Multi-language DFA minimisation

Question

I'm interested in a slight generalisation of DFA. As usual we have state-set $Q$, finite alphabet $\Sigma$, a $\Sigma^*$-action defined on $Q$ by $\delta : Q\times\Sigma\rightarrow Q$, and initial state $q_0$; but instead of the usual terminal set, we take a family $(T_i)_{i\in 1..n}$ of subsets of $Q$. A Multi-language DFA $M$ is then the tuple

$(Q, \Sigma, \delta, q_0, (T_i))$

and $L \subseteq \Sigma^*$ is recognised by $M$ iff $L = \{s\in\Sigma^*|q_0s\in T_i\}$ for some $i\in 1..n$ . Define $(L_i(M))_{i\in 1..n}$ to be the family of languages recognised by M, if you like.

Okay, now for my question: given a family of regular languages $(L_i)_{i\in 1..n}$ , I want to find the minimal Multi-language DFA $M$ as described above such that $L_i = L_i(M)$ for all $i\in 1..n$ , that is, such that $|Q|$ is minimised over all such machines. My question is, are there any known efficient ways of doing this, perhaps analogous to standard DFA minimisation theory? Conversely, is there any evidence that this problem might be hard?

Answer 1

Short answer. Given a finite family of regular languages $\mathcal{L} = (L_i)_{1 \leqslant i \leqslant n}$, there is a unique minimal deterministic complete multi-automaton recognizing this family.

Details. The case $n = 1$ corresponds to the standard construction and the general case is not much different in spirit. Given a language $L$ and a word $u$, let $u^{-1}L = \{ v \in A^* \mid uv \in L \}$. Define an equivalence relation $\sim$ on $A^*$ by setting $$ u \sim v \iff \text{for each }L \in \mathcal{L},\ u^{-1}L = v^{-1}L $$ Since the $L_i$ are regular, this congruence has finite index. Further, it is easy to see that each $L_i$ is saturated by $\sim$ and that for each $a \in A$, $u \sim v$ implies $ua \sim va$. Let us denote by $1$ the empty word and by $[u]$ the $\sim$-class of a word $u$. Let $\mathcal{A}_\mathcal{L} = (Q, [1], \cdot, (F_i)_{1 \leqslant i \leqslant n})$ be the deterministic multi-automaton defined as follows:

1. $Q = \{ [u] \mid u \in A^*\}$,
2. $[u] \cdot a = [ua]$,
3. $F_i = \{ [u] \mid u \in L_i\}$.

By construction, $[1] \cdot u \in F_i$ if and only if $u \in L_i$ and hence $\mathcal{A}_\mathcal{L}$ accepts the family $\mathcal{L}$. It remains to prove that $\mathcal{A}_\mathcal{L}$ is minimal. It is actually minimal in a strong algebraic sense (which implies that it has the minimal number of states). Let $\mathcal{A} = (Q, q_-, \cdot, (F_i)_{1 \leqslant i \leqslant n})$ and $\mathcal{A}' = (Q', q'_-, \cdot, (F'_i)_{1 \leqslant i \leqslant n})$ be two multi-automata. A morphism $f: \mathcal{A} \to \mathcal{A}'$ is a surjective map from $Q$ onto $Q'$ such that

1. $f(q_-) = q'_-$,
2. for $1 \leqslant i \leqslant n$, $f^{-1}(F'_i) = F_i$,
3. for all $u \in A^*$ and $q \in Q$, $f(q \cdot u) = f(q) \cdot u$.

Then for any accessible deterministic multi-automaton $\mathcal{A}$ accepting $\mathcal{L}$, there is a morphism from $\mathcal{A}$ onto $\mathcal{A}_\mathcal{L}$. To prove this, one first verifies that if $q_- \cdot u_1 = q_- \cdot u_2 = q$, then $u_1 \sim u_2$. Now $f$ is defined by $f(q) = [u]$ where $u$ is any word such that $q_- \cdot u = q$. Then one can show that $f$ satisfies the three required properties.

The end is a bit sketchy, let me know if you need more details.