[Haskell-cafe] Left-factoring with Parsec

[Joel Reymont]

How does

expr = b a*

translate back into the grammar? Assuming that I had b, c, d...

expr = b <|> c <|> d <|> many (do { symbol ":"; expr; symbol ":";  
expr })

Like this?

	Thanks, Joel

On Apr 11, 2007, at 8:56 PM, Lennart Augustsson wrote:

> 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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