Sum-of-squares hierarchy lower bounds
for symmetric formulations††thanks: Supported by the Swiss National Science Foundation project 200020-144491/1 “Approximation Algorithms for Machine Scheduling Through Theory and Experiments” and by Sciex Project 12.311.
Abstract
We introduce a method for proving Sum-of-Squares (SoS)/ Lasserre hierarchy lower bounds when the initial problem formulation exhibits a high degree of symmetry. Our main technical theorem allows us to reduce the study of the positive semidefiniteness to the analysis of “well-behaved” univariate polynomial inequalities.
We illustrate the technique on two problems, one unconstrained and the other with constraints. More precisely, we give a short elementary proof of Grigoriev/Laurent lower bound for finding the integer cut polytope of the complete graph. We also show that the SoS hierarchy requires a non-constant number of rounds to improve the initial integrality gap of 2 for the Min-Knapsack linear program strengthened with cover inequalities.
1 Introduction
Proving lower bounds for the Sum-of-Squares (SoS)/Lasserre hierarchy [25, 31] has attracted notable attention in the theoretical computer science community during the last decade, see e.g. [7, 12, 13, 18, 19, 27, 28, 29, 34, 36]. This is partly because the hierarchy captures many of the best known approximation algorithms based on semidefinite programming (SDP) for several natural 0/1 optimization problems (see [28] for a recent result). Indeed, it can be argued that the SoS hierarchy is the strongest candidate to be the “optimal” meta-algorithm predicted by the Unique Games Conjecture (UGC) [22, 32]. On the other hand, the hierarchy is also one of the best known candidates for refuting the conjecture since it is still conceivable that one could show that the SoS hierarchy achieves better approximation guarantees than the UGC predicts (see [6] for discussion). Despite the interest in the algorithm and due to the many technical challenges presented by semidefinite programming, only relatively few techniques are known for proving lower bounds for the hierarchy. In particular, several integrality gap results follow from applying gadget reductions to the few known original lower bound constructions.
Indeed, many of the known lower bounds for the SoS hierarchy originated in the works of Grigoriev [18, 19].111More precisely, Grigoriev considers the positivstellensatz proof system, which is the dual of the SoS hierarchy considered in this paper. For brevity, we will use SoS hierarchy/proof system interchangeably. We defer the formal definition of the hierarchy for later and only point out that solving the hierarchy after rounds takes time. In [19] Grigoriev showed that random 3Xor or 3Sat instances cannot be solved even by rounds of the SoS hierarchy (some of these results were later independently rediscovered by Schoenebeck [34]). Lower bounds, such as those of [7, 36] rely on [19, 34] combined with gadget reductions. Another important lower bound was also given by Grigoriev [18] for the Knapsack problem (a simplified proof can be found in [20]), showing that the SoS hierarchy cannot prove within rounds that the polytope contains no integer point when is odd. Using essentially the same construction as in [20], Laurent [27] independently showed that rounds are not enough for finding the integer cut polytope of the complete graph with nodes, where is odd (this result was recently shown to be tight in [15]).222The two problems, Knapsack and Max-Cut in complete graphs, considered respectively in [18, 20] and in [27], are essentially the same and we will use Max-Cut to refer to both. By using several new ideas and techniques, but a similar starting point as in [20, 27], Meka, Potechin and Wigderson [29] were able to show a lower bound of for the Planted-Clique problem. Common to the works [20, 27] and [29] is that the matrix involved in the analysis has a large kernel, and they prove that a principal submatrix is positive definite by applying the theory of association schemes [16]. It is also interesting to point out that for the class of Max-CSPs, Lee, Raghavendra and Steurer [28] proved that the SoS relaxation yields the “optimal” approximation, meaning that SDPs of polynomial-size are equivalent in power to those arising from rounds of the SoS relaxations. Then, by appealing to the result by Grigoriev/Laurent [18, 27] they showed an exponential lower bound on the size of SDP formulations for the integer cut polytope. For different techniques to obtain lower bounds, we refer for example to the recent papers [5, 23, 24] (see also Section 5.4) and the survey [13] for an overview of previous results.
In this paper we introduce a method for proving SoS hierarchy lower bounds when the initial problem formulation exhibits a high degree of symmetry. Our main technical theorem (Theorem 1) allows us to reduce the study of the positive semidefiniteness to the analysis of “well-behaved” univariate polynomial inequalities. The theorem applies whenever the solution and constraints are symmetric, informally meaning that all subsets of the variables of equal cardinality play the same role in the formulation (see Section 3 for the formal definition). For example, the solution in [18, 20, 27] for Max-Cut is symmetric in this sense.
We note that exploiting symmetry reduces the number of variables involved in the analysis, and different ways of utilizing symmetry have been widely used in the past for proving integrality gaps for different hierarchies, see for example [8, 17, 19, 21, 24, 35]. An interesting difference of our approach from others is that we establish several lower bounds without fully identifying the formula of eigenvectors. More specifically, the common task in this context is to identify the spectral structure to get a simple diagonalized form. In the previous papers the moment matrices belong to the Bose-Mesner algebra of a well-studied association scheme, and hence one can use the existing theory. In this paper, instead of identifying the spectral structure completely, we identify only possible forms and propose to test all the possible candidates. This is in fact an important point, since the approach may be extended even if the underlying symmetry is imperfect or its spectral property is not well understood.
The proof of Theorem 1 is obtained by a sequence of elementary operations, as opposed to notions such as big kernel in the matrix form, the use of interlacing eigenvalues, the machinery of association schemes and various results about hyper-geometric series as in [18, 20, 27]. Thus Theorem 1 applies to the whole class of symmetric solutions, even when several conditions and machinery exploited in [18, 20, 27] cannot be directly applied. For example the kernel dimension, which was one of the important key property used to prove the results in [18, 20, 27], depends on the particular solution that is used and it is not a general property of the class of symmetric solutions. The solutions for two problems considered in this paper have completely different kernel sizes of the analyzed matrices, one large and the other zero.
We demonstrate the technique with two illustrative and complementary applications. First, we show that the analysis of the lower bound for Max-Cut in [18, 20, 27] simplifies to few elementary calculations once the main theorem is in place. This result is partially motivated by the open question posed by O’Donnell [30] of finding a simpler proof for Grigoriev’s lower bound for the Knapsack problem.
As a second application we consider a constrained problem. We show that after levels the SoS hierarchy does not improve the integrality gap of for the Min-knapsack linear program formulation strengthened with cover inequalities [10] introduced by Wolsey [37]. Adding cover inequalities is currently the most successful approach for capacitated covering problems of this type [1, 2, 3, 9, 11].
Our result is the first SoS lower bound for formulations with cover inequalities. In this application we demonstrate that our technique can also be used for suggesting the solution and for analyzing its feasibility.
2 The SoS hierarchy
Consider a optimization problem with linear constraints , for and . We are interested in approximating the convex hull of the integral points of the set with the SoS hierarchy defined in the following.
The form of the SoS hierarchy we use in this paper (Definition 1) is equivalent to the one used in literature (see e.g. [4, 25, 26]). It follows from applying a change of basis to the dual certificate of the refutation of the proof system [26] (see also [29] for discussion on the connection to the proof system). We use this change of basis in order to obtain a useful decomposition of the moment matrices as a sum of rank one matrices of special kind. This will play an important role in our analysis. We refer the reader to Appendix A for more details and for a mapping between the different forms.
For any , let denote the solution obtained by setting for , and for . We denote by the value of the constraint evaluated at . For each integral solution , where , in the SoS hierarchy defined below there is a variable that can be interpreted as the “relaxed” indicator variable for the solution . We point out that in this formulation of the hierarchy the number of variables is exponential in , but this is not a problem in our context since we are interested in proving lower bounds rather than solving an optimization problem.
Let be the collection of subsets of of size at most . For every , the -zeta vector is a vector with -th entry () equal to if and only if .333In order to keep the notation simple, we do not emphasize the parameter as the dimension of the vectors should be clear from the context. Note that is a rank one matrix and the matrices considered in Definition 1 are linear combinations of these rank one matrices.
Definition 1.
The -th round SoS hierarchy relaxation for the set , denoted by , is the set of values that satisfy
(1) | |||||
(2) | |||||
(3) |
where if (no linear constraints), otherwise .
It is straightforward to see that the SoS hierarchy formulation given in Definition 1 is a relaxation of the integral polytope. Indeed consider any feasible integral solution and set and the other variables to zero. This solution clearly satisfies Condition (1), Condition (2) because the rank one matrix is positive semidefinite (PSD), and Condition (3) since .
3 The main technical theorem
The main result of this paper (see Theorem 1 below) allows us to reduce the study of the positive semidefiniteness for matrices (2) and (3) to the analysis of “well-behaved” univariate polynomial inequalities. It can be applied whenever the solutions and constraints are symmetric, namely they are invariant under all permutations of the set : for all (equivalently when whenever ),444We define the set-valued permutation by . where is understood to denote either or . For example, the solution for Max-Cut considered by Grigoriev [18] and Laurent [27] belongs to this class.
Theorem 1.
For any , let be the set of all polynomials , for , that satisfy the following conditions:
(4) | ||||
(5) | ||||
(6) |
For any fixed set of values , if the following holds
(7) |
then matrix (8) is positive-semidefinite
(8) |
Note that polynomial in (6) is nonnegative in a real interval, and in (5) it is zero for a finite set of integers. Moreover, constraints (7) are trivially satisfied for .
Theorem 1 is a actually a corollary of a technical theorem that is not strictly necessary for the applications of this paper, and therefore deferred to a later section (see Theorem 7 in Section 6). The proof (given in Section 6) is obtained by exploiting the high symmetry of the eigenvectors of the matrix appearing in (8). Condition (7) corresponds to the requirement that the Rayleigh quotient being non-negative restricted to some highly symmetric vectors (which we show are the only ones we need to consider).
4 Max-Cut for the complete graph
In the Max-Cut problem, we are given an undirected graph and we wish to find a partition of the vertices (a cut) which maximizes the number of edges whose endpoints are on different sides of the partition (cut value). For the complete graph with vertices, consider any solution with vertices on one side and the remaining on the other side of the partition. This gives a cut of value . When is odd and for any , Grigoriev [18] and Laurent [27] considered the following solution (reformulated in the basis considered in Definition 1, see Appendix B):
(9) |
It is shown [18, 27] that (9) is a feasible solution for the SoS hierarchy of value , for any up to round . In particular for the cut value of the SoS relaxation is strictly larger than the value of the optimal integral cut (i.e. ), showing therefore an integrality gap at round .
We note that the formula for the solution (9) is essentially implied by the requirement of having exactly vertices on one side of the partition (see [18, 27] and [29] for more details) and the core of the analysis in [18, 27] is in showing that (9) is a feasible solution for the SoS hierarchy. By taking advantage of Theorem 1, the proof that (9) is a feasible solution for the SoS relaxation follows by observing the fact below.
Lemma 2.
For any polynomial of degree and defined in (9) we have
Proof.
By the polynomial remainder theorem , where is a unique polynomial of degree at most . It follows that
since for any polynomial of degree at most .
∎
5 Min-Knapsack with cover inequalities
The Min-Knapsack problem is defined as follows: we have items with costs and profits , and we want to choose a subset of items such that the sum of the costs of the selected items is minimized and the sum of the profits is at least a given demand . Formally, this can be formulated as an integer program . It is easy to see that the natural linear program , obtained by relaxing to in , has an unbounded integrality gap.
By adding the Knapsack Cover (KC) inequalities introduced by Wolsey [37] (see also [10]), the arbitrarily large integrality gap of the natural LP can be reduced to 2 (and it is tight [10]). The KC constraints are as follows: for all , where and . Note that these constraints are valid constraints for integral solutions. Indeed, in the “integral world” if a set of items is picked we still need to cover ; the remaining profits are “trimmed” to be at most and this again does not remove any feasible integral solution.
The following instance [10] shows that the integrality gap implied by KC inequalities is 2: we have items of unit costs and profits. We are asked to select a set of items in order to obtain a profit of at least . The resulting linear program formulation with KC inequalities is as follows (for , )
s.t. | (11) | ||||
Note that the solution is a valid fractional solution of value whereas the optimal integral solution has value . In the following we show that , with arbitrarily close to a logarithmic function of , admits the same integrality gap as the initial linear program () relaxation.
Theorem 3.
For any and sufficiently large , let , and . Then the following solution is feasible for with integrality gap of
(16) |
5.1 Overview of the proof
An integrality gap proof for the SoS hierarchy can in general be thought of having two steps: first, choosing a solution to the hierarchy that attains a superoptimal value, and second showing that this solution is feasible for the hierarchy. We take advantage of Theorem 1 in both steps. Here we describe the overview of our integrality gap construction while keeping the discussion informal and technical details minimal, the proof can be found in Appendix 5.2.
Choosing the solution.
We make the following simplifying assumptions about the structure of the variables : due to symmetry in the problem we set for each such that , and for every we set in order to satisfy (2) for free. Furthermore, in order to have an integrality gap (i.e. a small objective function value), we guess that forcing the other variables to be small due to (1).
We then show that satisfying (3) for every constraint follows from showing that
(17) |
for every choice of real variables . We get this condition by observing similarities in the structure of the constraints and applying Theorem 1, then expressing the polynomial in root form.555We show that the roots can be assumed to be real numbers. If we set , the only negative term in the sum corresponds to . Then, it is clear that we need at least non-zero variables , otherwise the roots can be set such that the positive terms in (17) vanish and the inequality is not satisfied. Therefore, we choose exactly of the to be strictly positive (and the rest 0 excluding ), and we distribute them “far away” from each other, so that no root can be placed such that the coefficient of two positive terms become small. To take this idea further, for one “very small” (logarithmic in ), we set positive and space out the rest evenly.
Proving that the solution is feasible.
We show that (17) holds for all possible with our chosen solution by analysing two cases. In the first case we assume that all of the roots are larger than . Then, we show that the “small” point we chose is enough to satisfy the condition. In the complement case, we assume that there is at least one root that is smaller than . It follows that one of the evenly spaced points is “far” from any remaining root, and can be used to show that the condition is satisfied.
5.2 Proof of Theorem 3
We start by proving the claimed integrality gap. The defined solution has an objective value that is arbitrarily close to 1 whereas the optimal integral value is 2. Indeed, the objective value of the relaxation is (see Appendix A):
The remaining part of the Theorem 3 follows by showing that the suggested solution satisfies (1), (2) and (3). Note that (1) is immediately satisfied by the definition of variables , and (2) is satisfied since and the rank one matrix is positive semidefinite for every . It remains to prove that the condition (3) is also satisfied for all the constraints (11) and (11). Note that the constraint (11) is not symmetric (one variable is missing and sets of variables of the same size do not play the same role with respect to this constraint). However, the following lemma shows how to solve this issue by reducing to the form (8) of Theorem 1.
Lemma 4.
Proof.
We first show that this implies that (3) holds for the constraint . Since for a large , for and , the condition (3) takes the following form
which is implied if (18) is satisfied (recall that ). Next, we will show that (18) also implies that (3) is satisfied for the cover constraint (the other cases are similar). For this constraint Condition (3) can be written as
which is also implied if (18) is satisfied.
∎
Now by Theorem 1, Condition (18) holds if we have for and every univariate polynomial of degree such that for and for .
Note that the only nontrivial case is for , since otherwise the above condition is immediately satisfied. Indeed, for , we have that and the only remaining terms in the sum are non-negative. Thus in order to complete the proof of Theorem 3 it is enough to show that the following is satisfied
(19) |
The following lemma (proved in Section 5.3) further reduces the interesting cases.
Lemma 5.
The fundamental theorem of algebra states that any univariate polynomial of degree has exactly complex roots. According to Lemma 5 we can focus to polynomial with real roots. We prove that the suggested solution satisfies (19) by expressing the generic univariate polynomial using its roots , so that (19) becomes
(20) |
To show that (20) is satisfied we separate two cases: when all of the roots of the polynomial are greater or equal to a fixed treshold and when at least one root is smaller than this treshold. In order to simplify the computations we denote .
- 1.
-
2.
There is at least one root such that . It can be shown by straightforward induction on the number of roots that if for at least one , , then there exists a point , such that and for all . Let be such a point. It is sufficient to show that we can satisfy .
We have and the estimates . Substituting these we get the condition which gives us the requirement that .
These two cases suggest that we fix as
The proof has now been reduced to showing that with this choice of we have , i.e., . Assume . Then , when , using the estimate which holds when . Furthermore, the same estimate yields when . Hence, we have the bound
The right-hand side goes to 0 if as . This is clearly the case for , for any .
Next, assume . Then which immediately yields the condition on and that we need as . Substituting and , for any , allows us to write this as . By a change of variables of the form we get
which tends to 0 as .
5.3 Proof of Lemma 5
Lemma 5. In order to prove that Solution (26) satisfies (19) it is sufficient to prove that (26) satisfies (19) for polynomials with the following properties:
-
(a)
all the roots of are real numbers,
-
(b)
all the roots of are in the range, for all ,
-
(c)
the degree of is exactly .
Proof.
First notice that (19) is equivalent to
(21) |
where for the fixed the right-hand side is constant.
-
(a)
Let be the univariate polynomial with complex roots (complex roots appear in conjugate pairs) i.e. , for and the rest real roots. Let be the polynomial with all real roots such that for and .
For any and , a simple calculation shows that
Hence,
-
(b)
Let be the univariate polynomial with all positive roots but one i.e. , for . Let be the univariate polynomial with all positive roots such that and . Since for any
and follows that,
Now, let be the univariate polynomial with and , for . Let be the univariate polynomial with and .
Since for any
and follows that,
Next, let be the univariate polynomial with for and , for . Let be the univariate polynomial with and .
Since for any
and follows that,
-
(c)
Let be the univariate polynomial with degree with all real roots. Let be the polynomial of degree with all real roots such that and for
For any , we have
Hence,
and finally
∎
5.4 Further Results
In a recent paper [23] the authors characterize the class of the initial 0/1 relaxations that are maximally hard for the SoS hierarchy. Here, maximally hard means those relaxations that still have an integrality gap even after rounds of the SoS hierarchy.666Recall that at level the integrality gap vanishes. An illustrative natural member of this class is given by the simple LP relaxation for the Min Knapsack problem, i.e.
In [23] it is shown that at level the integrality gap is , for any if and only if . A natural question is to understand if the SoS hierarchy is able to reduce the gap when is “small”.
This problem, for , was considered by Cook and Dash [14] as an example where the Lovasz-Schrijver hierarchy rank is . Laurent [26] showed that the Sherali-Adams hierarchy rank is also equal to and raised the open question to find the rank for the Lasserre hierarchy. She also showed that when , the Lasserre relaxation has an integrality gap at level 1, but leaves open whether or not this happens at level for a general . In [23] it is ruled out the possibility that the Lasserre/SoS rank is for .
The following lemma provides a feasible solution for with integrality gap arbitrarily close to when and for any . The proof is omitted since it is similar to the proof of Theorem 3.
Theorem 6.
For any and sufficiently large , let , and . Then the following solution is feasible for with integrality gap arbitrarily close to .
(26) |
6 Proof of Theorem 1
Theorem 1 is actually a corollary of a stronger statement (see Theorem 7 below) that provides necessary and sufficient conditions for the matrix (8) being positive-semidefinite.
Theorem 7 uses a special family of polynomials whose definition is deferred to a later section (see Definition 3 in Section 6.1). We postpone the definition because it will become natural in the flow of the proof of Theorem 7. Here we remark that the polynomials of Definition 3 satisfy the conditions (4), (5) and (6) of Theorem 1 (as shown in Lemma 14 to follow).
Theorem 7.
Let for . Then for any the following matrix is positive-semidefinite
(27) |
if and only if
(28) |
for every univariate polynomial of degree at most as defined in Definition 3.
By Lemma 14, Theorem 1 is a straightforward corollary of Theorem 7. In the following we provide a proof for the latter.
6.1 Proof of Theorem 7
We study when the matrix where is positive-semidefinite. Theorem 7 allows us to reduce the condition to inequalities of the form , where is a univariate polynomial of degree with some additional remarkable properties.
A basic key idea that is used to obtain such a characterization is that the eigenvectors of are “very well” structured. This structure is used to get with the claimed properties.
The structure of the eigenvectors.
Let denote the group of all permutations of the set , i.e. the symmetric group. Let be the permutation matrix of size corresponding to any permutation of set , i.e. for any vector we have for any (see Footnote 4). Note that .
Lemma 8.
For every we have or, equivalently, and commute .
Proof.
Let denote the vector with a in the -th coordinate and ’s elsewhere. Observe
Then, .
∎
Corollary 9.
If is an eigenvector of then is also an eigenvector of for any .
By using Corollary 9 we can show that the set of interesting eigenvectors have some “strong” symmetry properties that will be used in our analysis. In the simplest case, for any eigenvector we could take the vector and observe that the elements of have the form for each such that . If then and are two eigenvectors corresponding to the same eigenvalue. The latter implies that by considering only eigenvectors having the form for each we would consider the eigenvalue corresponding to the “unstructured” eigenvector as well. This is not the case in general, however, since it is possible that .
We overcome this obstacle by restricting the permutations in a way which guarantees to be non-zero. Before going into the details, we introduce some notation.
Definition 2.
For any , we denote by the permutation group that fixes the set in the following sense: .
Note that the definition is equivalent to saying that if and only if for every and for every .
Now, we choose a subset such that for each such that and . Such a set always exists, since otherwise is a zero vector, since if there is one non-zero entry in , we can take and the resulting is non-zero. The choice of is not unique, but we can always assume that it is the subset of the first elements from , i.e. . Indeed, if it is not the case, there exists a permutation that maps to the subset of the first elements from and such that is an eigenvector of by Lemma 9. Now it holds that and the vector is a unit eigenvector corresponding to the same eigenvalue as and has many elements that are equal to each other.
Lemma 10.
Let be a unit eigenvector of corresponding to eigenvalue , and be the smallest subset of such that . Then is also a unit eigenvector of corresponding to eigenvalue .
The following lemma shows the structure of eigenvectors obtained from summing the permutations of any “unstructured” eigenvector.
Lemma 11.
Let . Then the vector is invariant under the permutations of , namely for . Equivalently, for all such that .
Proof.
We have where the last but one equality follows since permutations are bijective. The claim follows by observing that for all such that there exists such that .
∎
Lemma 12.
For any eigenvalue of there exists an such that the following is an eigenvector corresponding to :
(29) |
where , and such that if and , otherwise.
By Lemma 12, we have that the positive semidefiniteness condition of follows by ensuring that for any we have , i.e.
In the following (Lemma 13) we show that the above values are interpolated by the univariate polynomial defined in Definition 3. In Lemma 14 we prove some remarkable properties of as claimed in Theorem 1.
Definition 3.
For any , let be a univariate polynomial defined as follows
(30) |
where and (for ).777Denote by the falling factorial (with the convention that ).
Lemma 13.
For every the following identity holds .
Proof.
We start noting that for every , we have 888Recall that for any positive integer .
Indeed
It follows that we have
Splitting the sum over by considering the intersections of sizes , we have
Finally, we shift the sum over by and thus justify the equality
Now, the sum over is a Newton polynomial that we denote by . Note that by definition, here . Furthermore, observe that
and writing yields the claim. ∎
It follows that for any unit eigenvector of the form (29) the corresponding eigenvalue is equal to . Theorem 7 requires that which implies that the eigenvalue is nonnegative. In the following section we complete the proof by showing that the polynomials of Definition 3 satisfy the conditions (4), (5) and (6) of Theorem 1 (as shown in Lemma 14).
6.2 Properties of the univariate polynomials
Lemma 14.
For any , the polynomials as defined in Definition 3 have the following properties:
-
(a)
is a univariate polynomial of degree at most ,
-
(b)
for
-
(c)
for every .
Proof of (a).
Note that by Vandermonde’s identity. We prove the theorem by showing that has degree not larger than .
By Lemma 15 below, the degree of is at most and thus the degree of is at most
Lemma 15.
The degree of is at most .
The claim follows by showing that the degree of is at most the degree of . The degree of is .
Recall that the forward difference of function with respect to variable is a finite difference defined by . Higher order differences are obtained by repeated operations of the forward difference operator. We will use to denote the -th forward difference evaluated at . We will us the following easy to check identity: .
First note that any polynomial of degree can written as linear combinations of polynomials with . It follows that the claim follows by showing that the degree of the following is at most the degree of
where we have used the linearity of the forward difference operator and the Vandermonde’s identity to derive the last equality. The claim follows by observing that the forward difference operator does not increase the degree of its argument and therefore has degree at most .
Proof of (b).
Let . We have
where for each . Therefore is a sum of non-negative numbers weighted by positive coefficients .
Proof of (c).
With this aim, we start noting that for every set we have that the permutation group is the same as . Moreover if then , otherwise we would obtain a set smaller than with (contradicting our assumption that is a set with the smallest set size having ).
Now consider any set such that with and . By the previous observations it follows that . Note that since the set is equal to , since is the permutation group that maps any element from to any other element from of the same size. It follows that . Using the latter we get
proving the claim.
Acknowledgements.
The authors would like to express their gratitude to Ola Svensson for helpful discussions and ideas regarding this paper.
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Appendix
Appendix A The SoS hierarchy
In this section we recall the usual definition of the SoS/Lasserre hierarchy [25] and justify Definition 1. Notice that the SDP hierarchy that we discuss here is the dual certificate of a refutation of the positivstellensatz proof system, for further information about the connection to the proof system we refer the reader to [29]. In our setting we restrict ourselves to problems with -variables and linear constraints. More precisely, we consider the following general optimization problem : given a multilinear polynomial
(31) |
where is a polytope defined by linear inequalities . Many basic optimization problems are special cases of . For example, any -ary boolean constraint satisfaction problem, such as Max-Cut, is captured by (31) where a degree function counts the number of satisfied constraints, and no linear constraints are present. Also any integer linear program is a special case of (31), where is a linear function.
Lasserre [25] proposed a hierarchy of SDP relaxations parameterized by an integer ,
(32) |
where is a linear map with denoting the ring restricted to polynomials of degree at most .999In [4], is written and called the “pseudo-expectation” of . Note that (32) is a relaxation since one can take to be the evaluation map for any optimal solution .
Relaxation (32) can be equivalently formulated in terms of moment matrices [25]. In the context of this paper, this matrix point of view is more convenient to use and it is described below. In our notation we mainly follow the survey of Laurent [26] (see also [33]).
Variables and Moment Matrix.
Let denote the set . The collection of all subsets of is denoted by . For any integer , let denote the collection of subsets of having cardinality at most . Let . For any nonnegative integer , let denote the matrix with -entry for all . Matrix is termed in the following as the t-moment matrix of . For a linear function , we define as a vector, often called shift operator, where the -th entry is . Let denote the vector of coefficients of polynomial (where is the coefficient of monomial in ).
Definition 4.
The -th round SoS (or Lasserre) relaxation of problem (31), denoted as , is the following
(33) |
where is the set of vectors that satisfy the following PSD conditions
(34) | |||||
(35) | |||||
(36) |
where if (no linear constraints), otherwise .
Change of variables.
A solution of the SoS hierarchy as defined in Definition 4 is given by a vector . Next we show we can make a change of basis and replace the variables with other variables that are indexed by all the subsets of . Variable can be interpreted as the “relaxed” indicator variable for the integral solution , i.e. the solution obtained by setting for , and for . We use this change of basis in order to obtain a useful decomposition of the moment matrix as a sum of rank one matrices of special kind. Here it is not necessary to distinguish between the moment matrix of the variables and constraints, hence in what follows we denote a generic vector by , where is either or .
Definition 5.
Let . For every , define a vector such that
Note that the inverse (for ) is
(37) |
To simplify the notation, we note that the moment matrix of the variables is structurally similar to the moment matrix of the constraints: if is a vector such that for some , then . Hence, the following lemma holds for the moment matrix of variables and constraints.
Lemma 16.
Let , and such that . Then
Proof.
Since , we have by the change of variables that
where is the 0-1 indicator function such that if and only if . On the other hand, if , and 0 otherwise. Therefore .
∎
By the previous lemma it follows that given a solution by using variables we can obtain a solution with variables . Viceversa, given any assignment of variables in we can find an assignment of variables in such that and . Indeed, set for every such that . For the remaining ones note that for the square matrix corresponding to the following equalities is invertible since it is upper triangular.
Lemma 17.
[26] Given , for the vector we have
(38) |
where is a linear function corresponding to the constraint , evaluated at such that if and otherwise.
Proof.
We need to show that this choice of yields . We plug in (38) to obtain
Here the term is if and 0 otherwise. Taking this into account and changing the order of the sums, the above becomes
which proves the claim.
∎
The above discussion together with the observation that implies that and justifies Definition 1. Finally, we remark the following
Lemma 18.
Let denote the vector of coefficients of polynomial of (31). Then the objective value of the solution is given by