When converting NFA to DFA, do I always get the minimal DFA or not?

  • 1
    $\begingroup$ Generally, no.$\,$ $\endgroup$ Commented Jan 23, 2018 at 17:34
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    – Raphael
    Commented Jan 23, 2018 at 18:06
  • $\begingroup$ @Raphael but it's not a problem out of context... I simply asked a question about NFA to DFA conversion $\endgroup$
    – Sepfins
    Commented Jan 23, 2018 at 18:12
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    – D.W.
    Commented Jan 23, 2018 at 18:14
  • $\begingroup$ Of course @RickDecker gave the answer, not in general. But curiously, if you determinize, then reverse the language, determinize and repeat, you will have the minimal automaton. See Brzozowski's algorithm for DFA minimization. $\endgroup$ Commented Jan 23, 2018 at 18:59

2 Answers 2


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. (This is a particularly strong example of the first case I mentioned. All states except subsets of size $1$ will be disconnected from the start state, and the states of size $1$ are a copy of the original DFA.)


If we take NFA with n states and convert into DFA with 2^n states then there are unreachable states in resulting DFA which gives the result as not a minimal DFA.so the NFA to DFA conversion thus not give the result ad minimal DFA.


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