Regular operations

1. Regular operations Sipser 1.1 (pages 44 – 47)

2. Building languages • If L is a language, then its complement is L’ = {w | w ∉ L} • Let A = {w | wis a string of 0s and 1s containing an odd number of 1s}. What is A' ? • Let B = {w | wis a string of over {a,b} that starts and ends with the same symbol}. What is B' ? L’ L

3. Which complements are regular? • A = {w | wis a string of 0s and 1s containing an odd number of 1s} • What about A’? • B = {w | wis a string over {a,b} that starts and ends with the same symbol} • What about B’?

4. Our first theorem • Theorem: The class of regular languages is closed under the complement operator • How do we prove it?

5. Our first proof • Theorem: The class of regular languages is closed under the complement operation • Proof: Let A be a regular language. By definition, there exists a finite automaton M=(Q, Σ, δ, q0, F) that recognizes A=L(M). Since N=(Q, Σ, δ, q0, F’) is a finite automaton that recognizes L(N)=A’, A’ is a regular language.

6. Union and Intersection • Let A and B be languages • Define their union A∪B = { w | w∈A orw∈B} • Define their intersection A∩B = { w | w∈Aand w∈B}

7. Example: union • A = {w | wis a string over {a,b} containing an odd number of as} • B = {w | wis a string over {a,b} that starts and ends with the same symbol} • What is A∪B ? • Is A∪B regular?

8. Can we build a recognizer for the union from previous machines?

9. Now do it in general… our second theorem • Theorem: The class of regular languages is closed under the union operation

10. The proof • Theorem: The class of regular languages is closed under the union operation • Let A and B be regular languages. By definition, there exists finite automata M1 =(Q1, Σ, δ1, q1, F1) and M2 =(Q2, Σ, δ2, q2, F2) that recognize A=L(M1) and B=L(M2), respectively. What’s left? Build a finite automaton M that recognizes L(M) = A∪B.

11. What about intersection? • Theorem: The class of regular languages is closed under the intersection operation