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Theory of Computation Unit 2
- 1. UNIT II REGULAR EXPRESSIONS AND LANGUAGES Mrs.D.Jena Catherine Bel, Assistant Professor, CSE, Velammal Engineering College Mrs.D.Jena Catherine Bel, AP/CSE, VEC 1
- 2. UNIT II REGULAR EXPRESSIONS AND LANGUAGES Regular Expressions – FA and Regular Expressions – Proving Languages not to be regular – Closure Properties of Regular Languages – Equivalence and Minimization of Automata. Mrs.D.Jena Catherine Bel, AP/CSE, VEC 2
- 5. Regular Expressions • The algebraic notations describes exactly the same language as finite automata • The language described by regular expression or a finite automata is known as regular languages • The regular expression operators are union, concatenation and closure Mrs.D.Jena Catherine Bel, AP/CSE, VEC 5
- 6. Let Regular expression R, then Regular language L(R) defined as follows: • L(Φ)= Φ • L()={} Let Alphabet ∑={a,b} • If R=a then L(a)={a} • If R=b then L(b)={b} Mrs.D.Jena Catherine Bel, AP/CSE, VEC 6
- 7. Example RE Representation RL a+b L(a)U L(b) -> {a} U {b} {a,b} ab+bc L(ab)U L(bc) -> {ab} U {bc} {ab,bc} (a+b).c L(a+b).L(c) -> {a,b}.{c} {ac,bc} a(0+1)b L(a).L(0+1).L(b) {a}.{0,1}.{b} {a0b,a1b} (a+b)(0+1) L(a+b).L(0+1) ->{a,b}.{0,1} {a0,a1,b0,b1} Mrs.D.Jena Catherine Bel, AP/CSE, VEC 7
- 8. Operators of Regular expression • Union • L = {01 , 10} • M = {ε ,00, 11} • L U M = {ε ,00, 11,01,10} • Concatenation • L = {01 , 10} • M = {ε ,00, 11} • L M = {01, 10, 0100, 1000, 0111, 1011} Mrs.D.Jena Catherine Bel, AP/CSE, VEC 8
- 9. • Closure • L = {01 , 10} • L * = {ε, 01, 10, 0101,010101,0110,1001,…….} • L+ = {01, 10, 0101,010101,0110,1001,…….} Mrs.D.Jena Catherine Bel, AP/CSE, VEC 9
- 10. Example • Give the regular expression for set of all strings starting with 0. • Give the regular expression for set of all strings ending with 0. • Mrs.D.Jena Catherine Bel, AP/CSE, VEC 10
- 11. Example • Give the regular expression for set of all strings ending with 00. • Give the regular expression for set of all strings starting with 0 and ending with 1 Mrs.D.Jena Catherine Bel, AP/CSE, VEC 11
- 18. FA to RE - Methods 1.Using Rij formula (Transitive closure method) 2. State Elimination Method 3. Arden’s Method Mrs.D.Jena Catherine Bel, AP/CSE, VEC 18
- 19. Rij Method • DFA states are defined as {1,2,3,4,. . . n} • k={0,1,2,3,4,. . . n} • Rij Formula : 𝐑𝐢𝐣 (𝐤) = 𝐑𝐢𝐣 (𝐤−𝟏) + 𝐑𝐢𝐤 𝐤−𝟏 𝐑𝐤𝐤 𝐤−𝟏 ∗ 𝐑𝐤𝐣 (𝐤−𝟏) • i is the initial state (1) , j is the final state ꞓ F • The required RE is sum of union of 𝑹𝒊𝒋 (𝒏) such that j is the accepting state 𝑹𝒊𝒋 (𝒏) is computed by, 𝑹𝒊𝒋 (𝒌=𝟎) , 𝑹𝒊𝒋 (𝒌=𝟏) . . . 𝑹𝒊𝒋 (𝒌=𝒏) Mrs.D.Jena Catherine Bel, AP/CSE, VEC 19
- 20. Properties of Regular Expression • Commutative: • R+S = S+R • Associative: • (R+S)+T = R+(S+T) • (RS)T = R(ST) • Identity: • R+Φ = R • R = R = R • Annihilator: • ΦR = RΦ = Φ ◦ Distributive: ◦ R(S+T) = RS + RT ◦ (R+S)T = RT+ST ◦ Idempotent: R+ R = R ◦ Involving Kleene closures: ◦ (R*)* = R* ◦ Φ* = ◦ * = ◦ R+ =RR* ◦ R? = +R ◦ +R* = R* Mrs.D.Jena Catherine Bel, AP/CSE, VEC 20
- 21. DFA to RE Ex-1 • DFA: Start state =1 Final state=3 • Total Number of nodes (n) = 3 & k={0,1,2,3} • The required regular expression is 1 2 3 a b 𝑹𝒊𝒋 (𝒏) 𝑹𝟏𝟑 (𝟑) 𝐑𝐢𝐣 (𝐤) = 𝐑𝐢𝐣 (𝐤−𝟏) + 𝐑𝐢𝐤 𝐤−𝟏 𝐑𝐤𝐤 𝐤−𝟏 ∗ 𝐑𝐤𝐣 (𝐤−𝟏) i.e i=1 , j=3 ,k=3 𝐑𝟏𝟑 (𝟑) = 𝐑𝟏𝟑 (𝟐) + 𝐑𝟏𝟑 𝟐 𝐑𝟑𝟑 𝟐 ∗ 𝐑𝟑𝟑 (𝟐) Mrs.D.Jena Catherine Bel, AP/CSE, VEC 21
- 22. Computation of K=0 1 2 3 a b 𝑹𝟏𝟏 (𝟎) 𝑹𝟏𝟑 (𝟎) 𝑹𝟏𝟐 (𝟎) 𝑹𝟐𝟏 (𝟎) 𝑹𝟐𝟐 (𝟎) 𝑹𝟐𝟑 (𝟎) 𝑹𝟑𝟏 (𝟎) 𝑹𝟑𝟐 (𝟎) 𝑹𝟑𝟑 (𝟎) Mrs.D.Jena Catherine Bel, AP/CSE, VEC 22
- 23. Computation of K=0 - cont. 1 2 3 a b 𝑹𝟏𝟏 (𝟎) = 𝑹𝟏𝟑 (𝟎) = Φ 𝑹𝟏𝟐 (𝟎) = 𝒂 𝑹𝟐𝟏 (𝟎) = Φ 𝑹𝟐𝟐 (𝟎) = 𝑹𝟐𝟑 (𝟎) = 𝒃 𝑹𝟑𝟏 (𝟎) = Φ 𝑹𝟑𝟐 (𝟎) = Φ 𝑹𝟑𝟑 (𝟎) = Mrs.D.Jena Catherine Bel, AP/CSE, VEC 23
- 24. Computation of K=1 𝑹𝟏𝟏 (𝟏) = 𝑹𝟏𝟏 (𝟎) + 𝑹𝟏𝟏 𝟎 (𝑹𝟏𝟏 𝟎 ) ∗ 𝑹𝟏𝟏 𝟎 = + ()∗ = 𝑹𝟏𝟑 (𝟏) = 𝑹𝟏𝟑 (𝟎) + 𝑹𝟏𝟏 𝟎 (𝑹𝟏𝟏 𝟎 ) ∗ 𝑹𝟏𝟑 𝟎 =Φ+ ()∗Φ=Φ 𝑹𝟏𝟐 (𝟏) = 𝑹𝟏𝟐 (𝟎) + 𝑹𝟏𝟏 𝟎 (𝑹𝟏𝟏 𝟎 ) ∗ 𝑹𝟏𝟐 𝟎 = 𝒂+ ()∗ 𝐚 = a+a=a 𝑹𝟐𝟏 (𝟏) = 𝑹𝟐𝟏 (𝟎) + 𝑹𝟐𝟏 𝟎 (𝑹𝟏𝟏 𝟎 ) ∗ 𝑹𝟏𝟏 𝟎 =Φ+ Φ()∗ =Φ 𝑹𝟐𝟑 (𝟏) = 𝑹𝟐𝟑 (𝟎) + 𝑹𝟐𝟏 𝟎 (𝑹𝟏𝟏 𝟎 ) ∗ 𝑹𝟏𝟑 𝟎 =b+Φ= 𝒃 𝑹𝟐𝟐 (𝟏) = 𝑹𝟐𝟐 (𝟎) + 𝑹𝟐𝟏 𝟎 (𝑹𝟏𝟏 𝟎 ) ∗ 𝑹𝟏𝟐 𝟎 = +Φ= 𝑹𝟑𝟑 (𝟏) = 𝑹𝟑𝟑 (𝟎) + 𝑹𝟑𝟏 𝟎 (𝑹𝟏𝟏 𝟎 ) ∗ 𝑹𝟏𝟑 𝟎 = +Φ= 𝑹𝟑𝟐 (𝟏) = 𝑹𝟑𝟐 (𝟎) + 𝑹𝟑𝟏 𝟎 (𝑹𝟏𝟏 𝟎 ) ∗ 𝑹𝟏𝟑 𝟎 =Φ+Φ=Φ 𝑹𝟑𝟏 (𝟏) = 𝑹𝟑𝟏 (𝟎) + 𝑹𝟑𝟏 𝟎 (𝑹𝟏𝟏 𝟎 ) ∗ 𝑹𝟏𝟏 𝟎 =Φ+ Φ =Φ a+a = {a,a} = {a} so that RE is a Mrs.D.Jena Catherine Bel, AP/CSE, VEC 24
- 25. Computation of K=2 and K=3 𝑹𝟏𝟑 (𝟐) = 𝑹𝟏𝟑 (𝟏) + 𝑹𝟏𝟐 𝟏 (𝑹𝟐𝟐 𝟏 ) ∗ 𝑹𝟐𝟑 𝟏 =Φ+ a()∗b=ab 𝑹𝟑𝟑 (𝟐) = 𝑹𝟑𝟑 (𝟏) + 𝑹𝟑𝟐 𝟏 (𝑹𝟐𝟐 𝟏 ) ∗ 𝑹𝟐𝟑 𝟏 = +Φ= 𝑹𝟏𝟑 (𝟑) = 𝑹𝟏𝟑 (𝟐) + 𝑹𝟏𝟑 𝟐 (𝑹𝟑𝟑 𝟐 ) ∗ 𝑹𝟑𝟑 𝟐 =ab+ab()∗=ab +ab =ab The Required Regular Expression is 𝑹𝟏𝟑 (𝟑) ab+ab = {ab,ab} = {ab} so that RE is ab Mrs.D.Jena Catherine Bel, AP/CSE, VEC 25
- 26. DFA to RE Ex-2 Mrs.D.Jena Catherine Bel, AP/CSE, VEC 26
- 27. Computation of k=0 and k=1 Mrs.D.Jena Catherine Bel, AP/CSE, VEC 27
- 30. Finite Automata to Regular Expressions • State Elimination Method Mrs.D.Jena Catherine Bel, AP/CSE, VEC 30
- 31. Finite Automata to RE Mrs.D.Jena Catherine Bel, AP/CSE, VEC 31
- 32. Finite Automata to RE - Cont. Mrs.D.Jena Catherine Bel, AP/CSE, VEC 32
- 33. Finite Automata to RE - Cont. Mrs.D.Jena Catherine Bel, AP/CSE, VEC 33
- 34. Finite Automata to RE - Cont. Mrs.D.Jena Catherine Bel, AP/CSE, VEC 34
- 35. Arden’s Theorem • Let P and Q be two regular expressions. • If P does not contain null string, then R = Q + RP has a unique solution that is R = QP* Mrs.D.Jena Catherine Bel, AP/CSE, VEC 35
- 40. Proving Languages not to be regular • Pumping Lemma Mrs.D.Jena Catherine Bel, AP/CSE, VEC 40
- 41. Applications of Pumping Lemma Pumping Lemma is to be applied to show that certain languages are not regular. It should never be used to show a language is regular. If L is regular, it satisfies Pumping Lemma. If L does not satisfy Pumping Lemma, it is non-regular. Mrs.D.Jena Catherine Bel, AP/CSE, VEC 41
- 42. Prove that L={anbn | n>=1} is not regular • L={ab,aabb,aaabbb,aaaabbbb,….} • RE : aa*bb* • FA: • L={ab,aaaabb,abbbbb,….} a a b b Proof by contradiction Let L is regular with n states, then, w=“aaabbb” ꞓ L According pumping lemma, w divided into xyz, Let X=aa : y=abb : z= b For k=2, compute xyk z =aa(abb)2b =aaabbabbb not belongs to L So that L is not regular Mrs.D.Jena Catherine Bel, AP/CSE, VEC 42