Hi pi,

Quote ”Have you tried doing the same exercice of finding repeating patterns inside the z408? ”

Yes, but it was not much needed. Because the Z480 is already cracked. So why looking for? The goal is the Z340 to crack. What we have in hand? A cracked (Z480) Code (-18 last letters) and three other Codes.

I Remember to have read that Mr Harden said the following:

Quote:” Mr.Harden Said he broke the code by looking for four-letter patterns Which would fit in with the word "kill” -” We felt that ‘kill’ would be used more than once …"

I’ve never forgotten. It sounded like a motivation to me to venture a solution.

After many of my ideas and try it all ended in a dead end. Like all other users / peoples ideas and try in the World, who have tried the 340 code to crack.
That means for me, all previous try, even by computer geeks or mathematicians, Scientists etc. are an absolute ZERO! (My first solution a year ago included!)
A Aplaus i can only give for their effort, but not for the previous solutions

So what i do? I’m just trying to see something which might not have seen other user. I can even so not much use, but I think and I believe that this work like User Quicktrader or someone other, to a Idea brings.
That’s all.

What I admire most now is the teamwork like of Mr. and Mrs. Harden. Teamwork is the magic word for a solution.
It’s time for other ways to search for the Z340 to solve.

As an example, why examined Mr. and Mrs. Harden for a four letter word?
Why not for three letter word, or five letter vord etc.? Why just for a FOUR letter word?
He searched for the word” KILL” and they thought this word would be expected to repeat itself.
That means for me, Mr. and Mrs. Harden had an IDEA befor they came to a solution. They also looked for the letter ‘I’, because they also thought the Z would have to be an egomaniac, who boasts of his doing. This in turn means Mr. and Mrs. Harden had a psychological profile of Z., ok. a small profile, but it was enough to approach to find a pattern to the code. And they had success with it!

In plain text: Mr. and Mrs. Harden had an excellent IDEA and in 20 hours they had cracked the code!

Now i ask: how many users go to the Z340 code approach with a psychological profile of Zodiac?
To reduce it to a simple question: What would a killer like Zodiac after his letter Z480, in his Z340 writing a message? Catch me if you can?

Zambac

The approach IS brilliant..trying various words, e.g. ‘because’, you can do at 340-7=333 different positions – or you start at where the letters U and C could occur twice..possibly showing up twice in such a repeating structure. Also, the (accidentially) present homophones one line above or below were representing the letter ‘I’, a frequent vowel. So you also can make assumptions that those (standing alone) homophones might represent a frequent letter, e.g. a vowel.

QT

Hi, I found this: the first structure in the second and third line above is now extended. ” +”,” S”. And the Picture No.04 above (from yesterday) is with renewed at new double structure in Z340.

And yet another structure is added.

And here we go with another one..this might be interesting as it’s in a row, therefore could represent the same word (equalling the symbols inbetween).

QT

Hi Zambac,

Yes, but it was not much needed.

What were your findings?

Because the Z480 is already cracked. So why looking for?

The z408 is often a good place to test out some ideas we have about the z340 since both ciphers were created by the same author and that they share at least some similarities. Therefore, if we think we have found a unique property or feature in the z340, we can turn to the z408 and see if we can find the same thing; this sometimes can help assess the uniqueness or significance of our findings.

For example, in this case, what I tried to illustrate was that you could find similar repeating 2D structures in the z408. This indicates that those structures do also appear in a plain homophonic substitution cipher and are probably not a defining feature of the z340 but, as Quicktrader pointed out, are probably indicative of underlying plaintext repetitions, just as repeating n-grams are.

So what i do? I’m just trying to see something which might not have seen other user.

My intent was definitely not to discourage you but to provide my perspective. I clearly encourage you to keep digging on this idea and new ones as well!

Hi Quicktrader,

your last structur, I extended.

I try all the time to get behind it to make a word … So many repetitions are not normal, or unintentionally. This is based on a particular calculus. These structures want to say something, the question is what exactly? …

Hi pi

Quote: ”What were your findings?”

As the below example. It can be found many more. But to seek the repetitive structures only in Z480 makes no sense.

Sorry! Maybe I misunderstood you when you wrote :

Quote: ””Based on all of the above, my opinion is that these intra and inter repeating structures look cool but are probably coincidental.”

I am of the opinion, the Zodiac has not been repeated in his second letter, or he has not used the same sentences in Z340.
It makes no sense to repeat itself. Single words yes, but not whole sentences. Like (as example) ” I give you not my name” or ” I like killing people”, etc.

For me, all these are just repeats some (little percent from Z480) words, or the same letters but form a different word, the only sounds the same. As example: Kill – Will or Some – Love etc.

I repeat : Why should the Zodiac be repeated in the second letter?

Quote: ” This indicates that those structures do also appear in a plain homophonic substitution cipher …”

Yes you are right, this is a point, it should not be omitted.

Anything is possible, perhaps these findings are here unintentionally or they follow a certain schema. I think it’s too early to give about a fesste opinion.

Quote: ”My intent was definitely not to discourage you but to provide my perspective. I clearly encourage you to keep digging on this idea and new ones as well!”

Its OK. The misunderstanding came from me. Thanks for the build and morality.!

Zambac

Finding two identical structures (length=5) in the 340:

1.) Variety of 5 fields positioned on a 340 layer:

340 x 339 x 338 x 337 x 336 = 4,411,278,668,160

2.) Probability of chosing 5 identical (predefined) fields out of such a 340 layer (‘lottery’ calculation):

1 / (340! / (5! x (340-5)!)) = 1:36,760,655,568

3.) Now let this structure be positioned on ANY position of the 340 layer:

340 / (340! / (5! x (340-5)!)) = 340:36,760,655,568 = 1:108,119,575

4.) Probability of 5 homophones complying with a certain combination of 5 predefined (out of the 408 cipher) homophones:

1 / (63^5) = 1 / 992,436,543 or 1:992,436,543

5.) Now combining both, the probability of an identical field structure on no matter what position of the 340 layer AND the probability of 5 identical, predefined homophones:

(1 / 108,119,575) x (1 / 992,436,543) = 1:107,301,817,243,629,225

6.) Considering the amount of existing varieties of 5 fields in a 340 layer:

4,411,278,668,160 / 107,301,817,243,629,225 = 1:24,324

So I’d say chances to find one identical structure INSIDE the 340 would be 1:24,324.

In fact, we’ve found multiple, even longer structures. Same, however, may be performed in comparison to the 408, by modifying the first calculation step (variety of structures in the 408):

Finding two identical structures (length=5) in the 340 and the 408:

1.) Variety of 5 fields positioned on a 408 layer:

408 x 407 x 406 x 405 x 404 = 11,031,053,584,320

6.) Considering the amount of existing varieties of 5 fields in a 408 layer:

11,031,053,584,320 / 107,301,817,243,629,225 = 1:9,727

which of course is a ‘better’ probability due to the greater length (and therefore higher variety of structures) in the 408 layer.

The longest structure found in the 340, however, was of 8 different homophones in a correct structure and a correct order. Therefore:

Finding two identical structures (length=8) in the 340:

1.) Variety of 8 fields positioned on a 340 layer:

340 x 339 x 338 x 337 x 336 x 335 x 334 x 333 = 164,361,464,070,080,659,200

2.) Probability of chosing 8 identical (predefined) fields out of a 340 layer (‘lottery’ calculation):

1 / (340! / (8! x (340-8)!)) = 1:4,076,425,200,150,810

3.) Positioned anywhere on the 340 layer:

340 / 4,076,425,200,150,810 = 11,989,485,882,796

4.) Probability of 8 homophones complying with a certain order of 8 predefined (out of the 340 cipher given) homophones:

1 / (63^8) = 1 / 248,155,780,267,521 or 1:248,155,780,267,521

5.) Now combining both, the probability of an identical field structure (length 8) on no matter what position of the 340 layer AND the probability of 8 identical, predefined homophones:

(1 / 11,989,485,882,796) x (1 / 248,155,780,267,521) = 1:2,975,260,224,251,669,213,731,468,716

6.) Considering the amount of existing varieties of 8 fields on a 340 layer:

164,361,464,070,080,659,200 / 2,975,260,224,251,669,213,731,468,716 = 1:18,101,933

So I’d say chances to find an identical structure of 8 homophones twice in the 340 is a probability of 1:18 million.

Conclusio:

Please be aware that two factors may influence these odds, also that step 3.) should actually be calculated with 339 instead of 340 (otherwise two structures would actually lay exactly on the same position, therefore wouldn’t be two structures anymore):

– the HOMOPHONES’ FREQUENCY is not distributed equally, therefore it is easier to find a structure with e.g. the ‘+’ symbol inside rather than structures with other symbols (can’t say yet how much this actually does influence the overall probability)

– a LANGUAGE is not distributed equally either, instead we have certain (quite strong) lingual structures which may support an (incalculatable) shift of the previously considered probabilities of finding structures

And finally: Finding one structure twice is one thing…finding all of those we already got is definitely another: This actually requires to calculate all probabilities for each structure found (ideally considering the homophone’s frequency, too) and multiplying them each other.

Even under consideration of the critical factors above, I’d say it is still unlikely that so many structures appear twice in the 340 (and the 408), but as the recent analysis of the 408 has shown, I could still be wrong with that assumption.

One last comment on that one: Z has used his homophones in SEQUENCES, this would – again – shift our probability calculation towards a higher chance of finding structures.

What I really can’t say (and even not estimate) is if LANGUAGE, HOMOPHONE FREQUENCY and HOMOPHONE SEQUENCES in combination all together make it possible that we do now find so many identical structures. Opinions welcome.

Update:
An example for finding 1 structure of length=8 AND a total of 5 structures of length=5 would lead to:

9,727^5 x 18,101,933 = 1:1,576,229,794,692,874,376,947,707,731

which is rather unlikely..this is what I meant when I had said someone might want to protect nuclear weapons with it..

QT

(pi, great find with those in the 408, they clearly show that language does influence, e.g. the ‘because’ section)

p.s:

As to say about the structures in the 408, they seem to be accidentially or lingually related (e.g. MOST or BECAUSE or soMETHing). Take one identical word or two similar ones, add some accidentially appearing vowels around it (e.g. ‘I’) and you get your 4-5 structure..guess I’ll focus on the structures appearing (mostly) in one line, to figure out frequent words appearing twice. Sadly ‘because’ doesn’t work with the one I’ve found in the 340.

Regarding these kinds of repeated patterns:

I’ve explored these in the past and concluded that they are arising purely by coincidence. You can confirm this by the following experiment:

1) Shuffle the symbols of the 340 in any random order
2) Shuffle the symbols of the 408 in any random order

If the patterns were non-random, then shuffling would keep them from appearing. But you will discover that similar repeated patterns will be shared between the two completely randomized cipher texts.

Trav posted about the phenomenon here: http://www.zodiackillerfacts.com/forum/viewtopic.php?f=50&t=1310&start=30

I also posted there about the "Alignment Analysis" feature of CryptoScope which automatically discovers such patterns (which are numerous). Read more about it here: http://oranchak.com/zodiac/webtoy/new_cryptoscope_features.html#align

It’s important to remember that even though a single pattern may look improbable, the fact is that there is an astronomically high number of chances for improbable things to occur. If you have a one in a million chance of something happening, but you try it a billion times, then that rare thing will happen many times! That is basically what is happening here – there are so many combinations of positions you can select from the cipher texts.

Another way to look at it is if you toss your Scrabble tiles on the floor, chances are you won’t see them spell any valid words. But if you repeat the experiment thousands or millions of times, you’ll eventually see many valid words.

Here is another post on this forum about this topic: http://zodiackillersite.com/viewtopic.php?p=8907#p8907
And here: http://www.zodiackillersite.com/viewtopic.php?t=255&p=1824 (Scroll to my post on Tue Mar 05, 2013 2:13 pm)

D., All true. *sigh*
I wondered (at one time) if some of the "sequences" which follow along in order over a series of lines continuously (rather than zig-zagging all over the place) were more likely to be made up of high-incidence characters (like the IEETT example highlighted in the 408 a couple of posts ago.) I still think that’s, uh, possibly the case.

Hi doranchak,

thank you for posting. wow ‘look more professional than my jpg pictures. Interesting how many of these are repeats …

Can you help me with a question that prepares me for a few days headache. I am looking for a fingerprint of Z. in his ciphers.

I came up with the idea that what I found in Z408 and Z340 in his normal written letters to search. Is this posible?
Can the same strucures are in Z408 and Z340, also in his other Letters which has sent to newspapers?

I’m just curious to find out if anyone has already tested previously. So I can save myself a lot of work.

Thanks in advance.

Zambac

Not sure what you are asking. Maybe you mentioned it before and I missed it. Do you mean you want to find similar repeated patterns in his written letters?

Qote ”Do you mean you want to find similar repeated patterns in his written letters?”

Exactly.

For example: The Z408 has 17 characters from left to right, and from top to bottom 24. If we a letter (same Z sent to newpapers) take from Z and write it in 17 x 24 constellation that it afterwards as 408 looks like.
What do you mean (thing), we would also like repeats, structures found in the original 408?

It’s important to remember that even though a single pattern may look improbable, the fact is that there is an astronomically high number of chances for improbable things to occur. If you have a one in a million chance of something happening, but you try it a billion times, then that rare thing will happen many times!

Well, I actually DID consider that one..nevertheless, language is different than random number…homophone frequencies are not equal either, etc..

So imo we simply – for sure – won’t ever be able to tell the ‘true’ expectation of such structures.

QT