Decomposing graphs into forests
Christian Reiher, Lisa Sauermann

TL;DR
This paper presents a straightforward graph-theoretic proof of Nash-Williams' theorem on covering graphs with forests, including a generalization allowing preassigned edges to specific forests, enhancing understanding of graph decompositions.
Contribution
The paper offers a simple proof of Nash-Williams' classical result and extends it to cases with preassigned edges, broadening its applicability.
Findings
Provided a simple proof of Nash-Williams' theorem
Extended the theorem to include preassigned edges
Enhanced understanding of graph decompositions into forests
Abstract
We give a simple graph-theoretic proof of a classical result due to C. St. J. A. Nash-Williams on covering graphs by forests. Moreover we derive a slight generalisation of this statement where some edges are preassigned to distinct forests.
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Nash-Williams’ theorem on decomposing graphs into forests
Christian Reiher
Fachbereich Mathematik, Universität Hamburg, Bundesstraße 55, 20144 Hamburg, Germany
and
Lisa Sauermann
Department of Mathematics, Stanford University, 450 Serra Mall, Building 380, Stanford CA 94305, USA
Abstract.
We give a simple graph-theoretic proof of a classical result due to C. St. J. A. Nash-Williams on covering graphs by forests. Moreover we derive a slight generalisation of this statement where some edges are preassigned to distinct forests.
Key words and phrases:
Nash-William’s theorem, forest covering, arboricity
1. Introduction
All graphs considered in this note are finite. The sets of vertices and edges of a graph are denoted by and , respectively. The restriction of a graph to a subset of , i.e., the graph on whose edges are precisely those members of both of whose ends belong to , is indicated by . When is clear from the context, we write for the number of edges of that graph. For any integer we set .
The following result is due to C. St. J. A. Nash–Williams (see [NW2], and the related articles [NW1, Tut] as well as [China] for another simple proof).
Theorem 1**.**
If is a graph, and is an integer such that for all nonempty subsets of one has , then there exists a partition such that is a forest for .
It is plain that the sufficient condition for such a partition to exist given here is also necessary. Also, the cases and of this statement are immediate, and the case was recently posed at the All-Russian Mathematical Olympiad, [Russen]. That case can be dealt with by some peculiar tricks not discussed here and that do not straightforwardly generalize to , but which nevertheless motivated us to reprove the general case independently. In fact, it turned out that our arguments for the case yielded slightly more, namely that for any two distinct edges of there exists such a partition in which one of the edges belongs to while the other one belongs to . Hence one might guess:
Corollary 2**.**
Given a graph , an integer such that for all nonempty subsets of one has , and moreover a sequence of distinct edges of , there exists a partition such that for and is a forest for all .
As we shall see in Section 3, this can in fact be derived from Theorem 1.
All statements and arguments contained in this article are valid irrespective of whether multiple edges are allowed to occur in our graphs or not.
2. Proving Theorem 1
In this section we give a simple proof of Theorem 1 that is, to the best of our knowledge, new. For this purpose we need some preparation. Let be a forest, an integer, and mutually vertex disjoint, connected subgraphs of . Obviously are trees and each of them is contained in exactly one component of . We call isolated in if there is no with , that is contained in the same component of as . Furthermore we call peculiar in if there is an edge e_{i}\in E(T)\smallsetminus\bigl{(}E(T_{1})\cup\dots\cup E(T_{k})\bigr{)} incident with a vertex of , such that is isolated in (the reader may notice that one gets an equivalent notion without demanding to be incident with a vertex of ).
Fact 3**.**
At least two of the subgraphs , , …, are isolated or peculiar in .
Proof.
Otherwise take a counterexample where has as few vertices as possible. If no component of contains two or more of the , then each them is isolated and we are done. In the remaining cases is a tree. Consider any leaf of . As we cannot produce a smaller counterexample by deleting thus, there has to be some consisting solely of , and the edge of incident with witnesses that this is peculiar. Because of the tree has at least two vertices and, consequently, at least two leaves. Applying the foregoing argument to any two leaves of we see that at least two of the trees , , …, and are peculiar. ∎
Proof of Theorem 1..
Arguing indirectly we choose a graph and an integer contradicting Theorem 1 with minimal. Then . Let be an arbitrary edge of .
Because of the choice of , there is at least one partition such that is a forest for . For each of these partitions we consider the component of containing the vertex . From now on let
[TABLE]
be one of these partitions with minimum. Let .
If , the partition would satisfy all conditions of Theorem 1, consequently and . Therefore
[TABLE]
Thus, there is an with . This implies, that is not connected. Because of the definition of we have , so we can w.l.o.g. assume .
Let , …, be the connected components of , where obviously . Thus, , …, are mutually vertex disjoint, connected subgraphs of the forest . We define isolation and peculiarity in as applying to these subgraphs. By Fact 3 at least two of the subgraphs , …, are isolated or peculiar in , and at most one of them contains . Let w.l.o.g. be isolated or peculiar in and .
If is peculiar in , there is an edge
[TABLE]
incident with a vertex of , such that is isolated in . Notice that
[TABLE]
yields , wherefore connects and a vertex not in .
If is isolated in , let be an arbitrary vertex of .
We consider the uniquely determined path from to in the tree . Since and , this path contains an edge connecting a vertex of with a vertex of some with .
First, we consider the case when is isolated in . Then the graph is a forest, because connects different components of . Obviously, the graph is also a forest and its component including is a subgraph of . But this subgraph does not contain and has therefore a number of vertices smaller than . This contradicts
[TABLE]
being one of the partitions considered at the beginning.
For the second case let now be peculiar in . Then is isolated in and the graph \bigl{(}V,(E_{2}^{\prime}\smallsetminus\{e_{1}\})\cup\{e_{d}\}\bigr{)} is therefore a forest. On the other hand the graph is also a forest and a subgraph of the forest . Because belongs to the component of , the graph has two components being subgraphs of , one of which contains and the other one . The edge connects and a vertex not in , consequently the graph \bigl{(}V,(E_{1}^{\prime}\smallsetminus\{e_{d}\})\cup\{e_{1}\}\bigr{)} is a forest and its component including is equal to the component of including . Thus, its number of vertices is smaller than . This contradicts
[TABLE]
being one of the partitions considered at the beginning.
Since we have obtained a contradiction in each of the two cases, our assumption must have been wrong and Theorem 1 is true. ∎
3. Deducing Corollary 2
The strengthening given by Corollary 2 will now be deduced from Theorem 1 by means of a short argument.
Proof of Corollary 2.
Let , , and be as in Corollary 2. We call an integer restrained if there is a partition such that for and is acyclic for all . By Theorem 1 the integer [math] is restrained. Corollary 2 is equivalent to being restrained. It is therefore sufficient to prove the following: if an integer with is restrained, then the integer is also restrained.
Let be a partition such that for and is acyclic for all . If , we are done. We can therefore assume with . Obviously we can assume that the two vertices of belong to the same component of .
The two vertices of belong to different components of the forest . Therefore the uniquely determined path between these two vertices in the forest contains vertices of different components of . Hence, there is an edge of this path connecting vertices of different components of . Then the graph \bigl{(}V,(E_{\ell}\smallsetminus\{e_{k+1}\})\cup\{e\}\bigr{)} is a forest. On the other hand the graph \bigl{(}V,(E_{k+1}\smallsetminus\{e\})\cup\{e_{k+1}\}\bigr{)} is by the definition of also a forest.
Thus, the partition gained from by substituting by and by fulfills all conditions for being restrained. ∎
References
