Reputation: 2087
S -> aB | lamda
B -> bB
B is a useless production. Now after its removal
S -> a | lamda
Is this correct ?
Upvotes: 2
Views: 64
Reputation: 17724
Boy it's been a long time since I've looked at CFG's. B produces an infinite series of b's, does it not (b*?)
Assuming b* means an infinite series of B's I think I would reduce to:
S -> ab* | λ
EDIT:
Yes, my answer above is wrong. The definition of a "useless production" is a production that is never used in the derivation of a terminal string. Since B is non-terminal it can be removed thus S -> λ.
+1 to the answer by user574183.
Upvotes: 1
Reputation: 716
production S -> aB does not terminate. Because B -> bB does not terminate. So the production S -> aB is useless.
the answer should be
S -> lambda
Upvotes: 2