This document discusses non-deterministic finite automata (NFAs). It provides examples of NFA transition graphs and explains how NFAs can accept input strings in a non-deterministic manner, meaning there may be multiple possible computations for a given input. It also defines the extended transition function for NFAs and explains that a string is accepted by an NFA if at least one computation of the NFA leads to an accepting state while consuming all input symbols. The language accepted by an NFA is the set of all strings that have an accepting computation. Finally, it notes that NFAs and deterministic finite automata (DFAs) have equivalent computational power since any NFA can be converted to an equivalent DFA.

1. Non
Deterministic
Automata
This ppt covers 2.2 ~ 2.4.
NFA 1

2. Nondeterministic FA
 As we know, any w *, then *(q0, w)
corresponds to a unique walk on the
transition graph of a DFA M.
 When we allow FA to act non-
deterministically, it implies *(q0, w) may
not correspond to a unique walk on the
transition graph of such FA anymore.
2

3. Nondeterministic Finite Accepter
(NFA)
= {a}
q1 a q2
a
q0
a
q3
3

4. Nondeterministic Finite Accepter
(NFA)
= {a}
More than one choices
q1 a q2
a
q0
a
q3 (q0, a ) = {q1, q3 }
4

5. Nondeterministic Finite Accepter
(NFA)
= {a} No transition!
q1 a q2
a
q0
a
q3
i.e., (q2, a ) = = (q3, a )
5

6. Nondeterministic Finite Accepter
(NFA)
Consider input string aa :
q1 a q2
a
q0
a
q3
There are two computations!
6

7. Example
q1 a q2 q1 a q2
a a
q0 q0
a a
q3 q3
All input is consumed & No transition for (q3, a),
stops at a final state. the automaton hangs, a
dead configuration
aa is “accepted” aa is “rejected”
7

8. An NFA accepts a string:
when there is one computation of the NFA
that accepts the string
such that
all the input is consumed AND
the automaton is in a final state
Thus aa is “accepted” by the NFA.
8

9. Therefore, an NFA rejects a string:
when there is NO computation of the NFA
that accepts the string:
• All the input is consumed and the
automaton is in a non final state
OR
• The input cannot be consumed
(dead configuration)
9

10. Example
a is rejected by the NFA:
“reject”
q1 a q2 q1 a q2
a a
q0 q0
a a
q3 “reject” q3
All possible computations lead to rejection
10

11. Another Example
aaa is rejected by the NFA:
No transition:
“reject” the automaton hangs
q1 a q2 q1 a q2
a a
q0 q0
a a
“reject”
q3 q3
All possible computations lead to rejection
11

12. Language accepted: L {aa}
q1 a q2
a
q0
a
q3
12

13. Another NFA Example
transition is allowed for NFA
q0 a q1 b q2 q3
13

14. transition is allowed for NFA
It implies that the NFA can move from
state qi to state qj without moving the
read head, i.e. without consuming any
input symbol.
14

15. Another NFA Example
Consider the input string ab:
q0 a q1 b q2 ab is “accepted”
q0 a q1 b q2 q3 “rejected”
q0 a q1 b q2 q3 q0 “rejected”
q0 a q1 b q2 q3
15

16. Another String
a b a b
q0 a q1 b q2 q3
16

17. a b a b
q0 a q1 b q2 q3 q0 a q1 b q2
“accept”
q0 a q1 b q2 q3
17

18. Language accepted
L ab, abab, ababab, ...
ab
q0 a q1 b q2 q3
18

19. Another NFA Example
0
q0 q1 0, 1 q2
1
19

20. Language accepted
L(M ) = {λ, 10, 1010, 101010, ...}
= {10}*
(redundant state,
0 i.e. useless state)
q0 q1 0, 1 q2
1
20

21. NFAs are interesting because we can
express languages easier than DFAs
Example: design a FA for {a} with ={a}.
Example: design a FA for
{awa: w {a, b}*} with ={a, b}.
21

22. Formal Definition of NFAs (p.49)
M Q, , , q0 , F
Q : Set of states, Remember:
The is not a symbol,
: Input alphabet, it never appears on the
: Transition function
q0 : Initial state
F : Final states
22

23. NFA’s Transition Function
:Qx( { }) 2Q
Note that Every DFA is an NFA.
When input is , the original state is always a
member of its transition output, i.e. q (q, )
23

24. NFA’s Transition Function
:Qx( { }) 2Q
q0 , 1 q1
0
q0 q1 0, 1 q2
1
24

25. NFA’s Transition Function
:Qx( { }) 2Q
(q1,0) {q0 , q2}
0
q0 q1 0, 1 q2
1
25

26. NFA’s Transition Function
:Qx( { }) 2Q
(q0 , ) {q0 , q2}
0
q0 q1 0, 1 q2
1
26

27. :Qx( { }) 2Q
(q2 ,1) . So as (q0, 0) & (q2, 0)
0
q0 q1 0, 1 q2
1
Remark: Every DFA is an NFA.
27

28. Extended Transition
Function *
Similar as DFA, extended transition function is
defined to work with inputs to be strings.
*
:Qx( { } )* 2Q
*
or simply :Qx * 2Q
When input string is empty ( ), the original state
is always a member of its extended transition, i.e.
q * q,
28

29. Extended Transition
Function *
* q0 , a q1
q4 q5
a a
q0 a q1 b q2 q3
29

30. * q0 , aa q4 , q5
q4 q5
a a
q0 a q1 b q2 q3
30

31. * q0 , ab ?
q4 q5
a a
q0 a q1 b q2 q3
31

32. * q0 , ab q2 , q3 , q0
q4 q5
a a
q0 a q1 b q2 q3
32

33. * q0 , b ?
q4 q5
a a
q0 a q1 b q2 q3
33

34. Consider the transition graph:
a * a
q0 q0
q1 q1
q2 q2
a
q0 q1 q2
34

35. Formally, for an NFA
There is a walk from qi to q j with label
w 1 2 ... k
qi w qj
if and only if q j * qi , w w 1 2 ... k
qi 1 2 k
qj
Def. 2.5 p.51
For an nfa, the extended transition function is
defined so that qj *( qi , w) iff there is a walk
in the transition graph from qi to qj labeled w.
35

36. Hw # 20 p.56
 Show that for any nfa
*
(q, wv) 
*
*
( p, v).
p ( q , w)
for all q Q, and all w, v *
36

37. The Language of an NFA M
aa
F q0 ,q5 q4 q5
a a
q0 a q1 b q2 q3
Consider the aa, or the walk from q0 with label
aa:
* q0 , aa q4 , q5 aa L(M )
F 37

38. ab
F q0 ,q5 q4 q5
a a
q0 a q1 b q2 q3
* q0 , ab q2 , q3 , q0 ab L M
F 38

39. abaa
F q0 ,q5 q4 q5
a a
q0 a q1 b q2 q3
* q0 , abaa q4 , q5 abaa L(M )
F 39

40. F q0 ,q5 q4 q5
a a
q0 a q1 b q2 q3
* q0 , aba q1 aba L M
F 40

41. q4 q5
a a
q0 a q1 b q2 q3
LM (ab) n aa : n 0
41

42. Formally
The language accepted by NFA M is:
LM w1, w2 , w3 ,...
where * (q0 , wm ) {qi , q j ,..., qk ,...}
and there is some qk F (final state)
42

43. w LM * (q0 , w)
qi
w
qk qk F
q0 w
w qj
L( M ) {w *: * ( q0, w) F }
43

44. An NFA Example
Consider the input string ab:
q0 a q1 b q2 * (q0 , ab) {q0 , q2 , q3}
q0 a q1 b q2 q3
q0 a q1 b q2 q3 q0
q0 a q1 b q2 q3
44

45. Consider the input string ab:
* (q0 , ab) {q0 , q2 , q3}
q0 a q1 b q2 ab is “accepted”
q0 a q1 b q2 q3
q0 a q1 b q2 q3 q0
q0 a q1 b q2 q3
45

46. Another String
a b a b
q0 a q1 b q2 q3
46

47. q0 a q1 b q2 q3 q0 a q1 b q2
As long as there is one computation that
consumes all input symbols and stops at a final
state, aabb is accepted.
“accept”
q0 a q1 b q2 q3
47

48. Language accepted
L ab, abab, ababab, ...
ab
q0 a q1 b q2 q3
48

49. NFAs are interesting because we can
express languages easier than DFAs
Example: design an FA for {a} with ={a, b}.
Example: design an FA for
{awa: w {a, b}*} with ={a, b}.
49

50. Consider the transition graph:
a * a
q0 q0
q1 q1
q2 q2
a
q0 q1 q2
50

51. Hw # 20 p.56
 Show that for any nfa
*
(q, wv) U
*
*
( p, v).
p ( q , w)
for all q Q, and all w, v *
51

52. Remarks for FA M = {Q, , , q0, F }:
•The symbol never appears on the
input alphabet
•Simple NFA:
M1 M2
q0 q0
L(M1 ) = {} L(M 2 ) = {λ}
52

53. The language accepted by NFA M is:
L( M ) {w *: * ( q0, w) F }
Is it true that for any NFA M ,
L(M ) {w *: * ( q0, w) F }
Is it true that for any NFA M,
L(M ) {w *: * (q0, w) (Q F ) }
Referring to Hw 2.3 # 5, 6. P.62 53

54. NFAs accept the
Regular Languages
Hw of 2.2(p.54) 1 – 19 not hand in
NFA 54

55. Equivalence of Machines
Definition for Automata:
Machine M1 is equivalent to machine M 2
if L M1 L M2
55

56. Example of equivalent
machines
0
L M1 {10} *
q0 q1
NFA M1 1
0,1
0
L M2 {10}* q1 1 q2
q0
1
DFA M2 0
56

57. Which is more powerful?
 Every dfa is an nfa.
 Thus, every language accepted by
a dfa is also accepted by an nfa.
 Is nfa more powerful than dfa?
57

58. We will prove:
Languages
Regular
accepted
Languages
by NFAs
Languages
accepted
by DFAs
NFAs and DFAs have the
same computation power
58

59. Step 1
Languages
Regular
accepted
Languages
by NFAs
Proof: Every DFA is trivially an NFA
Any language L accepted by a DFA
is also accepted by an NFA
59

60. Step 2
Languages
Regular
accepted
Languages
by NFAs
Proof: Any NFA can be converted to an
equivalent DFA
Any language L accepted by an NFA
is also accepted by a DFA 60

61. Convert NFA to DFA
a
NFA M N q0 a q1 q2
b
* a b
q0
q1
q2
An extended transition
function is very helpful
61

62. Convert NFA to DFA
NFA M N a
q0 a q1 q2
b
Let p0={q0} be the initial state for MD
DFA MD
q0
Back to Procedure NFA to DFA
62

63. Convert NFA to DFA
NFA M N a
q0 a q1 q2
b
Now, define D(p0,a) for a
DFA MD D(p0,a) = N*({q0},a) ={q1,q2} p1
q0 a
q1, q2
63

64. Convert NFA to DFA
NFA M N a
q0 a q1 q2
b
D(p0,b) = N*({q0},b) = p2
DFA MD
q0 a
q1, q2
b
Back to Procedure NFA to DFA
64

65. Convert NFA to DFA
NFA M N a
q0 a q1 q2
b
D(p1,a) = N*({q1,q2},a)= N*({q1},a) N*({q2},a)
= {q1,q2} = {q1,q2} = p1
DFA MD a
q0 a
q1, q2
b
65

66. Convert NFA to DFA
NFA M N a
q0 a q1 q2
b
D(p1,b)
= N*({q1,q2},b) = N*({q1},b) N*({q2},b)
= {q0} {q0} = {q0} = p0
DFA MD b a
q0 a
q1, q2
b
66

67. Convert NFA to DFA
NFA M N a
q0 a q1 q2
b
D(p2,a) = N*( ,a) = =p2= D(p2,b)
DFA MD b a
q0 a
q1, q2
b
Back to Procedure NFA to DFA
a, b
67

68. Convert NFA to DFA
NFA M N a
q0 a q1 q2
b
How about final state? Lastly, check if L(M)
b a
DFA MD
q0 a
q1, q2
b
a, b back
68

69. NFA to DFA: Remarks
 We are given an NFA MN
 We want to convert it
to an equivalent DFA MD
 With L M N L( M D )
69

70.  If the NFA has states
q0 , q1, q2 ,...
 the DFA has states in the powerset
, q0 , q1 , q1, q2 , q3 , q4 , q7 ,....
70

71. Procedure NFA to DFA p.59
1. Initial state of NFA: q0
Initial state of DFA: p0 q0
To Conversion Example
71

72. Procedure NFA to DFA
2. For every DFA’s state pk {qi , q j ,..., qm }
Compute in the NFA
* qi , a ,
* q j,a , union {qi , q j ,..., qm}
...
Add transition to DFA To Convertion Example
D pk , a D {qi , q j ,..., qm }, a {qi , q j ,..., qm }
72

73. Procedure NFA to DFA
Repeat Step 2 for all letters in alphabet,
until
no more transitions can be added.
Remark:
In DFA, every vertex must have exactly | | outgoing
edges, each labeled with a different element of .
To Convertion Example
73

74. Procedure NFA to DFA
3. For any DFA state pk {qi , q j ,..., qm }
If some q j is a final state in the NFA
Then, make p k a final state in the DFA
Remark: If is accepted in the NFA then
p0 q0 should be a final state.
To Convertion Example
74

75. Theorem
Take NFA MN
Apply procedure to obtain DFA MD
Then MN and MD are equivalent :
L(MN) = L(MD)
75

76. Proof
L(MN) = L(MD)
L(MN) L(MD
) AND L(MN) L(MD
)
76

77. First we show: L(MN) L(MD
)
Take arbitrary: w L(MN)
We will prove: w L(MD)
77

78. w L(MN)
MN q0 w qf
w 1 2 ... k
MN q0 1 2 k
qf
78

79. We will show that if w L(MN)
w 1 2 ... k
MN q0 1 2 k
qf
1 2 k
MD
{q0} {q f ,...}
w L(MD)
79

80. More generally, we will show that if in MN :
(arbitrary walk) v a1a2 ... an
a1 a2 an
MN q0 qi qj ql qm
a1 a2 an
MD
{q0} {qi ,...} {q j ,...} {ql ,...} {qm ,...}
80

81. Proof by induction on |v|
Induction Basis: v a1
a1
MN q0 qi
a1
MD
{q0} {qi ,...}
81

82. Induction hypothesis: 1 |v| k
v a1a2 ... ak
a1 a2 ak
MN q0 qi qj qc qd
a1 a2 ak
MD
{q0} {qi ,...} {q j ,...} {qc ,...} {qd ,...}
82

83. Induction Step: |v| k 1
v a1a2 ... ak ak 1 v ak 1
v
a1 a2 ak
MN q0 qi qj qc qd
v
a1 a2 ak
MD
{q0} {qi ,....} {q j ,...} {qc ,...} {qd ,...}
v 83

84. Induction Step: |v| k 1
v a1a2 ... ak ak 1 v ak 1
v
a1 a2 ak ak 1
MN q0 qi qj qc qd qe
v
a1 a2 ak ak 1
MD
{q0} {qi ,...} {q j ,...} {qc ,...} {qd ,...} {qe ,...}
v 84

85. Therefore if w L(MN)
w 1 2 ... k
MN q0 1 2 k
qf
1 2 k
MD
{q0} {q f ,...}
w L(MD)
85

86. We have shown: L(MN) L(MD
)
We also need to show: L(MN) L(MD
)
(proof is similar)
86

87. Example
a b
NFA
q0 , a, b q1 a, b q2
Convert NFA to DFA Simplified version of DFA
a a
a p1 p1
b a, b a, b
p0 b b
b p2 a, b p3 a p4 p2 a, b p3
Ex. 2.13 Fig. 2.14 (p.59)– in class (you
can eliminate the redundant state first) 87

88. Minimal DFAs
 A dfa M’ is minimal if M is also a dfa such
that L(M’)=L(M) then |QM | |QM’|
88

89. Minimal DFAs
 States p and q are called indistinguishable if
* ( p, w) F implies * (q, w) F
and
* ( p, w) F implies * (q, w) F
for all w *
 States p and q are called distinguishable if
w * such that * ( p, w) F and * (q, w) F
89

90. Minimal DFAs
 To obtain a minimal DFA M’ from a given DFA M:
 Remove all inaccessible states, i.e. not reachable from the
initial state.
 Reduce states until it has no more indistinguishable states.
 Read procedures: mark (p.64) & reduce (p.66), and
theorem 2.3 (p.65) & 2.4 (p.67) for details.
Remove all inaccessible states can be accomplished
by enumerating all simple paths of the graph of the
dfa starting at the initial state, any state not part of
some path is inaccessible. 90

91. Example: Minimal DFAs
q1 0,1
0 1
0 0
q0 q2 1 q4
0 1
1 0
q3 q5
1
91

92. Minimal DFAs
Prove or Disprove:
 If M = (Q, , , q0, F) is a minimal DFA for a
regular language L, then M’ = (Q, , , q0, Q-F)
is a minimal DFA for the complement of L.
92

93. Minimal DFAs
 Prove or Disprove:
 If M = (Q, , , q0, F) is a minimal DFA for a
regular language L, then M’ = (Q, , , q0,
Q-F) is a minimal DFA for the complement of
L.
93

94. Homework Discussion
 2.3 (p.62) Discuss 7, 8, 11, 14, 15
Homework
2.3 (p.62) 1 ~15 & 2.4 (p.68) 1, 2, 4, 6
Hand in: 3, 5, 6, 9, 12 (p.62) & 2ae (p.68)
94