'Pumping theorem' presentation slideshows

Poetry The Pumping Lemma

Poetry The Pumping Lemma. Any regular language L has a magic number p And any long-enough word in L has the following property: Amongst its first p symbols is a segment you can find Whose repetition or omission leaves x amongst its kind. So if you find a language L which fails this acid test,

394 views • 36 slides

Poetry The Pumping Lemma

Poetry The Pumping Lemma. Any regular language L has a magic number p And any long-enough word in L has the following property: Amongst its first p symbols is a segment you can find Whose repetition or omission leaves x amongst its kind. So if you find a language L which fails this acid test,

346 views • 33 slides

Bottom-up parsing Pumping Theorem for CFLs

MA/CSSE 474 Theory of Computation. Bottom-up parsing Pumping Theorem for CFLs. Recap: Going One Way. Lemma : Each context-free language is accepted by some PDA. Proof (by construction): The idea: Let the stack do the work. Two approaches: Top down Bottom up. Top-down VS Bottom-up.

281 views • 28 slides

Bottom-up parsing Pumping Theorem for CFLs

MA/CSSE 474 Theory of Computation. Bottom-up parsing Pumping Theorem for CFLs. Recap: Going One Way. Lemma : Each context-free language is accepted by some PDA. Proof (by construction): The idea: Let the stack do the work. Two approaches: Top down Bottom up. Example from Yesterday.

245 views • 24 slides

MA/CSSE 474

309 views • 20 slides

Use the pumping theorem for context-free languages to prove that

Use the pumping theorem for context-free languages to prove that L= { a n b a n b a p : n, p ? 0 , p ? n } is not context-free. Hint: For the pumping theorem, factor in all ways as u v x y z such that |v| +|y| ? 1 and for each case, argue that u v n x y n z is not in L for some n.

429 views • 14 slides

Proof of the Pumping Theorem for Regular Languages

Proof of the Pumping Theorem for Regular Languages. Richard Beigel EECS University of Illinois at Chicago. The Pumping Theorem for Regular Languages. If L is regular then ? N ? z such that z ? L and | z | ? N

213 views • 13 slides