The Voynich Ninja

*actually what I said before works a lot better with three columns.

I came here to say the same thing Koen!

A word like [naeyun] is perhaps [na-e-yun], and [Kimaei] is [Ki-m-aei]. But it feels as though the middle column is optional.

byatan: Can you state that [Tiig], [Tiiig], and [Tiiiig] are all valid words? And that either [Tiig] or [Tiiig] isn't a spelling mistake? Otherwise I think the cipher would be ambiguous.

I got it down to 23-19-22 unique values for first, second and third column, omitting just two really annoying words (maybe those are numerals). This takes into account most capitals as unique characters as well. I will try to paste the way I split them in spoiler tags.

A simple comparison with the frequency table of the English language is not possible, because five letters are missing:  C, D, J, S, W. It is possible that these have been overwritten with other letters.

I used Nablator's java class to compute mattr 20 / 200 for the cipher text and four English texts:

k King James' Genesis
s Shakespeare's Sonnets
m Moby Dick
a Alice in Wonderland

For the English texts, I defined tokens in three different ways:

WORDS: the first book of moses called genesis
BIGRAMS: th ef ir st bo ok of mo se sc ...
TRIGRAMS: the fir stb ook ofm ose sca lle dge nes ...

[attachment=5686]

Unless I messed up something, the plot seems to confirm the idea put forward by Koen and Emma.

A bigram-based Voynich-like cipher was discussed by Donald Fisk You are not allowed to view links. Register or Login to view..

A trigram-based Voynich-like cipher is discussed in Rene's "Cardan Grille" paper. I posted an example You are not allowed to view links. Register or Login to view..

While the resulting word structure can be made to be very similar to Voynichese, of course all the features of Voynichese "grammar" are missing (e.g. no or very little full or partial reduplication).

Marco, I'd also like to pull You are not allowed to view links. Register or Login to view. into this discussion. He'll have to speak for himself, but it's my understanding that David Jackson leans gently in the direction of the VMs text being meaningless and stochastically generated strings of glyphs. But it's also clear from his blog post he's not against the possibility of the volvelles he proposes being used to generate meaningful, in other words encoded, strings of glyphs.

So here's what I'm picturing at this point:

1. The plaintext is demarcated into trigrams [deciding whether to include a space between words as a separate character for this purpose].
   
2. A three-wheeled nesting volvelle is constructed. Each wheel is marked off into somewhere between 21 and 24 arcs of equal size, one for each letter of the alphabet used in the writing of the plaintext [plus or minus a value of "space" or "start a new word" for one of the segments]
   
3. Assign 21~24 unique cipher glyphs — or short strings of cipher glyphs — to each arc of each wheel. For best results, keep overlap of cipher glyph strings between wheels to a minimum. At the very least, a plaintext letter should never encode the same cipher glyph string on more than one wheel
   
4. For each demarcated trigram in turn, rotate the inner volvelle for the first letter, the middle for the second, and the outer for the third, to the window or pointer.
   
5. The three ciphertext strings are transcribed together as one... vord. Move along to the next trigram and repeat step 4.
   

The anatomy of a vord, if you will, has always been one of my main areas of fascination with the VMs. This conversation includes several researchers who've put forth compelling arguments for modeling the formula for vords as three independent variables. Does anyone here foresee the feasibility of defining a finite set of 21~24 possible values for each of the three segments of any vord?

Hi RenegadeHealer,
if I understand correctly, the system you describe is 100% equivalent to what Rene discusses in his paper and I exemplified You are not allowed to view links. Register or Login to view.. A table with three columns is the same as a volvelle with three wheels (check the "hidden text" to see the full table).

My example was somehow sloppy and does not fully satisfy the sensible requirements you list. Anyway, it seems clear that with this method you can get a good approximation of Voynich word structure, but Emma pointed out that the system does not generate enough short words.
As I said, another major problem is that you don't get Voynich "grammar" (i.e. how words are arranged, with repetitions, quasi-repetition, line effects etc).

Good to see all the activity. I'm trying to make a cipher that's "more" voynichy as this one is very superficial. 

(21-07-2021, 09:51 PM)nablator Wrote: You are not allowed to view links. Register or Login to view.Is that what deterministic means: only one cipher text is possible?

For this it means a given plaintext will always produce the same ciphertext.

Am I on the right track at all with this? 
1) Somehow divide the plaintext in groups of three characters
2) Use three conversion tables, one for each position