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A k-tree of a connected graph G is a spanning tree with maximum degree at most k. The rupture degree for a connected graph G is defined by , where and , respectively, denote the order of the largest component and number of components in . In this paper, we show that for a connected graph G, if for any cut-set , then G has a k-tree.

Throughout this paper, We consider only finite undirected graphs without loops and multiple edges. A graph

nonempty subset S of

A k-tree of a connected graph G is a spanning tree with maximum degree k. Clearly, if

A nonempty set S of independent vertices of G is called a frame of G, if

In [

Theorem A. If

Theorem B. Let

graph, then G has a k tree.

Further, Kyaw in [

Theorem C. Let G be a connected graph and

every

The rupture degree of a graph G is introduced in [

where

In this paper, we consider the rupture degree and existence of k-tree in a connected graph G and thus give a new sufficient condition for a graph to have k tree.

Any undefined terms can be found in the standard references on graph theory, including Bondy and Murty [

Let G be a connected graph and k an integrity with

Theorem 1. Let G be a connected graph but not a tree. If

Let H be an induced subgraph of G and with maximal order among all subgraphs containing k-tree, and let A be a set adjacent to some vertices in G but not in

Claim 1. Let T be a k-tree of H. Then

Proof. Let T be a k-tree of H, which has maximal order among all the induced subgraphs of G having a k-tree. On the contrary, suppose that if there exist some vertex

Let T be a k-tree of H and

Claim 2. If there exist an edge

Proof. Suppose that

Claim 3. Let T be a k-tree of H and M is subset of

Proof. Since A is nonempty and for every

Let T be a k-tree of H with as many k degree vertices as possible and

Claim 4. For m and n, with

Proof. Suppose that

By taking

Claim 5. Let T be a k-tree of H with maximal number of k degree vertices. Then, there is no edge of H joining any components of

Proof. Given our choice of T, M and N as above, we derive a contradiction by assuming that there is an edge yz of H with endpoints y and z joining two components of

Now we are ready for the proof of theorem 2.1.

The proof of theorem 2.1. Since A is nonempty and thus H is an induced proper subgraph of G, we have

On the other hand, since

and,

This is a contradiction. Therefore A is empty and the proof is completed.

We thank the editor and the referee for their comments. Research of Qingning Wang and Xiaoling Wang is funded by the National Science Foundation grant 11561056. This support is greatly appreciated.

Yinkui Li,Qingning Wang,Xiaoling Wang, (2016) The Rupture Degree of Graphs with k-Tree. Open Journal of Discrete Mathematics,06,105-107. doi: 10.4236/ojdm.2016.62011