Decomposing graphs into forests

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.

Equations

$$E\setminus \{e\}=E_1^{\prime }\cup E_2^{\prime }\cup \cdots \cup E_r^{\prime }$$

$$|E_1^{\prime }\cap \overline{C}|+\cdots +|E_r^{\prime }\cap \overline{C}|<|\overline{C}|=e(C)⩽r(|C|-1).$$

$$e_{1}\in E_{2}^{\prime}\smallsetminus\bigl{(}E(D_{1})\cup\dots\cup E(D_{k})\bigr{)}$$

$$E(D_1)\cup \cdots \cup E(D_k)=\overline{C}\cap E_2^{\prime }$$

$$E\setminus \{e\}=(E_1^{\prime }\setminus \{e_d\})\cup (E_2^{\prime }\cup \{e_d\})\cup \cdots \cup E_r^{\prime }$$

$$E\smallsetminus\{e\}=\bigl{(}(E_{1}^{\prime}\smallsetminus\{e_{d}\})\cup\{e_{1}\}\bigr{)}\cup\bigl{(}(E_{2}^{\prime}\smallsetminus\{e_{1}\})\cup\{e_{d}\}\bigr{)}\cup\dots\cup E_{r}^{\prime}$$

Full text

1. Introduction

All graphs considered in this note are finite. The sets of vertices and edges of a graph $G$ are denoted by $V(G)$ and $E(G)$, respectively. The restriction of a graph $G=(V,E)$ to a subset $X$ of $V$, i.e., the graph on $X$ whose edges are precisely those members of $E$ both of whose ends belong to $X$, is indicated by $G|X$. When $G$ is clear from the context, we write $e(X)$ for the number of edges of that graph. For any integer $r\geq 0$ we set $[r]=\{1,2,\ldots,r\}$.

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 $G=(V,E)$ is a graph, and $r\geq 0$ is an integer such that for all nonempty subsets $X$ of $V$ one has $e(X)\leq r(|X|-1)$, then there exists a partition $E=E_{1}\cup E_{2}\cup\ldots\cup E_{r}$ such that $(V,E_{i})$ is a forest for $i\in[r]$.

It is plain that the sufficient condition for such a partition to exist given here is also necessary. Also, the cases $r=0$ and $r=1$ of this statement are immediate, and the case $r=2$ 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 $r>2$, but which nevertheless motivated us to reprove the general case independently. In fact, it turned out that our arguments for the case $r=2$ yielded slightly more, namely that for any two distinct edges of $G$ there exists such a partition in which one of the edges belongs to $E_{1}$ while the other one belongs to $E_{2}$. Hence one might guess:

Corollary 2**.**

Given a graph $G=(V,E)$, an integer $r\geqslant 0$ such that for all nonempty subsets $X$ of $V$ one has $e(X)\leqslant r(|X|-1)$, and moreover a sequence $e_{1},e_{2},\ldots,e_{r}$ of distinct edges of $G$, there exists a partition $E=E_{1}\cup E_{2}\cup\ldots\cup E_{r}$ such that $e_{i}\in E_{i}$ for $i\in[r]$ and $(V,E_{i})$ is a forest for all $i\in[r]$.

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 $T$ be a forest, $k\geq 2$ an integer, and $T_{1},T_{2},\ldots,T_{k}$ mutually vertex disjoint, connected subgraphs of $T$. Obviously $T_{1},T_{2},\ldots T_{k}$ are trees and each of them is contained in exactly one component of $T$. We call $T_{i}$ isolated in $T$ if there is no $T_{j}$ with $j\neq i$, that is contained in the same component of $T$ as $T_{i}$. Furthermore we call $T_{i}$ peculiar in $T$ 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 $T_{i}$, such that $T_{i}$ is isolated in $T-e_{i}$ (the reader may notice that one gets an equivalent notion without demanding $e_{i}$ to be incident with a vertex of $T_{i}$).

Fact 3**.**

At least two of the subgraphs $T_{1}$, $T_{2}$, …,$T_{k}$ are isolated or peculiar in $T$.

Proof.

Otherwise take a counterexample where $T$ has as few vertices as possible. If no component of $T$ contains two or more of the $T_{i}$, then each them is isolated and we are done. In the remaining cases $T$ is a tree. Consider any leaf $x$ of $T$. As we cannot produce a smaller counterexample by deleting $x$ thus, there has to be some $T_{i}$ consisting solely of $x$, and the edge of $T$ incident with $x$ witnesses that this $T_{i}$ is peculiar. Because of $k\geq 2$ the tree $T$ has at least two vertices and, consequently, at least two leaves. Applying the foregoing argument to any two leaves of $T$ we see that at least two of the trees $T_{1}$, $T_{2}$, …, and $T_{k}$ are peculiar. ∎

Proof of Theorem 1..

Arguing indirectly we choose a graph $G=(V,E)$ and an integer $r\geqslant 0$ contradicting Theorem 1 with $|E|$ minimal. Then $|E|\neq 0$. Let $e=ab$ be an arbitrary edge of $G$.

Because of the choice of $G$, there is at least one partition $E\smallsetminus\{e\}=E_{1}^{\prime}\cup E_{2}^{\prime}\cup\dots\cup E_{r}^{\prime}$ such that $(V,E_{i}^{\prime})$ is a forest for $i=1,2,\dots,r$. For each of these partitions we consider the component $(C,E_{C})$ of $(V,E_{1}^{\prime})$ containing the vertex $a$. From now on let

[TABLE]

be one of these partitions with $|C|$ minimum. Let $\overline{C}=E(G|C)$.

If $b\not\in C$, the partition $E=(E_{1}^{\prime}\cup\{e\})\cup E_{2}^{\prime}\cup\dots\cup E_{r}^{\prime}$ would satisfy all conditions of Theorem 1, consequently $b\in C$ and $e\in\overline{C}$. Therefore

[TABLE]

Thus, there is an $i\in[r]$ with $|E_{i}^{\prime}\cap\overline{C}|<|C|-1$. This implies, that $(C,E_{i}^{\prime}\cap\overline{C})$ is not connected. Because of the definition of $(C,E_{C})$ we have $i\neq 1$, so we can w.l.o.g. assume $i=2$.

Let $D_{1}$, …, $D_{k}$ be the connected components of $(C,E_{2}^{\prime}\cap\overline{C})$, where obviously $k\geq 2$. Thus, $D_{1}$, …, $D_{k}$ are mutually vertex disjoint, connected subgraphs of the forest $(V,E_{2}^{\prime})$. We define isolation and peculiarity in $(V,E_{2}^{\prime})$ as applying to these subgraphs. By Fact 3 at least two of the subgraphs $D_{1}$, …, $D_{k}$ are isolated or peculiar in $(V,E_{2}^{\prime})$, and at most one of them contains $a$. Let w.l.o.g. $D_{1}$ be isolated or peculiar in $(V,E_{2}^{\prime})$ and $a\not\in V(D_{1})$.

If $D_{1}$ is peculiar in $(V,E_{2}^{\prime})$, there is an edge

[TABLE]

incident with a vertex $v_{1}$ of $D_{1}$, such that $D_{1}$ is isolated in $(V,E_{2}^{\prime}\smallsetminus\{e_{1}\})$. Notice that

[TABLE]

yields $e_{1}\not\in\overline{C}$, wherefore $e_{1}$ connects $v_{1}$ and a vertex not in $C$.

If $D_{1}$ is isolated in $(V,E_{2}^{\prime})$, let $v_{1}$ be an arbitrary vertex of $D_{1}$.

We consider the uniquely determined path from $a$ to $v_{1}$ in the tree $(C,E_{C})$. Since $a\not\in V(D_{1})$ and $v_{1}\in V(D_{1})$, this path contains an edge $e_{d}$ connecting a vertex of $D_{1}$ with a vertex of some $D_{i}$ with $i\neq 1$.

First, we consider the case when $D_{1}$ is isolated in $(V,E_{2}^{\prime})$. Then the graph $(V,E_{2}^{\prime}\cup\{e_{d}\})$ is a forest, because $e_{d}$ connects different components of $(V,E_{2}^{\prime})$. Obviously, the graph $(V,E_{1}^{\prime}\smallsetminus\{e_{d}\})$ is also a forest and its component including $a$ is a subgraph of $(C,E_{C})$. But this subgraph does not contain $v_{1}$ and has therefore a number of vertices smaller than $|C|$. This contradicts

[TABLE]

being one of the partitions considered at the beginning.

For the second case let $D_{1}$ now be peculiar in $(V,E_{2}^{\prime})$. Then $D_{1}$ is isolated in $(V,E_{2}^{\prime}\smallsetminus\{e_{1}\})$ 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 $(V,E_{1}^{\prime}\smallsetminus\{e_{d}\})$ is also a forest and a subgraph of the forest $(V,E_{1}^{\prime})$. Because $e_{d}$ belongs to the component $(C,E_{C})$ of $(V,E_{1}^{\prime})$, the graph $(V,E_{1}^{\prime}\smallsetminus\{e_{d}\})$ has two components being subgraphs of $(C,E_{C})$, one of which contains $a$ and the other one $v_{1}$. The edge $e_{1}$ connects $v_{1}$ and a vertex not in $C$, consequently the graph \bigl{(}V,(E_{1}^{\prime}\smallsetminus\{e_{d}\})\cup\{e_{1}\}\bigr{)} is a forest and its component including $a$ is equal to the component of $(V,E_{1}^{\prime}-\{e_{d}\})$ including $a$. Thus, its number of vertices is smaller than $|C|$. 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 $G=(V,E)$, $r\geq 0$, and $e_{1},\dots,e_{r}$ be as in Corollary 2. We call an integer $0\leqslant r^{\prime}\leqslant r$ restrained if there is a partition $E=E_{1}\cup E_{2}\cup\dots\cup E_{r}$ such that $e_{i}\in E_{i}$ for $i\in[r^{\prime}]$ and $(V,E_{i})$ is acyclic for all $i\in[r]$. By Theorem 1 the integer [math] is restrained. Corollary 2 is equivalent to $r$ being restrained. It is therefore sufficient to prove the following: if an integer $k$ with $0\leqslant k\leqslant r-1$ is restrained, then the integer $k+1$ is also restrained.

Let $E=E_{1}\cup E_{2}\cup\dots\cup E_{r}$ be a partition such that $e_{i}\in E_{i}$ for $i\in[k]$ and $(V,E_{i})$ is acyclic for all $i\in[r]$. If $e_{k+1}\in E_{k+1}$, we are done. We can therefore assume $e_{k+1}\in E_{\ell}$ with $\ell\neq k+1$. Obviously we can assume that the two vertices of $e_{k+1}$ belong to the same component of $(V,E_{k+1})$.

The two vertices of $e_{k+1}$ belong to different components of the forest $(V,E_{\ell}\smallsetminus\{e_{k+1}\})$. Therefore the uniquely determined path between these two vertices in the forest $(V,E_{k+1})$ contains vertices of different components of $(V,E_{\ell}\smallsetminus\{e_{k+1}\})$. Hence, there is an edge $e\in E_{k+1}$ of this path connecting vertices of different components of $(V,E_{\ell}\smallsetminus\{e_{k+1}\})$. 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 $e$ also a forest.

Thus, the partition gained from $E=E_{1}\cup E_{2}\cup\dots\cup E_{r}$ by substituting $E_{\ell}$ by ${(E_{\ell}\smallsetminus\{e_{k+1}\})\cup\{e\}}$ and $E_{k+1}$ by $(E_{k+1}\smallsetminus\{e\})\cup\{e_{k+1}\}$ fulfills all conditions for $k+1$ being restrained. ∎