curiousBen wants to know more about the 340. I think that is a great idea, Ben. Doranchak made this video recently.
I posted the talk here:
Correction there are only 48 highlighted cells at periods 16, 32 and 48 versus 26 cells at periods 18, 36 and 54. It may have been the font. If I use numbers instead of symbols, I get 48, which is still high when I try to make messages. EDIT: Some of them are bunched up around the pivots, like Ben said.
curiousBen wants to know more about the 340. I think that is a great idea, Ben. Doranchak made this video recently.
I posted the talk here:
Thanks Smokie, (and Doranchak for sharing all that information).
As I learn more of the details of ciphers and the z340 specifically, I am going to continue to focus on the pivots and their relationship to the surrounding data. I’m sure I’ll duplicate some work that has already been done by others, but I think it will help me get a feel for the way the characters behave when transposed and more about periods and n-gram counting.
My hypothesis is that the pivots occurred (visible to us) by accident due to the "pile up" of many of the same plain text character in that block of the cipher. Despite transposition and/or encoding with multiple characters, there were still enough of those letters represented in that region that a unique shape occurred that caught our attention. I want to look at more of the plain text pivots.
I would assume that if there were only one random plaintext pivot in the region, that the encoding with 63 characters would have obscured it. But if there were so many of the same characters in the region, the encoding method disguised many of them, but still left us with a unique pattern that is driving us crazy.
Smokie attached a picture of a sample message that was in the Poems we analyzed and there are others in this same batch of messages. Both from the poems and Brave New World. Within one 6×6 region their were three or four pivots in the plain text. Would one of them survive the encoding due to the limited number of characters assigned to the letters that are in that region.
It appears in this case that there are a lot of diagonals of the same character in the plaintext in that region. Could we assume this is true in the plain text of the z340? How would we go about using this info?
See the examples below. There are 4 different pivots in the plain text. With T’s, H’s, and E’s on the diagonals. (Poem 379 on 17 grid)
This one is similar there are 3 overlapping pivots with with a 4 count diagonal of H’s and 4 T’s grouped, also multiple E’s on the diagonal. (Poem 85 on 20 grid)
This another four pivots in plain text. Five I’s on diagonal, with eight I’s in the pile up. Also 3 N’s on diagonal an 8 S’s in the pile up. (Poem 11 on 20 grid)
A few More:
I guess the question is: Do normal plaintext messages have this concentration of diagonals and multiple letters within 5×5 or 6×6 regions across messages?
Download pivot detector.2019.3 to see what happens to pivots after encoding with a homophonic key.
The first sheet holds the 32 messages with two pivots, and will randomly select a message unless you specify. As usual, make entries in green cells.
The second sheet is the encoder.
Right now the encoder is set for manual key in cell CF32. The manual key has only one symbol for each letter. Symbols are numbers.
Go to the single pivot detector to see the pivots in the message encoded. Set calculation to manual, and click on calculate now a bunch of time to see new encoded messages. Instead of letters, you see the symbols, and you can see the pivots.
Now go back to the encoder, and delete the Y from cell CF32.
Click on calculate now to see a homophonic message encoded. The target symbol count is desired number of symbols, and the actual symbol count will be generally close to that.
Set key efficiency to 1-4. 1 will make the key proportionate to letter frequencies in the message, while 4 will flatten out the key so that there will be fewer symbols for letters like E, and more symbols for letters like G and C. This will affect your symbol count distribution, as well as other statistics.
You can add polyphones if you want to. Those are symbols that map to more than one letter. The + symbol has count of 24, and does not cycle with other symbols well. It could, maybe, be a polyphone, as the 408 did have a few of them. You can make messages without any polyphones by leaving the cells in CR32 to CY32 blank.
The grid in C6 to AP25 allows you to make the symbol selection within homophone groups more or less cyclic. Let’s say we have symbols 1, 2, 3, 4 and 5 for letter E. If all of these cells have a 0 in them, then the assignments will be 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 etc. and perfect cycles, similar to the first two parts of the 408. If you put a 10 in every cell, ten percent of the time the encoder will choose a random symbol from 1, 2, 3 4 or 5. The zodiac 340 is very cyclic, but has about 25% on average random symbol selection based on making many test messages and comparing with tools that measure cycle lengths. The 340 is actually more cyclic at the top of the message, and increasingly more random toward the bottom. You can also make 100% random messages, with no cycles at all, by putting a 100 in each cell.
Now go back to the pivot detector and click on calculate now a bunch of times. You probably won’t see the pivots because with about 63 symbols, the chances of the same symbol being used for the same letter in each position of a pivots are very unlikely. Experiment by changing the target symbol count. Try 30 symbols, look at the keys generated, and look at a bunch of messages. Every few messages a pivot will appear, but even with 30 symbols, many messages will show no pivots. They are diffused and won’t shine through. Try 40 symbols and you will still get pivots. You will still get a pivot every once in a while with 50 symbols, but to get two pivots with 63 symbols is pretty difficult.
Remove the polyphone in CR32 if you want to because we don’t know what the + is anyway. We just want to see what happens when we use multiple symbols for the letters. Experiment with key efficiency. Change the value from 1 to 4 to see the shape of the key change. Try more with 63 symbols and different settings. You will get one pivot once in a while, but probably not two.
I will work with it.
I emailed you the file I built. Yours does a lot more stuff and I want to use it. But, just to feel like I am "contributing" I wanted to share the file. Maybe some of the formula’s could be helpful in future spreadsheets.
Maybe you will be successful uploading it to the folder.
Thanks,
I checked out the file. I have rarely used the Offset function. My spreadsheet is cumbersome, and there is a more efficient way to do the same thing. I like to cast messages into different shapes and back and forth from two dimensions to one dimension. I have to go to work now.
In general, you get more pivots if you substitute less cyclically.
I set my encoder to 100% random symbol selection just now and am trying with two pivot messages and 63 symbols, all homophonic. I am actually able to see some single pivots now and then. Maybe one out of 20 messages or so.
I like how this offers a different perspective on the pivots… That maybe its something about the plaintext structure, i.e. a repetitive message like your mikado example…
I think it’s good to open up the possibilities of what the plaintext could be in this way.
I’ve seen documents from a certain suspect that seemingly show this L shape pattern being made. It was from the same era.
If true, the 6×6 box surround it is the cipher relevant
i’ll request if i can share it here.
Yes, dyslexia is probably my first undiagnosed language.
I am going to try to find some text from children’s rhymes or songs with a short repeating chorus.
Smokie, do you know of any databases where we could get messages like that? I am at work now, but I will try to google it as soon as I get a chance.
A quick question for you Cipher experts. Do you know if this has been looked at;-
Splitting the cipher into 4 separate parts, and then trying to decipher using a unique word in each quarter. Also is it possible than one of these quarters i.e. the knife section could flow in a different way than the rest of the cipher ?
Hi,
I didn’t have much time, but it was enough for a short test. Please do not reference the following results or consider them correct. As soon as I have time, I want to test and check the whole thing again. Sorry if the following has already been discussed…I’m not up to date in this thread.
So here’s a short update, I’ll give you exact numbers later ( when, I don’t know):
– If you only look at plain texts, the pivot texts usually consist of the letters that occur most frequently in the English language. That was to be expected.
– If one considers substituted ciphers (25% randomness), then the pivots usually consist of letters for which there are only a few substitutions. In my case these were O (2x), M, F and P. Key:
dDiZ94bekBHax3y+0PfgIEJF2phcC5mj:uU;Aq7RX=YOnGVto1zKMLw8sNrvSQl-TW AAAABBCCDDDEEEEEEEEEFGGHHHIIIIJKLLLMNNNNOOPQRRRRSSSSSTTTTTUUUVWXYZ
To conclude with caution, I would say (also cautiously) that the pivots in z340 actually consist of letters for which there are few cipher symbols. It is possible that Zodiac made the same mistake with z340 as with z408: Using general letter frequencies for the key instead of using plain text frequencies. A poorly chosen key combined with above-average letter frequencies could favor pivots.
Something we mustn’t forget about this whole thing: If z340 actually uses a transposition, then the type of plaintext (poetry, lyrics) plays only a minor role. In this case it is only important that frequently used letters use only a few substitutions.
But one cannot exclude strange ciphers which are based on crossword puzzles or similar.
Sorry that this post is so undeveloped and fast written. I’m also sorry that I only flew over your last postings and haven’t answered yet. I don’t have the time, but I don’t want to lose the train of thought.
Hi,
I didn’t have much time, but it was enough for a short test. Please do not reference the following results or consider them correct. As soon as I have time, I want to test and check the whole thing again. Sorry if the following has already been discussed…I’m not up to date in this thread.
So here’s a short update, I’ll give you exact numbers later ( when, I don’t know):
– If you only look at plain texts, the pivot texts usually consist of the letters that occur most frequently in the English language. That was to be expected.
– If one considers substituted ciphers (25% randomness), then the pivots usually consist of letters for which there are only a few substitutions. In my case these were O (2x), M, F and P. Key:
dDiZ94bekBHax3y+0PfgIEJF2phcC5mj:uU;Aq7RX=YOnGVto1zKMLw8sNrvSQl-TW AAAABBCCDDDEEEEEEEEEFGGHHHIIIIJKLLLMNNNNOOPQRRRRSSSSSTTTTTUUUVWXYZTo conclude with caution, I would say (also cautiously) that the pivots in z340 actually consist of letters for which there are few cipher symbols. It is possible that Zodiac made the same mistake with z340 as with z408: Using general letter frequencies for the key instead of using plain text frequencies. A poorly chosen key combined with above-average letter frequencies could favor pivots.
Something we mustn’t forget about this whole thing: If z340 actually uses a transposition, then the type of plaintext (poetry, lyrics) plays only a minor role. In this case it is only important that frequently used letters use only a few substitutions.
But one cannot exclude strange ciphers which are based on crossword puzzles or similar.
Sorry that this post is so undeveloped and fast written. I’m also sorry that I only flew over your last postings and haven’t answered yet. I don’t have the time, but I don’t want to lose the train of thought.
I think it was a good post.
Let’s talk about the pivots. There are possibilities. They could be actual pivots in text, whether transposed or not, and after substitution they shine through. They could be nulls, or this may not even be a message at all. Just a very, very clever hoax, considering the different patterns.
This could be some type of home grown cipher. The period 29 repeats spike is high, and the pivots are period 29 RLTB. Also, when you look at all of the symbols involved in period 16, 32, and 48 unigram repeats, which take up 48 total positions in the message, that’s a lot too. Largo, I am still planning on looking into this. When I make homophonic messages, I don’t get this many positions. Consider taking a look at this.
