Given two strings x and y, I want to build a minimum size DFA that accepts x and rejects y. One way to do this is brute force search. You enumerate DFA's starting with the smallest. You try each DFA until you find one that accepts x and rejects y.

I want to know if there is any other known way of finding or building a minimum size DFA that accepts x and rejects y. In other words, can we beat brute force search?

More Detail:

(1) I really do want an algorithm to find a minimum size DFA, not a near minimum size DFA.

(2) I don't just want to know how large or small the minimum DFA is.

(3) Right here, I'm only focused on the case were you have two strings x and y.


Additional information for the interested reader:

Suppose $x$ and $y$ are binary strings of length at most $n$. It is a known result that there is a DFA that accepts $x$ and rejects $y$ with at most $\sqrt{n}$ states. Notice that there are about $n^{\sqrt{n}}$ DFA's with a binary alphabet and at most $\sqrt{n}$ states. Therefore, the brute force approach wouldn't require us to enumerate through more than $n^{\sqrt{n}}$ DFA's. It follows that the brute force approach couldn't take much more than $n^{\sqrt{n}}$ time.

Slides that I found helpful: https://cs.uwaterloo.ca/~shallit/Talks/sep2.pdf

  • 2
    $\begingroup$ @AndrásSalamon Is it still NP-complete if the sets to be distinguished each consist of only one string? It feels to me like this should be reasonably tractable. $\endgroup$
    – mhum
    Jan 12, 2015 at 20:45
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    $\begingroup$ @mhum the problem that there are many different regular languages that separate the two strings — DFA minimization will find the best automaton for any one of these languages but will do nothing to compare it to automata for the other separating languages. $\endgroup$ Jan 13, 2015 at 7:01
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    $\begingroup$ If $x$ and $y$ are different lengths, with the larger of length $n$, it is easy to quickly find a DFA with $O(\log n)$ states that separates them: just use a cycle of length $p$, where $p$ doesn't divide $|x|-|y|$. Find $p$ by trying $2, 3, 5,\ldots$ in order until you find the appropriate $p$. If $x$ and $y$ are the same length, then the $O(\sqrt{n})$ construction of Robson, in a 1996 paper, gives a simple machine that can be found by a search of size $O(n)$. Neither construction is guaranteed to be the smallest DFA. $\endgroup$ Jan 13, 2015 at 21:09
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    $\begingroup$ Shallit's notes, linked above, include the useful observation that the worst case for the separation problem is when the alphabet is binary: it's always possible to partition larger alphabets into two subsets that still distinguish the two input words and search for a binary automaton that treats letters in one subset as 0's and letters in the other subset as 1's. But for seeking the minimum separating automaton this doesn't seem to help, because you might be able to use the extra information from the original alphabet to do better than you could with a mapping to a binary alphabet. $\endgroup$ Jan 13, 2015 at 23:14
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    $\begingroup$ a special case of this other recent question where in-set and out-set sizes equal 1. minimal finite automata given in-words and out-words. that answer lists some learning literature incl some heuristics. $\endgroup$
    – vzn
    Jan 14, 2015 at 4:05

1 Answer 1


If I had to do this in practice, I would use a SAT solver.

The question of whether there is a DFA with $k$ states that accepts $x$ and rejects $y$ can be easily expressed as a SAT instance. For instance, one way is to have $2k^2$ boolean variables: $z_{s,b,t}$ is true if the DFA transitions from state $s$ to state $t$ on input bit $b$. Then add some clauses to enforce that this is a DFA, and some variables and clauses to enforce that it accepts $x$ and rejects $y$.

Now use binary search on $k$ to find the smallest $k$ such that a DFA of this sort exists. Based on what I've read in papers on related problem, I would expect that this might be reasonably effective in practice.

Other encodings of this as SAT are possible. For instance, we can use a trace encoding:

  • If $x$ is of length $m$, you could add $m\lg k$ boolean variables: let $s_0,s_1,\dots,s_m$ be the sequence of states traversed on input $x$, and represent each $s_i$ using $\lceil \lg k \rceil$ boolean variables.

  • Now for each $i,j$ such that $x_i=x_j$, you have the constraint that $s_{i-1}=s_{j-1} \implies s_i=s_j$.

  • Next, extend this to handle $y$: let $t_0,\dots,t_n$ be the sequence of states traversed on input $y$, and represent each $t_j$ using $\lg k$ boolean variables. For each $i,j$ such that $y_i=y_j$, add the constraint that $t_{i-1}=t_{j-1} \implies t_i=t_j$.

  • Similarly, for each $i,j$ such that $x_i=y_j$, add the constraint that $s_{i-1}=t_{j-1} \implies s_i=t_j$.

  • Both traces must start from the same starting point, so add the requirement that $s_0=t_0$ (WLOG you can require $s_0=t_0=0$).

  • To ensure that the DFA uses only $k$ states, require that $0 \le s_i < k$ and $0 \le t_j <k$ for all $i,j$.

  • Finally, to encode the requirement that $x$ is accepted and $y$ is rejected, require that $s_m \ne t_n$.

All of these requirements can be encoded as SAT clauses.

As before, you'd use binary search on $k$ to find the smallest $k$ for which such a DFA exists.

  • 3
    $\begingroup$ note this will actually be superior to brute force search if there are certain symmetries in the problem and they are recognized by the solver, but can currently be difficult to identify/ isolate those (either for human or machine). there is also some newer/ related "technology" of satisfiability modulo theories and answer set programming some of which has "built-in" graph predicates or can support their definitions. $\endgroup$
    – vzn
    Jan 14, 2015 at 18:25

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