Introduction to Computational Linguistics

17. Using Finite State Transducers 62
0 −→ a:b 1 and then 1 −→ ε:b
1 and 1 −→ a:ε
1. Now, for the pair aaa and bbbb there are
several runs:
(163)
0 a:b −→ 1 a:ε
−→ 1 a:ε
−→ 1 ε:b
−→ 1 ε:b
−→ 1 ε:b
−→ 1
0 a:b −→ 1 a:ε
−→ 1 ε:b
−→ 1 a:ε
−→ 1 ε:b
−→ 1 ε:b
−→ 1
0 a:b −→ 1 ε:b
−→ 1 a:ε
−→ 1 a:ε
−→ 1 ε:b
−→ 1 ε:b
−→ 1
0 a:b −→ 1 ε:b
−→ 1 a:ε
−→ 1 ε:b
−→ 1 a:ε
−→ 1 ε:b
−→ 1
(There are more runs that shown.) In the case of the machine just shown, the
number of runs grows with the size of the input/output pair. There is therefore no
hope of fixing the number of runs a priori.
Despite all this, we want to give algorithms to answer the following questions:
☞ Given a transducer T and a pair ⃗x : ⃗y of strings, is there a run of T that
accepts this pair? (That is to say: is 〈⃗x,⃗y〉 ∈ R T ?)
☞ Given ⃗x, is there a string ⃗y such that 〈⃗x,⃗y〉 ∈ R T ?
☞ Given ⃗x, is there a way to enumerate R T (⃗x), that is to say, all ⃗y such that
〈⃗x,⃗y〉 ∈ R T ?
Let us do these questions in turn. There are two ways to deal with nondeterminism:
we follow a run to the end and backtrack on need (depth first analysis); or we
store all partial runs and in each cycle try to extend each run if possible (breadth
first search). Let us look at them in turn.
Let is take the second machine. Assume that the transitions are numbered:
(164)
(165)
(166)
t(0) = 〈0, a, 1, b〉
t(1) = 〈1, a, 1, ε〉
t(2) = 〈1, ε, 1, b〉
This numbering is needed to order the transitions. We start with the initial state
0. Let input aaa and output bbbb be given. Initially, no part of input or output
is read. (Output is at present a bit of a misnomer because it is part of the input.)
Now we want to look for a transition that starts at the state we are in, and matches