06 Nov 2014 22:16:38 GMT

Related articles |
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Can this be proven for LL(1) grammars? lhp+news@toft-hp.dk (Lasse =?iso-8859-1?q?Hiller=F8e?= Petersen) (2014-11-06) |

Re: Can this be proven for LL(1) grammars? kaz@kylheku.com (Kaz Kylheku) (2014-11-07) |

Re: Can this be proven for LL(1) grammars? slkpg4@gmail.com (SLK Mail) (2014-11-08) |

From: | Lasse =?iso-8859-1?q?Hiller=F8e?= Petersen <lhp+news@toft-hp.dk> |

Newsgroups: | comp.compilers |

Date: | 06 Nov 2014 22:16:38 GMT |

Organization: | SunSITE.dk - Supporting Open source |

Keywords: | parse, theory, question, LL(1) |

Posted-Date: | 07 Nov 2014 13:14:12 EST |

Given a context free grammar G that is LL(1), is the answer to the

following questions "yes"? I intuitively believe it is, except for (2),

but I am not sufficiently confident of my skills in proving this in full

generality, so I would much appreciate expert advice.

1. From G, for each nonterminal symbol R define a grammar GR, where the

start rule is changed to R. Rules that are no longer reachable are of

course useless and discarded. Is each GR also LL(1)?

2. From all the GR grammars from (1), define G' with a new start rule S:

GR1 | GR2 ... |GRn. Is G' also LL(1)? Given that there could be rules Gi

and Gj such that their FIRST sets intersect, I suppose the answer is no.

What if instead, S is defined only from a selection of rules with

disjoint FIRST sets?

3. From G, define a new grammar G', such that for each nonterminal R, a

rule R: "R" is added, where "R" is a new terminal symbol. Is G' also LL

(1)?

4. Do (1) and (3) hold for other types of CFG?

This isn't homework, and I have tried searching for an answer, but found

none - this may of course be because the proof is actually trivial. If

so, I apologize and hope someone will enlighten me anyway.

/Lasse

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