Here is a problem I am trying to solve:
S -> 0A0 | 1B1 | BB A -> C B -> S | A C -> S | e
I know that C is nullable (since it produces an epsilon) and A is nullable (since it produces C, which is nullable). My question is if B is nullable too since it produces A which produces C which are both nullable?
Asked By : ipoood
Answered By : D.W.
Yes, absolutely. We have the derivation $B \to A \to C \to \epsilon$, so $B$ is nullable.
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Question Source : http://cs.stackexchange.com/questions/33675
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