Proving a Language Is Not Regular: Pumping Lemma Demystified

1. The Pumping Lemma CS 130: Theory of Computation HMU textbook, Chapter 4 (sections 4.1 and 4.2)

2. A language that is not regular • ConsiderL = {w | w = anbn for all n >= 0} • There is no DFA that accepts L, no regular expression that describes L • How do we prove this? • Intuition: DFAs cannot “count”, that is, DFAs cannot remember that n a’s have been recognized so that n b’s should follow • Proof technique: by contradiction

3. Proving that a language isnot regular • Suppose there is a DFA thatrecognizes L, and let k be the number of states in that DFA • Consider recognizing the strings ab, a2b2, a3b3, …, akbk, ak+1bk+1 • Note what state the DFA is in as it reads the last a. For the k+1 examples, there should be two examples where they are in the same state (pigeonhole principle)

4. Proving that a language isnot regular • This means there are two strings: arbr, ar+pbr+p, such that, the prefixes ar and ar+p brings the DFA to the same state, say qt. That is,q0s qt for both s = ar and s = ar+p • This in turn means thatqtv qtfor v = ap • Note also thatqtw qf for both w = br and w = br+p • Implications: arbr+p and ar+2pbr+p are acceptable in this DFA. A contradiction.

5. Summarizing the strategy • We looked for a state that can be pumped so that substrings admissible by starting and ending with that state can be arbitrarily inserted in the acceptable string • We formalize this notion through • the Pumping Lemma

6. Pumping Lemma • Let L be a regular language • There exists an n such that for all z in L, |z| >= n, z can be broken down into three substrings: z = uvw,where, for all i, uviw is in L • Note: |uv| <= n and |v| >= 1 • Proof of this lemma follows the outline of our earlier argument

7. Using the pumping lemma inour example • L is all strings of the form apbp • We are guaranteed there is an n as specified in the pumping lemma • We choose anbn in L and express this as uvw. Since |uv| <=n, v consists entirely of a’s, say v=ak. This means, strings like an-kbn and an+kbn should be in L, which is a contradiction