Introduction to Computational Linguistics

19. Parsing and Recognition 73
Z → UXVX, we eliminate the first rule (X → ε); furthermore, we add all rules
obtained by replacing any number of occurrences of X on the right by the empty
string. Thus, we add the rules Z → UVX, Z → UXV and Z → UV. (Since other
rules may have X on the left, it is not advisable to replace all occurrences of X
uniformly.) We do this for all such rules. The resulting grammar generates the
same set of strings, with the same set of constituents, excluding occurrences of
the empty string. Now we are still left with unary rules, for example, the rule
X → Y. Let ρ be a rule having Y on the left. We add the rule obtained by replacing
Y on the left by X. For example, let Y → UVX be a rule. Then we add the rule
X → UVX. We do this for all rules of the grammar. Then we remove X → Y.
These two steps remove the rules that do not expand the length of a string. We
can express this formally as follows. If ρ = X → ⃗α is a rule, we call |⃗α| − 1 the
productivity of ρ, and denote it by p(ρ). Clearly, p(ρ) ≥ −1. If p(ρ) = −1 then
⃗α = ε, and if p(ρ) = 0 then we have a rule of the form X → Y. In all other cases,
p(ρ) > 0 and we call ρ productive.
Now, if ⃗η is obtained in one step from ⃗γ by use of ρ, then |⃗η| = |⃗γ| + p(ρ).
Hence |⃗η| > |⃗γ| if p(ρ) > 0, that is, if ρ is productive. So, if the grammar only
contains productive rules, each step in a derivation increases the length of the
string, unless it replaces a nonterminal by a terminal. It follows that a string of
length n has derivations of length 2n−1 at most. Here is now a very simple minded
strategy to find out whether a string is in the language of the grammar (and to find
a derivation if it is): let ⃗x be given, of length n. Enumerate all derivations of
length < 2n and look at the last member of the derivation. If ⃗x is found once,
it is in the language; otherwise not. It is not hard to see that this algorithm is
exponential. We shall see later that there are far better algorithms, which are
polynomial of order 3. Before we do so, let us note, however, that there are
strings which have exponentially many different constituents, so that the task of
enumerating the derivations is exponential. However, it still is the case that we
can represent them is a very concise way, and this again takes only exponential
time.
The idea to the algorithm is surprisingly simple. Start with the string ⃗x. Scan
the string for a substring ⃗y which occurs to the right of a rule ρ = X → ⃗y. Then
write down all occurrences C = 〈⃗u,⃗v〉 (which we now represent by pairs of positions
— see above) of ⃗y and declare them constituents of category X. There is an