On the Complexity of Some Facet-Defining Inequalities of the QAP-polytope
aa r X i v : . [ c s . CC ] O c t On the Complexity of Some Facet-DefiningInequalities of the QAP-polytope
Pawan Aurora and Hans Raj Tiwary IISER Bhopal, India [email protected] Department of Applied Mathematics, Charles University, Prague, Czech Republic [email protected]
Abstract.
The Quadratic Assignment Problem (QAP) is a well-knownNP-hard problem that is equivalent to optimizing a linear objective func-tion over the QAP polytope. The QAP polytope with parameter n -QAP n - is defined as the convex hull of rank-1 matrices xx T with x asthe vectorized n × n permutation matrices.In this paper we consider all the known exponential-sized families offacet-defining inequalities of the QAP-polytope. We describe a new fam-ily of valid inequalities that we show to be facet-defining. We also showthat membership testing (and hence optimizing) over some of the knownclasses of inequalities is coNP-complete. We complement our hardness re-sults by showing a lower bound of 2 Ω ( n ) on the extension complexity ofall relaxations of QAP n for which any of the known classes of inequalitiesare valid. The Quadratic Assignment Problem (QAP) is a fundamental combinatorial op-timization problem from the category of facility location problems [23,24]. QAPis defined as the following problem: given n facilities and n locations, distances d ij between all pairs of locations i, j ∈ [ n ], flows f ij between all pairs of facilities i, j ∈ [ n ] and costs c ij of opening facility i at location j , for all pairs i, j ∈ [ n ], findan assignment σ of the n facilities to the n locations so that the total cost givenby the function P i,j f ij d σ ( i ) σ ( j ) + P i c iσ ( i ) is minimized. The problem is knownas QAP since it can be modeled as optimizing a quadratic function over linearand binary constraints. However, several linearizations of the problem have beenproposed. For details refer to the book [9] and the citations therein.Given an instance of QAP, it is NP-hard to approximate the optimum withinany constant factor [31]. What makes QAP one of the “hardest” problems incombinatorial optimization is the fact that unlike most NP-hard combinatorialoptimization problems, it is practically intractable. It is generally consideredimpossible to solve to optimality QAP instances of size larger than 20 withinreasonable time limits [9].As is common with combinatorial obtimization problems, QAP can be viewedas the problem of optimizing a linear objective function over the convex hull ofall feasible solutions. To this end, the QAP polytope is defined as QAP n = Pawan Aurora and Hans Raj Tiwary conv (cid:0)(cid:8) yy T | y = vec ( P σ ) , σ ∈ S n (cid:9)(cid:1) , where P σ is the n × n permutation matrixcorresponding to the premutation σ and y = vec ( P σ ) is its vectorization. Follow-ing the notation of [1], we denote a vertex yy T as P [2] σ . Note that P [2] σ ( ij, kl ) = P σ ( i, j ) · P σ ( k, l ). Clearly, QAP n ⊂ R n × n . In fact QAP n can be embedded in R ( n + n ) / since each point in the polytope is a symmetric n × n matrix andwe could only store its upper (or lower) triangular part. However, in this paperwe would conveniently denote a point in QAP n by a n × n matrix.One of the methods for solving hard combinatorial optimization problemsis the method of branch-and-cut [27]. For this method to be effective for theQAP, it is important to identify new valid and possibly facet-defining inequalitiesfor the QAP-polytope and to develop the corresponding separation algorithms.Given that it is NP-hard to optimize over the QAP-polytope, it is probablyimpossible to characterize all its facets [28]. In [17,18,26], the authors obtainearly results on the combinatorial structure of the QAP-polytope and some ofits facet-defining inequalities. In [1] the authors list all the known facets of theQAP-polytope besides the equations that define its affine hull. In this paper weadd another exponential sized family to the list of known facets of the QAP-polytope. Optimizing the QAP objective function over any of the relaxationsgiven by these families can provide an approximate solution to the QAP, providedthe optimization problem can be efficiently solved. In this paper we also showthat optimizing over the relaxations given by some of these exponential sizedfamily of facet-defining inequalities is NP-hard. We do it by proving that thecorresponding membership testing problem is coNP-complete for the appropriateclasses of inequalities.Furthermore, we prove a lower bound of 2 Ω ( n ) on the extension complexity ofbounded relaxations of QAP n obtained by each of these families of inequalities.To summarize, our main contributions are as follows. – We identify a new family of valid inequalities for QAP n (Section 2) andprove that they are facet-defining (Section 3), – We prove that membership testing for three out of the five known families ofvalid inequalities for the QAP-polytope(including the new one we introduce)is coNP-complete (Section 4), and – We prove a lower bound of 2 Ω ( n ) for the extension complexity of any bounded relaxation of QAP n that has any of the known families as valid inequalities(Section 5). Let P ⊂ R n be a polytope. A polytope Q ⊂ R n + r is called an extension or an extended formulation of P if P = { x ∈ R n | ∃ y ∈ R r , ( x, y ) ∈ Q } . In fact, boundedness is not required for the results in Section 5. However, sincewe will rely on existing results, such as Theorem 1, that are published with theboundedness assumption, we will include this assumption.n the Complexity of Some Facet-Defining Inequalities of the QAP-polytope 3
Let size( P ) denote the number of facets of polytope P and let Q ↓ P denotethat Q is an extended formulation of P . Then, the extension complexity of apolytope P - denoted by xc( P ) - is defined to be min Q ↓ P size( Q ).Extended formulations are a very useful tool in combinatorial optimizationas they allow the possibilty of drastically reducing the size of a Linear Programby introducing new variables (See [32,33,21,11] for surveys). In the past decadelower bounds on the extension complexity of various polytopes have been stud-ied [29,14,2,30] and the notion generalized and studied in various settings suchas general conic extensions [16], semidefinite extensions [7,25], approximation[6,10,4,3], parameterization [8,15], generalized probabilistic theories in Physics[13], and information theoretic perspective [6,5].Superpolynomial lower bounds on extension complexity are known for poly-topes related to many NP-hard problems [14,2] as well as for the Matchingpolytope: the convex hull of characteristic vectors of all matchings in K n [30].High extension complexity of the Matching polytope highlights the fact that asuperpolynomial lower bound on the extension complexity cannot be taken tomean that the underlying optimization problem is not solvable in polynomialtime. However, these lowers bounds are unconditional and do not require stan-dard complexity theoretic assumption such as P = N P . Moreover, apart fromthe exception of Matching polytope, linear optimization over all known poly-topes with superpolynomial lower bound is infeasible. Either because the linearoptimization over the polytope is NP-hard [14,2] or the polytope is not explicitlygiven, as is the case for some matroid polytopes [29].For the purposes of this paper the most relevant characterization of extensioncomplexity is given by Faenza et al. [12] where the authors prove the equivalencebetween existence of an extended formulation of size r with the existence of acertain two-party communication game requiring an exchange of Θ (log r ) bits.We will describe this connection here and use it as a black box in our proofs oflower bounds on the extension complexity of the polytopes considered here. EF-protocols: Computing a matrix in expectation
Let M be an m × n matrix with non-negative entries. Consider a communication game between twoplayers: Alice and Bob. Alice and Bob both know the matrix M and can agreeupon any strategy prior to the start of the game. In each round of the game,Alice receives a row index i ∈ [ m ] and Bob a column index j ∈ [ n ]. Both Aliceand Bob have no restriction on the computations that they perform and canalso use (private) random bits. They can also exchange information by sendingsome bits to the other player. At some point one of them outputs a non-negativenumber and the round finishes.Since they are allowed the use of random bits, the output X ij when Alice andBob receive inputs i and j respectively, is a random variable. We says that theirstrategy is an EF-protocol for M if E [ X ij ] = M ij for all ( i, j ) ∈ [ m ] × [ n ], where E [ X ij ] is the expected value of the random variable X ij . The complexity of anEF-protocol is defined to be the maximum number of bits exchanged betweenAlice and Bob for any input i, j to Alice and Bob respectively. Pawan Aurora and Hans Raj Tiwary
Let P = { x | Ax b } = conv( { v , . . . , v n } ) be a polytope where A is an m × d real matrix, b ∈ R m , v i ∈ R d , and conv( S ) denotes the convex hull of thepoints in a set S . The slack matrix of P with respect to this representation -denoted by S ( P ) - is the m × n (non-negative) matrix whose entry at i -th rowand j -th column is b i − A i v j , where A i denotes the i -th row of the matrix A .Note that a polytope is not defined uniquely this way: one can always embedthe polytope in higher dimensional space, and add redundant inequalities andpoints to the descriptions. However, in what follows, none of that makes anydifference and one can choose any description that they like. This justifies thenotation S ( P ) for any slack matrix of P even though the particular descriptionof P that defines this matrix is completely ignored in the notation.The following connection – which we will use in Section 5 – was shown byFaenza et al. [12] between existence of an EF-protocol computing a slack matrixof polytope P and that of an extended formulation of P . Theorem 1. [12] Let M be a non-negative matrix such that any EF-protocolfor M has complexity at least c . Further, let P be a polytope such that M is asubmatrix of some slack matrix S ( P ) of P . Then, xc( P ) > c . n In the following, Y is a n × n variable matrix that is used to denote an arbitrarypoint in QAP n . Further, Y ij,kl refers to Y ( n ∗ ( i −
1) + j, n ∗ ( k −
1) + l ).The most general family of valid inequalities that includes all known familiesas special cases is the following: QAP1: X ijkl n ij n kl Y ij,kl − (2 β − X ij n ij Y ij,ij > − ( β −
12 ) (1)where β ∈ Z and n ij ∈ Z for all i, j ∈ [ n ]. These inequalities were introducedin [1] as a generalization of all known facet-defining inequalities for the QAP-polytope. QAP2: m X r =1 Y i r j r ,kl − Y kl,kl − X r , m >
3. These inequalities were introduced in [1] and proved to be aspecial case of QAP1 that are facet-defining for the QAP-polytope.
QAP3: ( β − X ( ij ) ∈ P × Q Y ij,ij − X ( ij ) , ( kl ) ∈ P × Qi 3, (ii) | P | + | Q | n − β ,(iii) β > 2. These inequalities were introduced by J¨unger and Kaibel [19,20] whoalso proved that they are facet-defining for the QAP-polytope. They are also aspecial case of QAP1 [1]. n the Complexity of Some Facet-Defining Inequalities of the QAP-polytope 5 QAP4: − ( β − X ( ij ) ∈ P × Q Y ij,ij + β X ( ij ) ∈ P × Q Y ij,ij + X i 3, (ii) | P | + | P | ≤ n − 3, (iii) | P | > min { , β + 1 } , (iv) | P | > min { , − β + 2 } , (v) || P | − | P | − β | ≤ n − | Q | − 4, (vi) if | P | = 1: | Q | > min {− β + 5 , β + 2 } ; if | P | > | Q | > min {− β + 5 , β + 3 } or | Q | > min {− β + 4 , β + 4 } . These inequalities werealso introduced by Kaibel [20] and shown to be facet-defining for the QAP-polytope. They are also a special case of QAP1.It is known [1] that the inequalities QAP2 , QAP3 and QAP4 are specialinstances of the QAP1 inequalities. QAP1 inequalities are in general not facet-defining for the QAP-polytope and so it is interesting to identify conditionsunder which they do define facets. We identify a new special case (5) of QAP1inequality and show in Section 3 that they are facet-defining. Inequality QAP5follows from QAP1 by setting β = 2, n i j = n i j = · · · = n i m ,j m = 1 for distinct i , . . . , i m ∈ [ n ] and distinct j , . . . , j m ∈ [ n ], and n ij = 0 for i ∈ [ n ] \ { i , . . . , i m } or j ∈ [ n ] \ { j , . . . , j m } . QAP5: m X r =1 Y i r j r ,i r j r − X r 7. These inequalities are new and we discuss them in the followingsection. In this section we prove that the inequalities QAP5 are facet-defining. Let S k de-note the set of those vertices of QAP n that correspond to the permutations hav-ing i r j r for r ∈ { r , r , . . . , r k } ⊆ [ m ] and i r j r for r ∈ [ m ] \{ r , r , . . . , r k } .Here i r , j r , m are as in the definition of QAP5. If V is the set of all the verticesof QAP n then clearly V = ∪ mi =0 S i , and S i ∩ S j = ∅ for all i = j ∈ { , , . . . , m } . Lemma 1. The sets S , S together constitute the vertices that satisfy the in-equality (5) with equality.Proof. Consider a P [2] σ ∈ S k for 1 k m . W.l.o.g. let σ ( i ) = j , . . . , σ ( i k ) = j k and σ ( i r ) = j r for r ∈ { k + 1 , . . . , m } . Substituting Y = P [2] σ in (5) we have P mr =1 P σ ( i r , j r ) − P r 1, for k = 1 , k − (cid:0) k (cid:1) = 1, for k = 2 , k − (cid:0) k (cid:1) = 1 and for 3 k m we have k − (cid:0) k (cid:1) < 1. Hence a vertex of QAP n satisfies the inequality (5) withequality if and only if it belongs to S or S . ⊓⊔ Pawan Aurora and Hans Raj Tiwary Let S = S ∪ S . We will show that any vertex in V \ S can be expressedas a linear combination of the vertices in S and a fixed vertex P [2] σ ∗ ∈ S . Thiswill establish that the dimension of the face containing S is one less than thedimension of the polytope and hence it must be a facet.The following lemma from [1] provides a useful tool to express certain verticesas a linear combination of others. Lemma 2. [1, Lemma 16] Let k , k , k , x, y ∈ [ n ] be distinct indices. Let Σ = { σ , . . . , σ } be a set of permutations of [ n ] such that σ i ( z ) = σ j ( z ) for all z ∈ [ n ] \ { k , k , k } and for every i, j ∈ { , . . . , } . Further, let Σ ′ = { σ ′ , . . . , σ ′ } where σ ′ i is a transposition of σ i on the indices x, y , for each i = 1 , . . . , . Then ∀ i, j, k, l ∈ [ n ] , P σ ∈ Σ ∪ Σ ′ sign ( σ ) P [2] σ ( ij, kl ) = 0 . The next lemma shows that the vertices in the sets S , . . . , S m can be ex-pressed as a linear combination of the vertices in the sets S , . . . , S .In what follows, when it is clear from the context, we use σ to refer to avertex P [2] σ . Lemma 3. For k > , any vertex in S k can be expressed as a linear combinationof vertices in S k − , S k − , S k − , and S k − .Proof. Let σ ∈ S k . Since k is at least 4, we must have indices i r , i r , i r , i r , r k ∈ [ m ] , k = 1 , . . . , 4, such that σ ( i r k ) = j r k , k = 1 , . . . , 4. Let k = i r , k = i r , k = i r , x = i r , where k , k , k , x are as defined in Lemma 2. ApplyingLemma 2 with y chosen as an index such that either y = i p for any p ∈ [ m ]or when y = i p , p ∈ [ m ] then σ ( i p ) = j p , we get σ ′ ∈ S k − , σ , σ , σ ∈ S k − , σ , σ , σ ′ , σ ′ , σ ′ ∈ S k − , σ ′ , σ ′ ∈ S k − with the property that σ is alinear combination of σ , . . . , σ , σ ′ , . . . , σ ′ . ⊓⊔ Next, we show that vertices in S can be expressed as linear combinations ofvertices in S and S as well. Lemma 4. Any vertex in S can be expressed as a linear combination of verticesin S and S .Proof. Let σ ∈ S . So we have indices i r , i r , i r , r k ∈ [ m ] , k = 1 , , 3, suchthat σ ( i r k ) = j r k , k = 1 , , 3. Let k = i r , k = i r , x = i r , where k , k , x are as defined in Lemma 2. Applying Lemma 2 with k , y chosen arbitrarilyfrom [ n ] \ { i r , i r , i r } , we get σ , . . . , σ , σ ′ , σ ′ , σ ′ ∈ S, σ ′ , σ ′ , σ ′ ∈ S with theproperty that σ is a linear combination of σ , . . . , σ , σ ′ , . . . , σ ′ . ⊓⊔ Now that we have established that the linear hull of S , S , S equals thelinear hull of the QAP-polytope, the only remaining task is to show that insteadof the entire set S , a fixed vertex in S suffices to generate the entire linearhull. We will show that in fact any arbitrary vertex in S is sufficient. We do thisby first showing that permutations in S define a connected graph if the edgesconnect permutations that are one transposition apart. Then, we show that any n the Complexity of Some Facet-Defining Inequalities of the QAP-polytope 7 vertex σ ∈ S is sufficient to generate all vertices in the connected componentof σ by linear combination with vertices in S , S .In the following lemma we show that it is possible to obtain the permutationcorresponding to a vertex in S from the permutation corresponding to anyother vertex in S via transpositions such that the vertices corresponding to theintermediate permutations also lie in S . Lemma 5. Consider the graph G = ( S , E ) where S is the set of vertices of theQAP-polytope that correspond to permutations for which i r j r for all r ∈ [ m ] and { P [2] σ , P [2] σ } ∈ E for some P [2] σ , P [2] σ ∈ S if σ and σ are transpositions ofeach other. Then G is connected.Proof. Consider an arbitrary vertex P [2] σ ∈ S such that σ ( i r ) = k r , k r = j r forall r ∈ [ m ]. Also, consider another vertex P [2] σ ′ ∈ S such that σ ′ ( i r ) = l r , l r = j r for all r ∈ [ m ]. We will show that there is a path from P [2] σ to P [2] σ ′ in G . Forsimplicity, we will use σ to refer to P [2] σ . Let σ, σ , σ , . . . , σ t , σ ′ be a path oflength t + 1 between σ and σ ′ . Consider a vertex σ p , p ∈ [ t ] such that σ p ( i r ) = l r for all r ∈ [ s ] , s < m . In the next step we will extend the path from σ p to somevertex σ q such that σ q ( i r ) = l r for all r ∈ [ s + 1]. If σ p ( i x ) = l s +1 such that σ p ( i s +1 ) = j x then we can swap σ p ( i x ) with σ p ( i s +1 ) to get the desired vertex σ q . Otherwise, in the first swap we can move l s +1 to some index i x ′ , x = x ′ ,such that l s +1 = j x ′ and then in the second swap get σ p ( i s +1 ) to map to l s +1 .Note that both the swaps result in vertices within S . After the first swap wehave σ p ′ ( i x ′ ) = l s +1 and σ p ′ ( i x ) = σ p ( i x ′ ) and after the second swap we get σ q ( i s +1 ) = l s +1 and σ q ( i x ′ ) = j x , both of which avoid a map from i x to j x and i x ′ to j x ′ . In case it is not possible to find a suitable i x ′ to move l s +1 , it shouldbe possible to move σ p ( i s +1 ) instead. Once we have obtained a permutation σ ′′ such that σ ′′ ( i r ) = l r for all r ∈ [ m ], there must exist a path from σ ′′ to σ ′ sincethe set of permutations having i r l r for all r ∈ [ m ], forms a group isomorphicto the symmetric group on n − m elements. ⊓⊔ The following lemma gives a sequence of four vertices of QAP n such thata specific linear combination of these vertices reduces the number of non-zeroentries in the resulting vector to a constant independent of n . This lemma will beused crucially in Lemma 7 to express the difference of two neighboring verticesin S in terms of the vertices in S . Lemma 6. Given a sequence of permutations over the set [ n ] , σ , σ , σ , σ ,such that σ is obtained from σ by a transposition that swaps the values of σ ( i ) , σ ( j ) ; σ is obtained from σ by a transposition that swaps the values of σ ( i ′ ) , σ ( j ′ ) ( i ′ = i, j ′ = j ); and σ is obtained from σ by a transposition thatswaps the values of σ ( i ) , σ ( j ) . Then ( P [2] σ − P [2] σ ) − ( P [2] σ − P [2] σ ) has a numberof non-zeroes that is independent of n .Proof. Recall that P [2] σ ( ab, xy ) = P σ ( a, b ) · P σ ( x, y ). Since σ and σ differ atonly i, j , we have ( P [2] σ − P [2] σ )( ab, xy ) = 0 for all a, b, x, y ∈ [ n ] \ { i, j } . Similarly, Pawan Aurora and Hans Raj Tiwary we have ( P [2] σ − P [2] σ )( ab, xy ) = 0 for all a, b, x, y ∈ [ n ] \ { i, j } . One can verify that( P [2] σ − P [2] σ )( ab, xy ) = 1 for all x, y such that σ ( x ) = y , when a = i, b = σ ( i )or when a = j, b = σ ( j ). Symmetrically, we have ( P [2] σ − P [2] σ )( ab, xy ) = − x, y such that σ ( x ) = y , when a = i, b = σ ( i ) or when a = j, b = σ ( j ).For the case when a, b / ∈ { i, j } , ( P [2] σ − P [2] σ )( ab, xy ) = 1 when x = i, y = σ ( i )or when x = j, y = σ ( j ) and ( P [2] σ − P [2] σ )( ab, xy ) = − x = i, y = σ ( i ) or when x = j, y = σ ( j ). Similar values follow for P [2] σ − P [2] σ . Note that P [2] σ − P [2] σ and P [2] σ − P [2] σ differ only at the indices i ′ , j ′ . So subtracting thelatter from the former we get, (( P [2] σ − P [2] σ ) − ( P [2] σ − P [2] σ ))( ab, xy ) = 0 for all a, b, x, y / ∈ { i, j, i ′ , j ′ } . The only non-zero entries that remain are the following:(i) a = i, b = σ ( i ) , x = i ′ , y = σ ( i ′ ), (ii) a = i, b = σ ( i ) , x = i ′ , y = σ ( i ′ ),(iii) a = j, b = σ ( j ) , x = i ′ , y = σ ( i ′ ), (iv) a = j, b = σ ( j ) , x = i ′ , y = σ ( i ′ ),(v) a = i, b = σ ( i ) , x = i ′ , y = σ ( i ′ ), (vi) a = i, b = σ ( i ) , x = i ′ , y = σ ( i ′ ),(vii) a = j, b = σ ( j ) , x = i ′ , y = σ ( i ′ ), (viii) a = j, b = σ ( j ) , x = i ′ , y = σ ( i ′ ).Another 8 non-zero entries correspond to the case when x = j ′ taking the totalto 16. 16 more entries follow from symmetry, by swapping a, b with x, y . Thus,we get a total of 32 non-zero entries in the resulting matrix. Half of these are+1 and the remaining half are − 1. Note that these entries depend only on theindices where the four permutations map the indices i, j, i ′ , j ′ and not on thevalue of n or where these permutations map the remaining indices. ⊓⊔ Lemma 7. For any P [2] σ ∈ S , P [2] σ − P [2] σ ′ lies in the linear hull of S for everyneighbor P [2] σ ′ ∈ S , provided m > .Proof. Let σ , σ , σ , σ be as defined in Lemma 6. Let σ , σ , σ , σ be fourpermutations different from σ , σ , σ , σ but related to each other just like σ , σ , σ , σ are. This means that σ is obtained from σ by the same trans-position that is used to obtain σ from σ , σ is obtained from σ by the sametransposition that is used to obtain σ from σ , and σ is obtained from σ bythe same transposition that is used to obtain σ from σ . So from Lemma 6,we have P [2] σ − P [2] σ = ( P [2] σ − P [2] σ ) + ( P [2] σ − P [2] σ ) − ( P [2] σ − P [2] σ ). Let σ = σ and σ = σ ′ . If we can find σ , . . . , σ as defined above such that the corre-sponding vertices lie in S , then we are done. Since σ, σ ′ ∈ S , we have someindex a, a = i r , r ∈ [ m ] such that σ ( a ) = σ ′ ( a ) = j r . Consider the case when a = i p , p ∈ [ m ] and σ ( i r ) = j p . The case when a = i p for any p ∈ [ m ] is similar.We can swap σ ′ ( a ) with σ ′ ( i r ) to get σ ( a ) = σ ′ ( i r ) , σ ( i r ) = j r which clearlylies in S . We obtain σ from σ by the transposition defined in Lemma 6 andclearly σ also lies in S . Next we select a permutation σ ∈ S that matches with σ at the four indices defined in Lemma 6 and also maps an index i r ′ , r = r ′ to j r ′ . So we have σ , σ ∈ S , σ ∈ S . Obtaining σ , σ , σ as outlined above,we have σ ∈ S , σ , σ ∈ S . Next, consider the case when a = i p , p ∈ [ m ] and σ ( i r ) = j p . This can happen when m is even and any transposition of σ eitherresults in a permutation in S or in S . It is not possible to get a permutationin S by a single transposition of σ . So we obtain σ , σ as before but this timethe vertices lie in S instead of S . Moreover, this time we select a permutation n the Complexity of Some Facet-Defining Inequalities of the QAP-polytope 9 σ ∈ S that matches with σ at the four indices defined in Lemma 6 but has apair of indices i x , i y such that σ ( i x ) = j y but σ ( i y ) = j x . Obtaining σ , σ , σ as above, we have σ , σ ∈ S , σ , σ ∈ S . We can now repeat the above argu-ment with σ = σ , σ ′ = σ . This time however, by the choice of σ ∈ S , we haveensured that we are in the first case where we could express the difference vectoras a combination of vertices only in S . Note that we need m to be at least 7for the above argument to work. This is so because the transposition of σ thatgives σ ′ can use upto four indices so that these indices are no longer available toget to S . Further, as in the second case above, two other indices swap with eachother to get to S . So that takes up a total of six indices that are not availableto get the desired σ . Now if m is at least 7, we are guaranteed to find a pair ofindices i x , i y such that σ ( i x ) = j y but σ ( i y ) = j x . ⊓⊔ Theorem 2. The following inequality: m X r =1 Y i r j r ,i r j r − X r Given a point x ∈ R n with x , it is coNP-complete todecide whether x satisfies all inequalities of QAP2.Proof. The problem is clearly in coNP since given a violated inequality it canbe checked quickly that it is indeed violated.For establishing NP-hardness we will reduce the max-clique problem to mem-bership testing for QAP2. Let the given instance of the max-clique problem be G = ( V, E ) where V = [ n ]. We construct a n -partite graph G ′ = ( V ′ , E ′ ) where V ′ = { ( ij ) } for i, j ∈ [ n ] and { i j , i j } ∈ E ′ if and only if { i , i } ∈ E .So if there is an edge { i, j } ∈ E then we get a complete bi-partite graph be-tween the partitions i and j , else there is no edge between these two partitions.Consider a clique C = { i , i , . . . , i k } of size k in G . Then the set of vertices { i j , i j , . . . , i k j k } where j r could be any arbitrary index in [ n ], forms a cliqueof size k in G ′ . Conversely, given a clique C = { i j , i j , . . . , i k j k } of size k in G ′ , the set of vertices { i , i , . . . , i k } forms a clique of size k in G . Fix a pair ofindices k, l ∈ [ n ] arbitrarily and add edges { kl, i r j r } for all i r ∈ { [ n ] \ { k }} to G ′ (if the edge is not already present). Now construct a point Y as follows: Y i j ,i j = , if ( kl ) / ∈ { ( i j ) , ( i j ) } and { i j , i j } ∈ E ′ n, if ( kl ) / ∈ { ( i j ) , ( i j ) } and { i j , i j } / ∈ E ′ , if ( i = i = k and j = j = l )or ( k = i = i and l = j = j ) t, if i = i = k and j = j = ln , if i = i and j = j and ( i = k or j = l )where t > Y satisfies all theinequalities of QAP2 if and only if αY satisfies them for all α > 0. Therefore Y can be scaled to satisfy 0 Y 1. We will ignore this scale factor and continueour argument with Y as constructed above to avoid cluttered equations.We claim that Y satisfies all the inequalities of QAP2 if and only if everyclique in the subgraph induced by the neighborhood of the vertex ( kl ) in G ′ , hassize at most t .Suppose that the largest clique C = { i j , i j , . . . , i t ′ j t ′ } such that { i r j r , kl } ∈ E ′ for r ∈ [ t ′ ], has size t ′ > t . Without loss of generality, we can assumethat i , . . . , i t ′ as well as j , . . . , j t ′ are distinct. Consider the inequality h C cor-responding to the choice of indices { i , . . . , i t ′ , k } , { j , . . . , j t ′ , l } . From the aboveconstruction, P t ′ r =1 Y i r j r ,kl = t ′ , Y kl,kl = t and P r 3, (ii) | P | + | Q | n − β ,(iii) β > Theorem 4. Given a point x ∈ R n with x > , it is coNP-complete to decidewhether x satisfies all inequalities of QAP3.Proof. Again we will reduce the max-clique problem to membership testing forQAP3. Given an instance of the max-clique problem, G = ( V, E ) with | V | = n ,we construct a point Y as follows: Y i j ,i j = , if { i , i } ∈ En , if { i , i } / ∈ E ( i = i )1 /t, if i = i and j = j = 10 , if i = i and j = j and j > t n − Y satisfies allinequalities of QAP3 if and only if G doesn’t contain a clique of size larger than t . This gives an algorithm to find the size of largest clique in G by increasing t gradually and computing the smallest value of t for which Y becomes feasible. If Y remains infeasible for t = n − G has size n − O ( n ) possible subsets of verticesof G .To prove the claim, let P be a clique in G with t < | P | n − 3. Define Q = { } , β = 2 and consider the inequality h defined by P, Q, β . It can bechecked that Y violates h .Conversely, let h be an inequality defined by P, Q, β that is violated by Y . We first observe that 1 ∈ Q . Suppose not, then for any ( i, j ) ∈ P × Q wehave Y ij,ij = 0. Since β − β > β > h cannot be violated by Y . It follows that if i, k ∈ P and i = k , then { i, k } ∈ E. Again, suppose not.Then, Y ij,kl = n for all j, l but ( β − P ( ij ) ∈ P × Q Y ij,ij ( n − · ( n − < n .So Y cannot violate h as ( β − P ( ij ) ∈ P × Q Y ij,ij − P ( ij ) , ( kl ) ∈ P × Q,i 2, that is, larger than t. ⊓⊔ Finally, recall that QAP5 is defined by the following set of inequalities: m X r =1 Y i r j r ,i r j r − X r Given a point x ∈ R n with x > , it is coNP-complete to decidewhether x satisfies all inequalities of QAP5.Proof. Given an instance of the max-clique problem, G = ( V, E ) with | V | = n ,we construct a point Y as follows: Y i j ,i j = , if { i , i } ∈ En/ , if i = i and { i , i } / ∈ E /t, if i = i and j = j , otherwisewhere, t > Y is infeasible if and only if there exists a clique in G of sizeat least t + 1.Suppose there exists a clique C = { p , . . . , p m } in G with m > t +1. Considerthe inequality h C defined by the indices i , . . . , i m and j , . . . , j m with i k = j k = p k for all k ∈ [ m ]. Then, Y violates h C because m/t > G has size at most t and let h be aviolated inequality defined by indices i , . . . , i m and j , . . . , j m . It must holdthat { i r , i s } ∈ E otherwise the left hand side in the inequality h with respectto Y is at most m/t − n/ t > m n and so Y cannot violate h . So for a violation, G must contain a clique of size m . Butthen Y i r j r ,i s j s = 0 for all distinct r, s ∈ [ m ] and so m/t > G contains no cliques of size larger than t .Therefore, given a graph G = ( V, E ) if Y is feasible for all values of t > t for which Y is infeasible equals the sizeof the largest clique in G minus one. ⊓⊔ n the Complexity of Some Facet-Defining Inequalities of the QAP-polytope 13 In this section we will prove that any relaxation of the QAP-polytope for whichany of the families of inequalities defined in Section 2 are valid, has superpolyno-mial extension complexity. A set of linear inequalities is said to be a relaxationof the QAP-polytope if it contains the QAP-polytope.We will use Theorem 1 to prove each of the results in this section. The generalargument would be as follows. First we will show that certain specific matricesrequire exchange of many bits in any EF-protocol for them. Then, supposing thatR n is any relaxation of QAP n for which inequalities H are valid, we will provethat any EF-protocol for the slack matrix of R n requires an exchange of manybits by showing that computing the slack entries corresponding to the validinequalities in H and the vertices of QAP n allows us to compute our specialmatrices by exchanging few extra bits. So the slack matrix of R n contains asubmatrix that is difficult to compute and thus the whole slack matrix is difficultto compute. Let n be natural numbers. Consider the 2 n × n sized matrices M n , N n , N n defined as follows. The rows and columns of these matrices are indexed by 0 / n . Let a, b ∈ { , } n . The entry with row index a and columnindex b for each of these matrices is defined as follows: M kn ( a, b ) := ( a ⊺ b − k ) N kn ( a, b ) := ( a ⊺ b − k ) · ( a ⊺ b − k − M n require Ω ( n ) bits to be exchanged in any EF-protocol andwere used to prove that the correlation polytope has superpolynomial extensioncomplexity by Fiorini et al. [14]. Later Kaibel and Weltge [22] gave a simplecombinatorial proof of the hardness of computing the same matrices M n . Inparticular, the following holds: Lemma 8. [14,22] Any EF-protocol for computing M n requies an exchange of Ω ( n ) bits. For our purposes we need to work with matrices N kn which we now show torequire Ω ( n − k ) bits for k > Lemma 9. There exists an EF-protocol for computing N n that requires an ex-change of ⌈ log n ⌉ bits.Proof. Alice and Bob have a, b ∈ { , } n respectively and they wish to output a ⊺ b · ( a ⊺ b − 1) in expectation. Alice selects two indices i = j ∈ [ n ] uniformlyat random. That is, the probability of any two indices being chosen is 1 / (cid:0) n (cid:1) . If a i = 0 or a j = 0, Alice outputs zero and the protocol ends. Otherwise a i = a j = 1and Alice sends the binary encoding of indices i, j to Bob using 2 ⌈ log n ⌉ bits. If b i = b j = 1, Bob outputs n ( n − 1) otherwise he outputs zero, and the protocolends.A non-zero value is output if and only if the indices selected by Alice satisfy a i = a j = b i = b j = 1. This happens with probability (cid:0) a ⊺ b (cid:1) / (cid:0) n (cid:1) . Therefore theexpected output is n ( n − · ( a ⊺ b )( n ) = N n ( a, b ) . ⊓⊔ Lemma 10. Any EF-protocol for computing N kn for k > requires Ω ( n − k ) bits to be exchanged.Proof. We first prove the result for the case of k = 1.For any real number α it holds that α ( α − α − α − 2) = 2( α − . Using α = a ⊺ b we get that for all a, b ∈ { , } n we have that M n ( a, b ) = N n ( a,b )+ N n ( a,b )2 .Now, suppose that there exists an EF-protocol for computing N n that re-quires c bits to be exchanged. To compute M n , Alice decides whether to com-pute N n or N n (by tossing a fair coin) and tells this to Bob using one bit.They then proceed with the appropriate protocol. The expected value is clearly N n ( a,b )+ N n ( a,b )2 which equals M n ( a, b ) and they need to exchange only 1 + max { ⌈ log n ⌉ , c } bits. From Lemma 8 we know that Ω ( n ) bits may need to be ex-changed for this. Therefore, c = Ω ( n ) and the lemma holds for k = 1.Now, let a, b ∈ { , } n − k +1 be two binary strings of length n − k + 1 forsome k > 2. Obtain a ′ , b ′ ∈ { , } n by appending k − a ′ ⊺ b ′ = a ⊺ b + k − a ′ ⊺ b ′ − k )( a ′ ⊺ b ′ − k − 1) = ( a ⊺ b − a ⊺ b − N n − k +1 is a submatrix of N kn . It then follows that any EF-protocol for N kn requires Ω ( n − k + 1) bits to be exchanged thus proving the lemma for all k > ⊓⊔ n Recall the inequalities (1) of QAP1: X ijkl n ij n kl Y ij,kl − (2 β − X ij n ij Y ij,ij > − ( β − ) As noted in Section 2 these are the most general family of valid inequalitiesfor the QAP-polytope. Therefore any lower bound on the extension complexityof a relaxation R n corresponding to any of the families QAP2-QAP5 also holdfor QAP1 and so we state our next theorem without proof. Theorem 6. Let R1 n be any bounded relaxation of the QAP-polytope such thatthe inequalities of QAP1 are valid for R1 n . Then xc(R1 n ) > Ω ( n ) . Now consider any relaxation of QAP n for which the inequalities of QAP2 arevalid. n the Complexity of Some Facet-Defining Inequalities of the QAP-polytope 15 Lemma 11. Let i , i , . . . , i m , k be distinct indices in [ n ] , and j , j , . . . , j m , l beall distinct indices in [ n ] as well. Let σ ∈ S n be such that q indices ( i r , j r ) satisfy P σ ( i r , j r ) = 1 . Then, the slack of m X r =1 Y i r j r ,kl − Y kl,kl − X r The slack is P σ ( k, l ) + (cid:0) q (cid:1) − qP σ ( k, l ). If P σ ( k, l ) = 0 then this equals (cid:0) q (cid:1) .If P σ ( k, l ) = 1 then this equals 1 + (cid:0) q (cid:1) − q = q ( q − − q = (cid:0) q − (cid:1) . ⊓⊔ Theorem 7. Let R2 n be any bounded relaxation of QAP n such that the inequal-ities of QAP2 are valid for R2 n . Then xc(R2 n ) > Ω ( n ) .Proof. Suppose there exists an EF-protocol for computing the slack matrix ofR2 n that requires at most c bits to be exchanged. That is if Alice is given anyvalid inequality for R2 n and Bob any feasible point in R2 n they can compute thecorresponding slack in expectation by exchanging at most c bits. We will showthat they can modify this EF-protocol to get an EF-protocol for matrix N n withat most O ( c + log n ) bits exchanged. By Lemma 10 this requires Ω ( n ) bits tobe exchanged and so c = Ω ( n ). Finally, applying Theorem 1 we will get thatxc(R2 n ) > Ω ( n ) .So suppose, Alice and Bob get a, b ∈ { , } n respectively and wish to compute N n ( a, b ) = ( a ⊺ b − a ⊺ b − 2) in expectation. We can assume that Alice receivesneither the all zero nor the all one vector. If a = (0 , . . . , 0) then she can outputzero and stop. If a = (1 , . . . , 1) then she can tell Bob this using one bit and Bobcan output the number of nonzero entries in b . Further, we can assume that thevector b contains at least three zero entries. Otherwise Bob can tell Alice usingat most 2 log n bits the indices where b is zero and Alice can output the correctvalue.Let p , . . . , p m be the indices where a is non-zero and let p be an arbi-trary index such that a p = 0. Alice creates the inequality corresponding to sets i , . . . , i m , k and j , . . . , j m , l where i = j = p , . . . , i m = j m = p m and k = l = p .Alice then sends the index p to Bob who sets b p = 1 if it is not already so. Bobthen creates any permutation σ b such that σ b ( i ) = i if b i = 1 and σ b ( i ) = i if b i = 0. This is clearly possible since b still contains at least two zeroes. Bobselects the vertex P [2] σ b of QAP n corresponding to this permutation. Clearly, a ⊺ b equals the number of index pairs ( i r , j r ) in the set created by Alice such that P [2] σ b ( i r , j r ) = 1.Using P σ ( k, l ) = P σ b ( p, p ) = 1 and q = a ⊺ b in Lemma 11 we see that the slackof Alice’s inequality with respect to Bob’s vertex of QAP n is exactly (cid:0) a ⊺ b − (cid:1) = N n ( a, b ) and hence they can just use the protocol for computing the slackmatrix of R2 n for computing N n by agreeing that every time they wish to outputsomething they would output twice as much. ⊓⊔ Next, consider any relaxation of QAP n for which the inequalities of QAP3are valid. Recall the inequalities (3) of QAP3:( β − X ( ij ) ∈ P × Q Y ij,ij − X ( ij ) , ( kl ) ∈ P × Qi Let h be an inequality in QAP3 given by sets P, Q ⊂ [ n ] and β ∈ N .Let σ ∈ S n such that q index pairs ( i, j ) in P × Q satisfy P σ ( i, j ) = 1 . Then, theslack of h with respect to P [2] σ is (cid:0) q − ( β − (cid:1) . Proof. The slack of h with respect to P [2] σ is β − β − ( β − q + (cid:18) q (cid:19) = β ( β − − q ( β − 1) + q ( q − β − β − q ) + q ( q − β )2= ( q − β )( q − β + 1)2= (cid:18) q − ( β − (cid:19) (6) ⊓⊔ Theorem 8. Let R3 n be any bounded relaxation of QAP n such that the inequal-ities of QAP3 are valid for R3 n . Then xc(R3 n ) > Ω ( n ) .Proof. Suppose Alice and Bob get a, b ∈ { , } m respectively and wish to com-pute N m ( a, b ) in expectation. Let n = 2 m + 1 and let p , . . . , p t be the in-dices where a is non-zero. Alice creates the inequality corresponding to sets P = Q = { p , . . . , p t } and β = 2. Note that | P | m = ( n − / | P | + | Q | =2 | P | n − n − β . So the value of β = 2 the choices of P, Q, β satisfy theconditions of QAP3 for QAP n .Bob creates the following permutation σ b ∈ S n σ b ( i ) = i if 1 i m and b i = 1 ,i + m if 1 i m and b i = 0 ,i − m if m < i m + 1 and b i = 0 ,i otherwise.and selects the vertex P [2] σ b of the QAP n . Since R3 n is a relaxation of QAP n , Bobhas picked a feasible point of R3 n . Notice that a ⊺ b equals the number of indexpairs ( i, j ) in the set P × Q such that P σ b ( i, j ) = 1. n the Complexity of Some Facet-Defining Inequalities of the QAP-polytope 17 Therefore using q = a ⊺ b and β = 2 in Lemma 12 we get that the slack ofAlice’s inequality with respect to Bob’s feasible point of R3 n is (cid:0) a ⊺ b − ( β − (cid:1) = a ⊺ b − a ⊺ b − = N m ( a, b ) and hence they can just use the protocol for comput-ing the slack matrix of R3 n for computing N m by agreeing that every time theywish to output something they would output twice as much. ⊓⊔ We would like to remark that in the proof of the previous Theorem onecould have as easily used a value of β other than two as long as β αn for some0 < α < 1. So the extension complexity for QAP3 remains superpolynomial evenfor any fixed value of β if β αn for some 0 < α < n for which the inequalities of QAP4are valid. Recall the inequalities (4) of QAP4: − ( β − X ( ij ) ∈ P × Q Y ij,ij + β X ( ij ) ∈ P × Q Y ij,ij + X i Consider an inequality h in QAP4 given by sets P , P , Q ⊂ [ n ] and β ∈ N . Let σ ∈ S n be such that q index pairs ( i, j ) in P × Q satisfy P σ ( i, j ) = 1 and q index pairs ( i, j ) in P × Q satisfy P σ ( i, j ) = 1 . Then, theslack of h with respect to P [2] σ is (cid:0) q − ( β − (cid:1) + q β + q ( q − − q q . Proof. The slack of h with respect to P [2] σ is − ( β − q + βq + (cid:18) q (cid:19) + (cid:18) q (cid:19) − q q − β (1 − β )2= q ( q − − q ( β − − q ( β − − β (1 − β )2 + 2 q β + q ( q − − q q q ( q − β ) − ( β − q − β )2 + 2 q β + q ( q − − q q 2= ( q − β + 1)( q − β )2 + 2 q β + q ( q − − q q (cid:18) q − ( β − (cid:19) + 2 q β + q ( q − − q q ⊓⊔ Theorem 9. Let R4 n be any bounded relaxation of QAP n such that the inequal-ities of QAP4 are valid for R4 n . Then xc(R4 n ) > Ω ( n ) . Proof. Suppose Alice and Bob get a, b ∈ { , } m respectively and wish to com-pute N m ( a, b ) in expectation. We can assume that a is not all one vector. Ifnot, she can send this information to Bob using a single bit who can output thenumber of ones in vector b . Let P = Q = { i | a i = 1 } and let P ⊆ [ m ] be suchthat P ∩ P = ∅ and | P | = 1. Clearly, i ∈ P = ⇒ a i = 0. Alice creates theinequality corresponding to the sets P , P , Q and β = 2. One can verify thatall the conditions of QAP4 are satisfied for the above choice of P , P , Q and β .Alice tells Bob the set P using log n bits.Bob creates the following permutation σ b ∈ S m +1 σ b ( i ) = i if i ∈ [ m ] \ P and b i = 1 ,i + m if i ∈ P ,i + m if 1 i m and b i = 0 ,i − m if m < i m + 1 and b i = 0 ,i − m if i − m ∈ P ,i otherwise.and selects the vertex P [2] σ b of the QAP n . Since R3 n is a relaxation of QAP n , Bobhas picked a feasible point of R3 n . Notice that a ⊺ b equals the number of indexpairs ( i, j ) ∈ P × Q such that P σ b ( i, j ) = 1. Also, notice that the number ofindex pairs ( i, j ) ∈ P × Q such that P σ b ( i, j ) = 1 is zero.Using q = a ⊺ b, q = 0 and β = 2 in Lemma 13 we get that the slack of Alice’sinequality with respect to Bob’s point of R4 m +1 is (cid:0) a ⊺ b − ( β − (cid:1) = (cid:0) a ⊺ b − (cid:1) = N m +1 ( a, b ). Therefore, Alice and Bob can just use the protocol for computingthe slack matrix of R4 m +1 for computing N m ( a, b ) by agreeing that every timethey wish to output something they would output twice as much. ⊓⊔ Once again we remark that the previous proof can be easily modified forany other value of β as long as β αn for some 0 < α < 1. So the extensioncomplexity lower bound also holds for each subfamily of inequalities obtainedfor any such fixed β .Finally, consider any relaxation of QAP n for which the inequalities of QAP5are valid. Recall the inequalities (5) of QAP5: m X r =1 Y i r j r ,i r j r − X r Let h be an inequality in QAP5 given by i , . . . , i m all distinct and j , . . . , j m all distinct. Let σ ∈ S n with q index pairs ( i r , j r ) such that P σ ( i r , j r ) =1 . Then, the slack of h with respect to P [2] σ is (cid:0) q − (cid:1) . Proof. The slack equals 1 − q + (cid:0) q (cid:1) = − q + q ( q − = ( q − q − . ⊓⊔ Theorem 10. Let R5 n be any bounded relaxation of QAP n such that the inequal-ities of QAP5 are valid for R5 n . Then xc(R5 n ) > Ω ( n ) . n the Complexity of Some Facet-Defining Inequalities of the QAP-polytope 19 Proof. Suppose Alice and Bob get a, b ∈ { , } n respectively and wish to com-pute N n ( a, b ) in expectation. We can assume that n > p , . . . , p m be the indices where a is non-zero. We can assume that m > a is non-zero using at most 6 log n bits and Bob can output the correctvalue of N n ( a, b ). So m, n > i , . . . , i m and j , . . . , j m where i = j = p , . . . , i m = j m = p m .We can also assume that b contains at least two zeroes otherwise Bob cantell Alice the index where b is zero using log n bits and Alice can output thecorrect value of N n ( a, b ). Now Bob creates the following permutation σ b ∈ S n such that σ b ( i ) = i if b i = 1 and σ b ( i ) = i if b i = 0. This is clearly possible since b contains at least two zeroes. Bob selects the vertex P [2] σ b of QAP n correspondingto this permutation. Clearly, a ⊺ b equals the number of index pairs ( i r , j r ) in theinequality created by Alice such that P [2] σ b ( i, i ) = 1.Setting q = a ⊺ b in Lemma 14 we get that the slack of Alice’s inequality withrespect to Bob’s point of R5 n is exactly N n ( a, b ) and hence they can just use theprotocol for computing the slack matrix of R5 n for computing N n by agreeing thatevery time they wish to output something they would output twice as much. ⊓⊔ Acknowledgement. Pawan Aurora is partially supported by grant MTR/2018/000861 of the Scienceand Engineering Research Board, Government of India. Hans Raj Tiwary waspartially supported by grant 17-09142S of GA ˇCR. References 1. Aurora, P., Mehta, S.K.: The qap-polytope and the graph isomorphism problem.J. Comb. Optim. (3), 965–1006 (2018)2. Avis, D., Tiwary, H.R.: On the extension complexity of combinatorial polytopes.Math. Program. (1), 95–115 (2015)3. Bazzi, A., Fiorini, S., Pokutta, S., Svensson, O.: No small linear program approx-imates vertex cover within a factor 2 - ε . Math. Oper. Res. (1), 147–172 (2019)4. Braun, G., Fiorini, S., Pokutta, S., Steurer, D.: Approximation limits of linearprograms (beyond hierarchies). Math. Oper. Res. (3), 756–772 (2015)5. Braun, G., Jain, R., Lee, T., Pokutta, S.: Information-theoretic approximations ofthe nonnegative rank. Comput. Complex. (1), 147–197 (2017)6. Braverman, M., Moitra, A.: An information complexity approach to extended for-mulations. In: Proc. STOC 2013, pp. 161–170 (2013)7. Bri¨et, J., Dadush, D., Pokutta, S.: On the existence of 0/1 polytopes with highsemidefinite extension complexity. Math. Program. (1), 179–199 (2015)8. Buchanan, A.: Extended formulations for vertex cover. Oper. Res. Lett. (3),374–378 (2016)9. Cela, E.: The Quadratic Assignment Problem: Theory and Algorithms. Combina-torial Optimization. Springer US (1997)10. Chan, S.O., Lee, J.R., Raghavendra, P., Steurer, D.: Approximate constraint sat-isfaction requires large LP relaxations. In: FOCS ’13, pp. 350–359 (2013)0 Pawan Aurora and Hans Raj Tiwary11. Conforti, M., Cornu´ejols, G., Zambelli, G.: Extended formulations in combinatorialoptimization. Annals OR (1), 97–143 (2013)12. Faenza, Y., Fiorini, S., Grappe, R., Tiwary, H.R.: Extended formulations, nonneg-ative factorizations, and randomized communication protocols. Math. Program. (1), 75–94 (2015)13. Fiorini, S., Massar, S., Patra, M.K., Tiwary, H.R.: Generalized probabilistic the-ories and conic extensions of polytopes. Journal of Physics A: Mathematical andTheoretical (2), 025,302 (2014)14. Fiorini, S., Massar, S., Pokutta, S., Tiwary, H.R., de Wolf, R.: Exponential lowerbounds for polytopes in combinatorial optimization. J. ACM (2), 17 (2015)15. Gajarsk´y, J., Hlinen´y, P., Tiwary, H.R.: Parameterized extension complexity ofindependent set and related problems. Discret. Appl. Math. , 56–67 (2018)16. Gouveia, J., Parrilo, P.A., Thomas, R.R.: Lifts of convex sets and cone factoriza-tions. Math. Oper. Res. (2), 248–264 (2013)17. J¨unger, M., Kaibel, V.: A basic study of the qap-polytope. Tech. rep., Institut F¨rrInformatik, Universit¨at zu K¨oln, Germany (1996)18. J¨unger, M., Kaibel, V.: On the sqap polytope. Tech. rep., Institut F¨rr Informatik,Universit¨at zu K¨oln, Germany (1996)19. J¨unger, M., Kaibel, V.: The qap-polytope and the star transformation. DiscreteApplied Mathematics (3), 283 – 306 (2001)20. Kaibel, V.: Polyhedral Combinatorics of the Quadratic Assignment Problem. Ph.D.thesis, Faculty of Mathematics and Natural Sciences, University of Cologne (1997)21. Kaibel, V.: Extended formulations in combinatorial optimization. Optima , 2–7(2011)22. Kaibel, V., Weltge, S.: A short proof that the extension complexity of the cor-relation polytope grows exponentially. Discret. Comput. Geom. (2), 397–401(2015)23. Koopmans, T., Beckmann, M.: Assignment problems and the location of economicactivities. Tech. rep., Cowles Foundation, Yale University (1955)24. Lawler, E.L.: The quadratic assignment problem. Management Science (4), 586–599 (1963)25. Lee, J.R., Raghavendra, P., Steurer, D.: Lower bounds on the size of semidefiniteprogramming relaxations. In: STOC ’15, pp. 567–576 (2015)26. Padberg, M., Rijal, M.: Location, Scheduling, Design and Integer Programming.International Series in Operations Research & Management Science. Springer,Boston, USA (1996)27. Padberg, M., Rinaldi, G.: A branch-and-cut algorithm for the resolution of large-scale symmetric traveling salesman problems. SIAM Review (1), 60–100 (1991)28. Pitowsky, I.: Quantum Probability — Quantum Logic, Lecture Notes in Physics ,vol. 321. Springer (1989)29. Rothvoß, T.: Some 0/1 polytopes need exponential size extended formulations.Math. Program. (1-2), 255–268 (2013)30. Rothvoß, T.: The matching polytope has exponential extension complexity. J.ACM (6), 41:1–41:19 (2017)31. Sahni, S., Gonzalez, T.: P-complete approximation problems. J. ACM (3),555–565 (1976)32. Vanderbeck, F., Wolsey, L.A.: Reformulation and decomposition of integer pro-grams. In: M.J. et al. (ed.) 50 Years of Integer Programming 1958-2008, pp.431–502. Springer (2010)33. Wolsey, L.A.: Using extended formulations in practice. Optima85
.Proof. From Lemma 3 and Lemma 4 we can conclude that any vertex in V \ ( S ∪ S ) can be expressed as a linear combination of the vertices in S ∪ S .What remains to be shown is that any vertex in S can be expressed as a linearcombination of the vertices in S and a fixed vertex σ ∗ ∈ S . From Lemma 5 weknow that it is possible to go from any vertex in S to any other vertex in S viatranspositions such that all the intermediate vertices are in S . Let us fix somearbitrary vertex in S as σ ∗ . So there is a path from every other vertex in S to σ ∗ . Consider a vertex σ ∈ S . Let σ, σ , . . . , σ t , σ ∗ be a path from σ to σ ∗ . FromLemma 7 we can express the difference of any vertex σ ∈ S with any other vertex σ ′ ∈ S such that σ ′ is a transposition of σ , as a linear combination of the verticesin S . So we have σ − σ ∈ span ( S ) , σ − σ ∈ span ( S ) , . . . , σ t − σ ∗ ∈ span ( S )which implies that σ − σ ∗ ∈ span ( S ) or σ ∈ span ( S ∪ { σ ∗ } ). ⊓⊔ In this section we consider the membership testing problem for each of the QAPrelaxations defined in Section 2. That is, for each of these relaxations we wish totest whether a given point x satisfies all the constraints. Note that the separationproblem where one wishes to identify a violated inequality in case the answer tomembership testing is negative is a harder problem. Typically for efficient use incutting plane methods one would like to solve the (harder) separation problem.We show that membership testing for inequalities QAP2, QAP3, or QAP5 iscoNP-complete.Recall that QAP2 is defined by the following set of inequalities: m X r =1 Y i r j r ,kl − Y kl,kl − X r , m > Theorem 3. t + P r t + P r Y i r j r ,i s j s = 0 for all distinct r, s ∈ [ m ] and so { i r j r , , i s j s } ∈ E ′ . Butthen, m > t giving a clique of size larger than t in the neighborhood of ( kl )contradicting the assumption that the every such clique has size at most t .Therefore, given a membership oracle for QAP2 we can compute the size ofthe largest clique in any graph except K n by calling such an oracle for variouschoices of k, l and t and outputting the largest value of t for which the aboveconstructed point satisfies all the inequalities. ⊓⊔ Next, recall that QAP3 is defined by the following set of inequalities:( β − X ( ij ) ∈ P × Q Y ij,ij − X ( ij ) , ( kl ) ∈ P × Qi Theorem 5.