Non-Deterministic
Finite Automata
Fall 2006
Costas Busch - RPI
1
Nondeterministic Finite Automaton (NFA)
Alphabet = {a}
a
q0
q1 a
q2
a
q3
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Costas Busch - RPI
2
Alphabet = {a}
Two choices
a
q0
q1 a
q2
a
q3
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Costas Busch - RPI
3
Alphabet = {a}
Two choices
a
q0
q1 a
No transition
a
q3
Fall 2006
q2
No transition
Costas Busch - RPI
4
First Choice
a a
a
q0
q1 a
q2
a
q3
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Costas Busch - RPI
5
First Choice
a a
a
q0
q1 a
q2
a
q3
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6
First Choice
a a
All input is consumed
a
q0
q1 a
q2
“accept”
a
q3
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Second Choice
a a
a
q0
q1 a
q2
a
q3
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Second Choice
a a
Input cannot be consumed
a
q0
a
q1 a
Automaton Halts
q3
Fall 2006
q2
“reject”
Costas Busch - RPI
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An NFA accepts a string:
if there is a computation of the NFA
that accepts the string
i.e., all the input string is processed and the
automaton is in an accepting state
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Costas Busch - RPI
10
aa is accepted by the NFA:
“accept”
a
q0
q1 a
q2
a
q0
a
q3
because this
computation
accepts aa
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q1 a
a
q3
q2
“reject”
this computation
is ignored
Costas Busch - RPI
11
Rejection example
a
a
q0
q1 a
q2
a
q3
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Costas Busch - RPI
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First Choice
a
“reject”
a
q0
q1 a
q2
a
q3
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Second Choice
a
a
q0
q1 a
q2
a
q3
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14
Second Choice
a
a
q0
q1 a
a
q3
Fall 2006
q2
“reject”
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Another Rejection example
a a a
a
q0
q1 a
q2
a
q3
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First Choice
a a a
a
q0
q1 a
q2
a
q3
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First Choice
a a a
Input cannot be consumed
a
q0
q1 a
a
q3
Fall 2006
q2
“reject”
Automaton halts
Costas Busch - RPI
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Second Choice
a a a
a
q0
q1 a
q2
a
q3
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Second Choice
a a a
Input cannot be consumed
a
q0
q1 a
Automaton halts
a
q3
Fall 2006
q2
“reject”
Costas Busch - RPI
20
An NFA rejects a string:
if there is no computation of the NFA
that accepts the string.
For each computation:
• All the input is consumed and the
automaton is in a non final state
OR
• The input cannot be consumed
Fall 2006
Costas Busch - RPI
21
a
is rejected by the NFA:
“reject”
a
q0
q1 a
q2
a
q0
a
q3
“reject”
q1 a
q2
a
q3
All possible computations lead to rejection
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Costas Busch - RPI
22
aaa
is rejected by the NFA:
“reject”
a
q0
q1 a
q2
a
q0
a
q3
q1 a
a
q3
q2
“reject”
All possible computations lead to rejection
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Language accepted:
a
q0
L  {aa}
q1 a
q2
a
q3
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Lambda Transitions
q0 a
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q1 
q2 a
Costas Busch - RPI
q3
25
a a
q0 a
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q1 
q2 a
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q3
26
a a
q0 a
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q1 
q2 a
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q3
27
input tape head does not move
a a
q0 a
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q1 
q2 a
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q3
28
all input is consumed
a a
“accept”
q0 a
String
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q1 
q2 a
q3
aa is accepted
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Rejection Example
a a a
q0 a
Fall 2006
q1 
q2 a
Costas Busch - RPI
q3
30
a a a
q0 a
Fall 2006
q1 
q2 a
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q3
31
(read head doesn’t move)
a a a
q0 a
Fall 2006
q1 
q2 a
Costas Busch - RPI
q3
32
Input cannot be consumed
a a a
Automaton halts
“reject”
q0 a
String
Fall 2006
aaa
q1 
q2 a
q3
is rejected
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Language accepted:
q0 a
Fall 2006
q1 
L  {aa}
q2 a
Costas Busch - RPI
q3
34
Another NFA Example
q0
a
b
q1
q2

q3

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a b
q0
a
b
q1
q2

q3

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a b
q0
a
b
q1
q2

q3

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a b
“accept”
q0
a
b
q1
q2

q3

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Another String
a b a b
q0
a
b
q1
q2

q3

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a b a b
q0
a
b
q1
q2

q3

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a b a b
q0
a
b
q1
q2

q3

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a b a b
q0
a
b
q1
q2

q3

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a b a b
q0
a
b
q1
q2

q3

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a b a b
q0
a
b
q1
q2

q3

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a b a b
“accept”
q0
a
b
q1
q2

q3

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Language accepted
L  ab, abab, ababab, ...

 ab
q0
a
b
q1
q2

q3

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Costas Busch - RPI
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Another NFA Example
0
q0
1
q1
0, 1 q2

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Language accepted
L(M ) = {λ, 10, 1010, 101010, ...}
= {10} *
0
q0
1
q1
0, 1 q2

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Costas Busch - RPI
(redundant
state)
48
Remarks:
•The  symbol never appears on the
input tape
•Simple automata:
Fall 2006
M1
q0
M2
L(M1) = {}
L(M 2 ) = {λ}
q0
Costas Busch - RPI
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•NFAs are interesting because we can
express languages easier than DFAs
NFA
q0
a
DFA
M1
q2
q1
a
q0
L( M1) = {a}
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a
M2
a
q1
L( M 2 ) = {a}
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50
Formal Definition of NFAs
M  Q, ,  , q0 , F 
Q : Set of states, i.e. q0 , q1, q2 
 : Input aplhabet, i.e. a, b
:
q0 :
 
Transition function
Initial state
F : Accepting states
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Costas Busch - RPI
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Transition Function 
 q , x   q1, q2,, qk 
q
x
x
q1
q1
x
resulting states with
following one transition
with symbol x
qk
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 q0 , 1  q1
0
q0
1
q1
0, 1 q
2

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 (q1,0)  {q0 , q2}
0
q0
1
q1
0, 1 q
2

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 (q0 , )  {q2 }
0
q0
1
q1
0, 1 q
2

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 (q2 ,1)  
0
q0
1
q1
0, 1 q
2

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Extended Transition Function 
Same with

*
but applied on strings
 q0, a   q1 
*
q5
q4
a
q0
a
a
b
q1
q2

q3

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 q0, aa   q4, q5 
*
q5
q4
a
q0
a
a
b
q1
q2

q3

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 q0, ab   q2, q3, q0 
*
q5
q4
a
q0
a
a
b
q1
q2

q3

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Special case:
for any state q
q   q ,  
*
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In general
q j   qi ,w  : there is a walk from qi to q j
with label w
*
w
qi
qj
w  1 2  k
qi
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1
k
2
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qj
61
The Language of an NFA M
The language accepted by
M
is:
LM   w1,w2,...wn 
where
 (q0 ,wm )  {qi ,...,qk ,, q j }
*
and there is some
Fall 2006
qk  F
Costas Busch - RPI
(accepting state)
62
wm  LM 
 (q0,wm )
*
wm
qi
q0 w
m
qk
wm
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qk  F
qj
63
F  q0 ,q5
q5
q4
a
a
q0
a
b
q1
q2

q3


 q0 , aa   q4 , q5
*
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
aa  L(M )
F
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F  q0 ,q5
q5
q4
a
a
q0
a
b
q1
q2

q3


 q0 , ab   q2, q3, q0
*
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
ab LM 
F
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F  q0 ,q5
q5
q4
a
q0
a
a
b
q1
q2

q3


 q0 , abaa   q4 , q5
*
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
aaba L(M )
F
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F  q0 ,q5
q5
q4
a
q0
a
a
b
q1
q2

q3

 * q0, aba   q1
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F
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aba LM 
67
q5
q4
a
q0
a
a
b
q1
q2

q3

LM   ab *  ab * {aa }
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NFAs accept the Regular
Languages
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Equivalence of Machines
Definition:
Machine
if
Fall 2006
M1
is equivalent to machine
M2
LM1   LM 2 
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Example of equivalent machines
NFA
LM1   {10}*
0
q0
q1
1
DFA
LM 2   {10}*
M1
M2
0,1
0
q0
1
q1
1
q2
0
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71
Theorem:
Languages
accepted
by NFAs

Regular
Languages
Languages
accepted
by DFAs
NFAs and DFAs have the same computation power,
accept the same set of languages
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72
Proof:
we only need to show
Languages
accepted
by NFAs

Regular
Languages
AND
Languages
accepted
by NFAs
Fall 2006

Costas Busch - RPI
Regular
Languages
73
Proof-Step 1
Languages
accepted
by NFAs

Regular
Languages
Every DFA is trivially an NFA
Any language L accepted by a DFA
is also accepted by an NFA
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Costas Busch - RPI
74
Proof-Step 2
Languages
accepted
by NFAs

Regular
Languages
Any NFA can be converted to an
equivalent DFA
Any language L accepted by an NFA
is also accepted by a DFA
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75
Conversion NFA to DFA
a
NFA M
a
 q
q
q
0
1
2
b
DFA
M
q0 
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76
NFA
 * (q0, a )  {q1, q2 }
a
M
q0
a

q1
q2
b
DFA
M
q0 
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a
q1, q2 
Costas Busch - RPI
77
NFA
 * (q0 , b )  
a
M
q0
a
empty set

q1
q2
b
DFA
M
q0 
a
q1, q2 

trap state
b
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78
 (q1, a )  {q1, q2 }
 * (q2, a )  
*
NFA
a
M
q0
a

q1
union
q2
q1, q2 
b
DFA
a
M
q0 
a
q1, q2 
b

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79
 (q1, b )  {q0 }
 * (q2, b )  {q0 }
*
NFA
a
M
q0
a

q1
union
q2
q0 
b
DFA
M
a
b
q0 
a
q1, q2 
b

Fall 2006
Costas Busch - RPI
80
NFA
a
M
q0
a

q1
q2
b
DFA
M
a
b
q0 
a
q1, q2 
b

Fall 2006
Costas Busch - RPI
a, b trap state
81
END OF CONSTRUCTION
NFA
a
M
q0
a

q1
q1  F
q2
b
DFA
M
a
b
q0 
a
q1, q2 
b

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Costas Busch - RPI
q1, q2  F 
a, b
82
General Conversion Procedure
Input: an NFA
M
Output: an equivalent DFA M 
with LM   L(M )
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83
The NFA has states
q0 , q1, q2 ,...
The DFA has states from the power set
, q0 , q1 , q0 , q1 , q1, q2, q3, ....
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84
Conversion Procedure Steps
step
1.
Fall 2006
Initial state of NFA:
q0
Initial state of DFA:
q0 
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85
Example
NFA
a
M
q0
a

q1
q2
b
DFA
M
q0 
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86
step
2. For every DFA’s state {qi , q j ,..., qm}
compute in the NFA
 * qi , a 
  * q j , a 
...
Union
 {qk , ql,...,qn }
  * qm , a 
add transition to DFA
 {qi , qj ,...,qm }, a   {qk , ql,...,qn }
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87
Example
NFA
 * (q0 , a)  {q1, q2}
a
M
a
q0

q1
q2
b
DFA
M
 q0 , a   q1, q2 
q0 
Fall 2006
a
q1, q2 
Costas Busch - RPI
88
step
3. Repeat Step 2 for every state in DFA and
symbols in alphabet until no more states
can be added in the DFA
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89
Example
NFA
a
M
q0
a

q1
q2
b
DFA
M
a
b
q0 
a
q1, q2 
b

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Costas Busch - RPI
a, b
90
step
4. For any DFA state {qi , q j ,..., qm}
if some
q j is accepting state in NFA
Then, {qi , q j ,..., qm }
is accepting state in DFA
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91
Example
NFA
a
M
q0
a

q1
q1  F
q2
b
DFA
M
a
b
q0 
a
q1, q2 
b

Fall 2006
Costas Busch - RPI
q1, q2  F 
a, b
92
Lemma:
If we convert NFA M to DFA M 
then the two automata are equivalent:
L  M   L M  
Proof:
We only need to show:
LM   LM 
AND
LM   LM 
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93
First we show:
LM   LM 
We only need to prove:
w L(M )
Fall 2006
w  L(M )
Costas Busch - RPI
94
NFA
Consider
w L(M )
w
q0
qf
symbols
w  1 2  k
q0
Fall 2006
1
2
Costas Busch - RPI
k
qf
95
symbol
qi
i
qj
denotes a possible sub-path like
qi
Fall 2006


symbol
i
Costas Busch - RPI

qj
96
We will show that if
w L(M )
w  1 2  k
NFA
M:
q0
1
k
2
qf
then
DFA
1
M:
{q0}
Fall 2006
state
label
2
w  L(M )
Costas Busch - RPI
k
{q f ,}
state
label
97
More generally, we will show that if in
(arbitrary string)
NFA
M:
q0
M:
v  a1a2 an
a1
qi
a2
qj
ql
an
qm
then
DFA
Fall 2006
M:
a1
a2
{q0} {qi ,} {q j ,}
Costas Busch - RPI
an
{ql ,} {qm ,}
98
Proof by induction on
v  a1
Induction Basis: |v | 1
NFA
M:
DFA
M:
q0
|v|
a1
qi
a1
{q0}
{qi ,}
is true by construction of M 
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Costas Busch - RPI
99
Induction hypothesis:
1 | v | k
v  a1a2 ak
Suppose that the following hold
NFA
M:
DFA
M:
q0
a1
{q0}
Fall 2006
a1
qi
a2
qj
a2
{qi ,} {q j ,}
Costas Busch - RPI
qc
ak
qd
ak
{qc ,} {qd ,}
100
| v | k  1
v  a1a2 ak ak 1  vak 1


Induction Step:
v
Then this is true by construction of M 
NFA
M:
q0
a1
qi
a2
qj
qc
ak
qd
ak 1
qe
v
DFA
M:
a1
{q0}
Fall 2006
ak
a2
{qi ,} {q j ,}
v
Costas Busch - RPI
ak 1
{qc ,} {qd ,}
{qe ,}
101
Therefore if
w L(M )
w  1 2  k
NFA
M:
1
q0
k
2
qf
then
DFA
M:
1
{q0}
Fall 2006
2
w  L(M )
Costas Busch - RPI
k
{q f ,}
102
We have shown:
With a similar proof
we can show:
Therefore:
LM   LM 
LM   LM 
LM   LM 
END OF LEMMA PROOF
Fall 2006
Costas Busch - RPI
103