Lecture 36: pumping lemma

• Statement of pumping lemma
• Using pumping lemma
• Proof of pumping lemma

The pumping lemma

If L is a regular language, then

• there exists a number n > 0 such that
• for all x ∈ L with ∣x∣ ≥ n
• there exists strings u, v, and w, such that
  • x = uvw
  • ∣uv∣ ≤ n
  • ∣v∣ > 0
  • for all k ≥ 0, uv^kw ∈ L.

Using the pumping lemma

The pumping lemma is a tool for proving that a language is not regular. For example:

Claim: L = {0ⁿ1ⁿ  ∣  n ≥ 0} is not regular.

Proof: assume for the sake of contradiction that L is regular. Then there exists a number n as in the pumping lemma. Let x = 0ⁿ1ⁿ. Clearly x ∈ L and ∣x∣ ≥ n, so we can split x into u, v and w as in the pumping lemma. Now, since ∣uv∣ ≤ n, v can consist only of 0's. Therefore xʹ = uv²w = 0^{n + ∣v∣}1ⁿ. Now the pumping lemma says that xʹ ∈ L, but clearly xʹ has more 0's than 1's, so it is not in L. This is a contradiction, so the assumption that L was regular must be false.

Be careful to pay attention to the quantifiers (for all and there exists) while using the pumping lemma. You don't get to pick n or u, v, or w, but you do get to pick x and k.

Proof of the pumping lemma

Proof sketch: if L is regular, there is a DFA M that recognizes L. n is the number of states. While processing the first n characters of x, we transition through n + 1 states, so by the pigeonhole principle, two of them must be the same. In other words, there is a loop. Let u be the part of x that leads to the start of the loop, v be the part of x that goes around the loop, and let w be the part of the string that takes that machine from the end of the loop to the final state.

While processing uv^kw the machine transitions to the loop (using u), follows the loop k times (using v^k) and then transitions to the final state (using w). Thus uv^kw is accepted by the machine.