘if‘ but ‘how long‘ it will take to crack the Z340.

For me, this is comfortable: Besides smaller modifications regarding eg. the bigrams used, the program is working perfectly fine. All I can do is wait and watch.

QT
– Cryptophilosopher –

Seven years and 36 pages of trying to force a homophonic substitution solution. Hasn’t this been pretty convincingly shown to likely not be a homophonic substitution cipher?

Seven years and 36 pages of trying to force a homophonic substitution solution. Hasn’t this been pretty convincingly shown to likely not be a homophonic substitution cipher?

Good question but it has not. First of all, it indeed took 7 years of ‘hobby’ cryptanalysis to get to the point we are now. Second, using Python started only a few years ago. The program itself, as it consists of today, exists since less than a month. Thus, the cracking process has just begun – everything else before were thoughts, learning to program Python, preparing the dictionary, implementing the Aho-Corasick algorithm, programming the tool itself (with countless trial and errors), configuration of the program to not get endless lists nor one blinking cursor etc.

So far, the program has ‘only’ covered S (for the plus symbol) in combination with 6 different trigrams on a basis of 30 repeating bigrams. Total computation time approximately 48 hours (I still have some real life, too..).

IMO the Z340 may assumed to be a homophone substitution:

First of all, most encryption methods do not consist of homophones (‘symbols’) but figures or alphabetical letters. Here, this is not the case. Thus it must be an encryption using symbols – even more than 26. Therefore, the cipher is assumed to be a polyalphabetic (homophone) substitution, this way or another. It even was figured out that – opposite to the Z408 – the sequences are NOT constant over the complete cipher text (e.g. the w symbol occurs often but only in some sections of the cipher). But still it is homophone substitution.

In addition to that, solid sequences – typical for homophone substitution – have been found (see Doranchaks presentation minute 23:28 viewtopic.php?f=95&t=5044). Also other structures can be seen, true.

Until the cipher is solved, it is just an assumption. But I see more than 26 symbols which makes me believe to be right on this one. Considering the effort to solve the Z340 this way does not require any additional encryption method (eg transposition). It simply is hard to solve as it is.

QT

Well, good luck QT.

Well, good luck QT.

I’ll let you know if the cipher starts with ‘I have been up to something.."

UPDATE:

Searching for any (!) cleartext matching the 340 cipher, two things have recently been modified:

1) Randomization of the search process

On a certain level of progress, all alternatives for homophones (A-Z) should be covered to get closer to a (complete) cleartext match. However, this cannot be done only by using a ‘brute force’ attack; due to the huge amount of potential variations (63^26).

Therefore, the pre-setting of the attack is now running completely randomized, saying that moreless the first seven homophones are actually ‘guessed’ (with some preconditions, eg. that the + symbol is a frequent letter). 

The advantage of this is the following: The first part of the attack is accelerated while the latter part of the attack – when a first cleartext is found – remains structured.

2) Setting ‘levels’

The 340 rather cannot be brute-forced. Thus, the (new) FCCP method is dealing with (partial) sections of the cipher (e.g. a string of length 11 symbols). Finding any cleartext in such section already elminates a huge part of the variations mentioned above – a first level of the decryption process has been ‘found’.

Continuing with a second, third, fourth etc. section, the decryption is therefore structured into ‘levels’. The more levels found (matching any cleartext), the better: Finding twenty cleartext sections, for example, would reveal a near-to-complete match. In my little Python program, those levels are given Greek letters: Alfa, Beta etc. 

So far, the program is running based on six different sections/levels covered by only (!) 24 homophones. Those sections have been chosen ‘wisely’ from line 1, 3, 11, 13, 17, 18.

We are talking about 24^26 or approx. 7.68 e+35 or  768,231,807,465,763,655,682,670,928,358,010,000 different variations of how those sections could be theoretically filled with cleartext letters. This still be ‘boring’, as long as the sections would not overlap – but they do: Currently, those six sections cover a total of 64 letters (or 18.8% of the cipher). However, due to overlapping, repeating homophones, only 24 homophones are required to cover all symbols of those sections. Those, of course, are then cross-checked with a dictionary on each position of the section.

The following is one example of some ‘result’ – not a final one (looks a bit weird but is correct in its structure):

327964 THETH AIN RAT NG SE A D C E T I S E
NGAINSTSU USITHETHS SENTERAT SEDTTTEASSETCETH STINTSEAUE ACRESGESUHS 

It should always be considered that most of those ‘sections’ consist of homophones occurring in other sections, too (otherwise it’d be ‘easy’..).

To run half a million different pre-settings is currently done in approximately 20 minutes.

For each pre-setting, 11^26 homophones or 1,191,817,653,772,720,942,460,132,761 variations are added and cross-checked with ~4,500 words – on each position possible. Thus:  

500,000 (pre-settings) x 5,363,179,441,977,244,241,070,597,424,500 variations “cross-checked” with the dictionary in 20 minutes only.

Being aware of the much greater complexity of the cipher, this still is scratching on the surface. But now scratching a bit deeper.

Above: Parts of the code
Below: Some results produced by the code

QT 

Hi QT! I believe the code could be close but as David O told me when I gave him the solution to the last 18 of the 408, he told me it checked out correctly, but because Zodiac used so many letters for each symbol, it could also be other words. That was when I thought OK, that also means the same thing for the solved 340? 

I was given the solution by someone who claimed to be Zodiac himself! The last 18 was:” it me the hipie Robeet”. It was put into my mailbox on a green card that was an ad for a flooring company named St Patrick’s in Pleasant Hill Ca. St Patrick’s day was Darlene Ferrin’s birthday!

For whatever reason, he thought she and I were friends? The name Robert is the first name of one of my suspects. It is also a name that the FBI has for Zodiac’s first name! R of course matches part of the desk at Riverside: RH. The FBI has 12 letters for the full name of Zodiac from one of his letters, Bob is short for Robert, my suspect fits the 12 letters if I use Bob instead of Robert. His last name is 9 letters long starting with H, which equals 12 letters!

The Hardens were very close to solving the last 18 they had: Robert Emmet the hippie. “Two” too many letters. The other clue given to me with the 18 letter solution was: 5 E’s the 5th letter and three I’s the 9th letter. ( There are 5 E’s and three I’s in the last 18! Harden’s had the correct 18 letters, they just had them in the wrong order.  

QT, I would like your feedback on this, please? 

@sandy-betts 

EEBEORIETEMETHHPITI – Those letters could be fillers to complete the line of the cipher or indeed mean something. ‘Robert the Hipie’ is not the approach I would follow. Not knowing what you have or have not received and by whom (and in which year). However, there is some potential for clues in those letters: 

For example, one might extract the word ‘mother’ from those letters.

MOTHER  EEBEIETEHPITI

or, alternatively, the words ‘meet’ or ‘met the other’

I MET THE OTHER – EEEBPII

which still is no solution. Same with   

MEET THE EERIE TO – BHPII

Many possibilities to play with those 18 letters..even found a ‘complete’ one:

I HIT THE PEER – MEET OBIE

Regarding multiple solutions for homophone ciphers: Yes, they could exist but is the 340/408 long enough to rather work with few if not only one, I guess (we may already have seen one of such for the 340 btw).

Finally, the first name of Z: He may have (had) two  first namea..my guess is his first is pretty sure ‘PAUL’. Following his clues, I’d say the odds for this to be correct is 99.9%. He (often) gave clues in his communication and it took decades to figure out what he was actually referring to (bottom lid of a scotch tape / St. Paul, Minnesota). Well, who knows. 

“Maybe you play chess with me” – maybe his name was PAUL KING or PAUL BISHOP. The clue with the scuba gear is still foggy to me.

QT

@quicktrader, Thanks for your input. The solution you gave was: I HIT THE PEER – MEET OBIE. That would work except there aren’t six E’s in the solution there are only five and you have one too many letters to be 18. 

My point is, that the solution I received from the man who claimed to be Zodiac,  works out to be correct and fits perfectly.  I agree it can be made to say something very different, but if this was actually from the horses mouth so to speak , it shouldn’t be ignored. I believe Zodiac told the truth when he said his name was in the cipher, except it was not his full name, he wouldn’t be stupid enough to do that! The name Robert was a common first name, and it also works with the first initial of the RH at Riverside.

I believe that his signing off on the desk poem with RH,  was the best clue to his real name and was his largest mistake! Let us not forget Darlene’s own words: He was an “ole hippie”. She told her babysitter his first name was either Bill or Bob. These were factual words spoken by the victim about the man she was followed by and who she was fearful of!  I gave the name Paul in Darlene’s police report, because of a man who had acted suspiciously back in 1970. He had a solid alibi and was cleared.

@quicktrader Thanks for an interesting read!

Considering that we now have a solution to z340, are you still working on this? I’m asking because I do accept the correctness of the published solution to z340 – more or less anyway – but I still have a sneaking suspicion that there’s more to these ciphers that we haven’t found yet.

It’s entirely possible that I’m simply overestimating the abilities of the “cipher letter writer” but there are just so many loose ends that are often brushed aside or used to support the argument that the author must have been an amateur. I, however, would not be surprised if some of the typos and transposition errors in the ciphers are actually clues or remnants of additional cipher elements that we haven’t found yet.

For instance, it’s weird to me that z340 is split into 3 sections of 9+9+2 lines, that we see “LIFEIS” somewhere in the middle, that there are so many typos, that a character on line 15 needs to be moved to the left for the period 19 transposition to work, that the words on the last two lines are written forwards and backwards, etc. Perhaps it’s all an attempt to complicate our lives but it’s just not very aesthetically pleasing…

In z408 we also have a bunch of typos and what appears to be a missing word; often assumed to be “people”. And then, of course, there are the last 18 characters which no one has been able to explain properly yet. They might just be filler but who knows.

I keep wondering if we’re almost there with z408 and z340 but are still overlooking something. Maybe it’s possible to produce slightly different plaintexts by inserting lines from one cipher into another, maybe some lines are inherently ambiguous and should be read in a mirror, maybe z13 or z32 are involved somehow?

Anyway, I wonder what you and other web sleuths think about these ideas?

P.S. since you seem to know python, I thought you might be interested in a script I wrote the other day to test if other Zodiac letters use the z340 enciphering scheme (none seem to so far)

@jay Yes, Python :D. Regarding the 340 one may or may not describe it like this: Whilst searching for the highest mountain of the world, this one found might be the K, it might be the ‘Annapurna’. Not as close to it as the K2 but not a flat find, either. 

Average word length, shifting, asthetics etc. – I leave that discussion up to others. Hopefully my prog sooner or later would produce all of those solutions (it does spit out plenty on parts of the cipher – but not the correct/complete one, yet).

For ya entatainment have uploaded a video (each figure counting covers a total of 361,431,457,920,000 words being checked on various positions of the cipher parts).

QT 

Posted by: @quicktrader

@jay Yes, Python :D. Regarding the 340 one may or may not describe it like this: Whilst searching for the highest mountain of the world, this one found might be the K, it might be the ‘Annapurna’. Not as close to it as the K2 but not a flat find, either. 

Yes, python indeed – nice language! 🙂 Anyway, what you’re describing sounds like an analogy for a hill-climbing algorithm that gets stuck in a local maximum. Is that what you mean? As I understand it, that particular problem can be mitigated somewhat with techniques such as simulated annealing (wasn’t that what Oranchak et al. used as well?)

Posted by: @quicktrader

Average word length, shifting, asthetics etc. – I leave that discussion up to others.

Since you mention that you don’t use shifting (a form of transposition), are you treating z340 like a pure substitution cipher with no transposition? I.e. do you think the encryption scheme is identical to that of z408 but just uses a different key? I’d be amazed if you managed to produce a consistent and meaningful plaintext in this way considering that one has already been found using substitution and transposition!

Posted by: @quicktrader

For ya entatainment have uploaded a video (each figure counting covers a total of 361,431,457,920,000 words being checked on various positions of the cipher parts).

Sounds interesting… Where would I find this video?