30-01-2018, 01:31 PM
(This post was last modified: 30-01-2018, 01:31 PM by davidjackson.)
Quote:I think you are ignoring the 3 to 1 ratios that I indicated.
OK, let us try the thought experiment that you suggest:
Quote:moving across from left to right, every three glyphs converts to one letter (or blank space or punctuation mark).Some maths gives us the answer. Let a,b,c be natural numbers >= 0. How many combinations of a,b,c be there so that a+b+c = 13?
As you say, 13 is M. There are dozens of ways of arriving at 13 with three glyphs (eg. 10+2+1, 2+10+1, 5+5+3, 4+8+1, etc. etc) but as long as you know the numerical value of each glyph, it can be done quickly.
Let’s now suppose the second letter is a A. There are three possibilities, 0+0+1, 0+1+0, and 1+0+0, to get 1. So there you have, in a matter of seconds, we have MA and continue on.
Since a,b,c can be repeated (ie, 9+2+2) then this is just a combination problem. The solution is given by C(13,3) = 455 combinations of glyphs.
First off, let me note that there are fewer than 30 Voynich glyphs - there are aprox (ignoring the weirdoes) as many glyphs as letters in a European alphabet.
So we need to assume that there is a one on one correspondence going on; that is, that both languages have the same gemetria system.
Voynichese glyphs are very highly positional aware - they tend to appear in the same positions and correspondences. I would say that it is impossible for you to find more than 400 different combinations to write the glyph which is equivalent to M.

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