SR conflict in LALR(1)

Question

$A→ b.a , \left \{ b \right \}$

   $B→ b.,\left \{ a \right \}$

It is a SR conflict in LALR(1). Now how reduce state operated on $a$ for non terminal $B$(and not $b$) ?

Comments

S -> .AB ($)
A -> .aAb (d)
A -> .a (d)

here too how lookahead for A is d??

got from this link https://stackoverflow.com/questions/14103199/lr1-item-dfa-computing-lookaheads

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S -> .AB ($)

because S is start symbol and lookahead of start symbol is $.

after that we'll get all production of A.

A-> .aAb, (d) because FIRST(B) is d. we decide lookahead on the basis of FIRST of next symbol(S-> .AB here all production of A will have the lookahead of FIRST(B)).

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how lookahead is calculated?

follow of A or first of B?

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by FIRST of B.

Answer 1

This productions are not from a LR (1) grammar. If after merging the states of LR(1) grammar we end up with shift reduce conflict then be sure that the grammar we have used was not LR(1).  We can only have reduce reduce conflict in LALR(1) which we form by merging states of LR(1) grammar