340..partially solved ;)

[Quicktrader]

"Polly wants some cracker..."

Quite certainly, the Z340 can be cracked. Diagram #1 shows the method: In each colored area, 'step outward' towards the outer ring of the diagram to define your overall pre-settings before starting the computation process (e.g. + = 'S'; variables A-Z; trigrams = top 1000 frequent ones, IoFBc section = frequent 5-gram etc.):

cracker2.jpg

After having done so, a multi-step computation may start across (meanwhile/at least) 7-9 different text strings of the cipher. All considering opposite homophone relations, of course. Those, as an example, are shown in diagram #2:

cracker.jpg

On each blue-lined area, at least one word (if not two in line 17) shall be found by using a dictionary in combination with a linguistic search algorithm.

Results vary, e.g. using 8 strings it is possible to find thousands of solutions (variations), cross-checking an additional 9th one leads to computation without any result, so far.

By doing so, one pre-setting after another can be computed until a solution is found (all assuming the cipher to be homophone substition).

https://www.youtube.com/watch?v=OzO5pL8GAPo
(0:45 min..)

QT

[masootz]

you guys are wild. a continuous stream of good ideas.

[Simplicity]

Quicktrader,
Any resemblance to Zodiac symbol is coincidental and irrelevant.
I suggest 3 symbols form 1 letter or 2 symbols forming one letter which is less likely but plausible.
I stole this idea from Harvey Colliver partially.

[Quicktrader]

Crickedicracke....Polly wants..

QrickedicrackaTiquracker..kraaah!

[Simplicity]

Guess not, It’s to easy to be true.

[Quicktrader]

I wish it was that easy..for the first, easiest, slightest step, my pc is computing since approximately 1-2 weeks, few hours per day.

Besides IoFBc (65 most frequent), there is

- 1 bigram (30)
- 3 variables A-Z (26^3)
- 1 variable frequent (11 - ETAOIN..)

to get the first section in line 17, thus:

65x30x26^3x11 = 377,005,200 x 4,500 x 4,500 words (2 words to be found!) = 1,696,523,400,000 or 1.7 trillion words to be placed at any position of the string (x 3-5...depending on word length found..).

And that's only the beginning: Computation continues only with results found, nevertheless in further steps there (currently) are considered an additional

- 2 bigrams (30)
- 2 variables A-Z (26^2)
- 1 trig (1000 frequent) and to make it easy, in further steps
- 2 additional variables A-Z (26^2)

leading us to a total of 53,672,571,501,120,000,000,000 or 53.7 sextillion different letter variations 'covered' with a total of nine different strings of the Z340 cipher. If anybody thinks calculating with less would make sense: No problem as long as you can handle 2 billion results..

All of the previous 53.7 sextillion variations are considered during the computation process (all..). However, "exclusion happens during computation", that means that e.g. if there are no two words found in string on line 17, the rest will not be computed (and so on, step by step over all nine sections of the cipher).

Nevertheless, computation takes days, if not weeks or months (or forever..) for each pre-set configuration. If one (repeating) bigram is not amongst the chosen 30, the whole computation process will remain without result.

temp.jpg

Not to think what if the IoFBc section represented a non-frequent 5-gram..there is a total of 1,490,116,119,384,765,625 variations for those five letters alone! And, so far, we deal with the most frequent 65 only! The only 'hope' is that due to IoF as well as FBc repeating in the cipher, the approach to start with the most frequent ones could be a good one..

QT

[Simplicity]

I’m sorry i said anything but i’m glad someone finds it intersting.

My view is that it is tri-gram where symbols are for either A,B,C,D. I know bigram looks to be more fitting but the methodology of which a human might have made this makes me believe it is tri-gram (also the low variety of symbols producing few codes.

also with a trigram you start to notice that many of the symbols are just artistic noise to distract - whatever that means but it’s what i notice.

anyway what i suggest is that the symbols follow this key or one very similar

AAA = A
AAB = B
AAC = C
AAD = D
ABA = E
ABB = F
ABC = G

Etc etc until Z where it starts over again until DDD.

Some letters may be a Bigram because the structure allows it for instance A can just be “AA” instead of “AAA”

Anyway i wish you success, The Tri gram is most interesting, if you want my notes it might help you start i made progress with many symbols and have narrowed things down using common sense and pen and paper. (made guesses that certain letters wouldn’t be in certain places there by narrowing protential values for symbols.

[Quicktrader]

Update...slight modifications in which lines, where they start, which order etc..meanwhile finding a total of eleven (11) words of length >4 on different sections of the Z340 cipher. Results look like this (first shows chosen 5-gram, 3-gram):

Code: Select all

..
THERE VEN SEEARTHISSHAREAE SEVENTHEES EHSTERNA TONESSREST SSETHEREE EUSANTAEHISE NRREPOLESHS REPULSESRPENT EITETOORESTS EEAFSHOSTS
THERE VEN SEEARTHISSHAREAE SEVENTHEES EHSTERNA TONESSREST SSETHEREE EUSANTAEHISE NRREPOLESHS REPULSESRPENT EITETOORESTS EEAFSCOSTS
THERE VEN SEEARTHISSHAREAE SEVENTHEES EHSTERNA TONESSREST SSETHEREE EUSANTAEHISE NRREPOLESHS REPULSESRPENT EITETOORESTS EEAFSPOSTS
THERE VEN SEEARTHISSHAREAE SEVENTHEES EHSTERNA TONESSREST SSETHEREE EUSANTAEHISE NRREPOLESHS REPULSESRPENT EITETOORESTS EEAPSHOSTS
THERE VEN SEEARTHISSHAREAE SEVENTHEES EHSTERNA TONESSREST SSETHEREE EUSANTAEHISE NRREPOLESHS REPULSESRPENT EITETOORESTS EEAPSCOSTS
THERE VEN SEEARTHISSHAREAE SEVENTHEES EHSTERNA TONESSREST SSETHEREE EUSANTAEHISE NRREPOLESHS REPULSESRPENT EITETOORESTS EEAPSPOSTS


Still finding 100,000+ different variations just on that pre-setting. Some might seem to be duplicates, however might be configured based on e.g. different bigrams (e.g. EN and AN both contain N as the second letter).

'Cleartext' strings already cover 223 out of 340 symbols, thus more than 65% of the total cipher (of course only searching one word in each section, not overall cleartext, so far).

Thus, now adding a twelvth word the program shall search for. We'll see.

QT

[Jarlve]

Hey Quicktrader,

Index of coincidence of such a string is 0.1359. Should be closer to 0.0667 for English. You may need frequency control.

Code: Select all

THEREVENSEEARTHISSHAREAESEVENTHEESEHSTERNATONESSRESTSSETHEREEEUSANTAEHISENRREPOLESHSREPULSESRPENTEITETOORESTSEEAFSPOSTS


[Quicktrader]

Thank you for the hint. The IoC is based on frequency of the (alphabetical) letters. However, we deal with homophones' frequencies (first), thus the IoC of an encrypted message (homophones = 63) is way different than that of the same message in decrypted status (alphabet = 26). Because different frequencies as well as overall amount of homophones lead to different IoC values of encrypted and decrypted texts.

From en- to decrypted messages, the IoC value might therefore be multiplicated by the average amount of homophones per alphabetical letters (in use). Should be aware that latter could be wrong as homophones are not necessarily used equally often for each letter..

It is possible to test the aspect of different IoCs regarding encrypted and decrypted messages: If you enter the encrypted message of the Z408 cipher into an IoC analysis tool, you should get an IoC value of about 0.0269 instead of 0.066 (for English language). Nevertheless, the Z408 - as a cleartext - is written in English, has a higher IoC value when decrypted:

54 homophones vs. 23 alphabetical letters in use ['in use' correct?] is 2.3478 homophones per alphabetical letter, thus the cleartext should actually have an IoC of

0.0269 x 2.3478 = IoC 0.063 (Z408 expected)
(IoC English language ~ 0.066)

which is actually very close to the English language, while the IoC of the encrypted Z408 message has only an IoC value of 0.0269. The Z408 cleartext has a IoC of 0.0634, thus the multiplication is actually a 'match'.

Same with the Z340, even more homophones, though.

It would be nice to engineer the computation process to a degree that it decides while computing on which IoC level the computation itself is actually deciding where to compute..the problem is. If there is a short python script for that, I'd love to implement it (based on multiple IF/FOR loops, eg. 'interrupting' or rowing back if the first loop has a bad IoC). I'd appreciate..maybe sometime in the future.

The problems are manifold..with a solid dictionary it is obviously possible to find 10 words of length >4 or more (like ZDK does often). But that is not necessarily the cleartext solution. As if that wasn't worse enough, the more 'steps' are performed, the more computational effort is actually required. To find three words in the cipher can be done in a second..to find more than 10-12, however, could take us into the year 3,000..

Please keep in mind that the encrypted IoC values are not 'wrong' - they just deal with a different sized alphabet (e.g. 54...like an alien language ). Of course that has to be a different situation than (cleartext) English language (with a max. of 26 letters). Most likely, a language with 54 letters actually has a IoC level of less than three.

IoC calculator
https://planetcalc.com/7944/

IoC based on frequencies
https://pages.mtu.edu/~shene/NSF-4/Tuto ... g-IOC.html

Because of the effects above, I wouldn't rely too much on the IoC for homophone ciphers, at least not without considering the differences between encrypted and decrypted text.

Based on the previous thoughts, with 63 homophones divided by a max. of 26 letters (2.423 homophones per letter), as well as an English language IoC of 0.066, we could reconstruct if the Z340 is English language or not:

0.066 / 2.423 = 0.0272 IoC encrypted (Z340 expected).

Entering the encrypted message in the IoC calculator (considering 63 homophones!) indeed leads us to an encrypted IoC value of 0.0277 for the Z340, which is a deviation of only 1.7% from our expectation.

Thus, in a range of <3%, the encoded message of Z340 represents English (or similar language) cleartext. Therefore, with a probability of with 98.3%, the Z340 is not a hoax either, by the way.

QT