Non-regular languages

Transcription

1 Non-regular languages (Pumping Lemma) Fall 2017 Costas Busch - RPI 1

2 Non-regular languages { a n n : n 0} { vv R : v{ a, }*} Regular languages a* * c a c( a )* etc... Fall 2017 Costas Busch - RPI 2

3 How can we prove that a language is not regular? L Prove that there is no DFA or NFA or RE that accepts L Difficult: this is not eas to prove (since there is an infinite numer of them) Solution: use the Pumping Lemma!!! Fall 2017 Costas Busch - RPI 3

4 The Pigeonhole Principle Fall 2017 Costas Busch - RPI 4

5 4 pigeons 3 pigeonholes Fall 2017 Costas Busch - RPI 5

6 A pigeonhole must contain at least two pigeons Fall 2017 Costas Busch - RPI 6

7 n pigeons... m pigeonholes n m... Fall 2017 Costas Busch - RPI 7

8 The Pigeonhole Principle n m pigeons pigeonholes n m There is a pigeonhole with at least 2 pigeons... Fall 2017 Costas Busch - RPI 8

9 The Pigeonhole Principle and DFAs Fall 2017 Costas Busch - RPI 9

10 Consider a DFA with 4 states a q1 2 a q q3 a a q 4 Fall 2017 Costas Busch - RPI 10

11 Consider the walk of a long string: (length at least 4) aaaa A state is repeated in the walk of a a a a q1 q2 q3 q2 q3 q4 aaaa a a q1 q2 q3 q 4 a Fall 2017 Costas Busch - RPI 11 a

12 The state is repeated as a result of the pigeonhole principle Pigeons: (walk states) Walk of a a a a aaaa q1 q2 q3 q2 q3 q4 Are more than Nests: (Automaton states) q1 2 q 3 q q4 Repeated state Fall 2017 Costas Busch - RPI 12

13 Consider the walk of a long string: (length at least 4) Due to the pigeonhole principle: A state is repeated in the walk of aa aa a a q1 q2 q3 q4 q4 a a q1 q2 q3 q 4 a Fall 2017 Costas Busch - RPI 13 a

14 The state is repeated as a result of the pigeonhole principle Pigeons: (walk states) a Walk of a aa q1 q2 q3 q4 q4 Are more than Nests: (Automaton states) q1 q2 q3 q4 Automaton States Repeated state Fall 2017 Costas Busch - RPI 14

15 In General: If w #states of DFA, the pigeonhole principle, a state is repeated in the walk w q Walk of... i w 1 2 k q i i1 j 1... j... q i k q z Aritrar DFA q Fall 2017 Costas Busch - RPI 15 q i Repeated state k q z

16 w #states of DFA m Pigeons: q 1 (walk states)... q i Walk of w q i q z Are more than Nests: q1 2 (Automaton states) q qm1 qm Fall 2017 Costas Busch - RPI 16 q i A state is repeated

17 The Pumping Lemma Fall 2017 Costas Busch - RPI 17

18 Take an infinite regular language (contains an infinite numer of strings) L There exists a DFA that accepts L m states Fall 2017 Costas Busch - RPI 18

19 Take string w L with w m (numer of states of DFA) then, at least one state is repeated in the walk of w Walk in DFA of w 1 2 k q... k Repeated state in DFA Fall 2017 Costas Busch - RPI 19

20 There could e man states repeated Take q to e the first state repeated One dimensional projection of walk : w First Second occurrence occurrence j i i1 j 1 q q k Unique states Fall 2017 Costas Busch - RPI 20

21 We can write w xz One dimensional projection of walk : 1 2 First Second occurrence occurrence j 1... j i q i1 q w k x 1 i i 1 j z j 1 k Fall 2017 Costas Busch - RPI 21

22 In DFA: w x z contains onl first occurrence of q... 1 j... i j1... q i1... k x Fall 2017 Costas Busch - RPI 22 z

23 Oservation: length x m numer of states of DFA... 1 x... i j q i1 Unique States Since, in x Fall 2017 Costas Busch - RPI 23 no state is repeated (except q)

24 Oservation: length 1 Since there is at least one transition in loop... j q i1 Fall 2017 Costas Busch - RPI 24

25 We do not care aout the form of string z z ma actuall overlap with the paths of x and... z... q x Fall 2017 Costas Busch - RPI 25

26 Additional string: The string is accepted x z Do not follow loop... 1 j... i j1... q i1... k x Fall 2017 Costas Busch - RPI 26 z

27 Additional string: The string is accepted x z Follow loop 2 times... 1 j... i j q i1 k x Fall 2017 Costas Busch - RPI 27 z

28 Additional string: The string is accepted x z Follow loop 3 times... 1 j... i j q i1 k x Fall 2017 Costas Busch - RPI 28 z

29 In General: The string is accepted i x i z 0,1, 2,... Follow loop i times... 1 j... i j q i1 k x Fall 2017 Costas Busch - RPI 29 z

30 Therefore: x i z L i 0,1, 2,... Language accepted the DFA... 1 j... i j q i1 k x Fall 2017 Costas Busch - RPI 30 z

31 In other words, we descried: The Pumping Lemma!!! Fall 2017 Costas Busch - RPI 31

32 The Pumping Lemma: Given a infinite regular language L there exists an integer m (critical length) for an string w L with length w m we can write w x z with x m 1 and such that: x i z L i 0,1, 2,... Fall 2017 Costas Busch - RPI 32

33 In the ook: Critical length m = Pumping length p Fall 2017 Costas Busch - RPI 33

34 Applications of the Pumping Lemma Fall 2017 Costas Busch - RPI 34

35 Oservation: Ever language of finite size has to e regular (we can easil construct an NFA that accepts ever string in the language) Therefore, ever non-regular language has to e of infinite size (contains an infinite numer of strings) Fall 2017 Costas Busch - RPI 35

36 Suppose ou want to prove that An infinite language L is not regular 1. Assume the opposite: is regular L 2. The pumping lemma should hold for L 3. Use the pumping lemma to otain a contradiction 4. Therefore, L is not regular Fall 2017 Costas Busch - RPI 36

37 Explanation of Step 3: How to get a contradiction 1. Let m e the critical length for 2. Choose a particular string which satisfies the length condition 3. Write w xz 4. Show that w x zl w L i for some i 1 5. This gives a contradiction, since from pumping lemma w m w x i zl L Fall 2017 Costas Busch - RPI 37

38 Note: It suffices to show that onl one string w L gives a contradiction You don t need to otain contradiction for ever w L Fall 2017 Costas Busch - RPI 38

39 Example of Pumping Lemma application Theorem: The language L { a : n 0} n n is not regular Proof: Use the Pumping Lemma Fall 2017 Costas Busch - RPI 39

40 L n n { a : n 0} Assume for contradiction that L is a regular language Since L is infinite we can appl the Pumping Lemma Fall 2017 Costas Busch - RPI 40

41 L n n { a : n 0} Let m e the critical length for L Pick a string w such that: w L and length w m We pick m m w a Fall 2017 Costas Busch - RPI 41

42 From the Pumping Lemma: we can write w a m m x z with lengths x m, 1 m m w xz a m m a... aa... aa... a... x z Thus: k a, 1 Fall 2017 Costas Busch - RPI 42 k m

43 x z m m a k a, 1 k m From the Pumping Lemma: x i z L i 0,1, 2,... Thus: x 2 z L Fall 2017 Costas Busch - RPI 43

44 x z m m a k a, 1 k m From the Pumping Lemma: x 2 z L m k m x 2 z a... aa... aa... aa... a... L x z Thus: a mk m L Fall 2017 Costas Busch - RPI 44

45 a mk m L k 1 BUT: L n n { a : n 0} a mk m L CONTRADICTION!!! Fall 2017 Costas Busch - RPI 45

46 Therefore: Our assumption that is a regular language is not true L Conclusion: L is not a regular language END OF PROOF Fall 2017 Costas Busch - RPI 46

47 Non-regular language { a : n 0} n n Regular languages * * L ( a ) Fall 2017 Costas Busch - RPI 47