Discussion
Let \(N\) be an NFA with \(n\) states. Let \(k\) be the number of states of a minimal DFA which is equivalent to \(N\). Which one of the following is necessarily true?
Answer: D
The statement that is necessarily true is \(k \le 2^{n}\).
The subset construction turns an NFA with \(n\) states into an equivalent DFA whose states are sets of NFA states, so it has at most \(2^{n}\) states. The minimal DFA has no more states than this DFA, so \(k \le 2^{n}\) always holds.
The other bounds can fail. An NFA may have redundant states, so the minimal DFA can be much smaller than \(n\), which rules out \(k \ge n\) and \(k \ge 2^{n}\). And some languages need the full exponential blow-up, such as 'the \(n\)th symbol from the end is a', which an NFA recognises with \(n + 1\) states but any DFA needs \(2^{n}\), so \(k \le n^{2}\) is not guaranteed.
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