Introduction to Computational Linguistics

17. Using Finite State Transducers 63
characters of the input and output strings. There is only one possibility: the transition
t(0). This results in the following: we are now in state 1, and one character
from both strings has been read. The conjunction of facts (i) is in state 1, (ii) has
read one character from in put, (iii) has read one character from output we call
a configuration. We visualize this configuration by 1, a|aa : b|bbb. The bar is
placed right after the characters that have been consumed or read.
Next we face a choice: either we take t(1) and consume the input a but no
output letter, or we take t(2) and consume the output b but no input letter. Faced
with this choice, we take the transition with the least number, so we follow t(1).
The configuration is now this: 1, aa|a : b|bbb. Now face the same choice again,
to do either t(1) or t(2). We decide in the same way, choosing t(1), giving 1, aaa| :
b|bbb. Now, no more choice exists, and we can only proceed with t(2), giving
1, aaa| : bbb|b. One last time we do t(2), and we are done. This is exactly the first
of the runs. Now, backtracking can be one for any reason. In this case the reason
is: we want to find another run. To do that, we go back to the last point where
there was an actual choice. This was the configuration 1, aa|a : b|bbb. Here,
we were able to choose between t(1) and t(2) and we chose t(1). In this case we
decide differently: now we take t(2), giving 1, aa|a : bb|bb. From this moment
we have again a choice, but now we fall back on our typical recipe: choose t(1)
whenever possible. This gives the third run above. If again we backtrack, it is
the last choice point within the run we are currently considering that we look at.
Where we chose t(1) we now choose t(2), and we get this run:
(167)
0 a:b −→ 1 a:ε
−→ 1 ε:b
−→ 1 ε:b
−→ 1 a:ε
−→ 1 ε:b
−→ 1
Here is how backtracking continues to enumerate the runs:
(168)
0 a:b −→ 1 a:ε
−→ 1 ε:b
−→ 1 ε:b
−→ 1 a:ε
−→ 1 ε:b
−→ 1
0 a:b −→ 1 a:ε
−→ 1 ε:b
−→ 1 ε:b
−→ 1 ε:b
−→ 1 a:ε
−→ 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
0 a:b −→ 1 ε:b
−→ 1 a:ε
−→ 1 ε:b
−→ 1 ε:b
−→ 1 a:ε
−→ 1
0 a:b −→ 1 ε:b
−→ 1 ε:b
−→ 1 a:ε
−→ 1 ε:b
−→ 1 a:ε
−→ 1
0 a:b −→ 1 ε:b
−→ 1 ε:b
−→ 1 ε:b
−→ 1 a:ε
−→ 1 a:ε
−→ 1