Introduction to Computational Linguistics

13. Complexity and Minimal Automata 49
Inductively, we define the following:
(131)
q ∼ i+1 q ′ :⇔q ∼ i q ′ and for all a ∈ A, for all r ∈ Q:
if q a → r there is r ′ ∼ i r such that q ′ a
→ r
′
if q ′ a
→ r there is r ′ ∼ i r such that q ′ a
→ r
Evidently, ∼ i+1 ⊆∼ i . Also, if q ∼ i+1 q ′ and q ∈ F then also q ′ ∈ F, since this already
holds for ∼ 0 . Finally, if ∼ i+1 =∼ i then ∼ i is a net. This suggests the following
recipe: start with ∼ 0 and construct ∼ i one by one. If ∼ i+1 =∼ i , stop the construction.
It takes only finitely many steps to compute this and it returns the largest net on
an automaton.
Definition 17 Let A be a finite state automaton. A is called refined if the only net
on it is the identity.
Theorem 18 Let A and B be deterministic, total and refined and every state in
each of the automata is reachable. Then if L(A) = L(B), the two are isomorphic.
Proof. Let A be based on the state set Q A and B on the state set Q B . For q ∈ Q A
write I(q) := {⃗x : q → ⃗x r ∈ F}, and similarly for q ∈ Q B . Clearly, we have
I(i A 0 ) = I(iB 0 ), by assumption. Now, let q ∈ QA and q → a r. Then I(r) = a\I(q).
Hence, if q ′ ∈ Q B and q ′ a
→ r ′ and I(q) = I(q ′ ) then also I(r) = I(r ′ ). Now we
construct a map h : Q A → Q B as follows. h(i A 0 ) := iB 0 . If h(q) = q′ , q → a r
and q ′ a
→ r ′ then h(r) := r ′ . Since all states are reachable in A, h is defined on
all states. This map is injective since I(h(q)) = I(q) and A is refined. (Every
homomorphism induces a net, so if the identity is the only net, h is injective.) It is
surjective since all states in B are reachable and B is refined.
□
Thus the recipe to get a minimal automaton is this: get a deterministic total
automaton for L and refine it. This yields an automaton which is unique up to
isomorphism.