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Date: Thu, 12 Oct 2017 04:34:43 -0700 (PDT)
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Subject: [lojban] Re: How many gismu possible?
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Em quarta-feira, 11 de outubro de 2017 14:17:10 UTC+3, Jeremy Hussell 
escreveu:
>
> 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.
>

Unfortunately, my algorithm was straightforward, without much 
optimizations, without any tricky math involved. The result is that the 
algorithm is slow even for one round of filling the free gismu space, 
(every possible order of candidate gismy forms will combinatorically ~ 
infinite time).


> If I understand your algorithm, and it's bug-free
>

Well, the code is included, comments are added to it, one can check 
anytime. I tried to make it as simple as possible to get results for one 
round of filling the space in a human-bearable time (several minutes on my 
pcs)

 
> , 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.
>

I simply included them as already existing official gismu.
 

> Do your results change if you leave all but one of them out of the list of 
> existing gismu?
>

Well, I doubt this would change significantly. I was surprised that any 
reshuffling resulting in a new arbitrary order of gismu doesn't change the 
resulting number of possible gismu +/-10. So removing brodV must change the 
number within ~10 gismu, I suppose.
 

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<div dir=3D"ltr"><br><br>Em quarta-feira, 11 de outubro de 2017 14:17:10 UT=
C+3, Jeremy Hussell  escreveu:<blockquote class=3D"gmail_quote" style=3D"ma=
rgin: 0;margin-left: 0.8ex;border-left: 1px #ccc solid;padding-left: 1ex;">=
<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></div></blockquote><div><br=
></div><div>Unfortunately, my algorithm was straightforward, without much o=
ptimizations, without any tricky math involved. The result is that the algo=
rithm is slow even for one round of filling the free gismu space, (every po=
ssible order of candidate gismy forms will combinatorically ~ infinite time=
).</div><div><br></div><blockquote class=3D"gmail_quote" style=3D"margin: 0=
;margin-left: 0.8ex;border-left: 1px #ccc solid;padding-left: 1ex;"><div di=
r=3D"ltr"><br>If I understand your algorithm, and it&#39;s bug-free</div></=
blockquote><div><br></div><div>Well, the code is included, comments are add=
ed to it, one can check anytime. I tried to make it as simple as possible t=
o get results for one round of filling the space in a human-bearable time (=
several minutes on my pcs)</div><div><br></div><div>=C2=A0</div><blockquote=
 class=3D"gmail_quote" style=3D"margin: 0;margin-left: 0.8ex;border-left: 1=
px #ccc solid;padding-left: 1ex;"><div dir=3D"ltr">, I think you&#39;ve pro=
ved that it&#39;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 numb=
er possible. If you did indeed get a high count of 17,903 then you&#39;ve a=
lso proved that the trivial upper bound can be achieved.<br><br>It&#39;s po=
ssible the broda-brode-brodi-brodo-brodu group, which are an exception to t=
he &quot;can&#39;t differ only in the final vowel&quot; rule, are causing s=
ome trouble.</div></blockquote><div><br></div><div>I simply included them a=
s already existing official gismu.</div><div>=C2=A0</div><blockquote class=
=3D"gmail_quote" style=3D"margin: 0;margin-left: 0.8ex;border-left: 1px #cc=
c solid;padding-left: 1ex;"><div dir=3D"ltr"> Do your results change if you=
 leave all but one of them out of the list of existing gismu?<br></div></bl=
ockquote><div><br></div><div>Well, I doubt this would change significantly.=
 I was surprised that any reshuffling resulting in a new arbitrary order of=
 gismu doesn&#39;t change the resulting number of possible gismu +/-10. So =
removing brodV must change the number within ~10 gismu, I suppose.</div><di=
v>=C2=A0</div></div>

<p></p>

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