sure, but you could also just say "he used 18 letters = r, plus the zodiac symbol = 19 = s". why bother with the 6 iterations?

I guess I just wondered if there was anything to the letter E, or possibly C, besides poor ciphering because the second any letter is itself, and all letters of the alphabet are used, regardless of other symbol usage, if it doesn’t always happen?

-m

Your observation is interesting.

One situation where it wouldn’t always lead to E is when the mapping goes back to the original letter without reaching E.

For example, you could have these cipher-to-plaintext mappings:

A -> B
B -> C
C -> D
D -> A
E -> E
F -> K
..etc…

Thus, for letters A B C and D, you’d cycle back to A without reaching E. But the 408’s key doesn’t seem to do that. I verified that each normal letter cycles like this when alternately treated as cipher symbol and plain text:

A: W E
B: L T O N E
C: (no mapping)
D: N E
E
F: S A W E
G: A W E
H: T O N E
I: T O N E
J: F S A W E
K: S A W E
L: T O N E
M: H T O N E
O: N E
P: I T O N E
Q: F S A W E
R: G A W E
S: A W E
T: O N E
U: I T O N E
V: B L T O N E
W: E
X: O N E
Y: U I T O N E
Z: E

In case anyone’s confused about what’s being done here, consider the cipher symbol "F". The 408 decodes that to plaintext letter "S". But, if we treat "S" as a cipher symbol, it decodes to plaintext letter "A". Repeat like this, and you end up with this sequence: F S A W E.

They all end in E since E decodes to itself.

I’m very curious about whether or not this is an intentional feature of the key.

It’s a curious find indeed! It makes sense that all of the letters end in E because E maps to itself. So it in a sense short-circuits any further linking. If it was a different letter that mapped to itself, they would all likely end in that letter. But what’s curious is that there are no isolated loops that don’t end in E. For example, why there is no X -> Y -> Z -> X -> and around it goes? Was Zodiac careful to shuffle all letters around as much as possible? Or did it just happened that way by sheer luck?

Sounds like a key-shuffling experiment is in order.

I wonder if there’s some key construction method that produces these kinds of non-cycling assignments.

Hmm, yes, perhaps there is a standard way of building the key for a homophonic substitution cipher described in one of the books, that has this property.

So you know doranchak, it was your interesting info on patterns of letter usage in the thread "Detailed Analysis of the 408 solution" that gave me the masochistic idea to check for why the letter patterns break down in the third part of the cipher that inadvertently led me to it! I can’t really say how, I just tend to notice odd things sometimes.

-m

OK I ran the shuffle experiment. Here’s the key and assignment cycles for the 408:

   cipher:!#%()+/56789=@ABDEFGHIJKLMNOPQRSTUVWXYZ^_cdefjklpqrtz
    plain:OLLNHEKTESAIPSWLNESATTFSTHENIFGAOIBEOUERNYVOCDXIAEMRRD
sequences: DNE E FSAWE GAWE AWE BLTONE LTONE MHTONE NE ONE HTONE ITONE JFSAWE KSAWE UITONE TONE WE VBLTONE QFSAWE PITONE SAWE RGAWE YUITONE XONE ZE 
totals for each terminating letter: {E=25}

"cipher" indicates the cipher symbols. "plain" indicates the letters assigned to each cipher symbol. "sequences" are the assignment sequences that occur when you decode a alphabetic cipher symbol to a plaintext letter and then keep feeding the plaintext back in as a cipher symbol until the sequence cycles or terminates. Then the counts per termination symbols are shown.

Here’s the result of shuffling the cipher key 200 times: http://pastebin.com/raw.php?i=eM1ynY3q

The 408 had a max count of 25, corresponding to E as the "ending letter". In the 200 shuffles, only 3 of them had the same number (25) for the max count.

This suggests the chances of accidentally seeing the behavior we see in the 408 might be around 3 in 200, or 1.5%.

Then I ran another test of 10,000 shuffles, resulting in 282 shuffles that had a max count of 25. In that experiment, the chances went up to 2.82%.

So, the "all roads lead to E" feature is indeed unusual, but there’s an almost 3% chance that it just appeared completely at random. Add it to the pile of interesting and perplexing features.

Thanks for running the experiment! Seems to suggest it was done on purpose, considering only 1 in 35 chance of it happening, well, by chance. 🙂 Although I still can’t think of the reason why this "tree with no loops" homophone structure would be in any way beneficial to the cipher, or would hinder the deciphering somehow. It might be an artifact of how Z build the key maybe? Say, he thought, let’s start with the most frequent letter, E. Assign several letters/symbols to it. Then take the letters you just assigned to E, and assign some other homophones to them. And so on, so forth. That would definitely make it into a tree, with the root being E. But I don’t know why anyone would do it that way. I’ll definitely keep an eye on any books that might suggest a similar way of building the key.

Cool find marie, I wonder if this could be related (somehow) to the keyboard find?

I’m going to be a voice of dissension here, a bit. Although it is typical to make a homophonic cipher using – first – the letters A <-> Z and then using shapes and symbols for the homophones, Zodiac very well could have made both the 408 and 340 ciphers using symbols only.

He did not, of course, but he could have. And, certainly, it is much easier to start with the alphabet, since that gives you 26 ready-to-use symbols without having to create any new ones, just as one would make a simple substitution cipher like this:
GOYGLROYGNVJJXIVBYGNGLBIXXFBO

In a cipher like this, one does not try to read the letters themselves, anagram them, or treat them as anything other than substitutes. I could even make this into a homophonic by replacing one letter with a symbol, I will use "*"
GOY*LROYGNVJJXIVBY*NGLBIXXFBO

Or I could replace two more letters with homophones, I will use "#" and "$"
G$Y*LROYGNV#JXIVBY*NGLBIXXFB$

These are all the same cipher, with some using letters as substitutes, and some using symbols. I could make them all symbols, OFC.

My point then, is that I think – and this is my opinion – trying to treat the "letter-like" symbols in Zodiac’s ciphers AS LETTERS is probably misguided and the wrong approach. Like finding "GyKE" in the cipher, etc. It’s almost certainly mere coincidence.

-glurk

My point then, is that I think – and this is my opinion – trying to treat the "letter-like" symbols in Zodiac’s ciphers AS LETTERS is probably misguided and the wrong approach. Like finding "GyKE" in the cipher, etc. It’s almost certainly mere coincidence.

Sounds right to me.

It’s also a question of, well, intention – or plausible/likely intention on the part of the person who created the ciphers. As I’ve said before it’s quite possible that Z left us a clue to his identity (even purposely, yes) in his letters. But I don’t see it as likely that someone who went to considerable trouble constructing ciphers opted for leaving such clues in the plain text.

It’s tempting to read the plain text as – well – text, letters (as glurk says), because it’s easy to do so and it bypasses the problem of actually solving the cryptograms…and it yields all sorts of eye catching results. My name Kane – and so forth. The latter is a clear indication that this route is very problematic, however. You can pretty much read whatever you want into it.

I agree that seeing plaintext-like words and phrases directly in the cipher symbols is problematic.

I am curious about how he constructed the key. One thing I’ve always wondered is if he used a keyword or some other non-random procedure to "seed" the initial alphabet key (before extending it with symbols).

For example, write down the plaintext alphabet. Come up with some word, remove its duplicate letters, then write the word + the unused letters of the alphabet, underneath the plaintext alphabet:

ZODIAC SPEAKING ==> (remove duplicates) ==> ZODIACSPEKNG. Unused letters are BFHJLMQRTUVWXY. String them together under the regular alphabet:

ABCDEFGHIJKLMNOPQRSTUVWXYZ
ZODIACSPEKNGBFHJLMQRTUVWXY

The cipher text there is not meant to be interpreted as plaintext, but a normal phrase was used to generate the mappings. Extend that with symbols, and you’ve got your homophonic key.

Is there some key-initialization process that might tend to produce the unusual "terminates with E" behavior that we see in the 408’s key?

(Incidentally, I should run a brute force test that looks for possible initialization keywords, in case he did do something like that to start off the 408’s key).

Also, note in my example that the leftover letters tend to produce sequential assignments, so that when you sort the plaintext alphabet from A-Z, some of the cipher symbols form adjacent sequences:

ABCDEFGHIJKLMNOPQRSTUVWXYZ
ZODIACSPEKNGBFHJLMQRTUVWXY

The adjacent sequences in the key are: LM, QR, and TUVWXY

Looking at the 408’s key, similar adjacent sequences appear:

Plaintext C and D map to backwards E and F, respectively.
F and G map to Q and R, respectively.
S and T map to K and L, respectively.

That makes me curious.