On the Mysteries of MAX NAE-SAT
OOn the Mysteries of MAX NAE-SAT
Joshua Brakensiek ∗ Neng Huang † Aaron Potechin ‡ Uri Zwick § Abstract
MAX NAE-SAT is a natural optimization problem, closely related to its better-known relative MAXSAT. The approximability status of MAX NAE-SAT is almost completely understood if all clauses havethe same size k , for some k ≥
2. We refer to this problem as MAX NAE- { k } -SAT. For k = 2, it isessentially the celebrated MAX CUT problem. For k = 3, it is related to the MAX CUT problem ingraphs that can be fractionally covered by triangles. For k ≥
4, it is known that an approximation ratioof 1 − k − , obtained by choosing a random assignment, is optimal, assuming P (cid:54) = NP . For every k ≥ can be obtained for MAX NAE- { k } -SAT. There was some hope,therefore, that there is also a -approximation algorithm for MAX NAE-SAT, where clauses of all sizesare allowed simultaneously.Our main result is that there is no -approximation algorithm for MAX NAE-SAT, assuming theunique games conjecture (UGC). In fact, even for almost satisfiable instances of MAX NAE- { , } -SAT(i.e., MAX NAE-SAT where all clauses have size 3 or 5), the best approximation ratio that can beachieved, assuming UGC, is at most √ − ≈ . { } -SAT. We obtain an optimal algorithm, assuming UGC, for MAX NAE- { } -SAT, slightly im-proving on previous algorithms. The approximation ratio of the new algorithm is ≈ . { k } -SAT for every k ≥
2. Interestingly, the rounding function used bythis optimal algorithm is the solution of an integral equation.We complement our theoretical results with some experimental results. We describe an approximationalgorithm for almost satisfiable instances of MAX NAE- { , } -SAT with a conjectured approximationratio of 0.8728, and an approximation algorithm for almost satisfiable instances of MAX NAE-SAT witha conjectured approximation ratio of 0.8698. We further conjecture that these are essentially the bestapproximation ratios that can be achieved for these problems, assuming the UGC. Somewhat surprisingly,the rounding functions used by these approximation algorithms are non-monotone step functions thatassume only the values ± In a seminal paper, Goemans and Williamson [13] introduced the paradigm of obtaining an approximationalgorithm for a constraint satisfaction problem (CSP) by first solving a Semidefinite Programming (SDP)relaxation of the problem, and then rounding the solution. They used this paradigm to obtain an α GW ≈ . α GW + ε for MAX CUT, for any ε >
0, thus showing that the Goemans-Williamson MAX CUT algorithmis probably optimal. ∗ Stanford University, supported in part by an NSF Graduate Research Fellowship † University of Chicago, supported in part by NSF grant CCF:2008920 ‡ University of Chicago, supported in part by NSF grant CCF:2008920 § Blavatnik School of Computer Science, Tel Aviv University, Israel a r X i v : . [ c s . CC ] S e p mproved approximation algorithms for MAX DI-CUT and MAX 2-SAT were obtained by Feige and Goemans[10], Matuura and Matsui [24], and then by Lewin et al. [22] who gave a 0 . . k we let MAX { k } -SAT and MAX NAE- { k } -SAT be the versions of MAX SAT and MAXNAE-SAT in which all clauses are of size exactly k , and by MAX [ k ]-SAT and MAX NAE-[ k ]-SAT theversions in which clauses are of size at most k . Note that MAX NAE- { } -SAT is a natural generalizationof MAX CUT in which negations are allowed. The Goemans-Williamson algorithm also achieves a ratio of α GW for MAX NAE- { } -SAT.H˚astad [15], relying on the PCP theory developed by Arora et al. [3, 2], and others, proved that for every k ≥
3, obtaining an approximation ratio of (1 − k ) + ε for MAX { k } -SAT, for any ε >
0, is NP-hard.Similarly, for every k ≥
4, obtaining a ratio of (1 − k − ) + ε for MAX NAE- { k } -SAT is also NP-hard.Intriguingly, these ratios are obtained, in expectation, by simply choosing a random assignment.Thus, as mentioned in the abstract, while it is known that an approximation ratio of at least can beobtained for each individual clause size, for both MAX SAT and MAX NAE-SAT, it is not known whetheran approximation ratio of can be obtained for all clause sizes simultaneously (e.g., [23] ).(For MAX NAE-SAT we resolve this problem.)For MAX [3]-SAT (and MAX NAE-[4]-SAT), Karloff and Zwick [18] obtained a -approximation algorithmwhose computer assisted analysis appears in [36]. A -approximation algorithm is not known even for MAX { , } -SAT, i.e., when all clauses are either of size 1 or 4. (See Halperin and Zwick [14].)Avidor et al. [7], improving results of Andersson and Engebretsen [1], Asano and Williamson [4] and Zhang etal. [33], obtained a MAX SAT algorithm with a conjectured approximation ratio of 0 . . satisfiable instanced of MAX NAE-SAT, Zwick [35] gave an algorithm with a conjectured approximationratio of 0 . equal to the integrality ratio of a natural SDP relaxation of the problem. Furthermore, Raghavendra[29, 30] shows that the optimal approximation ratio can be obtained using a rounding procedure selectedfrom a specific family of rounding functions. (For more on finding almost optimal rounding procedures, seeRaghavendra and Steurer [31].)Raghavendra’s result [29, 30] does not solve the MAX SAT and MAX NAE-SAT problems, as it does nottell us what are the optimal approximation ratios achievable for these problems. It only tells us which SDPrelaxations are sufficient to obtain the optimal results and that the optimal ratios are equal to the integralityratios of these relaxations. This is certainly important and useful information, but it does not tell us whatthe integrality ratios are. [23] cites that the best known upper bound on the approximation ratio of MAX NAE-SAT is that of MAX CUT, ≈ . / / { } -SAT, which follows from H˚astad’s 7 / { } -SAT [15]. { k } -SAT Optimal ratio Algorithm Hardness k = 2 ≈ . k = 3 ≈ . (cid:63) ] [ (cid:63) ] k ≥ − k − Random assignment H˚astad [15]Table 1: Optimal approximation ratios for MAX NAE- { k } -SAT. Our first and main result is that under UGC, there is no -approximation algorithm for MAX NAE-SAT. Theproblem for MAX SAT is still open. Furthermore, assuming UGC, no polynomial time algorithm can achievea ratio of more than √ − ≈ . almost satisfiable instances of MAX NAE- { , } -SAT, i.e.,instances of MAX NAE-SAT in which all clauses are of size 3 or 5, and there is an assignment that satisfiesa 1 − ε (weighted) fraction of all the clauses, for an arbitrarily small ε >
0. We obtain the result by explicitlyconstructing instances of MAX NAE- { , } -SAT that exhibit an integrality ratio of √ − ≈ . { } -SAT, assuming UGC. Theapproximation ratio of this algorithm is ≈ . ≈ . { } -SAT [35]. This means that the optimal approximation ratios of MAX NAE- { k } -SAT, for every k ≥
2, are now known see Table 1. The rounding function used by the optimal MAXNAE- { } -SAT algorithm is the solution of an integral equation . The integral equation is obtained using a Calculus of Variations approach. The integral equation does not seem to have a closed-form solution, butthe optimal rounding function can be approximated to any desired accuracy. We show that this can be doneby solving a system of linear equations.A similar integral equation can be used to characterize the optimal rounding function for MAX CUT witha given completeness, giving an alternative description of the optimal rounding functions that are describedby O’Donnell and Wu [25].We next experiment with approximation algorithms for MAX NAE-SAT, as well as some restricted versionsof the problem. For a set K ⊆ { , , . . . } , we let MAX NAE- K -SAT be the restriction of MAX NAE-SAT toinstances in which each clause is of size k , for some k ∈ K .For MAX NAE- { , } -SAT we obtain an algorithm with a conjectured ratio of 0 . √ − ≈ . . . { , , } -SAT is as hard as MAX NAE-SAT.The exact approximation ratios achievable for MAX NAE- { , } -SAT, MAX NAE- { , , } -SAT and MAXNAE-SAT are not of great importance. What we believe is important is the nature of the rounding proceduresused to obtain what we believe are optimal, or close to optimal, results.All our algorithms, as well as most previous approximation algorithms, round the solution of the SDPrelaxation using the RPR (Random Projection followed by Randomized Rounding) technique introducedby Feige and Langberg [11]. This technique employs a rounding function f : ( −∞ , ∞ ) → [ − , technique see Section 2.2.) For MAX NAE- { } -SAT, the optimal rounding function f is thesolution of the integral equation mentioned above. What is intriguing about the best rounding functions wefound for versions of MAX NAE-SAT that involve clauses of at least two different sizes is that they are stepfunctions that only attain the values ±
1. We have some possible explanations for this phenomenon usingHermite expansion (see Section 5.3). 3 .3 Technical Overview
RPR and Raghavendra’s Theorem. Raghavendra [29, 30] showed that assuming the Unique Games Conjecture, to approximate CSPs it issufficient to consider a Basic SDP which assigns a unit vector v i to each variable x i . The idea behind thesevectors is that the SDP “thinks” there is a distribution of solutions where for all i and j , E [ x i x j ] = v i · v j .The value of the SDP is the expected number of constraints that the SDP “thinks” are satisfied by thisdistribution. We formally state the Basic SDP and make this precise in Section 2.In order to obtain an actual assignment from this Basic SDP, we need a rounding function which takes thesevectors { v i : i ∈ [ n ] } and outputs values { X i : i ∈ [ n ] } where each X i ∈ {− , } . For this, there is a canonicalmethod known as multi-dimensional RPR . Multi-dimensional RPR involves taking a rounding function f : R d → [ − , x i with probability f ( r · v i , . . . , r d · v i ) + 12 , where r , . . . , r d ∼ N (0 , I n ) are randomly sampled n -dimensional Gaussian variables, common to all thevariables.Raghavendra’s theorem [29, 30] formally states that if you can demonstrate a MAX CSP instance whose SDPvalue is at least c but the best integral assignment to the x i ’s satisfies at most s fraction, then distinguishinginstances which have a ( c − ε )-fraction satisfying assignment and not having a ( s + ε )-fraction satisfyingassignment in NP-hard by the Unique Games Conjecture. One can also prove that if this CSP instance ischosen to minimize s (for a fixed value of c ) then given any CSP with a ( c + ε )-fraction satisfying assignmentone can find a ( s − ε )-fraction satisfying assignment. By a suitable minimax argument (see, e.g., Section 7of [8]), such an algorithm can be attained by applying multidimensional RPR rounding, where the roundingfunction is sampled from a suitable distribution.In our proofs, we analyze the performance of various RPR rounding rules, by looking at the low-degreemoments . For instance, the second moment when d = 1 is F [ f ]( ρ ) = E r ∼N (0 ,I n ) [ f ( r · u ) f ( r · v )] , where u , v unit vectors with dot product ρ .In particular, for NAE- { k } -SAT, one can show it suffices to look at moments of 2 (cid:96) variables for 2 (cid:96) ≤ k . F [ f ]( ρ ) has a number of nice properties. It is an increasing function of ρ , it is an odd function, and it isconvex on nonnegative inputs. These properties play a crucial role in our results. √ − ≈ . UGC-hardness of MAX NAE-SAT.
In Section 3, we show that assuming the unique games conjecture, MAX NAE-SAT does not admit a 7 / near-satisfiable instances of MAX NAE- { , } -SAT that we cannotachieve a 7 / { } -SAT, adifficult triple ( v , v , v ) of unit vectors to round has pairwise dot product v i · v j = − / i (cid:54) = j .Likewise for NAE- { } -SAT, a difficult quintuple ( v , v , v , v , v ) has v i · v j = 1 / ≤ i < j ≤ v i · v = 0 for all 1 ≤ i ≤ . If we write out the expected value of our rounding rule, we derive thatbest integrality ratio ≤ min p ∈ [0 , max rounding rules (cid:20) (1 − p ) 3 + 3 F (1 / p − F (1 / − F (1 / (cid:21) , where F (1 /
3) = F (1 / , / , / , / , / , /
3) is the fourth moment when all of the pairwise biases are 1 / − p and p are the relative weights of the NAE- { } -SAT and NAE- { } -SAT clause types, respectively.We prove that F (1 / ∈ [0 , /
3] and F (1 / ≥ F (1 / . These together imply that for p = √ , the aboveexpression is at most √ − ≈ . . n optimal algorithm for MAX NAE- { } -SAT (assuming UGC). In Section 4, we tackle the problem of finding an optimal approximation algorithm for MAX NAE- { } -SAT.Our approach follows a similar template to that of [25] for MAX CUT. Like [25], we consider the max-minproblem of finding an RPR rounding function f which achieves at least the optimal approximation ratiofor any distribution D of triples of unit vectors (pairs of unit vectors for MAX CUT). In the case of MAXCUT, [25] showed that the “most difficult” D is a distribution over pairs of identical unit vectors, and pairsof unit vectors with dot product ρ <
0. Likewise, we show (see Theorem 4.8) that the most difficult D comefrom triples of vectors that are have pairwise dot products ( ρ, ρ,
1) or ( ρ , ρ , ρ ), where ρ ∈ ( − ,
0] and ρ = max( ρ, − /
3) or ρ = 1.Once we fix a distribution D , the performance of f can be expressed using a double integral. Using calculusof variations , we can then produce an integral equation that must be satisfied by the optimal f , known asa Fredholm integral equation of the second kind (c.f., [27]).If we discretize f as a step function, the integral equation becomes a system of linear equations (c.f.,Section 6 of the full version of [25]). Thus, like in [25], we can efficiently compute an arbitrarily closeapproximation of the optimal function for MAX NAE- { } -SAT. We found that the optimal approximationratio is approximately 0 . s -linear function,that is of the form f ( x ) = max( − , min(1 , sx )), but there is a nontrivial error term on the order of O ( x ). Conjectured near-optimal algorithm for MAX NAE-SAT with near-perfect completeness.
In Section 5, we give better experimental algorithms for MAX NAE-SAT and various restrictions of it, fornearly satisfiable instances (1 − ε completeness, for ε > ± monotone rounding functions f : R → [ − , { } -SAT, the optimal rounding functions are provably monotone.On the other hand, for even a simple extension like MAX NAE- { , } -SAT, it turns out the optimal roundingfunction is very likely not monotone, but rather a non-monotone step function. In fact, we conjecture thatthe function f ( x ) = − x < − α x ∈ [ − α, − x ∈ (0 , α ]1 x > α where α ≈ . { , } -SAT. We have further experimental results for near satisfiable instances of MAX NAE-SAT. Definition 2.1 (MAX CSP( F )) . Let F be a set of Boolean functions. Each f ∈ F is a function f : {− , } k ( f ) → { , } , where k ( f ) is the number of arguments of f . An instance of MAX CSP ( F ) is composedof a set of variables V = { x , x , . . . , x n } and a collection C of weighted clauses. Each clause C ∈ C is a tuple ( w, f, i , i , . . . , i k , b , . . . , b k ) , where w > , f ∈ F , k = k ( f ) , i , i , . . . , i k are distinct indices from [ n ] = { , , . . . , n } , and b , b , . . . , b k ∈ {− , } . Such a clause represents the constraint f ( b x i , b x i , . . . , b k x i k ) =5 . (Here b j x i j denotes multiplication.) The goal is to find an assignment α : V → {− , } that maximizesthe sum of the weights of the satisfied constraints. The above definition defines Boolean constraint satisfaction problems. (For technical reasons, the valuesassumed by the variables are − i , i , . . . , i k to be distinct. (If some of the indices are equal, this isequivalent to having a constraint on a smaller number of variables.) We allow constraints to be applied toboth variables and their negations. (This is equivalent to requiring the set F of allowed constrained to beclosed under negations of some of the arguments.) For more on general constraint satisfaction problems, andtheir maximization variants, see [9, 34]. Definition 2.2.
For every integer k ≥ , the Not-All-Equal predicate on k variables is defined as NAE k ( x , . . . , x k ) = (cid:26) if x = x = · · · = x k , otherwise. We remark that the Not-All-Equal predicates are even predicates: a collection of boolean variables are notall equal if and only if their negations are not all equal.In this paper, we only consider problems of the form MAX CSP( F ) where F ⊆ {
NAE k | k ≥ } . For a set K ⊆ { , , . . . } , we let MAX NAE- K -SAT be a shorthand for MAX CSP( { NAE k | k ∈ K } ).For a CSP instance with variable set V = { x , . . . , x n } and clause set C , we can define the Basic SDP asfollows. We maintain a unit vector v i ∈ R n +1 for each variable x i and a special unit vector v , and for eachclause C a probability distribution p C ( α ) over A ( C ), the set of all assignments on variables in C . Here z i stands for a literal which is either x i or − x i , and α ( x i ) is the value α assigns to x i . We use the notation z i ∈ C to denote that z i appears in the clause C .max (cid:88) C ∈C w C (cid:88) α ∈A ( C ) p C ( α ) C ( α ) ∀ i ∈ { , , , . . . , n } , v i · v i = 1 ∀ C ∈ C , ∀ z i , z j ∈ C, v i · v j = (cid:88) α ∈A ( C ) α ( x i ) α ( x j ) p C ( α ) ∀ C ∈ C , ∀ z i ∈ C, v i · v = (cid:88) α ∈A ( C ) α ( x i ) p C ( α ) ∀ C ∈ C , ∀ α ∈ A ( C ) , p C ( α ) ≥ . Note that the first two constraints imply that (cid:80) α ∈A ( C ) p C ( α ) = 1. Let b i = v · v i and b i,j = v i · v j for i (cid:54) = j ∈ [ n ]. We call b i biases and b i,j pairwise biases. Informally speaking, b i is intended for E [ x i ] and b i,j is intended for E [ x i x j ], and the SDP constraints are saying that local assignments should agree with thesebiases and pairwise biases. Note that for a CSP with even predicates as in our case, v (and therefore thebiases) is not useful, as we can always combine a solution and its negation to get a new solution with 0 biaseswhile preserving the objective value. Definition 2.3.
For a CSP instance Φ , we define its completeness, denoted c (Φ) , to be its SDP value, andits soundness, denoted s (Φ) , to be the value of the optimal integral solution to Φ . For a MAX CSP problem Λ , define its integrality gap curve to be the function S Λ ( c ) : c (cid:55)→ inf { s (Φ) | Φ ∈ Λ , c (Φ) = c } . The Unique Games Conjecture, introduced by Khot [19], is a central conjecture in the study of approximationalgorithms. One version of the conjecture is as follows.
Definition 2.4.
In the unique games problem with R labels, we are given a graph G = ( V, E ) and a set ofpermutations { π e : [ R ] → [ R ] | e ∈ E } . An assignment α : V → [ R ] is said to satisfy an edge e = ( u, v ) if π e ( α ( u )) = α ( v ) . Our goal is to find an assignment that maximizes the number of satisfied edges. onjecture 1 (Unique Games Conjecture) . For any constant ε > , there exists R ∈ N such that for uniquegames problem with R labels, it is NP-hard to distinguish between the following two cases: • there exists an assignment that satisfies − ε fraction of edges; • no assignment can satisfy more than ε fraction of edges. We say that a problem is unique games hard if it is NP-hard assuming the Unique Games Conjecture.Raghavendra, in his breakthrough paper [29, 30], showed the following result which exhibited the closerelation between the integrality gap curve of a CSP problem and its unique game hardness.
Theorem 2.5 (Raghavendra) . For a MAX CSP problem Λ , Let U Λ ( c ) be the best polynomial-time computableintegral value on instances with optimal value c assuming the Unique Games Conjecture. Then we have1. For every constant η > and c ∈ [0 , , U Λ ( c ) ≤ S Λ ( c + η ) + η .2. For every constant η > and c ∈ (0 , , there exists a polynomial time algorithm that on any inputwith completeness c , outputs an integral solution with value at least U Λ ( c − η ) − η . We will describe Raghavendra’s algorithm in the following subsection. For now we remark that this theoremessentially says that the integrality gap of the Basic SDP is the Unique Games Hardness threshold of theCSP, so in order to show unique games hardness, it suffices to construct an integrality gap instance for theBasic SDP.
The approximation constant of MAX NAE-SAT has multiple interpretations in the literature [7, 23]. Themost common one is to assume that any instance Ψ consists of an arbitrarily large clauses, in particularthey grow as a function of the number of variables of the instance. Note that Raghavendra’s result does notapply in this case, as the Basic SDP for an instance with clauses of length Ω( n ) has exponential size.A secondary interpretation is to consider the limiting behavior of MAX NAE-[ k ]-SAT as k → ∞ . In thiscase, each k is a fixed value so the Basic SDP has polynomial size and Raghavendra’s theorem applies.It turns out these two views are essentially identical in the limit, and as such we assume the latter in theremainder of the main body of the paper. See Appendix E for more details. rounding technique RPR (Random Projection followed by Random Rounding) is an algorithm for rounding SDP solutionsproposed by Feige and Langberg [11]. It generalizes hyperplane rounding and outward rotation techniquesand has played an important role in designing SDP-based approximation algorithms. The RPR techniquechooses a function f : R → [ − ,
1] and performs the following action: • Choose a random r ∼ N ( , I n ), where n is the length of the SDP solution vectors and N ( , I n ) is the d -dimensional standard normal distribution. • For each variable x i , compute the inner product t i = r · v i . (Random projection) • For every i , independently assign x i = 1 with probability f ( t i )2 , x i = − − f ( t i )2 (Random Rounding)Some previously used functions for f include the sign function (in hyperplane rounding) and piecewise linearfunctions. 7n [29, 30], Raghavendra showed an algorithm which achieves the integrality gap of Basic SDP for any CSPwithin an arbitrary precision, assuming UGC. His algorithm makes use of the following procedure (denoted Round f in [29, 30]), which is essentially a multi-dimensional version of RPR : • Choose a function f : R d → [ − , d random normal vectors r (1) , . . . , r ( d ) ∼ N ( , I n ). • For each variable x i and 1 ≤ j ≤ d , let t ( j ) i = v i · r ( j ) . • Let p i = f ( t (1) i , . . . , t ( d ) i ). Assign x i = 1 with probability p i and x i = − − p i .When d = 1, Round f is the usual RPR procedure. We will refer to this procedure as RPR d if f is a d -dimensional function. Raghavendra’s algorithm runs Round f for every f in a pre-computed ε -net andpicks the best f . It also has a preprocessing “smoothing” step, which in our case corresponds to scaling thepairwise biases by a common factor. It turns out that for Round f procedures, it suffices to consider odd f ,due to the following lemma. Lemma 2.6.
Let f : R d → [ − , and f (cid:48) ( x ) = ( f ( x ) − f ( − x )) / be its odd part. For any CSP, the worstcase performance of Round f (cid:48) is at least as good as Round f .Proof. Consider an arbitrary CSP and let Φ be a worst case instance for
Round f (cid:48) . Observe that the Round f (cid:48) procedure is equivalent to the following: independently for every variable x i , with probability 1/2 apply therounding function f on v i , and with probability 1/2 apply the rounding function f on − v i and flip the result.Observe further that by replacing v i with − v i and flipping the outcome, we are essentially applying f to anew instance with − x i in place of x i in Φ. This implies that the value of Round f (cid:48) on Φ is an average of 2 n values ( n is the number of variables in Φ) where each value is Round f evaluated on an instance obtainedby flipping some variables in Φ. It follows that in some of these instances Round f has a value as bad as thevalue of Round f (cid:48) on Φ.In the later sections, we will only consider Round f with odd f . Given a rounding function, we can view its outputs as random variables X , . . . , X n , where X i is the valueassigned to x i . An important property of RPR technique, either one-dimensional or multi-dimensional, isthat the k -wise moments of these random variables are determined entirely by the pairwise dot products ofthe vectors in the SDP solution. In other words, E [ X i · · · X i k ] for any group of variables { X i , . . . , X i k } isa function on b i ,i , b i ,i , . . . , b i k − ,i k . This inspires the following definition. Definition 2.7 (e.g., [28, 16]) . For rounding procedure
Round f , define F k [ f ]( b , , b , , . . . , b k − ,k ) to be the k -wise moment E [ X · · · X k ] where X , . . . , X k are random variables obtained by applying Round f to vectors v , . . . , v k such that v i · v j = b i,j for every ≤ i < j ≤ k . We will usually omit the argument f and write F k for F k [ f ] unless there are multiple rounding functions inquestion. The following observation is immediate: Proposition 2.8.
For every odd function f and odd integer k > , F k [ f ] = 0 .Proof. Let X , . . . , X k be a group of random variables that Round f outputs on SDP vectors v , . . . , v k . Nowtake the negation of each of these vectors, then each X i is also negated because f is odd, while the pairwise We are stating Raghavendra’s algorithm for even CSPs in which the assignment ( x , . . . , x n ) has the same weight asthe assignment ( − x , . . . , − x n ) for all assignments to ( x , . . . , x n ) (such as MAX CUT and MAX NAE-SAT). Otherwise,Raghavendra considers a more general algorithm which incorporates v · v i for all i . E [ X · · · X k ] = F k [ f ]( v · v , . . . , v k − · v k )= F k [ f ](( − v ) · ( − v ) , . . . , ( − v k − ) · ( − v k ))= E [( − X ) · · · ( − X k )]= ( − k E [ X · · · X k ] . The proposition then follows since k is odd.The first non-trivial moment function F has been studied in previous work and is relatively well understood.By definition, F ( x ) is the expected value of the product of a pair of variables whose SDP solution vectorshave inner product x . Observe that if r ∼ N (0 , I n ) is a standard normal vector and v ∈ R n is a unit vector,then their inner product r · v has the standard normal distribution N (0 , v i and v j areunit vectors with inner product x , then r · v i and r · v j are standard normal variables with correlation x . Itfollows that F [ f ]( x ) is equal to the Gaussian noise stability of f at x , defined as E u , v [ f ( u ) f ( v )] where u , v is a pair of d -dimensional x -correlated standard Gaussians. From here the standard analytic tools apply,and we refer to [26] for a more thorough treatment. Here we collect a few facts that are crucial to theunderstanding of MAX NAE- { } -SAT. These facts on F can also be found in O’Donnell and Wu [25]. Proposition 2.9 (Proposition 4.7 in [25]) . When f is odd, F [ f ]( x ) is a power series in x with nonnegativecoefficients, odd powers only and radius of convergence at least . In particular, F [ f ]( − x ) = − F [ f ]( x ) and F [ f ] is convex on [0 , , concave on [ − , . Theorem 2.10 (Theorem 4.4 in [25]) . For every f , there exists a Gaussian rearrangement f ∗ of f which isa one-dimensional, odd, increasing function with the property that for every x ∈ [0 , , F [ f ]( x ) ≤ F [ f ∗ ]( x ) and F [ f ](1) = F [ f ∗ ](1) . The power of these moment functions is best seen when combined with Fourier analysis. We have thefollowing proposition on the Fourier expansion of Not-all-equal predicates:
Proposition 2.11.
The Fourier expansion of
NAE k : {− , } k → { , } is given by NAE k ( x , . . . , x k ) = 12 k − k − − − (cid:88) i
0) for
NAE . Weshow that these biases “fool” the Basic SDP (has completeness 1) but are in fact very difficult to round.9 emma 3.1. Let Φ be a MAX NAE- { , } -SAT instance whose -clauses all have pairwise biases ( b , , b , ,b , ) = ( − , − , − ) and -clauses all have pairwise biases ( b , , b , , . . . , b , ) = ( , , , , , , , , , ,then Φ has completeness .Proof. It suffices to show that for every clause, there exists a distribution of satisfying assignments thatagrees with the global (pairwise) biases. • { (1 , , − , (1 , − , , ( − , , } has the same pairwise biases. • X X X X X − − − − − Proposition 3.2.
Let f be an odd function, v , . . . , v k be a group of unit vectors with v i · v j = b i,j . If b i,k = 0 for every ≤ i < k , then F k [ f ]( b , , b , , . . . , b k − ,k ) = 0 . Proof.
Let X , . . . , X k be ± Round f on v , . . . , v k . Since b i,k = 0for every 1 ≤ i < k , we have that v , . . . , − v k have the same pairwise products as v , . . . , v k . Since f isodd, if we run Round f on v , . . . , − v k , the distribution of the outputs will be the same as X , . . . , − X k . Itfollows that E [ X X · · · X k ] = F k [ f ]( b , , b , , . . . , b k − ,k ) = E [ X X · · · ( − X k )] = − E [ X X · · · X k ] , which implies that F k [ f ]( b , , b , , . . . , b k − ,k ) = 0 . Lemma 3.3.
Let x ∈ [0 , . For any rounding scheme, F ( x, x, x, x, x, x ) ≥ F ( x ) .Proof. Let v , . . . , v n be a group of unit vectors such that each pair has inner product x . This can be doneby having the vectors share a common component of length √ x and be orthogonal beyond that. Apply Round f procedure to these vectors and let X i be the value obtained by rounding v i . By Definition 2.7, wehave F ( x ) = E [ X i X j ] and F ( x ) = E [ X i X j X k X l ] for distinct i, j, k, l ∈ [ n ]. We have the inequality E (cid:32) n (cid:88) i =1 X i (cid:33) ≥ E (cid:32) n (cid:88) i =1 X i (cid:33) .
10y expanding out the inequality and using the fact that X i = 1, we have3 n − n + n ( n − n − F + n ( n − n − n − F ≥ ( n + n ( n − F ) . The lemma then follows by dividing both sides by n and letting n go to infinity. Theorem 3.4. A . -approximation for MAX NAE- { , } -SAT (clauses of size and ) is unique gameshard, even when the instance has completeness − ε , for ε > arbitrarily small.Proof. Recall that we have the Fourier expansions:
NAE ( x , x , x ) = 3 − x x − x x − x x , NAE ( x , x , x , x , x ) = 15 − (cid:80) ≤ i NAE ( x , x , x ) where b , = b , = b , = − then E [ NAE ( X , X , X )] = 3 + 3 F ( )42. If we have a 5-clause NAE ( x , x , x , x , x ) where b , = b , = b , = b , = b , = b , = and b , = b , = b , = b , = b , = 0 then E [ NAE ( X , X , X , X , X )] = 15 − F ( ) − F ( , , , , , )16The theorem now follows from the following lemma. Lemma 3.5. If the following conditions hold1. The average performance on -clauses is F 2. The average performance on -clauses is − F − F F ≥ F then taking the distribution where we choose a random -clause with probability − √ and choose a random -clause with probability √ , the average performance on this distribution is at most √ − < . .Proof. If we take a random 5-clause with probability p and take a random 3-clause with probability 1 − p ,the average performance on the clauses is(1 − p ) 3 + 3 F p − F − F 16 = 12 + 3 p + (12 − p ) F − pF F ≥ F , this is at most12 + 3 p + (12 − p ) F − pF 16 = 12 + 3 p + (6 − p ) p − p (cid:16) F − (6 − p ) p (cid:17) ≤ p + p − p , this is minimized when (84 − p ) = 0, which happens when p = √ . When p = √ , 84 p + p − √ 21 + 12 √ − √ − .2 Explicit integrality gap In this subsection, we explicitly construct a family of gap instances whose integrality ratio tends to √ − <. { e i | i ∈ [ n ] } be the canonical basis of R n . Consider the subset of R n containing vectors thathave exactly three nonzero coordinates, each being 1 / √ − / √ 3, namely V = (cid:26) b e i + b e j + b e k √ (cid:12)(cid:12)(cid:12)(cid:12) b , b , b ∈ {− , } , ≤ i < j < k ≤ n (cid:27) . For every v ∈ V , we assign a Boolean variable x v ∈ {− , } such that x − v = − x v . Our goal is to define aCSP instance Φ with variables x v such that assigning v to x v is an SDP solution with perfect completeness,while any integral solution has value at most √ − as n tends to infinity. Definition 3.6. We define C to be the set of -clauses of the form NAE ( x v , x v , x v ) where1. v = √ ( s e i − s e i + s e i ) v = √ ( s e i − s e i + s e i ) v = √ ( s e i − s e i + s e i ) for some distinct indices i , . . . , i ∈ [ n ] and signs s , . . . , s ∈ {− , } . Definition 3.7. We define C to be the set of -clauses of the form NAE ( x v , x v , x v , x v , x v ) where1. For all j ∈ { , , , } , v j = √ ( s e i + s j e i j + s j +1 e i j +1 ) v = √ ( s e i + s e i + s e i ) for some distinct indices i , . . . , i ∈ [ n ] and signs s , . . . , s ∈ {− , } . Remark 3.8. These sets of clauses are designed so that the pairwise biases for the -clauses are ( − , − , − ) and the pairwise biases for the -clauses are ( , , , , , , , , , . Definition 3.9. Let Φ be the MAX NAE- { , } -instance with variable set { X v | v ∈ V } and clause set C ∪ C , where every clause in C has weight − √ |C | and every clause in C has weight √ |C | . Theorem 3.10. For any integral solution to Φ , the weight of the satisfied clauses is at most √ − + O ( n ) .Proof. To analyze the weight of the satisfied constraints for a given solution, we consider the followingdistributions. Definition 3.11. For every k < n/ , we define D k to be the following distribution over V k :1. Sample k + 1 distinct indices i , i , . . . , i k +1 ∈ [ n ] uniformly at random.2. Sample k + 1 independent random coin flips b , . . . , b k +1 ∈ {− , } .3. For every j ∈ [ k ] , let v j = √ ( b e i + b j e i j + b j +1 e i j +1 ) . Return the k -tuple ( v , v , . . . , v k ) . Informally speaking, this distribution samples k vectors from V of “sunflower shape” in the sense that all ofthem share exactly one index on which they are nonzero. Definition 3.12. Given an assignment, we let F = E ( v , v ) ∼D [ x v x v ] , F = E ( v , v , v , v ) ∼D [ x v x v x v x v ] . emark 3.13. Here F and F come from an actual assignment rather than a rounding scheme, but theyplay the same role in the argument. Proposition 3.14. Given an assignment, the proportion of -clauses which are satisfied is F and theproportion of -clauses which are satisfied is − F − F Proof sketch. This can be shown by expanding out each constraint as a polynomial.By Lemma 3.5, if we had that F ≥ F then we would have that the total weight of the satisfied clauses is atmost √ − . Instead, we show that F ≥ F − O ( n ). Adapting the argument in Lemma 3.5 accordingly,this implies that the total weight of the satisfied clauses is at most √ − + O ( n ). Lemma 3.15. For any assignment, F ≥ F − O (cid:18) n (cid:19) . Proof. Let k = (cid:98) n/ (cid:99) − < n/ 2. Sample ( v , . . . , v k ) ∼ D k . Note that the marginal distribution of any pairof these vectors is exactly D and any 4 vectors exactly D . Now let X = (cid:80) ki =1 x v i . We have the inequalityVar[ X ] = E [ X ] − (cid:0) E [ X ] (cid:1) ≥ . We have that E (cid:2) X (cid:3) = E (cid:32) k (cid:88) i =1 X v i (cid:33) = k (cid:88) i =1 E [ X v i ] + (cid:88) i (cid:54) = j E [ X v i X v j ] = k + k ( k − F . Here we used the fact that X v i ∈ {− , } and X v i = 1. Similarly we can compute E (cid:2) X (cid:3) = E (cid:32) k (cid:88) i =1 X v i (cid:33) = 3 k − k + k ( k − k − F + k ( k − k − k − F . Plugging in these two expressions to the inequality above, we get3 k − k + k ( k − k − F + k ( k − k − k − F − ( k + k ( k − F ) ≥ . Our lemma follows by shifting terms. dividing both sides by k ( k − k − k − k = Θ( n ).A natural question is whether there exists an assignment such that the weight of the satisfied constraintsis at least ( √ − ) . If no such assignment exists, then it would be possible to further improve the upperbound. However, we show that for this set of constraints, our analysis is tight and there exists an assignmentsuch that the weight of the satisfied constraints is at least ( √ − ) . That said, there may be another set ofconstraints which gives a better upper bound. Theorem 3.16. There is an assignment that satisfies ( √ − ) -fraction of the clauses in Φ .Proof. It suffices to show there exists a probability distribution that satisfies ( √ − ) -fraction of the clausesin Φ in expectation. From the proof of Lemma 3.5 it can be seen that in order for a solution to achieve ( √ − ) it suffices to have F = 2 √ − F = F . We verify this as follows: given that F = 2 √ − F = F , on 3-clauses we achieve a ratio of3 + 3 F · (2 √ − (cid:0) √ − (cid:1) , − F − F 16 = 15 − √ − − (2 √ − 16 = 15 − √ − − (165 − √ (cid:0) √ − (cid:1) . Let us now consider the following rounding algorithm: if v has 3 positive coordinates, round X v to 1 withprobability p and − − p ; if v has exactly 2 positive coordinates, round X v to 1 withprobability p and − − p ; if v has less than 2 positive coordinates, we let X v = − X − v .We need to analyze the quantity E v ,..., v k ∼D k [ X v · · · X v k ] for k = 2 and k = 4.Let ( v , . . . , v k ) ∼ D k . Recall that these vectors have a “sunflower” shape: they all share a common non-zerocoordinate and each vector has a “petal” of two non-zero coordinates. Without loss of generality, assumethat their common coordinate is positive. Then, every vector independently has 3 positive coordinates withprobability 1 / 4, 2 positive coordinates with probability 1 / / X v i is rounding to 1 independently with probability14 · p + 12 · p + 14 · (1 − p ) = p − p + 14 . So we have E [ X v i ] = p − p + 14 − (cid:18) − p − p + 14 (cid:19) = p − p − F = (cid:18) p − p − (cid:19) , F = (cid:18) p − p − (cid:19) = F . Note that the range of F is [0 , . 25] and 2 √ − ≈ . p and p appropriately we canalso have F = 2 √ − 9. This completes the proof. { } -SAT In this section, we give a precise analytical condition for the optimal RPR rounding function for MAXNAE- { } -SAT using what is known as Fredholm integral equation of the second kind . To help illustrate ourtechniques, we also perform this analysis for the simpler case of MAX-CUT, giving an alternative descriptionfor the optimal rounding function found by O’Donnell and Wu [25]. Recall from Section 2.3 that the expected value of an NAE clause with pairwise bias b , is − F ( b , )2 andthat of an NAE clause with pairwise biases b , , b , , b , is given by − F ( b , ) − F ( b , ) − F ( b , )4 . In thissubsection, we use the fact that F is odd and convex on [0 , 1] to derive the hardest distribution for MAXNAE- { } -SAT. The idea is to maximize the sum of F while preserving the sum of the pairwise biases. Notethat a similar analysis was done for MAX CUT by O’Donnell and Wu [25]. We include this analysis herefor completeness and to help illustrate the techniques. x ≥ x ≥ Proposition 4.1. If F is convex for x ≥ , ≤ x ≤ x ≤ x ≤ x , and x + x = x + x then F ( x ) + F ( x ) ≤ F ( x ) + F ( x ) .Proof. If x = x = x = x then the result is trivial so we can assume that x < x . Writing x = ax + (1 − a ) x and x = bx + (1 − b ) x where a, b ∈ [0 , x + x = ( a + b ) x + (2 − a − b ) x = x + x + (1 − a − b )( x − x )Since x + x = x + x , we must have that b = 1 − a . Since F is convex we have that14. F ( x ) ≤ aF ( x ) + (1 − a ) F ( x ).2. F ( x ) ≤ bF ( x ) + (1 − b ) F ( x ) = (1 − a ) F ( x ) + aF ( x ).Adding these inequalities together we have that F ( x ) + F ( x ) ≤ F ( x ) + F ( x ), as needed.Our main tool is the following lemma: Lemma 4.2. If F is an odd function which is convex for x ≥ then1. For all x, x (cid:48) such that − x ≥ | x (cid:48) | , F (cid:16) x + x (cid:48) (cid:17) ≥ F ( x ) + F ( x (cid:48) ) .2. For all x, x (cid:48) such that x ≥ | x (cid:48) | and all y ≥ , F ( x + y ) + F ( x (cid:48) − y ) ≥ F ( x ) + F ( x (cid:48) ) .Proof. If − x ≥ | x (cid:48) | then there are two cases to consider:1. If x (cid:48) ≤ F ( x ) is convex for x ≥ 0, 2 F (cid:16) − x + x (cid:48) (cid:17) ≤ F ( − x ) + F ( − x (cid:48) ). Using the fact that F ( − x ) = − F ( x ) and rearranging, 2 F (cid:16) x + x (cid:48) (cid:17) ≥ F ( x ) + F ( x (cid:48) ).2. If x (cid:48) ≥ F ( x ) is convex for x ≥ F ( − ( x + x (cid:48) )) + F ( x (cid:48) ) ≤ F ( − x ) + F (0) and2 F (cid:16) − x + x (cid:48) (cid:17) ≤ F ( − ( x + x (cid:48) )) + F (0). Adding these two inequalities together, using the fact that F ( − x ) = − F ( x ), and rearranging, 2 F (cid:16) x + x (cid:48) (cid:17) ≥ F ( x ) + F ( x (cid:48) ).Similarly, if | x (cid:48) | ≤ x then there are three cases to consider:1. If x (cid:48) ≤ F ( x ) is convex for x ≥ F ( x ) + F ( y − x (cid:48) ) ≤ F ( − x (cid:48) ) + F ( x + y ). Rearrangingand using the fact that F ( − x ) = − F ( x ), F ( x + y ) + F ( x (cid:48) − y ) ≥ F ( x ) + F ( x (cid:48) ).2. If 0 ≤ x (cid:48) ≤ y then since F ( x ) is convex for x ≥ F ( x ) + F ( x (cid:48) ) ≤ F (0) + F ( x + x (cid:48) ) and F ( x + x (cid:48) ) + F ( y − x (cid:48) ) ≤ F ( x + y ) + F (0). Adding these two inequalities together, using the factthat F ( − x ) = − F ( x ), and rearranging, F ( x + y ) + F ( x (cid:48) − y ) ≥ F ( x ) + F ( x (cid:48) ).3. If x (cid:48) ≥ y then since F ( x ) is convex for x ≥ F ( x ) + F ( x (cid:48) ) ≤ F ( x + y ) + F ( x (cid:48) − y ). Remark 4.3. Technically, we need to make sure that we are making progress when we apply Lemma 4.2.To ensure this, we apply Lemma 4.2 so that1. All x stay in the range [ − , .2. We never change the value of any x to − unless it was − already.3. Whenever we apply the second statement (or more precisely, make a sequence of up to applicationsof the second statement), we change at least one x which is less than to .Under these conditions, our applications of Lemma 4.2 reduce (cid:80) x φ ( x ) for the following potential function φ ( x ) : φ ( x ) = x − x =1 + 1 x = − This implies that we make progress and can converge to a distribution where Lemma 4.2 cannot be applied. .1.2 The Hardest Distributions for MAX CUT/MAX-NAE- { } -SATTheorem 4.4 ([25]) . For MAX CUT/MAX-NAE- { } -SAT with any completeness, the most difficult distri-butions to round are when the constraints are supported on {− ρ, } for some ρ ∈ [0 , .Proof. To prove this theorem, we show that any distribution which is not of this form can be improved. Forthis, we use a sequence of lemmas. Lemma 4.5. For MAX CUT/MAX-NAE- { } -SAT ,1. If there are two constraints with pairwise biases x and x (cid:48) where | x (cid:48) | ≤ − x ≤ then replacing x and x (cid:48) with two copies of x + x (cid:48) does not affect the SDP value and can only decrease the performance of therounding scheme.2. If there are two constraints with pairwise biases x and x (cid:48) where < | x (cid:48) | ≤ x ≤ then replacing x and x (cid:48) with x + x (cid:48) − and does not affect the SDP value and can only decrease the performance ofthe rounding scheme.Proof. The SDP value is linear in x and x (cid:48) so it is not affected by these adjustments. To show that theperformance of the rounding scheme can only decrease, we make the following observations:1. The first statement follows directly from the first statement of Lemma 4.2.2. The second statement follows from applying the second statement of Lemma 4.2 to x and x (cid:48) with y = 1 − x .It is not hard to verify that the only distributions which cannot be improved using this lemma are distributionswhere the pairwise biases are supported on {− ρ, } for some ρ ∈ [0 , In order to analyze the hardest distribution of triples of pairwise biases for MAX-NAE- { } -SAT, we needto consider which triples of pairwise biases are possible for a set of three ± Lemma 4.6 (e.g., [10]) . For distributions on three ± variables x , x , x , the polytope of possible pairwisebiases is given by the following inequalities:1. b , + b , + b , ≥ − b , ≥ b , + b , − b , ≥ b , + b , − b , ≥ b , + b , − Remark 4.7. Observe that the inequalities − ≤ b , ≤ , − ≤ b , ≤ , and − ≤ b , ≤ are impliedby these inequalities. To see this, note that adding the first two inequalities and dividing by gives theinequality b , ≥ − . Adding the second and third inequality and dividing by gives the inequality b , ≤ .By symmetry, the other inequalities can be derived in a similar way.Proof. The possible pairwise biases for integral assignments are as follows.1. (1 , , , − , − − , , − 1) 16. ( − , − , b , = b , = b , = 1 with probability c +++ = b , + b , + b , +14 b , = 1 and b , = b , = − c + −− = b , − b , − b , +14 b , = 1 and b , = b , = − c − + − = b , − b , − b , +14 b , = 1 and b , = b , = − c −− + = b , − b , − b , +14 Since the inequalities are satisfied, we have that c +++ , c + −− , c − + − , c −− + ∈ [0 , c +++ + c + −− + c − + − + c −− + = 12. E [ x x ] = c +++ + c + −− − c − + − − c −− + = b , E [ x x ] = c +++ − c + −− + c − + − − c −− + = b , E [ x x ] = c +++ − c + −− − c − + − + c −− + = b , so this distribution matches the given pairwise biases. { } -SATTheorem 4.8. For MAX-NAE- { } -SAT with any completeness, the hardest distributions are of the followingforms.1. The distribution of the pairwise biases for the constraints is supported on { ( − ρ, − ρ, − ρ ) , ( − ρ, − ρ, } for some ρ ∈ [0 , ] .2. The distribution of the pairwise biases for the constraints is supported on { ( − , − , − ) , ( − ρ, − ρ, } for some ρ ∈ [ , .3. The distribution of the pairwise biases for the constraints is supported on { ( − ρ, − ρ, , (1 , , } for some ρ ∈ [ − , . Remark 4.9. The reason why we cannot have triples of pairwise biases ( − ρ, − ρ, − ρ ) where ρ > is becausefor any triple ( b , , b , , b , ) of pairwise biases, b , + b , + b , ≥ − .Proof. To prove this theorem, we show that any distribution which is not of this form can be improved. Forthis, we use the following lemmas. Lemma 4.10. For a single MAX NAE- { } -SAT constraint,1. If the pairwise biases are x, x (cid:48) , x (cid:48)(cid:48) where | x (cid:48) | ≤ − x ≤ then replacing x and x (cid:48) with two copies of x + x (cid:48) does not affect the SDP value and can only decrease the performance of the rounding scheme.2. If the pairwise biases are x, x (cid:48) , x (cid:48)(cid:48) where max {| x (cid:48) | , | x (cid:48)(cid:48) |} ≤ x ≤ , and x (cid:48)(cid:48) < x (cid:48) then replacing x and x (cid:48) with x + x (cid:48) − x (cid:48)(cid:48) and x (cid:48)(cid:48) does not affect the SDP value and can only decrease the performance of therounding scheme. Observe that x (cid:48)(cid:48) ≥ x + x (cid:48) − so x + x (cid:48) − x (cid:48)(cid:48) ≤ .3. If the pairwise biases are x, x (cid:48) , x (cid:48)(cid:48) where x (cid:48)(cid:48) = x (cid:48) and | x (cid:48) | < x ≤ then replacing x, x (cid:48) , x (cid:48)(cid:48) with , x (cid:48) + x − , and x (cid:48)(cid:48) + x − does not affect the SDP value and can only decrease the performance of therounding scheme. Observe that since x > | x (cid:48) | , x (cid:48) + x − = x + x (cid:48) + x (cid:48) − > x (cid:48) − ≥ − . emark 4.11. Whenever case holds and x + x (cid:48) − x (cid:48)(cid:48) < , we immediately apply case on ( x + x (cid:48) − x (cid:48)(cid:48) , x (cid:48)(cid:48) , x (cid:48)(cid:48) ) .This sequence changes x to , satisfying the conditions of Remark 4.3.Proof. The SDP value is linear in x, x (cid:48) , x (cid:48)(cid:48) so it is not affected by these adjustments. To show that theperformance of the rounding scheme can only decrease, we make the following observations:1. The first statement follows directly from the first statement of Lemma 4.2.2. The second statement follows from applying the second statement of Lemma 4.2 to x and x (cid:48) with y = x (cid:48) − x (cid:48)(cid:48) .3. The third statement follows from applying the second statement of Lemma 4.2 twice with y = − x .The first time we apply it to x and x (cid:48) and the second time we apply it to x +12 and x (cid:48)(cid:48) . Corollary 4.12. The hardest distributions have triples of pairwise biases of the form ( x, x, x ) where x ∈ [ − , or of the form ( x (cid:48) , x (cid:48) , where x (cid:48) ∈ [ − , . In order to prove Theorem 4.8, we need to show that there can only be one value of x and only one valueof x (cid:48) which is not equal to 1. For this, we use the following lemma. Lemma 4.13. For MAX-NAE- { } -SAT ,1. If there are two constraints with pairwise biases ( x, x, x ) and ( x (cid:48) , x (cid:48) , x (cid:48) ) where x, x (cid:48) ∈ [ − , thenreplacing these triples of pairwise biases with two copies of ( x + x (cid:48) , x + x (cid:48) , x + x (cid:48) ) does not affect the SDPvalue and can only decrease the performance of the rounding scheme.2. If there are two constraints with pairwise biases ( x, x, and ( x (cid:48) , x (cid:48) , where | x (cid:48) | ≤ − x ≤ thenreplacing these triples of pairwise biases with two copies of ( x + x (cid:48) , x + x (cid:48) , does not affect the SDPvalue and can only decrease the performance of the rounding scheme.3. If there are two constraints with pairwise biases ( x, x, and ( x (cid:48) , x (cid:48) , where | x (cid:48) | < x ≤ then replacingthese triples of pairwise biases with (1 , , and ( x + x (cid:48) − , x + x (cid:48) − , does not affect the SDP valueand can only decrease the performance of the rounding scheme.Proof. Again, the SDP value is linear in x and x (cid:48) so it is not affected by these adjustments. To show thatthe performance of the rounding scheme can only decrease, we make the following observations:1. The first and second statements follow directly from the first statement of Lemma 4.2.2. The third statement follows from applying the second statement of Lemma 4.2 to x and x (cid:48) with y = 1 − x .This implies that we can take the triples of pairwise biases to be supported on { ( x, x, x ) , ( x (cid:48) , x (cid:48) , , (1 , , } for some x ∈ [ − , 0] and some x (cid:48) ∈ [ − , x = x (cid:48) ∈ [ − , 0] or take x = − and x (cid:48) ∈ [ − , − ]. To show this, we use the following lemma. Lemma 4.14. For MAX-NAE- { } -SAT ,1. If there are two constraints with pairwise biases ( x, x, x ) and ( x (cid:48) , x (cid:48) , where x ∈ [ − , , x (cid:48) ∈ [ − , ,and x +2 x (cid:48) ≥ − then replacing these triples of pairwise biases with ( x +2 x (cid:48) , x +2 x (cid:48) , x +2 x (cid:48) ) and ( x +2 x (cid:48) , x +2 x (cid:48) , does not affect the SDP value and can only decrease theperformance of the rounding scheme.2. If there are two constraints with pairwise biases ( x, x, x ) and ( x (cid:48) , x (cid:48) , where x ∈ [ − , , x (cid:48) ∈ [ − , − ] ,and x +2 x (cid:48) ≤ − then replacing these triples of pairwise biases with ( − , − , − ) and ( x (cid:48) + x +12 , x (cid:48) + x +12 , does not affect the SDP value and can only decrease the performance of the rounding scheme.Proof sketch. This can be shown using the fact that F is an odd function and F ( x ) is convex for x ≥ x, x, x ) and (1 , , 1) then we can get rid of one of them. To see this,observe that as far as the SDP value and the performance of the rounding scheme are concerned, havingconstraints with pairwise biases ( x, x, x ) and (1 , , 1) with weights and respectively is the same as havinga constraint with pairwise biases ( x, x, x, x, x ) and (1 , , 1) we can convert theminto ( x, x, 1) until one of them is exhausted. Having derived the hardest distribution for MAX NAE- { } -SAT, we conclude this subsection with thefollowing optimality result. Definition 4.15. Let f : R k → [ − , be an RPR k rounding function. • For a distribution D of biases, define s ( f, D ) := − E b ∼D [ F [ f ]( b )]2 . • For a distribution D of triples of biases, define s ( f, D ) := − E ( b , ,b , ,b , ∼D [ F [ f ]( b , )+ F [ f ]( b , )+ F [ f ]( b , )]4 . Claim 4.16 ([25]) . For any f : R k → [ − , , there exists a monotone, odd, one-dimensional function g : R → [ − , such that • s ( g, D hard , ) ≥ s ( f, D hard , ) for any distribution of biases D hard , described by Theorem 4.4. • s ( g, D hard , ) ≥ s ( f, D hard , ) for any distribution of triples of biases D hard , described by Theorem 4.8.Proof. If we apply Theorem 2.10 and let g = f ∗ be the Gaussian rearrangement of f , then g is a monotone,odd, one-dimensional function. Furthermore, F [ g ](1) = F [ f ](1) and F [ g ]( x ) ≤ F [ f ]( x ) for all x ∈ [ − , { } -SAT is in [ − , ∪ 1. Thus, E b ∼D hard , [ F [ g ]( b )] ≤ E b ∼D hard , [ F [ f ]( b )] and E ( x ,x ,x ) ∼D hard , [ F [ g ]( x )+ F [ g ]( x )+ F [ g ]( x )] ≤ E ( x ,x ,x ) ∼D hard , [ F [ f ]( x )+ F [ f ]( x )+ F [ f ]( x )] . Therefore, s ( g, D hard , ) ≥ s ( f, D hard , ) and s ( g, D hard , ) ≥ s ( f, D hard , ). Let f : R → [ − , 1] be an odd, monotone RPR rounding function for MAX CUT. By Theorem 4.4, fora given completeness c ∈ [1 / , D are a combination of equal vectorsand vectors with dot product ρ . Let α ∈ [0 , 1] be the relative frequency of these two dot products, that is c = α · − ρ . The performance of the rounding scheme will be s ( f, D ) = (1 − α ) · − F (1)2 + α · − F ( ρ )2= 12 − − α (cid:90) R f ( x ) φ ( x ) dx − α (cid:90) R f ( x ) f ( y ) φ ρ ( x, y ) dxdy , where φ ( x ) = √ π exp( − x / 2) is the density of a standard normal variable, and φ ρ ( x, y ) = exp (cid:0) − ( x, y ) T Σ − ( x, y ) (cid:1) π (cid:112) − ρ = exp (cid:16) − x − ρxy + y − ρ ) (cid:17) π (cid:112) − ρ is the density function of a two-dimensional normal random variable with mean µ = (0 , 0) covariance matrixΣ = (cid:18) ρρ (cid:19) so that Σ − = − ρ (cid:18) − ρ − ρ (cid:19) . 19hus, maximizing s ( f, D ) is equivalent to minimizing L ( f ) = 1 − αα (cid:90) ∞−∞ f ( x ) φ ( x ) dx + (cid:90) R f ( x ) f ( y ) φ ρ ( x, y ) dxdy . We know that f is monotone and that it attains the values − | x | is sufficiently large. Let a := sup { x : f ( x ) < } . Since f is increasing and odd, we know that a ≥ | f ( x ) | < − a, a ).Note that it may be the case that a = ∞ .We prove in Appendix C that there is an ‘optimal’ f . That is, there exists a functions f which globallyminimizes L ( f ) among all valid RPR rounding functions. This means that any adjustment to f , notviolating the condition (cid:107) f (cid:107) ∞ ≤ a (cid:48) ∈ (0 , a ) and anymeasurable h : ( − a (cid:48) , a (cid:48) ) → [ − , 1] (not necessarily odd nor monotone). Note that for some ε sufficientlysmall, for all s ∈ [ − ε, ε ], (cid:107) f + sh (cid:107) ∞ ≤ 1. Then, in order for there to be a local minimum of L at s = 0, wemust have that 0 = 12 ∂∂s L ( f + sh ) (cid:12)(cid:12)(cid:12)(cid:12) s =0 = 1 − αα (cid:90) a (cid:48) − a (cid:48) f ( y ) h ( y ) φ ( y ) dy + (cid:90) ∞−∞ (cid:90) a (cid:48) − a (cid:48) f ( x ) h ( y ) φ ρ ( x, y ) dxdy = (cid:90) a (cid:48) − a (cid:48) h ( y ) (cid:18) − αα f ( y ) φ ( y ) + (cid:90) ∞−∞ f ( x ) φ ρ ( x, y ) dx (cid:19) dy . Since h is an arbitrary measurable function on ( − a (cid:48) , a (cid:48) ), we have that for almost every y ∈ ( − a (cid:48) , a (cid:48) ),1 − αα f ( y ) φ ( y ) + (cid:90) ∞−∞ f ( x ) φ ρ ( x, y ) dx = 0 . (1)Since a (cid:48) ∈ (0 , a ). The above holds for all x ∈ ( − a, a ). Further define K ( x, y ) := φ ρ ( x, y ) φ ( y ) = exp (cid:16) − ( x − ρy ) − ρ ) (cid:17)(cid:112) π (1 − ρ ) . and g ( y ) := α − α (cid:18)(cid:90) − a −∞ K ( x, y ) dx − (cid:90) −∞ a K ( x, y ) dx (cid:19) . Then, (1) becomes f ( y ) + α − α (cid:90) a − a f ( x ) K ( x, y ) dx = g ( y ) , − a ≤ y ≤ a . (2)This is a Fredholm integral equation of the second kind (c.f., [27]). Let s ( f, D ) | y be the expected size of the cut produced by using the rounding funtion f conditioned on thevalue of y , i.e., s ( f, D ) | y = 12 − − α f ( y ) − α (cid:90) ∞−∞ f ( x ) f ( y ) φ ρ ( x, y ) φ ( y ) dx . Feige and Langberg [11] argue that if f is optimal, then for every y we have s ( f, D ) | y ≥ , and if − 1, then s ( f, D ) | y = . Their intuitive argument is that if s ( f, D ) | y < then the cut produced bythe rounding procedure is not locally optimal in expectation and can thus be improved.If − < f ( y ) < s ( f, D ) | y = which, by moving sides and dividing by f ( y ), is equivalent to1 − αα f ( y ) + (cid:90) ∞−∞ f ( x ) φ ρ ( x, y ) φ ( y ) dx = 0 . This is exactly the integral equation we got above. 20 .3 Analogous Condition for MAX NAE- { } -SAT Let f : R → [ − , 1] be an odd, increasing RPR rounding function for MAX NAE- { } -SAT. By Theorem 4.8,for a given completeness c ∈ [3 / , D have ( ρ , ρ , ρ ) with probablity α and (1 , ρ, ρ ) with probability 1 − α , where ρ is either max( ρ, − / 3) or ρ = 1.We first consider the case ρ = max( ρ, − / s ( f, D ) = (1 − α ) · − F (1) − F ( ρ )4 + α · − F ( ρ )4= 34 − − α (cid:90) R f ( x ) φ ( x ) dx − − α (cid:90) R f ( x ) f ( y ) φ ρ ( x, y ) dxdy − α (cid:90) R f ( x ) f ( y ) φ ρ ( x, y ) dxdy . Thus, maximizing s ( f, D ) is equivalent to minimizing L ( f ) = 1 − α α (cid:90) R f ( x ) φ ( x ) dx + 2 − α α (cid:90) R f ( x ) f ( y ) φ ρ ( x, y ) dxdy + (cid:90) R f ( x ) f ( y ) φ ρ ( x, y ) dxdy . We know that f is monotone and that it attains the values − | x | is sufficiently large. Let a := sup { x : f ( x ) < } . Since f is increasing and odd, we know that a ≥ | f ( x ) | < − a, a ).Note that it may be the case that a = ∞ .We prove in Appendix C that there is an ‘optimal’ f . That is, there exists a functions f which globallyminimizes L ( f ) among all valid RPR rounding functions. This means that any adjustment to f , notviolating the condition (cid:107) f (cid:107) ∞ ≤ a (cid:48) ∈ (0 , a ) and anymeasurable h : ( − a (cid:48) , a (cid:48) ) → [ − , 1] (not necessarily odd nor monotone). Note that for some ε sufficientlysmall, for all s ∈ [ − ε, ε ], (cid:107) f + sh (cid:107) ∞ ≤ 1. Then, in order for there to be a local minimum of L at s = 0, wemust have that0 = 12 ∂∂s L ( f + sh ) (cid:12)(cid:12)(cid:12)(cid:12) s =0 = 1 − α α (cid:90) a (cid:48) − a (cid:48) f ( y ) h ( y ) φ ( y ) dy + 2 − α α (cid:90) ∞−∞ (cid:90) a (cid:48) − a (cid:48) f ( x ) h ( y ) φ ρ ( x, y ) dxdy + (cid:90) ∞−∞ (cid:90) a (cid:48) − a (cid:48) f ( x ) h ( y ) φ ρ ( x, y ) dxdy = (cid:90) a (cid:48) − a (cid:48) h ( y ) (cid:18) − α α f ( y ) φ ( y ) + 2 − α α (cid:90) ∞∞ f ( x ) φ ρ ( x, y ) dx (cid:90) ∞−∞ f ( x ) φ ρ ( x, y ) dx (cid:19) dy . Since h is an arbitrary measurable function on ( − a (cid:48) , a (cid:48) ), we have that for almost every y ∈ ( − a (cid:48) , a (cid:48) ),1 − α α f ( y ) φ ( y ) + 2 − α α (cid:90) ∞∞ f ( x ) φ ρ ( x, y ) dx (cid:90) ∞−∞ f ( x ) φ ρ ( x, y ) dx = 0 . (3)Since a (cid:48) ∈ (0 , a ). The above holds for all x ∈ ( − a, a ). Further define K ( x, y ) := 2 − α α · φ ρ ( x, y ) φ ( y ) + φ ρ ( x, y ) φ ( y ) = 2 − α α · exp (cid:16) − ( x − ρy ) − ρ ) (cid:17)(cid:112) π (1 − ρ ) + exp (cid:16) − ( x − ρ y ) − ρ ) (cid:17)(cid:112) π (1 − ρ ) . and g ( y ) = 3 α − α (cid:18)(cid:90) − a −∞ K ( x, y ) dx − (cid:90) −∞ a K ( x, y ) dx (cid:19) . Then, (3) becomes f ( y ) + 3 α − α (cid:90) a − a f ( x ) K ( x, y ) dx = g ( y ) , − a ≤ y ≤ a . (4)This is a Fredholm integral equation of the second kind.21n the other case, where ρ = 1, we have that s ( f, D ) = (1 − α ) · − F (1) − F ( ρ )4 + α · − F ( ρ )4= 34 − − α (cid:90) R f ( x ) φ ( x ) dx − − α (cid:90) R f ( x ) f ( y ) φ ρ ( x, y ) dxdy − α (cid:90) R f ( x ) φ ( x ) dx = 34 − − α (cid:90) R f ( x ) φ ( x ) dx − − α (cid:90) R f ( x ) f ( y ) φ ρ ( x, y ) dxdy. By applying a similar argument, we have a := sup { x : | f ( x ) | < } and for all y ∈ ( − a, a ),1 − α − α f ( y ) φ ( y ) + (cid:90) ∞−∞ f ( x ) φ ρ ( x, y ) dx = 0 . (5)We can then define K ( x, y ) := φ ρ ( x, y ) φ ( y ) = exp (cid:16) − ( x − ρy ) − ρ ) (cid:17)(cid:112) π (1 − ρ ) . and g ( y ) := 1 − α − α (cid:18)(cid:90) − a −∞ K ( x, y ) dx − (cid:90) −∞ a K ( x, y ) dx (cid:19) . Then, (5) becomes f ( y ) + 1 − α − α (cid:90) a − a f ( x ) K ( x, y ) dx = g ( y ) , − a ≤ y ≤ a . (6) A Fredholm integral equation of the second kind has the following standard form: f ( x ) − λ (cid:90) ba K ( x, y ) f ( y ) dy = g ( x ) , a ≤ x ≤ b . Perhaps the simplest way of solving such integral equations that do not have a closed-form solution is themethod of successive approximations. We construct a sequence of functions { f n ( x ) } that hopefully convergeto a solution of the integral equation. The sequence { f n ( x ) } is defined as follows: f ( x ) = g ( x ) .f n ( x ) = g ( x ) + λ (cid:90) ba K ( x, y ) f n − ( y ) dy , n > , Since we have more structure in our kernels for MAX CUT and MAX NAE- { } -SAT, we can compute adiscretization of f by solving a linear system. Recall that we sought to maximize a functional of the form L ( f ) = c − λ (cid:90) R f ( x ) φ ( x ) dx − λ (cid:90) R f ( x ) M ( x, y ) f ( y ) dy, (7)22or some suitable function M ( x, y ). Let N ≥ R into −∞ = a < a < · · · < a N = ∞ such that for all i ∈ { , , . . . , N } , (cid:90) a i a i − φ ( x ) dx = 1 N . Assume that f is piecewise constant, taking on value f i in the interval ( a i − , a i ). We let f denote the vectorof f i ’s. Note that we may still assume that f is odd and monotone. Define for all i, j ∈ { , . . . , N } defineˆ M i,j = (cid:90) a j a j − (cid:90) a i a i − M ( x, y ) dx dy. Then (7) becomes L ( f ) = c − λ N N (cid:88) i =1 f i + N (cid:88) i =1 N (cid:88) j =1 f i ˆ M i,j f j = c − f T ( λ I + λ ˆ M ) f . For all i such that | f i | < 1, we must have that0 = − ∂∂f i L ( f ) = 2 λ f i + λ n (cid:88) j =1 ( ˆ M i,j + ˆ M ji ) f j . In our situations, ˆ M is a symmetric matrix, so we have for all i such that | f i | < f i + λ λ n (cid:88) j =1 ˆ M i,j f j = 0 . Let λ := λ /λ and let i a be the largest index such that f i = − 1. Then, we can define for i ∈ { i a +1 , . . . , N − i a } , g i := λ i a (cid:88) j =1 ˆ M i,j − ˆ M i,N − j +1 . Let f (cid:48) be f restricted to { i a + 1 , . . . , N − i a } and ˆ M (cid:48) be ˆ M restricted to { i a + 1 , . . . , N − i a } . We then havethe linear system ( I + λ ˆ M (cid:48) ) f (cid:48) = g . This is a discrete Fredholm equation of the second kind. We can directly solve the linear system. Notethat we need to guess the value of i a , so we need to solve multiple linear systems. Empirically, we can use abinary search to find the value of i a .For any ε > 0, if we pick N = O (1 /ε ) and try α and ρ in a grid of size O (1 /ε ). For each choice, we shall getan optimal step function for that distribution. By trying all of these functions on any input and taking thebest expected result (e.g., [30]), we can compute optimal approximation factor to within an additive ε O (1) . Using MATLAB, we implemented a search to find the optimal ˆ f , for various choices of α ∈ [0 , ρ ∈ [ − , { } -SAT) ρ ∈ { max( − / , ρ ) , } . For MAX CUT, our results reproduced the resultsfound by [25].For MAX NAE- { } -SAT, we started with a coarse search. In particular, we did a grid search over 500 values of α and ρ and considered step functions with N = 100 steps. From this, we computed a numer-ical approximation of the completeness/soundness tradeoff curve (Figure 1). Using these calculations, weestimated that the approximation ratio of MAX NAE- { } -SAT is 0.9089 to four digits of precision. Note that I + λ ˆ M (cid:48) can only be non-invertible for O ( N ) values of λ . In the experiments (next section), this was onlyproblematic when α = 1 where I + λ ˆ M (cid:48) has rank 1. A discussion about invertibility of similar linear systems for MAX CUT isgiven in Section 6 of [25]. Proof of this follows by suitably modifying the arguments of [25]. .75 0.8 0.85 0.9 0.95 1 c s Figure 1: This plot shows the tradeoff between completeness ( x -axis) and soundness ( y -axis) for MAXNAE- { } -SAT (c.f., figure of [25]).With a more fine-grained search around the hardest points, we found that the most difficult point is α ≈ . ρ ≈ − . ρ = − / 3. At this point, we computed a step function with 600 steps whichattains an approximation ratio of ≈ . f , if we assume ρ = 1, then s ( D ) − . · c ( f, D )is a affine function in α and a convex function in ρ . This is also the case if ρ = − / ρ = ρ . Thus, foreach of α ∈ { , } and ρ ∈ { , − / , ρ } , we did a ternary search to check that f achieves an approximationof at least . − in the choice of ρ ).Figure 2: (left) Near-optimal rounding function for MAX NAE- { } -SAT in terms of approximation factor.(right) Approximate deviation of near-optimal rounding function for MAX NAE- { } -SAT from the best-fit s -linear function.This step function is essentially indistinguishable from a s -linear function of the form max(min( sx, , − s ≈ . f from an s -linear function in the region where f is strictly between − s -linear is smaller than 10 − .24 Approximation algorithms for satisfiable MAX NAE- K -SAT In this section we experiment with approximation algorithms for MAX NAE-SAT, as well as some restrictionsof it, such as MAX NAE- { , } -SAT and MAX NAE- { , , } -SAT. We focus in this paper on approximationalgorithms for satisfiable, or almost satisfiable, instances of these problems. When considering satisfiable, oralmost satisfiable, instances, we may assume that there are no clauses of size 2. Our proposed approximation algorithms round the solution of the basic SDP using the RPR method with acarefully chosen rounding function f : [ −∞ , ∞ ] → [ − , α K ( f ) of an algorithmfor MAX NAE- K -SAT, for some finite set K , that uses RPR with rounding function f is α K ( f ) = min k ∈ K min v , v ,..., v k relax ( v , v ,..., v k )=1 prob f ( v , v , . . . , v k ) , where v , v , . . . , v k are assumed to be unit vectors that can be written as a convex combination of integralsolutions, i.e., relax ( v , v , . . . , v k ) = 1, and prob f ( v , v , . . . , v k ) is the probability that rounding the vectorsusing f yields an assignment that satisfies the corresponding NAE k clause.Even for a given f , this is a fairly difficult optimization problem. For k = 9, for example, this is essentially a36-dimensional problem. (Here 36 = (cid:0) (cid:1) .) What makes the problem even harder is that even the computationof prob f ( v , v , . . . , v k ) for given vectors v , v , . . . , v k is a non-trivial task, as it essentially amounts tocomputing a k -dimensional integral.In view of these difficulties, the approximation ratios of the algorithms we consider for MAX NAE- K -SAT,where max K > 3, are only conjectured. We believe that we know which configuration v , v , . . . , v k attainsthe minimum in the above expression, and thus determines the approximation ratio, but we are not able, atthe moment, to prove it rigorously.It is, in principal, possible to make the analysis of the proposed approximation algorithms rigorous, as donefor MAX [3]-SAT and MAX NAE-[4]-SAT in [36], but this would require a lot of effort. We note that evenfor MAX [2]-SAT and MAX DI-CUT, this is a non-trivial task. (See Sj¨ogren [32].)Even though the approximation ratios that we obtain are only conjectured, we believe that they are usefulguides for further theoretical investigations of the MAX NAE-SAT and MAX SAT problems. The experiments we did with approximation algorithms for satisfiable, or almost satisfiable, instances ofMAX NAE- K -SAT, for various sets K , lead us to make the following conjectures: Conjecture 2. The hardest configuration v , v , . . . , v k , where k ∈ K and k ≥ , for the optimal RP R rounding function f K for satisfiable, or almost satisfiable, instances of MAX NAE- K -SAT is the symmetricconfiguration in which v i · v j = 1 − k , for every i (cid:54) = j . Our intuition for this conjecture is that this configuration comes from taking the uniform distribution oversatisfying assignments where all but one of the X i are the same, which we expect are the hardest satisfyingassignments to distinguish from the unsatisfying assignments.A conjecture similar to Conjecture 2 was also made in Avidor et al. [7]. Note that for K = { } , the conjectureis true and is a corollary of Theorem 4.8, as the only hard point with completeness 1 is ( − , − , − ). When K = { , } or K = { } , the conjecture is also true as hyperplane rounding gives a approximation.However, for K which contain larger k , the situation is more subtle. When k ≥ 5, this point is not thehardest configuration for all rounding functions because it is not a local maximum for (cid:80) i MAXNAE- K -SAT is the (one-dimensional) RP R procedure with an appropriate rounding function f = f K . Our intuition for this conjecture is that if the hardest configuration is the symmetric configuration in which v i · v j = 1 − k , for every i (cid:54) = j then these vectors can be split into a common component and a componentwhich is orthogonal to everything else. For more details, see Section 5.4. Our conjecture is that onedimensional rounding schemes are most effective for interacting with this common component.Note that Conjecture 3 does not follow from Ragavendra [29, 30] and UGC, as Ragavendra [29, 30] uses ahigh-dimensional version of RP R . Conjecture 4. Furthermore, if | K | ≥ and min K = 3 , then the optimal RP R rounding function f K for MAX NAE- K -SAT is a step function that only assumes the values +1 and − . Our intuition for this conjecture is that for RP R rounding functions, we can describe its performance interms of Hermite coefficients (for details of how this works for F , see Appendix A). If it is the case that it’smost important to optimize the first few Hermite coefficients, this is accomplished by a ± K = 3 in Conjecture 4 is important. If min K > 3, then it follows fromH˚astad [15] that the optimal rounding procedure is simply choosing a random assignment, which is equivalentto using RP R with the function f ( x ) = 0, which does not assume only ± We give a motivation for an improved rounding scheme which does well for MAX NAE- { , } -SAT withperfect completeness. A tool used by [25] in studying RPR rounding functions is the Hermite expansion . Define the n th normalized Hermite polynomial to be [12] H n ( x ) = 1 √ n ! (cid:98) n/ (cid:99) (cid:88) (cid:96) =0 ( − (cid:96) m (cid:96) ( K n ) x n − (cid:96) . n = 0 , , , . . . Here m (cid:96) ( K n ) is the number of l -matchings in the complete graph of order n . These have the property thatthey form an orthonormal basis with respect to the Gaussian measure on R . That is, (cid:90) ∞−∞ H i ( x ) H j ( x ) φ ( x ) dx = (cid:40) i = j i (cid:54) = j i, j = 0 , , , . . . , where φ ( x ) = √ π e − x / .Note that any rounding scheme can be described by its Hermite expansion: f ( x ) = ∞ (cid:88) i =1 odd c i H i ( x ) , where, c i = (cid:104) f ( x ) , H i ( x ) (cid:105) := (cid:90) ∞−∞ f ( x ) H i ( x ) φ ( x ) dx . For a positive integer k ≥ 1, let P k := { (cid:0) (cid:104) f ( x ) , H ( x ) (cid:105) , (cid:104) f ( x ) , H ( x ) (cid:105) , . . . , (cid:104) f ( x ) , H k − ( x ) (cid:105) (cid:1) | ∀ f : R → [ − , } . Note that P k is a convex set. See Figure 4 for a picture of P . [25] consider the Hermite expansion for multivariate polynomials as well, but we only consider single variable polynomialsin this section. - - - - Figure 3: The first few normalized Hermite polynomials of odd degree.Figure 4: The tradeoff between c and c for extreme RPR rounding functionals. The marked points arethe boundary of P . Claim 5.1. Every ( c , . . . , c k ) ∈ P k is attained by a step function with at most k steps.Proof. Since P k is convex, every point in P k can be expressed as a convex combination of k extreme points.Thus, it suffices to show that each extreme point of P k can be attained by a step function with at most 2 k steps.Fix an extreme point ( c , c , . . . , c k ) ∈ P k . By definition of being an extreme point there is exists ( α , . . . , α k )such that (cid:80) ki =1 α i x i is maximized in P k at ( c , c , . . . , c k ). That is, we seek to maximizemax f : R → [ − , k (cid:88) i =1 (cid:104) f ( x ) , α i H i − ( x ) (cid:105) = max f : R → [ − , (cid:42) f ( x ) , k (cid:88) i =1 α i H i − ( x ) (cid:43) . This expression is maximized when f = sign (cid:16)(cid:80) ki =1 α i H i − ( x ) (cid:17) . Since (cid:80) ki =1 α i H i − ( x ) is a degree 2 k − k steps.Thus, if we conjecture that optimizing the first few Hermite coefficients suffices to optimize the approximationratio of MAX NAE-SAT to high precision, it suffices to look at step functions.Using MATLAB, we searched through step functions achieving points in P , and as a result, we came across27he following RPR rounding function for almost satisfiable instances of MAX NAE- { , } -SAT: f α ( x ) = − x < − α x ∈ [ − α, − x ∈ (0 , α ]1 x > α where α ≈ . - - - - Figure 5: The conjectured optimal rounding function f for MAX NAE- { , } -SAT. Plotted alongside arefirst four terms of its Hermite expansion, and the sum of these terms (in brown).Through numerical experiments, this rounding function achieves a soundness of approximately 0 . Consider a collection of k unit vectors v , v , . . . , v k such that v i · v j = ρ , for every i (cid:54) = j . We say that sucha configuration is symmetric . Let p f ( k, ρ ) = prob f ( v , v , . . . , v k ) be the probability that a NAE k clausecorresponding to these vectors v , v , . . . , v k is satisfied when using rounding function f , then: p f ( k, ρ ) = 12 k − (cid:32) (2 k − − − (cid:88) i even (cid:18) ki (cid:19) F i ( ρ ) (cid:33) , where F i ( ρ ) = F i ( ρ, ρ, . . . , ρ ). Recall that F i is the i -wise moment function defined in Section 2.3.When the pairwise biases are all equal, we can evaluate F (cid:96) , where (cid:96) is a positive integer, using two integralsinstead of an 2 (cid:96) -dimensional integral. The key idea is to decompose the covariance into a weighted sum ofthe all-one matrix J and the identity matrix I . Assume that we have 2 (cid:96) unit vectors v , . . . , v (cid:96) ∈ R n withinner product ρ > r ∼ N (0 , I n )and let x i = r · v i be the inner product between r and v i , then x i ∼ N (0 , 1) and Cov[ x i , x j ] = ρ for i (cid:54) = j .The covariance matrix can thus be written as ρJ + (1 − ρ ) I .The same distribution can be obtained by the following procedure: first sample x ∼ N (0 , ρ ), and thensample 2 (cid:96) independent variables ε i ∼ N (0 , − ρ ) where i = 1 , , . . . , (cid:96) , and finally let x i = x + ε i . Using28his, F (cid:96) can be computed as follows: F (cid:96) ( ρ ) = E x ∼ N (0 ,ρ ) ,ε i ∼ N (0 , − ρ ) (cid:34) (cid:96) (cid:89) i =1 f ( x + ε i ) (cid:35) = E x ∼ N (0 ,ρ ) (cid:34) (cid:96) (cid:89) i =1 E ε i ∼ N (0 , − ρ ) [ f ( x + ε i )] (cid:35) = E x ∼ N (0 ,ρ ) (cid:34) (cid:96) (cid:89) i =1 E ε i ∼ N (0 , − ρ ) (cid:20) f (cid:18) √ ρ · x √ ρ + (cid:112) − ρ · ε i √ − ρ (cid:19)(cid:21)(cid:35) = E x ∼ N (0 ,ρ ) (cid:34) (cid:96) (cid:89) i =1 U √ ρ f (cid:18) x √ ρ (cid:19)(cid:35) = E x ∼ N (0 , (cid:104)(cid:0) U √ ρ f ( x ) (cid:1) (cid:96) (cid:105) = ∞ (cid:90) −∞ (( U √ ρ f )( x )) l φ ( x ) dx . Here f is the RPR function used and U η , where η ∈ [0 , η f ( x ) = E ε ∼ N (0 , (cid:104) f (cid:16) ηx + (cid:112) − η ε (cid:17)(cid:105) = ∞ (cid:90) −∞ f (cid:18) ηx + (cid:112) − η y (cid:19) φ ( y ) dy . We note that if f has the Hermite expansion (cid:80) ∞ i =0 c i H i , then U η f has the Hermite expansion (cid:80) ∞ i =0 c i η i H i .Further observe that p f ( k, ρ ) = 12 k − (cid:32) (2 k − − − (cid:88) i even (cid:18) ki (cid:19) F i ( ρ ) (cid:33) = 1 − k ∞ (cid:90) −∞ (cid:18)(cid:0) U √ ρ f )( x ) (cid:1) k + (cid:0) − ( U √ ρ f )( x ) (cid:1) k (cid:19) φ ( x ) dx. Formula for step functions For every a = ( a , a , . . . , a (cid:96) ), where 0 < a < a < · · · < a (cid:96) , and − ≤ b , b , . . . , b (cid:96) ≤ 1, let f ( x ) = f a , b ( x )be the function such that f ( x ) = b i , if a i ≤ x < a i +1 , for i = 0 , , . . . , (cid:96) , where a = 0 and a (cid:96) +1 = ∞ . Also, f ( x ) = − f ( − x ), if x < 0. We say that f a , b ( x ) is a ( (cid:96) + 1)-step function, only counting steps to the right ofthe origin. It is easy to check that( U ρ f )( x ) = (cid:96) (cid:88) i =0 b i (cid:32) Φ (cid:18) a i +1 − ρx (cid:112) − ρ (cid:19) − Φ (cid:18) a i − ρx (cid:112) − ρ (cid:19) − Φ (cid:18) − a i − ρx (cid:112) − ρ (cid:19) + Φ (cid:18) − a i +1 − ρx (cid:112) − ρ (cid:19)(cid:33) , where Φ( x ) = (cid:82) x −∞ φ ( y ) dy is the cumulative distribution function of the standard normal distribution. For every a = ( a , a , . . . , a (cid:96) ), where 0 < a < a < · · · < a (cid:96) , and − ≤ b , b , . . . , b (cid:96) ≤ 1, let f ( x ) = f a , b ( x ),as above, be the function such that f ( x ) = b i , if a i ≤ x < a i +1 , for i = 0 , , . . . , (cid:96) , where a = 0 and a (cid:96) +1 = ∞ . Also, f ( x ) = − f ( − x ), if x < 0. We say that f a , b ( x ) is a ( (cid:96) + 1)-step function, only countingsteps to the right of the origin. We also let f a ( x ) be the function f a ( x ) = f a , b ( x ), where b i = ( − i +1 , for i = 0 , , . . . , (cid:96) . Note that hyperplane rounding uses a 1-step function f () , (1) ( x ) with (cid:96) = 0 and b = 1.Motivated by the results of Sections 5.3 (see also Appendix D), we did extensive numerical experiments withstep functions. For a given set K , and a given number k + 1 of steps, we solved a numerical optimization29igure 6: The best 4-step rounding function for K = { , } (left) and for K = { , , } (right).problem in which the variables are a , a , . . . , a (cid:96) and b , b , . . . , b (cid:96) . The objective function maximized was α K ( f a , b ) = min k ∈ K α k ( f a , b ), where a = ( a , a , . . . , a (cid:96) ) and b = ( b , b , . . . , b (cid:96) ). When computing α k ( f a , b )we considered only the conjectured hardest configuration of Conjecture 2. As this configuration is symmetric,the probability α k ( f a , b ) = prob f a , b ( v , . . . , v k ) = p f a , b ( k, − k ), where v i · v j = 1 − k , for i ≤ j , when k ≥ 3, can be numerically computed using the formula, which involves integration, given in Section 5.4.Most of the experiments were carried out in Mathematica using numerical optimization tools, taking ad-vantage of the possibility of performing numerical calculations with arbitrary precision. The optimizationproblem of finding the optimal a , a , . . . , a (cid:96) and b , b , . . . , b (cid:96) is a fairly difficult optimization problem, asthe objective function α K ( f a , b ) = min k ∈ K α k ( f a , b ) is far from being a convex function. However, as thenumber of variables is relatively small, we were able to repeat the optimization attempts many times, fromdifferent initial points. This gives us some confidence that the best step functions found are close to beingthe optimal ones.No counterexamples to the other two conjectures of Section 5.2 were found. Most of our experiments weredevoted to confirming Conjecture 4, i.e., that the optimal RP R rounding function f K for MAX NAE- K -SAT, where | K | ≥ K ≥ 3, is a ± v , . . . , v k , where k ∈ K , for the optimal rounding function f K is the symmetric configurationwith v i · v j = 1 − k , for every i (cid:54) = j .The best step functions we found for satisfiable, or almost satisfiable, instances of MAX NAE- K -SAT, where K = { , k } , for k = 5 , , . . . , 8, and for K = { , , } are given in Table 2. What is interesting in theseexperiments is that when the number of allowed steps is high enough, the optimal step function is a ± K = { , } the best 2-step function found is already a ± K = { , } , the best step function found is a 3-step functions. For K = { , } and K = { , } , the beststep functions used are 4-step functions. For K = { , } and K = { , } , not shown in the table, the bestbest results are obtained using 5-step functions.Again, we could not improve on these functions by allowing more steps. It is possible, that tiny improvementscan be obtained by allowing more steps, but the improvements obtained in the approximation ratio, if any,are likely to be less than 10 − .Let α K be the best ratio found for MAX NAE- K -SAT. We currently have78 = α , > α , > α , > α , < α , < · · · Thus, if we just look at the mixture of two clause sizes, it seems that { , } is the hardest.However, it seems that α , , < α , < α , . The best rounding function found for MAX NAE- { , , } -SATis a 4-step ± K = { , } and K = { , , } are shown in Figure 6.30 = { , } a b b − K = { , } a a b b b − − − K = { , } a a a b b b b − − . − − − − K = { , } a a a b b b b − − . − − − − { } a a a b b b b − − . − − − − K -SAT with a given number of steps. The first column givesthe approximation ratio. The other columns give the vectors a and b .When using the best function found for K = { , , } , the satisfaction probabilities of clauses of size 3,7 and8 are the same, up to numerical error. The satisfaction probability of every clause size k > k (cid:54) = 7 , { , , } -SAT is as hard as approximating satisfiable, or almostsatisfiable, instances of MAX NAE-SAT. We thus conjecture that 0 . . 863 for satisfiable instances of MAX NAE-SAT is conjectured in Zwick [35].This conjectured approximation ratio is obtained using outward rotation, which is a special case of RP R rounding. Our new conjectured value of 0 . K = { , , } shown in Table 2 and Figure 6 is essentiallythe best rounding function for satisfiable, or almost satisfiable, instances of MAX NAE-SAT.If Conjecture 4 is true, i.e., the optimal rounding function is a step function, an interesting theoreticalquestion would be whether the optimal function has a finite or or an infinite number of steps. We speculatethat a finite number of steps is enough.We have made attempts to verify that the best step functions we found are at least local maxima, and thatthey cannot be improved by adding more steps.Figure 7 shows the effect of trying to add a second breakpoint, i.e., a third step, for K = { , } . The firstbreakpoint is held fixed at a ≈ . a . (The approximation ratio obtained using the best 2-step function is subtracted from theseprobabilities. (Both graphs show the same functions.) The two probabilities do cross each other around31igure 7: The effect of trying to add a second breakpoint for { , } . The two functions shown in the twofigures are the satisfaction probabilities of clauses of size 3 and 5, assuming Conjecture 2, as a function of a . The graph is the right is a close-up on a ≈ K = { , , } . a ≈ 7, but the two functions are increasing at that point, suggesting that the optimal value of a is + ∞ ,i.e., it is not beneficial to add a second breakpoint and a third step.Graphs showing the effect of adding the fourth step, and possibly a fifth step around a ≈ . K = { , , } , are shown in Figure 8. In the graph on the left, a a are fixed to the optimal settingfor a 3-step function, and the probabilities for clauses of sizes 3,7 and 8 are shown as a function of a . Thegraph on the right fixes a , a and a and shows the probabilities as a function of a . The graph on the rightseems to suggest that a really tiny improvement can be obtained by adding a fifth step. (Further numericalexperiments, that higher precisions, need to be done to verify it.) We presented the first improved hardness of approximation result for MAX NAE-SAT in nearly two decades,showing that no -approximation algorithm can be obtained for the problem, assuming UGC. We alsopresented an optimal algorithm for MAX NAE- { } -SAT, again assuming UGC. Finally, we presented analgorithm for (almost) satisfiable instances of MAX NAE-SAT that we conjecture to be nearly optimal.What we find striking is the contrast between MAX CUT and MAX NAE- { } -SAT, in which the optimalrounding functions are smooth (except for near-perfect completeness, where we get hyperplane rounding),monotone functions obtained as a solution of an integral equations, and MAX NAE- { , } -SAT, MAX NAE- { , , } -SAT and NAX NAE-SAT in which, at least for almost satisfiable instances, the apparent optimalrounding functions are non-monotone step functions that only assume ± F relatively well and used it to analyze N AE clauses of length 2 and3. However, higher moments functions are not well understood and several natural questions can be askedabout them. For example, we know that F ( x ) is convex in x for x ≥ 0. Does F (cid:96) ( x ) := F (cid:96) ( x, . . . , x ) havethe same property for (cid:96) ≥ 2? Having a better understanding of these higher moments can lead to a morerigorous analysis for clauses of length greater than 3.Finally, it would be interesting to see if some of the ideas introduced in this paper can be used to decidewhether there is a -approximation algorithm for MAX SAT. What makes MAX SAT potentially easierthan MAX NAE-SAT is that we can take advantage of individual biases. On the other hand, the search foroptimal rounding functions becomes harder. Acknowledgments We thank Ryan O’Donnell for sharing the code used to generate the experimental results in [25]. References [1] Gunnar Andersson and Lars Engebretsen. Better approximation algorithms for set splitting and not-all-equal SAT. Information Processing Letters , 65(6):305–311, 1998.[2] Sanjeev Arora, Carsten Lund, Rajeev Motwani, Madhu Sudan, and Mario Szegedy. Proof verificationand the hardness of approximation problems. Journal of the ACM (JACM) , 45(3):501–555, 1998.[3] Sanjeev Arora and Shmuel Safra. Probabilistic checking of proofs: A new characterization of np. Journalof the ACM (JACM) , 45(1):70–122, 1998.[4] Takao Asano and David P Williamson. Improved approximation algorithms for max sat. Journal ofAlgorithms , 42(1):173–202, 2002.[5] Per Austrin. Balanced max 2-SAT might not be the hardest. In Proceedings of the thirty-ninth annualACM symposium on Theory of computing , pages 189–197, 2007.[6] Per Austrin. Towards sharp inapproximability for any 2-CSP. SIAM Journal on Computing , 39(6):2430–2463, 2010.[7] Adi Avidor, Ido Berkovitch, and Uri Zwick. Improved approximation algorithms for MAX NAE-SATand MAX SAT. In Approximation and Online Algorithms, Third International Workshop, WAOA 2005 ,volume 3879 of Lecture Notes in Computer Science , pages 27–40. Springer, 2005.[8] Jonah Brown-Cohen and Prasad Raghavendra. Combinatorial optimization algorithms via polymor-phisms. CoRR , abs/1501.01598, 2015.[9] Andrei A Bulatov, Andrei A Krokhin, and Peter Jeavons. Constraint satisfaction problems and finitealgebras. In International Colloquium on Automata, Languages, and Programming , pages 272–282.Springer, 2000.[10] Uriel Feige and Michel Goemans. Approximating the value of two prover proof systems, with applicationsto MAX 2SAT and MAX DICUT. In Proceedings Third Israel Symposium on the Theory of Computingand Systems , pages 182–189. IEEE, 1995.[11] Uriel Feige and Michael Langberg. The RPR rounding technique for semidefinite programs. Journalof Algorithms , 60(1):1–23, 2006.[12] Chris D Godsil. Hermite polynomials and a duality relation for matchings polynomials. Combinatorica ,1(3):257–262, 1981. 3313] Michel X Goemans and David P Williamson. Improved approximation algorithms for maximum cut andsatisfiability problems using semidefinite programming. Journal of the ACM , 42(6):1115–1145, 1995.[14] Eran Halperin and Uri Zwick. Approximation algorithms for MAX 4-SAT and rounding procedures forsemidefinite programs. Journal of Algorithms , 40(2):184–211, 2001.[15] Johan H˚astad. Some optimal inapproximability results. Journal of the ACM (JACM) , 48(4):798–859,2001.[16] Neng Huang and Aaron Potechin. On the approximability of presidential type predicates. arXiv preprintarXiv:1907.04451 , 2019.[17] Leon Isserlis. On a formula for the product-moment coefficient of any order of a normal frequencydistribution in any number of variables. Biometrika , 12(1/2):134–139, 1918.[18] Howard Karloff and Uri Zwick. A 7/8-approximation algorithm for max 3sat? In Proceedings 38thAnnual Symposium on Foundations of Computer Science , pages 406–415. IEEE, 1997.[19] Subhash Khot. On the power of unique 2-prover 1-round games. In Proceedings of the thiry-fourthannual ACM symposium on Theory of computing , pages 767–775, 2002.[20] Subhash Khot, Guy Kindler, Elchanan Mossel, and Ryan O’Donnell. Optimal inapproximability resultsfor max-cut and other 2-variable csps? SIAM Journal on Computing , 37(1):319–357, 2007.[21] Giovanni Leoni. A first course in Sobolev spaces . American Mathematical Soc., 2017.[22] Michael Lewin, Dror Livnat, and Uri Zwick. Improved rounding techniques for the max 2-sat and maxdi-cut problems. In International Conference on Integer Programming and Combinatorial Optimization ,pages 67–82. Springer, 2002.[23] Konstantin Makarychev and Yury Makarychev. Approximation Algorithms for CSPs. In Andrei Krokhinand Stanislav Zivny, editors, The Constraint Satisfaction Problem: Complexity and Approximability ,volume 7 of Dagstuhl Follow-Ups , pages 287–325. Schloss Dagstuhl–Leibniz-Zentrum fuer Informatik,Dagstuhl, Germany, 2017.[24] Shiro Matuura and Tomomi Matsui. New approximation algorithms for max 2sat and max dicut. Journal of the Operations Research Society of Japan , 46(2):178–188, 2003.[25] Ryan O’Donnell and Yi Wu. An optimal SDP algorithm for Max-Cut, and equally optimal long codetests. In Proceedings of the fortieth annual ACM symposium on Theory of computing , pages 335–344,2008.[26] Ryan O’Donnell. Analysis of Boolean Functions . Cambridge University Press, USA, 2014.[27] Andrei D Polyanin and Alexander V Manzhirov. Handbook of integral equations . CRC press, 2008.[28] Aaron Potechin. On the approximation resistance of balanced linear threshold functions. In Proceedingsof the 51st Annual ACM SIGACT Symposium on Theory of Computing , pages 430–441, 2019.[29] Prasad Raghavendra. Optimal algorithms and inapproximability results for every CSP? In Proceedingsof the fortieth annual ACM symposium on Theory of computing , pages 245–254, 2008.[30] Prasad Raghavendra. Approximating NP-hard Problems - Efficient Algorithms and their Limits . PhDthesis, University of Washington, 2009.[31] Prasad Raghavendra and David Steurer. How to round any CSP. In , pages 586–594. IEEE, 2009.[32] Henrik Sj¨ogren. Rigorous analysis of approximation algorithms for MAX 2-CSP. Master’s thesis, KTHRoyal Institute of Technology, 2009. 3433] Jiawei Zhang, Yinyu Ye, and Qiaoming Han. Improved approximations for max set splitting and maxNAE SAT. Discrete Applied Mathematics , 142(1-3):133–149, 2004.[34] Dmitriy Zhuk. A proof of csp dichotomy conjecture. In , pages 331–342. IEEE, 2017.[35] Uri Zwick. Outward rotations: A tool for rounding solutions of semidefinite programming relaxations,with applications to MAX CUT and other problems. In Proceedings of the Thirty-First Annual ACMSymposium on Theory of Computing , pages 679–687. ACM, 1999.[36] Uri Zwick. Computer assisted proof of optimal approximability results. In SODA , pages 496–505, 2002. A Hermite Decomposition of F F in terms of the Hermite coefficients of the rounding function, as is previouslydone for F . We recall the following theorem from probability theory. Theorem A.1 (Isserlis, e.g., [17]) . Let X , X , . . . , X n be jointly gaussian random variables such that E [ X i ] = 0 for every i ∈ [2 n ] , then E [ X X · · · X n ] = (cid:88) M (cid:89) { i,j }∈ M E [ X i X j ] , E [ X X · · · X n − ] = 0 . Here the summation (cid:80) M runs over every perfect matching M of the complete graph K n . Corollary A.2. Let i, j, k, l be nonnegative integers, X , X , X , X be jointly gaussian zero-mean randomvariables such that E [ X s X t ] = ρ st for every ≤ s < t ≤ , then E [ H i ( X ) H j ( X ) H k ( X ) H l ( X )] = (cid:88) a,b,c,d,e,f : a + b + c = ia + d + e = jb + d + f = kc + e + f = l √ i ! j ! k ! l ! a ! b ! c ! d ! e ! f ! · ρ a ρ b ρ c ρ d ρ e ρ f . Proof. By linearity of expectation, we have E [ H i ( X ) H j ( X ) H k ( X ) H l ( X )]= (cid:98) i/ (cid:99) (cid:88) t =0 (cid:98) j/ (cid:99) (cid:88) t =0 (cid:98) k/ (cid:99) (cid:88) t =0 (cid:98) l/ (cid:99) (cid:88) t =0 E (cid:34) ( − t m t ( K i ) X i − t · ( − t m t ( K j ) X j − t · ( − t m t ( K k ) X k − t · ( − t m t ( K l ) X l − t √ i ! j ! k ! l ! (cid:35) = 1 √ i ! j ! k ! l ! (cid:98) i/ (cid:99) (cid:88) t =0 (cid:98) j/ (cid:99) (cid:88) t =0 (cid:98) k/ (cid:99) (cid:88) t =0 (cid:98) l/ (cid:99) (cid:88) t =0 ( − t + t + t + t m t ( K i ) m t ( K j ) m t ( K k ) m t ( K l ) E (cid:104) X i − t X j − t X k − t X l − t (cid:105) . We can apply Isserlis’ theorem to E (cid:104) X i − t X j − t X k − t X l − t (cid:105) and express it in terms of matchings. Ifwe look at the expression so obtained for E [ H i ( X ) H j ( X ) H k ( X ) H l ( X )] combinatorially, we are doing thefollowing thing. We have a complete graph whose vertex set is partitioned into 4 parts, V , V , V , V with | V | = i, | V | = j, | V | = k, | V | = l . We first pick some partial matching M consisting of edges whose twoendpoints are in the same part, and we get a factor of ( − | M | . We then match up the remaining verticesarbitrarily and get a partial matching M . Each edge in M with one endpoint in V s and one endpoint in V t will contribute a factor of E [ X s X t ]. M ∪ M is a perfect matching of our complete graph, and note that aperfect matching may be counted multiple times in this procedure since M is arbitrary.We claim that if a perfect matching M contains an edge whose endpoints are from the same part, then it iscounted multiple times and the contributions from each time sum up to zero. Indeed, if M contains an edge35hose both endpoints are in, say V , then this edge can be included in M , the partial matching produced inthe first step, and contribute − 1. or it can be included in M and contribute E [ X ] = 1. So the contributionscancel out. On the other hand, if M does not contain edges whose endpoints are from the same part, thenit is counted exactly once, for M must be empty, and its contribution is ρ a ρ b ρ c ρ d ρ e ρ f where a is thenumber of edges between V and V , b is the number of edges between V and V , etc.Now we need to count the number of perfect matchings given the numbers of edges across any two differentparts. It is easy to see that if a is the number of edges between V and V , b is the number of edges between V and V , etc, then the number of such matchings is given by (cid:18) ia, b, c (cid:19) · (cid:18) ja, d, e (cid:19) · (cid:18) kb, d, f (cid:19) · (cid:18) lc, e, f (cid:19) · a ! b ! c ! d ! e ! f ! = i ! j ! k ! l ! a ! b ! c ! d ! e ! f ! . Hence the corollary follows.We can then use this corollary to obtain an expression for F . However, unlike the expression for F , theexpression that we obtain does not converge for all valid pairwise biases. B An Example Where F is Negative on Positive Inputs The idea for this example is to choose vectors v , v , v , v such that v + v + v = (2 − δ ) v for somesmall δ > v , v , v which are orthogonal to v have negative inner products. Wecan do this by choosing three orthonormal vectors e , e , e and taking1. v = e v = (2 − δ )3 e + √ δ − δ e v = (2 − δ )3 e + √ δ − δ (cid:16) − e + √ e (cid:17) v = (2 − δ )3 e + √ δ − δ (cid:16) − e − √ e (cid:17) This gives the following pairwise biases:1. ∀ i ∈ { , , } , b ,i = v · v i = − δ .2. ∀ i < j ∈ { , , } , b i,j = v i · v j = (2 − δ ) − δ − δ = − δ + δ .Note that these pairwise biases are all positive as long as δ ∈ (0 , − √ ε > 01. Choose a random vector u .2. If | v i · u | ∈ [ ε, . ε ), take x i = sign ( v i · u ). Otherwise, choose x i by flipping a coin. Lemma B.1. For this rounding scheme, if x , x , x , x are all determined without flipping a coin, whichhappens with nonzero probability, then x x x x = − .Proof. Since v + v + v = (2 − δ ) v , we have that ( v · u ) + ( v · u ) + ( v · u ) = (2 − δ )( v · u ). If x , x , x are all determined without flipping a coin then ∀ i ∈ { , , } , | v i · u | ∈ [ ε, . ε ). We have the following casesfor ( v · u ), ( v · u ), and ( v · u ):1. If ( v · u ) , ( v · u ) , ( v · u ) are all in [ ε, . ε ) then ( v · u ) ∈ [ − δ ε, . − δ ε ) so x is determined by a coinflip.2. Similarly, if ( v · u ) , ( v · u ) , ( v · u ) are all in ( − . ε, − ε ] then x is determined by a coin flip.36. If two of the inner products ( v · u ) , ( v · u ) , ( v · u ) are in [ ε, . ε ) and the other inner product is in( − . ε, − ε ] then ( v · u ) ∈ ( . − δ ε, − δ ε ) so either x is determined by a coin flip or x = 1. If x = 1,which happens with nonzero probability, then x x x x = − v · u ) , ( v · u ) , ( v · u ) are in ( − . ε, − ε ] and the other innerproduct is in [ ε, . ε ) then either x is determined by a coin flip or x = − 1. If x = − x x x x = − Corollary B.2. For this rounding scheme, taking b ,i = − δ for all i ∈ { , , } and taking b i,j = (2 − δ ) − δ − δ = − δ + δ for all i < j ∈ { , , } , F ( b , , b , , b , , b , , b , , b , ) < . Remark B.3. The reason why this example works is that although the pairwise biases are all positive, oncewe consider the components of v , v , v which are orthogonal to v , their inner products are negative. Weconjecture that if the inner products remain positive throughout the Gram-Schmidt process then F must benon-negative. C Proof of Minimizer In this appendix, we justify that there exists f : R → [ − , 1] with (cid:107) f (cid:107) ∞ ≤ L ( f ) = 1 − αα (cid:90) ∞−∞ f ( x ) φ ( x ) dx + (cid:90) R f ( x ) f ( y ) φ ρ ( x, y ) dxdy , where ρ > − 1. The functional we consider for MAX NAE- { } -SAT are justified by near-identical logic.Consider a sequence of functions g i with (cid:107) g (cid:107) ∞ ≤ i →∞ L ( g i ) = inf { L ( f ) : (cid:107) f (cid:107) ∞ ≤ } . Note that by Claim 4.16, each g i can be assumed to be monotone. We shall use the topology of the spaceof functions to construct a function g for which L ( g ) equals this limit. In that case, g will be the minimizerwe desire. Such proofs are standard in functional analysis (e.g., [21]).Consider the Hilbert space L ( φ ), where φ is Gaussian measure on R . Note that every f such that (cid:107) f (cid:107) ∞ ≤ g (cid:96) i ofthe g i ’s which is weakly convergent to a function g . In particular, for any other h ∈ L ( φ ) we have thatlim i →∞ (cid:90) R g (cid:96) i ( x ) h ( x ) φ ( x ) dx = (cid:90) R g ( x ) h ( x ) φ ( x ) dx. By taking h to be the the indicator an an interval [ a, b ] and letting a approach b , we have that g is thepointwise limit of the g (cid:96) i almost everywhere. Thus, since the g (cid:96) i are bounded, on every compact interval[ a, b ], g (cid:96) i strongly converge to g in L ([ a, b ]). Therefore, for every interval [ a, b ], we have thatlim i →∞ (cid:34) − αα (cid:90) ba g (cid:96) i ( x ) φ ( x ) dx + (cid:90) [ a,b ] g (cid:96) i ( x ) g (cid:96) i ( y ) φ ρ ( x, y ) dxdy (cid:35) = 1 − αα (cid:90) ba g ( x ) φ ( x ) dx + (cid:90) [ a,b ] g ( x ) g ( y ) φ ρ ( x, y ) dxdy. For for every ε > 0, 1 − ε of the mass of φ ( x ) as well as φ ρ ( x, y ) is in a bounded rectangle. Thus, we maysend a → −∞ and b → −∞ to infer that lim i →∞ L ( g i ) = L ( g ) . Note that we need the g (cid:96) i are bounded and monotone to make this deduction. Heuristic argument for the optimality of ± functions In this appendix, we give a ‘heuristic argument’ in support of Conjecture 4, stating that the optimal roundingfunction f K , when | K | ≥ K = 3, is a ± K = { , k } , for k > f : ( −∞ , ∞ ) → [ − , 1] is locally optimal for MAX NAE- K -SAT if there exists ε > g : ( −∞ , ∞ ) → [ − ε, ε ] such that f + g : ( −∞ , ∞ ) → [ − , 1] and α K ( f + g ) > α K ( f ). (See Section 5.1 for the definition of α K ( f ).) The optimal function f K must of coursebe locally optimal.Suppose that f = f K does not assume the values ± a , b ] and [ a , b ] such that − δ ≤ f ( x ) ≤ − δ for every x ∈ [ a , b ] ∪ [ a , b ].Let g ( x ) = 1, if x ∈ [ a , b ], and g ( x ) = 0, otherwise. Define g ( x ) similarly for the interval [ a , b ]. For − δ < ε , ε < δ , the function f + ε g + ε g is a function from ( −∞ , ∞ ) to [ − , d k,i = dα k ( f + εg i ) dε be the derivative with respect to adding a small multiple of g i , for i = 1 , 2, and where k can also be 3. Forsmall enough ε , ε , we have α k ( f + ε g + ε g ) ≈ α k ( f ) + d k, ε + d k, ε .We expect d , and d k, , and also d , and d k, , to have opposite signs, if they are not 0, as otherwise f isclearly not locally optimal, as can be seen by adding a small multiple of g , or a small multiple of g .What is more interesting is the possibility of improving f by adding both a small multiple of g and asmall multiple of g . We thus ask whether there exists small enough ε , ε such that d , ε + d , ε ≥ d k, ε + d k, ε ≥ 0. A sufficient condition for the existence of such ε , ε is that the matrix D = (cid:18) d , d , d k, d k, (cid:19) is non-singular.In other words, if f is locally optimal and does not assume ± D = D [ a ,b ] , [ a ,b ] must be singular for every two intervals [ a , b ] and [ a , b ] in which f assume intermediatevalues. We believe that this condition cannot be satisfied. But, as conceded, we have no rigorous proof.The main reason we believe that this condition cannot be satisfied for K = { , k } is that the functions α ( f )and α k ( f ), for k > 3, seem to be ‘pulling’ in opposing directions.If this argument can be made rigorous, the proof would need to rely on the condition min K = 3, as ifmin K > 3, the optimal function is random assignment, i.e., f K ( x ) = 0, for every x ∈ ( −∞ , ∞ ). E Equivalence of Uniform and Non-uniform MAX NAE-SAT As discussed in the preliminaries, there are two natural definitions of MAX NAE-SAT: uniform MAX NAE-SAT, where the clause lengths can grow with the number of variables, and non-uniform MAX NAE-SAT,where the clause lengths can be an arbitrarily large constant (that is the limit of MAX NAE-[ k ]-SAT as k goes to infinity). In this appendix, we show that in terms of approximation, these two problems areessentially equivalent.Let U k ( c ) be the best soundness one can achieve in polynomial time for MAX NAE-[ k ]-SAT (assumingUGC). Let U ( c ) be the corresponding quantities for MAX NAE-SAT. Note the following is true. Proposition E.1. For any CSP Λ , and c, c (cid:48) , t ∈ [0 , , such that − t + tc ≥ c (cid:48) ≥ tc , we have − t + tU Λ ( c ) ≥ U Λ ( c (cid:48) ) .Proof. Consider an instance formed by taking the conjunction of a formula Ψ ( x ) with total weight t andcompleteness c and a second formula Ψ ( y ) with total weight 1 − t and completeness c (cid:48) − tc − t . Let A be apolynomial time algorithm which takes as input Ψ and Ψ and outputs an assignment ( x, y ). By definitionof U Λ ( c ), for any η > 0, there must be an input (Ψ (cid:48) , Ψ (cid:48) ) such that ( x, y ) := A (Ψ (cid:48) , Ψ (cid:48) ) satisfies at most U Λ ( c ) + η of the clauses of Ψ (cid:48) . Thus, the output of A satisfies at most 1 − t + t ( U Λ ( c ) + η ) fraction of theclauses. This gives the upper bound on U Λ (1 − t + tc ).38he following is the main result of this appendix. Lemma E.2. For all c ∈ [0 , , U ( c ) = lim k →∞ U k ( c ) . Proof. It suffices to prove the following two inequalities for all (cid:15) > U ( c ) ≤ lim k →∞ U k ( c ) (8)lim k →∞ U k ( c ) ≤ U ( c ) + (cid:15). (9)The inequality (8) follows from the fact that any instance of MAX NAE-[ k ]-SAT is an instance of uniformMAX NAE-SAT.To prove (9), it suffices to describe an algorithm for MAX NAE-SAT for any (cid:15) > 0. Let Φ be an instance ofMAX NAE-SAT.Define P k ( δ ) := 1 − − δ ) k + (1 − δ ) k and Q k ( δ ) := (1 − δ ) k .Run the following algorithm.1. Guess k = O ε (1) and set δ = Q − k (1 − (cid:15)/ 2) = − (1 − (cid:15)/ /k .2. Let Φ (cid:48) ⊂ Φ be the instance consisting of all clauses having length at most k . Run the optimalpolynomial-time approximation algorithm for this instance and get a solution x .3. For each variable x i , i ∈ [ n ]. With probability δ , set y i = 1 and with probability δ , set y i = − y i = x i .4. Output y .It suffices to prove that the above algorithm works for some k = O ε (1) as then we have a polynomial timealgorithm by enumerating over all k at most O ε (1).First, assume k is fixed. To analyze the algorithm, fix an optimal solution z to Φ. For each i ≥ 1, let w i the relative weight of clauses of Φ with size i (so that (cid:80) i w i = 1) and let c i be the fraction of clauses of Φsatisfied by x . Now define c (cid:48) k = (cid:80) ki =1 w i c i (cid:80) ki =1 w i , that is the completeness of Φ (cid:48) . By definition, the solution x will be a U k ( c (cid:48) k ) − (cid:15)/ (cid:48) .Observe that for a clause of length i , it is satisfied by (3) with probability 1 − − δ ) i + (1 − δ ) i = P i ( δ ) . Further, the probability that a clause of length k is ”untouched” (that is, no variables are changed) by (3)is (1 − δ ) k = Q k ( δ ).For a given k and δ such that Q k ( δ ) = 1 − (cid:15)/ 2, let k (cid:48) be the smallest integer such that P k (cid:48) ( δ ) ≥ − (cid:15)/ k (cid:48) is satisfied with probability at least 1 − (cid:15)/ 2. Thus, the fraction of39lauses satisfied by our algorithm is at least (cid:16) − (cid:15) (cid:17) n (cid:88) i = k (cid:48) w i + U k ( c (cid:48) k ) k (cid:88) i =1 w i − (cid:15) ≥ n (cid:88) i = k (cid:48) w i + U k ( c (cid:48) k ) k (cid:88) i =1 w i − (cid:15) n (cid:88) i = k +1 w i + U k ( c (cid:48) k ) k (cid:88) i =1 w i − k (cid:48) (cid:88) i = k +1 w i − (cid:15) ≥ U k (cid:32) n (cid:88) i =1 c i w i (cid:33) − k (cid:48) (cid:88) i = k +1 w i − (cid:15) U k ( c ) − k (cid:48) (cid:88) i = k +1 w i − (cid:15) . To finish, it suffices to show there exists k = O (cid:15) (1) such that (cid:80) k (cid:48) i = k +1 w i ≤ (cid:15) .To see why, fix k = 3 and then consider the sequence k (cid:48) , k (cid:48)(cid:48) , k (cid:48)(cid:48)(cid:48) , . . . k (5 /(cid:15) ) . (Where k ( i +1) is k (cid:48) for k = k ( i ) .)Since (cid:80) ni =1 w i = 1, there must exist i such that (cid:80) k (cid:48) i = k +1 w i ≤ (cid:15) for k = k ( i ))