[lojban] Re: How many gismu possible?

[jeremyhussell@gmail.com]

I'm willing to bet that the highest count you got for experimental gismu is exactly the number of possible 4-letter rafsi, less the number of official gismu. So, 19,295 - 1,392 = 17,903 by my count.

Why? 1) No two gismu can have the same 4-letter rafsi, otherwise they would differ only in the final vowel, which is one of the blocking rules. 2) Two gismu with different 4-letter rafsi can only block each other if they have the same final vowel and differ by exactly one consonant, because two differences is enough to unblock. 3) The blocking rules have disconnected groups of consonants which only block each other (bfpv, cjsz, dt, gkx, and lmnr). Therefore, any group of gismu which differ in the same consonant and block each other can be unblocked by making the final vowels different, since the largest such group will contain at most 4 gismu and there are 5 possible final vowels.

The above argument strongly suggests (but doesn't prove) that it's possible to pick gismu in an optimal way, such that they use all the possible 4-letter rafsi.

If I understand your algorithm, and it's bug-free, I think you've proved that it's possible - and probable - to pick gismu in a sub-optimal way, such that as many as 15 fewer would be available than the highest number possible. If you did indeed get a high count of 17,903 then you've also proved that the trivial upper bound can be achieved.

It's possible the broda-brode-brodi-brodo-brodu group, which are an exception to the "can't differ only in the final vowel" rule, are causing some trouble. Do your results change if you leave all but one of them out of the list of existing gismu?

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