[split] Verbose cipher?

[Koen G]

When one starts combining individual characters in the Voynich text, one can therefore never be sure if one is resolving a verbose cipher element, or compressing a frequent bigram.

Definitely, and I think there is no way to solve this issue. If I were to write English "a" as "aµ", there would be no way to distinguish this from English "qu"; however, one is a verbose cipher element while the other is a frequent bigram. Since "qu" is basically the only way "q" appears, it acts very much the same.

About the character set, I have the same ideas. However, the only test I did so far is change [y] into [a]. This increases h2 with a decent amount. This may be counter-intuitive since you remove variation in characters, but this variation was extremely predictable. So you end up with one new character that appears in more contexts.

The texts in my comparison corpus have had capitalization removed, but if this weren't the case, their h0 would be much higher as well. The solution here would also be to equate characters (upper case to lower case) which I did in pre-processing. 

However, I see no way this can be tested in Voynichese. The exercise would be more "random" than merging n-grams by frequency.