regular pumping examples

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  • More Applications

    of

    the Pumping Lemma

    Costas Busch - RPI

  • The Pumping Lemma: Given a infinite regular language

    there exists an integer (critical length)

    for any string with length

    we can write

    with and

    such that:

    Costas Busch - RPI

  • Regular languagesNon-regular languages

    Costas Busch - RPI

  • Theorem:The languageis not regularProof:Use the Pumping Lemma

    Costas Busch - RPI

  • Assume for contradictionthat is a regular languageSince is infinitewe can apply the Pumping Lemma

    Costas Busch - RPI

  • We pickLet be the critical length forPick a string such that: lengthand

    Costas Busch - RPI

  • we can write:with lengths:From the Pumping Lemma: Thus:

    Costas Busch - RPI

  • From the Pumping Lemma:Thus:

    Costas Busch - RPI

  • From the Pumping Lemma:Thus:

    Costas Busch - RPI

  • BUT:CONTRADICTION!!!

    Costas Busch - RPI

  • Our assumption thatis a regular language is not trueConclusion:is not a regular languageTherefore:END OF PROOF

    Costas Busch - RPI

  • Regular languagesNon-regular languages

    Costas Busch - RPI

  • Theorem:The languageis not regularProof:Use the Pumping Lemma

    Costas Busch - RPI

  • Assume for contradictionthat is a regular languageSince is infinitewe can apply the Pumping Lemma

    Costas Busch - RPI

  • We pickLet be the critical length of Pick a string such that: lengthand

    Costas Busch - RPI

  • We can writeWith lengthsFrom the Pumping Lemma:

    Thus:

    Costas Busch - RPI

  • From the Pumping Lemma:Thus:

    Costas Busch - RPI

  • From the Pumping Lemma: Thus:

    Costas Busch - RPI

  • BUT:CONTRADICTION!!!

    Costas Busch - RPI

  • Our assumption thatis a regular language is not trueConclusion:is not a regular languageTherefore:END OF PROOF

    Costas Busch - RPI

  • Regular languagesNon-regular languages

    Costas Busch - RPI

  • Theorem:The languageis not regularProof:Use the Pumping Lemma

    Costas Busch - RPI

  • Assume for contradictionthat is a regular languageSince is infinitewe can apply the Pumping Lemma

    Costas Busch - RPI

  • We pickLet be the critical length ofPick a string such that: length

    Costas Busch - RPI

  • We can writeWith lengthsFrom the Pumping Lemma:

    Thus:

    Costas Busch - RPI

  • From the Pumping Lemma:Thus:

    Costas Busch - RPI

  • From the Pumping Lemma: Thus:

    Costas Busch - RPI

  • Since:There must exist such that:

    Costas Busch - RPI

  • However:forfor any

    Costas Busch - RPI

  • BUT:CONTRADICTION!!!

    Costas Busch - RPI

  • Our assumption thatis a regular language is not trueConclusion:is not a regular languageTherefore:END OF PROOF

    Costas Busch - RPI

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