Introduction to Computational Linguistics

15. Finite State Transducers 56
(Here, A ε := A ∪ {ε}, and ε := B ∪ {ε}.) We write q → a:b
q ′ if δ(a, q, b, q ′ ) ∈ δ. We
say in this case that T makes a transition from q to q ′ given input a, and that it
outputs b. Again, it is possible to define this notation for strings. So, if q −→ ⃗u:⃗v
q ′
and q ′ ⃗x:⃗y
−→ q ′′ then we write q ⃗u⃗x:⃗v⃗y
−→ q ′′ .
It is not hard to show that a finite state transducer from A to B is equivalent to
a finite state automaton over A × B. Namely, put
(146) A := 〈A × B, Q, i 0 , F, θ〉
where q ′ ∈ θ(q, 〈a, b〉) iff 〈a, q, b, q ′ 〉 ∈ δ. Now define
(147) 〈⃗u,⃗v〉 · 〈⃗x,⃗y〉 := 〈⃗u⃗x, ⃗x⃗y〉
⃗x:⃗y
Then one can show that q −→ q ′ in T if and only if q −→ 〈⃗x,⃗y〉
q ′ in A. Thus, the
theory of finite state automata can be used here. Transducers have been introduced
as devices that translate languages. We shall see many examples below. Here,
we shall indicate a simple one. Suppose that the input is A := {a, b, c, d} and
B := {0, 1}. We want to translate words over A into words over B in the following
way. a is translated by 00, b by 01, c by 10 and d by 11. Here is how this is done.
The set of states is {0, 1, 2}. The initial state is 0 and the accepting states are {0}.
The transitions are
(148)
〈0, a, 1, 0〉, 〈0, b, 2, 0〉, 〈0, c, 1, 1〉, 〈0, d, 2, 1〉,
〈1, ε, 0, 0〉, 〈2, ε, 0, 1〉
This means the following. Upon input a, the machine enters state 1, and outputs
the letter 0. It is not in an accepting state, so it has to go on. There is only one
way: it reads no input, returns to state 0 and outputs the letter 0. Now it is back
to where is was. It has read the letter a and mapped it to the word 00. Similarly it
maps b to 01, c to 10 and d to 11.
Let us write R T for the following relation.
(149) R T := {〈⃗x,⃗y〉 : i 0
⃗x:⃗y
−→ q ∈ F}
Then, for every word ⃗x, put
(150) R T (⃗x) := {⃗y : ⃗x R T ⃗y}