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Examples. Applying Pumping Lemma. Proof by contradiction: Let be accepted by a k -state DFA. Choose For all prefixes of length show there exists such that i.e.,. Choose

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Examples

Examples

Applying Pumping Lemma

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Proof by contradiction:

  • Let be accepted by a k-state DFA.

  • Choose

  • For all prefixes of length

  • show there exists such that

  • i.e.,

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  • 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.

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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.

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  • 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.

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  • 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.

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Source of the problem

Source of the problem?

Regular (ultimately periodic)

Prime (sparse)

Composite(dense)

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Summary of proof techniques employed
Summary of Proof Techniques Employed

  • Counter Examples

  • Constructions/Simulations

    • Induction Proofs

  • Impossibility Proofs

    • Proofs by Contradiction

  • Reduction Proofs : Closure Properties

  • L11PLEG


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