Sender: lojban@googlegroups.com
Date: Tue, 10 Oct 2017 08:25:14 -0700 (PDT)
From: jeremyhussell@gmail.com
To: lojban <lojban@googlegroups.com>
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Subject: [lojban] Re: How many gismu possible?
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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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<div dir=3D"ltr">I&#39;m willing to bet that the highest count you got for =
experimental gismu is exactly the number of possible 4-letter rafsi, less t=
he number of official gismu. So, 19,295 - 1,392 =3D 17,903 by my count.<br>=
<br>Why? 1) No two gismu can have the same 4-letter rafsi, otherwise they w=
ould 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 g=
roups of consonants which only block each other (bfpv, cjsz, dt, gkx, and l=
mnr). Therefore, any group of gismu which differ in the same consonant and =
block each other can be unblocked by making the final vowels different, sin=
ce the largest such group will contain at most 4 gismu and there are 5 poss=
ible final vowels.<br><br>The above argument strongly suggests (but doesn&#=
39;t prove) that it&#39;s possible to pick gismu in an optimal way, such th=
at they use all the possible 4-letter rafsi.<br><br>If I understand your al=
gorithm, and it&#39;s bug-free, I think you&#39;ve proved that it&#39;s pos=
sible - and probable - to pick gismu in a sub-optimal way, such that as man=
y as 15 fewer would be available than the highest number possible. If you d=
id indeed get a high count of 17,903 then you&#39;ve also proved that the t=
rivial upper bound can be achieved.<br><br>It&#39;s possible the broda-brod=
e-brodi-brodo-brodu group, which are an exception to the &quot;can&#39;t di=
ffer only in the final vowel&quot; rule, are causing some trouble. Do your =
results change if you leave all but one of them out of the list of existing=
 gismu?<br><br><br></div>

<p></p>

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