Decompositions into isomorphic rainbow spanning trees
Stefan Glock, Daniela K\"uhn, Richard Montgomery, Deryk Osthus

TL;DR
This paper proves that large complete graphs with optimal edge colourings can be decomposed into isomorphic rainbow spanning trees, confirming longstanding conjectures in graph theory.
Contribution
It establishes the existence of such decompositions in large graphs, resolving conjectures by Brualdi--Hollingsworth and Constantine.
Findings
Decomposition of large complete graphs into isomorphic rainbow spanning trees is always possible.
Confirms conjectures for sufficiently large graphs.
Provides a new structural understanding of rainbow decompositions.
Abstract
A subgraph of an edge-coloured graph is called rainbow if all its edges have distinct colours. Our main result implies that, given any optimal colouring of a sufficiently large complete graph , there exists a decomposition of into isomorphic rainbow spanning trees. This settles conjectures of Brualdi--Hollingsworth (from 1996) and Constantine (from 2002) for large graphs.
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Decompositions into isomorphic rainbow spanning trees
Stefan Glock, Daniela Kühn, Richard Montgomery and Deryk Osthus
School of Mathematics, University of Birmingham, Edgbaston, Birmingham, B15 2TT, United Kingdom
[s.glock,d.kuhn,r.h.montgomery,d.osthus]@bham.ac.uk
Abstract.
A subgraph of an edge-coloured graph is called rainbow if all its edges have distinct colours. Our main result implies that, given any optimal colouring of a sufficiently large complete graph , there exists a decomposition of into isomorphic rainbow spanning trees. This settles conjectures of Brualdi–Hollingsworth (from 1996) and Constantine (from 2002) for large graphs.
This project has received partial funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (grant agreement no. 786198, D. Kühn and D. Osthus). The research leading to these results was also partially supported by the EPSRC, grant nos. EP/N019504/1 (S. Glock and D. Kühn) and EP/S00100X/1 (D. Osthus), as well as the Royal Society and the Wolfson Foundation (D. Kühn).
1. Introduction
Given an edge-coloured graph , we say a subgraph is rainbow if all the edges of have distinct colours. Moreover, we say that decompose if are edge-disjoint subgraphs of covering all the edges of .
In this paper, we address the problem of decomposing an optimally edge-coloured complete graph into (isomorphic) rainbow spanning trees. The study of rainbow decomposition problems can be traced back to the work of Euler, who investigated for which one can find a pair of orthogonal Latin squares of order . That is, equivalently, for which does there exist an optimally edge-coloured which can be decomposed into rainbow perfect matchings? Euler gave a construction for all and conjectured that these are the only admissible values. His conjecture was disproved by Parker, Bose and Shrikhande who provided constructions for the missing values, except for (which corresponds to Euler’s famous ‘36 officers problem’, for which the non-existence had already been shown by Tarry in 1901).
On the other hand, given an arbitrary optimally edge-coloured , a decomposition into rainbow perfect matchings need not exist. In fact, there are examples of such colourings that do not admit a single rainbow perfect matching. (An important conjecture widely attributed to Ryser–Brualdi–Stein postulates that there always exists a rainbow matching of size .)
Perfect matchings are, in some sense, very rigid objects, and it is natural to ask analogous questions for other types of subgraphs. In particular, several natural conjectures arose concerning decompositions into rainbow spanning trees. Here, the most notable are the Brualdi–Hollingsworth conjecture, Constantine’s conjecture and the Kaneko–Kano–Suzuki conjecture. Our main result implies the first two of these.
1.1. Decompositions into rainbow spanning trees
Note that if is optimally edge-coloured, then the colour classes form a -factorization, that is, a decomposition of into perfect matchings. We will here use the term -factorization synonymously with an edge-colouring whose colour classes form a -factorization. Note that if a -factorization of exists, then is even. We now state the Brualdi–Hollingsworth conjecture.
Conjecture 1** (Brualdi and Hollingsworth, [6]).**
For all and any -factorization of , there exists a decomposition of into rainbow spanning trees.
Note that the condition that is necessary. Suppose a -factorization of is given. Clearly, there always exists one rainbow spanning tree, for instance the star at any vertex. Brualdi and Hollingsworth in their original paper [6] showed that one can find two edge-disjoint rainbow spanning trees. Shortly afterwards, Krussel, Marshall and Verrall [17] were able to find three such trees. Later, Horn [13] significantly improved on this by finding edge-disjoint rainbow spanning trees. Very recently, Montgomery, Pokrovskiy, and Sudakov [21] proved Conjecture 1 approximately by showing that one can guarantee edge-disjoint rainbow spanning trees.
Several related problems have been studied, in two main directions. Firstly, we may wish to strengthen the conditions on the trees in the decomposition, most commonly by requiring the trees in the decomposition to be isomorphic. Secondly, we may wish to weaken the conditions on the colouring, most commonly by allowing non-optimal proper edge-colourings. (Naturally, those two directions can also be combined.)
A decomposition of an edge-coloured into isomorphic rainbow spanning trees is also known in the literature as a multicoloured tree parallelism (MTP). It turns out that the problem of finding an MTP is non-trivial even if one is allowed to choose the -factorization. Partial results were obtained by Constantine [9]. Akbari, Alipour, Fu and Lo [2] then proved that for all with , there exists a -factorization of which admits an MTP. Moreover, Constantine conjectured that in fact an MTP should exist for any given -factorization, thus generalizing the Brualdi–Hollingsworth conjecture.
Conjecture 2** (Constantine, [9, 10]).**
For all , any -factorization of admits a decomposition into isomorphic rainbow spanning trees.
Unsurprisingly, for this conjecture, much less was known than for the Brualdi–Hollingsworth conjecture. In [12] it was shown that three isomorphic rainbow spanning trees can be guaranteed. In [23], Pokrovskiy and Sudakov showed that one can find edge-disjoint rainbow spanning trees all isomorphic to a so-called -spider (which is even independent of the given -factorization). Montgomery, Pokrovskiy, and Sudakov [21], and independently, Kim, Kühn, Kupavskii and Osthus [16], proved a weak asymptotic version of the conjecture by showing that there are edge-disjoint rainbow paths each of length .
We now discuss results on more general colourings. Intuitively it might seem that dealing with an optimal colouring is the hardest case, as having more colours should make finding rainbow subgraphs easier. However, non-optimal colourings seem genuinely harder to deal with than -factorizations. Kaneko, Kano and Suzuki proved that for any proper colouring of , there exist three edge-disjoint rainbow spanning trees, and also generalized the Brualdi–Hollingsworth conjecture as follows.
Conjecture 3** (Kaneko, Kano, and Suzuki, 2002).**
For all , every properly edge-coloured contains edge-disjoint rainbow spanning trees.
Note that any proper colouring is -bounded, that is, every colour appears on at most edges. Under the weaker assumption that the colouring is -bounded, Akbari and Alipour [1] showed that one can guarantee two edge-disjoint rainbow spanning trees, and this was significantly improved by Carraher, Hartke, and Horn [8] who showed that such trees exist. For proper colourings, a linear number of rainbow spanning trees was independently obtained by Pokrovskiy and Sudakov [23] and by Balogh, Liu and Montgomery [5], where in the former work, the trees are even isomorphic. Finally, the aforementioned result from [21] on Conjecture 1 also applies to proper colourings, thus proving Conjecture 3 approximately.
We now state our main theorem, which implies the Brualdi–Hollingsworth conjecture and Constantine’s conjecture for large . This is the first general exact rainbow decomposition result for spanning subgraphs, where each subgraph in the decomposition has to use all the colours.
Theorem 4**.**
For all sufficiently large , there exists a tree on vertices such that for any -factorization of , there exists a decomposition into rainbow subgraphs each isomorphic to .
Note that whereas Constantine’s conjecture says that given a -factorization, one can decompose into isomorphic rainbow spanning trees, we actually show that one can use the same tree for any -factorization. This tree is made up of a path of length , with short paths attached to it (see Definition 5). By modifying our proof slightly, we can even ensure that . This is best possible in the sense that there exist -factorizations which do not admit a single rainbow Hamilton path [18].
Our argument relies upon the fact that the colouring is a -factorization. It would be very interesting to prove the result for more general colourings, in particular proper colourings.
It would also be interesting to investigate the -bounded setting further. The best known bound is the one from [8] mentioned earlier, which provides edge-disjoint rainbow spanning trees. A natural question is to ask for the maximum number of such trees that can be guaranteed. It seems unlikely that a decomposition can be obtained, but it would be interesting to see whether is possible or not. It is also natural to impose further local conditions on the colouring, e.g. that the colouring is locally -bounded, which means that the maximum degree of each colour class is at most . For instance, in [16] it is shown that for any -bounded colouring which is locally -bounded, there exists an approximate decomposition into almost spanning rainbow cycles (and thus into almost spanning paths).
1.2. Related problems
We now discuss some related results concerning rainbow decompositions. Let us first revisit the perfect matching case. As mentioned earlier, there exist proper optimal colourings of which do not contain a rainbow perfect matching. However, by imposing slightly stronger boundedness conditions on the colouring, one can obtain strong results. For example, Alon, Spencer and Tetali [3] showed that if is a power of and the edge-colouring is -bounded (and not necessarily proper), there exists a decomposition into rainbow perfect matchings. Montgomery, Pokrovskiy and Sudakov [21] showed that any proper edge-colouring of , where at most colours appear more than times, contains edge-disjoint rainbow perfect matchings. This implied a conjecture of Akbari and Alipour in a strong form (which was proved independently by Keevash and Yepremyan [15]) and a conjecture of Barat and Nagy approximately, both for large . Kim, Kühn, Kupavskii and Osthus [16] proved that for any -bounded and locally -bounded edge-colouring of , there exist edge-disjoint rainbow perfect matchings. The authors of both [16] and [21] also obtain analogous results (in their respective settings) on approximate decompositions of into rainbow Hamilton cycles. Furthermore, [16] contains results for approximate decompositions of into rainbow -factors (for any given graph ).
A further tantalizing problem concerning rainbow tree decompositions is the following special case of Rota’s basis conjecture. Let be spanning trees on a common vertex set of size , each monochromatic in a different colour. Then their union (allowing multiple edges) can be decomposed into rainbow spanning trees. The general version of Rota’s conjecture concerns the rearrangement of bases of a matroid into disjoint transversal bases. Recently, Bucić, Kwan, Pokrovskiy and Sudakov [7] showed that disjoint transversal bases can be found.
2. Notation
Given a graph with edge colouring , we say a subgraph is -rainbow if is rainbow and . We refer to an edge with colour as a -edge, and is a -neighbour of . For each colour , is the set of -edges in . For each vertex of , we let denote the set of all edges of incident to . For any , is the common neighbourhood in of the vertices in . For any and , is the number of neighbours of in . We denote by the graph obtained from by deleting the edges of .
For a hypergraph , let denote its maximum codegree, that is, the maximum number of edges containing any two fixed vertices.
Given a set and , a -random subset is a random subset which is obtained by including each element of independently with probability . If not otherwise stated, we always assume that such random subsets are independent. For instance, if we say that is a -random subset of and is a -random subset of , we implicitly assume that these random choices are made independently. Similarly, if is a graph, then a -random subgraph is the random graph with vertex set and a -random subset of as edge set.
On the other hand, we often split a random subset further into disjoint subsets. For instance, if is a -random subset of , we might say that we split into a -random set and a -random set , by which we mean that for each independently, we include in with probability and into otherwise. Note that then is indeed a -random subset of and is a -random subset of , but they are obviously not independent. To split into more sets, we use the following notation: By splitting randomly as
[TABLE]
we mean that for every element in independently, we choose an index according to the probability distribution , and put this element into the corresponding set .
We say that a random event holds with high probability if the probability that it holds tends to as tends to infinity (where is usually the number of vertices and the event depends on ).
We write . For , we write whenever . For , we write in our statements to mean that there are increasing functions such that whenever and , then the subsequent result holds.
3. Proof sketch
Our proof is based on hypergraph matching results and new absorption techniques. Suppose we are given a 1-factorization of the complete graph with colour set . We build the rainbow trees simultaneously, beginning with our absorbing structures and then gradually extending these structures to cover all the vertices and edges. For this, we further develop a recent ‘distributive’ form of the absorption method: we form an absorption structure along with a reservoir, such that, given any subset (of given size) from the reservoir we can distribute the elements of this subset among the different parts of the absorbing structure to always obtain a copy of the same tree. More precisely, we create a ‘global’ reservoir of edges, as well as ‘local’ reservoirs of colours and vertices (as explained below, ‘local’ refers to the fact that there is one such reservoir for each tree, while the ‘global’ reservoir is common to all trees). The structure of these absorbers and the corresponding reservoirs is described in more detail in Section 3.1.
Already, however, we can outline our proof strategy, as follows.
- (1)
Create an edge absorption structure and a global edge reservoir. 2. (2)
For each tree, create a colour absorption structure and a colour reservoir. 3. (3)
For each tree, create a vertex absorption structure and a vertex reservoir. 4. (4)
Find edge-disjoint almost spanning rainbow paths covering most of the remaining vertices. 5. (5)
Link up the absorbers and paths to form rainbow forests and thereby cover all non-reservoir vertices. 6. (6)
Cover non-reservoir edges by adding each such edge to one of the forests . 7. (7)
Incorporate non-reservoir colours for each forest, by adding a suitable edge from the edge reservoir. 8. (8)
Absorb the uncovered reservoir vertices into each forest, using edges and colours from the reservoirs. 9. (9)
Absorb the uncovered reservoir colours into each forest, using the colour absorption structure. 10. (10)
Absorb the uncovered edges from the global edge reservoir by distributing them among the forests to complete these forests into rainbow spanning trees .
To find all of the structures we use, we apply results on matchings in certain auxiliary hypergraphs, as described in Section 3.2. This allows the structures we find to look random-like, which in turn means that at each stage of the construction of the trees , the currently unused sets are also random-like. In particular, this means that the leftover sets which need to be absorbed are sufficiently small and well-distributed (again, the sets we track here are vertices and edges as well as colour sets).
The main difficulty in our proof lies in obtaining a decomposition into spanning trees. The property that these trees are isomorphic (even to some fixed in advance) can be achieved with only a little extra care. We comment more on this in Section 3.3. In Section 4, we list the tools that we use in our proof.
The above strategy is implemented in Section 5.6, following the proof of several lemmas allowing some of these tasks. In Section 5.1, we find a set of almost spanning rainbow paths. In Section 5.2, we find our colour absorption structure. In Section 5.3, we find our edge absorption structure. In Section 5.4, we show how we will connect these structures together. In Section 5.5, we find suitable rainbow matchings which we will use to absorb vertices.
3.1. Designing absorbers
The absorbing method has its roots in work by Erdős, Gyárfás, and Pyber, as well as Krivelevich, before its general codification by Rödl, Ruciński and Szemerédi. The key novelty in our work is to construct a ‘nested’ absorbing structure for the edges, colours and vertices. As the edges of a tree define its colours and vertices, we start by building an edge absorption structure and an accompanying edge reservoir (i.e. the edges in the reservoir are those which can later be absorbed).
Edge absorbers via monochromatic matchings. We create an edge absorption structure for a set of reservoir edges as follows, where is a small constant. (Recall that is the number of trees in our decomposition.) For each , we construct a rainbow forest (where we will have ) and matchings . Each will consist of edges of colour , and ranges over all elements of some colour set , where . The matchings may overlap but are edge-disjoint from , and, for each matching , any one of its edges can be added to to obtain a rainbow tree. More precisely, we have the following ‘local’ edge absorption property for each :
- (P)
If one edge is chosen from each matching , then is a rainbow tree with vertex set .
Note that since the are monochromatic, the colour set of does not depend on the choice of . See Figure 1 for our construction of such a subgraph and the matchings . We think of as being (the essential part of) an absorber which is able to ‘absorb’ exactly one of the edges it contains. The chosen edge is then added to to become part of the tree .
Since the will be small (of size 256) and monochromatic, the requirement that exactly one edge from each is to be added to is very restrictive. However, by carefully choosing how edges appear in different matchings , we can combine these to create the following ‘global’ edge absorption property for two suitable subgraphs and of . (Here , and the forests , , will be edge-disjoint.)
- (Q)
For any subset which consists of precisely edges of each colour , we can label as so that for each and .
Properties (P) and (Q) mean that, given any set with the right number of edges of each colour, we can absorb these edges (along with those in the ‘buffer set’ ) into the forests to obtain rainbow trees which span some pre-determined vertex set and colour set (these sets are different for each tree). In fact, the equidistribution condition on will be naturally satisfied as the resulting trees must contain exactly one edge of each colour. Thus altogether, the local edge absorption structures give rise to a global edge reservoir (namely ), for which we can absorb a leftover edge set into the existing forests.
To choose the matchings , we consider a set of auxiliary graphs (called ‘robustly matchable bipartite graphs’), introduced in [20] and already a standard technique in the construction of absorbers. As the name suggests, these graphs have the property that one can find a perfect matching even after the removal of an arbitrary set of vertices (of given size) from the larger vertex class, say. We will consider one such robustly matchable bipartite graph for each colour , where . The neighbourhood in of each vertex will correspond to some matching , where is such that . Thus adjacencies in encode the possible absorber matchings (and thus the possible trees) that a reservoir edge can be assigned to. A matching in saturating (where is the set of -edges in ) gives an assignment of these ‘leftover’ edges of colour to absorbers and thus to the trees . Carrying this out for all allows us to absorb all the leftover edges from the edge reservoir and the buffer edges (i.e. those in ).
The robustly matchable graphs are discussed in more detail in Section 4.2 and the properties of the edge absorption structure are described in Lemma 19.
Colour absorbers via rainbow matchings. The above properties allow us to use part of the edge reservoir to create separate colour absorbers for each tree. This means that for the th tree we have a reservoir of colours with the property that any ‘leftover’ (i.e. so far unused) set of colours of given size can be absorbed into the th rainbow forest so that the result is still a (larger) rainbow forest.
More precisely, for the th tree, we find a rainbow forest which is vertex- and colour-disjoint from , along with small rainbow matchings which are edge-disjoint from , as well as colour sets (the ‘colour reservoir’) and (the ‘buffer set’), such that the following ‘local colour absorption’ property holds for each (where , and ).
- (P*′*)
Given any set of colours, we can choose one edge from each so that is a -rainbow tree with vertex set .
For each colour appearing on an edge in , we think of as (part of) an absorber which can absorb colour into the th tree (and for each , we will provide several of these absorbers). The edges of the will lie in the edge reservoir . Crucially, this means that when absorbing a colour , it does not matter which edges/absorbers are actually involved in this colour absorption step – we can absorb any unused ones later. This means that the colour absorption step is less delicate than the edge absorption step. See Figure 1 for our construction of such an . (In the main proof, we actually construct the forests and simultaneously, and denote them .)
The matchings will be small edge-disjoint rainbow matchings, where the colours of each matching are chosen according to some auxiliary robustly matchable bipartite graph. We will consider one such auxiliary graph for each tree , with the larger vertex class consisting of the colour reservoir together with the buffer set . The edges of connect each colour to some indices . The colour set of will consist of precisely those colours in . For any set of size , a matching saturating absorbs all the ‘leftover’ colours, as required. The details are given in Lemma 18.
Vertex absorbers. We then use part of both the edge reservoir and the colour reservoir to create vertex absorbers. This construction is relatively simple, and the resulting vertex reservoir consists of some vertices unused by the th tree so far. For each , we take a small random set of vertices and connect them into a rainbow vertex absorbing path, while reserving a further random set of vertices that is slightly smaller than . When we reach Step (8), the set of uncovered vertices will be a subset of and contain almost all vertices of . (So one can view as a vertex reservoir.) We will match those vertices in which are still uncovered onto the vertex absorbing path. The randomness of and allows us to do this with a rainbow matching between and .
Covering outside the reservoirs. By construction, the edge and colour absorbing structures can only deal with edges/colours within the respective reservoirs. Thus, after we construct the th forest which covers almost all the colours, we must extend it slightly so that it now uses every colour outside its reservoir, and that collectively the resulting forests use all the edges outside of the global edge reservoir. We achieve this as follows: To cover an edge outside the global edge reservoir (in Step (6)), we include as an edge between and for some suitable . Similarly, to cover a colour outside the th colour reservoir (in Step (7)), we choose a suitable -edge between and , again from the edge reservoir. We can carry this out in such a way that these edges form a relatively small -matching, thus enabling us to carry out the vertex absorption procedure described above with only minor modifications.
3.2. Almost-packing random subgraphs
We will find the different structures for the strategy outlined above by defining (for each of these structures) an auxiliary hypergraph in which a large matching corresponds to the desired structure. The hypergraph will be roughly regular, with small codegrees, and thus the existence of this matching will follow from standard results (see Theorem 7 in Section 4.1). In each case, the hypergraph is defined in a similar way, but to give a concrete example we will sketch how to find almost-spanning rainbow paths in any optimally coloured . (Note that this construction as described below is already present in [16]. We repeat it informally here, as it forms a template for several more involved applications in this paper.)
To simplify further, we note that by randomly reserving edges, colours and vertices, we can greedily connect a given set of long disjoint rainbow paths together via very short paths (which use their own set of reserved edges, colours and vertices) into a single rainbow path. Thus, to cut to the main part of the argument, let us suppose we want to find the following, where is a large constant, and , for some small .
- Aim:
To find in , for each , a set of vertex-disjoint colour-disjoint rainbow cycles of length , so that all the cycles in are edge-disjoint.
The key is to construct a hypergraph in which a large matching corresponds to the required cycles (where each matching edge directs us to include some cycle into some set ). We take vertices for as follows. We need all the cycles we find to be edge-disjoint, so each edge in will appear as a vertex of . All the cycles in must be vertex-disjoint, so we wish to represent the vertices of as vertices in . However, different cycles in different sets are permitted to share vertices. Thus, for each , we include a copy of by including the vertices in in . Similarly, we represent the colours for cycles by including for each . We define the hyperedges of as follows. For each rainbow cycle of length and , we include the hyperedge
[TABLE]
Suppose then we had a matching in . Then, for each , let be the set of cycles with . Note that we have the following properties.
- •
If are distinct, we have the following.
- –
As , and , we have that and are vertex-disjoint.
- –
As , and , we have that and are colour-disjoint.
- •
For any and with , we have , , and , so and are edge-disjoint.
That is, each is a set of vertex- and colour-disjoint rainbow cycles, and the cycles in are edge-disjoint, as required in the above aim.
In the actual proof we will find the required structures within prescribed (randomly chosen) vertex, edge and colour sets, with parameters carefully chosen so that the construction uses almost all of the available sets each time. Together, this has the advantage that the overall leftover after removing these structures is also randomly distributed and sufficiently small so that it can be absorbed.
3.3. Isomorphic trees
The main achievement of our techniques is to find a decomposition into (any) spanning rainbow trees. However, by taking care at several points in our argument, the trees we construct can be kept isomorphic. The key point here is to observe that in Figure 1 the resulting structure from the absorber is the same regardless of which edges are used from the reservoir.
In fact, we not only find isomorphic trees, but we find copies of the same fixed tree, regardless of the -factorization of . We define this tree as follows (see Figure 1).
Definition 5**.**
Given such that , we define the tree as follows: Take a path of length . For all , add paths of length to (i.e. will become an endvertex of these paths), and add paths of length to each of and . Moreover, take a set of new vertices and add a perfect matching between and .
The set corresponds to the set in the vertex absorption structure. For each there will be an integer so that for each , the ‘middle’ edge on the path between and will be an edge of some ‘absorber-matching’ or . Note that and . We will prove Theorem 4 with , where and are small but linear in . So contains an almost spanning path. After proving Theorem 4 in Section 5.6, we describe how this construction can be slightly modified to achieve that .
4. Tools
4.1. Hypergraph matchings
We make frequent use of the existence of large matchings in almost regular hypergraphs with small codegrees (such matchings are constructed via semi-random nibble methods pioneered by Rödl [25]). Moreover, we wish to have a matching which is ‘well-distributed’ across a number of vertex subsets. To make this precise, we use the following definition.
Definition 6**.**
Given a hypergraph and a collection of subsets of , we say a matching in is -perfect if for each , at most vertices of are left uncovered by .
Pippenger and Spencer [22] showed that in an almost regular hypergraph with small codegrees there are many large edge-disjoint matchings. Alon and Yuster [4] observed that by randomly splitting into many parts, and applying the Pippenger–Spencer theorem to each induced subhypergraph and then selecting a matching in each of these subhypergraphs at random, one can obtain an almost perfect matching of that is ‘well-distributed’ in the sense of Definition 6. We will use the following consequence of Theorem 1.2 in [4].
Theorem 7** ([4]).**
Suppose . Let be an -uniform hypergraph on vertices such that for some , we have for all , and . Suppose that is a collection of subsets of such that . Then there exists a -perfect matching in .
We apply Theorem 7 to several different hypergraphs in our proof, each time checking the appropriate degree and codegree bounds. We comment here generally why our hypergraphs are almost regular with small codegree. Indeed, roughly speaking, in each hypergraph we define (see Section 3.2), estimating vertex degrees will correspond to counting the number of rainbow copies of a certain graph in with one fixed characteristic (e.g. one fixed vertex/edge/colour). The symmetry in our choice of random subsets and subgraphs will mean that for each characteristic, the vertex degrees in are roughly the same. Our choice of edge, colour and vertex probabilities then results in an almost regular hypergraph. (Here, it is also useful that we consider -factorizations rather than proper colourings.) Counting codegrees corresponds roughly to counting the number of copies of the same subgraph but with two characteristics fixed. This means that the codegrees are small in comparison to the degrees, giving the additional condition we need to apply Theorem 7.
4.2. Robustly matchable bipartite graphs
As noted in Section 3.1, we use robustly matchable bipartite graphs as auxiliary graphs to tell us how to distribute edges during the absorbing steps. These graphs are defined as follows.
Definition 8**.**
Given pairwise disjoint vertex sets , an is a bipartite graph with bipartition and the following crucial property: for any set with , the subgraph has a perfect matching.
We also refer to as an with parts .
Such graphs were introduced in [20] in order to find given spanning trees in random graphs.
Lemma 9** ([20, Lemma 10.7]).**
For all sufficiently large , there exists an with maximum degree at most .
We say that a bipartite graph with bipartition is -regular if all the vertices in have degree and all the vertices in have degree . Using the Max-Flow-Min-Cut-theorem, it is straightforward to find a supergraph of an RMBG from Lemma 9 which is appropriately regular.
Corollary 10**.**
For all fixed and sufficiently large , there exists a -regular .
**Proof. **Let be an with parts , and and maximum degree at most , as in Lemma 9. Take new vertices and let be the directed graph obtained from the complete bipartite graph between and (with all edges directed towards ) by removing the edges of and adding all edges from to and from to . An edge receives capacity , and an edge receives capacity . All edges in receive capacity . We claim that and are minimal -cuts. Indeed, first note that the capacity of these cuts is . Now, let be any -cut. Let , , and . The capacity of the cut satisfies
[TABLE]
Thus, if or , then , as desired. Therefore, assume that and . Then, (1) implies that , where the last inequality uses . This proves the claim.
By the Max-Flow-Min-Cut-theorem, there exists an -flow in with value . This yields a subgraph such that for all and for all . Thus, is the desired -regular .
4.3. Probabilistic tools
In order to show various properties of random subgraphs and subsets, we will use common concentration inequalities, as follows.
Lemma 11** (see [14, Corollary 2.3, Corollary 2.4 and Theorem 2.8]).**
Let be the sum of independent Bernoulli random variables. Then the following hold.
- (i)
For all , . 2. (ii)
If , then .
Throughout, we will refer to (i) as ‘Chernoff’s bound’. Often, we will use this in conjunction with an implicit union bound to show that several properties hold altogether with high probability.
Fact 12** (cf. [24, Lemma 8]).**
Let be Bernoulli random variables such that for all , we have . Let and . Then for all .
One important tool to prove concentration of our random variables is McDiarmid’s inequality. Let be independent random variables taking values in . Let be a function of such that
[TABLE]
for all , . If this holds, we say that affects by at most .
Theorem 13** (McDiarmid’s inequality, see [19, Lemma 1.2]).**
Let , and be as stated above. Then, for all ,
[TABLE]
For our purposes, we will have , and the will be indicator variables of certain events. We will often use different indicator variables (which are not necessarily independent) to compute , and then use McDiarmid’s inequality to prove concentration.
5. Proof
5.1. Approximate decomposition
The main result in this Section is Lemma 16, which implies the existence of an approximate decomposition into rainbow almost spanning paths for any given -factorization of . As noted earlier, this result was already proved in [16, 21].
However, we need to strengthen the result somewhat – in particular, we need to constrain the paths to use given (randomly chosen) vertex and colour sets, and we need the paths to be well-behaved towards given subsets of these sets (the latter is encapsulated in the concept of ‘boundedness’ defined below).
The proof of Lemma 16 relies on ideas from Theorem 1.5 and Lemma 2.14 in [16] (and simplifies some aspects of that argument). As described in Section 4.1, the strategy is to first find for each an almost spanning collection of vertex-disjoint long rainbow cycles. Then we delete an edge from each such cycle and connect them into a long rainbow path via Lemma 15.
Definition 14** (-bounded).**
Let be a -factorization of the complete graph with vertex set and colour set . Given a subgraph , vertex sets and colour sets , we say that is -bounded if the following hold:
- (B1)
for all , ; 2. (B2)
for all , we have and ; 3. (B3)
for all , and .
Here, we think of as being ‘leftovers’ that we want to be ‘well-behaved’ in the above sense.
The following lemma allows us to embed rooted graphs in a rainbow fashion. We will often apply it to find the desired rainbow subgraphs that were missed by an application of Theorem 7.
Lemma 15**.**
Suppose and . Let be a graph on vertices and a proper edge colouring. Moreover, let and be such that for each and any set with , we have that
[TABLE]
For each , let be a graph with and , and let be an injection, where is independent. Assume that for all , there are at most indices for which .
Then, there exist embeddings , , such that, for each , is rainbow with colours in , for all and for all , and such that are edge-disjoint.
**Proof. **We find the embeddings successively and greedily. For and a vertex , let be the number of indices for which . By assumption, .
Suppose that, for some , we have already found suitable embeddings such that, additionally,
[TABLE]
Now, we find a suitable embedding such that (3) holds with replaced by . Let be the set of all vertices whose degree in is larger than . Since , observe that .
We can now greedily embed while avoiding the vertices in . For all , define . Order the remaining vertices of arbitrarily and embed them one by one into as follows. When we consider , let be the set of images of the neighbours of which have already been embedded. We would like to choose an image for from .
First, note that by (2) with , and as , we have . At most vertices of are blocked because they have already been chosen as an image for , and at most vertices are blocked because contains a colour that has already been used. Moreover, invoking (3), at most vertices are blocked because for some . Hence, there exists a suitable image for . Thus, we can finish the embedding of in this way.
Clearly, is rainbow with colours in , and edge-disjoint from . Moreover, for any vertex , the degree of in is at most . In particular, if , then the degree of in is at most . Moreover, if , then if and only if , in which case the degree of in is at most . Thus, (3) holds with replaced by , which completes the proof.
Lemma 16**.**
Suppose and let and . Let be a -factorization of the complete graph with vertex set and colour set . For every , let be a -random subset of , and let be a -random subset of . Moreover, let be a -random subgraph of .
Then with high probability, there exist edge-disjoint rainbow paths in such that
- (P1)
, ; 2. (P2)
, , is -bounded; 3. (P3)
for all , the number of for which and the subpath from to one of the endvertices of has length at most , is at most .
**Proof. **Choose new constants and such that .
For each , split into a -random subset and a -random subset , and split into a -random subset and a -random subset . We also split into a -random subgraph and a -random subgraph . We claim that with high probability, we have the following:
- (a)
for each and with ,
[TABLE] 2. (b)
for each and with ,
[TABLE] 3. (c)
for each , ; 4. (d)
for each and , we have that ; 5. (e)
for each , ; 6. (f)
is -bounded.
Indeed, using Chernoff’s bound, it is straightforward to check that (c)–(f) hold with high probability. For (a) and (b), we can apply McDiarmid’s inequality, since of the at most edges incident with , each has an effect of at most , each vertex has an effect of at most , and each colour has an effect of at most .
Henceforth, assume that these random choices have been made and satisfy the above properties.
For all , let be the collection of all rainbow cycles of length in for which and . (Note that the ’s are not necessarily disjoint.) For , and , we let , and denote the set of all with , and , respectively. Using (a), we can now count that, for all ,
[TABLE]
and, for all with ,
[TABLE]
We define an auxiliary hypergraph as follows. The vertex set of consists of three parts. The first part is simply . The second part is the set of all pairs with and . The third part is the set of all pairs with and .
Now, we define the edge set of . For each and , we add the hyperedge
[TABLE]
Hence, is -uniform.
Clearly, using (4), we have for each that
[TABLE]
Moreover, we have for each that
[TABLE]
and for all that
[TABLE]
(Note that no hyperedge is counted more than once since each rainbow cycle contains at most one -edge.)
{NoHyper}
Claim 1*:*
.
*Proof of claim: *Recall, from (6), that each hyperedge of is uniquely fixed by some and . Note that for a set of vertices and , the number of with is at most . This easily implies that codegrees of pairs in , and are at most .
Next, consider distinct . For each , the number of with is at most . Summing over all yields the desired bound.
Now, take and . Each with will contain some -edge . We distinguish two cases for . If is incident to , there is only one choice for , and then at most choices left. If is not incident to , there are at most choices for , and then at most choices left. Thus, in total, there are at most choices.
Similarly, we check that the codegree of and for distinct is at most . We have to choose a -edge and a -edge and again distinguish two cases. If and share a vertex , there are at most choices for (which determines and ), and then at most choices left. If and form a matching, there are at most ways to choose and , and then at most choices left.
For each , let be the set of all pairs with and . For each colour , let be the set of all pairs with and . Let
[TABLE]
Using (7), (8), (9), and Claim 1, we now apply Theorem 7 to obtain a -perfect matching in . For , let be the collection of all for which . For distinct , and , and therefore, as is a matching, and are vertex-disjoint. Similarly, and are colour-disjoint. Thus, is a collection of vertex-disjoint -cycles in whose union is rainbow with colours in . Moreover, as for each and , , all these cycles are edge-disjoint.
For each , we will now randomly break each cycle in into a path, before joining all these paths together into a single cycle. For each and all , choose an edge uniformly at random. For each , let and . We claim that, with high probability, we have
[TABLE]
Indeed, fix a vertex . We have for all , and those events are independent. Similarly, for a fixed colour , we have for all , and those events are independent too. Thus, the claim follows with Lemma 11(ii) and a union bound.
Now, assume that (10) holds. For each , let be the graph obtained as follows: Give every edge an orientation, and (cyclically) enumerate these edges. Now, for each edge, add a path of length between its head and the tail of the next edge, using a new vertex as the internal vertex each time. (So consists of the union of all these paths of length , but does not contain the edges . In particular, is independent in .) Observe that .
By (b) and (10), we can apply Lemma 15 (with taking the place of ,,) to find for each an embedding such that is rainbow with colours in and for all and for all , and such that are edge-disjoint.
For , let
[TABLE]
We have that are edge-disjoint rainbow cycles in , where is rainbow with colours in and . Moreover, by the definition of , (f) and (10) and the fact that for all , we have that , , is -bounded.
Finally, choose for each an edge uniformly at random and let . For a vertex , let be the set of indices for which and the subpath from to one of the endvertices of has length at most . Note that, for each , as , the cycle has length at least by (e), implying , and these events are independent. Thus, with Lemma 11(ii), we conclude that (P3) holds with high probability. Similarly, we can deduce that with high probability, for every , the number of for which is incident with , is at most , and for every , the number of for which , is at most . This implies that (P2) still holds with high probability.
5.2. Matchings for colour absorption
In this subsection, we find the rainbow matchings which form the crucial part of the colour absorption structure (see Lemma 18). The following lemma prepares the ground for this.
Lemma 17**.**
Suppose and let , and . Let be a -factorization of with vertex set and colour set . For every , let be a -random subset of , and let be a -random subset of . Moreover, let be an -random subgraph of . Then, with high probability, there exist edge-disjoint matchings in such that the following hold:
- (i)
* for all ;* 2. (ii)
for all , consists of -edges for each ; 3. (iii)
for every vertex , the number of for which is covered by is .
Later on some edges of will be used to construct the th tree of the decomposition.
**Proof. **Choose a new constant such that . For each , we randomly split into a -random set and a -random set . Similarly, we split into an -random subgraph and an -random subgraph . For and , let denote the number of -edges in .
We define a (random) auxiliary hypergraph as follows. The vertex set of consists of three parts: The first part is simply . The second part is the set of all pairs with and . The third part of is the set of all triples with , and . For all , and , we add the hyperedge
[TABLE]
if and only if , and . Thus, is a -uniform hypergraph.
{NoHyper}
Claim 1*:*
With high probability, for each , .
*Proof of claim: *Fix an edge and assume . For , let be the indicator variable of the event that and . Note that . Since for each and the ’s are independent, we can easily deduce from Chernoff’s bound that the claim holds.
{NoHyper}
Claim 2*:*
With high probability, for each , .
*Proof of claim: *Fix and and assume . For every vertex , let be the indicator variable of the event that , and . Note that . Since for each and the ’s are independent, we can easily deduce from Chernoff’s bound that the claim holds.
{NoHyper}
Claim 3*:*
With high probability, for each , , and, for each and , .
*Proof of claim: *Fix , and and assume that . For every -edge , let be the indicator variable of the event that and . Note that . Since for each and the ’s are independent, we can easily deduce from Chernoff’s bound that the claim holds. (A similar argument works for .)
{NoHyper}
Claim 4*:*
With high probability, we have and for all .
*Proof of claim: *This is an easy consequence of Chernoff’s bound.
{NoHyper}
Claim 5*:*
.
*Proof of claim: *Clearly, the codegree of pairs in and is [math]. Moreover, the codegree of any pair in and is at most , and the codegree of any pair in is at most . Finally, consider a pair in , say and . If , then the codegree is [math], so assume . If , then the codegree is also zero, so assume otherwise and let be the colour of . Then the codegree is at most .
We now assume that the properties stated in Claims 1–5 are satisfied. By our choice of , we have that for all . For every vertex , let be the set of all pairs with and . For every , let be the set of all with and . Let
[TABLE]
Thus, we can apply Theorem 7 to find a -perfect matching in . For , let be the set of all edges such that for some .
Hence, by definition of , we have that are edge-disjoint matchings in , and, for each , we have and consists of at most 192 edges with colour , for each .
For each and , let . Thus, is the number of -edges that are missing in in order to satisfy (ii). Since is -perfect, we have for each and that
[TABLE]
Moreover, for each vertex , the number of for which but is not covered by , is at most . Since by Claim 4, this implies that the number of for which is covered by is .
Now, we wish to find edge-disjoint matchings in such that, for each , and contains precisely -edges for each . This can be done in order greedily using Claim 3 and (12). Indeed, suppose we want to add -edges to . By (12), we added at most -edges to previous matchings , , and at most edges to . Thus, at most -edges are blocked (since every edge in might block -edges). Since by Claim 3, we can find suitable -edges in and add them to .
Note that, by Claim 4, for every vertex , the number of for which is covered by , is at most . Finally, for each , let . It is easy to see that are the desired matchings.
Lemma 18**.**
Suppose and let , and . Suppose with and . Let be a -factorization of with vertex set and colour set . Let be an -random subgraph of . For every , let be a -random subset of , and let , be disjoint -random subsets of . Split further into an -random set and a -random subset .
Then with high probability, for each , there exist such that and and vertex-disjoint rainbow matchings in , such that altogether the following hold:
- (i)
for each , ; 2. (ii)
for all , ; 3. (iii)
for each , consists of -edges for each ; 4. (iv)
for each and any subset of size , there exists such that is -rainbow and contains exactly one edge from each of ; 5. (v)
the matchings are edge-disjoint, and for all ; 6. (vi)
for every vertex , the number of for which is covered by is .
Here, the crucial property is (iv), which will allow us to use some colours of flexibly before assigning the remaining colours (i.e. those in ) together with the ‘buffer’ in such a way that each matching contributes exactly one edge to which will be part of .
**Proof. **Clearly, we may assume that . We choose the random colour sets according to the following procedure: For each , let be a -random subset of , and let be a random function such that and . For and , let and , and let . Then are as in the statement.
Now, we first expose all random choices except the functions . By Lemma 17, with high probability, there exist edge-disjoint matchings in such that the following hold:
- (a)
for all ; 2. (b)
for all , consists of -edges for each ; 3. (c)
for every vertex , the number of for which is covered by is .
Henceforth, assume that these random choices have been made and satisfy the above properties. It remains to expose the functions .
With high probability, we have for all and that
[TABLE]
With high probability, we also have for all that
[TABLE]
For , let us call unreliable for if is covered by via an edge whose colour is in . Then, also with high probability, for all ,
[TABLE]
From now on, assume that (13)–(15) hold. For each and , note that by (13) we have and thus, again by (13), we can choose such that , and define
[TABLE]
Then (i) clearly holds and (ii) follows from (14). Moreover, let
[TABLE]
for each . Observe that (c) and (15) imply (vi).
We now use RMBG’s to break each into small rainbow matchings. For each , let be a -regular with parts and , which exist by Corollary 10. For each , partition into matchings , such that, for each , is an -rainbow matching. This can be done greedily since to do so we need precisely -edges of each colour , which contains. Clearly, , and thus (iii) and (v) hold.
Finally, we check that the crucial property (iv) holds. Consider and suppose has size . Since is an RMBG with parts , and , there exists a perfect matching in between and . Now, for each , we select the -edge from and include it in . (Here we view as the colour matched to in the matching , and we use that is -rainbow.) Clearly, is as desired.
5.3. Matchings for edge absorption
We now find the monochromatic matchings which form the crucial ingredients for the edge absorption process.
Lemma 19**.**
Suppose and let , , , and suppose with and . Let be a -factorization of with vertex set and colour set . Let be edge-disjoint -random subgraphs of , and split further into an -random subgraph and a -random subgraph . For each , let be a -random subset of and let be a -random subset of .
Then, with high probability, there exist and such that and with and, for each , there exists of size and vertex-disjoint matchings in , where consists of -edges, such that altogether the following hold:
- (i)
for each , and ; 2. (ii)
for any subset which consists of precisely edges of each colour , there exists a partition of into sets , such that, for each , contains exactly one edge from each of ; 3. (iii)
every vertex is covered by of the matchings .
Here, the crucial property is (ii), which will allow us to use some edges of the global edge reservoir flexibly before assigning the remaining edges (i.e. those in ) together with the ‘buffer’ in such a way that each matching contributes exactly one edge to . will then be assigned to the th tree .
**Proof. **We may clearly assume that . We also split further into a -random subgraph and an -random subgraph . We first expose . Using Chernoff’s bound, it is easy to see that, with high probability, we have for all , and that
[TABLE]
Henceforth, we assume that are fixed with the above properties, and expose the other random sets.
By (16), we have for and that . Therefore, by (16) again, for each , we can choose such that for all , and define
[TABLE]
Clearly, this choice of and satisfies the first part of (i). Moreover, from (17), we can infer that , as desired, and that
[TABLE]
As indicated in the proof sketch, the key to obtaining (ii) is to use an RMBG for each colour which matches the -edges of to ‘absorbers’. Let be a -regular with parts , which exists by Corollary 10. We identify and with and . We carry out this identification randomly in order to obtain a codegree condition in some hypergraph which we will define later. (This codegree condition will be needed when applying Theorem 7 to .) For each colour , pick random bijections and , all independently. Obtain a copy of by identifying with according to and with according to .
For two vertices , we define as the number of colours for which , where and are the unique -edges at and , respectively. (In particular, if or is not contained in , then contributes [math] to .)
{NoHyper}
Claim 1*:*
With positive probability, for all distinct vertices .
*Proof of claim: *Fix two distinct vertices . For , let be the indicator variable of the event that there exist -edges at and , respectively (which are unique if existent), and . Fix and let be as above. We claim that . Note that are distinct. Let be such that and . Thus,
[TABLE]
Hence, , and since the ’s are independent, Chernoff’s bound implies that the probability that is smaller than . A union bound then implies the claim.
From now on, fix RMBG’s for which the conclusion of Claim 1 holds. Let . For each , we define . We refer to as an absorber and will sometimes identify with . Note that is a matching consisting of -edges.
By our choice of RMBG’s, we have that for any two distinct vertices ,
[TABLE]
We will now assign to each absorber an index .
The assignment will be obtained as follows: We first define an auxiliary hypergraph , in which we will find an almost perfect matching that provides an almost complete assignment. For the remaining absorbers not yet assigned, we will greedily pick images from a reserve.
In order to set aside this ‘reserve’, we randomly split and further as follows. For each , split into a -random set and a -random set , and split into a -random set and a -random set .
We can now define the (random) auxiliary hypergraph as follows. The vertex set of consists of three different parts: The first part is simply the set which represents all the absorbers. The second part is the set of all pairs with and . The third part is the set of all pairs with and .
Now, we define the edge set of . For every and every absorber , we add the hyperedge
[TABLE]
if and only if and . Hence, is -uniform. (Recall that is a matching consisting of -edges.)
Moreover, for each absorber , we define the random set of indices for which and . We aim to apply Theorem 7 to . For this, we first establish the following properties.
{NoHyper}
Claim 2*:*
With high probability, for each , and .
*Proof of claim: *Fix an absorber . For , let be the indicator variable of the event that and and let be the indicator variable of the event that and . Note that and . For each , we have that and . Thus, and . Since the ’s are independent, and similarly, the ’s are independent, we can deduce with Chernoff’s bound that the claim holds.
{NoHyper}
Claim 3*:*
With high probability, for each , .
*Proof of claim: *Fix . For , let be the indicator variable of the event that . Note that . For each we have . Thus, .
Moreover, is determined by the independent random variables . Since affects by at most , the claim follows by an application of McDiarmid’s inequality.
{NoHyper}
Claim 4*:*
With high probability, for all , .
*Proof of claim: *Fix . For each edge at in , say with colour , has neighbours in , and for each of those we have iff and the other vertices of are contained in . Thus, by (18).
Moreover, is determined by the independent random variables . The effect of on is at most . Moreover, for each , by (19), affects by at most . The claim now follows from an application of McDiarmid’s inequality.
{NoHyper}
Claim 5*:*
.
*Proof of claim: *Clearly, the codegree of pairs in and is [math]. Moreover, the codegree of pairs in and is at most . It is also easy to see that the codegree of a pair in is at most .
Finally, consider a pair in , say and . If , then the codegree is [math], so assume . Crucially, by (19), the codegree of and is at most .
We now assume that the properties stated in Claims 2–5 are satisfied. Using Chernoff’s bound, we can assume that the following simple properties hold as well:
[TABLE]
By our choice of , we have that for all . In combination with Claim 5, we can thus apply Theorem 7 to find an almost perfect matching in . In order to gain control over the leftover vertices in , we define the following vertex sets. For each vertex , let be the set of all absorbers for which . Note that
[TABLE]
Define
[TABLE]
Now, apply Theorem 7 to find a
[TABLE]
Our goal is to define a map . Let be the set of absorbers which are not covered by . For each , the absorber is covered by a (unique) hyperedge , and we define . For all uncovered absorbers, we now use the ‘reserve’ sets and to pick suitable images.
For all , we successively define as follows: when we consider , let be the set of all previously considered with or . By (23), we have that
[TABLE]
Recall from Claim 2 that . Thus, there is and we define .
Altogether, we have found a map , which we show has the following properties:
- (a)
and for all ; 2. (b)
whenever ; 3. (c)
for all and , there is at most one with .
Here, (a) clearly holds by the definitions of , and . To see (b), suppose . If , then we have since is a matching and as such covers every vertex at most once. If and , then and . Finally, suppose and assume that we defined after . If , then (with notation as above) and hence , a contradiction.
For (c), fix and . Suppose for some . We consider two cases. In the first case, we have and . In particular, there is at most one which satisfies this and we must have . In the second case, we must have and , and there can only be one which satisfies this by definition of above. Since and are disjoint, (c) follows.
Now, for every , define and . By (c), is a bijection. For all , define
[TABLE]
For all , we have , so the second part of (i) holds too. Observe that if , then and hence there exists some for which . By (a), we have . Thus, . In particular, we have by (21). Moreover, since , at least elements of are covered by , which means that for at least colours in , we have and therefore . Thus, by (21). Hence, , as required.
Furthermore, for all and , let
[TABLE]
Clearly, is a matching consisting of -edges in . Using (a), we can also see that . Moreover, for fixed , all the matchings are vertex-disjoint by (b). To check (iii), consider any vertex . Clearly, the number of matchings covering is at most by (22). Moreover, since covers all but at most absorbers in , we obtain a lower bound of , as desired.
It remains to show the crucial property (ii). Suppose consists of precisely edges of each colour . For each , let be the set of -edges in . Since is an RMBG with parts , and , there exists a bijection such that for all .
We can now define the desired partition of as follows. Let . Let be the colour of and . Thus, we have , where . Note that and hence . Assign to . Clearly, this defines a partition of into . Consider . By construction, every edge belongs to some with . Moreover, for fixed , only one edge of is included in because and are bijective.
5.4. Connecting lemma
The following lemma will be used to efficiently connect up the (edges from the) matchings produced by Lemmas 18 and 19 of the trees .
Given a -uniform matching , we say that a graph is an -connector if is obtained from the empty graph on by adding, for every , new vertices , a perfect matching between and and all edges from to .
Lemma 20**.**
Suppose and let and and and suppose . Let be a -factorization of the complete graph with vertex set and colour set . Let be a -random subgraph of . For every , let be disjoint subsets of that are -random and -random, respectively, and let be a -random subset of .
Then, the following holds with high probability: Let be any -uniform (multi-)hypergraph which is the union of matchings such that and for all , and such that for all . Then, for each , there exists an -connector in such that the following hold:
- (i)
* are edge-disjoint;* 2. (ii)
for each , is rainbow with colours in ; 3. (iii)
* is -bounded.*
In the proof, we will find most of the required connections via Theorem 7 (which allows us to do this ‘efficiently’) and the remaining ones via Lemma 15.
**Proof. **Choose a new constant such that . Split further into a -random subgraph and a -random subgraph . Moreover, for each , split into a -random subset , a -random subset and a -random subset . Split into a -random subset and a -random subset . We will now establish a few properties concerning the random sets which hold with high probability. From these properties, we can then (deterministically) find the desired connections for any admissible .
For , let be the spanning subgraph of with all -edges, and let be the spanning subgraph of with all -edges.
For each edge , let be the set of for which and intersects both , and let be the set of for which and intersects both .
For and , let be the set of -edges in , and let be the set of -edges in .
We claim that the following hold with high probability:
- (a)
for all , , , and ; 2. (b)
for all , we have ; 3. (c)
for all and , , and ; 4. (d)
for all and distinct , ; 5. (e)
for all and with , we have ; 6. (f)
for all , and ; 7. (g)
for all and , and ; 8. (h)
is -bounded.
Indeed, (a), (b), (c), (f), (g) and (h) follow easily from Chernoff’s bound. For (d) and (e), we use McDiarmid’s inequality, as follows. Consider and distinct . Clearly, . Moreover, of the at most edges incident with either or , each has an effect of at most . Each vertex has an effect of at most , and each colour has an effect of at most , and so McDiarmid’s inequality applies. A similar argument works for (e).
Now assume that (a)–(h) hold. Let be given arbitrarily as in the lemma statement. Let . By (a) and since , we have that
[TABLE]
Moreover, for every vertex , it follows from (b) and since that
[TABLE]
From (c) and (24) we infer that
[TABLE]
For an edge , let be the set of for which intersects . From (f) and (25), we deduce that
[TABLE]
For all and , let be the set of -edges in . By (g) and (24), we have
[TABLE]
We now define an auxiliary hypergraph whose vertex set is the union of five parts. The first part is simply . The second part is the set of all pairs such that . The third part is the set of all pairs with . The fourth part is the set of all pairs with . The fifth part is the set of all pairs with .
We now define the edge set of . For disjoint , , and a bijection , let denote the graph on with edge set . Note that is an -connector.
For all , , and bijections , we add the hyperedge
[TABLE]
to if and only if is a rainbow subgraph of . Note that is -uniform since and hence . We will apply Theorem 7 to . For this, we first check that is roughly regular.
For each and , let
[TABLE]
and let for each and .
{NoHyper}
Claim 1*:*
For each and , we have
[TABLE]
*Proof of claim: *First, assume with and . There is a unique with . By (c), there are choices for . For each in turn, by (d), we have choices for while avoiding previously chosen vertices and previously used colours. We deduce that .
Next, assume with and . By (26), there are choices for , which yields a unique with . Using (d) as above, we conclude that .
Clearly, in any other case, we have .
We will use Claim 1 below without explicit reference.
{NoHyper}
Claim 2*:*
For all , we have .
*Proof of claim: *First, consider . We have
[TABLE]
as .
Next, consider . By (a), there are choices for . For each in turn, by (d), we have choices for while avoiding previously chosen vertices and previously used colours. We deduce that
[TABLE]
Now, consider . We have
[TABLE]
Next, consider . By assumption, we have choices for . For each in turn, by (d), we have choices for while avoiding previously chosen vertices and previously used colours. We deduce that
[TABLE]
Finally, consider . Note that . Hence,
[TABLE]
since .
{NoHyper}
Claim 3*:*
.
*Proof of claim: *Clearly, the codegrees of pairs in and are [math]. Moreover, by Claim 1, we have for all and . This implies that the codegrees of pairs in , , and are at most , as required. It is also easy to see that the codegrees of pairs in , , and are at most , since for fixed , we always have at most choices for and at most choices for each remaining vertex.
Consider distinct . There are choices for . If and with and , then there are at most choices for and at most choices for . Otherwise, we may assume that for and is incident to a vertex . Now determines and there are at most choices for . Altogether, we conclude that the codegree of is at most .
Next, consider with . We have to provide an upper bound for the number of for which and contains a -edge and a -edge . To count these possibilities, we distinguish some cases regarding how intersect . First, assume that . In this case, there are at most choices for and then at most choices for . Moreover, since must form a matching, two vertices of are determined. This leaves at most choices for the remaining vertices, which yields a total of choices in this case. Next, assume that . In this case, there are at most choices for , which then determines and and thus two vertices from . There are at most choices for and at most choices for the remaining vertices of . Finally, assume that . We divide this case into two subcases. First, assume that share their endpoint in . Then we have at most choices for , which determines , which in turn determines and , and leaves at most choices for the vertices in . On the other hand, if form a matching, then we have at most choices for , which determines and as before and leaves at most choices for the remaining vertices in . Thus, altogether, the codegree of is at most .
Next, consider and . We have to choose a -edge . If , there are at most choices for , which also fixes one vertex of , and leaves at most choices for the remaining vertices. If , then there are at most choices for , which fixes and one vertex from , and leaves at most choices for the remaining vertices. Thus, and have codegree at most .
Finally, consider and . We have to choose a -edge . If is incident with , then there is only one choice for . This either fixes , in which case at most choices are left for the remaining vertices, or it fixes another vertex from , in which case there are at most choices for and at most choices for the remaining vertices. If is not incident with , then there are at most choices for . However, this either fixes and leaves at most choices for the remaining vertices, or it fixes two more vertices, which leaves at most choices for and at most choices for the remaining vertices. Thus, and have codegree at most .
For , let be the set of all with , and let be the set of all pairs with and . For a colour , let be the set of all pairs with and . Let
[TABLE]
Now, apply Theorem 7 to obtain a -perfect matching in . For each , let be the set of all with . Since , we have . For each , there is a unique edge in which covers . Let and define . By construction of , are edge-disjoint subgraphs of , and, for each , we have that is a rainbow -connector with colours in , and and . Moreover, observe that
[TABLE]
Indeed, (B1) holds since for every . Similarly, (B2) holds since for every . Finally, (B3) holds since for every .
We will find the missing connectors using Lemma 15. Let be an -connector. Clearly, is an independent set in , and . Moreover, for every vertex , the number of for which , is at most .
Using (e), we can thus apply Lemma 15 (with playing the roles of ) to find for each , an embedding such that is rainbow with colours in and for all and for all , and such that are edge-disjoint.
Finally, let . Clearly, is a rainbow -connector in with colours in , and are edge-disjoint. Moreover, (iii) follows from (h), (29), (24) and (25).
5.5. Rainbow perfect matchings
Given a bipartite graph with vertex classes , we say that is -quasirandom if for all and distinct , we have and .
The next lemma follows easily from a result of Coulson and Perarnau [11, Lemma 6]. Indeed, it is well known that the above notion of quasirandomness implies super-regularity, which is sufficient for the existence of a perfect matching. Moreover, using regularity, it is straightforward to count the number of ‘switchable edges’, as required in the general statement in [11].
Lemma 21**.**
Suppose . Let be a bipartite graph with vertex classes such that and is -quasirandom. Then, given any edge-colouring of where each colour appears at most times, there exists a rainbow perfect matching of .
We now use Lemma 21 to obtain several edge-disjoint rainbow perfect matchings.
Lemma 22**.**
Suppose and let . Let be a vertex set of size and assume that are subsets of such that for all , for all distinct , and, for every , the number of for which is at most . For each , suppose is partitioned into equal-sized sets and , and is a -quasirandom bipartite graph with vertex classes . Assume that is edge-coloured and each colour appears at most times in .
Then there exist edge-disjoint such that is a rainbow perfect matching of for each .
We find using a randomised greedy algorithm. **Proof. **Let . Suppose that we have already found for some . We now define as follows. Let and let . If , then is -quasirandom. Thus, by Lemma 21 used repeatedly, we can find edge-disjoint rainbow perfect matchings of . Otherwise, if , let be empty graphs on . In either case, pick uniformly at random and let . The lemma clearly follows if the following holds with positive probability:
[TABLE]
For and , let be the set of indices such that , so that , and for , let be the indicator variable of the event that for some . Observe that
[TABLE]
Now, fix and . Crucially, for any , since , at most of the matchings that we picked in contain an edge incident to in (regardless of the previous choices). Let be the enumeration of in increasing order. By the above, for all , we have
[TABLE]
Let . Since , we have . Using Fact 12 and Lemma 11(ii), we infer that
[TABLE]
Finally, a union bound implies that (30) holds with high probability.
5.6. Proof of Theorem 4
We are now ready to prove our main theorem.
**Proof of Theorem 4. ** Choose new constants such that
[TABLE]
and let
[TABLE]
Let be a -factorization of with vertex set and colour set . We will obtain a decomposition into rainbow copies of (cf. Definition 5). Hence, and are essentially determined by and , respectively, and are best thought of as error parameters.
In order to apply the lemmas that we have proven in this section without interference, we will split , and into random subsets each reserved for the application of the relevant lemma. For convenience, we now define the relevant constants in one place (where the letters represent vertex, colour and edge probabilities, respectively).
.
Note that, as , and and hence and .
{NoHyper}
Step 1*:*
Random splitting
Split vertices. For each , we split randomly as follows:
[TABLE]
We split and further as follows:
[TABLE]
[TABLE]
Split colours. Moreover, for each , we split randomly as follows:
[TABLE]
We split further as follows:
[TABLE]
Let . Hence, is a -random set. Moreover, let .
Split edges. We split randomly as follows:
[TABLE]
Split further as follows:
[TABLE]
Let . Thus, is a -random subgraph. Moreover, let .
Create the edge reservoir. By Lemma 19 (with in place of ), with high probability, there exist such that
[TABLE]
with and, for each , there exists of size and vertex-disjoint matchings in , where consists of -edges, such that altogether the following hold:
- (M1)
and for all ; 2. (M2)
for any subset which consists of precisely edges of each colour , there exists a partition of into sets , such that for each , contains exactly one edge from each of ; 3. (M3)
every vertex is covered by of the matchings .
Create colour reservoirs. We apply Lemma 18 (with in place of ) to see that with high probability, for each , there exist
[TABLE]
and vertex-disjoint rainbow matchings in , such that altogether the following hold:
- (R1)
for each , ; 2. (R2)
for all , ; 3. (R3)
for each , consists of -edges for each ; 4. (R4)
for each and any subset of size , there exists such that is -rainbow and contains exactly one edge from each of ; 5. (R5)
the matchings are edge-disjoint, and for all ; 6. (R6)
for every vertex , the number of for which is covered by is .
Create short paths for the vertex absorption. By Lemma 16 (with playing the roles of ), with high probability, there exist edge-disjoint rainbow paths in such that
- (Q1)
for each , we have and ; 2. (Q2)
, , is -bounded; 3. (Q3)
for each , the number of for which and the subpath from to one of the endvertices of has length at most , is at most .
Properties for vertex absorption and covering non-reservoir edges/colours/vertices. For , let be the subgraph of containing precisely the -edges. In addition to the above, with high probability, the following hold:
- (A1)
for all and , there are at most -edges between and ; 2. (A2)
for all distinct , we have ; 3. (A3)
for every , the number of for which , is at most ; 4. (A4)
for all , and is -quasirandom; 5. (A5)
for all and all with , we have that
[TABLE] 6. (A6)
for all , the number of for which intersects and , and , is at least ; 7. (A7)
for all and , the number of -edges in , is at least ; 8. (A8)
for all , and ; 9. (A9)
for all , and .
Here, to deal with the sizes of the common neighbourhoods in (A4) and (A5), we use McDiarmid’s inequality. For all other claims, Chernoff’s bound suffices.
Find almost-spanning paths. We now find an approximate decomposition of (and thus ) into almost spanning rainbow paths. By Lemma 16 (with playing the roles of ), with high probability, there exist edge-disjoint rainbow paths in such that
- (P1)
for each , we have and ; 2. (P2)
, , is -bounded; 3. (P3)
every vertex is an endvertex of at most paths.
Establish connection properties. By Lemma 20 (with playing the roles of ), with high probability, the following is true:
- (C)
Let be any -uniform (multi-)hypergraph which is the union of matchings such that and for all , and such that for all . Then, for each , there exists an -connector in such that the following hold:
- (C1)
are edge-disjoint; 2. (C2)
for each , is rainbow with colours in ; 3. (C3)
is -bounded.
Henceforth, we assume that all random choices have been made and satisfy the above properties.
{NoHyper}
Step 2*:*
Connecting the pieces
For each , we now aim to connect the matchings and . For this, we will define a -uniform matching , which consists of vertices of one matching and vertices of the next matching, and then apply (C).
To make this more precise, for each , let
[TABLE]
So . Note that and hence . Also note that since and are disjoint, all the matchings in are vertex-disjoint, and recall that each matching consists of edges. For each , find two distinct such that altogether,
[TABLE]
This can clearly be done greedily. Now, for each , choose an arbitrary bijection such that and , and partition for each matching the vertices arbitrarily into a ‘tail set’ and a ‘head set’ such that is a perfect matching between and . Define
[TABLE]
Hence, is a -uniform matching in . Note that
[TABLE]
Let . By (R6), (M3) and (32), we have that for all . Hence, applying (C), for each , there exists an -connector in such that altogether the following hold:
- (C1*′*)
are edge-disjoint; 2. (C2*′*)
for each , is rainbow with colours in ; 3. (C3*′*)
is -bounded.
For each , let
[TABLE]
We will eventually have for all . Note that are edge-disjoint rainbow forests in . Moreover, for each , and . Let
[TABLE]
We think of the above as leftover sets. The following claim asserts that this leftover is well-behaved.
{NoHyper}
Claim 1*:*
* is -bounded.*
*Proof of claim: *Observe that
[TABLE]
Recall that and for all . Thus, (R1) and (A8) imply that, for all ,
[TABLE]
and (R2), (M1) and (A9) imply that, for all ,
[TABLE]
Hence, the claim follows together with (P2), (Q2) and (C3*′*).
We now use Lemma 15 to join the pieces of each together. Moreover, since the sets have different sizes, we artificially add some structure that will ensure that ultimately, all trees are isomorphic to (cf. below). In this process we can cover all remaining vertices outside the vertex reservoir .
For , let be the endvertices of , and let be the endvertices of . Let
[TABLE]
We now define a graph in which is independent and all other vertices are new vertices. Take new vertices . For each , add a path of length between and , and for each , add a path of length between and . For each , add a path of length between and . For each , add further paths of length onto (so will be an endvertex of those paths of length ), and add paths of length onto each of and . Connect and by a path which contains such that is an independent set. Finally, add a path of length between and . Clearly, . Note that since , and by (A4), we can choose in such a way that
[TABLE]
Also, for every , by (32), (P3), (Q3) and Claim 1, the number of for which , is at most .
By (A5), we can now apply Lemma 15 (with taking the place of ,,) to obtain, for each , an embedding such that is rainbow with colours in , for all , for all and such that are edge-disjoint.
For each , let
[TABLE]
Note that is rainbow as . By (36), we have that . Moreover, . Let be the set of the last vertices on , containing , so that . From (Q2), (Q3) and (A4), we deduce that, for each ,
[TABLE]
and, for each ,
[TABLE]
Crucially, observe that for each ,
any graph obtained from by adding a perfect matching between and and exactly one edge from each of the matchings in , is isomorphic to .
In particular,
[TABLE]
{NoHyper}
Step 3*:*
Final absorption
We will find the perfect matchings between and using Lemma 22, and then select exactly one edge from each of the matchings in using (R4) and (M2). For the last step to work, we need to ensure that all leftover colours are in and all leftover edges are in . Thus, prior to applying Lemma 22, we greedily deal with the colours in and the edges in .
Cover the remaining non-reservoir edges. First, find a partition of into rainbow matchings such that , and . This can be done greedily. Indeed, suppose we want to assign to an index . Let and . By (A6) and (38), the number of for which and , is at least . By Claim 1, we have that and . Thus, there exists a suitable such that no other -edge of has been assigned to , and does not yet cover or and contains at most edges so far. Finally, by (35) we have that
[TABLE]
Cover the remaining non-reservoir colours. Next, find edge-disjoint matchings in such that, for each , , , and consists of exactly one -edge for each . (Hence, .) This can also be done greedily. Indeed, suppose we want to add a -edge to . By (A7) and (37), there are at least -edges in . By Claim 1, we have that and . Also recall that . Hence, there exists a suitable -edge which has not been used by another matching , and whose endvertices are not covered by or yet by . Hence, for each , we have by (34) that
[TABLE]
Absorb the uncovered vertices. We now extend into a spanning forest by adding a matching . For each , let and . We aim to apply Lemma 22 (with playing the roles of ). (Recall that was defined just before (A1).) Clearly, . In particular, by (A4), and thus is -quasirandom. Finally, since and , the remaining conditions for Lemma 22 follow immediately from (A1), (A2) and (A3). Therefore, we can find edge-disjoint such that is a rainbow perfect matching of for each .
Absorb the uncovered colours. Now, for each , let
[TABLE]
Note that by (41), is a rainbow perfect matching between and . Similarly, by (41) and (42), is rainbow and . Also, is edge-disjoint from (since ). Since has edges by (39), we deduce that , implying that by (31) and (R1). Therefore, using (R4), there exists such that is -rainbow and contains exactly one edge from each of . Note that the are edge-disjoint from each other by (R5) and also edge-disjoint from by (40).
Absorb the uncovered edges. Finally, let
[TABLE]
We claim that contains precisely -edges for every . Note that, by (40), we have . Moreover, for each , is -rainbow, implying that for each the number of -edges in is by (M1). Thus, the number of -edges in is , which implies the claim, using (M1) again.
Thus, by (M2), there exists a partition of into sets , such that for each , contains exactly one edge from each of . In particular, is -rainbow.
Let . By , is a rainbow spanning tree isomorphic to , and decompose , as desired.
Finally, we briefly mention how the proof can be adapted to prove Theorem 4 with . The only necessary change is in how we connect the matchings in by using Lemma 20. Suppose that in Step 2 in the proof of Theorem 4, we want to connect the ‘head set’ with the ‘tail set’ for two consecutive . In the current proof, we find a vertex and internally disjoint paths of length from to each vertex in . Instead, we could also connect as follows: let be a binary tree with root and leaves , and let be a binary tree with root and leaves , and such that . (Recall that .) Let be the graph obtained from by adding a path of length between and , and then subdividing every edge once. Clearly, , and this construction ensures that still, the tree is always the same, independent of which edge of is ultimately selected for . To find all the required connections , one could still employ Lemma 20, here repeatedly, with . However, this necessitates to split , and into even more subsets, so for clarity, we omitted this from the proof.
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