# Coverings and Matchings in r-partite Hypergraphs

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1 Coverings and Matchings in r-partite Hypergraphs Douglas S. Altner Department of Mathematics, United States Naval Academy, Annapolis, Maryland, J. Paul Brooks Department of Statistical Sciences and Operations Research, Virginia Commonwealth University, Richmond, Virginia, June 10, 010 Abstract Ryser s conjecture postulates that, for r-partite hypergraphs, τ (r 1)ν where τ is the covering number of the hypergraph and ν is the matching number. Although this conjecture has been open since the 1960 s, researchers have resolved it for special cases such as for intersecting hypergraphs where r 5. In this paper, we prove several results pertaining to matchings and coverings in r-partite intersecting hypergraphs. First, we prove that finding a minimum cardinality vertex cover on an r-partite intersecting hypergraph is NP-hard. Second, we note Ryser s conjecture for intersecting hypergraphs is easily resolved if a given hypergraph does not contain a particular sub-hypergraph, which we call a tornado. We prove several bounds on the covering number of tornados. Finally, we prove the integrality gap for the standard integer linear programming formulation of the maximum cardinality r-partite hypergraph matching problem is at least r k where k is the smallest positive integer such that r k is a prime power. covering, matching, Ryser s conjecture, r-partite hypergraphs, intersecting hyper- Key words. graphs 1 Introduction Finding a maximum cardinality matching in a r-partite hypergraph is a well-studied problem in combinatorics, combinatorial optimization, and computer science that is known as the r-dimensional Matching Problem (rdm). rdm is NP-hard when r 3 . A related problem is that of finding a minimum cardinality vertex cover (rdvc) in an r-partite hypergraph. This problem is also NP-hard for r 3 . The relationship between the size of a minimum cardinality vertex cover τ and the size of a maximum cardinality matching ν in an r-partite hypergraph is the subject of an open conjecture due to Ryser, which is stated as follows: Ryser s Conjecture. For an r-partite hypergraph, τ (r 1)ν. 1

2 The case when r = is the subject of König s Theorem  for bipartite covering and matching. Aharoni  proves the conjecture when r = 3. The conjecture remains open for r 4. Tuza  proves the conjecture is true for intersecting hypergraphs when r = 4 and r = 5. The conjecture remains open for intersecting hypergraphs when r 6. There are other cases when Ryser s conjecture is known to be true. Berge [] establishes that any polytope defined by: {x R n : Ax b; x 0} or {x R n : Ax b; x 0} where A is a {0, 1} matrix has integral vertices if and only if A is a balanced matrix and b is a vector of positive integers. An implication of Berge s theorem is that rdm and rdvc can be solved with linear programming if the incidence matrix of the underlying hypergraph is a balanced matrix. If this is the case, then the strong duality theorem of linear programming implies τ = ν for hypergraphs with balanced incidence matrices. Thus, Ryser s conjecture is easily resolved for hypergraphs with balanced incidence matrices. Since rdm and rdvc can occur on hypergraphs having incidence matrices that are not balanced, the standard integer linear programming (ILP) formulation of rdm has a nontrivial integrality gap. Note that the LP relaxations of the standard ILP of rdm and of rdvc form a primal-dual LP pair. Therefore, a cover for a given hypergraph provides an upper bound on the integrality gap for the rdm ILP formulation. If Ryser s conjecture is true it would provide an upper bound of r 1 on the integrality gap for rdm. However, this upper bound is already known to be true due to the following result of Füredi: Füredi s Theorem  For an r-partite hypergraph, τ (r 1)ν. In this context, τ refers to the cardinality of the minimum fractional vertex cover in the hypergraph. Füredi s Theorem is tight for the case when r is a prime power. This can be shown by constructing an instance of rdm on the hypergraph that results from deleting a vertex and all incident hyperedges from the projective plane of order r [1, 3, 1]. There are also known bounds on the integrality gaps of the standard formulations of rdvc. Lovász  proves the ratio obtained by dividing the cardinality of the minimum vertex cover in a plain (i.e., not necessarily r-partite) hypergraph by the cardinality of the minimum fractional vertex cover is at most 1 + log d where d is the maximum degree in the hypergraph. Naturally, this result also extends to the integrality gap of rdvc. Lovász  also proves the integrality gap of rdvc is at most r. This paper makes several contributions related to covers and matchings in r-partite hypergraphs. First, we prove that rdvc in intersecting hypergraphs is NP-hard for r 3. We observe that an r-partite intersecting hypergraph with τ ν must contain a certain sub-hypergraph, which we call a tornado. We prove several upper bounds on the size of minimum cardinality covers for tornados. Finally, we present a proof that the integrality gap of the standard ILP formulation for rdm is at least r k, where k is the smallest positive integer such that r k is a prime power. Our last contribution is an extension of a known result; specifically, there exists an intersecting hypergraph with τ = r 1 if r is a prime power (e.g., Mansour et al. ). In addition, our proof is a novel construction that uses Latin squares while the proof of the previously-known result uses

3 projective planes. Preliminaries In this section we overview preliminary concepts that are used throughout this paper. Specifically, we present preliminary concepts for edge colorings, hypergraphs, integer programming, and Latin squares..1 Hypergraphs Let H = (V, E) denote an undirected hypergraph with vertex set V and hyperedge set E. A hyperedge e E is a subset of the vertices. That is, e V for each e E. A hypergraph is r-partite if the vertices are partitioned into r disjoint subsets V k, k = 1,..., r, and each hyperedge contains exactly one vertex from each of the r subsets. All hypergraphs discussed in this paper are r-partite. An incidence matrix A of a hypergraph is a {0, 1} matrix where every vertex has exactly one corresponding row in A, every hyperedge has exactly one corresponding column in A and the entry in the row corresponding to vertex v and the column corresponding to hyperedge e is 1 if and only if v is contained in e. An incidence matrix is balanced [] if it does not contain a submatrix that is equivalent to an incidence matrix of a graph that is an odd cycle. A hypergraph with a balanced incidence matrix is a balanced hypergraph. A hypergraph is intersecting if each pair of hyperedges has a non-empty intersection. A matching in a hypergraph is a subset of hyperedges M E such that each vertex appears in at most one of the hyperedges in M. A vertex cover in a hypergraph is a subset of the vertices C V such that every hyperedge contains at least one of the vertices in C. A fractional vertex cover C f is an assignment of weights to vertices such that every vertex receives a weight in the closed interval from 0 to 1 and the sum of the weights of vertices on each hyperedge is greater than or equal to one. C f denotes the cardinality of a fractional vertex cover C f, which is the sum of the weights in the fractional vertex cover.. Edge Colorings of Graphs Let G = (V, E) be a graph. An edge coloring is a partition of the edges into subsets where each subset uniquely corresponds to a color. An edge that is in the subset corresponding to color c is colored with color c. An edge coloring is proper if no two edges that are incident to the same vertex are colored with the same color. Let K n be the complete graph on n vertices. The following theorem is well-known: Theorem 1.  There exists a proper edge coloring of K n that uses n 1 colors. 3

4 .3 Integer Programming Given an instance I of an ILP formulation of a combinatorial optimization problem where the objective is to maximize a function, let zip (I) denote its optimal objective value. Let z LP (I) denote the optimal objective value of a corresponding linear programming (LP) relaxation, the linear program obtained by replacing the integrality restrictions on the variables with nonnegativity constraints. The integrality gap of this particular instance, (for this particular ILP and LP relaxation pair) is z LP (I) zip (I). Similarly, the integrality gap of this ILP (and a corresponding LP relaxation) equals: sup I I z LP (I) z IP (I), where I is the set of all possible instances of the problem. Note the integrality gap of any ILP is always greater than or equal to one. An ILP with an integrality of exactly one is said to have no integrality gap..4 Latin Squares Let [n] = {1,,..., n}. A Latin square of dimension n is an n n matrix where each number in [n] occurs in each row exactly once and each number in [n] occurs in each column exactly once. Such a Latin square has size n. Given two square matrices L 1 and L of size n, their paired matrix is a square matrix L 1, of dimension n such that the entry of the ith row and the jth column of L 1, is an ordered pair (a 1, a ) where a 1 is the element in the ith row and the jth column of L 1 and a is the element in the ith row and the jth column of L. Two Latin squares are orthogonal if their corresponding paired matrix contains each of the ordered pairings in the set [n] [n] exactly once. A set of pairwise orthogonal Latin squares are called mutually orthogonal Latin squares (MOLS). A well-known result concerning the existence of MOLS is the following: Theorem.  A set of n 1 distinct MOLS of dimension n exists if n is a prime power. 3 Covering r-partite Intersecting Hypergraphs is NP-hard Let IrDVC denote the special case of rdvc on intersecting hypergraphs. We provide a nontrivial reduction of rdvc to IrDVC to prove IrDVC is NP-hard. First, we need the following definitions: Definition 1. Let f : V [n 1] be a proper edge coloring of K n. Then a tournament function π : V [n 1] V is a function that, given a vertex u and a color c as input, outputs the vertex v such that the edge (u, v) has color c when the edge coloring f is applied to K n. 4

5 We use the name tournament function because if we were to schedule a single round-robin tournament between the vertices (players) using the well-known circle method of Kirkman , π would indicate the opponent of vertex u during the week that corresponds to color c. Definition. Let H = (V, E) be a hypergraph and let C V be a vertex cover of H. A vertex v C minimally covers a hyperedge e E if the vertex set C\{v} does not cover e. Theorem 3. IrDVC is NP-hard for r 5. Proof. Consider any instance of rdvc where r 5 and let H = ( r i=1 V i, E) be the corresponding hypergraph. We show how to transform this instance into an instance of IˆrDVC, which is on a hypergraph Ĥ = ( ˆr ˆV i=1 i, Ê) where ˆr = r + ( E 1). Assume for now that E is even. For each i {1,,..., r}, ˆV i = V i. In addition, for each hyperedge e i E and for each vertex subset V j such that j > r, create a vertex v i,j in partition ˆV j. For each hyperedge e i = (v (i,1), v (i,),..., v (i,r) ) in E, create the following two hyperedges in Ê: ê (i,1) = (v (i,1), v (i,),..., v (i,r), v (ψ(i,r+1),r+1), v (ψ(i,r+),r+),..., v (ψ(i,ˆr),ˆr) ), ê (i,) = (v (i,1), v (i,),..., v (i,r), v (ω(i,r+1),r+1), v (ω(i,r+),r+),..., v (ω(i,ˆr),ˆr) ), where { ( π i, p ψ(i, r + p) := ) if π ( i, p ) < i and p is odd i otherwise { i if ψ(i, r + p) = π(i, p ω(i, r + p) := ) π ( i, p ), otherwise and π is a tournament function on the complete graph K E such that:, Each hyperedge in H corresponds to a vertex in K E. The pth color on edge (i, j) in K E indicates each of the two hyperedges corresponding to e i intersects one of the two hyperedges corresponding to e j in the (p 1)st partition of Ĥ and intersects the other one in the (p)th partition of Ĥ. By construction, Ĥ is an intersecting, ˆr-partite hypergraph where ˆV = V + E ( E 1) and Ê = E. Thus, the input size of Ĥ is a polynomial function of r, V, and E, which makes this a polynomial reduction. What remains to show is that τ(h) = τ(ĥ). Clearly, τ(ĥ) τ(h) since the minimum vertex cover of H has a corresponding cover in Ĥ. We now show τ(h) τ(ĥ). Let ˆV L = r ˆV i=1 i and let ˆV R = ˆr ˆV i=r+1 i. Let Ĉ be a minimum vertex cover of Ĥ that is minimal with respect to the cardinality of Ĉ ˆV R. Every vertex v Ĉ ˆV R must minimally cover exactly two hyperedges since if it only minimally covered one hyperedge ê (i,j), then this would contradict the minimality of Ĉ since a new minimum vertex cover could be formed from Ĉ that replaces v with any vertex in ˆV L ê (i,j), which would use fewer vertices from ˆV R. 5

6 Furthermore, for any i, hyperedge ê (i,1) is minimally covered by a vertex in Ĉ ˆV R if and only if hyperedge ê (i,) is minimally covered by a vertex in Ĉ ˆV R. This is because a vertex in Ĉ ˆV L covers ê (i,1) if and only if it also covers ê (i,). Therefore, the vertices in Ĉ ˆV R must collectively minimally cover a set of hyperedges of the form: {ê (p(1),1), ê (p(1),), ê (p(),1), ê (p(),),..., ê (p(mr ),1), ê (p(mr ),)} where m R = Ĉ ˆV R. This means there exists an alternative minimum vertex cover that can be constructed by removing all of the vertices in Ĉ ˆV R from Ĉ and replacing them with one vertex from each of the following vertex sets: ê (p(1),1) V L, ê (p(),1) V L,..., ê (p(mr ),1) V L. However, this contradicts the minimality of Ĉ. Thus, any minimum vertex cover for Ĥ is also a minimum vertex cover for H, which implies that τ(ĥ) = τ(h). Finally, if E is odd, Ĥ can be constructed as follows: 1. Create a phantom r-partite hyperedge e E +1 on an arbitrary set of vertices to ensure the number of hyperedges is even.. Construct Ĥ as described above. 3. Remove the two hyperedges extended from e E +1 in Ĥ. The same reasoning shows that τ(ĥ) = τ(h) when E is odd. For illustrative purposes, we describe a small example of constructing Ĥ. Example of Construction: Suppose H = (V, E) where E consists of the following four hyperedges: 1. e 1 = (v (1,1), v (1,),..., v (1,r) ),. e = (v (,1), v (,),..., v (,r) ), 3. e 3 = (v (3,1), v (3,),..., v (3,r) ), 4. e 4 = (v (4,1), v (4,),..., v (4,r) ). Note that it is possible for any of these four hyperedges to intersect in H (i.e., v (i1,j) = v (i,j) for hyperedges e i1 and e i and some subset j), but they need not intersect. Then we can construct Ĥ by including the original r subsets of vertices, creating six new subsets of vertices, and including the following eight hyperedges: 1. ê 1,1 = (v (1,1), v (1,),..., v (1,r), v (1,r+1), v (1,r+), v (1,r+3), v (1,r+4), v (1,r+5), v (1,r+6) ),. ê 1, = (v (1,1), v (1,),..., v (1,r), v (,r+1), v (,r+), v (3,r+3), v (3,r+4), v (4,r+5), v (4,r+6) ), 6

9 vertices. Choosing C = {v 1, v 3, v 5,..., v E } is such a cover. The next proposition proves the upper bound from Proposition 5 is tight for tornados where E r+1 1. Proposition 6. Given r, there exist tornados with E = 3, 5,..., r+1 1 such that τ = E +1. Proof. Consider a proper edge coloring of the complete graph K m = ( ˆV m 1 {ˆv m }, Ê) where m = r+1 with m 1 colors. Such a proper edge coloring exists because m is even. Construct a hypergraph H = (V, E) where V = m 1 k=1 V k and E = {e i : i = 1,,..., E } such that there is a bijection between vertices in ˆV m 1 and hyperedges in E. For each edge (ˆv i, ˆv j ) K m with i, j m and color k, create a vertex in V k with hyperedges e i and e j incident to it. For each edge (ˆv i, ˆv m ) K m with color k, create a vertex in V k with only hyperedge e i incident to it. The hypergraph is (m 1)-partite because in K m every vertex has degree m 1 and for each edge in K m there is a vertex. The hypergraph is intersecting by construction because for any pair of hyperedges there is an edge in K m. The hypergraph is a tornado with eye {v 1,, v,3,..., v m,m 1, v m 1,1 } where v i,i+1 is the unique vertex that is contained in hyperedges e i and e i+1. A cover for H is the subset of vertices {v i,i+1 : i odd} v m 1,1, so τ E +1. This cover is of minimum size because a set of vertices of size E 1 or less can cover at most E 1 = E 1 1 hyperedges with τ = E +1. From H we can hyperedges. Therefore, H is a tornado with r+1 construct a tornado with any odd number of hyperedges h by taking the sub-hypergraph formed by edges {e i : i = 1,..., h} where h 3. The next proposition proves the upper bound on the covering number for r-partite tornados can be slightly tighter than that in Proposition 5 if E > r. Proposition 7. For an r-partite tornado with E > r, τ E 1. Proof. It suffices to construct a vertex cover C that contains at most E 1 vertices. Let e 1 be any hyperedge in the tornado. e 1 intersects at least E other hyperedges at vertices that are not in the eye of the tornado, which means these intersections occur in at most r subsets of the r-partition of vertices. If E > r, then E > r, and there is at least one vertex v where e 1 intersects two other hyperedges. If v is in the vertex cover, then the remaining E 3 hyperedges must be covered somehow. However, since E is odd, E 3 is even and there exists a vertex that is contained in any pair of these hyperedges. Thus, the remaining E 3 hyperedges can be covered with E 3 vertices, which means we have constructed a vertex cover of size E = E 1. 9

10 1,,3,4,..., E 1,,3,4,..., E 3,4 4,5 1,,3 1,,3,4,..., E 1 3,4 3 E,1 E 1, E 4 Figure : Illustration of a proof of Proposition 9. The black vertices correspond to vertices in V eye. The labels on the vertices represent the incident hyperedges. Each column of vertices corresponds to a subset of the 3-partition. Corollary 8. An r-partite tornado with τ = r 1, if one exists, has at least r 1 hyperedges. Proof. Suppose that an r-partite tornado with τ = r 1 and E < r 1 exists. If E r, then by Proposition 5, τ E +1 < r 1, which is a contradiction. Similarly, if r < E < r 1, then by Proposition 7, τ E 1 < r 1, which is a contradiction. Proposition 9. Every 3-partite tornado has exactly 3 hyperedges. Proof. Suppose that H = (V, E) is a 3-partite tornado with E 5. Without loss of generality, let v 1 V 1 and v V. Then e intersects hyperedges e 4,... e E in V 3 V \V eye, and all of the vertices in V eye are in V 1 and V. Then v E V 1, which means e 1 has two vertices in V 1, which contradicts the fact that H is 3-partite. Figure presents a schematic of this proof. We now present the final result of this section. Theorem 10. For a 4-partite tornado with E 5, τ = 1. Proof. Suppose, for the sake of contradiction, we are given a 4-partite tornado H T with the minimum number of hyperedges with τ and E 5. Without loss of generality, let v E V 1, v 1 V, and v V 3. We may assume the vertices of V eye are contained in at least 3 subsets of the 4-partition since otherwise the vertices in V eye correspond to an odd cycle in a bipartite graph, which is impossible. This setup implies e 1 and e 3 intersect at a vertex in V 4 V \V eye and e and e E intersect at a vertex in V 4 V \V eye. Figures 3, 4, and 5 present schematics for the proofs of Cases I, II, and III that follow. We refer to these schematics throughout our proofs. Case I: v 3 V. The first box in Figure 3 illustrates the starting point for Case I. 10

11 From this setup, we can infer that the following pairs of hyperedges must intersect at a vertex in V 4 V \V eye : e 1 and e 3, e and e E, and e 3 and e E. Therefore, e 1, e, e 3, and e E intersect at a single vertex in V 4 V \V eye. This situation is depicted in the second box in Figure 3. If E = 5, then v 4 V 3 and e 1 and e 4 intersect at a vertex in V 4 V \V eye. This means all hyperedges intersect at a single vertex in V 4 V \V eye, which contradicts our assumption that τ. Therefore, we may assume E 7. We can remove e 1, e, and e 3 and add a new hyperedge e a that contains the vertices (e 1 (V 1 V 3 V 4 )) (e 3 V ). This creates a 4-partite tornado H Ta with E hyperedges. This situation is depicted in the third box in Figure 3. Since we are assuming H T is a 4-partite tornado with the fewest number of hyperedges such that τ, then we know τ(h Ta ) = 1. If the minimum vertex cover of H Ta is comprised of the vertex in e a V 4, then that vertex also covers H T, which is a contradiction. Therefore, the vertex in e a V 3 is a cover for H Ta. This situation is depicted in the fourth box in Figure 3. Moreover, this setup implies v E 1 V. This situation is depicted in the fifth box in Figure 3. Further still, we can remove e a and restore hyperedges e 1, e and e 3. This situation is depicted in the sixth box in Figure 3. Now, we can remove hyperedges e E, e 1, and e and replace them with e b, where e b contains the vertices (e (V 1 V 3 V 4 )) (e E V ). Note this creates another tornado H Tb that has E hyperedges. This situation is depicted in the seventh box in Figure 3. By our assumption, we know τ(h Tb ) = 1. If the vertex in e b V 4 is a cover for H Tb then it is a cover for H T, so the vertex in e b V 1 must be the cover for H Tb. This setup is depicted in the eighth box in Figure 3 and the setup where e b is removed and e E, e 1 and e are restored is depicted in the ninth box in Figure 3. This setup implies v j V for every odd value of j and v j V 4 for every even value of j. This implies v E 1 V 4, which contradicts the fact that v E 1 V. This situation is depicted in the tenth box in Figure 3. Case II: v 3 V 1 and v E 1 V. The first box in Figure 4 illustrates the starting point for Case II. From this setup, we can infer the following pairs of hyperedges must intersect at a vertex in V 4 V \V eye : e 1 and e 3, e and e E, and e 3 and e E. Therefore, e 1, e, e 3, and e E intersect at a single vertex in V 4 V \V eye. This situation is depicted in the second box in Figure 4. If E = 5, then e and e 4 intersect at a vertex in V 4 V \V eye, which implies all five hyperedges intersect in at a single vertex in V 4 V \V eye, which contradicts our assumption. Thus, we may assume E 7. We can remove hyperedges e E, e 1, and e and replace them with a hyperedge e a that contains vertices (e (V 1 V 3 V 4 )) (e E V ). Note this creates a 4-partite tornado H Ta that has E hyperedges. This situation is depicted in the third box in Figure 4. We know τ(h Ta ) = 1 since we assumed H T is a 4-partite tornado with τ(h T ) that has the fewest number of hyperedges. The unique vertex in the minimum vertex cover for H Ta must be in V 4 since the vertices in V eye ensure this vertex cannot be in the other three vertex subsets. Specifically, the vertex that covers H Ta must be the same vertex in V 4 that is contained in hyperedges e a and 11

12 1, E 1,,3 1, E 1,,3 1,,3, E a, E a, E 3,4 3,4 a,4 1 3 b,3 b,3 1, E 1,,3 1,,3, E a, E a, E 3,4 4,5,..., E 1 3,4 1,4,5,..., E a,4 a,4,5,..., E E 1,b E 1, E b,3 b,3 1, E,3 1,,3, E 1, E 1,,3 1,,3, E,3,..., E 1 3,4 4,5,..., E 1,3,..., E 1 3,4 4,5,..., E 1,3,..., E 1 3,4 4,5,..., E 4,5 E 1,b E 1, E 7 8 E, E 1 E 3, E E 1, E 9 Figure 3: Illustration of a proof of Case I of Theorem 10. The black vertices correspond to vertices in V eye. The labels on the vertices represent the incident hyperedges. Each column of vertices corresponds to a subset of the 4-partition. 1

13 1, E 1,,3 1, E 1,,3 a,3 3,4 E 1, E 3,4 E 1, E 1,,3, E 3,4 E 1,a a, , E 1,,3 a,3 3,4 E 1, E 1,,..., E 3,4 E 1,a a,3,4,..., E Figure 4: Illustration of a proof of Case II of Theorem 10. The black vertices correspond to vertices in V eye. The labels on the vertices represent the incident hyperedges. Each column of vertices corresponds to a subset of the 4-partition. e 3. This situation is depicted in the fourth box in Figure 4. After removing e a and restoring hyperedges e E, e 1, and e, we note these three hyperedges are also incident to the vertex that covers H Ta, which means this vertex also covers H T. However, this is a contradiction since we assumed τ(h T ). This situation is depicted in the fifth box in Figure 4. Case III: v 3 V 1 and v E 1 V 3. The first box in Figure 5 illustrates the starting point for Case III. This setup implies the following pairs of hyperedges must intersect at a vertex in V 4 V \V eye : e 1 and e 3, e and e 4, and e and e E. If E = 5, then e 1 and e 4 intersect at a vertex in V 4 V \V eye, which implies all hyperedges intersect at a single vertex in V 4 V \V eye. Thus, we may assume E 7. If v 4 V 3, then we can remove hyperedges e, e 3, and e 4, replace them with a hyperedge e a that contains the vertices (e (V 1 V 3 V 4 )) (e 4 V 3 ). This creates a 4-partite tornado H Ta that has E hyperedges. We know τ(h Ta ) = 1 since we assumed H T is a 4-partite tornado with τ(h T ) that has the fewest number of hyperedges. Specifically, the vertex that covers H Ta can only be in V 4. However, this vertex would also be a cover for H T, which is a contradiction. Thus, we see v 4 / V 3 and we can conclude v 4 V. This situation is depicted in the second box in Figure 5. Then the following pairs of hyperedges intersect at the same vertex in V 4 V \V eye : e 3 and e 5, e 5 and e E, and e E 1 and e 4. Thus, e 1, e, e 3, e 4, e 5, e E 1, and e E intersect at the same vertex v V 4 V \V eye. This situation is depicted in the third box in Figure 5. If E = 7, then all hyperedges intersect at v, which is a contradiction. We now prove by induction on the number of remaining vertices in V eye that all hyperedges intersect at v for tornados with E > 7. 13

14 1, E 1,,3 1, E 1,,3,4, E 1, E 1,,3,4, E 3,4 E 1, E 3,4 E 1, E 1,3 3,4 4,5 E 1, E 1, , E 1,,3 3,4 4,5 E 1, E 1, E 1,,3 1,,...,j 1, E 1, E 3,4 4,5 E 1, E 1, E 1,,3 1,,3,4,5, E 1, E 3,4 4,5 E 1, E 1,,...,j 1, E 1, E 5 4 j 1,j or j 1,j j 1,j or 6 1, E 1,,3 1,,...,j, E 1, E 3,4 4,5 E 1, E 7 1, E 1,,3 3,4 4,5 E 1, E 1,,..., E 8 Figure 5: Illustration of a proof of Case III of Theorem 10. The black vertices correspond to vertices in V eye. The labels on the vertices represent the incident hyperedges. Each column of vertices corresponds to a subset of the 4-partition. Assume v i 1 (V 1 V V 3 ) for each i { E, 1,,..., j 1} and that hyperedges e E 1, e E, e 1,..., e j 1 all intersect at the same vertex v in V 4 V \V eye. We now show this implies v j 1 (V 1 V V 3 ) and e j also contains v. First, note that hyperedges v j 1 / V 4 because this contradicts the fact that e j 1 already contains vertex v V 4 V \V eye. Furthermore, note that if 1. v j 1 V 1, then e intersects e j at a vertex in V 4 V \V eye, which can only occur at v.. v j 1 V, then e 3 intersects e j at a vertex in V 4 V \V eye, which can only occur at v. 3. v j 1 V 3, then e 1 intersects e j at a vertex in V 4 V \V eye, which can only occur at v. Thus, in any case, e j contains v, and v is a cover for the tornado, which is a contradiction. 14

15 5 Integrality Gap for an r-dimensional Matching Formulation In this section, we provide an alternate proof that the integrality gap of the standard ILP formulation of rdm is exactly r 1 when r 1 equals a prime number raised to a power. Whereas the results of previous sections are concerned with intersecting hypergraphs, the main result of this section applies to the general rdm problem. Other researchers, such as Mansour et al. , have proven this result before using projective planes. Our proof is a construction that uses MOLS. (We also note there is a known correspondence between projective planes of order r and a complete set of r 1 r-dimensional MOLS .) In addition, we extend the result to show the integrality gap is at least r k where k is the smallest positive integer such that r k is a prime power. Let x e equal 1 if hyperedge e is in our matching and 0 otherwise. The standard ILP for rdm is the following: Integer linear programming formulation of rdm (rdm-ilp): Maximize e E x e, Subject to e E:v e x e 1 v V, x e {0, 1} e E. The LP relaxation of rdm-ilp is the ILP except with the binary constraints replaced by nonnegativity constraints. The LP relaxation of rdm-ilp shall henceforth be abbreviated as rdm-lp. Note that we need not constrain each of the LP decision variables to be less than or equal to 1, since these constraints are implied by the matching constraints. Lemma 11. The integrality gap of rdm-ilp is at most r 1. Proof. Let I be any instance of rdm so that ν is the objective value of an optimal solution to the instance as formulated by rdm-ilp. Since the underlying hypergraph of any instance of rdm is r-partite, we know from Füredi s Theorem that we can construct a fractional vertex cover C f with a cardinality of (r 1)ν. Note that such a cover is a feasible solution to the dual of rdm-lp and has an objective value of (r 1)ν. Thus, by duality theory, we know the objective value of an optimal solution to rdm-ilp of I is bounded above by (r 1)ν. To prove that the integrality gap of rdm-ilp is at least r 1 whenever r 1 is a prime power, it suffices to construct a family of instances where the optimal objective value to rdm-ilp is z r and the optimal objective value to rdm-lp is at least (r 1)z r. Theorem provides that for any prime power r 1, there exists a set of r MOLS L 1, L,..., L r, each of which have dimension r 1. We use this set of MOLS to construct an instance of rdm. In addition, define L r 1 to be the square matrix of dimension r 1 where every entry in row i is the number i, for each i in {1,,..., r 1}. (Note L r 1 is not a Latin square.) Let L k (i, j) denote the element in row i and column j of matrix L k. From the matrices L 1, L,..., L r 1, construct another set of (r 1)-dimensional square matrices M = {M 1, M,..., M r 1 } where 15

16 M k (i, j) = L(i, j) + (r 1)(k 1) for each i, j, k {1,,... r 1}. From M construct an r-partite hypergraph H r = ( r j=1 V j, E) as follows: 1. Let each subset V 1, V,..., V r contain r 1 vertices. Let v i,j be the ith vertex in subset V j.. For each k and for each element a contained in matrix M k, create a hyperedge e a that contains the vertex v r,k and the vertex v i,j for each pair (i, j) such that M k (i, j) = a. Example Construction: For illustrative purposes, we describe how our method constructs H r for the case when r = 4. First, we construct two MOLS of size 3: L 1 = Next, we construct the matrix L 3 : L 3 = L = Afterwards, we use L 1, L and L 3 to construct three matrices in M: M 1 = M = M 3 = This, allows us to construct H 4 = ( 4 j=1 V j, E) where V j = {v 1,j, v,j, v 3,j } for each j in {1,, 3, 4} and E = (v 1,1, v 3,, v,3, v 1,4 ), (v,1, v 1,, v 3,3, v 1,4 ), (v 3,1, v,, v 1,3, v 1,4 ), (v 1,1, v,, v 3,3, v,4 ), (v,1, v 3,, v 1,3, v,4 ), (v 3,1, v 1,, v,3, v,4 ), (v 1,1, v 1,, v 1,3, v 3,4 ), (v,1, v,, v,3, v 3,4 ), (v 3,1, v 3,, v 3,3, v 3,4 ) We now prove several properties about H r. Lemma 1. H r is an intersecting hypergraph. Proof. Note that any hyperedge in H r is created using one matrix in M. Any two hyperedges created using the same matrix in M intersect because they both contain the same vertex from V r. Now consider two hyperedges e a and e b that are created from two distinct matrices M a and M b and recall their corresponding matrices L a and L b. Let L a,b be the paired matrix that results from pairing L a with L b.. 16

17 Every hyperedge created using M a intersects with every hyperedge created using M b if L a,b contains every ordered pair from the set [r 1] [r 1]. First, consider the case when a = r 1. For each element i in [r 1], every element in row i from L a is the number i since L a = L r 1. Similarly, every element from [r 1] appears in exactly one entry in row i of L b since L b is a Latin square. Thus, for each i in [r 1], row i in L a,b contains (i, j) for each j in [r 1]. Therefore, every hyperedge created using M a intersects with every hyperedge that is created using M b. The same argument applies for the case when b = r 1. Now consider the case where a r 1 and b r 1. This means both L a and L b are MOLS, which means that their paired matrix contains (i, j) exactly once, for each (i, j) in [r 1] [r 1]. Therefore, in this case every hyperedge created using M a intersects with every hyperedge created using M b. Lemma 13. The optimal objective value to rdm-lp for H r is at least r 1. Proof. To prove this lemma, we construct a feasible solution to rdm-lp that has objective value r 1. Specifically, set x e = 1 r 1 for each e E. For each vertex v r j=1 V j, the corresponding constraint is satisfied because each vertex has a degree of r 1. This is because for each vertex there is one hyperedge created using each matrix in M and M = r 1. Thus, this solution is feasible. In addition, there are (r 1) hyperedges in H r since each matrix in M is used to create r 1 hyperedges and M = r 1. Thus, the objective value of this solution is (r 1) (r 1) = r 1. Theorem 14. The integrality gap of rdm-ilp when r 1 is a prime power is exactly r 1. Proof. By Lemma 11, the integrality gap of rdm-ilp is at most r 1. By Lemma 1, H r is intersecting and has matching number 1, so that the optimal objective value of rdm-ilp is 1. By Lemma 13, the objective value of rdm-lp is at least r 1. Therefore, the integrality gap for instances H r is exactly r 1. Corollary 15. The integrality gap of rdm-ilp is at least r k where k is the smallest positive integer such that r k is a prime power. Proof. Assume k > 1 since the case when k = 1 is handled by Theorem 14. Since r k is a prime power, the hypergraph H r k+1 that is defined using the aforementioned construction exists. This hypergraph is (r k+1)-partite and each subset of the partition contains r k+1 vertices. Construct a r-partite hypergraph H r from H r k+1 as follows. For each vertex in H r k+1, create a corresponding vertex in H r. In addition, create r k vertices v i,r k+, v i,r k+3,..., v i,r for each i {r k +, r k +3,..., r} where v i,r j V i for each j {r k +, r k +3,..., r}. Lastly, for each hyperedge e = (v i,j,..., v l,r k+1 ) in H r k+1 create the hyperedge ẽ = (v i,j,..., v l,r k+1, v l,r k+, v l,r k+3,..., v l,r ) in hypergraph H r. H r has a maximum cardinality matching of exactly one hyperedge since it is an intersecting hypergraph. We can obtain a solution to rdm-lp with objective value r k by setting x e = 1 r k for each hyperedge e in H r. 17

18 6 Conclusions and Future Work In this paper, we use ideas common in the math programming community to gain further insight into two well-studied combinatorics problems. First, we show that the r-dimensional Vertex Cover Problem is NP-hard for intersecting hypergraphs. This result is interesting because the r-dimensional Matching Problem is trivial for intersecting hypergraphs. The construction in our proof uses a well-known algorithm for obtaining a proper edge coloring in a complete graph (which is also a well-known algorithm for scheduling a single round-robin tournament). We note that Ryser s conjecture is easily resolved for intersecting hypergraphs that do not contain a sub-hypergraph that we call a tornado. A tornado-free intersecting hypergraph can be covered with a single vertex. We prove several bounds on the covering number of tornados. Extending these results to a possible proof of Ryser s conjecture for intersecting hypergraphs would involve characterizing how tornados can be arranged in a hypergraph. We provide an alternate proof of a previous result concerning the integrality gap for the r-dimensional Matching Problem. Our proof is unique in that it uses of Latin squares instead of projective planes. Our result also provides a provides a bound for any value of r, whereas the previous result only provides bounds for the case when r 1 is a prime power. References  R. Aharoni. Ryser s conjecture for tripartite 3-graphs. Combinatorica, 1:1 4, 001. [] C. Berge. Balanced matrices. Mathematical Programming, :19 31, 197.  Y. H. Chan and L. C. Lau. On linear and semidefinite programming relaxations for hypergraph matching. Proceedings of the Twenty-First Annual ACM-SIAM Symposium on Discrete Algorithms, pages , 010.  C. J. Colbourn and J. H. Dinitz. Handbook of Combinatorial Designs. Chapman & Hall/CRC, second edition, 006.  Z. Füredi. Maximum degree and fractional matchings in uniform hypergraphs. Combinatorica, 1:155 16,  G. Gottlob and P. Senellart. Schema mapping discovery from data instances. Journal of the ACM, 57, 010.  R. M. Karp. Reducibility among combinatorial problems. Complexity of Computer Computations, pages , 197.  D. Kőnig. Gráfok és mátrixok. Matematikai és Fixikai Lapok, 38: ,  T.P. Kirkman. On a problem in combinatorics. Cambridge and Dublin Mathematical Journal, :191 04,  L. Lovász. On minimax theorems of combinatorics. Matemathikai Lapok, 6:09 64, (in Hungarian). 18

19  L. Lovász. On the ratio of optimal integral and fractional covers. Discrete Mathematics, 13: ,  T. Mansour, C. Song, and R. Yuster. A comment on Ryser s conjecture for intersecting hypergraphs. Graphs and Combinatorics, 5: , 009.  Z. Tuza. Ryser s conjecture on transversals of r-partite hypergraphs. Ars Combinatoria, 16: 01 09,  D. B. West. Introduction to Graph Theory. Pearson Education,

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