On Wed, Aug 1, 2012 at 9:38 PM, Jorge Llambías
<jjllambias@gmail.com> wrote:
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.
I see what you mean. It would be interesting to attempt this computationally; the "transitivity" if you will of the blocking rules makes this seem like it could be subtler than either of us are seeing. You're closer to right than I was, though, if not actually right.
> 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.
Good point, I didn't think of that possibility. I actually just wrote some code that puts down a gismu list with all letters distinct. It's pretty readily extensible; in particular I can easily change the order of the nesting, which significantly alters the words that actually get generated. A variant in which initial consonants vary first, then the cluster, then the initial vowel, then the final vowel, with no initial clusters whatsoever, results in
http://pastebin.com/SYfHQf8p, which I found pretty interesting.
> 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.
Well, in the slightly naive computation earlier, it's 9/(191+48) = about 3.8%.
mu'o mi'e latros