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NFA with exponential number of states when determinized

Submitted by: @import:stackexchange-cs··
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Problem

How can I build an example of a regular language where the minimal DFA has $2^n$ states and the minimal NFA has $n$ states? Obviously the DFA's state-set should contain all subsets of the the NFA's state-set, but I don't know how to start. Any suggestions to put me on the right track?

Solution

The standard example is the language $L$ of all words over an alphabet $A$ of size $n$ that don't contain all the different letters. There is an NFA accepting $L$ with $n+1$ states (or $n$ states if you allow multiple starting states): first guess a letter $a$ which is missing, then go (with an $\epsilon$-move) to an accepting state with self-loops for all letters other than $A$.

Any DFA for $L$ requires at least $2^n$ states. This can be seen using the Myhill-Nerode theorem. Let $S_1,S_2$ be two different subsets of $A$, and $w(S_1),w(S_2)$ words which contain all and only the letters in $S_1,S_2$, respectively. Without loss of generality, suppose $a \in S_1 \setminus S_2$, and let $w = w(A-a)$. Then $w(S_1)w \notin L$ while $w(S_2)w \in L$.

Context

StackExchange Computer Science Q#3381, answer score: 21

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