Related articles |
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Is the language: { (a^n b^m) | n>m>0 } LL(k) ? LR(k) ? avi.tal@altavista.net (Avi Tal) (2000-04-05) |
Re: Is the language: { (a^n b^m) | n>m>0 } LL(k) ? LR(k) ? torbenm@diku.dk (2000-04-11) |
Re: Is the language: { (a^n b^m) | n>m>0 } LL(k) ? LR(k) ? feriozi@my-deja.com (2000-04-14) |
Re: Is the language: { (a^n b^m) | n>m>0 } LL(k) ? LR(k) ? feriozi@my-deja.com (2000-04-14) |
Re: Is the language: { (a^n b^m) | n>m>0 } LL(k) ? LR(k) ? AlexanderS@tidex.co.il (Alexander Sherman) (2000-04-29) |
From: | Avi Tal <avi.tal@altavista.net> |
Newsgroups: | comp.compilers |
Date: | 5 Apr 2000 22:34:31 -0400 |
Organization: | Compilers Central |
Keywords: | parse, question |
Is the language: { (a^n b^m) | n>m>0 } LL(k) ? LR(k) ?
An example of a CFG for this grammar is : S -> aaXb , X -> aXb | epsilon .
Thanks.
Avi.
[Feed that grammar to yacc, it'll generate a parser, which is as good
a constructive proof as any that it's LALR and LR(1). -John]
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