Generalizations and strengthenings of Ryser's conjecture

Ryser's conjecture says that for every $r$-partite hypergraph $H$ with matching number $\nu(H)$, the vertex cover number is at most $(r-1)\nu(H)$. This far reaching generalization of K\"onig's theorem is only known to be true for $r\leq 3$, or $\nu(G)=1$ and $r\leq 5$. An equivalent formulation of Ryser's conjecture is that in every $r$-edge coloring of a graph $G$ with independence number $\alpha(G)$, there exists at most $(r-1)\alpha(G)$ monochromatic connected subgraphs which cover the vertex set of $G$. We make the case that this latter formulation of Ryser's conjecture naturally leads to a variety of stronger conjectures and generalizations to hypergraphs and multipartite graphs. Regarding these generalizations and strengthenings, we survey the known results, improving upon some, and we introduce a collection of new problems and results.

A hypergraph H is intersecting if every pair of edges has non-empty intersection; equivalently H is intersecting if ν(H) = 1. The most well-studied special case of Ryser's conjecture is that for every r-partite intersecting hypergraph H, τ (H) ≤ r − 1.
Finally, we note that in [38], Ryser's conjecture was not orginally formulated in the way we have stated above. The original, equivalent, formulation is as follows: Let r ≥ 2 and let A be a r-dimensional 0, 1-matrix. The term rank of A, denoted ν(A), is the maximum number of 1's, such that no pair is in the same (r − 1)-dimensional hyperplane. The covering number of A, denoted τ (A), is the minimum number of (r − 1)-dimensional hyperplanes which contain all of the 1's of A. In this language, Ryser's conjecture says that if A is an r-dimensional 0, 1-matrix, then τ (A) ≤ (r − 1)ν(A). All three fractional versions of Ryser's conjecture are known to be true (that is, replacing at least one of τ or ν with τ * or ν * respectively):

Duality
We say that a hypergraph H is connected if for all u, v ∈ V (H) there exists e 1 , . . . , e k ∈ E(H) such that u ∈ e 1 , v ∈ e k and e i ∩ e i+1 = ∅ for all i ∈ equivalently, e ∈ E i . In an r-colored hypergraph H, a monochromatic cover of H is a set T of monochromatic connected subgraphs of H such that V (H) = ∪ T ∈T V (T ). For a positive integer t, a monochromatic t-cover of H is a monochromatic cover of order at most t. Let tc r (H) be the minimum integer t such that in every r-coloring of the edges of H, there exists a monochromatic t-cover of H. Note that since every connected subgraph contains a spanning tree, we can think of the connected subgraphs in a monochromatic cover as trees; this explains the notation "tc" which stands for "tree cover." In this language, the well known remark of Erdős and Rado that a graph or its complement is connected (see [11]), can be formulated as tc 2 (K n ) = 1.
Gyárfás [29] noted that Ryser's conjecture is equivalent to the following statement about edge colored graphs.  To see why this equivalence holds, given an r-colored graph G, we let H be a hypergraph where the vertex set is the set of monochromatic components in G which is naturally partitioned into r parts depending on the color of the component, and a set of vertices in H forms an edge if the corresponding set of components has non-empty intersection in G and is maximal with respect to this property. One can see that an independent set of order m in G will correspond to a matching of order m in H, and a monochromatic cover of G will correspond to a vertex cover of H.
On the other hand, given an r-partite hypergraph H with vertex set partitioned as {V 1 , . . . , V r }, we let G be a graph with V (G) = E(H) and we put an edge of color i between e, f ∈ V (G) if and only if e ∩ f ∩ V i = ∅. Since edges from H can intersect in more than one set, G will be an r-colored multigraph (or a r-multicolored graph) in which every monochromatic component is a clique. Note that a matching of order m in H will correspond to an independent set of order m in G, and a vertex cover of H will correspond to a monochromatic cover of G (in which all of the monochromatic components are cliques).
We have now seen that there are at least three equivalent ways of stating Ryser's conjecture. For the remainder of the paper we focus on these two.
(R2) For every graph G, tc r (G) ≤ (r − 1)α(G). Now suppose we have two r-colored graphs G and G on the same vertex set V such that for all i ∈ [r], the components of color i in G and the components of color i in G give the same partition of V . In the above discussion, we see that from G and G , we will derive the exact same r-partite hypergraph H. On the other hand, given an r-partite hypergraph H, we will only derive a single r-colored graph G.
This brings us to one of the main themes of this paper. It is certainly true that (R1) feels most natural in that it directly generalizes the well known König's theorem. However, by stating Ryser's conjecture in terms of (R2), we can access a whole host of interesting strengthenings and generalizations which have no analogue in the r-partite hypergraph setting. For instance we can ask if there is a monochromatic cover T in which every subgraph in T has small diameter, or whether T can be chosen so that subgraphs in T are pairwise disjoint (i.e. T forms a partition rather than just a cover). Furthermore, we can generalize the problem to settings like complete multipartite graphs and hypergraphs. In Section 8 we give many more such examples.
Finally, we make note of the following trivial upper bound on tc r (G) in the (R2) language.

Lower bounds
A projective plane of order q is a (q + 1)-uniform hypergraph on q 2 + q + 1 vertices and q 2 + q + 1 edges such that each pair of vertices is contained in exactly one edge. A truncated projective plane of order q is a (q + 1)-uniform hypergraph on q 2 + q vertices and q 2 edges obtained by deleting one vertex v from a projective plane of order q and removing the q + 1 edges which contained v. An affine plane of order q is a q-uniform hypergraph on q 2 vertices and q 2 + q edges obtained by deleting one edge e from a projective plane of order q and removing the q + 1 vertices which are contained in e. Note that truncated projective planes and affine planes are duals of each other in the geometric sense where the roles of lines and points are switched. It is well known that a projective plane of order q exists whenever q is a prime power (and it is unknown whether there exists a projective plane of non-prime power order). Also it is clear that a truncated projective plane of order q and an affine plane of order q exist if and only if a projective plane of order q exists.
A truncated projective plane H of order r − 1 is an intersecting r-uniform hypergraph with vertex cover number r − 1 and if we take ν vertex disjoint copies of H, we have an r-uniform hypergraph with matching number ν and vertex cover number (r −1)ν. Thus Ryser's conjecture is tight for a given value of r whenever a truncated projective plane H of order r − 1 exists.
An affine plane H of order r − 1 is an (r − 1)-uniform hypergraph with edge chromatic number r and edge cover number r − 1. From α vertex disjoint affine planes of order r − 1, we can create an r-colored graph G with independence number α such that tc r (G) = (r − 1)α.
So we have the following fact.
Fact 1.6. Let r ≥ 2 and α ≥ 1 be integers. If there exists an affine plane of order r − 1, then for all n ≥ (r −1) 2 α there exists a graph G on n vertices with α(G) = α such that tc r (G) ≥ (r −1)α.
Finding matching lower bounds when affine plane of order r − 1 does not exist is an active area of research with some interesting recent results ( [3], [2], [5], [37]); however, it is still unknown whether for all r ≥ 2 and α ≥ 1, there exists a graph G with α(G) = α such that tc r (G) ≥ (r − 1)α. The best general result is due to Haxell and Scott [37] who show that for all r ≥ 5, there exists a graph G such that tc r (G) ≥ (r − 4)α(G).
Finally, we note that most efforts to improve the lower bound have focused on the case α(G) = 1 because if one can prove that tc r (K) ≥ r − 1 for a complete graph K, then by taking α disjoint copies of K, we obtain a graph G such that tc r (G) ≥ (r − 1)α(G). It was shown in [36] that for r = 3, this is essentially the only such example. However, it was shown in [1] that for r = 4, there is an example which is different than two disjoint 4-colored complete graphs and a more general example was given in [13].

Large monochromatic components
We now briefly discuss the related problem of finding large monochromatic components in rcolored graphs. Theorem 1.7 (Füredi [26] (see Gyárfás [31])). In every r-coloring of the edges of a graph G with n vertices, there exists a monochromatic component of order at least n (r−1)α(G) . In the dual language, in every r-partite hypergraph H with n edges, there exists a vertex of degree at least n (r−1)ν(H) .
Let G be a graph. For sets A, B ⊆ V (G), an A, B-edge is an edge with one endpoint in A and the other in B. If A = {v}, we write v, B-edge instead of {v}, B-edge. We write δ(A, B) for min{|N (v) ∩ B| : v ∈ A}. If A and B are disjoint, we let [A, B] be the bipartite graph induced by the sets A and B. Given k ≥ 2, we let K k be the family of all complete k-partite graphs.
Given a set S we say that G is S-colored if the edges of G are colored from the set S. Given an integer r, we say that G is r-colored if the edges of G are colored with r colors (unless otherwise stated, the set of colors will be [r]). Given an r-coloring of G, say c :

Overview of the paper
In this section we give a detailed overview of the results in the paper. In addition, we discuss a variety of other generalizations and strengthenings of Ryser's conjecture in Section 8 and we collect some observations about a hypothetical minimal counterexample to Ryser's conjecture in Appendix A.

Monochromatic covers with restrictions on the colors
We begin with a few conjectures which can be stated both in terms of intersecting r-partite hypergraphs (R1) and in terms of r-colored complete graphs (R2).
The results mentioned here are proved in Section 3 and Section 5. In the dual (R1) language, Conjecture 2.1 says that for every r-partite intersecting hypergraph with vertex partition {V 1 , . . . , V r } and every S ⊆ [r], there is a vertex cover of order r − 1 which is contained in We prove Conjecture 2.1 for r ≤ 4. In the process of doing so, we formulate three other conjectures (all of which imply the α = 1 case of Ryser's conjecture).

Conjecture 2.2.
For all integers r ≥ 2 and all K ∈ K r , tc r−1 (K) ≤ r − 1. In particular, this implies that for every r-coloring of a complete graph K and every color i ∈ [r], either there is a monochromatic (r − 1)-cover consisting entirely of subgraphs of color i, or entirely of subgraphs which don't have color i.
In the dual (R1) language, Conjecture 2.2 says that for every r-partite intersecting hypergraph, if some part V i has at least r vertices, then there is a vertex cover of order at most r − 1 which uses no vertices from V i .
Note that Conjecture 2.2 implies the α = 1 case of Ryser's conjecture, but we will actually prove the following stronger conjecture for r ≤ 4. Conjecture 2.3. For all integers r ≥ 3 and all K ∈ K r−1 , tc r−1 (K) ≤ r − 1. In particular, this implies that for every r-coloring of a complete graph K and every color i ∈ [r], either there is a monochromatic (r − 2)-cover consisting entirely of subgraphs of color i, or a monochromatic (r − 1)-cover consisting entirely of subgraphs which don't have color i.
A special case of Conjecture 2.1 obtained by setting |S| = r/2 is the following. Conjecture 2.4. For all integers r ≥ 2, in every r-coloring of a complete graph K there exists a monochromatic (r−1)-cover such that the monochromatic subgraphs have at most r/2 different colors.
In the dual (R1) language, in every r-partite intersecting hypergraph, there is a vertex cover of order at most r − 1 which is made up of vertices from at most r/2 parts.
We give an example to show that r/2 cannot be reduced in Conjecture 2.4.
Example 2.5. For all r ≥ 3 and n ≥ r r r/2 +1 , there exists an r-coloring of K n , such that every monochromatic cover of K n with at most r − 1 components consists of components of at least r/2 different colors.
r/2 +1 (that is the family of subsets of [r] with just over half the elements). Now let V be a set of at least r r r/2 +1 vertices and let {V X : X ∈ A} be a partition of V into sets of order at least r which are indexed by the elements in A. For all u ∈ V X , v ∈ V Y , let uv be an edge of some arbitrarily chosen color i ∈ X ∩ Y (which is possible since X ∩ Y = ∅ for all X, Y ∈ A). We now have an r-colored complete graph K on vertex set V . Suppose for contradiction that there exists S ⊆ [r] with |S| = r/2 − 1 and that K has a monochromatic (r − 1)-cover T such that all of the subgraphs in T have a color in S. Since r − ( r/2 − 1) = r/2 + 1, there exists X ∈ A such that X = [r] \ S. This means that there are no edges having a color from S which are incident with a vertex in V X . Since there are at most r − 1 components in T all having colors from S and there are at least r vertices in V X , this contradicts the fact that T was the desired monochromatic cover.

Monochromatic covers with subgraphs of bounded diameter
Now we move on to some results which can only be stated in terms of r-colored graphs (R2).
The results mentioned here are proved in Section 4. Let G be a graph. For vertices u, v ∈ V (G), let d(u, v) denote the length of the shortest u, v- The radius of G, denoted rad(G), is the smallest integer r such that there exists u ∈ V (G) such that d(u, v) ≤ r for all v ∈ V (G).

6
It is well known that a graph or its complement has diameter at most 3; in other words, in every 2-coloring of a complete graph K, there is a spanning monochromatic subgraph of diameter at most 3.
Milićević conjectured an extension of this to r-colors which strengthens the α = 1 case of Ryser's conjecture. Conjecture 2.6 (Milićević [48]). For all r ≥ 2, there exists d = d(r) such that in every r-coloring of a complete graph K, there exists a monochromatic (r − 1)-cover consisting of subgraphs of diameter at most d.
Milićević proved that in every 3-coloring of a complete graph K, there is a monochromatic 2-cover consisting of subgraphs of diameter at most 8 [47], and in every 4-coloring of a complete graph K, there is a monochromatic 3-cover consisting of subgraphs of diameter at most 80 [48].
For the case r = 3, we improve the upper bound on the diameter from 8 to 4. In the case r = 4, we improve the upper bound on the diameter from 80 to 6 while at the same time giving a significantly simpler proof.
(i) In every 3-coloring of a complete graph K, there is a monochromatic 2-cover consisting of trees of diameter at most 4.
(ii) In every 4-coloring of a complete graph K, there is a monochromatic 3-cover consisting of subgraphs of diameter at most 6.
We also conjecture a generalization of Ryser's conjecture for graphs with arbitrary independence number.
Conjecture 2.8. For all α ≥ 1, there exists d = d(α) such that for all r ≥ 2, if G is a graph with α(G) = α, then in every r-coloring of G, there exists a monochromatic (r − 1)α-cover consisting of subgraphs of diameter at most d.
Note that in Conjecture 2.6, it is conjectured that d depends on r. We speculate that it is even possible to choose a d which is independent of both r and α, but we have no concrete evidence to support this.
Theorem 2.9. Let G be a graph with α(G) = 2. In every 2-coloring of G there is a monochromatic 2-cover consisting of subgraphs of diameter at most 6.
Gyárfás raised the following problem which would strengthen Theorem 1.7 in the case α = 1.
Problem 2.10 (Gyárfás [31]). In every r-coloring of the edges of K n , there exists a monochromatic subgraph of diameter at most 3 on at least n r−1 vertices. Perhaps the subgraph can even be chosen to be a tree of diameter at most 3 (which is necessarily a double star).
Theorem 2.11 (Letzter [44]). In every r-coloring of the edges of K n , there exists a monochromatic tree of diameter at most 4 (in fact, the tree can be chosen to be a triple star) on at least n r−1 vertices.
Note that Theorem 2.7(i) implies Letzter's result in the case r = 3 (except we can't guarantee that both of the trees are triple stars).

Monochromatic covers of complete multipartite graphs
The results mentioned here are proved in Section 4 and Section 5.
Gyárfás and Lehel made the following conjecture which would be tight if true.
Chen, Fujita, Gyárfás, Lehel, and Tóth [16] proved this for r ≤ 5. Also note that for all K ∈ K 2 , a trivial upper bound is tc r (K) ≤ 2r − 1 (by considering a pair of vertices u, v on opposite sides of the bipartition and the union of the monochromatic components containing u and v).
We now mention the following generalization of Conjecture 2.12 for which we don't even have a conjecture. The first interesting test case (outside the scope of Conjectures 2.2 and 2.3) is k = 3 and r = 4. Problem 2.13. Let k and r be integers with k, r ≥ 2. Determine an upper bound on tc r (K) which holds for all K ∈ K k .
We also make the following strengthening of Conjecture 2.12 and prove it for r = 2 and r = 3 (the r = 2 case is an improvement of a result of Milićević [47]).
Conjecture 2.14. There exists d such that for all r ≥ 2, if K ∈ K 2 , then in every r-coloring of K, there exists a monochromatic (2r − 2)-cover consisting of subgraphs of diameter at most d.
(i) In every 2-coloring of K, there is a monochromatic 2-cover consisting of trees of diameter at most 4.
(ii) In every 3-coloring of K, there is a monochromatic 4-cover consisting of subgraphs of diameter at most 6.

Partitioning into monochromatic connected subgraphs
The results mentioned here are proved in Section 6. For positive integers t and r, a monochromatic t-partition of an r-colored hypergraph H is a monochromatic t-cover T of H such that V (T ) ∩ V (T ) = ∅ for all T, T ∈ T . Let tp r (H) be the minimum integer t such that in every r-coloring of the edges of H, there exists a monochromatic t-partition of H.
Erdős, Gyárfás, and Pyber made the following conjecture and proved it for r = 3.
Later Fujita, Furuya, Gyárfás, and Tóth made the following conjecture and proved it for r = 2. Note that this is a significant strenghtening of Ryser's conjecture.
Haxell and Kohayakawa [35] proved tp r (K n ) ≤ r for sufficiently large n (in fact, they proved that there is a monochromatic r-partition consisting of trees of radius at most 2). The bound on n was improved in [8]. In Section 6, we discuss why the bound on n essentially cannot be improved any further using this approach, and in the process find an interesting connection to a different problem.
We also raised the question of determining an upper bound on tp r (K) for K ∈ K k . Surprisingly we found that in contrast to the cover version of the problem, no such upper bound (which depends only on k) is possible.
Theorem 2.18. For all k ≥ 2 and all functions f : Z + → R there exists K ∈ K k such that tp 2 (K) > f (k).

Monochromatic covers of hypergraphs
The results mentioned here are proved in Section 7.
Denote the complete r-uniform hypergraph on n vertices by K r n . Again, the well known remark of Erdős and Rado, which says tc 2 (K 2 n ) = 1, was generalized by Gyárfás [29] who proved that for all r ≥ 2, tc r (K r n ) = 1. Király [41] proved that for all k ≥ 3, tc r (K k n ) = r/k . In the dual (R1) language, this means that for k ≥ 3 if we have an r-partite hypergraph H in which every set of k edges has a common non-empty intersection, then τ (H) ≤ r/k .
We begin the study of a much more general setting in which we allow for different notions of connectivity in hypergraphs. Given an k-uniform hypergraph H, say that H is tightly connected if for every pair of vertices u, v ∈ V (H), there exists edges e 1 , . . . , e p ∈ E(H) such that u ∈ e 1 , v ∈ e p , and |e i ∩ e i+1 | = k − 1 for all i ∈ [p − 1]. We prove a generalization of Király's theorem, but we delay the statement until Section 7.
One of our main (and easiest to state) conjectures in this setting is the following strengthening of Gyárfás' result, which we prove for r = 3. Problem 2.20. Let r, k ≥ 2 be integers. Given an arbitary r-coloring of K k n , determine an upper bound on the number of monochromatic tightly connected subgraphs needed to cover V (K k n ).
The following tables given in Table 1 recap what is known about the various generalizations and strengthenings of Ryser's conjecture discussed so far, using green to indicate previously known results and yellow to indicate new or improved results that we will show in the following sections.
Finally, the case (r, α) = (4, 2) is claimed in [57] and [58], but no proof is given. Note that the cases (r, α) ∈ {(3, 1), (3,2), (3,3), (3,4)} are superseded by Aharoni's theorem [4], but Tuza's proof may still be of some interest because of its elementary nature. In all cases, Tuza's proofs are given in the dual (R1) language of vertex covers of r-partite hypergraphs. The objective of this section is to both reprove all of these results in the language of monochromatic covers of edge colored graphs and do so in such a way that we can use these results to prove Conjecture 2.4 for r ≤ 5 which in turn, together with the results in Section 5, allow us to prove Conjecture 2.1 for r ≤ 4. Also, since the r = 5 case is unpublished, we feel that this may be of some benefit to others who would like to understand Tuza's proof of this case. One of the original goals of this project was to explore the possibility of extending Tuza's methods to prove the case (r, α) = (6, 1). While we were unsuccessful in this goal, we were able to classify the (many) special cases which would need to be dealt with in order to prove such a result. More specifically, when (r, α) = (5, 1), Tuza's proof goes by making some general observations which, out of 37 possible cases, leaves two special cases each of which can be dealt with in an ad-hoc manner. In trying to extend this to the case (r, α) = (6, 1), we make analogous observations which, out of 560 possible cases, leaves 173 special cases (most of which do not seem to have an analogously easy ad-hoc proof).
We will prove the following. We begin with some general observations. The closure of a graph G with respect to a given coloring is a multigraphĜ on V (G) with edge set defined as follows: there is an edge of color i between u and v inĜ if and only if there is a path of color i between u and v in G.
Let the edges of a graph G be r-colored. Take the closure of G with respect to this coloring. Note that tc r (G) = tc r (Ĝ), since given a monochromatic cover ofĜ, the corresponding monochromatic components of G form a monochromatic cover.
Observation 3.2. In proving an upper bound on tc r (G) we will instead prove an upper bound on tc r (Ĝ); that is, we will assume that every monochromatic component in the r-edge (multi)coloring of G is a clique.
Let G i,j be the subgraph of G induced by the edges of colors i and j. By Theorem 1.4, Thus we have the following useful observation.  and let X = {x 1 , x 2 , x 3 } be an independent set in G 2,3 . This means every edge in X has color 1. Also since X is independent in G 2,3 , then by Observation 3.2, every vertex sends at most one edge of color 2 and at most one edge of color 3 to X. Thus every vertex sends an edge of color 1 to X and thus there is monochromatic cover consisting of a single component of color 1. We now claim that {A 1 , B 1 , B 2 } is the desired monochromatic 3-cover. If v ∈ A 1 , then v sends no edges of color 1 to X, at most one edge of color 3, at most one edge of color 4, and consequently at least two edges of color 2. Thus v must be in either B 1 or B 2 .

r = 5, α = 1
Let G be an r-colored graph and let X ⊆ V (G). For all i ∈ [r], the i-signature of X, denoted σ i [X], is the integer partition (n i 1 , . . . , n i t i ) of |X| such that the graph G i [X] induced by edges of color i in the set X has components of order n i Now let n and p be positive integers and let σ = {σ 1 , . . . , σ p } be a set of integer partitions of n. We say that σ is a valid signature, if there exists a p-coloring of a graph F on n vertices such that the [p]-signature of V (F ) is σ (note that a valid signature may be realized by nonisomorphic colored graphs). For example, {(4, 1), (3, 1, 1), (2, 2, 1)} is not a valid signature since there is no way to 3-color a K 5 so that there are components of order 4 and 1 in color 1, components of order 3, 1, and 1 in color 2, and components of order 2, 2, and 1 in color 3.
While we don't have a characterization of all valid signatures, the following is a useful necessary condition (and the above example shows that it is not sufficient), which follows simply by counting the number of possible edges.
, then there exists a monochromatic 4-cover in which all of the subgraphs have colors from [3].
Since the conditions in Lemma 3.5 are a bit hard to parse at first sight, note that (i) says that the number of components of color i or j plus the number of components of order at least 3 of color k in the graph induced by X is at most 4, and (ii) says that the number of components of color i plus the number of components of order at least 2 of color j or k in the graph induced by X is at most 4. For example, {(5), (3, 2), (3, 2)} and {(4, 1), (3, 1, 1), (3, 1, 1)} are valid signatures to which Lemma 3.5(i) and Lemma 3.5(ii), respectively, apply.
Proof. Let T denote the set of at most four monochromatic components which intersect X as described in one of the two cases. Suppose for contradiction that T is not a monochromatic 4-cover and let v be an uncovered vertex. In either case, this implies v ∈ X. First note that since G 4,5 [X] is an independent set, v sends at most one edge of color 4 and at most one edge of color 5 to X. Thus v sends at least three edges of color 1, 2, or 3 to X ( ).

(i) Without loss of generality we can assume T contains all components of colors 2 and 3
which intersect X and all components of of color 1 which intersect X in at least 3 vertices.
Then v sends no edges of color 2 or 3 to X, and at most 2 edges of color 1 to X; a contradiction to ( ).
(ii) Without loss of generality we can assume T contains all components of color 3 which intersect X and all components of colors 1 or 2 which intersect X in at least 2 vertices.
Then v sends no edges of color 3 to X, at most one edge of color 1, and at most one edge of color 2; a contradiction to ( ).
By direct inspection, one can see that there are only two valid signatures which do not meet the conditions of Lemma 3.5: {(3, 2), (3, 2), (3, 2)} and {(4, 1), (3,2), (3, 2)}. In both cases there are two components of each color which intersect X. Let A 1 , A 2 be the components of color 1 which intersect X, let B 1 , B 2 be the components of color 2 which intersect X, and let C 1 , C 2 be the components of color 3 which intersect X. Suppose that We now deal with these two cases separately.
Without loss of generality, we must have the following situation: then u must send one edge of color 4 and one edge of color 5 to {x 4 , x 5 } and three edges of color 1 to Without loss of generality, we must have the following situation: Suppose that neither {A 1 , A 2 , B 1 , B 2 } nor {A 1 , A 2 , C 1 , C 2 } are monochromatic 4-covers of K. Note that any vertex u which is not in A 1 ∪ A 2 ∪ B 1 ∪ B 2 must send one edge of color 4 and one edge of color 5 to {x 1 , x 2 } and three edges of color 3 to {x 3 , x 4 , x 5 } (so u ∈ C 1 ). Likewise any vertex v which is not in A 1 ∪ A 2 ∪ C 1 ∪ C 2 must send one edge of color 4 and one edge of color 5 to {x 3 , x 4 } and must send three edges of color 2 to {x 1 , The only possible color for the edge uv is color 1. Let A 3 be the component of color 1 which contains u and v. We now claim that {A 1 , A 2 , A 3 , B 1 } is a monochromatic cover. We establish this claim by showing that if w ∈ A 1 ∪ A 2 ∪ B 1 , then w must send an edge of color 1 to either u or v. So let w be such that w ∈ A 1 ∪ A 2 ∪ B 1 and suppose for contradiction that w does not send an edge of color 1 to {u, v}. If w sends an edge of color 3 to {x 1 , x 2 }, then w must send an edge of color 2 to {x 3 , x 4 } which further implies that w must send an edge of color 4 or 5, say color 5, to {x 5 }. Now w can only send edges of color 4 to {u, v}, but then this causes u and v to be in the same component of color 4, a contradiction. So suppose w does not send an edge of color 3 to {x 1 , x 2 }, which means w must send an edge of color 4 to {x 1 , x 2 } and an edge of color 5 to {x 1 , x 2 }, consequently w must send an edge of color 3 to {x 3 , x 4 , x 5 } (so w ∈ C 1 ). Now w is forced to send an edge of color 1 to v. This completes the case.
3.4 What we know for r = 6, α = 1 Let S ⊆ [6] with |S| = 2 and without loss of generality, suppose S = {5, 6}. If α(G 5,6 ) ≤ 5, then we are done by Observation 3.3; so suppose α(G 5,6 ) ≥ 6 and let We now split into cases depending on the [4]-signature of X. There are 1001 possible signatures, 560 of which are valid. The following two lemmas deal with 387 of the 560 cases 1 .
then there exists a monochromatic 5-cover in which all of the subgraphs have colors from [4].
Proof. Let T denote the set of at most five monochromatic components which intersect X as described in the three cases. Suppose for contradiction that T is not a monochromatic 5-cover and let v be an uncovered vertex. First note that since G 5,6 [X] is an independent set, v sends at most one edge of color 5 and at most one edge of color 6 to X, unless v ∈ X in which case v sends no edges of color 5 and no edges of color 6 to X. Thus in any case v sends at least four edges of color 1, 2, 3, or 4 to X ( ).
(i) Without loss of generality we can assume T contains all components of color 4 which intersect X and all components of color 1, 2, or 3 which intersect X in at least 2 vertices.
Thus v sends no edges of color 4 to X, and at most one edge of colors 1,2, or 3; a contradiction to ( ).
(ii) Without loss of generality we can assume T contains all components of color 3 and 4 which intersect X, all components of color 2 which intersect X in at least 3 vertices, and all components of color 1 which intersect X in at least 2 vertices. Thus v sends no edges of color 3 or 4 to X, at most two edges of color 2 to X, and at most one edge of color 1 to X; a contradiction to ( ).
(iii) Without loss of generality, we can assume T contains all components of colors 2, 3, and 4 which intersect X, and all components of color 1 which intersect X in at least 4 vertices.
Thus v sends no edges of color 2, 3 or 4 to X and at most three edges of color 1 to X; a contradiction to ( ).
Lemma 3.7. Let W ⊆ X and let then there exists a monochromatic 5-cover in which all of the subgraphs have colors from [4].
Proof. Let T denote the set of at most five monochromatic components which intersect W as described in the three cases. Suppose for contradiction that T is not a monochromatic 5-cover and let v be an uncovered vertex. First note that since G 5,6 [X] is an independent set, v sends at most one edge of color 5 and at most one edge of color 6 to X, unless v ∈ X in which case v sends no edges of color 5 and no edges of color 6 to X ( ).
(i) Note that T contains all components of colors 1, 2, 3, and 4 which intersect W , so v sends no edges of color 1, 2, 3, or 4 to W which together with ( ) and the fact that |W | = 3 is a contradiction.
(ii) Without loss of generality we can assume T contains all components of color 2, 3, and 4 which intersect W , and all components of color 1 which intersect W in at least 2 vertices.
Thus v sends at most one edge of color 1 to W which together with ( ) and the fact that |W | = 4 is a contradiction.
(iii) Without loss of generality, we can assume T contains all components of colors 3 and 4 which intersect W , and all components of color 1 or 2 which intersect W in at least 2 vertices. Thus v sends no edges of color 3 or 4 to W , at most one edge of color 1, and at most one edge of color 2, which together with ( ) and the fact that |W | = 5 is a contradiction.
(iv) Without loss of generality, we can assume T contains all components of colors 2, 3, and 4 which intersect W , and all components of color 1 which intersect W in at least 3 vertices.
Thus v sends no edges of color 2, 3, or 4 to W , at most two edges of color 1, which together with ( ) and the fact that |W | = 5 is a contradiction.
We are left with are 173 valid signatures for which an ad-hoc proof is needed (see Table 2).

Covering with monochromatic subgraphs of bounded diameter
The following is a well-known fact (see [60, Theorem 2.1.11]).
Also note that Proposition 4.1 is best possible. To see this, partition V as {V 1 , V 2 , V 3 , V 4 } and color all edges from V i to V i+1 with color 1 for all i ∈ [3] and color all other edges with color 2. Both G 1 and G 2 have diameter 3.
Let dc δ r (G) be the smallest integer t such that in every r-coloring of the edges of G, there exists a monochromatic t-cover T such that for all T ∈ T , diam(T ) ≤ δ. For r ≥ 1 and a graph G, let D r (G) be the smallest δ such that dc δ r (G) ≤ tc r (G). For instance, Proposition 4.1 implies dc 3 2 (K) = 1 for all complete graphs K (i.e. D 2 (K) = 3). Erdős and Fowler [22] proved that there exists a 2-coloring of K n such that every monochromatic subgraph of diameter at most 2 has order at most (3/4 + o(1))n and thus dc 2 2 (K n ) ≥ 2. Also by considering the edges incident with any vertex, we clearly have dc 2 2 (K) = 2 for all complete graphs K. In this language, Milićević conjectured the following strengthening of Ryser's conjecture.
We make the following more general conjecture which is also stronger in the sense that δ doesn't depend on r (we note that perhaps δ doesn't even depend on α).
Sometimes we will make the distinction between whether the subgraphs in our monochromatic cover are trees or not. Let tdc δ r (G) be the smallest integer t such that in every r-coloring of the edges of G, there exists a monochromatic t-cover T such that for all T ∈ T , T is a tree and diam(T ) ≤ δ. For r ≥ 1 and a graph G, let T D r (G) be the smallest δ such that tdc δ r (G) ≤ tc r (G). The following fact implies that tdc 2δ Note that by considering a random 2-coloring of K n , there is no monochromatic spanning tree of diameter 3, so T D 2 (K) ≥ 4. It is well-known (see [60, Exercise 2.1.49] and [11, Theorem 2.1]) that tdc 4 2 (K) = 1 for all complete graphs K and thus T D 2 (K) = 4. The following theorem summarizes the relevant results from [47] and [48]. We improve the bounds in each item of Theorem 4.5 and give a simpler proof for (iii).   We also prove analogous results in some new cases.    (Banach [9]). Every contracting operator on a complete metric space has a fixed point.
Austin [7] conjectured that every commuting contracting family {f 1 , f 2 , . . . , f r } of operators 2 on a complete metric space (M, d) has a common fixed point and proved it for r = 2. Milićević proved the case r = 3.
Theorem 4.12 (Milićević [47]). Every commuting contracting family {f 1 , f 2 , f 3 } of operators on a complete metric space has a common fixed point.
In the course of proving Theorem 4.12, Milićević requires a lemma which says that that there exists some δ (δ = 8 suffices) such that dc δ 3 (K) ≤ 2 for the countably infinite complete graph K.
We note that Milićević's proofs and our proofs apply equally well to finite or infinite (countable or uncountable) graphs.

Examples
Given a graph G, a blow-up of G is a graph obtained by replacing each vertex of G with an independent set and replacing each edge of G with a complete bipartite graph between the corresponding independent sets. A closed blow-up of G is a graph obtained by replacing each vertex of G with a clique and each edge of G with a complete bipartite graph between the corresponding cliques.  Proof. Take a red P 7 and note that its bipartite complement is a blue P 7 . Now take a blow-up of this example on n vertices coloring the edges between the sets according to the original coloring. If {H 1 , H 2 } is a monochromatic 2-cover, then for some i ∈ [2], H i contains vertices from at least four different sets which implies diam(H i ) ≥ 3.
Another example (provided m ≥ 4) comes from taking a red C 8 and noting that its bipartite complement is a blue C 8 . Proof. Trees of diameter 3 are double-stars. In a random 3-coloring of K n (for sufficiently large n), no two monochromatic double stars will cover the vertex set. Proof. As above, in a random 2-coloring of K m,n (for sufficiently large m, n), no two monochromatic double stars will cover the vertex set.

Complete graphs, r = 3
Proof of Theorem 4.6. Let x ∈ V (G). For i ∈ [3], let A i be the neighbors of x of color i. If For distinct i, j ∈ [3], define B ij to be the set of vertices v ∈ A i such that v sends no edge of color j to A j . Next, suppose there exist distinct i, j, k ∈ [3] such that B ij \ B ik = ∅. Without loss of generality say B 12 \ B 13 = ∅ and let z ∈ B 12 \ B 13 . Then there is a vertex u ∈ A 3 such that zu is color 3. Since every z, B 21 -edge is color 3, there are trees Note that it may be possible to improve the previous result by covering with two monochromatic subgraphs of diameter at most 3, but we cannot hope to cover with two monochromatic trees of diameter at most 3 (see Example 4.15).

Complete bipartite graphs, r = 2
Lemma 4.18. Let G be a complete bipartite graph with vertex classes X and Y . In any 2coloring of G, one of the following properties holds: (P1) There exists x 1 , x 2 ∈ X such that every edge incident with x i has color i or there exists y 1 , y 2 ∈ Y such that every edge incident with y i has color i. In this case, G can be covered by a color i tree of diameter at most 3 and color (3 − i) tree of diameter at most 2 for all i ∈ [2].
In this case, G i can be covered by two color i trees of diameter at most 3, for all i ∈ [2].
(P3) There exists i ∈ [2] such that G i has diameter at most 6 and G has a monochromatic 2-cover consisting of trees of diameter at most 4.
Proof. First suppose there exists x 1 , x 2 ∈ X such that every edge incident with x i has color i. Let y ∈ Y and note that the tree consisting of all color 1 edges incident with x 1 or y has diameter at most 3 and covers Y . The star consisting of all color 2 edges incident with y covers the remaining vertices in X and has diameter at most 2. So suppose that every vertex is incident with, say, a color 1 edge. If there exists x ∈ X such that every edge incident with x has color 1, then since every vertex is incident with an edge of color 1, we have that G contains a spanning tree of color 1 and diameter at most 4 in which case we are in (P3). Looking ahead to a potential improvement of the upper bound on the diameter in this result, we note that in this particular case (where there exists x ∈ X such that every edge incident with x has color 1) we can say more than just that we are in (P3). For any y ∈ Y , the tree consisting of all color 1 edges incident with x or y has diameter at most 3 and covers Y and the star consisting of all color 2 edges incident with y covers the remaining vertices in X and has diameter at most 2. So for the rest of the proof, suppose every vertex is incident with edges of both colors. Suppose both G 1 and G 2 are disconnected. Let {X 1 , X 2 } and {Y 1 , Y 2 } be partitions of X and Y respectively such that there are no color 2 edges from X 1 to Y 2 and no color 2 edges from X 2 to Y 1 . Note that X i = ∅ and Y i = ∅ for all i ∈ [2] since every vertex is incident with a color 2 edge. Thus [X 1 , Y 2 ] and [X 2 , Y 1 ] are complete bipartite graphs of color 1. Since we are assuming G 1 is disconnected, both [X 1 , Y 1 ] and [X 2 , Y 2 ] are complete bipartite graphs of color 2 and thus we have (P2).
Finally, suppose that at least one of G 1 and G 2 is connected and recall that every vertex is incident with edges of both colors. If diam(G i ) = 2 for some i ∈ Note that in the above proof, the only case in which we are not able to get a monochromatic 2-cover consisting of subgraphs of diameter at most 3 is the case where say G 1 has diameter 4 and G 2 can be covered by at most two subgraphs (trees), each of diameter at most 4. So if it were the case that D 2 (G) ≥ 4 for some G ∈ K 2 , then the example would have the property that both G 1 and G 2 have diameter exactly 4.
Note that Theorem 4.7 is a direct corollary of Lemma 4.18.
Later we will want to use a simpler version of Lemma 4.18 which doesn't make reference to the diameter of the subgraphs in the specific cases.
Lemma 4.19. Let G be a complete bipartite graph with vertex classes X and Y . In any 2coloring of G, one of the following properties holds: (P1 ) There exists x 1 , x 2 ∈ X such that every edge incident with x i has color i (in which case we say Y is the double covered side) or there exists y 1 , y 2 ∈ Y such that every edge incident with y i has color i (in which case we say that X is the double covered side).

Graphs with α(G) = 2, r = 2
Proof of Theorem 4.9. Let {x, y} be an independent set in G. Since α(G) = 2, every vertex in . Notice that if any A x , A 21 -edge were color 1, then So assume that every A x , A 21 -edge is color 2. Likewise we can assume that every A y , A 21 -edge is color    Figure 6: Set-up for the proof of Theorem 4.8 Since u sends a color k edge to A k and a color l edge to A l , Note that since every edge in [B ik , B ki ] has color j or l, we may suppose without loss of generality that u i u k has color j. Also every edge from u k to B ik has color j or l. Since u k ∈ B kl , u k sends an edge of color l to A l and since u i ∈ B ij , u i sends an edge of color j to A j . Thus letting   So suppose neither F 1 nor F 2 satisfy (P3). If both F 1 and F 2 satisfy (P2), then F 1 ∪ F 2 can be covered with at most two monochromatic subgraphs of color 4. If F 1 ∪ F 2 can be covered with exactly one monochromatic subgraph of color 4, let H 2 be this subgraph and we have the desired cover of V (K) with diam(H 1 ) ≤ 4 and diam(H 2 ) ≤ 6. If F 1 ∪ F 2 must be covered with two monochromatic subgraphs of color 4, let H 2 , H 3 be these subgraphs and we have the desired cover of V (K) with diam(H 1 ) ≤ 4, diam(H 2 ) ≤ 4, and diam(H 3 ) ≤ 2. Now suppose say F 2 satisfies (P2). If F 1 satisfies (P1 ) where B 12 is the double covered side, then letting H 2 be the nontrivial color 3 subgraph of F 1 and letting H 3 be the color 4 subgraph which covers the rest of F 1 ∪ F 2 , we have the desired cover of V (K) with diam(H 1 ) ≤ 4, diam(H 2 ) ≤ 3, and diam(H 3 ) ≤ 4. If F 1 satisfies (P1 ) where B 21 is the double covered side, then letting H 2 be the color 4 subgraph that covers B 21 along with the the color 4 subgraph of F 2 which it intersects and letting H 3 be the other color 4 subgraph of F 2 , we have the desired cover of V (K) with diam(H 1 ) ≤ 4, diam(H 2 ) ≤ 4, and diam(H 3 ) ≤ 3.
We may now suppose both F 1 and F 2 satisfy (P1 ). If say B 21 is the double covered side of

Complete graphs, r = 5
We note that in order to generalize of proof of Theorem 4.8 to prove Conjecture 4.2 for r = 5, it would be helpful to solve the following problem which is analogous to the last (main) case in the proof of Theorem 4.8.
] is a subgraph of color 3 which has diameter at most 6 and has We where X i is the double covered side. Since for any x ∈ X i \ (D 2 ∪ D 4 ), for every y ∈ Y 0 , the edge xy exists and is color 1 or 2, let which is in both the color 1 and color 2 component of [X 2 , Y 2 ], then we are done since every vertex in Y 0 is adjacent to x in color 1 or 2. So let x ∈ X 2 \ D 4 and suppose without loss of generality that x is only in the color 1 component. If x sends a color 1 edge to Y 0 , then we are done. Otherwise, x only sends color 2 edges to Y 0 . So if x sends a color 2 edge to Y 0 , then we are done by using the two color 2 subgraphs from [X 1 , Y 1 ], as one of these subgraphs has now been extended to cover all of Y 0 ; or else x only sends color 1 edges to Y 0 , in which case we are done by using the two color 1 subgraphs from [X 1 , Y 1 ], one of which contains Y 0 , together with the color 1 subgraph from [X 2 , Y 2 ], which contains Y 0 , and the color 2 subgraph from [X 2 , Y 2 ].
Lastly, assume (P2) holds for both [X 1 , If any vertex in Y 0 sends an edge of color 1 to X 2 , then we are done. Otherwise there is a vertex in X 2 which only sends color 2 edges to Y 0 in which case we are done.

Monochromatic covers of complete multipartite graphs
In this section, we prove Conjecture 2.3 for r ≤ 4. Let K k be the family of complete k-partite graphs. Lemma 4.19 implies the following (which was already known by [16], and was almost certainly a folklore result before that).
Proof. Let {V 1 , V 2 , V 3 } be the tripartition of K (we may assume K is 3-partite). First suppose there exists a monochromatic component C, say of color 3, which covers, say V 3 . Then either C covers all of V (K) and we are done, or K[(V 1 ∪ V 2 ) \ V (C), V 3 ] is a complete 2-colored bipartite graph and thus can be covered by two monochromatic components and we are done. So suppose for the remainder of the proof that for all monochromatic components C and all Proof. Let C be a monochromatic component and without loss of generality, suppose V (C) ∩ be the induced 2-colored complete bipartite graph, say the colors are red and blue. We apply Lemma 4.19 to K . By ( ), K cannot satisfy (P3 ). If (P1 ) is the case, then by ( ), it cannot be that V 3 is the double covered side, so (V 1 ∪ V 2 ) ∩ V (C) is the double covered side and thus we have a monochromatic component which has nontrivial intersection with all three parts. So finally, suppose (P2 ) is the case, and let {X 1 , X 2 } be the corresponding bipartition of V 3 and let {Y 1 , Y 2 } be the corresponding bipartition of ( then we have a monochromatic component which has nontrivial intersection with all three parts. Otherwise we have, without loss of generality, is say blue and every edge in [X 2 , Y 2 ] is then red. Since every edge from Y 2 to V 1 \ Y 1 is either red or blue, this gives us a monochromatic component which has nontrivial intersection with all three parts. Now by Claim 5.3, there exists a monochromatic component C so that and note that by ( ), all of these sets are non-empty. Note that the sets X 1 , Y 2 , Z 1 , X 2 , Y 1 , Z 2 form a 2-colored (say red and blue) blow-up of a C 6 . In the remainder of the proof, we implicity prove the general result that a 2-colored blow-up of a C 6 can be covered by at most 3 monochromatic components.
Claim 5.4. If there exists a monochromatic component covering any of Proof. Suppose without loss of generality that there is a monochromatic component covering is a 2-colored complete bipartite graph, we are done by Theorem 5.1.
We begin by focusing on the 2-colored (say red and blue) complete bipartite graphs K 1 = [Z 1 , X 2 ∪ Y 2 ] and K 2 = [Z 2 , X 1 ∪ Y 1 ], but note that [X 1 , Y 2 ] and [X 2 , Y 1 ] are also 2-colored complete bipartite graphs colored with red and blue. We apply Corollary 4.19 to each of K 1 and K 2 . Case 1 (K 1 or K 2 satisfies (P3 )) Say K 1 satisfies (P3 ). Since K 2 can be covered by at most two monochromatic components, we are done; thus we may assume that (P3 ) is never the case. Case 2 (K 1 or K 2 satisfies (P2 )) Without loss of generality, say K 1 satisfies (P2 ). This means there are two red components covering K 1 and there are two blue components covering K 1 .
Case 2.1 (K 2 satisfies (P2 )) There are two red components covering K 2 and there are two blue components covering K 2 . Using the fact that [X 1 , Y 2 ] is a 2-colored complete bipartite graph, there is a, say, red edge from X 1 to Y 2 . This red edge joins one of the red components covering K 1 to one of the red components covering K 2 and thus there are at most three red components covering K.
Case 2.2 (K 2 satisfies (P1 ) and X 1 ∪ Y 1 is the double covered side) So there is a red component R and a blue component B which together cover K 2 . Using the fact that [X 1 , Y 2 ] is a 2-colored complete bipartite graph, there is a, say, red edge from X 1 to Y 2 . This red edge joins one of the red components covering K 1 to the red component R and thus there are two red components and one blue component (B) which together cover K. Case 2.3 (K 2 satisfies (P1 ) and Z 2 is the double covered side) So there is a red component R and a blue component B which together cover K 2 . Using the fact that both [X 1 , Y 2 ] and [X 2 , Y 1 ] are 2-colored complete bipartite graphs, we either have that there is a blue edge from B to X 2 ∪ Y 2 in which case there are two blue components and one red component (R) which together cover K, or every edge from B to X 2 ∪ Y 2 is red and thus there are three red components which cover K. Case 3 (K 1 and K 2 both satisfy (P1 )) In K 1 we have that Z 1 is the double covered side or X 2 ∪ Y 2 is the double covered side, and in K 2 we have that Z 2 is the double covered side or X 1 ∪ Y 1 is the double covered side. For all i ∈ [2], let R i and B i be, respectively, the red and blue components covering K i .
We will split into two subcases. Case 3.1 (X 1 ∪ Y 1 is the double covered side or X 2 ∪ Y 2 is the double covered side) Without loss of generality, say X 2 ∪ Y 2 is the double covered side in K 1 . If there is a blue edge from B 2 to X 2 ∪ Y 2 , then B 1 and B 2 are contained together in a single blue component B and thus B, R 1 , R 2 forms a monochromatic cover of K. So suppose every edge from B 2 to X 2 ∪ Y 2 is red. So there is a red component R which covers R 1 and B 2 ∩ (V 1 ∪ V 2 ). Thus R, B 1 , R 2 forms a monochromatic cover of K.
Case 3.2 (Z 1 is the double covered side and Z 2 is the double covered side) If there is a blue edge between B 1 and B 2 or a red edge between R 1 and R 2 , we would have three monochromatic components which cover K, so suppose every edge between B 1 and B 2 is red and every edge between R 1 and R 2 is blue ( ). For Without loss of generality, suppose X 2 (R) = ∅ (which implies Y 2 (R) = ∅). If Y 1 (R) = ∅, then we must have X 1 (R) = ∅ and thus by ( ), we have that the complete bipartite graph [Y 2 (R), X 1 (R)] colored in blue, together with B 1 and B 2 form a monochromatic cover. So suppose Y 1 (R) = ∅.
If Y 1 (B) = ∅, then either every vertex in X 2 (B) sends a red edge to Y 1 (R) and thus there is red component covering X 2 ∪ Y 1 ∪ Z 2 and we are done by Claim 5.4, or there is a vertex in X 2 (B) which only sends blue edges to Y 1 (R) and thus there is a blue component covering Z 1 ∪ X 2 ∪ Y 1 and we are again done by Claim 5.4. So suppose Y 1 (B) = ∅.
By ( ), we have that every edge in [X 2 , Y 1 (B)] is red. So if there is a red edge from X 2 to Y 1 (R), then there is a red component covering X 2 ∪ Y 1 ∪ Z 2 and we are done by Claim 5.4. So suppose every edge from X 2 (B) to Y 1 (R) is blue, which implies there is a blue component B covering B 1 and Y 1 (R). Now either X 1 (R) = ∅ in which case every edge in [Y 2 (R), X 1 (R)] is blue and thus B, [Y 2 (R), X 1 (R)], B 2 forms the desired monochromatic cover, or X 1 (R) = ∅ in which case B, R 1 , B 2 forms the desired monochromatic cover.
We have by ( ) that every edge from Y 2 (R) to X 1 (R) is blue and every edge from Y 2 (B) to X 1 (B) is red. Suppose without loss of generality that there is a red edge from X 1 (R) to Y 2 (B) in which case there is a red component R covering V (R 1 ) ∪ Y 2 (B) ∪ X 1 (B) and thus R, R 2 and the red component covering X 2 (B) ∪ Y 1 (B) is the desired monochromatic cover.
The following example shows that in general, Conjecture 2.3 is best possible if true.
Example 5.5. For all k, r ≥ 2, there exists K ∈ K k such that tc r (K) ≥ r.
Proof. Let K ∈ K k in which one of the parts X has order at least r, and let x 1 , . . . , x r be r distinct vertices in X. For all i ∈ [r], color all edges incident with x i with color i and color all other edges arbitrarily.
The following is another example which shows that Theorem 5.2 is best possible with the additional property that all of the parts have order 2.
Finally, we see that Conjecture 2.1 holds for r = 3 by either Theorem 3.1 with r = 3 or Theorem 5.1, and Conjecture 2.1 holds for r = 4 by combining Theorem 3.1 with r = 4 and Theorem 5.2.
Note that if Conjecture 2.2 was true for r = 5 (or, even stronger, if Conjecture 2.3 is true for r − 1 = 4), then together with Theorem 3.1 with r = 5 this would imply Conjecture 2.1 is true for r = 5. 28 Erdős, Gyárfás, and Pyber proved the following theorem which strengthens the α = 1, r = 3 case of Ryser's conjecture.
Interestingly, no proof is known for infinite graphs (although the existence of a proof via personal communication is referenced in [23]). On the other hand, Hajnal proved [34, Theorem 1] the weaker result that tp r (K) ≤ r for all infinite (countable or uncountable) complete graphs K (in fact it can be specified that the trees have distinct colors and radius at most 2). Problem 6.2. Let K be a (countably) infinite complete graph. Prove tp 3 (K) = 2.
We know that tc 2 (G) ≤ α(G). Let R 1 , . . . , R p be the red components and let B 1 , . . . , B q be the blue components in such a monochromatic cover. Note that Since p + q + p + q ≤ 2α(G), we have say p + q ≤ α(G). So C 1 , . . . , C p , B 1 , . . . , B q is the desired monochromatic partition.
Haxell and Kohayakawa proved a weaker version of Conjecture 2.16 (but stronger in the sense that the subgraphs have bounded radius). Theorem 6.4 (Haxell,Kohayakawa [35]). Let r ≥ 2. If n ≥ 3r 4 r! ln r (1−1/r) 3(r−1) , then tp r (K n ) ≤ r. Furthermore, it can be specified that the trees have radius at most 2 and have distinct colors.
Given this result, it would be interesting to prove a bounded diameter strengthening of Theorem 6.1.
Problem 6.5. Does there exist a constant d such that in every 3-coloring of K n there exists a monochromatic 2-partition consisting of subgraphs of diameter at most d?
The lower bound on n in Theorem 6.4 was slightly improved by Bal and DeBiasio [8] to n ≥ 3r 2 r! ln r. The proofs in [35] and [8] go as follows: Construct a set X = {x 1 , . . . , x r } and disjoint set Y so that for all i ∈ [r], x i only sends edges of color i to Y . Then letting Z = V (K n ) \ (X ∪ Y ), we have an r-colored complete bipartite graph [Y, Z]. We say that Y has a good partition if there exists an integer 1 ≤ k ≤ r and a partition {Y 1 , . . . , Y k } of Y (allowing for parts of the partition to be empty) such that for all z ∈ Z, there exists i ∈ [k] and y ∈ Y i such that zy has color i. Then it is shown that if Y is large enough (equivalently, Z is small enough) then Y has a good partition. If Y has a good partition, then there exists a partition {Z 1 , . . . , Z k } of Z (allowing for parts of the partition to be empty) such that for all i ∈ [k] and z ∈ Z i , z sends an edge of color i to Y i . Thus for all i ∈ [k], the graph of color i induced on {x i } ∪ Y i ∪ Z i can be covered by a tree of radius at most 2.
The following lemma is a slight modification of the relevant lemma in [8].
Lemma 6.6. Let r ≥ 2 and let G ∈ K 2 with parts Y and Z, where Y is finite if r ≥ 3. If |Z| < ( r r−1 ) |Y | , then for every r-coloring of the edges of G there exists a good partition of Y . Proof. We say that a partition {Y 1 , . . . , Y k } of Y is good for z ∈ Z if there exists i ∈ [k] and y ∈ Y i such that zy has color i; otherwise we say that the partition of Y is bad for z.
For all z ∈ Z, there are (r − 1) |Y | partitions of Y which are bad for z. Since there exists a partition of Y which is good for every vertex in Z.
Our original intention was to come up with a new proof of Theorem 6.1 which would allow us to solve Problem 6.2 or Problem 6.5, but in the process we found an example to show that Lemma 6.6 is tight when r = 2 or Y is infinite. For each z ∈ Z b and y ∈ Y , color zy with b(y) (so the colors are 0 and 1). We also have an example which shows that Lemma 6.6 is close to tight when Y is finite and r ≥ 3. Example 6.8. Let r ≥ 3 and let G ∈ K 2 with parts Y and Z (where Y and Z are finite). If |Z| > 4|Y | ln r( r r−1 ) |Y | , then there exists an r-coloring of the edges of G such that Y does not have a good partition. This is obtained by showing that with positive probability, a random r-coloring of G does not have a good partition. However, we don't give the details here since this result will be superseded by an upcoming result with a better constant term.
With regards to Problem 6.2, Lemma 6.6 has the following consequence.
Corollary 6.9. Let K be a 3-colored complete graph on a set V . If there exists a maximal monochromatic component C (that is a monochromatic component which is not properly contained in a monochromatic component of another color) such that |V (C)| < 2 |V \V (C)| , then there exists a 2-partition of K. In particular, if V is countably infinite, then there is a 2-partition of K unless every maximal monochromatic component is cofinite.
Proof. By the assumption, let C be a maximal monochromatic component with |V (C)| < 2 |V \V (C)| and without loss of generality, suppose C is green. Set Z = V (C) and Y = V \ V (C).
By maximality of C, we may suppose that all edges between Y and Z are either red or blue. We apply Lemma 4.19 and note that we are done unless (P1 ) holds where Y is the double covered side (again by maximality of C). Now since |Z| < 2 |Y | , we can apply Lemma 6.6 to get a good partition {Y 1 , Y 2 } of Y and a corresponding partition The following corollary provides a proof of Theorem 2.18.
In particular, for all integers t ≥ 1 and k ≥ 2, there exists G ∈ K k such that tp 2 (G) > t.
Proof. We let Z = V i and Y = V (G) \ V i and color the edges between Y and Z as in Example 6.7 (where we partition Z into as equal sized sets as possible so that each part of the partition has at least |Z|/2 |Y | elements). Regardless of the edges inside the set Y , no matter how the set Y gets partitioned into red and blue subgraphs, there will be a part of the partition of Z which sends blue edges to the red subgraphs and red edges to the blue subgraphs.
The following question essentially asks whether the situation described in Corollary 6.10 is the only way to avoid having a 2-partition of a 2-colored multipartite graph. Problem 6.11. Is the following true? Let k ≥ 2 be an integer and let G ∈ K k with vertex partition Given an r-colored graph G and a color i ∈ [r], let G cross (i) be the multipartite graph consisting of the edges going between the components of color i. So if there are k components of color i, then G cross (i) is a k-partite graph colored with [r] \ {i}. Problem 6.12. Is the following true? There exists a 3-coloring of a complete graph such that for all i ∈ [3], there are at least three components of color i and there is no partition of G cross (i) into two monochromatic connected subgraphs.
Encouraged by the exact answer for r = 2 (from Lemma 6.6 and Example 6.7), we attempted to obtain a precise answer for r ≥ 3 (even though it wouldn't help improve the lower bound in Theorem 6.4 by any significant amount). Towards this end, for all integers r ≥ 2 and d ≥ 1, let Z(r, d) be the smallest positive integer z such that if G is a complete bipartite graph with parts Y and Z with |Y | = d and |Z| = z, then there exists an r-coloring of G in which there is no good partition of Y . In this language, we know from Lemma 6.6, Example 6.7, and Example 6.8 that Z(2, d) = 2 d for all d ≥ 1 and Problem 6.13. For all r ≥ 3 and d ≥ 1, determine Z(r, d).
We begin with a few simple observations. Observation 6.14. For all r ≥ 2 and d ≥ 1, (i) If r ≥ r, then Z(r , d) ≤ Z(r, d).
(iv) Z(r, r) ≤ r + r 2 + 1 Proof. Let G be a complete bipartite graph with parts Y and Z with |Y | = d and |Z| = z. (i) If there exists an r-coloring of G such that every partition of Y is bad, then since r ≥ r the r-coloring of G is an r -coloring of G such that every partition of Y is bad. (iv) Suppose z = r + r 2 + 1. Label the vertices of Z as v 1 , . . . , v r , u 1 , . . . , u r 2 +1 and label the vertices of Y as y 1 , . . . , y r . Consider the following coloring of G. For all i ∈ [r], all edges incident with v i get color i. For all i ∈ [ r 2 + 1], the edges from u i to {y 1 , . . . , y r 2 } are colored with i and the edges from u i to Y \{y 1 , . . . , y r 2 } are colored with i+1, except that the edges from u r/2 +1 to Y \{y 1 , . . . , y r 2 } are colored with 1. If there is a good partition {Y 1 , . . . , Y r } of Y , it must be the case that all sets in the partition are singletons because otherwise one of v 1 , . . . , v r would witness a bad partition. Also there is exactly one vertex u i ∈ {u 1 , . . . , u r 2 +1 } which is not satisfied by a vertex from {y 1 , . . . , y r 2 }; however, the only color that u i sends to Y \ {y 1 , . . . , y r 2 } has already been used on {y 1 , . . . , y r 2 }.
We were able to compute some small values of Z(r, d) using an integer linear program. Surprisingly, we didn't even have enough computing power to determine Z(4, 4) or Z (3,5).
Note  We now show that Z(r, d) is equivalent to two other well-studied parameters whose bounds seem to be difficult to improve in general.
Let Theorem 6.15. For all r ≥ 2 and d ≥ 1, Z(r, d) = γ t (K ×d r ) = τ (H(r, d)) Proof. Let r ≥ 2 and d ≥ 1 be given. First note that γ t (K ×d r ) = τ (H(r, d)) since the vertex sets of H(r, d) and K ×d r correspond to each other, and the edges of H(r, d) correspond to the neighborhoods of vertices in K ×d r . Clearly a transversal in H(r, d) corresponds to a total dominating set in K ×d r . To see that Z(r, d) = γ t (K ×d r ), suppose that we have a total dominating set T of order z in K ×d r , each vertex of which is a vector of length d over the alphabet {0, . . . , r − 1}. Now let Z be a set of z vertices and for each vertex in Z, color the edges according to the corresponding vertex (vector) from T . Every partition of Y now corresponds to a vertex (x 1 , . . . , x d ) in V (K ×d r ) and since T is a total dominating set (and the definition of K ×d r ), there exists a vertex (x 1 , . . . , x d ) ∈ T such that x i = x i for all i ∈ [d] which means (x 1 , . . . , x d ) is a bad partition of Y . On the other hand if Z is a set of z − 1 vertices, then since every set T of z − 1 vertices in K ×d r is not a total dominating set, there exists a vertex (x 1 , . . . , x d ) in V (K ×d r ) which is not adjacent to anything in T and this vertex corresponds to a good partition of Y .
The following is a known fact about the total domination number of a graph (see [39]).
Proposition 6.18. Let G be a graph on n vertices with minimum degree δ and maximum degree ∆. Then n ∆ ≤ γ t (G) ≤ 1+ln δ δ n We have |V (K ×d r )| = r d and δ(K ×d r ) = ∆(K ×d r ) = (r − 1) d and thus we have the following corollary.
Note that by Theorem 6.15, Corollary 6.17 and Corollary 6.19 can be derived from each other; however, it is interesting to note that they can be derived independently using the known bounds from Proposition 6.16 and Proposition 6.18 respectively.

Monochromatic covers of hypergraphs
The α = 1 case of Ryser's conjecture says tc r (K 2 n ) ≤ r − 1. Király [41] surprisingly gave a very simple proof that for all k ≥ 3, tc r (K k n ) = r/k . Earlier, Aharoni and Ziv [6] proved that for k ≥ 3, tc r (K k n ) ≤ r−1 k−1 (they proved this in the dual language of r-partite hypergraphs in which every k edges intersect). Part of the reason determining tc r (K k n ) is so much easier for k ≥ 3 than k = 2 seems to come down to the very weak notion of connectivity typically used for hypergraphs. Inspired by some recent results ( [17], [18], [19], [27]), we propose a more general problem which allows for stronger notions of connectivity in hypergraphs.
Let c, , k be positive integers with k ≥ 2 and c, ≤ k − 1 and let H be a k-uniform hypergraph. Say that a pair of c-sets S, S ∈ V (H) Let tc c, r (H) be the smallest integer t such that in every r-coloring of the edges of H, there exists a set of at most t monochromatic (c, )-components C (that is, each C ∈ C is a component in When c = , we write tc r (H) to mean tc , r (H).
In this language, we can state Király's result as follows. We will also give Király's proof of the upper bound.
Proof. If r = 1, the result is trivial, so let r ≥ 2 and suppose that tc 1 r−1 (K k n ) ≤ (r − 1)/k . If there exists a set S of k − 1 vertices such that S is contained in edges of at most r/k colors, then we are done. So for every set S ⊆ V (K) of order k − 1, S is contained in edges of at least r/k + 1 colors. For every edge e of color r, there are k distinct k − 1 sets contained in e and thus there are distinct S, S ⊆ e with |S| = k −1 = |S | and i ∈ [r −1] such that S and S are contained in a component of color i which implies that e is contained in a component of color i. Since all of the edges of color r are contained in a component of color i ∈ [r − 1], we actually have an (r − 1)-coloring of K and thus by induction there is a monochromatic (r − 1)/k -cover (which is of course a r/k -cover).
We propose the following general problem.
Problem 7.2. Let r, c, , k be positive integers such that c, ≤ k − 1. Determine the value of tc c, r (K k n ).
We prove the following results.
Theorem 7.3. Let r, c, , k be positive integers such that 1 ≤ ≤ c ≤ k/3. Then Note that this gives Theorem 7.1 when c = 1 = .
In the case when r = 2, we are essentially able to give a complete answer.
Theorem 7.4. Let c, , k be positive integers such that , c ≤ k − 1. Then The case c < is harder to analyze, but we are able to determine one interesting case exactly.

Lower bounds
The following example generalizes Király's example (which corresponds to the case c = = 1) and provides the lower bound in Theorem 7.3. Example 7.6. For all c, ≥ 1, r ≥ 2, k ≥ 3 and n ≥ c · r r k/c −1 , tc c, r (K k n ) ≥ r k/c .
Proof. Set t := k/c and q := r/t − 1. Let K = K k n and partition V (K) into m := r q sets V x 1 , . . . , V xm of order at least c, where x 1 , . . . , x m represent each of the subsets of [r] of order q. For each edge e ∈ E(K), let φ(e) = i:|e∩Vx i |≥c x i . Since |e| = k < ( k/c + 1)c = (t + 1)c, e intersects at most t of the sets V x 1 , . . . , V xm in at least c vertices, so |φ(e)| ≤ tq < r and thus [r] \ φ(e) = ∅. Color e with the smallest j ∈ [r] \ φ(e). Now let A ⊆ [r] with |A| = q and note that there exists i such that A = x i . Note that no (c, )-component having a color in A contains any of the c-sets from V x i and thus the number of (c, )-components needed to cover V (K) c is more than q; i.e. tc c, r (K k n ) ≥ q + 1 = r k/c .
The next example provides the lower bound in the last case of Theorem 7.4.
Proof. Set t := n/c , let K = K k n , and choose t disjoint sets x 1 , . . . , x t ⊆ V (K) each of order c. Let X = {x 1 , . . . , x t }. First note that since c > k − (1 − 1/r) we have r(c + − k) > r(c + − (c + (1 − 1/r) )) = . (2) For each -set y ∈ V (K) let (equivalently I y = {i ∈ [t] : y ∪ x i ⊆ e ∈ E(K)}) and note that |I y | ≤ r − 1 as otherwise = |y| ≥ i∈Iy |y ∩ x i | ≥ r( + c − k), contradicting (2). Let φ y be an injective function from I y to [r − 1] and for all i ∈ I y , color all edges containing y ∪ x i with color φ y (i). Now color all other edges with color r. Since c > 2k, no edge in E(K) contains more than one element of x as a subset. So by this fact and the way in the which the coloring was defined, no pair of c-sets from X is in the same monochromatic (c, )component.

c ≥
Observation 7.9. Let k−1 ≥ c ≥ ≥ 1, r ≥ 2, and let H be an r-colored k-uniform hypergraph. Let 2 ≤ k ≤ k c and let G be a k -uniform hypergraph on vertex set V (H) c where {S 1 , . . . , S k } ∈ E(G) if and only if there exists e ∈ E(H) such that S 1 ∪ · · · ∪ S k ⊆ e. Furthermore, color {S 1 , . . . , S k } ∈ E(G) with the color of the edge e ∈ E(H) such that S 1 ∪ · · · ∪ S k ⊆ e, and note that {S 1 , . . . , S k } may receive more than one color.
Proof. Let K := K k n with a given r-coloring of the edges. Suppose k/2 < c ≤ k −(1−1/r) . First note that given a c-set A and an -set B, there exists e ∈ E(K) such that A ∪ B ⊆ e if and only if k ≥ |A ∪ B| = |A| + |B| − |A ∩ B| = c + − |A ∩ B|; i.e. |A ∩ B| ≥ c + − k.
Given any family of r + 1 c-sets X = {X 1 , . . . , X r+1 }, since r(c + − k) ≤ r(c + − (c + (1 − 1/r) )) = , for every set of r elements of X, there exists an -set which is contained in an edge with each of the r elements. Furthermore, since c ≥ 2 /r (using c ≥ and r ≥ 2) we have (r + 1)(c + − k) ≤ k, and we can choose a family of r + 1 -sets Y = {Y 1 , . . . , Y r+1 } such that Y i is contained in an edge with every element in X \ {X i } and |Y 1 ∪ · · · ∪ Y r+1 | ≤ k which implies that every pair Y i , Y j is contained in the same edge of K of some color, say r. So if any two sets X i , X j are both contained in an edge of color r with an element in Y , there would be a -walk of color r between X i and X j in K. So suppose that at most one element from X, say X r+1 , is contained in an edge of color r with some element of Y . However, now Y r+1 is contained in an edge with every element in X \ {X r+1 } and since there are only r − 1 colors used on such edges, there is a monochromatic -walk between some X i and X j in K. Altogether, this implies that there is a monochromatic -walk between some pair of distinct elements from X, so the closure of G (the auxiliary graphĜ with an edge of color i between any two vertices (c-sets) which have an -walk of color i between them) has independence number at most r and thus by Observation 3.2 and Fact 1.5, we have tc r (G) ≤ tc r (Ĝ) ≤ rα(Ĝ) ≤ r 2 . If r = 2, then Theorem 1.4 applies and we have tc 2 (G) ≤ tc 2 (Ĝ) ≤ α(Ĝ) ≤ 2. If r = 3, then Aharoni's theorem [4] applies (in the dual language) and we have tc 3 (G) ≤ tc 3 (Ĝ) ≤ 2α(Ĝ) ≤ 6.
We now prove an upper bound on tc c, r (K k n ) when c ≤ k/2 and c ≥ . In particular, when c ≤ k/3, this provides the upper bound for Theorem 7.3.
Note that previous observation in particular implies tc 1,k−1 r (K k n ) ≤ tc 1 r (K 2 n ) = tc r (K n ). The first interesting test case for c < is tc 1,2 3 (K 3 n ). We have tc 1,2 3 (K 3 n ) ≤ tc 1 3 (K 2 n ) = 2 from above, but perhaps, tc 1,2 3 (K 3 n ) = 1? We now show that this is indeed the case by more carefully considering the possible structures in the 3-colored link graph of a vertex.
The following Lemma appears in [20], but we reproduce it here for completeness.
Lemma 7.14. Let K be a complete graph. For every 3-coloring of K, either (i) there exists a monochromatic connected subgraph on n vertices, or Proof of Theorem 7.5. Let K = K 3 n and let u ∈ V (K). If the link graph K(u) is connected in any color, then we are done (as in the proof of Observation 7.8(iv)). So by Lemma 7.14, there are two cases (type (ii) and type (iii)). We will consider how the edges which do not contain u (which have order 3) interact with the link graph K(u) (which is a 2-uniform hypergraph).
Claim 7.15. (i) Let H be a connected color i subgraph in the link graph K(u) for some i ∈ [3].
If for all c ∈ V (K(u)) \ V (H), there exists ab ∈ E(H) such that abc is a color i edge of K, then there is a monochromatic spanning tight component of K. Problem 7.17. Determine mc c, r (K k n ). In particular, determine mc 1,2 r (K 3 n ) for r ≥ 4.