Regular_Pumping-1

Regular_Pumping-1 - Non-regular languages (Pumping Lemma)...

Info iconThis preview shows pages 1–16. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Non-regular languages (Pumping Lemma) Regular languages b a * a c b + * ... etc * ) ( b a c b + + Non-regular languages } : { n b a n n }*} , { : { b a v vv R How can we prove that a language is not regular? L Prove that there is no DFA or NFA or RE that accepts L Difficulty: this is not easy to prove ( since there is an infinite number of them) Solution: use the Pumping Lemma !!! The Pigeonhole Principle pigeons pigeonholes 4 3 A pigeonhole must contain at least two pigeons ........... ........... pigeons pigeonholes n m m n The Pigeonhole Principle ........... pigeons pigeonholes n m m n There is a pigeonhole with at least 2 pigeons The Pigeonhole Principle and DFAs Consider a DFA with states 4 1 q 2 q 3 q a b 4 q b a b b a a Consider the walk of a long string: 1 q 2 q 3 q a b 4 q b b b a a a aaaab 1 q 2 q 3 q 2 q 3 q 4 q a a a a b A state is repeated in the walk of (length at least 4) aaaab aaaab 1 q 2 q 3 q 2 q 3 q 4 q a a a a b 1 q 2 q 3 q 4 q Pigeons: Nests: (Automaton states) Are more than Walk of The state is repeated as a result of the pigeonhole principle (walk states) Repeated state Consider the walk of a long string: 1 q 2 q 3 q a b 4 q b b b a a a aabb 1 q 2 q 3 q 4 q 4 q a a b b A state is repeated in the walk of (length at least 4) Due to the pigeonhole principle: aabb aabb 1 q 2 q 3 q 4 q Automaton States Pigeons: Nests: (Automaton states) Are more than Walk of The state is repeated as a result of the pigeonhole principle (walk states) 1 q 2 q 3 q 4 q 4 q a a b b Repeated state i q ..........
View Full Document

This note was uploaded on 12/02/2011 for the course AR 107 taught by Professor Gracegraham during the Fall '11 term at Montgomery College.

Page1 / 47

Regular_Pumping-1 - Non-regular languages (Pumping Lemma)...

This preview shows document pages 1 - 16. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online