Chapter1Section4_F11

Chapter1Section4_F11 - 91.304 Foundations of (Theoretical)...

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91.304 Foundations of (Theoretical) Computer Science Chapter 1 Lecture Notes (Section 1.4: Nonregular Languages) David Martin dm@cs.uml.edu With some modifications by Prof. Karen Daniels, Fall 2011 This work is licensed under the Creative Commons Attribution-ShareAlike License. To view a copy of this license, visit http://creativecommons.org/licenses/by- 1 sa/2.0/ or send a letter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA.
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§ 1.4 Nonregular languages ± For each possible language L, ² L. So is the smallest language. nd regular And is regular ² L Σ * . So Σ * is the largest language. nd * regular And Σ is regular ± Yet there are languages in between ese two extremes that are ot these two extremes that are not regular 2
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A nonregular language = { 0 n n n } B { 0 1 | n 0 } = { ε , 01, 0011, 000111, L } not regular is not regular ± Why? ² Q: how many bits of memory would a DFA need in order to recognize B? ² A: there appears to be no single number of bits that's big enough to work for every element of B ± Remember, the DFA needs to reject all strings that are not in B 3
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Other examples = { w 0 1} * n ) = n ) } ± C = { w {0,1} | n 0 (w) = n 1 (w) } ² Needs to count a potentially unbounded number of '0's. .. so nonregular ub e o0 s s oo e g u a ± D = { w {0,1} * | n 01 (w) = n 10 (w) } ² Needs to count a potentially unbounded number of '01' substrings. .. so ?? ± Need a technique for establishing nonregularity that is more formal and. .. less intuitive? 4
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Proving nonregularity o rove at a language is nonregular you ± To prove that a language is nonregular, you have to show that no DFA whatsoever recognizes the language ot just the DFA that is your best effort at ² Not just the DFA that is your best effort at recognizing the language ± The pumping lemma can be used to do that he pumping lemma says that every ± The pumping lemma says that every regular language satisfies the "regular pumping property " ( RPP ) ² Given this, if we can show that a language like B doesn't satisfy the RPP, then it's not regular ² B = { 0 n 1 n | n 0 } 5
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Pumping lemma, informally oughly: "if a regular ± Roughly: if a regular language contains any 'long' strings, then it contains infinitely many strings" ± Start with a regular nguage and suppose that language and suppose that some DFA M=(Q, Σ , δ ,q 0 ,F) for it has |Q|=10 states.
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This note was uploaded on 02/13/2012 for the course CS 91.304 taught by Professor Staff during the Fall '11 term at UMass Lowell.

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Chapter1Section4_F11 - 91.304 Foundations of (Theoretical)...

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