Examples Applying Pumping Lemma cs 466Prasad L 11
- Slides: 8
Examples Applying Pumping Lemma cs 466(Prasad) L 11 PLEG 1
Proof by contradiction: • Let be accepted by a k-state DFA. • Choose • For all prefixes of length • show there exists such that • i. e. , cs 466(Prasad) L 11 PLEG 2
• Choose (For this specific problem happens to be independent of j, but that need not always be the case. ) • is non-regular because it violates the necessary condition. cs 466(Prasad) L 11 PLEG 3
Proof : (For this example, choice of initial string is crucial. ) • For this choice of s, the pumping lemma cannot generate a contradiction! • However, let instead. cs 466(Prasad) L 11 PLEG 4
• For • Thus, by pumping the substring containing a’s 0 times (effectively deleting it), the number of a’s can be made smaller than the number of b’s. • So, by pumping lemma, L is non-regular. cs 466(Prasad) L 11 PLEG 5
• Proof by contradiction: – If is regular, then so is , the complement of – But which is known to be nonregular. – So, cannot be regular. • Proving to be non-regular using pumping lemma may be difficult/impossible. cs 466(Prasad) L 11 PLEG 6
Source of the problem? Regular (ultimately periodic) … Prime (sparse) … Composite(dense) … cs 466(Prasad) L 11 PLEG 7
Summary of Proof Techniques Employed • Counter Examples • Constructions/Simulations • Induction Proofs • Impossibility Proofs • Proofs by Contradiction • Reduction Proofs : Closure Properties cs 466(Prasad) L 11 PLEG 8
