# Re: Re: VMs: voynich dice game ... sunday thoughts

Hi,
I think that the only TRUE conclusion that can be picked from my space(ship?) investigation ('till now :-) is that:

Theorem:
spaces are not randomly inserted over a plain unencoded text (or over a text encoded with a "space unaware" schema):

```Proof:
if they were inserted randomly the frequencies distributions of these sets:```

A={the characters in VM}, B = {the first characters of each word}, C = {the last characters of each word}

should be almost the same ... but they are completely different. CVD

```:-)))
Bye,
Marzio```

At 01.21 13/09/04, you wrote:
Hello Rene,

======= At 2004-09-12, 06:04:00 you wrote: =======

```>This really is not correct. The word length
>distribution depends on two main questions:
>- which transcription alphabet is used?  . . .```

There is apparently more things to consider, since we really do not know the function of the
spaces yet. Let's cathegorize:

1) The spaces are used as delimiters of words as in natural language - that is I believe the
case you described.

2) The spaces have been inserted as nulls ( in plain language, the spaces between words
are actually nulls as well - except being delimiters - but that is not what I mean here :-).
Say if I write the whole VM without spaces - as one word - and then insert spaces by
random, I get text, where the solver needs only to eliminate the spaces and guess the
proper spaces by the content. I believe it will give him say 97 percent certainty, if we are
talking about plain, simple text and provided he can read the script and knows the language.
However, with limited number of "word lengths" used ( say by throwing multi-sided dice)
he will not get the VM "word frequency curve" - all word lengths will have about the same
frequency, right?

3) Same as 2), but more complicated - say length of the next "word" will be function of the
dice throw and of the length of "preceding word". There are many combinations and the
results may vary.

4) The "word length" is the function of some encrypting algorithm - but at the same time
it will simulate the bell curve. I cannot imagine off-hand how it can be done.

5) Other possibilites - but hard to specify in detail.

In none of above cases is the "word length frequency curve" the result of PURE
probability, since the VM has similar shape as for other natural languages. What we seem
to know for sure is that the longer "words" in the VM are somehow missing in
comparison with majority of natural languages - and the bell curve is skewed or simply
nonsymetrical. That still does not eliminate the use of natural language. Interestingly
enough, the AVERAGE "length of the VM words" according to our statistics is only
about 1 character shorter than for some other languages. Can it indicate that the total
number of the spaces is about the same, they are just misplaced?

Now talking about Cardan grille - would Ruggs gibberish have spaces and how would they
be encoded? The use of grill will certainly mess them up :-). Or would the grill provide new
spaces with the same word frequency curve as the VM? And how?

Regards,

Jan

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