That is what I’ve been talking about since years…you are right.

Currently able to break down that number to a few (tens of) thousands results.

QT

I would like to point out my observation on the magnitude of the search space. I’m sure others have a better idea of this than I do, but here goes.

26^63 is the starting point if it’s a straightforward left-to-right top-to-bottom homophonic substitution cipher.

But the evidence suggests that there is some type of mutation of the ciphertext and/or plaintext. For example it could be mutated using columnar transposition, route cipher, four square, along with many others or even something the zodiac made up himself. And for each of those there are any number of keys.

For example with a four square cipher, you’ve got (25!^4) keys. So the search space really is closer to (26^63) * (25!^4). And since we don’t know what type of mutation(s) he decided to use, the true search space is yet some multiple of that, so in reality closer to (26^63) * (25!^4) * N, where N is the number of mutation types. Essentially the search space is still something approximating infinity.

Eliminating the useless keys such as those containing all consonants is still a good heuristic! I don’t mean to downplay that by any means.

One of the guys who programs homophonic substitution solvers needs to answer this question about number of possible keys.

I suppose I technically fall into that category, at least since a few weeks ago. I released this project (free, open source, Java) which successfully solves the z408:

https://github.com/beldenge/zenith

26^63 can be confirmed, imo.

Under this given clompexity I don’t see any need or use of shuffling lines/rows etc.

Each cipher structure, however, reduces the amount of potential keys/variations; e.g. it is very unlikely that the letter Q occured 7.5% often.

What makes the cipher so hard to crack, imo, is the circumstance that sequences have not, at least not strictly been used. Also, that even if you consider all existing cipher structures, the amount of data is still way too large – even if you focus only on well-structured parts of the cipher.

QT

Until the z340 is solved we cannot assume that the upper limit is 26^63. We have a growing number of tools that can all successfully solve the z408 which had 26^54 possible keys. It seems to follow that if our tools that solve the z408 cannot solve the z340 then there must be another wrinkle in it.

My earlier four square example was a poor choice, as it’s simpler than that. If we entertain the hypothesis that perhaps the z340 is actually inscribed in reverse, then right there we already have doubled the number of keys, i.e. (26^63) * 2. Expand that to every possible mutation and we end up with (26^63) * N. That’s all I really meant to convey in my earlier comment, so my apologies for making it unnecessarily complicated.

The reason I think this is important is that the N above unfortunately approximates infinity. If we hope to solve the z340, it would be of great help for people to keep offering ideas of how it might have been mutated, perhaps based on some clue in another of the zodiac mailings or just by attempting to get in the mind of the zodiac a little bit. Then all of those of us who are able can quickly try those ideas programmatically.

True…63-54 = 9 or 1,000,000,000 or 1B times more complicated than the Z408. However: Shuffled homophones (no sequences) plus almost no cipher structures (e.g. no repeating 4-gram) plus last but not least a shorter cipher.

Easy stuff to solve on a lazy Sunday afternoon :p

QT

So, how many plausible keys are there for Z340? Can we reduce the number of combinations or is it still up there around 1.xE89?

That’s a very interesting question. However, it assumes that z340 is a text with average statistics. In other words, a text that corresponds to an ordinary continuous text. If -and only if- this is the case, you might consider reducing the search space. z340 might just as well be a text with highly repeated passages. If you reduce the search space too much, exotic solutions can no longer be found. Even a completely random plaintext is not excluded (this only applies to the plaintext, not to the substitution). In addition, there are many other possibilities, such as two keys or polyphones.

Since current solvers achieve extremely good results even with more than 63 symbols, I wouldn’t focus at all on reducing the search space of the key. It would be much more important to find out what z340 is really about. In my opinion, only one thing can be excluded: A straight forward substitution as in z408. I think this is only likely if the plaintext is something really confusing (many repetitions, fillers, skips, etc.). Otherwise, one of our solvers would have already found the solution.
If z340 is actually based on a transposition, then the whole search space should be 26^63 * 340! if I am not mistaken. If you consider fillers, skips, plaintext letter ! A-Z (e.g. numbers), then the search space strives towards infinity.

Quicktrader: Do you speak German? If yes, I would like to send you a PM in German. Is easier for me than in English =)

Translated with http://www.DeepL.com/Translator

Ja logisch.

z340 might just as well be a text with highly repeated passages.

Disagreed, it would show up with many reps in the cipher text as well.

Ja logisch.

Klasse! Ich schreib morgen =)

z340 might just as well be a text with highly repeated passages.

Disagreed, it would show up with many reps in the cipher text as well.

Agreed, my fault. I still had transposition in mind.

I didn’t really ask the question as I thought about it. I know that it may not be a homophonic cipher. But if it is, can we limit the search space and prove that it is or that it is not. And the answer is no, not really, I guess.

I had a weird idea –
What if the plaintext was actually just a drawing or pattern made with regular letters?
Something like this simple 17×20 example, but maybe with more detail to account for the variety of symbols:

You’d have a bunch of homophones for the dots and a bunch for the X’s.

If you permitted this possibility, it would greatly influence the key space.

What if the plaintext was actually just a drawing or pattern made with regular letters?

Cool idea! Do you happen to have a text version?