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2 Answers

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4 votes

S->Bb/a
B->Bc/Sd/e
there is both direct (B->Bc)as well as indirect recursion(B->Sd,S->Bb)
step i-
S->Bb/a
B->Bc/Bbd/e/ad
step 2
S->Bb/a
B->eB' /adB'
B'->cB' /bdB
'/ε

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jayanth asked Jun 16, 2022
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Compute the FIRST and FOLLOW for the postfix grammar after removing the left recursion:$ S → S S+/ S S* /a$