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In DFA, does each state need to have transition on "EACH" input alphabet?

The answer was given "False" but I dont think so. Can anyone explain?

Because if this statement is False, then there is no use of "Dead State"
asked in Theory of Computation by Active (1k points) | 67 views
i think it should be true.In DFA any state has transition for all input symbols.
@Subham Nagar,Only a single transition for a given alphabet. If there is more than one transition,in that case it's an NFA.:). Please correct me if I am wrong.

@Devshree, yes that is evident, but what the concerned doubt over here was that "whether for every state, there should be transition on EACH alphabet". 

For example if $\sum {a,b}$  , then a state Qi should have a defined transition on  both a and b

Yes it should be corect statement.
Made Easy and Ace Academy are known for irrelevant and incorrect solutions as well. The given statement is true.

1 Answer

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I think its true , if anyone finds counter example then please explain it too.
answered by Junior (657 points)

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