On the Hosoya Indices of Bicyclic Graphs with Small Diameter

Wu, Tingzeng; Yu, Yong

doi:https://doi.org/10.1155/2021/5555700

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Volume 2021 | Article ID 5555700 | https://doi.org/10.1155/2021/5555700

On the Hosoya Indices of Bicyclic Graphs with Small Diameter

Tingzeng Wu¹and Yong Yu¹

Academic Editor: Muhammad Faisal Nadeem

Received27 Feb 2021

Revised07 Jun 2021

Accepted17 Aug 2021

Published03 Sept 2021

Abstract

Let be a graph. The Hosoya index of , denoted by , is defined as the total number of its matchings. The computation of is NP-Complete. Wagner and Gutman pointed out that it is difficult to obtain results of the maximum Hosoya index among tree-like graphs with given diameter. In this paper, we focus on the problem, and a sharp bound of Hosoya indices of all bicyclic graphs with diameter of 3 is determined.

1. Introduction

Hosoya index is an important topological index introduced by Hosoya [1]. It was found that Hosoya index is related to a variety of physicochemical properties of alkanes (= saturated hydrocarbons). In particular, the boiling points of alkanes are well correlated with Hosoya index. Another series of researches revealed the applicability of Hosoya index in the theory of conjugated -electron systems [2, 3]. Jerrum [4] showed that the computing complexity of Hosoya index is NP-Complete. The Hosoya index got much attention by many researchers in the past decades. They have been interested in identifying the extremal value of Hosoya index for various classes of graphs, such as trees [5–7], unicyclic graphs [8–12], bicyclic graphs [13], and -graphs [14, 15]. Wagner and Gutman [16] gave an exhaustive survey for Hosoya index, and they pointed out some open problems (also see [17]) as follows:•It seems difficult to obtain results of the maximum Hosoya index among trees with a given number of leaves or given diameter. However, partial results are available, so the problem might not be totally intractable, and results in this direction would definitely be interesting.•If the aforementioned questions can be answered for trees, then it is also natural to consider the analogous questions for tree-like graphs.

According to the open problems, Liu et al. [17] discussed the problem in which unicyclic graph with diameter of 3 or 4 has the maximum Hosoya index. In this paper, we focus on similar problems to the above. That is, which bicyclic graph with diameter of 3 has the maximum Hosoya index? We give an answer of the problem as follows.

Theorem 1. and ; each of the following holds:

The rest of this paper is organized as follows. In Section 2, we shall present some definitions and lemmas. In Section 3, we will prove Theorem 1. Furthermore, some upper bounds for Hosoya index of some special classes of bicyclic graphs with diameter of 3 are also determined.

2. Preliminaries

In this paper, we only consider finite and simple graphs. Let be a graph with vertices and edges. The neighborhood of vertex in a graph , denoted by , is the set of vertices adjacent to . The degree of , denoted by , is the number of neighbors of in . The distance of two vertices is the length of a shortest path from to , denoted by . We will use to represent the graph after deleting the vertex . The diameter of is max .

Let be the set of all bicyclic graphs with vertices and diameter of 3. It is easy to verify that the structure of graph must be isomorphic to , where . The resulting graph can be seen in Figure 1.

Let be the number of -matchings of . It is convenient to denote and for . The Hosoya index of , denoted by , is defined as the sum of all the numbers of its matchings; namely,

Lemma 1 (see [16]). Let be a graph and let be a vertex of graph . Then(i)(ii), where is a component of

Let be a graph obtained by joining the centers of two stars and , denoted by . By Lemma 1, we obtain the following result.

Lemma 2. (i)(ii)

3. The Proof of Theorem 1

In order to prove Theorem 1, we first give some lemmas.

Lemma 3. Let be a graph with vertices. Then

Proof. Consider that vertex of degree 1 is adjacent to in . By Lemmas 1 and 2, we have . In the following, we use the method of Lagrange multipliers to find the sharp bound of Hosoya index of . First, we make auxiliary function as follows: , where , , and at most one of them is 1. Taking the partial derivatives of , and in , we can obtain the following equations:Solving the equations as above, we obtain that , , and . Because is a unique stable point, must be a unique extreme point. Since , , and are integers, we consider three cases. Case 1: assume that . We know that , , and ; , , and ; or , , and . Thus, . It is easy to verify that when , , and ; , , and ; or , , and . This implies that and the equality holds iff . Case 2: suppose that . We have , , and . So . To simplify the calculation, we know that when , , and . This means that and the equality holds iff . Case 3: assume that . Then , , and ; or , , and . Thus, . It is easy to check that when , , and ; or , , and . This indicates that and the equality holds if and only if .

Lemma 4. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, we have . By the method of Lagrange multipliers, we make an auxiliary function , where , , and . Taking the partial derivatives of , , and in , we get the following equations:Solving the equations as above, we obtain that and . Because is a unique stable point, must be the unique extreme point. Since and are integers, in the following we consider two cases: Case 1: suppose that is even. Then and . Thus, . To simplify calculation, we know that when and . This implies that and the equality holds iff . Case 2: assume that is odd. We obtain and , or and . So, . It is easy to check that when and , or and . This means that and the equality holds iff .

Lemma 5. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, we get that . According to the method of Lagrange multipliers, we make an auxiliary function , where , , and . Taking the partial derivatives of , and in , we obtain the following equations:Solving the equations as above, we obtain that and . Because is a unique stable point, must be the unique extreme point. Since and are integers, we only consider two cases. Case 1: assume that is even. We obtain and , or and . Thus, . It is easy to check that when and , or and . This implies that and the equality holds if and only if . Case 2: consider that is odd. We have and . So, . It is easy to verify that when and . This means that and the equality holds if and only if .

Lemma 6. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, we get . According to the method of Lagrange multipliers, we make an auxiliary function , where , , and . Taking the partial derivatives of , and in , we can obtain the following equations:Solving the equations as above, we obtain that and . Because is a unique stable point, must be the unique extreme point. Since and are integers, we only consider two cases: Case 1: consider that is even. We have and . Thus . To simplify calculation, we get when and . This implies that and the equality holds iff . Case 2: assume that is odd. Then and , or and . So . It is easy to check that when and , or and . This implies that and the equality holds if and only if .

Lemma 7. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, we get . By the method of Lagrange multipliers, we make an auxiliary function , where , , and . Taking the partial derivatives of , and in , we can obtain the following equations:Solving the equations as above, we obtain that and . Because is a unique stable point, must be the unique extreme point. Since and are integers, in the following we only consider two cases. Case 1: consider that is even. We have and . Thus . To simplify calculation, we get when and . This implies that and the equality holds iff . Case 2: assume that is odd. Then and , or and . So . It is easy to check that when and , or and . This implies that and the equality holds if and only if .

Lemma 8. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, direct computing yields . By the method of Lagrange multipliers, we make auxiliary function , where , , and at most one of them is 1. Taking the partial derivatives of , , , and in , we can obtain the following equations:Solving the equations as above, we get . Since is an integer, we know that or . By and , we can obtain that , , and . Similarly, by and , we also can obtain that , , and . Thus, we discuss two cases as follows: Case 1: , , and . Because , , and is a unique root of (14), , , and must be the unique extreme point. Since , , and are integers, we consider three subcases: Subcase 1: assume that . Then , , and , or , , and . Thus . It is easy to verify that when , , and , or , , and . This implies that and the equality holds if and only if . Subcase 2: suppose that . Then , , and . Thus . It is easy to check that when , , and . This implies that and the equality holds iff . Subcase 3: consider that . Then , , and . Thus . To simplify calculation, we know that when , , and . This implies that and the equality holds iff . Case 2: , , and . Similar to the proof of Case 1, because , , and is a unique root of (14), , , and must be the unique extreme point. Since , , and are integers, we also consider the three following subcases: Subcase 1: set . Then , , and , or , , and . Thus . It is easy to verify that when , , and , or , , and . This implies that and the equality holds if and only if . Subcase 2: suppose that . Then , , and . Thus . It is easy to check that when , , and . This implies that and the equality holds iff . Subcase 3: consider that . Then , , and . Thus . To simplify calculation, we know that when , , and . This implies that and the equality holds iff .

Lemma 9. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, direct computing yields . By the method of Lagrange multipliers, we make auxiliary function , where , , and . Taking the partial derivatives of , , and in , we can obtain the following equations:Solving the equations as above, we get . Since is an integer, we know that or . By and , we can obtain that and . Similarly, by and , we also can obtain that and . So, we discuss two cases as follows: Case 1: and . Because and is a unique root of (16), and must be the unique extreme point. Since and are integers, we consider two subcases: Subcase 1: assume that is even. Then and ; thus . It is easy to verify that when and . This implies that and the equality holds if and only if . Subcase 2: suppose that is odd. We know that and . Thus . It is easy to verify that when and . This implies that and the equality holds iff . Case 2: and . Similar to the proof of Case 1, because and is a unique root of (2), and must be the unique extreme point. Since and are integers, we also consider two subcases. Subcase 1: consider that is even. Then we get and ; thus . It is easy to verify that when and . This implies that and the equality holds if and only if . Subcase 2: assume that is odd. We know that and . Thus . It is easy to verify that when and . This implies that and the equality holds iff .

Lemma 10. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, we get . By the method of Lagrange multipliers, we make an auxiliary function , where , , and at most one of them is 1. Taking the partial derivatives of , and in , we can obtain the following equations:Solving the equations as above, we get . Since is an integer, we know that or . By and , we can obtain that and . Similarly, by and , we also can obtain that and . So, we discuss two cases as follows: Case 1: and . Because and is a unique root of (18), and must be the unique extreme point. Since and are integers, we consider two subcases. Subcase 1: consider that is even. We have and . Thus . To simplify calculation, we get when and . This implies that and the equality holds iff . Subcase 2: assume that is odd. Then and . So . It is easy to check that when and . This implies that and the equality holds if and only if . Case 2: and . Similar to the proof of Case 1, because and is a unique root of (18), and must be the unique extreme point. Since and are integers, we consider two subcases. Subcase 1: consider that is even. We have and . Thus . To simplify calculation, we get when and . This implies that and the equality holds iff . Subcase 2: assume that is odd. Then and . So . It is easy to check that when and . This implies that and the equality holds if and only if .

Lemma 11. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, we get . By the method of Lagrange multipliers, we make an auxiliary function , where , , and at most one of them is 1. Taking the partial derivatives of , and in , we can obtain the following equations:Solving the equations as above, we obtain that and . Because is a unique stable point, must be the unique extreme point. Since and are integers, we consider the two following cases. Case 1: consider that is even. We have and , or and . Thus . To simplify calculation, we get when and , or and . This implies that and the equality holds iff . Case 2: assume that is odd. We know that and . Thus . It is easy to check that when and . This implies that and the equality holds if and only if .

Lemma 12. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, direct computing yields . By the method of Lagrange multipliers, we make auxiliary function , where , , and at most one of them is 1. Taking the partial derivatives of , , , and in , we can obtain the following equations:Solving the equations as above, we get . Since is an integer, we know that or . By and , we can obtain that , , and . Similarly, by and , we also can obtain that , , and . Thus, we discuss two cases as follows: Case 1: , , and . Because , , and is a unique root of (22), , , and must be the unique extreme point. Since , , and are integers, we consider three subcases as follows: Subcase 1: assume that . Then , , and . Thus . It is easy to verify that when , , and . This implies that and the equality holds if and only if . Subcase 2: suppose that . Then , , and , or , , and . Thus . It is easy to check that when , , and , or , , and . This implies that and the equality holds iff . Subcase 3: consider that . Then , , and . Thus . To simplify calculation, we know that when , , and . This implies that and the equality holds iff . Case 2: , , and . Similar to the proof of Case 1, because , , and is a unique root of (22), , , and must be the unique extreme point. Since , , and are integers, after taking the integer, we also consider three subcases as follows: Subcase 1: assume that . Then , , and . Thus . It is easy to verify that when , , and . This implies that and the equality holds if and only if . Subcase 2: suppose that . Then , , and , or , , and . Thus . It is easy to check that when , , and , or , , and . This implies that and the equality holds iff . Subcase 3: consider that . Then , , and . Thus . To simplify calculation, we know that when , , and . This implies that and the equality holds iff .

Lemma 13. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, we get . By the method of Lagrange multipliers, we make an auxiliary function , where , , and . Taking the partial derivatives of , and in , we can obtain the following equations:Solving the equations as above, we obtain that and . Because is a unique stable point, must be the unique extreme point. Since and are integers, in the following we consider two cases. Case 1: consider that is even. We have and . Thus . To simplify calculation, we get when and . This implies that and the equality holds iff . Case 2: assume that is odd. We know that and , or and . Thus . It is easy to check that when and , or and . This implies that and the equality holds if and only if .

Lemma 14. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, direct computing yields . Using the method of Lagrange multipliers, we make auxiliary function , where , , and . Taking the partial derivatives of , , and in , we can obtain the following equations:Solving the equations as above, we get . Since is an integer, we know that or . By and , we can obtain that and . Similarly, by and , we also can obtain that and . So, we discuss two cases as follows: Case 1: and . Because and is a unique root of (26), and must be the unique extreme point. Since and are integers, we consider two subcases. Subcase 1: assume that is even. Then and . Thus . It is easy to verify that when and . This implies that and the equality holds if and only if . Subcase 2: suppose that is odd. We know that and . Thus . It is easy to check that when and . This implies that and the equality holds iff . Case 2: and . Similar to the proof of Case 1, because and is a unique root of (26), and must be the unique extreme point. Since and are integers, we consider two subcases. Subcase 1: assume that is even. Then and . Thus . It is easy to verify that when and . This implies that and the equality holds if and only if . Subcase 2: suppose that is odd. We know that and . Thus . It is easy to check that when and . This implies that and the equality holds iff .

Lemma 15. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, we get that . According to the method of Lagrange multipliers, we make an auxiliary function , where , , and . Taking the partial derivatives of , and in , we obtain the following equations:Solving the equations as above, we obtain that and . Because is a unique stable point, must be the unique extreme point. Since and are integers, we only consider two cases. Case 1: assume that is even. We obtain and . Thus . It is easy to check that when and . This implies that and the equality holds if and only if . Case 2: consider that is odd. We have and , or and . So, . It is easy to verify that when and , or and . This means that and the equality holds if and only if .

Lemma 16. Let be a graph with vertices. Then

Proof. Similar to the proof of Lemma 3, by Lemmas 1 and 2, we obtain that . According to the method of Lagrange multipliers, we make an auxiliary function , where , , and . Taking the partial derivatives of , and in , we can obtain the following equations:Solving the equations as above, we obtain that and . Because is a unique stable point, must be the unique extreme point. Since and are integers, we only consider two cases. Case 1: consider that is even. We have and , or and . Thus . To simplify calculation, we get when and , or and . This implies that and the equality holds iff . Case 2: assume that is odd. Then and . So . It is easy to check that when and . This implies that and the equality holds if and only if .

Proof of Theorem 1. By Lemmas 3–16, we know the maximum Hosoya index in . Employing Mathematica 12.0 to compute the difference of the maximum Hosoya indices of and , it yields directly the result in Theorem 1.

4. Concluding Remark

In this paper, we characterize the sharp upper bound of Hosoya indices of all graphs in . Furthermore, we also determine the upper bound of every type of bicyclic graphs. There exists an interesting problem:

Problem 1. Which graph has the minimum Hosoya index?

We attempted to find a solution to the problem; however, the problem is very difficult for us.

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this study.

Acknowledgments

This work was supported by the National Natural Science Foundation of China (no. 11761056), Natural Science Foundation of Qinghai Province (no. 2020-ZJ-920), and the Scientific Research Innovation Team in Qinghai Nationalities University.

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Copyright © 2021 Tingzeng Wu and Yong Yu. 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.

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