{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Fall2004-2005 CSI 301 1 More Applications of the Pumping Lemma
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
Fall2004-2005 CSI 301 3 Regular languages Non-regular languages *} : { Σ = v vv L R
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Fall2004-2005 CSI 301 4 Theorem: The language is not regular Proof: Use the Pumping Lemma *} : { Σ = v vv L R } , { b a = Σ
Image of page 4
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
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 6
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:
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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 =
Image of page 8
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 +
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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!!!
Image of page 10
Image of page 11
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern