Ryser's Conjecture for t -intersecting hypergraphs
aa r X i v : . [ m a t h . C O ] N ov Ryser’s Conjecture for t -intersecting hypergraphs Anurag Bishnoi ∗ Shagnik Das †‡ Patrick Morris †§ Tibor Szab´o †¶ November 30, 2020
Abstract
A well-known conjecture, often attributed to Ryser, states that the cover number of an r -partite r -uniform hypergraph is at most r − t -intersecting, conjecturing that the cover number τ ( H ) of such a hypergraph H is at most r − t . Inthese papers, it was proven that the conjecture is true for r ≤ t −
1, but also that it need not besharp; when r = 5 and t = 2, one has τ ( H ) ≤ t ≥ r ≤ t −
1, we prove atight upper bound on the cover number of these hypergraphs, showing that they in fact satisfy τ ( H ) ≤ ⌊ ( r − t ) / ⌋ + 1. Second, we extend the range of t for which the conjecture is known tobe true, showing that it holds for all r ≤ t −
5. We also introduce several related variations onthis theme. As a consequence of our tight bounds, we resolve the problem for k -wise t -intersectinghypergraphs, for all k ≥ t ≥
1. We further give bounds on the cover numbers of strictly t -intersecting hypergraphs and the s -cover numbers of t -intersecting hypergraphs. We define an r -uniform hypergraph H to be r -partite if one can partition the vertex set into r parts,say V ( H ) = P ⊔ · · · ⊔ P r , such that for all e ∈ E ( H ) and all j ∈ [ r ], we have | e ∩ P j | = 1. For1 ≤ t ≤ r −
1, an r -uniform hypergraph is t -intersecting if | e ∩ f | ≥ t for all e, f ∈ E ( H ). We will referto r -uniform r -partite t -intersecting hypergraphs as ( r, t ) -graphs throughout.For an r -uniform hypergraph H , a set of vertices C ⊂ V ( H ) is a cover of the hypergraph if C ∩ e = ∅ for all e ∈ E ( H ). The cover number of the hypergraph, denoted τ ( H ), is the smallest cardinality of acover for the hypergraph. Our goal is to bound the cover numbers of ( r, t )-graphs. Definition 1.1.
We define the following extremal function:Ryser( r, t ) = max { τ ( H ) : H is an ( r, t )-graph } . The problem is motivated by an old unsolved conjecture of Ryser from around 1970, first appearingin the PhD thesis of his student Henderson [20] (see [4] for more on the history of this conjecture), whichclaims that the cover number of any r -uniform r -partite hypergraph H satisfies τ ( H ) ≤ ( r − ν ( H ).Here, ν ( H ) denotes the matching number of the hypergraph, that is, the size of the largest set ofpairwise disjoint edges. When r = 2, this is a statement about bipartite graphs, and is equivalent ∗ Department of Applied Mathematics, Technische Universiteit Delft, Delft, Netherlands. E-mail: [email protected] . Research supported in part by a Humboldt Research Fellowship for Postdoctoral Re-searchers and by a Discovery Early Career Award of the Australian Research Council (No. DE190100666). † Institut f¨ur Mathematik, Freie Universit¨at Berlin, 14195 Berlin, Germany. ‡ E-mail: [email protected] . Research supported in part by GIF grant G-1347-304.6/2016 and by theDeutsche Forschungsgemeinschaft (DFG) - Project 415310276. § E-mail: [email protected] . Research supported in part by a Leverhulme Trust Study Abroad Studentship(SAS-2017-052 \
9) and by the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation) under Germany’sExcellence Strategy - The Berlin Mathematics Research Center MATH+ (EXC-2046/1, project ID: 390685689). ¶ E-mail: [email protected] . Research supported in part by GIF grant G-1347-304.6/2016.
1o the classic theorem of K¨onig [25]. Although Ryser’s Conjecture has attracted significant attentionover the years, the only other resolved case is r = 3. This was proven via topological methods byAharoni [3], with the extremal hypergraphs classified by Haxell, Narins and Szab´o [18].In this latter result, it was shown that the extremal hypergraphs with ν ( H ) = ν ≥ ν extremal intersecting hypergraphs. Thus, much research in this direction hasfocussed on intersecting hypergraphs. Here we have ν ( H ) = 1, and Ryser’s Conjecture assertsRyser( r, ≤ r −
1. Further motivation for considering the intersecting case comes from a connectionwith a conjecture of Gy´arf´as [17] which states that the vertices of any r -edge-coloured clique can becovered by at most r − r ≤ r, ≥ r − r − √ r ) many non-isomorphic minimal examples whenever r − r , Haxell and Scott [19] construct nearly-extremal intersecting hypergraphs;more precisely, they show Ryser( r, ≥ r − r large enough.This led Bustamante and Stein [6] and, independently, Kir´aly and T´othm´er´esz [23] to investigatewhat occurs when we impose the stricter condition of the hypergraph H being t -intersecting. Inthis case, any subset of r − t + 1 vertices from an edge must form a cover, and so we trivially have τ ( H ) ≤ r − t + 1. While one can construct r -uniform t -intersecting hypergraphs attaining this bound,it was conjectured that, as in Ryser’s Conjecture, one can do better when the hypergraph is also r -partite; that is, when considering ( r, t )-graphs. Conjecture 1.2 (Bustamante–Stein [6], Kir´aly–T´othm´er´esz [23]) . For all ≤ t ≤ r − , we have Ryser( r, t ) ≤ r − t. Note that while Ryser’s Conjecture for intersecting hypergraphs is a special case ( t = 1), it infact implies Conjecture 1.2. Indeed, Ryser( r, t ) ≤ Ryser( r − t + 1 ,
1) since deleting t − r − t + 1 , r − t .Therefore, one might hope to be able to make progress on Conjecture 1.2 for larger values of t ,and indeed, results have been obtained when t is linear in r . Bustamante and Stein [6] proved theconjecture for r ≤ t + 2, with Kir´aly and T´othm´er´esz [23] extending this to r ≤ t −
1. With regardsto lower bounds on Ryser( r, t ), the conjecture is trivially tight for t = r −
1, and Bustamante andStein [6] showed that it is also tight for t = r −
2. However, they demonstrated that it is not alwaysbest possible by proving Ryser(5 ,
2) = 2. More generally, they proved Ryser( r, t ) ≥ Ryser( ⌊ rt ⌋ ,
1) byobserving that replacing every vertex of an ( r ′ , t vertices gives an ( r ′ t, t )-graph.Given the aforementioned results on Ryser’s conjecture, this shows Ryser( r, t ) ≥ ⌊ r/t ⌋ − r, t ), and Bustamante and Stein suggested this lower bound should be closer to the truth thanthe upper bound of Conjecture 1.2. Our first result shows that the lower bound of Bustamante and Stein is, in fact, far from optimal.Indeed, we provide a construction, valid for all t and r , that greatly improves on the previous lowerbound when t ≥ Theorem 1.3.
For all ≤ t ≤ r we have Ryser( r, t ) ≥ (cid:22) r − t (cid:23) + 1 . Although recent constructions of Abu-Khazneh [1], for ν = 2 and r = 4, and Bishnoi and Pepe [5], for ν ≥ r ≥ r −
2e next prove a matching upper bound when t is large, showing that when r is less than thrice t ,the true value of Ryser( r, t ) is half the bound of Conjecture 1.2. Theorem 1.4.
For t, r ∈ N such that t + 1 ≤ r ≤ t − , we have Ryser( r, t ) = (cid:22) r − t (cid:23) + 1 . Theorem 1.3 gives the lower bound needed for Theorem 1.4, and hence all that is required isa matching upper bound. In fact, using different arguments, we are able to prove a few upperbounds on Ryser( r, t ). The theorem below collects the best upper bounds (excluding the trivialRyser( r, t ) ≤ r − t + 1) that we have in various ranges of the parameters. Theorem 1.5.
Let ≤ t ≤ r . Then Ryser( r, t ) ≤ (cid:4) r − t (cid:5) + 1 if t ≤ r ≤ t − , r − t + 2 if t ≤ r ≤ t , (cid:4) r − t (cid:5) + 2 if t ≤ r ≤ t − , (cid:4) r − t (cid:5) + 3 if t − ≤ r ≤ t − . In particular, the first case of the theorem gives the upper bound needed for Theorem 1.4. Tovisualise our results, it helps to focus on the asymptotics when t is linear in r . To this end, we definethe function f ( α ) := lim r →∞ Ryser( r,αr ) r . Theorem 1.5 can then be seen as a piecewise linear upper boundon f ( α ). Figure 1 summarises our knowledge of f ( α ): we know it exactly for α ≥ , while we can stillstrongly restrict f ( α ) for smaller values of α . αf ( α )
736 13 r − t + 1)Upper bound of Theorem 1.5Lower bound of Theorem 1.3Possible values of f ( α )Figure 1: The asymptotics of Ryser( r, αr ).In particular, we extend the results of Bustamante and Stein [6] and Kir´aly and T´othm´er´esz [23]by showing Conjecture 1.2 continues to hold for smaller values of t , and that it is not tight in most ofthese cases. Corollary 1.6.
Conjecture 1.2 holds for all but eight pairs ( r, t ) satisfying r ≤ t − . Furthermore,the conjecture is not tight (that is, Ryser( r, t ) ≤ r − t − ) for all but pairs where t + 3 ≤ r ≤ t − . The exceptional pairs, for which Conjecture 1.2 remains open, are (12 , , (13 , , (16 , , (17 , , (18 , , (22 , , (23 , , t − comes from comparing when the upper bound on Ryser( r, t ) in the fourthrange of Theorem 1.5 falls strictly below r − t + 1, using the fact that the cover number must be aninteger. This comparison describes exactly when our bounds become trivial for all large values of r and t . However, for small values of r and t , we may fall in other ranges and so have to check a finitenumber of cases for exceptions. In doing so, we note that for certain values of ( r, t ) we must appeal tothe previously mentioned result of Kir´aly and T´othm´er´esz [23] who proved the conjecture whenever r ≤ t −
1. The range and exceptions for when the conjecture is not tight are also calculated in asimilar manner, using Theorem 1.5.These results beg the question of what the true value of Ryser( r, t ) should be; we discuss thisfurther in Section 3, and propose a new conjecture in Conjecture 3.1. k -wise intersecting hypergraphs In the above results, we require that all pairwise intersections of the edges of the hypergraphs have sizeat least t . A natural stronger condition is to impose the same restriction on all k -wise intersectionsof edges, rather than just pairwise. This setting has often been studied in the extremal combinatoricsliterature. Frankl [11] first studied such hypergraphs, determining the maximum number of edgespossible when all k -wise intersections are non-empty. S´os [28] then raised the problem of finding thelargest hypergraphs where the sizes of all k -wise intersections lie in some set L , and various results inthis direction were obtained by F¨uredi [13], Vu [31, 32], Grolmusz [15], Grolmusz and Sudakov [16],F¨uredi and Sudakov [14] and Szab´o and Vu [29].We say a hypergraph H is k -wise t -intersecting if, for any edges e , e , . . . , e k ∈ E ( H ), we have |∩ ki =1 e i | ≥ t . Following Ryser’s Conjecture, we study how much smaller a cover one is guaranteedto find in r -uniform r -partite hypergraphs satisfying the more restrictive condition of being k -wise t -intersecting. In a stroke of serendipity, the range of intersection sizes for which Theorem 1.4 holdsis precisely what is needed to give an exact answer in this setting. Theorem 1.7.
Let H be an r -uniform r -partite k -wise t -intersecting hypergraph. If k ≥ and t ≥ ,or k = 2 and t > r , then τ ( H ) ≤ (cid:22) r − tk (cid:23) + 1 , and this bound is best possible. Organisation of the paper.
We prove the above theorems in the following section: the lowerbound of Theorem 1.3 is derived in Section 2.1, the upper bounds of Theorem 1.5 are proven inSection 2.2, and Theorem 1.7 is deduced in Section 2.3. Thereafter we suggest several directions forfurther research in Section 3 by presenting initial results on variants of the problem where we requirethe hypergraphs to be strictly t -intersecting or we try to cover each edge of an ( r, t )-graph at least s times. In this section we prove our main results, Theorems 1.3, 1.5 and 1.7, by establishing lower (Section 2.1)and upper (Section 2.2) bounds on the extremal function Ryser( r, t ), and then extending them to k -wise t -intersecting hypergraphs (Section 2.3). To obtain the lower bound, and thereby prove Theorem 1.3, we need to construct ( r, t )-graphs withlarge cover numbers. The hypergraphs we consider are of the following form.
Definition 2.1.
For 0 ≤ ℓ ≤ r −
1, we define H rℓ to be the following r -uniform r -partite hypergraph.Let m = (cid:0) rr − ℓ (cid:1) and fix some ordering (cid:0) [ r ] r − ℓ (cid:1) = { S , . . . , S m } of the ( r − ℓ )-subsets of [ r ]. We define V ( H rℓ ) := { , , . . . , m } × [ r ] and E ( H rℓ ) := { e i : i ∈ [ m ] } , e i = { (0 , j ) : j ∈ S i } ∪ { ( i, j ) : j ∈ [ r ] \ S i } , for each i ∈ [ m ].Note that H rℓ is indeed r -partite with parts P j = { ( i, j ) : 0 ≤ i ≤ m } for 1 ≤ j ≤ r . We now showthat choosing ℓ appropriately gives a construction verifying Theorem 1.3. Proposition 2.2.
For ≤ ℓ ≤ ⌊ r − ⌋ , H rℓ is ( r − ℓ ) -intersecting with τ ( H rℓ ) = ℓ + 1 .Proof. To see that H rℓ is ( r − ℓ )-intersecting, observe that each edge of H rℓ misses exactly ℓ verticesfrom the set L = { (0 , j ) : j ∈ [ r ] } . It then follows that, for any two edges e i , e i ′ ∈ E ( H rℓ ), we have | e i ∩ e i ′ | ≥ | e i ∩ e i ′ ∩ L | ≥ | L | − ℓ = r − ℓ. We now establish the cover number of H rℓ . Assume for a contradiction that H rℓ has a cover C ⊂ V ( H rℓ ) of size c ≤ ℓ . First we show that we may assume that C ⊂ L . Indeed, if v = ( i, j ) ∈ C forsome i ≥
1, then the only edge that could contain v is e i . If we replace v with any vertex in e i ∩ L ,the modified set C has not increased in size and still covers H rℓ .Since | C | = c ≤ ℓ , we have | L \ C | ≥ r − ℓ and hence there exists an i ∗ ∈ [ m ] such that { } × S i ∗ ⊆ L \ C . Then e i ∗ ∩ C = ∅ , contradicting the fact that C is a cover. Thus τ ( H rℓ ) ≥ ℓ + 1.To see we have equality, note that any subset of ℓ + 1 vertices in L forms a cover.The key property needed in the above proof is that, for each subset S ⊆ L of size r − ℓ , thereis an edge of H rℓ intersecting L exactly at the vertices of S . Our construction is edge-minimal withrespect to this key property, and further ensures that all vertices not in L have degree at most one,leading to an easy proof of the cover number.However, as long as the key property is maintained, there is great flexibility in how the rest ofthe hypergraph is constructed. For instance, one can instead make it vertex-minimal, having partsof size ℓ + 1 rather than (cid:0) rr − ℓ (cid:1) + 1, so that each part P j also forms a minimum vertex cover. Weomit the details of this construction for the sake of brevity, as we already have all we need to proveTheorem 1.3. Proof of Theorem 1.3.
Set ℓ = ⌊ r − t ⌋ . By Proposition 2.2, H rℓ is ( r − ℓ )-intersecting, and as r − ℓ ≥ t ,it follows that H rℓ is an ( r, t )-graph. The proposition further asserts that τ ( H rℓ ) = ℓ + 1, and thusRyser( r, t ) ≥ τ ( H rℓ ) = (cid:22) r − t (cid:23) + 1 . In this section we prove Theorem 1.5. These upper bounds are derived from a sequence of resultsobtained by considering configurations of two or three edges of the hypergraph. Before proceeding,we fix some notation that will be useful in what follows.
Definition 2.3.
Let H be an r -uniform r -partite hypergraph with parts P j for 1 ≤ j ≤ r . Supposethat e , . . . , e k ∈ E ( H ). Then for v ∈ V ( H ), we define d ( v ; e , . . . , e k ) = |{ i ∈ [ k ] : v ∈ e i }| to be the degree of v with respect to the k edges e , . . . , e k . Also, given a vertex subset C ⊆ V ( H )not wholly containing any part P j , we define∆ H ( C ; e , . . . , e k ) = r X j =1 max v ∈ P j \ C d ( v ; e , . . . , e k )to be the maximum sum of degrees (with respect to e , . . . , e k ) when we take one vertex from eachpart and avoid C .The utility of this definition comes from the following easy observation, which we use repeatedlyin the subsequent proofs. 5 bservation 2.4. Suppose that H is an r -uniform r -partite hypergraph, C ⊂ V ( H ) and e , . . . , e k ∈ E ( H ) are edges of H . Now if f ∈ E ( H ) and f ∩ C = ∅ , we have k X i =1 | f ∩ e i | ≤ ∆ H ( C ; e , . . . , e k ) . (1)Consequently, if H is t -intersecting and ∆ H ( C ; e , . . . , e k ) ≤ kt −
1, then C is a cover for H . Indeed,there can be no f ∈ E ( H ) disjoint from C , as by (1) and the pigeonhole principle, there would besome i ∈ [ k ] for which | e i ∩ f | ≤ t −
1, contradicting H being t -intersecting.Armed with this observation, we can prove upper bounds on the cover numbers of ( r, t )-graphs.In the following lemma, we begin by constructing a cover consisting of vertices lying in two edges ofsuch a hypergraph. Lemma 2.5.
Let H be an ( r, t ) -graph, let e , e ∈ E ( H ) , and set t ′ = | e ∩ e | ≥ t . Then τ ( H ) ≤ (j r − t ′ k + t ′ − t + 1 if r − t + 1 ≤ t ′ ≤ r, r − t − t ′ + 2 if t ≤ t ′ ≤ r − t. Proof.
Let the parts of H be P j for j ∈ [ r ]. If t ′ ≥ r − t + 1 then s = ⌊ r − t ′ ⌋ + t ′ − t + 1 satisfies1 ≤ s ≤ t ′ . We claim that an arbitrary set C of s vertices from e ∩ e is a cover. Indeed,∆ H ( C ; e , e ) = X j ∈ [ r ]: e ∩ P j = e ∩ P j ,e ∩ P j / ∈ C X j ∈ [ r ]: e ∩ P j = e ∩ P j | ( e ∩ e ) \ C | + r − t ′ = r + t ′ − s ≤ t − , and thus the conclusion follows from Observation 2.4.Now consider the case where t ′ ≤ r − t . Without loss of generality, we may assume the intersectionof e and e is contained in the first t ′ parts, labelling the vertices of e as { u , . . . , u r } and of e as { u , . . . , u t ′ , v t ′ +1 , . . . , v r } , where u j , v j ∈ P j for all j . Letting C = { u , . . . , u t ′ } ∪ r − t +1 [ j = t ′ +1 { u j , v j } , we have ∆ H ( C ; e , e ) = 2 t −
1. By Observation 2.4, we can again deduce that τ ( H ) ≤ | C | =2( r − t + 1) − t ′ = 2 r − t − t ′ + 2.Lemma 2.5 will suffice to prove the first two parts of Theorem 1.5. For the latter parts, we shallneed to consider covers consisting of vertices lying in three edges instead. Before proceeding, though,we present a reformulation of Lemma 2.5 that will be more convenient for later proofs. Corollary 2.6.
Let η ∈ N and let H be an ( r, t ) -graph such that τ ( H ) ≥ η + 1 . Then, for all e, f ∈ E ( H ) , we have that either ( i ) | e ∩ f | ≥ η + 2 t − r , or ( ii ) | e ∩ f | ≤ r − t − η + 1 .Proof. Let e, f ∈ E ( H ) be an arbitrary pair of edges of H and take t ′ = | e ∩ f | . If t ′ ≥ r − t + 1, thenwe claim that in fact t ′ ≥ η + 2 t − r . Indeed, if t ′ ≤ η + 2 t − r −
1, then Lemma 2.5 implies that τ ( H ) ≤ (cid:22) r − t ′ (cid:23) + t ′ − t + 1 ≤ r + t ′ − t + 1 ≤ η + 12 , a contradiction.Hence, if ( i ) is not satisfied for t ′ = | e ∩ f | , we must have t ′ ≤ r − t . By Lemma 2.5, it followsthat η + 1 ≤ τ ( H ) ≤ r − t − t ′ + 2, from which we deduce that t ′ ≤ r − t − η + 1.6he following lemma is the analogue of Lemma 2.5 when constructing covers from vertices thatlie in three fixed edges, as opposed to only using two edges. Lemma 2.7.
Let r ≥ t , let H be an ( r, t ) -graph, and let e , e , e ∈ E ( H ) . Set t = | e ∩ e ∩ e | and t = | e ∩ e \ e | + | e ∩ e \ e | + | e ∩ e \ e | . Then τ ( H ) ≤ (2 t + t + r − t + 3) if r − t + 1 + t ≤ t ≤ r,r − t + 1 + t if r − t + 1 − t ≤ t ≤ r − t + 1 + t , r − t − t − t + 3 if ≤ t ≤ r − t + 1 − t . Proof.
Observe that t counts the number of vertices that are in all three edges, while t counts thenumber of vertices in precisely two of the three edges. Let T = e ∩ e ∩ e be the set of t verticescontained in all three edges and let D be the set of t vertices contained in exactly two of the threeedges. When building a transversal of the parts that intersects e , e and e as much as possible, itis optimal to choose as many vertices from T as possible, followed by vertices from D . Therefore, tominimise ∆ H ( C ; e , e , e ), we will choose C so as to first block the vertices in T , followed by thosein D .Let us first consider the case when t ≥ r − t + 1 + t . Setting s = t − (cid:4) ( t − t − r + 3 t − (cid:5) ≤ (2 t + t + r − t + 3), note that 1 ≤ s ≤ t . Taking C to be an arbitrary subset of T of size s , wehave ∆ H ( C ; e , e , e ) = 3 | ( e ∩ e ∩ e ) \ C | + 2 t + ( r − t − t ) ≤ t − , since we cannot select a vertex of e ∪ e ∪ e in the s parts spanned by C . It thus follows fromObservation 2.4 that C is a cover for H , giving the claimed bound on τ ( H ).Next, suppose r − t + 1 ≤ t ≤ r − t + 1 + t , and set s = r − t + 1 + t − t , noting that0 ≤ s ≤ t . Take C = T ∪ S , where S is a subset of D of size s . Consider a transversal ofthe parts that is disjoint from C . There are t − s parts (those intersecting D \ S ) in which thetransversal could intersect up to two of the three edges e , e and e . In all other parts the transversalcan intersect at most one of the three edges and there are t parts (those that intersect T ) in whichthe transversal must be disjoint from all three edges. Thus∆ H ( C ; e , e , e ) = 2( t − s ) + ( r − t − ( t − s )) = 3 t − , and so, by Observation 2.4, C covers H , showing τ ( H ) ≤ | C | = r − t + 1 + t .In the range r − t + 1 − t ≤ t ≤ r − t , set s = r − t + 1 − t , whence 1 ≤ s ≤ t . Wedefine D ′ to be the t vertices which are contained in exactly one of the edges e , e and e and liein parts which intersect D . Now take C = T ∪ D ∪ S where S is an arbitrary subset of D ′ of size s . A transversal disjoint from C can then only meet the edges e , e and e in the t − s verticesof D ′ \ S , as well as in the r − t − t parts where the three edges are pairwise disjoint. We thereforehave ∆ H ( C ; e , e , e ) = ( t − s ) + ( r − t − t ) = 3 t − , and so Observation 2.4 implies C is a cover of the stated size.Finally, we are left with the case when t ≤ r − t + 1 − t , for which we set s = r − t + 1 − t − t .Take C = T ∪ D ∪ D ′ ∪ S , where D and D ′ are defined as above and S consists of the 3 s verticesof e ∪ e ∪ e from s of the parts where the edges e , e and e are pairwise disjoint. Then anytransversal disjoint from C can only meet the three edges in the r − t − t − s parts not spannedby C , from which it follows that ∆ H ( C ; e , e , e ) = 3 t −
1. By Observation 2.4, C is a cover of H ,and so τ ( H ) ≤ | C | = t + 2 t + 3 s = 3 r − t − t − t + 3.By applying Lemma 2.7 in conjunction with Corollary 2.6, we will prove the following upper boundson the cover numbers of ( r, t )-graphs. Proposition 2.8.
For all t ≥ and r ≥ t , we have Ryser( r, t ) ≤ ((cid:6) r − t +24 (cid:7) + (cid:6) t − r − (cid:7) if t ≤ r ≤ t − , (cid:6) r − (cid:7) + (cid:6) r − t +138 (cid:7) if t − ≤ r ≤ t − . Proof of Theorem 1.5.
We prove the theorem by induction on r − t . For the base case, we have t = r .Trivially, an ( r, r )-graph can have at most one edge, and thus can be covered by a single vertex. ThusRyser( r, r ) = 1, as stated in the first case of the theorem.For the induction step, suppose r − t ≥
1, and let H be an ( r, t )-graph of maximum cover number,so that Ryser( r, t ) = τ ( H ). As previously stated, we shall use Lemma 2.5 to prove the first two cases ofthe theorem. To this end, let e , e ∈ E ( H ) be a pair of edges with the smallest intersection. We mayassume that | e ∩ e | = t , as otherwise H is in fact an ( r, t + 1)-graph, and we are done by induction(as our upper bound on Ryser( r, t ) is decreasing in t ).Applying Lemma 2.5 with the edges e and e , we have t ′ = t , from which it follows thatRyser( r, t ) = τ ( H ) ≤ ((cid:4) r − t (cid:5) + 1 if r − t + 1 ≤ t ≤ r, r − t + 2 if t ≤ r − t. Simplifying the ranges for which these bounds hold, we see that the first is valid when r ≤ t −
1, asrequired for the first case of the theorem, while the second bound above is valid provided r ≥ t .The latter two cases of the theorem are direct consequences of Proposition 2.8, which we can applywhenever r ≥ t . To simplify the bounds, we estimate the ceiling terms, obtaining (cid:24) r − t + 24 (cid:25) + (cid:24) t − r − (cid:25) ≤ r − t + 54 + 6 t − r + 68 = 9 r − t (cid:24) r − (cid:25) + (cid:24) r − t + 138 (cid:25) ≤ r + 24 + 9 r − t + 208 = 15 r − t . That is, Proposition 2.8 implies that whenever r ≥ t , we haveRyser( r, t ) ≤ ( r − t + 2 if 3 t ≤ r ≤ t − , r − t + 3 if 5 t − ≤ r ≤ t − . This matches the latter two cases of Theorem 1.5. Finally, we note that the bound r − t + 2 improvesthe earlier bound of 2 r − t + 2 whenever r ≥ t , justifying the endpoints of the ranges of the secondand third cases in the theorem.All that remains is the proof of Proposition 2.8, which we delay no further. Proof of Proposition 2.8.
In the first case, let r and t be such that 3 t ≤ r ≤ t −
2, and define x = (cid:24) r − t + 24 (cid:25) and z = (cid:24) t − r − (cid:25) . Note that 1 ≤ z ≤ t and 0 ≤ x ≤ r − t, (2)using here that 3 t ≤ r ≤ t −
2. Suppose for contradiction that H is an ( r, t )-graph with τ ( H ) ≥ x + z +1.Applying Corollary 2.6, any pair of edges e, f ∈ E ( H ) must satisfy | e ∩ f | ≥ x + 2 z + 2 t − r or | e ∩ f | ≤ r − t − x − z + 1 . (3)Now take e , e ∈ E ( H ) to be two edges such that | e ∩ e | = t (again, we may assume such a pairexists, as otherwise H is an ( r, t + 1)-graph and we obtain a stronger bound on τ ( H )). Let Z ⊆ e ∩ e be a set of z vertices and X ⊆ e \ e a set of x vertices, noting that this is possible due to (2). Wetake Y = X ∪ Z and note that Y intersects both e and e (as z ≥
1) and is not a cover as it has sizeexactly x + z and we assumed that τ ( H ) ≥ x + z + 1. Therefore, there exists an edge e ∈ E ( H ) suchthat e ∩ Y = ∅ . We define a, b, c ∈ N as follows: a = | e ∩ e \ e | , b = | e ∩ e \ e | and c = | e ∩ e \ e | . | e ∩ e ∩ e | = t − a . Using the fact that e ∩ Y = ∅ , we can derive some bounds onthe parameters a, b and c . Indeed, since e is disjoint from Z , a ≥ z . As | e ∩ e | = b + ( t − a ) ≥ t ,we must further have b ≥ a , while considering | e ∩ e | similarly shows c ≥ a . Finally, as e is disjointfrom X , we must have b ≤ r − t − x . Putting this all together, we have z ≤ a ≤ b ≤ r − t − x and a ≤ c. (4)A further restriction on the parameters comes from considering | e ∩ e | = ( t − a ) + c . We havefrom (4) that b ≥ z and, since H is r -partite, in each of the b parts which contain vertices of e ∩ e \ e ,there are no vertices which lie in e ∩ e . Moreover e and e are also disjoint in the z parts whichhost vertices of Z . Thus we can conclude that t + c − a = | e ∩ e | ≤ r − b − z ≤ r − z. Due to the fact that 2 x + 2 z + 2 t − r ≥ r − t + 34 > r − t + 14 ≥ r − z, (5)it follows from (3) that we must in fact have t + c − a = | e ∩ e | ≤ r − t − x − z + 1 . (6)We now look to apply Lemma 2.7 to the three edges e , e and e to show that τ ( H ) ≤ x + z , thusreaching a contradiction. To this end, note that in the notation of Lemma 2.7, we have t = t − a and t = a + b + c . First suppose that our parameters fall into the first range given by the upper boundin Lemma 2.7. That is, t − a ≥ r − t + 1 + a + b + c or, rearranging,2 a + b + c ≤ t − r − . (7)We then have that τ ( H ) ≤ r + b + c − a − t + 33 ≤ t − a + 23 ≤ t + 23 ≤ x + z, (8)using (7) in the second inequality, the fact that a ≥ r ≥ t in the lastinequality.Now we turn to the second case of Lemma 2.7 and observe that the given bound is τ ( H ) ≤ r − t + 1 + a + b + c ≤ r − t + 1 + a + b + (2 r − t − x − z + 1 + a ) ≤ r − t + 2 − x − z + 3 b ≤ r − t + 2 − x − z + 3( r − t − x )= 6 r − t − x − z + 2 , where we used (6) to bound c in the second inequality, and the bounds a ≤ b and b ≤ r − t − x from (4)in the third and fourth inequalities respectively. One has that5 x + 2 z ≥ (cid:18) r − t + 24 (cid:19) + 2 (cid:18) t − r − (cid:19) ≥ r − t + 2 , (9)and hence τ ( H ) ≤ x + z in this case too.Finally, in the third case of Lemma 2.7, we have τ ( H ) ≤ r − t + 3 + a − b − c ≤ r − t + 3 − a ≤ r − t + 3 − z, using (4) to bound b, c ≥ a in the second inequality and a ≥ z in the last inequality. As x + 2 z ≥ r − t + 24 + 6 t − r −
14 = r − t + 14 ≥ r − t + 3 (10)9or all r ≤ t −
2, we can conclude that τ ( H ) ≤ x + z in this case as well. Therefore, we have shown τ ( H ) ≤ x + z , providing the contradiction needed to complete the proof of the first bound.The proof of the second bound is almost identical to that of the first and so we omit the details.The difference here comes as we define x and z as follows: x = (cid:24) r − (cid:25) and z = (cid:24) r − t + 138 (cid:25) . The rest of the proof goes through verbatim and one simply has to check that the inequalities (2), (5),(8), (9) and (10) all hold. One has to use the fact that 5 t − ≤ r ≤ t − in order to prove (2), (5)and (8). The lower bound on r is necessary for (5) to hold, whilst the upper bound on r is necessary sothat the upper bound on z in (2) always holds. Given this, we can again conclude τ ( H ) ≤ x + z . k -wise intersecting hypergraphs As we shall show now, our exact results for Ryser( r, t ) allow us to obtain tight bounds on the covernumbers of k -wise t -intersecting r -uniform r -partite hypergraphs. Proof of Theorem 1.7.
We prove the upper bound by induction on k . The base case, when k = 2 (and t > r ), is the first case of Theorem 1.5.For the induction step, we have k ≥ t ≥
1. Given k − e , . . . , e k − ∈ E ( H ), let U = ∩ k − i =1 e i . The k -wise intersection condition implies that every edge meets U in at least t elements.Thus, if B is obtained by removing t − U , B must be a cover for H .If there are k − (cid:4) r − tk (cid:5) + t , we are done. We maytherefore assume H is ( k − t ′ -intersecting, where t ′ = (cid:4) r − tk (cid:5) + t + 1. Note that if k = 3, then t ′ = (cid:22) r − t (cid:23) + t + 1 ≥ r + 2 t > r . Hence, by induction, τ ( H ) ≤ j r − t ′ k − k + 1. Define integers a, b such that 0 ≤ b ≤ k − r − t = ak + b and note that it follows from the definition of t ′ that t ′ = a + t + 1. We then have (cid:22) r − t ′ k − (cid:23) + 1 = (cid:22) r − a − t − k − (cid:23) + 1 = (cid:22) a ( k −
1) + b − k − (cid:23) + 1 ≤ a + 1 = (cid:22) r − tk (cid:23) + 1 , completing the induction.To finish, we show that the bound is best possible. Setting ℓ = (cid:4) r − tk (cid:5) , consider the hypergraph H rℓ from Definition 2.1. To see that H rℓ is k -wise t -intersecting, observe that each edge misses ℓ verticesof the form (0 , j ). Hence, in the intersection of k edges, we can miss at most kℓ of these r vertices,and thus the k edges must intersect in at least r − kℓ ≥ t vertices, as required. By Proposition 2.2, τ ( H rℓ ) = ℓ + 1 = (cid:4) r − tk (cid:5) + 1, matching the upper bound. In this paper, we have studied Ryser’s Conjecture for t -intersecting hypergraphs. In particular, wehave shown Ryser( r, t ) = ⌊ r − t ⌋ + 1 whenever r ≤ t −
1, and have proved Conjecture 1.2 for all butfinitely many pairs ( r, t ) satisfying r ≤ t − . Given these results, it is natural to ask what happenswhen r is larger with respect to t .Since the upper bounds of Theorem 1.5 are obtained by considering configurations of two andthree edges (Lemmas 2.5 and 2.7 respectively), the obvious next step is to prove an analogous resultfor configurations of four edges. However, as one increases the number of edges in the configuration,the number of variables (representing the intersections of these edges) grows exponentially and onehas much less control over the values that these sizes of intersections can have. Indeed, even with justfour edges, we could not see a way to channel our ideas to get a stronger upper bound.Another approach to understanding the behaviour of Ryser( r, t ) is to try and determine the value ofthe function for small values of r and t , using this as a testing ground for new ideas to give more general10roofs. We considered the smallest open cases: 3 ≤ Ryser(6 , ≤ ≤ Ryser(7 , ≤
5, where thelower bounds follow from Theorem 1.3, the upper bound on Ryser(6 ,
2) follows from Theorem 1.5 andthe upper bound on Ryser(7 ,
2) follows from the work of Kir´aly and T´othm´er´esz [23]. We managedto improve these upper bounds, showing that Ryser(6 ,
2) = 3 and Ryser(7 , ≤
4. Unfortunately, thearguments for these two new bounds required ad hoc methods and, as we doubt such arguments willlead to a significantly wider range of results, we have chosen to omit these proofs.With regards to the broader picture, we concede that it may be challenging to resolve Conjecture 1.2in full, since the case t = 1 is the intersecting case of Ryser’s Conjecture itself. As discretion is thebetter part of valour, one might restrict one’s attention to t ≥ r, t ). In this range, given the lackof a better construction, Theorem 1.4, and the k -wise result of Theorem 1.7, we propose the followingconjecture. Conjecture 3.1.
For all ≤ t ≤ r , Ryser( r, t ) = (cid:22) r − t (cid:23) + 1 . If we are to be honest, it is only a proper (but non-empty) subset of the authors that fully believesin this conjecture. That said, we are all happy to pose it, in the hopes of provoking the communityinto finding a proof or a counterexample. Should the conjecture be true, it would represent a markeddifference between the intersecting and t -intersecting ( t ≥
2) versions of Ryser’s Conjecture. Thoughthis may be surprising at first sight, such discrepancies are not unheard of in extremal combinatorics.At the very least, the determination of the asymptotic behaviour of Ryser( r, t ) when t is linear in r is an intriguing question in its own right, and even just reducing the grey area in Figure 1 seemsto require new ideas. This further motivates the pursuit of Ryser-type problems for various otherclasses of hypergraphs commonly studied in the field, some of which we outline below. We believethat the techniques and constructions used in answering these questions could shed further light onConjecture 3.1 and perhaps even on Ryser’s Conjecture. t -intersecting hypergraphs One advantage of the construction of Bustamante and Stein [6], in which each vertex of the truncatedprojective plane is replaced by a set of t vertices, is that it is regular. On the other hand, in ourconstruction for Theorem 1.3, while the majority of vertices are in at most one edge, some verticeshave very large degree. This begs the question of whether or not one can find a regular constructionmatching our bound, but, as we shall now show, the great irregularity is necessary for the cover numberof the hypergraph to be large.To start, observe that if H is a d -regular ( r, t )-graph, and V i is any one of the r parts, then H hasexactly d | V i | edges, since each edge meets V i in exactly one vertex. Since any set S ⊂ V ( H ) can coverat most d | S | edges, it follows that τ ( H ) ≥ | V i | ; that is, the parts are minimum covers. Therefore,maximising the cover number of d -regular ( r, t )-graphs is equivalent to maximising the number ofvertices in such graphs.An upper bound was provided by Frankl and F¨uredi [12], with a short proof later given byCalderbank [7]: they proved that any regular t -intersecting r -uniform hypergraph can have at most( r − r + t ) /t vertices. In the r -partite setting, it follows that we have a part of size at most( r − /t + 1 /r < r/t , and hence this is an upper bound on the cover number of any regular ( r, t )-graph.Note that for t ≥ r, t )-graphs.Frankl and F¨uredi [12] and Calderbank [7] further showed that the hypergraphs achieving equalityin their bound are precisely the symmetric 2-( v, r, t ) designs, a class of hypergraphs we now define. For instance, our proof that Ryser(6 ,
2) = 3 reduced the problem to the strictly 2-intersecting case, and our proofthat Ryser(7 , ≤ efinition 3.2. Given v, r, t ∈ N , a 2 - ( v, r, t ) design is an r -uniform hypergraph on v vertices withthe property that any two vertices share exactly t common edges. The design is symmetric if it hasexactly v edges.Note that designs are never r -partite, since two vertices in the same part could not have anycommon edges. One might therefore hope for an even smaller upper bound if the hypergraph is also r -partite, but the construction of Bustamante and Stein shows that there can be regular ( r, t )-graphswith cover number r/t −
1, and so there is not much room for improvement in general. Still, when it comes to ( r, t )-graphs, our next result shows that one can obtain strong upper boundson the cover number even if the condition of regularity is weakened to just having some control overthe minimum and maximum degrees.
Lemma 3.3.
Let ∆ be the maximum degree and δ the minimum degree of an ( r, t ) -graph H . Then τ ( H ) ≤ (cid:18) ∆ − δ (cid:19) rt − ∆ − δ − δ . Proof.
Let m be the total number of edges in H , and let e be one such edge. Double-counting pairs( v, f ) where f ∈ E ( H ) \ { e } and v ∈ e ∩ f , we get r (∆ − ≥ ( m − t , or(∆ − rt + 1 ≥ m. (11)Let u be a vertex of maximum degree ∆ and let P be the part of the r -partition that contains u .Since every edge is incident to a unique vertex in P , by looking at the edges incident to each vertexin P we get m ≥ ( | P | − δ + ∆ ≥ ( τ ( H ) − δ + ∆ , (12)where the final inequality follows from the fact that P is a vertex cover. Combining the upper andlower bounds on m then gives the desired result.In particular, this restricts the cover number of d -regular ( r, t )-graphs, and, if d < r , the boundwe obtain is smaller than that derived from Frankl and F¨uredi [12] and Calderbank [7]. As with theirresults, we can characterise the hypergraphs achieving equality, for which we require a couple moredesign-theoretic definitions. Definition 3.4.
Given a hypergraph H , the dual hypergraph H D has V ( H D ) = E ( H ) and E ( H D ) = {{ e ∈ E ( H ) : u ∈ e } : u ∈ V ( H ) } ;that is, we transpose the incidence relation between vertices and edges. Also, we say a 2-( v, r, t ) designis resolvable if its edges can be partitioned into perfect matchings.Now we can state our result for the regular setting. Corollary 3.5. If H is a d -regular ( r, t ) -graph, then τ ( H ) ≤ rt − rdt + 1 d , with equality if and only if H is the dual of a resolvable - ( v, d, t ) design. For example, for a prime power q and dimensions 1 ≤ k < n , the k -dimensional affine subspacesin F nq form a resolvable 2-( q n , q k , (cid:0) n − k − (cid:1) q ) design, whose dual is therefore a tight construction for Using the truncated projective plane for some prime power q , the Bustamante-Stein construction gives d -regular( r, t )-graphs with cover number τ ( H ) = d = r/t − q . For other values of the parameters d, r and t , it may be possibleto obtain better upper bounds, for instance in Corollary 3.5, which gives a stronger bound when d < r/t − Where (cid:0) n − k − (cid:1) q = ( q n − − q n − − ... ( q n − k +1 − q k − − q k − − ... ( q − is the Gaussian binomial coefficient, which counts the number of k -dimensional spaces that contain a given pair of points. In fact, we have a rich and storied variety of extremal constructions, as the studyof resolvable designs dates back to Kirkman’s famous schoolgirl problem [24] from 1857, which askedfor resolvable 2-(15 , ,
1) designs. This was greatly generalised by Ray-Chaudhuri and Wilson [26, 27],who showed the existence of resolvable designs of all uniformities r whenever v is sufficiently largeand the trivial divisibility conditions are satisfied. More recently, Keevash [21] resolved some long-standing conjectures by extending these results to designs of greater strength, while results of Ferberand Kwan [9] suggest that, when v ≡ v ) many 2-( v, , Proof of Corollary 3.5.
The upper bound follows immediately from Lemma 3.3 by substituting ∆ = δ = d .For the characterisation of equality, first observe that the inequality (11) is always tight if andonly if any two edges of H share exactly t vertices, in which case we say H is strictly t -intersecting.Next, we note that the inequalities in (12) are always tight in the regular setting; the first becauseall degrees are equal to d , and the second because, as argued at the beginning of Section 3.1, a partis always a minimum cover in a regular r -partite r -uniform hypergraph. Thus we see that we haveequality if and only if H is strictly t -intersecting.Now suppose H is d -regular ( r, t )-graph that is strictly t -intersecting, and consider the dual hyper-graph H D . Since every vertex of H has degree d , every edge of H D contains d vertices. Furthermore,as every pair of edges in H shares t vertices, every pair of vertices in H D have t common edges. Thus H D is a 2-( v, d, t ) design, where v is the number of edges in H . Finally, since each edge of H containsexactly one vertex from any of the r parts, a part corresponds to a perfect matching in H D , with everyvertex covered exactly once. Hence, since H is r -partite, the edges of H D can be partitioned into r perfect matchings; that is, H D is resolvable.Conversely, the same reasoning shows that the dual H of a resolvable 2-( v, d, t ) design gives astrictly t -intersecting d -regular r -partite r -uniform hypergraph with v edges, where r is the numberof perfect matchings in the resolution of the design. From our above remarks, this implies the dualachieves equality in the upper bound on τ ( H ).In the above proof, we saw that for a regular ( r, t )-graph to have as large a cover number aspossible, it must be strictly t -intersecting, with every pair of edges meeting in exactly t vertices. Itis therefore natural to ask what happens when we drop the condition of regularity, and only requirethe ( r, t )-graph be strictly t -intersecting. Such study has previously been carried out in the setting ofRyser’s Conjecture for (1-)intersecting families.Recall that Ryser’s Conjecture for intersecting r -partite hypergraphs was proved by Tuza [30] forall r ≤
5. Franceti´c, Herke, McKay and Wanless [10] showed that if we restrict ourselves to linear(that is, strictly 1-intersecting) hypergraphs, then the conjecture is true for all r ≤
9. Inspired by this,we prove Conjecture 1.2 for strictly t -intersecting hypergraphs for a much wider range of parameters r and t than covered by Corollary 1.6. Theorem 3.6.
Let t ≥ and t < r ≤ t + 3 t − be integers. If H is a strictly t -intersecting r -partitehypergraph, then τ ( H ) ≤ r − t .Proof. For the sake of contradiction, suppose τ ≥ r − t + 1, and, for the sake of convenience, let δ = δ ( H ), ∆ = ∆( H ) and τ = τ ( H ).Let v be an arbitrary vertex of H and let e be an edge through v . Since τ ≥ r − t + 1, we have thatfor every set S of r − t vertices in e \ { v } , there exists an edge f of H with e ∩ f ⊆ e \ S . Since H is t -intersecting and | e \ S | = t , we must have e ∩ f = e \ S , which in particular implies that v ∈ f . Thisshows that d ( v ) ≥ (cid:0) r − r − t (cid:1) + 1, and, since v was arbitrary, we get δ ≥ (cid:0) r − r − t (cid:1) + 1. Solving the inequalityin Lemma 3.3 for ∆, we obtain ∆ ≥ δ ( τ − r/t − , When n = 2, the dual is the truncated projective plane, and thus this construction generalises the classic tightconstruction for Ryser’s Conjecture. δ ≥ (cid:0) r − r − t (cid:1) + 1 and τ ≥ r − t + 1, yields∆ ≥ t (cid:18) r − r − t (cid:19) + t + 1 . Now let u be a vertex of degree d ( u ) = ∆ and let f be an edge through u . There are (cid:0) r − t − (cid:1) = (cid:0) r − r − t (cid:1) choices of t -subsets of f containing u , and every edge f ′ = f through u intersects f in one of thesesets. Since there are at least t (cid:0) r − r − t (cid:1) + t edges through u other than f , by the pigeonhole principlethere must exist a t -subset S of f containing u for which there are at least t + 1 edges f , . . . , f t +1 with f ∩ f i = S for all i . Since H is strictly t -intersecting, we further have f i ∩ f j = S for all i = j .We claim that S is a vertex cover. If not, there exists an edge e such that e ∩ f i ⊆ f i \ S for all0 ≤ i ≤ t + 1. Since f \ S, . . . , f t +1 \ S are disjoint sets, and e can only contain one vertex from eachpart of the r -partition, we get t ( t + 2) ≤ r − t , contradicting our upper bound on r .Thus, we have a vertex cover S of size t . Since Conjecture 1.2 is known for r ≤ t , it follows that τ ≤ r − t , contradicting our original supposition that τ ≥ r − t + 1.Although the restriction of being strictly t -intersecting allows us to prove the bound from Con-jecture 1.2 for a wider range of parameters ( r, t ), we believe this is far from tight. Indeed, in thissetting, we even lack constructions that come close to the smaller bound of Theorem 1.3. The bestconstructions we have found thus far are the duals of resolvable designs, as given in Corollary 3.5. Asthese are also regular, their cover numbers are smaller than rt , significantly smaller than the upperbound of Theorem 3.6. Problem 1.
Prove that τ ( H ) ≤ rt for any strictly t -intersecting r -partite r -uniform hypergraph H ,or find constructions with larger cover numbers. s -covers Another new direction is to ask for more from our vertex covers – rather than just intersecting eachedge, we could ask for a set that meets every edge in many vertices.
Definition 3.7.
Let H be an ( r, t )-graph. For s ≥
1, we define an s -cover of H to be a set B ⊆ V ( H )such that | B ∩ e | ≥ s for every e ∈ E ( H ). We further define τ s ( H ) = min {| B | : B is an s -cover of H} . Observe that τ ( H ) = τ ( H ). We can then generalise Ryser’s Conjecture (in the intersecting case)by asking for the maximum of τ s ( H ) over all ( r, t )-graphs. If s ≤ t , then, since every pair of edgesintersects in at least t vertices, any edge e ∈ E ( H ) is an s -cover, and so we always have τ s ( H ) ≤ r .However, the problem is ill-posed if s > t . For arbitrary n ∈ N , we can take H be the complete r -partite r -graph with parts V , . . . , V r , where | V | = . . . = | V t | = 1 and | V t +1 | = . . . = | V r | = n . Itis then easy to see that τ s ( H ) = ( s − t ) n + t , and therefore τ s ( H ) is unbounded for ( r, t )-graphs. Weshall thus require 1 ≤ s ≤ t ≤ r . Definition 3.8.
Given integers 1 ≤ s ≤ t ≤ r , defineRyser s ( r, t ) = max { τ s ( H ) : H is an ( r, t )-graph } . The case s = 1 is obviously what we have been talking about all along, and the following lemmashows how we can leverage our constructions from that case to obtain lower bounds when s ≥ Lemma 3.9.
For all ≤ s ≤ t ≤ r and a ≥ , Ryser s + a ( r + a, t + a ) ≥ Ryser s ( r, t ) + a. Proof.
Let H ′ be an ( r, t )-graph with τ s ( H ) = Ryser s ( r, t ). Form H by adding the same set S of a vertices to each edge of H ′ . H is then an ( r + a, t + a )-graph. Let B be a smallest ( s + a )-cover of H . By removing a elements from B , including all members of B ∩ S , we obtain an s -cover B ′ of H ′ .Hence we must have | B ′ | ≥ τ s ( H ′ ) = Ryser s ( r, t ), and thus Ryser s + a ( r + a, t + a ) ≥ τ s + a ( H ) = | B | ≥ Ryser s ( r, t ) + a . 14he next proposition extends Theorem 1.4 to the case when s ≥ Proposition 3.10. If r ≤ t − s , then Ryser s ( r, t ) = (cid:22) r − t (cid:23) + s. Proof.
The lower bound, valid for all 1 ≤ s ≤ t ≤ r , is an easy consequence of Lemma 3.9 andTheorem 1.3: Ryser s ( r, t ) ≥ Ryser ( r − s + 1 , t − s + 1) + s − ≥ (cid:22) r − t (cid:23) + s. For the upper bound, let t ≥ r +2 s . Let H be an ( r, t )-graph, and let t ′ ≥ t be the minimum sizeof an intersection of two edges. We will show τ s ( H ) ≤ τ ′ = j r − t ′ k + s . Let e and e be two edgesintersecting in exactly t ′ elements, and let S = e ∩ e . By our bound on t , we have t ′ ≥ τ ′ .Let B be a set of τ ′ elements from S . We claim that B is an s -cover. Indeed, suppose there wassome e ∈ E ( H ) with | B ∩ e | ≤ s −
1. Then e can contain at most s − t ′ − τ ′ elements from S .In all parts outside S , e intersects e ∪ e in at most one vertex. Thus | e ∩ e | + | e ∩ e | ≤ s − t ′ − τ ′ ) + r − t ′ = r + 2 s + t ′ − τ ′ − ≤ t ′ − , which contradicts e intersecting both e and e in at least t ′ elements.Observe that the lower bound was proved using Lemma 3.9, reducing the problem to the s = 1case via Ryser s ( r, t ) ≥ Ryser( r − s + 1 , t − s + 1). When s = t , this also reduces the problem to theclassic setting, where we have much stronger lower bounds. Here we know that, whenever r − t is aprime power, Ryser( r − t + 1 , ≥ r − t , and thus Ryser t ( r, t ) ≥ r −
1. The constructions of Haxelland Scott [19] further show Ryser t ( r, t ) ≥ r − r .Our final result uses the construction of Bustamante and Stein [6] to significantly improve thelower bound for large r , whenever s > t . Proposition 3.11.
Let ≤ s ≤ t and suppose r = t ( q + 1) for some prime power q . Then Ryser s ( r, t ) ≥ s (cid:16) rt − (cid:17) . Proof.
Let H ′ be the truncated projective plane of order q , which is a q -regular ( q + 1 , q edges, and let H be the hypergraph obtained by replacing each vertex of H ′ with a set of t vertices. H is then a q -regular ( t ( q + 1) , t )-graph with q edges.Since each vertex covers q edges, to cover all of the edges at least s times, we require at least sq q = sq = s (cid:0) rt − (cid:1) vertices. Thus Ryser s ( r, t ) ≥ τ s ( H ) ≥ s (cid:0) rt − (cid:1) .It remains an open problem to find matching upper bounds in these ranges. Problem 2.
Determine Ryser s ( r, t ), at least asymptotically, when r is large and 1 < s < t . Note added in proof
During the publication of our manuscript, it was brought to our attentionthat the t = 1 case of Theorem 1.7 (when k ≥
3) was earlier proven by Kir´aly [22], in the context ofmonochromatic components in edge-coloured hypergraphs. As observed by DeBiasio (personal com-munication), Kir´aly’s theorem can be used as a base case for induction on t , providing an alternativeproof of Theorem 1.7 for k ≥
3. For more information on various generalisations of Ryser’s Conjecture,we refer the reader to the recent survey of DeBiasio, Kamel, McCourt and Sheats [8].15 eferences [1] A. Abu-Khazneh,
Matchings and covers of multipartite hypergraphs , PhD Thesis, The LondonSchool of Economics and Political Science (2016), http://etheses.lse.ac.uk/3360/ .[2] A. Abu-Khazneh, J. Bar´at, A. Pokrovskiy and T. Szab´o,
A family of extremal hypergraphs forRyser’s conjecture , J. Comb. Theory A (2019), 164–177.[3] R. Aharoni,
Ryser’s conjecture for tripartite -graphs , Combinatorica (2001), 1–4.[4] D. Best and I. M. Wanless, What did Ryser conjecture? , arXiv:1801.02893 (2018).[5] A. Bishnoi and V. Pepe,
Non-intersecting Ryser hypergraphs , SIAM J. Discrete Math. (2020), 230–240.[6] S. Bustamante and M. Stein,
Monochromatic tree covers and Ramsey numbers for set-colouredgraphs , Discrete Math. (2018), 266–276.[7] A. R. Calderbank,
Symmetric designs as the solution of an extremal problem in combinatorialset theory , Eur. J. Combin. (1987), 171–173.[8] L. DeBiasio, Y. Kamel, G. McCourt and H. Sheats, Generalizations and strengthenings of Ryser’sconjecture , arXiv:2009.07239 (2020).[9] A. Ferber and M. Kwan,
Almost all Steiner triple systems are almost resolvable , arXiv::1907.0674(2019).[10] N. Franceti´c, S. Herke, B. D. McKay and I. M. Wanless,
On Ryser’s conjecture for linear inter-secting multipartite hypergraphs , Eur. J. Combin. (2017), 91–105.[11] P. Frankl, On Sperner families satisfying an additional condition , J. Comb. Theory Ser. A (1976), 1–11.[12] P. Frankl and Z. F¨uredi,
Finite projective spaces and intersecting hypergraphs , Combinatorica (1986), 335–354.[13] Z. F¨uredi,
On finite set-systems whose every intersection is a kernel of a star , Discrete Math. (1983), 129–132.[14] Z. F¨uredi and B. Sudakov, Extremal set systems with restricted k -wise intersections , J. Comb.Theory Ser. A (2004), 143–159.[15] V. Grolmusz, Set-systems with restricted multiple intersections , Electron. J. Comb. (2002), R8.[16] V. Grolmusz and B. Sudakov, On k -wise set-intersections and k -wise Hamming-distances , J.Comb. Theory Ser. A (2002), 180–190.[17] A. Gy´arf´as, Partition coverings and blocking sets in hypergraphs (in Hungarian) , Communi-cations of the Computer and Automation Institute of the Hungarian Academy of Sciences (1977), 62.[18] P. Haxell, L. Narins and T. Szab´o, Extremal hypergraphs for Ryser’s Conjecture , J. Comb. TheoryA (2018), 492–547.[19] P. Haxell and A. Scott,
A note on intersecting hypergraphs with large cover number , Electron. J.Comb. (2017), P3.26.[20] J. R. Henderson, Permutation Decompositions of (0 , -matricesand decomposition transversals , PhD Thesis, Caltech (1971), http://thesis.library.caltech.edu/5726/1/Henderson_jr_1971.pdf .1621] P. Keevash, The existence of designs II , arXiv:1802.05900 (2018).[22] Z. Kir´aly,
Monochromatic components in edge-colored complete uniform hypergraphs , ElectronicNotes in Discrete Mathematics (2011), 517–521.[23] Z. Kir´aly and L. T´othm´er´esz, On Ryser’s conjecture for t -intersecting and degree-bounded hyper-graphs , Electron. J. Comb. (2017), P4.40.[24] T. P. Kirkman, On a problem in combinatorics , Cambridge and Dublin Math. J. , (1847),191–204.[25] D. K¨onig, Gr´afok ´es m´atrixok , Matematikai ´es Fizikai Lapok (1931), 116–119.[26] D. K. Ray-Chaudhuri and R. M. Wilson, Solution of Kirkman’s school girl problem , Proc. Symp. inPure Mathematics (1971), 187–203.[27] D. K. Ray-Chaudhuri and R. M. Wilson, The existence of resolvable designs, in: A survey ofcombinatorial theory , North-Holland, Amsterdam, 1973.[28] V. T. S´os,
Remarks on the connection of graph theory, finite geometry and block designs , TheorieCombinatorie, Proc. Colloq., Rome, Vol. II, Atti dei Convegni Lincei (1973), 223–233. Accad.Naz. Lincei, Rome.[29] T. Szab´o and V. H. Vu, Exact k -wise intersection theorems , Graph. Combinator. (2005),247–261.[30] Zs. Tuza, Ryser’s conjecture on transversals of r -partite hypergraphs , Ars Combin. (1983),201–209.[31] V. H. Vu, Extremal set systems with upper bounded odd intersections , Graph. Combinator. (1997), 197–208.[32] V. H. Vu,
Set systems with weakly restricted intersections , Combinatorica19.4