1. CSCI 2670 Introduction to Theory of Computing
September 6, 2005

2. Announcements Homework due next Tuesday (9/13)
1.7(d,h), 1.16b, 1.21b (show the GNFA steps),1.19a, 1.46c, 1.55(c,f,j)
Old text numbers 1.5(d,g), 1.12b, 1.16b, 1.14a, 1.23d, 1.38(same instructions on (0010*1*, , and Σ*)
If you ask me a question on a homework problem by , please me the problem in question
Quiz tomorrow
Section 1.2 & Formal definition of RE

3. Agenda Last week Today This week Section 1.2 NFA’s
Regular expression intro (Section 1.3)
Today
Continue regular expressions
This week
Sections 1.3 and 1.4

4. RE inductive definition
R is a regular expression if R is
a for some a   ε 
R1  R2 where R1 and R2 are both regular expressions
R1  R2 where R1 and R2 are both regular expressions
(R1*) where R1 is a regular expression
Abuse of notation. These should be sets!

5. RE’s and regular languages
Theorem: A language is regular if and only if some regular expression describes it.
i.e., every regular expression has a corresponding DFA and vice versa

6. RE’s and regular languages
Lemma: If a language is described by a regular expression, then it is regular.
find an NFA corresponding to any regular expression
use inductive definition of RE’s

7. 1. R=a for some a a q1 q2
N = {{q1,q2},,,q1,{q2}} where (q1,a)={q2} and (r,x)= whenever r=q2 or x≠a

8. 2. R=ε q1
N = {{q1},,,q1,{q1}} where (q1,x)= for all x

9. 3. R= q1
N = {{q1},,,q1,} where (q1,x)= for all x

10. Remaining constructions
R = R1R2
R = R1R2
R = R1*
These were all shown to be regular operators
We know we can construct NFA’s for R provided they exist for R1 and R2

11. Example
R = 1
R = (01)1
R3 = 01 1 ε
R1 = 0 1
R2 = 1 1
R = Σ1 ε

12. Example2 R = 1(0ε)* ε ε R = 1(0ε)* 1 R1 = 1 0,1 R3 = * ε R2 = 0εε
R = 1(0ε)* 1 ε
0,1

13. Equivalence of RE’s and DFA’s
We have seen that every RE has a corresponding NFA
Therefore, every RE has a corresponding DFA
I.e, every RE describes a regular language
We need to show that every regular language can be described by a RE
Begin by converting all DFA’s into GNFA’s
Generalized Non-deterministic Finite Automata

14. GNFA’s A GNFA is an NFA with the following properties:
The start state has transition arrows going to every other state, but no arrows coming in from any other state
There is exactly one accept state and there is an arrow from every other state to this state, but no arrows to any other state from the accept state
The start state is not the accept state

15. GNFA’s (continued)
Except for the start and accept states, one arrow goes from every state to every other state and also from each state to itself
Instead of being labeled with symbols from the alphabet, transitions are labeled with regular expressions

16. Example GNFA 
01  1 10

17. Equivalence of DFA’s and RE’s
First show every DFA can be converted into a GNFA that accepts the same language
Then show that any GNFA has a corresponding RE that accepts the same language

18. Converting a DFA into a GNFA
Add two new states
New start state with an ε jump to the original DFA’s start state
New accept state with an ε jump from each of the original DFA’s accept states
This new state will be the only accept state
All transition labels with multiple labels are relabeled with the union of the previous labels
All pairs of states without transitions get a transition labeled 

19. Converting a DFA to a GNFA1 q1 q2 q3 q4
0,1 qs qt ε ε
Add two new states

20. Converting a DFA to a GNFAq2 1 qs qt q1 1 ε ε q3 q4
0,1
01
0,1
01
All transition labels with multiple labels are relabeled with the union of the previous labels

21. Converting a DFA to a GNFAq2 1 qs qt q1 1 ε ε q3 q4
01
01
All pairs of states without transitions get a transition labeled 

22. Converting a DFA to a GNFAq2 1 qs qt q1 1 ε ε q3 q4
01
01
The resulting state diagram is a GNFA
All GNFA properties are satisfied

23. Converting a DFA to a GNFAq2 1 qs qt q1 1 ε ε q3 q4
01
01
No step changed the strings accepted by the machine

24. 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

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