Ryser's Conjecture for t-intersecting hypergraphs

the of ) the the for all r ≤ 367 t − 5. We also introduce several related variations on this theme. As a consequence of our tight bounds, we resolve the problem for k -wise t -intersecting hypergraphs, for all k ≥ 3 and t ≥ 1. We further give bounds on the cover numbers of strictly t -intersecting hypergraphs and the s -cover numbers of t -intersecting hypergraphs.


Introduction
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 1 ⊔ • • • ⊔ P r , such that for all e ∈ E(H) and all j ∈ [r], we have |e ∩ P j | = 1.For 1 ≤ t ≤ r − 1, an r-uniform hypergraph is t-intersecting if |e ∩ f | ≥ t for all e, f ∈ E(H).We will refer to 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 a cover 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 appearing in the PhD thesis of his student Henderson [19] (see [4] for more on the history of this conjecture), which claims that the cover number of any r-uniform r-partite hypergraph H satisfies τ (H) ≤ (r − 1)ν(H).Here, ν(H) denotes the matching number of the hypergraph, that is, the size of the largest set of pairwise disjoint edges.When r = 2, this is a statement about bipartite graphs, and is equivalent to the classic theorem of König [24].Although Ryser's Conjecture has attracted significant attention over the years, the only other resolved case is r = 3.This was proven via topological methods by Aharoni [3], with the extremal hypergraphs classified by Haxell, Narins and Szabó [17].
In this latter result, it was shown that the extremal hypergraphs with ν(H) = ν ≥ 2 can essentially be decomposed into ν extremal intersecting hypergraphs.Thus, much research in this direction has focussed on intersecting hypergraphs. 1 Here we have ν(H) = 1, and Ryser's Conjecture asserts Ryser(r, 1) ≤ r − 1.Further motivation for considering the intersecting case comes from a connection with a conjecture of Gyárfás [16] which states that the vertices of any r-edge-coloured clique can be covered by at most r − 1 monochromatic trees.Indeed, this conjecture in the setting of coloured complete graphs is in fact equivalent to Ryser's conjecture for intersecting hypergraphs, see e.g.[21].Not much more is known even in this simpler setting; Tuza [29] proved the conjecture for r ≤ 5, but it remains otherwise open.The apparent difficulty of this conjecture is perhaps explained by the abundance of extremal constructions: the classic example of truncated projective planes shows Ryser(r, 1) ≥ r − 1 whenever r − 1 is a prime power, while Abu-Khazneh, Barát, Pokrovskiy and Szabó [2] construct exponentially (in √ r) many non-isomorphic minimal examples whenever r − 2 is a prime power.For general r, Haxell and Scott [18] construct nearly-extremal intersecting hypergraphs; more precisely, they show Ryser(r, 1) ≥ r − 4 for all r large enough.
This led Bustamante and Stein [6] and, independently, Király and Tóthmérész [22] to investigate what occurs when we impose the stricter condition of the hypergraph H being t-intersecting.In this 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.
Note that while Ryser's Conjecture for intersecting hypergraphs is a special case (t = 1), it in fact implies Conjecture 1.2.Indeed, Ryser(r, t) ≤ Ryser(r − t + 1, 1) since deleting t − 1 parts and removing the deleted vertices from each edge leaves us with an (r − t + 1, 1)-graph, which, by Ryser's Conjecture, should have a cover of size at most 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 the conjecture for r ≤ 2t + 2, with Király and Tóthmérész [22] extending this to r ≤ 4t − 1.With regards to lower bounds on Ryser(r, t), the conjecture is trivially tight for t = r − 1, and Bustamante and Stein [6] showed that it is also tight for t = r − 2. However, they demonstrated that it is not always best possible by proving Ryser(5, 2) = 2.More generally, they proved Ryser(r, t) ≥ Ryser(⌊ r t ⌋, 1) by observing that replacing every vertex of an (r ′ , 1)-graph with a set of t vertices gives an (r ′ t, t)-graph.Given the aforementioned results on Ryser's conjecture, this shows Ryser(r, t) ≥ ⌊r/t⌋ − 1 for many pairs (r, t), and Bustamante and Stein suggested this lower bound should be closer to the truth.

Our results
Our first result shows that this is far from the case, providing a construction, valid for all t and r, that greatly improves the previous lower bound.
Theorem 1.3.For all 1 ≤ t ≤ r we have We 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 1.3 gives the lower bound needed for Theorem 1.4, and hence all that is required is a matching upper bound.In fact, using different arguments, we are able to prove a few upper bounds on Ryser(r, t).The theorem below collects, in a simplified form, the best upper bounds (excluding the trivial Ryser(r, t) ≤ r − t + 1) that we have in various ranges of the parameters.
Theorem 1.5.Given t and r with t = αr for some α ∈ 9 52 , 1 , we have Ryser(r, t) ≤ f (α)r + 3, where A more precise version of our results, with careful attention paid to the additive constants, can be found in Theorem 2.3.In particular, the first case of that theorem gives the upper bound needed for Theorem 1.4.Although we fall short of determining Ryser(r, t) exactly when r ≥ 3t, the latter three cases strongly restrict how large the cover number of (r, t)-graphs can be.In particular, we extend the results of Bustamante and Stein [6] and Király and Tóthmérész [22] by showing Conjecture 1.2 continues to hold for smaller values of t, and that it is not tight in most of these cases.

k-wise intersecting hypergraphs
In the above results, we require that all pairwise intersections of the edges of the hypergraphs have size at least t.A natural stronger condition is to impose the same restriction on all k-wise intersections of edges, rather than just pairwise.This setting has often been studied in the extremal combinatorics literature.Frankl [10] first studied such hypergraphs, determining the maximum number of edges possible when all k-wise intersections are non-empty.Sós [27] then raised the problem of finding the largest hypergraphs where the sizes of all k-wise intersections lie in some set L, and various results in this direction were obtained by Füredi [12], Vu [30,31], Grolmusz [14], Grolmusz and Sudakov [15], Füredi and Sudakov [13] and Szabó and Vu [28].
We say a hypergraph H is k-wise t-intersecting if, for any edges e 1 , e 2 , . . ., e k ∈ E(H), we have |∩ k i=1 e i | ≥ t.Following Ryser's Conjecture, we study how much smaller a cover one is guaranteed to 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

Proofs of the main results
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 kwise t-intersecting hypergraphs (Section 2.3).

Lower bound construction
To obtain the lower bound, and thereby prove Theorem 1.3, we need to construct (r, t)-graphs with large cover numbers.The hypergraphs we consider are of the following form.
Proof.To see that H r ℓ is (r − 2ℓ)-intersecting, observe that each edge of H r ℓ misses exactly ℓ vertices from the set L 0 = {(0, j) : j ∈ [r]}.It then follows that, for any two edges e i , e i ′ ∈ E(H r ℓ ), we have We now establish the cover number of H r ℓ .Assume for a contradiction that H r ℓ has a cover C ⊂ V (H r ℓ ) of size c ≤ ℓ.First we show that we may assume that C ⊂ L 0 .Indeed, if v = (i, j) ∈ C for some i ≥ 1, then the only edge that could contain v is e i .If we replace v with a vertex in e i ∩ L 0 , the modified set C has not increased in size and still covers H r ℓ .Since |C| = c ≤ ℓ, we have |L 0 \ C| ≥ r − ℓ and hence there exists an i * ∈ [m] such that {0} × S i * ⊆ L 0 \ C. Then e i * ∩ C = ∅, contradicting the fact that C is a cover.Thus τ (H r ℓ ) ≥ ℓ + 1.To see we have equality, note that any subset of ℓ + 1 vertices in L 0 forms a cover.
The key property needed in the above proof is that, for each subset S ⊆ L 0 of size r − ℓ, there is an edge of H r ℓ intersecting L 0 exactly at the vertices of S. Our construction is edge-minimal with respect to this key property, and further ensures that all vertices not in L 0 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 of the hypergraph is constructed.For instance, one can instead make it vertex-minimal, having parts of size ℓ + 1 rather than r r−ℓ + 1, so that each part P j also forms a minimum vertex cover.We omit the details of this construction for the sake of brevity, as we already have all we need to prove Theorem 1.3.

Upper bounds
In this section we prove the following upper bounds on Ryser(r, t), from which Theorem 1.5 easily follows. .
These bounds are derived from a sequence of results obtained 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 2.4.Let H be an r-uniform r-partite hypergraph with parts P j for 1 ≤ j ≤ r.Suppose that e 1 , . . ., e k ∈ E(H).Then for v ∈ V (H), we define to be the degree of v with respect to the k edges e 1 , . . ., e k .Also, given a vertex subset C ⊆ V (H) not wholly containing any part P j , we define to be the maximum sum of degrees (with respect to e 1 , . . ., e k ) when we take one vertex from each part and avoid C.
The utility of this definition comes from the following easy observation, which we use repeatedly in the subsequent proofs.
Consequently, if and ∆ H (C; e 1 , . . ., 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 be some 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 of such a hypergraph.Lemma 2.6.Let H be an (r, t)-graph, let e 1 , e 2 ∈ E(H), and set Proof.Let the parts of H be P j for j ∈ and thus the conclusion follows from Observation 2.5.Now consider the case where t ′ ≤ r−2t.Without loss of generality, we may assume the intersection of e 1 and e 2 is contained in the first t ′ parts, labelling the vertices of e 1 as {u 1 , . . ., u r } and of e 2 as {u 1 , . . ., u t ′ , v t ′ +1 , . . ., v r }, where u j , v j ∈ P j for all j.Letting Lemma 2.6 suffices to prove the first two parts of Theorem 2.3.For the latter parts, we shall need to consider covers consisting of vertices lying in three edges instead.Before proceeding, though, we present a reformulation of Lemma 2.6 that will be more convenient for later proofs.
Corollary 2.7.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 Proof.Let e, f ∈ E(H) be an arbitrary pair of edges of H and take The following lemma is the analogue of Lemma 2.6 when constructing covers from vertices that lie in three fixed edges, as opposed to only using two edges.
Lemma 2.8.Let r ≥ 3t, let H be an (r, t)-graph, and let e 1 , e 2 , e 3 ∈ E(H).Set Proof.Observe that t 1 counts the number of vertices that are in all three edges, while t 2 counts the number of vertices in precisely two of the three edges.Let T = e 1 ∩ e 2 ∩ e 3 be the set of t 1 vertices contained in all three edges and let D be the set of t 2 vertices contained in exactly two of the three edges.Finally, in the t 2 parts spanned by D, let D ′ be the set of vertices contained in the third (disjoint) edge.When building a transversal of the parts that intersects e 1 , e 2 and e 3 as much as possible, it is optimal to choose as many vertices from T as possible, followed by vertices from D. Therefore, to minimise ∆ H (C; e 1 , e 2 , e 3 ), we will choose C so as to first block the vertices in T , followed by those in D.
Let us first consider the case when , where S 2 is a subset of D of size s 2 .When choosing a transversal of the parts that is disjoint from C 2 , one can at best select the t 2 − s 2 vertices from D \ S 2 that are in two of the edges, as well as the s 2 vertices from D ′ .One can also select a vertex in a single edge from the r − t 1 − t 2 parts where the three edges are pairwise disjoint.Thus and so, by Observation 2.5, In the range r We now take C 3 = T ∪ D ∪ S 3 where S 3 is an arbitrary subset of D ′ of size s 3 .A transversal disjoint from C 3 can then only meet the edges e 1 , e 2 and e 3 in the t 2 − s 3 vertices of D ′ \ S 3 , as well as in the r − t 1 − t 2 parts where the three edges are pairwise disjoint.We therefore have and so Observation 2.5 implies C 3 is a cover of the stated size.
Finally, we are left with the case when t 1 ≤ r−3t+1−t 2 , for which we set , where S 4 consists of the 3s 4 vertices of e 1 ∪ e 2 ∪ e 3 from s 4 of the parts where the edges e 1 , e 2 and e 3 are pairwise disjoint.Then any transversal disjoint from C 4 can only meet the three edges in the r−t 1 −t 2 −s 4 parts not spanned by C 4 , from which it follows that ∆ H (C 4 ; e 1 , e 2 , e 3 ) = 3t−1.By Observation 2.5, C 4 is a cover of H, and so τ (H) By applying Lemma 2.8 in conjunction with Corollary 2.7, we will prove the following upper bounds on the cover numbers of (r, t)-graphs.Proposition 2.9.For all t ≥ 1 and r ≥ 3t, we have .
Before proving this proposition, we observe that we have the necessary bounds to establish our main result.
Proof of Theorem 2.3.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.Thus Ryser(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 2.6 to prove the first two cases of the theorem.To this end, let e 1 , e 2 ∈ E(H) be a pair of edges with the smallest intersection.We may assume that |e 1 ∩ e 2 | = 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 2.6 with the edges e 1 and e 2 , we have t ′ = t, from which it follows that Simplifying the ranges for which these bounds hold, we see that the first is valid when r ≤ 3t − 1, as required for the first case of the theorem, while the second bound above is valid provided r ≥ 3t.
The latter two cases of the theorem are direct consequences of Proposition 2.9, which we can apply whenever r ≥ 3t.To simplify the bounds, we estimate the ceiling terms, obtaining This matches the latter two cases of Theorem 2.3.Finally, we note that the bound 9r−14t 8 + 2 improves the earlier bound of 2r − 5t + 2 whenever r ≥ 26  7 t, justifying the endpoints of the ranges of the second and third cases in the theorem.
All that remains is the proof of Proposition 2.9, which we delay no further.
Proof of Proposition 2.9.In the first case, let r and t be such that 3t ≤ r ≤ 5t − 2, and define using here that 3t ≤ r ≤ 5t−2.Suppose for contradiction that H is an (r, t)-graph with τ (H) ≥ x+z+1.Applying Corollary 2.7, any pair of edges e, f ∈ E(H) must satisfy Now take e 1 , e 2 ∈ E(H) to be two edges such that |e 1 ∩ e 2 | = t (again, we may assume such a pair exists, as otherwise H is an (r, t + 1)-graph and we obtain a stronger bound on τ (H)).Let Z ⊆ e 1 ∩ e 2 be a set of z vertices and X ⊆ e 1 \ e 2 a set of x vertices, noting that this is possible due to (2).We take Y = X ∪ Z and note that Y intersects both e 1 and e 2 (as z ≥ 1) and is not a cover as it has size exactly x + z.Therefore, there exists an edge A further restriction on the parameters comes from considering |e 2 ∩ e 3 | = (t − a) + c.We have from ( 4) that b ≥ z and, since H is r-partite, in each of the b parts which contain vertices of e 1 ∩ e 3 \e 2 , there are no vertices which lie in e 2 ∩ e 3 .Moreover e 3 and e 2 are also disjoint in the z parts which host vertices of Z. Thus we can conclude that it follows from (3) that we must in fact have We now look to apply Lemma 2.8 to the three edges e 1 , e 2 and e 3 to show that τ (H) ≤ x + z, thus reaching a contradiction.To this end, note that in the notation of Lemma 2.8, we have t 1 = t − a and t 2 = a+b+c.First suppose that our parameters fall into the first case.That is, t−a ≥ r−3t+1+a+b+c or, rearranging, 2a We then have that using (7) in the second inequality, the fact that a ≥ 0 in the third and the fact that r ≥ 3t in the last inequality.Now we turn to the second case of Lemma 2.8 and observe that the given bound is 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 that and hence τ (H) ≤ x + z in this case too.Finally, in the third case of Lemma 2.8, we have ) to bound b, c ≥ a in the second inequality and a ≥ z in the last inequality.As for all r ≤ 5t − 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: 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 5t − 1 ≤ r ≤ 52t−13 9 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 so that the upper bound on z in (2) always holds.Given this, we can again conclude τ (H) ≤ x + z.
As can be seen in the statement of Proposition 2.9, the general upper bounds stated in Theorem 2.3 come from the worst-case estimates with respect to the rounding of the ceiling terms.For specific values of r and t, one often can get a better upper bound by taking the actual values of these terms.In fact, in many cases one can improve the upper bound by 1 by defining either x or z to be the floor of the same function of r and t, as opposed to the ceiling.One simply has to check that defining x and z in this way still satisfies the inequalities (2), ( 5), ( 8), ( 9) and (10) for such a fixed r and t.

k-wise intersecting hypergraphs
As we shall show now, our exact results for Ryser(r, t) allow us to obtain tight bounds on the cover numbers 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 3 ), is the first case of Theorem 2.3.
For the induction step, we have k ≥ 3 and The k-wise intersection condition implies that every edge meets U in at least t elements.Thus, if B is obtained by removing t − 1 elements from U , B must be a cover for H.
If there are k − 1 edges whose intersection has size at most r−t k + t, we are done.We may therefore assume H is (k − 1)-wise t ′ -intersecting, where completing the induction.
To finish, we show that the bound is best possible.Setting ℓ = r−t k , consider the hypergraph H r ℓ from Definition 2.1.To see that H r ℓ is k-wise t-intersecting, observe that each edge misses ℓ vertices of 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 r ℓ ) = ℓ + 1 = r−t k + 1, matching the upper bound.

Further variants and open problems
In this paper, we have studied Ryser's Conjecture for t-intersecting hypergraphs.In particular, we have shown Ryser(r, t) = ⌊ r−t 2 ⌋ + 1 whenever r ≤ 3t − 1, and have proved Conjecture 1.2 for all but finitely many pairs (r, t) satisfying r ≤ 36t−17

7
. Given these results, it is natural to ask what happens when r is larger with respect to t.
Since the upper bounds of Theorem 1.5 are obtained by considering configurations of two and three edges (Lemmas 2.6 and 2.8 respectively), the obvious next step is to prove an analogous result for 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 one has much less control over the values that these sizes of intersections can have.Indeed, even with just four 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 of the function for small values of r and t, using this as a testing ground for new ideas to give more general proofs.We considered the smallest open cases: 3 ≤ Ryser(6, 2) ≤ 4 and 3 ≤ Ryser(7, 2) ≤ 5, where the lower bounds follow from Theorem 1.3, the upper bound on Ryser(6, 2) follows from Theorem 2.3 and the upper bound on Ryser(7, 2) follows from the work of Király and Tóthmérész [22].We managed to improve these upper bounds, showing that Ryser(6, 2) = 3 and Ryser(7, 2) ≤ 4. Unfortunately, the arguments for these two new bounds required ad hoc methods and, as we doubt such arguments will lead 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 1.2 in full, since the case t = 1 is Ryser's Conjecture itself.As discretion is the better part of valour, one might restrict one's attention to t ≥ 2 and seek in these cases to prove the conjecture and, with a bit more ambition, to fully determine Ryser(r, t).In this range, given the lack of a better construction, Theorem 1.4, and the k-wise result of Theorem 1.7, we propose the following conjecture.Conjecture 3.1.For all 2 ≤ t ≤ r, If we are to be honest, it is only a proper (but non-empty) subset of the authors that fully believes in this conjecture.That said, we are all happy to pose it, in the hopes of provoking the community into finding a proof, a counterexample, or both.Should the conjecture be true, it would represent a marked difference between the intersecting and t-intersecting (t ≥ 2) versions of Ryser's Conjecture.Though this 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 seems to require new ideas.This further motivates the pursuit of Ryser-type problems for various other classes of hypergraphs commonly studied in the field, some of which we outline below.We believe that the techniques and constructions used in answering these questions could shed further light on Conjecture 3.1 and perhaps even on Ryser's Conjecture. 3

Strictly t-intersecting hypergraphs
One advantage of the construction of Bustamante and Stein [6], in which each vertex of the truncated projective plane is replaced by a set of t vertices, is that it is regular.On the other hand, in our construction for Theorem 1.3, while the majority of vertices are in at most one edge, some vertices have very large degree.This begs the question of whether or not one can find a regular construction matching our bound, but, as we shall now show, the great irregularity is necessary for the cover number of 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 has exactly d|V i | edges, since each edge meets V i in exactly one vertex.Since any set S ⊂ V (H) can cover at 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 of vertices in such graphs.
An upper bound was provided by Frankl and Füredi [11], with a short proof later given by Calderbank [7]: they proved that any regular t-intersecting r-uniform hypergraph can have at most (r 2 − r + t)/t vertices.In the r-partite setting, it follows that we have a part of size at most (r −1)/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 ≥ 3 this is significantly smaller than the lower bound of Theorem 1.3, showing that the added condition of regularity considerably restricts the cover number of (r, t)-graphs.
Frankl and Füredi [11] and Calderbank [7] further showed that the hypergraphs achieving equality in their bound are precisely the symmetric 2-(v, r, t) designs, a class of hypergraphs we now define.Definition 3.2.Given v, r, t ∈ N, a 2-(v, r, t) design is an r-uniform hypergraph on v vertices with the property that any two vertices share exactly t common edges.The design is symmetric if it has exactly v edges.
Note that designs are never r-partite, since two vertices in the same part could not have any common 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)-graphs with cover number r/t − 1, and so there is not much room for improvement in general 4 .
Still, when it comes to (r, t)-graphs, our next result shows that one can obtain strong upper bounds on the cover number even if the condition of regularity is weakened to just having some control over the minimum and maximum degrees.Lemma 3.3.Let ∆ be the maximum degree and δ the minimum degree of an (r, t)-graph H. Then 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(∆ − 1) ≥ (m − 1)t, or 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 vertex in where the final inequality follows from the fact that P is a vertex cover.Combining the upper and lower 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 bound we obtain is smaller than that derived from Frankl and Füredi [11] and Calderbank [7].As with their results, we can characterise the hypergraphs achieving equality, for which we require a couple more design-theoretic definitions.that is, we transpose the incidence relation between vertices and edges.Also, we say a 2-(v, r, t) design is resolvable if its edges can be partitioned into perfect matchings.Now we can state our result for the regular setting.For example, for a prime power q and dimensions 1 ≤ k < n, the k-dimensional affine subspaces in F n q form a resolvable 2-(q n , q k , n−1 k−1 q ) design, 5 whose dual is therefore a tight construction for Corollary 3.5. 6In fact, we have a rich and storied variety of extremal constructions, as the study of resolvable designs dates back to Kirkman's famous schoolgirl problem [23] from 1857, which asked for resolvable 2-(15, 3, 1) designs.This was greatly generalised by Ray-Chaudhuri and Wilson [25,26], who showed the existence of resolvable designs of all uniformities r whenever v is sufficiently large and the trivial divisibility conditions are satisfied.More recently, Keevash [20] resolved some longstanding conjectures by extending these results to designs of greater strength, while results of Ferber and Kwan [8] suggest that, when v ≡ 3 (mod 6), almost all of the exponentially (in v 2 ) many 2-(v, 3, 1) designs should be resolvable. 4Using 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 − 1 = q.For other values of the parameters d, r and t, it may be possible to obtain better upper bounds, for instance in Corollary 3.5, which gives a stronger bound when d < r/t − 1. 5 Where n−1 k−1 q = (q n−1 −1)(q n−2 −1)...(q n−k+1 −1) is the Gaussian binomial coefficient, which counts the number of k-dimensional spaces that contain a given pair of points.
For the characterisation of equality, first observe that the inequality ( 11) is always tight if and only 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 because all degrees are equal to d, and the second because, as argued previously, a part is always a minimum cover in a regular r-partite r-uniform hypergraph.Thus we see that if have equality 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 hypergraph 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 contains exactly one vertex from any of the r parts, a part corresponds to a perfect matching in H D , with every vertex 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 a strictly t-intersecting d-regular r-partite r-uniform hypergraph with v edges, where r is the number of perfect matchings in the resolution of the design.From our above remarks, this implies the dual achieves 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 as possible, it must be strictly t-intersecting, with every pair of edges meeting in exactly t vertices.It is therefore natural to ask what happens when we drop the condition of regularity, and only require the (r, t)-graph be strictly t-intersecting.Such study has previously been carried out in the setting of Ryser's Conjecture for (1-)intersecting families.
Recall that Ryser's Conjecture for intersecting r-partite hypergraphs was proved by Tuza [29] for all r ≤ 5. Francetić, Herke, McKay and Wanless [9] 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 ≥ 1 and t < r ≤ t 2 + 3t − 1 be integers.If H is a strictly t-intersecting r-partite hypergraph, then τ (H) ≤ r − t.
Let v be an arbitrary vertex of H and let e be an edge through v.For 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 .This shows that d(v) ≥ r−1 r−t + 1, and, since v was arbitrary, we get δ ≥ r−1 r−t + 1. Solving the inequality in Lemma 3.3 for ∆, we obtain which, upon substituting δ ≥ r−1 r−t + 1 and τ ≥ r − t + 1, yields Now let u be a vertex of degree d(u) = ∆ and let f 0 be an edge through u.There are r−1 t−1 = r−1 r−t choices of t-subsets of f 0 containing u, and every edge f ′ = f 0 through u intersects f 0 in one of these sets.Since there are at least t r−1 r−t + t edges through u other than f 0 , by the pigeonhole principle there must exist a t-subset S of f 0 containing u for which there are at least t + 1 edges f 1 , . . ., f t+1 with f 0 ∩ 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 all 0 ≤ i ≤ t + 1.Since f 0 \ S, . . ., f t+1 \ S are disjoint sets, and e can only contain one vertex from each part 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 ≤ 2t, 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 Conjecture 1.2 for a wider range of parameters (r, t), we believe this is far from tight.Indeed, in this setting, we even lack constructions that come close to the smaller bound of Theorem 1.3.The best constructions we have found thus far are the duals of resolvable designs, as given in Corollary 3.5.As these are also regular, their cover numbers are smaller than r t , significantly smaller than the upper bound of Theorem 3.6.
Problem 1. Prove that τ (H) ≤ r t 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 each edge, 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 1 (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 edges intersects in at least t vertices, any edge e ∈ E(H) is an s-cover, and so we always have τ s (H) ≤ r.
The case s = 1 is obviously what we have been talking about all along, and the following lemma shows how we can leverage our constructions from that case to obtain lower bounds when s ≥ 2. Lemma 3.9.For all 1 ≤ s ≤ t ≤ r and a ≥ 1, 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.
The next proposition extends Theorem 1.4 to the case when s ≥ 2.

e 3 ∈
E(H) such that e 3 ∩ Y = ∅.We define a, b, c ∈ N as follows: a = |e 1 ∩ e 2 \ e 3 |, b = |e 1 ∩ e 3 \ e 2 | and c = |e 2 ∩ e 3 \ e 1 |.Observe that |e 1 ∩ e 2 ∩ e 3 | = t − a.Using the fact that e 3 ∩ Y = ∅, we can derive some bounds on the parameters a, b and c.Indeed, since e 3 is disjoint from Z, a ≥ z.As |e 1 ∩ e 3 | = b + (t − a) ≥ t, we must further have b ≥ a, while considering |e 2 ∩ e 3 | similarly shows c ≥ a. Finally, as e 3 is disjoint from X, we must have b ≤ r − t − x.Putting this all together, we have z ≤ a ≤ b ≤ r − t − x and a ≤ c.
holds is precisely what is needed to give an exact answer in this setting.