Every Permutation CSP of arity 3 is Approximation Resistant
Moses Charikar∗ Venkatesan Guruswami† Rajsekar Manokaran‡ Dept. of Computer Science Dept. of Computer Science & Engg. Dept. of Computer Science Princeton University University of Washington Princeton University Princeton, NJ 08540. Seattle, WA 98195. Princeton, NJ 08540. [email protected] [email protected] [email protected]
Abstract—A permutation constraint satisfaction problem Max 3LIN, Unique Games, etc. are examples of CSPs. A (permCSP) of arity k is specified by a subset Λ ⊆ Sk of CSP of arity k over domain [q] = {0, 1, . . . , q − 1} (called permutations on {1, 2, . . . , k}. An instance of such a permCSP a q-ary kCSP) is specified by a predicate P :[q]k → {0, 1}. consists of a set of variables V and a collection of constraints each of which is an ordered k-tuple of V . The objective is An instance of such a CSP consists of a set of variables to find a global ordering σ of the variables that maximizes V and a collection C of constraints each of which applies the number of constraint tuples whose ordering (under σ) P to a k-tuple of literals (which are variables or their follows a permutation in Λ. This is just the natural extension “negations,” i.e., translates modulo q). The objective is to of constraint satisfaction problems over finite domains (such find an assignment A : V → [q] of values to the variables as Boolean CSPs) to the world of ordering problems. The simplest permCSP corresponds to the case when Λ that maximizes the number of constraints of C that are consists of the identity permutation on two variables. This is satisfied. just the Maximum Acyclic Subgraph (MAS) problem. It was For most CSPs, it is NP-hard to find an optimal as- recently shown that the MAS problem is Unique-Games hard to signment with maximum number of satisfied constraints. approximate within a factor better than the trivial 1/2 achieved Therefore, one settles for approximation algorithms that by a random ordering [6]. Building on this work, in this paper we show that for every permCSP of arity 3, beating the random deliver provable guarantees. Ideally, for each CSP, we would ordering is Unique-Games hard. The result is in fact stronger: like to know its approximation threshold, i.e., a value α < 1 we show that for every Λ ⊆ Π ⊆ S3, given an instance of for which we can give a polynomial time algorithm that permCSP(Λ) that is almost-satisfiable, it is hard to find an satisfies at least α times the optimum number of constraints |Π| 1 ordering that satisfies more than 6 + ε of the constraints on every instance of that CSP , together with a matching even under the relaxed constraint Π (for arbitrary ε > 0). hardness result that rules out a factor (α +ε)-approximation A special case of our result is that the Betweenness problem is hard to approximate beyond a factor 1/3. Interestingly, for for arbitrary constant ε > 0. For a CSP, we define its random satisfiable instances of Betweenness, a factor 1/2 approximation assignment threshold to be the fraction ρ of assignments that algorithm is known. satisfy its predicate P . A trivial algorithm that ignores the Thus, every permutation CSP of arity up to 3 resists structure of the instance, and simply picks an assignment approximation beyond the trivial random ordering threshold. to variables randomly and independently, satisfied an ex- In contrast, for Boolean CSPs, there are both approximation resistant and non-trivially approximable CSPs of arity 3. pected fraction ρ of the constraints. (This algorithm can be derandomized using standard methods.) The approximation Keywords-hardness of approximation; betweenness; permu- tation constraint satisfaction problems; approximation resis- threshold is thus always at least the random assignment tance threshold. Approximation Resistance. The discovery of the PCP the- I.INTRODUCTION orem and semi-definite programming based approximation Constraint satisfaction problems (CSPs) are a rich class of algorithms in the early 90’s has led to a rich body of optimization problems that arise naturally in many settings. work which has identified the approximation threshold for A large number of well-studied problems such as Max 3SAT, many important CSPs. The breakthrough paper of Hastad˚ [8] Max Cut, Max k-colorable subgraph, Max k-set splitting, proved the striking result that many important CSPs are approximation resistant: it is NP-hard to approximate them ∗Supported by NSF ITR grants CCF-0205594 and CCF-0426582, NSF CAREER award CCF-0237113, MSPA-MCS award 0528414, and NSF better than their random assignment threshold, and thus their expeditions award 0832797. approximation thresholds equals their random assignment †Work done while on leave at the Computer Science Department, threshold ! The list of such approximation resistant CSPs Carnegie Mellon University, Pittsburgh, PA 15213, and the School of Mathematics, Institute for Advanced Study, Princeton, NJ 08540. Research include Max 3SAT, Max 3LIN (whose predicate stipulates supported in part by a Packard Fellowship, and NSF grants CCR-0324906 that the parity of 3 literals is 0), and in fact any binary and CCF-0835814. ‡Supported by NSF grants 0830673, 0832797, 528414. 1For convenience, we are focusing on maximization problems only. 3CSP whose predicate is implied by the parity constraint powerful techniques from the analysis of Boolean functions x ⊕ y ⊕ z = 0, Max k-set splitting for k > 4, etc. can be directly employed to establish strong results, whereas This motivates the natural question of classifying ap- permutations tend to resist such an analysis. proximation resistant predicates. Posed in this generality, In a recent work [6], a tight inappproximability result was this is a very challenging goal. But we now know fairly shown for the Maximum Acyclic Subgraph (MAS) problem broad classes of CSPs which are approximation resistant, as (assuming the UGC). In the MAS problem, one is given a well as those that are not. We now discuss some of these collection of u < v constraints (for variables u, v ∈ V ) results. Complementing Hastad’s˚ hardness result for 3-CSPs, and the goal is to permute V so that a maximum number Zwick [16] gave approximation algorithms outperforming a of these local constraints are met. A random permutation random assignment for every 3-ary predicate not implied will satisfy an expected fraction 1/2 of constraints, and the by parity, thereby leading to a precise classification of result of [6] shows that beating this is Unique Games-hard. approximation resistant binary 3CSPs. The situation for arity In other words, MAS is approximation resistant. This was 4 and higher gets more complicated as one might imagine. the first such result, and indeed the first tight hardness of Hast succeeds in characterizing 355 out of 400 different approximation result, for an ordering problem. predicate types for binary 4CSPs [7]. Hastad˚ [9] showed In this paper, we study problems with more general that most predicates (a fraction 1−o1(k) of k-ary predicates) ordering constraints. A permutation constraint satisfaction are approximation resistant (this relies on the Unique Games problem (permCSP) of arity k is specified by a subset conjecture). Λ ⊆ Sk of permutations on {1, 2, . . . , k}. An instance of It is known that every 2CSP, even over non-binary do- such a permCSP consists of a set of variables V and a mains, can be approximated better than the random assign- collection of constraints each of which is an ordered k- ment threshold [5], [4], [8]. The approximation threshold of tuple of V . The objective is to find a global ordering σ 2CSPs (such as Max Cut) remained a fascinating mystery of the variables that maximizes the number of constraint until recent progress based on the Unique Games conjecture tuples whose ordering (under σ) follows a permutation in (UGC) tied it to the integrality gap of semi-definite program- Λ. This is just the natural extension of CSPs to the world of ming (SDP) relaxations [11], [1], [15]. In fact, under the ordering problems. As with CSPs, we say a permCSP Λ is UGC, Raghavendra showed the striking result [15] that for approximation resistant if its approximation threshold equals every CSP, the approximation threshold equals the integrality |Λ| k! , which is the expected fraction of constraints satisfied by gap of a natural SDP. Thus believing the UGC is equivalent a random permutation of the variables. to believing that SDPs are the “best algorithms out there” Note that in this language, MAS corresponds to the for approximating CSPs. simplest permCSP: the arity 2 permCSP where Λ consists Note that this result does not pinpoint the value of the of the identity permutation on two variables. Also note that approximation threshold, but only ties it to the integrality MAS is the only non-trivial permCSP of arity 2. gap. For various CSPs, identifying the integrality gap itself Our main result is that every permCSP of arity 3 is ap- is a challenging task (this is especially so for arity 3 and proximation resistant. Specifically, for every such permCSP higher). So this connection per se does not advance the outperforming the trivial approximation ratio achieved by classification of approximation resistant predicates. random ordering is Unique Games-hard. The result is in Permutation CSPs. As discussed above, there have been fact stronger: significant advances on the approximability of CSPs, both Theorem 1. For every ε > 0, Λ ⊆ Π ⊆ S , the following in algorithms and hardness results. However, a rich class 3 is Unique Games hard: given an instance of permCSP(Λ) of CSP-like problems that have resisted such progress are that is (1 − ε)-satisfiable, it is hard to find an ordering that ordering (or permutation) problems. At a high level, an satisfies more than |Π|/6 + ε of the constraints even under ordering problem seeks a permutation of the variables of the relaxed constraint Π. a CSP (or vertices of a graph) to optimize some objective function. The list of ordering problems that have been A special case of our result is that the Betweenness studied in the literature include Bandwidth, Minimum Linear problem is hard to approximate beyond a factor 1/3. The Arrangement, Feedback arc set, Maximum acyclic subgraph, Betweenness problem consists of constraints of the form “j Betweenness, etc. lies between i and k” corresponding to the subset {123, 321} Our understanding of approximability of ordering prob- of S3. Interestingly, for satisfiable instances of Betweenness, lems lags that of CSPs significantly. In terms of algorithms, it is possible to outperform the random ordering: using a for CSPs, one typically rounds a SDP/linear program solu- semidefinite programming relaxation, Chor and Sudan [3] tion into a small number of values (0/1 for binary CSPs), gave a factor 1/2 approximation algorithm. whereas for ordering problems, the rounding algorithm has Thus, every permutation CSP of arity up to 3 resists to pick one of an unbounded number of values. In terms approximation beyond the trivial random ordering threshold. of hardness results, due to the bounded range for CSPs, In contrast, as mentioned above for binary CSPs, there are both approximation resistant and non-trivially approximable In light of this difficulty, the approach taken in [6] is to CSPs of arity 3 (and we even precisely know which ones relax the requirements on the gap instance. Start with a gap are in each category). instance on m vertices that has good SDP value, but much Our work suggests the intriguing possibility that this smaller t-ordering value, for t m. Thus, the instance approximation resistance holds for larger arities as well, is only required to perform poorly on t-orderings and not and that every permutation CSP might be approximation on normal orderings (i.e., m-orderings). In fact, given the resistant. Such a result would be striking for its generality difference in objectives (m vs. t-orderings), we can even — it would highlight the fundamental difficulty of satisfying hope for an instance with a “c vs. s” gap between optimum ordering constraints in a powerful way, and be in sharp integral m-orderings (not SDP vectors) and t-orderings. We contrast to the situation for binary and other bounded domain can plug in such a gap instance into the above-mentioned CSPs. UG reduction. This will yield an instance of permCSP(Λ) with m-ordering value (and hence also optimum ordering II.PROOF OVERVIEW AND RELATION TO [15], [6] value) at least c − ε for Yes instances of UG, and t-ordering value at most s + ε for No instances of UG. In the latter In this section, we describe the high level ideas behind case, using Gaussian noise stability bounds, the optimum inapproximability results for ordering problems, and place ordering value will also be close to s (for t chosen large it in context with two closely related works [15], [6]. The enough). We thus get a “c vs s” inapproximability result for underlying theme in these works is to convert some kind of permCSP(Λ). combinatorial “gap instance” into a reduction from Unique For the MAS problem, the authors of [6] showed how Games (UG) which preserves this gap. After describing this to use a construction of a directed acyclic graph with low framework, we then highlight the main technical contribu- cut-norm due to Charikar, Makarychev, and Makarychev [2] tion that is new to this work. to obtain the required gap instance. The gap instance is a As explained above, Raghavendra [15] showed how to directed graph on m vertices which is nearly acyclic, and convert an integrality gap instance of a natural SDP for yet for t m the MAS value of any t-ordering is close to an arbitrary CSP into a matching hardness result. Consider 1/2. We will refer to this graph as the CMM instance in a permCSP problem specified by Λ ⊆ Sk. An instance the sequel. of permCSP(Λ) with m variables can be viewed as an Our technical contribution in this work is to construct instance of a CSP with domain size m, and viewed this an appropriate gap instance for arity 3 permutation CSPs way admits a similar SDP relaxation. Suppose there is a with the completeness and soundness properties listed below. “c vs s” gap instance for this SDP that has m variables. In Plugging this instance into the above UG-reduction frame- other words, the SDP optimum is at least c but the permCSP work of [6] then gives our inapproximability result. (Specif- instance admits no ordering satisfying more than fraction s ically, this gives a strong hardness for permCSP(123), or of constraints. Using the techniques of [15], one can then get the 3-ary monotone ordering problem, where the constraints a UG-hardness for an m-ordering version of permCSP(Λ). are of the form i < j < k. Our general result for all 3-ary In this version, the goal is to map the variables to m ordered permCSPs follows by a simple reduction.) buckets (the variables inside a bucket are not ordered), and We now describe the properties of the gap instance. For the payoff of a constraint is the probability that it is satisfied integer parameters t < m, and η > 0, the gap instance if each bucket is randomly and independently permuted. For consists of a distribution D on [m] × [m] × [m] with the example, if Λ = 123, the payoff for a constraint i < j < k following properties: when i, j are placed in the same bucket and k is placed in • Completeness: Pr(i,j,k)∈D i < j < k > 1 − η. a later bucket equals 1/2. In fact, one can conclude something stronger. Using • Soundness: For every permutation π ∈ S3 and every Gaussian noise stability bounds, and the notion of influential t-ordering Ot of [m], the probability over random linear coordinates for orderings put forth in [6], one can show extensions of Ot that a sample (i, j, k) ∈ D is ordered according to π is at most 1 + η. the following fact in the soundness case: For the instance 6 of permCSP(Λ) produced by the reduction, the value of We obtain such a distribution over triples by first sampling the best ordering cannot be much higher than the best m- 3-tuples from the circle S1 according to a suitable con- ordering. Thus one can conclude a “c vs s” inapproximabil- tinuous distribution, and then discretizing the distribution. ity result for permCSP(Λ). We use Fourier analysis over the circle S1 to prove the So why are we not done? The problem is that we do not soundness property. The continuous view also gives an know strong integrality gaps to use as starting point for the alternate, somewhat simpler, explanation of why the CMM above reduction. In fact, even for the simplest permCSP, instance works in the MAS setting. The extension to the the MAS problem, tight integrality gaps were obtained only arity 3 case presents additional challenges and technical from the UG-hardness in [6]! difficulties compared to the CMM analysis, and working in the continuous setting makes it possible to overcome these follow immediately from the definition of the quantities with a reasonable economy of analysis. involved. A gap instance with similar properties for larger arity k Observation 2. For all 3-MO instances G, and integers would show that every permutation CSP is approximation 0 π π π t t and all π, Val (G) Val 0 (G) Val (G). resistant (under the UGC) via the same reduction framework. 6 t 6 t 6 However, we do not know whether such gap instances Observation 3. For all 3-MO instances G and integers t, for exist for arities larger than 3. Our proof technique for any ordering O of the points in G, P Valπ(G, O) = 1. π∈S3 t triples proceeds by some sort of “reduction” to the CMM A. Fourier analysis over S1 construction for arity 2. It is not clear if such a method can be extended to the larger arity case. We review the basis facts that we will use about Fourier analysis of real valued functions defined on the circle S1 = III.PRELIMINARIES R/Z. A great reference for this topic is Korner’s¨ book [13]. 1 For a positive integer t, Simt denotes the the t dimensional For a Riemann integrable function f : S → R, define the simplex. We will use boldface letters z to denote vectors Fourier coefficients of f by (1) (R) z = (z , . . . , z ). Let oτ (1) denote a quantity that tends Z 1 to zero as τ → 0, while keeping all other parameters fixed. fˆ(n) = f(t)e−2πintdt , For an integer m, we denote by [m] the set {1, 2, . . . , m}. 0 For notational convenience, an ordering O of a set of points for integers n ∈ Z. Under fairly general conditions, for f V is represented by a map O : V → Z. On the other hand, example if is continuous everywhere and has a continuous a t-ordering O consists of a map O : V → [t]. For a t- bounded derivative except at finitely many points, then the PM ˆ 2πinx ordering, the map need not be injective or surjective, but an Fourier series n=−M f(n)e converges uniformly to ordering is required to be injective. f(x) as M → ∞, and we can write 3-ary Monotone Ordering (3-MO) ∞ An instance of the X problem, G, is a set of points V along with a (weighted) f(x) = fˆ(n)e2πinx . set E of 3-tuples of V .A 3-tuple (u, v, w) in E is said to n=−∞ be satisfied by an ordering O if O(u) < O(v) < O(w). We will only encounter such functions in our application. The quantity Val(G) refers to the maximum over orderings In this case, for such functions f and g, Parseval’s identity of V the (weighted) fraction of satisfied constraints. We will says that use Val(G, O) to denote the fraction satisfied by a particular ∞ Z X ordering O. f(x)g(x)dx = fˆ(n)gˆ(n) , (1) 1 The problem can be naturally extended to t-orderings as S n=−∞ follows. A t-ordering O can be extended to a (total) ordering, which in particular implies that if |f(x)| 1 for every x, O0, by ordering within the t partitions randomly, while 6 then P∞ |fˆ(n)|2 1. If f, g : S1 → are continuous retaining the natural order amongst the partitions. Define n=−∞ 6 R functions, then their convolution f ∗ g is a continuous the payoff of a tuple (u, v, w) in E in the t-ordering O function given by is to be the probability that O0(u) < O0(v) < O0(w). For Z example, if O(u) = O(v) < O(w), then the tuple is ordered (f ∗ g)(r) = f(r − u)g(u) du . correctly with probability half and hence the payoff is 1/2. S1 The quantity Val (G) refers to the maximum expected total t Its Fourier coefficients are given by f[∗ g(n) = fˆ(n)ˆg(n). payoff (where the expectation is taken over choice of a The Fourier coefficients of the function h(x) = f(−x) are random tuple from G) over all t-orderings. given by hˆ(n) = fˆ(−n). We will be interested in general arity 3 ordering problems where the constraints may accept an arbitrary subset of B. Noise Operators and Influences the permutations of 3 elements. For a permutation π of Let Ω denote the finite probability space correspond- 3 elements, the payoff of a tuple (u, v, w) with respect ing to the uniform distribution over [m]. Let {χ0 = to π in a t-ordering O is the probability that the random 0 1, χ1, χ2, . . . , χm−1} be an orthonormal basis for the space extension O orders the tuple according to π. The quantity R Q (k) L2(Ω). For σ ∈ [m] , define χσ(z) = χσ (z ). Valπ(G) refers to the maximum expected payoff with respect k∈[R] i t Every function F :ΩR → can be expressed as a to π in a t-ordering. The quantity Val (G), then, is a short- R t multilinear polynomial as F(z) = P Fˆ(σ)χ (z). The L form for Valid(G) where id is the identity permutation. As σ σ 2 t norm of F in terms of the coefficients of the multilinear before, we will use Valπ(G, O) and Val (G, O) to denote t t polynomial is ||F||2 = P Fˆ2(σ). the value obtained by a particular ordering O. Finally, the 2 σ π R quantity Val (G) is simply the maximum expected payoff Definition 4. For a function F :Ω → R, define Infk(F) = P 2 with respect to π in any ordering. The following observations E [Var (k) [F]] = Fˆ (σ). z z σk6=0 Here Varz(k) [F] denotes the variance of F(z) over the D. Influential variables for orderings th (k) choice of the k coordinate z . We will use the notion of influence for orderings first R introduced in [6]. Definition 5. For a function F :Ω → R, define the function TρF as follows: Definition 10. Given an ordering O of vertices V , its t- coarsening is a t-ordering O∗ obtained by dividing O into X |σ| ˆ TρF(z) = E[F(z˜) | z] = ρ F(σ)χσ(z) t contiguous blocks, and assigning label i to vertices in the σ∈[m]R ith block. Formally, if M = |V |/t then (k) (1) (R) j|{v|O(v) < O(u)}|k where each coordinate z˜ of z˜ = (˜z ,..., z˜ ) is equal O∗(u) = + 1 to z(k) with probability ρ and with the remaining probability, M z˜(k) is a random element from the distribution Ω. For an ordering O of points in [m]R, Define functions [p,q] R The following result is established in [6] using the Majority FO :[m] → {0, 1} for integers p, q as follows: is Stablest theorem (see Theorem 4.4 in [14]). ( [p,q] 1 if O(x) ∈ [p, q] FO (x) = Lemma 6. For every ε > 0, there exists a µ0 > 0 0 otherwise for which the following holds for all µ µ . Suppose 6 0 [p,q] [p,q] F, G :[m]R → [0, 1] are two functions with E[F] = We will omit the subscript and write F instead of FO , when it is clear. E[G] = µ, and Infk(T1−εF) 6 τ, Infk(T1−εG) 6 τ R for all k ∈ [R]. Let x, y be random vectors in [m] Definition 11. For an ordering O of [m]R, define the set of R whose marginal distributions are uniform over [m] but are influential coordinates Sτ (O) as follows: arbitrarily correlated. Then [p,q] Sτ (O) = {k | Infk(T1−εF ) 6 τ for some p, q ∈ Z} 1+ε/2 Ex,y[T1−2εF(x)T1−2εG(y)] µ + oτ (1) . 6 An ordering O is said to be τ-pseudo-random if Sτ (O) is empty. Lemma 7. Given a function F :[m]R → [0, 1], then PR 1 1 k=1 Infk(T1−εF) 6 e ln 1/(1−ε) 6 ε Lemma 12 (Few Influential Coordinates). [6] For any R 400 ordering O of [m] , we have |Sτ (O)| 6 ετ 3 C. Unique Games Claim 13. For any τ-pseudo-random ordering O of [m]R, Definition 8. An instance of Unique Games represented as its t-coarsening O∗ is also τ-pseudo-random. Υ = (A ∪ B,E, Π, [R]), consists of a bipartite graph over [·,·] Proof: {F ∗ } node sets A,B with the edges E between them. Also part of Since the functions O are a subset of the [·,·] ∗ the instance is a set of labels [R] = {1,...,R}, and a set of functions {FO }, Sτ (O ) ⊆ Sτ (O). permutations πab :[R] → [R] for each edge e = (a, b) ∈ E. IV. COARSENING GAP INSTANCES FOR 3-MONOTONE An assignment Λ of labels to vertices is said to satisfy an ORDERING edge e = (a, b), if π (Λ(a)) = Λ(b). The objective is to ab Definition 14 ((η, t)-Coarsening Gap). A (weighted) 3-MO find an assignment Λ of labels that satisfies the maximum instance over [m], G = ([m],E) is a (η, t)-coarsening gap number of edges. if: For a vertex a ∈ A ∪ B, we shall use N(a) to denote • Completeness: Pr(i,j,k)∈E[i < j < k] > 1 − η. In other its neighborhood. For the sake of convenience, we shall use words, Val(G) > 1 − η and in particular is obtained by the following version of the Unique Games Conjecture [12] the obvious ordering of [m]. which is equivalent to the original conjecture [10]. • Soundness: For all permutations π of 3 elements, π 1 Conjecture 9. (Unique Games Conjecture [12], [10]) For |Valt (G) − 6 | 6 η all constants δ > 0, there exists large enough constant R Note that it is easy to modify the construction to obtain a such that given a bipartite unique games instance Υ = (A∪ coarsening gap instance with perfect completeness—where B,E, Π = {π :[R] → [R]: e ∈ E}, [R]) with number of e Val(G) is 1—by simply throwing away the η fraction of bad labels R, it is NP-hard to distinguish between the following tuples. This however does not help in the hardness reduction two cases: as the unique games instance has imperfect completeness. • (1−δ)-satisfiable instances: There exists an assignment Moreover, the construction is more natural if allowed to have Λ of labels such that for 1−δ fraction of vertices a ∈ A, imperfect completeness. all the edges incident at a are satisfied. The main result of this section is the following theorem • Instances that are not δ-satisfiable: No assignment which constructs (η, t)-coarsening gap for arbitrarily large t satisfies more than a δ-fraction of the constraints Π. and arbitrarily small η. Theorem 15. For every integer t > 0 and η > 0, there The following lemma and Lemma 20 are at heart of argu- exists an m = m(η, t) such that there is a (η, t)-coarsening ing the soundness property (Definition 16) of the coarsening gap instance for 3-MO over [m]. gap instance for t-orderings. This may not be apparent, but will be in the proof of Lemma 23. A. Continuous gap Instances over S1 We will prove Theorem 15 by first constructing a con- Lemma 19. For large enough T , for all continuous func- tions f, g, h : S1 → [0, 1] with continuous derivatives except tinuous analogue over the unit circle S1 = R/Z. We will later discretize the instance to obtain the necessary result. In at finitely many points: particular, we will analyze the following distribution over 3- h i 1 tuples of points on S1. E f(x) g(y)h(z) − h(y)g(z) O (x,y,z)∈DT 6 ln T Definition 16. For every T > 0, let DT be the following distribution over S1 × S1 × S1: Proof: We have 1 • Pick x ∈ S uniformly at random E[f(x){g(y)h(z) − g(z)h(y)}] • Pick ∆ ∈ [−T, 0] uniformly at random Z 1 "Z 0 Z 0 ln T ∆−T es ∆ • Pick s ∈ [− , 0] uniformly at random = f(x) g(x + e )h(x + e ) 2 ln T 0 −T − ∆−T es ∆ 2 • Output (x, x + e , x + e ) s d∆ 2ds − h(x + e∆−T e )g(x + e∆ dx Remark 17 (Relation to CMM instance). A directed graph T ln T with m vertices with properties similar to the CMM instance, namely a large gap between the MAS value for m-orderings We will bound the inner integrals for every x. Define the and t-orderings for t m, can be obtained by a suitable functions g1, h1 : R → [0, 1] as discretization (see Section IV-B below) of the following T (y−1) T (z−1) distribution on S1 × S1: Pick x ∈ S1 uniformly at random, g1(y) = g(x + e ) and h1(z) = h(x + e ) . pick s ∈ [− ln T, 0] at random, and output (x, y = x + es). This instance has roughly the same mass in all (geometric) (Note these are defined with domain R and not S1.) Now, “scales” of the jump between x and y. The fact that t- setting u = 1 + ∆/T and r = es, for each x the inner orderings have MAS value close to 1/2 follows from the integral can be written as: bound Z 1 Z 1 2dr 1 Ix = g1(u−r)h1(u)−g1(u)h1(u−r) du s s 1 Ex,s g(x)h(x + e ) − h(x + e )g(x) 6 O (2) 0 − √ r ln T ln T T 1 for continuous functions g, h : S → [0, 1], via an argument 1 Define functions g0, h0 : S → [0, 1] as g0(θ) = g1(θ) similar to (and in fact simpler than) Lemma 23 below. A and h0(θ) = h1(θ) where the argument θ in the right hand bound similar to (2) is at the heart of the proof of Lemma 19 side (i.e., for g1, h1) is treated as a real in [0, 1). Let below. The distribution in Definition 16 behaves similar to a Z Z 1 0 2dr I = [g0(u−r)h0(u)−g0(u)h0(u−r)]du . CMM instance even on fixing x (see Lemma 19). Intuitively, x 1 u∈S1 − √ r ln T this means that any t-ordering of the discretized instance T cannot order the tuples significantly more or less often as 0 i < j < k i < k < j In other words, Ix is the integral of the same integrand as than as . Interestingly, as shown 1 in the harder Lemma 20, a similar property is true even in Ix but with operations over the unit circle S . Note that when fixing z. These two properties are enough to argue the expressions are different exactly when u − r < 0. Since about every permutation of the tuples (i, j, k) as is shown each integrand is at most 1 in absolute value, we can bound in Lemma 23. Z 1 Z 1 0 2dr |Ix − I | 2 du x 6 1 The proof of the following integral is in Appendix A. 0 max{− √ ,u} r ln T T √ Lemma 18. Let m be an integer and a, b be such that 0 < Z 1/ T Z 1 + + 4 a < b < 1. Let ψ : R → R be a continuous, non- 6 2 du + √ − ln u du increasing function that is at least 1 in the interval [a, b]. 0 ln T 1/ T Then, 2 4 5 6 √ + 6 (3) Z b sin(2πmx) T ln T ln T
I = dx 6 14 . a ψ(x)x 0 for large enough T . To bound Ix, we use Fourier analysis over S1 and proceed as follows: wrap-around the unit interval, the tuple (i, j, k) will be ZZ 1 ordered as i < j < k. Hence, 0 2dr Ix = g0(u − r)h0(u) − g0(u)h0(u − r) du √1 r ln T Wt(i < j < k) = Wt(i < k) > Pr [x < z] T (x,y,z)←DT Z 1 ∞ X 2πinr −2πinr 2dr (As i < k implies i < j < k in the construction) = g0(−n)hc0(n) e − e √1 b r ln T Z 0 −T n=−∞ ∆ d∆ 1 − e 1 T > 1 − e = 1 − > 1 − (using transform identity for convolution) −T T T T 4i X Z 1 sin(2πnr) = g0(−n)hc0(n) dr . ln T b √1 r n T Lemma 23. For every positive integer t, for large enough m, T , the following holds for every π ∈ S : Using Cauchy Schwarz, Parseval’s identity, and Lemma 18, 3 we get t3 |Valπ(Gm) − 1 | O s t T 6 6 ln T 0 56 X 2 X 2 1 |I | |g0(n)| |hc0(n)| O (4) x 6 ln T b 6 ln T n n Proof: Fix a t-ordering O. We will bound the differ- π1 m π2 m ence between Valt (GT , O) and Valt (GT , O) for every Combining (3) and (4), we have the desired bound: two permutations π1, π2. Z 1 Let P = {P ,P ...P } be the partition of [m] into t 1 2 t E f(x) g(y)h(z) − g(z)h(y) = f(x) · Ix dx pieces induced by O. Abusing notation, we will use P (i) 0 to denote the partition to which i belongs and P to denote Z 1 1 a 6 |f(x)||Ix|dx 6 O the indicator of the part Pa. 0 ln T m 123 m Valt(GT , O) = Valt (GT , O) = Pr[P (i) < P (j) < P (k)] 1 + Pr[P (i) = P (j) < P (k)] + Pr[P (i) < P (j) = P (k)] The proof of the following lemma is more involved than 2 the above proof, and is deferred to Appendix A. 1 + Pr[P (i) = P (j) = P (k)] Lemma 20. For all large enough T , for all continuous 6 1 X 1 X functions f, g, h : S → [0, 1] with continuous derivatives = E[P (i)P (j)P (k)] + E[P (i)P (j)P (k)] a b c 2 a a c except at finitely many points: a
E(i,j,k) Pa(i)Pb(j)Pc(k) − Pa(i)Pb(k)Pc(j) and Lemma 26 below, whose proofs parallel those of similar statements in [6]. = E(x,y,z)←D fa(x) fb(y)fc(z) − fc(y)fb(z) T Lemma 25. For every ε > 0, there exists sufficiently large 1 m, t such that : For any τ-pseudo-random ordering O of O 6 ln T [m]R and permutation π,
123 m 132 m ε We therefore conclude |Val (G , O)−Val (G , O)| π π ∗ − 2 t T t T 6 Val (O) 6 Valt (O ) + O(t ) + oτ (1) t3 O ln T . An almost identical proof bounds the differences 213 m 312 m 231 m where O∗ is the t-coarsening of O. |Valt (GT , O) − Valt (GT , O)| and |Valt (GT , O) − 321 m Val (G , O)|. Using a similar argument and em- ∗ π ∗ t T Proof: As O is a coarsening of O, clearly Val (O ) 6 ploying Lemma 20 instead of Lemma 19, the differ- Valπ(O) 123 m 213 m 132 m t . Note that the loss due to coarsening, is because ences |Valt (GT , O)−Valt (GT , O)|, |Valt (GT , O)− e = (u, v, w) 231 m 312 m 321 m some tuples which are oriented correctly in Valt (GT , O)|, |Valt (GT , O)−Valt (GT , O)| can also O O∗(u) 3 , fall into the same block during coarsening, i.e , t ∗ ∗ be bounded by O ln T from above. O (v), O (w) are not all distinct. Thus we can write π1 m π2 m Thus, for all π1, π2, |Valt (GT , O) − Valt (GT , O)| 6 3 t Val(O) Val (O∗) + Pr O∗(z˜ ) = O∗(z˜ ) O ln T . Now, Observation 3 gives the required result. 6 t u v 3 ∗ ∗ ∗ ∗ t + Pr O (z˜u) = O (z˜w) + Pr O (z˜v) = O (z˜w) Given (η, t), setting T = exp(Ω( η )) and setting m = t3 exp(exp(Ω( η ))) and using Lemma 22 and Lemma 23 proves Theorem 15. ∗ ∗ Pr O (z˜u) = O (z˜v) V. DICTATORSHIP TEST X h i i i = E Ez ,z Ez˜ ,z˜ F (z˜u) ·F (z˜v) Let G = (V,E) be a (η, t)-coarsening gap instance for e=(u,v) u v u v i∈[t] the 3-MO. Without loss of generality, V = [m] for an X h i i i m divisible by t. Using G, construct a (weighted) 3-MO = Ee=(u,v)Ezu,zv T1−2εFu(zu) · T1−2εFv(zv) R dictatorship test instance DICTG on orderings O of [m] as i∈[t] follows: As O is a t-coarsening of O, for each value i ∈ [t], there 1 z O∗(z) = i DICT are exactly t fraction of for which . Hence for G Test: i 1 each i ∈ [t], Ez[Fu(z) = ]. Further, since the ordering • Pick a tuple e = (u, v, w) ∈ E at random. t O∗ is τ-pseudo-random, for every k ∈ [R] and i ∈ [t], • Pick ze = (zu, zv, zw) to be k random shifts of e. i Infk(T1−εFa) 6 τ. Hence using for sufficiently large t, the In other words, pick k random integers r1, r2 . . . rk −1− ε i i i above probability is bounded by t · t 2 + t · oτ (1) = from [m] and set zu = u + ri, zv = v + ri, zw = − ε O(t 2 ) + o (1). The same bound holds for the other two w + r (mod m). τ i probabilities too, hence giving the required result. • Obtain z˜u, z˜v, z˜w by perturbing each coordinate of zu, zv, and zw independently. Specifically, sample Lemma 26. For every choice of m, t, ε, and any τ-pseudo- th (k) (k) ∗ R π ∗ π the k coordinates z˜u , z˜v as follows: With prob- random t-ordering O of [m] , Valt (O ) 6 Valt (G) + (k) (k) (k) ability (1 − 2ε), z˜u = zu , and otherwise z˜u is oτ (1). a new sample from V . Proof: Deferred to Appendix A. • Output tuple (z˜u, z˜v, z˜w).
VI.HARDNESS REDUCTION
Theorem 24. (Soundness Analysis) For every ε > 0, for Let G = (V,E) be a (η, t)-coarsening gap instance, and any τ-pseudo-random ordering O of [m]R, let m = |V |. Let Υ = (A ∪ B,E, Π = {πe :[R] → [R]|e ∈ ε π π − 2 E}, [R]) be a bipartite unique games instance. Towards Val (O) 6 Valt (G) + O(t ) + oτ (1) constructing a 3-MO instance G = (V, E) from Υ, we shall for all permutations π, where oτ (1) → 0 as τ → 0 keeping introduce a long code for each vertex in B. Specifically, the all other parameters fixed. set of vertices V of G is indexed by B × [m]R. Hardness Reduction: Theorem 29 (Main). Conditioned on the Unique Games Input : Unique games instance Υ = (A ∪ B,E, Π = conjecture, for every Λ ⊆ Π ⊆ S3, given an instance of {πe :[R] → [R]|e ∈ E}, [R]) and a (η, t) coarsening permCSP(Λ) that is 1 − ε satisfiable, it is hard to find an gap instance G = (V,E). ordering that satisfies more than |Π|/6+ε of the constraints Output : 3-MO instance G = (V, E) with set of vertices : even under the relaxed constraint Π (for arbitrary ε > 0). V = B ×[m]R and edges E given by the following verifier: VII.ACKNOWLEDGMENTS • Pick a random vertex a ∈ A. Choose three neighbors b, b0, b00 ∈ B independently at random. Let π, π0, π00 We thank Prasad Raghavendra for several useful discus- denote the permutations on the edges (a, b), (a, b0) sions and collaboration during early stages of this research. and (a, b00). REFERENCES • Pick a tuple e = (u, v, w) ∈ E at random from the coarsening gap instance G. [1] P. Austrin. Towards sharp inapproximability for any 2-csp. • Sample ze = {zu, zv, zw} using k random shifts of In FOCS, pages 307–317. IEEE Computer Society, 2007. z as in the dictatorship test. e [2] M. Charikar, K. Makarychev, and Y. Makarychev. On the • Obtain z˜u, z˜v, z˜w by perturbing each coordinate of zu, advantage over random for maximum acyclic subgraph. In zv, and zw independently. FOCS, pages 625–633. IEEE Computer Society, 2007. 0 0 00 00 • Output tuple ((b, π(z˜u)), (b , π (z˜v)), (b , π (z˜w))). [3] B. Chor and M. Sudan. A geometric approach to betweenness. The proof of the following theorem follows from the sound- SIAM Journal on Discrete Mathematics, 11(4):511–523, Nov. ness analysis (Theorem 24) of the dictatorship test and using 1998. techniques similar to [6] and is omitted. [4] L. Engebretsen and V. Guruswami. Is constraint satisfaction Theorem 27. For every γ > 0, there exists choice of over two variables always easy? Random Struct. Algorithms, 25(2):150–178, 2004. parameters ε, η, t, δ such that: • COMPLETENESS: If Υ is a (1 − δ)-satisfiable instance [5] M. X. Goemans and D. P. Williamson. Improved ap- of Unique Games, then there is an ordering O for the proximation algorithms for maximum cut and satisfiability graph G with value at least (1−γ). i.e. Val(G) 1−γ problems using semidefinite programming. Journal of the 6 ACM, 42(6):1115–1145, 1995. • SOUNDNESS: If Υ is not δ-satisfiable, then no ordering 1 to G has value more than 6 + γ with respect to any [6] V. Guruswami, R. Manokaran, and P. Raghavendra. Beating π permutation π, i.e, for all permutations π, Val (G) 6 the random ordering is hard: Inapproximability of maximum 1 + γ acyclic subgraph. In FOCS, pages 573–582. IEEE Computer 6 Society, 2008. Given a hard instance G with the above properties, we can give an easy reduction to a hard instance for permCSP(Λ). [7] G. Hast. Beating a Random Assignment. PhD thesis, Royal In fact, in the soundness case we can even allow a more Institute of Technology, Stockholm, Sweden, 2005. relaxed set of constraints Π ⊇ Λ, and it will still be hard to [8] J. Hastad.˚ Every 2-csp allows nontrivial approximation. In outperform a random ordering. H. N. Gabow and R. Fagin, editors, STOC, pages 740–746. ACM, 2005. Lemma 28. For every Λ ⊆ Π ⊆ S3, there is a polynomial time reduction from 3-MO instance, G to permCSP(Λ) [9] J. Hastad.˚ On the approximation resistance of a random instance H such that: predicate. In M. Charikar, K. Jansen, O. Reingold, and J. D. P. Λ Rolim, editors, APPROX-RANDOM, volume 4627 of Lecture • COMPLETENESS: Val (H) > Val(G). π 1 γ Notes in Computer Science, pages 149–163. Springer, 2007. • SOUNDNESS: If, for some γ, Val (G) 6 6 + 6 for all Π |Π| permutations π, then Val (H) 6 6 + γ. [10] S. Khot. On the power of unique 2-prover 1-round games. In IEEE Conference on Computational Complexity, page 25, Proof: Given Λ ⊆ Π ⊆ S3 and a 3-MO instance G, 2002. pick an arbitrary permutation π ∈ Λ and permute the tuples of G by π to obtain the instance H. It is easy to see that for [11] S. Khot, G. Kindler, E. Mossel, and R. O’Donnell. Optimal 0 0 every permutation π0, Valπ◦π (H) = Valπ (G). In particular, inapproximability results for max-cut and other 2-variable Λ π csps? SIAM J. Comput., 37(1):319–357, 2007. since π ∈ Λ, Val (H) > Val (H) > Val(G). Similarly, if π0 1 γ 0 Π Val (G) 6 6 + 6 for all permutations π , then Val (H) = [12] S. Khot and O. Regev. Vertex cover might be hard to P π0 |Π| approximate to within 2-epsilon. J. Comput. Syst. Sci., π0∈Π Val (H) 6 + γ 6 74(3):335–349, 2008. Our final result follows immediately follows from Theo- rem 27 and Lemma 28. [13] T. Korner.¨ Fourier Analysis. Cambridge University Press, 1988. [14] E. Mossel, R. O’Donnell, and K. Oleszkiewicz. Noise as follows: stability of functions with low influences: invariance and 1 optimality. In FOCS, pages 21–30. IEEE Computer Society, k+ 2 Z k+1 sin(2πmx) Z m 1 2005. m dx = sin(2πmx) k xψ(x) k xψ(x) m m [15] P. Raghavendra. Optimal algorithms and inapproximability 1 − dx results for every csp? In STOC ’08: Proceedings of the 40th ψ(x + 1 )(x + 1 ) ACM Symposium on Theory of Computing, 2008. 2m 2m 1 k+ Z 2 m 1 1 [16] U. Zwick. Approximation algorithms for constraint satisfac- − dx 6 k xψ(x) ψ(x + 1 )(x + 1 ) tion problems involving at most three variables per constraint. m 2m 2m In SODA, pages 201–210, 1998. 1 k+ 2 Z m dx 1 1 max , 6 k 2 1 2mx ψ(x) ψ(x + 2m ) APPENDIX m 1 k+ 2 Z m dx 1 A. Omitted proofs 6 k 2mx2 6 4k2 m Lemma 30 (Lemma 18). Let m be an integer and a, b 1 P∞ 1 π2 We conclude that I 6 4π + 4 k=1 k2 6 4π + 24 6 14. be such that 0 < a < b < 1. Let ψ : R+ → R+ be a continuous, non-increasing function that is at least 1 in the Lemma 31 (Lemma 20). For all large enough T , for all interval [a, b]. Then, continuous functions f, g, h : S1 → [0, 1] with continuous derivatives except at finitely many points: Z b sin(2πmx) h i 1 E h(z) f(x)g(y) − f(y)g(x) O . I = dx 6 14 . (x,y,z)∈DT 6 ln T a ψ(x)x ∆ Proof: The distribution Dm is equivalent to (z−e , z− ∆ ∆−T es 1 Proof: w.l.o.g., we can assume m is positive. Set l = e + e , z) where z ∈ S is chosen uniformly at dmae and n = bmbc. If l > n, the interval [a, b] does not random, and ∆, s are chosen as in Dm. Thus, the expectation can be written as follows: contain any multiple of 1/m and thus b − a 6 1/m. In this case the integral is at most h i E h(z) f(x)g(y) − g(x)f(y) Z b Z b sin(2πmx) sin(2πmx) Z 1 "Z 0 Z 0 I = dx dx s xψ(x) 6 xψ(x) = h(z) f(z − e∆)g(z − e∆ + e∆−T e ) a a ln T 0 −T − b 2 Z |2πmx| s d∆ 2ds 6 dx 6 2πm(b − a) 6 2π − g(z − e∆)f(z − e∆ + e∆−T e ) dz a xψ(x) T ln T As in the previous lemma, we will bound the inner integral Therefore, assume l 6 n. The absolute value of the integral ∆ −T es ∆−rT can be bounded as: for every z. Write e (1 − e ) as e . For large T , s 0 − ln T r r ≈ e−T /T as varies√ from to 2 , varies from 0 to − T l k+1 r1 ≈ e /T . Z n−1 Z s m sin(2πmx) X m sin(2πmx) Differentiating e−rT = 1 − e−T e , we can compute the I 6 dx + dx ψ(x)x k ψ(x)x probability density function of r, and it is given by a k=l m Z b 2T e−rT dr 2ds sin(2πmx) −p(r) dr = = + dx −rT 1 ln T n ψ(x)x ln T (1 − e ) ln( 1−e−rT ) m
n−1 Z k+1 (The negative sign in front of p(r)dr is because r decreases X m sin(2πmx) 4π + dx when s increases.) 6 k ψ(x)x k=l m We will bound the inner integral above for every z. Now, as in the proof of Lemma 19, define functions f1, g1 : R → [0, 1] by Using the identity sin(π +x) = − sin(x), we can split the T (x−1) T (y−1) interval [k/m, (k + 1)/m] into two halves and group terms f1(x) = f(z − e ) and g1(y) = g(z − e ) 1 (Note these are defined with domain R and not S .) Setting where the argument θ in the right hand side (i.e., for f1, g1) u = 1 + t/T , the inner integral can be written as is treated as a real between 0 and 1. Now let 1 r Z Z 1 ZZ r1 2 f0(u)g0(u − r) − g0(u)f0(u − r) dudr 00 Iz = f1(u)g1(u − r) − g1(u)f1(u − r) du p(r)dr Iz = 1 0 r 0 r0 ln(T )r ln( rT ) √ −T − T To bound I00, as in the proof of Lemma 19, we use Further, r varies from r0 ≈ e /T to r1 ≈ e /T ; z 1 hence for large enough T , 1−e−rT ' rT . More formally, we Fourier analysis over S , transform identity for convolutions, will approximate p(r) by a simpler function for the purpose Lemma 18, Cauchy-Schwarz, and Parseval’s theorem, and of the analysis. Setting h = rT , we see that, proceed as follows: Z Z r1 −h 00 e 1 |Iz | = [f0(u)g0(u − r) − g0(u)f0(u − r)]du − S1 r −h 1 1 0 (1 − e ) ln( 1−e−h ) h ln( h ) 2 dr e−h e−h ln(T )r ln( 1 ) rT − ∞ 6 −h 1 1 Z r1 (1 − e ) ln( 1−e−h ) h ln( 1−e−h ) 4i X sin(2πnr) = g0(−n)fb0(n) 1 dr −h −h −h ln T b r ln( ) e e e 1 n=−∞ r0 rT + − + − 1 1 1 1 s h ln( 1−e−h ) h ln( h ) h ln( h ) h ln( h ) 56 X 2 X 2 1 |fb0(n)| |g0(n)| O 6 ln T b 6 ln T e−h 1 1 n n − 6 1 1 − e−h h √ ln( 1−e−h ) − ∆ Finally, since r is at most e /T , the√ difference between −h ! −h 00 0 − ∆ e 1 1 e − 1 the integrals I and I is also at most e /T as for a fixed + − + z z h 1 1 1 r, u < r with probability at most r. We therefore reach our ln( 1−e−h ) ln( h ) h ln( h ) desired conclusion h i Z 1 For our setting of parameters, we have h/2 1 − e−h h E h(z) f(x)g(y) − g(x)f(y) = h(z) Iz dz −h 6 6 0 1−e 1 and | ln( h )| 6 5h. Hence, Z 1 |h(z)| |I | dz O 6 z 6 ln T e−h 1 0 − −h 1 1 (1 − e ) ln( 1−e−h ) h ln( h ) e−h h − 1 − e−h Lemma 32 (Lemma 26). For every choice of m, t, ε, and 5 ∗ R π ∗ 6 1 2 any τ-pseudo-random t-ordering O of [m] , Valt (O ) 6 ln( h ) h π Valt (G) + oτ (1). −h 1−e−h ! −h e ln( h ) e − 1 10 + + 6 Proof: We will look at the case where π is the identity h ln2( 1 ) h ln( 1 ) ln( 1 ) h h h permutation. The proof for the other orderings is almost Thus, identical. The t-ordering problem is a CSP over a finite domain, and is thus amenable to techniques of [15]. Specif- 2 40T 40T P :[t]3 → [0, 1] p(r) − . ically, consider the payoff function defined 1 6 1 6 ln(T )r ln( rT ) ln T ln( rT ) ln T by: P (i, j, k) = 1 for i < j < k, P (i, j, k) = 1/2 for i = j < k and i < j = k, P (i, j, k) = 1/6 for i = j = k Hence, it is enough to bound the simpler integral and P (i, j) = 0 otherwise. The t-ordering problem (with respect to the identity permutation) is a Generalized CSP(see Z 1 Z 2 f1(u)g1(u − r) − g1(u)f1(u − r) dudr Definition 3.1, [15]) with the payoff function P . I0 = z 1 For the sake of exposition, let us pretend that t = m. 0 ln(T )r ln( rT ) The dictatorship test we construct essentially picks a random More precisely, the difference between the two integrals is integral solution from the m different integral solutions at most: (corresponding to the m cyclic shifts of 1 . . . m) to perform 200 √ 1 the test. This is exactly the same as the t-ordering CSP |I − I0 | e− T O z z 6 ln T 6 ln T dictatorship DICTG obtained by running the reduction of [15] on the trivial SDP solution corresponding to the convex As in the proof of the previous lemma, define functions combination of the integral solutions. Further, all the SDP 1 f0, g0 : S → [0, 1] as f0(θ) = f1(θ) and g0(θ) = g1(θ) constraints are satisfied as the SDP solution is integral. ∗ A t-ordering solution O for the dictaDICTG corresponds Hence the above lemma is just a restatement of Corollary R naturally to a function F :[t] → Simt. Thus, we have the 2.2 of [15] for the specific generalized CSP – t-Ordering – following observations: albeit in the language of τ-pseudo-random orderings.
• The t-ordering instance DICTG is identical to the Recall that the actual case of interest here satisfies t < m. dictatorship test described in this section when t = m. Unfortunately, in this case, a black box application of the ∗ • For a τ-pseudo-random t-ordering O , for every k ∈ result from [15] does not suffice. However, the proof in [R] and i ∈ [t], the corresponding function F sat- [15] can be easily adopted without any new technical ideas. i In fact, many of the technical difficulties encountered in isfies Infk(T1−εF ) 6 τ. In the terminology of [15](Definitions 4.1 and 4.2), this is equivalent to [15] can be avoided here. For instance, the SDP solution the function F = (F 1,..., F t) being “(γ, τ)-pseudo- associates with each vertex u, the uniform probability dis- random” with γ = 0. tribution over {1 . . . m} (due to the cyclic shifts), unlike [15] • By Corollary 2.2 in [15], for a (γ, τ)-pseudo-random where there are several arbitrary probability distributions to function F, its probability of acceptance on the dicta- deal with. With the value of m fixed, this removes the need torship test is at most Valt(G) + oγ,τ (1). for smoothing the SDP solution (Lemma 3.4 in [15]).