In this paper, we obtain the quantitative calculation formula of the degree-based topological indices of four standard product for the path and regular graphs, which unify to solve the question on product of these basic graphs without having to deal with it one by one separately. As applications, we give corresponding calculation formula of the general Randić index, the first general Zagreb index and the general sum-connectivity index.
Throughout the article, \(G\) is a simple undirected connected graph with vertex set \(V\left(G\right)\) and edge set \(E\left(G\right)\). The number of vertices and edges of \(G\) is called order and size, respectively. If the vertices \(u\) and \(v\) are adjacent, then we write \(u\sim v\). For \(v\in V\left(G\right)\), \(d_v=d_G\left(v\right)\) denotes the degree of vertex \(v\) in \(G\). Denote by \(P_n\) and \(K_{1,\,n-1}\) the path and star with \(n\) vertices, respectively.
Cheminformatics is a new interdiscipline composed of chemistry, mathematics and information science, which contributes a major role in the field of chemical sciences by implementing graph theory to mathematical modeling of chemical occurrence. In cheminformatics, the topological indices play a significant role in predicting the biological activities and properties of chemical compounds due to the fact that the numerical characteristics of topological indices reflect certain physico-chemical properties of chemical compounds, such as boiling point, stability, strain energy etc. A large number of topological indices have been studied in the models of Quantitative structure-activity relationships (QSAR) and structure-property relationships (QSPR), such as Wiener index, Randic index, Zagreb index, ABC index and so on.
The study on degree-based topological indices has been one of the hotspots in cheminformatics [1]. Let \(K=\{\left(i,\,j\right) \in \mathbb{N} \times \mathbb{N}: 1\leq i\leq j \leq n-1\}\) and \(m_{i,\, j}=m_{i,\, j}\left(G\right)\) be the number of edges in \(G\) joining vertices of degree \(i\) and \(j\). For any set of numbers \(\{\varphi_{i,\,j }\}_{\left(i,\, j\right)\in K}\), the general formula of degree-based topological indices is \[DTI\left(G\right)=\sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j}.\] In particular, we obtain the first Zagreb index and the second Zagreb index when \(\varphi_{i,\,j}=i+j\) and \(\varphi_{i,\,j}=ij\), respectively.
In 1998, the general Randic index of a graph \(G\), introduced by Bollobás and Erdos [2], is defined as \[R^{t}=R^{t}\left(G\right)=\sum\limits_{v_iv_j\in E\left(G\right)}\left(d_id_j\right)^t, \quad t\in \mathbb{R}.\] Clearly, we have that \(R^{0}\) is the number of edges, \(R^{-\frac{1}{2}}\) is the Randic index [3], \(R^{-1}\) is the modified second Zagreb index [3], \(R^{\frac{1}{2}}\) is the reciprocal Randic index [4], \(R^{2}\) is the second Hyper-Zagreb index [4], \(R^{1}\) is the second Zagreb index [5], etc.
In 2005, the first general Zagreb index of a graph \(G\) was introduced by Li and Zheng [6] and is defined as \[Z^{t}=Z^{t}\left(G\right)=\sum\limits_{v_i\in V\left(G\right)}d_i^t=\sum\limits_{v_iv_j\in E\left(G\right)}\left(d_i^{t-1}+d_j^{t-1}\right), \quad t\in \mathbb{R}.\] It is easy to see that \(Z^{0}\) is the number of vertices, \(Z^{1}\) is twice the number of edges, \(Z^{2}\) is the first Zagreb index [5], \(Z^{3}\) is the forgotten topological index [7], etc.
In 2010, Zhou and Trinajstic [8] proposed the general sum-connectivity index of a graph \(G\) as follows: \[\chi^{t}=\chi^{t}\left(G\right)=\sum\limits_{v_iv_j\in E\left(G\right)}\left(d_i+d_j\right)^t, \quad t\in \mathbb{R}.\] It is not difficult to find that \(2\chi^{-1}\) is the harmonic index [9], \(\chi^{-\frac{1}{2}}\) is the sum-connectivity index [10], \(\chi^{\frac{1}{2}}\) is the reciprocal sum-connectivity index [11], etc.
The product graphs are useful in constructing many important structural models with regularities [12], especially the following four standard product graphs which are widely used in network design [13], multiprocessor system [14], automata theory [15] and other fields. Let \(G_1\) and \(G_2\) be two graphs with disjoint vertex sets \(\{u_1, \ldots, u_m\}\) and \(\{v_1,\ldots, v_n\}\), respectively. The Cartesian product of \(G_{1}\) and \(G_{2}\), denoted by \(G_1\Box G_2\) is the graph, where \(\left(u_i, v_j\right)\sim \left(u_r, v_s\right)\) if either (\(u_i = u_r\) and \(v_j\sim v_s\) in \(G_2\)) or (\(u_i\sim u_r\) in \(G_1\) and \(v_j = v_s\)). The direct product or Kronecker product of \(G_1\) and \(G_2\), denoted by \(G_1\otimes G_2\), is the graph where \(\left(u_i, v_j\right)\sim \left(u_r, v_s\right)\) if \(u_i\sim u_r\) in \(G_1\) and \(v_j\sim v_s\) in \(G_2\). The strong product of \(G_{1}\) and \(G_{2}\), denoted by \(G_{1}\boxtimes G_{2}\), is graph where \(\left(u_{i},u_{j}\right)\sim\left(u_{r},u_{s}\right)\) if either (\(u_{i}=u_{r}\) and \(u_{j}\sim u_{s}\) in \(G_{2}\)) or (\(u_{i}\sim u_{r}\) in \(G_{1}\) and \(u_{j}=u_{s}\)) or (\(u_{i}\sim u_{r}\) in \(G_{1}\) and \(u_{j}\sim u_{s}\) in \(G_{2}\)). The lexicographic product of \(G_{1}\) and \(G_{2}\), denoted by \(G_1[G_2]\), is the graph where \(\left(u_i, v_j\right)\sim\left(u_r, v_s\right)\) if either (\(u_i\sim u_r\) in \(G_1\)) or (\(u_i = u_r\) and \(v_j\sim v_s\) in \(G_2\)).
In this paper, we give a unified approach to solve the computational problems of degree-based topological indices of standard product graphs for the path and regular graphs, which is generalization of many specific degree-based topological indices. As applications, the corresponding calculation formulas of the general Randic index, the first general Zagreb index and the general sum-connectivity index are obtained.
Theorem 1. Let \(P_{n_{1}}\) and \(P_{n_{2}}\) be two path graphs of order \(n_{1}\) and \(n_{2}\), respectively. Then \[DTI\left(P_{n_{1}}\Box P_{n_{2}}\right) = 8\varphi_{2,3}+2\left(n_{1}+n_{2}-6\right)\varphi_{3,3}+2\left(n_{1}+n_{2}-4\right)\varphi_{3,4} +\left(2n_{1}n_{2}-5n_{1}-5n_{2}+12\right)\varphi_{4,4}\] for \(n_{1}\geq n_{2}\geq3\).
Proof. By the definition of Cartesian product, we obtain the basic information on \(P_{n_{1}}\Box P_{n_{2}}\) in the Table 1.
\(m_{2,3}\) | \(m_{3,3}\) | \(m_{3,4}\) | \(m_{4,4}\) |
---|---|---|---|
\(8\) | \(2\left(n_{1}+n_{2}-6\right)\) | \(2\left(n_{1}+n_{2}-4\right)\) | \(2n_{1}n_{2}-5n_{1}-5n_{2}+12\) |
Thus we have
\[ DTI\left(P_{n_{1}}\Box P_{n_{2}}\right) = \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j} =8\varphi_{2,3}+2\left(n_{1}+n_{2}-6\right)\varphi_{3,3}+2\left(n_{1}+n_{2}-4\right)\varphi_{3,4} +\left(2n_{1}n_{2}-5n_{1}-5n_{2}+12\right)\varphi_{4,4}. \] This completes the proof.Corollary 1. Let \(P_{n_{1}}\) and \(P_{n_{2}}\) be two path graphs of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(P_{n_{1}}\Box P_{n_{2}}\right) & = 8\cdot6^{t}+2\cdot9^{t}\left(n_{1}+n_{2}-6\right)+2\cdot12^{t}\left(n_{1}+n_{2}-4\right) +16^{t}\left(2n_{1}n_{2}-5n_{1}-5n_{2}+12\right),\\ Z^{t}\left(P_{n_{1}}\Box P_{n_{2}}\right) & = 8\left(2^{t-1}+3^{t-1}\right)+4\cdot 3^{t-1}\left(n_{1}+n_{2}-6\right)+2\left(n_{1}+n_{2}-4\right)\left(3^{t-1}+4^{t-1}\right) +2\cdot4^{t-1}\left(2n_{1}n_{2}-5n_{1}-5n_{2}+12\right),\end{align*}\begin{align*} \chi^{t}\left(P_{n_{1}}\Box P_{n_{2}}\right) & = 8\cdot5^{t}+2\cdot 6^{t}\left(n_{1}+n_{2}-6\right)+2\cdot7^{t}\left(n_{1}+n_{2}-4\right) +8^{t}\left(2n_{1}n_{2}-5n_{1}-5n_{2}+12\right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Theorem 2. Let \(P_{n_{1}}\) and \(G_{r}\) be a path and a \(r\)-regular graph of order \(n_{1}\) and \(n_{2}\), respectively. Then \[ DTI\left(P_{n_{1}}\Box G_{r}\right) = rn_{2}\varphi_{r+1,r+1}+ 2n_{2}\varphi_{r+1,r+2}+\frac{1}{2}[rn_{2}\left(n_{1}-2\right)+2n_{1}n_{2}-6n_{2}]\varphi_{r+2,r+2} \] for \(n_{1}\geq n_{2}\geq2\).
Proof. By the definition of Cartesian product, we obtain the basic information on \(P_{n_{1}}\Box G_{r}\) in the following Table 2.
\(m_{r+1,r+1}\) | \(m_{r+1,r+2}\) | \(m_{r+2,r+2}\) |
---|---|---|
\(rn_{2}\) | \(2n_{2}\) | \(\dfrac{rn_{2}\left(n_{1}-2\right)}{2}+n_{1}n_{2}-3n_{2}\) |
Thus we have
\[DTI\left(P_{n_{1}}\Box G_{r}\right) = \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j} = rn_{2}\varphi_{r+1,r+1}+ 2n_{2}\varphi_{r+1,r+2}+\frac{1}{2}[rn_{2}\left(n_{1}-2\right)+2n_{1}n_{2}-6n_{2}]\varphi_{r+2,r+2}.\] This completes the proof.Corollary 2. Let \(P_{n_{1}}\) and \(G_{r}\) be a path and a \(r\)-regular graph of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(P_{n_{1}}\Box G_{r}\right) & = rn_{2}\left(r+1\right)^{2t}+ 2n_{2}\left(r+1\right)^{t}\left(r+2\right)^{t}+\left[\dfrac{rn_{2}\left(n_{1}-2\right)}{2}+n_{2}\left(n_{1}-3\right)\right]\left(r+2\right)^{2t},\\ Z^{t}\left(P_{n_{1}}\Box G_{r}\right) & = 2rn_{2}\left(r+1\right)^{t-1}+ 2n_{2}\left[\left(r+1\right)^{t-1}+\left(r+2\right)^{t-1}\right] +2\left(r+2\right)^{t-1}\left[\dfrac{rn_{2}\left(n_{1}-2\right)}{2}+n_{2}\left(n_{1}-3\right)\right],\\ \chi^{t}\left(P_{n_{1}}\Box G_{r}\right) & = 2^{t}rn_{2}\left(r+1\right)^{t}+ 2n_{2}\left(2r+3\right)^{t}+2^{t}\left(r+2\right)^{t}\left[\dfrac{rn_{2}\left(n_{1}-2\right)}{2}+n_{2}\left(n_{1}-3\right)\right] \end{align*} for \(n_{1}\geq n_{2}\geq2\).
Theorem 3. Let \(G_{r}\) and \(P_{n_{2}}\) be a \(r\)-regular and a path of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{eqnarray*} DTI\left(G_{r}\Box P_{n_{2}}\right) & = & rn_{1}\varphi_{r+1,r+1}+ 2n_{1}\varphi_{r+1,r+2}+ \frac{1}{2}\left[rn_{1}\left(n_{2}-2\right)+2n_{1}n_{2}-6n_{1}\right]\varphi_{r+2,r+2} \end{eqnarray*} for \(n_{1}\geq n_{2}\geq2\).
Proof. By the definition of Cartesian product, we obtain the basic information on \(G_{r}\Box P_{n_{2}}\) in the following Table 3.
\(m_{r+1,r+1}\) | \(m_{r+1,r+2}\) | \(m_{r+2,r+2}\) |
---|---|---|
\(rn_{1}\) | \(2n_{1}\) | \(\dfrac{rn_{1}\left(n_{2}-2\right)}{2}+n_{1}n_{2}-3n_{1}\) |
Thus we have
\[DTI\left(G_{r}\Box P_{n_{2}}\right) = \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j} = rn_{1}\varphi_{r+1,r+1}+ 2n_{1}\varphi_{r+1,r+2}+ \frac{1}{2}\left[rn_{1}\left(n_{2}-2\right)+2n_{1}n_{2}-6n_{1}\right]\varphi_{r+2,r+2}.\] This completes the proof.Corollary 3. Let \(G_{r}\) and \(P_{n_{2}}\) be a \(r\)-regular and a path of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(G_{r}\Box P_{n_{2}}\right) & = rn_{1}\left(r+1\right)^{2t}+ 2n_{1}\left(r+1\right)^{t}\left(r+2\right)^{t}+\left[\dfrac{rn_{1}\left(n_{2}-2\right)}{2}+n_{1}\left(n_{2}-3\right)\right] \left(r+2\right)^{2t},\\ Z^{t}\left(G_{r}\Box P_{n_{2}}\right) & = 2rn_{1}\left(r+1\right)^{t-1}+ 2n_{1}\left[\left(r+1\right)^{t-1}+\left(r+2\right)^{t-1}\right] +2\left(r+2\right)^{t-1}\left[\dfrac{rn_{1}\left(n_{2}-2\right)}{2}+n_{1}\left(n_{2}-3\right)\right],\\ \chi^{t}\left(G_{r}\Box P_{n_{2}}\right) & = 2^{t}rn_{1}\left(r+1\right)^{t}+ 2n_{1}\left(2r+3\right)^{t}+2^{t}\left(r+2\right)^{t}\left[\dfrac{rn_{1}\left(n_{2}-2\right)}{2}+n_{1}\left(n_{2}-3\right)\right] \end{align*} for \(n_{1}\geq n_{2}\geq2\).
Theorem 4. Let \(G_{1}\) and \(G_{2}\) be a \(r_{1}\)-regular graph and a \(r_{2}\)-regular graph with order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} DTI\left(G_{1}\Box G_{2}\right) = & \frac{n_{1}n_{2}\left(r_{1}+r_{2}\right)}{2}\varphi_{r_{1}+r_{2},r_{1}+r_{2}} \end{align*} for \(n_{1}\geq n_{2}\geq2\).
Proof. By the definition of Cartesian product, we have \(G_{1}\Box G_{2}\) is a \(\left(r_{1}+r_{2}\right)\)-regular graph with \(\frac{n_{1}n_{2}\left(r_{1}+r_{2}\right)}{2}\) edges. Thus \[DTI\left(G_{1}\Box G_{2}\right) = \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j} = \frac{n_{1}n_{2}\left(r_{1}+r_{2}\right)}{2}\varphi_{r_{1}+r_{2},r_{1}+r_{2}}.\] This completes the proof.
Corollary 4. Let \(G_{1}\) and \(G_{2}\) be a \(r_{1}\)-regular graph and a \(r_{2}\)-regular graph with order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(G_{1}\Box G_{2}\right) & = \dfrac{n_{1}n_{2}\left(r_{1}+r_{2}\right)^{2t+1}}{2},\\ Z^{t}\left(G_{1}\Box G_{2}\right) & = n_{1}n_{2}\left(r_{1}+r_{2}\right)^{t},\\ \chi^{t}\left(G_{1}\Box G_{2}\right) & = 2^{t-1}n_{1}n_{2}\left(r_{1}+r_{2}\right)^{t+1} \end{align*} for \(n_{1}\geq n_{2}\geq2\).
Theorem 5. Let \(P_{n_{1}}\) and \(P_{n_{2}}\) be two path graphs of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} DTI\left(P_{n_{1}}\otimes P_{n_{2}}\right) & = 4\varphi_{1,4}+4\varphi_{2,2}+4\left(n_{1}+n_{2}-6\right)\varphi_{2,4}+2\left(n_{1}-3\right)\left(n_{2}-3\right)\varphi_{4,4} \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Proof. By the definition of direct product, we obtain the basic information on \(P_{n_{1}}\otimes P_{n_{2}}\) in the following Table 4.
\(m_{1,4}\) | \(m_{2,2}\) | \(m_{2,4}\) | \(m_{4,4}\) |
---|---|---|---|
\(4\) | \(4\) | \(4\left(n_{1}+n_{2}-6\right)\) | \(2\left(n_{1}-3\right)\left(n_{2}-3\right)\) |
Thus we have
\[ DTI\left(P_{n_{1}}\otimes P_{n_{2}}\right) = \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j} = 4\varphi_{1,4}+4\varphi_{2,2}+4\left(n_{1}+n_{2}-6\right)\varphi_{2,4}+2\left(n_{1}-3\right)\left(n_{2}-3\right)\varphi_{4,4}. \] This completes the proof.Corollary 5. Let \(P_{n_{1}}\) and \(P_{n_{2}}\) be two path graphs of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(P_{n_{1}}\otimes P_{n_{2}}\right) = & 8\cdot4^{t}+2\cdot8^{t}\left(n_{1}n_{2}-n_{1}-n_{2}-3\right),\\ Z^{t}\left(P_{n_{1}}\otimes P_{n_{2}}\right) = & 4\left[1+2^{t}+\left(n_{1}+n_{2}-6\right)\left(2^{t-1}+4^{t-1}\right)\right]+4^{t}\left[1+\left(n_{1}-3\right)\left(n_{2}-3\right)\right],\\ \chi^{t}\left(P_{n_{1}}\otimes P_{n_{2}}\right) = & 4\left[4^{t}+5^{t}+6^{t}\left(n_{1}+n_{2}-6\right)\right]+2^{3t+1}\left(n_{1}-3\right)\left(n_{2}-3\right) \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Theorem 6. Let \(P_{n_{1}}\) and \(G_{r}\) be a path and a \(r\)-regular of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} DTI\left(P_{n_{1}}\otimes G_{r}\right) = & 2rn_{2}\varphi_{r,2r}+rn_{2}\left(n_{1}-3\right)\varphi_{2r,2r} \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Proof. By the definition of direct product, we obtain the basic information on \(P_{n_{1}}\otimes G_{r}\) in the following Table 5.
\(m_{r,2r}\) | \(m_{2r,2r}\) |
---|---|
\(2rn_{2}\) | \(rn_{2}\left(n_{1}-3\right)\) |
Thus we have
\[DTI\left(P_{n_{1}}\otimes G_{r}\right) = \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j}= 2rn_{2}\varphi_{r,2r}+rn_{2}\left(n_{1}-3\right)\varphi_{2r,2r}.\] This completes the proof.Corollary 6. Let \(P_{n_{1}}\) and \(G_{r}\) be a path and a \(r\)-regular of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(P_{n_{1}}\otimes G_{r}\right) = & 2^{t+1}r^{2t+1}n_{2}+2^{2t}r^{2t+1}n_{2}\left(n_{1}-3\right),\\ Z^{t}\left(P_{n_{1}}\otimes G_{r}\right) = & \left(2+2^{t}\right)r^{t}n_{2}+2^{t}r^{t}n_{2}\left(n_{1}-3\right),\\ \chi^{t}\left(P_{n_{1}}\otimes G_{r}\right) = & 2n_{2}\cdot3^{t}\cdot r^{t+1}+4^{t}r^{t+1}n_{2}\left(n_{1}-3\right) \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Theorem 7. Let \(G_{r}\) and \(P_{n_{2}}\) be a \(r\)-regular and a path of order \(n_{1}\) and \(n_{2}\), respectively. Then \[DTI\left(G_{r}\otimes P_{n_{2}}\right) = 2rn_{1}\varphi_{r,2r}+rn_{1}\left(n_{2}-3\right)\varphi_{2r,2r}\] for \(n_{1}\geq n_{2}\geq3\).
Proof. By the definition of direct product, we obtain the basic information on \(G_{r}\otimes P_{n_{2}}\) in the following Table 6.
\(m_{r,2r}\) | \(m_{2r,2r}\) |
---|---|
\(2rn_{1}\) | \(rn_{1}\left(n_{2}-3\right)\) |
Thus we have
\[DTI\left(G_{r}\otimes P_{n_{2}}\right) = \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j}= 2rn_{1}\varphi_{r,2r}+rn_{1}\left(n_{2}-3\right)\varphi_{2r,2r}.\] This completes the proof.Corollary 7. Let \(G_{r}\) and \(P_{n_{2}}\) be a \(r\)-regular and a path of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(G_{r}\otimes P_{n_{2}}\right) = & 2^{t+1}r^{2t+1}n_{1}+2^{2t}r^{2t+1}n_{1}\left(n_{2}-3\right),\\ Z^{t}\left(G_{r}\otimes P_{n_{2}}\right) = & r^{t}n_{1}\left(2+2^{t}\right)+2^{t}r^{t}n_{1}\left(n_{2}-3\right),\\ \chi^{t}\left(G_{r}\otimes P_{n_{2}}\right) = & 2\cdot3^{t}\cdot r^{t+1}n_{1}+4^{t}r^{t+1}n_{1}\left(n_{2}-3\right) \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Theorem 8. Let \(G_{1}\) and \(G_{2}\) be a \(r_{1}\)-regular graph and a \(r_{2}\)-regular graph with order \(n_{1}\) and \(n_{2}\), respectively. Then \[DTI\left(G_{1}\otimes G_{2}\right)=\frac{r_{1}r_{2}n_{1}n_{2}}{2}\varphi_{r_{1}r_{2},r_{1}r_{2}}\] for \(n_{1}\geq n_{2}\geq2\).
Proof. By the definition of direct product, we have \(G_{1}\otimes G_{2}\) is a \(r_{1}r_{2}\)-regular graph with \(\frac{r_{1}r_{2}n_{1}n_{2}}{2}\) edges. Thus \[DTI\left(G_{1}\otimes G_{2}\right)= \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j}=\frac{r_{1}r_{2}n_{1}n_{2}}{2}\varphi_{r_{1}r_{2},r_{1}r_{2}}.\] This completes the proof.
Corollary 8. Let \(G_{1}\) and \(G_{2}\) be a \(r_{1}\)-regular graph and a \(r_{2}\)-regular graph with order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(G_{1}\otimes G_{2}\right) = &\dfrac{n_{1}n_{2}\left(r_{1}r_{2}\right)^{2t+1}}{2},\\ Z^{t}\left(G_{1}\otimes G_{2}\right) = &n_{1}n_{2}\left(r_{1}r_{2}\right)^{t},\\ \chi^{t}\left(G_{1}\otimes G_{2}\right) = &2^{t-1}n_{1}n_{2}\left(r_{1}r_{2}\right)^{t+1} \end{align*} for \(n_{1}\geq n_{2}\geq2\).
Theorem 9. Let \(P_{n_{1}}\) and \(P_{n_{2}}\) be two path graphs of order \(n_{1}\) and \(n_{2}\), respectively. Then \[DTI\left(P_{n_{1}}\boxtimes P_{n_{2}}\right) = 8\varphi_{3,5}+ 4\varphi_{3,8}+ 2\left(n_{1}+n_{2}-4\right)\varphi_{5,5}+\left(6n_{1}+6n_{2}-32\right)\varphi_{5,8} +[4n_{1}n_{2}-11\left(n_{1}+n_{2}\right)+30]\varphi_{8,8}\] for \(n_{1}\geq n_{2}\geq 3\).
Proof. By the definition of strong product, we obtain the basic information on \(P_{n_{1}}\boxtimes P_{n_{2}}\) in the following Table 7.
\(m_{3,5}\) | \(m_{3,8}\) | \(m_{5,5}\) | \(m_{5,8}\) | \(m_{8,8}\) |
---|---|---|---|---|
\(8\) | \(4\) | \(2\left(n_{1}+n_{2}\right)-8\) | \(6\left(n_{1}+n_{2}\right)-32\) | \(4n_{1}n_{2}-11\left(n_{1}+n_{2}\right)+30\) |
Thus we have
\begin{align*} DTI\left(P_{n_{1}}\boxtimes P_{n_{2}}\right) = & \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j}\\ = & 8\varphi_{3,5}+ 4\varphi_{3,8}+ 2\left(n_{1}+n_{2}-4\right)\varphi_{5,5}+\left(6n_{1}+6n_{2}-32\right)\varphi_{5,8} +[4n_{1}n_{2}-11\left(n_{1}+n_{2}\right)+30]\varphi_{8,8}. \end{align*} This completes the proof.Corollary 9. Let \(P_{n_{1}}\) and \(P_{n_{2}}\) be two path graphs of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(P_{n_{1}}\boxtimes P_{n_{2}}\right) = & 8\cdot15^{t}+ 4\cdot24^{t}+ 25^{t}\cdot[2\left(n_{1}+n_{2}-4\right)]+40^{t}\cdot\left(6n_{1}+6n_{2}-32\right)\\ & +64^{t}\cdot[4n_{1}n_{2}-11\left(n_{1}+n_{2}\right)+30],\\ Z^{t}\left(P_{n_{1}}\boxtimes P_{n_{2}}\right) = & 8\cdot\left(3^{t-1}+ 5^{t-1}\right)+4\cdot\left(3^{t-1}+ 8^{t-1}\right)+4\cdot5^{t-1}[\left(n_{1}+n_{2}\right)-4]\\ & +\left(6n_{1}+6n_{2}-32\right)\cdot\left(5^{t-1}+8^{t-1}\right)+2\cdot8^{t-1}[4n_{1}n_{2}-11\left(n_{1}+n_{2}\right)+30],\\ \chi^{t}\left(P_{n_{1}}\boxtimes P_{n_{2}}\right) = & 8^{t+1}+4\cdot11^{t}+10^{t}\cdot[2\left(n_{1}+n_{2}\right)-8]+13^{t}\cdot\left(6n_{1}+6n_{2}-32\right)\\ &+16^{t}\cdot[4n_{1}n_{2}-11\left(n_{1}+n_{2}\right)+30] \end{align*} for \(n_{1}\geq n_{2}\geq 3\).
Theorem 10. Let \(P_{n_{1}}\) and \(G_{r}\) be a path and a \(r\)-regular of order \(n_{1}\) and \(n_{2}\), respectively. Then \[DTI\left(P_{n_{1}}\boxtimes G_{r}\right) = rn_{2}\varphi_{2r+1,2r+1}+2\left(r+1\right)n_{2}\varphi_{2r+1,3r+2} +\frac{1}{2}\left[n_{1}n_{2}\left(3r+2\right)-2n_{2}\left(4r+3\right)\right]\varphi_{3r+2,3r+2}\] for \(n_{1}> n_{2}\geq2\).
Proof. By the definition of strong product, we obtain the basic information on \(P_{n_{1}}\boxtimes G_{r}\) in the following Table 8.
\(m_{2r+1,2r+1}\) | \(m_{2r+1,3r+2}\) | \(m_{3r+2,3r+2}\) |
---|---|---|
\(rn_{2}\) | \(2\left(r+1\right)n_{2}\) | \(n_{1}n_{2}\left(\frac{3r}{2}+1\right)-n_{2}\left(4r+3\right)\) |
Thus we have
\begin{align*} DTI\left(P_{n_{1}}\boxtimes G_{r}\right) = &\sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j}\\ =& rn_{2}\varphi_{2r+1,2r+1}+2\left(r+1\right)n_{2}\varphi_{2r+1,3r+2} +\frac{1}{2}\left[n_{1}n_{2}\left(3r+2\right)-2n_{2}\left(4r+3\right)\right]\varphi_{3r+2,3r+2}. \end{align*} This completes the proof.Corollary 10. Let \(P_{n_{1}}\) and \(G_{r}\) be a path and a \(r\)-regular of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(P_{n_{1}}\boxtimes G_{r}\right) = & rn_{2}\left(2r+1\right)^{2t}+2n_{2}\left(r+1\right)\left(2r+1\right)^{t}\left(3r+2\right)^{t} +\left[n_{1}n_{2}\left(\frac{3r}{2}+1\right)-n_{2}\left(4r+3\right)\right]\left(3r+2\right)^{2t},\\ Z^{t}\left(P_{n_{1}}\boxtimes G_{r}\right) = & 2rn_{2}\left(2r+1\right)^{t-1}+2n_{2}\left(r+1\right)\left[\left(2r+1\right)^{t-1}+\left(3r+2\right)^{t-1}\right] +2\left(3r+2\right)^{t-1}\left[n_{1}n_{2}\left(\frac{3r}{2}+1\right)-n_{2}\left(4r+3\right)\right],\\ \chi^{t}\left(P_{n_{1}}\boxtimes G_{r}\right) = & 2^{t}rn_{2}\left(2r+1\right)^{t}+2n_{2}\left(r+1\right)\left(5r+3\right)^{t} +2^{t}\left(3r+2\right)^{t}\left[n_{1}n_{2}\left(\frac{3r}{2}+1\right)-n_{2}\left(4r+3\right)\right] \end{align*} for \(n_{1}> n_{2}\geq2\).
Theorem 11. Let \(G_{r}\) and \(P_{n_{2}}\) be a \(r\)-regular and a path of order \(n_{1}\) and \(n_{2}\), respectively. Then \[ DTI\left(G_{r}\boxtimes P_{n_{2}}\right) = rn_{1}\varphi_{2r+1,2r+1}+2\left(r+1\right)n_{1}\varphi_{2r+1,3r+2} +\frac{1}{2}\left[n_{1}n_{2}\left(3r+2\right)-2n_{1}\left(4r+3\right)\right]\varphi_{3r+2,3r+2} \] for \(n_{1}\geq n_{2}\geq3\).
Proof. By the definition of strong product, we obtain the basic information on \(G_{r}\boxtimes P_{n_{2}}\) in the following Table 9.
\(m_{2r+1,2r+1}\) | \(m_{2r+1,3r+2}\) | \(m_{3r+2,3r+2}\) |
---|---|---|
\(rn_{1}\) | \(2\left(r+1\right)n_{1}\) | \(n_{1}n_{2}\left(\frac{3r}{2}+1\right)-n_{1}\left(4r+3\right)\) |
Thus we have
\begin{align*}DTI\left(G_{r}\boxtimes P_{n_{2}}\right) &= \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j} \\&= rn_{1}\varphi_{2r+1,2r+1}+2\left(r+1\right)n_{1}\varphi_{2r+1,3r+2} +\frac{1}{2}\left[n_{1}n_{2}\left(3r+2\right)-2n_{1}\left(4r+3\right)\right]\varphi_{3r+2,3r+2}.\end{align*} This completes the proof.Corollary 11. Let \(G_{r}\) and \(P_{n_{2}}\) be a \(r\)-regular and a path of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(G_{r}\boxtimes P_{n_{2}}\right) = & rn_{1}\left(2r+1\right)^{2t}+2n_{1}\left(r+1\right)\left(2r+1\right)^{t}\left(3r+2\right)^{t} +\left(3r+2\right)^{2t}\left[n_{1}n_{2}\left(\frac{3r}{2}+1\right)-n_{1}\left(4r+3\right)\right],\\ Z^{t}\left(G_{r}\boxtimes P_{n_{2}}\right) = & 2rn_{1}\left(2r+1\right)^{t-1}+2n_{1}\left(r+1\right)\left[\left(2r+1\right)^{t-1}+\left(3r+2\right)^{t-1}\right] +2\left(3r+2\right)^{t-1}\left[n_{1}n_{2}\left(\frac{3r}{2}+1\right)-n_{1}\left(4r+3\right)\right],\\ \chi^{t}\left(G_{r}\boxtimes P_{n_{2}}\right) = & 2^{t}rn_{1}\left(2r+1\right)^{t}+2n_{1}\left(r+1\right)\left(5r+3\right)^{t} +2^{t}\left(3r+2\right)^{t}\left[n_{1}n_{2}\left(\frac{3r}{2}+1\right)-n_{1}\left(4r+3\right)\right] \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Theorem 12. Let \(G_{1}\) and \(G_{2}\) be a \(r_{1}\)-regular graph and a \(r_{2}\)-regular graph with order \(n_{1}\) and \(n_{2}\), respectively. Then \[DTI\left(G_{1}\boxtimes G_{2}\right)= \frac{n_{1}n_{2}\left(r_{1}r_{2}+r_{1}+r_{2}\right)}{2}\varphi_{r_{1}r_{2}+r_{1}+r_{2},r_{1}r_{2}+r_{1}+r_{2}}\] for \(n_{1}\geq n_{2}\geq2\).
Proof. By the definition of strong product, we have \(G_{1}\boxtimes G_{2}\) is a \(\left(r_{1}r_{2}+r_{1}+r_{2}\right)\)-regular graph with \(\frac{n_{1}n_{2}\left(r_{1}r_{2}+r_{1}+r_{2}\right)}{2}\) edges. Thus \[DTI\left(G_{1}\boxtimes G_{2}\right) =\sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j}= \frac{n_{1}n_{2}\left(r_{1}r_{2}+r_{1}+r_{2}\right)}{2}\varphi_{r_{1}r_{2}+r_{1}+r_{2},r_{1}r_{2}+r_{1}+r_{2}}.\] This completes the proof.
Corollary 12. Let \(G_{1}\) and \(G_{2}\) be a \(r_{1}\)-regular graph and a \(r_{2}\)-regular graph with order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(G_{1}\boxtimes G_{2}\right) = & \dfrac{n_{1}n_{2}\left(r_{1}r_{2}+r_{1}+r_{2}\right)^{2t+1}}{2},\\ Z^{t}\left(G_{1}\boxtimes G_{2}\right) = & n_{1}n_{2}\left(r_{1}r_{2}+r_{1}+r_{2}\right)^{t},\\ \chi^{t}\left(G_{1}\boxtimes G_{2}\right) = & 2^{t-1}n_{1}n_{2}\left(r_{1}r_{2}+r_{1}+r_{2}\right)^{t+1} \end{align*} for \(n_{1}\geq n_{2}\geq2\).
Theorem 13. Let \(P_{n_{1}}\) and \(P_{n_{2}}\) be two path graphs of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} DTI\left(P_{n_{1}}[P_{n_{2}}]\right)& = 4\varphi_{n_{2}+1,n_{2}+2}+8\varphi_{n_{2}+1,2n_{2}+1}+4\left(n_{2}-2\right)\varphi_{n_{2}+1,2n_{2}+2}+2\left(n_{2}-3\right)\varphi_{n_{2}+2,n_{2}+2} +4\left(n_{2}-2\right)\varphi_{n_{2}+2,2n_{2}+1}\\ & +2\left(n_{2}-2\right)^{2}\varphi_{n_{2}+2,2n_{2}+2} +4\left(n_{1}-3\right)\varphi_{2n_{2}+1,2n_{2}+1}+[2\left(n_{1}-2\right)+4\left(n_{1}-3\right)\left(n_{2}-2\right)]\varphi_{2n_{2}+1,2n_{2}+2}\\ & +[\left(n_{1}-2\right)\left(n_{2}-3\right)+\left(n_{1}-3\right)\left(n_{2}-2\right)^{2}]\varphi_{2n_{2}+2,2n_{2}+2} \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Proof. By the definition of lexicographic product, we obtain the basic information on \(P_{n_{1}}[P_{n_{2}}]\) in the following Table 10.
\(m_{n_{2}+1,n_{2}+2}\) | \(4\) |
---|---|
\(m_{n_{2}+1,2n_{2}+1}\) | \(8\) |
\(m_{n_{2}+1,2n_{2}+2}\) | \(4\left(n_{2}-2\right)\) |
\(m_{n_{2}+2,n_{2}+2}\) | \(2\left(n_{2}-3\right)\) |
\(m_{n_{2}+2,2n_{2}+1}\) | \(4\left(n_{2}-2\right)\) |
\(m_{n_{2}+2,2n_{2}+2}\) | \(2\left(n_{2}-2\right)^{2}\) |
\(m_{2n_{2}+1,2n_{2}+1}\) | \(4\left(n_{1}-3\right)\) |
\(m_{2n_{2}+1,2n_{2}+2}\) | \(2\left(n_{1}-2\right)+4\left(n_{1}-3\right)\left(n_{2}-2\right)\) |
\(m_{2n_{2}+2,2n_{2}+2}\) | \(\left(n_{1}-2\right)\left(n_{2}-3\right)+\left(n_{1}-3\right)\left(n_{2}-2\right)^{2}\) |
Thus we have
\begin{align*} DTI\left(P_{n_{1}}[P_{n_{2}}]\right) = &\sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j}\\ = & 4\varphi_{n_{2}+1,n_{2}+2}+8\varphi_{n_{2}+1,2n_{2}+1}+4\left(n_{2}-2\right)\varphi_{n_{2}+1,2n_{2}+2}\\ & +2\left(n_{2}-3\right)\varphi_{n_{2}+2,n_{2}+2}+4\left(n_{2}-2\right)\varphi_{n_{2}+2,2n_{2}+1}+2\left(n_{2}-2\right)^{2}\varphi_{n_{2}+2,2n_{2}+2}\\ & +4\left(n_{1}-3\right)\varphi_{2n_{2}+1,2n_{2}+1}+[2\left(n_{1}-2\right)+4\left(n_{1}-3\right)\left(n_{2}-2\right)]\varphi_{2n_{2}+1,2n_{2}+2}\\ & +[\left(n_{1}-2\right)\left(n_{2}-3\right)+\left(n_{1}-3\right)\left(n_{2}-2\right)^{2}]\varphi_{2n_{2}+2,2n_{2}+2}. \end{align*} This completes the proof.Corollary 13. Let \(P_{n_{1}}\) and \(P_{n_{2}}\) be two path graphs of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(P_{n_{1}}[P_{n_{2}}]\right) = & 4[\left(n_{2}+1\right)\left(n_{2}+2\right)]^{t}+8[\left(n_{2}+1\right)\left(2n_{2}+1\right)]^{t}+4\left(n_{2}-2\right)[\left(n_{2}+1\right)\left(2n_{2}+2\right)]^{t}\\ & +2\left(n_{2}-3\right)\left(n_{2}+2\right)^{2t}+4\left(n_{2}-2\right)[\left(n_{2}+2\right)\left(2n_{2}+1\right)]^{t}\\ &+2\left(n_{2}-2\right)^{2}[\left(n_{2}+2\right)\left(2n_{2}+2\right)]^{t}+4\left(n_{1}-3\right)\left(2n_{2}+1\right)^{2t}\\ & +[2\left(n_{1}-2\right)+4\left(n_{1}-3\right)\left(n_{2}-2\right)][\left(2n_{2}+1\right)\left(2n_{2}+2\right)]^{t}\\ & +[\left(n_{1}-2\right)\left(n_{2}-3\right)+\left(n_{1}-3\right)\left(n_{2}-2\right)^{2}]\left(2n_{2}+2\right)^{2t},\\ Z^{t}\left(P_{n_{1}}[P_{n_{2}}]\right)= & 4[\left(n_{2}+1\right)^{t-1}+\left(n_{2}+2\right)^{t-1}]+8[\left(n_{2}+1\right)^{t-1}+\left(2n_{2}+1\right)^{t-1}]\\ &+4\left(n_{2}-2\right)[\left(n_{2}+1\right)^{t-1}+\left(2n_{2}+2\right)^{t-1}]+4\left(n_{2}-3\right)\left(n_{2}+2\right)^{t-1}\\ & +4\left(n_{2}-2\right)[\left(n_{2}+2\right)^{t-1}+\left(2n_{2}+1\right)^{t-1}]+2\left(n_{2}-2\right)^{2}[\left(n_{2}+2\right)^{t-1}\\ & +\left(2n_{2}+2\right)^{t-1}]+8\left(n_{1}-3\right)\left(2n_{2}+1\right)^{t-1}\\ &+[2\left(n_{1}-2\right)+4\left(n_{1}-3\right)\left(n_{2}-2\right)][\left(2n_{2}+1\right)^{t-1}+\left(2n_{2}+2\right)^{t-1}]\\ & +2\left(2n_{2}+2\right)^{t-1}[\left(n_{1}-2\right)\left(n_{2}-3\right)+\left(n_{1}-3\right)\left(n_{2}-2\right)^{2}],\\ \chi^{t}\left(P_{n_{1}}[P_{n_{2}}]\right) = & 4\left(2n_{2}+3\right)^{t}+8\left(3n_{2}+2\right)^{t} +8\cdot3^{t}\cdot\left(n_{2}-2\right)\left(n_{2}+1\right)^{t}+2\left(n_{2}-3\right)\left(2n_{2}+4\right)^{t}\\ & +2\left(n_{2}-2\right)^{2}\left(3n_{2}+4\right)^{t}+4\left(n_{1}-3\right)\left(4n_{2}+2\right)^{t}\\ &+[2\left(n_{1}-2\right)+4\left(n_{1}-3\right)\left(n_{2}-2\right)]\left(4n_{2}+3\right)^{t}\\ & +4^{t}\left(n_{2}+1\right)^{t}[\left(n_{1}-2\right)\left(n_{2}-3\right)+\left(n_{1}-3\right)\left(n_{2}-2\right)^{2}] \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Theorem 14. Let \(P_{n_{1}}\) and \(G_{r}\) be a path and a r-regular of order \(n_{1}\) and \(n_{2}\), respectively. Then \[DTI\left(P_{n_{1}}\left[G_{r}\right]\right) = rn_{2}\varphi_{r+n_{2},r+n_{2}}+ 2n_{2}^{2}\varphi_{r+n_{2},2n_{2}+r}+\frac{1}{2}\left[rn_{2}\left(n_{1}-2\right)+2\left(n_{1}-3\right)n_{2}^{2}\right]\varphi_{2n_{2}+r,2n_{2}+r}\] for \(n_{1}> n_{2}\geq2\).
Proof. By the definition of lexicographic product, we obtain the basic information on \(P_{n_{1}}[G_{r}]\) in the following Table 11.
\(m_{r+n_{2},r+n_{2}}\) | \(m_{r+n_{2},2n_{2}+r}\) | \(m_{2n_{2}+r,2n_{2}+r}\) |
---|---|---|
\(rn_{2}\) | \(2n_{2}^{2}\) | \(\dfrac{rn_{2}\left(n_{1}-2\right)}{2}+n_{2}^{2}\left(n_{1}-3\right)\) |
Corollary 14. Let \(P_{n_{1}}\) and \(G_{r}\) be a path and a r-regular of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(P_{n_{1}}\left[G_{r}\right]\right) = & rn_{2}\left(r+n_{2}\right)^{2t}+ 2n_{2}^{2}\left(r+n_{2}\right)^{t}\left(2n_{2}+r\right)^{t} +\left[\dfrac{rn_{2}\left(n_{1}-2\right)}{2}+\left(n_{1}-3\right)n_{2}^{2}\right]\left(2n_{2}+r\right)^{2t},\\ Z^{t}\left(P_{n_{1}}\left[G_{r}\right]\right) = & 2rn_{2}\left(r+n_{2}\right)^{t-1}+ 2n_{2}^{2}\left[\left(r+n_{2}\right)^{t-1}+\left(2n_{2}+r\right)^{t-1}\right] +\left[rn_{2}\left(n_{1}-2\right)+2\left(n_{1}-3\right)n_{2}^{2}\right]\left(2n_{2}+r\right)^{t-1},\\ \chi^{t}\left(P_{n_{1}}\left[G_{r}\right]\right) = & 2^{t}rn_{2}\left(r+n_{2}\right)^{t}+ 2n_{2}^{2}\left(2r+3n_{2}\right)^{t} +2^{t}\left[\dfrac{rn_{2}\left(n_{1}-2\right)}{2}+\left(n_{1}-3\right)n_{2}^{2}\right]\left(2n_{2}+r\right)^{t} \end{align*} for \(n_{1}> n_{2}\geq2\).
Theorem 15. Let \(G_{r}\) and \(P_{n_{2}}\) be a \(r\)-regular and a path of order \(n_{1}\) and \(n_{2}\), respectively. Then \[ DTI\left(G_{r}[P_{n_{2}}]\right) = 2rn_{1}\varphi_{rn_{2}+1,rn_{2}+1}+2n_{1}\left[1+r\left(n_{2}-2\right)\right]\varphi_{rn_{2}+1,rn_{2}+2} +\frac{1}{2}\left[2n_{1}\left(n_{2}-3\right)+rn_{1}\left(n_{2}-2\right)^{2}\right]\varphi_{rn_{2}+2,rn_{2}+2} \] for \(n_{1}\geq n_{2}\geq3\).
Proof. By the definition of lexicographic product, we obtain the basic information on \(G_{r}[P_{n_{2}}]\) in the following Table 12.
\(m_{rn_{2}+1,rn_{2}+1}\) | \(m_{rn_{2}+1,rn_{2}+2}\) | \(m_{rn_{2}+2,rn_{2}+2}\) |
---|---|---|
\(2rn_{1}\) | \(2n_{1}[1+r\left(n_{2}-2\right)]\) | \(n_{1}\left(n_{2}-3\right)+\frac{rn_{1}\left(n_{2}-2\right)^{2}}{2}\) |
Corollary 15. Let \(G_{r}\) and \(P_{n_{2}}\) be a \(r\)-regular and a path of order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(G_{r}\left[P_{n_{2}}\right]\right) = & 2rn_{1}\left(rn_{2}+1\right)^{2t}+2n_{1}\left[1+r\left(n_{2}-2\right)\right]\left[\left(rn_{2}+1\right)\left(rn_{2}+2\right)\right]^{t}\\ & +\left[n_{1}\left(n_{2}-3\right)+\dfrac{rn_{1}\left(n_{2}-2\right)^{2}}{2}\right]\left(rn_{2}+2\right)^{2t},\\ Z^{t}\left(G_{r}\left[P_{n_{2}}\right]\right) = & 4rn_{1}\left(rn_{2}+1\right)^{t-1}+2n_{1}\left[1+r\left(n_{2}-2\right)\right]\left[\left(rn_{2}+1\right)^{t-1}+\left(rn_{2}+2\right)^{t-1}\right]\\ & +\left[2n_{1}\left(n_{2}-3\right)+rn_{1}\left(n_{2}-2\right)^{2}\right]\left(rn_{2}+2\right)^{t-1},\\ \chi^{t}\left(G_{r}\left[P_{n_{2}}\right]\right) = & 2^{t+1}rn_{1}\left(rn_{2}+1\right)^{t}+2n_{1}\left[1+r\left(n_{2}-2\right)\right]\left(2rn_{2}+3\right)^{t}\\ & +2^{t}\left[n_{1}\left(n_{2}-3\right)+\dfrac{rn_{1}\left(n_{2}-2\right)^{2}}{2}\right]\left(rn_{2}+2\right)^{t} \end{align*} for \(n_{1}\geq n_{2}\geq3\).
Theorem 16. Let \(G_{1}\) and \(G_{2}\) be a \(r_{1}\)-regular graph and a \(r_{2}\)-regular graph with order \(n_{1}\) and \(n_{2}\), respectively. Then \[DTI\left(G_{1}[G_{2}]\right)=\frac{1}{2}n_{1}n_{2}\left(r_{2}+r_{1}n_{2}\right)\varphi_{r_{1}n_{2}+r_{2},r_{1}n_{2}+r_{2}}\] for \(n_{1}\geq n_{2}\geq2\).
Proof. By the definition of lexicographic product, we have \(G_{1}[G_{2}]\) is a \(\left(r_{1}n_{2}+r_{2}\right)\)-regular graph. Thus \[ DTI\left(G_{1}[G_{2}]\right) = \sum\limits_{\left(i,\, j\right)\in K}m_{i,\,j}\left(G\right)\varphi_{i,\,j}=\frac{1}{2}n_{1}n_{2}\left(r_{2}+r_{1}n_{2}\right)\varphi_{r_{1}n_{2}+r_{2},r_{1}n_{2}+r_{2}}.\] This completes the proof.
Corollary 16. Let \(G_{1}\) and \(G_{2}\) be a \(r_{1}\)-regular graph and a \(r_{2}\)-regular graph with order \(n_{1}\) and \(n_{2}\), respectively. Then \begin{align*} R^{t}\left(G_{1}[G_{2}]\right) = & \dfrac{n_{1}n_{2}\left(r_{2}+r_{1}n_{2}\right)^{2t+1}}{2},\\ Z^{t}\left(G_{1}[G_{2}]\right) = & n_{1}n_{2}\left(r_{2}+r_{1}n_{2}\right)^{t},\\ \chi^{t}\left(G_{1}[G_{2}]\right) = & 2^{t-1}n_{1}n_{2}\left(r_{2}+r_{1}n_{2}\right)^{t+1} \end{align*} for \(n_{1}\geq n_{2}\geq2\).