|Can this be proven for LL(1) grammars? email@example.com (Lasse =?iso-8859-1?q?Hiller=F8e?= Petersen) (2014-11-06)|
|Re: Can this be proven for LL(1) grammars? firstname.lastname@example.org (Kaz Kylheku) (2014-11-07)|
|Re: Can this be proven for LL(1) grammars? email@example.com (SLK Mail) (2014-11-08)|
|From:||Lasse =?iso-8859-1?q?Hiller=F8e?= Petersen <firstname.lastname@example.org>|
|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
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.
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