1. CSCI 2670 Introduction to Theory of Computing September 13, 2005

2. Announcement Homework due next Tuesday 9/20 –Use the pumping lemma to prove the following language is not regular A = {ww R | w  {a,b} * } –1.54, 2.1 c & d, 2.4 c & e, 2.6 b & d Note in 2.6 d, it is possible that i = j

3. Agenda Last week –Proved correspondence between regular languages and regular expressions This week –Learn how to prove a language is not regular –Introduce a new class of language –Introduce a new type of state machine

4. Nonregular languages So far, we have explored several ways to identify regular languages –DFA’s, NFA’s, GNFA’s, RE’s There are many nonregular languages –{0 n 1 n | n  0} –{101,101001,1010010001,…} –{w | w has the same number of 0s and 1s} How can we tell if a language is not regular?

5. Property of regular languages All regular languages can be generated by finite automata States must be reused if the length of a string is greater than the number of states If states are reused, there will be repetition

6. The pumping lemma Theorem: If A is a regular language, then there is a number p where, if s is any string in A of length at least p, then s may be divided into three pieces, s = xyz, satisfying the following conditions 1.for each i  0, xy i z is in A 2.|y| > 0, and 3.|xy|  p p is called the pumping length

7. Proof idea Pumping length is equal to the number of states in the DFA whose language is A –p = |Q| If A accepts a word w with |w| > p, then some state must be entered twice while processing w –Pigeonhole principle

8. Proof idea 1.for each i  0, xy i z is in A 2.|y| > 0, and 3.|xy|  p x y z

9. Using the pumping lemma We can use the pumping lemma to prove a language B is not regular Proof by contradiction –Assume B is regular with pumping length p –Find a string in w  B with |w| ≥ p –Show that the pumping lemma is not satisfied Show that any xyz satisfying the properties of the pumping lemma cannot be pumped You can choose a specific w, but you cannot choose a specific xyz!

10. Example B={wbbw | w  {a,b} * } Assume B is regular and p is the pumping length of B Consider the string w = a p bba p w  B and |w| ≥ p so the pumping lemma aplies –w = xyz, |xy| ≤ p, |y| > 0, xy i z  B for all i

11. Example Consider the string w = a p ba p w  B and |w| ≥ p so the pumping lemma aplies –w = xyz, |xy| ≤ p, |y| > 0, xy i z  B for all i Since |xy| ≤ p and w begins with xy, xy = a k for some k ≤ p –y = a j for some j = 1, 2, …, k Therefore xy 2 z = a p+j bba p  B –Pumping lemma is contradicted so B is not regular

12. Proof of Pumping Lemma Let A be any regular language Find DFA M=(Q, , ,q 0,F) with L(M)=A Let p=|Q| Let s=s 1 s 2 s 3 …s n be any string in A with |s| = n ≥ p –What if no such s is in A?

13. Proof (cont.) Let r 1, r 2, r 3, …, r n+1 be the sequence of states entered while processing s –r 1 = q 0 –r n+1  F –r i+1 =  (r i, s i )

14. Proof (cont.) Consider the first p+1 elements of this sequence –p+1 states must contain a repeated state Let r k be the first state to be repeated and let r t be the second occurance of this state –t  p+1

15. Proof (cont.) Let x=s 1 s 2 …s k-1, y=s k s k+1 …s t-1, z=s t s t+1 …s n –x takes M from r 1 to r k If k = 1, then x =  –y takes M from r k to r t –z takes M from r t to r n+1, which is an accept state Since r k and r t are the same state, M must accept xy i z for any i=0, 1, 2, …

16. Proof (cont.) Have we satisfied the conditions of the theorem? 1.for each i  0, xy i z is in A 2.|y| > 0, and 3.|xy|  p Have we satisfied the conditions of the theorem? 1.for each i  0, xy i z is in A Yes 2.|y| > 0, and Yes since t > k and y=s k s k+1 …s t-1 3.|xy|  p Yes since t  p+1 and xy = s 1 s 2 …s t-1

17. Regular languages -- Summary Let R be any language. The following are equivalent 1.R is a regular language 2.R = L(M) for some finite automata M, where M is a DFA, an NFA, or a GNFA 3.R is describe by some regular expression If R can be shown not to have a finite pumping length, then R is not regular