Theory of computation lecture NFA

1. Formal Languages
NFAs Accept the Regular
Languages
Prof. Charbel Boustany
Msc Systems Engineering
PHD Genetic Algorithms (current)
IEEE Member

2. EQUIVALENCE OF MACHINES
 Definition:
 Machine is equivalent to machine
 if

2
1
M 2
M
   
2
1 M
L
M
L 

3. EXAMPLE OF EQUIVALENT MACHINES

3
0
q 1
q
0
1
0
q 1
q 2
q
0
1
1
0
1
,
0
NFA
FA
  *
}
10
{
1 
M
L
  *
}
10
{
2 
M
L
1
M
2
M

4. 4
We will prove:
Languages
accepted
by NFAs
Regular
Languages

NFAs and FAs have the
same computation power
Languages
accepted
by FAs

5. 5
Languages
accepted
by NFAs
Regular
Languages

Languages
accepted
by NFAs
Regular
Languages

We will show:

6. 6
Languages
accepted
by NFAs
Regular
Languages

Proof-Step 1
Proof?

7. 7
Languages
accepted
by NFAs
Regular
Languages

Proof-Step 1
Proof: Every FA is trivially an NFA
Any language accepted by a FA
is also accepted by an NFA
L

8. 8
Languages
accepted
by NFAs
Regular
Languages

Proof-Step 2
Proof: Any NFA can be converted to an
equivalent FA
Any language accepted by an NFA
is also accepted by a FA
L

9. CONVERT NFA TO FA

9
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q
M
M 

10. CONVERT NFA TO FA

10
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q  
2
1,q
q
a
M
M 

11. CONVERT NFA TO FA

11
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q  
2
1,q
q
a

b
M
M 

12. CONVERT NFA TO FA

12
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q  
2
1,q
q
a

b
a
M
M 

13. CONVERT NFA TO FA

13
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q  
2
1,q
q
a

b
a
b
M
M 

14. CONVERT NFA TO FA

14
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q  
2
1,q
q
a

b
a
b
b
a,
M
M 

15. CONVERT NFA TO FA

15
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q  
2
1,q
q
a

b
a
b
b
a,
M
M 
  )
(M
L
M
L 


16. NFA TO FA CONVERSION
 We are given an NFA
 We want to convert it
 to an equivalent FA
 With
16
M
M 
  )
(M
L
M
L 


17. WHAT WE NEED TO CONSTRUCT
 Finite Automaton (FA)
17
 
F
q
Q
M ,
,
,
, 0



Q


0
q
F
: set of states
: input alphabet
: transition function
: initial state
: set of accepting states

18.  If the NFA has states
 the FA has states in the power set

18
,...
,
, 2
1
0 q
q
q
    ,....
,
,
,
,
,
,
, 7
4
3
2
1
1
0 q
q
q
q
q
q
q


19. PROCEDURE NFA TO FA

1. Initial state of NFA:

 Initial state of FA:
19
0
q
 
0
q

20. EXAMPLE

20
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q
M
M 

21. PROCEDURE NFA TO FA
 2. For every FA’s state
 Compute in the NFA

 Add transition to FA
21
}
,...,
,
{ m
j
i q
q
q
 
 
...
,
,
*
,
,
*
a
q
a
q
j
i


}
,...,
,
{ m
j
i q
q
q 


  }
,...,
,
{
},
,...,
,
{ m
j
i
m
j
i q
q
q
a
q
q
q 






22. EXAMPLE

22
a
b
a

0
q 1
q 2
q
NFA
 
0
q  
2
1,q
q
a
FA
}
,
{
)
,
(
* 2
1
0 q
q
a
q 

 
   
2
1
0 ,
, q
q
a
q 

M
M 

23. PROCEDURE NFA TO FA
 Repeat Step 2 for all letters in alphabet,
 until
 no more transitions can be added.
23

24. EXAMPLE

24
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q  
2
1,q
q
a

b
a
b
b
a,
M
M 
Done?

25. PROCEDURE NFA TO FA
 3. For any FA state
 If is accepting state in NFA
 Then,
 is accepting state in FA
25
}
,...,
,
{ m
j
i q
q
q
j
q
}
,...,
,
{ m
j
i q
q
q

26. EXAMPLE

26
a
b
a

0
q 1
q 2
q
NFA
FA
 
0
q  
2
1,q
q
a

b
a
b
b
a,
F
q 
1
  F
q
q 

2
1,
M
M 

27. THEOREM

27
Take NFA M
Apply procedure to obtain FA M 
Then and are equivalent :
M M 
   
M
L
M
L 


28. 28
Proof (slide 28 – 40) (reading recommended)
   
M
L
M
L 

   
M
L
M
L 
    
M
L
M
L 

AND

29. 29
   
M
L
M
L 

First we show:
)
(M
L
w
Take arbitrary:
We will prove: )
(M
L
w 


30. 30
)
(M
L
w
w
k
w 

 
2
1

0
q f
q
:
M
:
M 0
q f
q
1
 2
 k


31. 31
i

i

 
denotes

i
q j
q
i
q j
q

32. 32
0
q f
q
k
w 

 
2
1

1
 2
 k

:
M
}
{ 0
q
1
 2
 k

:
M
}
,
{ 
f
q
)
(M
L
w
)
(M
L
w 

We will show that if
then

33. 33
0
q m
q
n
a
a
a
v 
2
1

1
a 2
a n
a
:
M
}
{ 0
q
1
a 2
a n
a
:
M
}
,
{ 
i
q
i
q j
q l
q
}
,
{ 
j
q }
,
{ 
l
q }
,
{ 
m
q
More generally, we will show that if in :
M
(arbitrary string)
then

34. 34
0
q 1
a
:
M
}
{ 0
q
1
a
:
M
}
,
{ 
i
q
i
q
Proof by induction on
Induction Basis: 1
a
v 
|
| v
Is true by construction of :
M

35. 35
Induction hypothesis: k
v 
 |
|
1
0
q d
q
1
a 2
a k
a
:
M i
q j
q c
q
}
{ 0
q
1
a 2
a k
a
:
M
}
,
{ 
i
q }
,
{ 
j
q }
,
{ 
c
q }
,
{ 
d
q
k
a
a
a
v 
2
1


36. 36
Induction Step: 1
|
| 
 k
v
0
q d
q
1
a 2
a k
a
:
M i
q j
q c
q
}
{ 0
q
1
a 2
a k
a
:
M
}
,
{ 
i
q }
,
{ 
j
q }
,
{ 
c
q }
,
{ 
d
q
1
1
2
1 




 k
k
v
k a
v
a
a
a
a
v






v
v

37. 37
Induction Step: 1
|
| 
 k
v
0
q d
q
1
a 2
a k
a
:
M i
q j
q c
q
}
{ 0
q
1
a 2
a k
a
:
M
}
,
{ 
i
q }
,
{ 
j
q }
,
{ 
c
q }
,
{ 
d
q
1
1
2
1 




 k
k
v
k a
v
a
a
a
a
v






v
v
e
q
1

k
a
1

k
a
}
,
{ 
e
q

38. 38
0
q f
q
k
w 

 
2
1

1
 2
 k

:
M
}
{ 0
q
1
 2
 k

:
M
}
,
{ 
f
q
)
(M
L
w
)
(M
L
w 

Therefore if
then

39. 39
We have shown:    
M
L
M
L 

We also need to show:    
M
L
M
L 

(proof is similar)

40. 40
Induction Step: 1
|
| 
 k
v
0
q d
q
1
a 2
a k
a
:
M i
q j
q c
q
}
{ 0
q
1
a 2
a k
a
:
M
}
,
{ 
i
q }
,
{ 
j
q }
,
{ 
c
q }
,
{ 
d
q
1
1
2
1 




 k
k
v
k a
v
a
a
a
a
v






v
v
e
q
1

k
a
1

k
a
}
,
{ 
e
q
All cases covered?

41. SINGLE ACCEPTING
STATE FOR NFAS
41

42.  Any NFA can be converted
 to an equivalent NFA
 with a single accepting state
42

43. 43
a
b
b
a NFA
Equivalent NFA
with single
accepting state?
Example

44. 44
a
b
b
a NFA
Equivalent NFA


a
b
b
a
Example

45. 45
NFA
In General
Equivalent NFA
Single
accepting
state




46. 46
Extreme Case
NFA without accepting state
Add an accepting state
without transitions

47. PROPERTIES OF
REGULAR LANGUAGES
47

48. 48
1
L 2
L
2
1L
L
Concatenation:
*
1
L
Star:
2
1 L
L 
Union:
Are regular
Languages
For regular languages and
we will prove that:
1
L
2
1 L
L 
Complement:
Intersection:
R
L1
Reversal:

49. 49
We say: Regular languages are closed under
2
1L
L
Concatenation:
*
1
L
Star:
2
1 L
L 
Union:
1
L
2
1 L
L 
Complement:
Intersection:
R
L1
Reversal:

50. 50
1
L
Regular language
  1
1 L
M
L 
1
M
Single accepting state
NFA 2
M
2
L
Single accepting state
  2
2 L
M
L 
Regular language
NFA

51. EXAMPLE
51
}
{
1 b
a
L n

a
b
1
M
 
ba
L 
2
a
b
2
M
0

n

52. UNION
 NFA for
52
1
M
2
M
2
1 L
L 



53. EXAMPLE

53
a
b
a
b


}
{
1 b
a
L n

}
{
2 ba
L 
}
{
}
{
2
1 ba
b
a
L
L n



NFA for

54. CONCATENATION
 NFA for
54
2
1L
L
1
M 2
M
 

55. EXAMPLE

 NFA for
55
a
b a
b
}
{
1 b
a
L n

}
{
2 ba
L 
}
{
}
}{
{
2
1 bba
a
ba
b
a
L
L n
n


 

56. 56
How do we construct automata for the
remaining operations?
2
1L
L
Concatenation:
*
1
L
Star:
2
1 L
L 
Union:
1
L
2
1 L
L 
Complement:
Intersection:
R
L1
Reversal:

57. STAR OPERATION
 NFA for
57
*
1
L
1
M


*
1
L


 

58. EXAMPLE
 NFA for
58
*
}
{
*
1 b
a
L n

a
b
}
{
1 b
a
L n





1
2
1
L
w
w
w
w
w
i
k

 

59. REVERSE
59
R
L1
1
M
NFA for

1
M
1. Reverse all transitions
2. Make initial state accepting state
and vice versa
1
L

60. 60
Example
}
{
1 b
a
L n

a
b
1
M
}
{
1
n
R
ba
L 
a
b

1
M

61. 61
Complement
1. Take the FA that accepts 1
L
1
M
1
L 
1
M
1
L
2. Make final states non-final,
and vice-versa
Why not NFA?

62. 62
Example
}
{
1 b
a
L n

a
b
1
M
b
a,
b
a,
}
{
*
}
,
{
1 b
a
b
a
L n

 a
b

1
M
b
a,
b
a,

63. INTERSECTION
63
1
L regular
2
L regular
We show 2
1 L
L 
regular

64. 64
DeMorgan’s Law: 2
1
2
1 L
L
L
L 


2
1 , L
L regular
2
1 , L
L regular
2
1 L
L  regular
2
1 L
L  regular
2
1 L
L  regular

65. 65
Example
}
{
1 b
a
L n

}
,
{
2 ba
ab
L 
regular
regular
}
{
2
1 ab
L
L 

regular

66. 66
1
L
for for 2
L
FA
1
M
FA
2
M
Construct a new FA that accepts
Machine Machine
M 2
1 L
L 
M simulates in parallel and
1
M 2
M
Another Proof for Intersection Closure

67. 67
States in M
j
i p
q ,
1
M 2
M
State in State in

68. 68
1
M 2
M
1
q 2
q
a
transition
1
p 2
p
a
transition
FA FA
1
1, p
q a
transition
M
FA
2
2, p
q

69. 69
0
q
initial state
0
p
initial state
Initial state
0
0, p
q
1
M 2
M
FA FA
M
FA

70. 70
i
q
accept state
j
p
accept states
accept states
j
i p
q ,
k
p
k
i p
q ,
1
M 2
M
FA FA
M
FA
Both constituents must be accepting states

71. 71
M simulates in parallel and
1
M 2
M
M accepts string w if and only if
accepts string and
w
1
M
accepts string w
2
M
)
(
)
(
)
( 2
1 M
L
M
L
M
L 


72. 72
Example: what are the DFAs for the
Following:
}
{
1 b
a
L n

a
b
1
M
0

n
}
{
2
n
ab
L 
b
b
2
M
0
q 1
q 0
p 1
p
0

n
2
q 2
p
a
a
b
a, b
a,
b
a,

73. 73
Construct machine for intersection

74. 74
0
0, p
q
Automaton for intersection
}
{
}
{
}
{ ab
ab
b
a
L n
n



1
0, p
q
a
2
1, p
q
b
a
b 1
1, p
q
2
0, p
q
a
1
2, p
q
2
2, p
q
b
b
a,
a
b
b
a,
b
a

75. 75
under union, star, concatenation with
NFAs. Would be much harder with
DFAs.