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I am seeking a minimal Deterministic Finite Automaton (DFA) that determines the number of distinct symbols in a given word, modulo 3, over an alphabet of size N. Specifically, the DFA should have the following properties:

Input: A word w over the alphabet of size N. Output: The DFA should enter one of three final states: Q[0], Q[1], or Q[2]. Acceptance condition: The DFA enters state Q[k] if and only if the number of distinct symbols in w is congruent to k modulo 3.

My current approach involves constructing a DFA with 2^(3N) states, where each state represents a unique subset of distinct elements encountered. However, this solution is clearly far from optimal. I would appreciate guidance on constructing a minimal DFA for this problem, or insights into more efficient approaches to reduce the state space.

Avi Tal
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I expect that the best you can do is an automaton that has close to $2^N$ states, one for each unique subset of elements that have been encountered so far. (I'm not sure where your $2^{3N}$ came from, but $2^N$ is the correct number of such subsets.) You can do slightly better for some $N$: if $N=3k+2$, then I expect $2^{N} - N$ states suffice, as all the states corresponding to a subset with $3k+1$ or $3k+2$ unique elements can be merged into a single state. However I doubt you can do better than that.

I expect you can prove this using the Myhill-Nerode theorem that it's not possible to do significantly better. In particular, I suggest you identify all of the equivalence classes; then the number of such classes is the size of the minimal DFA.

D.W.
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