CSCI 2670 Introduction to Theory of Computing

1. CSCI 2670Introduction to Theory of Computing September 7, 2004

2. Agenda • Last week • Section 1.2 • NFA’s • Today • Start Section 1.3 • Regular expressions • This week • Sections 1.3 and 1.4 • May not complete Section 1.4

3. Announcement • Homework due next Tuesday (9/14) • 1.5e, 1.12a, 1.16a (show the GNFA steps), 1.17b, 1.39 • Quiz tomorrow • Section 1.2 & Formal definition of RE

4. Regular expressions (RE’s) • So far we have had to describe languages either with finite automata or with words • Potentially clumsy or imprecise • Today we learn precise expression to describe regular languages • Example: All strings with at least one 1 becomes *{1}*, or more simply *1*

5. Where have you seen RE’s? • Grep • Awk • Perl • Search expressions within emacs or vi

6. 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*) R1 is a regular expression

7. Examples • 0*10*10* • {w | w contains exactly two 1’s} • 11 • {w | w contains two consecutive 1’s} • 1(0ε)1 • {w | w contains two 1’s separated by at most one 0} • (0ε)(1ε) • {0,1,01,ε}

8. 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

9. 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

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

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

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

13. 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

14. 0 R1 = 0 1 R2 = 1 R = (01)1 0 0 ε R3 = 01 ε 1 ε 1 ε ε 1 ε Example • R = 1 • R = (01)1

15. 1 R1 = 1 0,1 R3 = * ε R = 1(0ε)* 0 R2 = 0ε ε 0 ε ε ε 0,1 1 ε ε ε ε ε ε Example2 • R = 1(0ε)*

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

17. 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

18. 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