[PDF] CFL Pumping Lemma

PUMPING LEMMA FOR CFL’S
CS 516
1. Pumping Lemma for CFL’s
Theorem 1. Let L be a CFL. There exists a natural number N (depending
on L) such that if z ∈ L and |z| ≥ N, then there exists strings u, v, w, x, y
satisfying the following conditions:
(1) z = uvwxy
(2) v ≠ ɛ or x ≠ ɛ
(3) |vwx| ≤ N
(4) ∀i ≥ 0, uv i wx i y ∈ L.
Proof.
(1) Let G = (V, Σ, P, S) be the CNF grammar for L. The parse tree for
any z ∈ L using G will be a binary tree. Let N = 2 |V |+1 .
(2) If |z| ≥ N, the height of the parse tree for z is ≥ |V | + 1. Therefore
there is a path from the root to a leaf that contains |V | + 1 edges
and uses |V | + 1 internal nodes. By the Pigeon Hole Principle, two
of these nodes must be labeled with the same nonterminal, say A.
S
A
A
u v w x
y
(3) The following three derivations can occur yielding z = uvwxy ∈ Σ ∗ :
(a) S ∗ ⇒ uAy (assume this is the penultimate A in the path)
(b) A ∗ ⇒ vAx (assume this is the last A from the root in the path)
(c) A ∗ ⇒ w
(4) G contains no null productions and all internal nodes have exactly
two children. Therefore v and x can not both be empty.
(5) The height of the upper A is at most |V | + 1 so |vwz| ≤ 2 |V |+1 = N.
(6) By applying the first derivation above once, the second derivation i
times, and the third once we get:
S ∗ ⇒ uAy ∗ ⇒ uv i Ax i y ∗ ⇒ uv i wx i y. □
Date: September 15, 2009.
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2 CS 516
2. Proving some languages are not context-free
The sole purpose of the Pumping Lemma is to prove that certain languages
are not context-free. The proofs generally follow this pattern:
(1) Assume L is context-free (with the hopes of yielding a contradiction)
and thus has a “pumping number” N.
(2) Pick a cleverly crafted string z such that z ∈ L and |z| ≥ N. We
don’t know N in general so z is always defined in terms of N.
(3) Since z ∈ L and |z| ≥ N, all four conditions of the lemma must
apply.
(4) We deduce “something useful” about v and x from the first three
conditions. Often we must consider several possible cases.
(5) Using our knowledge of what v and x must look like, we pick a value
i such that uv i wx i y ∉ L contradicting the last condition.
For example, we can show L = {a n b n c n | n ≥ 0} is not context-free as
follows.
(1) Assume L is context-free and thus has a “pumping number” N.
(2) Let z = a N b N c N .
(3) Since z ∈ L and |z| = 3N ≥ N, the four conditions of the Pumping
Lemma must apply.
(4) Therefore, there exists strings u, v, w, x, y such that
z = uvwxy = a } .{{ . . a}
b } .{{ . . b}
c } .{{ . . c}
.
N N N
(5) Since |vwx| ≤ N we have
v, x ∈ a ∗ ∪ b ∗ ∪ c ∗ ∪ a ∗ b ∗ ∪ b ∗ c ∗
(6) If either v or x is in the set a + b + ∪b + c + then uv 2 wx 2 y ∉ a ∗ b ∗ c ∗ ⊃ L.
(7) So v, x ∈ a ∗ ∪ b ∗ ∪ c ∗ but v and x are not both empty. Therefore
uv 2 wx 2 y does not contain an equal number of a’s, b’s, and c’s and
thus can not be in L. □
Note that L is context-sensitive:
S → aSBC | aBC
CB → BC
aB → ab
bB → bb
bC → bc
cC → cc
Notes:
• Regular languages are “1-pumpable.”
• Context-free languages are “2-pumpable.”
• As before, you only get to pick z and i. The nature of the strings
u, v, w, x, y are deduced from the first three conditions.