Non-Regular Languages and The Pumping Lemma

Transcription

1 CS240 Lnguge Theory nd Automt Fll 2011 Non-Regulr Lnguges nd The Pumping Lemm Non-Regulr Lnguges Not every lnguge is regulr lnguge. However, there re some rules tht sy "if these lnguges re regulr, so is this one derived from them" There is lso powerful technique -- the pumping lemm -- tht helps us prove lnguge not to e regulr. Key tool: Since we know RE's, DFA's, NFA's, NFAε's ll define exctly the regulr lnguges, we cn use whichever representtion suits us when proving something out regulr lnguge. The Pumping Lemm If L is regulr lnguge, then there exists constnt n such tht every string w in L, of length n or more, cn e written s w = xyz, where: 0 < y xy <= n For ll i 0, xy i z is lso in L Note y i = y repeted i times; y 0 = ε. Intuitive Explntion The utomton elow hs n sttes nd no loops. Expressed in terms of n, wht is the longest string this utomton cn ccept? Generlly, in n utomton (grph) with n sttes (vertices), ny "wlk" of length n or greter must repet some stte (vertex)--tht is, it must contin cycle 1

2 Proof of Pumping Lemm Since we clim L is regulr, there must e DFA A such tht L = L(A). Let A hve n sttes; choose this n for the pumping lemm Let w e string of length n in L, sy w = 1 2 m, where m n. Let q i e the stte A is in fter reding the first i symols of w. q 0 = strt stte, q 1 = δ(q 0, 1 ), q 2 = ˆ δ (q 0, 1 2 ), etc. Since there re only n different sttes, two of q 0, q 1, q n must e the sme; sy q i = q j, where 0 <= i < j <= n. Let x = 1 i ; y = i+1 j ; z = j +1 m. Then y repeting the loop from q i to q j with lel i +1 j zero times, once, or more, we cn show tht xy i z is ccepted y A. y! i+1 j! 1! q j!. i q i! j +1 m! z! x! Exmple DFA with 6 sttes tht ccepts n infinite lnguge , Any string of length 6 or more contins circuit Some strings with length < 6 lso contin circuit (), Pth tht this string tkes through the DFA cn e decomposed into three stges: 1. x prt : goes from strt stte to eginning of the first circuit 2. y prt : circuit 3. z prt : the rest x y z 2

3 PL gets its nme ecuse the repeted string is "pumped" Note tht ecuse of the nture of FAs, we cnnot control the numer of times it is pumped So, regulr lnguge with strings of length n is lwys infinite! PL is only interesting for infinite lnguges ut works for finite lnguges, which re lwys regulr--for finite lnguges n is lrger thn the longest string, so nothing cn e pumped PL Use We use the PL to show lnguge L is not regulr. Strt y ssuming L is regulr. Then there must e some n tht serves s the PL constnt. We my not know wht n is, ut we cn work the rest of the "gme" with n s prmeter. We choose some w tht is known to e in L. Typiclly, w depends on n. Applying the PL, we know w cn e roken into xyz, stisfying the PL properties. Agin, we my not know how to rek w, so we use x, y, z s prmeters. We derive contrdiction y picking i (which might depend on n, x, y, nd/or z) such tht xy i z is not in L. Exmple Consider the lnguge i i This lnguge is not regulr! Intuitive explntion: Imgine n FA to ccept this lnguge Since the numer of s must e equl to the numer of s, must hve some wy to rememer how mny s were seen, nd ccept if the rest of the string contins the sme numer of s 3

4 How mny sttes re needed? 1 2 s etc. 1, 1 2 s, 2 s Using the PL to prove L = i i is not regulr Suppose L is regulr. Then there is constnt n stisfying the PL conditions. Consider the string w = n n Then w = xyz, where xy <= n nd y ε, nd we cn rek this string into xyz where for ny j 0 xy j z is in L But ecuse xy <= n nd y >0, the string y hs to consist of s only. So no mtter wht segment of the string xy covers, pumping y dds to the numer of s nd hence there re more s thn s There is NO WAY to segment w into xyz such tht pumping will not led to string tht is not in the lnguge! CONTRADICTION! L is therefore not regulr Importnt point It is necessry to show there is no segmenttion of the chosen string tht won t led to contrdiction This mens considering every possile mpping of xy onto the first n symols in the chosen string We chose our string to mke this esy, since very possile segmenttion consists of s only Pumping therefore disrupts the equivlence of the numer of s nd s Exmple Consider the set of strings of 's whose length is squre; formlly, L = { i i is squre}. We clim L is not regulr. Suppose L is regulr. Then there is constnt n stisfying the PL conditions. Consider w = n2, which is surely in L. Then w = xyz, where xy <= n nd y ε. 4

5 By PL, xyyz (xy 2 z) is in L. The length of xyyz is greter thn n 2 nd no greter thn n 2 + n. (Why?) However, the next perfect squre fter n 2 is (n+1) 2 = n 2 + 2n + 1. Thus, xyyz is not of squre length nd is not in L. Since we hve derived contrdiction, the only unproved ssumption -- tht L is regulr -- must e t fult, nd we hve "proof y contrdiction" tht L is not regulr. The PL "gme" Gol: win the PL gme ginst our opponent y estlishing contrdiction of the PL, while the opponent tries to foil us. Four steps: 1.The numer of sttes in the utomton is n. Note tht we don't hve to know wht n is, since we use the vrile to define our string. 2. Given n, we pick string w in L of length equl to or greter thn n. We re free to choose ny w, suject to w L nd w n. We usully define the string in terms of n. 3. Our opponent chooses the decomposition xyz, suject to xy <= n, y We try to pick i (the power fctor in xy i z) in such wy tht the pumped string w i is not in L. If we cn do so, we win the gme! Exmple 1 Σ = {,}; L = {ww R w Σ*} Whtever n is, we cn lwys choose w s follows: n n n n x y Becuse of this choice nd the requirement tht xy <= n, the opponent is restricted in step 3 to choosing y tht consists entirely of s. In step 4, we use i=2.the string xy 2 z hs more 's on the left thn on the right, so it cnnot e of form ww R. So L is not regulr. z 5

6 Exmple 2 L = {w w hs n equl numer of 1's nd 0's} Given n, we choose the string (01) n We need to show splitting this string into xyz where xy i z is in L is impossile But it is possile! If x = ε, y = 01, nd z = (01) n-1, xy i z is in L for every vlue of i. Are we out of luck? First lw of PL use: If your string does not succeed, try nother! Let's try 1 n 0 n. Agin, we need to show splitting this string into xyz where xy i z is in L is impossile But it is possile! If x nd z re the empty string nd y is 1 n 0 n, then xy i z lwys hs n equl numer of 0's nd 1's. Are we still in troule? Not this time the PL sys tht our string hs to e divided so tht xy <= n nd y. If xy <= n then y must consist only of 0's, so xyyz L. Contrdiction! We win! Exmple 3 L = {ww w Σ*} We choose the string n n, where n is the numer of sttes in the FA. We now show tht there is no decomposition of this string into xyz where for ny j 0 xy j z is in L. Agin, it is crucil tht the PL insists tht xy <= n, ecuse without it we could could pump the string if we let x nd z e the empty string. With this condition, it's esy to show tht the PL won't pply ecuse y must consist only of 's, so xyyz is not in L. 6

7 In the previous exmple s efore, the choice of string is criticl: hd we chosen n n (which is memer of L) insted of n n, it wouldn't work ecuse it cn e pumped nd still stisfy the PL. MORAL Choose your strings wisely.! Exmple 4 L = {0 i 1 j i > j} Given n, choose s = 0 n+1 1 n. Split into xyz etc. Becuse y the PL xy <= n, y consists only of 0's. Is xyyz in L? The PL sttes tht xy i z is in L even when i = 0 So, consider the string xy 0 z Removing string y decreses the numer of 0's in s s hs only one more 0 thn 1 Therefore, xz cnnot hve more 0's thn 1's, nd is not memer of L. Contrdiction! This strtegy is clled pumping down Exmple 5 L = { i i is prime} Let n e the pumping lemm vlue nd let k e prime greter thn n. If L is regulr, PL implies tht k cn e decomposed into xyz, y > 0, such tht xy i z is in L for ll i 0. Assume such decomposition exists. The length of w = xy k+1 z must e prime if w is in L. But length(xy k+1 z) = length(xyzy k ) = length(xyz) + length(y k ) = k + k(length(y) = k (1 + length(y)) The length of xy k+1 z is therefore not prime, since it is the product of two numers other thn 1. So xy k+1 z is not in L. Contrdiction! 7

8 Exmple 6 L = {w Σ* n (w) > n (w)} Let n e the pumping lemm constnt. Then if L is regulr, PL implies tht s = n n+1 cn e decomposed into xyz, y > 0, xy n, such tht xy i z is in L for ll i 0. Since the length of xy n, y consists of ll s Then xy 2 z = k-j j n-k n+1, where the length of of y = j. We know j > 0 so the length of the pumped string contins t lest s mny s s s, nd is not in L. Contrdiction! Exmple 7 L = { 3 m c m-3 m > 3} Let n e the pumping lemm constnt. Then if L is regulr, PL implies tht s = 3 n c n-3 cn e decomposed into xyz, y > 0, xy n, such tht xy i z is in L for ll i 0. Since the length of xy n, there re three wys to prtition s: 1. y consists of ll s Pumping y will led to string with more thn 3 s -- not in L 2. y consists of ll s Pumping y will led to string with more thn m s, nd leve the numer of c s untouched, such tht there re no longer 3 fewer c s thn s -- not in L 3. y consists of s nd s Pumping y will led to string with s efore s, -- not in L There is no wy to prtition 3 n n-3 so tht pumped strings re still in L. Contrdiction! Rememer You need to find only ONE string for which the PL does not hold to prove lnguge is not regulr But you must show tht for ANY decomposition of tht string into xyz the PL holds This sometimes mens considering severl different cses The Pumping Lemm Poem Any regulr lnguge L hs mgic numer p And ny long-enough word in L hs the following property: Among its first p symols is segment you cn find Whose repetition or omission leves x mong its kind. So if you find lnguge L which fils this cid test, And some long word you pump ecomes distinct from ll the rest, By contrdiction you hve shown tht lnguge L is not A regulr guy, resilient to the dmge you hve wrought. But if, upon the other hnd, x stys within its L, Then either L is regulr, or else you chose not well. For w is xyz, nd y cnnot e null, And y must come efore p symols hve een red in full. As mthemticl postscript, n ddendum to the wise: The sic proof we outlined here does certinly generlize. So there is pumping lemm for ll lnguges context-free, Although we do not hve the sme for those tht re r.e. 8

9 Proving lnguge non-regulr without the pumping lemm The pumping lemm isn't the only wy we cn prove lnguge is non-regulr Other techniques: show tht the desired DFA would require infinite sttes to model the intended lnguge use closure properties to relte to other non-rl lnguges DFA Method Consider the lnguge { i i i >= 0} nd DFA to recognize it For ny i, let i e the stte entered fter processing i, i.e., ˆ δ (q 0, i ) = i. Consider ny i nd j such tht i j. ˆ δ (q 0, i i ) ˆ δ (q 0, j i ), since the former is ccepting, nd the ltter is rejecting. ˆ δ (q 0, i i ) = ˆ δ ( ˆ δ (q 0, i ), i ) = ˆ δ ( i, i ), y definition of ˆ δ nd definition of i, respectively. ˆ (q 0, j i ) = ˆ δ δ ( ˆ δ (q 0, j ), i ) = ˆ δ ( j, i ), y the sme resoning. Since i nd j led to different sttes on the sme input, i j. Since i nd j were ritrry, nd since there re n infinite numer of wys to pick them, there must e n infinite numer of sttes. Thus, there is no DFA to recognize this lnguge, nd the lnguge is non-regulr. 9

10 Closure Properties Certin opertions on regulr lnguges re gurnteed to produce regulr lnguges Closure properties cn lso e used to prove lnguge non-regulr (or regulr) Regulr lnguges re closed under common set opertions Union : L 1 L 2 Intersection : L 1 L 2 Conctention : L 1 L 2 Complementtion : Str-closure : L 1 * L 1 Other closures Difference: If L 1 nd L 2 re regulr, then L 1 - L 2 is lso regulr Proof : Set difference is defined s L 1 - L 2 = L 1 L 2 We know tht if L 2 is regulr, so is L 2. We lso know regulr lnguges re closed under intersection. Therefore, we know tht L 1 L 2 is regulr. Reversl : If L 1 is regulr, then L R is lso regulr. Proof: Suppose L is regulr lnguge. We cn therefore construct n NFA with single finl stte tht ccepts L. We cn then mke the strt stte of this NFA the finl stte, mke the finl stte the strt stte, nd reverse the direction of ll rcs in the NFA. The modified NFA ccepts string w R if nd only if the originl NFA ccepts w. Therefore the modified NFA ccepts L R. Difference sometimes notted s L 1 \ L 2 10

11 Using regulr lnguge closure properties Showing lnguge is regulr show tht y using two or more known regulr lnguges nd one or more of the opertions over which regulr lnguges re closed, you cn produce tht lnguge Lnguges known to e regulr Bsic templte L REG1 [OP] L REG2 = L REG3 Lnguge to prove regulr where OP is one of the opertions over which regulr lnguges re closed L REG3 is the lnguge in question (i.e., the one we need to prove is regulr) L REG1 nd L REG2 re known regulr lnguges If the two lnguges on the left side of the opertor re regulr then so too must e the one on the right side NB: Cnnot ssume tht if the lnguge on the right is regulr, so too must e oth lnguges on the left Exmple If L is regulr lnguge, is L 1 = {uv u L, v = 2} lso regulr? We know L is regulr Every string in L 1 consists of string from L conctented to string of length 2 The set of strings of length 2 (cll it L 2 ) over ny lphet is finite, nd therefore this is regulr lnguge since ll finite lnguges re regulr. Therefore we hve L [conctention] L 2 = L 1 [ L REG1 ] [OP] [ L REG2 ] = [ L REG3 ] Since L 1 is the conctention of two regulr lnguges, L 1 must lso e regulr Exmple Prove the lnguge { n m n,m > 3} is regulr Show tht this lnguge cn e produced using regulr lnguge closure properties on known regulr lnguges L 1 = {**}, L 2 = {,, }, L 3 = {,,} s follows: conctenttion: L 4 = L 2 L 3 complementtion: L 5 = L 4 intersection : L 6 = L 5 L 1 = { n m n,m > 3} 11

12 Using regulr lnguge closure properties Showing lnguge is not regulr Use the sme templte L REG1 [OP] L REG2 = L REG3 However: the lnguge in question is plugged into the templte in the position of L REG1 wnt to use known regulr lnguge for L REG2 If we cn show tht L REG3 is not regulr, then it must e the cse tht L REG1 is not regulr Exmple Show L = {w w in {,}* w hs equl numer of 's nd 's} is non-regulr Use the templte: L ** = { n n n 0} If oth lnguges on the left side of the = re regulr, the lnguge on the right side is regulr (closure of regulr lnguges over intersection) { n n n 0} esily proved non-regulr using the pumping lemm We know ** is regulr Therefore L must e non-regulr Exmple Given L 1 is regulr L 1 L 2 is regulr L 2 is non-regulr Is L 1 L 2 regulr? Use sme strtegy s previous exmple: Mke the unknown lnguge (L 1 L 2 ) one of the lnguges on the left side in templte Mke the other left side lnguge known regulr lnguge Show tht the lnguge on the right side is not regulr The unknown lnguge is it more complicted ecuse it is the union of two other lnguges, ut this doesn t chnge nything Fill the templte: Use one of our known RLs on left side Use known non-rl on right So we hve (L 1 L 2 ) OP [known regulr lnguge] = L 2 Need to put in n opertion nd known regulr lnguge tht we know yields L 2 To do this, get L 2 isolted from L 1 L 2 Given L 1 is regulr L 1 L 2 is regulr L 2 is non-regulr Is L 1 L 2 regulr? L 1 L 2 Σ* 12

13 Cn t extrct L 2 from L 1 L 2 using only L 1 or L 1 L 2, since tking the difference of L 1 L 2 nd L 1 gives us only wht is left of L 2 tht is not in L 1 Hve to remove nything tht is in L 1 L 2 from L 1, then sutrct the result (everything in L 1 tht is not lso in L 2 ) from L 1 L 2 difference nd union re closed for regulr lnguges fter doing this we know we still hve regulr lnguge to sutrct from L 1 L 2 Result: (L 1 L 2 ) (L 1 - (L 1 L 2 )) = L 2 Templte L REG1 is (L 1 L 2 ) [OP] is - (difference) L REG2 is (L 1 - (L 1 L 2 )) L REG3 is L 2 We know L REG2 is regulr produced y pplying the closure opertions on two known regulr lnguges (L 1 nd L 1 L 2 ) if L REG1 is regulr, so is L REG3 But we were given the fct tht L 2 is non-regulr We cn conclude tht L REG1 = L 1 L 2 is non-regulr s well Divide nd Conquer L = {w {,}* w contins n even numer of s nd n odd numer of s nd ll s come in runs of three} Regulr: L = L 1 L 2, where L 1 = {w {,}* w {,}* w contins n even numer of s nd n odd numer of s} nd L 2 = {w {,}* ll s come in runs of three} To prove it, uild n FSA for ech Esier thn FSA for the originl lnguge L 1 : L 2 : Even s Even s Even s Odd s Odd s Even s Odd s Odd s Becuse we cn uild n FSA for ech of these lnguges, they re oth regulr We get L y intersecting these two lnguges RLs re closed under intersection Therefore, L is regulr! 13

14 Wht the Closure Theorem for Union Does Not Sy Closure theorem for union sys : If L 1 nd L 2 re regulr, then L = L 1 L 2 is regulr. Wht hppens if (for exmple) L is regulr? Does tht men tht L 1 nd L 2 re lso? Mye. Exmple We know + is regulr Consider two cses for L 1 nd L = { n n > 0 nd n is prime} { n n > 0 nd n is not prime} + = L 1 L 2 Neither L 1 nor L 2 is regulr! 2. + = { n n > 0 nd n is even} { + = { n n > 0 nd n is odd} + = L 1 L 2 Both L 1 nd L 2 re regulr! Wht the Closure Theorem for Conctention Does Not Sy Closure Theorem for Conctention sys : If L 1 nd L 2 re regulr, then L = L 1 L 2 is regulr. Wht hppens (for exmple) if L 2 is not regulr? Does tht men tht L isn t regulr? Mye. Consider two exmples: 1. { n n n 0} = {} { n n n 0} L = L 1 L 2 L 2 is not regulr! 2. { * } = {*} { n n is prime} L = L 1 L 2 L 2 is not regulr, ut L is! 14

15 True or Flse? If L 1 L 2 nd L 1 is not regulr, then L 2 is not regulr. Flse! {,}* is regulr, nd it hs non-regulr suset { n n n 0} If L 1 L 2 nd L 2 is not regulr, then L 1 is not regulr. Flse! Non-regulr lnguges hve finite susets, nd finite lnguges re regulr Hints When you need known regulr lnguge, rememer tht Σ*,ε, *, **, etc. re regulr When you need known non-regulr lnguge, use n n or ny lnguge with similr dependency 15