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LALR parser
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=== LL parsers === The LALR(''j'') parsers are incomparable with [[LL parser|LL(''k'') parsers]]: for any ''j'' and ''k'' both greater than 0, there are LALR(''j'') grammars that are not [[LL grammar|LL(''k'') grammars]] and vice versa. In fact, it is undecidable whether a given LL(1) grammar is LALR(''k'') for any <math>k > 0</math>.<ref name=chapman/> Depending on the presence of empty derivations, a LL(1) grammar can be equal to a SLR(1) or a LALR(1) grammar. If the LL(1) grammar has no empty derivations it is SLR(1) and if all symbols with empty derivations have non-empty derivations it is LALR(1). If symbols having only an empty derivation exist, the grammar may or may not be LALR(1).<ref>{{harv|Beatty|1982}}</ref>
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