Then the left side is clearly \an y sequence of 0's and - PDF document

Extended RE's UNIX pioneered the use of additional op erators and notation for RE's: � E ? = 0 or 1 o ccurrences of E = � + E . + � = 1 or more o ccurrences of = . E E E E � � � Char acter classes [ a z GX ] = the union of all (ASCI I) c haracters from a to z , plus the c haracters G and X , for example. Algebraic La ws for RE's If t w o expressions E and F ha v e no v ariables, then = means that L ( E ) = L ( F ) (not that and E F E are iden tical expressions). F + � Example: 1 = 11 � . If and are RE's with v ariables, then = E F E ( E is ) means that whatev er F e quivalent to F languages w e substitute for the v ariables (pro vided w e substitute the same language ev erywhere the same v ariable app ears), the resulting expressions denote the same language. � Example: + = . R RR � With t w o notable exceptions, w e can think of ; union (+) as if it w ere addition with in place of the iden tit y 0, and concatenation, with � in place of the iden tit y 1, as m ultiplication. � + and concatenation are b oth asso ciativ e. � + is comm utativ e. � La ws of the iden tities hold for b oth. � ; is the annihilator for concatenation. � The exceptions: 1. Concatenation is comm utativ e: ab 6 = not ba . 2. + is idemp otent : E + E = E for an y expression E . Chec king a La w Supp ose w e are told that the la w ( R + ) = S � ( R ) holds for RE's. Ho w w ould w e c hec k that � S � � this claim is true? � Think of R and S as if they w ere single sym b ols, rather than placeholders for languages, i.e., = f 0 g and = f 1 g . R S ✦ Then the left side is clearly \an y sequence of 0's and 1's. 1

✦ The righ t side also denotes an y string of 0's and 1's, since 0 and 1 are eac h in L ( 0 � 1 � ). � That test is (i.e., if the test fails, ne c essary then the la w do es not hold. ✦ W e ha v e particular languages that serv e as a coun terexample. � But is it (if the test succeeds, the su�cient la w holds)? Pro of of Su�ciency The b o ok has a fairly simple argumen t for wh y , when the \concretized" expressions denote the same language, then the languages w e get b y substituting an y languages for the v ariables are also the same. � But if y ou think that's ob vious, the b o ok also has an example of \RE's with in tersection" where the same statemen t is false. � Also | is it clear that w e can tell whether t w o RE's without v ariables denote the same language? ✦ Algorithm to do so will b e co v ered. Closure Prop erties � Not ev ery language is a regular language. � Ho w ev er, there are some rules that sa y \if these languages are regular, so is this one deriv ed from them. � There is also a p o w erful tec hnique | the pumping lemma | that helps us pro v e a language to b e regular. not � Key to ol: Since w e kno w RE's, DF A's, NF A's, � -NF A's all de�ne exactly the regular languages, w e can use whic hev er represen tation suits us when pro ving something ab out a regular language. Pumping Lemma If L is a regular language, then there exists a constan t n suc h that ev ery string w in L , of length or more, can w e written as = , where: n w xy z 1. 0 j y j . < j xy j � 2. n . 2

i 3. F or all � 0, is also in L . i w y z ✦ i 0 Note = rep eated times; = � . y y i y � The alternating quan ti�ers in the logical statemen t of the PL mak es it v ery complex: ( 8 L )( 9 n )( 8 w )( 9 x; y ; z )( 8 i ). Pro of of Pumping Lemma � Since w e claim is regular, there m ust b e a L DF A suc h that = L ( A ). A L � Let ha v e states; c ho ose this for the A n n pumping lemma. � Let b e a string of length � in L , sa y = w n w � � � , where � n . a a a m 1 2 m � Let b e the state is in after reading the q A i �rst sym b ols of . i w ✦ = start state, = ( q ), = q q � ; a q 0 1 0 1 2 ^ ( q ), etc. � ; a a 0 1 2 � Since there are only di�eren t states, t w o of n m ust b e the same; sa y = , q ; q ; : : : ; q q q 0 1 n i j where 0 � � n . i < j � Let = � � � ; = � � � ; = x a a y a a z 1 i i +1 j � � � . a a j +1 m � Then b y rep eating the lo op from to with q q i i lab el � � � zero times once, or more, w e a a i +1 j i can sho w that is accepted b y A . xy z PL Use W e use the PL to sho w a language is L not regular. � Start b y assuming L is regular. � Then there m ust b e some that serv es as the n PL constan t. ✦ W e ma y no kno w what is, but w e can n w ork the rest of the \game" with n as a parameter. � W e c ho ose some w that is kno wn to b e in L . ✦ T ypically , w dep ends on n . � Applying the PL, w e kno w can b e brok en w in to xy z , satisfying the PL prop erties. ✦ Again, w e ma y not kno w ho w to break w , so w e use x; y ; z as parameters. � W e deriv e a con tradiction b y pic king i (whic h migh t dep end on n , x , y , and/or z ) suc h that i xy z is not in L . 3

Example Consider the set of strings of 0's whose length is a p erfect square; formally = f 0 i j is a square g . L i � W e claim L is not regular. � Supp ose is regular. Then there is a constan t L satisfying the PL conditions. n 2 � n Consider w = 0 , whic h is surely in L . � Then = , where j xy j � and 6 = � . w xy z n y � By PL, is in L . But the length of xy y z xy y z is greater than 2 and no greater than 2 + n . n n 2 � Ho w ev er, the next p erfect square after n is 2 2 ( n + 1) = n + 2 n + 1. � Th us, is not of square length and is not xy y z in L . � Since w e ha v e deriv ed a con tradiction, the only unpro v ed assumption | that L is regular | m ust b e at fault, and w e ha v e a \pro of b y con tradiction" that L is not regular. Closure Prop erties Certain op erations on regular languages are guaran teed to pro duce regular languages. � Example: the union of regular languages is regular; start with RE's, and apply + to get an RE for the union. Substitu ti on � T ak e a regular language L o v er some alphab et �. � F or eac h in �, let b e a regular language. a L a � Let s b e the substitution de�ned b y s ( a ) = L a for eac h a . ✦ Extend to strings b y s ( a � � � ) = s a a n 1 2 s ( a ) s ( a ) � � � s ( a ); i.e., concatenate the 1 2 n languages L L � � � L . a a a 1 2 n ✦ Extend to languages b y s ( M ) = [ s w in M s ( w ). � Then s ( L ) is regular. Pro of That Substitu ti on of Regular Languages In to a Regular Language is Regular � Let R b e a regular expression for language L . 4

� Let b e a regular expression for language R a s ( a ) = L , for all sym b ols a in �. a � Construct a RE E for s ( L ) b y starting with R and replacing eac h sym b ol a b y the RE L . a � Pro of that L ( E ) = s ( L ) is an induction on the heigh t of (the expression tree for) RE R . : is a single sym b ol, a . Then = , Basis R E R a = f a g , and s ( L ) = s ( f a g ) = L ( R ). L a � Cases where R is � or ; easy . Induction : There are three cases, dep ending on whether R = R + R , R = R R , or R = R � . 1 2 1 2 1 W e'll do only R = R R . 1 2 � = , where = L ( R ) and = L L L L L 1 2 1 1 2 L ( R ). 2 � Let E b e R , with eac h a replaced b y R , 1 1 a and E similarly . 2 � By the IH, L ( E ) = s ( L ) and L ( E ) = s ( L ). 1 1 2 2 � Th us, L ( E ) = s ( L ) s ( L ) = s ( L ). 1 2 Application s of the Substitu ti on Theorem � If L and L are regular, so is L L . 1 2 1 2 ✦ Let s ( a ) = L and s ( b ) = L . Substitute 1 2 in to the regular language f ab g . � So is [ . L L 1 2 ✦ Substitute in to f a; b g . � Ditto L � . 1 ✦ Substitute in to L ( a � ). � Closure under = substitution homomorphism of one string for eac h sym b ol. ✦ Sp ecial case of a substitution. Example: Homomorphism Let = L ( 0 ), and let b e a homomorphism L � 1 � h de�ned b y h (0) = and h (1) = � . aa � � � Then h ( L ) = aa ) = all strings of an ev en L � n um b er of a 's. Closure Under In v erse Homomorphism � � 1 f w j L g . h ( L ) = h ( w ) is in 5