1 / 22

Nonregularity Proofs

Nonregularity Proofs. Regular Languages: Grand Unification. (Parallel Simulation) (Rabin and Scott’s work). (Collapsing graphs; Structural Induction) (S. Kleene’s work). (Construction) (Solving linear equations). Role of various representations for Regular Languages.

neal
Download Presentation

Nonregularity Proofs

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Nonregularity Proofs Pumping Lemma

  2. Regular Languages: Grand Unification (Parallel Simulation) (Rabin and Scott’s work) (Collapsing graphs; Structural Induction) (S. Kleene’s work) (Construction) (Solving linear equations) Pumping Lemma

  3. Role of various representations for Regular Languages • Closure under complemention. (DFAs) • Closure under union, concatenation, and Kleene star. (NFA-ls, Regular expression.) • Consequence: Closure under intersection by De Morgan’s Laws. • Relationship to context-free languages. (Regular Grammars.) • Ease of specification. (Regular expression.) • Building tokenizers/lexical analyzers. (DFAs) Pumping Lemma

  4. Consider pairs of strings: IfL were regular,then there exists a DFA M accepting L with the following property: Pumping Lemma

  5. CLAIM: JUSTIFICATION: Otherwise, from the definition of DFA, which contradicts the earlier conclusion. In order to satisfy the machine M must have a unique state for every i. Thus, M must have infinite number of states, if L is regular. This violates the definition of DFA. So, L must be non-regular. Pumping Lemma

  6. Using Closure Properties • Regular languages are closed under set-intersection. • Note that regularity is a property of a collection, and not a property of an individual string in the collection. L1=bit strings with even parity L2=bit strings with number of 1’s divisible by 3 L=bit strings with number of 1’s a multiple of 6 Pumping Lemma

  7. If R is a regular language and C is context-free, then may not be regular. • Proof: • Show that • is not regular. • Proof: If L were regular, ought to be regular. However, is known to be non-regular. Hence, L cannot be regular. Pumping Lemma

  8. Prelude to Pumping Lemma • Is 46551 divisible by 46? • Is 46554 divisible by 46? • Is 46552 divisible by 46? Necessary vs sufficient condition Pumping Lemma

  9. Pumping Lemma for Regular Languages • It is a necessary condition. • Every regular language satisfies it. • If a language violates it, it is not regular. • RL => PL not PL => not RL • It is not a sufficient condition. • Not every non-regular language violates it. • not RL =>? PL or not PL (no conclusion) Pumping Lemma

  10. Basic Idea: b a q0 a q1 b b a q2 q3 a,b Pumping Lemma

  11. Note, So, Pumping Lemma

  12. Fundamental Observation • Given a “sufficiently” long string, the states of a DFA must repeat in an accepting computation. These cycles can then be used to predict (generate) infinitely many other strings in (of) the language. Pigeon-Hole Principle Pumping Lemma

  13. Pumping Lemma • Let L be a regular language that is accepted by a DFA M with k states. Let z be any string in L with . Then z can be decomposed as uvw with Pumping Lemma

  14. For all sufficiently long strings (z) There exists non-null prefix (uv) and substring (v) For all repetitions of the substring (v), we get strings in the language. Pumping Lemma

  15. Proving non-regularity • If there exists an arbitrarily long string s L, and for each decomposition s = uvw, there exists an i such that , then L is non-regular. Negation of the necessary condition: Pumping Lemma

  16. Examples Applying Pumping Lemma Pumping Lemma

  17. Proof by contradiction: • Let be accepted by a k-state DFA. • Choose • For all prefixes of length • show there exists such that • i.e., Pumping Lemma

  18. Choose (For this specific problem happens to be independent of j, but that need not always be the case.) • is non-regular because it violates the necessary condition. Pumping Lemma

  19. Proof : (For this example, choice of initial string is crucial.) • For this choice of s, the pumping lemma cannot generate a contradiction! • However, let instead. Pumping Lemma

  20. For • Thus, by pumping the substring containing a’s 0 times (effectively deleting it), the number of a’s can be made smaller than the number of b’s. • So, by pumping lemma, L is non-regular. Pumping Lemma

  21. Proof by contradiction: • If is regular, then so is , the complement of • But which is known to be non-regular. • So, cannot be regular. Pumping Lemma

  22. Summary: Proof Techniques • Counter Examples • Constructions/Simulations • Induction Proofs • Impossibility Proofs • Proofs by Contradiction • Reduction Proofs : Closure Properties Pumping Lemma

More Related