Graph Expansion and the Unique Games Conjecture

∗ Prasad Raghavendra David Steurer MSR New England Princeton University Cambridge, MA Princeton, NJ

ABSTRACT General Terms The edge expansion of a subset of vertices S ⊆ V in a graph Algorithms, Theory. G measures the fraction of edges that leave S. In a d-regular Keywords graph, the edge expansion/conductance Φ(S) of a subset S ⊆ V is defined as Φ(S) = |E(S,V \S)| . Approximating the Graph Expansion, Unique Games Conjecture, Hardness of d|S| Approximation, Spectral profile. conductance of small linear sized sets (size δn) is a natural optimization question that is a variant of the well-studied Sparsest Cut problem. However, there are no known al- 1. INTRODUCTION gorithms to even distinguish between almost complete edge The phenomenon of vertex and edge expansion in graphs expansion (Φ(S) = 1 − ε), and close to 0 expansion. has been a subject of intense study with applications pervad- In this work, we investigate the connection between Graph ing almost all branches of theoretical computer science. From Expansion and the Unique Games Conjecture. Specifically, an algorithmic standpoint, approximating expansion or lack we show the following: thereof (finding good cuts or separators) is a fundamental optimization problem with numerous applications. Yet, the –We show that a simple decision version of the problem of computational complexity of detecting and approximating approximating small set expansion reduces to Unique expansion in graphs is not very well understood. Games. Thus if approximating edge expansion of small Among the two notions of expansion, this work will con- sets is hard, then Unique Games is hard. Alternatively, cern mostly with edge expansion. For simplicity, let us first a refutation of the UGC will yield better algorithms to consider the case of a d-regular graph G = (V,E). The edge approximate edge expansion in graphs. expansion of a subset of vertices S ⊆ V measures the fraction This is the first non-trivial “reverse” reduction from a of edges that leave S. Formally, the edge expansion Φ(S) of natural optimization problem to Unique Games. a subset S ⊆ V can be defined as, –Under a slightly stronger UGC that assumes mild expansion |E(S, V \ S)| ΦG(S) = , of small sets, we show that it is UG-hard to approximate d|S| small set expansion. where E(S, V \ S) denotes the set of edges with one endpoint –On instances with sufficiently good expansion of small sets, in S and the other endpoint in V \ S. The conductance we show that Unique Games is easy by extending the or the Cheeger’s constant associated with the graph G is techniques of [4]. the minimum of Φ(S) over all sets S with at most half the vertices, i.e.,

Categories and Subject Descriptors ΦG = min ΦG(S) . |S|6n/2 F.2.2 [Theory of Computation]: Analysis of Algorithms The definitions of conductance of sets Φ (S) and the graph and Problem Complexity—Nonnumerical Algorithms and G Φ can be extended naturally to non-regular graphs, and Problems G finally to arbitrary weighted graphs. For the sake of sim- ∗Supported by NSF grants 0830673, 0832797, 528414. Part plicity, we restrict our attention to regular graphs here, and of this work done while at Microsoft Research New England. defer the discussion of general weighted graphs to the full version. Henceforth, we will use the notation µ(S) to denote the normalized set size µ(S) = |S|/n. The problem of approximating ΦG also referred to as the Permission to make digital or hard copies of all or part of this work for the uniform Sparsest Cut (equivalent within a factor of personal or classroom use is granted without fee provided that copies are 2), is among the fundamental problems in approximation not made or distributed for profit or commercial advantage and that copies algorithms. Efforts towards approximating ΦG have led to bear this notice and the full citation on the first page. To copy otherwise, to a rich body of work with strong connections to spectral republish, to post on servers or to redistribute to lists, requires prior specific techniques and metric embeddings. permission and/or a fee. STOC’10, June 5–8, 2010, Cambridge, Massachusetts, USA. The first approximation for conductance was obtained by Copyright 2010 ACM 978-1-4503-0050-6/10/06 ...$10.00. discrete analogues of the Cheeger inequality [9] shown by Alon-Milman [2] and Alon [1]. Specifically, they show the Unique Games Conjecture. following: The Unique Games Conjecture (UGC) of Khot [13] is among the central open problems in hardness of approximation, and Theorem 1.1 (Cheeger’s Inequality). If λ2 denotes has fueled many developments in the area in recent years. the second largest eigenvalue of the suitably normalized adja- The UGC is shown to imply optimal inapproximability results cency matrix of a graph G then, for classic problems like Max Cut [14], Vertex Cover [15] 1 − λ and Sparsest Cut [16] and constraint satisfaction problems 2 Φ p2(1 − λ ) . 2 6 G 6 2 [19]. For the sake of concreteness, we formally state the unique Since the second eigenvalue λ can be efficiently computed, 2 games conjecture here. The problem is Cheeger’s inequality yields an approximation algorithm for Unique Games defined as follows: ΦG, indeed one that is used heavily in practice for graph partitioning. However, the approximation for ΦG obtained Definition 1. An instance of Unique Games represented via Cheeger’s inequality is poor in terms of a approximation as Υ = (V, E, Π, [R]) consists of a graph over vertex set V with ratio, especially when the value of ΦG is small (λ2 is close the edges E between them. Also part of the instance is a set to 1). of labels [R] = {1,...,R}, and a set of permutations πv←w : An O(log n) approximation algorithm for ΦG was obtained [R] → [R] for each edge e = (w, v) ∈ E. An assignment A by Leighton and Rao [17]. Later work by Linial et al. [18] of labels to vertices is said to satisfy an edge e = (w, v), if and Aumann and Rabani [6] established a strong connection πv←w(A(w)) = A(v). The objective is to find an assignment between the Sparsest Cut problem and the theory of metric A of labels that satisfies the maximum number of edges. spaces, in turn spurring a large and rich body of literature. As is customary in hardness of approximation, one defines a More recently,√ in a breakthrough result Arora et al. [5] obtained an O( log n) approximation for the problem using gap-version of the Unique Games problem as follows: semidefinite programming techniques. Problem 2. (Unique Games (R, 1−ε, η)) Given a Unique Games instance Υ = (V, E, Π = {πv←w :[R] → [R] | e = Small Set Expansion. (w, v) ∈ E}, [R]) with number of labels R, distinguish be- Note that the ΦG is a fairly coarse measure of edge expan- tween the following two cases: sion, in that it is the worst case edge expansion over sets S of all sizes. In a typical graph (say a random d-regular –(1 − ε)- satisfiable instances: There exists an assignment A graph), smaller sets of vertices expand to a larger extent of labels that satisfies a 1 − ε fraction of edges. than sets with half the vertices. For instance, all sets S of –Instances that are not η-satisfiable: No assignment satisfies 2 δn vertices in a random d-regular graph have Φ(S) ≈ 1 − d more than a η-fraction of the edges E. with very high probability, while the conductance ΦG of the 1 The Unique Games Conjecture asserts that the above decision entire graph is roughly 2 . Moreover, the stronger expansion exhibited by small sets has numerous applications in graph problem is NP-hard when the number of labels is large enough. theory. Formally, A natural finer measure of the edge expansion of a graph Conjecture 1.2 (Unique Games Conjecture [13]). is its expansion profile. Specifically, for a regular graph G For all constants ε, η > 0, there exists large enough constant 1 the expansion profile is given by the curve R such that Unique Games (R, 1 − ε, η) is NP-hard. 1 ΦG(δ) = min Φ(S) ∀δ ∈ [0, /2] . While the implications of the conjecture are well-understood, µ(S)=δ there has been much slower progress towards its resolution. The problem of approximating the expansion profile is seem- Results supporting the truth of UGC have been especially ingly far-less tractable than approximating ΦG itself. For difficult to show. In particular, it is unknown whether many instance, there is no known algorithm for the following easily of the implications of UGC are equivalent to the conjecture. stated decision problem concerning the expansion profile: In other words, it is entirely consistent with existing litera- ture that all the implications of UGC on problems like Max Problem 1. Gap-Small-Set Expansion (η, δ) Given a Cut and Vertex Cover hold, but the conjecture itself is graph G and constants η, δ > 0, distinguish whether false. More precisely, although the Unique Games prob- lem is known to efficiently reduce to classic problems like ΦG(δ) > 1 − η or ΦG(δ) 6 η . Max Cut and Vertex Cover, there are no known “reverse” Spectral techniques fail in approximating the expansion reductions from these problems back to Unique Games. of small sets in graphs. On one hand, even with the largest The only reverse reduction towards which there is any possible spectral gap, the Cheeger’s inequality cannot yield literature is the reduction from Max Cut to Unique Games. 1 a lower bound greater than /2 for the conductance ΦG(δ). Note that the Max Cut problem is a special case of Unique More importantly, there exists graphs such as hypercube Games over the binary alphabet {0, 1}. Hence a Max Cut where there are sets S of half the vertices with small conduc- instance is readily reduced to a Unique Games instance by tance (Φ(S) < η), yet every sufficiently small set S satisfies using parallel repetition. With a sufficiently strong parallel Φ(S) > 1 − η. This implies that ΦG (and the second eigen repetition theorem (conjectured by Feige et al. [11]), this value λ2) do not yield any information about ΦG(δ) for small would yield a reverse reduction from Max Cut to Unique δ. Games. Unfortunately, a strong parallel repetition theorem of this nature was shown not to hold by Raz [24]. Subsequent 1 In irregular graphs, it is more convenient to permit sets S within a range of sizes say [δ, 10δ], since in an arbitrary (non- work by Barak et al. [7] almost entirely ruled out this regular) graph there could be no sets satisfying µ(S) = δ. approach to reduce Max Cut to Unique Games. Expansion and Unique Games. refutation of UGC was not known to formally imply any Vertex and edge expansion in graphs appear to be closely new algorithmic technique. An important implication of the tied to the hard instances for linear and semidefinite program- above result is that a refutation of the UGC would yield an ming relaxations. Many integrality gap instances have been algorithm for approximating edge expansion (only in a cer- constructed for problems like Vertex Cover or Max Cut tain regime) in graphs – a basic optimization problem. Now against the linear programming hierarchies such as Lov´asz– we state the main result of the paper from this algorithmic Schriver and Sherali–Adams hierarchies (see [8, 25] and the standpoint. To this end, we formally state a hypothesis that references therein). Not only do most of these instances would be emerge from a refutation of UGC. consist of expanding graphs but the arguments rely crucially on either vertex or edge expansion. Hypothesis 1.5 (Unique Games is easy). There ex- The situation is little bit more subtle in case of semidefinite ists a constant ε > 0 and a function f : N → N such that programming. Semidefinite programs can approximate Max given a Unique Games instance Υ with n vertices and k la- Cut well on instances that have very good conductance bels, it is possible distinguish between the cases opt(Υ) > 1−ε f(k) (spectral gap). Hence, the SDP integrality gap instances and opt(Υ) 6 ε in time n . known are graphs where small sets expand well, while the larger sets do not. Indeed, SDP integrality gap constructions Theorem 1.6. Suppose the above hypothesis ( Unique for Max Cut [10, 16, 21], Vertex Cover [12], Unique Games is easy) holds for a constant ε0 and a function f, Games [16, 20] and Sparsest Cut [16] all have near-perfect then there exists a function g : [0, 1] → N such that given edge expansion for small sets. any graph G with n vertices, and it is possible to distinguish In case of Unique Games, not only do all known inte- whether ΦG(δ) 6 ε1 or ΦG(δ) > 1 − ε1 for some absolute g(δ) grality gap instances have near-perfect edge expansion of constant ε1 in time n . small sets, even the analysis relies directly on this property. Furthermore, it is known that the best possible soundness A somewhat stronger consequence of the Unique Games is for Unique Games with label size R and completeness 1 − ε easy hypothesis follows using the parallel repetition theorem. is a constant η(R, ε) which is roughly 1 . The constant Rε/2 The following corollary is an immediate consequence of the 1 parallel repetition theorem of Rao [23] and our reduction 1 − η arises directly from the expansion of sets of size R in a certain graph defined over the Gaussian space. from Gap-Small-Set Expansion to Unique Games. While this suggests that Unique Games is closely tied to expansion of small sets in graphs, somewhat contrastingly, Corollary 1.7. Suppose the hypothesis – Unique Games Arora et al. [4] show that Unique Games is easy when the instance is easy) holds for a constant ε0 and a function constraint graph involved is a good spectral expander, i.e., f. Then, given a graph G with n vertices and parameters has a non-trivial spectral gap for the Laplacian. ε, δ such that ε < ε1 for some absolute constant ε1, we can distinguish the following cases in time ng(ε,δ) for some Motivated by the above reasons, we investigate the connec- 2 tion between graph expansion and Unique Games in this g : [0, 1] → N. work. 1. There exists S ⊆ V with µ(S) = δ and Φ(S) 6 ε 1.1 Results 2. Every√ set S ⊆ V with µ(S) 6 1500δ/ε satisfies Φ(S) > The main result of this work is a reduction from the prob- 1500 ε. lem of approximating expansion of small sets to the Unique Games problem. The main implication of the result can be 1.1.1 Towards an Equivalence succinctly stated as follows. A natural question that arises from Theorem 1.4 is whether Let us consider the following hardness assumption about the Unique Games Conjecture is equivalent to the Gap- the complexity of the Gap-Small-Set Expansion problem. Small-Set Expansion conjecture. More specifically, could it be true that Unique Games Conjecture implies the Gap- Conjecture 1.3. (Gap-Small-Set Expansion Conjecture) Small-Set Expansion conjecture. For every η > 0, there exists δ such that the problem Gap- Showing a result of this nature amounts to obtaining a Small-Set Expansion (η, δ) is NP-hard. reduction from Unique Games to Gap-Small-Set Expan- sion. Despite the large number of UG reductions and a fairly Then, an immediate consequence of the reduction pre- thorough understanding of how to construct reductions from sented in this work is, Unique Games, obtaining reductions to graph expansion Theorem 1.4. The Gap-Small-Set Expansion conjec- problems is often problematic. ture implies the Unique Games Conjecture. The main issue is that hardness reductions via local gadgets do not alter the global structure of the graph. For example, To the best of our knowledge, this is the first non-trivial if the Unique Games instance is disconnected, the resulting “reverse” reduction from a natural combinatorial optimization graph produced by a gadget reduction is also disconnected. problem to Unique Games. On one hand, it connects the Hence, to show a UG-hardness result for a global property somewhat non-standard problem of Unique Games to the such as expansion, it often seems necessary to assume the much well-studied problem of approximating graph expan- corresponding global property on the constraint graph of the sion. Furthermore, the result makes concrete the conspicious Unique Games. More specifically, in our case we require that presence of small set expansion in SDP integrality gaps for the Unique Games instance has good local expansion, in Unique Games and related problems. that sufficiently small sets have conductance close to 1. The While a confirmation of UGC was known to imply opti- formal statement of the modified Unique Games Conjecture mal inapproximability results for fundamental problems, a with mild expansion on small sets is given below: Conjecture 1.8. (Unique Games with Small Set Ex- Unlike the combinatorial small set expansion ΦG(δ), the pansion Conjecture) For every ε > 0, there exists δ such quantity Λ˜ G(δ) is efficiently computable, thus making it that for all η > 0 the following problem is NP-hard with a easy to recognize graphs that satisfy the required expansion sufficiently large R = R(ε, η): property. Raghavendra et al. [22] show that this parameter Given an instance Υ = (V, E, Π = {πv←w :[R] → [R]}, [R]) approximates the combinatorial expansion profile ΦG(δ) as of Unique Games with alphabet size R > R0 distinguish follows: between the following cases: 1/2 ˜ “˜ ” ΛG(δ) 6 ΦG(δ) 6 O ΛG(δ/2) log(1/δ) . –Completeness There exists an assignment A : V → [R] that satisfies at least 1 − ε fraction of edges. In this work, we show that Unique Games is easy if the ˜ –Soundness Every assignment A : V → [R] satisfies at most constraint graph G has sufficiently large ΛG(δ). The formal η fraction of edges and the constraint graph (V, E) sat- statement of the result is as follows: isfies the following expansion property: Theorem 1.11. Let Υ be unique game with SDP value –(Local Expansion Property) For every set S with 1 − ε and constraint graph G. Suppose there exists δ > 0 µ(S) 6 δ, Φ(S) > 1 − ε. such that ˜ 6 1 The above conjecture assumes fairly mild expansion in that ΛG(δ) > 10 ε log( /εδ) , sufficiently small sets have conductance close to 1. In fact, then the integral value of Υ is at least δ/10. existing SDP integrality gap instances for Unique Games (see [16, 20]) satisfy the above local expansion property. Corollary 1.12. If the constraint graph G associated 6 1 Under this stronger Unique Games Conjecture stated above, with a unique games instance Υ satisfies Λ˜ G(δ) > 10 ε log( /εδ), we show the following hardness result for Gap-Small-Set then it is easy to distinguish whether instance has value 1 − ε Expansion. or less than δ.

Theorem 1.9. The Unique Games with Small Set Ex- Independent of this work, Arora et al. [3] also show that pansion conjecture implies the Gap-Small-Set Expansion Unique Games is easy if the underlying constraint graph conjecture. has sufficient local expansion using different techniques. Due As an immediate consequence of the UG hardness reduc- to space constraints, we defer the proof of Theorem 1.11 to the full version. tion we also obtain super-constant hardness of approximation −1/ε for the k-Densest Subgraph Problem. Relation with Conjecture 1.8: For δ < 2 the quantity 6 1 ˜ 10 ε log( /εδ) > 1, while ΛG(δ) 6 1 for all graphs G and all Theorem 1.10. Assuming the Unique Games with Small δ > 0. Hence, Theorem 1.11 is not applicable on graphs with Set Expansion conjecture, it is NP-hard to approximate the expansion of sets of size δ < 2−1/ε. In Conjecture 1.8, there k-Densest Subgraph Problem to any constant factor. is no lower bound on the size δ of sets with local expansion. Consequently, Theorem 1.11 only yields an upper bound on Using standard techniques, the UG-hardness reductions the choice of the set size δ in Conjecture 1.8. can be used to construct SDP integrality gaps for the Gap- Small-Set Expansion and the k-Densest Subgraph Prob- 2. PRELIMINARIES lem. Due to space constraints, we defer the proofs of Theo- rem 1.9 and Theorem 1.10 to the full version. Notation: Unless otherwise stated, all graphs considered henceforth can be assumed to be regular unweighted graphs. 1.2 Unique Games with Small Set Expansion The proofs and results generalize to arbitrary weighted are Easy graphs, but we defer its discussion to the full version. Further exploring the connection between Unique Games Given a graph G = (V,E), we will use v ∼ V to denote and small set expansion, we show that the Unique Games a vertex sampled from the probability distribution given by problem is easy when there is sufficiently high expansion degrees deg(v) (in this case, the uniform distribution). Fur- of small sets. This result generalizes the work of [4] which thermore, for a set S ⊆ V , µ(S) = Pv∼V {v ∈ S} . Similarly, showed that Unique Games is easy when the associated the notation e ∼ E will denote an edge sampled from a distri- constraint graph is a good spectral expander. bution proportional to the weights (the uniform distribution We will use an analytic notion of a graph being a small set in unweighted graphs). We denote by E(A, B) the set of expander for this purpose. This is akin to spectral expansion edges with one endpoint in A and the other endpoint in B. 0 0 of a graph instead of combinatorial expansion. For a subset of edges E ⊆ E, |E | will denote the sum of the edge weights within E0. Therefore for a subset of edges ˜ 0 0 |E0| 0 Definition 2. For a graph G and δ > 0, define ΛG(δ) as E , Pe∼E [e ∈ E ] = /|E| = 2|E | . the minimum of Definition 3. For a vertex subset S ⊆ V , we define 2 kXu − Xvk E def def X (u,v)∼G ∂(S) = |E(S, V \ S)| and µ(S) = deg(s) . over all collection of vectors {Xv}v∈V that satisfy the con- s∈S straints Definition 4. (Conductance) The conductance/Cheeger’s hXu, Xvi δ , (1.1) constant associated with a subset S ⊆ V is given by Φ(S) = u,vE∼V 6 ∂(S) 2 µ(S) . The conductance/Cheeger’s constant for the graph G E kXuk = 1 . (1.2) u∼V is ΦG = min 1 Φ(S). µ(S)6 2 Due to space constraints we omit the proofs of the following –The provers are expected to pick one of the vertices of the simple facts. tuple they receive. Specifically, the answer set of the first prover is {u1, ... , uR}, while the second prover’s ˛ 2 2˛ Fact 2.1. For all a, b ∈ [0, 1], ˛a − b ˛ 6 2a + 2b − 4ab. answer set is {v1, . . . , vR}. Lemma 2.2 (Glorified Markov Inequality). Let Ω –The provers win if they pick two vertices (ui, vi) correspond- ing to some edge in the set M. be a probability space and let X,Y : Ω → R+ be two jointly distributed non-negative random variables over Ω . Sup- Completeness: Let us suppose there exists a set S of δn pose E X 6 γ E Y . Then, there exists ω ∈ Ω such that vertices such that Φ(S) ε. In this case, the provers can use X(ω) 6 2γY (ω) and Y (ω) > E Y/2. 6 the following simple strategy: If exactly one vertex from S Partial 2-prover games: A general 2-prover game Γ is appears in the tuple, return that vertex, else refuse to answer specified by a vertex set V , an edge set E, an alphabet Σ, the question. R 1 and a collection of predicates {Πu,v} indexed by vertex pairs The set S is of size δn and a question U ∈ V has d δ e u, v ∈ V . The value of the game Γ is defined as the maximum vertices. Therefore, with constant probability (at least 0.01) success probability exactly one of the vertices from S appears in the question n o U, and the provers answer the question. Consequently, the (F (u),F (v)) ∈ Πu,v , above strategy is a valid partial strategy. Observe that if uvP∈E the set S has small conductance Φ(S) 6 ε, a random edge over all strategies F : V → Σ. incident on S is with high probability completely contained Given a 2-prover game Γ, a partial game is one in which within S. In other words, if ui ∈ S then with very high the provers are permitted to refuse to answer the question. probability its neighbour vi also belongs to S. This implies The provers win only if both of them refuse to answer the that whenever the first prover decides to answer the vertex ui, question or else they both answer correctly as per the game Γ. the second prover also answers vi with very high probability. To preclude the trivial strategy that always refuses to answer Hence a small non-expanding set S translates directly in to questions, we require that the provers answer at least an α- a good partial strategy for the unique game. fraction of the questions for some constant α > 0. Formally, Soundness: Let F : V R → [R] be a strategy for the two we define the α-partial value of a 2-prover game Γ as the 1 provers that succeeds with probability at least 2 . For a maximum success probability R−1 R vertex sequence U ∈ V , let U +i x ∈ V denote the n o n o vertex sequence obtained by inserting x at index i. (F (u),F (v)) ∈ Πu,v / F (u) 6= ⊥ , R−1 uvP∈E u∼PV For every vertex sequence U ∈ V and an index i ∈ [R], (i) over all α-partial strategies F : V → Σ ∪ {⊥} . Here, ⊥ let us define a {0, 1}-valued function FU as follows: is a designated symbol (not a member of Σ) and an α- F (i)(x) = 1 if F (U + x) = i else it is 0 partial strategy is an assignment F : V → Σ ∪ {⊥} such that U i {F (u) 6= ⊥} α. As usual, the notation v ∼ V means (i) Pu∼V > Specifically, F is the indicator function of the set of vertices that v is sampled with probability proportional to its degree. U x such that the strategy decides to pick up the vertex x, when Organization: In the remainder of the paper, we present it is inserted at the ith location in U. the reduction from to Gap-Small-Set Expansion Unique Notice that, for the strategy proposed in the completeness Games and its analysis. For conceptual clarity, we subdivide case, for every setting of U, the set F (i) is either the non- the reduction in to two parts: In the first part (Section 3), U expanding set S or the empty set. Extrapolating from here, we reduce Gap-Small-Set Expansion to a partial Unique games, and then in Section 4 exhibit a generic reduction it is natural to look for non-expanding sets by considering (i) R−1 from partial 2-prover games to a corresponding traditional the set FU for a a random choice of U ∈ V and i. This 2-prover game. Finally, in Section 5 we wrap up the proof of is the basic intuition behind the soundness analysis presented the main result of this work - Theorem 1.4. in Section 3.1. Over a random choice of U ∈ V R−1 and i, the expected size of F (i) is indeed roughly Θ( 1 ) = Θ(δ). However, it 3. FROM GAP-SMALL-SET-EXPANSION TO U R could be possible that with very high probability over the PARTIAL UNIQUE GAMES (i) choice of U and i, FU is either too large or too small a In this section, we outline the key ideas of the reduction set. To rule out this possibility, we modify the above unique from Gap-Small-Set Expansion to Partial Unique Games. game by including random noise in to the questions. More specifically, the referee changes each vertex of the questions First Attempt. U, V independently with probability ε, before sending it to Let G = (V,E) be an instance of the Gap-Small-Set the provers. In the modified game, we bound the second (i) Expansion (η, δ) problem. For the sake of simplicity, let G moment of the random variable – the size of FU over a be a d-regular unweighted graph. We define the following random choice of U and i. unique game: Formal Reduction. –Fix R = d 1 e. The referee/verifier picks R edges M = δ Here we present the details of the reduction from Gap- {(u1, v1),..., (uR, vR)} uniformly at random from G. Small-Set Expansion problem to Partial Unique Games. –The referee sends a random permutation of the tuple U = For a vertex v ∈ V and ε > 0, we define a distribution (u1, ... , uR) to the first prover, and a random permu- Nε(v) on V as follows: with probability 1 − ε, we output tation of the tuple V = (v1, . . . , vR) to second prover. u := v and with probability ε, we output u ∼ V . For a vertex R−1 sequence v1, . . . , vR, the notation Nε(v1, . . . , vR) refers to the For M ∈ E , we define a function fM : V → [0, 1] as R product distribution Nε(v1) ⊗ · · · ⊗ Nε(vR) on V . follows For an edge e ∈ E, we define D(e) to be the distribution def ˜ fM (x) = E E f(U, x˜) . (3.3) that picks a random endpoint of e (uniformly). For a edge U∼D(M) (U,˜ x˜)∼N (U,x) R ε sequence e1, ... , eR, the distribution D(e1, ... , eR) on V We establish the following three relatively straight-forward is the product of the distributions D(e1), ... , D(eR). For a permutation π :[R] → [R] and a sequence A ∈ V R, we write properties of the functions fM . A0 = π.A to denote the permutation of A according to π, 0 Proposition 3.2. Suppose that F has value 1 − η for the i.e., A = Ai for all i ∈ [R]. π(i) game Υ and that F differs from ⊥ on exactly an α fraction Let R ∈ N. We define a Unique Games instance Υ = R of the inputs, i.e., α := R {F (X) 6= ⊥} . Then the ΥR,ε(G) with vertex set V and alphabet Σ = [R]. The edge PX∼V constraints in Υ are given by the tests performed by the following hold, following verifier. α E E fM (x) = R , (3.4) M∈ER−1 x∼V SSE-to-UG Reduction 2 R−1 E fM (x) 6 εR ∀M ∈ E (3.5) Let F : V R → Σ be an assignment to the instance Υ. The x∼V 2 value of the assignment is the success probability of the “ ” α E E E fM (x) = (1 − η) R . (3.6) following test: M∈ER−1 e∈E x∼D(e)

1. Sample R random edges e1, ... , eR from E and let Before proving the proposition, we will first show that the M := (e1, . . . , eR). existence of such functions {fM } as outlined in the proposi- tion, suffices to finish the soundness analysis. 2. Sample A ∼ D(M) and B ∼ D(M). Lemma 3.3. Suppose there exists a collection of functions 3. Sample A˜ ∼ Nε(A) and B˜ ∼ Nε(B). 1 {fM } satisfying the three conditions of 3.2 for some η 6 24 . Then there exists a set S ⊆ V with µ(S) ∈ ˆ α , 6 ˜ and 4. Sample two permutations π1, π2 :[R] → [R]. 4R εR Φ(S) 6 96η 5. Output Success if Proof. Let X and Y be the following non-negative ran- −1 “ ˜ ” −1 “ ˜ ” dom variables over the probability space ER−1, π1 F (π1.A) = π2 F (π2.B) . (3.1) X(M) := E fM (x)fM (y) Y (M) := E fM (x) . In the remainder of the section, we will show the following xy∈E x∼V theorem. 0 Since X 6 Y (pointwise), X = Y − X is also a non- R−1 1 negative random variable over E . The conditions (3.6) Theorem 3.1. For η < , given a graph G = (V,E), the 0 24 and (3.4) imply that X 2η Y and Y = α/R. Hence, 3 reduction produces an instance Υ of Unique Games such E 6 E E Lemma 2.2 (Glorified Markov Inequality) asserts that there that: ∗ R−1 0 ∗ exists an edge sequence M ∈ E such that X (M ) 6 ∗ ∗ ˆ 1 1 ˜ 4ηY (M ) and Y (M ) > E Y/2. In other words, –Completeness: If Φ(S) 6 η for some S with µ(S) ∈ 20R , R then there exists a 1/10e-partial strategy of value at least E fM∗ (x)fM∗ (y) > (1 − 4η) E fM∗ (x) , (3.7) 1 − 40eη − 4ε. xy∈E x∼V α E fM∗ (x) . (3.8) –Soundness: If there exists a α-partial strategy with value 1−η x∼V > 2R for the Unique Games instance Υ, then there exists a h i By condition (3.5), we can also upper bound the expectation set S ⊆ V with Φ(S) 96η and µ(S) ∈ α , 6 . 6 4R εηR of fM∗ , 2 E fM∗ (x) . (3.9) Due to space limitations, we defer the somewhat straightfor- x∼V 6 εR ward proof of the completeness case to the full version. The soundness claim of the above theorem is proven in the next Lemma 3.4 shows that we can round the function fM∗ to a ∗ section (Lemma 3.3). cut S that satisfies Φ(S∗) 96η , and µ(S∗) ∈ ˆ α , 6 ˜ . 3.1 Soundness 6 4R εR Let F : V R → [R] ∪ {⊥} be a partial assignment for the unique game Υ. For U ∈ V R−1 and x ∈ V , we let f(U, x) ∈ Lemma 3.4. Let G be a graph with vertex set V and edge [0, 1] denote the probability that F selects the coordinate of set E. Suppose f : V → [0, 1] is a bounded function on V x after we place it at a random position of U and permute such that the sequence randomly, i.e., f(x)f(y) (1 − η) f(x) , (3.10) def xyE∼E > x∼EV f(U, x) = P {F (π.(U +i x)) = π(i)} . (3.2) i∈[R] 1 π for η < 6 . Let β := Ex∼V f(x). Then, there exists a vertex ∗ ∗ ˆ β ˜ ∗ R set S ⊆ V such that µ(S ) ∈ 2 , 3β and Φ(S ) 6 24η. Here, U +i x denotes the vertex sequence in V obtained from U by inserting x as the i-th coordinate (the original Proof. Define a distribution over sets S given by the coordinates i, . . . , R − 1 of U are moved by one to the right). following sampling procedure: 1. sample a threshold t ∈ [1/9, 4/9] uniformly at random, strategy with value (1 − η) · α on the game Υ. Expressing this fact we get, ˘ 2 ¯ 2. output the set S := x | f (x) > t . α (1 − η) R = Since f = β the set S always satisfies n ˜ ˜ o E E P F (π1.A) = π1(r) ∧ F (π2.B) = π2(r) M∈ER,A,B∼D(M) π1,π2 E[f] A˜∼Nε(A), B˜∼Nε(B) µ(S) 6 = 3β . (3.11) r∈[R] 1/3 “ n o ”2 = F (π.A˜) = π(r) Note that condition (3.10) implies E R E E P M∈E , A∼D(M) A˜∼Nε(A) π r∈[R] 1 ` 2 2 ´ 2 (1−η)β 6 E f(x)f(y) 6 E 2 f (x)+f (y) = E f . 2 xy∼E xy∼E x∼V “ n ˜ o ” = E E P F (π.(U +r x˜)) = π(r) M∈ER, x∼D(Mr ) π Therefore we have Ex∼V f(1 − f) 6 ηβ. Rewriting the r∈[R] U∼D(M−r ) previous inequality, (U,˜ x˜)∼Nε(U,x) “ ”2 ηβ [f < 2/3] [f(1 − f)|f < 2/3] = f(U,˜ x˜) > P E E R E E −r E x∼V M∈E , x∼D(Mr ) U∼D(M ) (U,˜ x˜)∼Nε(U,x) 1 r∈[R] 2 2 > P[f < /3] E [f|f < /3] . 2 3 x∼V “ ” = E E fM−r (x) M∈ER, x∼D(Mr ) This implies a lower bound on the size of the set S as follows, r∈[R] 2 2 2 2 “ ” µ(S) [f /3] [f /3] [f|f /3] 0 > P > > P > x∼EV > = E E E fM (x) . M0∈ER−1 e∈E x∼D(e) = E[f] − P[f < 2/3] E [f|f < 2/3] > β(1 − 3η) . x∼V In the second equality, we use the fact that the events (3.12) F (π1.A˜) = π1(r) and F (π2.B˜) = π2(r) are independent given M ∈ ER and r ∈ [R]. In the third equality, e denotes Finally, we estimate the boundary of the set S. r the edge in the r-th coordinate of M. 1 1 ˆ ˛ 2 2 ˛˜ ∂(S) = E [| S (x) − S (y)|] 6 E 3 ˛f (x) − f (y)˛ xy∼E xy∼E Lemma 3.7. Suppose εR is sufficiently large. Then for R−1 12 f − 12 f(x)f(y)( 0 f 1 and Fact 2.1) every M ∈ E , 6 Ex xyE∼E ∵ 6 6 2 E fM (x) 6 . 6 12ηβ (3.13) x∼V εR

The claim follows from (3.11),(3.12) and (3.13). Proof. As before, we first unroll the definition of fM (x) ˜ E fM (x) = E E E f(U +x, ˜ x˜) 3.2 Proof of Proposition 3.2 x∼V U∼D(M) x∼V (U,˜ x˜)∼Nε(U,x) R Let F : V → [R]∪{⊥} be a partial strategy of value 1−η n ˜ o = E E E P F (π.(U +i x˜))) = π(i) for the game Υ. Suppose F differs from ⊥ on exactly an α U∼D(M) x∼V (U,˜ x˜)∼Nε(U,x) i∈[R] π fraction of inputs, i.e., PX∼V {F (X) ∈ [R]} = α. We prove (3.14) the three conditions of 3.2 in the following lemmas. To estimate this expectation, we generate random variables α ˜ Lemma 3.5. EM∈ER−1 Ex∼V fM (x) = R . π.(U +i x˜) and π(i) in an different fashion. Let ~ denote a placeholder symbol (not a member of V ). Consider the We unroll the definition of f (x), Proof. M following randomized procedure generating A ∈ V R, r ∈ [R]:

fM (x) ER−1 E M∈E x∼V 1. Sample U ∼ D(M), a permutation π :[R] → [R] and n ˜ o an index i ∈ [R]. Set B := π.(U + ). = E P F (π.(U +i x˜))) = π(i) i ~ M∈ER−1,U∼D(M) i∈[R] ˜ π 0 R x∼V,(U,x˜)∼Nε(U,x) 2. Generate B ∈ (V ∪ {~}) by changing each co- α ordinate of B randomly to ~ with probability ε. = P {F (X) = r} = R . 0 X∼V R, r∈[R] Let I ⊆ [R] denote the set of coordinates of B that contain . In the last equality, we use that the joint distribution of ~ R π.(U˜ +i x˜) and π(i) is the same as the joint distribution of 3. Generate A ∈ V by replacing all the placeholders in X and r. B0 by vertices sampled from V .

“ ”2 4. Sample r ∈ I uniformly at random. Lemma 3.6. EM∈ER−1 Ee∈E Ex∼D(e) fM (x) = (1 − α We claim that the joint distribution of A and r is the same η) R . as the joint distribution of π.(U˜ +i x˜) and π(i) in (3.14). Proof. For an edge sequence M ∈ ER and r ∈ [R], the By construction, the joint distribution of A and π(i) is the −r R−1 notation M refers to the edge sequence in E obtained same as the joint distribution of π.(U˜ +i x˜) and π(i) in (3.14) from M by removing the r-th coordinate. Similarly, Mr (with an obvious coupling). From the description of the denotes the r-th coordinate of M. Recall that F is a partial construction, it is also clear that the r-th coordinate of A has the same distribution as the π(i)-th coordinate of A (both Reduction from a partial 2-prover game Γ to a total are sampled from V ). 2-prover game Γ0 Since the pair (A, r) is identical in distribution to the pair 1.Sample c vertex pairs (u1, v1),..., (uc, vc) ∼ E . (π.(U˜ +i x), π(i)), we can finish the proof of the lemma as follows 2.Send the vertex sequence u1, . . . , uc to the first prover and the sequence v1, . . . , vc to the second prover. n ˜ o E fM (x) = E P F (π.(U +i x˜))) = π(i) x∼V U∼D(M),x∼V i∈[R] 3.Let (i, a) and (j, b) be their answers. π (U,˜ x˜)∼Nε(U,x) 4.The provers win if i = j and (a, b) ∈ Πui,vj . (by (3.14)) We observe that the reduction preserves the uniqueness prop- n o = F (A) = r (A and r generated as above) erty. A,rP n o Observation 4.2. If Γ is a unique game, then Γ0 is a = E P F (A) = r (A and I generated as above) A, I r∈I unique game as well. 1 1 = E {F (A)∈I} A, I |I| We will show Theorem 4.1 in two parts: Lemma 4.3 and 1 Lemma 4.4. E ( |I| ∼ 1 + Binom(R − 1, ε)) 6 t∼Binom(R−1,ε) t+1 ∵ 1.5 Lemma 4.3 (Completeness). If the α-partial value of + P {t + 1 εR/1.5} 0 −α·c 6 εR t∼Binom(R−1,ε) 6 Γ is at least 1−ε, then the value of Γ is at least 1−ε−e . 1.5 + 2−Ω(εR) (using Chernoff bound) Proof. Let F : V → Σ ∪ {⊥} be an α-partial strategy of 6 εR 0 0 2 value 1 − ε. We consider the following strategy F for Γ . 6 εR (for sufficiently large εR) ( 0 (i, a) if (F (ui) ∈ Σ) ∧ (∀j < i, F (uj ) = ⊥) , F (u1, . . . , uc) := (1, 1) if F (u1) = ... = F (uc) = ⊥ .

4. FROM PARTIAL 2-PROVER GAMES TO (In words, the prover answers with the first answer in the TOTAL 2-PROVER GAMES list F (u1), ... ,F (uc) (ignoring ⊥). If the partial strategy We show a general reduction from a partial 2-prover game refuses to answer on all inputs u1, ... , uc, then the prover Γ to a corresponding 2-prover game Γ0. returns an arbitrary answer.) Recall that in the partial game Γ, the provers are allowed to Let u1v1, ... , ucvc be a sequence of c vertex pairs, inde- 0 pendently drawn from E . The probability of the event refuse to answer questions. However, in the game Γ no such c −αc choice should be available. To achieve this, the referee gives F (u1) = ... = F (uc) is at most (1 − α) 6 e . Let us multiple questions from Γ simultaneously and the provers condition on the complementary event, i.e., the event that have to answer any question of their choice. F (ui) 6= ⊥ for at least one coordinate i ∈ [c]. Let i0 be the 0 first coordinate such that F (u ) 6= ⊥ or F (u ) 6= ⊥. The Specifically, the questions in Γ consists of a sequence of i0 i0 c questions from the game Γ. The provers are required to winning probability of the provers (conditioned on the event choose one of the c questions to answer, and return the that F (ui) 6= ⊥ for at least one coordinate i ∈ [c]) is equal answer. The provers win only if they pick the corresponding to pair of questions (say the ith question), and also answer the n o P (F (ui0 ),F (vi0 )) ∈ Πui ,vi question correctly as per the game Γ. Formally, we will show 0 0 the following theorem in the remainder of the section. n ˛ o = (F (u),F (v)) ∈ Πu,v ˛ F (u) ∈ Σ ∨ F (v) ∈ Σ uv∼PE

Theorem 4.1. For all positive integers c, given a 2-prover Puv∼E {(F (u),F (v)) ∈ Πu,v} game Γ with n vertices, there is a reduction to another 2- = Puv∼E {F (u) ∈ Σ ∨ F (v) ∈ Σ} prover game Γ0 running in time nO(c) such that: Puv∼E {(F (u),F (v)) ∈ Πu,v} = . –Completeness If the α-partial value of Γ is at least 1 − ε, 2 u∼V {F (u) ∈ Σ} − uv∼E {F (u),F (v) ∈ Σ} P P then the value of Γ0 is at least 1 − ε − e−α·c. Without loss of generality, we may assume u∼V {F (u) ∈ Σ} = 0 P –Soundness If the value of Γ is at least 1 − η, then the α. Since the partial strategy F 0 has value 1 − ε, we have 1 3c 1 1 / -partial value of Γ is at least 1 − 12η. uv∈E {(F (u),F (v)) ∈ Πu,v} /2 ·(1−ε)α+ /2 ·α. There- P > ε fore, also uv∈E {F (u),F (v) ∈ Σ} (1 − /2)α. We can P > Furthermore, the reduction preserves the uniqueness property conclude of the games. n ˛ o (F (u),F (v)) ∈ Πu,v ˛ F (u) ∈ Σ ∨ F (v) ∈ Σ uv∼PE We will assume that the edge set E of the game Γ does not contain any self-loops. We denote by E the following (1−ε/2)α > (1+ε/2)α > 1 − ε . distribution over vertex pairs (u, v) ∈ V 2: sample a ran- dom edge e ∈ E and let u and v be (independent) random It follows that the value of the strategy F 0 for the game Γ0 endpoints of e (in particular, u = v with probability 1/2). is at least 1 − ε − e−αc. We add predicates Πu,u = {(a, a) | a ∈ Σ} for all self-loops 2 0 (u, u) ∈ V to the collection {Πu,v}. For a parameter c ∈ N, Lemma 4.4 (Soundness). If the value of Γ is at least 0 let Γ be the following 2-prover game: 1 − η, then the 1/3c-partial value of Γ is at least 1 − 12η. Proof. Let F 0 : V c → [c]×Σ be a strategy for Γ0 of value For the reader’s convenience, we repeat the (short) argu- 0 0 1 − η. We first construct a partial strategy for Γ that uses ment: Let η 6 3η be such that 1 − η = (1 − η)/(1 + 2η). c−1 0 shared randomness. For i ∈ [c] and u ∈ V , we define a Let Good be the event Val > (1 − 2η )Vol and let Bad be partial strategy Fi,u : V → Σ ∪ {⊥} as the complementary event. We denote by 1Good and 1Bad the indicator variables of these events. We can relate 1GoodVol ( 0 E a if F (u +i u) = (i, a) , and Vol as follows Fi,u(u) := E ⊥ otherwise. 0 0 6 E Val − (1 − η ) E Vol (using (4.5) and (4.6)) Here, u + u denotes the vertex sequence u0 ∈ V c obtained 1 0 1 0 i 6 E GoodVol + (1 − 2η ) E BadVol − (1 − η ) E Vol from u by inserting u as the i-th coordinate (the original 0 1 0 coordinates i, . . . , c − 1 of u are moved by one to the right). = 2η E GoodVol − η E Vol .

It is clear that Hence, we can find i, u, and v such that 1Good(i, u, v) = 1 n o and Vol(i, u, v) E Vol/2, i.e., P Fi,u(u) ∈ Σ = 1/c . (4.1) > i∈[c], u∈V c−1, u∼V n o (Fi,u(u),Fi,v(v)) ∈ Πu,v 0 uv∼PE Since the value of F is 1 − η, we have n o 0 n o (1 − 2η ) · P Fi,u(u) ∈ Σ ∨ Fi,v(u) ∈ Σ , (4.7) (1 − η)/c = P (Fi,u(u),Fi,v(v)) ∈ Πu,v > u∼V i∈[c], uv∼Ec−1, uv∼E n o Fi,u(u) ∈ Σ ∨ Fi,v(u) ∈ Σ (1 − 2η)/2c . (4.8) 1 n o u∼PV > = 2 P (Fi,u(u),Fi,v(v)) ∈ Πu,v i∈[c], uv∈Ec−1, uv∈E We claim that the two conditions (4.7) and (4.8) allows 1 n o us to construct an O(1/c)-partial strategy of value 1 − O(η). + 2 P Fi,u(u) = Fi,v(u) ∈ Σ i∈[c], uv∈Ec−1, u∼V More concretely, we can modify Fi,u and Fi,v in such a way

that Fi,u(u) = Fi,v(u) for all u ∈ V while maintaining the From (4.1), it follows that both probabilities in the previous conditions (4.7) and (4.8). (Here, we use the fact that with 1 equation are at most 1/c. Hence, for their average to be probability /2, we test Fi,u(u) = Fi,v(u) ∈ Σ.) In this way, (1 − η)/c, both of them have to be at least (1 − 2η)/c, we get an assignment F ∗ : V → Σ∪{⊥} such that F ∗(u) = ⊥ ∗ n o if Fi,u(u) = Fi,v = ⊥ and F (u) ∈ {Fi,u(u),Fi,v(u)} ∩ Σ P (Fi,u(u),Fi,v(v)) ∈ Πu,v > (1 − 2η)/c , otherwise. Furthermore, the assignment F ∗ satisfies i∈[c], uv∈Ec−1, uv∈E

n ∗ o n o (4.2) P F (u) ∈ Σ = P Fi,u(u) ∈ Σ ∨ Fi,v(u) ∈ Σ , n o u∼V u∼V P Fi,u(u) = Fi,v(u) ∈ Σ > (1 − 2η)/c . n ∗ ∗ o i∈[c], uv∈Ec−1, u∼V (F (u),F (v)) ∈ Πu,v uv∼PE

(4.3) n o (Fi,u(u),Fi,v(v)) ∈ Πu,v . > uv∼PE To further analyze the partial strategies, we define two ran- dom variables ∗ Let α = u∼V {F (u) ∈ Σ}. We have n o P Vol(i, u, v) := P Fi,u(u) ∈ Σ ∨ Fi,v(u) ∈ Σ , 0 n ∗ ∗ o u∼V (1 − 2η ) · α (F (u),F (v)) ∈ Πu,v 6 uv∼PE n o Val(i, u, v) := (Fi,u(u),Fi,v(v)) ∈ Πu,v . uv∼PE 1 n ∗ ∗ o 1 = (F (u),F (v)) ∈ Πu,v + α 2 uvP∈E 2 The measure on (i, u, v) is as follows: We choose i ∈ [c] ∗ ∗ 0 uniformly at random, and sample uv from Ec−1. It is clear It follows that Puv∈E { (F (u),F (v)) ∈ Πu,v} > (1−4η )·α. ∗ We can conclude that F is an (1−2η)/2c-partial strategy of that Val 6 Vol (pointwise) and value at least 1 − 4η0 1 − 12η for Γ. h > E Vol = E P {Fi,u(u) ∈ Σ} (4.4) i∈[c], uv∼Ec−1 u∼V 5. WRAPPING UP n o i + P {Fi,v(u) ∈ Σ} − P Fi,u(u),Fi,v(u) ∈ Σ The main result of this work namely Theorem 1.4 is an u∼V u∼V immediate consequence of the following theorem. n o = 2/c − P Fi,u(u),Fi,v(u) ∈ Σ i∈[c], uv∼Ec−1, u∼V 5.1. Given a regular graph G of size n and Theorem parameters ε, δ > 0 such that δ = m/n for some integer 6 (1 + 2η)/c (using (4.3)) . (4.5) m, we can compute in polynomial time a Unique Games 0 On the other hand, since the value of F is 1 − η, E Val is instance Υ with (n/ε)O(1/δ) variables and with poly(1/εδ) given by, labels such that the following holds with C = 1500: n o P (Fi,u(u),Fi,v(v)) ∈ Πu,v = (1−η)/c . –Completeness: For all η > 0 if i∈[c], uv∼Ec−1, uv∼E

(4.6) ΦG(δ) 6 η =⇒ opt(Υ) > 1 − C(η + ε) At this point, we could use Lemma 2.2 (Glorified Markov Inequality) to argue that there exists a triple (i, u, v) such –Soundness: that Val(i, u, v) > (1 − O(η))Vol(i, u, v) and Vol(i, u, v) > 2 2 2 1 ΦG(δ) > 1 − ε /2C log (1/ε) =⇒ opt(Υ) 6 1 − E Vol/2 > 1/2c. 2C Proof. Execute the SSE-to-UG reduction with param- [9] J. Cheeger. A lower bound on smallest eigenvalue of a 1 eters R = b δ c and ε on the input graph G = (V,E). laplacian. Problems in Analysis (Papers dedicated to This yields a Unique Games instance Υ∗ on alphabet size Salomon Bochner), pages 195–199, 1970. R. Apply the reduction from partial Unique Games to [10] Feige and Schechtman. On the optimality of the Unique Games presented in Section 4 with parameter c = random hyperplane rounding technique for MAX CUT. 40e log(1/ε), to get the Unique Games instance Υ. RSA: Random Structures Algorithms, 20, 2002. Completeness If ΦG(δ) 6 η then there exists a subset S [11] U. Feige, G. Kindler, and R. O’Donnell. Understanding such that µ(S) ∈ [δ/10, δ] and Φ(S) 6 ε. By choice of R, we parallel repetition requires understanding foams. In 1 1 have µ(S) ∈ [ 20R , R ]. Hence by Theorem 3.1, there exists IEEE Conference on Computational Complexity, pages ∗ a 1/20e-partial strategy with value (1 − 40eη − 4ε) for Υ . 179–192, 2007. Finally, by Theorem 4.1 there exists a strategy for the game [12] K. Georgiou, A. Magen, T. Pitassi, and I. Tourlakis. 2 Υ succeeding with probability at least 1 − 40eη − 4ε − ε . Integrality gaps of 2 - o(1) for vertex cover SDPs in the 1 Soundness Let β = 2C . Suppose not, let us say opt(Υ) > lov´esz-schrijver hierarchy. In FOCS, pages 702–712. 1 − β. By Theorem 4.1, if opt(Υ) > 1 − β then there exists IEEE Computer Society, 2007. a 1 -partial strategy for Υ∗ with value at least 1 − 12β. 3c [13] S. Khot. On the power of unique 2-prover 1-round Applying Theorem 3.1, we get that there exists a set S with games. In STOC, pages 767–775. ACM, 2002. Φ(S) 1152β and 6 [14] S. Khot, G. Kindler, E. Mossel, and R. O’Donnell. » δ 6δ – Optimal inapproximability results for max-cut and µ(S) ∈ , . 480e log(1/ε) ε other 2-variable csps? SIAM J. Comput., 37(1):319–357, 2007. If µ(S) < δ then pad it with arbitrary vertices to con- [15] S. Khot and O. Regev. Vertex cover might be hard to 0 0 struct a set S such that µ(S ) = δ. On the other hand, approximate to within 2-ε. J. Comput. Syst. Sci., 0 if µ(S) > δ, randomly subsample a subset S ⊆ S of vol- 74(3):335–349, 2008. 0 ume δ. In either case, it is easy to check that Φ(S ) 6 2 2 2 2 2 2 [16] S. Khot and N. K. Vishnoi. The unique games 1 − (1 − 1152β)ε /C log (1/ε) 6 1 − ε /2C log (1/ε). conjecture, integrality gap for cut problems and embeddability of negative type metrics into l . In Acknowledgments: We thank Boaz Barak for suggesting 1 FOCS, pages 53–62. IEEE Computer Society, 2005. the notion of partial games, which helped to improve the presentation of the results. 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