Abstract

We provide a structural property of trees, which is applied to show that if a plane graph contains two edge-disjoint spanning trees, then its dual graph has the vertex-arboricity at most 2. We also show that every maximal plane graph of order at least 4 contains two edge-disjoint spanning trees.

1. Introduction

All graphs considered in this paper are finite simple graphs. Given graph , let , , , and denote its vertex set, edge set, vertex number, and edge number, respectively. For a vertex , let denote the degree of in . Moreover, let and denote the maximum degree and minimum degree of , respectively. A tree is a connected graph without cycles. A plane graph is a particular drawing of a planar graph on the Euclidean plane.

A subgraph of is called a spanning one if . A plane graph is called maximal if every face of is a triangle. A connected Eulerian graph is a connected one that contains no vertices of degree odd. The dual, denoted by , of a plane graph is a plane graph whose vertices correspond to the faces of and edges correspond to the edges of in this way: if is an edge of incident to faces , then the endpoints of the dual edge are vertices of that represent the faces of of . The vertex-arboricity of a graph is the minimum number of subsets into which can be partitioned so that each subset induces a forest.

The vertex-arboricity of a graph was first introduced by Chartrand et al. [1], named as point-arboricity. Among other things, they proved that the vertex-arboricity of planar graphs is at most 3. Chartrand and Kronk [2] showed that this bound is sharp by presenting a planar graph of the vertex-arboricity 3. More generally, Kronk [3] showed that if is a surface with Euler genus , then and if is the sphere or the Klein bottle, then . Hara et al. [4] extended partially Kronk’s result by proving that if is the projective plane or the torus, and if is the Klein bottle. Here the vertex-arboricity of surface is defined to be the maximum of the vertex-arboricity of all graphs embeddable into . Other results about the vertex-arboricity of embedded graphs in the surface are referred to in [5–8].

The following theorem, due to Stein [9], characterizes completely maximal plane graphs with vertex-arboricity 2.

Theorem 1. Let be a maximal plane graph with . Then if and only if is Hamiltonian.

Hakimi and Schmeichel [10] extended Stein’s theorem to the following form.

Theorem 2. Let be a plane graph. Then if and only if contains a connected Eulerian spanning subgraph.

Note that determining whether a graph to have a connected Eulerian spanning subgraph is quite difficult. Thus a natural question is as follows: which plane graphs have a connected Eulerian spanning subgraph? In this paper, we provide a sufficient condition about the problem.

2. Structural Property

Let denote a path from vertex to vertex . We call a vertex of degree 1 a leaf in a tree. The following lemma is of interest by itself.

Theorem 3. Let be a tree with and be an integer. If with , then can be partitioned into two sets and such that contains edge-disjoint paths .

Proof. The proof is proceeded by induction on the order of . If , then is a path of length 1 or 2. Note that in this case, so the theorem holds trivially. Suppose that is a tree with . Let be a subset of with . If contains a leaf which is not in , then . Let . Then is a tree of order and satisfies . By the induction hypothesis, can be partitioned into two sets and such that contains edge-disjoint paths . Clearly, also is a required partition of .
So assume that contains all the leaves of . If there exist two leaves and adjacent to a common vertex , we let and . Then is a tree of order and satisfies . By the induction hypothesis, can be partitioned into two sets and such that contains edge-disjoint paths , . Defining , for , , and , we obtain a desired partition of .
Now suppose that every vertex of is adjacent to at most one leaf. Let be a longest path in . Then it is easy to see that and are leaves. If is adjacent to some vertex, , different from and , then is not a leaf by the assumption. There exists a vertex , other than , adjacent to . However, is a path whose length is greater than that of , contradicting the choice of . Hence it follows that . Similarly, we derive that .
If , we define and . Then satisfies . By the induction hypothesis, admits a required partition with so that contains edge-disjoint paths . In , we define , , and .
If , then . Let and . Then is a tree of order and with . By the induction hypothesis, can be partitioned into two sets and such that contains edge-disjoint paths . Since is a leaf of , is an end of some path . Furthermore, . Without loss of generality, suppose that and . In , we let , , , and for . A desired partition of in is established. This proves the theorem.

By means of Theorem 3, we can give a simple proof for the following result.

Theorem 4. Let be a graph with . If contains two edge-disjoint spanning trees, then contains a connected Eulerian spanning subgraph.

Proof. Suppose that and are two edge-disjoint spanning trees of . Then . Let denote the subset of vertices of degree odd in . Then is even. Since contains at least two leaves, . Since , by Theorem 3, can be partitioned into two sets and such that contains edge-disjoint paths , .
Let denote the subgraph of induced by the edge set . Let . is connected and spanning, so is . Let . It is easy to observe that . If , then both and are even, and hence is even. If , then both and are odd, and so is even. It follows that is Eulerian. Thus is a connected Eulerian spanning subgraph of . The proof of the theorem is complete.

Combining Theorems 2 and 4, we obtain the main result in this paper.

Theorem 5. Let be a plane graph. If contains two edge-disjoint spanning trees, then .

3. Spanning Trees in Maximal Plane Graphs

Let be a maximal plane graph. Then the following properties (a)–(d) hold:(a);(b) if ;(c)Each of the faces in is of degree 3;(d)If , then is isomorphic to ; if , then is isomorphic to ; if , then is isomorphic to , where is any edge of . In particular, when , we have that .

Theorem 6. Every maximal plane graph with contains two edge-disjoint spanning trees.

Proof. We prove the theorem by induction on the vertex number . Since , it follows that . If , then is isomorphic to , and can be easily edge-partitioned into two edge-disjoint spanning trees. Hence the basis step of induction is established. Let be a maximal plane graph with . So . By the above property (b), contains a vertex with . Set . Let denote the neighbors of in in clockwise order and denote the incident faces of in in clockwise order with as two boundary edges of for . Here all the indices are taken modulo . Note that are all 3 faces.
The proof is split into the following three cases, depending on the size of .
Case 1 . Let . Then is a maximal plane graph with and . By the induction hypothesis, contains two edge-disjoint spanning trees and . Let and . Obviously, is a spanning tree of for , and . Thus, the theorem holds in this situation.
Case 2 . In view of the planarity of , at least one of and does not belong to . Without loss of generality, assume that . Let . Then is a maximal plane graph with and . By the induction hypothesis, contains two edge-disjoint spanning trees and . If , then we define for . If , then we define and . If , then we define and . It is easy to inspect that and are two edge-disjoint spanning trees of in each of the above three cases.
Case 3 . Again, by the planarity of , there exists a vertex , say , such that . Let . Then is a maximal plane graph with and . By the induction hypothesis, contains two edge-disjoint spanning trees and . We have to consider the following subcases.
Case 3.1. At least one of and does not belong to , say .
If , then we set for . If , then we set and . If , then we set and .
Case 3.2 . We need to deal with the following two subcases by symmetry.
Case 3.2.1 for some . Without loss of generality, assume that . It is enough to set and .
Case 3.2.2 and . Let and denote the two components of with and , and and denote the two components of with and . Since is a cut edge of , it follows that belongs to exactly one of and . Similarly, belongs to exactly one of and .
If , then we define and .
If , then we define and .
Now assume that and . This implies that and . If , then we define and . If , then we define and . Otherwise, and . It suffices to define and .
It is easy to inspect that and are two edge-disjoint spanning trees of in every possible case above. This completes the proof of the theorem.

The following consequence follows immediately from Theorems 4 and 6.

Corollary 7. Every maximal planar graph with contains a connected Eulerian spanning subgraph.

Conflicts of Interest

The authors declare that they have no conflicts of interest.