Intro to Theory of Computation

Transcription

1 Intro to Theory of Computation 1/26/2016 LECTURE 5 Last time: Closure properties. Equivalence of NFAs, DFAs and regular expressions Today: Conversion from NFAs to regular expressions Proving that a language is not regular: pumping lemma Sofya Raskhodnikova Sofya Raskhodnikova; based on slides by Nick Hopper 5.1

2 NFAs to regular expressions Theorem. Every NFA has an equivalent regular expression. Proof idea: Transform NFA to a regular expression by removing states and relabeling the arrows with regular expressions. 0 01* /26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L4.2

3 I-clicker problem (frequency: AC) What is the regular expression that generates all strings that take this machine from q s to q a? b q s a q c q a A. ab c d B. ab c d C. abc d D. ab cd E. None of the above. d

4 Generalized NFAs Each transition is labeled with a regular expression Unique and distinct start and accept states No transitions to the start state No transitions from the accept state 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L4.4

5 Generalized NFAs a b q s a*b q a q a G accepts R(q w if s,q) it finds = a*b q 0 q 1 q k, w 1 w k : w i is generated by R(q i,q i+1 ) R(q a,q) = Ø w = w 1 w 2 w k R(q,q q 0 s =q ) = s Ø, q k =q a

6 NFA to GNFA NFA Add a new start state with no incoming arrows. Make a unique accept state with no outgoing arrows. 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L4.6

7 GNF to regular expression NFA While machine has more than 2 states: Pick an internal state, rip it out and relabel the arrows with regular expressions to account for the missing state. 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L4.7

8 a a,b a*b(a*b)(a b)* b q 0 q 1 q 2 q 3 R(q 0,q 3 ) = (a*b)(a b)*

9 bb b a q 1 b b a a q 2 q 3 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

10 bb b a q 1 b a a ba b a q 2 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

11 bb (a bb ba)b*a = R(q 1,q 1 ) b a q 1 a ba q 2 b (a ba)b* 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

12 bb (a ba)b*a = R(q 1,q 1 ) q 1 q s b (a ba)b* = R(q 1,q a ) q a 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper

13 Convert the NFA to a regular expression q 1 a, b q 2 b (a b)b*b (a b)b*b(bb*b)*a b bb*b a b q 3 (a b)b*b(bb*b)*

14 Convert the NFA to a regular expression (a b)b*b(bb*b)*a ((a b)b*b(bb*b)*a )*( (a b)b*b(bb*b)*) (a b)b*b(bb*b)*

15 NFA to regular expression Add q start and q accept to create GNFA G. Run CONVERT(G) CONVERT(G): If #states = > 2 return the expression on the arrow going from q start to q accept 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper 5.15

16 NFA to regular expression Add q start and q accept to create GNFA G. Run CONVERT(G) CONVERT(G): If #states > 2 Build G from G: select q rip Q different from q start and q accept define Q = Q {q rip } define R as: R (q i,q j ) = R(q i,q rip )R(q rip,q rip )*R(q rip,q j ) R(q i,q j ) return CONVERT(G ) 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper 5.16

17 Conversion procedures DFA NFA definition Regular Language Regular Expression 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper 5.17

18 REGULAR OR NOT? Design an NFA for the language: {0 n 1 n 0 < n 2} 0 n 1 n 0 < n k} {0 n 1 n n > 0}? n (For R a regexp, R 2 means RR, and R n means RR R) 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.18

19 SOME LANGUAGES ARE NOT REGULAR! B = {0 n 1 n n 0} is NOT regular! 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.19

20 Proof (by contradiction) Let M be a k-state DFA that recognizes B. Consider the path M takes on 0 k 1 k : q k q 2k F q 0 q 1 q 2 q i q i+1 q j There must be i < j k such that q i = q j M accepts 0 k-(j-i) 1 k B! So M does not recognize the language B. 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.20

21 REGULAR OR NOT? C = { w w has equal number of 1s and 0s} NOT REGULAR D = { w w has equal number of occurrences of 01 and 10} (0Σ*0) (1Σ*1) 1 0 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.21

22 THE PUMPING LEMMA Let L be a regular language with L = Then there exists a length p such that if w L and w p then w can be split into three parts w=xyz where: 1. y > 0 2. xy p 3. xy i z L for all i 0 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.22

23 THE PUMPING LEMMA Example: Let L = 0*1* ; p = 1 w = 011 x = if w L and w p y = 0 then w = xyz, where: z = y > 0 Let L = (0 1)2*; p = 2. xy p w = 12 x = 1 3. xy i z L for all y i= 20 z = 2 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.23

24 Proof of the pumping lemma Let M be a DFA that recognizes L. Let p be the number of states in M. Assume w L is such that w p. We show w = xyz 1. y > 0 2. xy p 3. xy i z L for all i 0 x y z q 0 q i q j q w There must be j > i such that q i = q j. 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.24

25 USING THE PUMPING LEMMA Use the pumping lemma to prove that B = {0 n 1 n n 0} is not regular Hint: Assume B is regular, and try pumping w = 0 P 1 P If B is regular, w can be split into w = xyz, where 1. y > 0 2. xy p 3. xy i z B for all i 0 y is all 0s: xyyz has more 0s than 1s Contradiction! 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.25

26 GENERAL STRATEGY Proof by contradiction: assume L is regular. Then there is a pumping length p. Find a string w L with w p. Show that no matter how you choose xyz, w cannot be pumped! Conclude that L is not regular. 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.26

27 USING THE PUMPING LEMMA PALINDROMES = { ww R w {0,1}* } is not regular. Proof: Assume pumping length p Find a w PALINDROMES longer than p 0 P 1 P 1 P 0 P Show that w cannot be pumped: w= p 2p y must be in this part xyyz = p > p 2p p 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.27

28 USING THE PUMPING LEMMA PALINDROMES = { ww R w {0,1}* } is not regular. Proof: Assume pumping length p Find a w PALINDROMES longer than p 0 P 1 P 1 P 0 P Show that w cannot be pumped: If w = xyz with xy p then y = 0 J for some J>0. Then xyyz = 0 P+J 1 2P 0 P PALINDROMES Contradiction! 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.28

29 PUMPING DOWN Prove C = { 0 i 1 j i > j 0} is not regular. Proof: Assume pumping length p Find a w C longer than p 0 P+1 1 P Show that w cannot be pumped: p+1 p w= y must be in this part xyyz = xz = > p+1 P p P 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.29

30 PUMPING DOWN Prove C = { 0 i 1 j i > j 0} is not regular. Proof: Assume pumping length p Find a w C longer than p 0 P+1 1 P If w = xyz with xy p then y = 0 J for some J 1. Then xy 0 z = xz = 0 P+1-J 1 P C Contradiction! xz = /26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.30

31 Pumping lemma as a game 1. YOU pick the language L to be proved nonregular. 2. ADVERSARY picks p, but doesn't reveal to YOU what p is; YOU must devise a play for all possible p's. 3. YOU pick w, which should depend on p and which must be of length at least p. 4. ADVERSARY divides w into x, y, z, obeying conditions stipulated in the pumping lemma: y > 0 and xy n. Again, ADVERSARY does not tell YOU what x, y, z are, although they must obey the constraints. 5. YOU win by picking i, which may be a function of p, x, y, z, such that xy i z is not in L. 1/26/2016 Sofya Raskhodnikova; based on slides by Nick Hopper L5.31