Regular Expressions Kleene s Theorem Equation-based alternate construction. Regular Expressions. Deepak D Souza

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1 Regular Expressions Deepak D Souza Department of Computer Science and Automation Indian Institute of Science, Bangalore. 11 August 2011

2 Outline 1 Regular Expressions 2 Kleene s Theorem 3 Equation-based alternate construction

3 Examples of Regular Expressions Expressions built from a, b, ǫ, using operators +,, and. (a + b ) c Strings of only a s or only b s, followed by a c. (a + b) abb(a + b) contains abb as a subword. (a + b) b(a + b)(a + b) 3rd last letter is a b. (b ab a) b

4 Examples of Regular Expressions Expressions built from a, b, ǫ, using operators +,, and. (a + b ) c Strings of only a s or only b s, followed by a c. (a + b) abb(a + b) contains abb as a subword. (a + b) b(a + b)(a + b) 3rd last letter is a b. (b ab a) b Even number of a s.

5 Examples of Regular Expressions Expressions built from a, b, ǫ, using operators +,, and. (a + b ) c Strings of only a s or only b s, followed by a c. (a + b) abb(a + b) contains abb as a subword. (a + b) b(a + b)(a + b) 3rd last letter is a b. (b ab a) b Even number of a s. Ex. Give regexp for Every 4-bit block of the form w[4i,4i + 1,4i + 2,4i + 3] has even parity.

6 Examples of Regular Expressions Expressions built from a, b, ǫ, using operators +,, and. (a + b ) c Strings of only a s or only b s, followed by a c. (a + b) abb(a + b) contains abb as a subword. (a + b) b(a + b)(a + b) 3rd last letter is a b. (b ab a) b Even number of a s. Ex. Give regexp for Every 4-bit block of the form w[4i,4i + 1,4i + 2,4i + 3] has even parity. ( ) (ǫ )

7 Formal definitions Syntax of regular expresions over an alphabet A: where a A. r ::= a r + r r r r Semantics: associate a language L(r) A with regexp r. L( ) = {} L(a) = {a} L(r + r ) = L(r) L(r ) L(r r ) = L(r) L(r ) L(r ) = L(r).

8 Formal definitions Syntax of regular expresions over an alphabet A: where a A. r ::= a r + r r r r Semantics: associate a language L(r) A with regexp r. L( ) = {} L(a) = {a} L(r + r ) = L(r) L(r ) L(r r ) = L(r) L(r ) L(r ) = L(r). Question: Do we need ǫ in syntax?

9 Formal definitions Syntax of regular expresions over an alphabet A: where a A. r ::= a r + r r r r Semantics: associate a language L(r) A with regexp r. L( ) = {} L(a) = {a} L(r + r ) = L(r) L(r ) L(r r ) = L(r) L(r ) L(r ) = L(r). Question: Do we need ǫ in syntax? No. ǫ.

10 Example: Semantics of regexp (a + b ) c + a b c

11 Example: Semantics of regexp (a + b ) c + {a} a b {b} c {c}

12 Example: Semantics of regexp (a + b ) c + {ǫ, a, aa,...} {ǫ, b, bb,...} {a} a b {b} c {c}

13 Example: Semantics of regexp (a + b ) c {ǫ, a, b, aa, bb,...} + {ǫ, a, aa,...} {ǫ, b, bb,...} {a} a b {b} c {c}

14 Example: Semantics of regexp (a + b ) c {c, ac, bc, aac, bbc,...} {ǫ, a, b, aa, bb,...} + {ǫ, a, aa,...} {ǫ, b, bb,...} {a} a b {b} c {c}

15 Kleene s Theorem: RE = DFA Class of languages defined by regular expressions coincides with regular languages. Proof RE DFA: Use closure properties of regular languages. DFA RE:

16 DFA RE: Kleene s construction Let A = (Q,s,δ,F) be given DFA. Define L pq = {w A δ(p,w) = q}. Then L(A) = f F L sf. For X Q, define L X pq = {w A δ(p,w) = q via a path that stays in X except for first and last states} X p q Then L(A) = f F LQ sf.

17 DFA RE: Kleene s construction r X p q Advantage: L X {r} pq = L X pq + L X pr (L X rr) L X rq.

18 DFA RE: Kleene s construction (2) Method: Begin with L Q sf for each f F. Simplify by using terms with strictly smaller X s: L X {r} pq For base terms, observe that L {} pq = = L X pq + L X pr (L X rr) L X rq. { {a δ(p,a) = q} if p q {a δ(p,a) = q} {ǫ} if p = q. Exercise: convert NFA/DFA s below to RE s: a, b b a b b s f a

19 DFA RE: Kleene s construction (2) Method: Begin with L Q sf for each f F. Simplify by using terms with strictly smaller X s: L X {r} pq For base terms, observe that L {} pq = = L X pq + L X pr (L X rr) L X rq. { {a δ(p,a) = q} if p q {a δ(p,a) = q} {ǫ} if p = q. Exercise: convert NFA/DFA s below to RE s: a, b b a b b s f e a o

20 DFA RE using system of equations Aim: to construct a regexp for L q = {w A δ(q,w) F }. Note that L(A) = L s. Example: b a b Set up equations to capture L q s: x e = b x e + a x o x o = a x e + b x o + ǫ. Solution is a RE for each x, such that languages denoted by LHS and RHS RE s coincide. a

21 DFA RE using system of equations Aim: to construct a regexp for L q = {w A δ(q,w) F }. Note that L(A) = L s. Example: b a b Set up equations to capture L q s: e a o x e = b x e + a x o x o = a x e + b x o + ǫ. Solution is a RE for each x, such that languages denoted by LHS and RHS RE s coincide.

22 Solving a system of equations L q s are a solution to the system of equations In general there could be many solutions to equations. Consider x = A x. In this case, L q s can be seen to the least solution to the equations.

23 Least solution to the system of equations Equations can be viewed as: [ b a [ x e x o ] = a b ] [ x e x o ] + [ ǫ ] System of equations have the general form: X = AX T + B. Claim: A B is the least solution to the equations above. [See Kozen, Supplementary Lecture A]. Definition of A : [ ] [ a b (a + bd = c) (a + bd c) bd ] c d (d + ca b) ca (d + ca b)