Re: LR(1) resolving SLR(1) reduce/reduce conflict

  Hans Aberg <haberg@math.su.se> wrote:
>Is it possible for LR(1) to resolve a reduce/reduce conflict appearing in
>SLR(1)?




The following grammar is LR(1) but not SLR(1) or LALR(1)


        S -> A x B x
        S -> B y A y
        A -> w
        B -> w


In the LR(0) machine you have a state that looks like this:


        A -> w .
        B -> w .


SLR(1) uses follow(A) = {x,y} as the lookahead for the first item
                and follow(B) = {x,y} as the lookahead for the second,
                    so ther's a reduce-reduce conflict.


LR(1) has two different states:


      A -> w . { x }
      A -> w . { y }


and


      A -> w . { y }
      A -> w . { x }


Neither state has a conflict.


LALR(1) merges these two LR(1) states, because they have the same
core. The net results is the same as for SLR(1) - a reduce-reduce conflict.

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