Complete Description of Matching Polytopes with One Linearized Quadratic Term for Bipartite Graphs

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1 Complete Description of Matching Polytopes with One Linearized Quadratic Term for Bipartite Graphs Matthias Walter RWTH Aachen University: July 7, 06 Abstract We consider, for complete bipartite graphs, the convex hulls of characteristic vectors of ings, extended by a binary number indicating whether the ing contains two specific edges. This polytope is associated to the quadratic ing problem with a single linearized quadratic term. We provide a complete irredundant inequality description, which settles a conjecture by Klein (Ph.D. thesis, TU Dortmund, 05). Introduction Let K m,n = (A B, E) be the complete bipartite graph with the node partition A B, A = m and B = n for m, n. The maximum weight ing problem is to maximize the sum w(m) = e M w e over all ings M in K m,n for given edge weights w Q E. Note that we generally abbreviate j J v j as v(j) for vectors v and subsets J of their index sets. Following the usual approach in polyhedral combinatorics, we identify the ings M with their characteristic vectors χ(m) {0, } E, which satisfy χ(m) e = if and only if e M holds. The maximum weight ing problem is then equivalent to the problem of maximizing the linear objective w over the ing polytope, i.e., the convex hull of all characteristic vectors of ings. In order to use linear programming techniques, one requires a description of that so-called polytope in terms of linear inequalities. Such a description is well-known and consists of the following constraints []: x e 0 for all e E () x(δ(v)) for all v A B () For general (nonbipartite) graphs, Edmonds [4, 5] proved that adding the following Blossom Inequalities is sufficient to describe the ing polytope: x(e[s]) ( S ) for all S V, S odd His result was based on a primal-dual approximation algorithm, which also proved that the weighted ing problem can be solved in polynomial time. Later, Schrijver [] gave a direct (and more geometric) proof of the polyhedral result. Note that one also often considers the special case of perfect ings, which are those ings that every node of the graph. The associated perfect ing polytope is the face of the ing polytope obtained by requiring that all Inequalities () are satisfied with equality: x(δ(v)) = for all v A B. (3) For more background on ings and the ing polytopes we refer to Schrijver s book []. For a basic introduction into polytopes and linear programmg we recommend to read [3]. In this paper, we consider the more general quadratic ing problem for which we have, in addition to w, a set Q ( E ) and weights p Q Q for the edge-pairs in Q. The objective is now to maximize w(m) + q Q,q M p q, again over all ings M. A special case of the problem (by

2 requiring the ings to be perfect) is the quadratic assignment problem, a problem that is not just NP-hard [], but also hard to solve in practice (see [0] for a survey). A common strategy is then to linearize this quadratic objective function by introducing additional variables y e,f = x e x f for all {e, f} Q. Usually, the straight-forward linearization of this product equation is very weak, and one seeks to find (strong) inequalities that are valid for the associated polytope. One way of finding such inequalities, recently suggested by Buchheim and Klein [], is the socalled one term linearization technique. The idea is to consider the special case of Q in which the optimization problem is still polynomially solvable. By the equivalence of separation and optimization [8], one can thus hope to characterize all (irredundant) valid inequalities and develop separation algorithms. These inequalities remain valid when more than one monomial are present, and hence one can use the results of this special case in the more general setting. Buchheim and Klein suggested this for the quadratic spanning-tree problem and conjectured a complete description of the associated polytope. This conjecture was later confirmed by Fischer and Fischer [6] and Buchheim and Klein [3]. Fischer et al. [7] recently generalized this result to matroids and multiple monomials, which must be nested in a certain way. In her dissertation [9], Klein considered several other combinatorial polytopes, in particular the quadratic assignment polytope that corresponds to the quadratic perfect ing problem in bipartite graphs. We extend this to the non-perfect ings and our setup is as follows: Consider two disjoint edges e = {a, b } and e = {a, b } (with a i A and b i B for i =, ) in K m,n and denote by V = {a, a, b, b } the union of their node sets. Our polytopes of interest are the convex hulls of all vectors (χ(m), y) for which M is a ing in G, y {0, } and one of the relationships between M and y holds: ˆ P Q Q = P (K m,n, e, e ): y = implies e, e M. ˆ P Q Q = P (K m,n, e, e ): y = 0 implies e M or e M. ˆ P Q Q = P (K m,n, e, e ): y = if and only if e, e M. Note that P Q Q Q (resp. P ) is the downward (resp. upward) monotonization of P respect to the y-variable, and that P Q Q = conv(p P Q (ZE Z)) holds. Clearly, Constraints () and () as well as the bound constraints are valid for all three polytopes. Additionally, the two inequalities with 0 y (4) y x ei i =, (5) are also valid for P Q Q and P (and belong to the standard linearization of y = x e x e ). Klein [9] introduced two more inequality classes, and proved them to be facet-defining (see Theorems 6.. and 6..3 in [9]). They read x(e[s]) + y ( S ) for all S S and (6) x(e[s]) + x e + x e y S for all S S, (7) where the index sets (see Figure ) are defined as follows: S = {S A B S odd and either S V = {a, a } and S A = S B + or S V = {b, b } and S B = S A + holds}. S = {S A B S A = S B and either S V = {a, b } or S V = {a, b }.

3 a e b a e e b a e b a b a 3 b 3 a 3 b 3 a 4 b 4 a 4 b 4 a 5 b 5 a 5 b 5 (a) A set S S indexing Inequality (6). (b) A set S S indexing Inequality (7). Figure : Node Sets Indexing Additional Facets. Klein [9] even conjectured, that Constraints () (7), where Inequalities () are replaced by equations, completely describe the mentioned face of P Q. We will confirm this conjecture in Corollary 8. In contrast to the two proofs for the one-quadratic-term spanning-tree polytopes [6, 3], our proof technique is not based on linear programming duality. In fact, the two additional inequality families presented above introduce two sets of dual multipliers, which seem to make this proof strategy hard, or at least quite technical. Instead, we were heavily inspired by Schrijver s direct proof [] for the ing polytope. The paper is structured as follows: In Section we present our main results together with their proofs, which are based on two key lemmas, one for P Q Q and one for P. Then, in Section 3 we focus on perfect ings and establish the corresponding results for this special case. The proofs for the two key lemmas are similiar with respect to the general strategy, but are still quite different due to the specific constructions they depend on. Hence, each lemma has its own dedicated section. Although Klein already proved that the new inequalities are facet-defining, she only did so for the case of perfect ings. Hence, for the sake of completeness, we do the same for the general case in Section 6. We conclude this paper with a small discussion on our proof strategy and on a property of P Q. Main Results We will prove our result using two key lemmas, each of which is proved within its own section. Lemma. Let (ˆx, ŷ) Q E Q satisfy Constraints (), (), (4), (5) and (6). Let furthermore (ˆx, ŷ) satisfy at least one of the Inequalities (5) for i {, } or (6) for a set S S with equality. Then (ˆx, ŷ) is a convex combination of vertices of P Q. Lemma will be proved in Section 4. Lemma. Let (ˆx, ŷ) Q E Q satisfy Constraints (), (), (4), and Inequalities (7) for all S S. Let furthermore (ˆx, ŷ) satisfy at least one of the Inequalities(7) for a set S S with equality. Then (ˆx, ŷ) is a convex combination of vertices of P Q. Lemma will be proved in Section 5. We continue with the consequences of the two lemmas. Theorem 3. P Q is equal to the set of (x, y) RE R that satisfy Constraints (), (), (4), (5) and (6). Proof. Let P be the polytope defined by Contraints () (6). We first show P Q P by showing (χ(m), y) P for all feasible pairs (χ(m), y), i.e., ings M in G and suitable y. Clearly, χ(m) satisfies Constraints () and (), and y satisfies (4). 3

4 Let S S, define S = S {a, a, b, b }, and observe that S is odd. If y =, then e, e M holds, i.e., the Constraint (5) is satisfied. Hence, only nodes in S can be ed to other nodes in S, and there are at most S / = ( S 3)/ of them. If y = 0 holds, then the validity follows from the fact that S has odd cardinality. This shows that Constraint (6) is always satisfied. To show P P Q, we consider a vertex (ˆx, ŷ) of P. Note that since P is rational we have (ˆx, ŷ) Q E Q. If it satisfies at least one of the Inequalities (5) for some i {, } or (6) for some S S, Lemma yields that (ˆx, ŷ) is a convex combination of vertices of P Q, which are vertices of P Q. Hence, (ˆx, ŷ) is even a vertex of the polytope defined only by the Constraints (), () and (4). Thus, ŷ {0, } and ˆx = χ(m) holds for some ing M in K m,n. Since Inequalities (5) are strictly satisfied, we must have ŷ = 0, which concludes the proof. Theorem 4. P Q is equal to the set of (x, y) RE R that satisfy Constraints (), (), (4) and (7). Proof. Let P be the polytope defined by Contraints () (4) and (7). We first show P Q P by showing (χ(m), y) P for all feasible pairs (χ(m), y), i.e., ings M in G and suitable y. Clearly, χ(m) satisfies Constraints () and (), and y satisfies (4). For S S, M contains at most ( S e e ) = S + edges in E[S] e e. Thus, if y = holds, Constraint (7) is satisfied. If y = 0 holds and e, e M, then it is trivially satisfied. Otherwise, i.e., if y = 0 holds and M contains exactly one of the two edges, we can assume w.l.o.g. that e δ(m) and e δ(m) hold. Since S e has odd cardinality and at most S / edges of M can have both endnodes in S, the constraint is also satisfied in this case. To show P P Q, we consider a vertex (ˆx, ŷ) of P. Note that since P is rational we have (ˆx, ŷ) Q E Q. If it satisfies at least one of the Inequalities(7) for some S S, Lemma yields that (ˆx, ŷ) is a convex combination of vertices of P Q Q, which are vertices of P. Hence, (ˆx, ŷ) is even a vertex of the polytope defined only by the Constraints (), () and (4). Thus, ŷ {0, } and ˆx = χ(m) holds for some ing M in K m,n. If ŷ = 0 holds, then Inequality (7) for S = {a, b } reads ˆx e + ˆx a,b + ˆx e, and thus implies e M or e M, which concludes the proof. Theorem 5. P Q is equal to the set of (x, y) RE R that satisfy Constraints (), (), (4), (5), (6) and (7), i.e., P Q = P Q P Q. Proof. Let P be the polytope defined by Contraints () (7). By Theorems (3) and (4) we have P Q P Q P Q = P. To show P P Q, we consider a vertex (ˆx, ŷ) of P. Note that since P is rational we have (ˆx, ŷ) Q E Q. If it satisfies at least one of the Inequalities (5) for some i {, } (6) for some S S, Lemma yields that (ˆx, ŷ) is a convex combination of vertices of P Q. If it satisfies at least one of the Inequalities(7) for some S S, Lemma yields that (ˆx, ŷ) is a convex combination of vertices of P Q. Hence, (ˆx, ŷ) is even a vertex of the polytope defined only by the Constraints (), () and (4). Thus, ŷ {0, } and ˆx = χ(m) holds for some ing M in K m,n. Inequalities (5) and Inequality (7) for S = {a, b } imply that y = holds if and only if e, e M holds, which concludes the proof. 3 Perfect Matchings Throughout this section we assume m = n, since otherwise, K m,n does not contain perfect ings. Since the formulations for perfect ings are obtained by replacing Inequalities () by Equations (3), the corresponding polytopes are faces of the ones defined in the Section, and we immediately obtain the following results from the corresponding theorems in Section : Corollary 6. The convex hull of all (χ(m), y) {0, } E {0, }, for which M is a perfect ing M in K n,n and y = implies e, e M, is equal to the set of (x, y) R E R that satisfy Constraints (), (4), (5), (6) and (3). 4

5 Corollary 7. The convex hull of all (χ(m), y) {0, } E {0, }, for which M is a perfect ing M in K n,n and y = 0 implies e M or e M, is equal to the set of (x, y) R E R that satisfy Constraints (), (4), (7) and (3). Corollary 8. The convex hull of all (χ(m), y) {0, } E {0, }, for which M is a perfect ing M in K n,n and y = holds if and only if e, e M, is equal to the set of (x, y) R E R that satisfy Constraints (), (4), (5), (6), (7) and (3). 4 Proofs for the Downward Monotonization This section contains the proof of Lemma. We first introduce relevant objects which are fixed for the rest of this section, and then present the main proof. To improve readability, the proofs of several claims are deferred to the end of this section. Let (ˆx, ŷ) Q E Q be as stated in the lemma. Let Ḡ = (A B, Ē) be the graph K m,n with the additional edges e a = {a, a } and e b = {b, b }, i.e., Ē = E {e a, e b }. Define the vector x QĒ as follows (see Figure ): ˆ x e = ˆx e for all e E {e, e }. ˆ x ei = ˆx ei ŷ for i =,. ˆ x ea = x eb = ŷ. a ˆx e ŷ b ŷ ŷ a ˆx e ŷ b Figure : Graph Ḡ and vector x in the Proof of Lemma. Claim 9. x is in the ing polytope of Ḡ. By Claim 9, and since x is rational, it can be written as a convex combination of characteristic vectors of ings using only rational multipliers. Multiplying with a sufficiently large integer k, we obtain that x = k k j= χ( M j ) holds for ings M,..., M k in G, where ings may occur multiple times. Let J a = {j [k] e a M j } and J b = {j [k] e b M j }, and observe that J a = ŷk = J b holds. We may assume that the convex combination is chosen such that J a J b is minimum. Claim 0. The convex combination satisfies J a = J b. By Claim 0 we can write J = J a = J b. We construct ings ˆM j for j [k] that are related to the corresponding M j. To this end, let C = {e, e, e a, e b } and define ˆM j = M j C for all j J and ˆM j = M j for all j [k] J. All ˆMj are ings in Ḡ since for all j J, the ings M j contain both edges e a and e b. In fact, none of the ˆM j contains these edges, and hence they are even ings in K m,n. Claim. We have ˆx = k k j= χ( ˆM j ). 5

6 Together with ŷk = J, Claim yields (ˆx, ŷ) = k (χ( ˆM j ), ) + j J j [k] J (χ( ˆM j ), 0), and it remains to prove that all participating vectors are actually feasible for P Q. For the first sum, this is easy to see, since for all j J, the ings ˆM j contain both edges e and e by construction. The ings in the second sum are considered in two claims, depending (ˆx, ŷ). Claim. Let (ˆx, ŷ) satisfy Inequality (5) for some î {, } with equality. Then ˆM j contains at most one of the two edges e, e for all j [k] J. Claim 3. Let (ˆx, ŷ) satisfy Inequality (6) for some Ŝ S with equality. Then most one of the two edges e, e for all j [k] J. ˆM j contains at Since, by the assumptions of Lemma, the premise of at least one of the Claims or 3 is satisfied, (ˆx, ŷ) is indeed a convex combination of vertices of P Q, which concludes the proof of Lemma. Before actually proving the claims of this section, we list further valid inequalities. Proposition 4. Let (ˆx, ŷ) satisfy Constraints () (6) and define S ext = {S A B S is odd and S V {{a, a }, {b, b }}}. Then (ˆx, ŷ) even satisfies Inequality (6) for all S S ext. Proof of Proposition 4. We only have to prove the statement for S S ext S. W.l.o.g. we assume that S V = {a, a }, since the other case is similar. Let A = S A and B = S B. If A < B + holds, we have A B since S is odd. Then the sum of x(δ(a)) for all a A plus the sum of x e 0 for all e δ(a ) (E[S] {e }) reads x(e[s])+x e A ( S ). Adding y x e yields the desired inequality. If A > B + holds, we have A B + 3 since S is odd. Then the sum of x(δ(b)) for all b B plus the sum of x e 0 for all e δ(b ) E[S] reads x(e[s]) B ( S ). Adding y yields the desired inequality, which concludes the proof. Proof of Claim 9. From ˆx O and (5) we obtain that also x O holds. By construction and since ˆx satisfies (), x(δ(v)) holds for every node v A B. Suppose, for the sake of contradiction, that x(e[s]) > ( S ) holds for some odd-cardinality set S A B. From ˆx(E[S]) ( S ) we deduce x(e[s]) > x(e[s]), i.e., E[S] contains at least one of the edges {e a, e b }, since only for these edges the x-value is strictly greater than the corresponding ˆx-value. Observe that E[S] also must contain at most one of these edges, since otherwise it would also contain the two edges edges e, e, which yielded x(e[s]) = ˆx(E[S]) ( S ). Hence, we have S S ext, and thus x(e[s]) = ˆx(E[S]) + ŷ ( S ) by Proposition 4. This proves that x is in the ing polytope of Ḡ. Proof of Claim 0. Suppose, for the sake of contradiction, that J a J b holds. Let j a J a J b and let j b J b J a. Consider the ings M ja and M jb and note that M ja M jb contains both edges e a and e b. Let C a and C b be (the edge sets of) the connected components of Mja M jb that contain e a and e b, respectively. We claim that C a and C b are not the same component. Assume, for the sake of contradiction, that C = C a = C b is a connected component (i.e., an alternating cycle or path) of M M that contains a and b. Consider a path P C {e a, e b } that connects an endnode of e a with an endnode of e b (if C is an alternating cycle, there exist two such paths and we pick one arbitrarily). On the one hand, (A B, Ē {e a, e b }) is bipartite and thus P must have odd length. On the other hand, e a M ja and e b M jb hold, and hence P must have even length, yielding a contradiction. Define two new ings M j a = M ja C a and M j b = M jb C a, and note that χ( M ja ) + χ( M jb ) = χ( M j a )+χ( M j b ) holds, i.e., we can replace M ja and M jb by M j a and M j b in the convex combination. The fact that M ja contains none of the two edges e a and e b, while M j b contains both, yields a contradiction to the assumption that J a J b is minimum. 6

7 Proof of Claim. Consider the vector d = k j=(χ( ˆM j ) χ( M j )). By construction of the ˆM j, we have d e = 0 for all e C and d e = d e = d ea = d eb = J = kŷ. A simple comparison with the construction of x from ˆx concludes the proof. Proof of Claim. From ˆx ei = ŷ we obtain that x ei = 0 and hence e i M j holds, which concludes the proof. Proof of Claim 3. From ˆx(E[S ])+ŷ = ( S ) and the construction of x we obtain x(e[s ]) = ( S ). But since x(e[s ]) ( S ) is valid for all χ( M j ), equality must hold for all j [k]. Thus, M j {e, e } M j δ(s ) holds for all j, which concludes the proof. 5 Proofs for the Upward Monotonization This section contains the proof of Lemma. The setup is similar to that of the previous section, starting with the relevant objects. Let (ˆx, ŷ) Q E Q be as stated in the lemma. Let Ḡ = ( V, Ē) be the graph K m,n with two additional nodes u and v, i.e., V = A B {u, v}, and edge set Ē = E {{u, v}, {a, u}, {a, v}, {b, v}, {b, u}}. Define two vectors x, x RĒ as follows (see Figure 3): ˆ x e = ˆx e and x e = ˆx e for all e E {e, e }. ˆ x ei = ˆx ei and x ei = ŷ for i =,. ˆ x {u,v} = and x {u,v} = ˆx e ˆx e + ŷ. ˆ x a,u = x b,v = 0 and x {a,u} = x {b,v} = ˆx e ŷ. ˆ x a,v = x b,u = 0 and x {a,v} = x {b,u} = ˆx e ŷ. The vector x is essentially a trivial lifting of ˆx into RĒ by setting the value for edge {u, v} to and the values for the other new edges to 0. It is easy to see that x is in the ing polytope of Ḡ. The vector x is a modification of x on the edges of the following two cycles: C = {{a, u}, {u, v}, {v, b }, {b, a }}, C = {{a, v}, {v, u}, {u, b }, {b, a }}. The values on the two opposite (in C ) edges {a, b } and {u, v} are decreased by ˆx e ŷ, and increased by the same value on the other two edges. Similarly, the values on the edges {a, b } and {u, v} are decreased by ˆx e ŷ, while they are increased by the same value on the other two edges of C. ŷ a ˆx e ŷ u b ˆx e ˆx e + ŷ ˆx e ŷ C a ˆx e ŷ v ˆx e ŷ b C ŷ Figure 3: Graph Ḡ, vector x and cycles C and C in the Proof of Lemma. Claim 5. x is in the ing polytope of Ḡ. 7

8 By Claim 5, and since x is rational, it can be written as a convex combination of characteristic vectors of ings using only rational multipliers. Multiplying with a sufficiently large integer k, we obtain that x = k k j= χ( M j ) holds for ings M,..., M k in G, where ings may occur multiple times. We define the index sets J a = {j [k] {a, u}, {a, v} M j }, J b = {j [k] {b, v}, {b, u} M j }, J = {j [k] {a, u}, {b, v} M j }, J = {j [k] {a, v}, {b, u} M j }, J = {j [k] {a, b } M j }, N = {j [k] {u, v} M j }. J = {j [k] {a, b } M j } and We assume that the convex combination is chosen such that J a + J b is minimum. Using the assumption from the lemma, that (ˆx, ŷ) satisfies Inequality (7) for some set S S with equality, we can derive the following statement. Claim 6. Let j [k]. The ing M j contains at most one of the edges e, e, or {u, v}. It furthermore es u and v (not necessarily to each other). Claim 7. The convex combination satisfies J a = J b =, and J J N is a partitioning of [k]. Claim 8. We have J i J for i =, and thus J and J are disjoint. Claim 9. We have J J = ŷk. We construct ings M j and ˆM j for j [k] that are related to the corresponding M j. Define M j = M j C for all j J, Mj = M j C for all j J. By Claim 7, all remaining indices are the j N, and for those we define M j = M j. All Mj are ings in Ḡ since for all j J i (i =, ) the cycle C i is an M j -alternating cycle. We define ˆM j = M j {u, v} for all j [k], which are ings in K m,n since {u, v} M j holds for all j [k], Claim 0. We have x = k k j= χ( M j ) and ˆx = k k j= χ( ˆM j ). Claims 9 and 0 yield (ˆx, ŷ) = k (χ( ˆM j ), ) + (χ( ˆM j ), ) + (χ( ˆM j ), 0) j J j J j [k] (J J ), and it remains to prove that all participating vectors are actually feasible for P Q. To this end, let j J and observe that {a, b } M j and, by Claim 8, {a, u}, {b, v} M j holds. Thus, the symmetric difference with C yields {a, b }, {a, b } ˆM j. Similarly, {a, b }, {a, b } ˆM j holds for all j J. Let j [k] (J J ). First, Mj contains none of the edges {a, b }, {a, b }. Second, the construction of ˆMj from M j adds at most one of the two edges {a, b }, {a, b }, which proves that ˆM j does not contains both of them. This concludes the proof. Before actually proving the claims of this section, we list further valid inequalities. Proposition. Let (ˆx, ŷ) satisfy Constraints () (4) and (7). Then (ˆx, ŷ) satisfies the following inequalities: (a) x e + x e y. (b) x(e[s]) + x ei ŷ S for i {, }, S A B with S even and e i δ(s). (c) Inequality (7) is satisfied even for all S S ext S, defined as follows: S ext = {S A B S is even and S V {{a, b }, {a, b }}}. If (ˆx, ŷ) also satisfies Inequality (7) for some S S with equality, then the following inequalities hold as well: (d) y x e and y x e. 8

9 Proof of Proposition. We prove validity for each inequality individually: (a) The inequality is the sum of Inequality (7) for S = {a, b } and x {a,b } 0. (b) Since S e i is odd (and since K m,n is bipartite), the usual Blossom Inequality x(e[s e i ]) S is implied by Constraints () and (). Adding x e 0 for all e E[S e ] {e } and y 0 yields the desired inequality. (c) We only have to prove the statement for S S ext S. W.l.o.g. we assume that S V = {a, b } and S A < S B, since the other cases are similar. Let A = S A and B = S B and observe that A B holds because S is even. Then the sum of x(δ(a)) for all a A plus the sum of x e 0 for all e δ(a ) (E[S] {e }) reads x(e[s]) + x e A = S. Adding x e and y 0 yields the desired inequality, which concludes the proof. (d) Let i {, } and j = 3 i. Similar to the proof of (b) we have that x(e[s ]) + x ej S is implied by Constraints () and (). Subtracting this from the equation x(e[s ])+x e +x e y = S yields x ei y 0, which concludes the proof. Proof of Claim 5. Since (ˆx, ŷ) O holds, Parts (a) and (d) of Propositon yield x O. The degree constraints are also satisfied, since x(δ(v)) = ˆx(δ(v)) holds for the nodes v V and since x(δ(u)) = x(δ(v)) = holds. Suppose, for the sake of contradiction, that x(e[ S]) > ( S ) holds for some odd-cardinality set S V. Clearly, x(e[ S]) ( S ) holds, i.e., ( x x)(e[ S]) > 0. This implies that E[ S] must intersect some C i (i =, ) in such a way that the sum of the respective modifications (increase or decrease by ˆx ei ŷ) is positive. Similar to the proof of Claim 9, we conclude that S must touch one of the cycles in precisely two nodes, whose connecting edge e satisfies x e > x e. Hence, (at least) one of the following four conditions must be satisfied: () S {a, b, u, v} = {a, u}, () S {a, b, u, v} = {b, v}, (3) S {a, b, u, v} = {a, v}, (4) S {a, b, u, v} = {b, u}. We define V = {a, a, b, b, u, v} and S = S {u, v}. Note that we always have S = S since each of the four conditions implies that either u or v is contained in S. We now make a case distinction, based on S V : Case : S V is equal to {a, u}, {a, u, a }, {a, u, a, b }, {b, v}, {b, v, b } or {b, v, b, a }. In this case x(e[ S]) = x(e[ S]) + ˆx e ŷ = ˆx(E[S]) + ˆx e ŷ S = ( S ) holds by Proposition (b), which yields a contradiction. Case : S V is equal to {a, v}, {a, v, a }, {a, v, a, b }, {b, u}, {b, u, b }, {b, u, b, a }. In this case x(e[ S]) = x(e[ S]) + ˆx e ŷ = ˆx(E[S]) + ˆx e ŷ S = ( S ) holds by Proposition (b), which yields a contradiction. Case 3: S V is equal to {a, u, b } or {a, v, b }. In this case x(e[ S]) = x(e[ S]) + ˆx e + ˆx e ŷ = ˆx(E[S]) + ˆx e + ˆx e ŷ S = ( S ) holds by Proposition (c), which yields a contradiction. Proof of Claim 6. Let S S be such that (ˆx, ŷ) satisfies Inequality (7) with equality. If a, b S holds, then we define S = S {u}, and otherwise S = S {v}. A simple calculation shows that x(e[ S]) = ˆx(E[S ]) + ˆx e + ˆx e ŷ = S = ( S ) holds, i.e., x satisfies the Blossom Inequality induced by S with equality. Furthermore, x satisfies the degree inequalities () for nodes u and v with equality. This implies that all characteristic vectors χ( M j ) satisfy these three inequalities with equality, i.e., we have M j E[ S] = ( S ) and M j δ(u) = M j δ(v) =. From the first equation we derive M j δ( S). This, together with the second equation proves the claimed properties. Proof of Claim 7. First, the set J a, J b, J, J and N are disjoint since the indexed ings all nodes u and v in different ways. Second, their union is equal to [k] due to the second part of Claim 6. From this we obtain J a + J = x a,uk = (ˆx e y) k = x b,vk = J b + J, and conclude that J a = J b holds. 9

10 Now suppose, for the sake contradiction, that J a (and thus J b = J a ) holds. Let j J a and j J b and let C be the connected component (i.e., an alternating cycle or path) of Mj M j that contains {a, v}. We claim that {a, u} C holds. Assuming the contrary, there must exist an odd-length (alternating) path in K m,n that connects either a with a or b with b or there must exist an even-length (alternating) path in K,n that connects either a with b or a with b. Since K m,n is bipartite, none of these paths exist, which proves {a, u} C. Define two new ings M j = M j C and M j = M j C, and note that χ( M j ) + χ( M j ) = χ( M j ) + χ( M j ) holds, i.e., we can replace M j and M j by M j and M j in the convex combination. The fact that M j contains the edges {a, u} and {b, v} and that M j contains the edges {b, u} and {a, v} contradicts the assumption that the convex combination was chosen with minimum J a + J b. Hence, J a = J b = holds. Proof of Claim 8. Let j J. Using {a, b } M j, Claim 6 shows that {u, v} M j, and thus (since a and b are already ed to each other) that M j contains the two edges {a, u} and {b, v}, i.e., j J holds. The proof for J is similar. From Claim 7 we have J J =, and hence J J = holds as well. Proof of Claim 9. By Claim 8, we have J J =. The statement now follows from the fact that x e = x e = ŷ holds. Proof of Claim 0. Similar to the proof of Claim, we consider the vector d = k j=(χ( M j ) χ( M j )). By construction of the M j, we have ˆ d e = 0 for all e C C, ˆ d e = d {a,u} = d {b,v} = J = (ˆx e ŷ)k, ˆ d e = d {a,v} = d {b,u} = J = (ˆx e ŷ)k, and ˆ d {u,v} = d e + d e = (ˆx e + ˆx e ŷ)k. A simple comparison with the construction of x and x from ˆx proves the first part. The construction of ˆMj from M j by removing edge {u, v} corresponds to the fact that ˆx is the orthogonal projection of x onto R E, which proves that second part. 6 Facet Proofs Proposition. The polytopes P Q, P Q Q and P are full-dimensional. Proof. The point (χ( ), 0), the points (χ({e}), 0) for all e E and the point (χ({e, e }), ) are E + affinely independent points that are contained in all three polytopes. This proves the statement. Proposition 3. Let e E. Then Inequality () defines a facet for P Q. Furthermore, it defines a facet for P Q Q (and thus for P ) if and only if e {e, e } holds. Proof. Consider the point (x, 0) R E R defined via x e = and x e = 0 for all e E {e }. Clearly, (x, 0) satisfies Inequalities () for e e, but satisfies it for e = e. Since x e appears in the -Inequalities (), (4), (6) and (7) with a nonnegative coefficient, and since O P Q holds, all these are satisfied by (x, 0). If e {e, e } holds, then the above also holds for Inequalities (5), which proves the statement for this case and for P Q, since there Inequalities (5) are not present. If e = e i holds for some i {, }, then x e 0 is clearly implied by 0 y and y x ei, proving that it is not facet-defining for P Q Q and P. This concludes the proof. Proposition 4. Let v A B and let k = δ(v ). Then Inequality () is facet-defining for ˆ P Q in any case, for ˆ P Q if and only if k 3 or v V holds, and for 0

11 ˆ P Q if and only if k 3 holds. Proof. First note that k = m or k = n holds, since we consider the complete bipartite graph, and thus k holds. If k = and v V holds, then let S = δ(v ), i.e., S contains v and its neighbors. Inequality () is the sum of Inequality (6) for S and y 0. Both inequalities are valid for P Q If k = and v V holds, then let S be equal to {a, b } or equal to {a, b } such that v S is satisfied. Inequality () is the sum of Inequality (7) for S and Inequality (5) for i {, } such and P Q, hence Inequality () for v = v cannot be facet-defining for those polytopes. that e i δ(s) holds. Both inequalities are valid for P Q, hence Inequality () for v = v cannot be facet-defining for this polytope. We now prove non-redundancy of the inequality in the remaining cases. Let ε > 0 be such that εk + ε holds. We consider the two points (x, 0), (x, ε) R E R defined via x e = k + ε for all e δ(v ) and x e = 0 for all e E δ(v ). We will prove that (x, 0) satisfies all inequalities, except for the one in question, that are valid for P Q Q and P if the respective conditions on k and v are satisfied. Furthermore, we will prove that (x, ε) satisfies all inequalities, except for the one in question, that are valid for P Q. Both points clearly satisfy Constraints () and (4) and violate Inequality () for v = v. They furthermore satisfy Inequalities () for v v since at most one of the edges in the support (in the x-space) of such an inequality can be in δ(v ). The first point (x, 0) also clearly satisfies (5). Suppose Inequality (6) is violated by (x, 0) for a set S S. Since x = + kε holds, we must have S = 3 and the right-hand side of the inequality equals. Thus, in order to be violated, δ(v ) E[S] must hold. It is easy to see that in this case, we have k = and v must be the unique node in S V. Suppose Inequality (7) is violated by (x, 0) or (x, ε) for a set S S. Since x + ε holds, we must have S =, i.e., S = {a, b } or S = {a, b }, and the right-hand side is equal to. In order to be violated, δ(v ) {e, e } E[S] must hold, which implies v S V. We prove that (x, ε) is not violated and assume that, w.l.o.g., S = {a, b } and v = b holds. Now observe that x (E[S]) = x a,b and we have x (E[S])+x e +x e ε = ( +ε)+( +ε)+0 ε =. This concludes the proof. Proposition 5. The inequality y 0 is facet-defining for P Q, P Q Q and P, while y is facet-defining for P Q Q Q, but not for P and P. Proof. For fixed value k {0, }, The point (χ( ), k) and the points (χ({e}), k) for all e E are E + affinely independent points. For k = 0, they are contained in all three polytopes and satisfy y 0 with equality, which proves the first statement. For k =, they are contained in P Q and satisfy y with equality, which proves one direction of the second statement. For the reverse direction, observe that y is the sum of Inequality (5) for i {, } and x ei, which concludes the proof. Proposition 6. For i =,, Inequalities (5) define facets for P Q Q and P. Proof. Let i {, }. We consider the point (x, y ) defined via x e i = y = and x e = 0 for e E {e i } Since x + y holds, the point clearly satisfies Constraints (), (), (4), (6), (7), and Inequality (5) for i i, but violates the latter for i = i. This concludes the proof. For the remaining two proofs we will consider a set S A B of nodes and denote by A = S A and B = S B the induced sides of the bipartition. For a ing M in K m,n we denote by y(m) {0, } its corresponding y-value, i.e., y(m) = if and only if e, e M. Note that this implies (χ(m), y(m)) P Q. Another concept from ing theory also turns out to be useful: We say that a ing is near-perfect in a set of nodes if it es all nodes but one of this set. Proposition 7. For all S S, Inequalities (6) define facets for P Q Q and P. Proof. Let S S. We will assume w.l.o.g. that A + = B holds, since the proof for B = A + is similar. Let M denote the set of ings M in K m,n that are either near-perfect in S or are near-perfect in S V and contain edges e and e. In the first case we have M E[S ] = ( S ) and y(m) = 0, and in the second case we have M E[S ] = ( S 3) and y(m) =. Hence, for all M M, the vector (χ(m), y(m)) satisfies Inequality (6) with equality.

12 Let c, x + γy δ dominate Inequality (6) for S = S, i.e., it is valid for P Q and the vectors (χ(m), y(m)) and for all M M we have c, χ(m) +γy(m) = δ. We now analyze the coefficients and the right-hand side of the inequality. ˆ Let e E E[S ]. If e intersects S, then let v e S be its endnode in S, otherwise let v S be arbitrary. If v A, then let M be a perfect ing in S {v} (which exists due to A {v} = B ). Otherwise, let M be a perfect ing in S {v, a, a } (which exists due to A {a, a } = B {v} ), and extend it to the ing M = M {e, e }. Then e does not intersect M and thus M = M {e} is also a ing that satisfies y(m ) = y(m ). By construction we have M, M M, and hence c, χ(m ) + γy(m ) = δ = c, χ(m ) + γy(m ) holds. This proves c e = 0. ˆ Let u B and let e = {u, v} and f = {u, w} be two incident edges with endnodes v, w S. Let M be a perfect ing in S {v} that uses edge f. Then M = (M {f}) {e} is perfect in S {w}. Clearly, M, M M holds by construction. Thus, c, χ(m ) +γy(m ) = δ = c, χ(m ) + γy(m ) holds, i.e., c f = c e holds. ˆ If B, then also A 3 holds. Let v, w B be two nodes, let u A {a, a }, and let e = {u, v} and f = {u, w}. Let M be a perfect ing in S {a, a, u, v, w} (which exists due to A {a, a, u} = B {v, w} ). Define ings M = M {e, e, e } and M = M {f, e, e } and observe that M, M M and y(m ) = = y(m ) holds. Thus, c, χ(m ) + γy(m ) = δ = c, χ(m ) + γy(m ) holds, i.e., c e = c f holds. ˆ Let M be a perfect ing in S {a } and let e M be the edges that es a. Define ing M = (M {e}) {e, e }, and note that M, M M holds. Again, c, χ(m ) + γy(m ) = δ = c, χ(m ) + γy(m ) holds, i.e., c e = γ holds. The arguments above already fix (c, γ) up to multiplication with a scalar. Hence we can assume that γ = holds, which proves that (c, γ) is equal to the coefficient vector of Inequality (6) for S = S. Since there always exists a near-perfect ing M in S, and since such a ing has cardinality M = ( S ), we derive δ = ( S ). This concludes the proof. Proposition 8. For all S S, Inequalities (7) define facets for P Q Q and P. Proof. Let S S. We will assume w.l.o.g. that a, b S holds, since the proof for a, b S is similar. Let M denote the set of ings M in K m,n that are either perfect in S or contain exactly one edge e {e, e } and are near-perfect in S e or contain both, e and e, and are perfect in S {a, b }. In the first two cases we have M (E[S ] {e, e }) = S and y(m) = 0, and in the third case we have M (E[S ] {e, e }) = ( S ) + and y(m) =. Hence, for all M M, the vector (χ(m), y(m)) satisfies Inequality (7) with equality. Let c, x + γy δ dominate Inequality (7) for S = S, i.e., it is valid for P Q and the vectors (χ(m), y(m)) and for all M M we have c, χ(m) +γy(m) = δ. We now analyze the coefficients and the right-hand side of the inequality. ˆ Let e E (E [ S ] {e, e }). If e intersects S, then let v e S be its endnode in S, otherwise let v S be arbitrary. If v A, then let M be a perfect ing in S {v, b } (which exists due to A {v} = B {b } ), and extend it to the ing M = M {e }. Otherwise, let M be a perfect ing in S {v, a } (which exists due to A {a } = B {v} ), and extend it to the ing M = M {e }. Then e does not intersect M and thus M = M {e} is also a ing that satisfies y(m ) = 0 = y(m ). By construction we have M, M M, and hence c, χ(m ) +γy(m ) = δ = c, χ(m ) +γy(m ) holds. This proves c e = 0. ˆ Let u S {a, b } and let e = {u, v} and f = {u, w} be two incident edges with endnodes v, w S. W.l.o.g. we can assume u A, since the case of u B is similar. Let M be a perfect ing in S {a, u, v, w} (which exists due to A {a, u} = B {v, w} ). Define the two ings M = M {e, e} and M = M {e, f}, and observe that M, M M and y(m ) = 0 = y(m ) holds. From c, χ(m ) + γy(m ) = δ = c, χ(m ) + γy(m ) we obtain that c e = c f holds.

13 ˆ Let M be a perfect ing in S {a, b }. Define the ings M = M {a, b }, M = M {e }, M 3 = M {e } and M 4 = M {e, e }. By construction we have M, M, M 3, M 4 M, y(m ) = y(m ) = y(m 3 ) = 0 and y(m 4 ) =. Thus, c, χ(m i ) + γy(m i ) = δ holds for i =,, 3, 4, which proves c {a,b } = c e = c e = c e + c e γ. The arguments above already fix (c, γ) up to multiplication with a scalar. Hence we can assume that γ = holds, which proves that (c, γ) is equal to the coefficient vector of Inequality (7) for S = S. Since there always exists a perfect ing M in S, and since such a ing has cardinality M = S, we derive δ = S. This concludes the proof. 7 Discussion The observation from Section that P Q = P Q P Q polytopes. In fact, this is a property of convex sets: holds, is not specific to ing Proposition 9. Let C R n be a convex set and let C = {x + λe x C, λ 0} and C = {x λe x C, λ 0} its respective up- and downward monotonization of the first variable. Then C = C C holds. Proof. Clearly, C C, C and thus C C C holds. In order to prove the reverse direction, let x C C. By definition, there exist x (), x () C and λ, λ 0 such that x () + λ e = x = x () λ e holds. If λ holds, then x = x () C holds, and we are done. Otherwise, the equation λ x () λ + x () = λ (x λ e) + λ (x + λ e) = x λ + λ λ + λ λ + λ proves that x is a convex combination of two points in C, i.e., x C holds, which concludes the proof. In the case of ing polytopes we intersect the up- and downward monotonizations with the 0/-cube, but this does not interfere with the arguments provided above. In fact Proposition 9 does not generalize to the simultaneous monotonization of several variables. To see this, consider P = conv {(0, 0), (, ) }. Its upward-monotonization w.r.t. two both variables is P + R + = R +, its downward-monotonization is P R + = (, ) + R, but their intersection is equal to [0, ] P. Hence, this is a purely one-dimensional phenomenon. Descriptions of monotonizations. A second property is specific, at least to polytopes arising from one-term linearizations: we can obtain the complete description for P Q from the one for P Q by omitting the -inequalities that have a negative y-coefficient. Similarly, we obtain the complete description for P Q Q from the one for P by omitting the -inequalities that have a positive y-coefficient and adding y (which is not facet-defining for P Q, see Proposition 5). The reason turns out to be that all facets of the projection of P Q onto the x-variables are projections of facets of P Q. The arguments for the upward-monotonization are as follows: Let P R n+ be a polytope. After normalizing, we can write its outer description as P = {(x, y) R n R Ax b, Bx + y c, Cx y d}. We assume that P s projection onto the x-variables is the polytope defined by Ax b only. P s upward-monotonization can be obtained by projecting the extended formulation {(x, y, y ) R n R R (x, y) P, y y 0} onto the (x, y )-variables. Fourier-Motzkin elimination yields that this projection is described by Ax b, Cx y d and inequalities that are the sum of an inequality from Bx + y c and an inequality from Cx y d. Since the last type of inequalities are already valid for P s projection onto the x-variables, these are already present in Ax b. Proof technique. The technique we applied in Sections 4 and 5 in order to prove our results can be summarized as follows: 3

14 . Consider a fractional (extreme) point (of the polytope P in question) that satisfies a certain inequality with equality.. Modify that point in a way such that it lies in a face F of a polytope Q that we have under control. 3. Write the modified point as a convex combination of vertices of F and derive structural properties that are implied by the fact that they are vertices of that face. 4. Revert the modifications by replacing some of the vertices in the convex combination by others. Clearly, this technique does not work for arbitrary polytopes. In fact it heavily depends on the fact that P is very related to Q, e.g., a subpolytope. Clearly, the more complicated the modifications are, the more involved the proof will probably be. Thus, on the one hand we believe that the applicability of the technique is quite limited. On the other hand, it does not require LP duality, and hence it could be useful when duality-based methods become unattractive because of many inequality classes. Acknowledgements. The author is grateful to Mirjam Friesen and Volker Kaibel for valueable discussions. References [] Garrett Birkhoff. Tres observaciones sobre el algebra lineal. Revista Facultad de Ciencias Exactas, Puras y Aplicadas Universidad Nacional de Tucuman, Serie A (Matematicas y Fisica Teorica), 5:47 5, 946. [] Christoph Buchheim and Laura Klein. The Spanning Tree Problem with One Quadratic Term. In Kamiel Cornelissen, Ruben Hoeksma, Johann Hurink, and Bodo Manthey, editors, th Cologne-Twente Workshop on Graphs and Combinatorial Optimization CTW 03, pages 3 34, 03. [3] Christoph Buchheim and Laura Klein. Combinatorial optimization with one quadratic term: Spanning trees and forests. Discrete Applied Mathematics, 77(0):34 5, 04. [4] Jack Edmonds. Maximum Matching and a Polyhedron with 0,-Vertices. Journal of Research of the National Bureau of Standards B, 69:5 30, 965. [5] Jack Edmonds. Paths, trees, and flowers. Canad. J. Math., 7: , 965. [6] Anja Fischer and Frank Fischer. Complete description for the spanning tree problem with one linearised quadratic term. Operations Research Letters, 4(6):70 705, 03. [7] Anja Fischer, Frank Fischer, and S. Thomas McCormick. Matroid Optimisation Problems with Nested Non-linear Monomials in the Objective Function. NAM Preprint 06-, Institute for Numerical and Applied Mathematics, University of Goettingen, 06. [8] Martin Grötschel, László Lovász, and Alexander Schrijver. The ellipsoid method and its consequences in combinatorial optimization. Combinatorica, ():69 97, 98. [9] Laura Klein. Combinatorial Optimization with One Quadratic Term. PhD thesis, TU Dortmund, 04. [0] Eliane Maria Loiola, Nair Maria Maia de Abreu, Paulo Oswaldo Boaventura-Netto, Peter Hahn, and Tania Querido. A survey for the quadratic assignment problem. European Journal of Operational Research, 76(): , 007. [] Sartaj Sahni and Teofilo Gonzalez. P-Complete Approximation Problems. J. ACM, 3(3): , 976. [] A Schrijver. Short Proofs on the Matching Polyhedron. Journal of Combinatorial Theory, Series B, 34():04 08, 983. [3] Alexander Schrijver. Theory of Linear and Integer Programming. John Wiley & Sons, Inc., New York, NY, USA,