# On the Reformulated Second Zagreb Index of Graph Operations

Authors: Durbar Maji; Ganesh Ghorai; Yaé Ulrich Gaba
Journal: Journal of Chemistry
 (2021)

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## Abstract

Topological indices (TIs) are expressed by constant real numbers that reveal the structure of the graphs in QSAR/QSPR investigation. The reformulated second Zagreb index (RSZI) is such a novel TI having good correlations with various physical attributes, chemical reactivities, or biological activities/properties. The RSZI is defined as the sum of products of edge degrees of the adjacent edges, where the edge degree of an edge is taken to be the sum of vertex degrees of two end vertices of that edge with minus 2. In this study, the behaviour of RSZI under graph operations containing Cartesian product, join, composition, and corona product of two graphs has been established. We have also applied these results to compute RSZI for some important classes of molecular graphs and nanostructures.


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## Body

## 1. Introduction

In the whole study, we only consider the molecular graph [1, 2], a graphical representation of molecular structure, in which every vertex corresponds to the atoms and the edges to the bonds between them. Assume J as a simple (molecular) graph with VJ vertex set and EJ edge set. The notations VJ and EJ represent the number of elements of J in VJ and EJ, respectively. Also, dJx denotes the degree of a vertex (x) in J and is defined as the number of edges incident to x.TIs can be expressed by real numbers related to graphs. There exist many applications as tools for modelling chemical and other properties of molecules for TIs. They determine the correlation between the specific properties of molecules and the biological activity with their configuration in the study of quantitative structure-activity relationships (QSARs) and quantitative structure-property relationships (QSPRs) [3]. To develop the scientific knowledge in 20th century, the concept of molecular structure plays an important role in chemical graph theory, a branch of mathematical chemistry which is closely related to chemical graph. The molecular structure descriptor, namely, topological index expresses the numerical value obtained from the molecular graph that represents its topology and is necessarily invariant under the automorphism of graphs.The Zagreb indices, namely, first Zagreb index [4] and second Zagreb index [5] were introduced by Gutman et al. in 1972 and 1975, respectively. These two indices are, respectively, defined for molecular graph (J) as(1)M1J=∑x∈VJdJ2x=∑xy∈EJdJx+dJy,(2)M2J=∑xy∈EJdJxdJy.Letx, y, and z be its three vertices of J forming a path of length two. If x and y are two adjacent vertices of J, we express as x∼y and similarly for y and z as y∼z. The edge connecting these vertices will be denoted by xy∼yz. Then, the three auxiliary Zagreb-type indices are introduced by Basavanagoud et al. [6] in 2015. They are denoted as follows:(3)M3J=∑xy∼yz∈EJdJ2y,(4)M4J=∑xy∼yz∈EJdJx+2dJy+dJz,(5)M5J=∑xy∼yz∈EJdJxdJy+dJydJz+dJzdJx.In 2004, Milicevic et al. [7] reformulated the Zagreb indices by replacing vertex degree with edge degree, and the edge degree of an edge is defined as de=dx+dy−2. The first and second reformulated Zagreb indices [8] of a graph J are defined as(6)EM1J=∑e∈EJd2e,(7)EM2J=∑e∼fdedf,where e∼f means that the edges e and f share a common end vertex is, and e and f are adjacent. In 2015, Furtula and Gutman [9] introduced forgotten index (F-index) and is defined as(8)FJ=∑x∈VJdJ3x=∑xy∈EJdJ2x+dJ2y.In mathematical chemistry, graph operations perform a significant role in the formation of new classes of graphs. By different graph operations on some general or particular graphs, some chemically interesting graphs can be obtained. In [10], Khalifeh et al. computed the first and second Zagreb indices under some graph operations. Some explicit formulae of Zagreb coindices under some graph operations were presented by Ashrafi et al. [11]. In [12], Das et al. derived multiplicative Zagreb indices of different graph operations. In [13], De et al. computed the reformulated first Zagreb index under some graph operations. Recently, the analytical expressions for various topological indices under some binary graph operations have been discussed in [14–16]. We also refer to [17–23] in this regard for interested readers.


## 2. Main Results

In the following, we study different binary graph operations such as join, Cartesian product, composition, and corona product of two molecular graphs and compute some exact formulae for RSZI with respect to those operations separately. SupposeJ1 and J2 be two molecular graphs with the vertex sets VJ1, VJ2, such that VJ1=n1, VJ2=n2, and the edge sets EJ1, EJ2, such that EJ1=m1, EJ2=m2, respectively. We consider the notations Pn, Cn, and Kn for path, cycle, and complete graph with n vertices. To establish the main results, we also follow equations (1)–(8).


### 2.1. Join

The join [24] of two graphs J1 and J2, denoted by J1+J2, contains the vertex set VJ1+J2=VJ1∪VJ2 and the edge set EJ1+J2=EJ1∪EJ2∪xy:x∈VJ1,y∈VJ2.In the following theorem, we compute RSZI for join of two graphs.Theorem 1.
LetJ1+J2=J be the join of J1 and J2 graphs. Then, RSZI of J is given by(9)EM2J=EM2J1+EM2J2+2VJ2M4J1+2VJ1M4J2+2VJ2VJ2−2M1J1−2EJ1+4EJ1M1J2+4EJ2M1J1+2VJ1VJ1−2M1J2−2EJ2+VJ1FJ2+VJ2FJ1+2VJ12|VJ1+VJ2−3M1J2+2VJ2VJ1+2VJ2−3M1J1+VJ224EJ12−M1J1+VJ124EJ22−M1J2+CVJ12M1J2+CVJ22M1J1+12VJ1VJ2VJ1+VJ2−23+12EJ1EJ2VJ1+VJ2−2+2VJ1M2J2+2VJ2M2J1+2VJ1EJ2VJ1+VJ2−23VJ1+VJ2−4+2VJ2EJ1VJ1+VJ2−2VJ1+3VJ2−4.Proof.
ConsiderdJx+dJy−2dJy+dJz−2=X. Let J1 and J2 be two graphs with VJ1, VJ2 vertices and EJ1, EJ2 edges, respectively. Then, by Table 1, we obtain(10)EM2J=∑xy∼yz∈EJX=∑xy∼yz∈EJ1X+∑xy∼yz∈EJ2X+∑x∈VJ1,y∈VJ2,yz∈EJ2X+∑x∈VJ2,y∈VJ1,yz∈EJ1X+∑x,z∈VJ1,y∈VJ2X+∑x,z∈VJ2,y∈VJ1X=D1+D2+D3+D4+D5+D6say,respectively.
First part:(11)D1=∑L1:xy∼yz∈EJ1X=∑L1dJ1x+dJ1y+2VJ2−2dJ1y+dJ1z+2VJ2−2=∑xy∼yz∈EJ1dJ1x+dJ1y−2dJ1y+dJ1z−2+4∑xy∼yz∈EJ1VJ22+2VJ2∑xy∼yz∈EJ1dJ1x+2dJ1y+dJ1z−8∑xy∼yz∈EJ1VJ2=M2LJ1+4VJ2212M1J1−EJ1+2VJ2M4J1−8VJ212M1J1−EJ1=EM2J1+2VJ2M4J1+2VJ2VJ2−2M1J1−2EJ1.
Second part:(12)D2=∑L2:xy∼yz∈EJ2X=∑L2dJ2x+dJ2y−2+2VJ1dJ2y+dJ2z−2+2VJ1=∑xy∼yz∈EJ2dJ2x+dJ2y−2dJ2y+dJ2z−2+4∑xy∼yz∈EJ2VJ12+2VJ1∑xy∼yz∈EJ2dJ2x+2J2y+dJ2z−8∑xy∼yz∈EJ2VJ1=M2LJ2+4VJ1212M1J2−EJ2+2VJ1M4J1−8VJ112M1J2−EJ2=EM2J2+2VJ1M4J2+2VJ1VJ1−2M1J2−2EJ2.
Third part:(13)D3=∑L3:x∈VJ1,y∈VJ2,yz∈EJ2X=∑L3dJ1x+dJ2y+VJ1+VJ2−2dJ2y+dJ2z+2VJ1−1=2∑x∈VJ1dJ1x∑yz∈EJ2dJ2y+dJ2z+4VJ1−1∑yz∈EJ2∑x∈VJ1dJ1x+∑x∈VJ1∑yz∈EJ2dJ22y+dJ22z+2∑x∈VJ1∑yz∈EJ2dJ2ydJ2z+3VJ1+VJ2−4∑x∈VJ1∑yz∈EJ2dJ2y+dJ2z+VJ1+VJ2−2∑x∈VJ1∑yz∈EJ2dJ2y+dJ2z+4VJ1−1VJ1+VJ2−2∑x∈VJ1∑yz∈EJ21=4EJ1M1J2+8EJ1EJ2VJ1−1+VJ1FJ2+2VJ1M2J2+2VJ12VJ1+VJ2−3M1J2+4EJ2VJ1VJ1−1VJ1+VJ2−2.
Fourth part:(14)D4=∑L4:x∈VJ2,y∈VJ1,yz∈EJ1X=∑L4dJ2x+dJ1y+VJ1+VJ2−2dJ1y+dJ1z+2VJ2−1=2∑x∈VJ2dJ2x∑yz∈EJ1dJ2y+dJ2z+4VJ2−1∑yz∈EJ1∑x∈VJ2dJ2x+∑x∈VJ2∑yz∈EJ1dJ12y+dJ12z+∑x∈VJ2∑yz∈EJ1dJ1ydJ1z+3VJ2+VJ1−4∑x∈VJ2∑y∈VJ1dJ1y+VJ1+VJ2−2∑x∈VJ2∑z∈VJ1dJ1z+4∑x∈VJ2∑yz∈EJ1VJ1−1VJ1+VJ2−2=4EJ2M1J1+8EJ1EJ2VJ2−1+VJ2FJ1+2VJ2M2J1+2VJ2VJ1+2VJ2−3M1J1+4EJ1VJ2VJ2−1VJ1+VJ2−2.
Fifth part:(15)D5=∑x,z∈VJ1,y∈VJ2X=∑x,z∈VJ1,y∈VJ2dJ1x+dJ2y+VJ1+VJ2−2dJ1y+dJ2z+VJ1+VJ2−2=∑y∈VJ2∑xizj∈CVJ1xizjdJ1xidJ1zj+∑y∈VJ2dJ2y∑xizj∈CVJ1xizjdJ1xi+dJ1zj+∑xizj∈CVJ1xizj∑y∈VJ2dj22y+VJ1+VJ2−2∑y∈VJ2∑xizj∈CVJ1xizjdJ1xi+dJ1zj+2VJ1+VJ2−2∑xizj∈CVJ1xizj∑y∈VJ2dJ2y+VJ1+VJ2−22∑xizj∈CVJ1xizj∑y∈VJ21=VJ224EJ12−M1J1+4VJ1−1|EJ1||EJ2|+CVJ12M1J2+2VJ2VJ1−1VJ1+VJ2−2EJ1+4CVJ12EJ2VJ1+VJ2−2+CVJ12VJ2VJ1+VJ2−22.
Sixth part:(16)D6=∑L6:x,z∈VJ2,y∈VJ1X=∑L6dJ2u+dJ1v+VJ1+VJ2−2dJ1v+dJ2w+VJ1+VJ2−2=∑y∈VJ1∑xizj∈CVJ2xizjdJ2xidJ2zj+∑y∈VJ1dJ1y∑xizj∈CVJ2xizjdJ2xi+dJ2zj+∑xizj∈CVJ2xizj∑y∈VJ1dJ12y+VJ1+VJ2−2∑y∈VJ1∑xizj∈CVJ2xizjdJ2xi+dJ2zj+2VJ1+VJ2−2∑xizj∈CVJ2xizj∑y∈VJ1dJ1y+VJ1+VJ2−22∑xizj∈CVJ2xizj∑y∈VJ11=VJ124EJ22−M1J2+4VJ2−1EJ2EJ1+CVJ22M1J1+2EJ2VJ1VJ2−1VJ2+VJ1−2+4CVJ22VJ2+VJ1−2EJ1+CVJ22|VJ1|VJ1+VJ2−22.
By adding∑i=16Di, we get the desired result.Table 1
The degree distribution of joinJ=J1+J2.
x∈VJ1x∈VJ2dJ1x+VJ2dJ2x+VJ1


### 2.2. Applications

The suspension of a graph H is the join or sum of H with a single vertexK1.Corollary 1.
The RSZI of suspension of H that containVJ=n and EJ=m can be expressed as EM2H+K1=EM2H+2nM4H+FH+2M2H+1/24n−9M1H+2m2+4m+2mn−12+n/2n−13.Example 1.
The cone graphCm,n is defined as Cm+Kn. Then, EM2Cm,n=2m11n2+5mn+2+1/2mnm+n−2m+n2+42n−3.Example 2.
The RSZI of suspension of graph such asCn,K¯n−1,Pn,mK2 are expressed in Table 2.Table 2
The results are calculated using Corollary1.
Suspension ofNamelyJoin ofEM2 valuesCnWheel graphWnCn+K1n/2n3+n2+47n−15K¯n−1Star graphSnK¯n−1+K11/2n−1n−23PnFan graphFnPn+K11/2n+4n−13+26n2−46n−17mK2Flower graphmK2+K14m22m2−m+2Example 3.
The complete bipartite graphKp,q is the join of Kp+Kq. Using Theorem 1, EM2Kp,q=1/2pqp+q−23.


### 2.3. The Cartesian Product

The Cartesian product (CP) [25] of J1 and J2, denoted by J1×J2, is a graph with VJ1×J2=VJ1×VJ2 and any f,pg,q∈EJ1×J2 if and only if fg∈EJ1,p=q or pq∈EJ1,f=g. Also, dJ1×J2fi,pj=dJ1fi+dJ2pj;i=1,2,…,VJ1 and j=1,2,…,VJ2.Now, we obtain RSZI for Cartesian product of two graphs.Theorem 2.
IfJ1×J2=J be the CP of J1 and J2 graphs, then RSZI of J is(17)EM2J=VJ2EM2J1+VJ1EM2J2+4EJ2M4J1+4EJ1M4J2+4M1J1M1J2+4EJ2FJ1+2M2J1−2M1J1+4EJ1FJ2+2M2J2−2M1J2+4M1J1−2EJ1M1J2−3EJ2+4M1J2−2EJ2M1J1−3EJ1.Proof.
By definition of RSZI, from equation (7) and the degree distribution for CP of two graphs, we have(18)EM2J=∑f,pg,q∼g,qh,r∈EJdJf,p+dJg,q−2dJg,q+dJh,r−2=∑f∈VJ1,pq∼qr∈EJ2dJf,p+dJf,q−2dJf,q+dJf,r−2+∑fg∼gh∈EJ1,p∈VJ2dJf,p+dJg,p−2dJg,p+dJh,p−2+∑fg∈EJ1,pq∈EJ2dJf,p+dJf,q−2dJf,q+dJg,q−2=U1+U2+U3say.
The notationsU1,U2, and U3 represent the sum of above terms in order.Step 1:(19)U1=∑f∈VJ1,pq∼qr∈EJ2dJf,p+dJf,q−2dJf,q+dJf,r−2=∑f∈VJ1∑pq∼qr∈EJ22dJ1f+dJ2p+dJ2q−22dJ1f+dJ2q+dJ2r−2=∑f∈VJ1∑pq∼qr∈EJ2dJ2p+dJ2q−2dJ2q+dJ2r−2+4∑pq∼qr∈EJ2∑f∈VJ1dJ1f2+2∑f∈VJ1dJ1f∑pq∼qr∈EJ2dJ2p+2dJ2q+dJ2z−8∑pq∼qr∈EJ2∑f∈VJ1dJ1f=VJ1M2LJ2+4EJ1M4J2+412M1J2−EJ2M1J1−1612M1J2−EJ2EJ1=VJ1EM2J2+4EJ1M4J2+2M1J2−2EJ2M1J1−4EJ1.Step 2:(20)U2=∑fg∼gh∈EJ1,p∈VJ2dJf,p+dJg,p−2dJg,p+dJh,p−2=∑p∈VJ2∑fg∼gh∈EJ12dJ2J+dJ1f+dJ1g−22dJ2p+dJ1g+dJ1h−2=∑p∈VJ2∑fg∼gh∈EJ1dJ1f+dJ1g−2dJ1g+dJ1h−2+2∑p∈VJ2dJ2p∑fg∼gh∈EJ1dJ1f+2dJ1g+dJ1h+4∑fg∼gh∈EJ1∑p∈VJ2dJ22p−8∑fg∼gh∈EJ1∑p∈VJ2dJ2p=VJ2M2LJ1+4EJ2M4J1+412M1J1−EJ1M1J2−812M1J1−EJ12EJ2=VJ2EM2J1+4EJ2M4J1+2M1J1−2EJ1M1J2−4EJ2.Step 3:(21)U3=∑fg∈EJ1,pq∈EJ2dJf,p+dJf,q−2dJf,q+dJg,q−2=∑fg∈EJ1∑pq∈EJ22dJ1f+dJ2p+dJ2q−22dJ2q+dJ1f+dJ1g−2=4∑fg∈EJ1∑pq∈EJ2dJ1fdJ2q+2∑fg∈EJ1∑pq∈EJ2dJ2p+dJ2q−2dJ2q+2∑pq∈EJ2∑fg∈EJ1d1fdJ1f+dJ1g−2+∑fg∈EJ1∑pq∈EJ1dJ2p+dJ2q−2dJ1f+dJ1g−2=4∑fg∈EJ1dJ1f+dJ1g∑pq∈EJ2dJ2p+dJ2q+4∑fg∈EJ1∑pq∈EJ2dJ22p+dJ22q+8∑fg∈EJ1∑pq∈EJ2dJ2pdJ2q−8∑fg∈EJ1∑pq∈EJ2dJ2p+dJ2q+4∑pq∈EG2∑fg∈EJ1dJ12f+dJ12g+8∑pq∈EJ2∑fg∈EJ1dJ1fdJ1g−8∑pq∈EJ2∑fg∈EJ1dJ1f+dJ1g+4∑fg∈EJ1dJ1f+dJ1g−2∑pq∈EJ2dJ2x+dJ2y−2=4M1J1M1J2+4EJ1FJ2+2M2J2−2M1J2+4EJ2FJ1+2M2J1−2M1J1+4M1J1−2EJ1M1J2−2EJ2.By addingU1,U2,U3, we get the required result.


### 2.4. Applications

LetP, Q, R, and S be the grids Pn×Pm=P, rook’s graph Kn×Km=Q, C4-nanotorus TC4n,m=Cn×Cm=R, and C4-nanotube TUC4n,m=Pn×Cm=S. Then, by Theorem 2, we get the following results.Example 4.
The RSZI for P is given byEM2P=454mn−89m+n+132, for m,n≥4.Example 5.
The RSZI forQ is given by EM2Q=2mnm−1m−22mn+3m−2n−4+n−1n−22mn+3n−2m−4+n−1m−16mn−11m−11n+22, for m,n≥2.Example 6.
The RSZI forR is given by EM2R=216mn.Example 7.
The RSZI for S is given byEM2S=4m54n−89.


### 2.5. Lexicographic Product

The lexicographic product (LP) or composition [26] of two graphs J1 and J2 is denoted by J1J2, and any two vertices u1,u2 and v1,v2 are adjacent if and only if u1,v1∈EJ1 or u1=v1 and u2v2∈EJ2. The vertex set of J1J2 is VJ1×VJ2, and the degree of a vertex a,b∈J1J2 is given by dJ1J2a,b=VJ2dJ1a+dJ2b.In the following theorem, we compute RSZI for composition of two graphsJ1 and J2.Theorem 3.
LetJ1J2=J be the composition of J1 and J2. The RSZI of J is given by(22)EM2J=VJ1EM2J2+VJ25M5J1+VJ25M3J1+4EJ1VJ2M4J2+VJ23VJ22−VJ2+4EJ2FJ1+2M2J1+2EJ1VJ2FJ2+2M2J2+172VJ22M1J1M1J2+2VJ232EJ2−VJ2M4J1+2VJ24EJ2VJ22−2VJ23+3VJ22−18EJ2VJ2+7EJ22M1J1−2EJ111VJ2−4EJ2M1J2−4EJ1VJ22−12EJ2VJ2+6EJ22.Proof.
By using the definition of RSZI and from the equation (7), we have(23)EM2J=∑p,xq,y∼q,yr,z∈EJdJp,x+dJq,y−2dJq,y+dJr,z−2=∑p∈VJ1∑xy∼yz∈EJ2dJp,x+dJp,y−2dJp,y+dJp,z−2+∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJp,x+dJp,y−2dJp,y+dJq,z−2+∑pq∼qr∈EJ1∑x∈VJ2∑y∈VJ2∑z∈VJ2dJp,x+dJq,y−2dJq,y+dJr,z−2+∑pq∈EJ1∑x∈VJ2∑y∈VJ2dJp,x+dJq,x−2dJq,x+dJp,y−2+∑xz∈CVJ2−yxz∑y∈VJ2∑pq∈EJ1dJp,x+dJq,y−2dJq,y+dJp,z−2=T1+T2+T3+T4+T5say,respectively.
Now,(24)T1=∑p∈VJ1∑xy∼yz∈EJ2dJp,x+dJp,y−2dJp,y+dJp,z−2=∑p∈VJ1∑xy∼yz∈EJ2dJp,x+dJp,y−2dJp,y+dJp,z−2=∑p∈VJ1∑xy∼yz∈EJ22VJ2dJ2p+dJ2x+dJ2y−22VJ2dJ2p+dJ2y+dJ2z−2=∑p∈VJ1∑xy∼yz∈EJ2dJ2x+dJ2y−2dJ2y+dJ2z−2+2VJ2∑p∈VJ1dJ1p∑xy∼yz∈EJ2dJ2x+2dJ2y+dJ2z+4VJ22∑xy∼yz∈EJ2∑p∈VJ1dJ12p−8VJ2∑xy∼yz∈EJ2∑p∈VJ1dJ1p=VJ1M2LJ2+4EJ1VJ2M4J2+4|VJ2212M1J2−EJ2M1J1−16VJ212M1J2−EJ2EJ1=VJ1EM2J2+4EJ1VJ2M4J2+2VJ2VJ2M1J1−4EJ1M1J2−2EJ2.
Next,(25)T2=∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJp,x+dJp,y−2dJp,y+dJq,z−2=∑pq∈EJ1∑xy∈EJ2∑z∈VJ22VJ2dJ1p+dJ2x+dJ2y−2VJ2dJ1p+dJ1q+dJ2y+dJ2z−2=2VJ2∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJ1pdJ2y+dJ2z+∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJ2x+dJ2y−2dJ2y+dJ2z+2VJ2∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJ1pVJ2dJ1p+VJ2dJ1q−2+∑z∈VJ2∑xy∈EJ2dJ2x+dJ2y−2∑pq∈EJ1VJ2dJ1p+dJ1q−2=2VJ2∑pq∈EJ1dJ1p+dJ1q∑xy∈EJ2VJ2dJ2x+dJ2y+4EJ2+2∑pq∈EJ1∑xy∈EJ2dJ2x+dJ2y−2VJ2dJ2x+dJ2y+4EJ2+4VJ2∑z∈VJ2∑xy∈EJ2∑pq∈EJ1dJ1p+dJ1qVJ2dJ1p+dJ1q−2+∑z∈VJ2∑xy∈EJ2dJ2x+dJ2y−2∑pq∈EJ1VJ2dJ1p+dJ1q−2=2VJ2M1J1VJ2M1J2+4EJ22+2EJ1VJ2FJ2+4EJ12EJ2−VJ2M1J2−16EJ22EJ1+4EJ1VJ2M2J2+4EJ2VJ23FJ1+8EJ2VJ23M2J1−8EJ2VJ22M1J1+4VJ2M1J2−2EJ2VJ2M1J1−2EJ1,T3=∑Ω:pq,qr∈EJ1∑α:x∈VJ2∑β:y∈VJ2∑γ:z∈VJ2dJp,x+dJq,y−2dJq,y+dJr,z−2=∑Ω∑α∑β∑γVJ2dJ1p+dJ1q+dJ2x+dJ2y−2VJ2dJ1p+dJ1r+dJ2y+dJ2z−2=VJ22∑α∑β∑γ∑ΩdJ1p+dJ1qdJ1q+dJ1r+VJ2∑ΩdJ1p+dJ1q∑α∑β∑γdJ2y+dJ2z−2+VJ2∑ΩdJ1q+dJ1r∑γ∑β∑αdJ2x+dJ2y−2+∑Ω∑β∑α∑γdJ2x+dJ2y−2dJ2y+dJ2z−2=VJ24∑y∈VJ2∑pq∼qr∈EJ1dJ1pdJ1q+dJ1qdJ1r+dJ1pdJ1r+dJ12q+VJ22∑ΩdJ1p+dJ1q∑α∑γ∑βdJ2y+∑β∑γdJ2z−∑γ∑β2+VJ22∑ΩdJ1q+dJ1r∑γ∑β∑αdJ2x+∑α∑βdJ2y−∑α∑β2+∑Ω∑γ∑α∑βdJ2y−22+∑βdJ2y−2∑γ∑αdJ2x+∑α∑γdJ2z+∑Ω∑β∑αdJ2x∑γdJ2z=VJ24∑β∑ΩdJ1pdJ1q+dJ1qdJ1r+dJ1pdJ1r+VJ24∑β∑ΩdJ12q+VJ22∑ΩdJ1p+2dJ1q+dJ1r∑α∑γ∑βdJ2y+∑β∑γdJ2z−∑γ∑β2+∑Ω∑γ∑α∑βdJ2y−22+∑βdJ2y−2∑γ∑αdJ2x+∑α∑γdJ2z+∑Ω∑β∑αdJ2x∑γdJ2z=VJ25M5J1+VJ25M3J1+2VJ232EJ2−VJ2M4J1+12M1J1−EJ1VJ22M1J2−8EJ2+4VJ2+2EJ2−VJ24EJ2VJ2+4EJ22VJ2=VJ25M5J1+VJ25M3J1+2VJ232EJ2−VJ2M4J1+VJ22M1J2+12VJ2EJ22−16VJ22EJ2+4VJ2312M1J1−EJ1.
For(26)T4=∑pq∈EJ1∑x∈VJ2∑y∈VJ2dJp,x+dJq,x−2dJq,x+dJp,y−2=∑pq∈EJ1∑x∈VJ2∑y∈VJ2VJ2dJ1p+dJ1q+2dJ2x−1VJ2dJ1q+dJ1r+dJ2x+dJ2y−2,=PartA−PartB (say), where(27)PartA=2VJ22∑x∈VJ2∑y∈VJ2∑pq∈EJ1dJ1p+dJ1q2+2VJ2∑x∈VJ2∑y∈VJ23dJ2x+dJ2y−4∑pq∈EJ1dJ1p+dJ1q+4∑pq∈EJ1∑x∈VJ2∑y∈VJ2dJ2x−1dJ2x+dJ2y−2=2VJ24FJ1+2M2J1+8VJ222EJ2−VJ2M1J1+4EJ1VJ2M1J2+16EJ1EJ22+8EJ1VJ22−32EJ1EJ2VJ2,PartB=∑pq∈EJ1∑x∈VJ2VJ2dJ1p+dJ1q+2dJ2x−12=2VJ22∑x∈VJ2∑pq∈EJ1dJ1p+dJ1b2+8VJ2∑x∈VJ2dJ2x−1∑pq∈EJ1dJ1p+dJ1q+8∑pq∈EJ1∑x∈VJ2dJ2x−12=2VJ23FJ1+2M2J1+8VJ22EJ2−VJ2M1J1+8EJ1M1J2+8EJ1VJ2−32EJ1EJ2.
Therefore, we get(28)T4=partA−partB=2VJ23VJ2−1FJ1+2M2J1+8VJ2VJ2−12EJ2−VJ2M1J1+4EJ1VJ2−2M1J2+16EJ1EJ22+8EJ1VJ2−1VJ2−4EJ2.
Lastly,(29)T5=∑xz∈CVJ2−yxz∑y∈VJ2∑pq∈EJ1dJp,x+dJq,y−2dJq,y+dJp,z−2=∑xz∈CVJ2−yxz∑y∈VJ2∑pq∈EJ1VJ2dJ1p+dJ1q+dJ2x+dJ2y−2VJ2dJ1p+dJ1q+dJ1z+dJ1y−2=VJ22∑xz∈CVJ2−yxz∑y∈VJ2∑pq∈EJ1dJ1p+dJ1q2+VJ2∑pq∈VJ2dJ1p+dJ2q∑xz∈CVJ2−yxz∑y∈VJ22dJ2y+dJ2x+dJ2z+∑pq∈EJ1∑xz∈CVJ2−yxz∑y∈VJ2dJ2x+dJ2y−2dJ2y+dJ2z−2=VJ23VJ2−1VJ2−2FJ1+2M2J1+4VJ2VJ2−1VJ2−22EJ2−VJ2M1J1+3EJ1VJ2−24EJ22−M1J2+EJ1VJ2−1VJ2−2M1J2+4EJ1VJ2−1VJ2−2VJ2−4EJ2.
By taking the summation of the five casesT1,T2,T3,T4, and T5 and after simplification, we get the desired result.


### 2.6. Applications

The fence graph is the composition ofPn and P2.Example 8.
From Theorem3, RSZI of PnP2 is calculated as EM2PnP2=1608n−17, where n>3.The closed fence graph is the composition ofCn and P2.Example 9.
The RSZI ofCnP2 is given by EM2CnP2=1280n.


### 2.7. Corona Product

For the corona product (COP) [27] of J1 and J2, denoted by J1∘J2, the degree of a vertex r∈J1∘J2 is given in Table 3.Table 3
The degree distribution of COP forJ and J2,i is the ith copy of the graph J2.
r∈VJ1r∈VJ2,i,i=1,2,…,n1dJ1r+VJ2dJ2r+1Now, we obtain the explicit expression of RSZI for corona product of two graphs.Theorem 4.
The RSZI ofJ1∘J2 is given by(30)EM2J1∘J2=EM2J1+VJ2EM1J1+2VJ2M4J1+VJ1M5J2+M3J2+23VJ22−2VJ2+2EJ2M1J1−2EJ1+VJ1FJ2+2VJ1M2J2+2VJ1VJ2−1+2EJ1EJ2VJ2−1+M1J2+VJ124EJ22−M1J2+CVJ22M1J1+2EJ1VJ23VJ22−2VJ2+4EJ2+1+12VJ1VJ2VJ2−13.Proof.
From definition of RSZI and from equation (7) and Table 3, we have(31)EM2J1∘J2=∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑rs∼st∈EJ1dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+VJ1∑rs∼st∈EJ2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+∑rs∈EJ1,t∈VJ2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+∑t∈VJ1,rs∈VJ2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+∑s∈VJ1,rs∈VJ2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+I1+I2+I3+I4+I5sayin order.
First,(32)I1=∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑rs∼st∈EJ1dJ1r+dJ1s−2+2n2dJ1s+dJ1t−2+2n2=∑rs∼st∈EJ1dJ1r+dJ1s−2dJ1s+dJ1t−2+2n2∑rs∼st∈EJ1dJ1r+2dJ1s+dJ1t+∑rs∼st∈EJ14n2n2−2=M2LJ1+2n2M4J1+4n2n2−212M1J1−EJ1=EM2J1+2VJ2M4J1+2VJ2VJ2−2M1J1−2EJ1.
Second,(33)I2=VJ1∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=VJ1∑rs∼st∈EJ2dJ2r+1+dJ2s+1−2dJ2s+1+dJ2t+1−2=VJ1∑rs∼st∈EJ2dJ2rdJ2s+dJ2sdJ2t+dJ2rdJ2t+VJ1∑rs∼st∈EJ2dJ22s=n1M5J2+M3J2=VJ1M5J2+M3J2.
Third,(34)I3=∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑rs∈EJ1dJ1r+n2+dJ1s+n2−2dJ1s+n2+dJ2t+1−2=∑rs∈EJ1dJ1r+dJ1s+2n2−1∑t∈VJ2dJ1s+dJ2t+n2−1=∑rs∈EJ1dJ1r+dJ1s+2n2−1n2dJ1rdJ1s+4m2+2n2n2−1=n2∑rs∈EJ1dJ1r+dJ1s−22+4m2+n22=∑rs∈EJ1dJ1r+dJ1s−2+∑rs∈EJ14n22m2+n22=n2EM1J1+4m2+n22M1J1−2M1+4n22m2+n22m1=VJ2EM1J1+4EJ2+VJ22M1J1−2EJ1+4EJ1VJ22EJ2+VJ22.
Fourth,(35)I4=∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑t∈VJ1,rs∈EJ2dJ2r+1+dJ2s+1−2dJ2s+1+dJ1t+n2−2=∑w∈VJ1∑rs∈EJ2dJ22r+dJ22s+2∑t∈VJ1∑rs∈EJ2dJ2rdJ2s+2∑t∈VJ1dJ1t∑rs∈EJ2dJ2r+dJ2s+2n2−1∑t∈VJ1∑rs∈EJ2dJ2r+dJ2s=VJ1FJ2+2VJ1M2J2+4EJ1M1J2+2VJ1VJ2−1M1J2=VJ1FJ2+2VJ1M2J2+22EJ1+VJ1VJ1−1M1J2.
Last,(36)I5=∑rs∼st∈EJ1°J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑s∈VJ1;r,t∈VJ2dJ2r+dJ1s+n2−1dJ1s+dJ2t+n2−1=∑s∈VJ1dJ1s∑ritj∈CVJ2ritjdJ2ri+dJ2tj+∑s∈VJ1∑ritj∈CVJ2ritjdJ2ri+dJ2tj+∑ritj∈CVJ2ritj∑s∈VJ1dJ22s+2n2−1∑ritj∈CVJ2ritj∑s∈VJ1dJ1s+n2−1∑s∈VJ1∑ritj∈CVJ2ritjdJ2ri+dJ2tj+n2−12∑ritj∈CVJ2ritj∑s∈VJ11=4EJ1EJ2VJ2−1+VJ124EJ22−M1J2+CVJ22M1J1+4VJ2−1EJ1CVJ22+2VJ1EJ2VJ2−1VJ2−1+VJ1VJ2−12CVJ22=4EJ1EJ2VJ2−1+VJ124EJ22−M1J2+CVJ22M1J1+2EJ1VJ2VJ2−12+2EJ2VJ1VJ2−12+12VJ1VJ2VJ2−13.
By simplifying the sum∑i=15Ii, we get the required result.


### 2.8. Applications

The t-thorny graph of a graphJ, denoted as Jt, is obtained by joining t-number of thorns (pendent edges) to each vertex of J. It is defined as the corona product of J and complement of complete graph Kt. To know more about the thorn graphs, it may be followed in [28]. By using Theorem 4, we have the following results.Example 10.
The RSZI ofJt is given by EM2Jt=EM2J+tEM1J+2tM4J+2t3t−2+Ct2M1J+2mtt−13t−5+n/2tt−13.Example 11.
The RSZI of t-thorny ofPn is given by EM2Pnt=1/2tt−122n−3+n−1t−1t+3+4t+12t+1+t+1n−4+t4n−10+4t24n−6+4n−1t2t−2.Example 12.
The RSZI ofCn is given by EM2Cnt=2n3t3+5t2−2t+1+1/2ntt−13.The bottleneck graph of a graphJ is defined as the corona product of K2 and J.Example 13.
The RSZI is given byEM2K2∘J=2M5J+2M3J+2FJ+4n−1M1J+4M2J+n2+3n+2mn2+2m−2mn.


## 2.1. Join

The join [24] of two graphs J1 and J2, denoted by J1+J2, contains the vertex set VJ1+J2=VJ1∪VJ2 and the edge set EJ1+J2=EJ1∪EJ2∪xy:x∈VJ1,y∈VJ2.In the following theorem, we compute RSZI for join of two graphs.Theorem 1.
LetJ1+J2=J be the join of J1 and J2 graphs. Then, RSZI of J is given by(9)EM2J=EM2J1+EM2J2+2VJ2M4J1+2VJ1M4J2+2VJ2VJ2−2M1J1−2EJ1+4EJ1M1J2+4EJ2M1J1+2VJ1VJ1−2M1J2−2EJ2+VJ1FJ2+VJ2FJ1+2VJ12|VJ1+VJ2−3M1J2+2VJ2VJ1+2VJ2−3M1J1+VJ224EJ12−M1J1+VJ124EJ22−M1J2+CVJ12M1J2+CVJ22M1J1+12VJ1VJ2VJ1+VJ2−23+12EJ1EJ2VJ1+VJ2−2+2VJ1M2J2+2VJ2M2J1+2VJ1EJ2VJ1+VJ2−23VJ1+VJ2−4+2VJ2EJ1VJ1+VJ2−2VJ1+3VJ2−4.Proof.
ConsiderdJx+dJy−2dJy+dJz−2=X. Let J1 and J2 be two graphs with VJ1, VJ2 vertices and EJ1, EJ2 edges, respectively. Then, by Table 1, we obtain(10)EM2J=∑xy∼yz∈EJX=∑xy∼yz∈EJ1X+∑xy∼yz∈EJ2X+∑x∈VJ1,y∈VJ2,yz∈EJ2X+∑x∈VJ2,y∈VJ1,yz∈EJ1X+∑x,z∈VJ1,y∈VJ2X+∑x,z∈VJ2,y∈VJ1X=D1+D2+D3+D4+D5+D6say,respectively.
First part:(11)D1=∑L1:xy∼yz∈EJ1X=∑L1dJ1x+dJ1y+2VJ2−2dJ1y+dJ1z+2VJ2−2=∑xy∼yz∈EJ1dJ1x+dJ1y−2dJ1y+dJ1z−2+4∑xy∼yz∈EJ1VJ22+2VJ2∑xy∼yz∈EJ1dJ1x+2dJ1y+dJ1z−8∑xy∼yz∈EJ1VJ2=M2LJ1+4VJ2212M1J1−EJ1+2VJ2M4J1−8VJ212M1J1−EJ1=EM2J1+2VJ2M4J1+2VJ2VJ2−2M1J1−2EJ1.
Second part:(12)D2=∑L2:xy∼yz∈EJ2X=∑L2dJ2x+dJ2y−2+2VJ1dJ2y+dJ2z−2+2VJ1=∑xy∼yz∈EJ2dJ2x+dJ2y−2dJ2y+dJ2z−2+4∑xy∼yz∈EJ2VJ12+2VJ1∑xy∼yz∈EJ2dJ2x+2J2y+dJ2z−8∑xy∼yz∈EJ2VJ1=M2LJ2+4VJ1212M1J2−EJ2+2VJ1M4J1−8VJ112M1J2−EJ2=EM2J2+2VJ1M4J2+2VJ1VJ1−2M1J2−2EJ2.
Third part:(13)D3=∑L3:x∈VJ1,y∈VJ2,yz∈EJ2X=∑L3dJ1x+dJ2y+VJ1+VJ2−2dJ2y+dJ2z+2VJ1−1=2∑x∈VJ1dJ1x∑yz∈EJ2dJ2y+dJ2z+4VJ1−1∑yz∈EJ2∑x∈VJ1dJ1x+∑x∈VJ1∑yz∈EJ2dJ22y+dJ22z+2∑x∈VJ1∑yz∈EJ2dJ2ydJ2z+3VJ1+VJ2−4∑x∈VJ1∑yz∈EJ2dJ2y+dJ2z+VJ1+VJ2−2∑x∈VJ1∑yz∈EJ2dJ2y+dJ2z+4VJ1−1VJ1+VJ2−2∑x∈VJ1∑yz∈EJ21=4EJ1M1J2+8EJ1EJ2VJ1−1+VJ1FJ2+2VJ1M2J2+2VJ12VJ1+VJ2−3M1J2+4EJ2VJ1VJ1−1VJ1+VJ2−2.
Fourth part:(14)D4=∑L4:x∈VJ2,y∈VJ1,yz∈EJ1X=∑L4dJ2x+dJ1y+VJ1+VJ2−2dJ1y+dJ1z+2VJ2−1=2∑x∈VJ2dJ2x∑yz∈EJ1dJ2y+dJ2z+4VJ2−1∑yz∈EJ1∑x∈VJ2dJ2x+∑x∈VJ2∑yz∈EJ1dJ12y+dJ12z+∑x∈VJ2∑yz∈EJ1dJ1ydJ1z+3VJ2+VJ1−4∑x∈VJ2∑y∈VJ1dJ1y+VJ1+VJ2−2∑x∈VJ2∑z∈VJ1dJ1z+4∑x∈VJ2∑yz∈EJ1VJ1−1VJ1+VJ2−2=4EJ2M1J1+8EJ1EJ2VJ2−1+VJ2FJ1+2VJ2M2J1+2VJ2VJ1+2VJ2−3M1J1+4EJ1VJ2VJ2−1VJ1+VJ2−2.
Fifth part:(15)D5=∑x,z∈VJ1,y∈VJ2X=∑x,z∈VJ1,y∈VJ2dJ1x+dJ2y+VJ1+VJ2−2dJ1y+dJ2z+VJ1+VJ2−2=∑y∈VJ2∑xizj∈CVJ1xizjdJ1xidJ1zj+∑y∈VJ2dJ2y∑xizj∈CVJ1xizjdJ1xi+dJ1zj+∑xizj∈CVJ1xizj∑y∈VJ2dj22y+VJ1+VJ2−2∑y∈VJ2∑xizj∈CVJ1xizjdJ1xi+dJ1zj+2VJ1+VJ2−2∑xizj∈CVJ1xizj∑y∈VJ2dJ2y+VJ1+VJ2−22∑xizj∈CVJ1xizj∑y∈VJ21=VJ224EJ12−M1J1+4VJ1−1|EJ1||EJ2|+CVJ12M1J2+2VJ2VJ1−1VJ1+VJ2−2EJ1+4CVJ12EJ2VJ1+VJ2−2+CVJ12VJ2VJ1+VJ2−22.
Sixth part:(16)D6=∑L6:x,z∈VJ2,y∈VJ1X=∑L6dJ2u+dJ1v+VJ1+VJ2−2dJ1v+dJ2w+VJ1+VJ2−2=∑y∈VJ1∑xizj∈CVJ2xizjdJ2xidJ2zj+∑y∈VJ1dJ1y∑xizj∈CVJ2xizjdJ2xi+dJ2zj+∑xizj∈CVJ2xizj∑y∈VJ1dJ12y+VJ1+VJ2−2∑y∈VJ1∑xizj∈CVJ2xizjdJ2xi+dJ2zj+2VJ1+VJ2−2∑xizj∈CVJ2xizj∑y∈VJ1dJ1y+VJ1+VJ2−22∑xizj∈CVJ2xizj∑y∈VJ11=VJ124EJ22−M1J2+4VJ2−1EJ2EJ1+CVJ22M1J1+2EJ2VJ1VJ2−1VJ2+VJ1−2+4CVJ22VJ2+VJ1−2EJ1+CVJ22|VJ1|VJ1+VJ2−22.
By adding∑i=16Di, we get the desired result.Table 1
The degree distribution of joinJ=J1+J2.
x∈VJ1x∈VJ2dJ1x+VJ2dJ2x+VJ1


## 2.2. Applications

The suspension of a graph H is the join or sum of H with a single vertexK1.Corollary 1.
The RSZI of suspension of H that containVJ=n and EJ=m can be expressed as EM2H+K1=EM2H+2nM4H+FH+2M2H+1/24n−9M1H+2m2+4m+2mn−12+n/2n−13.Example 1.
The cone graphCm,n is defined as Cm+Kn. Then, EM2Cm,n=2m11n2+5mn+2+1/2mnm+n−2m+n2+42n−3.Example 2.
The RSZI of suspension of graph such asCn,K¯n−1,Pn,mK2 are expressed in Table 2.Table 2
The results are calculated using Corollary1.
Suspension ofNamelyJoin ofEM2 valuesCnWheel graphWnCn+K1n/2n3+n2+47n−15K¯n−1Star graphSnK¯n−1+K11/2n−1n−23PnFan graphFnPn+K11/2n+4n−13+26n2−46n−17mK2Flower graphmK2+K14m22m2−m+2Example 3.
The complete bipartite graphKp,q is the join of Kp+Kq. Using Theorem 1, EM2Kp,q=1/2pqp+q−23.


## 2.3. The Cartesian Product

The Cartesian product (CP) [25] of J1 and J2, denoted by J1×J2, is a graph with VJ1×J2=VJ1×VJ2 and any f,pg,q∈EJ1×J2 if and only if fg∈EJ1,p=q or pq∈EJ1,f=g. Also, dJ1×J2fi,pj=dJ1fi+dJ2pj;i=1,2,…,VJ1 and j=1,2,…,VJ2.Now, we obtain RSZI for Cartesian product of two graphs.Theorem 2.
IfJ1×J2=J be the CP of J1 and J2 graphs, then RSZI of J is(17)EM2J=VJ2EM2J1+VJ1EM2J2+4EJ2M4J1+4EJ1M4J2+4M1J1M1J2+4EJ2FJ1+2M2J1−2M1J1+4EJ1FJ2+2M2J2−2M1J2+4M1J1−2EJ1M1J2−3EJ2+4M1J2−2EJ2M1J1−3EJ1.Proof.
By definition of RSZI, from equation (7) and the degree distribution for CP of two graphs, we have(18)EM2J=∑f,pg,q∼g,qh,r∈EJdJf,p+dJg,q−2dJg,q+dJh,r−2=∑f∈VJ1,pq∼qr∈EJ2dJf,p+dJf,q−2dJf,q+dJf,r−2+∑fg∼gh∈EJ1,p∈VJ2dJf,p+dJg,p−2dJg,p+dJh,p−2+∑fg∈EJ1,pq∈EJ2dJf,p+dJf,q−2dJf,q+dJg,q−2=U1+U2+U3say.
The notationsU1,U2, and U3 represent the sum of above terms in order.Step 1:(19)U1=∑f∈VJ1,pq∼qr∈EJ2dJf,p+dJf,q−2dJf,q+dJf,r−2=∑f∈VJ1∑pq∼qr∈EJ22dJ1f+dJ2p+dJ2q−22dJ1f+dJ2q+dJ2r−2=∑f∈VJ1∑pq∼qr∈EJ2dJ2p+dJ2q−2dJ2q+dJ2r−2+4∑pq∼qr∈EJ2∑f∈VJ1dJ1f2+2∑f∈VJ1dJ1f∑pq∼qr∈EJ2dJ2p+2dJ2q+dJ2z−8∑pq∼qr∈EJ2∑f∈VJ1dJ1f=VJ1M2LJ2+4EJ1M4J2+412M1J2−EJ2M1J1−1612M1J2−EJ2EJ1=VJ1EM2J2+4EJ1M4J2+2M1J2−2EJ2M1J1−4EJ1.Step 2:(20)U2=∑fg∼gh∈EJ1,p∈VJ2dJf,p+dJg,p−2dJg,p+dJh,p−2=∑p∈VJ2∑fg∼gh∈EJ12dJ2J+dJ1f+dJ1g−22dJ2p+dJ1g+dJ1h−2=∑p∈VJ2∑fg∼gh∈EJ1dJ1f+dJ1g−2dJ1g+dJ1h−2+2∑p∈VJ2dJ2p∑fg∼gh∈EJ1dJ1f+2dJ1g+dJ1h+4∑fg∼gh∈EJ1∑p∈VJ2dJ22p−8∑fg∼gh∈EJ1∑p∈VJ2dJ2p=VJ2M2LJ1+4EJ2M4J1+412M1J1−EJ1M1J2−812M1J1−EJ12EJ2=VJ2EM2J1+4EJ2M4J1+2M1J1−2EJ1M1J2−4EJ2.Step 3:(21)U3=∑fg∈EJ1,pq∈EJ2dJf,p+dJf,q−2dJf,q+dJg,q−2=∑fg∈EJ1∑pq∈EJ22dJ1f+dJ2p+dJ2q−22dJ2q+dJ1f+dJ1g−2=4∑fg∈EJ1∑pq∈EJ2dJ1fdJ2q+2∑fg∈EJ1∑pq∈EJ2dJ2p+dJ2q−2dJ2q+2∑pq∈EJ2∑fg∈EJ1d1fdJ1f+dJ1g−2+∑fg∈EJ1∑pq∈EJ1dJ2p+dJ2q−2dJ1f+dJ1g−2=4∑fg∈EJ1dJ1f+dJ1g∑pq∈EJ2dJ2p+dJ2q+4∑fg∈EJ1∑pq∈EJ2dJ22p+dJ22q+8∑fg∈EJ1∑pq∈EJ2dJ2pdJ2q−8∑fg∈EJ1∑pq∈EJ2dJ2p+dJ2q+4∑pq∈EG2∑fg∈EJ1dJ12f+dJ12g+8∑pq∈EJ2∑fg∈EJ1dJ1fdJ1g−8∑pq∈EJ2∑fg∈EJ1dJ1f+dJ1g+4∑fg∈EJ1dJ1f+dJ1g−2∑pq∈EJ2dJ2x+dJ2y−2=4M1J1M1J2+4EJ1FJ2+2M2J2−2M1J2+4EJ2FJ1+2M2J1−2M1J1+4M1J1−2EJ1M1J2−2EJ2.By addingU1,U2,U3, we get the required result.


## 2.4. Applications

LetP, Q, R, and S be the grids Pn×Pm=P, rook’s graph Kn×Km=Q, C4-nanotorus TC4n,m=Cn×Cm=R, and C4-nanotube TUC4n,m=Pn×Cm=S. Then, by Theorem 2, we get the following results.Example 4.
The RSZI for P is given byEM2P=454mn−89m+n+132, for m,n≥4.Example 5.
The RSZI forQ is given by EM2Q=2mnm−1m−22mn+3m−2n−4+n−1n−22mn+3n−2m−4+n−1m−16mn−11m−11n+22, for m,n≥2.Example 6.
The RSZI forR is given by EM2R=216mn.Example 7.
The RSZI for S is given byEM2S=4m54n−89.


## 2.5. Lexicographic Product

The lexicographic product (LP) or composition [26] of two graphs J1 and J2 is denoted by J1J2, and any two vertices u1,u2 and v1,v2 are adjacent if and only if u1,v1∈EJ1 or u1=v1 and u2v2∈EJ2. The vertex set of J1J2 is VJ1×VJ2, and the degree of a vertex a,b∈J1J2 is given by dJ1J2a,b=VJ2dJ1a+dJ2b.In the following theorem, we compute RSZI for composition of two graphsJ1 and J2.Theorem 3.
LetJ1J2=J be the composition of J1 and J2. The RSZI of J is given by(22)EM2J=VJ1EM2J2+VJ25M5J1+VJ25M3J1+4EJ1VJ2M4J2+VJ23VJ22−VJ2+4EJ2FJ1+2M2J1+2EJ1VJ2FJ2+2M2J2+172VJ22M1J1M1J2+2VJ232EJ2−VJ2M4J1+2VJ24EJ2VJ22−2VJ23+3VJ22−18EJ2VJ2+7EJ22M1J1−2EJ111VJ2−4EJ2M1J2−4EJ1VJ22−12EJ2VJ2+6EJ22.Proof.
By using the definition of RSZI and from the equation (7), we have(23)EM2J=∑p,xq,y∼q,yr,z∈EJdJp,x+dJq,y−2dJq,y+dJr,z−2=∑p∈VJ1∑xy∼yz∈EJ2dJp,x+dJp,y−2dJp,y+dJp,z−2+∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJp,x+dJp,y−2dJp,y+dJq,z−2+∑pq∼qr∈EJ1∑x∈VJ2∑y∈VJ2∑z∈VJ2dJp,x+dJq,y−2dJq,y+dJr,z−2+∑pq∈EJ1∑x∈VJ2∑y∈VJ2dJp,x+dJq,x−2dJq,x+dJp,y−2+∑xz∈CVJ2−yxz∑y∈VJ2∑pq∈EJ1dJp,x+dJq,y−2dJq,y+dJp,z−2=T1+T2+T3+T4+T5say,respectively.
Now,(24)T1=∑p∈VJ1∑xy∼yz∈EJ2dJp,x+dJp,y−2dJp,y+dJp,z−2=∑p∈VJ1∑xy∼yz∈EJ2dJp,x+dJp,y−2dJp,y+dJp,z−2=∑p∈VJ1∑xy∼yz∈EJ22VJ2dJ2p+dJ2x+dJ2y−22VJ2dJ2p+dJ2y+dJ2z−2=∑p∈VJ1∑xy∼yz∈EJ2dJ2x+dJ2y−2dJ2y+dJ2z−2+2VJ2∑p∈VJ1dJ1p∑xy∼yz∈EJ2dJ2x+2dJ2y+dJ2z+4VJ22∑xy∼yz∈EJ2∑p∈VJ1dJ12p−8VJ2∑xy∼yz∈EJ2∑p∈VJ1dJ1p=VJ1M2LJ2+4EJ1VJ2M4J2+4|VJ2212M1J2−EJ2M1J1−16VJ212M1J2−EJ2EJ1=VJ1EM2J2+4EJ1VJ2M4J2+2VJ2VJ2M1J1−4EJ1M1J2−2EJ2.
Next,(25)T2=∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJp,x+dJp,y−2dJp,y+dJq,z−2=∑pq∈EJ1∑xy∈EJ2∑z∈VJ22VJ2dJ1p+dJ2x+dJ2y−2VJ2dJ1p+dJ1q+dJ2y+dJ2z−2=2VJ2∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJ1pdJ2y+dJ2z+∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJ2x+dJ2y−2dJ2y+dJ2z+2VJ2∑pq∈EJ1∑xy∈EJ2∑z∈VJ2dJ1pVJ2dJ1p+VJ2dJ1q−2+∑z∈VJ2∑xy∈EJ2dJ2x+dJ2y−2∑pq∈EJ1VJ2dJ1p+dJ1q−2=2VJ2∑pq∈EJ1dJ1p+dJ1q∑xy∈EJ2VJ2dJ2x+dJ2y+4EJ2+2∑pq∈EJ1∑xy∈EJ2dJ2x+dJ2y−2VJ2dJ2x+dJ2y+4EJ2+4VJ2∑z∈VJ2∑xy∈EJ2∑pq∈EJ1dJ1p+dJ1qVJ2dJ1p+dJ1q−2+∑z∈VJ2∑xy∈EJ2dJ2x+dJ2y−2∑pq∈EJ1VJ2dJ1p+dJ1q−2=2VJ2M1J1VJ2M1J2+4EJ22+2EJ1VJ2FJ2+4EJ12EJ2−VJ2M1J2−16EJ22EJ1+4EJ1VJ2M2J2+4EJ2VJ23FJ1+8EJ2VJ23M2J1−8EJ2VJ22M1J1+4VJ2M1J2−2EJ2VJ2M1J1−2EJ1,T3=∑Ω:pq,qr∈EJ1∑α:x∈VJ2∑β:y∈VJ2∑γ:z∈VJ2dJp,x+dJq,y−2dJq,y+dJr,z−2=∑Ω∑α∑β∑γVJ2dJ1p+dJ1q+dJ2x+dJ2y−2VJ2dJ1p+dJ1r+dJ2y+dJ2z−2=VJ22∑α∑β∑γ∑ΩdJ1p+dJ1qdJ1q+dJ1r+VJ2∑ΩdJ1p+dJ1q∑α∑β∑γdJ2y+dJ2z−2+VJ2∑ΩdJ1q+dJ1r∑γ∑β∑αdJ2x+dJ2y−2+∑Ω∑β∑α∑γdJ2x+dJ2y−2dJ2y+dJ2z−2=VJ24∑y∈VJ2∑pq∼qr∈EJ1dJ1pdJ1q+dJ1qdJ1r+dJ1pdJ1r+dJ12q+VJ22∑ΩdJ1p+dJ1q∑α∑γ∑βdJ2y+∑β∑γdJ2z−∑γ∑β2+VJ22∑ΩdJ1q+dJ1r∑γ∑β∑αdJ2x+∑α∑βdJ2y−∑α∑β2+∑Ω∑γ∑α∑βdJ2y−22+∑βdJ2y−2∑γ∑αdJ2x+∑α∑γdJ2z+∑Ω∑β∑αdJ2x∑γdJ2z=VJ24∑β∑ΩdJ1pdJ1q+dJ1qdJ1r+dJ1pdJ1r+VJ24∑β∑ΩdJ12q+VJ22∑ΩdJ1p+2dJ1q+dJ1r∑α∑γ∑βdJ2y+∑β∑γdJ2z−∑γ∑β2+∑Ω∑γ∑α∑βdJ2y−22+∑βdJ2y−2∑γ∑αdJ2x+∑α∑γdJ2z+∑Ω∑β∑αdJ2x∑γdJ2z=VJ25M5J1+VJ25M3J1+2VJ232EJ2−VJ2M4J1+12M1J1−EJ1VJ22M1J2−8EJ2+4VJ2+2EJ2−VJ24EJ2VJ2+4EJ22VJ2=VJ25M5J1+VJ25M3J1+2VJ232EJ2−VJ2M4J1+VJ22M1J2+12VJ2EJ22−16VJ22EJ2+4VJ2312M1J1−EJ1.
For(26)T4=∑pq∈EJ1∑x∈VJ2∑y∈VJ2dJp,x+dJq,x−2dJq,x+dJp,y−2=∑pq∈EJ1∑x∈VJ2∑y∈VJ2VJ2dJ1p+dJ1q+2dJ2x−1VJ2dJ1q+dJ1r+dJ2x+dJ2y−2,=PartA−PartB (say), where(27)PartA=2VJ22∑x∈VJ2∑y∈VJ2∑pq∈EJ1dJ1p+dJ1q2+2VJ2∑x∈VJ2∑y∈VJ23dJ2x+dJ2y−4∑pq∈EJ1dJ1p+dJ1q+4∑pq∈EJ1∑x∈VJ2∑y∈VJ2dJ2x−1dJ2x+dJ2y−2=2VJ24FJ1+2M2J1+8VJ222EJ2−VJ2M1J1+4EJ1VJ2M1J2+16EJ1EJ22+8EJ1VJ22−32EJ1EJ2VJ2,PartB=∑pq∈EJ1∑x∈VJ2VJ2dJ1p+dJ1q+2dJ2x−12=2VJ22∑x∈VJ2∑pq∈EJ1dJ1p+dJ1b2+8VJ2∑x∈VJ2dJ2x−1∑pq∈EJ1dJ1p+dJ1q+8∑pq∈EJ1∑x∈VJ2dJ2x−12=2VJ23FJ1+2M2J1+8VJ22EJ2−VJ2M1J1+8EJ1M1J2+8EJ1VJ2−32EJ1EJ2.
Therefore, we get(28)T4=partA−partB=2VJ23VJ2−1FJ1+2M2J1+8VJ2VJ2−12EJ2−VJ2M1J1+4EJ1VJ2−2M1J2+16EJ1EJ22+8EJ1VJ2−1VJ2−4EJ2.
Lastly,(29)T5=∑xz∈CVJ2−yxz∑y∈VJ2∑pq∈EJ1dJp,x+dJq,y−2dJq,y+dJp,z−2=∑xz∈CVJ2−yxz∑y∈VJ2∑pq∈EJ1VJ2dJ1p+dJ1q+dJ2x+dJ2y−2VJ2dJ1p+dJ1q+dJ1z+dJ1y−2=VJ22∑xz∈CVJ2−yxz∑y∈VJ2∑pq∈EJ1dJ1p+dJ1q2+VJ2∑pq∈VJ2dJ1p+dJ2q∑xz∈CVJ2−yxz∑y∈VJ22dJ2y+dJ2x+dJ2z+∑pq∈EJ1∑xz∈CVJ2−yxz∑y∈VJ2dJ2x+dJ2y−2dJ2y+dJ2z−2=VJ23VJ2−1VJ2−2FJ1+2M2J1+4VJ2VJ2−1VJ2−22EJ2−VJ2M1J1+3EJ1VJ2−24EJ22−M1J2+EJ1VJ2−1VJ2−2M1J2+4EJ1VJ2−1VJ2−2VJ2−4EJ2.
By taking the summation of the five casesT1,T2,T3,T4, and T5 and after simplification, we get the desired result.


## 2.6. Applications

The fence graph is the composition ofPn and P2.Example 8.
From Theorem3, RSZI of PnP2 is calculated as EM2PnP2=1608n−17, where n>3.The closed fence graph is the composition ofCn and P2.Example 9.
The RSZI ofCnP2 is given by EM2CnP2=1280n.


## 2.7. Corona Product

For the corona product (COP) [27] of J1 and J2, denoted by J1∘J2, the degree of a vertex r∈J1∘J2 is given in Table 3.Table 3
The degree distribution of COP forJ and J2,i is the ith copy of the graph J2.
r∈VJ1r∈VJ2,i,i=1,2,…,n1dJ1r+VJ2dJ2r+1Now, we obtain the explicit expression of RSZI for corona product of two graphs.Theorem 4.
The RSZI ofJ1∘J2 is given by(30)EM2J1∘J2=EM2J1+VJ2EM1J1+2VJ2M4J1+VJ1M5J2+M3J2+23VJ22−2VJ2+2EJ2M1J1−2EJ1+VJ1FJ2+2VJ1M2J2+2VJ1VJ2−1+2EJ1EJ2VJ2−1+M1J2+VJ124EJ22−M1J2+CVJ22M1J1+2EJ1VJ23VJ22−2VJ2+4EJ2+1+12VJ1VJ2VJ2−13.Proof.
From definition of RSZI and from equation (7) and Table 3, we have(31)EM2J1∘J2=∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑rs∼st∈EJ1dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+VJ1∑rs∼st∈EJ2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+∑rs∈EJ1,t∈VJ2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+∑t∈VJ1,rs∈VJ2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+∑s∈VJ1,rs∈VJ2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2+I1+I2+I3+I4+I5sayin order.
First,(32)I1=∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑rs∼st∈EJ1dJ1r+dJ1s−2+2n2dJ1s+dJ1t−2+2n2=∑rs∼st∈EJ1dJ1r+dJ1s−2dJ1s+dJ1t−2+2n2∑rs∼st∈EJ1dJ1r+2dJ1s+dJ1t+∑rs∼st∈EJ14n2n2−2=M2LJ1+2n2M4J1+4n2n2−212M1J1−EJ1=EM2J1+2VJ2M4J1+2VJ2VJ2−2M1J1−2EJ1.
Second,(33)I2=VJ1∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=VJ1∑rs∼st∈EJ2dJ2r+1+dJ2s+1−2dJ2s+1+dJ2t+1−2=VJ1∑rs∼st∈EJ2dJ2rdJ2s+dJ2sdJ2t+dJ2rdJ2t+VJ1∑rs∼st∈EJ2dJ22s=n1M5J2+M3J2=VJ1M5J2+M3J2.
Third,(34)I3=∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑rs∈EJ1dJ1r+n2+dJ1s+n2−2dJ1s+n2+dJ2t+1−2=∑rs∈EJ1dJ1r+dJ1s+2n2−1∑t∈VJ2dJ1s+dJ2t+n2−1=∑rs∈EJ1dJ1r+dJ1s+2n2−1n2dJ1rdJ1s+4m2+2n2n2−1=n2∑rs∈EJ1dJ1r+dJ1s−22+4m2+n22=∑rs∈EJ1dJ1r+dJ1s−2+∑rs∈EJ14n22m2+n22=n2EM1J1+4m2+n22M1J1−2M1+4n22m2+n22m1=VJ2EM1J1+4EJ2+VJ22M1J1−2EJ1+4EJ1VJ22EJ2+VJ22.
Fourth,(35)I4=∑rs∼st∈EJ1∘J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑t∈VJ1,rs∈EJ2dJ2r+1+dJ2s+1−2dJ2s+1+dJ1t+n2−2=∑w∈VJ1∑rs∈EJ2dJ22r+dJ22s+2∑t∈VJ1∑rs∈EJ2dJ2rdJ2s+2∑t∈VJ1dJ1t∑rs∈EJ2dJ2r+dJ2s+2n2−1∑t∈VJ1∑rs∈EJ2dJ2r+dJ2s=VJ1FJ2+2VJ1M2J2+4EJ1M1J2+2VJ1VJ2−1M1J2=VJ1FJ2+2VJ1M2J2+22EJ1+VJ1VJ1−1M1J2.
Last,(36)I5=∑rs∼st∈EJ1°J2dJ1∘J2r+dJ1∘J2s−2dJ1∘J2s+dJ1∘J2t−2=∑s∈VJ1;r,t∈VJ2dJ2r+dJ1s+n2−1dJ1s+dJ2t+n2−1=∑s∈VJ1dJ1s∑ritj∈CVJ2ritjdJ2ri+dJ2tj+∑s∈VJ1∑ritj∈CVJ2ritjdJ2ri+dJ2tj+∑ritj∈CVJ2ritj∑s∈VJ1dJ22s+2n2−1∑ritj∈CVJ2ritj∑s∈VJ1dJ1s+n2−1∑s∈VJ1∑ritj∈CVJ2ritjdJ2ri+dJ2tj+n2−12∑ritj∈CVJ2ritj∑s∈VJ11=4EJ1EJ2VJ2−1+VJ124EJ22−M1J2+CVJ22M1J1+4VJ2−1EJ1CVJ22+2VJ1EJ2VJ2−1VJ2−1+VJ1VJ2−12CVJ22=4EJ1EJ2VJ2−1+VJ124EJ22−M1J2+CVJ22M1J1+2EJ1VJ2VJ2−12+2EJ2VJ1VJ2−12+12VJ1VJ2VJ2−13.
By simplifying the sum∑i=15Ii, we get the required result.


## 2.8. Applications

The t-thorny graph of a graphJ, denoted as Jt, is obtained by joining t-number of thorns (pendent edges) to each vertex of J. It is defined as the corona product of J and complement of complete graph Kt. To know more about the thorn graphs, it may be followed in [28]. By using Theorem 4, we have the following results.Example 10.
The RSZI ofJt is given by EM2Jt=EM2J+tEM1J+2tM4J+2t3t−2+Ct2M1J+2mtt−13t−5+n/2tt−13.Example 11.
The RSZI of t-thorny ofPn is given by EM2Pnt=1/2tt−122n−3+n−1t−1t+3+4t+12t+1+t+1n−4+t4n−10+4t24n−6+4n−1t2t−2.Example 12.
The RSZI ofCn is given by EM2Cnt=2n3t3+5t2−2t+1+1/2ntt−13.The bottleneck graph of a graphJ is defined as the corona product of K2 and J.Example 13.
The RSZI is given byEM2K2∘J=2M5J+2M3J+2FJ+4n−1M1J+4M2J+n2+3n+2mn2+2m−2mn.


## 3. Conclusion

In this study, we have executed the explicit expressions for RSZI under several graph operations such as join, Cartesian product, lexicographic product, and corona product. By applying these results, RSZI is also computed for some classes of graphs by specializing the components of graph operations. As a future work, we want to generalize the above theorems forn graphs. These results will also be helpful for further development using remaining graph operations.



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*Source: 9289534-2021-12-15.xml*