Subject: [lojban] Re: valfendi algorithm

From: Robin Lee Powell <[email-masked]>
Date: Thu, 23 Jan 2003 17:22:48 -0800

On Thu, Jan 23, 2003 at 06:32:18PM -0500, Pierre Abbat wrote:
> On Thursday 23 January 2003 15:13, Robin Lee Powell wrote:
> > Hoo-boy.
> >
> > What, exactly, do the proofs demonstrate?
> 
> Hopefully, that all well-formed Lojban words can be lexed correctly by
> the algorithm, regardless of what Lojban words precede and follow
> them, as long as the syllables are stressed correctly and pauses are
> inserted where required.

Seems to me that you'd need agreement on the algorithm before
constructing the proof.

> > I hereby offer to attempt to debug any reasonably rigorous proof you
> > come up with.
> 
> Lemma: All brivla contain a consonant followed two letters later,
> ignoring apostrophes, by a vowel. If a brivla contains 'y', it
> contains such a consonant after the 'y'.
> 
> Proof: If a brivla contains no 'y', it contains two adjacent
> consonants in the first five letters, ignoring apostrophes. Find the
> first vowel after this consonant cluster. Two letters before it is a
> consonant. If a brivla contains 'y', there is at least one rafsi after
> 'y'. Consider the last rafsi. Either it is a CVV or CCV rafsi of a
> gismu, in which case the first and last letters of the rafsi are the
> consonant and the vowel two letters later, or it is the final long
> rafsi of a gismu or fu'ivla, in which case, being identical to the
> selrafsi, it has such a consonant by the first part.

Granted.

> Theorem: If two lerpoi R and S which both lack 'y' are such that for
> all i R[i:i+1] 

That's the string of lerpoi in R from index i to index i+1, yes?

> is a valid initial consonant pair, valid consonant pair, valid lujvo
> diphthong, fa'u 

Seems like .a would work there just fine.

> valid fu'ivla diphthong iff S[i:i+1], 

I'm sorry, it stopped making sense for me there. iff what about
S[i:i+1]?

> and R[i] is a vowel, consonant,
> fa'u y'ybu iff S[i] is, then R is a valid brivla iff S is, regardless
> of whether for some i R[i] is 'n', 'r', or 'l' and S[i] isn't.
> 
> The proof of this was discussed on the list, and I think it's right,
> but haven't written it down as a proof yet.

All right.

> Please send me any test data you can think of. This should be Lojban
> or not-quite-Lojban text with words run together and stress indicated,

I don't personally feel up to the challenge of constructing things that
will trip *you* up.

-Robin

-- 
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