2.4 Nondeterministic Finite State Automata
Definition 6 : A nondeterministic finite state automaton(NFA)
M = (, Q, , q 0, F), where
 is a finite set alphabet,
Q is a finite set of states,
 : Q  {}  2 Q is a transition function, where 2 Q
is a power set of Q,
q 0  Q is the initial state,
F  Q is the set of final states.
Example 1: A nondeterministic finite state automaton(NFA)
M = (, Q, , q 0, F), where
 = {0, 1},
Q = {q 0, q 1, q 2, q 3},
q 0  Q is the initial state,
F = {q 3},
 : Q {}  2 Q is defined as follows.

q0
q1
0
{q 0, q 1}

1
{q 0}
{q 2}
q2
{q 3}

q3


0, 1
q0
0
q1
1
q2
0
q3
The transition (q 0, 0) = {q 0, q 1} says that at the state q 0 after
reading the symbol 0 the machine can enter either the state q 0 or
the state q 1.
Each move of the machine may have more than one state to select.
Now, we get a problem that how to determine wether an input
string is accepted by the machine or not.
Consider the example 1 as the machine reading the input 01010.
The NFA M may have the following possible moves.
0, 1
q0
0
q1
1
q2
0
q3
Case 1: q 0 01010 |-- 0 q 1 1010 |-- 0 1 q 2 010 |-- 0 10 q 3 10 |--?
Case 2: q 0 01010 |-- 0 q 0 1010 |-- 0 1 q 0 010 |-- 0 10 q 0 10 |-0 10 1 q 0 0 |-- 0 10 1 0 q 0
Case 3: q 0 01010 |-- 0 q 0 1010 |-- 0 1 q 0 010 |-- 0 10 q 0 10 |-0 10 1 q 0 0 |-- 0 10 1 0 q 1
Case 4: q 0 01010 |-- 0 q 0 1010 |-- 0 1 q 0 010 |-- 0 10 q 1 10 |-0 10 1 q 2 0 |-- 0 10 1 0 q 3
There are 4 different ways for M to read the input 01010.
Although 3 ways do not lead to a final state, but there is one
way, i.e., in case 4, that the machine enters a final state q 3.
Since there is a way for M to enter a final state q 3 after reading
the input 01010, we say that the machine M accepts 01010.
Definition 7 : A string  is accepted by an NFA M = (, Q, , q 0,
F), if there is some way that q 0  |--*  q f , where q f  F.
The language accepted by the machine M is L(M) = {  * | if
there is some way that q 0  |--* q f , where q f  F. }
The language L accepted by the following NFA M is the set of
strings ending with 010.
0, 1
q0
0
q1
1
q2
0
q3
Compare the design of the DFA accepting language L in
section 2.3 with the design of the above NFA. It is quite
obvious that the design of an NFA is much easier than that of
a DFA, since we take less care of interaction between states.
Althogh the design of an NFA is much easier than that of a
DFA, later we will show that NFA‘s and DFA’s are of
same expressive power, i.e., if L = L(M) for some NFA M,
then there is a DFA M‘ such that L = L(M’), and vise
versa.
Therefore, we may first design an NFA, then convert to a DFA.
Example 2: Design an NFA M to accept strings of 0’s and 1’s such
that the 3rd symbol from the right end is 0.
Solution :
0, 1
0
q0
0, 1
q1
An equivalent DFA is as follows.
1
0
q0
1
q1
0
q4
1
q5
q2
q3
0
0
q3
q 0= q 111 q 1= q 110
1
q 2= q 100 q 3= q 000
1
1
0
q2
0, 1
0
0
q6
1
q7
0
1
q 4= q 011 q 5= q 101
q 6= q 010 q 7= q 001
An NFA with -moves is much easier to design and to understand,
and more flexible and important to use.
If we allow an NFA M to have -moves, then we allow M to
change states without reading any input symbol.
Because of -moves, it is quite easy to group a substructure to
form a module. And the connection between modules is much
easier to handle.
Example 3: Design an NFA M to accept strings of 0’s and 1’s
such that the strings either contain a substring 101 or end with the
symbols 00.
0, 1
0, 1

q0
q1
1
q2
0
q3

q5
0, 1
0
q6
0
q7
1
q4
The transition function  is as follows.

0
1

q0


{q 1, q 5}
q1
{q 1}
{q 1, q 2}

q2
{q 3}


q3

{q 4}

q4
{q 4}
{q 4}

q5
{q 5, q 6}
{q 5}

q6
{q 7}


q7
{q 7}
{q 7}

For the input 10100, M has the following ways to accept it.
Case 1: q 0 10100 |-- q 110100 |--1q 20100 |-- 10q 3100 |-- 101q 400
|-- 1010q 40 |-- 10100q 4
Case 2: q 0 10100 |-- q 510100 |--1q 50100 |-- 10q 5100 |-- 101q 500
|-- 1010q 60 |-- 10100q 7
q0


0, 1
0, 1
q1 1
q5
0, 1
0
q2 0
q6
0
q3 1
q7
q4
Example 4: Design an NFA M to accept strings over {0, 1} so that
the evaluation of the string  according to the following operation
is same as the evaluation of  R.
0
1
0
0
0
1
1
0
Solution :
Let M = ( ={0, 1}, Q, , q 0, F ), where
each state q  Q and q  q 0, q is of the form [f, b, z],
f, b, z {0, 1, },
F = { [z, , z] | z = 0 or 1},
if [g, b, z]  ([f, c, z], a), then g = fa and c = ba, and
(q 0, )={[, a, a] | a = 0 or 1}.
Each state q  Q and q  q 0, q is of the form [f, b, z], where
f is the evaluation of the substring from the initial state up to
current state,
z is a fixed value used to indicate the evaluation of the reversal
of a string ,
and b is the value of the coming substring evaluated in
reverse direction.
The idea is as follows.
(1) If [g, b, z]  ([f, c, z], a), then g = fa, c = ba.
In another word, g = fa is evaluated forward, and c = ba is
evaluate backward.
(2) [z, , z] is a final state which can not proceed any more.
Since to evaluate forward equal to the backward evaluation z, it
should be a final state.
(3) The start state should be of the form [, z, z]. Although
we do not know what will be the input , we can assume the
evaluation of the string  R is z. And it is possible to trace
back.
(4) ([1, 0, 0], 1) = {[0, 0, 0], [0, 1, 0]}, since
considering the first component 0 of [0, 0, 0] and [0, 1, 0],
the forward evaluation 11 = 0, and
considering the second components of [0, 0, 0] and [0, 1, 0],
the backward evaluation 10 R = 01 = 0, and 11R = 11 = 0.
(5) ([, 0, 0], 0) = {[0, , 0], [0, 0, 0]}, where [0, , 0] is a
final state.
(6) An NFA M is listed below.
1
1
0, 1
[0,1,0]
[0,0,0]
0
0
0
1
[0,,0]
1
0
0
[,0,0]
[1,0,0]
1

1
[1,1,0]
q0
0
0

[1,,1]
0
1
1
[,1,1]
0
[0,1,1]
1
[0,,1]
[1,,0]
The expressive power of NFA’s seems more powerful than that of
DFA’s. It is easier to design an NFA than to design a DFA to
accept the same language.
Now we have a problem that whether or not NFA’s are really more
powerful than DFA’s. Is there any language L accepted by an NFA
but not DFA’s?
By the following theorem, we shall see that NFA’s are equivalent to
DFA’s in the sense that if there is a language L accepted by an
NFA, then there is a DFA that accepts L, and vise versa.
We are going to use constructive method to prove the following
theorem , and that is the method we use to convert an NFA to an
equivalent DFA.
First, we need to talk about -closure of a set of states.
Definition 8 : An -closure of a set P of states is the set
 -closure( P ) defined recursively by
(1) P  -closure( P ).
(2) If q  -closure( P ), then ( q,  )  -closure( P ).
Example 5: Considering the following transition diagram,
find -closure( {q 0, q 2} ).
0, 1
0, 1
q0

q1
1
q2

q3
1
q4


q5

q6
0
q7
0, 1
Solution :
( q 0,  ) = {q 1, q 5}, ( q 2,  ) = {q 3, q 7}
( q 1,  ) =  = ( q 3,  ) = ( q 7,  ), ( p 5,  ) = {q 6}
Therefore, -closure( {q 0, q 2} ) = {q 0, q 2, q 1, q 5 , q 6 , q 3, q 7}
Theorem 1: L = L( M 1 ) for some NFA M 1 iff L = L( M 2 )
for some DFA M 2.
Proof :
(1) Suppose that L = L( M 2 ) for some DFA M 2 .
Since a DFA M 2 can be treated as an NFA M 1, where M 1
= M 2, we have that L = L( M 1 ) for some NFA M 1.
(2) Suppose that L = L( M 1 ) for some NFA M 1 = ( , Q,
, q 0, F).
Construct a DFA M 2 = ( , Q’, ’, q 0 ’, F ’), where
Q’=2 Q,
q 0 ’ = -closure( {q 0} ),
F ’ = { P Q’ | P  F  },
’(P, a) =  q P -closure((q, a)),
Then for any input   *, *(q 0 , ) = the set of all
possible states that can be reached after reading the input
 by machine M 1= ’*(q 0 ’, )= the state for the
machine M 2 to enter after reading the input .
We can prove the above statement by induction.
First, if = , we have that *(q 0 , ) = *(q 0 , ) =
-closure( {q 0 } ) = ’*(q 0 ’, ) = q 0 ’.
Second, if || = n and *(q 0 , ) = ’*(q 0 ’, ) = P,
then *(q 0 , a) =  q  P -closure((q, a)) = ’(P,
a) = ’*(q 0 ’, a) for a  .
And it is obvious that L = L(M 1) = L(M 2).
Example 6: Design a DFA M 2 = ( , Q’, ’, q 0 ’, F ’) to accept
the set L of strings of 0’s and 1’s such that the strings end with the
symbols 010.
Solution :
The following is the design of an NFA M 1 = ( , Q, , q 0, F)
accepting L.
0, 1
q0
0
q1
1
q2
0
q3
We now convert the NFA M 1 to an equivalent DFA M 2 . There is
no -move in the machine. Therefore, we do not need to worry
about -closure of a set of states.
First, we have the state q 0 ’ = [q 0] and the transitions
’([q 0] , 0) = [q 0, q 1]) and ’([q 0] , 1) = [q 0]).
0, 1
q0 0
q1 1
M1
M2
1
[q 0]
0
[q 0, q 1]
q2 0
q3
Second, consider the transitions ’([q 0, q 1] , 0) = [q 0, q 1]) and
’([q 0, q 1] , 1) = [q 0 , q 2]).
0, 1
q0 0
q1 1
q2 0
q3
M1
M2
0
1
[q 0]
0
[q 0, q 1]
1
[q 0, q 2]
Third, consider the transitions ’([q 0, q 2] , 0) = [q 0, q 1 , q 3]) and
’([q 0, q 2] , 1) = [q 0]).
0, 1
q0 0
q1 1
q2 0
q3
M1
M2
0
1
[q 0]
0
[q 0, q 1]
1
1
[q 0, q 2]
0
[q 0,q 2,q3]
Fourth, consider the transitions ’([q 0, q 1 , q 3] , 0) = [q 0, q 1]) and
’([q 0, q 1 , q 3] , 1) = [q 0 , q 2]), done!
0, 1
q0 0
q1 1
q2 0
q3
M1
M2
0
1
[q 0]
0
[q 0, q 1]
1
1
[q 0, q 2]
1
0
[q 0,q 1,q3]
0
Example 7: Design a DFA M 2 = ( , Q’, ’, q 0 ’, F ’) to accept
the set L of strings of 0’s and 1’s such that the strings either
contain a substring 101 or end with the symbols 00.
Solution :
The following is the design of an NFA M 1 = ( , Q, , q 0, F)
accepting L.
0, 1
0, 1

q0
q1
1
q2
0
q3
1

q5
0
q6
0
q7
0, 1
We now convert the NFA M 1 to an equivalent DFA M 2 .
q4
First, according to a part of transition diagram as follows
we can get the state q 0 ’ = [q 0, q 1, q 5].
q0
M1


0, 1
0, 1
q1 1
q5
0, 1
M2
[q 0, q 1, q 5]
0
q2 0
q6
0
q3 1
q7
q4
Second, ’ (q 0 ’, 0) = [q 1, q 5 , q 6] and ’ (q 0 ’, 1) = [q 1, q 2, q 5].
q0


M1
0, 1
0, 1
q1 1
q5
0, 1
M2
0
q2 0
q6
0
q3 1
q7
0
[q 1, q 5 , q 6]
1
[q 1, q 2 , q 5]
[q 0, q 1, q 5]
q4
Third, ’ ( [q 1, q 5 , q 6], 0 )= [q 1, q 5 , q 6 , q 7], and
’ ([q 1, q 5 , q 6], 1) = [q 1, q 2, q 5], .
q0


M1
0, 1
0, 1
q1 1
q5
0, 1
0
q2 0
q6
0
q3 1
q4
q7
0
0
M2
[q 1, q 5 , q 6]
1
[q 0, q 1, q 5]
1
[q 1, q 2 , q 5]
[q 1, q 5 ,q 6 , q 7]
Fourth, ’ ( [q 1, q 2 , q 5], 0 )= [q 1, q 3 , q 5 , q 6], and
’ ([q 1, q 2 , q 5], 1) = [q 1, q 2, q 5], .
q0


M1
0, 1
0, 1
q1 1
q5
0, 1
0
q2 0
q6
0
q3 1
q4
q7
0
0
M2
[q 1, q 5 , q 6]
1
[q 0, q 1, q 5]
1
[q 1, q 5 ,q 6 , q 7]
0
[q 1, q 2 , q 5]
1
[q 1 , q 3, q 5 ,q 6]
Fifth, ’ ( [q 1, q 3 , q 5 , q 6], 0 )= [q 1, q 5 , q 6 , q 7], and
’ ([q 1, q 3 , q 5 , q 6], 1) = [q 1, q 2, q 4 , q 5 ], .

q0

0, 1
0, 1
q1 1
q2 0
q5
0, 1
M1
q6
0
q3 1
q7
0
0
M2
0
[q 0, q 1, q 5]
[q 1, q 5 , q 6]
[q 1, q 2 , q 5]
1
[q 1, q 5 ,q 6 , q 7]
0
1
1
q4
0
[q 1 , q 3, q 5 ,q 6]
1
[q1,q2,q4 ,q5]
Sixth, ’ ( [q 1, q 5 , q 6 , q 7], 0 )= [q 1, q 5 , q 6 , q 7], and
’ ([q 1, q 5 , q 6 , q 7], 1) = [q 1, q 2, q 5 ], .

q0

M1
0, 1
0, 1
q1 1
q5
0
0, 1
q2 0
q6
0
M2
0
[q 0, q 1, q 5]
1
[q 1, q 5 , q 6]
1
1
q4
q7
0
0
[q 1, q 5 ,q 6 , q 7]
1
0
[q 1, q 2 , q 5]
q3 1
0
[q 1 , q 3, q 5 ,q 6]
1
[q1,q2,q4 ,q5]
Seventh, ’ ( [q 1, q 2 , q 4 , q 5], 0 )= [q 1, q 3 , q 4 , q 5 , q 6], and
’ ([q 1, q 2 , q 4 , q 5], 1) = [q 1, q 2 , q 4, q 5 ], .

q0

M1
0, 1
0, 1
q1 1
q5
0
0, 1
q2 0
q6
0
M2
0
[q 0, q 1, q 5]
1
[q 1, q 5 , q 6]
1
1
q4
q7
0
0
[q 1, q 5 ,q 6 , q 7]
1
0
[q 1, q 2 , q 5]
q3 1
[q1,q3,q4 ,q5 ,q6]
0
[q 1 , q 3, q 5 ,q 6]
0
1
[q1,q2,q4 ,q5]
1
Eighth, ’ ( [q 1, q 3 , q 4 , q 5 , q 6], 0 )= [q 1, q 4 , q 5 , q 6 , q 7], and
’ ([q 1, q 3 , q 4 , q 5 , q 6], 1) = [q 1, q 2 , q 4, q 5 ], .

q0

M1
0, 1
0, 1
q1 1
q5
0
0, 1
q2 0
q6
0
M2
0
[q 0, q 1, q 5]
1
[q 1, q 5 , q 6]
1
1
q4
q7
0
0
[q1,q4,q5 ,q6 ,q7]
[q 1, q 5 ,q 6 , q 7]
1
0
[q 1, q 2 , q 5]
q3 1
0
[q1,q3,q4 ,q5 ,q6]
0
[q 1 , q 3, q 5 ,q 6]
0
1
[q1,q2,q4 ,q5]
1
1
Ninth, ’ ( [q 1, q 4 , q 5 , q 6 , q 7], 0 )= [q 1, q 4 , q 5 , q 6 , q 7], and
’ ([q 1, q 4 , q 5 , q 6 , q 7], 1) = [q 1, q 2 , q 4, q 5 ], .

q0

M1
0, 1
0, 1
q1 1
q5
0
0, 1
q2 0
q6
0
M2
0
[q 0, q 1, q 5]
1
[q 1, q 5 , q 6]
1
1
q4
q7
0
0
0
[q1,q4,q5 ,q6 ,q7]
[q 1, q 5 ,q 6 , q 7]
1
0
[q 1, q 2 , q 5]
q3 1
0
[q1,q3,q4 ,q5 ,q6]
0
[q 1 , q 3, q 5 ,q 6]
0
1
[q1,q2,q4 ,q5]
1
1
1
After rewrite the machine M2, we come up with the following result.
0
q0
0
q1
1
1
1
0
q3
q2
0
q4
0
0
0
q6
1
0
1
1
q5
1
The above machine can be simplified as follows.
0
q0
q1
1
1
1
q3
1
0
0
q2
0
q4
q7
0
0, 1
1
q5
Example 8: Design a DFA M to accept strings over {0, 1} so that
the evaluation of the string  according to the following operation
is same as the evaluation of  R.
0
1
0
0
0
1
1
0
Solution : Convert the previous NFA in example 4 to the
following DFA.
0
0
q0
1
q4
0, 1
1
q1
q2
1
0
q5
1
0
0, 1
q3