Introduction to Computational Linguistics

19. Parsing and Recognition 74
actual derivation that defines this constituent structure:
(192) ⃗uX⃗v, ⃗u⃗y⃗v = ⃗x
We scan the string for each rule of the grammar. In doing so we have all possible
constituents for derivations of length 1. Now we can discard the original string,
and work instead with the strings obtained by undoing the last step. In the above
case we analyze ⃗uX⃗v in the same way as we did with ⃗x.
In actual practice, there is a faster way of doing this. All we want to know
is what substrings qualify as constituents of some sort. The entire string is in
the language if it qualifies as a constituent of category S for some S ∈ Σ. The
procedure of establishing the categories is as follows. Let ⃗x be given, length n.
Constituents are represented by pairs [i, δ], where i is the first position and i+δ the
last. (Hence 0 < δ ≤ n.) We define a matrix M of dimension (n + 1) × n. The entry
m(i, j) is the set of categories that the constituent [i, j] has given the grammar G.
(It is clear that we do not need to fill the entries m(i, j) where i + j > n. They
simply remain undefined or empty, whichever is more appropriate.) The matrix
is filled inductively, starting with j = 1. We put into m(i, 1) all symbols X such
that X → x is a rule of the grammar, and x is the string between i and i + 1. Now
assume that we have filled m(i, k) for all k < j. Now we fill m(i, j) as follows. For
every rule X → ⃗α check to see whether the string between the positions i and i + k
has a decomposition as given by ⃗α. This can be done by cutting the string into
parts and checking whether they have been assigned appropriate categories. For
example, assume that we have a rule of the form
(193) X → AbuXV
Then ⃗x = [i, j] is a string of category X if there are numbers k, m, n, p such that
[i, k] is of category A, [i + k, m] = bu (so m = 2), [i + k + m, n] is of category X and
[i + k + m + n, p] is of category V, and, finally k + m + n + p = k. This involves
choosing three numbers, k, n and p, such that k + 2 + n + p = j, and checking
whether the entry m(i, k) contains A, whether m(i+k+2, n) contains X and whether
m(i+k+2+n+ p) contains V. The latter entries have been computed, so this is just
a matter of looking them up. Now, given k and n, p is fixed since p = j − k − 2 − n.
There are O(k 2 ) ways to choose these numbers. When we have filled the relevant
entries of the matrix, we look up the entry m(0, n). If it contains a S ∈ Σ the string
is in the language. (Do you see why?)
The algorithm just given is already polynomial. To see why, notice that in each
step we need to cut up a string into a given number of parts. Depending on the