1. Principles of Computing – UFCFA3-30-1
Week-5
Instructor : Mazhar H Malik
Global College of Engineering and Technology

2. NON-DETERMINISTIC FINITE AUTOMATA (NFA)1

3. Kleene Closure/Plus

4. Kleene Closure

5. Kleene Star – Example 2

6. Kleene Plus

7. Kleene Plus…

8. NON-DETERMINISTIC FINITE AUTOMATA (NFA)
Automata that are non-deterministic (NFA) can be in several states at once
from a state q on input a it can go to several different states this is "non-determinism"
Deterministic Finite Automata (DFA) can go only to one state in such
situations
it also has one start state and several final states
so when there are choices where to go, a NFA tries all 8

9. NFA Transition Function
Unlike in DFAs, δ(q,a) returns a set of states δ(q,a)={p1,...,pm} and it can return an empty set ∅ if there's nothing to go next 9

10. Definition: NON-DETERMINISTIC FINITE AUTOMATA (NFA)10

11. NON-DETERMINISTIC FINITE AUTOMATA (NFA)
In a nondeterministic finite automaton (NFA), for each state there can be zero, one, two, or more transitions corresponding to a particular symbol.
If NFA gets to state with more than one possible transition corresponding to the input symbol, we say it branches.
If NFA gets to a state where there is no valid
transition, then that branch dies. 11

12. 11

13. Example: An NFA for a++(ab)+
Input : - 12

14. Example: An NFA for a++(ab)+
Input : a 13

15. Example: An NFA for a++(ab)+
Input : a 14

16. Example: An NFA for a++(ab)+
Input : a b 15

17. Example: An NFA for a++(ab)+
Input : a b (Accepted) 16

18. Example: An NFA for a++(ab)+
Input : a b a (Rejected) 17

19. Example: An NFA for a++(ab)+
Input : a a (Rejected) 18

20. 19

21. Nondeterministic Finite Accepter (NFA)
Alphabet = {a} q1 q2 q0 q3

22. q1 q2 q0 q3 Nondeterministic Finite Accepter (NFA) Alphabet = {a}
Two choices q1 q2 q0 q3

23. q3 No transition q1 q2 q0 Nondeterministic Finite Accepter (NFA)
Alphabet = {a}
Two choices q1 q2
No transition q0
q3 No transition

24. First Choice q1 q2 q0 q3

25. First Choice q1 q2 q0 q3 24

26. First Choice q1 q2 q0 q3 25

27. First Choice q1 q2
“accept” q0 q3

28. Second Choice q1 q2 q0 q3

29. Second Choice q1 q2 q0 q3

30. Second Choice q1 q2 q0
No transition:
the automaton hangs q3

31. Second Choice q1 q2 q0 q3
“reject”

32. Observation An NFA accepts a string if
there is a computation of the NFA that accepts the string

33. aa is accepted by the NFA:
Example
aa is accepted by the NFA: q1 q2 q0 q3

34. Design NFA by observing the Output Language Given in side window33

35. 34

36. Example: Moves on a Chessboard
States = squares.
Inputs = r (move to an adjacent red square) and b (move to an adjacent black square).
Start state, final state are in opposite
corners. 1 2 3 4 5 6 7 8 9 35

37. Example: Chessboard – (2)1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8 1 2 3 4 5 6 7 8 9 r b b 2 4 1 3 5 7 5 1 3 7 9 1 *
Accept, since final state reached 36

38. Example 37

39. Language of an NFA A string w is accepted by an NFA if δ(q0, w)
contains at least one final state.
The language of the NFA is the set of
strings it accepts. 39

40. Example: Language of an NFA1 2 3 4 5 6 7 8 9
Example: Language of an NFA
For our chessboard NFA we saw that rbb is
accepted.
If the input consists of only b’s, the set of accessible states alternates between {5} and {1,3,7,9}, so only even-length, nonempty strings of b’s are accepted.
What about strings with at least one r? 40

41. 40

42. Lambda Transitions q0 q1 q1 q3

43. q0 q1 q2 q3

44. q0 q1 q2 q3

45. (read head doesn’t move)q0 q1 q2 q3

46. q0 q1 q2 q3

47. “accept” q3 q0 q1 q2
String aa is accepted

48. Language accepted: L  {aa}q0 q1 q2 q3

49. Another NFA Example a b q1 q2 q3 q0

50. a b a b q1 q2 q3 q0

51. a b a b q1 q2 q3 q0

52. a b a b q1 q2 q3 q0

53. a b
“accept” q2 a b q1 q3 q0

54. Another String a b a b q0 q1 q2 q3

55. a b a b q0 q1 q2 q3

56. a b a b q0 q1 q2 q3

57. a b a b q0 q1 q2 q3

58. a b a b q0 q1 q2 q3

59. a b a b q0 q1 q2 q3

60. a b a b q0 q1 q2 q3

61. a b
“accept” q2 a b q0 q1 q3

62. Language accepted
L  ab, abab, ababab, ...
 ab a b q1 q2 q3 q0

63. Another NFA Example
0, 1 q1 q2 q0 1

64. Language accepted
L  , 10, 1010, , ...
 10*
0, 1
q1 q2 q0 1

65. q0 : Initial state  : Q : M  Q, ,  , q0 , F   : F :
Formal Definition of NFAs
M  Q, ,  , q0 , F 
Q :
 :
 :
Set of states, i.e. q0 , q1, q2
Input aplhabet, i.e. a, b
Transition function
q0 : Initial state
F :
Final states

66. 65

67. Transition Function
 q0 , 1  q1
0, 1 q0 q1 q2 1

68.  (q1,0)  {q0 , q2}
0, 1 q0 q1 q2 1

69.  (q0 ,)  {q0 , q2}
0, 1 q0 q1 q2 1

70.  (q2 ,1)  
0, 1 q0 q1 q2 1

71. 70

72. Extended Transition Function  *
 * q0 , a  q1 q4 q5
a a q1 a b q0 q2 q3

73.  * q0 , aa  q4 , q5 q4 q5
a a q1 a b q0 q2 q3

74.  * q0 , ab  q2 , q3, q0 q4 q5
a a q1 a b q0 q2 q3

75. q j   *qi , w Formally It holds if and only if
there is a walk from qi
with label w
to q j

76. Your Turn 75

77. Example NFA.12
Design NFA for the language L = all string over {0,1} that have at least two consecutive 0’s or 1’s
Solution: By the analysis, it is clear that NFA will accept the string of the patern 00, 11, , 76

78. Example NFA.13 Example: Design NFA for language L = (ab ∪ aba)*
Solution: Let us first analyze the given language i.e. L = (ab
∪ aba)* means either ab, aba or any combination of should be accepted by the NFA. Null string (€) is also accepted. 77

79. Example NFA.13 Example: Design NFA for language L = (ab ∪ aba)*
Solution: Let us first analyze the given language i.e. L = (ab
∪ aba)* means either ab, aba or any combination of should be accepted by the NFA. Null string (€) is also accepted. 78

80. Example NFA.13 Example: Design NFA for language L = (ab ∪ aba)*
Solution: Let us first analyze the given language i.e. L = (ab
∪ aba)* means either ab, aba or any combination of should be accepted by the NFA. Null string (€) is also accepted. 79

81. Example NFA.13 Example: Design NFA for language L = (ab ∪ aba)*
Solution: Let us first analyze the given language i.e. L = (ab
∪ aba)* means either ab, aba or any combination of should be accepted by the NFA. Null string (€) is also accepted. 80

82. Example NFA.13 Example: Design NFA for language L = (ab ∪ aba)*
Solution: Let us first analyze the given language i.e. L = (ab
∪ aba)* means either ab, aba or any combination of should be accepted by the NFA. Null string (€) is also accepted. 81

83. Example NFA.14
Example: Draw NFA for language L = (ab)* (ba) ∪ aa*

84. Example NFA.14 Example: Draw NFA for language L = (ab)* (ba) ∪ aa*
Solution: We can construct NFA for the language L in two
paths i.e.
L = L1 ∪ L2
L1 = (ab)* (ba)
L2 = aa*
First we can construct NFA L1

85. Example NFA.14
Secondly we construct NFA for L2

86. Example NFA.14
Now combine L2 and L2 85

87. Example NFA.14
Input: Empty 86

88. Example NFA.14
Input: Empty 87

89. Example NFA.14
Input: Empty 88

90. Example NFA.14
Input: aa 89

91. Example NFA.14
Input: aa 90

92. NFA TO DFA

93. 92

94. Non Deterministic Features of NFA
There are three main cases of non- determinism in NFAs:
Transition to a state without consuming any input.
Multiple transitions on the same input symbol.
No transition on an input symbol.
To convert NFAs to DFAs we need to get rid of non-determinism from NFAs.

95. NFA to DFA 94

96. Example: Subset Construction1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8 r b
{1}
{2,4}
{5}
{2,4}
{5}
Alert: What we’re doing here is the lazy form of DFA construction, where we only construct a state
if we are forced to. * 96

97. Example: Subset Construction1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8 r b
{1}
{2,4}
{5}
{2,4,6,8}
{1,3,5,7}
{2,4} {5}
{2,4,6,8} {1,3,5,7} * 97

98. Example: Subset Construction1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8 r b
{1}
{2,4}
{5}
{2,4,6,8}
{1,3,5,7}
* {1,3,7,9}
{2,4} {5}
{2,4,6,8} {1,3,5,7}
{2,4,6,8} {1,3,7,9} * 98

99. Example: Subset Construction1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8 r b
{1}
{2,4}
{5}
{2,4,6,8}
{1,3,5,7}
* {1,3,7,9}
* {1,3,5,7,9}
{2,4} {5}
{2,4,6,8} {1,3,5,7}
{2,4,6,8} {1,3,7,9}
{2,4,6,8} {1,3,5,7,9} * 99

100. Example: Subset Construction1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8 r b
{1}
{2,4}
{5}
{2,4,6,8}
{1,3,5,7}
* {1,3,7,9}
* {1,3,5,7,9}
{2,4} {5}
{2,4,6,8} {1,3,5,7}
{2,4,6,8} {1,3,7,9}
{2,4,6,8} {1,3,5,7,9} *
100

101. Example: Subset Construction1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8 r b
{1}
{2,4}
{5}
{2,4,6,8}
{1,3,5,7}
* {1,3,7,9}
* {1,3,5,7,9}
{2,4} {5}
{2,4,6,8} {1,3,5,7}
{2,4,6,8} {1,3,7,9}
{2,4,6,8} {1,3,5,7,9}
{2,4,6,8} {5} *
101

102. Example: Subset Construction1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8 r b
{1}
{2,4}
{5}
{2,4,6,8}
{1,3,5,7}
* {1,3,7,9}
* {1,3,5,7,9}
{2,4} {5}
{2,4,6,8} {1,3,5,7}
{2,4,6,8} {1,3,7,9}
{2,4,6,8} {1,3,5,7,9}
{2,4,6,8} {5} *
102

103. Example: Subset Construction1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8 r b
{1}
{2,4}
{5}
{2,4} {5}
{2,4,6,8} {1,3,5,7}
{2,4,6,8} {1,3,7,9} A B C
{2,4,6,8} D {2,4,6,8} {1,3,5,7,9}
{1,3,5,7} E
{2,4,6,8} {1,3,5,7,9}
F {2,4,6,8} {5}
* {1,3,7,9}
* {1,3,5,7,9}
G {2,4,6,8} {1,3,5,7,9} *
103

104. Example: Subset Construction1
2,4 5 2
4,6
1,3,5 3
2,6 4
2,8
1,5,7
2,4,6,8
1,3,7,9 6
3,5,9 7
4,8 8
5,7,9 9
6,8
r b
B C
D E
D F
D G
F D C
G D G *
104

105. Example: Subset Construction
r b
B D D D D D D
C E
F G G C G
A B C D E
* F
* G
105

106. 105

107. Subset Construction Method
Fig1. NFA without λ-transitions 3
a b a a a 2
a,b b 1 5 a
a,b 4 b

108. Subset Construction Method
Fig1. NFA without λ-transitions 3
a b
Step1
Construct a transition table showing all reachable states for every state for every input signal. a a a 2
a,b b 1 5 a
a,b 4 b

109. Subset Construction Method
Fig1. NFA without λ-transitions 3
Fig2. Transition table
a b a a a 2
a,b b 1 5 a
a,b 4 b

110. Subset Construction Method
Fig1. NFA without λ-transitions 3
Fig2. Transition table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5
a b a a a 2
a,b b 1 5 a
a,b 4 b

111. Subset Construction Method
Transition from stateFqig2. TraTnrasnitsiiotionn tfarobmlestate q with input a with input b
Fig1. NFA without λ-transitions q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 3
a b
Starts here a a a 2
a,b b 1 5 a
a,b 4 b

112. Subset Construction Method
Fig2. Transition table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5
Step2
The set of states resulting from every transition function constitutes a new state. Calculate all reachable states for every such state for every input signal.

113. Fig3. Subset Construction table
Fig2. Transition table
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
Starts with
Initial state

114. ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 Fig2. Transition table 1 {4,5} 2
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
Starts with Initial state

115. ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) Step3 4 5 Fig2. Transition table 1
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
Starts with Initial state
Step3
Repeat this process(step2) until no more new states are reachable.

116. ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 Fig2. Transition table 1 {4,5} 2
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5}

117. ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5} 5 4

118. ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5} 5 4
{3,5}

119. Fig3. Subset Construction table
Fig2. Transition table
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5} 5 4
{3,5} ∅

120. ∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5} 5 4
{3,5} ∅
We already got 4 and 5.
So we don’t add them again.

121. ∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5} 5 4
{3,5} ∅ 2

122. ∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5} 5 4
{3,5} ∅ 2

123. ∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5} 5 4
{3,5} ∅ 2 3

124. ∅ ∅ q δ(q,a) δ(q,b) q δ(q,a) δ(q,b) 4 5 5 4 Fig2. Transition table 1
Fig3. Subset Construction table q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5} 2
{3}
{5} 3 ∅
{2} 4
{4} 5 q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5} 5 4
{3,5} ∅ 2 3
Stops here as there are
no more reachable states

125. Fig4. Resulting FA after applying Subset Construction to fig1
Fig3. Subset Construction table a q
δ(q,a)
δ(q,b) 1
{1,2,3,4,5}
{4,5}
{2,4,5} 5 4
{3,5} ∅ 2 3 b a
12345
245 35 a
a,b a b b ∅
a,b a 1 3 b a b 2 a 45 5 b b a 4 b

126. Fig4. Resulting FA after applying Subset Construction to fig1
Fig3. Subset Construction table a q
δ(q,a)
δ(q,b) I 1
{1,2,3,4,5}
{4,5} II
{2,4,5}
III 5 4 IV
{3,5} V ∅ VI
VII 2
VIII IX 3 X b a II IV
VII a
a,b a b b a I b
VIII X b
a,b a IX a
III V b b a VI b

127. Summary
DFA’s, NFA’s, and ε–NFA’s all accept exactly the same set of languages: the regular languages.
The NFA types are easier to design and may have exponentially fewer states than a DFA.
But only a DFA can be implemented!
126

128. AI: Chapter 2: Intelligent 127
Agents