All graphs considered in this paper are finite, loopless, and without multiple edges. For a graph G, we refer to and as the vertex set and the edge set, respectively. The cardinality of is called the order of G, denoted by. The (open) neighborhood of a vertex x is the set of vertices adjacent to x in G, and the close neighborhood is. A vertex x is said to be a leaf if. A vertex v of G is a support vertex if it is adjacent to a leaf in G. Two distinct vertices u and v are called duplicated if. Note that u and v are duplicated vertices in a tree, and then they are both leaves. The n-path is the path of order. For a subset, the induced subgraph induced by A is the graph with vertex set A and the edge set, the deletion of A from G is the graph by removing all vertices in A and all edges incident to these vertices and the complement of A is the set. For notation and terminology in graphs we follow [1] in general.

A set is an independent set of G if no two vertices of I are adjacent in G. The independence number of G is the maximum cardinality among all independent sets of G. If I is an independent set of G with cardinality, we call I an -set of G. If I is an -set of G containing all leaves of G, we call I an -set of G.

The independence problem is to find an -set in G. The problem is known to be NP-hard in many special classes of graphs. Over the past few years, several studies have been made on independence (see [2] -[6] ). For

any tree T of order, it is easy to see that. In addition, if there are duplicated leaves in a tree, then these duplicated leaves are all lying in every maximum independent set. In this paper, we will show that if T is a tree of order without duplicated leaves, then. Moreover, we constructively

characterize the extremal trees T of order, which are without duplicated leaves, achieving these upper bounds.

In this section, we will show a sharp upper bound on the independence number of a tree T without duplicated leaves.

Lemma 1 If H is an induced subgraph of G, then.

Proof. If S is an -set of H, then S is an independent set of G. It follows that.

Lemma 2 ([4] ) If T is a tree of order, then.

Lemma 3 ([5] ) If T is a tree of order, then there exists an -set of T.

Lemma 4 For an integer,.

Proof. It is straightforward to check that for. Let. Since P_n is a tree of order, by Lemma 2, we have that. Suppose that there exists an independent set I of P_n with, then there exists i, , such that and. This is a contradiction, therefore we obtain that.

Theorem 1 If T is a tree of order without duplicated leaves, then.

Proof. We prove it by induction on. By Lemma 4 and T is a tree without duplicated leaves, it’s true for all. For all we assume that the assertion is true for all. Suppose that T is a tree of order without duplicated leaves and x is a leaf lying on a longest path of T. Let. Since T has no duplicated leaves, this implies that, say. Let, then T' is a tree of order. For the case in which T' has no duplicated leaves, by induction hypothesis, we have that

. Since an -set of T', together with, form an -set of T. Therefore we obtain that. For the other case in which T' has duplicated leaves z and, then is a tree of order without duplicated leaves. By induction hypothesis, we have that. Since an -set of, together with, form an -set of T. Therefore, we obtain that. Hence we conclude that

.

Note that the result in Theorem 1 is sharp and some such T are illustrated below.

Let be the class of all trees T of order without duplicated leaves such that.

We will constructively characterize these extremal trees. Let and, respectively, denote the collections of all leaves and all support vertices of T. First, we define four operations on a tree T of order as follows, where I is an -set of T.

Operation O1. Join a vertex of to a vertex of such that, where (mod 3).

Operation O2. Join a vertex of to a vertex of such that, where (mod 3).

Operation O3. Join a vertex u of to a leaf of (say) such that, where (mod 3).

Operation O4. Join a vertex of T to a leaf v₃ of P₃ (say) such that.

Lemma 5 Suppose that for.If I is an -set of T, then

Proof. It’s true for all. So we assume that. Since I is an -set of T, this implies that. By Theorem 1, we have that

Hence we obtain that. Let. Note that and

for every i. In addition, , these imply that. Thus we obtain that. It follows that

This completes the proof.

Lemma 6 Let be a tree of order with an -set I. Suppose that T' is obtained from T by Operation O1, then is a tree of order and is an -set of T'.

Proof. Suppose that is a tree of order (mod 3) with an -set I, by Lemma 5, then. Let T' be the tree obtained from T by Operation O1. Since, this implies that u is not a support vertex of T and T' is a tree of order without duplicated leaves. On the other hand, is an independent

set of with such that, where (mod 3). Hence. In conclusion, is a tree of order with an -set I_O1.

Lemma 7 Let be a tree of order with an -set I such that. If is obtained from T by Operation O2, then is a tree of order and is an -set of.

Proof. Note that such a tree T exists, as, for instance, the tree in Figure 1 is as desired. If is a tree of order (mod 3) with an -set I such that. Let T' be the tree obtained from T by Operation O2. Since u is not a support vertex of T, this implies that T' is a tree of order without duplicated leaves. And is an independent set of T' with such that

, where (mod 3). Hence

. In conclusion, is a tree of order with an -set I_O2.

Lemma 8 Let be a tree of order with an -set I. If T' is obtained from T by Operation O3, then is a tree of order and I_O3 is an -set of T'.

Proof. Note that T' is a tree of order n + 2 without duplicated leaves. And I_O3 is an independent set of T' with

such that, where (mod 3). Hence. In conclusion, is a tree of order with an - set I_O3.

Lemma 9 Let be a tree of order with an -set I. If T' is obtained from T by Operation O4, then is a tree of order and I_O4 is an -set of T'.

Proof. Note that T' is a tree of order without duplicated leaves. And I_O4 is an independent set of T' with

such that. Hence

. In conclusion, is a tree of order with an -set I_O4.

Let be the class of all trees obtained from or by a finite sequence of Operations O1-O4. Suppose that, we will show that.

Theorem 2 T is in if and only if T is in.

Proof. If T is in, by Lemmas 6, 7, 8 and 9, then T is in. Now, we want to show the converse by contradiction. Suppose to the contrary that there exists a tree and such that is as small as possible. We can see that. Let be a longest path of T. Then and is a tree of order. We consider two cases.

Case 1. T' has no duplicated leaves.

For an -set I of T, is an independent set of T', this implies that. By

Theorem 1, we have that. Then and (mod 3). This follows that, where

(mod 3), by hypothesis,. Note that T can be obtained from by Operation O3, this implies that, which is a contradiction.

Case 2. T' has duplicated leaves z and z'.

Let. Then is a tree of order. Since z' is a leaf of T, this implies that z and z' are in every -set of T. For an -set I of T, is an independent set of, thus . By Theorem 1, we have that

. Then. This fol-

lows that, where, by hypothesis,. Note that T can be obtained from by Operation O4, this implies that, which is a contradiction.

By Cases 1 and 2, we conclude that T is in, then T is in.

Now, we obtain the main theorem in this paper.

Theorem 3 Suppose that T is a tree of order without duplicated leaves, then. Furthermore, the equality holds if and only if.

Min-JenJou,Jenq-JongLin, (2015) Independence Numbers in Trees. Open Journal of Discrete Mathematics,05,27-31. doi: 10.4236/ojdm.2015.53003