1. Introduction

Topological indices play a significant role mainly in chemistry, pharmacology, etc. (see [1] - [7] ). Many of the topological indices of current interest in mathematical chemistry are defined in terms of vertex degrees of the molecular graph. Two of the most famous topological indices of graphs are the first and second Zagreb indices which have been introduced by Gutman and Trinajstic in [8] , and defined as $M_1\left(G\right)={\displaystyle {\sum }_{u\in V\left(G\right)}}{\left(d\left(u\right)\right)}^2$ and $M_2\left(G\right)={\displaystyle {\sum }_{uv\in E\left(G\right)}}d\left(u\right)d\left(v\right)$, respectively. The Zagreb indices have been studied extensively due to their numerous applications in the place of existing chemical methods which need more time and increase the costs. Many new reformulated and extended versions of the Zagreb indices have been introduced for several similar reasons (cf. [9] - [17] ).

One of the present authors Saleh [18] has recently introduced a new matrix representation for a graph G by defining the locating matrix $Lo\left(G\right)$ over G. We will redefine this representation as in the following.

Definition 1 ( [18] ) Let $G=\left(V,E\right)$ be a connected graph with vertex set $V=\left\{v_1,v_2,\cdots ,v_n\right\}$. A locating function of G denoted by $L\left(G\right)$ is a function $L\left(G\right)\mathrm{<mo>\; :\; </mo>}V\left(G\right)\to {\left({\mathbb{Z}}^{+}\cup \left\{0\right\}\right)}^n$ such that $L\left(v_i\right)=v_i=\left(d\left(v_1\mathrm{,}{v}_i\right)\mathrm{,}d\left(v_2\mathrm{,}{v}_i\right)\mathrm{,}\cdots \mathrm{,}d\left(v_n\mathrm{,}{v}_i\right)\right)$, where $d\left(v_i\mathrm{,}{v}_j\right)$ is the distance between the vertices $v_i$ and $v_j$ in G. The vector $v_i$ is called the locating vector corresponding to the vertex $v_i$, where $v_i\cdot v_j$ is actually the dot product of the vectors $v_i$ and $v_j$ in the integers space ${\left({\mathbb{Z}}^{+}\cup \left\{0\right\}\right)}^n$ such that $v_i$ is adjacent to $v_j$.

The above locating function and huge applications of Zagreb indices motivated us to introduce two new topological indices, namely first and second locating indices, based on the locating vectors.

Definition 2. Let $G=\left(V,E\right)$ be a connected graph with a vertex set $V=\left\{v_1,v_2,\cdots ,v_n\right\}$ and an edge set $E\left(G\right)$. Then we define the first and second locating indices as

$M_1^{\mathcal{L}}\left(G\right)={\displaystyle \underset{v_i\in V\left(G\right)}{\sum }}{\left(v_i\right)}^2\text{and}{M}_2^{\mathcal{L}}\left(G\right)={\displaystyle \underset{v_i{v}_j\in E\left(G\right)}{\sum }}{v}_i\cdot v_j\mathrm{,}$

respectively.

All graphs in this paper will be assumed simple, undirected and connected unless stated otherwise. For graph theoretical terminologies, we refer [19] to the readers.

2. Some Exact Values in Terms of Locating Indices

In this section, by considering Definition 2, we determine the first and second locating indices for the standard graphs $K_n$, $C_n$, $K_{n\mathrm{,}m}$, $W_n$, $P_n$, and also for the join graph $G\cong G_1+G_2$ such that $G_1$ and $G_2$ are both connected graphs with diameter 2 and G will be assumed as $C_3$, $C_5$ -free graphs.

Theorem 3. Let $G\cong K_n$ be the complete graph with a vertex set $V\left(G\right)=\left\{v_1,v_2,\cdots ,v_n\right\}$, where $n\ge 2$. Then $M_1^{\mathcal{L}}\left(K_n\right)=n\left(n-1\right)$ and $M_2^{\mathcal{L}}\left(K_n\right)=\frac{n\left(n-1\right)\left(n-2\right)}{2}$.

Proof. Let $v_i$ be a locating vector corresponding to the vertex $v_i\in V\left(G\right)$. Then $v_i=\left(a_1,a_2,\cdots ,a_n\right)$ such that $a_i=0$ and $a_{i+1}=1$. Thus ${\left(v_i\right)}^2=n-1$. But we have total n vertices in $V\left(G\right)$, and so $M_1^{\mathcal{L}}\left(K_n\right)=n\left(n-1\right)$, as required. On the other hand, for any two locating vectors $v_i$ and $v_j$, where $i\ne j$, we

definitely have $v_i\cdot v_j=n-2$. Hence $M_2^{\mathcal{L}}\left(K_n\right)=\frac{n\left(n-1\right)\left(n-2\right)}{2}$.+

In the next two Theorems, we investigate the cycle $C_n$ depends on the status of n.

Theorem 4. For an even integer $n\ge 2$, let $G\cong C_n$. Then $M_1^{\mathcal{L}}\left(C_n\right)=\frac{n\left(n^2+2\right)}{12}$ and $M_2^{\mathcal{L}}\left(C_n\right)=\frac{n^2{\left(n-2\right)}^2}{12}$.

Proof. By labeling the vertices of the cycle $C_n$ as $\left\{v_1\mathrm{,}{v}_2\mathrm{,}\cdots \mathrm{,}{v}_n\right\}$ in the anticlockwise direction, we obtain

$v_1=\left(0,1,2,3,\cdots ,\frac{n}{2},\frac{n}{2}-1,\frac{n}{2}-2,\cdots ,1\right),$

$v_2=\left(1,0,1,2,\cdots ,\frac{n}{2}-1,\frac{n}{2},\frac{n}{2}-1,\cdots ,2\right),$

$v_3=\left(2,1,0,1,\cdots ,\frac{n}{2}-2,\frac{n}{2}-1,\frac{n}{2},\cdots ,3\right),$

$⋮$

$v_n=\left(1,2,3,\cdots ,\frac{n}{2},\frac{n}{2}-1,\frac{n}{2}-2,\frac{n}{2}-2,\cdots ,0\right),$

and hence $v_i^2=2{\displaystyle {\sum }_{i=1}^{\frac{n}{2}}}{i}^2-\frac{n^2}{4}$. It is not difficult to see that each $v_i$ has the same components within different location, and so each $v_i^2$ has the same sum as the form of $v_i^2=\frac{n\left(n+1\right)\left(n+2\right)-3n^2}{12}$. Therefore $M_1^{\mathcal{L}}\left(C_n\right)=\frac{n\left(n^2+2\right)}{12}$. In addition, by the symmetry,

$v_i\cdot v_{i+1}=2{\displaystyle \underset{i=2}{\overset{\frac{n}{2}}{\sum }}}i\left(i-1\right)=2\left(\frac{\frac{n}{2}\left(\frac{n}{2}+1\right)\left(n+1\right)}{6}-1\right)-2\left(\frac{\frac{n}{2}\left(\frac{n}{2}+1\right)}{2}-1\right)=\frac{n{\left(n-2\right)}^2}{12}$

which gives $M_2^{\mathcal{L}}\left(C_n\right)=\frac{n^2{\left(n-2\right)}^2}{12}$.+

Theorem 5. For an odd integer $n\ge 3$, let $G\cong C_n$. Then $M_1^{\mathcal{L}}\left(C_n\right)=\frac{n^2\left(n^2-1\right)}{12}$ and $M_2^{\mathcal{L}}\left(C_n\right)=\frac{n\left(n-1\right)\left(n-2\right)\left(n+3\right)}{12}$.

Proof. With a similar procedure as in the proof of Theorem 4, we get

$v_1=\left(0,1,2,3,\cdots ,\frac{n-1}{2},\frac{n-1}{2}-1,\frac{n-1}{2}-2,\cdots ,1\right),$

$v_2=\left(1,0,1,2,\cdots ,\frac{n-1}{2}-1,\frac{n-1}{2},\frac{n-1}{2}-1,\cdots ,2\right),$

$v_3=\left(2,1,0,1,\cdots ,\frac{n-1}{2}-2,\frac{n-1}{2}-1,\frac{n-1}{2},\cdots ,3\right),$

$⋮$

$v_n=\left(1,2,3,\cdots ,\frac{n-1}{2},\frac{n-1}{2}-1,\frac{n-1}{2}-2,\cdots ,0\right)$

which implies

$v_i^2=2{\displaystyle \underset{i=1}{\overset{\frac{n-1}{2}}{\sum }}}{i}^2=\frac{n\left(n^2-1\right)}{12},$

and so $M_1^{\mathcal{L}}\left(C_n\right)=\frac{n^2\left(n^2-1\right)}{12}$. Also, by the symmetry,

$\begin{array}{l}{v}_i\cdot v_{i+1}=2{\displaystyle \underset{i=2}{\overset{\frac{n-1}{2}}{\sum }}}i\left(i-1\right)+\frac{{\left(n-1\right)}^2}{4}\\ =\left(2\frac{\frac{n-1}{2}\left(\frac{n-1}{2}+1\right)\left(2\frac{n-1}{2}+1\right)}{6}\right)-1-\left(2\frac{\frac{n-1}{2}\left(\frac{n-1}{2}+1\right)}{2}-1\right)+\frac{{\left(n-1\right)}^2}{4}\\ =\frac{\left(n-1\right)\left(n-2\right)\left(n+3\right)}{12}\end{array}$

which gives the exact value of $M_2^{\mathcal{L}}\left(C_n\right)$ as depicted in the statement of theorem.+

Now we will take into account the complete bipartite graphs to determine the locating indices.

Theorem 6. Let $G\cong K_{n\mathrm{,}m}$, where $1\le n\le m$. Then $M_1^{\mathcal{L}}\left(K_{n,m}\right)=4\left(n^2+m^2\right)-4\left(n+m\right)+2nm$ and $M_2^{\mathcal{L}}\left(K_{n,m}\right)=2nm\left(n+m-2\right)$.

Proof. For all $1\le i\le n$ and $1\le j\le m$, by labeling the adjacent vertices $v_i$ and $v_{n+j}$ of $K_{n\mathrm{,}m}$, the locating vectors $v_i$ of $v_i$ are given by:

$\begin{array}{l}{v}_1=\left(0,\stackrel{n-1}{\stackrel{︷}{2,\cdots ,2}},\stackrel{m}{\stackrel{︷}{1,1,\cdots ,1}}\right),v_2=\left(2,0,\stackrel{n-2}{\stackrel{︷}{2,\cdots ,2}},\stackrel{m}{\stackrel{︷}{1,1,\cdots ,1}}\right),\\ v_3=\left(2,2,0,\stackrel{n-3}{\stackrel{︷}{2,\cdots ,2}},\stackrel{m}{\stackrel{︷}{1,1,\cdots ,1}}\right),\cdots ,v_n=\left(\stackrel{n-1}{\stackrel{︷}{2,\cdots ,2}},0,\stackrel{m}{\stackrel{︷}{1,1,\cdots ,1}}\right),\\ v_{n+1}=\left(\stackrel{n}{\stackrel{︷}{1,\cdots ,1}},0,\stackrel{m-1}{\stackrel{︷}{2,2,\cdots ,2}}\right),v_{n+2}=\left(\stackrel{n}{\stackrel{︷}{1,\cdots ,1}},2,0,\stackrel{m-2}{\stackrel{︷}{2,\cdots ,2}}\right),\cdots ,\\ v_{n+m}=\left(\stackrel{n}{\stackrel{︷}{1,\cdots ,1}},\stackrel{m-1}{\stackrel{︷}{2,\cdots ,2}},0\right).\end{array}$

In here, for any $i=1,2,\cdots ,n$, we have $v_i^2=4\left(n-1\right)+m$ and for any $i=n+1,n+2,\cdots ,n+m$, we get $v_i^2=4\left(m-1\right)+n$. Therefore

$\begin{array}{c}{M}_1^{\mathcal{L}}\left(K_{n,m}\right)=n\left(4\left(n-1\right)+m\right)+m\left(4\left(m-1\right)+n\right)\\ =4\left(n^2+m^2\right)-4\left(n+m\right)+2nm.\end{array}$

On the other hand, for any two consecutive locating vertices $v_i\mathrm{,}{v}_{i+1}$ in $K_{n\mathrm{,}m}$, since $v_i\cdot v_{i+1}=2\left(n+m-2\right)$, we obtain $M_2^{\mathcal{L}}\left(K_{n,m}\right)=2nm\left(n+m-2\right)$.+

Since the following consequences of Theorem 6 are very special cases and clear, we will omit their proofs.

Corollary 7. Let $G\cong K_{n\mathrm{,}n}$, where $n\ge 1$. Then $M_1^{\mathcal{L}}\left(K_{n,n}\right)=2n\left(5n-4\right)$ and $M_2^{\mathcal{L}}\left(K_{n,n}\right)=2n^2\left(2n-2\right)$.

Corollary 8. Let $G\cong K_{\mathrm{1,}m}$. Then $M_1^{\mathcal{L}}\left(K_{1,m}\right)=2m\left(2m-1\right)$ and $M_2^{\mathcal{L}}\left(K_{1,m}\right)=2m\left(m-1\right)$.

The case for wheel graphs will be investigated in the following result.

Theorem 9. Let us consider G as the wheel graph $W_n$ ( $n\ge 4$ ) with $n+1$ vertices. Then we have $M_1^{\mathcal{L}}\left(W_n\right)=4n\left(n-2\right)$ and $M_2^{\mathcal{L}}\left(W_n\right)=n\left(6n-15\right)$.

Proof. With a similar approximation as in the previous results, by labeling the vertices of $V\left(G\right)$ in the anticlockwise direction as $v_1\mathrm{,}{v}_2\mathrm{,}\cdots \mathrm{,}{v}_n\mathrm{,}{v}_{n+1}$ such that $v_{n+1}$ is the center of the wheel, we obtain

$\begin{array}{l}{v}_1=\left(0,1,\stackrel{n-3}{\stackrel{︷}{2,2,\cdots ,2}},1,1\right),v_2=\left(1,0,1,\stackrel{n-3}{\stackrel{︷}{2,2,\cdots ,2}},1\right),\\ v_3=\left(2,1,0,1,\stackrel{n-2}{\stackrel{︷}{2,\cdots ,2}},1\right),\cdots ,v_n=\left(1,\stackrel{n-3}{\stackrel{︷}{2,2,\cdots ,2}},1,0,1\right),\\ v_{n+1}=\left(\stackrel{n}{\stackrel{︷}{1,1,\cdots ,1}},0\right).\end{array}$

Now for any locating vector $v_i$ corresponding to a vertex $v_i$ $\left(i\in \left\{\mathrm{1,2,}\cdots \mathrm{,}n\right\}\right)$, we have $v_i^2=4n+9$ and $v_{n+1}^2=n$. Hence $M_1^{\mathcal{L}}\left(W_n\right)=4n\left(n-2\right)$.

For $M_2^{\mathcal{L}}\left(W_n\right)$, by labeling the vertices as above, we have

$\begin{array}{l}{v}_1=\left(0,1\mathrm{,}\underset{n-3}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,1,}1\right)\mathrm{,}{v}_2=\left(1,0,1\mathrm{,}\underset{n-3}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,1}\right)\mathrm{,}\\ v_3=\left(\mathrm{2,}1,0,1\mathrm{,}\underset{n-2}{\underset{︸}{\mathrm{2,}\cdots \mathrm{,2}}}\mathrm{,1}\right)\mathrm{,}{v}_4=\left(\mathrm{2,2,}1,0,1\mathrm{,}\underset{n-1}{\underset{︸}{\mathrm{2,}\cdots \mathrm{,2}}}\mathrm{,1}\right)\mathrm{,}\\ v_5=\left(\mathrm{2,2,2,}1,0,1\mathrm{,}\underset{n}{\underset{︸}{\mathrm{2,}\cdots \mathrm{,2}}}\mathrm{,1}\right)\mathrm{,}\cdots \mathrm{,}{v}_n=\left(\mathrm{1,}\underset{n-3}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}1,0,1\right)\mathrm{,}\\ v_{n+1}=\left(\underset{n}{\underset{︸}{\mathrm{1,1,}\cdots \mathrm{,1}}}\mathrm{,0}\right)\mathrm{.}\end{array}$

Bearing in mind the permutation of components $1,0,1$ in each vector $v_i$, where $i=1,2,\cdots ,n$, it is easy to see that any two adjacent vertices $v_i$ and $v_j$ $\left(i\mathrm{,}j\in \left\{\mathrm{1,2,}\cdots \mathrm{,}n\right\}\right)$ satisfy $v_i\cdot v_j=4n-11$ and $v_i\cdot v_{i+1}=2n-4$ for $i=1,2,\cdots ,n$. Hence $M_2^{\mathcal{L}}\left(W_n\right)=n\left(6n-15\right)$.+

The result for determining of locating indices on path graphs can be given as in the following.

Theorem 10. Let $G\cong P_n$ $\left(n\ge 3\right)$. Then

$M_1^{\mathcal{L}}\left(P_n\right)={\displaystyle \underset{j=1}{\overset{n-1}{\sum }}}\frac{\left(n-j\right)\left(n-j+1\right)\left(2n-2j+1\right)}{3},$

and

$M_2^{\mathcal{L}}\left(P_n\right)=2{\displaystyle \underset{j=1}{\overset{n-1}{\sum }}}\frac{\left(n-j\right)\left(n-j+1\right)\left(n-j-1\right)}{3}.$

Proof. Assume that G is the graph $P_n$ ( $n\ge 3$ ). By labeling the vertices from left to right as $v_1\mathrm{,}{v}_2\mathrm{,}\cdots \mathrm{,}{v}_n$ according to the locating function, the corresponding vector for each vertex $v_i\in V\left(G\right)$ ( $i=1,\cdots ,n$ ) will be the form of

$\begin{array}{l}{v}_1=\left(0,1,2,3,\cdots ,n-1\right),v_2=\left(1,0,1,2,\cdots ,n-2\right),\cdots ,\\ v_{n-1}=\left(n-2,n-1,\cdots ,0,1\right),v_n=\left(n-1,n-2,n-3,\cdots ,0\right).\end{array}$

By applying the symmetry on components between the vector pairs $v_1\mathrm{,}{v}_n$ and $v_2\mathrm{,}{v}_{n-1}\mathrm{,}\cdots$ and so on, we can see that

$M_1^{\mathcal{L}}\left(P_n\right)=2{\displaystyle \underset{j=1}{\overset{n-1}{\sum }}}{\displaystyle \underset{i=1}{\overset{n-j}{\sum }}}{i}^2={\displaystyle \underset{j=1}{\overset{n-1}{\sum }}}\frac{\left(n-j\right)\left(n-j+1\right)\left(2n-2j+1\right)}{3}.$

For $M_2^{\mathcal{L}}\left(P_n\right)$, we see that

$v_1\cdot v_2=\left(0\cdot 1\right)+\left(1\cdot 0\right)+\cdots +\left(n-1\right)\left(n-2\right)={\displaystyle \underset{i=1}{\overset{n-1}{\sum }}}i\left(i-1\right),$

$v_2\cdot v_3=\left(1\cdot 2\right)+\left(0\cdot 1\right)+\cdots +\left(n-2\right)\left(n-3\right)={\displaystyle \underset{i=1}{\overset{n-1}{\sum }}}i\left(i-1\right),$

$v_3\cdot v_4={\displaystyle \underset{i=1}{\overset{n-1}{\sum }}}i\left(i-1\right),$

$⋮$

However, by the symmetry between the components of the vectors as mentioned above, we get

$M_2^{\mathcal{L}}\left(P_n\right)=2{\displaystyle \underset{j=1}{\overset{n-1}{\sum }}}{\displaystyle \underset{i=1}{\overset{n-j}{\sum }}}i\left(i-1\right)=2{\displaystyle \underset{j=1}{\overset{n-1}{\sum }}}\left({\displaystyle \underset{i=1}{\overset{n-j}{\sum }}}{i}^2-{\displaystyle \underset{i=1}{\overset{n-j}{\sum }}}i\right)$

which can be rewritten as in the form

$\begin{array}{c}{M}_2^{\mathcal{L}}\left(P_n\right)=2{\displaystyle \underset{j=1}{\overset{n-1}{\sum }}}\left(\frac{\left(n-j\right)\left(n-j+1\right)\left(2n-2j+1\right)}{6}-\frac{\left(n-j\right)\left(n-j+1\right)}{2}\right)\\ =2{\displaystyle \underset{j=1}{\overset{n-1}{\sum }}}\frac{\left(n-j\right)\left(n-j+1\right)\left(n-j-1\right)}{3}.\end{array}$

This complete the proof.+

It is known that from the elementary textbooks the join $G=G_1+G_2$ of graphs $G_1$ and $G_2$ with disjoint vertex sets $V_1$ and $V_2$ and edge sets $E_1$ and $E_2$ is the graph union $G_1\cup G_2$ together with all the edges joining $V_1$ and $V_2$. In the following theorem we find first and second locating indices for the join graph G.

Theorem 11. Let $G\cong G_1+G_2$ such that $G_1$ and $G_2$ are both connected graphs with diameter 2 and G is a $C_3$ or $C_5$ -free graph. Assume that $G_1$ has $n_1$ vertices and $m_1$ edges while $G_2$ has $n_2$ vertices and $m_2$ edges. Then

$M_1^{\mathcal{L}}\left(G\right)=2n_1{n}_2+4\left(n_1^2+n_2^2-n_1-n_2\right)-6\left(m_1+m_2\right),$

and

$M_2^{\mathcal{L}}\left(G\right)=2\left(m_1{n}_1+m_2{n}_2\right)-4\left(m_1+m_2\right)+2n_1{n}_2\left(n_1+n_2-2\right).$

Proof. Assume that G satisfies the conditions in the statement of theorem. Let us label the vertices of the graph G as

$v_1\mathrm{,}{v}_2\mathrm{,}\cdots \mathrm{,}{v}_{n_1}\mathrm{,}{v}_{n_1+1}\mathrm{,}{v}_{n_1+2}\mathrm{,}\cdots \mathrm{,}{v}_{n_1+n_2}\mathrm{,}$

where $v_1\mathrm{,}{v}_2\mathrm{,}\cdots \mathrm{,}{v}_{n_1}\in V\left(G_1\right)$ and $v_{n_1+1}\mathrm{,}{v}_{n_1+2}\mathrm{,}\cdots \mathrm{,}{v}_{n_1+n_2}\in V\left(G_2\right)$. Also let $v$ be the locating vector corresponding to the vertex v such that $v\in V\left(G_1\right)$ :

$v=\left(\mathrm{0,}\underset{\mathrm{deg}\left(v\right)}{\underset{︸}{\mathrm{1,}\cdots \mathrm{,1}}}\mathrm{,}\underset{n_1-\mathrm{deg}\left(v\right)-1}{\underset{︸}{\mathrm{2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{n_2}{\underset{︸}{\mathrm{1,}\cdots \mathrm{,1}}}\right)\mathrm{.}$

Then $v^2=n_2+4n_1-4-3\mathrm{deg}\left(v\right)$.

Similarly, for any vertex $w\in V\left(G_2\right)$, the locating vector $w$ corresponding to w:

$w=\left(\underset{n_1}{\underset{︸}{\mathrm{1,}\cdots \mathrm{,1}}}\mathrm{,0,}\underset{\mathrm{deg}\left(w\right)}{\underset{︸}{\mathrm{1,}\cdots \mathrm{,1}}}\mathrm{,}\underset{n_2-\mathrm{deg}\left(w\right)-1}{\underset{︸}{\mathrm{2,}\cdots \mathrm{,2}}}\right)\mathrm{.}$

So $w^2=n_1+4n_2-4-3\mathrm{deg}\left(w\right)$. Therefore, by the above equalities on $v^2$ and $w^2$, we obtain

$\begin{array}{c}{M}_1^{\mathcal{L}}\left(G\right)=n_1\left(n_2+4n_1-4\right)-6m_1+n_2\left(n_1+4n_2-4\right)-6m_2\\ =2n_1{n}_2+4\left(n_1^2+n_2^2-n_1-n_2\right)-6\left(m_1+m_2\right).\end{array}$

Now, let us make partition to the set of vertices of G as

$A=\left\{u\cdot v:u,v\in V\left(G_1\right)\right\},$

$B=\left\{u\cdot v:u,v\in V\left(G_2\right)\right\},$

$C=\left\{u\cdot v\mathrm{<mo>\; :\; </mo>}u\in V\left(G_1\right)\mathrm{,}v\in V\left(G_2\right)\right\}\mathrm{.}$

Hence $M_2^{\mathcal{L}}\left(G\right)$ can be written as ${\displaystyle {\sum }_{u\cdot v\in A}}u\cdot v+{\displaystyle {\sum }_{u\cdot v\in B}}u\cdot v+{\displaystyle {\sum }_{u\cdot v\in C}}u\cdot v$. To get ${\displaystyle {\sum }_{u\cdot v\in A}}u\cdot v$ for any two adjacent vertices $u\mathrm{,}v\in V\left(G_1\right)$, let us consider

$u=\left(0,\underset{\mathrm{deg}\left(u\right)}{\underset{︸}{1,\cdots ,1}},\underset{n_1-\mathrm{deg}\left(u\right)-1}{\underset{︸}{2,\cdots ,2}},\underset{n_2}{\underset{︸}{1,\cdots ,1}}\right)$

$v=\left(\mathrm{1,0,}\underset{\mathrm{deg}\left(u\right)-1}{\underset{︸}{\mathrm{2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{n_1-\mathrm{deg}\left(u\right)-1}{\underset{︸}{\mathrm{1,}\cdots \mathrm{,1}}}\mathrm{,}\underset{n_2}{\underset{︸}{\mathrm{1,}\cdots \mathrm{,1}}}\right)\mathrm{.}$

We then have

$u\cdot v=2\left(\mathrm{deg}\left(u\right)-1\right)+2\left(n_1-\mathrm{deg}\left(u\right)-1\right)+n_2=n_2+2n_1-4$

which implies ${\displaystyle {\sum }_{u\cdot v\in A}}u\cdot v=m_1\left(n_2+2n_1-4\right)$. With a similar calculation, we get ${\displaystyle {\sum }_{u\cdot v\in B}}u\cdot v=m_2\left(n_1+2n_2-4\right)$.

Next, we need to calculate ${\displaystyle {\sum }_{u\cdot v\in C}}u\cdot v$. To do that let us take $u\in V\left(G_1\right)$ and $v\in V\left(G_2\right)$, and then labeling as

$u=\left(0,\underset{\mathrm{deg}\left(u\right)}{\underset{︸}{1,\cdots ,1}},\underset{n_1-\mathrm{deg}\left(u\right)-1}{\underset{︸}{2,\cdots ,2}},\underset{n_2}{\underset{︸}{1,\cdots ,1}}\right)$

$v=\left(\underset{\mathrm{deg}\left(u\right)+1}{\underset{︸}{\mathrm{1,}\cdots \mathrm{,1}}}\mathrm{,}\underset{n_1-\mathrm{deg}\left(u\right)-1}{\underset{︸}{\mathrm{1,}\cdots \mathrm{,1}}}\mathrm{,0,}\underset{\mathrm{deg}\left(v\right)}{\underset{︸}{\mathrm{1,}\cdots \mathrm{,1}}}\mathrm{,}\underset{n_2-\mathrm{deg}\left(v\right)-1}{\underset{︸}{\mathrm{2,}\cdots \mathrm{,2}}}\right)\mathrm{.}$

Hence we get

$\begin{array}{c}u\cdot v=\mathrm{deg}\left(u\right)+2\left(n_1-\mathrm{deg}\left(u\right)-1\right)+\mathrm{deg}\left(v\right)+2\left(n_2-\mathrm{deg}\left(v\right)-1\right)\\ =2\left(n_1+n_2\right)-4-(\mathrm{deg}\left(u\right)-\mathrm{deg}(v))\end{array}$

and so ${\displaystyle {\sum }_{u\cdot v\in C}}u\cdot v=n_1{n}_2\left(2\left(n_1+n_2\right)-4\right)-2n_2{m}_1-2n_1{m}_2$.

After all above calculations, we finally obtain

$\begin{array}{c}{M}_2^{\mathcal{L}}\left(G\right)=m_1\left(n_2+2n_1-4\right)+m_2\left(n_1+2n_2-4\right)\\ +n_1{n}_2\left(2\left(n_1+n_2\right)-4\right)-2n_2{m}_1-2n_1{m}_2\\ =2\left(m_1{n}_1+m_2{n}_2\right)-4\left(m_1+m_2\right)+2n_1{n}_2\left(n_1+n_2-2\right)\mathrm{.}\end{array}$

Hence the result.+

3. Locating Indices of Firefly Graphs

We recall that a firefly graph $F_{s\mathrm{,}t\mathrm{,}n-2s-2t-1}$ ( $s\ge \mathrm{0,}t\ge 0$ and $n-2s-2t-1\ge 0$ ) is a graph of order n that consists of s triangles, t pendant paths of length 2 and $n-2s-2t-1$ pendent edges that are sharing a common vertex (cf. [20] ). Let ${\mathcal{F}}_n$ be the set of all firefly graphs $F_{s\mathrm{,}t\mathrm{,}n-2s-2t-1}$. Note that ${\mathcal{F}}_n$ contains the stars $S_n\left(\cong F_{\mathrm{0,0,}n-1}\right)$, stretched stars $\left(\cong F_{\mathrm{0,}t\mathrm{,}n-2t-1}\right)$, friendship graphs $\left(\cong F_{\frac{n-1}{2}\mathrm{,0,0}}\right)$ and butterfly graphs $\left(\cong F_{s\mathrm{,0,}n-2s-1}\right)$.

In the next theorem we present the first and second locating indices for the firefly graph. In our calculations, for simplicity, we denote $n-2s-2t-1$ by a single letter l.

Theorem 12. Let $G\cong F_{s\mathrm{,}t\mathrm{,}l}$ ( $s\ge \mathrm{0,}t\ge 0$ and $l\ge 0$ ) be a firefly graph of order n. Then

$M_1^{\mathcal{L}}\left(G\right)=4l^2+16ls+26lt-2l+16s^2+52st-10s+38t^2-28t\mathrm{,}$

and

$M_2^{\mathcal{L}}\left(G\right)=2l^2+16ls+13lt-2l+24s^2+52st-20s+22t^2-17t\mathrm{.}$

Proof. Let $G\cong F_{s\mathrm{,}t\mathrm{,}l}$ ( $s\ge \mathrm{0,}t\ge 0$ and $l\ge 0$ ) is a firefly graph of order n. Let us label the vertices with clockwise direction as

$\begin{array}{l}{v}_1\mathrm{,}{v}_2\mathrm{,}\cdots \mathrm{,}{v}_{2s+1}\mathrm{,}{v}_{2s+2}\mathrm{,}{v}_{2s+3}\mathrm{,}\cdots \mathrm{,}{v}_{2s+l+1}\mathrm{,}{v}_{2s+l+2}\mathrm{,}\\ v_{2s+l+3}\mathrm{,}\cdots \mathrm{,}{v}_{2s+l+t+1}\mathrm{,}{v}_{2s+l+t+2}\mathrm{,}{v}_{2s+l+t+3}\mathrm{,}\cdots \mathrm{,}{v}_{2s+2t+l+1}\mathrm{,}\end{array}$

where $v_1$ is the center of the firefly graph and

$\underset{2s}{\underset{︸}{v_2\mathrm{,}{v}_3\mathrm{,}\cdots \mathrm{,}{v}_{2s+1}}}\mathrm{<mo>\; :\; </mo>}\text{verticesoftriangles,}$

$\underset{l}{\underset{︸}{v_{2s+2}\mathrm{,}{v}_{2s+3}\mathrm{,}\cdots \mathrm{,}{v}_{2s+l+1}}}\mathrm{<mo>\; :\; </mo>}\text{verticesofpendentedges,}$

$\underset{t}{\underset{︸}{v_{2s+l+2}\mathrm{,}{v}_{2s+l+3}\mathrm{,}\cdots \mathrm{,}{v}_{2s+l+t+1}}}\mathrm{<mo>\; :\; </mo>}\text{verticesofpendentpathoflength1,}$

$\underset{t}{\underset{︸}{v_{2s+l+t+2}\mathrm{,}{v}_{2s+l+t+3}\mathrm{,}\cdots \mathrm{,}{v}_{2s+2t+l+1}}}\mathrm{<mo>\; :\; </mo>}\text{verticesofpendentpathoflength2}\text{.}$

Now we calculate the corresponding vectors $v_i$ for each vertex $v_i\in V\left(G\right)$, where $i=1,2,\cdots ,2s+2t+l+1$, as in the following:

$v_1=\left(0,\underset{2s}{\underset{︸}{1,1,\cdots ,1}},\underset{l}{\underset{︸}{1,1,\cdots ,1}},\underset{t}{\underset{︸}{1,1,\cdots ,1}},\underset{t}{\underset{︸}{2,2,\cdots ,2}}\right),$

$v_2=\left(1,0,1,\underset{2s-2}{\underset{︸}{2,2,\cdots ,2}},\underset{l}{\underset{︸}{2,2,\cdots ,2}},\underset{t}{\underset{︸}{2,2,\cdots ,2}},\underset{t}{\underset{︸}{3,3,\cdots ,3}}\right),$

$v_3=\left(1,1,0,\underset{2s-2}{\underset{︸}{2,2,\cdots ,2}},\underset{l}{\underset{︸}{2,2,\cdots ,2}},\underset{t}{\underset{︸}{2,2,\cdots ,2}},\underset{t}{\underset{︸}{3,3,\cdots ,3}}\right),$

$v_4=\left(1,2,2,0,1,\underset{2s-4}{\underset{︸}{2,2,\cdots ,2}},\underset{l}{\underset{︸}{2,2,\cdots ,2}},\underset{t}{\underset{︸}{2,2,\cdots ,2}},\underset{t}{\underset{︸}{3,3,\cdots ,3}}\right),$

$v_5=\left(\mathrm{1,2,2,1,0,}\underset{2s-4}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{l}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\right)\mathrm{,}$

$⋮$

$v_{2s+1}=\left(\mathrm{1,}\underset{2s-2}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,1,0,}\underset{l}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\right)\mathrm{,}$

$v_{2s+2}=\left(\mathrm{1,}\underset{2s}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,0,}\underset{l-1}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\right)\mathrm{,}$

$v_{2s+3}=\left(\mathrm{1,}\underset{2s}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,2,0,}\underset{l-2}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\right)\mathrm{,}$

$⋮$

$v_{2s+l+1}=\left(\mathrm{1,}\underset{2s}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{l-1}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,0,}\underset{t}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\right)$

$v_{2s+l+2}=\left(\mathrm{1,}\underset{2s}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{l}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,0,}\underset{t-1}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,1,}\underset{t-1}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\right)$

$⋮$

$v_{2s+l+t+1}=\left(\mathrm{1,}\underset{2s}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{l}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,}\underset{t-1}{\underset{︸}{\mathrm{2,2,}\cdots \mathrm{,2}}}\mathrm{,0,}\underset{t-1}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,1}\right)\mathrm{,}$

$v_{2s+l+t+2}=\left(\mathrm{2,}\underset{2s}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,}\underset{l}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,1,}\underset{t-1}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,0,}\underset{t-1}{\underset{︸}{\mathrm{4,4,}\cdots \mathrm{,4}}}\right)\mathrm{,}$

$v_{2s+l+t+3}=\left(\mathrm{2,}\underset{2s}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,}\underset{l}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,3,1,}\underset{t-2}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,0,}\underset{t-2}{\underset{︸}{\mathrm{4,4,}\cdots \mathrm{,4}}}\right)\mathrm{,}$

$⋮$

$v_{2s+l+2t+1}=\left(\mathrm{2,}\underset{2s}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,}\underset{l}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,}\underset{t-1}{\underset{︸}{\mathrm{3,3,}\cdots \mathrm{,3}}}\mathrm{,1,}\underset{t-1}{\underset{︸}{\mathrm{4,4,}\cdots \mathrm{,4}}}\mathrm{,0}\right)\mathrm{.}$

Suppose that $A\mathrm{,}B\mathrm{,}C\mathrm{,}D\subset V\left(G\right)$ such that

$\begin{array}{l}A=\left\{v_2,v_3,\cdots ,v_{2s+1}\right\},C=\left\{v_{2s+l+2},v_{2s+l+3},\cdots ,v_{2s+l+t+1}\right\},\\ B=\left\{v_{2s+2},v_{2s+3},\cdots ,v_{2s+l+1}\right\},D=\left\{v_{2s+l+t+2},v_{2s+l+t+3},\cdots ,v_{2s+2t+l+1}\right\}.\end{array}$

Therefore we can write

$M_1^{\mathcal{L}}\left(G\right)={\displaystyle \underset{v\in A}{\sum }}{v}^2+{\displaystyle \underset{v\in B}{\sum }}{v}^2+{\displaystyle \underset{v\in C}{\sum }}{v}^2+{\displaystyle \underset{v\in D}{\sum }}{v}^2.$

For the calculation of ${\displaystyle {\sum }_{v\in A}}{v}^2$, we have the cases $v_1^2=2s+l+t+4t=2s+l+5t$ and

$v_i^2=2+4\left(2s-2\right)+4l+4t+9t=4l+8s+13t-6,$

where $i=2,3,\cdots ,2s+1$. Hence

$\begin{array}{c}{\displaystyle \underset{v\in A}{\sum }}{v}^2=2s+l+5t+2s\left(4l+8s+13t-6\right)\\ =2s+l+5t+8sl+16s^2+26st-12s\\ =l-10s+5t+8ls+26st+16s^2\mathrm{.}\end{array}$

On the other hand, for the calculation of ${\displaystyle {\sum }_{v\in B}}{v}^2$, we have

$v_i^2=1+4\left(2s\right)+4\left(l-1\right)+4t+9t=4l+8s+13t-\mathrm{3,}$

where for $i=2s+1,2s+2,\cdots ,2s+l+1$. Thus

${\displaystyle \underset{v\in B}{\sum }}{v}^2=l\left(4l+8s+13t-3\right)=4l^2+8ls+13lt-3l.$

Thirdly to calculate ${\displaystyle {\sum }_{v\in C}}{v}^2$, we have

$v_i^2=2+4\left(2s\right)+4l+4\left(t-1\right)+9\left(t-1\right)=4l+8s+13t-\mathrm{11,}$

where $i=2s+l+2,1,2s+l+3,\cdots ,2s+l+t+1$, and so

${\displaystyle \underset{v\in C}{\sum }}{v}^2=t\left(4l+8s+13t-11\right)=4tl+8ts+13t^2-11t.$