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On Wed, Aug 1, 2012 at 9:50 PM, Ian Johnson <blindbravado@gmail.com> wrote:
> 5 vowels
> 17 consonants
> 48 initial clusters
> {11 unvoiced consonants, 10 voiced consonants, (m, l, n, and r get double
> counted) 9 "special exceptions": cs sc jz zj cx kx xc xk mz are all
> forbidden, and no double consonants}->191 internal clusters
> gives exactly 20315 gimtai. This avoids only collisions of final vowels, and
> permits every other apparent "collision", including all those listed in the
> CLL. These restrictions, to within a very crude approximation, probably
> divide the number given above by something between 2 and 3, giving a number
> between 6700 and 10000, probably closer to the former given that most of the
> collisions involve two letters.

I don't think so. "blanu" for example will block "planu", but it won't
block "plana", so by playing with the final vowel you can probably
accomodate all 20315 four-letter initial forms.

> An interesting computation: assuming every combination of 2 vowels and 3
> consonants can serve as exactly one gismu (which lets words differ only in
> their final vowel; not doing so makes the current number of gismu just
> barely fit), you get 6800 possible words.

Three _different_ consonants and two _different_ vowels, right? But
you can have gismu with two repeated consonants or vowels (e.g.
"nanba"). So I think the number for different combinations comes out
as 14280.

> Almost all such combinations work,
> too: ignoring the 9 special exceptions, any 2 vowels obviously work, and any
> group of 3 consonants will have at least 2 with have the same voicing.

I would guess around 5-10% are blocked by the special exceptions.

mu'o mi'e xorxes

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