(T/F and explain) An NFA accepts the empty string if and only if its start state is a final state

Question

Does an NFA accept an empty string if and only if its start state is a final state? Is this true?

Please explain why.

This question is related to Automata and NFAs and DFAs.

This seems like a better fit for cs.stackexchange.com. – ggorlen Sep 19 '19 at 00:04 — ggorlen, Sep 19 '19 at 00:04

score 1 · Answer 1 · answered Sep 19 '19 at 09:24

This is false. Consider a two-state NFA with a non-accepting initial state leading to an accepting state by means of a lambda- (or epsilon-, or empty) transition. The empty string is accepted by this NFA by traversing the transition, but the initial state is non-accepting.

If the claim were about DFAs, then it would be true, since lambda- (or epsilon-, or empty) transitions wouldn't be available.

score 0 · Answer 2 · answered Feb 02 '21 at 03:16

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Yes it is true. By defualt NFA means NFA without Epsilon transition. If it is an Epsilon NFA (NFA which can change state without consuming the input symbol) then the answer is false.

answered Feb 02 '21 at 03:16

Jithin Parakka

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score -2 · Answer 3 · answered Jan 15 '20 at 23:34

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Yes, it is true. When the start state become the final state, then without reading any string you reached the final state. So the string accepted is empty. Refer https://www.udemy.com/course/introduction-to-theory-of-computations/

answered Jan 15 '20 at 23:34

Jithin Parakka

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(T/F and explain) An NFA accepts the empty string if and only if its start state is a final state

3 Answers3