# Maximum Reciprocal Degree Resistance Distance Index of Unicyclic Graphs

Authors: Gai-Xiang Cai; Xing-Xing Li; Gui-Dong Yu
Journal: Discrete Dynamics in Nature and Society
 (2020)

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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 Un denote the set of unicyclic graphs with n vertices. We study the graph with maximum reciprocal degree resistance distance index among all graphs in Un and characterize the corresponding extremal graph.


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## Body

## 1. Introduction

LetG=V,E be a simple connected graph of order n with vertex set V=VG=v1,v2,…,vn and edge set E=EG. For any u∈VG, dGu is the degree of vertex u, and the distance between vertices of u and v, denoted by dGu,v, is the length of a shortest path between them. Topological indices are numbers associated with molecular structures which serve for quantitative relationships between chemical structures and properties. The first such index was published by Wiener [1], but the name topological index was invented by Hosoya [2]. Many of them are based on the graph distance [3] and the vertex degree [4]. In addition, several graph invariants are based on both the vertex degree and the graph distance [5].One of the most intensively studied topological indices is the Wiener index. The Wiener index was introduced by American chemist Wiener in [1], defined as(1)WG=∑u,v⊆VGdGu,v.Another distance-based graph invariant Harary index has been introduced by Plavšić et al. [6] and independently by Ivanciuc et al. [7] in 1993 for the characterization of molecular graphs. The Harary index HG of graph G is defined as(2)HG=∑u,v⊆VG1dGu,v.For more results related to Harary index, refer to [8, 9–17].The resistance distanceru,v (if more than one graphs are considered, we write rGu,v in order to avoid confusion) between vertices u and v in G is defined as the effective resistance between the two nodes of the electronic network obtained so that its nodes correspond to the vertices of G and each edges of G is replaced by a resistor of unit resistance, which is compared by the methods of the theory of resistive electrical networks based on Ohm’s and Kirchoff’s laws.The Kirchhoff indexKfG of a graph G is defined as [18, 19](3)KfG=∑u,v⊆VGrGu,v.As a new structure descriptor, the Kirchoff index is well studied (see recent papers [20–31]). In 2017, Chen et al. [32] introduced a new graph invariant reciprocal to Kirchoff index, named Resistance-Harary index:(4)RHG=∑u,v⊆VG1rGu,v.For more results related to Resistance-Harary index, refer to [32–34].The first and the second Zagreb indices are defined as(5)M1=M1G=∑u∈VGdGu2=∑uv∈EGdGu+dGv,M2=M2G=∑uv∈EGdGudGv,respectively. These are the oldest [35, 36] and best studied degree-based topological indices (see the reviews [4], recent papers [37, 38], and the references cited therein).Dobrynin and Kochtova [39] and Gutman [40] independently proposed a vertex-degree-weighted version of Wiener index called degree distance, which is defined for a connected graph G as(6)DDG=12∑u,v∈VGdGu+dGvdGu,v.The reciprocal degree distance [41] is defined as(7)RDDG=12∑u,v∈VGdGu+dGvdGu,v.Hua and Zhang [42] have obtained lower and upper bounds for the reciprocal degree distance of graph in terms of other graph invariants. The chemical applications and mathematical properties of the reciprocal degree distance are well studied in [41, 43].Analogous to the relationship between degree distance and reciprocal degree distance, we introduce here a new graph invariant based on both the vertex degree and the graph distance, named the reciprocal degree resistance distance index:(8)RDRG=∑u,v⊆VGdGu+dGvrGu,v.A unicyclic graph is a connected graph withn vertices and n edges. Let Un denote the set of unicyclic graphs with n vertices. In this paper, we determine the graph with maximum reciprocal degree resistance distance index among all graphs in Un and characterize the corresponding extremal graph.


## 2. Preliminaries

In this section, we will introduce some useful lemmas and three transformations.LetCg=v1v2⋯vgv1 be the cycle on g≥3 vertices. For any two vertices vi,vj∈VCg with i<j, by Ohm’s law, one has(9)rCgvi,vj=j−ig+i−jg.Lemma 1 (see [19]).
Letx be a cut vertex of a connected graph G and let a and b be vertices occurring in different components which arise upon deletion of x. Then,(10)rGa,b=rGa,x+rGx,b.


### 2.1. Edge-Lifting Transformation

LetG1 and G2 be two graphs with n1≥2 and n2≥2 vertices, respectively. G is the graph obtained from G1 and G2 by adding an edge between a vertex u0 of G1 to a vertex v0 of G2. And G′ is the graph obtained by identifying u0 of G1 to a vertex v0 of G2 and adding a pendant edge to u0v0. We say that G′ is obtained from G by an edge-lifting transformation at e=u0,v0 (see Figure 1).Figure 1
The edge-lifting transformation. (a)G. (b) G′.
(a)(b)Lemma 2.
IfG′ can be obtained from G by an edge-lifting transformation, then RDRG<RDRG′.Proof.
ConsiderG and G′ shown in Figure 1. By the definition of RDRG,(11)RDRG=∑x,y∈VG1\u0dGx+dGyrGx,y+∑x,y∈VG2\v0dGx+dGyrGx,y+∑x∈VG1\u0dGu0+dGxrGu0,x+∑x∈VG2\v0dGu0+dGxrGv0,x+1+∑x∈VG1\u0dGv0+dGxrGu0,x+1+∑x∈VG2\v0dGv0+dGxrGv0,x+∑x∈VG1\u0,y∈VG2\v0dGx+dGyrGx,u0+1+rGv0,y+dGu0+dGv0rGu0,v0,RDRG′=∑x,y∈VG1\u0dG′x+dG′yrG′x,y+∑x,y∈VG2\v0dG′x+dG′yrG′x,y+∑x∈VG1\u0dG′u0+dG′xrG′u0,x+∑x∈VG2\v0dG′v0+dG′xrG′v0,x+∑x∈VG1\u0dG′w0+dG′xrG′u0,x+1+∑x∈VG2\v0dG′w0+dG′xrG′v0,x+1+∑x∈VG1\u0,y∈VG2\v0dG′x+dG′yrG′x,u0+rG′v0,y+dG′u0+dG′w0rG′u0,w0.(i)
Note thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for x,y∈VG1\u0 or x,y∈VG2\v0, and dGu0+dGv0=dG′u0+dG′w0, rGu0,v0=rG′u0,w0; then, we have(12)∑x,y∈VG1\u0dGx+dGyrGx,y=∑x,y∈VG1\u0dG′x+dG′yrG′x,y,∑x,y∈VG2\v0dGx+dGyrGx,y=∑x,y∈VG2\v0dG′x+dG′yrG′x,y,dGu0+dGv0rGu0,v0=dG′u0+dG′w0rG′u0,w0.(ii)
Note thatdGx+dGy=dG′x+dG′y, rGx,u0+rGv0,y=rG′x,u0+rG′v0,y for x∈VG1\u0,y∈VG2\v0; then, we have(13)∑x∈VG1\u0,y∈VG2\v0dGx+dGyrGx,u0+1+rGv0,y<∑x∈VG1\u0,y∈VG2\u0dG′x+dG′yrG′x,u0+rG′v0,y.(iii)
Note thatrG′u0,x=rGu0,x for any x∈VG1\u0, and dG′u0=dGu0+dGv0−1; then, we have(14)∑x∈VG1\u0dG′u0+dG′xrG′u0,x−∑x∈VG1\u0dGu0+dGxrGu0,x+∑x∈VG1\u0dG′w0+dG′xrG′u0,x+1−∑x∈VG1\u0dGv0+dGxrGu0,x+1=∑x∈VG1\u0dGv0−1rGu0,x+∑x∈VG1\u01−dGv0rGu0,x+1>0.(iv)
Note thatrG′v0,x=rGv0,x, dG′x=dGx for any x∈VG2, and dG′v0=dGu0+dGv0−1; then, we have(15)∑x∈VG2\v0dG′v0+dG′xrG′v0,x−∑x∈VG2\v0dGv0+dGxrGv0,x+∑x∈VG2\v0dG′w0+dG′xrG′v0,x+1−∑x∈VG2\v0dGu0+dGxrGv0,x+1=∑x∈VG2\v0dGu0−1rGv0,x+∑x∈VG2\v01−dGu0rGv0,x+1>0.
Thus, by (i)–(iv), we getRDRG′−RDRG>0.


### 2.2. Cycle-Lifting Transformation

LetG be a graph as shown in Figure 2. Take a cycle C in G, say C=v1v2⋯vpv1; G can be viewed as a graph obtained by coalescing C with a number of star subgraphs of G, say G1,G2,…,Gp, by identifying vi with the center of Gi for all i1≤i≤p, denoted as EGi=si and VGi=si+1. Deleting all edges in Gi and joining v1 to all pendant vertices of Gi2≤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 C.Figure 2
The cycle-lifting transformation. (a)G. (b) G′.
(a)(b)Lemma 3.
IfG′ can be obtained from G by a cycle-lifting transformation, then RDRG<RDRG′.Proof.
ConsiderG and G′ shown in Figure 2. By the definition of RDRG,(16)RDRG=∑i=1p∑x,y∈VGi\vidGx+dGyrGx,y+∑i=1p∑x∈VGi\vidGvi+dGxrGvi,x+∑i,j=1i≠jp∑x∈VGi\viy∈VGj\vjdGx+dGyrGx,vi+rGvi,vj+rGvj,y+∑i=1p∑x∈VGi\viy∈VGj\vjdGx+dGyrGx,y+∑x,y∈VCpdGx+dGyrGx,y,RDRG′=∑i=1p∑x,y∈VGi\v1dG′x+dG′yrG′x,y+∑i=1p∑x∈VGi\v1dG′v1+dG′xrG′v1,x+∑i,j=1i≠jp∑x∈VGi\v1y∈VGj\v1dG′x+dG′yrG′x,v1+rG′v1,y+∑i=1p∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y+∑x,y∈VCpdG′x+dG′yrG′x,y.(i)
Note thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for x,y∈VGi\vi in G, and x,y∈VGi\v1 in G′; then, we have(17)∑i=1p∑x,y∈VGi\vidGx+dGyrGx,y=∑i=1p∑x,y∈VGi\v1dG′x+dG′yrG′x,y.(ii)
LetrGvi,vj=rG′vi,vj=rt; when j−i=t, then rt=rp−t. dGvi=si+2 for any vi∈VCp in G and dG′v1=s1+⋯+sp+2 for v1∈VCp in G′; then, we have(18)∑x,y∈VCpdGx+dGyrGx,y=∑i=2pdGv1+dGvirGv1,vi+∑i=3pdGv2+dGvirGv2,vi+⋯+dGvp−1+dGvprGvp−1,vp.Thus,(19)∑i=2pdGv1+dGvirGv1,vi=s1+s2+4r1+s1+s3+4r2+⋯+s1+sp+4rp−1,∑i=3pdGv2+dGvirGv2,vi=s2+s3+4r1+⋯+s2+sp+4rp−2,⋯dGvp−1+dGvprGvp−1,vp=sp−1+sp+4r1.Thus,(20)∑x,y∈VCpdGx+dGyrGx,y=s1+⋯+sp−1+s2+⋯+sp+4p−1r1+s1+⋯+sp−2+s3+⋯+sp+4p−2r2+⋯+s1+sp+4rp−1.Similarly,(21)∑x,y∈VCpdG′x+dG′yrG′x,y=s1+⋯+sp+4p−1r1+s1+⋯+sp+4p−2r2+⋯+s1+⋯+sp+4rp−1.Then,(22)∑x,y∈VCpdGx+dGyrGx,y−∑x,y∈VCpdG′x+dG′yrG′x,y=s2+⋯+sp−1r1+s3+⋯+sp−2r2+⋯+−s2+⋯+sp−1rp−1=0.(iii)
Note thatdGx+dGy=dG′x+dG′y, rGx,vi+rGvj,y=rG′x,v1+rG′v1,y for x∈VGi\vi,y∈VGj\vj in G and for x∈VGi\v1,y∈VGj\v1 in G′; then, we have(23)∑i,j=1i≠jp∑x∈VGi\viy∈VGj\vjdGx+dGyrGx,vi+rGvi,vj+rGvj,y<∑i,j=1i≠jp∑x∈VGi\v1y∈VGj\v1dG′x+dG′yrG′x,v1+rG′v1,y.(iv)
Note thatrGvi,x=rG′v1,x=1, dGvi+dGx=si+3 for x∈VGi\vi in G, and dG′v1+dG′x=s1+⋯+sp+3 for x∈VGi\v1 in G′; then, we have(24)∑i=1p∑x∈VGi\vidGvi+dGxrGvi,x=s1s1+3+s2s2+3+⋯+spsp+3=∑i=1psisi+3,∑i=1p∑x∈VGi\v1dG′v1+dG′xrG′v1,x=s1s1+⋯+sp+3+⋯+sps1+⋯+sp+3=∑i=1psis1+⋯+sp+3.Thus, we have(25)∑i=1p∑x∈VGi\v1dG′v1+dG′xrG′v1,x−∑i=1p∑x∈VGi\vidGvi+dGxrGvi,x=∑i,j=1i≠jpsi⋅sj.(v)
Note thatdGx+dGy=3+si,dG′x+dG′y=3, rGx,y=rG′x,y for any x∈VCp\vi,y∈VGi\vi in G and for x∈VCp\v1,y∈VGi\v1 in G′; when si+t>sp1≤t≤p−1, then si+t=si+t−p; we have(26)∑i=1p∑x∈VCp\viy∈VGi\vidGx+dGyrGx,y=∑x∈VCp\v1y∈VG1\v1dGx+dGyrGx,y+∑x∈VCp\v2y∈VG1\v2dGx+dGyrGx,y+⋯+∑x∈VCp\vpy∈VGp\vpdGx+dGyrGx,y.
Thus,(27)∑x∈VCp\v1y∈VG1\v1dGx+dGyrGx,y=s13+s2r1+1+s13+s3r2+1+⋯+s13+sprp−1+1,∑x∈VCp\v2y∈VG1\v2dGx+dGyrGx,y=s23+s3r1+1+s23+s4r2+1+⋯+s23+s1rp−1+1,⋯∑x∈VCp\vpy∈VGp\vpdGx+dGyrGx,y=sp3+s1r1+1+sp3+s2r2+1+⋯+sp3+sp−1rp−1+1,and we have(28)∑i=1p∑x∈VCp\vjy∈VGp\vjdGx+dGyrGx,y=∑i=1psi3+si+1modpr1+1+∑i=1psi3+si+2modpr2+1+⋯+∑i=1psi3+si+p−1modprp−1+1.
Thus,(29)∑i=1p∑x∈VCp\viy∈VGp\vidGx+dGyrGx,y=∑i,j=1i≠jpsi⋅3+sjrGvi,vj+1.
Similarly,(30)∑i=1p∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y+∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y+⋯+∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y,∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=3s1r1+1+3s1r2+1+⋯+3s1rp−1+1,∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=3s2r1+1+3s2r2+1+⋯+3s2rp−1+1,⋯∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=3spr1+1+3spr2+1+⋯+3sprp−1+1.
Thus, we have(31)∑i=1p∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=∑i=1p3sir1+1+∑i=1p3sir2+1+⋯+∑i=1p3sirp−1+1.
Then,(32)∑i=1p∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y−∑i=1p∑x∈VCp\viy∈VGi\vidGx+dGyrGx,y=−∑i,j=1i≠jpsi⋅sjrGvi,vj+1.
Thus, by (i)–(v), we getRDRG′−RDRG>0.


### 2.3. Cycle-Shrinking Transformation

Denote bySnp the unicyclic graph obtained from cycle Cp by attaching n−p pendant edges to a vertex v1 of Cp (see Figure 3). Let G=Snpp≥4; deleting the edges v2v3,…,vp−1vp and adding the edges v2vp,v3v1,…,vp−1v1, we obtain the graph G′=Sn3. This operation is called cycle-shrinking transformation. Denote by WW′,resp the set of pendant vertices of GG′,resp. Let W=k, W′=k′. It is clear that p+k=n and 3+k′=n. Then, k′=k+p−3; thus, W′ can be partitioned into two subsets. One has k vertices, which is naturally corresponding to W. So, we also denote it by W; another has p−3 vertices, denoted by W˜.Figure 3
The cycle-shrinking transformation. (a)G. (b) G′.
(a)(b)Lemma 4.
LetG be an unicyclic graph of order n and a cycle Cp with p≥4; if G′ can be obtained from G by a cycle-shrinking transformation, then RDRG<RDRG′.Proof.
ConsiderG and G′ shown in Figure 3. By the definition of RDRG,(33)RDRG=∑x,y∈WdGx+dGyrGx,y+∑x∈WdGv1+dGxrGv1,x+∑x∈Wy∈v2,vpdGx+dGyrGx,y+∑x∈Wy∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x∈v2,vpdGv1+dGxrGv1,x+∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,x+∑x∈VCp\v1,v2,vpy∈v2,vpdGx+dGyrGx,y+∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y+dGv2+dGvprGv2,vp,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∈W,y∈W˜dG′x+dG′yrG′x,y+∑x∈v2,vpdG′v1+dG′xrG′v1,x+∑x∈W˜dG′v1+dG′xrG′v1,x+∑x∈Wy∈v2,vpdG′x+dG′yrG′x,y+∑x,y∈W˜dG′x+dG′yrG′x,y+dG′v2+dG′vprG′v2,vp.(i)
Note thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for any x,y∈W; then, we have(34)∑x,y∈WdGx+dGyrGx,y=∑x,y∈WdG′x+dG′yrG′x,y.(ii)
Note 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′; then, we have(35)∑x∈Wy∈v2,vpdG′x+dG′yrG′x,y−∑x∈Wy∈v2,vpdGx+dGyrGx,y≥0.(iii)
Note thatdGv1+dGx=k+3, dG′v1+dG′x=k+p, rGv1,x=rG′v1,x=1, for x∈W in G and G′, respectively; then, we have(36)∑x∈WdG′v1+dG′xrG′v1,x−∑x∈WdGv1+dGxrGv1,x=kk+p−kk+3=kp−3.(iv)
Note 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˜; then, we have(37)∑x∈Wy∈W˜dG′x+dG′yrG′x,y−∑x∈Wy∈VCp\v1,v2,vpdGx+dGyrGx,y≥kp−3−3kp−32=−kp−32.Combingiii−iv, we have(38)∑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.(v)
Note thatdGv1+dGx=k+4,rGv1,x=p−1/p, and dG′v1+dG′x=k+p+1,rG′v1,x=2/3, for x∈v2,vp in G and G′, respectively; then, we have(39)∑x∈v2,vpdG′v1+dG′xrG′v1,x−∑x∈v2,vpdGv1+dGxrGv1,x=3k+p+1−2pk+4p−1=k+3p+1p−3p−1.(vi)
Note thatdGv1+dGx=k+4,rGv1,x=rGv1,vi≥13≤i≤p−1 for x∈VCp\v1,v2,vp and dG′v1+dG′x=k+p,rG′v1,x=1 for x∈W˜; then, we have(40)∑x∈W˜dG′v1+dG′xrG′v1,x−∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,x≥p−3k+p−p−3k+4=p−3p−4.(vii)
Note thatdGx+dGy=3 for x∈VCp\v1,v2,vp, y∈v2,vp and dG′x+dG′y=4, rG′x,y=rG′v2,v1+1=2/3+1=5/3 for x∈W˜, y∈v2,vp; by the symmetry between v2 and vp, we have(41)∑x∈W˜y∈v2,vpdG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈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 2p−2/p≥5/3, then p≥12.Then,(42)2∑i=4p−14rGv2,vi≤24p−45.Thus,(43)∑x∈W˜y∈v2,vpdG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈v2,vpdGx+dGyrGx,y≥18p−35−8pp−1+24p−45.(viii)
Note thatdGx+dGy=4, rGx,y=rGvi,vj3≤i<j≤p−1 for x,y∈VCp\v1,v2,vp and dG′x+dG′y=2, rG′x,y=2 for x,y∈W˜; then, we have(44)∑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.Then,(45)∑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, rGvi,vj=rt=rp−t; thus,(46)∑i=4pdGv3+dGvirGv3,vi=4r1+4r2+⋯+4rp−4,∑i=5pdGv4+dGvirGv4,vi=4r1+⋯+4rp−5,⋯dGvp−2+dGvp−1rGvp−2,vp−1=4r1,and(47)∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y=4p−4r1+4p−5r2+⋯+4rp−4.Sincer1=p−1/p, rt≥r2=2p−2/p2≤t≤p−4, if 2p−2/p≥5/3, then p≥12.Then,(48)∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y≤4pp−4p−1+6p−4p−55.Thus,(49)∑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 v−viii, we have(50)∑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∈W˜y∈v2,vpdG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈v2,vpdGx+dGyrGx,y+∑x,y∈W˜dG′x+dG′yrG′x,y−∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y≥k+3p+1p−3p−1+3p−3p−42+18p−35−4pp−2p−1−6p−1p−45.=3p3−22p2+33p−54+k10p−3010p−1.Letfp=3p3−22p2+33p−54,p∈12,+∞; this function is strictly increasing in this interval; thus, fp≥f12=2358>0.Hence,(51)3p3−22p2+33p−54+k10p−3010p−1>0.(ix)
Note thatdGx+dGy=4, rGv2,vp=2p−2/p for x∈v2,vp in G and dG′x+dG′y=4, rG′v2,vp=2/3 for x∈v2,vp in G′; then, we have(52)dG′v2+dG′vprG′v2,vp−dGv2+dGvprGv2,vp=6−2pp−2=4p−12p−2>0.
Thus, by (i)–(ix), whenp≥12, we get RDRG′−RDRG>0.
When4≤p≤11, UnG0=Sn4,Sn5,Sn6,Sn7,Sn8,Sn9,Sn10,Sn11. By comparison of direct calculation, we can obtain RDRSn3>RDRSnp.
The proof is completed.


## 2.1. Edge-Lifting Transformation

LetG1 and G2 be two graphs with n1≥2 and n2≥2 vertices, respectively. G is the graph obtained from G1 and G2 by adding an edge between a vertex u0 of G1 to a vertex v0 of G2. And G′ is the graph obtained by identifying u0 of G1 to a vertex v0 of G2 and adding a pendant edge to u0v0. We say that G′ is obtained from G by an edge-lifting transformation at e=u0,v0 (see Figure 1).Figure 1
The edge-lifting transformation. (a)G. (b) G′.
(a)(b)Lemma 2.
IfG′ can be obtained from G by an edge-lifting transformation, then RDRG<RDRG′.Proof.
ConsiderG and G′ shown in Figure 1. By the definition of RDRG,(11)RDRG=∑x,y∈VG1\u0dGx+dGyrGx,y+∑x,y∈VG2\v0dGx+dGyrGx,y+∑x∈VG1\u0dGu0+dGxrGu0,x+∑x∈VG2\v0dGu0+dGxrGv0,x+1+∑x∈VG1\u0dGv0+dGxrGu0,x+1+∑x∈VG2\v0dGv0+dGxrGv0,x+∑x∈VG1\u0,y∈VG2\v0dGx+dGyrGx,u0+1+rGv0,y+dGu0+dGv0rGu0,v0,RDRG′=∑x,y∈VG1\u0dG′x+dG′yrG′x,y+∑x,y∈VG2\v0dG′x+dG′yrG′x,y+∑x∈VG1\u0dG′u0+dG′xrG′u0,x+∑x∈VG2\v0dG′v0+dG′xrG′v0,x+∑x∈VG1\u0dG′w0+dG′xrG′u0,x+1+∑x∈VG2\v0dG′w0+dG′xrG′v0,x+1+∑x∈VG1\u0,y∈VG2\v0dG′x+dG′yrG′x,u0+rG′v0,y+dG′u0+dG′w0rG′u0,w0.(i)
Note thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for x,y∈VG1\u0 or x,y∈VG2\v0, and dGu0+dGv0=dG′u0+dG′w0, rGu0,v0=rG′u0,w0; then, we have(12)∑x,y∈VG1\u0dGx+dGyrGx,y=∑x,y∈VG1\u0dG′x+dG′yrG′x,y,∑x,y∈VG2\v0dGx+dGyrGx,y=∑x,y∈VG2\v0dG′x+dG′yrG′x,y,dGu0+dGv0rGu0,v0=dG′u0+dG′w0rG′u0,w0.(ii)
Note thatdGx+dGy=dG′x+dG′y, rGx,u0+rGv0,y=rG′x,u0+rG′v0,y for x∈VG1\u0,y∈VG2\v0; then, we have(13)∑x∈VG1\u0,y∈VG2\v0dGx+dGyrGx,u0+1+rGv0,y<∑x∈VG1\u0,y∈VG2\u0dG′x+dG′yrG′x,u0+rG′v0,y.(iii)
Note thatrG′u0,x=rGu0,x for any x∈VG1\u0, and dG′u0=dGu0+dGv0−1; then, we have(14)∑x∈VG1\u0dG′u0+dG′xrG′u0,x−∑x∈VG1\u0dGu0+dGxrGu0,x+∑x∈VG1\u0dG′w0+dG′xrG′u0,x+1−∑x∈VG1\u0dGv0+dGxrGu0,x+1=∑x∈VG1\u0dGv0−1rGu0,x+∑x∈VG1\u01−dGv0rGu0,x+1>0.(iv)
Note thatrG′v0,x=rGv0,x, dG′x=dGx for any x∈VG2, and dG′v0=dGu0+dGv0−1; then, we have(15)∑x∈VG2\v0dG′v0+dG′xrG′v0,x−∑x∈VG2\v0dGv0+dGxrGv0,x+∑x∈VG2\v0dG′w0+dG′xrG′v0,x+1−∑x∈VG2\v0dGu0+dGxrGv0,x+1=∑x∈VG2\v0dGu0−1rGv0,x+∑x∈VG2\v01−dGu0rGv0,x+1>0.
Thus, by (i)–(iv), we getRDRG′−RDRG>0.


## 2.2. Cycle-Lifting Transformation

LetG be a graph as shown in Figure 2. Take a cycle C in G, say C=v1v2⋯vpv1; G can be viewed as a graph obtained by coalescing C with a number of star subgraphs of G, say G1,G2,…,Gp, by identifying vi with the center of Gi for all i1≤i≤p, denoted as EGi=si and VGi=si+1. Deleting all edges in Gi and joining v1 to all pendant vertices of Gi2≤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 C.Figure 2
The cycle-lifting transformation. (a)G. (b) G′.
(a)(b)Lemma 3.
IfG′ can be obtained from G by a cycle-lifting transformation, then RDRG<RDRG′.Proof.
ConsiderG and G′ shown in Figure 2. By the definition of RDRG,(16)RDRG=∑i=1p∑x,y∈VGi\vidGx+dGyrGx,y+∑i=1p∑x∈VGi\vidGvi+dGxrGvi,x+∑i,j=1i≠jp∑x∈VGi\viy∈VGj\vjdGx+dGyrGx,vi+rGvi,vj+rGvj,y+∑i=1p∑x∈VGi\viy∈VGj\vjdGx+dGyrGx,y+∑x,y∈VCpdGx+dGyrGx,y,RDRG′=∑i=1p∑x,y∈VGi\v1dG′x+dG′yrG′x,y+∑i=1p∑x∈VGi\v1dG′v1+dG′xrG′v1,x+∑i,j=1i≠jp∑x∈VGi\v1y∈VGj\v1dG′x+dG′yrG′x,v1+rG′v1,y+∑i=1p∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y+∑x,y∈VCpdG′x+dG′yrG′x,y.(i)
Note thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for x,y∈VGi\vi in G, and x,y∈VGi\v1 in G′; then, we have(17)∑i=1p∑x,y∈VGi\vidGx+dGyrGx,y=∑i=1p∑x,y∈VGi\v1dG′x+dG′yrG′x,y.(ii)
LetrGvi,vj=rG′vi,vj=rt; when j−i=t, then rt=rp−t. dGvi=si+2 for any vi∈VCp in G and dG′v1=s1+⋯+sp+2 for v1∈VCp in G′; then, we have(18)∑x,y∈VCpdGx+dGyrGx,y=∑i=2pdGv1+dGvirGv1,vi+∑i=3pdGv2+dGvirGv2,vi+⋯+dGvp−1+dGvprGvp−1,vp.Thus,(19)∑i=2pdGv1+dGvirGv1,vi=s1+s2+4r1+s1+s3+4r2+⋯+s1+sp+4rp−1,∑i=3pdGv2+dGvirGv2,vi=s2+s3+4r1+⋯+s2+sp+4rp−2,⋯dGvp−1+dGvprGvp−1,vp=sp−1+sp+4r1.Thus,(20)∑x,y∈VCpdGx+dGyrGx,y=s1+⋯+sp−1+s2+⋯+sp+4p−1r1+s1+⋯+sp−2+s3+⋯+sp+4p−2r2+⋯+s1+sp+4rp−1.Similarly,(21)∑x,y∈VCpdG′x+dG′yrG′x,y=s1+⋯+sp+4p−1r1+s1+⋯+sp+4p−2r2+⋯+s1+⋯+sp+4rp−1.Then,(22)∑x,y∈VCpdGx+dGyrGx,y−∑x,y∈VCpdG′x+dG′yrG′x,y=s2+⋯+sp−1r1+s3+⋯+sp−2r2+⋯+−s2+⋯+sp−1rp−1=0.(iii)
Note thatdGx+dGy=dG′x+dG′y, rGx,vi+rGvj,y=rG′x,v1+rG′v1,y for x∈VGi\vi,y∈VGj\vj in G and for x∈VGi\v1,y∈VGj\v1 in G′; then, we have(23)∑i,j=1i≠jp∑x∈VGi\viy∈VGj\vjdGx+dGyrGx,vi+rGvi,vj+rGvj,y<∑i,j=1i≠jp∑x∈VGi\v1y∈VGj\v1dG′x+dG′yrG′x,v1+rG′v1,y.(iv)
Note thatrGvi,x=rG′v1,x=1, dGvi+dGx=si+3 for x∈VGi\vi in G, and dG′v1+dG′x=s1+⋯+sp+3 for x∈VGi\v1 in G′; then, we have(24)∑i=1p∑x∈VGi\vidGvi+dGxrGvi,x=s1s1+3+s2s2+3+⋯+spsp+3=∑i=1psisi+3,∑i=1p∑x∈VGi\v1dG′v1+dG′xrG′v1,x=s1s1+⋯+sp+3+⋯+sps1+⋯+sp+3=∑i=1psis1+⋯+sp+3.Thus, we have(25)∑i=1p∑x∈VGi\v1dG′v1+dG′xrG′v1,x−∑i=1p∑x∈VGi\vidGvi+dGxrGvi,x=∑i,j=1i≠jpsi⋅sj.(v)
Note thatdGx+dGy=3+si,dG′x+dG′y=3, rGx,y=rG′x,y for any x∈VCp\vi,y∈VGi\vi in G and for x∈VCp\v1,y∈VGi\v1 in G′; when si+t>sp1≤t≤p−1, then si+t=si+t−p; we have(26)∑i=1p∑x∈VCp\viy∈VGi\vidGx+dGyrGx,y=∑x∈VCp\v1y∈VG1\v1dGx+dGyrGx,y+∑x∈VCp\v2y∈VG1\v2dGx+dGyrGx,y+⋯+∑x∈VCp\vpy∈VGp\vpdGx+dGyrGx,y.
Thus,(27)∑x∈VCp\v1y∈VG1\v1dGx+dGyrGx,y=s13+s2r1+1+s13+s3r2+1+⋯+s13+sprp−1+1,∑x∈VCp\v2y∈VG1\v2dGx+dGyrGx,y=s23+s3r1+1+s23+s4r2+1+⋯+s23+s1rp−1+1,⋯∑x∈VCp\vpy∈VGp\vpdGx+dGyrGx,y=sp3+s1r1+1+sp3+s2r2+1+⋯+sp3+sp−1rp−1+1,and we have(28)∑i=1p∑x∈VCp\vjy∈VGp\vjdGx+dGyrGx,y=∑i=1psi3+si+1modpr1+1+∑i=1psi3+si+2modpr2+1+⋯+∑i=1psi3+si+p−1modprp−1+1.
Thus,(29)∑i=1p∑x∈VCp\viy∈VGp\vidGx+dGyrGx,y=∑i,j=1i≠jpsi⋅3+sjrGvi,vj+1.
Similarly,(30)∑i=1p∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y+∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y+⋯+∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y,∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=3s1r1+1+3s1r2+1+⋯+3s1rp−1+1,∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=3s2r1+1+3s2r2+1+⋯+3s2rp−1+1,⋯∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=3spr1+1+3spr2+1+⋯+3sprp−1+1.
Thus, we have(31)∑i=1p∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y=∑i=1p3sir1+1+∑i=1p3sir2+1+⋯+∑i=1p3sirp−1+1.
Then,(32)∑i=1p∑x∈VCp\v1y∈VGi\v1dG′x+dG′yrG′x,y−∑i=1p∑x∈VCp\viy∈VGi\vidGx+dGyrGx,y=−∑i,j=1i≠jpsi⋅sjrGvi,vj+1.
Thus, by (i)–(v), we getRDRG′−RDRG>0.


## 2.3. Cycle-Shrinking Transformation

Denote bySnp the unicyclic graph obtained from cycle Cp by attaching n−p pendant edges to a vertex v1 of Cp (see Figure 3). Let G=Snpp≥4; deleting the edges v2v3,…,vp−1vp and adding the edges v2vp,v3v1,…,vp−1v1, we obtain the graph G′=Sn3. This operation is called cycle-shrinking transformation. Denote by WW′,resp the set of pendant vertices of GG′,resp. Let W=k, W′=k′. It is clear that p+k=n and 3+k′=n. Then, k′=k+p−3; thus, W′ can be partitioned into two subsets. One has k vertices, which is naturally corresponding to W. So, we also denote it by W; another has p−3 vertices, denoted by W˜.Figure 3
The cycle-shrinking transformation. (a)G. (b) G′.
(a)(b)Lemma 4.
LetG be an unicyclic graph of order n and a cycle Cp with p≥4; if G′ can be obtained from G by a cycle-shrinking transformation, then RDRG<RDRG′.Proof.
ConsiderG and G′ shown in Figure 3. By the definition of RDRG,(33)RDRG=∑x,y∈WdGx+dGyrGx,y+∑x∈WdGv1+dGxrGv1,x+∑x∈Wy∈v2,vpdGx+dGyrGx,y+∑x∈Wy∈VCp\v1,v2,vpdGx+dGyrGx,y+∑x∈v2,vpdGv1+dGxrGv1,x+∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,x+∑x∈VCp\v1,v2,vpy∈v2,vpdGx+dGyrGx,y+∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y+dGv2+dGvprGv2,vp,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∈W,y∈W˜dG′x+dG′yrG′x,y+∑x∈v2,vpdG′v1+dG′xrG′v1,x+∑x∈W˜dG′v1+dG′xrG′v1,x+∑x∈Wy∈v2,vpdG′x+dG′yrG′x,y+∑x,y∈W˜dG′x+dG′yrG′x,y+dG′v2+dG′vprG′v2,vp.(i)
Note thatdGx+dGy=dG′x+dG′y, rGx,y=rG′x,y for any x,y∈W; then, we have(34)∑x,y∈WdGx+dGyrGx,y=∑x,y∈WdG′x+dG′yrG′x,y.(ii)
Note 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′; then, we have(35)∑x∈Wy∈v2,vpdG′x+dG′yrG′x,y−∑x∈Wy∈v2,vpdGx+dGyrGx,y≥0.(iii)
Note thatdGv1+dGx=k+3, dG′v1+dG′x=k+p, rGv1,x=rG′v1,x=1, for x∈W in G and G′, respectively; then, we have(36)∑x∈WdG′v1+dG′xrG′v1,x−∑x∈WdGv1+dGxrGv1,x=kk+p−kk+3=kp−3.(iv)
Note 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˜; then, we have(37)∑x∈Wy∈W˜dG′x+dG′yrG′x,y−∑x∈Wy∈VCp\v1,v2,vpdGx+dGyrGx,y≥kp−3−3kp−32=−kp−32.Combingiii−iv, we have(38)∑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.(v)
Note thatdGv1+dGx=k+4,rGv1,x=p−1/p, and dG′v1+dG′x=k+p+1,rG′v1,x=2/3, for x∈v2,vp in G and G′, respectively; then, we have(39)∑x∈v2,vpdG′v1+dG′xrG′v1,x−∑x∈v2,vpdGv1+dGxrGv1,x=3k+p+1−2pk+4p−1=k+3p+1p−3p−1.(vi)
Note thatdGv1+dGx=k+4,rGv1,x=rGv1,vi≥13≤i≤p−1 for x∈VCp\v1,v2,vp and dG′v1+dG′x=k+p,rG′v1,x=1 for x∈W˜; then, we have(40)∑x∈W˜dG′v1+dG′xrG′v1,x−∑x∈VCp\v1,v2,vpdGv1+dGxrGv1,x≥p−3k+p−p−3k+4=p−3p−4.(vii)
Note thatdGx+dGy=3 for x∈VCp\v1,v2,vp, y∈v2,vp and dG′x+dG′y=4, rG′x,y=rG′v2,v1+1=2/3+1=5/3 for x∈W˜, y∈v2,vp; by the symmetry between v2 and vp, we have(41)∑x∈W˜y∈v2,vpdG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈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 2p−2/p≥5/3, then p≥12.Then,(42)2∑i=4p−14rGv2,vi≤24p−45.Thus,(43)∑x∈W˜y∈v2,vpdG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈v2,vpdGx+dGyrGx,y≥18p−35−8pp−1+24p−45.(viii)
Note thatdGx+dGy=4, rGx,y=rGvi,vj3≤i<j≤p−1 for x,y∈VCp\v1,v2,vp and dG′x+dG′y=2, rG′x,y=2 for x,y∈W˜; then, we have(44)∑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.Then,(45)∑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, rGvi,vj=rt=rp−t; thus,(46)∑i=4pdGv3+dGvirGv3,vi=4r1+4r2+⋯+4rp−4,∑i=5pdGv4+dGvirGv4,vi=4r1+⋯+4rp−5,⋯dGvp−2+dGvp−1rGvp−2,vp−1=4r1,and(47)∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y=4p−4r1+4p−5r2+⋯+4rp−4.Sincer1=p−1/p, rt≥r2=2p−2/p2≤t≤p−4, if 2p−2/p≥5/3, then p≥12.Then,(48)∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y≤4pp−4p−1+6p−4p−55.Thus,(49)∑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 v−viii, we have(50)∑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∈W˜y∈v2,vpdG′x+dG′yrG′x,y−∑x∈VCp\v1,v2,vpy∈v2,vpdGx+dGyrGx,y+∑x,y∈W˜dG′x+dG′yrG′x,y−∑x,y∈VCp\v1,v2,vpdGx+dGyrGx,y≥k+3p+1p−3p−1+3p−3p−42+18p−35−4pp−2p−1−6p−1p−45.=3p3−22p2+33p−54+k10p−3010p−1.Letfp=3p3−22p2+33p−54,p∈12,+∞; this function is strictly increasing in this interval; thus, fp≥f12=2358>0.Hence,(51)3p3−22p2+33p−54+k10p−3010p−1>0.(ix)
Note thatdGx+dGy=4, rGv2,vp=2p−2/p for x∈v2,vp in G and dG′x+dG′y=4, rG′v2,vp=2/3 for x∈v2,vp in G′; then, we have(52)dG′v2+dG′vprG′v2,vp−dGv2+dGvprGv2,vp=6−2pp−2=4p−12p−2>0.
Thus, by (i)–(ix), whenp≥12, we get RDRG′−RDRG>0.
When4≤p≤11, UnG0=Sn4,Sn5,Sn6,Sn7,Sn8,Sn9,Sn10,Sn11. By comparison of direct calculation, we can obtain RDRSn3>RDRSnp.
The proof is completed.


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

LetG be a connected graph with exactly one cycle, say Cp=v1v2⋯vpv1. G can be viewed as a graph obtained by coalescing Cp with a number of trees, denoted by G1,G2,…,Gp, by identifying vi with some vertex of Gi for all i1≤i≤p. Denote Sn3 by the unicyclic graph obtained from cycle C3 by attaching n−3 pendant edges to a vertex of C3. The next theorem discusses the maximum reciprocal degree resistance distance index in Un, which is the main result of our work.Theorem 1.
LetG be a unicyclic graph of order n; then, RDRG≤RDRSn3, and the equality holds if and only if G≅Sn3.Proof.
LetG be a unicyclic graph of order n such that ξG is as big as possible. By Lemmas 2–4, we have RDRG≤RDRSn3; the equality holds if and only if G≅Sn3.



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