# Maximum Reciprocal Degree Resistance Distance Index of Bicyclic Graphs

Authors: Gaixiang Cai; Xing-Xing Li; Guidong Yu
Journal: Discrete Dynamics in Nature and Society
 (2021)

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## Abstract

The reciprocal degree resistance distance index of a connected graphG is defined as RDRG=∑u,v⊆VGdGu+dGv/rGu,v, where rGu,v is the resistance distance between vertices u and v in G. Let ℬn denote the set of bicyclic graphs without common edges and with n vertices. We study the graph with the maximum reciprocal degree resistance distance index among all graphs in ℬn and characterize the corresponding extremal graph.


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## Body

## 1. Introduction

LetG=VG,EG be a simple connected graph of order n with vertex set VG=v1,v2,…,vn and edge set EG. For any vertex, u∈VG, denoted by dGu the degree of vertex u. The distance between vertices of u and v is defined as the length of the shortest path between them and denoted by dGu,v. The resistance distance between vertices of u and v, denoted by rGu,v, is the effective resistance between the two nodes of the electronic network obtained so that its nodes correspond to the vertices of G and each edge of G is replaced by a resistor of unit resistance.Topological indices are numbers related to molecular structures, used as quantitative relationships between chemical structures and properties. Some of them are based on the distance of graph [1], the vertex degree [2], or the resistance distance. The first topological index was published by Wiener [3]. From then, many topological indices are defined, such as the Harary index [4], Kirchhoff index [5, 6], the first and the second Zagreb indices [7, 8], and degree distance [9, 10].Recently, Alizadeh and Iranmanesh [11] proposed a new topological index, the reciprocal degree distance, which is defined as(1)RDDG=12∑u,v∈VGdGu+dGvdGu,v.Its lower and upper bounds were obtained in [12], and its chemical applications and mathematical properties have been well studied in [11, 13].Especially, analogous to the reciprocal degree distance indexRDDG, Cai et al. [14] introduced a new graph invariant based on both the vertex degree and the resistance, named the reciprocal degree resistance distance index, shown as follows:(2)RDRG=∑u,v⊆VGdGu+dGvrGu,v.A bicyclic (or unicyclic) graph is a connected graph of ordern with n+1 (or n) edges. Let ℬn be the set of bicyclic graphs without common edges of order n. Tree, unicyclic graph, and bicyclic graph are three kinds of graphs with a simple structure; the reciprocal degree resistance distance index of tree is the same as the reciprocal degree distance, Cai et al. [14] determined the graph with the maximum reciprocal degree resistance distance index among all unicyclic graph and characterized the corresponding extremal graph. In this paper, we determine the graph with the maximum reciprocal degree resistance distance index among all graphs in ℬn and characterize the corresponding extremal graph.This paper is organized as follows: in the second part, we give three types of transformation, edge-lifting transformation, cycle-lifting transformation, and cycle-shrinking transformation, to keep the reciprocal degree resistance distance index increasing. In the third part, we give the maximum reciprocal degree resistance distance index among all graphs inℬn.


## 2. Preliminaries

LetCg=v1v2,…,vgv1 be the cycle graph with girth g≥3. For any two vertices vi,vj∈VCg with i<j, by Ohm’s law, one has(3)rCgvi,vj=j−ig+i−jg.Lemma 1.
(See [6]). Let x be a cut vertex of a connected graph G and s and t be two vertices occurring in different components which arise upon the deletion of x.Then,(4)rGs,t=rGs,x+rGx,t.


### 2.1. Edge-Lifting Transformation

LetG′ be obtained from G by an edge-lifting transformation at e=u0,v0, see Figure 1. See [14] for specific description.Figure 1
The edge-lifting transformation.Lemma 2.
(See [14]). If G′ can be obtained from G by an edge-lifting transformation, see Figure 1, then RDRG<RDRG′.


### 2.2. Cycle-Lifting Transformation

LetGpi be the star graph K1,si1≤i≤p with order si+1 and Gqj be the star graph K1,tj2≤j≤q with order tj+1. Let C be a bicyclic graph with exactly two cycles, which are Cp=v1v2,…,vpv1 and Cq=v1u2,…,uqv1vi≠uj,1≤i≤p,2≤j≤q, and G is obtained from C by identifying vi with the center of Gpi for all i1≤i≤p and uj with the center of Gqj for all j2≤j≤q. Deleting all edges in Gpi2≤i≤p, joining v1 to all pendent vertices of Gpi2≤i≤p, we obtained a new graph, denoted by G′, see Figure 2. This operation is called a cycle-lifting transformation of G with respect to Cp.Figure 2
The cycle-lifting transformation.Lemma 3.
IfG′ can be obtained from G by a cycle-lifting transformation, see Figure 2, and ∑i=2psi>0, then RDRG<RDRG′.Proof. LetG and G′ be as shown in Figure 2. By the definition of RDRG,(5)RDRG=∑i=1p∑x,y∈VGpi\vidGx+dGyrGx,y+∑i=1p∑x∈VGpi\vidGvi+dGxrGvi,x+∑i=1p∑x∈VGpi\viy∈VCp\vidGx+dGyrGx,y+∑i,j=1i≠jp∑x∈VGpi\viy∈VGpj\vjdGx+dGyrGx,y+∑i=1p∑x∈VGpi\viy∈VCq\v1dGx+dGyrGx,y+∑i=1x∈VGpi\vip∑j=2y∈VGqj\ujqdGx+dGyrGx,y+∑x,y∈VCpdGx+dGyrGx,y+∑x∈VCpy∈VCq\v1dGx+dGyrGx,y+∑j=2q∑x∈VCpy∈VGpj\ujdGx+dGyrGx,y+∑x,y∈VCq\v1dGx+dGyrGx,y+∑j=2q∑x∈VCq\v1y∈VGpj\ujdGx+dGyrGx,y+∑j=2q∑x,y∈VGpj\ujdGx+dGyrGx,y+∑i,j=2i≠jq∑x∈VGqi\uiy∈VGqj\ujdGx+dGyrGx,y.(6)RDRG′=∑i=1p∑x,y∈VGpi\v1dG′x+dG′yrG′x,y+∑i=1p∑x∈VGpi\v1dG′x+dG′yrG′x,y+∑i=1p∑x∈VGpi\v1y∈VCp\v1dG′x+dG′yrG′x,y+∑i,j=1i≠jp∑x∈VGpi\v1y∈VGpj\v1dG′x+dG′yrG′x,y+∑i=1p∑x∈VGpi\v1y∈VCq\v1dG′x+dG′yrG′x,y+∑i=1x∈VGpi\v1p∑j=2y∈VGqj\ujqdG′x+dG′yrG′x,y+∑x,y∈VCpdG′x+dG′yrG′x,y+∑x∈VCpy∈VCq\v1dG′x+dG′yrG′x,y+∑j=2q∑x∈VCpy∈VGpj\ujdG′x+dG′yrG′x,y+∑x,y∈VCq\v1dG′x+dG′yrG′x,y+∑j=2q∑x∈VCq\v1y∈VGpj\ujdG′x+dG′yrG′x,y+∑j=2q∑x,y∈VGpj\ujdG′x+dG′yrG′x,y+∑i,j=2i≠jq∑x∈VGqi\uiy∈VGqj\ujdG′x+dG′yrG′x,y.(1)
Noting thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for x,y∈VGpi\vi of G and x,y∈VGpi\v1 of G′, we have(7)∑i=1p∑x,y∈VGpi\v1dG′x+dG′yrG′x,y−∑i=1p∑x,y∈VGpi\vidGx+dGyrGx,y=0.(2)
Noting thatrGvi,x=rG′v1,x=1, dGv1+dGx=s1+5 for x∈VGp1\v1 and dGvi+dGx=si+3 for x∈VGpi\vi2≤i≤p of G, and dG′v1+dG′x=s1+⋯+sp+5 for x∈VGpi\v1 of G′, we have(8)∑i=1p∑x∈VGpi\v1dG′v1+dG′xrG′v1,y−∑i=1p∑x∈VGpi\vidGvi+dGxrGvi,y=∑i,j=1i≠jpsi⋅sj+2∑i=2psi.(3)
Noting thatdGx+dGv1=s1+5, rGx,v1=1+rGvi,v1 for x∈VGpi\vi2≤i≤p of G, dGx+dGy=sj+3, rGx,y=1+rGvi,vj, for x∈VGpi\vi1≤i≤p,y=vj∈VCp\v1,vi of G, and dG′x+dG′y=3, rG′x,y=1+rG′v1,y for x∈VGpi\v1,y∈VCp\v1 of G′, we have(9)∑i=1p∑x∈VGpi\v1y∈VCp\v1dG′x+dG′yrG′x,y−∑i=1p∑x∈VGpi\viy∈VCp\vidGx+dGyrGx,y=−∑i,j=1i≠jpsi⋅sj1+rGvi,vj+2∑i=2psi1+rGvi,v1.(4)
Noting thatdGx+dGy=dG′x+dG′y, rGx,y=rGx,vi+rGvi,vj+rGvj,y>rG′x,v1+rG′v1,y=rG′x,y for x∈VGpi\vi,y∈VGpj\vj1≤i,j≤p,i≠j of G and x∈VGpi\v1,y∈VGpj\v11≤i,j≤p,i≠j of G′, similarly for x∈VGpi\vi1≤i≤p,y∈VCq\v1 and x∈VGpi\v11≤i≤p,y∈VCq\v1 of G and G′, respectively, and for x∈VGpi\vi1≤i≤p,y∈VGqj\uj2≤j≤q and x∈VGpi\v11≤i≤p,y∈VGqj\uj2≤j≤q of G and G′, respectively, we have the same conclusions; thus,(10)∑i,j=1i≠jp∑x∈VGpi\v1y∈VGpj\v1dG′x+dG′yrG′x,y−∑i,j=1i≠jp∑x∈VGpi\viy∈VGpj\vjdGx+dGyrGx,y+∑i=1p∑x∈VGpi\v1y∈VCq\v1dG′x+dG′yrG′x,y−∑i=1p∑x∈VGpi\viy∈VCq\v1dGx+dGyrGx,y+∑i=1x∈VGpi\v1p∑j=2y∈VGqj\ujqdG′x+dG′yrG′x,y−∑i=1x∈VGpi\vip∑j=2y∈VGqj\ujqdGx+dGyrGx,y>0presents.(5)
We note thatdGv1=s1+4, dGvi=si+22≤i≤p for any vi∈VCp of G and dG′v1=s1+⋯+sp+4, dG′vi=22≤i≤p for any vi∈VCp of G′. Let rGvi,vj=rG′vi,vj=rt, when j−i=t; then, rt=rp−t. Thus, we have(11)∑x,y∈VCpdGx+dGyrGx,y=∑i=2pdGv1+dGvirGv1,vi+∑i=3pdGv2+dGvirGv2,vi+⋯+dGvp−1+dGvprGvp−1,vp,and then, we have(12)∑i=2pdGv1+dGvirGv1,vi=s1+s2+6r1+s1+s3+6r2+⋯+s1+sp+6rp−1,∑i=3pdGv2+dGvirGv2,vi=s2+s3+4r1+⋯+s2+sp+4rp−2,…dGvp−1+dGvprGvp−1,vp=sp−1+sp+4r1.Thus,(13)∑x,y∈VCpdGx+dGyrGx,y=s1+⋯+sp−1+s2+⋯+sp+4p−2+6r1+s1+⋯+sp−2+s3+⋯+sp+4p−3+6r2+⋯+s1+sp+6rp−1.Similarly,(14)∑x,y∈VCpdG′x+dG′yrG′x,y=s1+⋯+sp+4p−2+6r1+s1+⋯+sp+4p−3+6r2+⋯+s1+⋯+sp+6rp−1.Then,(15)∑x,y∈VCpdG′x+dG′yrG′x,y−∑x,y∈VCpdGx+dGyrGx,y=s2+⋯+sp−1r1+s3+⋯+sp−2r2+⋯+−s2+⋯+sp−1rp−1=0.(6)
It is noted thatdGv1=s1+4, dGvi=si+22≤i≤p, dGuj=tj+22≤j≤q for vi∈VCp and uj∈VCq\v1 of G. Similarly, It is noted that dG′v1=∑i=1psi+4, dG′vi=22≤i≤p, dG′uj=tj+22≤i≤q, for vi∈VCp and uj∈VCq\v1 of G′. Let rGvi,v1=rpt, when i−1=t2≤i≤p,1≤t≤p−1. Let rGv1,uj=rqs, when j−1=s2≤j≤q,1≤s≤q−1. Then, rGx,y=rGvi,v1+rGv1,ui=rpt+rqs=rG′x,y for x∈VCp and y∈VCq\v1. Thus, we have(16)∑x∈VCpy∈VCq\v1dGx+dGyrGx,y=∑i=2qdGv1+dGuirGv1,ui+∑i=2qdGv2+dGuirGv2,ui+⋯+∑i=2qdGvp+dGuirGvp,ui,and then, we have(17)∑i=2qdGv1+dGuirGv1,ui=s1+t2+6rq1+s1+t3+6rq2+⋯+s1+tq+6rqq−1,∑i=2qdGv2+dGuirGv2,ui=s2+t2+4rp1+rq1+s2+t3+4rp1+rq2+⋯+s2+tq+4rp1+rqq−1,…∑i=2qdGvp+dGuirGvp,ui=sp+t2+4rpp−1+rq1+sp+t3+4rpp−1+rq2+⋯+sp+tq+4rpp−1+rqq−1.Similarly,(18)∑x∈VCpy∈VCq\v1dG′x+dG′yrG′x,y=∑i=2qdG′v1+dG′uirG′v1,ui+∑i=2qdG′v2+dG′uirG′v2,ui+⋯+∑i=2qdG′vp+dG′uirG′vp,ui,and then, we have(19)∑i=2qdG′v1+dG′uirG′v1,ui=s1+⋯+sp+t2+6rq1+s1+⋯+sp+t3+6rq2+⋯+s1+⋯+sp+tq+6rqq−1,∑i=2qdG′v2+dG′uirG′v2,ui=t2+4rp1+rq1+t3+4rp1+rq2+⋯+tq+4rp1+rqq−1,…∑i=2qdG′vp+dG′uirG′vp,ui=t2+4rpp−1+rq1+t3+4rpp−1+rq2+⋯+tq+4rpp−1+rqq−1.Thus,(20)∑x∈VCpy∈VCq\v1dG′x+dG′yrG′x,y=−∑x∈VCpy∈VCq\v1dGx+dGyrGx,y=∑i=2p∑j=1q−1sirqj−∑i=2p∑j=1q−1sirpi−1+rqj>0.(7)
Forx∈VCp,y∈VGpj\uj of G and G′, respectively, simulating the abovementioned calculation method of (6), we get the similar conclusion,(21)∑j=2q∑x∈VCpy∈VGpj\ujdG′x+dG′yrG′x,y−∑j=2q∑x∈VCpy∈VGpj\ujdGx+dGyrGx,y=∑i=2p∑j=2qsi⋅tjrqj−1+1−∑i=2p∑j=2qsi⋅tjrpi−1+rqj−1+1>0.(8)
Obviously,(22)∑x,y∈VCq\v1dGx+dGyrGx,y+∑j=2q∑x∈VCq\v1y∈VGpj\ujdGx+dGyrGx,y+∑j=2q∑x,y∈VGpj\ujdGx+dGyrGx,y+∑i,j=2i≠jq∑x∈VGqi\uiy∈VGqj\ujdGx+dGyrGx,y,=∑x,y∈VCq\v1dG′x+dG′yrG′x,y+∑j=2q∑x∈VCq\v1y∈VGpj\ujdG′x+dG′yrG′x,y+∑j=2q∑x,y∈VGpj\ujdG′x+dG′yrG′x,y+∑i,j=2i≠jq∑x∈VGqi\uiy∈VGqj\ujdG′x+dG′yrG′x,y.Thus, combing (1)–(8), we get RDRG′−RDRG>0, and the results are as follows.Corollary 1.
IfG′ can be obtained from G by a cycle-lifting transformation, see Figure 3, and ∑j=2qtj>0, then RDRG<RDRG′.Figure 3
The cycle-lifting transformation.


### 2.3. Cycle-Shrinking Transformation

LetG=Snp,q, which is obtained by coalescing Cp and Cq at the vertex v1 and attaching n−p−q+1 pendent edges to the vertex v1. We note that p,q≥3,p+q≤n+1. Deleting the edges v2v3,…,vp−1vp and adding the edges v3v1,v4v1,…,vp−1v1,v2vp, we get G′=Sn3,q This operation is called cycle-shrinking transformation. It is denoted by (ww′, resp), the set of pendent vertices of (GG′, resp). Let w=k≥0; w′ can be partitioned into two subsets: one has k vertices, which is naturally corresponding to w (also denoted by w), and another has p−3 vertices, denoted by w˜.Lemma 4.
LetG=Snp,qp≥4,p+q≤n+1; if G′ can be obtained from G by a cycle-shrinking transformation, see Figure 4, then RDRG<RDRG′.Figure 4
The cycle-shrinking transformation.Proof. LetG and G′ be as shown in Figure 4. By the definition of RDRG,(23)RDRG=∑x,y∈wdGx+dGyrGx,y+∑x∈wdGv1+dGxrGv1,y+∑x∈wy∈v2,vpdGx+dGyrGx,y+∑x∈wy∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x∈wy∈VCq\v1dGx+dGyrGx,y+∑x∈v2,vpdGv1+dGxrGv1,y+∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,y+∑x∈VCq\v1dGv1+dGxrGv1,x+dGv2+dGvprGv2,vp+∑x∈v2,vpy∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x∈v2,vpy∈VCq\v1dGx+dGyrGx,y+∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x∈VCp\v1,v2,vpy∈VCq\v1dGx+dGyrGx,y+∑x,y∈VCq\v1dGx+dGyrGx,y,RDRG′=∑x,y∈wdG′x+dG′yrG′x,y+∑x∈wdG′v1+dG′xrG′v1,x+∑x∈wy∈v2,vpdG′x+dG′yrG′x,y+∑x∈wy∈w˜dG′x+dG′yrG′x,y+∑x∈wy∈VCq\v1dG′x+dG′yrG′x,y+∑x∈v2,vpdG′v1+dG′xrG′v1,x+∑x∈w˜dG′v1+dG′xrG′v1,x+∑x∈VCq\v1dG′v1+dG′xrG′v1,x+dG′v2+dG′vprG′v2,vp+∑x∈v2,vpy∈w˜dG′x+dG′yrG′x,y+∑x∈v2,vpy∈VCq\v1dG′x+dG′yrG′x,y+∑x,y∈w˜dG′x+dG′yrG′x,y+∑x∈w˜y∈VCq\v1dG′x+dG′yrG′x,y+∑x,y∈VCq\v1dG′x+dG′yrG′x,y.(1)
Noting thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for any x,y∈w of G and G′, respectively, we have(24)∑x,y∈wdGx+dGyrGx,y=∑x,y∈wdG′x+dG′yrG′x,y.(2)
Noting thatdGv1+dGx=k+5, dG′v1+dG′x=k+p+2, rGv1,x=rG′v1,x=1 for x∈w in G and G′, respectively, we have(25)∑x∈wdG′v1+dG′xrG′v1,x−∑x∈wdGv1+dGxrGv1,x=kk+p+2−kk+5=kp−3.(3)
Noting thatdGx+dGy=dG′x+dG′y=3, rGx,y=p−1/p+1≥rG′x,y=2/3+1 for any x∈w,y∈v2,vp in G and x∈w,y∈v2,vp in G′, we have(26)∑x∈wy∈v2,vpdG′x+dG′yrG′x,y−∑x∈wy∈v2,vpdGx+dGyrGx,y>0.(4)
Noting thatdGx+dGy=3,rGx,y=rGvi,v1+1>23≤i≤p−1 for x∈w,y∈VCp\v1,v2,vp and dG′x+dG′y=2,rG′x,y=2 for x∈w,y∈w˜, we have(27)∑x∈wy∈w˜dG′x+dG′yrG′x,y−∑x∈wy∈VCp\v1,v2,vpdGx+dGyrGx,y≥kp−3−3kp−32=−kp−32.Thus, combing (2) and (4), we have(28)∑x∈wdG′v1+dG′xrG′v1,x−∑x∈wdGv1+dGxrGv1,x+∑x∈wy∈w˜dG′x+dG′yrG′x,y−∑x∈wy∈VCp\v1,v2,vpdGx+dGyrGx,y≥kp−32≥0.(5)
Noting thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for any x∈w,y∈VCq\v1 of G and G′, respectively, we have(29)∑x∈wy∈VCq\v1dGx+dGyrGx,y=∑x∈wy∈VCq\v1dG′x+dG′yrG′x,y.(6)
Noting thatdGv1+dGx=k+6,rGv1,x=p−1/p and dG′v1+dG′x=k+p+3,rG′v1,x=2/3 for x∈v2,vp of G and G′, respectively, we have(30)∑x∈v2,vpdG′v1+dG′xrG′v1,x−∑x∈v2,vpdGv1+dGxrGv1,x=3k+p+3−2pk+6p−1=k+3p+3p−3p−1.(7)
Noting thatdGv1+dGx=k+6,rGv1,x=rGv1,vi>13≤i≤p−1 for x∈VCp\v1,v2,vp of G and dG′v1+dG′x=k+p+2,rG′v1,x=1 for x∈w˜ of G′, we have(31)∑x∈w˜dG′v1+dG′xrG′v1,x−∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,x≥p−3k+p+2−p−3k+6=p−3p−4.(8)
Noting thatdGv1+dGx=k+6,dG′v1+dG′x=k+p+3,rGv1,x=rG′v1,x for x∈VCq\v1 of G and G′, respectively, we have(32)∑x∈VCq\v1dG′v1+dG′xrG′v1,x−∑x∈VCq\v1dGv1+dGxrGv1,x=p−3∑j=2q1rGv1,uj.(9)
Noting thatdGv2+dGvp=4, rGv2,vp=2p−2/p and dG′v2+dG′vp=4, rG′v2,vp=2/3, we have(33)dG′v2+dG′vprG′v2,vp−dGv2+dGvprGv2,vp=6−2pp−2=4p−12p−2>0.(10)
Noting thatdGx+dGy=4 for x∈v2,vp, y∈VCp\v1,v2,vp of G and dG′x+dG′y=3, rG′x,y=rG′v2,v1+1=2/3+1=5/3 for x∈v2,vp,y∈w˜ of G′, by the symmetry between v2 and vp, we have(34)∑x∈v2,vpy∈w˜dG′x+dG′yrG′x,y−∑x∈v2,vpy∈VCp\v1,v2,vpdGx+dGyrGx,y=2∑i=3p−13rG′v2,v1+1−2∑i=3p−14rGv2,vi,=18p−35−8pp−1+2∑i=4p−14rGv2,vi.SincerGv2,vi≥rGv2,v4=2p−2/p4≤i≤p−1, if p≥12, then 2p−2/p≥5/3, and(35)2∑i=4p−14rGv2,vi≤24p−45.Thus,(36)∑x∈v2,vpy∈w˜dG′x+dG′yrG′x,y−∑x∈v2,vpy∈VCp\v1,v2,vpdGx+dGyrGx,y≥18p−35−8pp−1+24p−45.(11)
Noting thatdGx+dGy=4 for x∈v2,vp, y∈VCq\v1 of G, dG′x+dG′y=4 for x∈v2,vp,y∈VCq\v1 of G′, and rGx,y=rGv2,v1+rGv1,ui=p−1/p+rGv1,ui≥rG′x,y=rG′v2,v1+rG′v1,ui=2/3+rG′v1,ui2≤i≤q, by symmetry between v2 and vp, we have(37)∑x∈v2,vpy∈VCq\v1dG′x+dG′yrG′x,y−∑x∈v2,vpy∈VCq\v1dGx+dGyrGx,y≥0.(12)
Noting thatdGx+dGy=4, rGx,y=rGvi,vj3≤i<j≤p−1 for x,y∈VCp\v1,v2,vp of G and dG′x+dG′y=2, rG′x,y=2 for x,y∈w˜ of G′, we have(38)∑x,y∈w˜dG′x+dG′yrG′x,y−∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y=p−3p−42−∑i,j=3i<jp−14rGvi,vj,∑i,j=3i<jp−14rGvi,vj=∑i=4pdGv3+dGvirGv3,vi+∑i=5pdGv4+dGvirGv4,vi+⋯+dGvp−2+dGvp−1rGvp−2,vp−1.Whenj−i=ti<j, we denote rGvi,vj=rt, and then, rt=rp−t=tp−t/p; thus,(39)∑i=4pdGv3+dGvirGv3,vi=4r1+4r2+⋯+4rp−4,∑i=5pdGv4+dGvirGv4,vi=4r1+⋯+4rp−5,…dGvp−2+dGvp−1rGvp−2,vp−1=4r1.Thus,(40)∑i,j=3i<jp−14rGvi,vj=4p−4r1+4p−5r2+⋯+4rp−4.Sincer1=p−1/p, rt≥r2=2p−2/p2≤t≤p−4. If p≥12, then 2p−2/p≥5/3, and(41)∑i,j=3i<jp−14rGvi,vj≤4pp−4p−1+6p−4p−55.Thus,(42)∑x,y∈w˜dG′x+dG′yrG′x,y−∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y≥p−3p−42−4pp−4p−1+6p−4p−55.Whenp≥12, combing (6), (7), (10), and (12), we have(43)∑x∈v2,vpdG′v1+dG′xrG′v1,x−∑x∈v2,vpdGv1+dGxrGv1,x+∑x∈w˜dG′v1+dG′xrG′v1,x−∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,x+∑x∈v2,vpy∈w˜dG′x+dG′yrG′x,y−∑x∈v2,vpy∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x,y∈w˜dG′x+dG′yrG′x,y−∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y≥k+3p+3p−3p−1+3p−3p−42+18p−35−4pp−2p−1−6p−1p−45=3p3−22p2+53p−114+k10p−3010p−1.(i)
Letfp=3p3−22p2+53p−114,p∈12,+∞; this function is strictly increasing in this interval; thus, fp≥f12=2538>0.Thus,(44)3p3−22p2+33p−54+k10p−3010p−1>0.(13)
Noting thatdGx+dGy=4, rGx,y=rGvi,v1+rGv1,uj≥1+rGv1,uj3≤i≤p−1,2≤j≤q for x∈VCp\v1,v2,vp,y∈VCq\v1 of G and dG′x+dG′y=3, rG′x,y=1+rGv1,uj2≤j≤q for x,y∈w˜ of G′, we have(45)∑x∈w˜y∈VCq\v1dG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈VCq\v1dGx+dGyrGx,y,=p−3∑j=2q31+rGv1,uj−∑i=3p−1∑j=2q4rGvi,v1+rGv1,uj≥−p−3∑j=2q11+rGv1,uj.Thus, combining (8) and (13), we have(46)∑x∈VCq\v1dGv1+dGxrGv1,x−∑x∈VCq\v1dGv1+dGxrGv1,x+∑x∈w˜y∈VCq\v1dG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈VCq\v1dGx+dGyrGx,y,=p−3∑j=2q1rGv1,uj+−p−3∑j=2q11+rGv1,uj>0.So, ifp≥12, we get RDRG′−RDRG>0, and the results are as follows.If4≤p≤11, BnG0=Sn4,q,Sn5,q,Sn6,q,Sn7,q,Sn8,q,Sn9,q,Sn10,q,Sn11,q. By comparison of direct calculation, we can obtain RDRSn3,q>RDRSnp,q.The proof is completed.Corollary 2.
IfG′ can be obtained from G by a cycle-shrinking transformation, see Figure 5, and q≥4, then RDRG<RDRG′.Figure 5
The cycle-shrinking transformation.


## 2.1. Edge-Lifting Transformation

LetG′ be obtained from G by an edge-lifting transformation at e=u0,v0, see Figure 1. See [14] for specific description.Figure 1
The edge-lifting transformation.Lemma 2.
(See [14]). If G′ can be obtained from G by an edge-lifting transformation, see Figure 1, then RDRG<RDRG′.


## 2.2. Cycle-Lifting Transformation

LetGpi be the star graph K1,si1≤i≤p with order si+1 and Gqj be the star graph K1,tj2≤j≤q with order tj+1. Let C be a bicyclic graph with exactly two cycles, which are Cp=v1v2,…,vpv1 and Cq=v1u2,…,uqv1vi≠uj,1≤i≤p,2≤j≤q, and G is obtained from C by identifying vi with the center of Gpi for all i1≤i≤p and uj with the center of Gqj for all j2≤j≤q. Deleting all edges in Gpi2≤i≤p, joining v1 to all pendent vertices of Gpi2≤i≤p, we obtained a new graph, denoted by G′, see Figure 2. This operation is called a cycle-lifting transformation of G with respect to Cp.Figure 2
The cycle-lifting transformation.Lemma 3.
IfG′ can be obtained from G by a cycle-lifting transformation, see Figure 2, and ∑i=2psi>0, then RDRG<RDRG′.Proof. LetG and G′ be as shown in Figure 2. By the definition of RDRG,(5)RDRG=∑i=1p∑x,y∈VGpi\vidGx+dGyrGx,y+∑i=1p∑x∈VGpi\vidGvi+dGxrGvi,x+∑i=1p∑x∈VGpi\viy∈VCp\vidGx+dGyrGx,y+∑i,j=1i≠jp∑x∈VGpi\viy∈VGpj\vjdGx+dGyrGx,y+∑i=1p∑x∈VGpi\viy∈VCq\v1dGx+dGyrGx,y+∑i=1x∈VGpi\vip∑j=2y∈VGqj\ujqdGx+dGyrGx,y+∑x,y∈VCpdGx+dGyrGx,y+∑x∈VCpy∈VCq\v1dGx+dGyrGx,y+∑j=2q∑x∈VCpy∈VGpj\ujdGx+dGyrGx,y+∑x,y∈VCq\v1dGx+dGyrGx,y+∑j=2q∑x∈VCq\v1y∈VGpj\ujdGx+dGyrGx,y+∑j=2q∑x,y∈VGpj\ujdGx+dGyrGx,y+∑i,j=2i≠jq∑x∈VGqi\uiy∈VGqj\ujdGx+dGyrGx,y.(6)RDRG′=∑i=1p∑x,y∈VGpi\v1dG′x+dG′yrG′x,y+∑i=1p∑x∈VGpi\v1dG′x+dG′yrG′x,y+∑i=1p∑x∈VGpi\v1y∈VCp\v1dG′x+dG′yrG′x,y+∑i,j=1i≠jp∑x∈VGpi\v1y∈VGpj\v1dG′x+dG′yrG′x,y+∑i=1p∑x∈VGpi\v1y∈VCq\v1dG′x+dG′yrG′x,y+∑i=1x∈VGpi\v1p∑j=2y∈VGqj\ujqdG′x+dG′yrG′x,y+∑x,y∈VCpdG′x+dG′yrG′x,y+∑x∈VCpy∈VCq\v1dG′x+dG′yrG′x,y+∑j=2q∑x∈VCpy∈VGpj\ujdG′x+dG′yrG′x,y+∑x,y∈VCq\v1dG′x+dG′yrG′x,y+∑j=2q∑x∈VCq\v1y∈VGpj\ujdG′x+dG′yrG′x,y+∑j=2q∑x,y∈VGpj\ujdG′x+dG′yrG′x,y+∑i,j=2i≠jq∑x∈VGqi\uiy∈VGqj\ujdG′x+dG′yrG′x,y.(1)
Noting thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for x,y∈VGpi\vi of G and x,y∈VGpi\v1 of G′, we have(7)∑i=1p∑x,y∈VGpi\v1dG′x+dG′yrG′x,y−∑i=1p∑x,y∈VGpi\vidGx+dGyrGx,y=0.(2)
Noting thatrGvi,x=rG′v1,x=1, dGv1+dGx=s1+5 for x∈VGp1\v1 and dGvi+dGx=si+3 for x∈VGpi\vi2≤i≤p of G, and dG′v1+dG′x=s1+⋯+sp+5 for x∈VGpi\v1 of G′, we have(8)∑i=1p∑x∈VGpi\v1dG′v1+dG′xrG′v1,y−∑i=1p∑x∈VGpi\vidGvi+dGxrGvi,y=∑i,j=1i≠jpsi⋅sj+2∑i=2psi.(3)
Noting thatdGx+dGv1=s1+5, rGx,v1=1+rGvi,v1 for x∈VGpi\vi2≤i≤p of G, dGx+dGy=sj+3, rGx,y=1+rGvi,vj, for x∈VGpi\vi1≤i≤p,y=vj∈VCp\v1,vi of G, and dG′x+dG′y=3, rG′x,y=1+rG′v1,y for x∈VGpi\v1,y∈VCp\v1 of G′, we have(9)∑i=1p∑x∈VGpi\v1y∈VCp\v1dG′x+dG′yrG′x,y−∑i=1p∑x∈VGpi\viy∈VCp\vidGx+dGyrGx,y=−∑i,j=1i≠jpsi⋅sj1+rGvi,vj+2∑i=2psi1+rGvi,v1.(4)
Noting thatdGx+dGy=dG′x+dG′y, rGx,y=rGx,vi+rGvi,vj+rGvj,y>rG′x,v1+rG′v1,y=rG′x,y for x∈VGpi\vi,y∈VGpj\vj1≤i,j≤p,i≠j of G and x∈VGpi\v1,y∈VGpj\v11≤i,j≤p,i≠j of G′, similarly for x∈VGpi\vi1≤i≤p,y∈VCq\v1 and x∈VGpi\v11≤i≤p,y∈VCq\v1 of G and G′, respectively, and for x∈VGpi\vi1≤i≤p,y∈VGqj\uj2≤j≤q and x∈VGpi\v11≤i≤p,y∈VGqj\uj2≤j≤q of G and G′, respectively, we have the same conclusions; thus,(10)∑i,j=1i≠jp∑x∈VGpi\v1y∈VGpj\v1dG′x+dG′yrG′x,y−∑i,j=1i≠jp∑x∈VGpi\viy∈VGpj\vjdGx+dGyrGx,y+∑i=1p∑x∈VGpi\v1y∈VCq\v1dG′x+dG′yrG′x,y−∑i=1p∑x∈VGpi\viy∈VCq\v1dGx+dGyrGx,y+∑i=1x∈VGpi\v1p∑j=2y∈VGqj\ujqdG′x+dG′yrG′x,y−∑i=1x∈VGpi\vip∑j=2y∈VGqj\ujqdGx+dGyrGx,y>0presents.(5)
We note thatdGv1=s1+4, dGvi=si+22≤i≤p for any vi∈VCp of G and dG′v1=s1+⋯+sp+4, dG′vi=22≤i≤p for any vi∈VCp of G′. Let rGvi,vj=rG′vi,vj=rt, when j−i=t; then, rt=rp−t. Thus, we have(11)∑x,y∈VCpdGx+dGyrGx,y=∑i=2pdGv1+dGvirGv1,vi+∑i=3pdGv2+dGvirGv2,vi+⋯+dGvp−1+dGvprGvp−1,vp,and then, we have(12)∑i=2pdGv1+dGvirGv1,vi=s1+s2+6r1+s1+s3+6r2+⋯+s1+sp+6rp−1,∑i=3pdGv2+dGvirGv2,vi=s2+s3+4r1+⋯+s2+sp+4rp−2,…dGvp−1+dGvprGvp−1,vp=sp−1+sp+4r1.Thus,(13)∑x,y∈VCpdGx+dGyrGx,y=s1+⋯+sp−1+s2+⋯+sp+4p−2+6r1+s1+⋯+sp−2+s3+⋯+sp+4p−3+6r2+⋯+s1+sp+6rp−1.Similarly,(14)∑x,y∈VCpdG′x+dG′yrG′x,y=s1+⋯+sp+4p−2+6r1+s1+⋯+sp+4p−3+6r2+⋯+s1+⋯+sp+6rp−1.Then,(15)∑x,y∈VCpdG′x+dG′yrG′x,y−∑x,y∈VCpdGx+dGyrGx,y=s2+⋯+sp−1r1+s3+⋯+sp−2r2+⋯+−s2+⋯+sp−1rp−1=0.(6)
It is noted thatdGv1=s1+4, dGvi=si+22≤i≤p, dGuj=tj+22≤j≤q for vi∈VCp and uj∈VCq\v1 of G. Similarly, It is noted that dG′v1=∑i=1psi+4, dG′vi=22≤i≤p, dG′uj=tj+22≤i≤q, for vi∈VCp and uj∈VCq\v1 of G′. Let rGvi,v1=rpt, when i−1=t2≤i≤p,1≤t≤p−1. Let rGv1,uj=rqs, when j−1=s2≤j≤q,1≤s≤q−1. Then, rGx,y=rGvi,v1+rGv1,ui=rpt+rqs=rG′x,y for x∈VCp and y∈VCq\v1. Thus, we have(16)∑x∈VCpy∈VCq\v1dGx+dGyrGx,y=∑i=2qdGv1+dGuirGv1,ui+∑i=2qdGv2+dGuirGv2,ui+⋯+∑i=2qdGvp+dGuirGvp,ui,and then, we have(17)∑i=2qdGv1+dGuirGv1,ui=s1+t2+6rq1+s1+t3+6rq2+⋯+s1+tq+6rqq−1,∑i=2qdGv2+dGuirGv2,ui=s2+t2+4rp1+rq1+s2+t3+4rp1+rq2+⋯+s2+tq+4rp1+rqq−1,…∑i=2qdGvp+dGuirGvp,ui=sp+t2+4rpp−1+rq1+sp+t3+4rpp−1+rq2+⋯+sp+tq+4rpp−1+rqq−1.Similarly,(18)∑x∈VCpy∈VCq\v1dG′x+dG′yrG′x,y=∑i=2qdG′v1+dG′uirG′v1,ui+∑i=2qdG′v2+dG′uirG′v2,ui+⋯+∑i=2qdG′vp+dG′uirG′vp,ui,and then, we have(19)∑i=2qdG′v1+dG′uirG′v1,ui=s1+⋯+sp+t2+6rq1+s1+⋯+sp+t3+6rq2+⋯+s1+⋯+sp+tq+6rqq−1,∑i=2qdG′v2+dG′uirG′v2,ui=t2+4rp1+rq1+t3+4rp1+rq2+⋯+tq+4rp1+rqq−1,…∑i=2qdG′vp+dG′uirG′vp,ui=t2+4rpp−1+rq1+t3+4rpp−1+rq2+⋯+tq+4rpp−1+rqq−1.Thus,(20)∑x∈VCpy∈VCq\v1dG′x+dG′yrG′x,y=−∑x∈VCpy∈VCq\v1dGx+dGyrGx,y=∑i=2p∑j=1q−1sirqj−∑i=2p∑j=1q−1sirpi−1+rqj>0.(7)
Forx∈VCp,y∈VGpj\uj of G and G′, respectively, simulating the abovementioned calculation method of (6), we get the similar conclusion,(21)∑j=2q∑x∈VCpy∈VGpj\ujdG′x+dG′yrG′x,y−∑j=2q∑x∈VCpy∈VGpj\ujdGx+dGyrGx,y=∑i=2p∑j=2qsi⋅tjrqj−1+1−∑i=2p∑j=2qsi⋅tjrpi−1+rqj−1+1>0.(8)
Obviously,(22)∑x,y∈VCq\v1dGx+dGyrGx,y+∑j=2q∑x∈VCq\v1y∈VGpj\ujdGx+dGyrGx,y+∑j=2q∑x,y∈VGpj\ujdGx+dGyrGx,y+∑i,j=2i≠jq∑x∈VGqi\uiy∈VGqj\ujdGx+dGyrGx,y,=∑x,y∈VCq\v1dG′x+dG′yrG′x,y+∑j=2q∑x∈VCq\v1y∈VGpj\ujdG′x+dG′yrG′x,y+∑j=2q∑x,y∈VGpj\ujdG′x+dG′yrG′x,y+∑i,j=2i≠jq∑x∈VGqi\uiy∈VGqj\ujdG′x+dG′yrG′x,y.Thus, combing (1)–(8), we get RDRG′−RDRG>0, and the results are as follows.Corollary 1.
IfG′ can be obtained from G by a cycle-lifting transformation, see Figure 3, and ∑j=2qtj>0, then RDRG<RDRG′.Figure 3
The cycle-lifting transformation.


## 2.3. Cycle-Shrinking Transformation

LetG=Snp,q, which is obtained by coalescing Cp and Cq at the vertex v1 and attaching n−p−q+1 pendent edges to the vertex v1. We note that p,q≥3,p+q≤n+1. Deleting the edges v2v3,…,vp−1vp and adding the edges v3v1,v4v1,…,vp−1v1,v2vp, we get G′=Sn3,q This operation is called cycle-shrinking transformation. It is denoted by (ww′, resp), the set of pendent vertices of (GG′, resp). Let w=k≥0; w′ can be partitioned into two subsets: one has k vertices, which is naturally corresponding to w (also denoted by w), and another has p−3 vertices, denoted by w˜.Lemma 4.
LetG=Snp,qp≥4,p+q≤n+1; if G′ can be obtained from G by a cycle-shrinking transformation, see Figure 4, then RDRG<RDRG′.Figure 4
The cycle-shrinking transformation.Proof. LetG and G′ be as shown in Figure 4. By the definition of RDRG,(23)RDRG=∑x,y∈wdGx+dGyrGx,y+∑x∈wdGv1+dGxrGv1,y+∑x∈wy∈v2,vpdGx+dGyrGx,y+∑x∈wy∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x∈wy∈VCq\v1dGx+dGyrGx,y+∑x∈v2,vpdGv1+dGxrGv1,y+∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,y+∑x∈VCq\v1dGv1+dGxrGv1,x+dGv2+dGvprGv2,vp+∑x∈v2,vpy∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x∈v2,vpy∈VCq\v1dGx+dGyrGx,y+∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x∈VCp\v1,v2,vpy∈VCq\v1dGx+dGyrGx,y+∑x,y∈VCq\v1dGx+dGyrGx,y,RDRG′=∑x,y∈wdG′x+dG′yrG′x,y+∑x∈wdG′v1+dG′xrG′v1,x+∑x∈wy∈v2,vpdG′x+dG′yrG′x,y+∑x∈wy∈w˜dG′x+dG′yrG′x,y+∑x∈wy∈VCq\v1dG′x+dG′yrG′x,y+∑x∈v2,vpdG′v1+dG′xrG′v1,x+∑x∈w˜dG′v1+dG′xrG′v1,x+∑x∈VCq\v1dG′v1+dG′xrG′v1,x+dG′v2+dG′vprG′v2,vp+∑x∈v2,vpy∈w˜dG′x+dG′yrG′x,y+∑x∈v2,vpy∈VCq\v1dG′x+dG′yrG′x,y+∑x,y∈w˜dG′x+dG′yrG′x,y+∑x∈w˜y∈VCq\v1dG′x+dG′yrG′x,y+∑x,y∈VCq\v1dG′x+dG′yrG′x,y.(1)
Noting thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for any x,y∈w of G and G′, respectively, we have(24)∑x,y∈wdGx+dGyrGx,y=∑x,y∈wdG′x+dG′yrG′x,y.(2)
Noting thatdGv1+dGx=k+5, dG′v1+dG′x=k+p+2, rGv1,x=rG′v1,x=1 for x∈w in G and G′, respectively, we have(25)∑x∈wdG′v1+dG′xrG′v1,x−∑x∈wdGv1+dGxrGv1,x=kk+p+2−kk+5=kp−3.(3)
Noting thatdGx+dGy=dG′x+dG′y=3, rGx,y=p−1/p+1≥rG′x,y=2/3+1 for any x∈w,y∈v2,vp in G and x∈w,y∈v2,vp in G′, we have(26)∑x∈wy∈v2,vpdG′x+dG′yrG′x,y−∑x∈wy∈v2,vpdGx+dGyrGx,y>0.(4)
Noting thatdGx+dGy=3,rGx,y=rGvi,v1+1>23≤i≤p−1 for x∈w,y∈VCp\v1,v2,vp and dG′x+dG′y=2,rG′x,y=2 for x∈w,y∈w˜, we have(27)∑x∈wy∈w˜dG′x+dG′yrG′x,y−∑x∈wy∈VCp\v1,v2,vpdGx+dGyrGx,y≥kp−3−3kp−32=−kp−32.Thus, combing (2) and (4), we have(28)∑x∈wdG′v1+dG′xrG′v1,x−∑x∈wdGv1+dGxrGv1,x+∑x∈wy∈w˜dG′x+dG′yrG′x,y−∑x∈wy∈VCp\v1,v2,vpdGx+dGyrGx,y≥kp−32≥0.(5)
Noting thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for any x∈w,y∈VCq\v1 of G and G′, respectively, we have(29)∑x∈wy∈VCq\v1dGx+dGyrGx,y=∑x∈wy∈VCq\v1dG′x+dG′yrG′x,y.(6)
Noting thatdGv1+dGx=k+6,rGv1,x=p−1/p and dG′v1+dG′x=k+p+3,rG′v1,x=2/3 for x∈v2,vp of G and G′, respectively, we have(30)∑x∈v2,vpdG′v1+dG′xrG′v1,x−∑x∈v2,vpdGv1+dGxrGv1,x=3k+p+3−2pk+6p−1=k+3p+3p−3p−1.(7)
Noting thatdGv1+dGx=k+6,rGv1,x=rGv1,vi>13≤i≤p−1 for x∈VCp\v1,v2,vp of G and dG′v1+dG′x=k+p+2,rG′v1,x=1 for x∈w˜ of G′, we have(31)∑x∈w˜dG′v1+dG′xrG′v1,x−∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,x≥p−3k+p+2−p−3k+6=p−3p−4.(8)
Noting thatdGv1+dGx=k+6,dG′v1+dG′x=k+p+3,rGv1,x=rG′v1,x for x∈VCq\v1 of G and G′, respectively, we have(32)∑x∈VCq\v1dG′v1+dG′xrG′v1,x−∑x∈VCq\v1dGv1+dGxrGv1,x=p−3∑j=2q1rGv1,uj.(9)
Noting thatdGv2+dGvp=4, rGv2,vp=2p−2/p and dG′v2+dG′vp=4, rG′v2,vp=2/3, we have(33)dG′v2+dG′vprG′v2,vp−dGv2+dGvprGv2,vp=6−2pp−2=4p−12p−2>0.(10)
Noting thatdGx+dGy=4 for x∈v2,vp, y∈VCp\v1,v2,vp of G and dG′x+dG′y=3, rG′x,y=rG′v2,v1+1=2/3+1=5/3 for x∈v2,vp,y∈w˜ of G′, by the symmetry between v2 and vp, we have(34)∑x∈v2,vpy∈w˜dG′x+dG′yrG′x,y−∑x∈v2,vpy∈VCp\v1,v2,vpdGx+dGyrGx,y=2∑i=3p−13rG′v2,v1+1−2∑i=3p−14rGv2,vi,=18p−35−8pp−1+2∑i=4p−14rGv2,vi.SincerGv2,vi≥rGv2,v4=2p−2/p4≤i≤p−1, if p≥12, then 2p−2/p≥5/3, and(35)2∑i=4p−14rGv2,vi≤24p−45.Thus,(36)∑x∈v2,vpy∈w˜dG′x+dG′yrG′x,y−∑x∈v2,vpy∈VCp\v1,v2,vpdGx+dGyrGx,y≥18p−35−8pp−1+24p−45.(11)
Noting thatdGx+dGy=4 for x∈v2,vp, y∈VCq\v1 of G, dG′x+dG′y=4 for x∈v2,vp,y∈VCq\v1 of G′, and rGx,y=rGv2,v1+rGv1,ui=p−1/p+rGv1,ui≥rG′x,y=rG′v2,v1+rG′v1,ui=2/3+rG′v1,ui2≤i≤q, by symmetry between v2 and vp, we have(37)∑x∈v2,vpy∈VCq\v1dG′x+dG′yrG′x,y−∑x∈v2,vpy∈VCq\v1dGx+dGyrGx,y≥0.(12)
Noting thatdGx+dGy=4, rGx,y=rGvi,vj3≤i<j≤p−1 for x,y∈VCp\v1,v2,vp of G and dG′x+dG′y=2, rG′x,y=2 for x,y∈w˜ of G′, we have(38)∑x,y∈w˜dG′x+dG′yrG′x,y−∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y=p−3p−42−∑i,j=3i<jp−14rGvi,vj,∑i,j=3i<jp−14rGvi,vj=∑i=4pdGv3+dGvirGv3,vi+∑i=5pdGv4+dGvirGv4,vi+⋯+dGvp−2+dGvp−1rGvp−2,vp−1.Whenj−i=ti<j, we denote rGvi,vj=rt, and then, rt=rp−t=tp−t/p; thus,(39)∑i=4pdGv3+dGvirGv3,vi=4r1+4r2+⋯+4rp−4,∑i=5pdGv4+dGvirGv4,vi=4r1+⋯+4rp−5,…dGvp−2+dGvp−1rGvp−2,vp−1=4r1.Thus,(40)∑i,j=3i<jp−14rGvi,vj=4p−4r1+4p−5r2+⋯+4rp−4.Sincer1=p−1/p, rt≥r2=2p−2/p2≤t≤p−4. If p≥12, then 2p−2/p≥5/3, and(41)∑i,j=3i<jp−14rGvi,vj≤4pp−4p−1+6p−4p−55.Thus,(42)∑x,y∈w˜dG′x+dG′yrG′x,y−∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y≥p−3p−42−4pp−4p−1+6p−4p−55.Whenp≥12, combing (6), (7), (10), and (12), we have(43)∑x∈v2,vpdG′v1+dG′xrG′v1,x−∑x∈v2,vpdGv1+dGxrGv1,x+∑x∈w˜dG′v1+dG′xrG′v1,x−∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,x+∑x∈v2,vpy∈w˜dG′x+dG′yrG′x,y−∑x∈v2,vpy∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x,y∈w˜dG′x+dG′yrG′x,y−∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y≥k+3p+3p−3p−1+3p−3p−42+18p−35−4pp−2p−1−6p−1p−45=3p3−22p2+53p−114+k10p−3010p−1.(i)
Letfp=3p3−22p2+53p−114,p∈12,+∞; this function is strictly increasing in this interval; thus, fp≥f12=2538>0.Thus,(44)3p3−22p2+33p−54+k10p−3010p−1>0.(13)
Noting thatdGx+dGy=4, rGx,y=rGvi,v1+rGv1,uj≥1+rGv1,uj3≤i≤p−1,2≤j≤q for x∈VCp\v1,v2,vp,y∈VCq\v1 of G and dG′x+dG′y=3, rG′x,y=1+rGv1,uj2≤j≤q for x,y∈w˜ of G′, we have(45)∑x∈w˜y∈VCq\v1dG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈VCq\v1dGx+dGyrGx,y,=p−3∑j=2q31+rGv1,uj−∑i=3p−1∑j=2q4rGvi,v1+rGv1,uj≥−p−3∑j=2q11+rGv1,uj.Thus, combining (8) and (13), we have(46)∑x∈VCq\v1dGv1+dGxrGv1,x−∑x∈VCq\v1dGv1+dGxrGv1,x+∑x∈w˜y∈VCq\v1dG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈VCq\v1dGx+dGyrGx,y,=p−3∑j=2q1rGv1,uj+−p−3∑j=2q11+rGv1,uj>0.So, ifp≥12, we get RDRG′−RDRG>0, and the results are as follows.If4≤p≤11, BnG0=Sn4,q,Sn5,q,Sn6,q,Sn7,q,Sn8,q,Sn9,q,Sn10,q,Sn11,q. By comparison of direct calculation, we can obtain RDRSn3,q>RDRSnp,q.The proof is completed.Corollary 2.
IfG′ can be obtained from G by a cycle-shrinking transformation, see Figure 5, and q≥4, then RDRG<RDRG′.Figure 5
The cycle-shrinking transformation.


## 3. Maximum Reciprocal Degree Resistance Distance Index of Bicyclic Graphs

LetG∈ℬn; applying a sequence of edge-lifting transformations, cycle-lifting transformations, and cycle-shrinking transformations to G, we can obtain a new graph G′, denoted by Sn3,3, see Figure 5. The next theorem discusses the maximum reciprocal degree resistance distance index in ℬn.Theorem 1.
LetG∈ℬn; then, RDRG≤RDRSn3,3, with equality holding if and only if G≅Sn3,3.Proof. LetG∈ℬn such that the RDRG is as big as possible. By Lemma 2–4, Corollary 2, we have RDRG≤RDRSn3,3, and equality holds if and only if G≅Sn3,3.


## 4. Conclusions

In this paper, we characterize the maximum reciprocal degree resistance distance index among all bicyclic graphs without common edges of ordern by using three types of transformation which keep the reciprocal degree resistance distance index increasing: edge-lifting transformation, cycle-lifting transformation, and cycle-shrinking transformation.



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*Source: 9998763-2021-06-19.xml*