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# On the atom-bond connectivity index and radius of connected graphs

*Journal of Inequalities and Applications*
**volume 2015**, Article number: 188 (2015)

## Abstract

Let *G* be a connected graph with edge set \(E (G)\). The atom-bond connectivity index (ABC index for short) is defined as \(ABC (G) = \sum_{uv \in E (G)} \sqrt{ \frac{d_{G} (u)+d_{G} (v)-2}{d_{G} (u) d_{G} (v)} }\), where \(d_{G} (u)\) denotes the degree of vertex *u* in *G*. The research of ABC index of graphs is active these years, and it has found a lot of applications in a variety of fields. In this paper, we will focus on the relationship between ABC index and radius of connected graphs. In particular, we determine the upper and lower bounds of the difference between ABC index and radius of connected graphs.

## Introduction

Let *G* be a connected graph with vertex set \(V (G)\) and edge set \(E (G)\). For \(v \in V (G)\), let \(d_{G} (v)\) denote the degree of vertex *v* in *G*.

In chemical graph theory, we usually use a graph to represent a molecule graph. One of the most useful tools to study and predict various properties of molecule graphs is the topological indices, which are used directly as simple numerical descriptors in quantitative structure property relationships (QSPR) and quantitative structure activity relationships (QSAR) [1].

In 1998, Estrada *et al.* [2] proposed a topological index based on the degrees of vertices of graphs, which is called the atom-bond connectivity index (ABC index for short). The ABC index of a graph *G* is defined as [2]

Initially, in the light of the close relationship between the ABC index and the heats of formation of alkanes, the ABC index became an efficient tool to model the thermodynamic properties of organic chemical compounds [2]. In 2008, Estrada [3] elaborated a novel quantum-theory-like justification for the ABC index and explained the stability of branched alkanes.

Since then, the research on the ABC index of graphs becomes rather active. Xing *et al.* [4] presented an upper bound for the ABC index of connected graphs with fixed number of vertices, number of edges and maximum degree. Das [5] established the lower and upper bounds on ABC index of graphs and trees and showed that these bounds are best possible. Fath-Tabar *et al.* [6] considered the effect on the ABC index of graphs under some graph operations. More results on ABC index of graphs can be found in [7–9], especially for trees [10–12].

In a connected graph *G*, for \(v \in V(G)\), the eccentricity of *v* in *G* is defined as the maximum distance from *v* to another vertex of *G*. The radius of *G*, denoted by \(r (G)\), is the minimum eccentricity among the vertices in *G*.

In this paper, we will study the relationship between the ABC index and radius of connected graphs. In particular, we determine the upper and lower bounds of \(ABC (G) - r (G)\) for connected graphs *G*.

## Preliminaries

For an edge subset *M* of the graph *G*, let \(G-M\) denote the graph obtained from *G* by deleting the edges in *M*, and for an edge subset \(M^{*}\) of the complement of *G*, let \(G+M^{*}\) denote the graph obtained from *G* by adding the edges in \(M^{*}\). In particular, if \(M = \{ u v \}\), then we write \(G - u v\) for \(G - \{ u v \}\), and if \(M^{*} = \{ u v \}\), then we write \(G + u v\) for \(G + \{ u v \}\).

The following result reveals that the deletion of an edge from a graph would decrease the ABC index.

### Lemma 1

([7], Theorem 2)

*Let*
*G*
*be a connected graph on*
\(n \ge3\)
*vertices*. *For*
\(uv \in E(G)\), *we have*

As an immediate application of Lemma 1, we know that the complete graph is the unique graph with maximum ABC index among connected graphs. Furthermore, we can deduce that the upper bound of \(ABC (G) - r (G)\) would be uniquely attained by the complete graph.

### Proposition 1

*Let*
*G*
*be a connected graph on*
\(n \ge2\)
*vertices*. *Then*

*with equality if and only if*
*G*
*is the complete graph on*
*n*
*vertices*.

In the following, we will consider the lower bound of \(ABC (G) - r (G)\) for connected graphs *G* and show that this lower bound would be uniquely attained by the path.

## Lemmas

First we give several additional lemmas which will be used in our proof later.

### Lemma 2

([7], Lemma 2)

*For integers*
\(x \ge2\)
*and*
\(y \ge1\), *we have*

A vertex of degree 1 is said to be a pendent vertex, while a vertex of degree at least 2 is said to be a non-pendent vertex.

### Lemma 3

*Let*
*H*
*be a nontrivial tree*, *where*
\(u,v \in V(H)\) (*possibly*
\(u=v\)). *Let*
*G*
*be the tree obtained from*
*H*
*by attaching a pendent vertex*
\(u_{1}\)
*to*
*u*, *and attaching a pendent vertex*
\(v_{1}\)
*to*
*v*; *see Figure *
1. *In particular*, *if*
\(u=v\), *then two pendent vertices*
\(u_{1}\), \(v_{1}\)
*are both attached to vertex*
\(v = u\)
*in*
*H*, *see Figure *
2. *Consider*
\(G_{1} = G - v v_{1} + u_{1} v_{1}\); *see Figures *
1
*and*
2. *If there are at most two non*-*pendent neighbors of*
*v*
*in*
*G*, *then*

### Proof

Denote by *a* the number of non-pendent neighbors of *v* in *G*. According to the hypothesis, \(a = 0,1,2\).

Clearly, \(d_{G} (v) \ge2\). For \(vx \in E (H)\), if \(d_{G} (x) = 1\), then

while if \(d_{G} (x) \ge2\), then by Lemma 2, we have

So it follows that

In the following we will split the proof into two cases.

*Case* 1. \(u \ne v\).

In this case, \(a = 1,2\) (if \(a=0\), then *G* is a star, which is impossible). Moreover, note that \(d_{G} (v) \ge a + 1 \ge2\). Now by (1), we have

On one hand, notice that \(\sqrt{ \frac{d_{G} (u) - 1}{d_{G} (u)} } = \sqrt { 1 - \frac{1}{d_{G} (u)} }\) is increasing on \(d_{G} (u) \ge2\). On the other hand, \(\sqrt{ \frac{d_{G} (v) - 1}{d_{G} (v)} } = \sqrt{ 1 - \frac {1}{d_{G} (v)} }\) and \(\sqrt{ \frac{1}{d_{G} (v)} } -\sqrt{ \frac{1}{d_{G} (v) - 1} }\) are both increasing on \(d_{G} (v) \ge a + 1\). So we have

*i.e.*, \(ABC (G) - ABC (G_{1}) > -0.5\).

*Case* 2. \(u = v\).

In this case, \(a = 0,1,2\). Moreover, note that \(d_{G} (v) \ge a + 2 \ge2\).

Similar to the arguments in Case 1, we have

*i.e.*, \(ABC (G) - ABC (G_{1}) > -0.5\).

Combining Cases 1 and 2, the result follows. □

The following lemma on the radius of a tree is clear.

### Lemma 4

*Let*
*G*
*be a tree*. *If the diameter of*
*G*
*is*
*d*, *then*
\(r(G) = \lfloor \frac{d+1}{2} \rfloor\).

In the following two lemmas, we will show that the path would attain the minimum value \(ABC(G) - r (G)\) among trees.

### Lemma 5

*Let*
*G*
*be a tree with a diametrical path*
*P*. *If*
\(ABC(G) - r (G)\)
*is minimum among trees*, *then there is at most one vertex outside*
*P*
*in*
*G*.

### Proof

Suppose to the contrary that there are at least two vertices outside *P* in *G*.

Assume that \(P = v_{0} v_{1} \cdots v_{d}\). Denote by \(T_{i}\) the component of \(G - E (P)\) containing \(v_{i}\), where \(0 \le i \le d\). Let \(x_{i}\) be a vertex in \(T_{i}\) such that

where \(0 \le i \le d\). Clearly, \(x_{i}\) is a pendent vertex if \(|V (T_{i})| \ge2\), and denote by \(x_{i}^{*}\) the unique neighbor of \(x_{i}\) in *G*.

Since there are some vertices outside *P* in *G*, thus there exists some index, say *k*, such that \(|V (T_{k})| \ge2\), where \(1 \le k \le d-1\). By the choice of \(x_{k}\), it is easily seen that there are at most two non-pendent neighbors of \(x_{k}^{*}\) in *G*.

Consider \(G_{1} = G - x_{k}^{*} x_{k} + v_{0} x_{k}\). By Lemma 3, we have

Clearly, \({x_{k}} \cup P = x_{k} v_{0} v_{1} \cdots v_{d}\) is a diametrical path of \(G_{1}\), thus the diameter of \(G_{1}\) is \(d+1\).

Observe that there are at least two vertices outside *P* in *G*, which implies that there are some vertices outside (diametrical path) \({x_{k}} \cup P\) in \(G_{1}\). Similar to the transformation from *G* to \(G_{1}\), we can construct another tree \(G_{2}\) based on \(G_{1}\), such that

and the diameter of \(G_{2}\) is \(d+2\).

Now by (2) and (3), it follows that

On the other hand, by Lemma 4, we have

Finally, by (4) and (5), we get

*i.e.*, \(ABC (G) - r (G) > ABC (G_{2}) - r (G_{2})\), which is a contradiction to the minimality of \(ABC (G) - r (G)\).

Then the result follows easily. □

Let \(P_{n}\) be the path on *n* vertices.

### Lemma 6

*Let*
*G*
*be a tree on*
\(n \ge4\)
*vertices with a diametrical path*
*P*. *If there is at most one vertex outside*
*P*
*in*
*G*, *then*

*with equality if and only if*
\(G \cong P_{n}\).

### Proof

From the hypothesis that there are at most one vertex outside *P* in *G*, we know that either *G* is the path \(P_{n}\), or *G* is a tree obtained from \(P_{n-1} = v_{1} v_{2} \cdots v_{n-1}\) by attaching a pendent vertex to \(v_{i}\), where \(2 \le i \le n - 2\).

If *G* is a tree obtained from \(P_{n-1} = v_{1} v_{2} \cdots v_{n-1}\) by attaching a pendent vertex to \(v_{2}\) or \(v_{n-2}\), then \(ABC (G) \geq \frac {n-3}{\sqrt{2}} + 2 \sqrt{\frac{2}{3}}\) with equality when \(n\geq 5\) and \(r (G) = \lfloor\frac{n - 1}{2} \rfloor\), and thus

If *G* is a tree obtained from \(P_{n-1} = v_{1} v_{2} \cdots v_{n-1}\) by attaching a pendent vertex to \(v_{i}\), where \(3 \le i \le n - 3\), then \(ABC (G)= \frac{n-2}{\sqrt{2}} + \sqrt{\frac{2}{3}}\) and \(r (G) = \lfloor\frac{n - 1}{2} \rfloor\), and thus

If \(G \cong P_{n}\), then \(ABC (P_{n}) = \frac{n-1}{\sqrt{2}}\) and \(r (P_{n}) = \lfloor\frac{n}{2} \rfloor\), and thus

It is easily verified that

Then the result follows easily. □

## The main result

Now we give the main result of this paper.

### Theorem 1

*Let*
*G*
*be a connected graph on*
\(n \ge2\)
*vertices*. *Then*

*with equality if and only if*
\(G \cong P_{n}\).

### Proof

Let *G* be a connected graph on \(n \ge2\) vertices.

If \(n = 2\), then \(G \cong P_{2}\), and thus the result holds trivially. If \(n = 3\), then either *G* is the path \(P_{3}\) or the triangle, and thus the result follows from a simple calculation.

Suppose in the following that \(n \ge4\).

*Case* 1. *G* is a tree.

Suppose that *G* is a tree with minimum value \(ABC (G) - r (G)\).

Let *P* be a diametrical path of *G*. By Lemma 5, there is at most one vertex outside *P* in *G*. Furthermore, by Lemma 6, we have

with equality if and only if \(G \cong P_{n}\).

*Case* 2. *G* is not a tree.

Let *T* be a spanning tree of *G*. By Lemma 1, we have \(ABC (G) > ABC (T)\). On the other hand, since the removal of edges potentially increases the eccentricities of some vertices, thus \(r (G) \le r (T)\) follows clearly. So we have

Now together with the arguments in Case 1, we get

Combining Cases 1 and 2, the result follows. □

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## Acknowledgements

This work was supported by the National Natural Science Foundation of China (Grant No. 11426199), Guangdong Provincial Natural Science Foundation of China (Grant No. 2014A030310277), and the Department of Education of Guangdong Province Natural Science Foundation of China (Grant No. 2014KQNCX224).

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### Cite this article

Du, Z. On the atom-bond connectivity index and radius of connected graphs.
*J Inequal Appl* **2015, **188 (2015). https://doi.org/10.1186/s13660-015-0714-4

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DOI: https://doi.org/10.1186/s13660-015-0714-4

### MSC

- 05C12
- 05C35
- 92E10

### Keywords

- atom-bond connectivity index
- degree of vertices
- radius