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Today

Today. Chapter 1: RE = Regular Languages, nonregular languages RL pumping lemma Chapter 2: Context-Free Languages (CFLs). Regular Expressions (Def. 1.26). Given an alphabet , R is a regular expression if: R = a, with a R =  R = 

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Today

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  1. Today • Chapter 1: • RE = Regular Languages, • nonregular languages • RL pumping lemma • Chapter 2: • Context-Free Languages (CFLs)

  2. Regular Expressions (Def. 1.26) • Given an alphabet , R is a regular expression if: • R = a, with a • R =  • R =  • R = (R1R2), with R1 and R2 regular expressions • R = (R1R2), with R1 and R2 regular expressions • R = (R1*), with R1 a regular expression

  3. Thm 1.28: RL ~ RE • We need to prove both ways: • If a language is described by a regular expression,then it is regular (Lemma 1.29)(Last week we saw how we can convert a regularexpression R into an NFA M such that L(R)=L(M)) • Today we do the second part:If a language is regular, then it can be described bya regular expression (Lemma 1.32)

  4. Generalized NFA • Generalized nondeterministic finite automaton M=(Q, , , qstart, qaccept) with • Q finite set of states •  the input alphabet • qstart the start state • qaccept the accept state • :(Q\{qaccept})(Q\{qstart})  R the transition function • (R is the set of regular expressions over )

  5. 0110 0  qS qA 01*  0*  11 Example GNFA

  6. Observation: This GNFA:recognizes the language L(R) RR qS qA Characteristics of GNFA’s  • :(Q\{qaccept})(Q\{qstart})  R The interior Q\{qaccept,qstart} is fully connected by  From qstart only ‘outgoing transitions’ To qaccept only ‘ingoing transitions’ Impossible qiqj transitions are “(qi,qj) = ”

  7. Proof Idea of Lemma 1.32 Proof idea (given a DFA M): Construct an equivalent GNFA M’ with k2 states Reduce one-by-one the internal states until k=2 This GNFA will be of the form This regular expression R will be such that L(R) = L(M) R qS qA

  8. q1 qS - Connect qstart to earlier q1: qj  qA - Connect old accepting states to qaccept  - Complete missing transitions by qi qj - Join multiple transitions: 1 becomes 01 qi qj qi qj 0 DFA M  Equivalent GNFA M’ Let M have k states Q={q1,…,qk} - Add two states qaccept and qstart

  9. R1 R2 qrip R4(R1R2*R3) qj qi qi R3 = R4 qj Remove Internal state of GNFA If the GNFA M has more than 2 states, ‘rip’internal qrip to get equivalent GNFA M’ by: - Removing state qrip: Q’=Q\{qrip} - Changing the transition function  by’(qi,qj) = (qi,qj)  ((qi,qrip)((qi,qj))*(qrip,qj)) for every qiQ’\{qaccept} and qjQ’\{qstart}

  10. Proof Lemma 1.32 Let M be DFA with k states Create equivalent GNFA M’ with k+2 states Reduce in k steps M’ to M’’ with 2 states The resulting GNFA describes a single regular expressions R The regular language L(M) equals the languageL(R) of the regular expression R

  11. Recap Regular Languages = Regular Expressions Let R be a regular expression, then there exists an NFA M such that L(R) = L(M) The language L(M) of a DFA M is equivalent toa language L(M’) of a GNFA = M’, which canbe converted to a two-state M’’ The transition qstartR qacceptof M’’ obeys L(R) = L(M’’) Hence: RE  NFA = DFA  GNFA  RE

  12. Nonregular Languages §1.4 • Which languages cannot be recognized by finite automata? • Example: L={ 0n1n | nN } • ‘Playing around’ with FA convinces you that the ‘finiteness’ of FA is problematic for “all nN” • The problem occurs between the 0n and the 1n • Informal: the memory of a FA is limited by the the number of states |Q|

  13. qj q1 qk Repeating DFA Paths Consider an accepting DFA M with size |Q| On a string of length p, p+1 states get visited For p|Q|, there must be a j such that the computational path looks like: q1,…,qj,…,qj,…,qk

  14. qj q1 qk Repeating DFA Paths The action of the DFA in qj is always the same. If we repeat (or ignore) the qj,…,qj part, the newpath will again be an accepting path

  15. Line of Reasoning • Proof by contradiction: • Assume that L is regular • Hence, there is a DFA M that recognizes L • For strings of length  |Q| the DFA M has to ‘repeat itself’ • Show that M will accept strings outside L • Conclude that the assumption was wrong Note that we use the simple DFA, not the more elaborate (but equivalent) NFA or GNFA

  16. Pumping Lemma (Thm 1.37) For every regular language L, there is a pumping length p, such that for any string sL and |s|p, we can write s=xyz with1) x yi z  L for every i{0,1,2,…}2) |y|  13) |xy|  p Note that 1) implies that xz  L 2) says that y cannot be the empty string Condition 3) is not always used

  17. Formal Proof of Pumping Lemma Let M = (Q,,,q1,F) with Q = {q1,…,qp} Let s = s1…snL(M) with |s| = n  p Computational path of M on s is thesequence r1,…,rn+1  Qn+1 withr1 = q1, rn+1F and rt+1= (rt,st) for 1tn Because n+1  p+1, there are two statessuch that rj = rk (with j<k and k  p+1) Let x = s1…sj–1, y = sj…sk–1, and z = sk…sn+1 x takes M from q1=r1 to rj, y takes M from rj to rj, and z takes M from rj to rn+1F As a result: xyiz takes M from q1 to rn+1F (i  0)

  18. Formal Proof of Pumping Lemma Let M = (Q,,,q1,F) with Q = {q1,…,qp} Let s = s1…snL(M) with |s| = n  p Computational path of M on s is thesequence r1,…,rn+1  Qn+1 withr1 = q1, rn+1F and rt+1= (rt,st) for 1tn Because n+1  p+1, there are two termssuch that rj = rk (with j<k and k  p+1) Let x = s1…sj–1, y = sj…sk–1, and z = sk…sn+1 x takes M from q1=r1 to rj, y takes M from rj to rj, and z takes M from rj to rn+1F As a result: xyiz takes M from q1 to rn+1F (i  0) |y|  1 and |xy|  p x yi z  L(M) for every i{0,1,2,…}

  19. Pumping 0n1n (Ex. 1.38) Assume that B = {0n1n | n0} is regular Let p be the pumping length, and s = 0p1p  B P.L.: s = xyz = 0p1p, with xyiz  B for all i0 Three options for y: 1) y=0k, hence xyyz = 0p+k1p  B 2) y=1k, hence xyyz = 0p1k+p  B 3) y=0k1l, hence xyyz = 0p1l0k1p  B Conclusion: The pumping lemma does not hold,the language B is not regular.

  20. F = { ww | w{0,1}* } (Ex. 1.40) Let p be the pumping length, and take s = 0p10p1 Let s = xyz = 0p10p1 with condition 3) |xy|p Only one option: y=0k, with xyyz = 0p+k10p1  F Without 3) this would have been a pain.

  21. Intersecting Regular Languages Let C = { w | # of 0s in w equals # of 1s in w} Problem: If xyzC with yC, then xyizC Idea: If C is regular and F is regular, then the intersection CF has to be regular as well Solution: Assume that C is regular Take the regular F = { 0n1m | n,mN}, then for the intersection: CF = { 0n1n | nN } But we know that CF is not regular Conclusion: C is not regular

  22. Pumping Down E = { 0i1j | ij } Problem: ‘pumping up’ s=0p1p with y=0k givesxyyz = 0p+k1p, xy3z = 0p+2k1p, which are all in E(hence do not give contradictions) Solution: pump down to xz = 0p–k1p. Overall for s = xyz = 0p1p (with |xy|p): y=0k, hence xz = 0p–k1p  E Contradiction: E is not regular End Chapter One

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