Lecture 9 Minimum DFA

1. Lecture 9 Minimum DFA

2. For each regular language, there is a unique minimum DFA. • How do we find it? • Applications.

3. For each language L, how many DFA accepts L? • Answer: infinity • If we don’t allow any useless state (i.e., a state not be visited), how many DFA accepts L? • Answer: still infinity • Is there one with minimum number of states? • Answer: yes!

4. Equivalent Relation • xRx • xRy => yRx • xRy, yRz => xRz • Equivalence class [x]R = {y | xRy}

5. Example 1 • For each DFA M, define x RMy if and only if x y [x] ↔ a state RM

6. Example 2 • For a language L, define x RL y if and only if for any w in Σ*, xw in L ↔ yw in L. [x] contains [x] if L=L(M) fot DFA M. RL RM x w y

7. Minimum DFA • Q = {[x] | x in Σ*} • δ([x] , a) = [xa] • s = [ε] • F = {[x] | x in L} RL RL RL RL RL

8. Why, well-defined? • x RL y => xa RL ya • # of [x] < infinity => For any w in Σ*, xw in L ↔ yw in L For any w in Σ*, xaw in L ↔ yaw in L RL

9. Example • L=(0+1)*00 • [ε] = (0*1)* = ε + (0+1)*1 • [0] = (0*1)*0 = 0 + (0+1)*10 • [00] = L RL RL RL 0 1 0 0 [ε] [0] [00] 1 1

10. General Way Given a regular language L, how to construct a minimum DFA for L? • Construct a DFA M for L=L(M). • Convert DFA to minimum DFA.

11. Convert DFA to minimum DFA • Consider a DFA M = (Q, Σ, δ, s, F). • RL induces an equivalent relation in Q. [x] RL [y] x RL y • Find all equivalence classes of RL in Q by using: x RL y xa RL ya for a in Σ xa RL ya x RL y RM RM

12. Example 1 0 0 1 0 1 0 1 0 0 0 1 0 1 1 1 0 1 0 1 1

13. Example 1 0 0 0 1 0 1 0 0 1 0 1 1 0 1 1 0 0 1 0 1 1 1

14. Example 1 0 0 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1 0 0 1 1 0 1 1 1

15. Let L = {x in (0+1)* | #0(x) ≡ 0 (mod 10), #1(x) ≡ 0 (mod 12)} . A DFA accepting L needs at least how many states: (a) 10 (b) 12 (c) 120 #0(x) ≡ 0 (mod 5), #1(x) ≡ 0 (mod 4) 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 1 1 1 1 1 1 1 1 0 0 0 0 1 1 0 1 1 1 1 0 0 0 0

16. irregularity • # of [x] = infinity => L is not regular. • xw RL yw => x RL y RL

17. Example 1 n n • L={0 1 | n > 0} is not regular. • [0], [0 ], [0 ], … are different classes. Proof. For i≠ j, 0 1 in L, 0 1 not in L. 2 3 i i j i

18. Example 2 R • L={xx | x in (0+1)* } is not regular. • Proof. Consider 0 1, n > 0. They all in different equivalence classes because for i ≠ j, 0 110 in L and 0 110 not in L. n i i j i

19. Example 3 + R • L = { wxx | w in (0+1)*, x in (0+1) } is not regular. • Proof. Consider (01) , n > 0. For i>j, (01) (10) in L, but (01) (10) not in L. n i i i j

20. Questions? R • Is {wxx | w, x in (0+1)*} regular? • Is {xx w | w in (0+1)*, x in (0+1) } regular? • Is {xwx | w in (0+1)*, x in (0+1) } regular? R + + R