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moreApponPumpingLemma

# moreApponPumpingLemma - M or e Appl i cati ons of the Pumpi...

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Fall2004-2005 CSI 301 1 More Applications of the Pumping Lemma

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Fall2004-2005 CSI 301 2 The Pumping Lemma: Given a infinite regular language L there exists an integer m for any string with length L w m w | | we can write z y x w = with and m y x | | 1 | | y such that: L z y x i ... , 2 , 1 , 0 = i
Fall2004-2005 CSI 301 3 Regular languages Non-regular languages *} : { Σ = v vv L R

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Fall2004-2005 CSI 301 4 Theorem: The language is not regular Proof: Use the Pumping Lemma *} : { Σ = v vv L R } , { b a = Σ
Fall2004-2005 CSI 301 5 Assume for contradiction that is a regular language L Since is infinite we can apply the Pumping Lemma L *} : { Σ = v vv L R

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Fall2004-2005 CSI 301 6 m m m m a b b a w = We pick Let be the integer in the Pumping Lemma Pick a string such that: w L w m w | | length m and *} : { Σ = v vv L R
Fall2004-2005 CSI 301 7 Write z y x a b b a m = it must be that length From the Pumping Lemma a ba bb ab a aa a xyz ... ... ... ... ... ... = x y z m m m m 1 | | , | | y m y x 1 , = k a y k Thus:

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Fall2004-2005 CSI 301 8 From the Pumping Lemma: L z y x i ... , 2 , 1 , 0 = i Thus: L z y x 2 1 , = k a y k m a b b a z y x =
Fall2004-2005 CSI 301 9 From the Pumping Lemma: L a ba bb ab a aa aa a z xy ... ... ... ... ... ... ... = 2 x y z k m + m m m 1 , = k a y k y L z y x 2 Thus: m a b b a z y x = L a b b a m m m k m +

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Fall2004-2005 CSI 301 10 L a b b a m m m k m + L a b b a m m m k m + BUT: CONTRADICTION!!!
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