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in Theory of Computation 613 views

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No, it is not in general minimal. The standard determinization procedure, the subset construction, always converts an NFA with n states to a DFA with 2^n states and there is no reason that this should be minimal.

In practical terms, if you convert an NFA, you'll often find that there are unreachable states in the resulting DFA, which immediately shows that the result is not minimal. For a more extreme example, take any DFA with n states, call it an NFA and determinize it. You now have a DFA with 2^n>n states that accepts the same language as it did before, so it's certainly not minimal.

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