Intersecting and 2 -intersecting hypergraphs with maximal covering number: the Erdős-Lovász theme revisited
IIntersecting and 2-intersecting hypergraphs with maximal coveringnumber: the Erd˝os-Lov´asz theme revisited
J´anos Bar´at ∗ University of Pannonia, Department of Mathematics8200 Veszpr´em, Egyetem utca 10., HungaryandMTA-ELTE Geometric and Algebraic Combinatorics Research GroupH–1117 Budapest, P´azm´any P. s´et´any 1/C, Hungary [email protected]
November 10, 2020
Abstract
Erd˝os and Lov´asz noticed that an r -uniform intersecting hypergraph H with maximal covering num-ber, that is τ ( H ) = r , must have at least r − q ( r ) denote the minimum number of edges in an inter-secting r -uniform hypergraph. It was known that q (3) = 6 and q (4) = 9. We obtain the following newresults: The extremal example for uniformity 4 is unique. Somewhat surprisingly it is not symmetricby any means. For uniformity 5, q (5) = 13, and we found 3 examples, none of them being some knowngraph. We use both theoretical arguments and computer searches. In the footsteps of Erd˝os and Lov´asz,we also consider the special case, when the hypergraph is part of a finite projective plane. We determinethe exact answer for r ∈ { , , , } . For uniformity 6, there is a unique extremal example.In a related question, we try to find 2-intersecting r -uniform hypergraphs with maximal coveringnumber, that is τ ( H ) = r −
1. An infinite family of examples is to take all possible r -sets of a (2 r − λ = 2.We determined that only 3 biplanes of the 18 known examples are extremal. Keywords: intersecting hypergraph, cover, projective plane, combinatorial design, biplane A hypergraph consists of vertices and edges, where edges are subsets of vertices. We use | H | or m to denotethe number of edges (also called lines) and | V ( H ) | or n the number of vertices (also called points) of H . Ahypergraph is r -uniform if every line has r points on it. A hypergraph is intersecting if any two edges havea vertex in common. In this paper we only consider intersecting, uniform hypergraphs. A projective plane is an r -uniform intersecting hypergraph on r − r + 1 vertices such that there is precisely one line throughany pair of points. In the standard terminology of projective planes, this has order r −
1. For instance,
P G (2 , r −
1) is such an example if r − k -cover of a hypergraph is a set of k verticesmeeting every edge of the hypergraph. The covering number τ ( H ) of a hypergraph H is the minimum k for which there is a k -cover of H . Since covering is our main subject, it makes no difference to repeat anedge. Therefore, we only consider simple hypergraphs. A hypergraph is r -partite if its vertex set V can bepartitioned into r sets V , . . . , V r , called the sides of the hypergraph, so that every edge contains preciselyone vertex from each side. In particular, r -partite hypergraphs are r -uniform. ∗ Supported by Sz´echenyi 2020 under the EFOP-3.6.1-16-2016-00015 and OTKA-ARRS Slovenian-Hungarian Joint ResearchProject, grant no. NN-114614 and the grant of the Hungarian Ministry for Innovation and Technology (Grant Number:NKFIH-1158-6/2019). a r X i v : . [ m a t h . C O ] N ov or intersecting hypergraphs, any edge is a cover. Therefore, if H is an r -uniform intersecting hyper-graph, then τ ≤ r . Erd˝os and Lov´asz initiated the study of extremal examples [7]. That is, the hypergraphswith τ = r . For every r , where r − r -element subsets of a 2 r − (cid:0) r − r (cid:1) , second the projective planesof uniformity r . Among other things, Erd˝os and Lov´asz asked the following extremal question. What isthe minimum number of edges q ( r ) that an intersecting r -uniform hypergraph H with τ ( H ) = r can have?They proved a lower bound of r − $ for a proof. Kahn confirmed that a linear example exists [11].Tripathi[16] gave a short proof that q (3) = 6 and settled q (4) = 9. In this direction, we show that theextremal example for uniformity 4 is unique. Somewhat surprisingly it is not symmetric by any means. Wedetermine the next value, and q (5) = 13. For uniformity 5, we found 3 examples, none of them being someknown graph. These properties might indicate why it is difficult to improve the Erd˝os-Lov´asz lower boundin general. Our arguments combine computation and human reasoning. The strategies are somewhatsimilar to that of Francetiˇc et al. [8].Erd˝os and Lov´asz asked the analogous problem in projective planes. We determine the exact answerfor r ∈ { , , , } . For uniformity 5, we determine the five non-isomorhic extremal examples. We couldnot recognize any of these as some obvious geometric construction . For uniformity 6, there is a uniqueextremal example. These results are falling out of a simple computer search.For intersecting, r -partite hypergraphs, Ryser conjectured that τ ≤ r −
1. This conjecture in fullgenerality appeared in [9]. A hypergraph is t - intersecting if any two edges have at least t common vertices.Recently, Kir´aly and T´othm´er´esz [12] and independently Bustamante and Stein [3] conjectured the following t -intersecting generalisation of Ryser’s conjecture: If H is an r -partite hypergraph and any two hyperedgesintersect in at least t vertices, then τ ( H ) ≤ r − t . Bishnoi et al.[2] proved the tight upper bound (cid:98) ( r − t ) / (cid:99) +1for the range t ≥ r ≤ t −
1. Also DeBiasio et al. [5] considers this problem in section 8.4 in theirsurvey-like paper.Inspired by the development described in the previous paragraph, we initiate the study of 2-intersectinghypergraphs , in particular those, for which τ = r −
1. Our goal is to determine the minimum number ofedges of a hypergraph that belongs to this class. We collect the fundamental properties in the next section.We pose a provocative question whether there are only finitely many sporadic examples apart from thetrivial infinite class. A geometric hint is to study so called biplanes [17]. There are 18 examples known.We found that only 3 of them have maximal covering number. We studied the 4- and 5-uniform case indetail and said the final word only for the 4-uniform case.For our computational tasks, it makes good sense to think of a hypergraph H as a bipartite graph G ( V, E ), where V is the vertex set of H and E is the edge set. A vertex v ∈ V and a vertex u ∈ E areadjacent in G if and only if v ∈ e in H . This is the Levi graph of hypergraph H . We use the n × m incidence matrix of G to describe the properties of H .We used the following (commercially available) configurations to execute our searches and calculate thecovering number:1. Intel Core i3-4150 CPU 3.50GHz x 4 with 8 GiB RAM and ubuntu 16.04 64-bit operating system.2. Intel Core i5-4200 CPU 1.6 GHz x 4 with 8 GiB RAM and ubuntu 16.04 64-bit operating system.At the end it was the storage capicity that made our further searches impossible. We remark that allsearches were exhaustive. We plan to implement some randomized searches in a different project. -intersecting hypergraphs Let us first try to modify our two standard extremal intersecting hypergraphs. Taking all possible r -sets ofa (2 r − τ = r −
1. On the other hand, it appearsharder to get a 2-intersecting r -uniform hypergraph from a projective plane, where τ = r −
1. For instance,we can consider two projective planes π and π of the same order q , and a bijection φ between the lines. We not necessarily r -partite!! q + 2-uniform hypergraph, whose edges are L ∪ φ ( L ) for every possible L ∈ π .However, any line of π is still a cover, hence τ remains q + 1, which is only r/
2. It is worth mentioning thefollowing construction, which gives very similar parameters. Let the vertex set be the points of an n × m grid. That is, V ( H ) = { ( i, j ) : 1 ≤ i ≤ n, ≤ j ≤ m } . Let the hyperedges be the crosses of this grid. Thatis, E ( H ) = { C ( i, j ) : 1 ≤ i ≤ n, ≤ j ≤ m } and C ( a, b ) = { ( a, j ) : 1 ≤ j ≤ m } ∪ { ( i, b ) : 1 ≤ i ≤ n } . Thishypergraph H is ( n + m − τ = min( n, m ).There is a geometric object that resembles the properties of a 2-intersecting hypergraph: a biplane isa symmetric 2-design with λ = 2; that is, every set of two points is contained in two blocks (lines), whileany two lines intersect in two points [17]. They are similar to finite projective planes, except that ratherthan two points determining one line (and two lines determining one point), two points determine two lines(respectively, points). A biplane of order n is a symmetric design, where blocks have k = n + 2 points.We might say that 2-intersecting uniform hypergraphs are generalisations of biplanes. There are only 18biplanes known [17].The order 1 biplane geometrically corresponds to the tetrahedron. The vertices are the 4 points and theedges are the 4 faces. It is 3-uniform and no vertex covers the opposite face, therefore the covering numberis 2.The order 2 biplane is the complement of the Fano plane. It has covering number 3. The order 3 biplanehas 11 points (and lines of size 5), and is also known as the Paley biplane. Using our small program, wechecked its covering number: 4. At this point, one might get excited to believe that all other biplanes havemaximum covering number.Let us examine the next case. There are three biplanes of order 4 (and 16 points, lines of size 6). We foundthe combinatorial description of two of them. We checked these by hand, and their covering number wassmaller than 5. Let us show one of them. The example is the Kummer configuration. Let the points be thenumbers from 1 to 16 and arrange them in a 4 × and each of them has covering number 4.We pose the following Problem 2.1.
Let H be a -intersecting r -uniform hypergraph, and let (cid:0) r − r (cid:1) be the hypergraph on (2 r − vertices and all possible r -sets as edges. The maximum value of τ ( H ) among all -intersecting r -uniformhypergraphs is taken if H = (cid:0) r − r (cid:1) . Are there infinitely many values of r such that this is the only example?Are there infinitely many other examples? We start investigating these questions. We partly search by computer and try to prove facts to reducethe search space. The following two observations help.
Lemma 2.2.
Let H be an intersecting hypergraph with e edges. If the maximum degree is ∆ , then τ ≤ (cid:100) e − ∆2 (cid:101) . If H is r -uniform, τ = r − , then ∆ ≤ e + 4 − r if e − ∆ is even, and ∆ ≤ e + 5 − r if e − ∆ is odd.Proof. We find a cover by taking a vertex of maximum degree and covering the other lines in pairs.
Lemma 2.3.
Let H be a -intersecting r -uniform hypergraph. If there exists a vertex of degree at most r − , then τ ( H ) ≤ r − . We used a simple algorithm going through all possible vertex sets to find a cover of prescribed size and implemented inpython. roof. Let v be a vertex of degree at most r −
1. Let e, e , . . . , e r − be the edges through v , and let theirunion be U . Since H was 2-intersecting, there is at least one vertex different from v in each of e ∩ e ,. . . , e r − ∩ e . Let these vertices be v , . . . , v r − (some of them may coincide). Now let x be a vertex in e different from each of v, v , . . . , v r − . We claim that C = e \ { v, x } is a cover of size at most r −
2. Indeed,any edge f / ∈ U intersects e in at least 2 vertices, one of them different from x . On the other hand, anyedge f ∈ U , where v, v i ∈ f , intersects C in v i for some i . Therefore any edge f intersects C .Since we are looking for a 2-intersecting hypergraph that satisfies τ = r −
1, therefore in a computersearch we may assume the minimum degree to be at least r .Recall that intersecting r -uniform hypergraphs with maximal covering number can have linearly manyedges by Kahn’s result [11]. Now this changes dramatically for 2-intersecting hypergraphs, if we insist thattwo edges cannot intersect in more than 2 vertices. Corollary 2.4. If H is a -intersecting r -uniform hypergraph with maximal covering number and no twoedges intersect in at least vertices, then | E ( H ) | ≥ r − r + 1 .Proof. Consider any edge e and the vertices { v , . . . , v r } on e . Since the covering number of H is r −
1, wecan apply Lemma 2.3. That is, the degree of every vertex v i is at least r . Therefore the number of edgesis at least 1 + ( r − r .We can also adapt the Erd˝os-Lov´asz idea for intersecting hypergraphs to settle a lower bound on thenumber of edges of a 2-intersecting hypergraph. Lemma 2.5.
Let H be a -intersecting r -uniform hypergraph. If k r + 1 < | E ( H ) | , then there is a vertexof degree at least k + 2 .Proof. We fix an edge e . Any other edge f intersects e in at least 2 vertices. Therefore f contributes atleast 2 to the sum of the degrees of the vertices on e . Since r edges add at least r to the total degree, theaverage degree raises by 1. Corollary 2.6.
For large enough r , the number of edges of a -intersecting r -uniform hypergraph H withmaximal covering number satisfies r < | E ( H ) | .Proof. We can cover the last r edges by r vertices of degree 2. Before that, we can cover r edges by r vertices of degree 3 etc. Since (cid:88) k <
1, we can greedily cover more than 5 r edges with less than r − (cid:88) k >
1. Notice for small r all the integer parts make a substantialcontribution.As a warm-up, we can clear the 3-uniform case. Let e and e be two different edges of a 2-intersecting3-uniform hypergraph H . Let { u, v } = e ∩ e . Let x be the point in e \ e and y the vertex in e \ e . Ifevery other edge intersects e in { u, v } , then the covering number is only 1. Since our target is coveringnumber 2, there exists another edge f that intersects e in 2 points, say x and u . Now f must intersect e in two points. Therefore f = { x, u, y } . Still u is a cover of size one. There must be an edge g intersecting e in x and v and not containing u . Now g must intersect f in 2 points. Therefore y ∈ g and g = { x, y, v } . Sofar, we constructed (cid:0) (cid:1) . Clearly, we cannot add any new edge to this hypergraph keeping all requirements. Proposition 2.7.
There is only one -uniform -intersecting hypergraph with maximum covering number,namely (cid:0) (cid:1) the biplane of order . -uniform case In this section, we try to find a 2-intersecting 4-uniform hypergraph different from (cid:0) (cid:1) that has coveringnumber 3. We may assume that n ≥
7, and consider a 4-uniform hypergraph H that is 2-intersecting andsatisfies τ = 3. We think of H as an incidence matrix, where the rows correspond to vertices and columnsto edges of H . Each column of the incidence matrix contains precisely four 1s. Since τ >
2, we notice thatfor every pair of rows r and r , there must be a column that contains 0 in row r and r . Therefore, wemay assume that the first four rows start as follows: x y z z (cid:48) y (cid:48) x (cid:48) x to x (cid:48) must contain two 1s below row4. Since x and x (cid:48) correspond to hyperedges, they must intersect in two vertices. We may assume that thecorresponding 1s lie in row 5 and 6. Similarly y and y (cid:48) must intersect in two vertices, so the correspondingcolumns contain two 1s in the same two rows below row 4. However, they must also intersect x and x (cid:48) inat least two vertices. There are two possibilities here, as shown in the next two incomplete matrices. x y z z (cid:48) y (cid:48) x (cid:48) x y z z (cid:48) y (cid:48) x (cid:48) z and z (cid:48) must contain 1s in the same two rows below row 4. Assume to the contrarythat these two rows are row 5 and 6. In that case, row i and j correspond to two vertices covering H forany 1 ≤ i ≤ ≤ j ≤
6. Therefore, we must add a column that contains 0 in those pair of rows. Toachieve a 2-intersecting hypergraph, all other entries must be 1s. In this way we get (cid:0) (cid:1) .Now we may assume that z and z (cid:48) contain 0 in row 5. As a consequence they contain 1 in row 6, andwe may assume that they have a 1 in row 7. At this point, vertex 6 and vertex i corresponds to a 2-coverof the current hypergraph for 1 ≤ i ≤
4. Let us fix i = 2. There must be another edge f corresponding tothe next column, which contains 0 in rows 2 and 6. Comparing y and f , we conclude that f must contain1 in row 4 and 5. Comparing now z and f we conclude the last 1s must be in row 3 and 7. Now x (cid:48) and f intersect only in 1 vertex, a contradiction. x y z z (cid:48) y (cid:48) x (cid:48) f y z z (cid:48) y (cid:48) x (cid:48) x y z z (cid:48) y (cid:48) x (cid:48) x y z z (cid:48) y (cid:48) x (cid:48) f contains precisely two 1s in the firstfour rows, then it must coincide with one of the six edges x to x (cid:48) . Therefore any new edge f must containthree 1s in the first three rows and 0 in row 4 and 5. Now the partial edge f intersects y and z (cid:48) in only 1vertex. However the remaining 1 of f cannot intersect both y and z (cid:48) , since they are disjoint below row 5.We conclude: Proposition 3.1.
There are precisely two non-isomorphic -uniform -intersecting hypergraphs that havecovering number , namely (cid:0) (cid:1) and the complement of the Fano plane. -uniform case In this section, we seek a 2-intersecting 5-uniform hypergraph H . We deduced a lower bound for | E ( H ) | for somewhat large r in Corollary 2.6. Here, r = 5 too small to apply the Corollary directly. Still we canuse the same idea: that is, relate the degrees to small coverings as follows.To start with, we create a 3-edge hypergraph such that it needs 2 vertices to cover all edges. Let V ( H ) = { , . . . , } and E ( H ) = { (1 , , , , , (4 , , , , , (7 , , , , } . Since 4 edges can be greedilycovered by taking the intersection of two pairs of edges, Lemma 2.5 can be first applied for a hypergraphwith 5 edges. For k = 1, it yields there must be a vertex of degree 3. The other two edges intersect, sowe have a 2-cover again. Next, for the study of 6 edges, we use the following 5-uniform hypergraph. Let V ( H ) = { , . . . , } and E ( H ) = { (1 , , , , , (4 , , , , , (7 , , , , } . This construction resemblesthe circular motif of three hares , see Figure 1. Now we can add the edges (9 , , , , , , , , , , , ,
8) to H to get the unique 2-intersecting hypergraph with 10 vertices and 6 edges. This hypergraphis 3-regular and cannot be covered with 2 vertices, hence τ = 3, see Figure 1. (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) Figure 1: The three hares. The smallest 2-intersecting 5-uniform hypergraph without a 2-cover.We next show that any 2-intersecting 5-uniform hypergraph with at most 10 edges can be covered byat most 3 vertices. If there are 7 edges, then there is a vertex of degree at least 3, and the leftover 4 edgescan be greedily covered by 2 vertices. If there are 8 edges, then there is a vertex of degree at least 4 byLemma 2.5 with k = 2. The remaining 4 edges can be greedily covered by 2 vertices. If there are 9 edges,6hen we apply Lemma 2.5 with k = 3, and find a vertex of degree at least 5. The remaining 4 edges canbe greedily covered by 2 vertices. If there are 10 edges, then we apply Lemma 2.5 with k = 3, and finda vertex of degree at least 5. Now if there are 5 edges left, then we find a vertex of degree at least 3.Therefore the last 2 edges can be covered by one of their intersection points.There are two examples with maximum covering number listed in the Preliminaries: the standardexample (cid:0) (cid:1) with 8 vertices and 56 edges and the Paley biplane with 11 vertices and 11 edges. One guessesthere might be more examples in between. That is, we expect the number of vertices to be 9 or 10 and thenumber of edges between 11 and 56. We know that our hypothetical example must satisfy the followingproperties: Every vertex has degree at least 5 and at most e − n \ e 11 12 13 14 15 169 134.252.82210 1 17 462 7965 196.514As one can see, there are many candidates. However, checking their covering number, none of themwas an extremal example. We continued the search with increasing number of edges, but it was impossibleto get an exhaustive result within reasonable time frame. We found the list of known biplanes on Gordon Royle’s Home Page [15]. He gives each biplane explicitlyas a list of blocks. We refer to each example with Royle’s notation. There are four biplanes of uniformity 9(order 7).B9A has a 7-cover { , , , , , , } .Its dual B9A ∗ has a 7-cover { , , , , , , } .B9B has a 7-cover { , , , , , , } .We could not manually find a 7-cover of the last biplane, called B9C in Royle’s list. However, ourprogram showed a cover in a split of a second. Namely { , , , , , , } . It took more time (4-5minutes) for the program to show that indeed 7 is the covering number.There are five biplanes of uniformity 11.B11A has a 9-cover { , , , , , , , , } .B11B has a 9-cover { , , , , , , , , } .B11C has a 9-cover { , , , , , , , , } .B11D has a 9-cover { , , , , , , , , } .B11E has a 9-cover { , , , , , , , , } .There is only one biplane of uniformity 13 known.B13A has an 11-cover { , , , , , , , , , , } .The conclusion is simple: the known biplanes of order 7, 9 and 11 do not have maximal covering number. Erd˝os and Lov´asz proved in [7] the following lower bound for intersecting r -uniform hypergraphs withmaximum covering number: q ( r ) ≥ r −
3, where q ( r ) denotes the minimum number of edges in anintersecting r -uniform hypergraph with maximal covering number. There has been no improvement onthis bound for 45 years. We use the following simple facts numerous times: Observation 6.1.
An intersecting r -uniform hypergraph with maximal covering number has minimumdegree at least . It took 28.1 hours to generate and 13 hours to check the covering number of the examples with 9 vertices and 15 edges It took 14 minutes to find an 11-cover by the cover calculator. roof. Assume to the contrary that e is a hyperedge containing a vertex v of degree 1. Now the verticesof e except v form a cover of size r − Observation 6.2.
Let H be an intersecting hypergraph with e edges. If the maximum degree is ∆ , then τ ≤ (cid:100) e − ∆2 (cid:101) .If H is r -uniform and τ = r , then ∆ ≤ e − r − if e − ∆ is even, and ∆ ≤ e − r −
1) + 1 otherwise.Proof.
We can build a cover using the vertex of maximum degree and greedily taking the intersection oftwo lines. -uniform case The lower bound is (cid:100) r − (cid:101) = 8. However, this bound cannot be achieved, as it was first shown by Tripathi[16]. He also constructed an example with 9 edges. In what follows, we argue that the example is unique.We searched for an intersecting, 4-uniform hypergraph with 9 edges and satisfying the necessary degreeconditions, and found the following: H with maximal covering number existed with9 edges and at least 14 vertices. Let us double count the intersecting ordered pairs of edges. There are9 ∗ d gives rise to d ( d − ∗ ∗
12 + 6 ∗ ∗ ∗ ∗ .2 The -uniform case The lower bound gives (cid:100) r − (cid:101) = 11. There must be at least 11 edges in H . However, this very case canbe excluded. Proposition 6.3.
Any -uniform intersecting hypergraph with edges can be covered by at most vertices.Proof. By Observation 6.2 we may assume the maximum degree is at most 4. Suppose to the contrary that H is a 5-uniform intersecting hypergraph with maximal covering number, 11 edges and a vertex v of degree4. Notice that H \ v has 7 edges, and therefore there is a vertex v of degree 3 in H \ v . However, theremaining 4 lines can be covered by 2 vertices, contradicting having maximal covering number. Therefore,we are seeking a 5-uniform, intersecting, maximum degree 3 hypergraph H with covering number 5. Notice,that there must be at least 19 vertices in H by the hand-shake lemma, since 18 ∗ < ∗
11. However, usingthe double counting argument on the intersecting pairs of edges gives a contradiction as before: There areat most 17 vertices of degree 3, since the sum of the degrees is 55. In this case, the number of pairs countedat the vertices is at most 17 ∗ ∗ ∗
10 = 110. Therefore, 11edges can be excluded.Next, we assume that H is an intersecting 5-uniform hypergraph with 12 edges and maximal coveringnumber. Just as in the previous paragraph, we notice that H cannot have a vertex of degree 5. The hand-shake lemma shows that maximum vertex degree 4 implies there must be at least 15 vertices. (Otherwise14 ∗ < ∗
18 12 1.814.037 136.052.775 1.733.108.268 019 12 7.483 598.640 7.594.138 22.5 days 020 12 39 1190 15.069 39 days 0We complement the above computer results by the following
Lemma 6.4.
Every -uniform, intersecting hypergraph with edges and at least vertices has coveringnumber at most .Proof. Suppose to the contrary that such a hypergraph with covering number 5 exists. We consider theLevi graph of H , and use the following double counting argument. Since the hypergraph is intersecting, wecan calculate the number of intersecting pairs of edges as 12 ∗
11 = 132. Here every pair is counted twice.On the other hand, if a vertex v in V has degree d , then there are d ( d −
1) pairs intersecting in that vertex.In our case, every vertex degree must be between 2 and 4 by Observation 6.1 and 6.2. More precisely, ifthere was a vertex v of degree at least 5, then we put v into a cover C . Now removing v and all edgescontaining v from H yields a 5-uniform, intersecting hypergraph H (cid:48) with 7 edges. This subhypergraphstill contains a vertex v of degree at least 3. We put v into C . The vertices v and v together cover 8edges. Therefore, we can greedily extend C into a cover of size 4, a contradiction.Since there are at least 21 vertices, we get the maximum number of intersecting pairs, if there are 9vertices of degree 4 and 12 vertices of degree 2. If we either have more vertices or lower vertex degrees, itresults in a decrease of the intersecting pairs (since the number of edges is fixed). Now simple calculation It took 2 days to check by the covering number calculator. ∗
12 + 12 ∗ | ) pairwise balanced design,and there is a unique such object the dual of AG (2 , L and the intersection of the two lines parallel to L .All these human and computer results together imply Corollary 6.5.
The smallest number of edges a -uniform intersecting hypergraph with maximum coveringnumber can have is at least . Next we set the number of edges to be 13. The number of vertices must be at least 17 by the hand-shakelemma, since 64 = 16 ∗ < ∗
13 = 65. We made nauty to look for an intersecting hypergraph with 17vertices and 13 edges. There were 670914 graphs found in 21.5 days. Our covering number calculatorfound in 5 hours that three of the candidates have covering number 5. We give the adjacency matrix ofthe first one, B say.11110000000001100000000110101010010000010000100010011000001110000010110100000001000101000100100000011001001101001000000100010010100010001000101000100010101000010001001010000111000000000010001001100001000011000000010010100For completeness, we argue that the covering number is maximal. Lemma 6.6.
The above intersecting hypergraph B has covering number 5.Proof. Suppose to the contrary that there is a cover C = { v , v , v , v } of size 4. Since the maximumvertex degree is 4, in principle, there might be two possibilities. Either the vertices cover 4 + 3 + 3 + 3 or4 + 4 + 3 + 2 edges in this order. In the former case, consider three vertices of degree 4 such that there areno two of them covering disjoint edges. Then either they cover at most 9 lines or v , v , v belong to thesame edge, in which case they cover 10 edges. Hence v must cover 3 edges and v , v , v , v belong to thesame edge. Also in the first case, there must be two vertices covering 4 edges each. Now we simply have tomake a list of pairs of vertices of degree 4 satisfying this. In the above matrix, there are 7 such pairs: 1-15,2-9, 3-8, 4-6, 7-11,10-13, 11-12. We can manually check that in each case the remaining 5 edges cannot becovered by two vertices. Theorem 6.7.
The smallest number of edges a -uniform intersecting hypergraph with maximal coveringnumber can have is . p.269 in the Handbook of Combinatorial Designs. Erd˝os-Lov´asz in projective planes
In the same paper [7], Erd˝os and Lov´asz studied subset of lines in projective planes also. They explicitlyposed the question whether one needs more than a linear number of lines to ensure the covering numberis maximal. Kahn [10] proved that indeed r log r is the correct order of magnitude. One may define m ( r )to be the minimum number of lines in P G (2 , r −
1) that cannot be covered by r − . Let us study m ( r ) for small values and compare it to q ( r ). r = 3: In Section 6, we defined the minimal example as part of the Fano plane. Therefore, q (3) = m (3) = 6.We recall our earlier result, which we use in the next proof. Lemma 7.1 (The oval construction [1]) . In P G (2 , r − , there exists a set of r + r lines with coveringnumber r . r = 4: Notice the example we found in subsection 6.1 contained two lines with intersection size two.Therefore, we need to study lines of P G (2 ,
3) more closely. What is the smallest subset of lines in
P G (2 , Lemma 7.2.
Any lines of P G (2 , can be covered by points.Proof. Let H be a set of 9 lines in P G (2 , v ofdegree 4 in H , then we can find a cover of size 3 as follows. Let I denote the hypergraph formed by theremaining 5 lines of H . We may assume the first 3 lines form a triangle ABC . How can we add 2 more linesto create I without a vertex of degree 3? We must use the remaining points on the sides of the triangle: A , A , B , B , C , C . There is essentially only one way to do it. Let the new lines be: A , B , C , D and A , B , C , D . However, this configuration is not part of a projective plane. We cannot add lines through A , C without violating the axioms.In what follows, we assume that H has maximum degree 3. How can one remove 4 lines of P G (2 ,
3) toensure that each vertex degree is decreasing? There is only one way to do so: we have to remove a pencilof lines through a fixed point F . What remains now is the dual of the affine plane AG (2 , F . Corollary 7.3. m (4) = 10 . r = 5: We have to work in P G (2 , P G (2 , q ) has q + q + 1lines and it has covering number q + 1. We perform the following three-step algorithm:(i) Delete an edge from the current hypergraph in all possible ways such that the minimum degree of eachvertex remains at least 2.(ii) Check isomorphism and only keep non-isomorphic examples.(iii) Determine the covering number. Keep only the maximal ones.We stop if τ < q + 1 for all examples.Clearly this determines m ( r ). Asymptotically it must be much better than the oval construction. Wewondered how far we can go with our resources.It takes only a few minutes to consider subhypergraphs of P G (2 , One may go even further and define a similar function for any projective plane of order r −
1. We do not consider thisline here and now. For these values, there are only Desarguesian planes. orollary 7.4. m (5) = 14 . Here we give the point-line incidence matrix for one example:110000000000000011000000000000001100000000000000111100000000000000111100000010001000001000010001000000100010001000010100010001000000001000010010000001001000000110000100000110010010001000100001010010100100101000100000010000011001001000001001000101000010011010100000010100000001100001010000100100 r = 6: We have to work in P G (2 , P G (2 ,
5) with 21 edges, 112 of them has maximal coveringnumber. In one more step, we found 178 subhypergraphs with 20 edges in 56 minutes, 99 of them hasmaximal covering number. Next, we found 207 subhypergraphs with 19 edges in 42 minutes, 23 of themhas maximal covering number. Finally, there is a unique example with 18 edges and maximal coveringnumber.
Corollary 7.5. In P G (2 , , there is a unique set of 18 lines that cannot be covered with fewer than points: m (6) = 18 . m (8) is smaller than +72 = 28. However, due to storage constraints, we could only go exhaustively toPG(2,7)-13 edges. To find a configuration smaller than 28 edges requires someone with more resources ora different idea. Final remarks
In some geometric problems, it makes better sense to define a blocking set of an r -uniform intersectinghypergraph as a cover, which does not contain an edge of H . There is a function similar to q ( r ) definedas follows. We define an r -uniform intersecting hypergraph H to be maximal , if there is no blocking set ofsize at most r . Every cover of size at most r is an edge. Let q ∗ ( r ) be the minimum number of edges in amaximal r -uniform intersecting hypergraph. The original Erd˝os-Lov´asz lower bound of r − q ∗ ( r ) ≥ r if r ≥
4. However, theirproof idea does not work for q ( r ). They also determined the following exact value: q ∗ (4) = 12, which is incontrast to q (4) = 9. Acknowledgement
We thank M´at´e B´ar´any for stimulating discussions and writing the covering number calculator for us. Wethank Brendan McKay for all help using nauty, Marston Conder and Gordon Royle for valuable remarks.We thank the two anonymous reviewers for pointing out reference [16], the table in p.269 of the Handbookof Combinatorial Designs, and their constructive criticism that improved the presentation.
References [1] R. Aharoni, J. Bar´at, I.M. Wanless. Multipartite hypergraphs achieving equality in Ryser’s conjecture.
Graphs Combin. (2016), 1–15.[2] A. Bishnoi, S. Das, P. Morris, T, Szab´o. Ryser’s Conjecture for t -intersecting hypergraphs. manuscript retrieved on 3 Nov 2020. https://arxiv.org/abs/2001.04132 [3] S. Bustamante and M. Stein. Monochromatic tree covers and Ramsey numbers for set-coloured graphs. Disc. Math. (2018) 266–276. http://arxiv.org/abs/1510.05190v3 retrieved on 3 Jul 2017.[4] DistanceRegular.org manuscript https://arxiv.org/pdf/2009.07239.pdf retrieved on 3 Nov 2020.[6] S.J. Dow, D.A. Drake, Z. F¨uredi and J.A. Larson. A lower bound for the cardinality of a maximalfamily of mutually intersecting sets of equal size.
Congressus Numerantium (1985), 47–48.[7] P. Erd˝os and L. Lov´asz. Problems and results on 3-chromatic hypergraphs and some related questions. Infinite and finite sets (Colloq., Keszthely, 1973; dedicated to P. Erd˝os on his 60th birthday), Vol. II,pp. 609–627; Colloq. Math. Soc. J´anos Bolyai, Vol. 10, North-Holland, Amsterdam, 1975.[8] N. Francetiˇc, S. Herke, B.D. McKay and I.M. Wanless. On Ryser’s conjecture for linear intersectingmultipartite hypergraphs.
Europ. J. Combin.
61 (2017), 91–105.[9] J.R. Henderson. Permutation Decompositions of (0,1)-matrices and decomposition transversals, Ph.D.Thesis, Caltech (1971), http://thesis.library.caltech.edu/5726 [10] J. Kahn. On a problem of Erd˝os and Lov´asz: random lines in a projective plane.
Combinatorica (1992), 417–423.[11] J. Kahn. On a problem of Erd˝os and Lov´asz. II. n ( r ) = O ( r ). J. Amer. Math. Soc. (1994), 125–143.[12] Z. Kir´aly and L. T´othm´er´esz. On Ryser’s conjecture for t -intersecting and degree-bounded hyper-graphs. Electron. J. Combin.
24 (2017), no. 4, Paper 4.40, 15 pp. http://arxiv.org/abs/1705.10024v2 retrieved on 9 Dec 2017.[13] B.D. McKay and A. Piperno. Practical graph isomorphism II.
J. Symbolic Comput. (2014), 94–112.[14] B.D. McKay, personal communication[15] G. Royle’s home page on Biplanes. https://staffhome.ecm.uwa.edu.au/~00013890/remote/biplanes/ [16] A. Tripathi. A result on intersecting families with maximum transversal size. manuscript https://arxiv.org/pdf/1409.4610.pdf [17] Wikipedia on Biplanes. https://en.wikipedia.org/wiki/Block_designhttps://en.wikipedia.org/wiki/Block_design