Tricyclic graphs with maximal revised Szeged index

Lily Chen, Xueliang Li, Mengmeng Liu

Center for Combinatorics, LPMC

Nankai University, Tianjin 300071, China

Email: lily60612@126.com, lxl@nankai.edu.cn, liumm05@163.com

Abstract

The revised Szeged index of a graph is defined as where and are, respectively, the number of vertices of lying closer to vertex than to vertex and the number of vertices of lying closer to vertex than to vertex , and is the number of vertices equidistant to and . In this paper, we give an upper bound of the revised Szeged index for a connected tricyclic graph, and also characterize those graphs that achieve the upper bound.

Keywords: Wiener index, Szeged index, Revised Szeged index, tricyclic graph.

AMS subject classification 2010: 05C12, 05C35, 05C90, 92E10.

## 1 Introduction

All graphs considered in this paper are finite, undirected and simple. We refer the reader to [2] for terminology and notation not given here. Let be a connected graph with vertex set and edge set . For , denotes the distance between and in , we use for short, if there is no ambiguity. The Wiener index of is defined as

This topological index has been extensively studied in the mathematical literature; see, e.g., [6, 8]. Let be an edge of , and define three sets as follows:

Thus, is a partition of the vertices of respect to . The number of vertices of , and are denoted by , and , respectively. A long time known property of the Wiener index is the formula [7, 16]:

which is applicable for trees. Motivated by the above formula, Gutman [5] introduced a graph invariant, named as the Szeged index, as an extension of the Wiener index and defined by

Randić [14] observed that the Szeged index does not take into account the contributions of the vertices at equal distances from the endpoints of an edge, and so he conceived a modified version of the Szeged index which is named as the revised Szeged index. The revised Szeged index of a connected graph is defined as

Some properties and applications of these two topological indices have been reported in [3, 4, 9, 10, 11, 12, 13, 15]. In [1], Aouchiche and Hansen showed that for a connected graph of order and size , an upper bound of the revised Szeged index of is . In [17], Xing and Zhou determined the unicyclic graphs of order with the smallest and the largest revised Szeged indices for , and they also determined the unicyclic graphs of order with the unique cycle of length , with the smallest and the largest revised Szeged indices. In [11], we identified those graphs whose revised Szeged index is maximal among bicyclic graphs. In this paper, we give an upper bound of the revised Szeged index for a connected tricyclic graph, and also characterize those graphs that achieve the upper bound.

###### Theorem 1.1

## 2 Main result

So, we are left to show that for any connected tricyclic graph of order , other than , . Using the fact that and , we have

For convenience, let , where We have

### 2.1 Proof for tricyclic graphs with connectivity 1

###### Lemma 2.1

Let be a connected tricyclic graph of order with at least one pendant edge. Then

Proof. Let be a pendant edge and . Then, for we have

Combining with equality , the result follows.

###### Lemma 2.2

Let be a connected tricyclic graph of order without pendant edges but with a cut vertex. Then, we have

Proof. Suppose that is a cut vertex. Since is a tricyclic graph without pendant edge, is composed of a bicyclic graph and a cycle and . It is obvious that . If is even, for every edge in , we have So

If is odd, for all edges in but the edge such that , we have So

If , then If , then , so .

Combining with equality , this completes the proof.

### 2.2 Proof for 2-connected tricyclic graphs

In this section, , then it must be one of the graphs depicted in Figure 2.2. The letters stand for the lengths of the corresponding paths between vertices of degree greater than 2. For the sake of brevity, we refer to these paths as , respectively. In the statement of the following lemmas, we call these four graphs in Figure 2.2 as and , respectively.

###### Lemma 2.3

Let be a -graph composed of four paths , and , and . Then if and only if is in the middle of an odd path of the four paths , , and .

Proof. Assume that belongs to , the th path connecting and . Then, with respect to and , there are three cases to discuss.

Case . are in different sets. We claim that

where (resp. ) is the distance between (resp. ) and the edge .

To see this, assume that . Then we have vertices more in than in on the path , but on each path , we have vertices more in than in . Hence

Case . are in the same set. We claim that

where is the length of the shortest cycle of that contains .

To see this, assume that . Thus all vertices from the paths are in . Therefore, , while . So

Case . One of is in . We claim that

with equality if and only if two paths of () have length , where is the length of a shortest path of the four paths ().

To see this, assume that , . Then the shortest cycle of that contains is odd. Let be the furthest vertex from such that . Then

From the above, we know that in Case . In Case , if two paths of () have length , which is impossible since is simple. So, if and only if are in different sets and , that is, is in the middle position of an odd path of ().

###### Lemma 2.4

If is a -graph of order . Then, we have

Proof. Without loss of generality, assume that , then Now consider the six edges which are incident with and but do not belong to . Let be one of them, by Lemma 2.3, . Similar thing is true for the other five edges. Hence

Combining with equality , this completes the proof.

###### Lemma 2.5

If is a -graph of order . Then, we have

Proof. Without loss of generality, let . In order to complete the proof, we consider the following four cases.

Case . .

Consider the two edges which belong to , then

Therefrom, we get

Since , If , then , so

If then . Now consider the edge . So

If since . Now consider the edge . So

If then . Now consider the edge . So

If if then , so If or Now consider the edge . So If , then . Now consider the edge . So

Case . .

Subcase . .

Consider two edges and ,

Therefrom, we get . So, .

If or ,

If , and , Now consider the edge . So

Subcase . .

Consider the edge , since . Let be the furthest vertex in such that , be the vertex incident with but not in . If the cycle is even, then , that is . If the cycle is odd, then , that is . So we have .

Then consider the edge , since . Let be the furthest vertex in and such that , be the vertex incident with but not in . If the cycle is even, then , . If the cycle is odd, then , . So we have .

From above, we have

unless . If , now consider the edge belonging to , so .

Subcase . .

Consider the edge , then .

If , then .

If , then , which is impossible since .

Case . .

First, we know that .

Subcase . .

Consider the edges and , then

Since and , .

If , then .

If , now consider the edge . If , then . If , then . If , then . So .

Subcase . .

Consider the edge , since , then . Let be the furthest vertex in such that , be the vertex incident with but not in . If the cycle is even, then , . If the cycle is odd, then , . So we have .

Similarly

where is the edge belonging to .

Since .

If , then .

If , that is , then . Now consider , then . So .

Case . .

Subcase . .

Consider the edge , then . Similarly for the other three edges incident with .

If , then .

If , since . Now consider the edges belonging to , so .

Subcase . .

Consider the edge , . For , we also have

If , then .

If , that is , then . Now consider , then . So .

Combining with equality , this completes the proof.

###### Lemma 2.6

If is a -graph of order . Then, we have

Proof. Without loss of generality, let . In order to complete the proof, we consider the following four cases.

Case . .

Consider the edge ,

Therefrom we get

Since

If , then .

If , that is . Now consider the edge then , so .

Case . .

Subcase . .

Consider the edge , , then

Therefrom, we get

If or or , then .

If , then consider the edge , we have , so .

Subcase . .

It’s similar to the Subcase 2.1.

Subcase . .

Consider the edge , then Since , then , so .

Case . .

Subcase . .

Consider the edge , Similarly where is the edge belonging to .

Since , otherwise is not simple, then

If , then .

If , then . Now consider the edge , so .

If , then . Now consider the edge , so .

If , then and . Now consider the edge , no matter or , we both have , so .

Subcase . .

Now consider the edge , then

Therefrom, we get .

Since , .

If then . So .

If that is , then . Now consider the edge , . So .

If that is , then . Now consider the edge , then we have . Since . Then we have unless . When , we can draw the graph exactly, we also have . So .

Case . .

We may assume that .

Subcase . .

Consider the edge , . For , we also have

Since