Closure for Spanning Trees with k-Ended Stems

Let T be a tree. The set of leaves of T is denoted by ( ) Leaf T . The subtree ( ) Leaf T T − of T is called the stem of T . A stem is called a k-ended stem if it has at most k-leaves in it. In this paper, we prove the following theorem. Let G be a connected graph and 2 k ≥ be an integer. Let u and v be a pair of nonadjacent vertices in G . Suppose that ( ) ( ) 1 G G N u N v G k ≥ − −  . Then G has a spanning tree with k-ended stem if and only if G uv + has a spanning tree with k-ended stem. Moreover, the condition on ( ) ( ) G G N u N v  is sharp.


Introduction
We consider simple graphs, which have neither loops nor multiple edges.For a graph G , let ( ) ( ) E G denote the set of vertices and the set of edges of G , respectively.We write G for the order of G (i.e., ( ) .In particular, ( ) ( ) An edge joining two vertices x and y is denoted by xy or yx .
Let T be a tree.A vertex of T with degree one is often called a leaf of T , and the set of leaves of T is denoted by ( ) Leaf T .The subtree ( ) of T is called the stem of T and denoted by ( ) Stem T .A spanning tree with specified stem was first considered in [1].
A tree having at most k leaves is called a k-ended tree.So a tree whose stem has at most k leaves in it is called a tree with k-ended stem.Notice that a tree with 2-ended stem is nothing but a caterpillar, whose stem is a path.We consider a spanning tree with k-ended stem.
We make a remark about spanning trees with k-ended stem from the point of view of dominating set.A subgraph H of a graph G is said to dominate G if every vertex of G not contained in H has a neighbor in H . Namely, H dominates G if every vertex . So a graph G has a spanning tree with k-ended stem if and only if G has a k-ended tree that dominates G .There are many researches on dominating cycles and dominating paths (for example, see [2] and [3] with stronger definition of domination).Thus the concept of spanning trees with k-ended stem can be also considered as a generalization of dominating paths.
For an integer 2 k ≥ and a graph G , ( ) denotes the minimum degree sum of k independent ver- tices of G .The following theorem gives a sufficient condition using for a graph to have a spanning tree with k-ended stem.
Theorem 1 (Tsugaki and Zhang [4]) Let G be a connected graph and 2 k ≥ be an integer.If ( ) then G has a spanning tree with k-ended stem.
Another result on spanning trees with k-ended stem is the following.Theorem 2 (Kano and Yan [5]) Let G be a connected graph and 2 k ≥ be an integer.If G satisfies one of the following conditions, then G has a spanning tree with k-ended stem. (1) ( ) operation is useful in the study of the existence of hamiltonian cycles, hamiltonian paths and other spanning subgraphs in graphs.It was first introduced by Bondy and Chvátal.
Theorem 3 (Bondy and Chval [6]) Let G be a graph and let u and v be two nonadjacent vertices of G . ( . Then G has a hamiltonian cycle if and only if G uv + has a hamiltonian cycle. ( Then G has a hamiltonian path if and only if G uv + has a hamiltonian path.
After [6], many researchers have defined other closure concepts for various graph properties.The following theorem gives a result on closure for spanning k-ended tree.
Theorem 4 (Broersma and Tuinstra [7]) Let 2 ≥ k be an integer, and let G be a graph.Let u and v be a pair of nonadjacent vertices of G with ( ) ( ) . Then G has a spanning k-ended tree if and only if G uv + has a spanning k -ended tree.Another type of closure theorem on spanning k-ended tree can be found in Fujisawa, Saito and Schiermeyer [8].The interested reader is referred to the survey [9] on closure concepts.
In this paper, we prove the following theorem.Theorem 5 Let G be a connected graph and 2 ≥ k be an integer.Let u and v be a pair of nonadjacent vertices of G such that ( ) ( ) Then G has a spanning tree with k-ended stem if and only if uv G + has a spanning tree with k-ended stem.
Before proving Theorem 5, we show that the condition (1) in Theorem 5 is sharp.We construct a graph G as follows.Let 2 ≥ k and 1 ≥ m be integers, and let m K be a complete graph of order m , which is a subgraph of G .Let , , , , , , , to u and i v by edges.Then the resulting graph is G (see Figure 1).It is immediate that G uv + has a spanning tree with k-ended stem, where all the vertices of m K and v are leaves of the spanning tree, and . However G has no spanning tree with k-ended stem.Therefore the condition (1) is sharp.Moreover, in Figure 1, Since m be an arbitrary integer, the degree sum of u and v can be arbitrarily great.This implies that we can not find a condition similar to Theorem 4 for spanning tree with k-ended stem.
Some results on spanning k-ended trees and other spanning trees with given properties can be found in [10], and many current results on spanning trees can be found in [11].

Proof of Theorem 5
In this section we prove Theorem 5. Without mentioning, we often use the fact that ( ) ) T .Proof of Theorem 5. Since the necessity of the theorem is trivial, we only prove the sufficiency.Assume, to the contrary, that uv G + has a spanning tree with k-ended stem but G does not have a spanning tree with k- ended stem.Let us denote * G G uv = + .Choose a spanning tree T with k-ended stem of * G so that (T1) ( )

( )
Leaf Stem T is as small as possible, and Stem T is as small as possible subject to (T1).
It is obvious that the edge uv is contained in T since otherwise T is a spanning tree with k-ended stem of G .Let ( ) , there exists a vertex ( ) , then X is an independent set of * G , and so is of G .Assume that 2 l = and the two leaves 1 x and 2 x of ( ) It is easy to see that 1 x and 2 x are not adjacent in

( )
Stem T since otherwise we can obtain a spanning tree with 2-ended stem of G from T .If u and v are both contained in ( )

T uv x x
− + is a spanning tree with 2-ended stem of G , which contradicts the assumption.Hence we may assume that u is a vertex of ( ) Stem T and v is a leaf of T by symmetry of u and v and by ( ) Stem T , then T uv wv − + is a spanning tree with 2-ended stem of G , a con- tradiction.If w is a leaf of T , let * w be the vertex of ( ) Stem T adjacent to w in T , and let z be a vertex of ( ) T uv w z x x wv − − + + is a spanning tree with 2-ended stem of G , a contradiction.Therefore, is an independent set of * G .Hence (1) of Claim 1 follows.Suppose that there exists a vertex s x , 1 s l ≤ ≤ , such that every leaf y adjacent to s x in T satisfies . Then for every leaf y adjacent to s x in T , remove the edge s yx from T and add an edge yz of * G , where ( ) T is a spanning tree of * G and satisfies ( ) , which contradicts the condition (T2).Therefore, (2) of Claim 1 holds.Hereafter, we take the vertices ( ) T uv T T − =  , where ( ) ( ) The claim holds when 2 k = , and so we assume that 3 ≤ − , then T uv ab − + is a spanning tree with k-ended stem of G , which contradicts the assumption.Next we consider the case where 1 l k = − .If either v is a leaf of T or the degree of v in

( )
Stem T is 1 or greater than 2, then T uv ab − + is a spanning tree with k-ended stem of G , which contradicts the assumption.Thus the degree of v in

( )
Stem T is 2. By symmetry of u and v , we may assume that the degree of u in

( )
Stem T is also 2, and it is clear that uv is an edge of ( ) + − is a spanning tree with k-ended stem of G , a contradiction.Thus no vertex of ( ) T contains at least two vertices of X .Then the path in ( ) x and u contains a vertex of degree at least 3 in ( ) Stem T .Let e be an edge of the path which is incident with a vertex with degree at least 3 in

( )
Stem T ′ .Then we can derive a contradiction by the same argument as in the above first paragraph.Therefore if a vertex x of X is adjacent to u in G , then 1 T contains exactly one vertex of X .Note that x may be adjacent to u but not to v .Since T contains at least two vertices of X , which means no vertex of X is adjacent to v in G by the same argument on 1 T and x given above.By the above fact and Claim (1), we obtain x be the leaves of ( ) Stem T where ( ) x V T ∈ and ( ) x V T ∈ .By the same argument as in the above paragraph, we that neither u and 2 x nor v and 1 x are adjacent in G .We shall show that u and x are not adjacent in G .
By the above fact and Claim 1, we obtain For a vertex v of G , the degree of v in G is denoted by ) ( deg v G , and the set of vertices adjacent to v is called the neighborhood of v and denoted by ) (v N G

Figure 1 .
Figure 1.A graph G having no spanning tree with k-ended stem.
tree with k-ended stem of * G , and u has degree at least 3 in are not adjacent in G .Assume that u and 1 x are adjacent in G .Let * a be the vertex adjacent to a in T if tree with 3-ended stem of G , a contradiction.Similarly, v and 2 1x