1 / 17

DFA vs NFA

DFA vs NFA. Deterministic Finite Automata. Non-Deterministic Finite Automata. Characterized as 5-tuple < S, A, T, s 0 , F > S is the set of states A is the alphabet T is the transition function: S x A  S s 0 is the start state F is the set of accepting states.

rooney-sparks
Download Presentation

DFA vs NFA

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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  0 0 {1} ø {3} a 1 ø {2} ø  2 ø ø ø 3 {3} ø {4}  4 3 a 4 {2} ø ø Waseem Besada

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

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

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

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

  10. Solution to exercise 5 NFA transition table DFA transition table T a b a b  0 Ø {1,2} {1}1 {2} Ø Ø 2 Ø {2} {1} {0,1} ? ? {1,2} {1,2} 0 {1,2} {1,2} {1,2} 1 a,b a,b 0 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 0 0 0 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 0 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} Transition table for DFA a {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   0 1 6 6 0 1   b 4 5  Waseem Besada

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

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

More Related