In this article, we compute closed forms of M-polynomial for three general classes of convex polytopes. From the M-polynomial, we derive degree-based topological indices such as first and second Zagreb indices, modified second Zagreb index, Symmetric division index, etc.
Definition 1.1. [3] The M-polynomial of G is defined as: \begin{equation*} M\left(G,x,y\right)=\sum _{\delta \le i\le j\le \Delta }m_{ij} \left(G\right)x^{i} y^{j} \end{equation*} where \(\delta ={\rm Min}\{ d_{v} |v\in {\rm V\; (G)}\} ,\) \(\Delta ={\rm Max}\{ d_{v} |v\in {\rm V\; (G)}\} ,\) and \(m_{ij} (G)\) is the edge \(vu\in E(G)\) such that \(\left\{d_{v} ,d_{u} \right\}=\left\{i,j\right\}.\)
The first topological index was introduced by Wiener [9] and it was named path number, which is now known as Wiener index. In chemical graph theory, this is the most studied molecular topological index due to its wide applications; see for details in [10, 11]. Rendić index, [12] denoted by \(R_{-1/2} (G)\) and introduced by Milan Rendić in 1975 is also one of the oldest topological index. The Rendić index is defined as \begin{equation*} R_{-1/2} (G)=\sum _{uv\in E(G)}\frac{1}{\sqrt{d_{u} d_{v} } }. \end{equation*} In 1998, working independently, Bollobás and Erdös, [13] and Amic et al. [14] proposed the generalized Rendić index which has been studied extensively by both chemists and mathematicians [15]. Many mathematical properties have been discussed [16]. For a detailed survey we refer the book [17].Topological Index | Derivation from \(M(G;x,y)\) |
---|---|
First Zagreb | \(\left(D_{x} +D_{y} \right)(M(G;x,y))_{x=y=1} \) |
Second Zagreb | \(\left(D_{x} D_{y} \right)(M(G;x,y))_{x=y=1} \) |
Second Modified Zagreb | \(\left(S_{x} S_{y} \right)(M(G;x,y))_{x=y=1} \) |
Rendić Index | \(\left(D_{x}^{\alpha } D_{y}^{\alpha } \right)(M(G;x,y))_{x=y=1} \) |
Inverse Rendić Index | \(\left(S_{x}^{\alpha } S_{y}^{\alpha } \right)(M(G;x,y))_{x=y=1} \) |
Symmetric Division Index | \(\left(D_{x} S_{y} +S_{x} D_{y} \right)(M(G;x,y))_{x=y=1} \) |
Harmonic Index | \(2S_{x} J(M(G;x,y))_{x=1} \) |
Inverse sum Index | \(S_{x} JD_{x} D_{y} (M(G;x,y))_{x=1} \) |
Augmented Zagreb Index | \(S_{x} ^{3} Q_{-2} JD_{x} ^{3} D_{y} ^{3} (M(G;x,y))_{x=1} \) |
Theorem 2.1. Assume we have a convex polytope \(T_{n}\), then the M-Polynomial of \(T_{n}\) is \begin{equation*} M\left(T_{n} ;x,y\right)=2nx^{3} y^{3} +2nx^{3} y^{6} +nx^{4} y^{4} +2nx^{4} y^{6} +nx^{6} y^{6}. \end{equation*}
Proof. Let \(G=T_{n}\) be a convex polytope. It is easy to see form Figure 1 that \begin{equation*} \left|V\left(T_{n} \right)\right|=4n, \end{equation*} \begin{equation*} \left|E\left(T_{n} \right)\right|=8n. \end{equation*} The vertex set of \(S_{n} \) has two partitions: \begin{equation*} V_{1} (T_{n} )=\left\{u\in V(T_{n} ):{\rm \; \; \; }d_{u} =3\right\}, \end{equation*} \begin{equation*} V_{2} (T_{n} )=\left\{u\in V(T_{n} ):{\rm \; \; \; }d_{u} =4\right\},\end{equation*} \begin{equation*}V_{4} (T_{n} )=\left\{u\in V(T_{n} ):{\rm \; \; \; }d_{u} =6\right\}, \end{equation*} such that \begin{equation*} \left|V_{1} (T_{n} )\right|=2n,\left|V_{2} (T_{n} )\right|=n,\left|V_{3} (T_{n} )\right|=n, \end{equation*} The edge set of \(T_{n} \) has three partitions: \begin{equation*}E_{1} (T_{n} )=\left\{e=uv\in E(T_{n} ):{\rm \; \; \; }d_{u} =d_{v} =3\right\},\end{equation*} \begin{equation*}E_{2} (T_{n} )=\left\{e=uv\in E(T_{n} ):{\rm \; \; \; }d_{u} =3,d_{v} =6\right\},\end{equation*} \begin{equation*}E_{3} (T_{n} )=\left\{e=uv\in E(S_{n} ):{\rm \; \; \; }d_{u} =d_{v} =4\right\},\end{equation*} \begin{equation*}E_{4} (T_{n} )=\left\{e=uv\in E(T_{n} ):{\rm \; \; \; }d_{u} =4,d_{v} =6\right\},\end{equation*} \begin{equation*}E_{5} (T_{n} )=\left\{e=uv\in E(T_{n} ):{\rm \; \; \; }d_{u} =d_{v} =6\right\}, \end{equation*} From Figure 1, \begin{equation*} \left|E_{1} (T_{n} )\right|=2n,\left|E_{2} (T_{n} )\right|=2n,\left|E_{3} (T_{n} )\right|=n,\left|E_{4} (T_{n} )\right|=2n,\left|E_{5} (T_{n} )\right|=n, \end{equation*} From the definition of the M-polynomial \begin{align*} M\left(T_{n} ;x\; ,y\right)&=\mathop{\sum }\limits_{i\le j} m_{ij} \left(T_{n} \right)x^{i} y^{j} \nonumber \& =\mathop{\sum }\limits_{3\le 3} m_{33} \left(T_{n} \right)x^{3} y^{3} +\mathop{\sum }\limits_{3\le 6} m_{36} \left(T_{n} \right)x^{3} y^{6} +\mathop{\sum }\limits_{4\le 4} m_{44} \left(T_{n} \right)x^{4} y^{4} \nonumber\& +\mathop{\sum }\limits_{4\le 6} m_{46} \left(T_{n} \right)x^{4} y^{6} +\mathop{\sum }\limits_{6\le 6} m_{66} \left(T_{n} \right)x^{6} y^{6} \nonumber\& =\mathop{\sum }\limits_{uv\in E_{1} } m_{33} \left(T_{n} \right)x^{3} y^{3} +\mathop{\sum }\limits_{uv\in E_{2} } m_{36} \left(T_{n} \right)x^{3} y^{6} +\mathop{\sum }\limits_{uv\in E_{3} } m_{44} \left(T_{n} \right)x^{4} y^{4} \nonumber\& +\mathop{\sum }\limits_{uv\in E_{4} } m_{46} \left(T_{n} \right)x^{4} y^{6} +\mathop{\sum }\limits_{uv\in E_{5} } m_{66} \left(T_{n} \right)x^{6} y^{6} \nonumber\& =\left|E_{1} \right|x^{3} y^{3} +\left|E_{2} \right|x^{3} y^{6} +\left|E_{3} \right|x^{4} y^{4} +\left|E_{4} \right|x^{4} y^{6} +\left|E_{5} \right|x^{6} y^{6} \nonumber\& =2nx^{3} y^{3} +2nx^{3} y^{6} +nx^{4} y^{4} +2nx^{4} y^{6} +nx^{6} y^{6} . \end{align*}
Now we compute some degree-based topological indices of double antiprism from this M-polynomial.Proposition 2.2. Let \(T_{n} \) be the double antiprism, then
Proof. Let \begin{equation*} M\left(T_{n} ;x,y\right)=f(x,y)=2nx^{3} y^{3} +2nx^{3} y^{6} +nx^{4} y^{4} +2nx^{4} y^{6} +nx^{6} y^{6} \end{equation*} Then \begin{equation*} D_{x} \left(f(x,y)\right)=6nx^{3} y^{3} +6nx^{3} y^{6} +4nx^{4} y^{4} +8nx^{4} y^{6} +6nx^{6} y^{6} ,\end{equation*} \begin{equation*}D_{y} \left(f(x,y)\right)=6nx^{3} y^{3} +12nx^{3} y^{6} +4nx^{4} y^{4} +12nx^{4} y^{6} +6nx^{6} y^{6} ,\end{equation*} \begin{equation*}\left(D_{y} D_{x} \right)\left(f(x,y)\right)=18nx^{3} y^{3} +36nx^{3} y^{6} +16nx^{4} y^{4} +48nx^{4} y^{6} +36nx^{6} y^{6} ,\end{equation*} \begin{equation*}S_{x} S_{y} (f(x,y))=\frac{2}{9} nx^{3} y^{3} +\frac{1}{9} nx^{3} y^{6} +\frac{1}{16} nx^{4} y^{4} +\frac{1}{12} nx^{4} y^{6} +\frac{1}{36} nx^{6} y^{6} ,\end{equation*} \begin{equation*}D_{x} ^{\alpha } D_{y} ^{\alpha } (f(x,y))=2\times 9^{\alpha } nx^{3} y^{3} +2\times 18^{\alpha } nx^{3} y^{6} +16^{\alpha } nx^{4} y^{4} +2\times 24^{\alpha } nx^{4} y^{6} +36^{\alpha } nx^{6} y^{6} ,\end{equation*} \begin{equation*}S_{x} ^{\alpha } S_{y} ^{\alpha } (f(x,y))=\frac{2n}{9^{\alpha } } x^{3} y^{3} +\frac{2n}{16^{\alpha } } x^{3} y^{6} +\frac{n}{16^{\alpha } } x^{4} y^{4} +\frac{2n}{24^{\alpha } } x^{4} y^{6} +\frac{n}{36^{\alpha } } x^{6} y^{6} ,\end{equation*} \begin{equation*}S_{y} D_{x} \left(f(x,y)\right)=2nx^{3} y^{3} +nx^{3} y^{6} +nx^{4} y^{4} +\frac{4n}{3} x^{4} y^{6} +nx^{6} y^{6} ,\end{equation*} \begin{equation*}S_{x} D_{y} \left(f(x,y)\right)=2nx^{3} y^{3} +4nx^{3} y^{6} +nx^{4} y^{4} +3nx^{4} y^{6} +nx^{6} y^{6} ,\end{equation*} \begin{equation*}S_{x} Jf(x,y)=\frac{n}{3} x^{6} +\frac{n}{4} x^{8} +\frac{2n}{9} x^{9} +\frac{n}{5} x^{10} +\frac{n}{12} x^{12} ,\end{equation*} \begin{equation*}S_{x} JD_{x} D_{y} \left(f(x,y)\right)==3nx^{6} +2nx^{8} +4nx^{9} +\frac{24}{5} nx^{10} +3nx^{12} ,\end{equation*} \begin{equation*}S_{x} ^{3} Q_{-2} JD_{x} ^{3} D_{y} ^{3} f(x,y)=\frac{1458}{64} nx^{4} +\frac{4096}{216} nx^{6} +\frac{11664}{343} nx^{7} +\frac{27648}{512} nx^{8} +\frac{46656}{1000} nx^{10} .\end{equation*} Now from Table 1
Theorem 2.3. Let \(A_{n}\) be the double antiprism, then the M-Polynomial of \(A_{n}\) is \begin{equation*} M\left(A_{n} \; ,\; x,y\right)=2nx^{4} y^{4} +4nx^{4} y^{6} +nx^{6} y^{6} \end{equation*}
Proof. Let \(G=A_{n}\) is double antiprism. It is easy to see form figure 2 that \begin{equation*} \left|V\left(A_{n} \right)\right|=3n, \end{equation*} \begin{equation*} \left|E\left(A_{n} \right)\right|=7n. \end{equation*} The vertex set of \(A_{n}\) has two partitions: \begin{equation*}V_{1} (A_{n} )=\left\{u\in V(A_{n} ):{\rm \; \; \; }d_{u} =4\right\}, \end{equation*} \begin{equation*} V_{2} (A_{n} )=\left\{u\in V(A_{n} ):{\rm \; \; \; }d_{u} =6\right\}, \end{equation*} such that \begin{equation*} \left|V_{1} (A_{n} )\right|=2n,\left|V_{2} (A_{n} )\right|=n. \end{equation*} The edge set of \(A_{n} \) has three partitions: \begin{equation*}E_{1} (A_{n} )=\left\{e=uv\in E(A_{n} ):{\rm \; \; \; }d_{u} =d_{v} =4\right\},\end{equation*} \begin{equation*}E_{2} (A_{n} )=\left\{e=uv\in E(A_{n} ):{\rm \; \; \; }d_{u} =4,d_{v} =6\right\},\end{equation*} \begin{equation*}E_{3} (A_{n} )=\left\{e=uv\in E(A_{n} ):{\rm \; \; \; }d_{u} =d_{v} =6\right\},\end{equation*} From Figure 2, \begin{equation*} \left|E_{1} (A_{n} )\right|=2n,\left|E_{2} (A_{n} )\right|=4n,\left|E_{3} (A_{n} )\right|=n, \end{equation*} Now from the definition of the M-polynomial \begin{align*} M\left(A_{n} ,\; x\; ,y\right)&=\mathop{\sum }\limits_{i\le j} m_{ij} \left(A_{n} \right)x^{i} y^{j} \nonumber \\ &=\mathop{\sum }\limits_{4\le 4} m_{44} \left(A_{n} \right)x^{4} y^{4} +\mathop{\sum }\limits_{4\le 6} m_{46} \left(A_{n} \right)x^{4} y^{6} {\rm \; }+\mathop{\sum }\limits_{5\le 5} m_{66} \left(A_{n} \right)x^{6} y^{6} \nonumber \\ &=\mathop{\sum }\limits_{uv\in E_{1} } m_{44} \left(A_{n} \right)x^{4} y^{4} +\mathop{\sum }\limits_{uv\in E_{2} } m_{46} \left(A_{n} \right)x^{4} y^{6} +\mathop{\sum }\limits_{uv\in E_{3} } m_{66} \left(A_{n} \right)x^{6} y^{6} \nonumber \\ &=\left|E_{1} \right|x^{4} y^{4} +\left|E_{2} \right|x^{4} y^{6} +\left|E_{3} \right|x^{6} y^{6} \nonumber \\ &=2nx^{4} y^{4} +4nx^{4} y^{6} +nx^{6} y^{6}. \end{align*}
Now we compute some degree-based topologcal indices of double antiprism from this M-polynomial.Proposition 2.4. Let \(A_{n}\) be the double antiprism, then
Theorem 2.5. Let \(S_{n}\) be the double antiprism, then the M-Polynomial of \(S_{n}\) is \begin{equation*} M\left(S_{n} ;\; x,y\right)=2nx^{3} y^{3} +2nx^{3} y^{5} +4nx^{5} y^{5} . \end{equation*}
Proof. Let \(G=S_{n}\) be the double antiprism. It is easy to see form Figure 3 that \begin{align*} \left|V\left(S_{n} \right)\right|&=4n, \\ \left|E\left(S_{n} \right)\right|&=8n. \end{align*} The vertex set of \(S_{n}\) has two partitions: \begin{align*} & V_{1} (S_{n} )=\left\{u\in V(S_{n} ):{\rm \; \; \; }d_{u} =3\right\}, \& V_{2} (S_{n} )=\left\{u\in V(S_{n} ):{\rm \; \; \; }d_{u} =5\right\}, \end{align*} such that \begin{align*} \left|V_{1} (S_{n} )\right|=2n,\left|V_{2} (S_{n} )\right|=2n. \end{align*} The edge set of \(A_{n}\) has three partitions: \begin{align*} & E_{1} (S_{n} )=\left\{e=uv\in E(S_{n} ):{\rm \; \; \; }d_{u} =d_{v} =3\right\},\& E_{2} (S_{n} )=\left\{e=uv\in E(S_{n} ):{\rm \; \; \; }d_{u} =3,d_{v} =5\right\},\& E_{3} (S_{n} )=\left\{e=uv\in E(S_{n} ):{\rm \; \; \; }d_{u} =d_{v} =5\right\}, \end{align*} From Figure 3, \begin{align*} \left|E_{1} (S_{n} )\right|=2n,\left|E_{2} (S_{n} )\right|=2n,\left|E_{3} (S_{n} )\right|=4n, \end{align*} Now from the definition of the M-polynomial \begin{align*} M\left(S_{n} ;x\; ,y\right)&=\mathop{\sum }\limits_{i\le j} m_{ij} \left(S_{n} \right)x^{i} y^{j}\nonumber \& =\mathop{\sum }\limits_{uv\in E_{1} } m_{33} \left(S_{n} \right)x^{3} y^{3} +\mathop{\sum }\limits_{uv\in E_{2} } m_{35} \left(S_{n} \right)x^{3} y^{5} +\mathop{\sum }\limits_{uv\in E_{3} } m_{55} \left(S_{n} \right)x^{5} y^{5} \nonumber \& =\left|E_{1} \right|x^{3} y^{3} +\left|E_{2} \right|x^{3} y^{5} +\left|E_{3} \right|x^{5} y^{5} \nonumber \& =2nx^{3} y^{3} +2nx^{3} y^{5} +4nx^{5} y^{5} . \end{align*}
Now we compute some degree-based topologcal indices of double antiprism from this M-polynomial.Proposition 2.6. Let \(A_{n}\) be the double antiprism, then