Example for LL(1) Grammar which is NOT LALR?

Question

I am learning now about parsers on my Theory Of Compilation course. I need to find an example for grammar which is in LL(1) but not in LALR. I know it should be exist. please help me think of the most simple example to this problem.

Answer 1

Some googling brings up this example for a non-LALR(1) grammar, which is LL(1):

S ::= '(' X 
    | E ']' 
    | F ')'
X ::= E ')' 
    | F ']'
E ::= A
F ::= A
A ::= ε

The LALR(1) construction fails, because there is a reduce-reduce conflict between E and F. In the set of LR(0) states, there is a state made up of

E ::= A . ;
F ::= A . ;

which is needed for both S and X contexts. The LALR(1) lookahead sets for these items thus mix up tokens originating from the S and X productions. This is different for LR(1), where there are different states for these cases.

With LL(1), decisions are made by looking at FIRST sets of the alternatives, where ')' and ']' always occur in different alternatives.

Answer 2

From the Dragon book (Second Edition, p. 242):

The class of grammars that can be parsed using LR methods is a proper superset of the class of grammars that can be parsed with predictive or LL methods. For a grammar to be LR(k), we must be able to recognize the occurrence of the right side of a production in a right-sentential form, with k input symbols of lookahead. This requirement is far less stringent than that for LL(k) grammars where we must be able to recognize the use of a production seeing only the first k symbols of what the right side derives. Thus, it should not be surprising that LR grammars can describe more languages than LL grammars.