CSCI 3130: Automata theory and formal languages

1. Fall 2011 The Chinese University of Hong Kong CSCI 3130: Automata theory and formal languages Non-regular languages Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130

2. A non-regular language • An example • We reason by contradiction: • Suppose we have managed to construct a DFA M for L • We argue something must be wrong with this DFA • In particular, M must accept some strings outsideL L = {0n1n: n ≥ 0} is not regular.

3. A non-regular language imaginary DFA for L with n states • What happens when we run M on input x = 0n+11n+1? • M better accept, because xL M x

4. A non-regular language • What happens when we run M on input x = 0n+11n+1? • M better accept, because xL • But since M has n states, it must revisit at least one of its states while reading 0n+1 0 M x 0 0 0 0 0r1n+1

5. Pigeonhole principle • Here, balls are 0s, bins are states: Suppose you are tossing n+ 1 balls into n bins. Then two balls end up in the same bin. If you have a DFA with n states and it reads n + 1 consecutive 0s, then it must end up in the same state twice.

6. A non-regular language • What happens when we run M on input x = 0n+11n+1? • M better accept, because xL2 • But since M has n states, it must revisitat least one of its states while reading 0n+1 • But then the DFA must contain a loop with 0s 0 M x 0 0 0 0 0r1n+1

7. A non-regular language • The DFA will then also accept strings that go around the loop multiple times • But such strings have more 0s than 1s, so they are not in L2! 0 M 0 0 0 0 0r1n+1

8. General method for showing non-regularity • Every regular language L has a property: • For every sufficiently long input z in L, there is a “middle part” in z that, even if repeated any number of times, keeps the input inside L an z an-1 a1 ak+1 ak … … an+1…am

9. Pumping lemma for regular languages • Pumping lemma: For every regular language L There exists a number nsuch that for every string z in L, we can write z = u v w where |uv| ≤ n |v| ≥ 1 For every i≥ 0, the string u vi w is in L. v z … … w u

10. Arguing non-regularity • If L is regular, then: • So to prove L is not regular, it is enough to show: There exists nsuch that for every z in L, we can write z = u v w where|uv| ≤ n, |v| ≥ 1 and  For every i≥ 0, the string u vi w is in L. For every nthere exists z in L, such that for every way of writing z = u v w where|uv| ≤ n and|v| ≥ 1, the string u vi w is not in L for some i≥ 0.

11. Proving non regularity For every nthere exists z in L, such that for every way of writing z = u v w where|uv| ≤ n and|v| ≥ 1, the string u vi w is not in L for some i≥ 0. • This is a game between you and an imagined adversary (say Donald) Donald choose nwrite z = uvw (|uv| ≤ n,|v| ≥ 1) you choose zLchoose iyou win if uviwL 1 2

12. Arguing non-regularity • You need to give a strategy that, regardless of what the adversary does, always wins you the game Donald choose nwrite z = uvw (|uv| ≤ n,|v| ≥ 1) you choose zLchoose iyou win if uviwL 1 2

13. Example Donald choose nwrite z = uvw (|uv| ≤ n,|v| ≥ 1) you choose zLchoose iyou win if uviwL 1 2 L = {0n1n: n ≥ 0} Donald you choose n z= 0n1n 1 write z = uvw i = 2 2 uv2w = 0n+k1nL 000000000000000111111111111111 u v w 0000000000000000000111111111111111 u v w v

14. Example LDUP = {0n10n1: n ≥ 0} Donald you choose n z= 0n10n1 1 write z = uvw i = 2 2 uv2w = 0n+k10n1 L 000000000000001000000000000001 u v w 0000000000000000001000000000000001 u v w v

15. Which of these are regular? S = {0, 1} L1 = {x: x has same number of 0s and 1s} L2 = {x: x = 0n1m, n > m≥ 0} L3 = {x: x has same number of patterns 01 and 10} L4 = {x: x has same number of patterns 01 and 10} L5 = {x: x has different number of 0s and 1s}

16. Example L1 = {x: x has same number of 0s and 1s} Donald you choose n z= 0n1n 1 write z = uvw i = 2 2 uv2w = 0n+k1nL3 000000000000000111111111111111 u v w 0000000000000000000111111111111111 u v w v

17. Example L2 = {x: x = 0m1n, m > n≥ 0} Donald you choose n z= 0n+11n 1 write z = uvw i = 0 2 uv0w = 0j+l1n+1 L2 000000000000000111111111111111 u v w 00000000000111111111111111 u w

18. Example L3 = {x: x has same number of 01s and 11s} Donald you choose n z= (01)n(11)n 1 n = 1 L4 z= 0111 has too many 11s What we have in mind: n = 1 z= 011 z= (01)n1n n = 2 z= 010111 has n01s and n11s n = 3 z= 010101111

19. Example L3 = {x: x has same number of 01s and 11s} Donald you win! choose n z= (01)n1n 1 write z = uvw i = 0 2 Taking out v will kill at least one 01, but it does not kill any 11s or 010101010101010111111111 010101010101010111111111 010101010101010111111111 so uv0w L3 or u u u v v v w w w

20. Example L4 = {x: x has same number of 01s and 01s} Donald you choose n z= (01)n(10)n 1 n = 1 z= 0110 n = 2 z= 01011010 n = 3 z= 010101101010

21. Example L4 = {x: x has same number of 01s and 10s} is regular! Donald you choose n z= (01)n(10)n 1 write z = uvw i = 2 i = 0 2 5 01 patterns 5 10 patterns 010101101010 u v w 6 01 patterns 4 01 patterns 6 10 patterns 4 10 patterns 01010101101010 0101101010 u v u v w w

22. Example one more 10 1 0 0 r1 r0 1 1 q0 one more 01 0 1 0 1 s1 s0 0 L4 = {x: x has same number of 01s and 10s}

23. Example L4 = {x: x has different number of 0s than 1s} Donald you choose n z= ? 1 there is an easier way! L1 = {x: x has same number of 0s and 1s} = L4 If L4 is regular, then L1= L4is also regular But L1 is not regular, so L4 cannot be regular

24. Extra example L5 = {1p: p is prime} Donald you choose n z= 1p: p > n is prime 1 = a + c = ? write z = uvw = 1a1b1c i 2 uviw = 1a1ib1c = 1(a+c)+ib = 1(a+c)+(a+c)b = 1(a+c)(b+1) = 1compositeL5 111111111111111111111111111111 u = 1a v = 1b w = 1c