1 Introduction
Given a graph , we denote by and by its vertex set and its edge set, respectively. For , denotes the induced subgraph of on . Unless we specify otherwise, we consider graphs to be simple graphs without loops and multigraphs to have multiple edges and loops.
For graphs and , we say that is decomposable if there exists a partition of such that every forms an isomorphic copy of . We then call an decomposition of . Note that if has an decomposition, divides . More generally, if there exists a partition of such that each forms an isomorphic copy of one of the graphs , we call a decomposition of .
In [bib:baratthomassen], Barát and Thomassen conjectured that for a fixed tree , every sufficiently edgeconnected graph with number of edges divisible by has a decomposition.
For any tree on edges, there exists an integer such that every edgeconnected graph with number of edges divisible by has a decomposition.
They also observed a relation between decompositions and Tutte’s conjecture, which states that every edgeconnected graph admits a nowherezero flow. Until recently it was not even known that any constant edgeconnectivity would suffice. Barát and Thomassen have shown that if every edgeconnected graph has a decomposition, then every edgeconnected graph has a nowherezero flow and, vice versa, Tutte’s flow conjecture would imply that every edgeconnected graph with number of edges divisible by 3 has a decomposition.
A series of results showing that Conjecture 1 holds for specific trees followed [bib:BT4, bib:thomassenpaths, bib:BTbistars, bib:BTdiam4, bib:BTpaths3, bib:BTpaths2, bib:BTpaths5, bib:BTbotlerallpaths, bib:thomassen3flow, bib:origpaths]. Recently, the conjecture was proven by Bensmail, Harutyunyan, Le, Merker and the second author [bib:BHLMT].
For any tree , there exists an integer such that every edgeconnected graph with number of edges divisible by has a decomposition.
In [bib:origpaths], the authors posed the following, strengthened version of the conjecture of Barát and Thomassen and they proved it for being a path.
There is a function such that, for any fixed tree with maximum degree , every edgeconnected graph with minimum degree at least and number of edges divisible by has a decomposition.
We define the length of a path as its number of edges. The following result from [bib:origpaths] answers the previous question for .
For every integer , there exists such that the edge set of every edgeconnected graph with minimum degree and number of edges divisible by has a decomposition into paths of length .
However, Conjecture 1 is not true in general. In Section 2, we show that it does not hold even for trees of maximum degree three.
In Section 3, we consider a variation of Conjecture 1 for forests. We call a tree proper if it has at least one edge. We call a forest proper if each of its connected components is a proper tree. Note that if is a forest which is not proper, for the purpose of finding an decomposition, one can disregard components with a single vertex, provided that the host graph has at least as many vertices as . Having this in mind, we state our result only for proper forests.
Theorem 1 can be easily extended to proper forests by gluing several copies of a forest together (see 3 for details). Moreover, one can replace edgeconnectivity requirement by minimum degree requirement in case of coprime forests. We call a forest coprime if there is no integer which divides the number of edges of all of its components.
For any proper coprime forest , there exists an integer such that every graph with minimum degree at least and number of edges divisible by the number of edges of has an decomposition.
Note that the requirement that the forest is coprime is necessary. For instance, if every component of
has even number of edges, a graph with a connected component that has an odd number of edges does not have any
decomposition.From Theorem 1, we can easily derive, in a sense, a relaxation of Conjecture 1. In particular, for two trees , with coprime numbers of edges, high minimum degree implies the existence of a decomposition.
For any two trees with coprime numbers of edges, there exists an integer such that every graph with minimum degree at least has a decomposition.
The core idea of the proof of Theorem 1 (which is used in Lemma 3) is to partition the host graph along edgecuts of bounded size into vertex disjoint parts that are highly edgeconnected. This is achieved by repeated applications of the following lemma.
Let be an integer and be a multigraph. Then, there is a cut in of order at most such that is edgeconnected or has only one vertex.
We then obtain decompositions of these parts using the result of [bib:BHLMT]. Finally, we construct a decomposition of the whole graph by combining decompositions of the parts.
2 Disproving Conjecture 1
In this section, we disprove Conjecture 1. We show that it does not hold even for trees of maximum degree three.
Assume that Conjecture 1 holds for some function . Let be the complete binary tree of depth (see Figure 1 for an example). The maximum degree of is three and the number of edges of is for every . Let be the set of possible numbers of edges contained in a component of for some edge of . Observe that the components of have edges and edges for some , for every edge . Thus, and then .
It follows that the sum , where for every , can attain at most different values. Therefore, there exists such that for every .
Fix . Then, there exists such that for any choice of values of .
Let and be edgeconnected graphs with minimum degree at least such that the number of edges of is congruent to modulo , the number of edges of is congruent to modulo and there are sets , of size such that the distance between every two distinct vertices of in , , is greater than (where the distance is the length of the shortest path between the two vertices). See Figure 2 for an example of a construction of with these properties, can be constructed in an analogous way.
Let be a graph obtained from the disjoint union of and by adding a matching of size between the vertices of and . Then, is edgeconnected, of minimum degree at least and with number of edges divisible by . Assume that there exists a decomposition of . Since the distance between any two vertices in and any two vertices in is greater than the distance between any two vertices in , every copy of in contains at most one vertex of and and therefore at most one edge of . Note that each copy of with an edge in contains edges of . Therefore, the number of edges of is , where is an integer and for every . This yields a contradiction with the choice of the number of edges of .
3 Decomposition into coprime trees
In this section, we prove Theorem 1 after introducing some tools.
Let be a set of graphs. We say that has a decomposition if its edge set can be partitioned into edge disjoint copies of graphs in . Decompositions are transitive in the following sense.
Let be a graph and let and be sets of graphs. If the edge set of can be partitioned in edgedisjoint subgraphs such that each of them has an decomposition, then has an decomposition. In particular, if every graph in has an decomposition and has a decomposition, has an decomposition.
Given a sequence of vertex disjoint proper trees , we define a chain as a graph obtained from by choosing two distinct leaves and in each and identifying and for . Note that a chain is a tree.
For a proper tree and an integer , we define a chain of to be a chain , where each is isomorphic to . We denote it by . Similarly, we define a chain of a proper forest with connected components as the tree .
Let be a proper forest and . Then has an decomposition.
Next, we argue that Theorem 1 can be easily extended to proper forests. In particular, given a fixed proper forest , if has number of edges divisible by and is edgeconnected, where is as in Theorem 1, then has an decomposition. Indeed, from Theorem 1 it follows that if is divisible by , then has an decomposition. The existence of an decomposition follows from Observation 3.
If is divisible by but not by , we can remove any copy of to make the number of edges of divisible by , decreasing edgeconnectivity by at most (to at least ) and then argue as before.
Let , and be trees such that the number of edges of is coprime with both the number of edges of and the number of edges of . Then, there exists an integer such that every graph with minimum degree at least and number of edges divisible by the greatest common divisor of and has a decomposition that contains the same number of copies of and .
Proof of Theorem 1.
Let be the connected components of the coprime forest . By Bézout’s lemma, there exist integers such that . We define and . Let be and be . Then, , i.e. the number of edges of is coprime with the number of edges of . Moreover, , thus the number of edges of is also coprime with number of edges of . Let be a prime number greater than and and define to be . Then, is coprime with both and .
Then, by Lemma 3, has a decomposition . Let be a subgraph of formed by the copies of and in . Since the graph has a decomposition by Observation 3 and 3, it also has an decomposition. Thus, to show that has an decomposition, it suffices to show that has an decomposition.
Note that has a decomposition into edge disjoint copies of and edge disjoint copies of for some integer . Let . We show that has an decomposition for every and thus, by Observation 3, has an decomposition. By construction, has a decomposition in which each appears times. Similarly, has a decomposition in which each appears times. Note that and are at least for every . We first construct copies of in combining trees from and in such a way that at the end of the process, there will be exactly unused copies of each in each of and . We construct one copy of at a time by greedily picking copies of one by one, always choosing it from that of and which contains the greatest number of unused copies of the component (in case of equality we choose arbitrarily). We argue that it is always possible to pick a copy which is vertex disjoint from all the previously picked components in the currently constructed copy of . Assume that we need to pick a copy of and assume that contains greater number of unused copies of  the argument for is analogous.
The total number of copies of in and is and therefore after using less than copies of , contains more than unused copies of . By construction, each copy of , in or intersects at most copies of in . Therefore, the already chosen components of the currently constructed copy of intersect at most copies of in , so there exist at least one copy of which is vertex disjoint. Note that after constructing copies of in this way, both and contain unused copies of each , . We denote these collections of unused copies by and respectively and argue that can be partitioned into copies of . Then, can be partitioned into copies of by analogous arguments.
We call a copy of , , in bad if it intersects a copy of in with . We call all other copies in good. Observe that by construction, there are at most bad copies in . Any collection of copies in containing one copy of for each such that at most one of the copies is bad forms a copy of the forest (i.e., the copies in the collection are pairwise disjoint). Thus, it is possible to construct copies of greedily by picking one bad copy and good copies of the remaining components of . Since contains copies of each , and , there are enough good copies of each component of for repeatedly creating copies of in this way, until all the bad copies in are used. Once all the bad copies are used, the remaining good copies in can be arbitrarily partitioned into copies of .
∎
4 Proof of Lemma 3
We start by showing that sufficiently highly edgeconnected graph can be decomposed into copies of two trees with coprime numbers of edges.
Let be trees with coprime numbers of edges. Then, there exists an integer such that every edgeconnected graph has a decomposition with less than copies of .
Proof.
Let , be numbers of edges of , , respectively, and let be as in Theorem 1 and let . Let be a edgeconnected graph and let be the smallest nonnegative integer such that . Since and are coprime, such exists and is smaller than by Bézout’s Lemma.
By the greedy algorithm, it is possible to find a collection of edgedisjoint copies of in . Then, is a edgeconnected graph with number of edges divisible by . The result follows from Theorem 1.
∎
For the purpose of the proof of Lemma 3, we extend the definition of a graph by allowing hyperedges of size one, which we call stubs and we call the resulting object a stub graph. Each vertex of a stub graph can be incident with arbitrarily many stubs and the degree of a vertex is the number of edges and stubs incident with it. Moreover, we assign a positive integer to each stub and we call the index of the stub . Intuitively, a stub can be viewed as a remainder of an edge after one of its endvertices has been removed from the graph. The index of the stub then contains some information about the removed endvertex.
We write to denote the set consisting of the edges and the stubs of a stub graph and we call it the edge set of , but we do not refer to stubs as edges otherwise. We denote the set of stubs of index in a stub graph by . A subgraph of a stub graph is also a stub graph and we say that two subgraphs are edgedisjoint if their edge sets are disjoint, i.e., they do not share any edge or stub.
Let be a set of trees (without stubs). We extend the definition of decomposition for graphs to stub graphs. Informally, in a decomposition of a stub graph, a stub incident with a vertex plays the role of a subtree of , such that the vertex is a leaf of this subtree.
A twig is a pair , where is a proper tree and is a leaf of . We say that is the root of the twig. Let be a stub graph, a stub incident with a vertex in and a twig disjoint from . A stub graph obtained from by expanding by is the stub graph obtained from and by identifying and .
An embedding of a tree in a stub graph is a subgraph of such that each stub in has different index and there exists a mapping assigning a twig to each stub in such that expanding every stub in by yields a copy of . See Figure 3 for an example.
We say that a stub graph has a decomposition if its edge set can be decomposed into disjoint sets such that each forms an embedding of some . Note that this definition coincides with the definition of a decomposition in the usual sense if has no stubs.
We denote the graph obtained from by removing all the stubs by and we say that is edgeconnected if is edgeconnected. The next observation asserts that a decomposition of can be easily extended to a decomposition of .
If has a decomposition , there exists a decomposition of such that . Moreover, given , there exists such that contains only embeddings of .
Proof.
It follows from the fact that a stub (with the vertex incident to it) forms an embedding of any proper tree . It is enough to let , where is some leaf of . ∎
In particular, a stub graph with no edges has a decomposition for any set of proper trees , since has a trivial (empty) decomposition. It follows that Lemma 4 holds for stub graphs as well.
Next, we introduce some more tools and terminology. Let be a multigraph. We call a partition of into two parts a cut in . We denote the set of edges of incident with a vertex in both and and call the order of the cut . We now prove Lemma 1. Note that the definition of a cut and the statement of the lemma trivially extend to stub graphs (by considering the multigraph instead of the stub graph ).
Proof of Lemma 1.
Let be a cut in of order at most such that is inclusionwise minimal. Assume that has more than one vertex and let be a cut in . By minimality of , we have that the cuts and have order greater than . Since , has order greater than . ∎
The following results of Czumaj and Strothmann were originally proven only for simple graphs, however, they easily extend to multigraphs.
[Czumaj, Strothmann [bib:czumaj], extended]
Every edgeconnected multigraph contains a spanning tree such that for every vertex of .
To find such a tree, it is enough to take an outbranching in a strongly connected balanced orientation of .
[Czumaj, Strothmann [bib:czumaj], extended]
Let be a positive integer. If a multigraph contains edgedisjoint spanning trees, then has a spanning tree such that for every vertex of .
The following Corollary 4 of Theorem 4 was essentially proven in [bib:thomassenpaths]. Our version differs in some details. Among other things, we use the following theorem of NashWilliams and Tutte to replace the requirement of having a collection of edgedisjoint spanning trees by edgeconnectivity.
[NashWilliams [bib:nashwilliams], Tutte [bib:tutte]] If a multigraph is edgeconnected, then contains edgedisjoint spanning trees.
For every and integer , there exists such that every edgeconnected stub graph with minimum degree at least , where , has edgedisjoint spanning trees such that
for every .
Proof.
By Theorem 4, the graph contains edgedisjoint spanning trees. Thus, contains edgedisjoint edgeconnected spanning subgraphs (formed by unions of the spanning trees). From Theorem 4 applied to these edgeconnected graphs, it follows that contains edgedisjoint spanning trees such that
for every . Moreover for sufficiently large. The result follows. ∎
Corollary 4 implies the following.
For any positive integers and , there exists such that the edge set of every edgeconnected stub graph with minimum degree at least can be decomposed into a edgeconnected graph and a stub graph of minimum degree at least .
Proof.
We need the following easy consequence of Bézout identity for the proof of Lemma 3. Let be positive integers and let be an integer such that and is divisible by the greatest common divisor of and . Then there exist nonnegative integers and such that satisfying .
Proof.
Let be the greatest common divisor of and and let be an integer such that . By Bézout identity there exist integers and satisfying , thus . Note that then also for any integer . Let be such that . Then and satisfy the claim of the observation, in particular, since and , must be positive. ∎
We now prove Lemma 3.
Proof of Lemma 3.
Let and we repeat the following recursive procedure, obtaining pairs of stub graphs until is empty. Let be the number of steps before is empty.
If is edgeconnected or has only one vertex, we let and . Assume that is not edgeconnected and has more than one vertex. Then, by Lemma 1, there exists a cut in of order at most such that is edgeconnected or has only one vertex.
Let and let be the set of edges with and . Let be the stub graph obtained from by adding a stub with index incident with for every edge in incident with .
Observe that . Moreover, the graphs and have the following properties.
For every , the following holds:

There are at most stubs with index created during the procedure (i.e., in total, there are at most stubs with index in the stub graphs ),

has minimum degree at least ,

is edgeconnected or has only one vertex,

has minimum degree at least (minimum of the empty set is ).
Since there is a onetoone correspondence between the stubs of index and the edges in and , there are at most stubs with index . Moreover, for every and therefore has minimum degree at least by induction.
By construction, is edgeconnected or has only one vertex and since has minimum degree at least , has minimum degree at least as required.
We call a decomposition balanced if the numbers of copies of and differ by at most . Since is empty, it has a balanced decomposition (the trivial one). Next, we proceed inductively, constructing a balanced decomposition of from a balanced decomposition of for . Moreover, in each step we increase the number of copies of in the decomposition by at most and keep the number of copies of the same, or the other way round, we increase the number of copies of in the decomposition by at most and keep the number of copies of the same.
In the last step, we construct a decomposition of from a balanced decomposition of in a similar way, ensuring that the numbers of copies of and in the constructed decomposition are the same.
Roughly speaking, each step of construction has two phases: first, we replace every stub in of index by a subtree in . Then we decompose the remaining part of using Lemma 4. In the last step, when i=1, we will proceed in a slightly different way to ensure that the resulting decomposition will contain the same number of copies of and .
More formally, given a balanced decomposition of , let if the number of copies of in is smaller than the number of copies of and otherwise. Recall that and .
If has more than one vertex, it is edgeconnected and thus, by Corollary 4, contains a spanning subgraph with minimum degree at least such that is edgeconnected. If has only one vertex, let and is an isolated vertex.
We replace every embedding of , or in which contains a stub by an embedding of , or in . This yields a partial decomposition of .
Moreover, is such that . Thus, the stub graph has a decomposition that contains at most copies of by Lemma 4 and by Observation 4, because is either edgeconnected or has only one vertex (and therefore no edges). Then, forms a decomposition of . Since there are at most copies of in , from the choice of it follows that if the difference between the number of copies of and in was at most , the difference in is also at most .
Let be an embedding of , or containing a stub with index (i.e., ). Note that contains at most one stub in . For each such , we construct such that in the following way. We assume that is an embedding of , the construction for and is analogous.
Let be the set of the indices of the stubs in . Let be the vertex incident with and let be the edge in corresponding to the stub . We find an embedding of such that

and it corresponds to the root of ,

and , and

no stub in has its index in .
Then, is an embedding of in .
Moreover, we ensure that and are disjoint for every two distinct stubs . Thus, the edge sets of the embeddings in will be mutually disjoint.
We construct an embedding of greedily. Starting from which consists of the root in and the edge , we add edges and stubs one by one. At the same time, we remove the used edges and stubs from , making sure that no edge and no stub is used in more than one embedding. Let be the set of the indices of the stubs in . Assume that is not yet an embedding of and let be a vertex of to which we need to add an edge or a stub. We argue that either has a neighbor in and therefore we can extend by the edge (removing from ) or is incident with a stub in such that its index is not in . This is indeed the case; at the beginning, had degree at least in , at most edges and stubs from were removed by embedding trees corresponding to stubs in and at most edges and stubs incident with were removed or cannot be used for extending because the other endpoint of the edge is already in . This leaves at least available edges and stubs incident with . Since , contains a stub incident with such that its index is not in , or an edge with .
At the last step, it remains to construct a decomposition of from a balanced decomposition of . Note that is not a single vertex and has no stubs. Let , be subgraphs of defined as before, in particular, is edgeconnected (since ).
Let , and be the numbers of embeddings of , and in respectively. Without loss of generality, we assume that .
Since is divisible by the greatest common divisor of and , is also divisible by the greatest common divisor of and . By Observation 4, there exists an integer such that is divisible by .
We greedily construct copies of and copies of in , denote the resulting partial decomposition by and remove its edges from . Note that the minimum degree of decreases by at most