Formal Languages, Automata and Models of Computation

Transcription

1 CDT314 FABER Formal Languages, Automata and Models of Computation Lecture 5 School of Innovation, Design and Engineering Mälardalen University

2 Content - More Properties of Regular Languages (RL) - Standard Representations of RL - Elementary Questions about RL - Non-Regular Languages - The Pigeonhole Principle - The Pumping Lemma - Applications of the Pumping Lemma - NFA-DFA repetition 2

3 More Properties of Regular Languages 3

4 We have proven Regular languages g are closed under Union Concatenation Star operation Reverse 4

5 Namely, for regular languages L1 and L2 : Union Concatenation Star operation Reverse L1 L 2 L 1 L 2 L 1 R L 1 Regular Languages 5

6 We will prove Regular languages are also closed under Complement Intersection 6

7 Namely, for regular languages L1 and L2 : Complement L1 Intersection L1 L 2 Regular Languages 7

8 Complement Theorem For regular language L the complement L is regular Proof Take DFA that accepts L and exchange: non-final states final states non-accepting states accepting states Resulting DFA accepts L 8

9 Example a L = L( a * b) a,b b a, b q0 q1 q2 L = L ( a + a b ( a + b )( a + b ) a ) a, b b a, b q 0 q 1 q 2 9

10 Intersection Theorem e For regular languages L 1 and L 2 the intersection L1 L 2 is regular Proof Apply DeMorgan s Law: L1 L2 = L1 L 2 10

11 L 1, L 2 1, regular L 1, L 2 L1 L 2 L1 L 2 L1 L 2 regular regular regular regular 11

12 Standard Representations of Regular Languages 12

13 Regular Language Representations Regular Expression Regular Language DFA NFA Regular Grammar 13

14 Elementary Questions about Regular Languages 14

15 Membership Question Question: Given regular language L and string w how can we check if w L? Answer: Take the DFA that accepts L and check if w is accepted 15

16 DFA w w L DFA w w L 16

17 Question: Given regular language how can we check L ( L = ) L if is empty:? Answer: Take the DFA that accepts L Check if there is a path from the initial state to a final state 17

18 DFA L DFA L = 18

19 Question: Given regular language how can we check L if is finite? L Answer: Take the DFA that accepts Check if there is a walk with cycle from the initial iti state t to a final state t L 19

20 DFA L is infinite it DFA L is finite 20

21 Question: L L2 Given regular languages 1 and how can we check if L 1 = L 2? Answer: Find if ( L1 L2 ) ( L1 L2 ) = 21

22 ( 2 L1 L2 ) ( L1 L ) = L 1 L 2 = and L 1 L 2 = L 1 L2 L 2 L 2 1 L L1 L1 L 2 L2 L1 L = 1 L 2 22

23 ( 2 L1 L2 ) ( L1 L ) L 1 L 2 or L 1 L 2 L 1 L L 2 2 L1 L1 L 2 L2 L1 L 1 L 2 23

24 Non-Regular Languages 24

25 Non-regular languages??? Regular languages a *b b * c + a b + c ( a + b)* etc... 25

26 How can we prove that a language L is not regular? a a a λ b λ λ b b 26

27 How can we prove that a language L is not regular? Prove that there is no DFA that accepts L a a a Problem: this is not easy to prove Solution: the Pumping Lemma! λ λ λ b b L n n = { a b : n 0} b 27

28 The Pigeonhole Principle i 28

29 The Pigeonhole Principle 29

30 4 pigeons 3 pigeonholes 30

31 A pigeonhole must contain at least two pigeons pg 31

32 n pigeons... m pigeonholes n > m... 32

33 The Pigeonhole Principle n pigeons m pigeonholes n > m There is a pigeonhole with at least 2 pigeons... 33

34 The Pigeonhole Principle and DFAs 34

35 DFA with 4 states b b b a b q 1 q 2 q 3 q b q 4 a a 35

36 In walks of strings: b b a aa aab no state is repeated b a a q1 q2 q3 a b a q 4 36

37 In walks of strings: aabb a state bbaa abbabb abbbabbabb... is repeated b b b a a q1 q2 q3 b q 4 a a 37

38 w w 4 If the walk of string has length then a state is repeated b b b a a q1 q2 q3 a b a q 4 38

39 Pigeonhole principle for any DFA: If in a walk of a string transitions then a state is repeated w states of DFA 39

40 In other words for a string w a q transitions are pigeons states t are pigeonholes 40

41 In general A string w has length number of states A state q must be repeated in the walk of w walk of w q 41

42 The Pumping Lemma for Regular Languages see also 42

43 Take an infinite regular language g L DFA that accepts L n states 43

44 Take string w with w L There is a walk with label w... walk w 44

45 If string w has length w m = n + ( n number of states) then, from the pigeonhole principle: 1 a state q is repeated in the walk w q walk w 45

46 Write w = x y z y q x z 46

47 Lengths: x y m (from pigeon pg principle, p as q is the first repetition in sequence) y 1 (there is a walk in the graph) y... q... x z 47

48 Observation The string x z is accepted y q x z 48

49 Observation The string x y y z is accepted y q x z 49

50 Observation The string is accepted Th t i x y y y z y... q... x z 50

51 Generally The string x y i z is accepted i = 0,1, 2,... y q x z 51

52 The Pumping Lemma Given an infinite regular language L there exists an integer m for any string w L with length w m we can write w = x y z with x y m and y 1 such hthat: t i x y i z L i = 0,1, 2,... 52

53 Applications of the Pumping Lemma 53

54 Theorem The language L n n = { a b : n 0} is not regular Proof Use the Pumping Lemma! 54

55 L n n = { a b : n 0} Assume to the contrary, that L is a regular language L Since is infinite we can apply the Pumping Lemma 55

56 L n n = { a b : n 0} Let m be the integer in the Pumping Lemma Pick a string such that: w w L length w m e.g. pick m b m w = a 56

57 Write: a m b m = x y z From the Pumping Lemma it must be that length x y m, y 1 Therefore: a m m m b m = a... aa... a... ab... b x y z y k = a, k 1 57

58 We have: x y z = m b m a y k = a, k 1 From the Pumping Lemma: x y i z i = 0,1, 2,... L We can choose i = 0 w = a m k b m L CONTRADICTION! 58

59 Therefore: Our assumption that L is a regular language is not true Conclusion L is not a regular language END OF PROOF 59

60 Non-regular languages n n { a b : n 0} Regular languages a *b b * c + a b + c ( a + b)* etc... 60

61 Theorem The language L R = { ww : w Σ*} Σ = { a, b} is not regular Proof Use the Pumping Lemma! 61

62 L R = { ww : w Σ*} Assume to the contrary, L that is a regular language L Since is infinite we can apply the Pumping Lemma 62

63 L R = { ww : w Σ *} Let m be the integer in the Pumping Lemma Pick a string w such that: w L length w m pick w = a m b m b m a m 63

64 Write a m b m b m a m = x y z From the Pumping Lemma it must be that length x y m, y 1 a m b m b m a m = m m m m a... aa... a... ab... bb... ba... a y k = a, k 1 x y z 64

65 a m b m b m a m = m m m m a... aa... a... ab... bb... ba... a y k, 1 = a k x y z We can choose i = 0 m k So we get a s as on the left, while m is on the right: CONTRADICTION! 65

66 Therefore: Our assumption that L is a regular language is not true Conclusion L is not a regular language END OF PROOF 66

67 n Non-regular languages g n { a b : n 0} { ww R : w Σ*} Regular languages a *b b * c + a b + c ( a + b)* etc... 67

68 Theorem The language L n = { a b c : n, l is not regular l n+ l 0} Proof Use the Pumping Lemma 68

69 L n l n+ l = { a b c : n, l 0} Assume to the contrary that L is a regular language L Since is infinite we can apply the Pumping Lemma 69

70 L n l n+l = { a b c : n, l 0} Let m be the integer in the Pumping Lemma Pick a string such that: w w L length w m Pick m m 2 m 2 w = a b c 70

71 Write a m b m c 2m = x y z From the Pumping Lemma it must be that length x y m, y 1 a m m 2m b c = m m 2m a... aa... aa... ab... bc... cc... c y k = a, k 1 x y z 71

72 m m m c b a z y x 2 = We have:, 1 = k a y k From the Pumping Lemma L z y x i... 2, = 0,1, i Thus: L z y x 0 L c b a z x z y x m m k m = =

73 Therefore: a m k m 2m b c L BUT: L n l n+ l = { a b c : n, l 0} a m k m 2m b c L CONTRADICTION! 73

74 Therefore: Our assumption that L is a regular language is not true Conclusion L is not a regular language END OF PROOF 74

75 Non-regular languages g { ww R : w Σ*} n n { a b : n 0} n l n+ l { a b c : n, l 0} Regular languages 75

76 Theorem The language L n = { a! : n 0} is not regular n! = 1 2L ( n 1 ) n Proof Use the Pumping Lemma Factorial of n, (n!) is the product of all positive integers less than or equal to n. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120 The value of 0! is 1. 76

77 L n = { a! : n 0} Assume to the contrary that L is a regular language L Since is infinite we can apply the Pumping Lemma 77

78 L n = { a! : n 0} Let m be the integer in the Pumping Lemma Pick a string ti w such hthat: t w L length w m Pick w = a m!! 78

79 Write a m! = x y z From the Pumping Lemma it must be that length x y m, y 1 m m!! m! = a m a... aa... aa... aa... aa... a x y z k y = a, 1 k m 79

80 We have: x y z = m! a k y = a, 1 k m From the Pumping Lemma: x y i z L i = 0,1, 2,... Thus: x y 2 z L x y 2 z = x y y z = a m!+ k L 80

81 Therefore: a m!+ k L 1 k m Since: L n = { a! : n 0} There is p m! + k = p!! 1 k m There is p 81

82 However for m > 1 m! +k m!+ m m!+ m! < m! m + m! = m! ( m + 1) = ( m +1)! m! < m! + k < ( m + 1)! m! + k p! for any p 82

83 Therefore: a m!+ k L BUT: L = { a! : n 0} n and 1 k m a m!+ k L CONTRADICTION! 83

84 Therefore: Our assumption that L is a regular language is not true Conclusion L is not a regular language END OF PROOF 84

85 Non-regular languages { ww R : w Σ*} { a b : n 0} n n n l n+ l { a b c : n, l 0} { a n! : n 0} Regular languages 85

86 Theorem The language L = i { a : i prime } is not regular Proof Use the Pumping Lemma 86

87 L = i { a : i prime } Assume to the contrary that L is a regular language Since is infinite L we can apply the Pumping Lemma 87

88 w L length w m From the Pumping Lemma it must be that length x y m, y 1 From the Pumping Lemma: x y i z L i = 0,1, 2,... k+ 1 The length of w = xy z = k > m must be prime for each string of L 88

89 But, choosing length ( xyz ) = k length( xy k + 1 z) = length( xyy k z) = length( xyz) + length( y k ) = k + k ( length ( y )) = k (1 + length( y)) which is not prime! ( k > m) Thus: w L CONTRADICTION! 89

90 Therefore: Our assumption that L is a regular language is not true Conclusion L is not a regular language END OF PROOF 90

91 Non-regular languages { ww R : w Σ*} { a b : n 0} n n n l n+ l { a b c : n, l 0} { a n! : n 0} Regular languages L i = { a : i prime} 91

92 An alternative variant of proof from slide 53. Theorem The language is not regular L n n = { a b : n 0} Proof Use the Pumping Lemma! 92

93 L n n = { a b : n 0} Assume to the contrary that L is a regular language L Since is infinite we can apply the Pumping Lemma 93

94 L n n = { a b : n 0} Let m be the integer in the Pumping Lemma Pick a string w such that: w L length w m Pick m b m w = a 94

95 Write: a m b m = x y z From the Pumping Lemma it must be that: length x y m, y 1 m m Therefore: a m b m = a... aa... a... ab... b y k = a, k 1 x y z 95

96 We have: x y z = a m b m y = a, k 1 k From the Pumping Lemma: x y i z L i = 0,1, 2,... Thus: x y 2 z L x y 2 z = x y y z = a m+ k b m L 96

97 m+k + b m Therefore: a b L BUT: L n n = { a b : n 0} a m+ k b m L CONTRADICTION 97

98 Therefore: Our assumption that L is a regular language is not true Conclusion: L is not a regular language END OF PROOF 98

99 Pumping Lemma, in short 99

100 Observation: Every language of finite size has to be regular (we can easily construct an NFA that accepts every string in the language) Therefore, every non-regular language g has to be of infinite size (contains an infinite number of strings) 100

101 Suppose you want to prove that An infinite language L is not regular 1. Assume the opposite: L is regular 2. The pumping lemma should hold for L 3. Use the pumping lemma to obtain a contradiction 4. Therefore, L is not regular 101

102 Explanation of Step 3: How to get a contradiction with pumping lemma w L i. Find a particular string which satisfies the conditions of the pumping p lemma ii. Write w = xyz iii. Show that w i for some i 1 = xy z L iv. This gives a contradiction, since from pumping lemma w = xy i z L 102

103 Note: It suffices to show that only one string w gives a contradiction L You don t need to obtain contradiction for every w L 103

104 Extra Examples on Regular Languages 104

105 FA RE 105

106 FA RE or 106

107 NFA DFA Subset Construction Delmängdkonstruktion 107

108 Convert an NFA to DFA using the subset construction 108

109 Each state t of the DFAis a set of states t of the NFA. Start by the initial state of the DFA which is the ε-closure of the initial state of the NFA (all the states you reach by ε-transitions from the initial state). CLOSE{0}={0, {0, 1, 3}=S 0 109

110 Determine the transition function of the DFA on all inputs. Begin with the initial state S 0, and determine the transition on input a. CLOSE{0} = {0, 1, 3} = S 0 δ(s 0, a) = CLOSE{1, 2} 110

111 The ε-closure of the set {1, 2} is {1, 2, 3} This is a new state in the DFA, call it S 1. CLOSE{0} = {0, 1, 3} = S 0 δ(s 0, a) = CLOSE{1, 2} = {1, 2, 3} = S 1 111

112 With the initial state S 0, determine the transition on input b. CLOSE{0} = {0, 1, 3} = S 0 δ(s 0, a) = CLOSE{1, 2} = {1, 2, 3} = S 1 δ(s0, b) = CLOSE{3} 112

113 The ε-closure of the set {3} is {1, 3}. This is a new state in the DFA, call it S 2. CLOSE{0}={0 {0, 1, 3}=S 0 δ(s 0, a) = CLOSE{1, 2} = {1, 2, 3} = S 1 δ(s0, b) = CLOSE{3} = {1, 3} = S 2 113

114 Determine the transition function of the DFA from state S 1 on inputs a and b. On a there is no where to go in the NFA, so we create a sink ( trap ) state for DFA. δ(s 0, a) = CLOSE{1, 2} = {1, 2, 3} = S 1 δ(s 0, b) = CLOSE{3} = {1, 3} = S 2 δ(s 1, a) = CLOSE{} = = S 3 114

115 Determine the transition function of the DFA from state S 1 on inputs a and b. On b there is a transition to an existing state S 2. δ(s 0, a) = CLOSE{1, 2} = {1, 2, 3} = S 1 δ(s 0, b) = CLOSE{3} = {1, 3} = S 2 δ(s 1, a) = CLOSE{} = = S 3 δ(s 1, b) = CLOSE{3} = {1, 3} = S2 115

116 Determine the transition from state S 2 on inputs a and b. δ(s 0, a) = CLOSE{1, 2} = {1, 2, 3} = S 1 δ(s 0, b) = CLOSE{3} = {1, 3} = S 2 δ(s 1, a) = CLOSE{} = = S 3 δ(s 1, b) = CLOSE{3} = {1, 3} = S 2 δ(s 2, a) = CLOSE{} = = S 3 δ(s 2, b) = CLOSE{3} = {1, 3} = S 2 116

117 Determining the transition from state S 2 on inputs a and b is easy; from the empty set of states there are no transitions in the NFA. In the DFA this is represented by a transition from the empty set back to itself. 117

118 The final states t of the DFA are determined d from the final states t of the NFA. State t 3 was the only final state in the NFA. (Any set of NFA states containing a final state is a final state in the DFA.) 118

119 NFA DFA RESULT 119

120 State elimination 0*1 q 0 λ+10* q 1 0*11 q 2 01 (λ+10*)(0*1)*0*11 q 0 q 2 01 (λ+10*)(0*1)*0*11 1) q 0 q 2

121 Example 7 Convert the following NFA to DFA. 1 0 q 0 0 s 0 ε ε 1 s r 0 q p ε r p r s p q, r, p 1

122 Converting the NFA to DFA s p 0 λ λ q λ 1 r p r s 1 q r p 0 1 q p r s

123 Application: Lex - lexical analyzer Lex RE NFA DFA Minimal DFA The final states of the DFA are associated with actions. 123