Journal of Applied Mathematics

Volume 2014, Article ID 316108, 9 pages

http://dx.doi.org/10.1155/2014/316108

## On Minimum Wiener Polarity Index of Unicyclic Graphs with Prescribed Maximum Degree

Department of Mathematics, Wuyi University, Jiangmen 529020, China

Received 6 April 2014; Accepted 15 July 2014; Published 9 November 2014

Academic Editor: Chongxin Liu

Copyright © 2014 Jianping Ou et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The Wiener polarity index of a connected graph is defined as the number of its pairs of vertices that are at distance three. By introducing some graph transformations, in different way with that of Huang et al., 2013, we determine the minimum Wiener polarity index of unicyclic graphs with any given maximum degree and girth, and characterize extremal graphs. These observations lead to the determination of the minimum Wiener polarity index of unicyclic graphs and the characterization of the extremal graphs.

#### 1. Introduction

All the graphs considered in this paper are connected and simple. The distance between two vertices and of a graph , denoted by or for short, is the length of a shortest path connecting these two vertices. The Wiener polarity index of a graph is defined as the number of its unordered pairs of vertices that has distance three [1]. Namely, the Wiener polarity index of a molecular graph is

In 1998, by using the Wiener polarity index, Lukovits and Linert in [2] demonstrated quantitative structure-property relationships in a series of acyclic and cycle-containing hydrocarbons. Besides, a physicochemical interpretation of was found by Rouvray and King [3]. M. Liu and B. Liu described the relations between Wiener polarity index, Zagreb index, Wiener index, and hyper-Wiener index in [4]; they also determined there the first two smallest Wiener polarity indices among all unicyclic graphs. Recently, Du et al. obtained the smallest and largest Wiener polarity indices together with the corresponding graphs among all trees with any given number of vertices [5]. In [6], Deng and Xiao, determined the maximum Wiener polarity index of trees with pendants and characterized the extremal graphs. In [7], the authors determined the Wiener polarity of fullerenes and hexagonal systems. In [8], Hou et al. determine the maximum wiener polarity index of unicyclic graphs and characterizes the corresponding extremal graphs. This work determines in differrent way with that of [9] the minimum Wiener polarity index of unicyclic graphs with any given maximum degree and girth; these observations lead to the determination of minimum wiener polarity index and the characterization of the corresponding extremal graphs.

Before proceeding, let us introduce some more symbols and terminology. Denote by , or for short, the neighborhood of vertex in graph , the set of vertices adjacent to . Let represent the degree of vertex in graph . If , vertex is called a pendent vertex. The girth of a unicyclic graph is the length of its unique cycle. As usual, let , , and represent a cycle, a star, and a path on vertices, respectively, and the set of unicyclic graphs with vertices and maximum degree . For any unicyclic graph and any vertex in its cycle , the union of vertex and the component of that has a vertex adjacent to induces a tree in , which is called a hanging tree on vertex and is denoted by . The union of all these hanging trees is denoted by . For other symbols and terminology not specified herein, we follow that of [10].

#### 2. Property of Graph Transformations

Let be two integers. After joining and vertices, each by an edge, to the two endpoints of a path ,, respectively, we obtain a new graph and denote it by [11]. For example, the path on two vertices may be regarded as , and the star with may be regarded as . Denote by the set of trees with order ( vertices) and maximum degree .

Let be a cycle with and let be nonnegative integers. We attach at first isolated vertices to vertex for every . And then past a 1-degree vertex of a star that has vertices to vertex to obtain a graph denoted by , where denotes the number of vertices of the obtained graph; or attach isolated vertices directly to vertex to obtain graph ; or past the 1-degree vertex of whose neighbor has degree two to to obtain graph . It is not difficult to see that for every with , for the second case and for the other two cases. Furthermore, either , or , or for some .

Finally, let denote the graph obtained by attaching at first isolated vertices to vertex and then pasting the 1-degree vertex of whose neighbor has degree two (if contains such pendent vertex) to . For clarity, we depict some of these graphs in Figure 1.

Lemma 1 (see [11]). *If and , then , with the equality holding if and only if , where .*

Lemma 2 (see [11]). *; .*

In what follows, for clarity, we denote by the unique cycle of any graph with girth .

Lemma 3. *Let and be one of its hanging tree. If and , then , where , is the maximum degree of , and is obtained from by transforming to with and the neighbor of in having degree two if contains such pendent vertices.*

*Proof. *Let and . If a unicyclic graph has cycle , then its wiener polarity index can be expressed as (refer to [4] for example)

Notice that if is a tree, then its wiener polarity index can be expressed as . According to the differences between and , it is not difficult to deduce (for simplicity the common terms are deleted, this method is also employed in the proof of other leammas) that
Since , it follows that . Combining this observation with , , Lemmas 1 and 2 and the fact that among trees with maximun degree and order graph has maximum wiener polarity index, we deduce that
Again by Lemmas 1 and 2, the inequalities in above formula become equalities if and only if for some positive integer and . So, and the lemma follows.

Lemma 4. *Let and be one of its hanging tree. If is obtained from by transforming to then , with equality holding if and only if with and the neighbor of in having degree two.*

*Proof. *Let and . According to the differences between and and formula (2), it is not difficult to deduce that
Since , and , from Lemmas 1 and 2 it follows that

Notice that implies . If diameter and is not a path, then , and . So,
If diameter but either or with , then , , and at least one of these two inequalities strictly holds. By formula (6), we also have . Hence, the lemma follows.

Lemma 5. *Let be a unicylic graph with girth . If its hanging trees are , and , then , where is obtained from by transforming to a hanging tree on any vertex with and the neighbor of in having degree two.*

*Proof. *Assume without loss of generality that and is a hanging tree on . After tranforming every hanging tree of to a path , we obtain a new graph . By Lemma 4 we have .

Let . If has a hanging tree on vertex , let , where is the neighbor of in . Since , , , and , it follows from formula (2) that when ,
when , since and , we have
By induction on the number of hanging trees on vertex , it follows that if we denote by the graph obtained from by transforming to then when has at least two hanging trees on vertex , where .

In the same way, one can show with ease that if one hanging tree of on vertex has at least three vertices, then the graph has less wiener polarity index than that of , where is the neighbor of in . And so, we leave its proof to the readers.

Now, we consider the case when every hanging tree of on vertices , , is a path of order two. Let , and assume without loss of generality that , where . Denote by the graph obtained from by transforming to a path of order . Then

In conclusion, if is the graph obtained from by transforming to a path of order for every , then .

Finally, if and has a hanging tree on vertex with , then let . Since the total contribution of and its neighbor in to is not less than the total contribution of them to , furthermore, the contribution of other vertex to is not less than the contribution of its corresponding vertex to , it follows that . And so, the lemma follows.

Lemma 6. *Let be a unicylic graph with girth . If its hanging trees are , , , and then , where is obtained from by transforming to a hanging tree on any vertex with and the neighbor of in having degree two.*

*Proof. *Assume without loss of generality that and is a hanging tree on . After tranforming every hanging tree of to a path , we obtain a new graph . By Lemma 4 we have .

Let . If has a hanging tree on vertex , let , where is the neighbor of in . Since , , , and , it follows from formula (2) that when ,

When , since and , we have
By induction on the number of hanging trees on vertex , it follows that if we denote by the graph obtained from by transforming to then when has at least two hanging trees on vertex , where .

In the same way, one can show with ease that if one hanging tree of on vertex has at least three vertices, then the graph has less wiener polarity index than that of , where is the neighbor of in . And so, we leave its proof to the readers.

Now, we consider the case when every hanging tree of on vertices , , is a path of order two. Let be a hanging tree of . If , say , let and , then from formula (2) we deduce that
The second equality holds since and . And so, the lemma follows.

Lemma 7. *Let be a unicylic graph with girth . If one of its hanging tree is and all others are paths, and then , where is obtained from by transforming to a hanging tree on any vertex with and the neighbor of in having degree two, and .*

*Proof. *Assume without loss of generality that is a hanging tree on . Consider at first the case when has at least two hanging trees on vertex . Let . If is a longest hanging tree of on vertex , let , where is the neighbor of in . Since , , and , it follows from formula (2) that when we have
and that when , since , and , we have
The above inequality becomes equality if and only if for every , and , namely, with and .

By induction on the number of hanging trees on vertex , it follows that if we denote by the graph obtained from by transforming to then , where .

Consider secondly the case when is the unique hanging tree of on vertex . By similar reasoning as above, one can show with ease that if one hanging tree of on vertex has at least three vertices, then the graph obtained from by transforming to a path of order has less wiener polarity index than that of . And so, we assume in what follows that every hanging tree of on vertices , , is a path of order two.

Let . Since , it follows that and . Let be the neighbor of in , be the neighbor of in , where and are postulated in the lemma. Noticing that if then and , and that if then , we deduce from formula (2) that
And so, the lemma follows.

Lemma 8. *Let . If every hanging tree is a path then , where is obtained from by transforming to with and .*

*Proof. *Assume without loss of generality that . Let , and let be a hanging tree on that has maximum length. If is another hanging tree on with , let , since it follows that
And so, the lemma follows by induction on the number of hanging trees on vertex that has length of at least three.

#### 3. Minimum Index and Extremal Graphs with Any Given Girth

This section charaterizes the extremal unicyclic graphs with any given girth and minimum wiener polarity index. To this end, we first consider the case when the maximum degree vertices are contained in the cycle and then compare the results with those obtained in the previous section. Since there is only one unicyclic graph that has maximum degree , girth , and vertices, we need to only consider the case when a unicyclic graph has at least vertices.

Lemma 9. *Let be a unnicylic graph with girth and vertices. If and every hanging tree of is a path then , with equality if and only if .*

*Proof. *The lemma is clearly true when since there are only five nonisomorphic unicyclic graphs in this subcase. So, assume in what follows that . Let , , and be the longest hanging tree on .

Let us consider at first the case when . In this case, every hanging tree on is a path on two vertices. To prove the lemma, we may assume in what follows that is an extremal unicyclic graph with minimum wiener polarity index. If the total number of hanging trees on vertices and is at least two, assume without loss of generality that and are two of these hanging trees. Let . When or , say , by formula (2) we have
Since and , it follows that in this subcase, which is a contradiction since is an extremal graph. By induction on the total number of hanging trees on vertices and , we deduce that in this subcase.

When every hanging tree on vertices or is a path on two vertices, which implies that , we deduce that or since . And so . Assume without loss of generality that and . Let . Similar to the reasoning employed in the proof of formula (18), we have
Now, becomes the first subcase. And so, the lemma is ture in the first case.

Secondly, we consider the case when . If has a hanging tree on or , say , let . As in the proof of formula (18), we have
Since , , it follows that
Since , , , and in this case, it follows that , with the equality holding if and only if , and . Therefore, . By formula (2) it is not difficult to show that . And so, the lemma follows.

Lemma 10. *Let be a unnicylic graph with girth and vertices. If and every hanging tree of is a path then , with equality if and only if .*

*Proof. *Since there are only two graphs that satisfy the postulated conditions, namely, , , and , the lemma follows.

Lemma 11. *Let be a unicyclic graph with girth and vertices. If the cycle of contians a maximum-degree vertex, say , then*(1)* when , with equality if and only if , where and when ;*(2)* when , with equality if and only if , where .*

*Proof. *Assume that is an extremal graph with minimum wiener polarity index and properties postulated in the lemma. By Lemma 4 we may assume at first that every hanging tree of is a path. And so, by Lemma 8, every of is either a path or with having maximum degree in for every , where . If and ; we move all the hanging trees on vertex to to obtain graph . Since , by the reasoning as is employed in the first three paragraphs of the proof of Lemma 6, one can deduce from formula (2) that . And so, we may assume in what follows that .

Let be the longest hanging tree of on vertex , . If , let ; when we have
when we have
These contradictions show that , namely, . Similarly, we have .

If , let , as in above paragraph one can show with ease that . This contradiction shows that and is a star with maximum degree .

If then
with the equality holding if and only if the union of hanging trees of on is a path. Now, combining this observation with Lemma 4 we deduce that the first statement of Lemma 11 is true since implies .

If , which implies , then
And so, the second statement of the lemma is also true.

*Let denote the set of all such unicyclic graph that has following properties: girth ; every hanging tree of is a path; is a star with center ; for every edge in the cycle of , either or is a vertex; at most two elements of is not a vertex; if two elements of are not vertices then both are paths of length two.*

*Theorem 12. Let be a unicyclic graph with girth . If then
When the equality holds and , when the equality holds if and only if .*

*Proof. *Since , it follows that the maximum-degree vertex is contained in the cycle of . Now, it is not difficult to see that if the lemma is not true then when , and when . But, from formula (2) we deduce that in the first case and in the second case. And so, the lemma follows.

*Lemma 13. Let be a unicyclic graph with girth and vertices. If and every hanging tree of is a path then
with equality holding if and only if .*

*Proof. *We need to only consider the case when has minimum wiener polarity index. By Lemma 4, it suffices to consider the case when is a path for every . If for some edge in the cycle of , assume without loss of generality that is a path and path is a hanging tree on vertex (it is possible that ), and let ; from formula (2) one can show with no difficulty that . This observation shows that for every edge in the cycle of , either or is a vertex.

Suppose that , , and are nontrivial, where . Let and . From formula (2) we can deduce with ease that . This contradiction shows that at most two elements of are not a vertex. Similarly, one can show that if two elements of are not vertices then both are paths of length two. Now, the wiener polarity indices of graphs in can be obtained by formula (2) as is listed in the lemma. And so, the lemma follows.

*Theorem 14. Let be a unicyclic graph with girth . If then
with equality holding if and only if ; if then
with equlity holding if and only if .*

*Proof. *Assume that is an extremal graph with minimum wiener polarity index and the postulated conditions. From Lemmas 3, 4, 5, and 13 we deduce that . From formula (2) it follows that if with then
Similarly, if then
Since , comparing these obsevations with the indices of obtained in Lemma 13 we complete the proof of this theorem.

*Theorem 15. Let be a unicyclic graph with girth . If then with equality if and only if , where ; if then , with equality if and only if either with and when , or with and .*

*Proof. *Assume that is an extremal graph with minimum wiener polarity index and the postulated conditions. From Lemmas 3, 4, 6, and 11 it follows that , where and . By formula (2) we have
Comparing this observation with the results of Lemma 11, we deduce that if then and , where ; if then and either with and when , or with and . So, the theorem follows.

*Theorem 16. Let be a unicyclic graph with girth . If then , with equality if and only . If then , the equality holds if and only if , *