Conditional Hardness for Approximate Coloring

Transcription

1 Condtonal Hardness for Approxmate Colorng Irt Dnur Elchanan Mossel Oded Regev Aprl 30, 2009 Abstract We study the APPROXCOLORING(q, Q) problem: Gven a graph G, decde whether χ(g) q or χ(g) Q. We present hardness results for ths problem for any constants 3 q < Q. For q 4, our result s based on Khot s 2-to-1 label cover, whch s conjectured to be NPhard [Khot02]. For q = 3, we base our hardness result on a certan < shaped varant of hs conjecture. Prevously no hardness result was known for q = 3 and Q 6. At the heart of our proof are tght bounds on generalzed nose-stablty quanttes, whch extend the recent work of Mossel et al. [MOO05] and should have wder applcablty. 1 Introducton The approxmate graph colorng problem, whch we descrbe next, s one of a few classcal optmzaton problems whose approxmablty behavor s stll qute mysterous, despte ncreasngly complex technques developed n the past 15 years. For an undrected graph G = (V, E), let χ(g) be ts chromatc number,.e., the smallest number of colors needed to color the vertces of G wthout monochromatc edges. Then the approxmate graph colorng problem s defned as follows. APPROXCOLORING(q, Q) : Gven a graph G, decde between χ(g) q and χ(g) Q. Ths problem also has a natural search varant, whch can be stated as follows: gven a graph G wth χ(g) q, color G wth less than Q colors. It s easy to see that the search varant s not easer than the orgnal decson varant, and hence for the purpose of showng hardness results t s enough to consder the decson varant. It s easy to solve the problem APPROXCOLORING(2, Q) for any Q 3 n polynomal tme as t amounts to checkng bpartteness. The stuaton wth APPROXCOLORING(3, Q) s much more nterestng, as there s a huge gap between the value of Q for whch an effcent algorthm s known and that for whch a hardness result exsts. Indeed, untl not long ago, the best known polynomaltme algorthm was due to Blum and Karger [9], who solve the problem for Q = Õ(n3/14 ) colors, where n s the number of vertces and the Õ notaton hdes poly-logarthmc factors (as s often A prelmnary verson of ths paper has appeared n the proceedngs of STOC Hebrew Unversty. Emal: dnur@cs.huj.ac.l. Supported by the Israel Scence Foundaton and by the Bnatonal Scence Foundaton. Statstcs, U.C. Berkeley. Emal: mossel@stat.berkeley.edu. Supported by a Mller fellowshp n Computer Scence and Statstcs, a Sloan fellowshp n Mathematcs, the Bnatonal Scence Foundaton, and NSF grants DMS and DMS (CAREER) Department of Computer Scence, Tel-Avv Unversty, Tel-Avv 69978, Israel. Supported by an Alon Fellowshp, by the Bnatonal Scence Foundaton, by the Israel Scence Foundaton, by the European Commsson under the Integrated Project QAP funded by the IST drectorate as Contract Number , and by a European Research Councl (ERC) Startng Grant. 1

2 the case, ther algorthm actually solves the search varant). Ther work contnues a long lne of research [37, 8, 27] and s based on a sem-defnte relaxaton. Very recently, Arora et al. [2] were able to mprove ths to Q = O(n ) colors, and the constant n the exponent can possbly be reduced even further f certan geometrc conjectures are proven. However, there s some ndcaton that ths lne of work s lmted by Q = n δ for some fxed δ > 0, see [17]. In contrast, the strongest known hardness result shows that the problem s NP-hard for Q = 5 [28, 23]. Thus, the problem s open for all 5 < Q < O(n ). In ths paper we prove a hardness result for any constant (.e., ndependent of n) value of Q. As we shall explan later, our hardness result s based on the conjectured NP-hardness of certan nstances of the label cover problem due to Khot [30]. The stuaton wth APPROXCOLORING(q, Q) for small values q 4 s smlar. The best known algorthm, due to Halpern et al. [24], solves APPROXCOLORING(q, Q) for Q = Õ(nα q ) where 0 < α q < 1 s some constant dependng on q. For example, α On the other hand, there are several known NP-hardness results. One of the strongest s due to Khot [29], who mproved on an earler result of Fürer [20] by showng that for any large enough constant q and Q = q log q 25, APPROXCOLORING(q, Q) s NP-hard. Notce that for any fxed q, Khot s result, as well as all other known hardness results, apply only up to some fxed Q. Our result holds for any Q > q 3. Hardness Results. One of the most successful approaches to dervng hardness proofs, whch s also the one we shall take here, s by a reducton from a certan combnatoral problem known as the label-cover problem [3]. The PCP theorem [5, 4] says that ths problem s NP-hard. In the labelcover problem, we are gven an undrected graph and a number R. Each edge s assocated wth a bnary relaton on {1,..., R} and we refer to t as a constrant. The goal s to label the vertces wth values from {1,..., R} such that the number of satsfed constrants s maxmzed, where a constrant s satsfed f the labels on the two ncdent vertces satsfy the relaton assocated wth t. Wthout gong nto the detals of the reducton (these detals are descrbed n Secton 4), we remark that for our reducton to work, the label-cover nstances we use must have constrants of a very specfc form. For example, we mght requre all constrants to be bjectons,.e., a bnary relaton n whch any labellng of one vertex determnes the other, and vce versa. We call ths specal case the 1 1-label-cover. The precse defnton of ths and other specal cases wll appear later. Unfortunately, these specal cases of the label-cover problem are not known to be NP-hard. Nevertheless, n hs semnal work [30] Khot conjectured that such problems are n fact NP-hard, although the tools necessary to prove ths conjecture seem to be beyond our current reach. Ths conjecture has snce been heavly scrutnzed [36, 13, 22, 14], and so far there s no evdence aganst the conjecture. Ths ssue s currently one of the central topcs n theoretcal computer scence. Khot s conjecture s known to mply many strong, and often tght, hardness results. Two examples are the NP-hardness of approxmatng the VERTEXCOVER problem to wthn factors below 2 [32], whch s tght by a smple greedy algorthm, and the NP-hardness of approxmatng the MAXCUT problem to wthn factors larger than roughly [31], whch s tght by the algorthm of Goemans and Wllamson [21]. Our result contnues ths lne of work by showng that (varants of) Khot s conjecture mply strong hardness results for another fundamental problem that of approxmate graph colorng. More specfcally, we present three reductons, each from a dfferent specal case of the label-cover problem. An exact descrpton of the three reductons wll be gven later. For now, we just state nformally one mplcaton of our reductons. Theorem 1.1 (nformal) If a certan specal case of the label-cover problem s NP-hard, then for any Q > 3, APPROXCOLORING(3, Q) s NP-hard. 2

3 Bounds on Blnear Forms At the heart of our hardness results are certan bounds on blnear forms descrbng the correlaton under nose between two functons f, g : [q] n R where [q] := {1,..., q}. Snce we beleve these bounds mght be useful elsewhere, we now descrbe them n some detal. Let T be a symmetrc Markov operator on [q] (equvalently, T s the random walk on a regular undrected weghted graph on vertex set [q] possbly wth self-loops). We study the expectaton E x,y [f(x)g(y)] where x [q] n s chosen unformly and y [q] n s obtaned from x by applyng T to each coordnate ndependently. Usng notaton we ntroduce later, ths expectaton can also be wrtten as f, T n g = E x [f(x) T n g(x)]. (1) We are nterested n the case where T and q are fxed, and n tends to nfnty. Our man techncal result provdes tght bounds on the blnear form f, T n g for bounded functons f, g, n terms of E[f], E[g], and ρ, where ρ s the second largest egenvalue n absolute value of T. To motvate ths result, consder the followng concrete example. Take q = 2 and f : {0, 1} n {0, 1} satsfyng E[f] = 1/2 (.e., f s balanced). Fx some ρ (0, 1), and let T ρ be the operator that flps each bt wth probablty (1 ρ)/2, T ρ = ρ ( ) ( 1 + (1 ρ) 2 We would lke to know how hgh the stablty of a balanced Boolean functon f can be, where the stablty of f (wth parameter ρ) s defned as the probablty that f(x) = f(y) where x s chosen unformly from {0, 1} n and y s obtaned from x by flppng each bt ndependently wth probablty (1 ρ)/2. It s easy to see that the stablty of f can be wrtten as 1 2 Pr [f(x) = f(y)] = 2E x,y[f(x)f(y)] = 2 f, T n x,y ρ f (2) and hence an upper bound on the stablty of a balanced Boolean functon would follow from an upper bound on (1). If the functon f depends on just one coordnate of ts nput, say f(x 1,..., x n ) = x 1, then ts stablty s smply (1 + ρ)/2, and t can be shown that ths s the hghest possble for any balanced functon f. We consder such cases degenerate and nstead study functons that do not depend too strongly on any one coordnate of ther nput (ths wll be made precse soon). An example of such a functon s the majorty functon, whose value s 1 f and only f more than half of ts nput bts are 1 (assume for smplcty n s odd). It can be shown that the stablty of ths functon approaches arcsn(ρ)/π as n goes to nfnty. But s majorty the most stable functon among those who do not depend too strongly on any one coordnate? The results of [33] mply that the answer s essentally yes. In the work presented here we generalze such stablty statements to cases where T s a general reversble Markov operator (and not just the specfc operator defned above) and relax the assumptons on nfluences as dscussed later. Functons wth low nfluences. The noton of the nfluence of a varable on a functon defned n a product space [26] played a major role n recent developments n dscrete mathematcs, see for example [34, 35, 19, 18, 11, 7, 12]. Consder the space [q] n equpped wth the unform measure. Then the nfluence of the th varable on the functon f : [q] n R s defned by ). I (f) := E[Var x [f(x) x 1,..., x 1, x +1,..., x n ]]. 3

4 In recent years, startng wth [7, 11], an effort has been made to study propertes of functons all of whose varables have low nfluences. In addton to a natural mathematcal nterest n functons that do not depend strongly on any one coordnate, the study of such functons s essental for proofs of hardness of approxmaton results, see for example [16, 30, 32, 31]. Man Techncal Result. Our man techncal result s a bound on (1) for bounded functons f, g that have no common nfluental coordnate. The upper and lower bounds are stated n terms of nner products of the form F η, U ρ F ν γ where γ denotes the standard Gaussan measure on R, F µ (s) = 1 s<t s an ndcator functon where t s chosen so that E γ [F µ ] = µ, and U ρ s the Ornsten- Uhlenbeck operator, U ρ G(x) = E y γ [G(ρx + 1 ρ 2 y)]. These nner products can be wrtten as certan double ntegrals, and we menton the easy bound 0 < F η, U ρ F ν γ < mn(η, ν) for all ν, η and ρ strctly between 0 and 1. Theorem 1.2 Let T be some fxed symmetrc Markov operator on a fnte state space [q] whose second largest egenvalue n absolute value s ρ = r(t ) < 1. Then for any ε > 0 there exsts a δ > 0 such that f f, g : [q] n [0, 1] are two functons satsfyng for all, then t holds that mn ( I (f), I (g) ) < δ F E[f], U ρ (1 F 1 E[g] ) γ ε f, T n g F E[f], U ρ F E[g] γ + ε. (3) Stated n the contrapostve, ths theorem says that f for some two functons f, g, (1) devates from a certan range, then there must exst a coordnate that s nfluental n both functons. The fact that we obtan a common nfluental coordnate s crucal n our applcatons, as well as n a recent applcaton of Theorem 1.2 to the characterzaton of ndependent sets n graph powers [15]. We remark that for any T, the bounds n the theorem are essentally tght (see Appendx B). Gong back to our earler example, consder a balanced functon f : {0, 1} n {0, 1} all of whose nfluences are small. Applyng the theorem wth q = 2 and T ρ, and usng (2), gves us an upper bound of essentally 2 F 0.5, U ρ F 0.5 γ on the stablty of f. A straghtforward calculaton shows that ths value equals arcsn(ρ)/π, and hence we obtan as a specal case of our man theorem that asymptotcally, the majorty functon s the most stable among all balanced functons wth low nfluences. As mentoned before, ths specal case s not new to our work. It was orgnally presented as a conjecture n the work of [31] on the computatonal hardness of the MAXCUT problem, and has snce been proven by Mossel et al [33], who refer to t as the majorty s stablest theorem. For the proof, [33] developed a very powerful nvarance prncple. Ths prncple allows one to translate questons on low-nfluence functons n the dscrete settng (such as the above queston on {0, 1} n ) to correspondng questons n other spaces, and n partcular Gaussan space. The advantage of ths s that one can then apply known (and powerful) results n Gaussan space (such as [10]). Our proof of Theorem 1.2 also reles on ths nvarance prncple, and can n fact be seen as an extenson of the technques n [33]. Our theorem mproves on the one from [33] n the followng two aspects: The analyss of [33] only consders a very partcular nose operator known as the Beckner operator. We extend ths to more general nose operators that are gven by an arbtrary symmetrc Markov operator. In the applcaton to hardness of colorng we apply the result to three dfferent operators. We remark that our man theorem can be easly extended to 4

5 reversble Markov operators, wth the statonary dstrbuton takng the place of the unform dstrbuton. Perhaps more mportantly, our Theorem 1.2 allows one to conclude about the exstence of a common nfluental coordnate. A more drect applcaton of [33] only mples the exstence of an nfluental varable n one of the functons (n other words, one would have a max nstead of the mn n the theorem). As mentoned above, ths dfference s crucal for our applcaton as well as to the recent results n [15]. Independent sets n graph powers To demonstrate the usefulness of Theorem 1.2, let us consder a queston n the study of ndependent sets n graph powers. Let G = ([q], E) be a regular, connected, non-bpartte graph on q vertces. Consder the graph G n on vertex set [q] n n whch vertces (x 1,..., x n ) and (y 1,..., y n ) are connected f and only f x s connected to y n G for all (ths s known as the n-fold weak product of G wth tself). Let T G be the symmetrc Markov operator correspondng to one step n a random walk n G. It s easy to verfy that the operator T n G corresponds to one step n Gn. Let f, g : [q] n {0, 1} be two Boolean functons, and thnk of them as beng the ndcator functons of two subsets of [q] n. Then, the blnear form n (1) gves the fracton of edges that are spanned between these two subsets n the graph G n. In partcular, f, T n G f = 0 f and only f f s the ndcator functon of an ndependent set n G n. Usng the lower bound n Theorem 1.2, we obtan that for any µ > 0 there exsts a δ > 0 such that any ndependent set of measure µ (.e., E[f] = µ) must have at least one coordnate wth nfluence at least δ. Less formally, ths says that any reasonably bg ndependent set n graph powers must have some structure (namely, have an nfluental coordnate). Our hardness result for approxmate graph colorng uses Theorem 1.2 n a smlar fashon. We remark that graph powers were studed n a smlar context n [1], where a smlar structure theorem was proved for the restrcted case that the ndependent set has nearly maxmal sze. Moreover, Theorem 1.2 was recently used n [15] to show that every ndependent set n a graph power s contaned (up to o(1)) n a nontrval set descrbed by a constant number of coordnates. 2 Prelmnares 2.1 Functons on the q-ary hypercube Let [q] denote the set {0,..., q 1}. For an element x of [q] n wrte x a for the number of coordnates k of x such that x k = a and x = a 0 x a for the number of nonzero coordnates. In ths paper we are nterested n functons from [q] n to R. We defne an nner product on ths space by f, g = 1 q x f(x)g(x). In our applcatons, we usually take q to be some constant (say, n 3) and n to be large. Defnton 2.1 Let f : [q] n R be a functon. The nfluence of the th varable on f, denoted I (f) s defned by I (f) = E[Var x [f(x) x 1,..., x 1, x +1,..., x n ]] where x 1,..., x n are unformly dstrbuted. Consder a sequence of vectors α 0 = 1, α 1,..., α q 1 R q formng an orthonormal bass of R q. Equvalently, we can thnk of these vectors as functons from [q] to R. These vectors can be used to form an orthonormal bass of the space of functons from [q] n to R, as follows. 5

6 Defnton 2.2 Let α 0 = 1, α 1,..., α q 1 be an orthonormal bass of R q. For x [q] n, wrte α x R qn for α x1 α x2 α xn. Equvalently, we can defne α x as the functon mappng y [q] n to α x1 (y 1 )α x2 (y 2 ) α xn (y n ). Clearly, any functon from [q] n to R can be wrtten as a lnear combnaton of α x for x [q] n. Ths leads to the followng defnton. Defnton 2.3 For a functon f : [q] n R, defne ˆf(α x ) = f, α x and notce that f = x ˆf(α x )α x. The followng standard clam relates the nfluences of a functon to ts decomposton. Notce that the clam holds for any choce of orthonormal bass α 0,..., α q 1 as long as α 0 = 1. Clam 2.4 For any functon f : [q] n R and any {1,..., n}, I (f) = ˆf 2 (α x ). x:x 0 Proof: Let us frst fx the values of x 1,..., x 1, x +1,..., x n. Then [ ] [ Var x [f] = Var x ˆf(α y )α y = Var x y y:y 0 ˆf(α y )α y ], where the last equalty follows from the fact that f y = 0 then α y s a constant functon of x. If y 0, then the expected value of α y wth respect to x s zero. Therefore, 2 [ ] Var x ˆf(α y )α y = E x ˆf(α y )α y = E x ˆf(α y ) ˆf(α z )α y α z. Thus, y:y 0 I (f) = E x as needed. y,z:y 0,z 0 y:y 0 ˆf(α y ) ˆf(α z )α y α z = We now defne the noton of low-level nfluence. y,z:y 0,z 0 y,z:y 0,z 0 ˆf(α y ) ˆf(α z )E x [α y α z ] = y:y 0 ˆf 2 (α y ), Defnton 2.5 Let f : [q] n R be a functon, and let k n. The low-level nfluence of the th varable on f s defned by I k (f) = ˆf 2 (α x ). It s easy to see that for any functon f, I k (f) = x: x k x:x 0, x k ˆf 2 (α x ) x k x ˆf 2 (α x ) = k f 2 2. In partcular, for any functon f obtanng values n [0, 1], I k (f) k. Moreover, let us menton that I k s n fact ndependent of the partcular choce of bass α 0, α 1,..., α q 1 as long as α 0 = 1. Ths follows by notng that I k s the squared length of the projecton of f on the subspace spanned by all α x wth x 0, x k, and that ths subspace can be equvalently defned n terms of tensor products of α 0 and α0. There s a natural equvalence between [q] 2n and [q 2 ] n. As ths equvalence s used often n ths paper, we ntroduce the followng notaton. 6

7 Defnton 2.6 For any x [q] 2n we denote by x the element of [q 2 ] n gven by x = ((x 1, x 2 ),..., (x 2n 1, x 2n )). For any y [q 2 ] n we denote by y the element of [q] 2n gven by y = (y 1,1, y 1,2, y 2,1, y 2,2,..., y n,1, y n,2 ). For a functon f on [q] 2n we denote by f the functon on [q 2 ] n defned by f(y) = f(y). Smlarly, for a functon f on [q 2 ] n we denote by f the functon on [q] 2n defned by f(x) = f(x). Clam 2.7 For any functon f : [q] 2n R, any {1,..., n}, and any k 1, I k (f) I 2k 2 1 (f) + I 2k 2 (f). Proof: Fx some bass α x of [q] 2n as above and let α x be the bass of [q 2 ] n defned by α x (y) = α x (y). Then, t s easy to see that ˆf(α x ) = ˆf(α x ). Hence, I k (f) = x:x (0,0), x k ˆ f 2 (α x ) where we used that x 2 x. x:x 2 1 0, x 2k ˆf 2 (α x ) + x:x 2 0, x 2k ˆf 2 (α x ) = I 2k 2 1 (f) + I 2k 2 (f) For the followng defnton, recall that we say that a Markov operator T s symmetrc f t s reversble wth respect to the unform dstrbuton,.e., f the transton matrx representng T s symmetrc. Defnton 2.8 Let T be a symmetrc Markov operator on [q]. Let 1 = λ 0 λ 1 λ 2 λ q 1 be the egenvalues of T. We defne r(t ) to be the second largest egenvalue n absolute value, that s, r(t ) = max{ λ 1, λ q 1 }. For T as above, we may defne a Markov operator T n on [q] n n the standard way. Note that f T s symmetrc then T n s also symmetrc and r(t n ) = r(t ). If we choose α 0,..., α q 1 to be an orthonormal set of egenvectors for T wth correspondng egenvalues λ 0,..., λ q 1 (so α 0 = 1), we see that and hence T n α x = ( a 0 λ x a a T n f = x ) α x. ( ) a 0 λ x a a ˆf(αx )α x. holds for any functon f : [q] n R. We now descrbe two operators that we use n ths paper. The frst s the Beckner operator, T ρ. For any ρ [ 1 q 1, 1], t s defned by T ρ(x x) = 1 q + (1 1 q )ρ and T ρ(x y) = 1 q (1 ρ) for any x y n [q]. It can be seen that T ρ s a Markov operator as n Defnton 2.8 wth λ 1 = = λ q 1 = ρ and hence r(t ρ ) = ρ. Another useful operator s the averagng operator, A S. For a subset S {1,..., n}, t acts on functons on [q] n by averagng over coordnates n S, namely, A S (f) = E xs [f]. Notce that the functon A S (f) s ndependent of the coordnates n S. 7

8 2.2 Functons n Gaussan space We let γ denote the standard Gaussan measure on R n wth densty (2π) n/2 e x 2 2 /2. We denote by E γ the expected value wth respect to γ and by, γ the nner product on L 2 (R n, γ). Notce that E γ [f] = f, 1 γ where 1 s the constant 1 functon. For ρ [ 1, 1], we denote by U ρ the Ornsten-Uhlenbeck operator, whch acts on L 2 (R, γ) by U ρ f(x) = E y γ [f(ρx + 1 ρ 2 y)]. Snce for x, y γ we have that ρx + 1 ρ 2 y s also dstrbuted accordng to the standard Gaussan dstrbuton, E x γ [U ρ f(x)] = E x γ [f(x)]. Fnally, for 0 < µ < 1, let F µ : R {0, 1} denote the functon F µ (x) = 1 x<t where t s chosen n such a way that E γ [F µ ] = µ. One useful quantty that wll appear later s F η, U ρ F ν γ, whch by defnton can also be wrtten as F η, U ρ F ν γ = Pr x,y γ [x < s and ρx + 1 ρ 2 y < t], where s and t are such that F η (x) = 1 x<s and F ν (x) = 1 x<t. It s not dffcult to see that for any ν, η > 0, and any ρ [ 1, 1], t holds that F η, U ρ F ν γ = F ν, U ρ F η γ (say, snce U ρ s self-adjont) and that F τ, U ρ F τ γ F η, U ρ F ν γ τ, where τ = mn(η, ν). Moreover, for all τ > 0 and ρ > 1 t holds that F τ, U ρ F τ γ > 0. 3 An Inequalty for Nose Operators The man analytc result of the paper, Theorem 3.1, s a generalzaton of the result of [33]. It shows that f the nner product of two functons f and g under some nose operator devates from a certan range then there must exst an ndex such that the low-level nfluence of the th varable s large n both f and g. Ths range depends on the expected value of f and g, and on r(t ). Note n partcular that Theorem 3.1 mples Theorem 1.2. Theorem 3.1 Let q be a fxed nteger and let T be a symmetrc Markov operator on [q] such that ρ = r(t ) < 1. Then for any ε > 0 there exst δ > 0 and k N such that f f, g : [q] n [0, 1] are two functons satsfyng for all, then t holds that and where µ = E[f], ν = E[g]. mn ( I k (f), I k (g) ) < δ f, T n g F µ, U ρ (1 F 1 ν ) γ ε (4) f, T n g F µ, U ρ F ν γ + ε (5) Note that (4) follows from (5). Indeed, apply (5) to 1 g to obtan f, T n (1 g) F µ, U ρ F 1 ν γ + ε 8

9 and then use f, T n (1 g) = f, 1 f, T n g = µ f, T n g = F µ, U ρ 1 γ f, T n g. From now on we focus on provng (5). Followng the approach of [33], the proof conssts of two powerful technques. The frst s an nequalty by Chrster Borell [10] on contnuous Gaussan space. The second s an nvarance prncple shown n [33] that allows us to translate our dscrete queston to the contnuous Gaussan space. Defnton 3.2 (Gaussan analogue of an operator) Let T be an operator as n Defnton 2.8. We defne ts Gaussan analogue as the operator T on L 2 (R q 1, γ) gven by T = U λ1 U λ2 U λq 1. For example, the Gaussan analogue of T ρ s Uρ (q 1). We need the followng powerful theorem by Borell [10]. It says that the functons that maxmze the nner product under the operator U ρ are the ndcator functons of half-spaces. Theorem 3.3 (Borell [10]) Let f, g : R n [0, 1] be two functons and let µ = E γ [f], ν = E γ [g]. Then f, U n ρ g γ F µ, U ρ F ν γ. The above theorem only apples to the Ornsten-Uhlenbeck operator. In the followng corollary we derve a smlar statement for more general operators. The proof follows by wrtng a general operator as a product of the Ornsten-Uhlenbeck operator and some other operator. Corollary 3.4 Let f, g : R (q 1)n [0, 1] be two functons and defne µ = E γ [f], ν = E γ [g]. Let T be an operator as n Defnton 2.8 and let ρ = r(t ). Then f, T n g γ F µ, U ρ F ν γ. Proof: For 1 q 1, let δ = λ /ρ. Note that δ 1 for all. Let S be the operator defned by Then, S = U δ1 U δ2 U δq 1. Uρ (q 1) S = U ρ U δ1 U ρ U δq 1 = U ρδ1 U ρδq 1 = T (ths s often called the sem-group property). It follows that T n = Uρ (q 1)n S n. Snce S n s an averagng operator, the functon S n g obtans values n [0, 1] and satsfes E γ [S n g] = E γ [g]. Thus the clam follows by applyng Theorem 3.3 to the functons f and S n g. Defnton 3.5 (Real analogue of a functon) Let f : [q] n R be a functon wth decomposton f = ˆf(αx )α x. Consder the (q 1)n varables z 1 1,..., z1 q 1,..., zn 1,..., zn q 1 and let Γ x = n =1,x 0 z x. We defne the real analogue of f to be the functon f : R n(q 1) R gven by f = ˆf(αx )Γ x. 9

10 Clam 3.6 For any two functons f, g : [q] n R and operator T on [q] n, f, g = f, g γ f, T n g = f, T n g γ where f, g denote the real analogues of f, g respectvely and T denotes the Gaussan analogue of T. Proof: Both α x and Γ x form an orthonormal set of functons hence both sdes of the frst equalty are ˆf(α x )ĝ(α x ). x For the second clam, notce that for every x, α x s an egenvector of T n and Γ x s an egenvector of T n and both correspond to the egenvalue a 0 λ x a a. Hence, both sdes of the second equalty are ( ) a 0 λ x a a ˆf(αx )ĝ(α x ). x Defnton 3.7 For any functon f wth range R, defne the functon chop(f) as f(x) f f(x) [0, 1] chop(f)(x) = 0 f f(x) < 0 1 f f(x) > 1 The followng theorem s proven n [33]. It shows that under certan condtons, f a functon f obtans values n [0, 1] then f and chop( f) are close. Its proof s non-trval and bulds on the man techncal result of [33], a result that s known as an nvarance prncple. In essence, t shows that the dstrbuton of values obtaned by f and that of values obtaned by f are close. In partcular, snce f never devates from [0, 1], t mples that f rarely devates from [0, 1] and hence f and chop( f) are close. See [33] for more detals. Theorem 3.8 ([33, Theorem 3.20]) There exsts a functon δ MOO (η, ε) such that for any η < 1 and ε > 0 the followng holds. For any functon f : [q] n [0, 1] such that d ˆf(α x ) 2 η d and I (f) < δ MOO (η, ε), t holds that x: x d f chop( f) 2 ε. We are now ready to prove the frst step n the proof of Theorem 3.1. It s here that we use the nvarance prncple and Borell s nequalty. Lemma 3.9 Let q be a fxed nteger and let T be a symmetrc Markov operator on [q] such that ρ = r(t ) < 1. Then for any ε > 0, η < 1, there exsts a δ > 0 such that for any functons f, g : [q] n [0, 1] satsfyng and t holds that d where µ = E[f], ν = E[g]. x: x d max (I (f), I (g)) < δ ˆf(α x ) 2 η d, d x: x d f, T n g F µ, U ρ F ν γ + ε 10 ĝ(α x ) 2 η d,

11 Proof: Let µ = E γ [chop( f)] and ν = E γ [chop( g)]. We note that F µ, U ρ F ν γ s a unformly contnuous functon of µ and ν. Let ε 1 be chosen such that f µ µ ε 1 and ν ν ε 1 then t holds that F µ, U ρ F ν γ F µ, U ρ F ν γ ε/2. Let ε 2 = mn(ε/4, ε 1 ) and let δ = δ MOO (η, ε 2 ) be the value gven by Theorem 3.8. Then, usng the Cauchy-Schwartz nequalty, µ µ = E γ [chop( f) f] = chop( f) f, 1 γ chop( f) f 2 ε 2 ε 1. Smlarly, we have ν ν ε 1. Now, f, T n g = f, T n g γ (Clam 3.6) = chop( f), T n chop( g) γ + chop( f), T n ( g chop( g)) γ + f chop( f), T n g γ chop( f), T n chop( g) γ + 2ε 2 F µ, U ρ F ν γ + 2ε 2 (Corollary 3.4) F µ, U ρ F ν γ + ε/2 + 2ε 2 F µ, U ρ F ν γ + ε where the frst nequalty follows from the Cauchy-Schwartz nequalty together wth the fact that chop( f) and g have L 2 norm at most 1 and that T n s a contracton on L 2. We complete the proof of Theorem 3.1 by provng: Lemma 3.10 Let q be a fxed nteger and let T be a symmetrc Markov operator on [q] such that ρ = r(t ) < 1. Then for any ε > 0, there exsts a δ > 0 and an nteger k such that f f, g : [q] n [0, 1] satsfy mn ( I k (f), I k (g) ) < δ (6) then where µ = E[f], ν = E[g]. f, T n g F µ, U ρ F ν γ + ε (7) Proof: Let f 1 = Tη n f and g 1 = Tη n g where η < 1 s chosen so that ρ j (1 η 2j ) < ε/4 for all j. Then f 1, T n g 1 f, T n g = ˆf(α x )ĝ(α x ) λ x a a (1 η 2 x ) x a 0 ρ x (1 η 2 x ) ˆf(α x )ĝ(α x ) ε/4 x where the last nequalty follows from the Cauchy-Schwartz nequalty. Thus, n order to prove (7) t suffces to prove f 1, T n g 1 F µ, U ρ F ν γ + 3ε/4. (8) Let δ(ε/4, η) be the value gven by Lemma 3.9 pluggng n ε/4 for ε. Let δ = δ(ε/4, η)/2. Let k be chosen so that η 2k < mn(δ, ε/4). Defne C = k/δ and δ = (ε/8c) 2 < δ. Let B f = { : I k (f) δ }, B g = { : I k (g) δ }. 11

12 We note that B f and B g are of sze at most C = k/δ. By (6), we have that whenever B f, I k (g) < δ. Smlarly, for every B g we have I k (f) < δ. In partcular, B f and B g are dsjont. Recall the averagng operator A. We now let f 2 = A Bf (f 1 ) = ˆf(α x )α x η x, g 2 = A Bg (g 1 ) = x:x Bf =0 x:x B g =0 ĝ(α x )α x η x. Clearly, E[f 2 ] = E[f] and E[g 2 ] = E[g], and for all x, f 2 (x), g 2 (x) [0, 1]. It s easy to see that I (f 2 ) = 0 f B f and I (f 2 ) I k (f) + η 2k < 2δ otherwse and smlarly for g 2. Thus, for any, max (I (f 2 ), I (g 2 )) < 2δ. We also see that for any d, x: x d ˆf 2 (α x ) 2 η d and the same for g 2. Thus, we can apply Lemma 3.9 to obtan that f 2, T n g 2 F µ, U ρ F ν γ + ε/4. In order to show (8) and complete the proof, we show that f 1, T n g 1 f 2, T n g 2 ε/2. Ths follows by f 1, T n g 1 f 2, T n g 2 = ˆf(αx )ĝ(α x ) x:x Bf Bg 0 η 2k ˆf(α x )ĝ(α x ) ε/4 + ε/4 + x: x k B f B g B f B g a 0 λ x a a η 2 x + { } ˆf(αx )ĝ(α x ) : x Bf B g 0, x k { } ˆf(αx )ĝ(α x ) : x 0, x k I k ε/4 + δ( B f + B g ) ε/4 + 2C δ = ε/2, (f) where the next-to-last nequalty holds because for each B f B g one of I k (f), I k (g) s at most δ and the other s at most 1. The fnal theorem of ths secton s needed only for the APPROXCOLORING(3, Q) result. Here, the operator T acts on [q 2 ] and s assumed to have an addtonal property. Before proceedng, t s helpful to recall Defnton 2.6. Theorem 3.11 Let q be a fxed nteger and let T be a symmetrc Markov operator on [q 2 ] such that ρ = r(t ) < 1. Suppose moreover, that T has the followng property. Gven (x 1, x 2 ) chosen unformly at random and (y 1, y 2 ) chosen accordng to T appled to (x 1, x 2 ) we have that (x 2, y 2 ) s dstrbuted unformly at random. Then for any ε > 0, there exsts a δ > 0 and an nteger k such that for any functons f, g : [q] 2n [0, 1] satsfyng that for = 1,..., n I k (g) mn ( I k 2 1 (f), I k 2 1 (g)) < δ, mn ( I k 2 1 (f), I k 2 (g)) < δ, and 12 mn ( I k 2 (f), I k 2 1 (g)) < δ

13 t holds that and where µ = E[f], ν = E[g]. f, T n g F µ, U ρ (1 F 1 ν ) γ ε (9) f, T n g F µ, U ρ F ν γ + ε (10) Proof: As n Theorem 3.1, (9) follows from (10) so t s enough to prove (10). Assume frst that n addton to the three condtons above we also have that for all = 1,..., n, mn ( I k 2 (f), I k 2 (g)) < δ. (11) Then t follows that for all, ether both I k 2 1 (f) and I k 2 (g) are smaller than δ. Hence, by Clam 2.7, we know that for all we have I k 2 ( mn I k/2 ) (f), I k/2 (g) < 2δ (f) are smaller than δ or both I k 2 1 (g) and and the result then follows from Lemma However, we do not have ths extra condton and hence we have to deal wth bad coordnates for whch mn(i k 2 (f), I k 2 (g)) δ. Notce that for such t must be the case that both I k 2 1 (f) and I k 2 1 (g) are smaller than δ. Informally, the proof proceeds as follows. We frst defne functons f 1, g 1 that are obtaned from f, g by addng a small amount of nose. We then obtan f 2, g 2 from f 1, g 1 by averagng the coordnates 2 1 for bad. Fnally, we obtan f 3, g 3 from f 2, g 2 by averagng the coordnate 2 for bad. The pont here s to mantan f, T n g f 1, T n g 1 f 2, T n g 2 f 3, T n g 3. The condton n Equaton 11 now apples to f 3, g 3 and we can apply Lemma 3.10, as descrbed above. We now descrbe the proof n more detal. We frst defne f 1 = Tη n f and g 1 = Tη n g where η < 1 s chosen so that ρ j (1 η 2j ) < ε/4 for all j. As n the prevous lemma t s easy to see that and thus t suffces to prove that f 1, T n g 1 f, T n g < ε/4 f 1, T n g 1 F µ, U ρ F ν γ + 3ε/4. Let δ(ε/2, η), k(ε/2, η) be the values gven by Lemma 3.10 wth ε taken to be ε/2. Let δ = δ(ε/2, η)/2. Choose a large enough k so that 128kη k < ε 2 δ and k/2 > k(ε/2, η). We let C = k/δ and δ = ε 2 /128C. Notce that δ < δ and η k < δ. Fnally, let { B = I k 2 (f) δ, I k 2 (g) δ }. We note that B s of sze at most C. We also note that f B then we have I k 2 1 (f) < δ and I k 2 1 (g) < δ. We clam that ths mples that I 2 1(f 1 ) δ + η k < 2δ and smlarly for g. To see that, take any orthonormal bass β 0 = 1, β 1,..., β q 1 of R q and notce that we can wrte f 1 = ˆf(βx )η x β x. x [q] 2n Hence, I 2 1 (f 1 ) = x [q] 2n x ˆf(β x ) 2 η 2 x < δ + η k 13 x [q] 2n x > k ˆf(β x ) 2 δ + η k

14 where we used that the number of nonzero elements n x s at least half of that n x. Next, we defne f 2 = A 2B 1 (f 1 ) and g 2 = A 2B 1 (g 1 ) where A s the averagng operator and 2B 1 denotes the set {2 1 B}. Note that f 2 f = f 2 f B I 2 1 (f 1 ) 2Cδ. and smlarly, Thus g 2 g = g 2 g Cδ. f 1, T n g 1 f 2, T n g 2 f 1, T n g 1 f 1, T n g 2 + f 1, T n g 2 f 2, T n g 2 2 2Cδ = ε/4 where the last nequalty follows from the Cauchy-Schwartz nequalty together wth the fact that f and also T n g Hence, t suffces to prove f 2, T n g 2 F µ, U ρ F ν γ + ε/2. We now defne f 3 = A 2B (f 2 ) and g 3 = A 2B (g 2 ). Equvalently, we have f 3 = A B (f 1 ) and g 3 = A B (g 1 ). We show that f 2, T n g 2 = f 3, T n g 3. Let α x, x [q 2 ] n, be an orthonormal bass of egenvectors of T n. Then f 3, T n g 3 = x,y [q 2 ] n,x B =y B =0 ˆ f 1 (α x )ĝ 1 (α y ) α x, T n α y. (12) Moreover, snce A s a lnear operator and f 1 can be wrtten as x [q 2 ] ˆf1 n (α x )α x and smlarly for g 1, we have f 2, T n g 2 = ˆf1 (α x )ĝ 1 (α y ) A 2B 1 (α x ), T n A 2B 1 (α y ). (13) x,y [q 2 ] n Frst, notce that when x B = 0, A 2B 1 (α x ) = α x snce α x does not depend on coordnates n B. Hence, n order to show that the expressons n (12) and (13) are equal, t suffces to show that A 2B 1 (α x ), T n A 2B 1 (α y ) = 0 unless x B = y B = 0. So assume wthout loss of generalty that B s such that x 0. The above nner product can be equvalently wrtten as E z,z [q 2 ] n[a 2B 1(α x )(z) A 2B 1 (α y )(z )] where z s chosen unformly at random and z s chosen accordng to T n appled to z. Fx some arbtrary values to z 1,..., z 1, z +1,..., z n and z 1,..., z 1, z +1,..., z n and let us show that E z,z [q2 ][A 2B 1 (α x )(z) A 2B 1 (α y )(z )] = 0. Snce B, the two expressons nsde the expectaton do not depend on z,1 and z,1 (where by z,1 we mean the frst coordnate of z ). Moreover, by our assumpton on T, z,2 and z,2 are ndependent. Hence, the above expectaton s equal to E z [q 2 ][A 2B 1 (α x )(z)] E z [q 2 ][A 2B 1 (α y )(z )]. 14

15 Snce x 0, the frst expectaton s zero. Ths establshes that f 2, T n g 2 = f 3, T n g 3. The functons f 3, g 3 satsfy the property that for every = 1,..., n, ether both I k 2 1 (f 3) and I k 2 (f 3) are smaller than δ or both I k 2 1 (g 3) and I k 2 (g 3) are smaller than δ. By Clam 2.7, we get that for = 1,..., n, ether I k/2 (f 3 ) or I k/2 (g 3 ) s smaller 2δ. We can now apply Lemma 3.10 to obtan f 3, T n g 3 F µ, U ρ F ν γ + ε/2. 4 Approxmate Colorng As mentoned n the ntroducton, one of the most successful approaches to dervng hardness proofs, whch s also the one we shall take here, s by a reducton from a combnatoral problem known as the label-cover problem. To recall, n the label-cover problem we are gven an undrected graph together wth a constrant (.e., a bnary relaton on {1,..., R}) for each edge. The goal s to label the vertces wth values from {1,..., R} such that the number of satsfed constrants s maxmzed, where a constrant s satsfed f the labels on the two ncdent vertces satsfy the relaton assocated wth t. It s known that n ths problem (as well as n many of ts varants), t s NP-hard to tell whether there exsts a way to label the vertces such that all constrants are satsfed, or whether any labelng satsfes at most, say, 0.01 fracton of the constrants. Our reducton follows the general paradgm of [6, 25]. Each vertex of the label-cover nstance s replaced wth a block of vertces, often known as a gadget. In our case, the gadget s smply a set of q R vertces, and we thnk of them as correspondng to elements of [q] R. We then add edges between these gadgets n a way that encodes the label-cover constrants. For the reducton to work, we need to have two propertes. Frst, f the label-cover s satsfable, then the resultng graph s q-colorable (ths s known as the completeness part). Ths property would follow mmedately from our constructon. The more dffcult part s to show that f there s no way to satsfy more than 0.01 fracton of the constrants n the label-cover nstance, then the resultng graph has chromatc number at least Q (ths s known as the soundness part). The way ths s shown s by assumng towards contradcton that there exsts a colorng wth less than Q colors, and then decodng t nto a labelng of the label-cover nstance that satsfes more than 0.01 of the constrants. It s ths part that s usually the most dffcult to establsh. In our case, we wll apply Theorem 1.2 to detect nfluental coordnates n each block based on the colorng gven to t. The above outlne hdes one very mportant fact: for our reducton to work, the label-cover nstances we use must have constrants of a very specfc form. For example, we mght requre all constrants to be bjectons,.e., a bnary relaton n whch any labellng of one vertex determnes the other, and vce versa. We call ths specal case 1 1-label-cover. We wll also consder two other restrctons of the label-cover problem, whch we call the 2 2-label-cover and the <-label-cover (read: alpha-label-cover). The precse defntons of these problems wll appear later. As already dscussed n the ntroducton, these specal cases of the label-cover problem are not known to be NP-hard. Nevertheless, Khot s unque games conjecture [30] asserts that such problems are n fact NP-hard. The conjecture has been heavly scrutnzed [36, 13, 22, 14], and so far there s no evdence aganst the conjecture. Our Hardness Results: We now descrbe our hardness results n more detal. In addton to APPROXCOLORING(q, Q), we consder the followng computatonal problem, defned for any ε > 0. 15

16 ALMOST3COLORING ε : Gven a graph G = (V, E), decde between There exsts a set V V, V (1 ε) V such that χ(g V ) 3 where G V s the graph nduced by V. Every ndependent set S V n G has sze S ε V. Observe that these two tems are mutually exclusve for ε < 1/4. We consder three conjectures: the 1 1 conjecture, the 2 2 conjecture, and the < conjecture. Roughly speakng, each conjecture says that n the correspondng label-cover nstances t s NPhard to dstngush between completely satsfable nstances, and nstances that are almost completely unsatsfable. The only excepton s the 1 1 conjecture: t s easy to see that checkng f a 1 1-label-cover s completely satsfable can be done n polynomal tme. Hence the 1 1 conjecture says that t s NP-hard to dstngush between almost completely satsfable and almost completely unsatsfable. Ths drawback of the 1 1 conjecture, often known as mperfect completeness, prevents us from usng t for provng the hardness of the approxmate colorng problem. Instead, we use t to show hardness of the (somewhat harder) problem ALMOST3COLORING. We present three reductons, each from a dfferent specal case of the label-cover problem. These reductons yeld the followng. For any constant ε > 0, the 1 1 conjecture mples the NP-hardness of ALMOST3COLORING ε. For any constant Q > 4, the 2 2 conjecture mples that APPROXCOLORING(4, Q) s NPhard. Ths also holds for APPROXCOLORING(q, Q) for any q 4. For any constant Q > 3, the < conjecture mples that APPROXCOLORING(3, Q) s NP-hard. Ths also holds for APPROXCOLORING(q, Q) for any q 3. We remark that Khot s orgnal conjectures actually refer to slghtly dfferent varants of the label-cover problem. Most notably, hs label-cover nstances are bpartte. However, as we shall show later, Khot s unque-games conjecture mples our 1 1 conjecture, and Khot s two-to-one conjecture mples our 2 2 conjecture. The < conjecture s, to the best of our knowledge, new, and seems to be not weaker than the 2 2 conjecture. Future work: Our constructons can be extended n several ways. Frst, usng smlar technques, one can show hardness of APPROXCOLORING(q, Q) based on the d-to-1 conjecture of Khot for larger values of d (and not only d = 2 as we do here). It would be nterestng to fnd out how q depends on d. Second, by strengthenng the current conjectures to sub-constant values, one can obtan hardness for Q that depends on n, the number of vertces n the graph. Agan, t s nterestng to see how large Q can be. Fnally, let us menton that n all our reductons we n fact show n the soundness case that there are no ndependent sets of relatve sze larger than ε for arbtrarly small constant ε (note that ths s somewhat stronger than showng that there s no Q- colorng). In fact, a more careful analyss can be used to obtan the stronger statement that there are no almost-ndependent sets of relatve sze larger than ε. Organzaton: In Secton 4.1, we descrbe the three conjectures along wth some defntons. We then prove the three reductons mentoned above. The three reductons are very smlar, each combnng a conjecture wth an approprately constructed nose operator. In Secton 4.2 we descrbe the three nose operators, and n Secton 4.3 we spell out the reductons. Then, n Sectons 4.4 and 4.5, we prove the completeness and soundness of the three reductons. 16

17 4.1 Label-cover problems Defnton 4.1 A label-cover nstance s a trple G = ((V, E), R, Ψ) where (V, E) s a graph, R s an nteger, and Ψ = {ψ } e {1,..., R} 2 e E s a set of constrants (relatons), one for each edge. For a gven labelng L : V {1,..., R}, let sat L (G) = Pr [(L(u), L(v)) ψ e], e=(u,v) E sat(g) = max L (sat L(G)). For t, R N let ( R t) denote the collecton of all subsets of {1,..., R} whose sze s at most t. Defnton 4.2 A t-labelng s a functon L : V ( R t) that labels each vertex v V wth a subset of values L(v) {1,..., R} such that L(v) t for all v V. A t-labelng L s sad to satsfy a constrant ψ {1,..., R} 2 over varables u and v ff there are a L(u), b L(v) such that (a, b) ψ. In other words, ff (L(u) L(v)) ψ. In the specal case of t = 1, a 1-labelng s essentally a labelng L : V {1,..., R} (except that some vertces mght get no label). Smlar to the defnton of sat(g), we also defne sat(g) ( nduced-sat ) to be the relatve sze of the largest set of vertces for whch there s a labelng that satsfes all of the nduced edges. sat(g) = max S { S V }. L : S {1,..., R} that satsfes all the constrants nduced by S V Let sat t (G) denote the relatve sze of the largest set of vertces S V for whch there s a t-labelng that satsfes all the constrants nduced by S. { ( ) } S sat t (G) = max R S V L : S that satsfes all the constrants nduced by S V. t We next descrbe three conjectures on whch our reductons are based. The man dfference between the three conjectures s n the type of constrants that are allowed. The three types are defned next, and also llustrated n Fgure 1. Defnton 4.3 (1 1-constrant) A 1 1 constrant s a relaton {(, π())} R =1, where π : {1,..., R} {1,..., R} s any arbtrary permutaton. The constrant s satsfed by (a, b) ff b = π(a). Defnton 4.4 (2 2-constrant) A 2 2 constrant s defned by a par of permutatons π 1, π 2 : {1,..., 2R} {1,..., 2R} and the relaton 2 2 = {(2, 2), (2, 2 1), (2 1, 2), (2 1, 2 1)} R =1. The constrant s satsfed by (a, b) ff (π 1 1 (a), π 1(b)) Defnton 4.5 ( <-constrant) An < constrant s defned by a par of permutatons π 1, π 2 : {1,..., 2R} {1,..., 2R} and the relaton The constrant s satsfed by (a, b) ff (π 1 1 < = {(2 1, 2 1), (2, 2 1), (2 1, 2)} R =1. (a), π 1(b)) <. 2 17

18 Fgure 1: Three types of constrants (top to bottom): 1 1, <, 2 2 Conjecture 4.6 (1 1 Conjecture) For any ε, ζ > 0 and t N there exsts some R N such that gven a label-cover nstance G = (V, E), R, Ψ where all constrants are 1 1-constrants, t s NP-hard to decde between sat(g) 1 ζ sat t (G) < ε It s easy to see that the above problem s n P when ζ = 0. Conjecture 4.7 (2 2 Conjecture) For any ε > 0 and t N there exsts some R N such that gven a label-cover nstance G = (V, E), 2R, Ψ where all constrants are 2 2-constrants, t s NP-hard to decde between sat(g) = 1 sat t (G) < ε The above two conjectures are no stronger than the correspondng conjectures of Khot. Namely, our 1 1 conjecture s not stronger than Khot s (bpartte) unque games conjecture, and our 2 2 conjecture s not stronger than Khot s (bpartte) two-to-one conjecture. The former clam was already proven by Khot and Regev n [32]. The latter clam s proven n a smlar way. For completeness, we nclude both proofs n Appendx A. We also make a thrd conjecture that s used n our reducton to APPROXCOLORING(3, Q). Ths conjecture seems stronger than Khot s conjectures. Conjecture 4.8 ( < Conjecture) For any ε > 0 and t N there exsts some R N such that gven a label-cover nstance G = (V, E), 2R, Ψ where all constrants are <-constrants, t s NP-hard to decde between sat(g) = 1 sat t (G) < ε Remark: The (strange-lookng) <-shaped constrants have already appeared before n [16]. There, t s essentally proven that for all ε, ζ > 0 gven a label-cover nstance G where all constrants are <-constrants, t s NP-hard to dstngush between sat(g) > 1 ζ 18

19 sat t=1 (G) < ε The man dfference between ther theorem and our conjecture s that n our conjecture we consder any constant t, whle n ther case t s 1. Another dfference s that n our conjecture we assume perfect completeness (.e., sat(g) = 1) Nose operators We now defne the nose operators correspondng to the 1 1-constrants, <-constrants, and 2 2-constrants. The nose operator that corresponds to the 1 1-constrants s the smplest, and acts on {0, 1, 2}. For the other two cases, snce the constrants nvolve pars of coordnates, we obtan an operator on {0, 1, 2} 2 and an operator on {0, 1, 2, 3} 2. See Fgure 2 for an llustraton. (a) (b) (c) Fgure 2: Three nose operators (edge weghts not shown) correspondng to: (a) 1 1, (b) <, and (c) 2 2. Lemma 4.9 There exsts a symmetrc Markov operator T on {0, 1, 2} such that r(t ) < 1 and such that f T (x y) > 0 then x y. Proof: Take the operator gven by 0 1/2 1/2 T = 1/2 0 1/2. 1/2 1/2 0 See Fgure 2(a). Lemma 4.10 There exsts a symmetrc Markov operator T on {0, 1, 2, 3} 2 such that r(t ) < 1 and such that f T ((x 1, x 2 ) (y 1, y 2 )) > 0 then {x 1, x 2 } {y 1, y 2 } =. Proof: Our operator has three types of transtons, wth transtons probabltes β 1, β 2, and β 3. 1 The man dea n ther constructon s to take an NP-hard label-cover as gven by the parallel repetton theorem appled to the PCP theorem, and to construct a new <-label-cover wth ( ) R X l varables correspondng to all subsets of sze l of X {1,..., R}, where l = cr for some large constant c. The number of labels s equal to the number of bnary strngs of length l whose Hammng weght s at least l/2r. Constrants are placed between any par of l-tuples for whch () ther ntersecton has sze l 1, and () the unque elements, one from each l-tuple, correspond to an nconsstency n the orgnal label-cover. These constrants check for agreement on ther ntersecton and that not both unque elements are 1, and are therefore essentally <-constrants. 19

20 Wth probablty β 1 we have (x, x) (y, y) where x y. Wth probablty β 2 we have (x, x) (y, z) where x, y, z are all dfferent. Wth probablty β 3 we have (x, y) (z, w) where x, y, z, w are all dfferent. These transtons are llustrated n Fgure 2(c). For T to be a symmetrc Markov operator, we need that β 1, β 2 and β 3 are non-negatve and 3β 1 + 6β 2 = 1, 2β 2 + 2β 3 = 1. It s easy to see that the two equatons above have solutons bounded away from 0 and that the correspondng operator has r(t ) < 1. For example, choose β 1 = 1 12, β 2 = 1 8, and β 3 = 3 8. Lemma 4.11 There exsts a symmetrc Markov operator T on {0, 1, 2} 2 such that r(t ) < 1 and such that f T ((x 1, x 2 ) (y 1, y 2 )) > 0 then x 1 / {y 1, y 2 } and y 1 / {x 1, x 2 }. Moreover, the nose operator T satsfes the followng property. Let (x 1, x 2 ) be chosen accordng to the unform dstrbuton and (y 1, y 2 ) be chosen accordng T appled to (x 1, x 2 ). Then the dstrbuton of (x 2, y 2 ) s unform. Proof: The proof resembles the prevous proof. Agan there are 3 types of transtons. Wth probablty β 1 we have (x, x) (y, y) where x y. Wth probablty β 2 we have (x, x) (y, z) where x, y, z are all dfferent. Wth probablty β 3 we have (x, y) (z, y) where x, y, z are all dfferent. For T to be a symmetrc Markov operator we requre β 1, β 2 and β 3 to be non-negatve and 2β 1 + 2β 2 = 1, β 2 + β 3 = 1. For the unformty property, assume (x 1, x 2 ) s chosen accordng to the unform dstrbuton and (y 1, y 2 ) s chosen accordng T appled to (x 1, x 2 ). It s not dffcult to verfy that each of the nne possble settngs of (x 2, y 2 ) s obtaned wth probablty ether 2β 3 /9 (f x 2 = y 2 ) or β 1 /9 + 2β 2 /9 (otherwse). Therefore, the unformty property amounts to the equaton β 1 + 2β 2 = 2β 3. It s easy to see that β 2 = β 3 = 1 2 and β 1 = 0 s the soluton of all equatons and that the correspondng operator has r(t ) < 1. Ths operator s llustrated n Fgure 2(b). 4.3 The three reductons The basc dea n all three reductons s to take a label-cover nstance and to replace each vertex wth a block of q R vertces, correspondng to the q-ary hypercube [q] R. The ntended way to q- color ths block s by colorng x [q] R accordng to x where s the label gven to ths block. One can thnk of ths colorng as an encodng of the label. We wll essentally prove that any other colorng of ths block that uses relatvely few colors, can be lst-decoded nto at most t labels from {1,..., R}. By properly defnng edges connectng these blocks, we can guarantee that the lsts decoded from two blocks can be used as t-labelngs for the label-cover nstance. In the rest of ths secton, we use the followng notaton. For a vector x = (x 1,..., x n ) and a permutaton π on {1,..., n}, we defne x π = (x π(1),..., x π(n) ). 20