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An NFA is nothing more than an ε-NFA with no ε transitions.

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An NFA is nothing more than an ε-NFA with no ε transitions.

(a) True

(b) False

This question was addressed to me by my school teacher while I was bunking the class.

My query is from The NFA with epsilon-moves to the DFA-2 topic in chapter Finite Automata and Regular Expression of Compiler

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The correct answer is (a) True

Easiest explanation: An NFA is nothing more than an ε-NFA with no ε transitions. Thus • δ (q, ε) for all states q = ∅.

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