CSCI 2670 Introduction to Theory of Computing September 9, 2004

Agenda Yesterday Today Proved correspondence between DFA’s and RE’s Began an example converting a DFA to an RE Today Complete example Learn how to prove languages are not regular

Next week New method for describing languages Context-free grammars Read Section 2.1 (pages 91 – 101)

Converting a GNFA to a RE If the GNFA has two states, then the label connecting the states is the RE Otherwise, remove one state at a time without changing the language accepted by the machine until the GNFA has two states

Removing one state from a GNFA q2 q1 q3 a12a13a33*a32 q1’ q2’

Accounting for loops a22a23a33*a32 a11a13a33*a31 a12a13a33*a32 q1’

Example 1 q1 q2

Example 1 1 q1 q2 ε ε qs qt Step 1: Add two new states

Example 101*0 1 1 q1 q2 ε ε qs qt 1*0 Step 2: Remove q1

Example 101*0 q2 ε qs qt 1*0 1*0(101*0)* Step 3: Remove q2

Example So this DFA 1 q1 q2 Is equivalent to the regular expression 1*0(101*0)*

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 {0n1n | 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?

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

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 for each i  0, xyiz is in A |y| > 0, and |xy|  p p is called the pumping length

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

Proof idea x y z for each i  0, xyiz is in A |y| > 0, and |xy|  p

Using the pumping lemma We can use the pumping lemma to prove a language B is not regular Assume B is regular Show that the pumping lemma is not satisfied Proof by contradiction

Example B={101,101001,1010010001,…} The jth and (j+1)th 1’s are separated by j 0’s Assume B is regular and let w be any string in B Let w=xyz, where |y|>0 We cannot make any assumptions on the value of p We want to show that no matter what y is, there is some i for which xyizB

Example (cont.) There are 3 possibilities for y If y=0b then xy2z  B y has no 1’s y = 0b for some b > 0 y has one 1 y = 0b10c for some b,c  0 y has two or more 1’s y = 0b10c1v for some b,c  0 and some v* If y=0b then xy2z  B The number of 1’s between the last 1 in x and the fist 1 in z has increased

Example (cont.) There are 3 possibilities for y y has no 1’s y = 0b for some b > 0 y has one 1 y = 0b10c for some b,c  0 y has two or more 1’s y = 0b10c1v for some b,c  0 and some v* If y = 0b10c then xy3z  B There are two consecutive 1’s with b+c 0’s between them

Example (cont.) There are 3 possibilities for y y has no 1’s y = 0b for some b > 0 y has one 1 y = 0b10c for some b,c  0 y has two or more 1’s y = 0b10c1v for some b,c  0 and some v* If y=0b10c1v then xy2z  B There are c 0’s between a pair of consecutive 1’s twice

Example (cont.) Therefore, the pumping lemma is not satisfied so B is not a regular language For every p and every y with |y| > 0, we can show that xyiz  B for some i

Have an excellent weekend