Re: CFG grammar inference problem

Thanks Theo Vosse. That formula is not a arbitrary one. It is based on firm
theory -- Minimum Description Length. We cannot just add a unjustified
factor.

I guess the reason my implementation did not work is that I did not find the
proper coding for Grammar.

Thanks anyway.

"Theo Vosse" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> > However, I found log(1/Prob(Data|Grammar)) and L(Grammar) are of totally
> > different magnitude. L(Grammar) is so big that makes
> > log(1/Prob(Data|Grammar)) has almost no influence to the goodness of the
> > grammar.  There must be something wrong. Could anyone help me to work it
> > out? Thanks!
>
> I don't see why there must be something wrong. If Prob(Data|Grammar)
> is very high, the length of the grammar starts being more imporant.
> This formula probably assumes that the Prob is very, very low. If you
> want to change that, you can add an arbitrary multiplication factor to
> either component to express your weighed interest in them, e.g. x *
> log(...) + y * L(Grammar). Making x much larger than y will change the
> way your greedy algorithm works, I guess...
>
>     Theo Vosse