1.4 Nonregular Languages

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1 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. 1

2 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 2

3 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

4 80 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. 81 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. 4

5 82 2. 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. S nd 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 ((( ))) = ((( ))) 83 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) 5

6 84 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 ( + ()) 85 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 6

7 86 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 )* 87 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 7

8 88 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 89 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. + T T + T F + T + T + F T + F T + F F + F F T + F + T + F + + The differences cused by vrying substitution order of vribles cn be bstrcted wy by exmining prse trees 8

9 90 T F + T F 91 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: + () 9

10 Chomsky Norml Form Definition 2.8 A context-free grmmr is in Chomsky norml form (CNF), if At most the strt vrible S derives the empty string, very rule is of the form A BC or A (except mybe S ), The strt vrible S does not pper in the right-hnd side of ny rule. Theorem 2.9 Any context-free lnguge is generted by contextfree grmmr in CNF. Proof We convert ny grmmr into CNF. The conversion hs three stges. First, we dd new strt vrible. Then, we eliminte ll rules nd unit rules. 10

11 94 liminting rules Lemm Any context-free lnguge cn be converted into n equivlent grmmr in which t most the strt vrible derives the empty string. Proof Let G = (V,, R, S). Computing the vribles of G tht derive the empty string: NULL = { A V A R } Repet until set NULL does not chnge ny more: NULL += { A V A B 1 B k R, B i NULL i = 1,, k } 95 ch rule A X 1 X k in G is replced by the set of ll such rules tht re of form A 1 k, where X i i X i if if X NULL i X NULL i In the end we remove ll rules tht hve the form A. If S belongs to the removed rules, we tke new strt vrible S' for the grmmr nd give it rules S S. 11

12 96 S A B A B B bab S A B A B B bab bb NULL = { A, B, S } S S S A B A B B bab bb 97 liminting unit rules A unit rule hs the form A B, where A nd B re vribles. Lemm Any context-free lnguge cn be converted into n equivlent grmmr which hs no unit rules. Proof Let G = (V,, R, S). Computing the unit followers for ech vrible in G: 1. F(A) = { B V A B R } 2. Until the F-sets do not chnge nymore F(A) += { F(B) A B R } In the end we remove ll unit rules in G nd replce them by ll rules of the form A, where B F(A) nd B. 12

13 98 S S S A B A B B bab bb S B bab bb S B bab bb A B B bab bb F(S') = { S, A, B } F(S) = { A, B } F(A) = F(B) = 99 Once ll rules nd unit rules hve been eliminted, ll rules hve form A, A X 1 X k, k 2, or S. For every we dd to the grmmr the vrible C nd rule C. A rule A X 1 X k, k 2, is replced by set of rules A X 1 A 1 A 1 X 2 A 2 A k-2 X k-2 A k-1 A k-1 X k-1 X k, where X X ' i C i if X V if X i i 13

14 100 S B bab bb S B bab bb A B B bab bb S C C S C b S 2 S 2 AC b S C b C b A C A 1 S' C S 1 S 1 BC S' C C S' C b S 2 S 2 AC b S' C b C b S' S C S 1 S 1 BC A 1 BC A C C B C b B 1 B 1 AC b B C b C b C C b b 101 Algorithm CYK The strings of context-free grmmr tht hs been converted into CNF cn be prsed in (n 3 ) time using the Cocke-Younger-Ksmi lgorithm In other words, context-free lnguges cn be efficiently recognized The operting principle of lgorithm CYK is dynmic progrmming For ech substring we tbulte those vribles from which the substring cn be derived from If in the end the strt vrible of the grmmr belongs to the set of vribles tht derive the whole string, the string t hnd belongs to the lnguge 14

15 Pushdown Automt Pushdown utomt re like NFAs, but hve n extr component: (n infinite) stck We cn write new symbol on the stck t the top by pushing it We cn red nd remove the top symbol from the stck by popping it In pushdown utomton the trnsitions lwys lso concern the stck The stck gives the utomton memory by which we cn void some of the limittions tht finite utomt hve 103 Definition 2.13 A pushdown utomton is 6-tuple M =(Q,,,, q 0, F), where Q is the finite set of sttes, is the input lphbet, is the stck lphbet, q 0 Q is the strt stte, F Q is the set of ccept sttes, nd is the set-vlued trnsition function: : Q P(Q ) 15

16 104 In generl pushdown utomt re nondeterministic: (r, x, ) = { (r 1, b 1 ),, (r k, b k ) } By reding the input symbol x nd stck symbol The utomton my trnsfer from stte r to one of the sttes r 1,, r k, nd Simultneously replce the top symbol of the stck by one of the symbols b 1,,b k. 1. If x =, the utomton trnsfers without reding n input symbol; 2. If =, the utomton does not red stck symbol, but writes new symbol t the top of the stck, leving the old top symbol s is (push); 3. If nd b i =, the top symbol of the stck is red nd removed, but no new symbol is not written in its sted (pop) 16

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