1. TK PrasadPumping Lemma1 Nonregularity Proofs

2. TK PrasadPumping Lemma2 Grand Unification Regular Languages: Grand Unification (Parallel Simulation) (Rabin and Scott’s work) (Collapsing graphs; Structural Induction) (S. Kleene’s work) (Construction) (Solving linear equations)

3. TK PrasadPumping Lemma3 Role of various representations for Regular Languages Closure under complemention. (DFAs) Closure under union, concatenation, and Kleene star. (NFA- s, 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)

4. TK PrasadPumping Lemma4 Consider pairs of strings: M If L were regular, then there exists a DFA M accepting L with the following property:

5. TK PrasadPumping Lemma5 JUSTIFICATION: Otherwise, from the definition of DFA, CLAIM: which contradicts the earlier conclusion. In order to satisfy the machine M must have a unique state for every i. infinite Thus, M must have infinite number of states, if L is regular. This violates the definition of DFA. So, L must be non-regular.

6. TK PrasadPumping Lemma6 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

7. TK PrasadPumping Lemma7 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. If R is a regular language and C is context-free, then may not be regular. Proof:

8. TK PrasadPumping Lemma8 Prelude to Pumping Lemma Is 46551 divisible by 46? Is 46554 divisible by 46? Is 46552 divisible by 46? Necessary vs sufficient condition

9. TK PrasadPumping Lemma9 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)

10. TK PrasadPumping Lemma10 Basic Idea: q0 b a a a,b b b a q2q3 q1

11. TK PrasadPumping Lemma11 Note, So,

12. TK PrasadPumping Lemma12 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

13. TK PrasadPumping Lemma13 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

14. TK PrasadPumping Lemma14 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.

15. TK PrasadPumping Lemma15 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:

16. TK PrasadPumping Lemma16 Examples Applying Pumping Lemma

17. TK PrasadPumping Lemma17 Proof by contradiction: Let be accepted by a k-state DFA. Choose For all prefixes of length show there exists such that i.e.,

18. TK PrasadPumping Lemma18 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.

19. TK PrasadPumping Lemma19 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.

20. TK PrasadPumping Lemma20 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.

21. TK PrasadPumping Lemma21 Proof by contradiction: –If is regular, then so is, the complement of – But which is known to be non-regular. –So, cannot be regular.

22. TK PrasadPumping Lemma22 Summary: Proof Techniques Counter Examples Constructions/Simulations Induction Proofs Impossibility Proofs Proofs by Contradiction Reduction Proofs : Closure Properties