Introduction to Computational Linguistics

13. Complexity and Minimal Automata 45
In Step 3 we collect into H 3 all the q 3 such that there exists a q 2 ∈ H 2 such that
x 3
q 2 → q 3 . And so on. It is easy to see that
(124) i 0
x 0 x 1 ...x j
−→ q j+1 iff q j+1 ∈ H j+1
Hence, ⃗x ∈ L(A) iff H n ∩ F ∅. For then there exists an accepting state in H n .
Here each step takes time quadratic in the number of states: for given H j , we
compute H j+1 by doing Q many lookups for every q ∈ H j . However, this number
is basically bounded for given A. So the time requirement is down to a constant
depending on A times the length of ⃗x. This is much better. However, in practical
terms this is still not good enough. Because the constant it takes to compute a
single step is too large. This is because we recompute the transition H j to H j+1 .
If the string is very long, this means that we recompute the same problem over
and over. Instead, we can precompute all the transitions. It turns out that we can
define an automaton in this way that we can use in the place of A.
Definition 10 Let A = 〈A, Q, i 0 , F, δ〉 be an automaton. Put F ℘ := {U ⊆ Q :
U ∩ F ∅}. And let
(125) δ ℘ = {〈H, a, J〉 : for all q ∈ H there is q ′ ∈ J: q a → q ′ }
Then
(126) A ℘ = 〈A, ℘(Q), {i 0 }, F ℘ , δ ℘ 〉
is called the exponential of A.
Definition 11 An automaton is deterministic if for all q ∈ Q and a ∈ A there is at
most one q ′ ∈ Q such that q a → q ′ .
It is clear that for a deterministic and total automaton, all we have to do is to
look up the next state in (??), which exists and is unique. For these automata,
recognizing the language is linear in the string.
Theorem 12 For every automaton A, the exponential A ℘ is total and deterministic.
Moreover, L(A ℘ ) = L(A).