Conditional Hardness for Approximate Coloring

Condtonal Hardness for Approxmate Colorng Irt Dnur Elchanan Mossel Oded Regev November 3, 2005 Abstract We study the APPROXIMATE-COLORING(q, Q) problem: Gven a graph G, decde whether χ(g) q or χ(g) Q (where χ(g) s the chromatc number of G). We derve condtonal hardness for ths problem for any constant 3 q < Q. For q 4, our result s based on Khot s 2-to-1 conjecture [Khot 02]. For q = 3, we base our hardness result on a certan < shaped varant of hs conjecture. We also prove that the problem ALMOST-3-COLORING ε s hard for any constant ε > 0, assumng Khot s Unque Games conjecture. Ths s the problem of decdng for a gven graph, between the case where one can 3-color all but a ε fracton of the vertces wthout monochromatc edges, and the case where the graph contans no ndependent set of relatve sze at least ε. Our result s based on boundng varous generalzed nose-stablty quanttes usng the nvarance prncple of Mossel et al [MOO 05]. Hebrew Unversty. Emal: dnur@cs.huj.ac.l. Supported by the Israel Scence Foundaton and by BSF grant 2004105 Statstcs, U.C. Berkeley. Emal: mossel@stat.berkeley.edu. Supported by a Mller fellowshp n Computer Scence and Statstcs, by a Sloan fellowshp n Mathematcs, by NSF grants DMS-0504245, DMS-0528488 and by BSF grant 2004105 Department of Computer Scence, Tel-Avv Unversty, Tel-Avv 69978, Israel. Supported by an Alon Fellowshp, by the Israel Scence Foundaton and by BSF grant 2004105

1 Introducton For a graph G = (V, E) we let χ(g) be the chromatc number of G,.e., the smallest number of colors needed to color the vertces of G wthout monochromatc edges. We study the followng problem, APPROXIMATE-COLORING(q,Q) : Gven a graph G, decde between χ(g) q and χ(g) Q. The problem APPROXIMATE-COLORING(3, Q) s notorous for the wde gap between the value of Q for whch an effcent algorthm s known and that for whch a hardness result exsts. The best known polynomaltme algorthm solves the problem for Q = Õ(n3/14 ) colors, where n s the number of vertces [3] (In fact, that algorthm solves the search problem of fndng a Q-colorng gven a q-colorable graph. Snce we are manly nterested n hardness results, we restrct our attenton to the decson verson of the problem). In contrast, the strongest known hardness result shows that the problem s NP-hard for Q = 5 [12, 8]. Thus, the problem s open for all 5 < Q < Õ(n3/14 ). In ths paper we gve some evdence that ths problem s hard for any constant value of Q. We remark that any hardness result for q = 3 mmedately carres over for all q > 3. The best algorthm known for larger values of q s due to Halpern et al. [9], mprovng on a prevous result of Karger et al [11]. Ther algorthm solves APPROXIMATE-COLORING(q, Q) for Q = n α q where 0 < α q < 1 s some functon of q. For example, α 4 0.37. Improvng on an earler result of Fürer [7], Khot has shown [13] that for any large enough constant q and Q = q log q 25, APPROXIMATE-COLORING(q, Q) s NP-hard. A related problem s that of approxmatng the chromatc number χ( ) of a gven graph. For ths problem, an napproxmablty result of n 1 o(1) s known [6, 13]. Constructons: Our constructons follow the standard composton paradgm ntated n [2, 10], whch has yelded numerous napproxmablty results by now. In our context, ths means that we show reductons from varants of a problem known as label-cover to approxmate graph colorng problems. In the label-cover 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 ts two ncdent vertces satsfy the relaton assocated wth t. As s the case wth other composton-based reductons, our reductons work by replacng each vertex of the label-cover nstance wth a block of vertces, known as a gadget. In other reductons, the gadget s often the bnary hypercube {0, 1} R, sometmes known as the long-code. In our case, the gadget s the q-ary hypercube, {1,...,q} R. We then connect the gadgets n a way that encodes the label-cover constrants. The dea s to ensure that any Q-colorng of the graph (where Q s some constant greater than q), can be decoded nto a labelng for the underlyng label-cover nstance that satsfes many label-cover constrants. We note that the dea of usng the q-ary hypercube as a gadget has been around for a number of years. Ths dea has been studed n [1] and some partal results were obtaned. The recent progress of [17] has provded the necessary tool for achevng our result. Conjectures: Let us now turn our attenton to the label-cover problem. None of the known NP-hard varants of the label-cover problem (or even more general PCP systems) seems sutable for composton n our settng. An ncreasngly popular approach s to rely on the Unque-Games conjecture of Khot [14]. The conjecture states that a very restrcted verson of label-cover s hard. The strength of ths restrcton s that n a sense, t reduces the analyss of the entre constructon, to the analyss of the gadget alone. However, ths conjecture suffers from nherent mperfect completeness, whch seems to prevent t from beng used n a reducton to approxmate colorng (although t s useful for almost approxmate colorng). Therefore, we consder some restrctons of label-cover that do have perfect completeness. Our approach s 1

to search for the least-restrcted such label-cover problem that would stll yeld the desred result. In all, we consder three varants, whch result n three dfferent reductons. We show that ALMOST-3-COLORING s as hard as Khot s Unque Games problem. We show that APPROXIMATE-COLORING(4, Q) s as hard as Khot s 2-to-1 problem for any constant Q > 4. Ths also holds for APPROXIMATE-COLORING(q, Q) for any q 4. We ntroduce a new conjecture, whch states that label-cover s hard when the constrants are restrcted to a form we call <-constrants (read: alpha constrants). We show that for any constant Q > 3, APPROXIMATE-COLORING(3, Q) s as hard as solvng the <-label-cover problem. We remark that <-constrants have already appeared n [5]. The plausblty of the Unque Games Conjecture, as well as that of other varants, s uncertan. Recently, Trevsan [18] showed that these conjectures are false when the parameters are pushed beyond certan subconstant values. Ths has been followed by addtonal algorthms [?,?] for varous sub-constant settngs. Hopefully, these results wll trgger more attempts to understand these type of constrant systems from both the algorthmc sde, and the napproxmablty sde. Technques: Our man technque s based on the recent progress of Mossel et al [17]. There, they present a powerful technque for boundng the stablty of functons under nose operators. For example, one specal case of ther result says that among all balanced Boolean functons that do not depend too strongly on any one coordnate, the one that s most stable under nose s the majorty functon. In other words, among all such functons, the majorty functon s least lkely to flp ts value f we flp each of ts nput bts wth some small constant probablty. In fact, ths specal case was presented as a conjecture n the work of [15] on MAXCUT and motvated the result of [17]. The technque of [17] s based on what s called an nvarance prncple. Ths prncple allows one to translate questons n the dscrete settng (such as the above queston on the Boolean hypercube) to correspondng questons n other spaces, and n partcular Gaussan space. One then apples known (and powerful) results n Gaussan space. In ths paper we extend ther approach s several respects. We consder more general nose operators that are gven by some arbtrary Markov operator. We then apply ths to three operators, one for each of the aforementoned constructons. We show that when the nner product under nose of two functons devates notably from that of two majorty functons, there must exst a varable that s nfluental n both functons (see Theorem 3.1). A drect applcaton of [17] only yelds a varable that s nfluental n one of the functons. Ths latter statement was enough for the applcaton to MAXCUT n [15]. We also present another result talored for the < constrants (see Theorem B.2). We beleve that the general framework developed here wll fnd many applcatons. It also demonstrates the flexblty of the approach of [17]. Future work: Our constructons can be extended n several ways. Frst, usng smlar technques, one can show hardness of APPROXIMATE-COLORING(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 2

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 are no almost-ndependent sets of relatve sze larger than ε. 2 Prelmnares 2.1 Approxmate colorng problems We now defne the colorng problems that we study n ths paper. For any graph G, let χ(g) be ts chromatc number. Namely, χ(g) s the smallest number of colors needed n order to color the vertces of G wthout monochromatc edges. For any 3 q < Q, we defne APPROXIMATE-COLORING(q, Q): Gven a graph G, decde between χ(g) q or χ(g) Q. For any ε > 0 we defne ALMOST-3-COLORING 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. 2.2 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 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, 3) and n to be n 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[V 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. 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 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 n [q] n R can be wrtten as a lnear combnaton of α x for x [q] n. Ths leads to the followng defnton. 3

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. Clam 2.4 For any functon f : [q] n R and any {1,...,n}, I (f) = ˆf 2 (α x ). x:x 0 We nclude the proof n Appendx F. Notce that ths clam holds for any choce of orthonormal bass α 0,...,α q 1 as long as α 0 = 1. 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 ). x:x 0, x k It s easy to see that for any functon f, I k (f) = ˆf x: x k 2 (α x ) x k ˆf x 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 can be verfed from the above defnton. 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. 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(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 where we used that x 2 x. ˆ f 2 (α x ) x:x 2 1 0, x 2k ˆf 2 (α x ) + 2 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), the spectral radus of T, by 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 ( ) T n α x = a 0 λ x a a α x. 4

and hence T n f = 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 ts spectral radus s 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. 2.3 Functons n Gaussan space We let γ denote the standard Gaussan measure on R n. 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)]. 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 value that wll appear later s F η, U ρ F ν γ. For our purposes t s useful to know that for any ν, η > 0, and any ρ [ 1, 1], t holds 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 result of ths secton, Theorem 3.1, s a generalzaton of the result of [17]. 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 the spectral radus of the operator T. 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 E[f] = µ,e[g] = ν and for all, then t holds that and mn ( I k (f), I k (g) ) < δ f, T n g F µ, U ρ (1 F 1 ν ) γ ε (1) f, T n g F µ, U ρ F ν γ + ε. (2) 5

Note that (1) follows from (2). Indeed, apply (2) to 1 g to obtan f, T n (1 g) F µ, U ρ F 1 ν γ +ε 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 (2). Followng the approach of [17], the proof conssts of two powerful technques. The frst s an nequalty by Chrster Borell [4] on contnuous Gaussan space. The second s an nvarance prncple shown n [17] 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 [4]. 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 [4]) 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 satsfyng E γ [f] = µ,e γ [g] = ν. 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 S = U δ1 U δ2... U δq 1. Then, 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. 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. Clam 3.6 For any two functons f, g : [q] n R and operator T on [q] n, f, g = f, g γ and 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 x )ĝ(α 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 x ( ) a 0 λ x a a ˆf(αx )ĝ(α x ). 6

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 [17]. 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 [17], a result that s known as an nvarance prncpal. 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 [17] for more detals. Theorem 3.8 ([17, Theorem 3.18]) For any η < 1 and ε > 0 there exsts a δ > 0 such that the followng holds. For any functon f : [q] n [0, 1] such that x ˆf(α x ) η x and I (f) < δ, then f chop( f) 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 E[f] = µ,e[g] = ν, max(i (f), I (g)) < δ and x ˆf(α x ) η x, ĝ(α x ) η x, t holds that f, T n g F µ, U ρ F ν γ + ε. 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(η, ε 2 ) be the value gven by Theorem 3.8 wth ε taken to be ε 2. Then, usng the Cauchy-Schwartz nequalty, µ µ = E γ [chop( f) f] = chop( f) f,1 γ chop( f) f 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. Due to lack of space, the rest of the proof s deferred to Appendx B. There, we complete the proof of the theorem by essentally showng how to replace the max(i (f), I (g)) by mn(i (f), I (g)), as n the statement of the theorem. Ths s based on the dea of usng two nfluence thresholds. We also present there a theorem talored to the < constrant, whch s somewhat more nvolved techncally. 7

4 Approxmate Colorng In ths secton we descrbe and prove reductons to the three problems descrbed n Secton 2, based on three conjectures on the hardness of label-cover. These conjectures, along wth some defntons, are descrbed n Secton 4.1. 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 constructons. Due to space lmtatons, Secton 4.3 contans only one of the three reductons (along wth proofs of completeness and soundness). The remanng two reductons are deferred to Appendx E. 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. For the specal case of t = 1, a 1-labelng s lke a labelng L : V {1,...,R} (except that some vertces get no label). In ths case, a constrant ψ over u, v s satsfed by L ff (L(u), L(v)) ψ. 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. { S sat(g) = max 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 now consder 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 llustrated n Fgure A.1. Due to lack of space, we only descrbe the frst conjecture and defer the remanng two to Appendx C. 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). Conjecture 4.4 (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 the case sat(g) 1 ζ and the case sat t (G) < ε. It s easy to see that the above problem s n P when ζ = 0. 8

4.2 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 A.2 for an llustraton. Lemma 4.5 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 The two remanng nose operators are descrbed n Appendx D. 4.3 The 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 ths secton we wll descrbe only the reducton for ALMOST-3-COLORING. We defer the two (smlar) remanng reductons (for APPROXIMATE-COLORING(4, Q) and for APPROXIMATE-COLORING(3, Q)) to Appendx E. 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) ). ALMOST-3-COLORING Let G = ((V, E), R, Ψ) be a label-cover nstance as n Conjecture 4.4. For v V wrte [v] for a collecton of vertces, one per pont n {0, 1, 2} R. Let e = (v, w) E, and let ψ be the 1 1-constrant assocated wth e. By Defnton 4.3 there s a permutaton π such that (a, b) ψ ff b = π(a). We now wrte [v, w] for the followng collecton of edges. We put an edge (x, y) for x = (x 1,...,x R ) [v] and y = (y 1,..., y R ) [w] ff {1,...,R}, T ( x y π() ) 0 where T s the nose operator from Lemma 4.5. In other words, x s adjacent to y whenever T R (x y π ) = R T ( ) x y π() 0. =1 The reducton outputs the graph [G] = ([V ], [E]) where [V ] s the dsjont unon of all blocks [v] and [E] s the dsjont unon of all collectons of edges [v, w]. 9

Completeness. If sat(g) 1 ε, then there s some S V of sze (1 ε) V and a labelng l : S R that satsfes all of the constrants nduced by S. We 3-color all of the vertces n v S [v] as follows. Let c : v S [v] {0, 1, 2} be defned as follows. For every v S, the color of x = (x 1,..., x R ) {0, 1, 2} R = [v] s defned to be c(x):=x, where = l(v) {1,...,R}. To see that c s a legal colorng on v S [v], observe that f x [v] and y [w] share the same color, then x = y j for = l(v) and j = l(w). Snce l satsfes every constrant nduced by S, t follows that f (v, w) s a constrant wth an assocated permutaton π, then j = π(). Snce T(z z) = 0 for all z {0, 1, 2}, there s no edge between x and y. Soundness. Before presentng the soundness proof, we need the followng corollary. It s smply a specal case of Theorem 3.1 stated n the contrapostve, wth ε playng the role of ν and µ. Here we use the fact that F ε, U ρ (1 F 1 ε ) γ > 0 whenever ε > 0. Corollary 4.6 Let q be a fxed nteger and let T be a reversble Markov operator on [q] such that r(t) < 1. Then for any ε > 0 there exst δ > 0 and k N such that the followng holds. For any f, g : [q] n [0, 1], f E[f] ε, E[g] ε, and f, T n g = 0, then {1,...,n}, I k (f) δ and I k (g) δ. We wll show that f [G] has an ndependent set S [V ] of relatve sze 2ε, then sat t (G) ε for a fxed constant t > 0 that depends only on ε. More explctly, we wll fnd a set J V, and a t-labelng L : J ( R t) such that J ε V and L satsfes all the constrants of G nduced by J. In other words, for every constrant ψ over an edge (u, v) E J 2, there are values a L(u) and b L(v) such that (a, b) ψ. Let J be the set of all vertces v V such that the fracton of vertces belongng to S n [v] s at least ε. Then, snce S 2ε [V ], Markov s nequalty mples J ε V. For each v J let f v : {0, 1, 2} R {0, 1} be the characterstc functon of S restrcted to [v], so E[f v ] ε. Select δ, k accordng to Corollary 4.6 wth ε and the operator T of Lemma 4.5, and set { } L(v) = {1,...,R} I k (f v ) δ. Clearly, L(v) k/δ because R =1 I k (f) k. Thus, L s a t-labelng for t = k/δ. The man pont to prove s that for every edge e = (v 1, v 2 ) E J 2 nduced on J, there s some a L(v 1 ) and b L(v 2 ) such that (a, b) ψ e. Ths would mply that sat t (G) J / V ε. Fx (v 1, v 2 ) E J 2, and let π be the permutaton assocated wth the 1 1 constrant on ths edge. (It may be easer to frst thnk of π = d.) Recall that the edges n [v 1, v 2 ] were defned based on π, and on the nose operator T defned n Lemma 4.5. Let f = f v1, and defne g by g(x π ) = f v2 (x). Snce S s an ndependent set, f(x) = f v1 (x) = 1 and g(y π ) = f v2 (y) = 1 mples that x, y are not adjacent, so by constructon T R (x y π ) = 0. Therefore, f, T R g = 3 R x f(x)t R g(x) = 3 R x f(x) y π T R (x y π )g(y π ) = x,y π 0 = 0. Also, by assumpton, E[g] ε and E[f] ε. Corollary 4.6 mples that there s some ndex {1,...,R} for whch both I k (f) δ and I k (g) δ. By defnton of L, L(v 1 ). Snce the -th varable n g s the π()-th varable n f v2, π() L(v 2 ). It follows that there are values L(v 1 ) and π() L(v 2 ) such that (, π()) satsfes the constrant on (v 1, v 2 ). Ths means that sat t (G) J / V ε. 10

5 Omtted proofs and dscussons We have omtted from ths extended abstract a comparson between the conjectures we have used and Khot s orgnal conjectures. A detaled comparson s gven n Appendx G. The rest of the man Fourer theorem s gven n Appendx B. The reductons for APPROXIMATE-COLORING are gven n Appendx E. Acknowledgements We thank Ur Zwck for hs help wth the lterature. References [1] N. Alon, I. Dnur, E. Fredgut, and B. Sudakov. Graph products, fourer analyss and spectral technques. GAFA, 14(5):913 940, 2004. [2] M. Bellare, O. Goldrech, and M. Sudan. Free bts, PCPs, and nonapproxmablty towards tght results. SIAM J. Comput., 27(3):804 915, 1998. [3] A. Blum and D. Karger. An Õ(n3/14 )-colorng algorthm for 3-colorable graphs. Inform. Process. Lett., 61(1):49 53, 1997. [4] C. Borell. Geometrc bounds on the Ornsten-Uhlenbeck velocty process. Z. Wahrsch. Verw. Gebete, 70(1):1 13, 1985. [5] I. Dnur and S. Safra. On the hardness of approxmatng mnmum vertex cover. Annals of Mathematcs, 2005. To appear. [6] U. Fege and J. Klan. Zero knowledge and the chromatc number. J. Comput. System Sc., 57(2):187 199, 1998. [7] M. Fürer. Improved hardness results for approxmatng the chromatc number. In Proc. 36th IEEE Symp. on Foundatons of Computer Scence, pages 414 421, 1995. [8] V. Guruswam and S. Khanna. On the hardness of 4-colorng a 3-colorable graph. In 15th Annual IEEE Conference on Computatonal Complexty (Florence, 2000), pages 188 197. IEEE Computer Soc., Los Alamtos, CA, 2000. [9] E. Halpern, R. Nathanel, and U. Zwck. Colorng k-colorable graphs usng relatvely small palettes. J. Algorthms, 45(1):72 90, 2002. [10] J. Håstad. Some optmal napproxmablty results. J. ACM, 48(4):798 859, 2001. [11] D. Karger, R. Motwan, and M. Sudan. Approxmate graph colorng by semdefnte programmng. Journal of the ACM, 45(2):246 265, 1998. [12] S. Khanna, N. Lnal, and S. Safra. On the hardness of approxmatng the chromatc number. Combnatorca, 20(3):393 415, 2000. [13] S. Khot. Improved napproxmablty results for MaxClque, chromatc number and approxmate graph colorng. In Proc. 42nd IEEE Symp. on Foundatons of Computer Scence, pages 600 609, 2001. 11

[14] S. Khot. On the power of unque 2-prover 1-round games. In Proceedngs of the thry-fourth annual ACM symposum on Theory of computng, pages 767 775. ACM Press, 2002. [15] S. Khot, G. Kndler, E. Mossel, and R. ODonnell. Optmal napproxmablty results for MAX-CUT and other 2-varable CSPs? In Proc. 45th IEEE Symp. on Foundatons of Computer Scence, pages 146 154, 2004. [16] S. Khot and O. Regev. Vertex cover mght be hard to approxmate to wthn 2 ε. In Proc. of 18th IEEE Annual Conference on Computatonal Complexty (CCC), pages 379 386, 2003. [17] E. Mossel, R. O Donnell, and K. Oleszkewcz. Nose stablty of functons wth low nfluences: nvarance and optmalty. Preprnt, 2005. [18] L. Trevsan. Approxmaton algorthms for unque games. In ECCCTR: Electronc Colloquum on Computatonal Complexty, techncal reports, 2005. A Fgures Fgure A.1: Three types of constrants (top to bottom): 1 1, <, 2 2 (a) (b) (c) Fgure A.2: Three nose operators (edge weghts not shown) correspondng to: (a) 1 1, (b) <, and (c) 2 2. B Rest of Proof of Theorem 3.1 We complete the proof of Theorem 3.1 by provng: 12

Lemma B.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 exsts a δ > 0 and an nteger k such that for any functons f, g : [q] n [0, 1] satsfyng E[f] = µ,e[g] = ν and mn ( I k (f), I k (g) ) < δ (3) then f, T n g F µ, U ρ F ν γ + ε. (4) Proof: Let f 1 = T n η 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 = x x ˆf(α x )ĝ(α x ) a 0λ x a a (1 η 2 x ) ρ x (1 η 2 x ) ˆf(α x )ĝ(α x ) ε/4 where the last nequalty follows from the Cauchy-Schwartz nequalty. Thus, n order to prove (4) t suffces to prove f 1, T n g 1 F µ, U ρ F ν γ + 3ε/4. (5) 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 }. We note that B f and B g are of sze at most C = k/δ. By (3), 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 (x) = A Bf (f 1 ) = ˆf(α x )α x η x, g 2 (x) = 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 x, ˆf 2 (α x ) η x 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 (5) and complete the proof, we show that f 1, T n g 1 f 2, T n g 2 ε/2. 13

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 APPROXIMATE-COLORING(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 B.2 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 E[f] = µ,e[g] = ν, and 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 mn ( I k 2 (f), I k 2 1 (g)) < δ t holds that and f, T n g F µ, U ρ (1 F 1 ν ) γ ε (6) f, T n g F µ, U ρ F ν γ + ε. (7) Proof: As n Theorem 3.1, (6) follows from (7) so t s enough to prove (7). 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. (8) Then t follows that for all, ether both I k 2 1 (f) and I k 2 (f) are smaller than δ or both I k 2 1 (g) and I k 2 (g) are smaller than δ. Hence, by Clam 2.7, we know that for all we have ( mn I k/2 ) (f), I k/2 (g) < 2δ and the result then follows from Lemma B.1. 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 for such t must be the case that both I k 2 1 (f) and I k (g) are smaller than δ. Informally, the proof proceeds as follows. We frst 2 1 14

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 8 now apples to f 3, g 3 and we can apply Lemma B.1, 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 B.1 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 Hence, I 2 1 (f 1 ) = x [q] 2n x 2 1 0 f 1 = x [q] 2n ˆf(βx )η x β x. ˆf(β x ) 2 η 2 x < δ + η k x [q] 2n x > k ˆf(β x ) 2 δ + η k 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 1 2 2 = f 2 f 1 2 2 B I 2 1 (f 1 ) 2Cδ. and smlarly, Thus g 2 g 1 2 2 = g 2 g 1 2 2 2Cδ. 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 1 2 1 and also T n g 2 2 1. Hence, t suffces to prove f 2, T n g 2 F µ, U ρ F ν γ + ε/2. 15

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. Moreover, snce A s a lnear operator and f 1 can be wrtten as x [q 2 ] n ˆf1 (α x )α x and smlarly for g 1, we have f 2, T n g 2 = x,y [q 2 ] n ˆf1 (α x )ĝ 1 (α y ) A 2B 1 (α x ), T n A 2B 1 (α y ). 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 two expressons above 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 )]. 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 B.1 to obtan f 3, T n g 3 F µ, U ρ F ν γ + ε/2. C The Two Remanng Conjectures Defnton C.1 (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)) 2 2. 2 16

Defnton C.2 ( <-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 Conjecture C.3 (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 the case sat(g) = 1 and the case sat t (G) < ε. The 1 1 conjecture and the above conjecture 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) 2 1 conjecture. The former clam was already proven by Khot and Regev n [16]. The latter clam s proven n a smlar way. For completeness, we nclude both proofs n Appendx G. We also make a thrd conjecture that s used n our reducton to APPROXIMATE-COLORING(3, Q). Ths conjecture seems stronger than Khot s conjectures. Conjecture C.4 ( < 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 the case sat(g) = 1 and the case sat t (G) < ε. Remark: The (strange-lookng) <-shaped constrants have already appeared before, n [5]. There, t s (mplctly) 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 the case sat(g) > 1 ζ and the case sat t=1 (G) < ε. The man dfference between ther case and our conjecture s that n our conjecture we consder any constant t, whle n ther case t s 1. D The Two Remanng Operators Lemma D.1 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. 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 A.2(c) wth red ndcatng β 1 transtons, blue ndcatng β 2 transtons, and black ndcatng β 3 transtons. For T to be 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. 17

Lemma D.2 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. Moreover, the last requrement of unformty of (x 2, y 2 ) amounts to the equaton β 1 /3 + 2β 2 /3 = 2β 3 /3. It s easy to see that β 2 = β 3 = 0.5 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 A.2(b). E The Two Remanng 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) ). APPROXIMATE-COLORING(4, Q): changes: Ths reducton s nearly dentcal to the one above, wth the followng The startng pont of the reducton s an nstance G = ((V, E), 2R, Ψ) as n Conjecture C.3. Each vertex v s replaced by a copy of {0, 1, 2, 3} 2R (whch we stll denote [v]). For every (v, w) E, let ψ be the 2 2-constrant assocated wth e. By Defnton C.1 there are two permutatons π 1, π 2 such that (a, b) ψ ff (π1 1 (a), π 1 2 (b)) 2 2. We now wrte [v, w] for the followng collecton of edges. We put an edge (x, y) for x = (x 1,...,x 2R ) [v] and y = (y 1,...,y 2R ) [w] f {1,...,R}, T((x π1 (2 1), x π1 (2)) (y π2 (2 1), y π2 (2))) 0 where T s the nose operator from Lemma D.1. Equvalently, we put an edge f T R (x π 1 y π 2) 0. As before, the reducton outputs the graph [G] = ([V ], [E]) where [V ] s the unon of all blocks [v] and [E] s the unon of collecton of the edges [v, w]. 18

APPROXIMATE-COLORING(3, Q): followng changes: Here agan the reducton s nearly dentcal to the above, wth the The startng pont of the reducton s an nstance of label-cover, as n Conjecture C.4. Each vertex v s replaced by a copy of {0, 1, 2} 2R (whch we agan denote [v]). For every (v, w) E, let π 1, π 2 be the permutatons assocated wth the constrant, as n Defnton C.2. Defne a collecton [v, w] of edges, by ncludng the edge (x, y) [v] [w] ff {1,...,R}, T((x π1 (2 1), x π1 (2)) (y π2 (2 1), y π2 (2))) 0 where T s the nose operator from Lemma D.2. As before, ths condton can be wrtten as T R (x π 1 y π 2 ) 0. As before, we look at the colorng problem of the graph [G] = ([V ], [E]) where [V ] s the unon of all blocks [v] and [E] s the unon of collecton of the edges [v, w]. E.1 Completeness APPROXIMATE-COLORING(4, Q): Let l : V {1,...,2R} be a labelng that satsfes all the constrants n G. We defne a legal 4-colorng c : [V ] {0, 1, 2, 3} as follows. For a vertex x = (x 1,...,x 2R ) {0, 1, 2, 3} 2R = [v] set c(x):=x, where = l(v) {1,...,2R}. To see that c s a legal colorng, fx any 2 2 constrant (v, w) E and let π 1, π 2 be the permutatons assocated wth t. Let = l(v) and j = l(w), so by assumpton (π1 1 (), π 1 2 (j)) 2 2. In other words there s some k {1,...,R} such that {π 1 (2k 1), π 1 (2k)} and j {π 2 (2k 1), π 2 (2k)}. If x [v] and y [w] share the same color, then x = c(x) = c(y) = y j. Snce x { x π 1 2k 1, } xπ 1 2k and y j { y π 2 2k 1, } yπ 2 2k we have that the above sets ntersect. Ths, by Lemma D.1, mples that T R (x π 1 y π 2) = 0. So the vertces x, y cannot be adjacent, hence the colorng s legal. APPROXIMATE-COLORING(3, Q): Here the argument s nearly dentcal to the above. Let l : V {1,...,2R} be a labelng that satsfes all of the constrants n G. We defne a legal 3-colorng c : [V ] {0, 1, 2} lke before: c(x):=x, where = l(v) {1,...,2R}. To see that c s a legal colorng, fx any edge (v, w) E and let π 1, π 2 be the permutatons assocated wth the <-constrant. Let = l(v) and j = l(w), so by assumpton (π1 1 (), π 1 2 (j)) <. In other words there s some k {1,...,R} such that {π 1 (2k 1), π 1 (2k)} and j {π 2 (2k 1), π 2 (2k)} and not both = π 1 (2k) and j = π 2 (2k). Assume, wthout loss of generalty, that = π 1 (2k 1), so x = x π 1 2k 1 and y j { y π 2 2k 1, yπ 2 2k}. If x [v] and y [w] share the same color, then x = c(x) = c(y) = y j, so By Lemma D.2 ths mples T((x π 1 and y. x π 1 2k 1 = x = y j { y π 2 2k 1, yπ 2 2k}. 2k 1, xπ 1 2k ) (yπ 2 2k 1, yπ 2 2k )) = 0, whch means there s no edge between x 19

E.2 Soundness APPROXIMATE-COLORING(4, Q): We outlne the argument and emphasze only the modfcatons. Assume that [G] contans an ndependent set S [V ] whose relatve sze s at least 1/Q and set ε = 1/2Q. Let f v : {0, 1, 2, 3} 2R {0, 1} be the characterstc functon of S n [v]. Defne the set J V as before and for all v J, defne { L(v) = {1,...,2R} I 2k (f v ) δ } 2 where k, δ are the values gven by Corollary 4.6 wth ε and the operator T of Lemma D.1. As before, J ε V and E[f v ] ε for v J. Now L s a t-labelng wth t = 4k/δ. Fx an edge (v, w) E J 2 and let π 1, π 2 be the assocated permutatons. Defne f, g by f(x π 1 ):=f v1 (x) and g(y π 2 ):=f v2 (y). Snce S s an ndependent set, f(x π 1 ) = f v1 (x) = 1 and g(y π 2 ) = f v2 (y) = 1 mples that x, y are not adjacent, so by constructon T(x π 1 y π 2 ) = 0. Therefore, f, Tg = 0. Now, recallng Defnton 2.6, consder the functons f, g : ({0, 1, 2, 3} 2 ) R {0, 1}. Applyng Corollary 4.6 on f, g we may deduce the exstence of an ndex {1,...,R} for whch both I k (f) δ and I k (g) δ. By Clam 2.7, δ I k (f) I 2k 2 1 (f) + I 2k 2 (f), so ether I 2k 2 1 (f) δ/2 or I 2k 2 (f) δ/2. Snce the j-th varable n f s the π 1 (j)-th varable n f v1, ths puts ether π 1 (2) or π 1 (2 1) n L(v 1 ). Smlarly, at least one of π 2 (2), π 2 (2 1) s n L(v 2 ). Thus, there are a L(v 1 ) and b L(v 2 ) such that (π1 1 (a), π 1 2 (b)) 2 2 so L satsfes the constrant on (v 1, v 2 ). We have shown that L satsfes every constrant nduced by J, so sat t (G) ε. APPROXIMATE-COLORING(3, Q): The argument here s smlar to the prevous one. The man dfference s n the thrd step, where we replace Corollary 4.6 by the followng corollary of Theorem B.2. The corollary follows by lettng ε play the role of µ and ν, and usng the fact that F ε, U ρ (1 F 1 ε ) γ > 0 whenever ε > 0. Corollary E.1 Let T be the operator on {0, 1, 2} 2 defned n Lemma D.2. For any ε > 0, there exsts δ > 0, k N, such that for any functons f, g : {0, 1, 2} 2R [0, 1] satsfyng E[f] ε,e[g] ε, there exsts some {1,...,R} such that ether mn ( I k 2 1 (f), I k 2 1 (g)) δ or mn ( I k 2 1 (f), I k 2 (g)) δ or mn ( I k 2 (f), I k 2 1 (g)) δ. Now we have functons f v : {0, 1, 2} 2R {0, 1}, and J s defned as before. Defne a labelng { } L(v) = {1,...,2R} I k (f v ) δ where k, δ are the values gven by Corollary E.1 wth ε. Then L s a t-labelng wth t = k/δ. Let us now show that L s a satsfyng t-labelng. Let (v 1, v 2 ) be a <-constrant wth assocated permutatons π 1, π 2. Defne f(x π 1 ) = f v1 (x), g(x π 2 ) = f v2 (x). We apply Corollary E.1 on f, g, and obtan an ndex {1,...,R}. Snce the j-th varable n f s the π 1 (j)-th varable n f v1, ths puts ether π 1 (2) or π 1 (2 1) n L(v 1 ). Smlarly, at least one of π 2 (2), π 2 (2 1) s n L(v 2 ). Moreover, we are guaranteed that ether π 1 (2 1) L(v 1 ) or π 2 (2 1) L(v 2 ). Thus, there are a L(v 1 ) and b L(v 2 ) such that (a), π 1 (π 1 1 2 (b)) < so L satsfes the constrant on (v 1, v 2 ). 20