[Haskell-cafe] Left-factoring with Parsec

[Lennart Augustsson]

I presume your grammar has other clauses for expr, otherwise the loop  
is inevitable.
Assuming you have other clauses you can always left-factor.

Here's how those of us with weak memory can remember how to do it:

Say that you have

   expr ::= expr ":" expr ":" expr
          | b
Let's call the part from the first ":" a, since it doesn't matter  
what it is.  So we have
   expr ::= expr a | b
Let's call expr x, and just change notation slightly
   x = x a + b
Now use high school algebra
   x = x*a + b
   x - x*a = b
   x*(1-a) = b
   x = b / (1-a)
   x = b * 1/(1-a)
Now you have to remember that the Taylor series expansion of 1/(1-a) is
   1/(1-a) = 1 + a + a^2 + a^3 + a^4 + ...

OK, now put your grammar hat back on.  What's
   1 | a | aa | aaa | aaaa | ...
it's just an arbitrary number of a:s, i.e., a* (or 'many a' in parsec).
So finally
   expr = b a*

	-- Lennart

On Apr 11, 2007, at 18:15 , Joel Reymont wrote:

> Suppose I have expr = expr ":" expr ":" expr.
>
> Can the above be left-factored to fail on empty input so that my  
> parser doesn't go into a loop?
>
> 	Thanks, Joel
>
> --
> http://wagerlabs.com/
>
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