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Given Some(DFA) = {|A is a DFA and L(A) is not empty and L(A) is not equal to Σ^(*)}

Show Some(DFA) is decidable.

I produced the following answer and wanted to check if I am correct

T="On input where A is a DFA: 1.Mark the start state of A 2.Repeat until no new state gets marked -Mark any state that has a transition coming into it from any state that is already marked 3.If every state is marked is an accepting state or no accepting state is marked. Reject. 4.Otherwise, accept input

Thanks for any help/advice in advance.

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Your answer takes care of the first part. See here for more detailed discussion on the topic "Decide whether a DFA accepts the empty language."

Now for the second part, minimize the given DFA. If it accepts $\Sigma^*$, then the minimized DFA would have a single state, which is both start and final, and all transitions on all input symbols are self-loops on that state. This can be verified easily.

If both of the above checks fail, you have a 'yes' candidate for your problem. If any one of the above checks is passed, you have a 'no' candidate.

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