1. CSC 3130: Automata theory and formal languages Andrej Bogdanov http://www.cse.cuhk.edu.hk/~andrejb/csc3130 The Chinese University of Hong Kong Closure properties Limitations of regular languages Fall 2009

2. Operations that preserve regularity We saw three operations that preserve regularity: –Union: If L, L’ are regular languages, so is L  L’ –Concatenation: If L, L’ are regular languages, so is LL’ –Star: If L is a regular language, so is L* Exercise: If L is regular, is L 4 also regular? Answer: Yes, because L 4 = ((LL)L)L

3. Example The language L of strings that end in 101 is regular How about the language L of strings that do not end in 101 ? (0+1)*101

4. Example Hint: A string does not end in 101 if and only if it ends in one of the following patterns: (or it has length 0, 1, or 2) So L can be described by the regular expression 000, 001, 010, 011, 100, 110, 111 (0+1)*(000+001+010+010+100+110+111) +  + (0 + 1) + (0 + 1)(0 + 1)

5. Complement The complement L of a language L is the set of all strings that are not in L Examples (  = {0, 1} ) –L 1 = all strings that end in 101 –L 1 = all strings that do not end in 101 = all strings end in 000, …, 111 or have length 0, 1, or 2 –L 2 = 1* = { , 1, 11, 111, …} –L 2 = all strings that contain at least one 0 = (0 + 1)*0(0 + 1)*

6. Closure under complement If L is a regular language, is L also regular? Previous examples indicate answer should be yes Theorem If L is a regular language, so is L.

7. Proof of closure under complement To argue this, we can use any of the equivalent definitions for regular languages: The DFA definition will be most convenient –We will assume L is accepted by a DFA, and show the same for L regular expression DFANFA

8. Proof of closure under complement Suppose L is regular, then it is accepted by a DFA M Now consider the DFA M’ with the accepting and rejecting states of M reversed

9. Proof of closure under complement Now for every input x   * : M accepts x After processing x, M ends in an accepting state After processing x, M’ ends in an rejecting state M’ rejects x Language of M’ is L L is regular

10. Intersection The intersection L  L’ is the set of strings that are in both L and L’ Examples: If L, L’ are regular, is L  L’ also regular? L = (0 + 1)*111L’ = 1* L  L’ = L = (0 + 1)*101L’ = 1* L  L’ = 1*111 ∅

11. Closure under intersection Theorem Proof: If L and L’ are regular languages, so is L  L’. L regular L’ regular L regular L’ regular L  L’ regular But L  L’ = L  L’ L  L’ regular L  L’ regular.

12. Reversal The reversal w R of a string w is w written backwards The reversal L R of a language L is the language obtained by reversing all its strings w = cavew R = evac L = {cat, dog}L R = {tac, god}

13. Reversal of regular languages L = all strings that end in 101 is regular How about L R ? This is the language of all strings beginning in 101 Yes, because it is represented by (0+1)*101 101(0+1)*

14. Closure under reversal Theorem Proof –We will use the representation of regular languages by regular expressions If L is a regular language, so is L R. regular expression DFANFA

15. Proof of closure under reversal If L is regular, then there is a regular expression E that describes it We will give a systematic way of reversing E Recall that a regular expression can be of the following types: –Special expressions  and  –Alphabet symbols a, b, … –The union, concatenation, or star of simpler expressions In each of these cases we show how to do a reversal

16. Proof of closure under reversal regular expression E   a (alphabet symbol) E 1 + E 2 reversal E R E1E2E1E2 E1*E1*   a E 1 R + E 2 R E2RE1RE2RE1R (E 1 R )*

17. A question If L is regular, is L DUP also regular? regular expression DFANFA ? L DUP = {ww: w  L} L = {cat, dog} Ex. L DUP = {catcat, dogdog}

18. A question Let’s try with regular expression: Let’s try with NFA: q0q0 q1q1  NFA for L   L DUP = LL L = {a, b} L DUP = {aa, bb} LL = {aa, ab, ba, bb}

19. An example Let’s try to design an NFA for L DUP L = 0*1 is regular L DUP = {11, 0101, 001001, 00010001,...} = {0 n 10 n 1: n ≥ 0} L DUP = {1, 01, 001, 0001,...}

20. An example L DUP = {11, 0101, 001001, 00010001,...} = {0 n 10 n 1: n ≥ 0} 0 000 0001 1 001 1 01 1 1 1

21. Non-regular languages

22. A non-regular language Another 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 outside L L = {0 n 1 n : n ≥ 0} is not regular.

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

24. A non-regular language What happens when we run M on input x = 0 n+1 1 n+1 ? –M better accept, because x  L –But since M has n states, it must revisit at least one of its states while reading 0 n+1 M x 0000 0 0 r 1 n+1

25. Pigeonhole principle Here, balls are 0 s, 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 0 s, then it must end up in the same state twice.

26. A non-regular language What happens when we run M on input x = 0 n+1 1 n+1 ? –M better accept, because x  L 2 –But since M has n states, it must revisit at least one of its states while reading 0 n+1 –But then the DFA must contain a loop with 0 s M x 0000 0 0 r 1 n+1

27. A non-regular language The DFA will then also accept strings that go around the loop multiple times But such strings have more 0 s than 1 s, so they are not in L 2 ! M 0000 0 0 r 1 n+1

28. 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 several times, keeps the input inside L z a1a1 a k+1 akak …… a n-1 anan a n+1 …a m

29. Pumping lemma for regular languages Pumping lemma: For every regular language L There exists a number n such 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 v i w is in L. z …… u v w

30. Proving non-regularity If L is regular, then: So to prove L is not regular, it is enough to show: There exists n such 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 v i w is in L. For every n there exists z in L, such that for every way of writing z = u v w where  |uv| ≤ n and  |v| ≥ 1, the string u v i w is not in L for some i ≥ 0.

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

32. Arguing non-regularity You need to give a strategy that, regardless of what the adversary does, always wins you the game adversary choose n write z = uvw ( |uv| ≤ n, |v| ≥ 1) you choose z  L choose i you win if uv i w  L 1 2

33. Example adversary choose n write z = uvw ( |uv| ≤ n, |v| ≥ 1) you choose z  L choose i you win if uv i w  L 1 2 adversaryyou 1 2 00000000000000011111111111111 1 u vw choose n write z = uvw z = 0 n+1 1 n+1 i = 2 uv 2 w = 0 j+2k+l 1 n+1 = 0 n+1+k 1 n+1  L L = {0 n 1 n : n ≥ 0} 0000000000000000000111111111111111 u vw v

34. Example adversaryyou 1 2 00000000000000100000000000000 1 u vw choose n write z = uvw z = 0 n+1 10 n+1 1 i = 2 uv 2 w = 0 j+2k+l 10 n+1 1 = 0 n+1+k 10 n+1 1  L L DUP = {0 n 10 n 1: n ≥ 0} 0000000000000000001000000000000001 u vw v

35. Which of these are regular? L 1 = {1 n : n is divisible by 3} L 2 = {1 n : n is prime}  = {1} L 3 = {x: x has same number of 0s and 1s} L 4 = {x: x has same number of patterns 01 and 10} L 5 = {x: x has more 0s than 1s} L 6 = {x: x has different number of 0s and 1s}  = {0, 1}

36. Example L 3 = {x: x has same number of 0s and 1s} adversaryyou 1 2 00000000000000011111111111111 1 u vw choose n write z = uvw z = 0 n+1 1 n+1 i = 2 uv 2 w = 0 j+2k+l 1 n+1 = 0 n+1+k 1 n+1  L 3 0000000000000000000111111111111111 u vw v

37. Example L 4 = {x: x has same number of 01s and 10s} adversaryyou 1 2 choose n write z = uvw z = (01) n+1 (10) n+1 i = 2 010101101010 u vw 5 01 patterns 5 10 patterns 01010101101010 u vw 6 01 patterns 6 10 patterns v 0101101010 u w 4 01 patterns 4 10 patterns is regular!

38. Example L 4 = {x: x has same number of 01s and 10s} r0r0 r1r1 1 0 0 1 s0s0 s1s1 0 1 1 0 q0q0 1 0 more 10 s more 01 s

39. Example L 5 = {x: x has more 0s than 1s} adversaryyou 1 2 00000000000000011111111111111 1 u vw choose n write z = uvw z = 0 n+1 1 n i = 0 uv 0 w = 0 j+l 1 n+1  L 5 00000000000111111111111111 u w

40. Example L 6 = {x: x has different number of 0s than 1s} adversaryyou 1 choose n z = ? there is an easier way! L 3 = {x: x has same number of 0s and 1s} = L 6 If L 6 is regular, then L 3 = L 6 is also regular But L 3 is not regular, so L 6 cannot be regular

41. Example L 2 = {1 p : p is prime} adversaryyou 1 2 11111111111111111111111111111 1 u = 1 a v = 1 b w = 1 c choose n write z = uvw = 1 a 1 b 1 c z = 1 n : n > p is prime i = a + c uv i w = 1 a 1 ib 1 c = 1 a+ib+c = 1 a+(a+c)b+c = 1 (a+c)(b+1) = 1 composite  L 2