A Note on SK, SK1, SK2 Indices of Interval Weighted Graphs
Abstract: In this study, the SK, SK1 and SK2 indices are defined on weighted graphs. Then, the SK, SK1 and SK2 indices are defined on interval weighted graphs. Their behaviors are investigated under some graph operations by using these definitions.

1. Introduction

A topological index of a chemical compound is an integer, derived following a certain rule, which can be used to characterize the chemical compound and predict certain physiochemical properties like boiling point, molecular weight, density, refractive index, and so forth [1].

Molecules and molecular compounds are often modeled by molecular graph. A molecular graph is a representation of the structural formula of a chemical compound in terms of graph theory, whose vertices correspond to the atoms of the compound and edges correspond to chemical bonds [2].

Let $G=\left(V,E\right)$ be a graph with the vertex set $V\left(G\right)$ and edge set $E\left(G\right)$ and ${v}_{G}=|V\left(G\right)|$ vertices and ${e}_{G}=|E\left(G\right)|$ edges. The degree ${d}_{v}$ of the vertex $v\in V\left(G\right)$ is the number of first neighbors of v. The edge of the graph G, connecting the verticesu and v, will be denoted by $e=uv$. Throughout this paper, the graphs considered are assumed to be connected. A connected graph is a graph such that there is a path between all pairs of vertices, see books [3] [4].

We now recall some graph operations we shall need in this paper.

Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple graphs. The sum ${G}_{1}+{G}_{2}$ of these two graphs is defined as the graph having the vertex set $V\left({G}_{1}+{G}_{2}\right)={V}_{1}\cup {V}_{2}$ and the edge set $E\left({G}_{1}+{G}_{2}\right)={E}_{1}\cup {E}_{2}\cup \left\{\left(u,v\right):u\in {V}_{1},v\in {V}_{2}\right\}$ [5].

The cartesian product ${G}_{1}×{G}_{2}$ is the graph with vertex set $V\left({G}_{1}×{G}_{2}\right)={V}_{1}×{V}_{2}$ ; the vertices $u=\left({u}_{1},{u}_{2}\right)$ and $v=\left({v}_{1},{v}_{2}\right)$ of ${G}_{1}×{G}_{2}$ are adjacent if and only if $\left[{u}_{1}={v}_{1},{u}_{2}{v}_{2}\in {E}_{2}\right]$ or $\left[{u}_{2}={v}_{2},{u}_{1}{v}_{1}\in {E}_{1}\right]$ [5].

Definition 1.1. ( [1], SK index). TheSK index of a graph $G=\left(V,E\right)$ is defined as

$\text{SK}\left(G\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }d\left(u\right)+d\left( v \right)$

where $d\left(u\right)$ and $d\left(v\right)$ are the degrees of the vertices u and v in G, respectively.

Definition 1.2. ( [1], SK1 index). TheSK1 index of a graph $G=\left(V,E\right)$ is defined as

${\text{SK}}_{1}\left(G\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }d\left(u\right)d\left( v \right)$

where $d\left(u\right)$ and $d\left(v\right)$ are the degrees of the vertices u and v in G, respectively.

Definition 1.3. ( [1], SK2 index). TheSK2 index of a graph $G=\left(V,E\right)$ is defined as

${\text{SK}}_{2}\left(G\right)=\frac{1}{4}\underset{uv\in E\left(G\right)}{\sum }{\left[d\left(u\right)+d\left(v\right)\right]}^{2}$

where $d\left(u\right)$ and $d\left(v\right)$ are the degrees of the vertices u and v in G, respectively.

2. Graph Operations on the SK, SK1, SK2 Indices of Weighted Graphs

In this section, we define the SK, SK1 and SK2 indices on weighted graphs. A weighted graph is a graph each edge of which has been assigned to a number called the weight of the edge. All the weight of the edges are assumed to be positive definite [6] [7].

Let G be a weighted graph with vertex set $V\left(G\right)=\left\{{v}_{1},{v}_{2},\cdots ,{v}_{n}\right\}$ and edge set E. Denote by ${w}_{ij}$ the positive definite weight matrix of order p of the edge ij and assume that ${w}_{ij}={w}_{ji}$. We write $i~j$ if vertices i and j are adjacent. Let ${w}_{i}={\sum }_{j:i~j}{w}_{ij}$ be the weight matrix of the vertex i [6] [7].

Definition 2.1. Let $G=\left(V,E\right)$ be a connected weighted graph having n vertices. Let each edge of G be weighted with positive real numbers. The weighted SK index $\text{SK}\left(G,w\right)$ of G is defined as follows:

$\text{SK}\left(G,w\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }w\left(u\right)+w\left( v \right)$

where $w\left(u\right)$ is the sum of the weights on u.

Definition 2.2. Let $G=\left(V,E\right)$ be a connected weighted graph having n vertices. Let each edge of G be weighted with positive real numbers. The weighted SK1 index ${\text{SK}}_{1}\left(G,w\right)$ of G is defined as follows:

${\text{SK}}_{1}\left(G,w\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }w\left(u\right)w\left( v \right)$

where $w\left(u\right)$ is the sum of the weights on u.

Definition 2.3. Let $G=\left(V,E\right)$ be a connected weighted graph having n vertices. Let each edge of G be weighted with positive real numbers. The weighted SK2 index ${\text{SK}}_{2}\left(G,w\right)$ of G is defined as follows:

${\text{SK}}_{2}\left(G,w\right)=\frac{1}{4}\underset{uv\in E\left(G\right)}{\sum }{\left[w\left(u\right)+w\left(v\right)\right]}^{2}$

where $w\left(u\right)$ is the sum of the weights on u.

Theorem 2.4. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected graphs. Then the SK, SK1 and SK2 indices of the sum of graphs ${G}_{1}$ and ${G}_{2}$ are respectively given by

$\begin{array}{c}\text{SK}\left({G}_{1}+{G}_{2}\right)=\frac{1}{2}\left[\underset{ij\in {E}_{1}}{\sum }\left({d}_{i}+{n}_{{G}_{2}}\right)+\left({d}_{j}+{n}_{{G}_{2}}\right)+\underset{ij\in {E}_{2}}{\sum }\left({d}_{i}+{n}_{{G}_{1}}\right)+\left({d}_{j}+{n}_{{G}_{1}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({d}_{i}+{n}_{{G}_{2}}\right)+\left({d}_{j}+{n}_{{G}_{1}}\right)\right]\end{array}$

$\begin{array}{c}{\text{SK}}_{1}\left({G}_{1}+{G}_{2}\right)=\frac{1}{2}\left[\underset{ij\in {E}_{1}}{\sum }\left({d}_{i}+{n}_{{G}_{2}}\right)\left({d}_{j}+{n}_{{G}_{2}}\right)+\underset{ij\in {E}_{2}}{\sum }\left({d}_{i}+{n}_{{G}_{1}}\right)\left({d}_{j}+{n}_{{G}_{1}}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({d}_{i}+{n}_{{G}_{2}}\right)\left({d}_{j}+{n}_{{G}_{1}}\right)\right]\end{array}$

$\begin{array}{c}{\text{SK}}_{2}\left({G}_{1}+{G}_{2}\right)=\frac{1}{4}\left[\underset{ij\in {E}_{1}}{\sum }{\left[\left({d}_{i}+{n}_{{G}_{2}}\right)+\left({d}_{j}+{n}_{{G}_{2}}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{ij\in {E}_{2}}{\sum }{\left[\left({d}_{i}+{n}_{{G}_{1}}\right)+\left({d}_{j}+{n}_{{G}_{1}}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }{\left[\left({d}_{i}+{n}_{{G}_{2}}\right)+\left({d}_{j}+{n}_{{G}_{1}}\right)\right]}^{2}\right]\end{array}$

Theorem 2.5. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected weighted graphs. Then the weighted SK, SK1 and SK2 indices of the sum of graphs ${G}_{1}$ and ${G}_{2}$ are respectively given by

$\begin{array}{c}\text{SK}\left({G}_{1}+{G}_{2},w\right)=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)+\left({w}_{j}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({w}_{i}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)+\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)+\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\right]\end{array}$

${\text{SK}}_{1}\left({G}_{1}+{G}_{2},w\right)=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)\left({w}_{j}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)$

$\begin{array}{c}\text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({w}_{i}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\right]\end{array}$

$\begin{array}{c}{\text{SK}}_{2}\left({G}_{1}+{G}_{2},w\right)=\frac{1}{4}\left[\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }{\left[\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)+\left({w}_{j}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)\right]}^{2}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }{\left[\left({w}_{i}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)+\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }{\left[\left({w}_{i}+{\sum }_{k\in {V}_{2}}{w}_{k}\right)+\left({w}_{j}+{\sum }_{t\in {V}_{1}}{w}_{t}\right)\right]}^{2}\right]\end{array}$

Theorem 2.6. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected graphs. Then the SK, SK1 and SK2 indices of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ are respectively given by

$\text{SK}\left({G}_{1}×{G}_{2}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(d\left({u}_{i}\right)+d\left({v}_{j}\right)\right)+\left(d\left({u}_{k}\right)+d\left({v}_{l}\right)\right)\right]$

${\text{SK}}_{1}\left({G}_{1}×{G}_{2}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(d\left({u}_{i}\right)+d\left({v}_{j}\right)\right)\left(d\left({u}_{k}\right)+d\left({v}_{l}\right)\right)\right]$

${\text{SK}}_{2}\left({G}_{1}×{G}_{2}\right)=\frac{1}{4}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }{\left[\left(d\left({u}_{i}\right)+d\left({v}_{j}\right)\right)+\left(d\left({u}_{k}\right)+d\left({v}_{l}\right)\right)\right]}^{2}\right]$

Theorem 2.7. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected weighted graphs. Then the weighted SK, SK1 and SK2 indices of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ are respectively given by

$\text{SK}\left({G}_{1}×{G}_{2},w\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(w\left({u}_{i}\right)+w\left({v}_{j}\right)\right)+\left(w\left({u}_{k}\right)+w\left({v}_{l}\right)\right)\right]$

${\text{SK}}_{1}\left({G}_{1}×{G}_{2},w\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(w\left({u}_{i}\right)+w\left({v}_{j}\right)\right)\left(w\left({u}_{k}\right)+w\left({v}_{l}\right)\right)\right]$

${\text{SK}}_{2}\left({G}_{1}×{G}_{2},w\right)=\frac{1}{4}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }{\left[\left(w\left({u}_{i}\right)+w\left({v}_{j}\right)\right)+\left(w\left({u}_{k}\right)+w\left({v}_{l}\right)\right)\right]}^{2}\right]$

3. Graph Operations on the SK, SK1, SK2 Indices of Interval Weighted Graphs

In this section, we define the SK, SK1 and SK2 indices on interval weighted graphs. An interval weighted graph (interval graph) is a weighted graph in which each edge is assigned an interval or an interval square matrix. All the interval square matrices are assumed to be of the same order and to be positive definite [8].

Let G be an interval graph on n vertices. Denote by ${\stackrel{˜}{w}}_{ij}$ the positive definite interval matrix of order p of the edge ij and assume that ${\stackrel{˜}{w}}_{ij}={\stackrel{˜}{w}}_{ji}$. We write $i~j$ if vertices i and j are adjacent. Let ${\stackrel{˜}{w}}_{i}={\sum }_{j:j~i}{\stackrel{˜}{w}}_{ij}$ be the weight interval matrix of the vertex i [8].

Definition 3.1. Let $G=\left(V,E\right)$ be a connected interval weighted graph having n vertices. Let weight each edge of G be an interval or an interval square matrix. The interval weighted SK index $\text{SK}\left(G,\stackrel{˜}{w}\right)$ of G is defined as follows:

$\text{SK}\left(G,\stackrel{˜}{w}\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }\text{ }\text{ }\stackrel{˜}{w}\left(u\right)+\stackrel{˜}{w}\left( v \right)$

where $\stackrel{˜}{w}\left(u\right)$ is the sum of the interval weights on u.

Definition 3.2. Let $G=\left(V,E\right)$ be a connected interval weighted graph having n vertices. Let weight each edge of G be an interval or an interval square matrix. The interval weighted SK1 index ${\text{SK}}_{1}\left(G,\stackrel{˜}{w}\right)$ of G is defined as follows:

${\text{SK}}_{1}\left(G,\stackrel{˜}{w}\right)=\frac{1}{2}\underset{uv\in E\left(G\right)}{\sum }\text{ }\text{ }\stackrel{˜}{w}\left(u\right)\stackrel{˜}{w}\left( v \right)$

where $\stackrel{˜}{w}\left(u\right)$ is the sum of the interval weights on u.

Definition 3.3. Let $G=\left(V,E\right)$ be a connected interval weighted graph having n vertices. Let weight each edge of G be an interval or an interval square matrix. The interval weighted SK2 index ${\text{SK}}_{2}\left(G,\stackrel{˜}{w}\right)$ of G is defined as follows:

${\text{SK}}_{2}\left(G,\stackrel{˜}{w}\right)=\frac{1}{4}\underset{uv\in E\left(G\right)}{\sum }{\left[\stackrel{˜}{w}\left(u\right)+\stackrel{˜}{w}\left(v\right)\right]}^{2}$

where $\stackrel{˜}{w}\left(u\right)$ is the sum of the interval weights on u.

Theorem 3.4. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by

$\begin{array}{c}\text{SK}\left({G}_{1}+{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right]\end{array}$

Proof. Let $V=V\left({G}_{1}\right)\cup V\left({G}_{2}\right)$, $E=E\left({G}_{1}\right)\cup E\left({G}_{2}\right)\cup \left\{\left({u}_{1},{u}_{2}\right):{u}_{1}\in V\left({G}_{1}\right),{u}_{2}\in V\left({G}_{2}\right)\right\}$. We partition the set of pairs of vertices of ${G}_{1}+{G}_{2}$ to obtain the following three sums denoted by ${K}_{1},{K}_{2},{K}_{3}$, respectively.

Firstly, for each sum, we consider ${\stackrel{˜}{w}}_{i}$ as the sum of the weights in each vertex i. In ${K}_{1}$, we collect all pairs of vertices i and j so that $i,j$ are in $V\left({G}_{1}\right)$ and ij is in $E\left({G}_{1}\right)$. Hence, i and j are adjacent vertices in $E\left({G}_{1}\right)$. For ${K}_{1}$, we obtain,

${K}_{1}=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\right].$

For the second sum ${K}_{2}$, we take the vertices i and j in $V\left({G}_{2}\right)$ so that ij is in $E\left({G}_{2}\right)$. Hence,

${K}_{2}=\frac{1}{2}\left[\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right].$

In the third sum ${K}_{3}$, i is taken in $V\left({G}_{1}\right)$ and j is taken in $V\left({G}_{2}\right)$. So,

${K}_{3}=\frac{1}{2}\left[\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right].$

The result now follows by adding the three contributions and simplifying the resulting expression.

Theorem 3.5. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK1 index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by

$\begin{array}{c}{\text{SK}}_{1}\left({G}_{1}+{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\left({\stackrel{˜}{w}}_{j}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right]\end{array}$

Proof. The proof is similarly done to the proof of Theorem 3.4.

Theorem 3.6. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK2 index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by

$\begin{array}{c}{\text{SK}}_{2}\left({G}_{1}+{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{4}\left[\underset{i,j\in {V}_{1},ij\in {E}_{1}}{\sum }{\left[\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i,j\in {V}_{2},ij\in {E}_{2}}{\sum }{\left[\left({\stackrel{˜}{w}}_{i}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right]}^{2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\underset{i\in {V}_{1},j\in {V}_{2}}{\sum }{\left[\left({\stackrel{˜}{w}}_{i}+{\sum }_{k\in {V}_{2}}{\stackrel{˜}{w}}_{k}\right)+\left({\stackrel{˜}{w}}_{j}+{\sum }_{t\in {V}_{1}}{\stackrel{˜}{w}}_{t}\right)\right]}^{2}\right]\end{array}$

Proof. The proof is similarly done to the proof of Theorem 3.4.

Theorem 3.7. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by

$\text{SK}\left({G}_{1}×{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(\stackrel{˜}{w}\left({u}_{i}\right)+w\left({v}_{j}\right)\right)+\left(\stackrel{˜}{w}\left({u}_{k}\right)+\stackrel{˜}{w}\left({v}_{l}\right)\right)\right]$

Proof. The set of vertices in the graph ${G}_{1}×{G}_{2}$ is $u=\left({u}_{1},{u}_{2}\right)$, $v=\left({v}_{1},{v}_{2}\right)\in {V}_{1}×{V}_{2}$ for ${u}_{1},{v}_{1}\in {V}_{1}$ and ${u}_{2},{v}_{2}\in {V}_{2}$. Also, $\stackrel{˜}{w}\left(u\right)$ is the interval weight of the vertex u. Thus, the interval weight of any vertex $\left({u}_{1},{u}_{2}\right)\in {V}_{1}×{V}_{2}$ in the graph ${G}_{1}×{G}_{2}$ is $\stackrel{˜}{w}\left({u}_{1}\right)+\stackrel{˜}{w}\left({u}_{2}\right)$.

The SK index is equal half of the sum of degrees of all adjacent vertex pairs of the graph. Since the degrees in an interval weighted graph will turn into interval weights, it is obtained

$\text{SK}\left({G}_{1}×{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(\stackrel{˜}{w}\left({u}_{i}\right)+w\left({v}_{j}\right)\right)+\left(\stackrel{˜}{w}\left({u}_{k}\right)+\stackrel{˜}{w}\left({v}_{l}\right)\right)\right].$

Theorem 3.8. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK1 index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by

${\text{SK}}_{1}\left({G}_{1}×{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{2}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }\left(\stackrel{˜}{w}\left({u}_{i}\right)+\stackrel{˜}{w}\left({v}_{j}\right)\right)\left(\stackrel{˜}{w}\left({u}_{k}\right)+\stackrel{˜}{w}\left({v}_{l}\right)\right)\right]$

Proof. The proof is similarly done to the proof of Theorem 3.7.

Theorem 3.9. Let ${G}_{1}=\left({V}_{1},{E}_{1}\right)$ and ${G}_{2}=\left({V}_{2},{E}_{2}\right)$ be two simple, connected interval weighted graphs. Then the interval weighted SK2 index of the cartesian product of graphs ${G}_{1}$ and ${G}_{2}$ is respectively given by

${\text{SK}}_{2}\left({G}_{1}×{G}_{2},\stackrel{˜}{w}\right)=\frac{1}{4}\left[\underset{\left({u}_{i},{v}_{j}\right)\left({u}_{k},{v}_{l}\right)\in E\left({G}_{1}×{G}_{2}\right)}{\sum }{\left[\left(\stackrel{˜}{w}\left({u}_{i}\right)+\stackrel{˜}{w}\left({v}_{j}\right)\right)+\left(\stackrel{˜}{w}\left({u}_{k}\right)+\stackrel{˜}{w}\left({v}_{l}\right)\right)\right]}^{2}\right]$

Proof. The proof is similarly done to the proof of Theorem 3.7.

Acknowledgements

The first author is supported in part by TÜBİTAK. This work is derived from the first author’s PH’s thesis.

Cite this paper: Nurkahlı, S. and Büyükköse, Ş. (2021) A Note on SK, SK1, SK2 Indices of Interval Weighted Graphs. Advances in Linear Algebra & Matrix Theory, 11, 14-20. doi: 10.4236/alamt.2021.111002.
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