As title says, can the number of states in DFA be greater than $2^n$ when language-equivalent NFA has $n$ states - that is if the NFA recognizes the same language as the DFA and has $n$ states, can DFA have more than $2^n$ states?

I think it is, and this is trivially. However Wikipedia seeems to write that it is not possible to have more than $2^n$ states. So comes the question.


closed as unclear what you're asking by D.W., David Richerby, Nicholas Mancuso, Juho, Rick Decker Feb 8 '15 at 19:47

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  • $\begingroup$ The person asking you the question probably means pairs $(A, f(A))$ where $f$ is the powerset construction for determinising NFA. $\endgroup$ – Raphael Feb 6 '15 at 8:01
  • $\begingroup$ The $2^n$ bound is for the minimal DFA. Of course you can increase the size arbitrarily by duplicating the set of states. $\endgroup$ – Klaus Draeger Feb 6 '15 at 13:43
  • $\begingroup$ Where does Wikipedia say that? Please provide a link and a quote. I suspect you are misunderstanding Wikipedia and your question is based on a faulty premise and a too-fast reading of what Wikipedia actually says. $\endgroup$ – D.W. Feb 6 '15 at 23:04

it surely can, but it needs not. Using the powerset construction you can convert any $n$-state NFA into a DFA with at most $2^n$ states.

  • $\begingroup$ @Juho My guess is the OP copied the question wrong. $\endgroup$ – Raphael Feb 6 '15 at 8:01
  • $\begingroup$ @Juho, it should answer the question. Starting from an NFA with n states, The MINIMAL equivalent DFA will have at most $2^n$, but an arbitrary DFA that accepts the same language may have more (redundant, useless) states. $\endgroup$ – Ran G. Feb 6 '15 at 14:08

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