Introduction to Computational Linguistics

17. Using Finite State Transducers 64
In general, backtracking works as follows. Every time we face a choice, we pick
the available transition with the smallest number. If we backtrack we go back in
the run to the point where there was a choice, and then we change the previously
chosen next transition to the next up. From that point on we consistently choose
the least numbers again. This actually enumerates all runs. It is sometimes a
dangerous strategy since we need to guarantee that the run we are following actually
terminates: this is no problem here, since every step consumes at least one
character.
Important in backtracking is that it does not give the solutions in one blow:
it gives us one at a time. We need to memorize only the last run, and then backtracking
gives us the next if there is one. (If there is none, it will tell us. Basically,
there is none if going back we cannot find any point at which there was a choice
(because you can never choose lower number than you had chosen).)
Now let us look at the other strategy. It consists in remembering partial runs,
and trying to complete them. A partial run on a pair of strings simply is a sequence
of transitions that sucessfully maps the machine into some state, consuming some
part of the input string and some part of the outpu string. We do not require that
the partial run is part of a successful run, otherwise we would require to know
what we are about to find out: whether any of the partial runs can be completed.
We start with a single partial run: the empty run — no action taken. In the first
step we list all transitions that extend this by one step. There is only one, t(0). So,
the next set of runs is t(0). It leads to the situation 1, a|aa : b|bbb. From here we
can do two things: t(1) or t(2). We do both, and note the result:
(169)
Step 2
t(0), t(1) : 1, aa|a : b|bbb
t(0), t(2) : 1, a|aa : bb|bb
In the next round we try to extend the result by one more step. In each of the two,
we can perform either t(1) or t(2):
(170)
Step 3
t(0), t(1), t(1) : 1, aaa| : b|bbb
t(0), t(1), t(2) : 1, aa|a : bb|bb
t(0), t(2), t(1) : 1, aa|a : bb|bb
t(0), t(2), t(2) : 1, a|aa : bbb|b