Chapter 9. Computational Complexity; P and NP. In the previous two chapters, we clarified what it means for a problem to be decidable or undecidable.

Transcription

1 Chaptr 9 Computational Complxity; P and NP 9.1 Th Class P In th prvious two chaptrs, w clarifid what it mans for a problm to b dcidabl or undcidabl. In principl, if a problm is dcidabl, thn thr is an algorithm (i.., a procdur that halts for vry input) that dcids vry instanc of th problm. Howvr, from a practical point of viw, knowing that aproblmisdcidablmaybuslss,ifthnumbrof stps (tim complxity) rquirdbythalgorithmisxcssiv, for xampl, xponntial in th siz of th input, or wors. 567

2 568 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP For instanc, considr th travling salsman problm, which can b formulatd as follows: W hav a st {c 1,...,c n } of citis, and an n n matrix D =(d ij )ofnonngativintgrs,thdistanc matrix, whr d ij dnots th distanc btwn c i and c j,which mans that d ii =0andd ij = d ji for all i j. Th problm is to find a shortst tour of th citis, that is, a prmutation π of {1,...,n} so that th cost C(π) =d π(1)π(2) + d π(2)π(3) + + d π(n 1)π(n) + d π(n)π(1) is as small as possibl (minimal). On way to solv th problm is to considr all possibl tours, i.., n! prmutations. Actually, sinc th starting point is irrlvant, w nd only considr (n 1)! tours, but this still grows vry fast. For xampl, whn n =40,itturnsoutthat39!xcds 10 45,ahugnumbr.

3 9.1. THE CLASS P 569 Considr th 4 4symmtricmatrixgivnby D =, and th budgt B =4. Th tour spcifid by th prmutation ( ) 1234 π = 1423 has cost 4, sinc c(π) =d π(1)π(2) + d π(2)π(3) + d π(3)π(4) + d π(4)π(1) = d 14 + d 42 + d 23 + d 31 = =4. Th citis in this tour ar travrsd in th ordr (1, 4, 2, 3, 1).

4 570 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Thus, to captur th ssnc of practically fasibl algorithms, w must limit our computational dvics to run only for a numbr of stps that is boundd by a polynomial in th lngth of th input. W ar ld to th dfinition of polynomially boundd computational modls. Dfinition 9.1. AdtrministicTuringmachinM is said to b polynomially boundd if thr is a polynomial p(x)sothatthfollowingholds: Forvryinputx Σ, thr is no ID ID n so that ID 0 ID 1 ID n 1 ID n, with n>p( x ), whr ID 0 = q 0 x is th starting ID. AlanguagL Σ is polynomially dcidabl if thr is a polynomially boundd Turing machin that accpts L. Thfamilyofallpolynomiallydcidabllanguagsis dnotd by P.

5 9.1. THE CLASS P 571 Rmark: Evn though Dfinition 9.1 is formulatd for Turing machins, it can also b formulatd for othr modls, such as RAM programs. Th rason is that th convrsion of a Turing machin into aramprogram(andvicvrsa)producsaprogram(or amachin)whossizispolynomialinthoriginaldvic. Th following lmma, although trivial, is usful: Lmma 9.1. Th class P is closd undr complmntation. Of cours, many languags do not blong to P. Onway to obtain such languags is to us a diagonal argumnt. But thr ar also many natural languags that ar not in P, althoughthismaybvryhardtoprovforsom of ths languags. Lt us considr a fw mor problms in ordr to gt a bttr fling for th family P.

6 572 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP 9.2 Dirctd Graphs, Paths Rcall that a dirctd graph, G, is a pair G =(V,E), whr E V V. Evry u V is calld a nod (or vrtx) and a pair (u, v) E is calld an dg of G. W will rstrict ourslvs to simpl graphs,thatis,graphs without dgs of th form (u, u); quivalntly, G =(V,E) is a simpl graph if whnvr (u, v) E, thnu v. Givn any two nods u, v V,apath from u to v is any squnc of n +1dgs(n 0) (u, v 1 ), (v 1,v 2 ),...,(v n,v). (If n =0,apathfromu to v is simply a singl dg, (u, v).)

7 9.2. DIRECTED GRAPHS, PATHS 573 AgraphG is strongly connctd if for vry pair (u, v) V V,thrisapathfromu to v. A closd path, or cycl, isapathfromsomnodu to itslf. W will rstrict out attntion to finit graphs, i.. graphs (V,E) whrv is a finit st. Dfinition 9.2. Givn a graph G, an Eulrian cycl is a cycl in G that passs through all th nods (possibly mor than onc) and vry dg of G xactly onc. A Hamiltonian cycl is a cycl that passs through all th nods xactly onc (not, som dgs may not b travrsd at all). Eulrian Cycl Problm: Givn a graph G, is thr an Eulrian cycl in G? Hamiltonian Cycl Problm: Givn a graph G, is thr an Hamiltonian cycl in G?

8 574 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP 9.3 Eulrian Cycls Th following graph is a dirctd graph vrsion of th Königsbrg bridg problm, solvd by Eulr in Th nods A, B, C, D corrspond to four aras of land in Königsbrg and th dgs to th svn bridgs joining ths aras of land. C A D B Figur 9.1: A dirctd graph modling th Königsbrg bridg problm Th problm is to find a closd path that crosss vry bridg xactly onc and rturns to th starting point. In fact, th problm is unsolvabl, as shown by Eulr, bcaus som nods do not hav th sam numbr of incoming and outgoing dgs (in th undirctd vrsion of th problm, som nods do not hav an vn dgr.)

9 9.3. EULERIAN CYCLES 575 It may com as a surpris that th Eulrian Cycl Problm dos hav a polynomial tim algorithm, but that so far, not such algorithm is known for th Hamiltonian Cycl Problm. Th rason why th Eulrian Cycl Problm is dcidabl in polynomial tim is th following thorm du to Eulr: Thorm 9.2. AgraphG =(V,E) has an Eulrian cycl iff th following proprtis hold: (1) Th graph G is strongly connctd. (2) Evry nod has th sam numbr of incoming and outgoing dgs. Proving that proprtis (1) and (2) hold if G has an Eulrian cycl is fairly asy. Th convrs is hardr, but not that bad (try!).

10 576 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Thorm 9.2 shows that it is ncssary to chck whthr agraphisstronglyconnctd.thiscanbdonbycomputing th transitiv closur of E, whichcanbdonin polynomial tim (in fact, O(n 3 )). Chcking proprty (2) can clarly b don in polynomial tim. Thus, th Eulrian cycl problm is in P. Unfortunatly, no thorm analogous to Thorm 9.2 is know for Hamiltonian cycls.

11 9.4. HAMILTONIAN CYCLES Hamiltonian Cycls AgaminvntdbySirWilliamHamiltonin1859uss argularsoliddodcahdronwhostwntyvrticsar labld with th nams of famous citis. Th playr is challngd to travl around th world by finding a closd cycl along th dgs of th dodcahdron which passs through vry city xactly onc (this is th undirctd vrsion of th Hamiltonian cycl problm).

12 578 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP In graphical trms, assuming an orintation of th dgs btwn citis, th graph D shown in Figur 9.2 is a plan projction of a rgular dodcahdron and w want to know if thr is a Hamiltonian cycl in this dirctd graph. Figur 9.2: A tour around th world. Finding a Hamiltonian cycl in this graph dos not appar to b so asy! AsolutionisshowninFigur9.3blow:

13 9.4. HAMILTONIAN CYCLES 579 v 20 v 1 v 19 v 9 v 8 v 2 v15 v 16 v 7 v 3 v 10 v 6 v 5 v 4 v 14 v 11 v 12 v13 v 18 v 17 Figur 9.3: A Hamiltonian cycl in D. Asolution!

14 580 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Rmark: W talkd about problms bing dcidabl in polynomial tim. Obviously, this is quivalnt to dciding som proprty of a crtain class of objcts, for xampl, finit graphs. Our framwork rquirs that w first ncod ths classs of objcts as strings (or numbrs), sinc P consists of languags. Thus, whn w say that a proprty is dcidabl in polynomial tim, w ar rally talking about th ncoding of this proprty as a languag. Thus, w hav to b carful about ths ncodings, but it is rar that ncodings caus problms.

15 9.5. PROPOSITIONAL LOGIC AND SATISFIABILITY Propositional Logic and Satisfiability W dfin th syntax and th smantics of propositions in conjunctiv normal form (CNF). Th syntax has to do with th lgal form of propositions in CNF. Such propositions ar intrprtd as truth functions, by assigning truth valus to thir variabls. W bgin by dfining propositions in CNF. Such propositions ar constructd from a countabl st, PV,ofpropositional (or boolan) variabls, say PV = {x 1,x 2,...,}, using th connctivs (and), (or) and (ngation).

16 582 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP W dfin a litral (or atomic proposition), L, as L = x or L = x, alsodnotdbyx, whrx PV. A claus, C, isadisjunctionofpairwisdistinctlitrals, C =(L 1 L 2 L m ). Thus, a claus may also b viwd as a nonmpty st C = {L 1,L 2,...,L m }. W also hav a spcial claus, th mpty claus, dnotd or (or {}). It corrsponds to th truth valu fals. A proposition in CNF, or boolan formula, P, is a conjunction of pairwis distinct clauss P = C 1 C 2 C n.

17 9.5. PROPOSITIONAL LOGIC AND SATISFIABILITY 583 Thus, a boolan formula may also b viwd as a nonmpty st P = {C 1,...,C n }, but this tim, th comma is intrprtd as conjunction. W also allow th proposition,and somtims th proposition (corrsponding to th truth valu tru). For xampl, hr is a boolan formula: P = {(x 1 x 2 x 3 ), (x 1 x 2 ), (x 2 x 3 ), (x 3 x 1 ), (x 1 x 2 x 3 )}. In ordr to intrprt boolan formula, w us truth assignmnts. W lt BOOL = {F, T}, thstoftruthvalus,whr F stands for fals and T stands for tru.

18 584 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP A truth assignmnt (or valuation), v, isanyfunction v : PV BOOL. Givn a truth assignmnt, v : PV BOOL, w dfin th truth valu, v(x), of a litral, claus, and boolan formula, X, usingthfollowingrcursivdfinition: (1) v( ) =F, v( ) =T. (2) v(x) =v(x), if x PV. (3) v(x) =v(x), if x PV, whrv(x) =F if v(x) =T and v(x) =T if v(x) =F. (4) v(c) =F if C is a claus and iff v(l i )=F for all litrals L i in C, othrwist. (5) v(p )=T if P is a boolan formula and iff v(c j )=T for all clauss C j in P,othrwisF.

19 9.5. PROPOSITIONAL LOGIC AND SATISFIABILITY 585 Dfinition 9.3. W say that a truth assignmnt, v, satisfis aboolanformula,p,if v(p )=T. In this cas, w also writ v = P. Aboolanformula,P,issatisfiabl if v = P for som truth assignmnt v, othrwis, it is unsatisfiabl. A boolan formula, P,isvalid (or a tautology) if v = P for all truth assignmnts v, inwhichcaswwrit = P. On should chck that th boolan formula P = {(x 1 x 2 x 3 ), (x 1 x 2 ), (x 2 x 3 ), (x 3 x 1 ), (x 1 x 2 x 3 )} is unsatisfiabl.

20 586 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP On may think that it is asy to tst whthr a proposition is satisfiabl or not. Try it, it is not that asy! As a mattr of fact, th satisfiability problm, tsting whthr a boolan formula is satisfiabl, also dnotd SAT, is not known to b in P. Morovr, it is an NP-complt problm. Most popl bliv that th satisfiability problm is not in P, buta proof still luds us! Bfor w xplain what is th class NP,ltusrmark that th satisfiability problm for clauss containing at most two litrals (2-satisfiability, or2-sat)issolvabl in polynomial tim.

21 9.5. PROPOSITIONAL LOGIC AND SATISFIABILITY 587 Th first stp consists in obsrving that if vry claus in P contains at most two litrals, thn w can rduc th problm to tsting satisfiability whn vry claus has xactly two litrals. Indd, if P contains som claus (x), thn any valuation satisfying P must mak x tru. Thn, all clauss containing x will b tru, and w can dlt thm, whras w can dlt x from vry claus containing it, sinc x is fals. Similarly, if P contains som claus (x), thn any valuation satisfying P must mak x fals. Thus, in a finit numbr of stps, ithr w gt th mpty claus, and P is unsatisfiabl, or w gt a st of clauss with xactly two litrals.

22 588 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Th numbr of stps is clarly linar in th numbr of litrals in P. For th scond stp, w construct a dirctd graph from P. Th nods of this graph ar th litrals in P,and dgs ar dfind as follows: (1) For vry claus (x y), thr is an dg from x to y and an dg from y to x. (2) For vry claus (x y), thr is an dg from x to y and an dg from y to x (3) For vry claus (x y), thr is an dg from x to y and an dg from y to x. Thn, it can b shown that P is unsatisfiabl iff thr is som x so that thr is a cycl containing x and x. As a consqunc, 2-satisfiability is in P.

23 9.6. THE CLASS NP, NP-COMPLETENESS Th Class NP,PolynomialRducibility, NP-Compltnss On will obsrv that th hard part in trying to solv ithr th Hamiltonian cycl problm or th satisfiability problm, SAT, is to find asolution,butthatchcking that a candidat solution is indd a solution can b don asily in polynomial tim. This is th ssnc of problms that can b solvd nondtrmistically in polynomial tim: A solution can b gussd and thn chckd in polynomial tim.

24 590 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Dfinition 9.4. AnondtrministicTuringmachinM is said to b polynomially boundd if thr is a polynomial p(x) sothatthfollowingholds: Forvryinput x Σ,thrisnoIDID n so that ID 0 ID 1 ID n 1 ID n, with n>p( x ), whr ID 0 = q 0 x is th starting ID. AlanguagL Σ is nondtrministic polynomially dcidabl if thr is a polynomially boundd nondtrministic Turing machin that accpts L. Th family of all nondtrministic polynomially dcidabl languags is dnotd by NP.

25 9.6. THE CLASS NP, NP-COMPLETENESS 591 Of cours, w hav th inclusion P NP, but whthr or not w hav quality is on of th most famous opn problms of thortical computr scinc and mathmatics. In fact, th qustion P NP is on of th opn problms listd by th CLAY Institut, togthr with th Poincaré conjctur and th Rimann hypothsis, among othr problms, and for which on million dollar is offrd as a rward! It is asy to chck that SAT is in NP,andsoisth Hamiltonian cycl problm.

26 592 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP As w saw in rcursion thory, whr w introducd th notion of many-on rducibility, in ordr to compar th dgr of difficulty of problms, it is usful to introduc th notion of rducibility and th notion of a complt st. Dfinition 9.5. Afunctionf :Σ Σ is polynomialtim computabl if thr is a polynomial p(x)sothatth following holds: Thr is a dtrministic Turing machin M computing it so that for vry input x Σ,thris no ID ID n so that ID 0 ID 1 ID n 1 ID n, with n>p( x ), whr ID 0 = q 0 x is th starting ID. Givn two languags L 1,L 2 Σ, apolynomialrduction from L 1 to L 2 is a polynomial-tim computabl function f :Σ Σ so that for all u Σ, u L 1 iff f(u) L 2.

27 9.6. THE CLASS NP, NP-COMPLETENESS 593 For xampl, on can construct a polynomial rduction from th Hamiltonian cycl problm to SAT. Rmarkably, vry languag in NP can b rducd to SAT. Intuitivly, if L 1 is a hard problm and L 1 can b rducd to L 2,thnL 2 is also a hard problm. Thus, SAT is a hardst problm in NP (Sinc it is in NP). Dfinition 9.6. AlanguagL is NP-hard if thr is a polynomial rduction from vry languag L 1 NPto L. AlanguagL is NP-complt if L NPand L is NP-hard.

28 594 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Thus, an NP-hard languag is as hard to dcid as any languag in NP. Th importanc of NP-complt problms stms from th following thorm: Thorm 9.3. Lt L b an NP-complt languag. Thn, P = NP iff L P. Nxt, w prov a famous thorm of Stv Cook and Lonid Lvin (provd indpndntly): SAT is NP-complt.

29 9.7. THE COOK-LEVIN THEOREM Th Cook Lvin Thorm: SAT is NP-Complt Instad of showing dirctly that SAT is NP-complt, which is rathr complicatd, w procd in two stps, as suggstd by Lwis and Papadimitriou. (1) First, w dfin a tiling problm adaptd from H. Wang (1961) by Harry Lwis, and w prov that it is NP-complt. (2) W show that th tiling problm can b rducd to SAT. W ar givn a finit st T = {t 1,...,t p } of til pattrns, for short, tils. Copisofthstilpattrnsmaybusd to til a rctangl of prdtrmind siz 2s s (s >1). Howvr, thr ar constraints on thwaythatthstils may b adjacnt horizontally and vrtically.

30 596 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Th horizontal constraints ar givn by a rlation H T T,andthvrtical constraints ar givn by a rlation V T T. Thus, a tiling systm is a tripl T =(T,V,H)withV and H as abov. Th bottom row of th rctangl of tils is spcifid bfor th tiling procss bgins. For xampl, considr th following til pattrns:

31 9.7. THE COOK-LEVIN THEOREM 597 a a d c, c, c,, a a b b c d c d, c d, d,, b c d c d, c d, d,, b c d Th horizontal and th vrtical constraints ar that th lttrs on adjacnt dgs match (blank dgs do not match). Lt us try to find a 6 3tilingwiththinitialrowshown on th nxt pag.

32 598 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP For s =3,givnthbottomrow a c b c d c d d d w hav th tiling shown blow: c a c d b d c d d a c a b c d b c d c d d d d a c b c d c d d d

33 9.7. THE COOK-LEVIN THEOREM 599 Th problm is thn as follows: Th Boundd Tiling Problm Givn any tiling systm (T,V,H), any intgr s>1, and any initial row of tils σ 0 (of lngth 2s) σ 0 : {1, 2,...,s,s+1,...,2s} T, find a 2s s-tiling σ xtnding σ 0,i..,afunction so that σ : {1, 2,...,s,s+1,...,2s} {1,...,s} T (1) σ(m, 1) = σ 0 (m), for all m with 1 m 2s. (2) (σ(m, n),σ(m +1,n)) H, forallm with 1 m 2s 1, and all n, with1 n s. (3) (σ(m, n),σ(m, n +1)) V,forallm with 1 m 2s, andalln, with1 n s 1.

34 600 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Formally, an instanc of th tiling problm is a tripl, ((T,V,H), ŝ, σ 0 ), whr (T,V,H)isatilingsystm,ŝ is th string rprsntation of th numbr s 2, in binary and σ 0 is an initial row of tils (th bottom row). For xampl, if s =1025(asadcimalnumbr),thnits binary rprsntation is ŝ = Thlngthof ŝ is log 2 s +1. Rcall that th input must b a string. This is why th numbr s is rprsntd by a string in binary. If w only includd a singl til σ 0 in position (s +1, 1), thn th lngth of th input ((T,V,H), ŝ, σ 0 )wouldb log 2 s + C +2forsomconstantC corrsponding to th lngth of th string ncoding (T,V,H).

35 9.7. THE COOK-LEVIN THEOREM 601 Howvr, th rctangular grid has siz 2s 2,whichisxponntial in th lngth log 2 s + C +2ofthinput ((T,V,H), ŝ, σ 0 ). Thus, it is impossibl to chck in polynomial tim that a proposd solution is a tiling. Howvr, if w includ in th input th bottom row σ 0 of lngth 2s, thnthsizofthgridisinddpolynomial in th siz of th input. Thorm 9.4. Th tiling problm dfind arlir is NP-complt. Proof. Lt L Σ b any languag in NP and lt u b any string in Σ.AssumthatL is accptd in polynomial tim boundd by p( u ).

36 602 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP W show how to construct an instanc of th tiling problm, ((T,V,H) L, ŝ, σ 0 ), whr s = p( u )+2,andwhr th bottom row ncods th starting ID, so that u L iff th tiling problm ((T,V,H) L, ŝ, σ 0 )hasasolution. First, not that th problm is indd in NP,sincw hav to guss a rctangl of siz 2s 2,andthatchcking that a tiling is lgal can indd b don in O(s 2 ), whr s is boundd by th th siz of th input ((T,V,H), ŝ, σ 0 ), sinc th input contains th bottom row of 2s symbols (this is th rason for including th bottom row of 2s tils in th input!).

37 9.7. THE COOK-LEVIN THEOREM 603 Th ida bhind th dfinition of th tils is that, in a solution of th tiling problm, th labls on th horizontal dgs btwn two adjacnt rows rprsnt a lgal ID, upav. In a givn row, th labls on vrtical dgs of adjacnt tils kp track of th chang of stat and dirction. Lt Γ b th tap alphabt of th TM, M. Asbfor,w assum that M signals that it accpts u by halting with th output 1 (tru). From M, wcratthfollowingtils: (1) For vry a Γ, tils a a

38 604 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP (2) For vry a Γ, th bottom row uss tils a, q 0,a whr q 0 is th start stat. (3) For vry instruction (p, a, b, R, q) δ, forvryc Γ, tils b p, a q, R, q, R q, c c

39 9.7. THE COOK-LEVIN THEOREM 605 (4) For vry instruction (p, a, b, L, q) δ, forvryc Γ, tils q, c c q, L, q, L b p, a (5) For vry halting stat, p, tils p, 1 p, 1 Th purpos of tils of typ (5) is to fill th 2s s rctangl iff M accpts u. Sinc s = p( u ) +2andthmachin runs for at most p( u ) stps,th2s s rctangl can b tild iff u L. Th vrtical and th horizontal constraints ar that adjacnt dgs hav th sam labl (or no labl).

40 606 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP If u = u 1 u k,thinitialbottomrowσ 0,oflngth2s, is: B q 0,u 1 u k B whr th til labld q 0,u 1 is in position s +1. Th xampl blow illustrats th construction: B B f,1 B... f,r f,r... B q, c 1 B B q, c 1 B... q, L q, L... B c p, a B B c p, a B... p, R p, R... B r, b a B

41 9.7. THE COOK-LEVIN THEOREM 607 It is not hard to chck that u = u 1 u k is accptd by M iff th tiling problm just constructd has a solution. This is bcaus s = p( u )+2andthmachinrunsfor at most p( u ) stps.soth2s s rctangl can b tild iff tils of typ (5) ar usd iff M accpts u (prints 1). Rmarks. (1) Th problm bcoms hardr if w only spcify a singl til σ 0 as input, instad of a row of lngth 2s. Ifs is spcifid in binary (or any othr bas, but not in tally notation), thn th 2s 2 grid has siz xponntial in th lngth log 2 s + C +2ofthinput((T,V,H), ŝ, σ 0 ), and this tiling problm is actually NEXP-complt! (2) If w rlax th finitnss condition and rquir that th ntir uppr half-plan b tild, i.., for vry s>1, thr is a solution to th 2s s-tiling problm, thn th problm is undcidabl. In 1972, Richard Karp publishd a list of 21 NP-complt problms. W finally prov th Cook-Lvin thorm.

42 608 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Thorm 9.5. (Cook, 1971, Lvin, 1973) Th satisfiability problm SAT is NP-complt. Proof. W rduc th tiling problm to SAT. Givn a tiling problm, ((T,V,H), ŝ, σ 0 ), w introduc boolan variabls x mnt, for all m with 1 m 2s, alln with 1 n s, and all tils t T. Th intuition is that x mnt = T iff til t occurs in som tiling σ so that σ(m, n) =t. W dfin th following clauss:

43 9.7. THE COOK-LEVIN THEOREM 609 (1) For all m, n in th corrct rang, as abov, for all p tils in T. (x mnt1 x mnt2 x mntp ), This claus stats that vry position in σ is tild. (2) For any two distinct tils t t T,forallm, n in th corrct rang, as abov, (x mnt x mnt ). This claus stats that a position may not b occupid by mor than on til.

44 610 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP (3) For vry pair of tils (t, t ) T T H, forallm with 1 m 2s 1, and all n, with1 n s, (x mnt x m+1 nt ). This claus nforcs th horizontal adjacncy constraints. (4) For vry pair of tils (t, t ) T T V,forallm with 1 m 2s, andalln, with1 n s 1, (x mnt x mn+1 t ). This claus nforcs th vrtical adjacncy constraints. (5) For all m with 1 m 2s, (x m1σ0 (m)). This claus stats that th bottom row is corrctly tild with σ 0.

45 9.7. THE COOK-LEVIN THEOREM 611 It is asily chckd that th tiling problm has a solution iff th conjunction of th clauss just dfind is satisfiabl. Thus, SAT is NP-complt. W sharpn Thorm 9.5 to prov that 3-SAT is also NPcomplt. This is th satisfiability problm for clauss containing at most thr litrals. W know that w can t go furthr and rtain NP-complttnss, sinc 2-SAT is in P. Thorm 9.6. (Cook, 1971) Th satisfiability problm 3-SAT is NP-complt.

46 612 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Proof. W hav to brak long clauss C =(L 1 L k ), i.., clauss containing k 4litrals,intoclausswith at most thr litrals, in such a way that satisfiability is prsrvd. For xampl, considr th following claus with k = 6 litrals: C =(L1 L2 L3 L4 L5 L6). W crat 3 nw boolan variabls y 1,y 2,y 3,andth4 clauss (L 1 L 2 y 1 ), (y 1 L 3 y 2 ), (y 2 L 4 y 3 ), (y 3 L 5 L 6 ). Lt C b th conjunction of ths clauss.

47 9.7. THE COOK-LEVIN THEOREM 613 W claim that C is satisfiabl iff C is. Assum that C is satisfiabl but C is not. If so, in any truth assigmnt v, v(l i )=F, fori =1, 2,...,6. To satisfy th first claus, w must hav v(y 1 )=T. Thn to satisfy th scond claus, w must hav v(y 2 )= T, andsimilarlysatisfyththirdclaus,wmusthav v(y 3 )=T. Howvr, sinc v(l 5 ) = F and v(l 6 ) = F, thonly way to satisfy th fourth claus is to hav v(y 3 )=F, contradicting that v(y 3 )=T. Thus, C is indd satisfiabl.

48 614 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Lt us now assum that C is satisfiabl. This mans that thr is a smallst indx i such that L i is satisfid. Say i =1,sov(L 1 )=T. Thnifwltv(y 1 )=v(y 2 )= v(y 3 )=F, wsthatc is satisfid. Say i =2,sov(L 1 )=F and v(l 2 )=T. Again if w lt v(y 1 )=v(y 2 )=v(y 3 )=F, wsthat C is satisfid. Say i =3,sov(L 1 )=F, v(l 2 )=F, andv(l 3 )=T. If w lt v(y 1 )=T and v(y 2 )=v(y 3 )=F, wsthat C is satisfid. Say i =4,sov(L 1 )=F, v(l 2 )=F, v(l 3 )=F, and v(l 4 )=T. If w lt v(y 1 )=T, v(y 2 )=T and v(y 3 )=F, ws that C is satisfid.

49 9.7. THE COOK-LEVIN THEOREM 615 Say i =5,sov(L 1 )=F, v(l 2 )=F, v(l 3 )=F, v(l 4 )= F, andv(l 5 )=T. If w lt v(y 1 )=T, v(y 2 )=T and v(y 3 )=T, ws that C is satisfid. Say i =6,sov(L 1 )=F, v(l 2 )=F, v(l 3 )=F, v(l 4 )= F, v(l 5 )=F, andv(l 6 )=T. Again, if w lt v(y 1 )=T, v(y 2 )=T and v(y 3 )=T, w s that C is satisfid. Thrfor if C is satisfid, thn C is satisfid in all cass. In gnral, for vry long claus (with k 4), crat k 3 nw boolan variabls y 1,...y k 3,andthk 2clauss (L 1 L 2 y 1 ), (y 1 L 3 y 2 ), (y 2 L 4 y 3 ),, (y k 4 L k 2 y k 3 ), (y k 3 L k 1 L k ). Lt C b th conjunction of ths clauss. W claim that C is satisfiabl iff C is.

50 616 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Assum that C is satisfiabl, but that C is not. Thn, for vry truth assignmnt v, whavv(l i )=F, for i =1,...,k. Howvr, C is satisfid by som v, andthonlywaythis can happn is that v(y 1 )=T, tosatisfythfirstclaus. Thn, v(y 1 )=F, andwmusthavv(y 2 )=T, tosatisfy th scond claus. By induction, w must hav v(y k 3 )=T, tosatisfyth nxt to th last claus. Howvr, th last claus is now fals, a contradiction. Thus, if C is satisfiabl, thn so is C.

51 9.7. THE COOK-LEVIN THEOREM 617 Convrsly, assum that C is satisfiabl. If so, thr is som truth assignmnt, v, sothatv(c) =T, andthus, thr is a smallst indx i, with1 i k, sothat v(l i )=T (and so, v(l j )=F for all j<i). Lt v b th assignmnt xtnding v dfind so that v (y j )=F if max{1,i 1} j k 3, and v (y j )=T, othrwis. It is asily chckd that v (C )=T. Anothr vrsion of 3-SAT can b considrd, in which vry claus has xactly thr litrals. W will call this th problm xact 3-SAT.

52 618 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Thorm 9.7. (Cook, 1971) Th satisfiability problm for xact 3-SAT is NP-complt. Proof. Aclausofthform(L) issatisfiabliffthfollowing four clauss ar satisfiabl: (L u v), (L u v), (L u v), (L u v). Aclausofthform(L 1 L 2 )issatisfiabliffthfollowing two clauss ar satisfiabl: (L 1 L 2 u), (L 1 L 2 u). Thus, w hav a rduction of 3-SAT to xact 3-SAT. W now mak som rmarks on th convrsion of propositions to CNF.

53 9.7. THE COOK-LEVIN THEOREM 619 Rcall that th st of propositions (ovr th connctivs,, and ) isdfindinductivlyasfollows: (1) Evry propositional lttr, x PV, isaproposition (an atomic proposition). (2) If A is a proposition, thn A is a proposition. (3) If A and B ar propositions, thn (A B) isaproposition. (4) If A and B ar propositions, thn (A B) isaproposition. Two propositions A and B ar quivalnt,dnotd A B, if for all truth assignmnts, v. v = A iff v = B

54 620 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP It is asy to show that A B iff th proposition is valid. ( A B) ( B A) Evry proposition, A, isquivalnttoaproposition,a, in CNF. Thr ar svral ways of proving this fact. On mthod is algbraic, and consists in using th algbraic laws of boolan algbra. First, on may convrt a proposition to ngation normal form, ornnf.apropositionisinnnfifoccurrncsof only appar in front of propositional variabls, but not in front of compound propositions.

55 9.7. THE COOK-LEVIN THEOREM 621 Any proposition can b convrtd to an quivalnt on in nnf by using th d Morgan laws: (A B) ( A B) (A B) ( A B) A A. Thn, a proposition in nnf can b convrtd to CNF, but th qustion of uniqunss of th CNF is a bit tricky. For xampl, th proposition has A =(u (x y)) ( u (x y)) A 1 =(u x y) ( u x y) A 2 =(u u) (x y) A 3 = x y, as quivalnt propositions in CNF!

56 622 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP W can gt a uniqu CNF quivalnt to a givn proposition if w do th following: (1) Lt Var(A) ={x 1,...,x m } b th st of variabls occurring in A. (2) Dfin a maxtrm w.r.t. Var(A) as any disjunction of m pairwis distinct litrals formd from Var(A), and not containing both som variabl x i and its ngation x i. (3) Thn, it can b shown that for any proposition A that is not a tautology, thr is a uniqu proposition in CNF quivalnt to A, whosclaussconsist of maxtrms formd from Var(A). Th abov dfinition can yild strang rsults. For instanc, th CNF of any unsatisfiabl proposition with m distinct variabls is th conjunction of all of its 2 m maxtrms! Th abov notion dos not cop wll with minimality.

57 9.7. THE COOK-LEVIN THEOREM 623 For xampl, according to th abov, th CNF of should b A =(u (x y)) ( u (x y)) A 1 =(u x y) ( u x y). Thr ar also propositions such that any quivalnt proposition in CNF has siz xponntial in trms of th original proposition. Hr is such an xampl: A =(x 1 x 2 ) (x 3 x 4 ) (x 2n 1 x 2n ). Obsrv that it is in DNF. W will prov a littl latr that any CNF for A contains 2 n occurrncs of variabls.

58 624 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP AnicmthodtoconvrtapropositioninnnftoCNFis to construct a tr whos nods ar labld with sts of propositions using th following (Gntzn-styl) ruls : and P, Q, (P Q), P, Q, (P Q), whr stands for any st of propositions (vn mpty), and th comma stands for union. Thus, it is assumd that (P Q) / inthfirstcas,andthat(p Q) / in th scond cas. Sinc w intrprt a st, Γ, of propositions as a disjunction, a valuation, v, satisfisγiffitsatisfissom proposition in Γ.

59 9.7. THE COOK-LEVIN THEOREM 625 Obsrv that a valuation v satisfis th conclusion of a rul iff it satisfis both prmiss in th first cas, and th singl prmis in th scond cas. Using ths ruls, w can build a finit tr whos lavs ar labld with sts of litrals. By th abov obsrvation, a valuation v satisfis th proposition labling th root of th tr iff it satisfis all th propositions labling th lavs of th tr. But thn, a CNF for th original proposition A (in nnf, at th root of th tr) is th conjunction of th clauss apparing as th lavs of th tr. W may xclud th clauss that ar tautologis, and w may discovr in th procss that A is a tautology (whn all lavs ar tautologis).

60 626 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP Going back to our bad proposition, A, by induction, w s that any tr for A has 2 n lavs. Howvr, it should b notd that for any proposition, A, wcanconstructinpolynomialtimaformula,a, in CNF, so that A is satisfiabl iff A is satisfiabl, by crating nw variabls. W procd rcursivly. Th trick is that w rplac by (C 1 C m ) (D 1 D n ) (C 1 y) (C m y) (D 1 y) (D n y), whr th C i s and th D j s ar clauss, and y is a nw variabl. It can b shown that th numbr of nw variabls rquird is at most quadratic in th siz of A.

61 9.7. THE COOK-LEVIN THEOREM 627 Warning: In gnral, th proposition A is not quivalnt to th proposition A. Ruls for daling for can also b cratd. In this cas, w work with pairs of sts of propositions, Γ, whr, th propositions in Γ ar intrprtd conjunctivly, and th propositions in ar intrprtd disjunctivly. W obtain a sound and complt proof systm for propositional logic (a Gntzn-styl proof systm, s Gallir s Logic for Computr Scinc).

62 628 CHAPTER 9. COMPUTATIONAL COMPLEXITY; P AND NP