1. CSCI 2670 Introduction to Theory of Computing
September 9, 2004

2. 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

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

4. 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

5. Removing one state from a GNFAq2 q1 q3
a12a13a33*a32
q1’
q2’

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

7. Example 1 q1 q2

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

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

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

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

12. 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, ,…}
{w | w has the same number of 0s and 1s}
How can we tell if a language is not regular?

13. 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

14. 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

15. 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

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

17. 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

18. Example
B={101,101001, ,…}
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

19. 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

20. 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

21. 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

22. 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

23. Have an excellent weekend