Intro to Theory of Computation

Intro to Theory of Computation • LECTURE5 • Theory of Computation • Conversion from NFAs to regular expressions • Proving that a language is not regular: pumping lemma CS 464 Sofya Raskhodnikova SofyaRaskhodnikova; based on slides by Nick Hopper

01*0 0 0 1 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. Sofya Raskhodnikova; based on slides by Nick Hopper

Generalized NFAs Sofya Raskhodnikova; based on slides by Nick Hopper 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

Generalized NFAs a[b a*b a qs q2 qa G accepts w if it finds q0q1…qk, w1…wk: wi2 R(qi,qi+1) w = w1w2…wk q0=qs, qk=qa R(qs,q2) = a*b R(qa,q) = Ø R(q,qs) = Ø Sofya Raskhodnikova; based on slides by Nick Hopper

NFA to GNFA ε ε ε NFA ε Add unique and distinct start and accept states Sofya Raskhodnikova; based on slides by Nick Hopper

GNF to regular expression ε ε ε NFA While machine has more than 2 states: Pick an internal state, rip it out and relabelthe arrows with regular expressions to account for the missing state Sofya Raskhodnikova; based on slides by Nick Hopper

b bb a q1 q2 a ε b ε b a ε q3 Sofya Raskhodnikova; based on slides by Nick Hopper

b bb a q1 q2 a  ba a ε b ε b a ε Sofya Raskhodnikova; based on slides by Nick Hopper

b = R(q1,q1) bb  (a  ba)b*a bb a q1 q2 a  ba ε ε b  (a  ba)b* b Sofya Raskhodnikova; based on slides by Nick Hopper

Convert the NFA to a regular expression a, b (a  b)b*b(bb*b)* b q1 q2 (a  b)b*b ε b (a  b)b*b(bb*b)*a bb*b a b ε q3 ε Sofya Raskhodnikova; based on slides by Nick Hopper

GNF to regular expression Formally: Add qstart and qaccept to create G Run CONVERT(G): If #states > 2 If #states = 2 return the expression on the arrow going from qstart to qaccept Sofya Raskhodnikova; based on slides by Nick Hopper

GNF to regular expression Formally: Add qstart and qaccept to create G Run CONVERT(G): If #states > 2 select qripQ different from qstart and qaccept define Q = Q – {qrip} define Ras: R(qi,qj) = R(qi,qrip)R(qrip,qrip)*R(qrip,qj)  R(qi,qj) return CONVERT(G) Sofya Raskhodnikova; based on slides by Nick Hopper

Conversion procedures NFA DFA definition Regular Language Regular Expression Sofya Raskhodnikova; based on slides by Nick Hopper

REGULAR OR NOT? Design a NFA for the language: {0n1n | 0 < n · 2} {0n1n | 0 < n · k} {0n1n | n > 0}? n (For R a regexp, R2 means RR, and Rnmeans RR…R) Sofya Raskhodnikova; based on slides by Nick Hopper

SOME LANGUAGES ARE NOT REGULAR! B = {0n1n | n ≥ 0} is NOT regular! Sofya Raskhodnikova; based on slides by Nick Hopper

qk … q2k 2 F qj 0.. 011111…11 Proof (by contradiction) Let M be a k-state DFA that recognizes B. Consider the path M takes on 0k1k: q0 q1 q2… qiqi+1 0000… 00.. There must be i < j ≤ k such that qi = qj M accepts 0k-(j-i)1k B! So M does not recognize the language B. Sofya Raskhodnikova; based on slides by Nick Hopper

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  ε Sofya Raskhodnikova; based on slides by Nick Hopper

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. xyiz L for all i ≥ 0 Sofya Raskhodnikova; based on slides by Nick Hopper

THE PUMPING LEMMA Example: Let L = 0*1* ; p = 1 w = 011 ε x = y = 0 if w  L and |w| ≥ p then w = xyz, where: z = 11 1. |y| > 0 2. |xy| ≤ p 3. xyiz L for all i ≥ 0 Let L = (0[1)2*; p = 2 2 w = 1 1 x = y = 2 z = ε Sofya Raskhodnikova; based on slides by Nick Hopper

q|w| 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. 1. |y| > 0 2. |xy| ≤ p 3. xyiz L for all i ≥ 0 We show w = xyz y x z … q0 qi qj There must be j > i such that qi = qj. Sofya Raskhodnikova; based on slides by Nick Hopper

USING THE PUMPING LEMMA Use the pumping lemma to prove that B = {0n1n | n ≥ 0} is not regular Hint: Assume B is regular, and try pumping w = 0P1P If B is regular, w can be split into w = xyz, where 1. |y| > 0 2. |xy| ≤ p 3. xyiz Bfor all i ≥ 0 xyyz has more 0s than 1s y is all 0s: Contradiction! Sofya Raskhodnikova; based on slides by Nick Hopper

GENERAL STRATEGY Proof by contradiction – assume L is regular Then there is a pumping length p Find a string w 2 L with |w| ¸p Show that no matter how you choose xyz, w cannot be pumped! Conclude that L is not regular Sofya Raskhodnikova; based on slides by Nick Hopper

USING THE PUMPING LEMMA PALINDROMES = { wwR | w 2 {0,1}* } is not regular. Proof: Assume … pumping length p Find a w 2 PALINDROMES longer than p 0P1P1P0P Show that … w cannot be pumped: 2p p p w= 00…0011…1100…00 y xyyz = 00…00011…1100…00 > p p 2p Sofya Raskhodnikova; based on slides by Nick Hopper

USING THE PUMPING LEMMA PALINDROMES = { wwR | w 2 {0,1}* } is not regular. Proof: Assume … pumping length p Find a w 2 PALINDROMES longer than p 0P1P1P0P Show that … w cannot be pumped: If w = xyz with |xy| ·p then y = 0Jfor some J>0. Then xyyz = 0P+J12P0P PALINDROMES Contradiction! Sofya Raskhodnikova; based on slides by Nick Hopper

PUMPING DOWN ProveC= { 0i1j | i > j ¸ 0} is not regular. Proof: Assume … pumping length p Find a w 2C longer than p 0P+11P Show that … w cannot be pumped: p p+1 w= 00…0011…11 y xz = 0…0011…11 xyyz = 00…00011…11 ·p > p+1 P P Sofya Raskhodnikova; based on slides by Nick Hopper

PUMPING DOWN ProveC= { 0i1j | i > j ¸ 0} is not regular. Proof: Assume … pumping length p Find a w 2C longer than p 0P+11P If w = xyz with |xy| ·p then y = 0J for some J¸ 1. Then xy0z = xz = 0P+1-J1P C Contradiction! xz = 0…0011…11 Sofya Raskhodnikova; based on slides by Nick Hopper

Intro to Theory of Computation