Lecture 8 From NFA to Regular Language

1. Lecture 8 From NFA to Regular Language

2. Induction on k= # of states other than initial and final states • K=0 a a* b c*a(d+bc*a)* a c d

3. k > 1 ac*e c ac*d a e bc*e d b bc*d

4. Example 1 0 1 0+10 1 0 0 Why? 10 0 1 1 (0+10)*0(1+10(0+10)*0)* 0+10 1 0 1 0+10 (0+10)*0 0 10(1+10)*0 How to get in the final state? 10 Cycles at the final state.

5. Example 2 0 1 0 1 0 0 1 ε 0 0 ε 1 1 1 1 0 1 0 0 0 0 0 1 1 1 1 Solution 1 (0+10)*0(1+10(0+10)*0)* + 0*(1+01*1)(00*(1+01*1))*

6. 0 0 1 0 01* ε 0 0 1 1 1+01*1 ε ε Solution 2 0+(1+01*1)0 (0+(1+01*1)0)*(01*+1+01*1) 01*+(1+01*1)

7. 0+10 1 0 1 0 ε+1 ε 0 10 1 1 1 ε Solution 3 0+10+01*10 (0+10+01*10)*(1+01*(ε+1)) 1+01*(ε+1) Different ways may give different regular expressions for the same language.

8. Theorem • A language is regular if and only if it can be accepted by an NFA if and only if it can be accepted by a DFA.

9. Closure Properties • If A is regular, so is its complement A. • If A and B are regular, then A∩B, A\B, A U B, AB are regular.

10. Quotient • L1/L2 = {x | there exists y in L2 s.t. xy in L1} • If L1 is regular, so is L1/L2 . L1=L(M), M=(Q,Σ,δ,s,F) L1/L2 = L(M’), M’ = (Q,Σ,δ,s, F’) where F’={q in Q | there exists y in L2 s.t. δ(q,y) in F}

11. Example L2 = 0*1+0*11 L1 = L(M) 0 0 1 0 1 1 0 1