Re: On the word length distribution

["John Grove" <John@xxxxxxxxxxxx>]

----- Original Message -----
From: "Jorge Stolfi" <stolfi@xxxxxxxxxxxxx>
To: <rene@xxxxxxxxxx>
Cc: <voynich@xxxxxxxx>
Sent: Wednesday, December 27, 2000 7:57 PM
Subject: Re: On the word length distribution


> Note that in fact the word counts for each length are actually about
> 12 times the pure binomial model.  So the dictionary has
> in fact about 6000 distinct words.
>
>     > If we take this further, there should be a set of 12 characters
>     > (nice number!) of which every Voynich word should have:
>     > - exactly one (simple scenario)
>     > - at least one (complicated scenario)
>
> Unfortunately that is not the case.  It looks more like some
> variant of the "diacritics" solution

Yikes - all this Math is hard on my head!!!  I find it difficult to believe
the author would have created such a cumbersome code as a 6000 word
dictionary - what if he misplaced his dictionary? Does this binomial system
account for the lack of doublets?

I don't know what the math would do with such a system, but what if you
created two 6x6 tables and filled them both with one 17 letter (plaintext)
alphabet, the numbers 0-9, and the most frequently used letters filling in
the remaining nine squares?  The first table would be labelled according to
the VMS character set found at the beginning of words, the second by word
final characters. When to switch from the first table to the other thus
bringing the encrypted word to a close could be random. Whenever the second
table is used a space is added to the encryption. Am I right in assuming
this would give you the binomial wordlength you've been looking at? You
could also ensure that doublets didn't occur by selecting one of the other
choices for most popular letters - or forcing a move to the second table.

Labelling in my mind would be done by the 'stroke-order' concept with the
initial stroke being the rows and the ligature being the columns. Perhaps a
blank row label could account for double ligatures. Gallows could indicate a
switch without a space - even backfilling a space so that a 'y' could
immediately precede a gallows.

Well, I'm not a cryptologist but I would like to think that a cipher system
sounds more logical than a codebook of 6000 words - and I still would rather
it turned out to be a natural language.

John