1. DFA vs NFA Deterministic Finite Automata
Non-Deterministic Finite Automata
Characterized as 5-tuple < S, A, T, s0, F >
S is the set of states
A is the alphabet
T is the transition function: S x A  S
s0 is the start state
F is the set of accepting states.
Characterized as 5-tuple < S, A, T, s0, F >
S is the set of states
A is the alphabet
T is the transition function: S x (A{})  PS
s0 is the start state
F is the set of accepting states.
Waseem Besada

2. From NFA To DFA 1 Transition function for NFA b a 2 T a b 
{1} ø {3} a
ø {2} ø 
ø ø ø
{3} ø {4}  4 3 a
{2} ø ø
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3. Calculating The Transition function for DFA
Transition function for NFA
Transition function for DFA
T a b 
T a b
{1} ø {3}
{0,3,4} ? ?
ø {2} ø
TD({0,3,4},a) = ({0},a)  ({3},a)  ({4},a)
= {1,3}  {3,4}  {2}
= {1,2,3,4}
ø ø ø
{3} ø {4}
TD({0,3,4},b) = ({0},b)  ({3},b)  ({4},b)
= ø  ø  ø
= ø
{2} ø ø
Waseem Besada

4. Calculating The Transition function for DFA (forts)
Transition function for NFA
T a b
T a b 
{0,3,4} {1,2,3,4} ø
{1} ø {3}
ø ? ? ø ø
ø {2} ø
{1,2,3,4} ? ?
{2,3,4}
{2}
ø ø ø
{3} ø {4}
TD({1,2,3,4},a) = ({1},a)  ({2},a)  ({3},a)  ({2},a)
= ø  ø  {3,4}  {2}
= {2,3,4}
Similarly,
TD({1,2,3,4},b) = {2}
{2} ø ø
Waseem Besada

5. Calculating The Transition function for DFA (forts)
Transition function for NFA
T a b
T a b 
{0,3,4} {1,2,3,4} ø
{1} ø {3}
ø ø ø
ø {2} ø
{1,2,3,4} {2,3,4} {2}
ø ø ø
{2,3,4} {2,3,4} {2}
{3} ø {4}
{2} ø ø
{2} ø ø
Waseem Besada

6. Finally 1 2 3 4 Transition function for DFA T a b ø ? ? ø ø b 4 b a,b1 2 3 4
{0,3,4} {1,2,3,4} ø 2 1
ø ? ? ø ø a b
{1,2,3,4} {2,3,4} {2} 3
{2,3,4} {2,3,4} {2} a
{2} ø ø
Waseem Besada

7. Exercise 2, solution
2. Construct directly (that is don’t use the transformation procedure given in the lecture) a DFA for each of the following REs.
a) a | b b) a | b* c) ab* | bc* b a
a,b
a,b a
a,c
a,b,c b
a,b a
a,b b
a,b
a,b c a) b c c) a b) b
Waseem Besada

8. Exercise 4, solution
4. Construct directly (that is don’t use the transformation procedure given in the lecture) a NFA for each of the following REs.
a) a*bc*| bc
b) a* | ab a c a  b  a c a b b) a)
Waseem Besada

9. Exercise 5
5. Construct a DFA table for the following NFA transition table The start state is 0 and there is one accepting state, 2.
a b 
0 Ø {1,2} {1} {2} Ø Ø
2 Ø {2} {1}  b 1 a  b 2 b
Waseem Besada

10. Solution to exercise 5 DFA transition table NFA transition table T a b
0 Ø {1,2} {1} {2} Ø Ø
2 Ø {2} {1}
{0,1} ? ?
{1,2}
{1,2}
{1,2}
{1,2}
{1,2} 1
a,b
a,b 1
Waseem Besada

11. Exercise 6 6. Consider the following NFA over the alphabet {a,b}:
a) Find a regular expression for the language accepted by the NFA.
b) Write down the transition table for the NFA.
c) Transform the NFA into a DFA
d) Draw a picture of the resulting DFA. a  b a 
Waseem Besada

12. Solution to exercise 6 (cont’d)
Transition table for NFA
T a b 
0 {1} Ø {1,2}
1 Ø {2} Ø
2 {2} Ø Ø a  b 1 2 a 
Waseem Besada

13. Solution to exercise 6 (cont’d)
Transition table for NFA
Transition table for DFA
T a b
{0,1,2} {1,2} {2}
{1,2} {2} {2}
{2} {2} Ø
Ø Ø Ø
T a b 
0 {1} Ø {1,2}
1 Ø {2} Ø
2 {2} Ø Ø
Waseem Besada

14. Solution to exercise 6(cont’d)
{0,1,2} a
Transition table for DFA
{1,2}
T a b
{0,1,2} {1,2} {2}
{1,2} {2} {2}
{2} {2} Ø
Ø Ø Ø b
a,b
{2} a
Waseem Besada

15. Exercise 7
7. Transform each of the following REs into a NFA, then into a DFA
a) a*b* b) a* | b* c) (a|b)*
Solution to a)  a b a      2 3   1 6 6 1   b 4 5 
Waseem Besada

16. Solution, exercise 6-a T a b Can be reduced to:
{0,1,2,3} {1,2,3} {2,3}
{1,2,3} {1,2,3} {2,3}
{2,3} Ø {2,3}
Ø Ø Ø a b    1 2 3 a a
Can be reduced to: a
a,b 1 b a b
a,b 2 3 b a 2 3 b b
Waseem Besada

17. It is your turn to solve the rest of the exercises
What to do next?
It is your turn to solve the rest of the exercises
Waseem Besada