1.3 Regular Expressions

Transcription

1 Regulr xpressions These hve n importnt role in describing ptterns in serching for strings in mny pplictions (e.g. wk, grep, Perl,...) All regulr expressions of lphbet re 1.Ønd re regulr expressions, 2.is regulr expression of for ll, 3. if R 1 nd R 2 re regulr expressions, then lso (R 1 R 2 ), (R 1 R 2 ) nd R 1 * re regulr expressions 57 ch regulr expression R of represents lnguge L(R) 1. L(Ø) = Ø, 2. L) = {}, 3. L() = {}, 4. L((R 1 R 2 )) = L(R 1 ) L(R 2 ), 5. L((R 1 R 2 )) = L(R 1 )L(R 2 ) nd 6. L(R 1 *) = (L(R 1 ))* Proper closure: R + is shorthnd for RR* (Kleene plus) Observe: R + = R* Let R k be shorthnd for the conctention of kr s with ech other. 1

2 58 xmples 0*10* = { w w contins single 1 } *001* = { w w contins the string 001 s substring } 1*(01 + )* = { w every 0 in w is followed by t lest one 1 } )* = { w w is string of even length } = { 01,10 } *0 *1 0 1 = { w w strts nd ends with the sme symbol } (0 )1* = 01* 1* (0 )(1 ) = {, 0, 1, 01 } 1*Ø = Ø Ø* = { } 59 For ny regulr expression R R Ø = R nd R = R However, it my hold tht R Rnd RR For exmple, the unsigned rel numbers tht cn be recognized using the previous utomton cn be expressed with the regulr expression d + (.d + )( (+ )d + ), where d = ( 0 9 ) 2

3 60 d d d. d q 0 q 1 q 2 q 3 d q 4 q 6 d +, - d q 5 61 Theorem 1.54 A lnguge is regulr if nd only if some regulr expression describes it. We stte nd prove both directions of this theorem seprtely. Lemm 1.55 If lnguge is described by regulr expression, then it is regulr. Proof. Any regulr expression cn be converted into finite utomton, which recognizes the sme lnguge s tht described by the regulr expression. There re only six rules by which regulr expressions cn be composed. The following pictures illustrte the NFA for ech of these cses. 3

4 62 r = Ø r = r = 63 r = s t N s N t 4

5 64 r = st N s N t 65 r = s* N s 5

6 66 r = ((b bb))* b b b b b b 67 Lemm 1.60 If lnguge is regulr, then it is described by regulr expression. Proof. By definition regulr lnguge cn be recognized with (nondeterministic) finite utomton, which cn be converted into generlized nondeterministic finite utomton (GNFA). The GNFA finlly yields regulr expression tht is equivlent with the originl utomton. Let R denote the set of regulr expressions over In GNFA the trnsition function is finite mpping : Q R P(Q) (q,w)(q', w') if q (q, r) for some r R s.t. w = zw', z L(r) 6

7 68 A GNFA M cn be reduced into regulr expression which describes the lnguge recognized by M 1. We compress M into GNFA with only 2 sttes (so tht the lnguge recognized remins equivlent) 1. The ccept sttes of M re replced by single one ( rrows) 2. We remove ll other sttes q except the strt stte nd finl stte. Let q i nd q j be the predecessor nd successor of q on some route pssing through q. Now we cn remove q nd renme the rrow between q i nd q j with new expression. 2. ventully the GNFA contins t most two sttes. It is esy to convert the lnguge recognized into regulr expression. 69 7

8 70 q i r q s q j q i rs q j t q i r q s q j q i rt*s q j 71 q i r s q j q i r s q j r r* r s t u r*s(t* ur*s)* 8

9 72 b b bb b b b b bb 73 b b b bb ( b)(b)*(bb ) (b ( b)(b)*(bb ))* 9

10 Nonregulr Lnguges The number of forml lnguges over ny lphbet (= decision/recognition problems) is uncountble On the other hnd, the number of regulr expressions (= strings) is countble Hence, ll lnguges cnnot be regulr Cn we find n intuitive exmple of nonregulr lnguge? The lnguge of blnced pirs of prentheses L prenth = { ( k ) k k 0 } 75 Theorem 1.70 (Pumping lemm) Let A be regulr lnguge. Then there exists p 1 (the pumping length) s.t. ny string s A, s p, my be divided into three pieces, s = xyz, stisfying the following conditions: xy p, y 1 nd xy i z A i = 0, 1, 2, Proof. Let M= (Q,,q 0,F)be DFA tht recognizes A s.t. Q = p. When the DFA is computing with input s A, s p, it must pss through some stte t lest twice when processing the first p chrcters of s. Let q be the first such stte. 10

11 76 Let us choose so tht: x is the prefix of s tht hs been processed when M enters q for the first time, y is tht prt of the suffix s tht gets processed by M before it reenters stte q, nd z is the rest of the string s. Obviously xy p, y 1 nd xy i z A for ll i = 0, 1, 2, x q y z Observe: The pumping lemm does not give us liberty to choose x nd y s we plese. 77 xmple Let us ssume tht L prenth is regulr lnguge. By the pumping lemm there exists some number p s.t. strings of L prenth of length t lest p cn be pumped. Let us choose s = ( p ) p. Then s = 2p > p. By Lemm 1.70 s cn be divided into three prts s = xyz s.t. xy p nd y 1. Therefore, it must be tht x = ( i i p 1, y = ( j j 1, nd z = ( p-(i+j) ) p. By our ssumption xy k z L prenth for ll k = 0, 1, 2,, but for exmple xy 0 z = xz = ( i ( p-(i+j) ) p = ( p-j ) p L prenth, becuse p j p since j 1. Hence, L prenth cnnot be regulr lnguge 11

12 78 The min limittion tht finite utomt hve is tht they hve no (externl) mens of keeping trck of n unlimited number of possibilities; i.e., to count Consider the following two lnguges C = { w w hs n equl number of 0s nd 1s } D= { w w hs n equl number of occurrences of 01 nd 10 s substrings } At first glnce recognizing mchine needs to count in ech cse The lnguge C contins { 0 k 1 k k 0 } s subset nd, hence, the nonregulrity of L prenth proves tht of C Surprisingly, D is regulr 79 An lterntive proof for nonregulrity of C The complement of regulr lnguge is regulr (xercises 1, question 5) The intersection of two regulr lnguges A nd B cn be expressed s A B A B Therefore, by Theorem 1.25, the intersection is lso regulr Lnguge L prenth cn be expressed s the intersection of C nd 0*1*, the ltter of which is regulr lnguge If C were regulr, then its intersection with the regulr lnguge 0*1* would lso be regulr However, we know tht L prenth is not regulr Therefore, C cnnot either be regulr 12

13 80 DFA for recognizing D xmple 1.75 Let F = { ww w { 0, 1 }* }. We show tht F is not regulr. Assume tht F is regulr. Let p be the pumping length given by the pumping lemm. Let s be the string 0 p 10 p 1. Becuse s is member of F nd it hs length more thn p, the pumping lemm gurntees tht s cn be split into pieces s = xyz, stisfying the three conditions of the lemm. We show tht this outcome is impossible. Becuse xy p, x nd y must consist only of 0s, so xyyz F. More exctly, x = 0 i, y = 0 j, nd z = 0 p (i+j) 10 p 1. Therefore, xy 2 z = xyyz = 0 i+j+j+p (i+j) 10 p 1 = 0 p+j 10 p 1 which does not belong to F since 0 p+j 1 hs more zeros thn 0 p 1 since by pumping lemm j 1. Hence, F is not regulr lnguge. 13

14 82 xmple 1.77 Let = { 0 i 1 j i > j }. We show tht is not regulr. Assume tht is regulr. Let p be the pumping length for given by the pumping lemm. Let s be the string 0 p+1 1 p. Then s cn be split into xyz stisfying the conditions of the pumping lemm. Becuse xy p, x nd y must consist only of 0s: x = 0 i nd y = 0 j Let us exmine the string xyyz to see whether it cn be in. Adding n extr copy of y increses the number of 0s. But contins ll strings in 0*1* tht hve more 0s thn 1s, so incresing the number of 0s will still give string in. We need to pump down: xy 0 z = xz = 0 i+p+1 (i+j) 1 p = 0 p+1 j 1 p since p+1 j p becuse by ssumption j 1. Hence, the clim follows Context-Free Lnguges The lnguge of blnced pirs of prentheses is not regulr one On the other hnd, it cn be described using the following substitution rules 1.Snd 2.S(S) These productions generte the strings of the lnguge L prenth strting from the strt vrible S S ² (S) ² ((S)) ² (((S))) ¹ ((())) = ((( ))) 14

15 84 The string being described is generted by substituting vribles one by one ccording to the given rules The string surrounding vrible does not determine the chosen production context-free grmmr One often bbrevites A w 1 w k to describe the lterntive productions ssocited with the vrible A A w 1,, A w k S (S) 85 Simple rithmetic expressions ( = expression, T = term nd F = fctor) + T T T T F F F () Genertion the expression ( + ()) T T F F F () F ( + T) F (T + T) F (F + T) F ( + T) F ( + F) F ( + ()) F ( + (T)) F ( + (F)) F ( + ()) F ( + ()) 15

16 86 Definition 2.2 A context-free grmmr is 4-tuple G = (V,, R, S), where V is finite set clled the vribles, is finite set, disjoint from V, clled the terminls V is the lphbet of G, R V (V )* is finite set of rules, nd S V is the strt vrible (A, w) R is usully denoted s A w 87 Let G = (V,, R, S), strings u, v, w (V )*, nd A w production in R uav yields string uwv in grmmr G, written uav G uwv String u derives string v in grmmr G, written u G v, if sequence u 1, u 2,, u k (V )* (k 0) exists s.t. u G u 1 G u 2 G G u k G v k = 0: u G u for ny u (V )* 16

17 88 u (V )* is sententil form of G if S G u A sententil form consisting of only terminls w * is sentence of G The lnguge of the grmmr G consists of sentences L(G) = { w * S G w } A forml lnguge L * is context-free, if it cn be generted using context-free grmmr 89 A context-free grmmr is right-liner if ll its productions re of type A or A B Theorem Any regulr lnguge cn be generted using rightliner context-free grmmr. Theorem Any right-liner context-free lnguge is regulr. Hence, right-liner grmmrs generte exctly regulr lnguges However, there re context-free lnguges which re not regulr; e.g., the lnguge of blnced pirs of prentheses L prenth Therefore, context-free lnguges re proper superset of regulr lnguges 17

18 90 Ambiguity The sequence of one-step derivtions leding from the strt vrible S to string w S w 1 w k w is clled the derivtion of w In the grmmr for rithmetic expressions the sentence + cn be derived in mny different wys: 1.+TT+TF+T+T+F+ 2.+T+FT+FF+FF++ 3.+T+F+T+F++ The differences cused by vrying substitution order of vribles cn be bstrcted wy by exmining prse trees 91 T F + T F 18

19 92 Context-free grmmr G is mbiguous if some sentence of G hs two (or more) distinct prse trees Otherwise the grmmr is unmbiguous Lnguge tht hs no unmbiguous context-free grmmr is inherently mbiguous.g. lnguge { i b j c k i = j j = k } is inherently mbiguous An lterntive grmmr for the simple rithmetic expressions: + ()