Chapter 11 - 1

1. Chapter 11 - 1 Regular Languages

2. Section 11.1 Regular Languages • Problem: Suppose the input strings to a program must be strings over the alphabet {a, b} that contain exactly one substring bb. In other words, the strings must be of the form xbby, where x and y are strings over {a, b} that do not contain bb, x does not end in b, and y does not begin with b. In a few minutes, we’ll see how to describe the strings formally.

3. regular language • A regular language over alphabet A is a language constructed by the following rules: • Фand {٨} are regular languages. • {a} is a regular language for all a ∈ A. • If M and N are regular language, then so are M ⋃ N, MN, and M*.

4. Example • Let A = {a, b}. Then the following languages are a sampling of the regular languages over A: Ф, {٨}, {a}, {b}, {a, b}, {ab}, {a}* = {٨, a, aa, aaa, …, an, …}.

5. regular expression • A regular expression over alphabet A is an expression constructed by the following rules: • Фand ٨ are regular expressions. • a is a regular expression for all a∈ A. • If R and S are regular expressions, then so are (R), R + S, R•S, and R*. The hierarchy in the absence of parentheses is, * (do it first), •, + (do it last). Juxtaposition will be used in place of •.

6. Example • Let A = {a, b}. Then the following expressions are a sampling of the regular expressions over A: Ф,٨, a, b, ab, a + ab, (a + b)*.

7. Regular expressions represent regular languages • Regular expressons represent regular languages by the following correspondence, where L(R) denotes the regular language of the regular expression R. L(Ф) = Ф L(٨) = {٨}, L(a) = {a} for all a ∈ A, L(R + S) = L(R) ⋃ L(S), L(RS) = L(R)L(S), L(R*) = L(R)*.

8. Example • The regular expression ab + a* represents the following regular language: L(ab + a*) = L(ab) ⋃ L(a*) = L(a)L(b) ⋃ L(a)* = {a}{b} ⋃ {a}* = {ab} ⋃ {٨, a, aa, aaa, …, an, …} ={ab, ٨, a, aa, aaa, …, an, …}.

9. Example • The regular expression (a + b)* represents the following regular language: L((a + b)*) = (L(a + b))* = {a, b}*, the set of all possible strings over {a, b}. • Back to the Problem: Suppose the input strings to a program must be strings over the alphabet {a, b} that contain exactly one substring bb. In other words, the strings must be of the form xbby, where x and y are strings over {a, b} that do not contain bb, x does not end in b, and y does not begin with b. How can we describe the set of inputs formally? • Solution: let x = (a + ba)* and y = (a + ab)*.

10. Quizzes • Quiz. Find a regular expresson for {abn | n ∈ N} ∪ {ban | n ∈ N}. • Answer. ab* + ba*. • Quiz. Use a sentence to describe the language of (b + ab)*(٨ + a). • Answer. All strings over {a, b} whose substrings of a’s have length 1.

11. The Algebra of Regular Expressions • Equality: Regular expressions R and S are equal, written R = S, when L(R) = L(S). • Examples. a + b = b + a, a + a = a, aa* = a*a, ab ≠ ba.

12. Properties of +, • and closure + is commutative, associative, Фis identity for +, and R + R = R. • is associative, ٨ is identity for • and Ф is zero for • • distributes over + (closure properties) Ф* = ٨* = ٨. R* = R*R* = (R*)* = R + R*. R* = ٨ + R* = ٨ + RR* = (٨ + R)* = (٨ + R)R*. R* = (R + R2 + …+ Rk)* = ٨ + R + R2 + …+ Rk-1 + RkR* for any k ≥ 1. R*R = RR*. (R + S)* = (R* + S*)* = (R*S*)* = (R*S)*R* = R*(SR*)*. R(SR)* = (RS)*R. (R*S)* = ٨ + (R + S)*S and (RS*)* = ٨ + R(R + S)*. Proof: All properties can be verified by showing that the underlying regular languages are equal as sets. QED.

13. Quizzes • Quiz. Explain each inequality. (1). (a + b)* ≠ a* + b*. (2) (a + b)* ≠ a*b*. • Answers. (1) ab ∈ L((a + b)*) - L(a* + b*). (2) ba ∈ L((a + b)*) - L(a*b*). • Quiz. Simplify the regular expression aa(b* + a) + a(ab* + aa). • Answer. aa(b* + a) + a(ab* + aa) = aa(b* + a) + aa(b* + a) • distributes over + = aa(b* + a) R + R = R.

14. Example/Quiz • Show that (a + aa)(a + b)* = a(a + b)*. • Proof: (a + aa)(a + b)* = (a + aa)a*(ba*)* (R + S)* = R*(SR*)* = a(٨ + a)a*(ba*)* R = R٨ and • distributes over + = aa*(ba*)* (٨ + R)R* = R* = a(a + b)* (R + S)* = R*(SR*)* QED.

15. Example/Quiz • Show that a*(b + ab*) = b + aa*b*. • Proof: a*(b + ab*) = (٨ + aa*)(b + ab*) R* = ٨ + RR* = b + ab* + aa*b + aa*ab* • distributes over + = b + (ab* + aa*ab*) + aa*b + is commutative and associative = b + (٨ + aa*)ab* + aa*b R = R٨ and • distributes over + = b + a*ab* + aa*b R* = ٨ + RR* = b + aa*b* + aa*b R*R = RR* = b + aa*(b* + b) • distributes over + = b + aa*b*. R* = R* + R QED.

16. Example • Show that a* + abb*a = a* + ab*a. • Proof: Starting on the RHS, we have a* + ab*a = a* + a(٨ + bb*)a R* = ٨ + RR* = a* + aa + abb*a • distributes over + = (٨ + aa*) + aa + abb*a R* = ٨ + RR* = ٨ + (aa* + aa) + abb*a + is associative = ٨ + a(a* + a) + abb*a • distributes over + = ٨ + aa* + abb*a R* = R + R* = a* + abb*a R* = ٨ + RR* QED.

17. Example • Show that (a + aa + … + an)(a + b)* = a(a + b)* for all n ≥ 1. • Proof (by induction): If n = 1, the statement becomes a(a + b)* = a(a + b)*, which is true. If n = 2, the statement becomes (a + aa)(a + b)* = a(a + b)*, which is true by a previous example. Let n > 2 and assume the statement is true for 1 ≤ k < n. We need to prove the statement is true for n. The LHS of the statement for n is (a + aa + … + an)(a + b)* = a(a + b)* + (aa + … + an)(a + b)* • distributes over + = a(a + b)* + a(a + aa + … + an-1)(a + b)* • distributes over + = a(a + b)* + aa(a + b)* induction assumption = (a + aa)(a + b)* • distributes over + = a(a + b)* induction assumption The last line is the RHS of the statement for n. So the statement is true for n. Therefore, the statement is true for all n ≥ 1. QED.

18. Example/Quiz • Use regular algebra to show that R + (R + S)* = (R + S)*. • Proof: R + (R + S)* = R + [٨ + (R + S) + (R + S)(R+ S)*] R* = ٨ + R + R2 + ... + Rk-1 + RkR* = ٨ + (R + R) + S + (R + S)(R+ S)* + is associative = ٨ + R + S + (R + S)(R+ S)* R + R = R = ٨ + (R + S) + (R + S)(R+ S)* + is associative = (R+ S)* R* = ٨ + R + R2 + ... + Rk-1 + RkR* QED. • Take-home quiz: Use regular algebra to show that (a + ab)*a = a(a + ba)*.

19. The End of Chapter 11 -1