Introduction to Computational Linguistics

15. Finite State Transducers 55
Consider again your computer. On an input of length 65536 (= 2 16 ) it takes one
second under the algorithm just described, while the naive algorithm would require
it run for 2159 seconds, which is more than half an hour.
In practice, one does not want to spell out in painful detail how many steps
an algorithm consumes. Therefore, simplifying notation is used. One writes that
a problem is in O(n) if there is a constant C such that from some n 0 on for an
input of length n the algorithm takes C · n steps to compute the solution. (One
says that the estimate holds for ‘almost all’ inputs if it holds only from a certain
point onwards.) This notation makes sense also in view of the fact that it is not
clear how much time an individual step takes, so that the time consumption cannot
not really be measured in seconds (which is what is really of interest for us). If
tomorrow computers can compute twice as fast, everything runs in shorter time.
Notice that O(bn + a) = O(bn) = O(n). It is worth understanding why. First,
assume that n ≥ a. Then (b + 1)n ≥ bn + n ≥ bn + a. This means that for almost all
n: (b+1)n ≥ bn+a. Next, O((b+1)n) = O(n), since O((b+1)n) effectively means
that there is a constant C such that for almost all n the complexity is ≤ C(b + 1)n.
Now put D := C(b + 1). Then there is a constant (namely D) such that for almost
all n the complexity is ≤ Dn. Hence the problem is in O(n).
Also O(cn 2 + bn + a) = O(n 2 ) and so on. In general, the highest exponent
wins by any given margin over the others. Polynomial complexity is therefore
measured only in terms of the leading exponent. This makes calculations much
simpler.
15 Finite State Transducers
Finite state transducers are similar to finite state automata. You think of them
as finite state automata that leave a trace of their actions in the form of a string.
However, the more popular way is to think of them as translation devices with
finite memory. A finite state transducer is a sextuple
(144) T = 〈A, B, Q, i 0 , F, δ〉
where A and B are alphabets, Q a finite set (the set of states), i 0 the initial state,
F the set of final states and
(145) δ ⊆ ℘(A ε × Q × B ε × Q)