Course 8 Properties of Regular Languages

Transcription

1 Course 8 Properties of Regular Laguages The structure ad the cotet of the lecture is based o

2 Topics ) How to prove whether a give laguage is ot regular? 2) Miimizatio of DFAs 2

3 Some laguages are ot regular Whe is a laguage is regular? if we are able to costruct oe of the followig: DFA or NFA or e -NFA or regular expressio Whe is it ot? If we ca show that o FA ca be built for a laguage 3

4 How to prove laguages are ot regular? What if we caot come up with ay FA? A) Ca it be laguage that is ot regular? B) Or is it that we tried wrog approaches? How do we decisively prove that a laguage is ot regular? The hardest thig of all is to fid a black cat i a dark room, especially if there is o cat! -Cofucius 4

5 Example of a o-regular laguage Let L = {w w is of the form, for all } Hypothesis: L is ot regular Ituitive ratioale: How do you keep track of a ruig cout i a FA? A more formal ratioale: Ø Ø By cotraditio, if L is regular the there should exist a DFA for L. Let k = umber of states i that DFA. Ø Cosider the special word w= k k => w Î L Ø DFA is i some state p i, after cosumig the first i symbols i w 5

6 Uses Pigeo Hole Priciple Ratioale Ø Ø Ø Ø Ø Ø Let {p,p, p k } be the sequece of states that the DFA should have visited after cosumig the first k symbols i w which is k But there are oly k states i the DFA! ==> at least oe state should repeat somewhere alog the path (by ++ Priciple) ==> Let the repeatig state be p i =p j for i < j ==> We ca fool the DFA by iputig (k-(j-i)) k ad still get it to accept (ote: k-(j-i) is at most k-). ==> DFA accepts strigs w/ uequal umber of s ad s, implyig that the DFA is wrog! 6

7 The Pumpig Lemma for Regular Laguages What it is? The Pumpig Lemma is a property of all regular laguages. How is it used? A techique that is used to show that a give laguage is ot regular. It ca ot be used to show that a give laguage is regular. 7

8 Pumpig Lemma for Regular Laguages Let L be a regular laguage The there exists some costat N such that for every strig w Î L s.t. w N, there exists a way to break w ito three parts, w=xyz, such that:. y e 2. xy N 3. For all k, all strigs of the form xy k z Î L This property should hold for all regular laguages. Defiitio: N is called the Pumpig Lemma Costat 8

9 Pumpig Lemma: Proof L is regular => it should have a DFA. Set N := umber of states i the DFA Ay strig wîl, s.t. w N, should have the form: w=a a 2 a m, where m N Let the states traversed after readig the first N symbols be: {p,p, p N } Ø Ø ==> There are N+ p-states, while there are oly N DFA states ==> at least oe state has to repeat i.e, p i = p j where i<j N (by PHP) 9

10 Pumpig Lemma: Proof Ø => We should be able to break w=xyz as follows: Ø x=a a 2..a i ; y=a i+ a i+2..a J ; z=a J+ a J+2..a m Ø Ø x s path will be p..p i y s path will be p i p i+..p j (but p i =p j implyig a loop) Ø Ø z s path will be p j p J+..p m Now cosider aother strig w k =xy k z, where k Ø Case k= y k (for k loops) p x z p i p m =p j Ø DFA will reach the accept state p m Ø Case k> Ø DFA will loop for y k, ad fially reach the accept state p m for z Ø I either case, w k Î L This proves part (3) of the lemma

11 Pumpig Lemma: Proof For part (): Sice i<j, y e y k (for k loops) p x z p i p m =p j For part (2): By PHP, the repetitio of states has to occur withi the first N symbols i w ==> xy N

12 Examples 2

13 Note: This N ca be aythig (eed ot ecessarily be the #states i the DFA.) Example Claim L = {w! = # $ % $, $ %} is ot regular Proof: By cotradictio, assume L be regular. The, P/L costat should exist; let N = that P/L costat Cosider iput w = N N By pumpig lemma, we should be able to break w=xyz, such that:. y e 2. xy N 3. For all k, the strig xy k z is also i L w ca have oe of the followig shapes: w = N N =... ==> #># x y x w = N N =... ==> #<# x y x w = N N =... ==>w has ot the required x y x 3

14 Example 2 Claim L eq = {w w is a biary strig with equal umber of s ad s} is ot regular. Assume L eq be regular. The there exists N (P.L. costat) such that for every strig w Î L s.t. w N, there exists a way to break w ito three parts, w=xyz, such that: ()y e, (2) xy N (3) For all k, all strigs of the form xy k z Î L. Take N=N*, ad w= N* N* ( w =2N* N*), w Î L eq. Proof proceeds like i Example. 4

15 Example 3 Prove L = { } is ot regular. Assume L eq be regular. The there exists N (P.L. costat) such that for every strig w Î L s.t. w N, there exists a way to break w ito three parts, w=xyz, such that: ()y e, (2) xy N (3) For all k, all strigs of the form xy k z Î L. Take N=N*, ad w= N* N* ( w =2N*+ N*). The w ca be divided ito 3 parts: x= (legth N*-2), y=,( xy = N*-2 +2 N*), z= (legth N*). The xy =N* N*. For k= we have xz= (o ). So ot i L eq. We foud a couterexample for which the PL does ot hold. Hece L eq is ot regular. 5

16 Example 4 Prove L = { is prime} is ot regular. Assume L eq be regular. The there exists N (P.L. costat) such that for every strig w Î L s.t. w N, there exists a way to break w ito three parts, w=xyz, such that: ()y e, (2) xy N (3) For all k, all strigs of the form xy k z Î L. Take N=p, ad w= p ( w =p p ad p - prime). The w ca be divided ito 3 parts: y =l (cod. () is satisfied, ad assume xy <=p s.t. (2) is satisfied). Tryig to prove (3): Let k=p+. We have xy p+ z = xyz + y p = p+p y =p(+ y ) which is ot always a prime umber, e.g. p=3, y =, xy p+ z =3(+)=6. 6

17 Equivalece & Miimizatio of DFAs 7

18 Applicatios of iterest Comparig two DFAs: L(DFA ) == L(DFA 2 )? How to miimize a DFA?. Remove ureachable states 2. Idetify & codese equivalet states ito oe 8

19 Whe to call two states i a DFA equivalet? Past does t matter - oly future does! AND Two states p ad q are said to be equivalet iff: i) Ay strig w accepted by startig at p is also accepted by startig at q; p q i) Ay strig w rejected by startig at p is also rejected by startig at q. è p q p q w w 9

20 Computig equivalet states i a DFA Table Fillig Algorithm A C E G B D F H A = B = = C x x = D x x x = E x x x x = F x x x x x = Pass # G x x x = x x =. Mark acceptig states o-acceptig states Pass # H x x = x x x x =. Compare every pair of states 2. Distiguish by oe symbol trasitio A B C D E F G H 3. Mark = or or blak (i.e. ca ot distiguish) Pass #2. Compare every pair of states 2. Distiguish by up to two symbol trasitios (util differet or same or tbd). (keep repeatig util table complete) How the table o the right was obtaied? Table Fillig Algorithm 2

21 Table Fillig Algorithm Recursive discovery of distiguishable states i a DFA Base case: If p is a acceptig state ad q is ot acceptig the the pair {p,q} is distiguishable. Iductio: Let p, q be states s.t. for some iput symbol a, r =! (p, a) ad s =! (q, a) are kow to be distiguishable. The the pair {p,q} is distiguishable. 2

22 Table Fillig Algorithm - step by step A C E G B D F H A = B = C = D = E = F = G = H = A B C D E F G H 22

23 Table Fillig Algorithm - step by step A C E G B D F H. Mark X betwee acceptig vs. o-acceptig state A = B = C = D = E X X X X = F X = G X = H X = A B C D E F G H 23

24 Table Fillig Algorithm - step by step A C E G B D F H. Mark X betwee acceptig vs. o-acceptig state 2. Look - hop away for distiguishig states or strigs A = B = C X = D X = E X X X X = F X = G X X = H X X = A B C D E F G H 24

25 Table Fillig Algorithm - step by step A C E G B D F H. Mark X betwee acceptig vs. o-acceptig state 2. Look - hop away for distiguishig states or strigs A = B = C X X = D X X = E X X X X = F X = G X X X = H X X X = A B C D E F G H 25

26 Table Fillig Algorithm - step by step A C E G B D F H. Mark X betwee acceptig vs. o-acceptig state 2. Look - hop away for distiguishig states or strigs A = B = C X X = D X X X = E X X X X = F X X = G X X X X = H X X = X = A B C D E F G H 26

27 Table Fillig Algorithm - step by step A C E G B D F H. Mark X betwee acceptig vs. o-acceptig state 2. Look - hop away for distiguishig states or strigs A = B = C X X = D X X X = E X X X X = F X X X = G X X X = X = H X X = X X = A B C D E F G H 27

28 Table Fillig Algorithm - step by step A C E G B D F H. Mark X betwee acceptig vs. o-acceptig state 2. Look - hop away for distiguishig states or strigs A = B = C X X = D X X X = E X X X X = F X X X = G X X X = X X = H X X = X X X = A B C D E F G H 28

29 Table Fillig Algorithm - step by step A C E G B D F H. Mark X betwee acceptig vs. o-acceptig state 2. Look - hop away for distiguishig states or strigs A = B = C X X = D X X X = E X X X X = F X X X = G X X X = X X = H X X = X X X X = A B C D E F G H 29

30 Table Fillig Algorithm - step by step A C E G B D F H A = B = =. Mark X betwee acceptig vs. o-acceptig state 2. Pass : Look - hop away for distiguishig states or strigs 3. Pass 2: Look -hop away agai for distiguishig states or strigs cotiue. C X X = D X X X = E X X X X = F X X X X X = G X X X = X X = H X X = X X X X = A B C D E F G H 3

31 Table Fillig Algorithm - step by step A C E G B D F H A = B = =. Mark X betwee acceptig vs. o-acceptig state 2. Pass : Look - hop away for distiguishig states or strigs 3. Pass 2: Look -hop away agai for distiguishig states or strigs cotiue. C X X = D X X X = E X X X X = F X X X X X = G X X X = X X = H X X = X X X X = A B C D E F G H Equivaleces: A=B C=H D=G 3

32 Table Fillig Algorithm - step by step A C E G B D F H {A,B} {C,H} E {D,G} F Retrai oly oe copy for each equivalece set of states Equivaleces: A=B C=H D=G 32

33 Table Fillig Algorithm special case A C E G B D F H Q) What happes if the iput DFA has more tha oe fial state? Ca all fial states iitially be treated as equivalet to oe aother? A = B = C = D = E? = F = G = H = A B C D E F G H 33

34 Puttig it all together How to miimize a DFA? Goal: Miimize the umber of states i a DFA Depth-first traversal from the start state Algorithm:. Elimiate states ureachable from the start state Table fillig algorithm 2. Idetify ad remove equivalet states 3. Output the resultat DFA 34

35 Summary How to prove laguages are ot regular? Pumpig lemma & its applicatios Simplificatio of DFAs How to remove ureachable states? How to idetify ad collapse equivalet states? How to miimize a DFA? 35