cs466(Prasad)L11PLEG1 Examples Applying Pumping Lemma

cs466(Prasad)L11PLEG2 Proof by contradiction: Let be accepted by a k-state DFA. Choose For all prefixes of length show there exists such that i.e.,

cs466(Prasad)L11PLEG3 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.

cs466(Prasad)L11PLEG4 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.

cs466(Prasad)L11PLEG5 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.

cs466(Prasad)L11PLEG6 Proof by contradiction: –If is regular, then so is, the complement of – But which is known to be non- regular. –So, cannot be regular. Proving to be non-regular using pumping lemma may be difficult/impossible.

cs466(Prasad)L11PLEG7 Source of the problem? Regular ( ultimately periodic ) … Prime (sparse) … … Composite (dense)

cs466(Prasad)L11PLEG8 Summary of Proof Techniques Employed Counter Examples Constructions/Simulations Induction Proofs Impossibility Proofs Proofs by Contradiction Reduction Proofs : Closure Properties