Self doubt in CLR(1)

Question

if a grammar is CLR(1) with no mergeable states, then it is LALR(1) ? I suppose it is “yes”, am I right?

Answer 1

In LALR(1) we merge states of CLR(1) if they have same reduction but different lookaheads. So if there is no mergeable states in CLR(1) then it is also  LALR(1).

Comments

Having different lookaheads is not compulsory, right?

It may or may not have different lookaheads, is this statement correct?

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If same productions have same lookaheads then why they will be in different states in CLR(1) ? In CLR(1) we generally get more states because same productions but different lookaheads are kept in different states , and we merge such states in LALR(1).

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Yes, agreed.

Answer 2

Just to add a few points,

1. If a grammar is not CLR(1) then it is definitely not LALR(1).
2. If a grammar is CLR(1), it may or may not be LALR(1) always.