1. Another NFA Example
0
q0 q1 0, 1 q2
1

Fall 2004 COMP 335 1

2. Language accepted
L(M ) = {λ, 10, 1010, 101010, ...}
= {10}*
0
q0 q1 0, 1 q2
1 (redundant
state)

Fall 2004 COMP 335 2

3. Remarks:
•The  symbol never appears on the
input tape
•Simple automata:
M1 M2
q0 q0
L(M1 ) = {} L(M 2 ) = {λ}
Fall 2004 COMP 335 3

4. •NFAs are interesting because we can
express languages easier than DFAs
NFA M1 DFA M2 a
q2
q0 a q1
a
q0 a q1
L( M1 ) = {a} L( M 2 ) = {a}
Fall 2004 COMP 335 4

5. Formal Definition of NFAs
M  Q, ,  , q0 , F 
Q : Set of states, i.e. q0 , q1, q2 
 : Input aphabet, i.e. a, b
: Transition function
q0 : Initial state
F : Final states
Fall 2004 COMP 335 5

6. Transition Function 
 q0 , 1  q1
0
q0 q1 0, 1 q
2
1

Fall 2004 COMP 335 6

7.  (q1,0)  {q0 , q2}
0
q0 q1 0, 1 q
2
1

Fall 2004 COMP 335 7

8.  (q0 ,  )  {q0 , q2}
0
q0 q1 0, 1 q
2
1

Fall 2004 COMP 335 8

9.  (q2 ,1)  
0
q0 q1 0, 1 q
2
1

Fall 2004 COMP 335 9

10. Extended Transition Function  *
 * q0 , a   q1
q4 q5
a a
q0 a q1 b q2  q3

Fall 2004 COMP 335 10

11.  * q0 , aa   q4 , q5
q4 q5
a a
q0 a q1 b q2  q3

Fall 2004 COMP 335 11

12.  * q0 , ab  q2 , q3 , q0 
q4 q5
a a
q0 a q1 b q2  q3

Fall 2004 COMP 335 12

13. Formally
q j   * qi , w : there is a walk from qi to q j
with label w
qi w qj
w  1 2  k
1 2 k
qi qj
Fall 2004 COMP 335 13

14. The Language of an NFA M
F  q0 ,q5
q4 q5
a a
q0 a q1 b q2  q3

 * q0 , aa   q4 , q5 aa  L(M )
F
Fall 2004 COMP 335 14

15. F  q0 ,q5
q4 q5
a a
q0 a q1 b q2  q3

 * q0 , ab   q2 , q3 , q0  ab LM 
F
Fall 2004 COMP 335 15

16. F  q0 ,q5
q4 q5
a a
q0 a q1 b q2  q3

 * q0 , abaa   q4 , q5 abaa L(M )
F
Fall 2004 COMP 335 16

17. F  q0 ,q5
q4 q5
a a
q0 a q1 b q2  q3

 * q0 , aba   q1 aba  LM 
F
Fall 2004 COMP 335 17

18. q4 q5
a a
q0 a q1 b q2  q3

LM   ab*  {ab}*{aa}
Fall 2004 COMP 335 18

19. Formally
The language accepted by NFA M is:
LM   w1, w2 , w3 ,...
where  * (q0 , wm )  {qi , q j ,..., qk ,}
and there is some qk  F (final state)
Fall 2004 COMP 335 19

20. w LM   * (q0 , w)
qi
w
q0
qk qk  F
w
w qj
Fall 2004 COMP 335 20

21. NFA accept Regular Languages
Fall 2004 COMP 335 21

22. Equivalence of FA
Definition:
An FA M1 is equivalent to FA M2
if LM1   LM 2 
that is if both accept the same language.
Fall 2004 COMP 335 22

23. Example of equivalent FA
NFA M1
LM1   {10} * 0
q0 q1
1
DFA M2 0,1
LM 2   {10}* 0
q0 q1 1 q2
1
0
Fall 2004 COMP 335 23

24. We will prove:

Languages
Regular
accepted
Languages
by NFA
Languages
accepted
by DFA
That is, NFA and DFA have
the same computation power
Fall 2004 COMP 335 24

25. Step 1
Languages
accepted  Regular
Languages
by NFA
Proof: Every DFA is trivially an NFA
Any language L accepted by a DFA
is also accepted by an NFA
Fall 2004 COMP 335 25

26. Step 2
Languages
accepted  Regular
Languages
by NFA
Proof: Any NFA can be converted into an
equivalent DFA
Any language accepted by an NFA
is also accepted by a DFA
Fall 2004 COMP 335 26

27. Convert NFA to DFA
NFA M
a
q a q 0
 q 1 2
b
DFA M
q0 
Fall 2004 COMP 335 27

28. Convert NFA to DFA
NFA M
a
q a q 0
 q 1 2
b
DFA M
q0  a
q1, q2 
Fall 2004 COMP 335 28

29. Convert NFA to DFA
NFA M
a
q a q 0
 q 1 2
b
DFA M
q0  a
q1, q2 
b

Fall 2004 COMP 335 29

30. Convert NFA to DFA
NFA M
a
q a q 0
 q 1 2
b
a
DFA M
q0  a
q1, q2 
b

Fall 2004 COMP 335 30

31. Convert NFA to DFA
NFA M
a
q a q 0
 q 1 2
b
a
DFA M b
q0  a
q1, q2 
b

Fall 2004 COMP 335 31

32. Convert NFA to DFA
NFA M
a
q a q 0
 q 1 2
b
a
DFA M b
q0  a
q1, q2 
b
 a, b
Fall 2004 COMP 335 32

33. Convert NFA to DFA
NFA M
a
q a q 0
 q 1 2
LM   L(M )
b
a
DFA M b
q0  a
q1, q2 
b
 a, b
Fall 2004 COMP 335 33

34. NFA to DFA: Remarks
We are given an NFA M
We want to convert it
into an equivalent DFA M
That is, LM   L(M )
Fall 2004 COMP 335 34

35. If the NFA has states
q0 , q1, q2 ,...
Then the DFA has states in the powerset
, q0 , q1, q1 , q2 ,....
Fall 2004 COMP 335 35

36. Procedure NFA to DFA
1. Initial state of NFA: q0
Initial state of DFA: q0 
Fall 2004 COMP 335 36

37. Example
NFA M a
q0 a q1  q2
b
DFA M
q0 
Fall 2004 COMP 335 37

38. Procedure NFA to DFA
2. For every DFA’s state {qi , q j ,..., qm}
Compute in the NFA
 * qi , a ,
 * q j , a ,  {qi , qj ,..., qm}

...
Add the following transition to the DFA
 {qi , q j ,..., qm}, a   {qi , qj ,..., qm}

Fall 2004 COMP 335 38

39. Example
NFA M a
q0 a q1  q2
b
 * (q0 , a )  {q1, q2 }
DFA M
q0  a
q1, q2 
 q0 , a   q1, q2 
Fall 2004 COMP 335 39

40. Procedure NFA to DFA
Repeat step 2 for all symbols in the alphabet
∑, until no more transitions can be added.
Fall 2004 COMP 335 40

41. Example
NFA M a
q0 a q1  q2
b
a
DFA M b
q0  a
q1, q2 
b
 a, b
Fall 2004 COMP 335 41

42. Procedure NFA to DFA
3. For any DFA state: {qi , q j ,..., qm}
If some qj is a final state in the NFA
Then, {qi , q j ,..., qm }
is a final state in the DFA
Fall 2004 COMP 335 42

43. Example
NFA M a
q0 a q1  q2 q1  F
b
a
DFA M b
q0  a
q1, q2 
b q1, q2  F 
 a, b
Fall 2004 COMP 335 43

44. Theorem
Take NFA M
Apply the procedure to obtain DFA M
Then, M and M  are equivalent:
LM   LM 
Fall 2004 COMP 335 44

45. Proof
LM   LM 
L  M   L M   AND L  M   L M  
Fall 2004 COMP 335 45

46. First we show: L  M   L M  
Take arbitrary string : w L (M )
We will prove: w  L(M )
Fall 2004 COMP 335 46

47. w L(M )
M: q0 w qf
w  1 2  k
1 2 k
M: q0 qf
Fall 2004 COMP 335 47

48. We will show that if w L(M )
w  1 2  k
1 2 k
M: q0 qf
1 2 k
M:
{q0} {q f ,}
w  L(M )
Fall 2004 COMP 335 48

49. More generally, we will show that if in M:
(arbitrary string) v  a1a2 an
a1 a2 an
M: q0 qi qj ql qm
M:
a1 a2 an
{q0} {qi ,} {q j ,} {ql ,} {qm ,}
Fall 2004 COMP 335 49

50. Proof by induction on |v|
The basis case: v  a1
a1
M: q0 qi
M:
a1
{q0} {qi ,}
Fall 2004 COMP 335 50

51. Induction hypothesis: 1 | v | k
v  a1a2 ak
a1 a2 ak
M: q0 qi qj qc qd
M:
a1 a2 ak
{q0} {qi ,} {q j ,} {qc ,} {qd ,}
Fall 2004 COMP 335 51

52. Induction Step: | v | k  1
v  a1a2 ak ak 1  vak 1
 
 
v
a1 a2 ak
M: q0 qi qj qc qd
v
M:
a1 a2 ak
{q0} {qi ,} {q j ,} {qc ,} {qd ,}
Fall 2004
v
COMP 335 52

53. Induction Step: | v | k  1
v  a1a2 ak ak 1  vak 1
 
 
v
a1 a2 ak ak 1
M: q0 qi qj qc qd qe
v
M:
a1 a2 ak ak 1
{q0} {qi ,} {q j ,} {qc ,} {qd ,} {qe ,}
Fall 2004
v
COMP 335 53

54. Therefore if w L(M )
w  1 2  k
1 2 k
M: q0 qf
1 2 k
M:
{q0} {q f ,}
w  L(M )
Fall 2004 COMP 335 54

55. We have shown: L  M   L M  
We also need to show: L  M   L M  
(proof is similar)
Fall 2004 COMP 335 55