Hey masootz,

In a nomenclator cipher a single symbol can stand for a word or multiple letters. Inversely, in a verbose cipher multiple symbols can stand for single letter. Verbose ciphers are typically heavy on n-gram repeats and the 340 does not have many so that can be ruled out.

AZdecrypt should be able to handle it if no more than about 20% of the cipher has such assignments. These assignments can also be seen as nulls and skips.

Nomenclator table:

I have been wondering about the same subject lately myself. Klaus Schmeh recently posted a cryptogram made by a priest that is a homophonic cipher, which someone solved. If you look closely at the key, which is really small and you have to zoom in, some of the assignments involve two symbols instead of one.

http://scienceblogs.de/klausis-krypto-k … le-priest/

http://scienceblogs.de/klausis-krypto-k … -Key-1.png

Jarlve, do you think that such a system could be responsible for P39? Too many double symbols for one letter would destroy P19, wouldn’t it? I think it would.

Looks like a homophonic + some polyalphabetic assignments here and there if you ask me.

Jarlve, do you think that such a system could be responsible for P39? Too many double symbols for one letter would destroy P19, wouldn’t it? I think it would.

This:

Inversely, in a verbose cipher multiple symbols can stand for single letter. Verbose ciphers are typically heavy on n-gram repeats and the 340 does not have many so that can be ruled out.

The priest used symbols that were shapes or partially closed shapes, and in some instances put a dot inside the shape and the symbol E either preceding or following the symbol with the dot.

I find this really interesting. How did the priest know to do this? Where did he get the idea? It is to make the cryptogram more difficult to solve and most of the symbols that he did this with map to low frequency plaintext: C D F G I L M P R S U V. It would make it look like the symbol E maps to a letter, but it doesn’t, and it makes for more symbols with the dot inside. It is a really interesting variation. I have never read the term verbose cipher before. Thanks.

The article says that the cryptograms were written in the 1990s. It also says that homophonic messages are typically quite difficult to solve. Every once in a while a homophonic message is posted on the site and someone solves it, by hand. I thought that you might be interested. Most of the messages are not in English.

Inversely, in a verbose cipher multiple symbols can stand for single letter. Verbose ciphers are typically heavy on n-gram repeats and the 340 does not have many so that can be ruled out.

thanks jarlve. could he have masked n-gram repeats, along the lines of smokie’s priest cipher? google calls it a "verbose homophonic cipher".

AB = "E"
BC = "E"
CD = "E"

maybe just for the vowels or higher frequency letters. is that testable or do we start getting into that grey area where that many variables would be hard to prove?

Hey masootz,

Anything substantially verbose will have allot of n-gram repeats. It can be ruled out.

masootz asked;
"are the current solvers (azdecrypt etc) efficient in the case where zodiac may have
intermingled some symbol pairs mapped to a letter? for example, ‘++’ = d but ‘+’ = h
or ‘>P’=r but ‘>’=j and ‘P’=t."

I can’t speak to AZdecrypt, but doranchak‘s Webtoy does not seem to allow for this.
You can understand that it would be representationally difficult to display a ‘symbol
solution’ such that a homophone symbol solver simultaneously showed two
different solved letters.
My view is that Zodiac did not go to the level of complication that your example uses,
so the Webtoy does not need to account for this – however, I do have the view that
Zodiac’s intention may be that some symbols are allowed to have two possible
choices for an outcome of which only one choice is the correct ‘solved letter’.
I give an explanation of the circumstance where that could occur here;
viewtopic.php?f=81&t=907&start=290#p68412

The Webtoy could conceivably be modified to account for the circumstance I cite
where there are two possible choices for an outcome of which only one choice
(for a specific position in the Z340) is the correct ‘solved letter’ .
To give an example, suppose the homophone symbol ‘I’=’T'(as usual solved plaintext),
but could also have an outcome where (for a specific position in the Z340) symbol ‘I’
is meant to be not actioned as a homophone but instead is read as simply being its
usual self (i.e. for a specific position in the Z340, a symbol ‘I’=’I’). If we look at
the first line of the Z340, we could take position Z340#11 as symbol ‘I’=’I’ whilst also
regarding other instances of the symbol ‘I’ as being ‘I’=’T’ .
The way the Webtoy could accomodate this circumstance would be to modify Webtoy
such that specific positions are able to be ‘locked‘ to a specific letter and Webtoy
does not act upon that specific position, whilst Webtoy does continue to act upon
other positions’ instances in the Z340 of the symbol ‘I’. Thereby, in this case,
position Z340#11 is set ‘locked’ to be ‘I’=’I’ and Webtoy does not attempt to ‘solve’ for
this position, whilst for other positions where the symbol ‘I’ occurs, because these
other positions are not ‘locked’, you can specify that the cipher symbol ‘I’=’T’ and
Webtoy will ‘solve’ by substituting ‘T’ into these other positions (-unless you decide
to also ‘lock’ some other singular position to also set ‘I’ to be ‘I’).
Cheers

Hey buyerninety,

To give an example, suppose the homophone symbol ‘I’=’T'(as usual solved plaintext),
but could also have an outcome where (for a specific position in the Z340) symbol ‘I’
is meant to be not actioned as a homophone but instead is read as simply being its
usual self (i.e. for a specific position in the Z340, a symbol ‘I’=’I’). If we look at
the first line of the Z340, we could take position Z340#11 as symbol ‘I’=’I’ whilst also
regarding other instances of the symbol ‘I’ as being ‘I’=’T’.

AZdecrypt does have support for something along these lines. In such that it can be configured to allow a symbol to represent multiple letters. Though these extra letters per symbol cannot be locked to a specific letter of the users choice yet, though that could be added in a later update.

The functionality is at the moment not super user friendly. It is on my to do list to improve that.

Open up a cipher (the 340) and go to Functions and then Symbols (use Update list if have to). Select a symbol in the first list, select "Set plaintext letters for selected symbol…", put 2 in the A1 input box and click on Apply manipulation. Do that for all symbols you want to represent 2 letters. To run the solver with those settings you need to select the "Substitution + polyphones [user]" solver and click on Solve. You probably want to use at least 6-grams also. These can be loaded using File and Load n-grams (location: AZdecrypt N-gram folder).

I could do that for you if you can tell me exactly which symbols may represent 2 letters.

My view is that Zodiac did not go to the level of complication that your example uses,
so the Webtoy does not need to account for this – however, I do have the view that
Zodiac’s intention may be that some symbols are allowed to have two possible
choices for an outcome of which only one choice is the correct ‘solved letter’.
I give an explanation of the circumstance where that could occur here;
https://www.zodiackillersite.com/viewto … 290#p68412

why do you think that would be more complicated a method than zodiac would try? just curious.

i try to think of how i would complicate a cipher if i were zodiac. i have a passing familiarity with different encryption techniques, i have read the basic information, i can follow along with these discussions although a lot of the math involved goes over my head – basically my guess is i have about the same knowledge as zodiac as far as creating ciphers. the 408 was simple, he took an ego hit when it was solved, and he was trying to make the 340 harder. he’s creating the "game" and it wouldn’t do for him to create an unsolvable cipher (his whole point is to make everyone else look dumb for not solving it).

i don’t think he spent weeks and months heavily engaged in learning new encryption techniques. i think the 340 is the 408 with some added step. not a complicated step, just an infuriating one – probably something the rest of us would think of as close to "cheating".

one way to easily (for him) complicate a homophonic substitution is to create a new rule – like "symbol @ = letter B but @ next to $, %, or & = letter E" (@$, @%, @& = E) so we’d end up with a whole assortment of possibilities of two symbols for one letter, plus possibly even additional individual symbols for that same letter. jarlve says we’d see more bigrams and, while i get that, i also think he could have gone to an extreme to mask that by making a great number of combinations.

he would consider it fair game, and it wouldn’t require any additional knowledge or time. i don’t think he double encrypted, changed directions, shuffled letters before encrypting, swapped rows, or anything like that – he was impatient and those things would require too much effort. he didn’t even bother recopying the 340 when he made the backwards "K" mistake. he’s lazy.

even considering i’m wrong about the technique above, the framework remains – what it some simple cheat he could do to a homophonic substitution that would make it hard to solve but still remain technically solvable so feels superior to the rest of us?

i’m making a cipher by hand similar to the 340 as an experiment (not being as versed as some of you has advantages if we assume zodiac didn’t have more than the average layman’s knowledge of encryption). i’m using the "encyclopedia of observations" you guys created to see what i might be missing as i attempt to fool azdecrypt. already i have accidentally bumped into several things he seemed to do (taken from encyclopedia of observations) –

"The top half of the cipher text, considered on its own, contains 9 repeated bigrams. However, the bottom half of the cipher text, considered on its own, contains only 1 repeated bigram. " – once the cipher maker encrypts the message he starts looking for bigrams that repeat too often. it’s much easier (if you’re lazy) to start swapping letters in a concentrated area. or at least i did the exact same thing he did by favoring replacing one character of a bigram in a more concentrated area than i would have if i wasn’t being lazy.

"The 340 cipher has 9 rows that each contain no repeated symbols" – not 9 exactly, but same thing. i was hurrying so i wouldn’t pay much attention at points, then realize i was reusing the same symbols too often so i’d make sure at least some rows didn’t have repeating symbols.

cycles – i was aware of cycles being used a clue for decrypting so i made sure to not use clean cycles although i think worrying about period 19 blah blah whatever is way beyond the scope of anything he’d be concerned about. i did use azdecrypt to find cycles and, while it reported several "perfect 2 symbol cycles" they were unintentional on my part and weren’t indicative of letters used more often.

columnar repeats – i didn’t account for this or try to prevent it at all.

some of the nonchalance in creating the cipher was 1) i’m not a murderer and 2) my cipher has a "trick" that, while fair, i think will fool azdecrypt. i don’t think there’s evidence that zodiac would have put revealing information in this cipher so he may similarly not have cared as much about the message being revealed as he did about fooling attempts to decode it. i mention this because he may have banged the whole thing out in an afternoon relying, like i did, on a trick in the cipher that would make it difficult to decode.

i think worrying about period 19 blah blah whatever is way beyond the scope of anything he’d be concerned about.

The Period 19 bigram repeats could be the result of a rather simple transposition scheme that ties in quite nicely with a 17 x 20 grid. Just add a column, and subtract a row. Now you have an 18 x 19 grid, consisting of 342 squares. Then don’t utilize the last 2 squares in column 18. Write out the 340 letter message in rows into the 18 x 19 grid, then take the letters from the grid in columns, filling them back into a 17 x 20 grid in rows. Now you’re transposed at P19 (I think I did that right).

But I don’t think he would have been worrying necessarily about transposing the message at P19 specifically. It’s more just about a slight modification to the shape of the grid, then working through what was described above. All in an effort to offset the message in an unusual way.