Spanning k-ended trees of 3-regular connected graphs

A tree is called k-ended tree if it has at most k leaves, where a leaf is a vertex of degree one. In this paper we prove that every 3-regular connected graph with n vertices such that n is greater than 8 has spanning sub tree with at most [(2n+4)/9]-ended tree. At the end we give a conjecture about spanning k-ended trees on 3-regular connected graphs.


Introduction
Throughout this article we consider only finite undirected labeled graphs without loops or multiple edges. The vertex set and edge set of graph G is denoted by V = V (G) and E = E(G), respectively. For u, v ∈ V , an edge joining two vertices u and v is denoted by uv or vu. The neighbourhood N G (v) or N (v) of vertex v is the set of all u ∈ V which are adjacent to v. The degree of a vertex v, denoted by deg G (u) = |N G (v)|.
The minimum degree of a graph G is denoted δ(G) and the maximum degree is denoted ∆(G). If all vertices of G have same degree k, then the graph G is called k-regular. The distance between vertices u and v, denoted by d G (u, v) or d(u, v), is the length of a shortest path between u and v. A Hamiltonian path of a graph is a path passing through all vertices of the graph. A graph is Hamiltonian-connected if every two vertices are connected with a Hamiltonian path. In graph G, an independent set is a subset S of V (G) such that no two vertices in S are adjacent. A maximum independent set is an independent set of largest possible size for a given graph G. This size is called the independence number of G, that denoted by α(G).
A vertex of degree one is called an end-vertex, and the set of end-vertices of G is denoted by End(G). If T is a tree, an end-vertex of a T is usually called a leaf of T and the set of leaves of T is denoted by leaf (T ). A spanning tree is called independence if End(G) is independent in G. For a positive integer k, a tree T is said to be a k-ended tree if |End(T )| ≤ k. We define σ k (G) = min{d By using σ 2 (G), Ore [4] obtain the following famous theorem on Hamiltonian path. Notice that a Hamiltonian path is spanning 2-ended tree. A Hamilton cycle can be interpreted as a spanning 1-ended tree. In particular, K 2 is hamiltonian and is a 1-ended tree.
The following theorem of Las Vergnas Broersma and Tuinstra [1] gives a similar sufficient condition for a graph G to have a spanning k-ended tree.
Let k ≥ 2 be an integer, and let G be a connected graph. If σ 2 (G) ≥ |G|−k +1, then G has a spanning k-ended tree.
Win [10] obtained a sufficient condition related to independent number for k-connected graph that confirms a conjecture of Las Vergnas Broersma and Tuinstra [1] gave a degree sum condition for a spanning k-ended tree. After [1], many researchers have defined other closure concepts for various graph properties. More on k-ended tree and spanning tree can be found in [6,7,8,9]. In this paper, we obtain sufficient conditions for spanning k-ended trees of 3-regular connected graphs and with construction sequence of graphs like G m , we will show this condition is sharp. At the end, we present a conjecture about spanning k-ended trees of 3-regular connected graphs.

Our results
Lemma 2.1. Let T be a tree with n vertices such that ∆(T ) ≤ 3. If |leaf (T )| = k and p be the number of vertices of degree 3 in T , then k = p + 2.
Proof. It is easy by the induction on p.
Lemma 2.2. Let G be a labelled graph and k ≥ 3 be the smallest integer such that G has a spanning tree T with k leaves. Then, no two leaves of T are adjacent in G.
Proof. Put S = {v 1 , v 2 , . . . , v k } be the set of all leaves of T . By contradiction, suppose that v 1 and v 2 are adjacent vertices in G. If T 1 = T + v 1 v 2 , then T 1 contains a unique cycle as Since k ≥ 3 then there exist vertex v s ∈ G such that it is not a vertex of C. Let P be the shortest path of vertex v s to the cycle C such that its intersection with cycle C is c j for 1 ≤ j ≤ ℓ. Now, we omit the edge Proof. For the graph T , we denote the vertices of degree 1 with the set A 1 , the vertices of degree 2 with the set A 2 and the vertices of degree 3 with the set A 3 .
If v ∈ A 3 then the two adjacent edges to v (those were in G but are not in T ), each one connects v to a vertex of A 2 in G, because by Lemma 2.2 it can not connect v to a member of A 1 . So, for each vertex in A 1 there exist two vertices in A 2 such that they are connected to v in G but not in T . Now, we have 2 × |A 1 | ≤ |A 2 |. Let |A 1 | = k, |A 2 | = s and |A 3 | = p. By Lemma 2.1 we have k = p + 2 and since 2|A 1 | ≤ |A 2 | then 2k ≤ s. We have Then k ≤ ⌊ n+2 4 ⌋.

Some concluding remarks
Now we construct the sequence G m of 3-regular graphs, For m = 1, Consider the graph G 1 as Figure 1.
Clearly G 1 has spanning subtree like T that has 3 leaves and G has no spanning subtree with less than 3 leaves. Every part of G 1 like subgraph induced by vertices {1, 2, 3, 4, 5} is called a branch, so G 1 has 3 branch. Let H be a branch of G 1 with vertices {1, 2, 3, 4, 5} and set of edges {12, 15, 23, 24, 34, 35, 45}. Since the edge {01} is a cut edge in G 1 , So T must has a vertex with degree one in H. Also in every other branches of G 1 , T must has a vertex with degree one. so G 1 is 3-ended tree and has no spanning tree with less than 3 leaves. Now, we counteract 3-regular graph www.ejgta.org    Clearly |G 2 | = 16 + 3 × 6 and minimum number leaves in every spanning subtree of G 2 is at least 2 × 3 and obviously G 2 has spanning subtree with 2 × 3 leaves. Let the number of vertices of G m is equal n and the number of branches of G m is equal k, then we have the It obvious for each m ∈ N if the number of vertices of G m is equal n and the number of branches of G m is equal k, then n+2 6 = k, and so G m is n+2 6 -ended tree (such that n+2 6 is the minimum number for that G m is n+2 6 -ended tree).
Conjecture 1. There exists n ∈ N such that each 3-regular graph with at least n vertices has a spanning ⌊ n+2 6 ⌋-ended tree.