NFA doesn’t require to show transitions for all input symbols in alphabet set over a statea.trueb.false
Question
NFA doesn’t require to show transitions for all input symbols in alphabet set over a statea.trueb.false
Solution
The statement is true. In a Non-deterministic Finite Automaton (NFA), it is not necessary to define transition functions for every input symbol on every state. This is one of the key differences between a Deterministic Finite Automaton (DFA) and an NFA. In a DFA, every state must have exactly one transition for each symbol in the alphabet. But in an NFA, a state can have zero, one, or more transitions for each symbol in the alphabet. So, the answer is (a) true.
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