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LALR parsers have an equal number of states as an LR(0) parser. LR(1) parsers have a number of states dependent on the lookahead. Often different states are identical, except they have a different lookahead. i.e.
Sample LR(1) item sets:
1)
A -> B.C,'x'
A -> E.C,'y'
2)
A -> B.C,'g'
A -> E.C,'h'
Equivalent LALR(1) item set:
1)
A -> B.C,'x'
A -> E.C,'y'
A -> B.C,'g'
A -> E.C,'h'
Fewer states means fewer shifts and reduces. And at least as importantly, a much smaller parse table.
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∙ 14y ago
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How do you show that a grammar is LALR but not SLR?
To show that a grammar is LALR but not SLR, you can construct a parsing table for the grammar and demonstrate that there are conflicts present in the SLR parsing table that are resolved in the LALR parsing table. Specifically, LALR parsers have larger look-ahead sets than SLR parsers which can resolve these conflicts. In other words, LALR parsers are able to distinguish between more parser states compared to SLR parsers, allowing them to handle more complex grammars.
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