Finite Automata-cont d

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1 Automt Theory nd Forml Lnguges Professor Leslie Lnder Lecture # 6 Finite Automt-cont d The Pumping Lemm WEB SITE: ~lnder/cs573.html Septemer 18, 2000 Exmple 1 Consider L = {ww R x : x,w (0 + 1) + }, where w R denotes the string w written in reverse If L is regulr there is n n s in the PL4RL Pick z = (01) n (10) n 1 = w 1 w 2 w n v, where u = nd w 1 = w 2 = = w n = 01, v = (10) n 1 There is loop somewhere in the pth Then there re i nd j so tht w 1 w 2 (w i+1 w j ) k w n v is in L for ll k CLASS 06-8 CLASS 06-9

2 Pump up However, w 1 w 2 (w i+1 w j ) 2 w n v = w 1 w 2 w j w i+1 w n v = (01) n + j - i (10) n 1 = longer shorter Conclusion Hence w 1 w 2 (w i+1 w j ) 2 w n v L Thus, L does not hve one of the fundmentl properties of regulr lnguge. So L cnnot e regulr The mid-point of the ww R prt needs 00 or 11 ut tht pttern occurs once nd is not ner the center CLASS CLASS Exmple 2 Pick the z Suppose we consider the lnguge L 1 = { 2j 3k : 0 <= j < k } We show tht L 1 is not regulr The nottion 2j indictes tht there is n even numer of s The nottion 3k indictes tht the numer of s is multiple of 3 CLASS We pick the z to e 2n 3n+3 Then z = w 1 w 2 w n v, where w 1 =w 2 = =w n = 2, v = 3n+3 Note tht we CANNOT tke z = 2n 3n+1 ecuse 2n 3n+1 L 1 If L 1 were regulr nd n were the numer from the PL4RL pplied to L then there would exist i, j with 0 < i < j < n nd w 1 w 2 (w i+1 w j ) k w n v L, k > 0 CLASS 06-13

3 Pump up one extr loop We consider w 1 w 2 (w i+1 w j ) 2 w n v p 0 p i = p j w 1 w j wj+1 wn p n v (w i +1 w j ) 2 Conclusion The string w 1 w 2 (w i +1 w j ) 2 w n v hs ONE extr copy of w i+1 w j = 2(j - i ) so the whole string is 2(n + j - i ) 3n+3, which is not in L 1 since n + j - i < n + 1 is flse. NOTE: We used w i = 2, so tht when we pumped up we were sure to keep n even numer of s CLASS CLASS Initil step for finding regulr expression Finding the regulr expression for the lnguge of n NFA Remove ll ded sttes from the DFA or NFA Remove ll unrechle sttes from the DFA or NFA If there is NO pth from the initil stte q 0 to stte q, then q is clled n unrechle stte We ssume ech stte in the DFA or NFA is on pth leding from q 0 to finl stte CLASS 06-17

4 Strt with n NFA Suppose we egin with the NFA from clss 3 q 5 q 1, q 3 q 0 q 4 q 6 Step 1: new single finl stte Add new stte s SINGLE finl stte q 5 with -trnsitions q 1, q 3 q 0 q 4 q 6 CLASS CLASS Remove sttes, one t time Remove stte nd replce it with new rcs leled y regulr expressions here we remove q 6 Remove q 5 Do not remove the initil or finl stte q 1, q 3 q 0 q 4 * CLASS q 5 q 1 q 0 * q 4, CLASS q 3

5 Remove q 1 Remove q 3 Be creful to put replcement for every trnsition tht is lost Wtch for loops + q 0 * q 4, q 3 q 0 * q 4 CLASS CLASS Now it gets tough Just one more... Remove q 4 + Remove This is the form we were iming for q 0 + * + q 0 + ( + )*( + ) + * + * + ( + )*( + *) CLASS CLASS 06-25

6 The regulr expression For wht it s worth [ + * + ( + )*( + *)]* [ + ( + )*( + )] Note tht ( + )*( + ) = ( + )* Sustitute [ + * + ( + )*( + *)]* [ + ( + )*( + )] = [ + * + ( + )*( + *)]* ( + )* since is prt of (+)* [ + ( + )*( + )] = + ( + )* = ( + )* since ( + )* includes CLASS CLASS Still some more Consider [ + * + ( + )*( + *)]* = [ + * + ( + )*]* since nything in ( + )** cn e otined from cycling through ( + )*, nd * You cn lso get * from [ + ( + )*]* At lest we get down to [ + ( + )*]*( + )* We ll stop simplifying here Now it looks like we might e le to reduce this to [ + ( + )*]*...ut this shorter expression contins ll strings k, k > 0, none of which is in [ + ( + )*]*( + )* ecuse the is missing However, s oserved in clss ( + )* [ + ( + )*]* ( + )* So we rech ( + )*( + )* CLASS CLASS 06-29

7 Principles Remove q We hve to uild n utomton with regulr expressions on the edges Strt y dding unique externl finl stte with -trnsitions from wht were finl sttes Go through process of removing internl sttes y comining regulr expressions Remove q q 1 r 1 r 2 q r r 3 r 4 q 3 q 4 Loops on sttes leled r ecome Kleene closures r * CLASS CLASS loops ecome closure opertions Continue until there re two sttes q 1 r 1 r *r 3 q 3 We re left with the initil nd finl sttes r 2 r *r 4 r 2 r *r 3 r 1 r *r 4 q 4 The finl regulr expression is r 1 *r 2 q 0 r1 r 2 CLASS CLASS 06-33

8 Union, Conctention, Kleene Closure Properties of regulr lnguges We know there re regulr expressions for the union, conctention nd Kleene closure of regulr expressions nd ll regulr expressions denote regulr lnguges, so The union, conctention nd Kleene closure of regulr lnguges re ll regulr CLASS Complement The complement of ny regulr lnguge is regulr: Tke regulr lnguge L nd uild FULL DFA M for L (it must e DFA nd ll sttes must e shown) Consider the complementry mchine M 1 where every finl stte of M is non-finl in M 1 nd every nonfinl stte in M is finl in M 1 The sttes nd trnsitions of M 1 re the sme s those of M L(M 1 ) is the complement of L(M) It is cler tht δ*(q 0, w) is finl stte of M 1 if nd only it is not finl stte of M Hence M 1 ccepts w if nd only if M does not ccept w _ Therefore M 1 ccepts exctly L CLASS CLASS 06-37

9 Intersection The intersection of two regulr lnguges is regulr L 1 L 2 = (L 1 L 2 ) The textook constructs DFA for the intersection Decidle properties of regulr lnguges CLASS CLASS Emptiness There is n lgorithm to decide if regulr lnguge is empty: If you re given regulr lnguge L over Σ, you cn otin DFA for the lnguge--ssume tht DFA hs n sttes Check if ny of the (k n -1)/(k - 1) strings of length <n in Σ* is in L, where Σ hs k elements If one does, L is not empty. If none re in L, then L is empty CLASS Proof Proof tht lgorithm works: Suppose L is not empty nd let z e the shortest string in L. If z > n then y the pumping lemm z = uvw nd uw is shorter string lso in L, which mens z ws not the shortest string Thus the shortest string in L must hve length < n, which is wht we look for in the lgorithm

10 Finiteness To check if the regulr lnguge L is finite, we check if there re ny strings z in L with n < z < 2n, where n is the numer of sttes in DFA for L If not, L is finite Proof - 1 Proof of the lgorithm: From the pumping lemm, if there is one string z = uvw in L with n < z, then L is infinite: it contins uv k w for ll k > 0 If there is string in tht rnge, L is infinite CLASS CLASS Proof - 2 We need to see tht there must e such z with z < 2n Let z e the shortest string in L with n < z, we ssert tht z < 2n If not, we know tht z = uvw, z > 2n nd v < n It follows tht uw is in L nd uw > n Hence z ws not the shortest with this property fter ll Minimiztion of the DFA CLASS 06-44

11 CLASS The construction The interesting thing out DFA s is tht every DFA for lnguge L cn e minimized to the sme miniml DFA (the sme up to renming sttes) Tke DFA for L with Q = {q 0, q 1,, q n } We hve to unify sttes tht cnnot e distinguished y δ* CLASS distinguished sttes Two sttes q i nd q j re distinguished y δ* if there is string w in Σ* such tht ONLY ONE of δ*(q i, w) nd δ*(q j, w) is finl If q i nd q j re not distinguished then they cn e unified in the DFA If ll such indistinguishle sttes re unified, we otin well-defined DFA tht recognizes the sme lnguge L Crete tle q 1 Begin with tringulr tle q 3... q n q 0 q 1... q n-2 q n-1 How will it finish We will mrk X where sttes re. distinguishle. If the is no X. for q i nd q j, q i X X X unify them. X. X. X... q j... CLASS CLASS 06-49

12 Round 1 To strt, mrk X for those sttes which re distinguished y The sttes q i nd q j re distinguished y if only one of δ*(q i, ) nd δ*(q j, ) is finl But these sttes re exctly q i nd q j Hence we put n X wherever one of q i nd q j is finl nd the other is not finl Round 2 For every q i nd q j nd every in Σ, consider q i ' = δ*(q i,) nd q j ' = δ*(q j,) IF there is n X for the pir q i ', q j ' [the order of the pir is NOT importnt], then put n X in the cell of the tle for the pir q i, q j Otherwise, write ij in the cell of the tle indexed y q i ' nd q j ' Do this for every pir q i nd q j CLASS CLASS Round 2½ Round 2½ - cont d All the wy through this process, wtch out to see if n X gets plced in cell contining some pirs ij, For exmple suppose q i ' = δ*(q i,) nd q j ' = δ*(q j,) nd we wrote ij in the cell of the tle indexed y q i ' nd q j ' Suppose lter we re considering δ*(q i ',) nd δ*(q j ',) nd we find n X for tht pir Tht will put n X for q i ' nd q j ', nd we use the ij in tht cell to go ck nd put n X in the cell indexed y q i nd q j CLASS CLASS 06-53

13 CLASS Round 2½ - cont d This process is trnsitive, i.e. if cell gets n X nd there is pir of indices mn in tht cell, then we lso dd n X in the cell indexed y q m nd q n CLASS Exmple - I (finl) q 0 q 6 q 1 (finl) q 1 q 3 q 6 (finl) q 8 q 3 q 0 q 9 q 4 q 9 q 4 (finl) q 5 q 9 (finl) q 6 q 0 q 5 q 7 q 9 q 7 (finl) q 8 q 3 q 0 q 9 q 6 q 3 Check your nswer Rechle sttes Note tht if you get interrupted y ny distrction, you will proly mke mistke To verify your nswer, you must drw out the minimized utomton nd mke sure ll the trnsitions re consistent We will see this in n exmple A stte q in DFA or NFA is rechle if there is w Σ* such tht δ*(q 0, w) = q (for the DFA) or q δ*(q 0, w) (for the NFA) CLASS CLASS 06-57

14 Check wht is REACHABLE FROM q 0 (finl) q 0 (Y) q 6 q 1 (finl) q 1 (Y) q 3 q 6 (finl) (Y) q 8 q 3 (Y) q 0 q 9 q 4 q 9 q 4 (finl) q 5 (Y) q 9 (finl) q 6 (Y) q 0 q 5 q 7 q 9 q 7 (finl) q 8 (Y) q 3 q 0 We otin q 1 Remove q 4 nd q 7 ; mke tle q 3 X X X q 5 q 6 q 8 X X X q 9 X X X X X X q 0 q 1 q 3 q 5 q 6 q 8 q 9 (Y) q 6 q 3 CLASS CLASS First column Work on column of q 0 q 1 08 Where numer 05 pir lnds on q 3 X X X n X, go q 5 06 X ck nd q X 08 write X q 8 02 X q 9 X X X X X 05 X q 0 q 1 q 3 q 5 q 6 q 8 Mrk X s Effect of X s on column of q 0 q 1 X q 3 X X X q 5 X 06 X q 6 02 X q 8 X 02 X q 9 X X X X X X q 0 q 1 q 3 q 5 q 6 q 8 CLASS CLASS 06-61

15 Column 2 Work on column of q 1 q 1 X q 3 X 16 X X 12 q 5 X 06 X q ,15 X 16 q 8 X 02 X 12 q 9 X X X 15 X X X q 0 q 1 q 3 q 5 q 6 q 8 Mrk X s Effect of X s on column of q 1 q 1 X DO NOT get X tempted to go q 3 X X X forwrd to q 5 X 06 X mrk X, q 6 18 X 02,15 X 16 e.g. t q 8 X 02 X 12 (6,8) q 9 X X X 15 X X X q 0 q 1 q 3 q 5 q 6 q 8 CLASS CLASS Third column Mrk X s Work on column of q 1 X 26 X q 3 X X X 28 q 5 X 06 X q 6 18 X 02,15 X 16 q 8 X X q 9 X X X X X X q 0 q 1 q 3 q 5 q 6 q 8 Effect on X s on columns of q 1 X 26 X q 3 X X X q 5 X 06 X X q 6 18 X 02,15 X 16 q 8 X 02 X X q 9 X X X 15 X X X q 0 q 1 q 3 q 5 q 6 q 8 CLASS CLASS 06-65

16 Fourth nd fifth Column Column of q 3 nd q 5 q 1 X 26, 58 X q 3 X X X q 5 X 06 X 56 X q 6 18,39 X 02,15 X 16 q 8 X 02 X X q 9 X 56 X X 15, 58 X X X q 0 q 1 q 3 q 5 q 6 q 8 Mrk X s Effect of X s on column of q 5 q 1 X if the cell for q 1, q 6 26, 58 X were not mrked, q 3 X X X we would mrk q 5 X 06 X X it now q 6 18,39 X 02,15 X X 16 q 8 X 02 X X q 9 X X X 15, 58 X X X q 0 q 1 q 3 q 5 q 6 q 8 CLASS CLASS Sixth column Mrk X s Work on column of q 6 q 1 X 26, 58 X q 3 X 68 X X q 5 X X X q 6 18,39 X 02,15 X X q 8 X 02 X X q 9 X X X 15, 58 X X X q 0 q 1 q 3 q 5 q 6 q 8 Effect of X s on column of q 6 q 1 X 26, 58 X q 3 X X X q 5 X 06 X X q 6 18,39 X 02,15 X X q 8 X 02 X X 26 X 12 q 9 X X X 15, 58 X X X q 0 q 1 q 3 q 5 q 6 q 8 CLASS CLASS 06-69

17 Finl tle Cler unused indices, unify: q q 1 X q nd q 3.9 X q 3 X X X q 5 X X X q 6 X X X q 8 X X X X q 9 X X X X X X q 0 q 1 q 3 q 5 q 6 q 8 CLASS CLASS Check consistency Is the finl trnsition tle (or grph) consistent with the initil tle (finl) q q q (finl) q q 3.9 q q 3.9 q q 3.9 q q q 3.9

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