1. CSCI 3130: Formal languages and automata theory Tutorial 1 Lee Chin Ho

2. About me Name: Chin Office: SHB 117 Office hour: Tuesday 11:00 – 12:00 Homework 1 is due on next Thursday

3. Finite Automata 1. Q - States 2.  - Alphabets 3.  - Transitions 4. q 0 - Initial state 5. F - Final state(s) / Accepting state(s)

4. Deterministic Finite Automata There must be a transition corresponding to each alphabet at each state

5. Nondeterministic Finite Automata Can be at multiple states at the same time Can go to multiple states on one alphabet May have no transitions on an alphabet (die)

6. Regular Expression The symbols  and  are regular expressions Every a in  is a regular expression If R and S are regular expressions, so are R+S, RS and R* Remember for R*, * could be 0

7. Regular Languages Languages that can be represented by a DFA / NFA / RE DFA = NFA = RE When asked if a language L is regular, represent L using a DFA / NFA / RE

8. Checklist 1. Always pay attention to  2. For DFA, out degree of each state must be equal |  | e.g.  = { 0, 1, 2 }, L = {w: w begins with 01 } 3. For NFA, pay attention to  -transitions 4. Remember to specify the initial state 5. Make sure you have considered the input  e.g. L = {w: w contains even number of 01 }

9. Example in Lecture q0q0 q1q1 0 1 1 0 q2q2 0, 1  = {0, 1} 1.q 2 is a “die” state 2.After a 1 has appeared, any 0 appears afterwards will die 3.0*1*

10. Exercise  = { 0, 1 } L = {w: w is non-empty and the sum of the digits in w is divisible by 5}

11. Idea Need to keep track of the sum of digits Arrive at the state q k when sum of digits = k q0q0 q1q1 0 0 1 q2q2 0 1 q3q3 0 1 1 q4q4 0 q5q5 0 1 …

12. Idea Need to keep track of the sum of digits Arrive at the state q k when sum of digits = k k divisible by 5 iff k mod 5 = 0 Keep track of the remainder instead q0q0 q1q1 0 0 1 q2q2 0 1 q3q3 0 1 1 q4q4 0 1

13. Answer Handle the input  q0q0 q1q1 0 0 1 q2q2 0 1 q3q3 0 1 1 q4q4 0 1 qsqs 0 1

14. Converting an NFA to a DFA Eliminate  for each state i for each alphabet x for each state j if i can reach j using one x and  (s) add an edge from i to j with label x q0q0 q1q1 0 q2q2  q0q0 q1q1  q2q2 1 q0q0 q1q1 0 q2q2 0 q0q0 q1q1 q2q2 1 1 q0q0 q1q1  q2q2 1 q0q0 q1q1 q2q2 1 1 q3q3  q3q3 11

15. Converting an NFA to a DFA Eliminate  Remember to consider the case when the initial state can reach accepting states on  (s) q0q0 q1q1  q2q2 1 q0q0 q1q1 q2q2 1 1

16. Converting an NFA to a DFA Every possible subsets of Q is a state in the DFA Going to multiple states at the same time in the NFA = going to a subset of Q, which is now a state in the DFA q0q0 q1q1 1 q2q2 1 {q 0 }{q 1, q 2 } 1

17. Converting an NFA to a DFA Empty set ∅ = die ∅  

18. Exercise q0q0 q1q1 0  0 q2q2 1 0, 1 1

19. Eliminating  -transitions q 0 can reach q 0 on 0 (q 0 -> q 1 -> q 0 ) q 0 can reach q 1 on 0 (q 0 -> q 1 ) q 1 can reach q 0 on 0 (q 1 -> q 0 ) q 1 can reach q 1 on 0 (q 1 -> q 1 ) The rest of the transitions remain unchanged 01 q0q0 q 0, q 1 q2q2 q1q1 q2q2 q1q1 q 1, q 2 q0q0 q1q1 0  0 q2q2 1 0, 1 1

20. Eliminating  -transitions q0q0 q1q1 0 0 q2q2 1 0, 1 1 0 0 01 q0q0 q 0, q 1 q2q2 q1q1 q2q2 q1q1 q 1, q 2 q0q0 q1q1 0  0 q2q2 1 0, 1 1 q 0 is the initial state q 0 can reach an accepting state on  (i.e. q 1 ) therefore q 0 is also an accepting state

21. Converting NFA to DFA 01 {q 0 }{q 0, q 1 }{q 2 } {q 1 }{q 0, q 1 } ∅ {q 2 }{q 1 }{q 1, q 2 } q0q0 q1q1 0 0 q2q2 1 0, 1 1 0 0

22. Converting NFA to DFA 01 {q 0 }{q 0, q 1 }{q 2 } {q 1 }{q 0, q 1 } ∅ {q 2 }{q 1 }{q 1, q 2 } Starting from initial state {q 0 } {q 0 } {q 0, q 1 } 0 {q 2 } 1

23. Converting NFA to DFA 01 {q 0 }{q 0, q 1 }{q 2 } {q 1 }{q 0, q 1 } ∅ {q 2 }{q 1 }{q 1, q 2 } Check if every state has |  | outgoing transitions No “Fix” {q 2 } and {q 0, q 1 } {q 0, q 1 } goes to {q 0, q 1 } ∪ {q 0, q 1 } = {q 0, q 1 } on 0 {q 0, q 1 } goes to {q 2 } ∪ ∅ = {q 2 } on 1 {q 0 } {q 0, q 1 } 0 {q 2 } 1 0 {q 1 } 0 {q 1, q 2 } 1 1

24. Converting NFA to DFA 01 {q 0 }{q 0, q 1 }{q 2 } {q 1 }{q 0, q 1 } ∅ {q 2 }{q 1 }{q 1, q 2 } Check if every state has |  | outgoing transitions No “Fix” {q 1 } and {q 1, q 2 } {q 1, q 2 } goes to {q 0, q 1 } ∪ {q 1 } = {q 0, q 1 } on 0 {q 1, q 2 } goes to ∅ ∪ {q 1, q 2 } = {q 1, q 2 } on 1 {q 0 } {q 0, q 1 } 0 {q 2 } 1 0 {q 1 } 0 {q 1, q 2 } 1 1 0 0 1 ∅ 1

25. Converting NFA to DFA 01 {q 0 }{q 0, q 1 }{q 2 } {q 1 }{q 0, q 1 } ∅ {q 2 }{q 1 }{q 1, q 2 } Check if every state has |  | outgoing transitions No “Fix” ∅ ∅ = die 0, 1 {q 0 } {q 0, q 1 } 0 {q 2 } 1 0 {q 1 } 0 {q 1, q 2 } 1 1 0 0 ∅ 1 1

26. Converting NFA to DFA 01 {q 0 }{q 0, q 1 }{q 2 } {q 1 }{q 0, q 1 } ∅ {q 2 }{q 1 }{q 1, q 2 } Check if every state has |  | outgoing transitions Yes Accepting states = states that contain q 0 or q 1 q0q0 q1q1 0 0 q2q2 1 0, 1 1 0 0 {q 0 } {q 0, q 1 } 0 {q 2 } 1 0 {q 1 } 0 {q 1, q 2 } 1 1 0 0 ∅ 1 1

27. Converting NFA to DFA Done q0q0 q1q1 0  0 q2q2 1 0, 1 1 {q 0 } {q 0, q 1 } 0 {q 2 } 1 0 {q 1 } 0 {q 1, q 2 } 1 1 0 0 ∅ 1 1