Related articles |
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Ambiguity and LR(k) leonardo@dcc.ufmg.br (Leonardo Teixeira Passos) (2006-10-03) |
Re: Ambiguity and LR(k) schmitz@i3s.unice.fr (Sylvain Schmitz) (2006-10-04) |
Re: Ambiguity and LR(k) debray@CS.Arizona.EDU (Saumya K. Debray) (2006-10-04) |
Re: Ambiguity and LR(k) int2k@gmx.net (Wolfram Fenske) (2006-10-06) |
From: | "Saumya K. Debray" <debray@CS.Arizona.EDU> |
Newsgroups: | comp.compilers |
Date: | 4 Oct 2006 16:47:14 -0400 |
Organization: | University of Arizona Computer Science Department |
References: | 06-10-013 06-10-017 |
Keywords: | LR(1), parse |
Posted-Date: | 04 Oct 2006 16:47:14 EDT |
Sylvain Schmitz wrote:
> Leonardo Teixeira Passos wrote:
>> I would like to know if a grammar is ambiguous then there isn't a
>> LR(k) syntax analyser that can be generated from it.
>
> Yes, this is true. You can find proofs of this for instance in Geller
> and Harrison, _On LR(k) Grammars and Languages_, TCS 4:245--276, 1977,
> or in most theory-oriented textbooks on parsing.
>
> Intuitively, you cannot generate a deterministic parser for an ambiguous
> grammar: if each parsing action done by the parser is chosen
> deterministically, then there is a unique way to recognize the entire
> input string, and the grammar is not ambiguous.
Well, just to be pedantic, "nondeterminism" isn't the same as
"ambiguity." For example, consider the language of palindromes,
L = { ww^R | w \in {0,1}* }, where w^R denotes the reverse of w.
This language is generated by the following context-free grammar:
S --> 0 S 0
S --> 1 S 1
S --> 0
S --> 1
S --> epsilon
This language can't be parsed by a deterministic parser, since for any
input string the parser has to "guess" when it has reached the midpoint
of the input. However, the grammar is not ambiguous: each string in the
language has exactly one parse tree (which is how "ambiguity" is defined).
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