Introduction to Computational Linguistics

23. Some Metatheorems 92
exists an accepting run for every string of the form ⃗u⃗v q ⃗w. For example,
(224)
(225)
(226)
(227)
q 0
⃗u
→ qi = q j
⃗w
→ qn
q 0
⃗u
→ qi
⃗v
→ q j = q i
⃗w
→ xn
⃗u ⃗v
⃗v
q 0 → qi → q j = q i → q j
q 0
⃗u
→ qi
⃗v
→ q j = q i
⃗w
→ xn
⃗v
→ q j = q i
⃗v
→ q j = q i
⃗w
→ xn
There are examples of this kind in natural languages. An amusing example is from
Steven Pinker. In the days of the arms race one produced not only missiles
but also anti missile missile, to be used against missiles; and then anti
anti missile missile missiles to attach anti missile missiles. And to attack
those, one needed anti anti anti missile missile missile missiles.
And so on. The general recipe for these expressions is as follows:
(228) (anti ⌢ □) n (missile ⌢ □) n⌢ missile
We shall show that this is not a regular language. Suppose it is regular. Then there
is a k satisfying the conditions above. Now take the word (anti ⌢ □) k (missile ⌢ □) k⌢ missile.
There must be a subword of nonempty length that can be omitted or repeated without
punishment. No such word exists: let us break the original string into a prefix
(anti ⌢ □) k of length 5k and a suffix (missile ⌢ □) k⌢ missile of length 8(k+1)−1.
The entire string has length 13k + 7. The string we take out must therefore have
length 13n. We assume for simplicity that n = 1; the general argument is similar.
It is easy to see that it must contain the letters of anti missile. Suppose we
decompose the original string as follows:
(229) ⃗u = (anti ⌢ □) k−1 ,⃗v = anti missile ⌢ □,
Then ⃗u⃗w is of the required form. Unfortunately,
⃗w = (missile ⌢ □) k−1⌢ missile
(230) ⃗u⃗v 2 ⃗w = (anti ⌢ □) k−n⌢ anti missile anti missile ⌢ □ ⌢
Similarly for any other attempt to divide the string.
(missile ⌢ □) k−2⌢ missile L