A basic introduction to single particles cryo-electron microscopy

Vittoria Raimondi; Alessandro Grinzato; Vittoria Raimondi; Alessandro Grinzato

doi:10.3934/biophy.2022002

AIMS Biophysics

2022, Volume 9, Issue 1: 5-20. doi: 10.3934/biophy.2022002

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A basic introduction to single particles cryo-electron microscopy

Vittoria Raimondi ¹,
Alessandro Grinzato ^{2,3
,
,}

1.
Veneto Institute of Oncology IOV – IRCCS, via Gattamelata 64, 35128, Padova, Italy
2.
Department of Biomedical Sciences, University of Padova, Via Ugo Bassi 58b 35131 Padova, Italy
3.
Present address: ESRF, The European Synchrotron, 71 Avenue des Martyrs, 38043 Grenoble, France

Received: 31 October 2021 Revised: 25 November 2021 Accepted: 02 December 2021 Published: 30 December 2021

In the last years, cryogenic-electron microscopy (cryo-EM) underwent the most impressive improvement compared to other techniques used in structural biology, such as X-ray crystallography and NMR. Electron microscopy was invented nearly one century ago but, up to the beginning of the last decades, the 3D maps produced through this technique were poorly detailed, justifying the term “blobbology” to appeal to cryo-EM. Recently, thanks to a new generation of microscopes and detectors, more efficient algorithms, and easier access to computational power, single particles cryo-EM can routinely produce 3D structures at resolutions comparable to those obtained with X-ray crystallography. However, unlike X-ray crystallography, which needs crystallized proteins, cryo-EM exploits purified samples in solution, allowing the study of proteins and protein complexes that are hard or even impossible to crystallize. For these reasons, single-particle cryo-EM is often the first choice of structural biologists today. Nevertheless, before starting a cryo-EM experiment, many drawbacks and limitations must be considered. Moreover, in practice, the process between the purified sample and the final structure could be trickier than initially expected. Based on these observations, this review aims to offer an overview of the principal technical aspects and setups to be considered while planning and performing a cryo-EM experiment.

Keywords:

Citation: Vittoria Raimondi, Alessandro Grinzato. A basic introduction to single particles cryo-electron microscopy[J]. AIMS Biophysics, 2022, 9(1): 5-20. doi: 10.3934/biophy.2022002

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Abstract

1. Introduction

Graph theory has provided the researcher with various useful tools, such as graph labeling, locating numbers and topological indices. Graph theory subject has many applications and implementations in different research subjects like chemistry, medicine and engineering. A graph can be recognized by a numeric value, a polynomial, a sequence of numbers or a matrix. The representation of the chemical compound in terms of diagram, known as its molecular graph, in which its atoms and the chemical bonding between them represent the nodes and edges, respectively. Recently, a new subject caught attention of the researchers was introduced, which is the combination of chemistry, information science and mathematics is called Cheminformatics, which studies QSAR/QSPR relationship, bioactivity and characterization of chemical compounds ^[1].

The topological index is a numeric value related with chemical compositions maintaining the correlation of chemical structures with many physico-chemical properties, chemical reactivity or biological activity. Topological indices are prepared on the grounds of the transformation of a chemical network into a number that describes the topology of the chemical network. Some of the main types of topological indices of graphs are distance-based topological indices, degree-based topological indices, and counting-related topological indices. Recently, numerous researchers have found topological indices for the study of fundamental properties of molecular graph or network. These networks have very motivating topological properties which have been considered in different characteristics in ^{[2,3,4,5,6,7,8,9]}.

Let $G = (V, E)$ be a simple connected graph, with $V$ be the vertex set and $E$ be the edge set of graph $G,$ with order $|V| = p$ , size $|E| = q.$ The number of edges incident with a vertex $\omega$ is known as the degree of $\omega$ , denoted by $\zeta(\omega)$ . The reverse vertex degree ( $\mathfrak{R}(\omega)$ ) was introduced by Kulli ^[10] defined as: $\mathfrak{R}(\omega) = 1-\zeta(\omega)+\Delta,$ where $\Delta$ denoted the maximum degree of the given graph. Let $\mathbf{E}_{\mathfrak{R}(\omega), \mathfrak{R}(\mu)}$ represents the edge partition of the given graph based on reverse degree of end vertices of an edge $\omega\mu\in E$ and $|\mathbf{E}_{\mathfrak{R}(\omega), \mathfrak{R}(\mu)}|$ represents its cardinality. There are detailed variations of topological indices mainly distance-based and degree-based indices, see ^{[11,12,13,14,15,16]}. Milan Randic ^[30] was the first who defined the degree-based indices and its reverse Randic index is defined as:

$\begin{equation} \mathfrak{R}\mathbb{R}_{\alpha}(G) = \sum\limits_{\omega\mu\in E(G)}\bigg( \mathfrak{R}(\omega)\times \mathfrak{R}(\mu) \bigg)^{\alpha}, \, \, \, \alpha = \frac{1}{2}, -\frac{1}{2}, 1, -1. \end{equation}$

(1.1)

Estrada et al. presented the atom bond connectivity ( $\mathbb{ABC}$ ) index in ^[18] and the reverse atom bond connectivity ( $\mathfrak{R}\mathbb{ABC}$ ) is defined as:

$\begin{equation} \mathfrak{R}\mathbb{ABC}(G) = \sum\limits_{\omega\mu\in E(G)}\sqrt{\frac{ \mathfrak{R}(\omega)+ \mathfrak{R}(\mu)-2}{ \mathfrak{R}(\omega)\times \mathfrak{R}(\mu)}} \end{equation}$

(1.2)

Vukicevic and Furtula defined the geometric arithmetic ( $\mathbb{GA}$ ) index in ^[19] and the reverse geometric arithmetic ( $\mathfrak{R}\mathbb{GA}$ ) is presented as:

$\begin{equation} \mathfrak{R}\mathbb{GA}(G) = \sum\limits_{\omega\mu\in E(G)}\frac{ 2\, \sqrt{\mathfrak{R}(\omega)\times \mathfrak{R}(\mu)}}{ \mathfrak{R}(\omega)+ \mathfrak{R}(\mu)} \end{equation}$

(1.3)

Gutman et al. ^[20,21] defined the first and second Zagreb and its reverse indices as:

$\begin{eqnarray} \mathfrak{R}\mathbb{M}_1(G) & = & \sum\limits_{\omega\mu\in E(G)}\bigg(\mathfrak{R}(\omega)+ \mathfrak{R}(\mu)\bigg) \end{eqnarray}$

(1.4)

$\begin{eqnarray} \mathfrak{R}\mathbb{M}_2(G) & = & \sum\limits_{\omega\mu\in E(G)}\bigg(\mathfrak{R}(\omega)\times \mathfrak{R}(\mu)\bigg) \end{eqnarray}$

(1.5)

Shirdel et al. ^[22] introduced hyper Zagreb index. We defined the reverse hyper Zagreb index as:

$\begin{equation} \mathfrak{R}\mathbb{HM}(G) = \sum\limits_{\omega\mu\in E(G)}\bigg( \mathfrak{R}(\omega)+ \mathfrak{R}(\mu) \bigg)^{2} \end{equation}$

(1.6)

Furtula and Gutman ^[23] accomplished the forgotten index and its reverse forgotten index as:

$\begin{equation} \mathfrak{R}\mathbb{F}(G) = \sum\limits_{\omega\mu\in E(G)}\bigg( \left(\mathfrak{R}(\omega)\right)^2+ \left(\mathfrak{R}(\mu)\right)^{2} \bigg) \end{equation}$

(1.7)

Augmented Zagreb index was introduced by Furtula et al. ^[24] and the reverse augmented Zagreb index as:

$\begin{equation} \mathfrak{R}\mathbb{AZI}(G) = \sum\limits_{\omega\mu\in E(G)}\bigg( \frac{\mathfrak{R}(\omega)\times \mathfrak{R}(\mu)}{\mathfrak{R}(\omega)+ \mathfrak{R}(\mu)-2} \bigg)^{3} \end{equation}$

(1.8)

Ranjini et al. ^[25] introduced the first redefined, second redefined and third redefined Zagreb indices. The reverse first redefined, second redefined and third redefined Zagreb indices are defined as:

$\begin{eqnarray} \mathfrak{R}\mathbb{ReZ}_1(G) & = & \sum\limits_{\omega\mu\in E(G)} \frac{\mathfrak{R}(\omega)+ \mathfrak{R}(\mu)}{\mathfrak{R}(\omega)\times \mathfrak{R}(\mu)} \end{eqnarray}$

(1.9)

$\begin{eqnarray} \mathfrak{R}\mathbb{ReZ}_2(G) & = & \sum\limits_{\omega\mu\in E(G)} \frac{\mathfrak{R}(\omega)\times \mathfrak{R}(\mu)}{\mathfrak{R}(\omega)+ \mathfrak{R}(\mu)} \end{eqnarray}$

(1.10)

$\begin{eqnarray} \mathfrak{R}\mathbb{ReZ}_3(G) & = & \sum\limits_{\omega\mu\in E(G)} \bigg(\mathfrak{R}(\omega)+ \mathfrak{R}(\mu)\bigg)\bigg(\mathfrak{R}(\omega)\times \mathfrak{R}(\mu)\bigg) \end{eqnarray}$

(1.11)

For latest results on topological indices see ^{[26,27,28,29,31,32,33,34,35]}. In this paper, we compute the exact results for all the above reverse indices.

2. Structure of third type hex-derived networks

With the help of complete graphs of order 3 ( $K_3$ ), Chen et al. ^[36] assembled a hexagonal mesh. In terms of chemistry, these $K_3$ graphs are also called oxide graphs. The is obtained by joining these $K_3$ graphs. Two dimensional mesh graph HX(2) (see ), is obtained by joining six $K_3$ graphs and three dimensional mesh graph HX(3) (see ) is obtained by putting $K_3$ graphs around all side of HX(2). Furthermore, repeating the same process by putting the $t$ $K_3$ graph around each hexagon, we obtained the $t^{th}$ hexagonal mesh. To be noted that the one dimensional hexagonal mesh graph does not exist.

Figure 1. Hexagonal meshes: (a)

$HX(2)$ and (b)

$HX(3)$ .

DownLoad: Full-Size Img PowerPoint

Simonraj et al. ^[37] created the new network which is named as third type of hex-derived networks. The graphically construction algorithm for third type of hexagonal hex-derived network $HHDN_3(t)$ (see ), triangular hex-derived network $THDN_3(t)$ (see ) and rectangular hex-derived network $RHDN_3(t)$ (see Figure 4) are defined in ^[38,39] and they determined some topological indices of these new derived networks. Some networks such as hexagonal, honeycomb, and grid networks, for instance, endure closeness to atomic or molecular lattice configurations. Related research that applies this theory and which could get additional advantages from the visions of the new research is found in ^{[40,41,42,43,44,45,46]}.

Figure 2. Third type of hexagonal hex-derived network

$HHDN_3(t)$ for

$t = 4$ .

DownLoad: Full-Size Img PowerPoint

Figure 3. Third type of triangular hex-derived network

$THDN_3(t)$ for

$t = 7$ .

DownLoad: Full-Size Img PowerPoint

Figure 4. Third type of rectangular hex-derived network

$RHDN_3(t)$ for

$t = 4$ .

DownLoad: Full-Size Img PowerPoint

3. Hexagonal hex-derived network $HHDN_3(t)$

Let $\Gamma_1 = HHDN_3(t)$ be the third type of hexagonal hex-derived network which is shown in , where $t \ge 4.$ The graph $\Gamma_1$ has $21t^2 -39t + 19$ vertices from which $18t^2-36t+18$ vertices of reverse degree $15$ , $4$ vertices of reverse degree $12$ , $6t-12$ vertices of reverse degree $9$ and $3t^2-9t+9$ vertices of reverse degree $1.$ There are $63t^2-123t+60$ number of edges of $\Gamma_1$ is partitioned into nine classes based on their reverse degrees which are given in Eq (3.1).

$\begin{equation} |\mathbf{E}_{\mathfrak{R}(\omega), \mathfrak{R}(\mu)}(\Gamma_1)| = \left\{ \begin{array}{ll} 9t^2-33t+30, & \hbox{for;}\ \mathfrak{R}(\omega) = 1, \mathfrak{R}(\mu) = 1 \\ 12t-24, & \hbox{for;}\ \mathfrak{R}(\omega) = 9, \mathfrak{R}(\mu) = 1 \\ 6t-18, & \hbox{for;}\ \mathfrak{R}(\omega) = 9, \mathfrak{R}(\mu) = 9 \\ 6, & \hbox{for;}\ \mathfrak{R}(\omega) = 12, \mathfrak{R}(\mu) = 1 \\ 12, & \hbox{for;}\ \mathfrak{R}(\omega) = 12, \mathfrak{R}(\mu) = 9 \\ 36t^2-108t+84, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 1 \\ 36t-72, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 9 \\ 24, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 12 \\ 18t^2-36t+18, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 15 \end{array} \right. \end{equation}$

(3.1)

In the next theorem, we will calculate the reverse Randic index, reverse Zagreb indices, reverse forgotten index for $\Gamma_1$ graph.

Theorem 3.1. Let $\Gamma_1$ be the third type of hexagonal hex-derived network, then

● $\mathfrak{R}\mathbb{R}_{\alpha}(\Gamma_1) = [9+36(15)^{\alpha}+18(225)^{\alpha}]t^2+[-33+12(9)^{\alpha}+6(81)^{\alpha}-108(15)^{\alpha}+36(135)^{\alpha}-36(225)^{\alpha}]t +30-24(9)^{\alpha}-18(81)^{\alpha}+6(12)^{\alpha}+12(108)^{\alpha}+84(15)^{\alpha}-72(135)^{\alpha}+24(180)^{\alpha}+18(225)^{\alpha}$

● $\mathfrak{R}\mathbb{M}_1(\Gamma_1) = 1134\, {t}^{2}-1782\, t+630$

● $\mathfrak{R}\mathbb{M}_2(\Gamma_1) = 4599\, {t}^{2}-4299\, t-366$

● $\mathfrak{R}\mathbb{HM}(\Gamma_1) = 25452\, {t}^{2}-36300\, t+11922$

● $\mathfrak{R}\mathbb{F}(\Gamma_1) = 16254\, {t}^{2}-27702\, t+12654$

Proof. Let $\Gamma_1$ be the third type of hexagonal hex-derived network which is shown in . The order of hexagonal hex derived network $\Gamma_1$ is $p = |\Gamma_1| = 21t^2 -39t + 19$ and size is $q = 63t^2-123t+60.$ The edge partitioned of $\Gamma_1$ based on their reverse degrees are shown in Eq (3.1). Reverse Randic index can be calculated by using Eq (3.1). Thus, from Eq (1.1), it follows,

$\mathfrak{R}\mathbb{R}_{\alpha}(\Gamma_1) = \left.(1)^{\alpha}|\mathbf{E}_{1, 1}(\Gamma_1)|+(9)^{\alpha}|\mathbf{E}_{9, 1}(\Gamma_1)| +(81)^{\alpha}|\mathbf{E}_{9, 9}(\Gamma_1)|+(12)^{\alpha}|\mathbf{E}_{12, 1}(\Gamma_1)|\right.\\ +(108)^{\alpha}|\mathbf{E}_{12, 9}(\Gamma_1)| +(15)^{\alpha}|\mathbf{E}_{15, 1}(\Gamma_1)| +(135)^{\alpha}|\mathbf{E}_{15, 9}(\Gamma_1)|+(180)^{\alpha}|\mathbf{E}_{15, 12}(\Gamma_1)| +(225)^{\alpha}|\mathbf{E}_{15, 15}(\Gamma_1)|.$

$= (9t^2-33t+30)+(9)^{\alpha}(12t-24) +(81)^{\alpha}(6t-18)+(12)^{\alpha}(6) +(108)^{\alpha}(12) +(15)^{\alpha}(36t^2-108t+84) +(135)^{\alpha}(36t-72)+(180)^{\alpha}(24) +(225)^{\alpha}(18t^2-36t+18).$

$= [9+36(15)^{\alpha}+18(225)^{\alpha}]t^2+[-33+12(9)^{\alpha}+6(81)^{\alpha}-108(15)^{\alpha}+36(135)^{\alpha}-36(225)^{\alpha}]t +30-24(9)^{\alpha}-18(81)^{\alpha}+6(12)^{\alpha}+12(108)^{\alpha}+84(15)^{\alpha}-72(135)^{\alpha}+24(180)^{\alpha}+18(225)^{\alpha}.$ Put $\alpha = 1$ and after some calculation, we get reverse second Zagreb index $\mathfrak{R}\mathbb{M}_2(\Gamma_1)$ as:

$\begin{equation} \mathfrak{R}\mathbb{M}_2(\Gamma_1) = 4599\, {t}^{2}-4299\, t-366. \end{equation}$

(3.2)

Using the Eq (1.4), we can determine the reverse first Zagreb index $\mathfrak{R}\mathbb{M}_1(\Gamma_1)$ as:

$\mathfrak{R}\mathbb{M}_1(\Gamma_1) = 2\times|\mathbf{E}_{1, 1}(\Gamma_1)|+10\times|\mathbf{E}_{9, 1}(\Gamma_1)| +18\times|\mathbf{E}_{9, 9}(\Gamma_1)|+13\times|\mathbf{E}_{12, 1}(\Gamma_1)|+21\times|\mathbf{E}_{12, 9}(\Gamma_1)| +16\times|\mathbf{E}_{15, 1}(\Gamma_1)| +24\times|\mathbf{E}_{15, 9}(\Gamma_1)|+27\times|\mathbf{E}_{15, 12}(\Gamma_1)| +30\times|\mathbf{E}_{15, 15}(\Gamma_1)|.$

By putting the values of from equation (3.1) and after simplification, we obtain:

$\begin{equation} \mathfrak{R}\mathbb{M}_1(\Gamma_1) = 1134\, {t}^{2}-1782\, t+630. \end{equation}$

(3.3)

Using the Eq (1.6), we can determine the reverse hyper Zagreb index $\mathfrak{R}\mathbb{HM}(\Gamma_1)$ as:

$\mathfrak{R}\mathbb{HM}(\Gamma_1) = 4\times|\mathbf{E}_{1, 1}(\Gamma_1)|+100\times|\mathbf{E}_{9, 1}(\Gamma_1)| +324\times|\mathbf{E}_{9, 9}(\Gamma_1)|+169\times|\mathbf{E}_{12, 1}(\Gamma_1)|+441\times|\mathbf{E}_{12, 9}(\Gamma_1)| +256\times|\mathbf{E}_{15, 1}(\Gamma_1)| +576\times|\mathbf{E}_{15, 9}(\Gamma_1)|+729\times|\mathbf{E}_{15, 12}(\Gamma_1)| +900\times|\mathbf{E}_{15, 15}(\Gamma_1)|.$

After simplification, we get

$\begin{equation*} \mathfrak{R}\mathbb{HM}(\Gamma_1) = 25452\, {t}^{2}-36300\, t+11922 \end{equation*}$

Using the Eq (1.7), we can determine the reverse forgotten index $\mathfrak{R}\mathbb{F}(\Gamma_1))$ as:

$\mathfrak{R}\mathbb{F}(\Gamma_1)) = 2\times|\mathbf{E}_{1, 1}(\Gamma_1)|+82\times|\mathbf{E}_{9, 1}(\Gamma_1)| +162\times|\mathbf{E}_{9, 9}(\Gamma_1)|+145\times|\mathbf{E}_{12, 1}(\Gamma_1)|+225\times|\mathbf{E}_{12, 9}(\Gamma_1)| +226\times|\mathbf{E}_{15, 1}(\Gamma_1)| +306\times|\mathbf{E}_{15, 9}(\Gamma_1)|+369\times|\mathbf{E}_{15, 12}(\Gamma_1)| +450\times|\mathbf{E}_{15, 15}(\Gamma_1)|.$

After simplification, we get

$\begin{equation*} \mathfrak{R}\mathbb{F}(\Gamma_1) = 16254\, {t}^{2}-27702\, t+12654 \end{equation*}$

In the next theorem, we will calculate the reverse atom bond connectivity index, reverse geometric arithmetic index for $\Gamma_1$ graph.

Theorem 3.2. Let $\Gamma_1$ be the third type of hexagonal hex-derived network, then

● $\mathfrak{R}\mathbb{ABC}(\Gamma_1) = \left({\frac {12\, \sqrt {210}}{5}}+{\frac {12\, \sqrt {7}}{5}} \right) {t}^{2}+ \left(8\, \sqrt {2}+\frac{8}{3}-{\frac {36\, \sqrt {210}}{5}} +\frac{4\, \sqrt {330}}{5}-{\frac {24\, \sqrt {7}}{5}} \right) t-8-16\, \sqrt {2} +\sqrt {33}+\frac{2\, \sqrt {57}}{3}+{\frac {28\, \sqrt {210}}{5}}-\frac{8\, \sqrt { 330}}{5}+4\, \sqrt {5}+{\frac {12\, \sqrt {7}}{5}}$

● $\mathfrak{R}\mathbb{GA}(\Gamma_1) = \left(27+\frac{9\, \sqrt {15}}{2} \right) {t}^{2}+ \left(-{\frac{279}{5}}-\frac{9\, \sqrt {15}}{2} \right) t+{\frac{78}{5}}+{\frac {792\, \sqrt {3}}{91}}-\frac{15\, \sqrt {15}}{2}+{\frac {32\, \sqrt {5}}{3}}.$

Proof. The reverse atom bond connectivity ( $\mathfrak{R}\mathbb{ABC}(\Gamma_1)$ ), can be determined by using Eq (1.2) and Eq (3.1), as follows:

$\mathfrak{R}\mathbb{ABC}(\Gamma_1) = 0\times|\mathbf{E}_{1, 1}(\Gamma_1)|+\sqrt{\frac{8}{9}}\times|\mathbf{E}_{9, 1}(\Gamma_1)| +\sqrt{\frac{16}{81}}\times|\mathbf{E}_{9, 9}(\Gamma_1)|+\sqrt{\frac{11}{12}}\times|\mathbf{E}_{12, 1}(\Gamma_1)| +\sqrt{\frac{19}{108}}\times|\mathbf{E}_{12, 9}(\Gamma_1)| +\sqrt{\frac{14}{15}}\times|\mathbf{E}_{15, 1}(\Gamma_1)| +\sqrt{\frac{22}{135}}\times|\mathbf{E}_{15, 9}(\Gamma_1)|+\sqrt{\frac{25}{180}}\times|\mathbf{E}_{15, 12}(\Gamma_1)| +\sqrt{\frac{28}{225}}\times|\mathbf{E}_{15, 15}(\Gamma_1)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ABC}(\Gamma_1) = \left({\frac {12\, \sqrt {210}}{5}}+{\frac {12\, \sqrt {7}}{5}} \right) {t}^{2}+ \left(8\, \sqrt {2}+\frac{8}{3}-{\frac {36\, \sqrt {210}}{5}} +\frac{4\, \sqrt {330}}{5}-{\frac {24\, \sqrt {7}}{5}} \right) t-8-16\, \sqrt {2} +\sqrt {33}+\frac{2\, \sqrt {57}}{3}+{\frac {28\, \sqrt {210}}{5}}-\frac{8\, \sqrt { 330}}{5}+4\, \sqrt {5}+{\frac {12\, \sqrt {7}}{5}}.$

The reverse geometric arithmetic ( $\mathfrak{R}\mathbb{GA}(\Gamma_1)$ ), can be determined by using Eq (1.3) and Eq (3.1), as follows:

$\mathfrak{R}\mathbb{GA}(\Gamma_1) = |\mathbf{E}_{1, 1}(\Gamma_1)|+\frac{2\, \sqrt{9}}{10}\times|\mathbf{E}_{9, 1}(\Gamma_1)| +\frac{2\, \sqrt{81}}{18}\times|\mathbf{E}_{9, 9}(\Gamma_1)|+\frac{2\, \sqrt{12}}{13}\times|\mathbf{E}_{12, 1}(\Gamma_1)| +\frac{2\, \sqrt{108}}{21}\times|\mathbf{E}_{12, 9}(\Gamma_1)| +\frac{2\, \sqrt{15}}{16}\times|\mathbf{E}_{15, 1}(\Gamma_1)| +\frac{2\, \sqrt{135}}{24}\times|\mathbf{E}_{15, 9}(\Gamma_1)|+\frac{2\, \sqrt{180}}{27}\times|\mathbf{E}_{15, 12}(\Gamma_1)| +\frac{2\, \sqrt{225}}{30}\times|\mathbf{E}_{15, 15}(\Gamma_1)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{GA}(\Gamma_1) = \left(27+\frac{9\, \sqrt {15}}{2} \right) {t}^{2}+ \left(-{\frac{279}{5}}-\frac{9\, \sqrt {15}}{2} \right) t+{\frac{78}{5}}+{\frac {792\, \sqrt {3}}{91}}-\frac{15\, \sqrt {15}}{2}+{\frac {32\, \sqrt {5}}{3}}.$ In the next theorems, we will calculate the reverse redefined Zagreb indices for $\Gamma_1$ graph.

Theorem 3.3. Let $\Gamma_1$ be the third type of hexagonal hex-derived network, then

● $\mathfrak{R}\mathbb{ReZ}_1(\Gamma_1) = {\frac {294\, {t}^{2}}{5}}-{\frac {2474\, t}{15}}+{\frac{3629}{30}}$

● $\mathfrak{R}\mathbb{ReZ}_2(\Gamma_1) = {\frac {693\, {t}^{2}}{4}}-{\frac {2949\, t}{20}}-{\frac{93907}{1820}}$

● $\mathfrak{R}\mathbb{ReZ}_3(\Gamma_1) = 130158\, {t}^{2}-142518\, t+24828$

Proof. Reverse redefined Zagreb indices can be calculated by using Eq (3.1), the $\mathfrak{R}\mathbb{ReZ}_1(\Gamma_1)$ by using Eq (1.9) as follows:

$\mathfrak{R}\mathbb{ReZ}_1(\Gamma_1) = 2\times|\mathbf{E}_{1, 1}(\Gamma_1)|+\frac{10}{9}\times|\mathbf{E}_{9, 1}(\Gamma_1)| +\frac{18}{81}\times|\mathbf{E}_{9, 9}(\Gamma_1)|+\frac{13}{12}\times|\mathbf{E}_{12, 1}(\Gamma_1)| +\frac{21}{108}\times|\mathbf{E}_{12, 9}(\Gamma_1)| +\frac{16}{15}\times|\mathbf{E}_{15, 1}(\Gamma_1)| +\frac{24}{135}\times|\mathbf{E}_{15, 9}(\Gamma_1)|+\frac{27}{108}\times|\mathbf{E}_{15, 12}(\Gamma_1)| +\frac{30}{225}\times|\mathbf{E}_{15, 15}(\Gamma_1)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ReZ}_1(\Gamma_1) = {\frac {294\, {t}^{2}}{5}}-{\frac {2474\, t}{15}}+{\frac{3629}{30}}.$

The $\mathfrak{R}\mathbb{ReZ}_2(\Gamma_1)$ can be determined by using Eq (1.10) as follows:

$\mathfrak{R}\mathbb{ReZ}_2(\Gamma_1) = \frac{1}{2}\times|\mathbf{E}_{1, 1}(\Gamma_1)|+\frac{9}{10}\times|\mathbf{E}_{9, 1}(\Gamma_1)| +\frac{81}{18}\times|\mathbf{E}_{9, 9}(\Gamma_1)|+\frac{12}{13}\times|\mathbf{E}_{12, 1}(\Gamma_1)| +\frac{108}{21}\times|\mathbf{E}_{12, 9}(\Gamma_1)| +\frac{15}{16}\times|\mathbf{E}_{15, 1}(\Gamma_1)| +\frac{135}{24}\times|\mathbf{E}_{15, 9}(\Gamma_1)|+\frac{108}{27}\times|\mathbf{E}_{15, 12}(\Gamma_1)| +\frac{225}{30}\times|\mathbf{E}_{15, 15}(\Gamma_1)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ReZ}_2(\Gamma_1) = {\frac {693\, {t}^{2}}{4}}-{\frac {2949\, t}{20}}-{\frac{93907}{1820}}.$

The $\mathfrak{R}\mathbb{ReZ}_3(\Gamma_1)$ can be calculated by using Eq (1.11) as follows:

$\mathfrak{R}\mathbb{ReZ}_3(\Gamma_1) = 2\times|\mathbf{E}_{1, 1}(\Gamma_1)|+90\times|\mathbf{E}_{9, 1}(\Gamma_1)| +1458\times|\mathbf{E}_{9, 9}(\Gamma_1)|+156\times|\mathbf{E}_{12, 1}(\Gamma_1)| +2268\times|\mathbf{E}_{12, 9}(\Gamma_1)| +240\times|\mathbf{E}_{15, 1}(\Gamma_1)| +3240\times|\mathbf{E}_{15, 9}(\Gamma_1)|+2916\times|\mathbf{E}_{15, 12}(\Gamma_1)| +6750\times|\mathbf{E}_{15, 15}(\Gamma_1)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ReZ}_3(\Gamma_1) = 130158\, {t}^{2}-142518\, t+24828.$

4. Triangular hex-derived network $THDN3(p)$

Let $\Gamma_2 = THDN_3(t)$ be the third type of triangular hex-derived network which is shown in , where $t \ge 4.$ The graph $\Gamma_2$ has $\frac{7t^2-11t+6}{2}$ vertices. There are $\frac{21t^2-39t+18}{2}$ number of edges of $\Gamma_2$ is partitioned into six classes based on their reverse degrees which are given in Eq (4.1). Now we calculated reverse degree based indices such as: reverse Randic index $\mathfrak{R}\mathbb{R}_{\alpha},$ reverse atom bond connectivity index $\mathfrak{R}\mathbb{ABC},$ reverse geometric arithmetic index $\mathfrak{R}\mathbb{GA},$ first reverse Zagreb index $\mathfrak{R}\mathbb{M}_1,$ second reverse Zagreb index $\mathfrak{R}\mathbb{M}_2$ , reverse forgotten index $\mathfrak{R}\mathbb{F},$ reverse hyper Zagreb index $\mathfrak{R}\mathbb{HM}$ and reverse redefined Zagreb indices for $\Gamma_2$ graph.

$\begin{equation} |\mathbf{E}_{\mathfrak{R}(\omega), \mathfrak{R}(\mu)}(\Gamma_2)| = \left\{ \begin{array}{ll} \frac{3t^2}{2}-\frac{21t}{2}+18, & \hbox{for;}\ \mathfrak{R}(\omega) = 1, \mathfrak{R}(\mu) = 1 \\ 6t-18, & \hbox{for;}\ \mathfrak{R}(\omega) = 9, \mathfrak{R}(\mu) = 1 \\ 3t-6, & \hbox{for;}\ \mathfrak{R}(\omega) = 9, \mathfrak{R}(\mu) = 9 \\ 6t^2-30t+36, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 1 \\ 18t-30, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 9 \\ 3t^2-6t+9, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 15 \end{array} \right. \end{equation}$

(4.1)

In the next theorem, we will calculate the reverse Randic index, reverse Zagreb indices, reverse forgotten index for $\Gamma_2$ graph.

Theorem 4.1. Let $\Gamma_2$ be the third type of triangular hex-derived network, then

● $\mathfrak{R}\mathbb{R}_{\alpha}(\Gamma_2) = \left(\frac{3}{2}+6\, (15)^{\alpha}+3\, (225)^{\alpha} \right) {t}^{2}+ \left(\frac{-21}{2}+6\, (9)^{\alpha}+3\, (81)^{\alpha}-30\, (15)^{\alpha}+\right. \\ \left. 18\, ( 135)^{\alpha}-6\, (225)^{\alpha} \right) t+18-18\, (9)^{\alpha}-6\, (81)^ {\alpha}+36\, (15)^{\alpha}-30\, (135)^{\alpha}+9\, (225)^{\alpha}.$

● $\mathfrak{R}\mathbb{M}_1(\Gamma_2) = 189\, {t}^{2}-135\, t-126$

● $\mathfrak{R}\mathbb{M}_2(\Gamma_2) = {\frac {1533\, {t}^{2}}{2}}+{\frac {1833\, t}{2}}-2115$

● $\mathfrak{R}\mathbb{HM}(\Gamma_2) = 4242\, {t}^{2}-1182\, t-3636$

● $\mathfrak{R}\mathbb{F}(\Gamma_2) = 2709\, {t}^{2}-3015\, t+594$

Proof. Let $\Gamma_2$ be the third type of triangular hex-derived network which is shown in . The order of triangular hex derived network $\Gamma_2$ is $p = |\Gamma_2| = \frac{7t^2-11t+6}{2}$ and size is $q = \frac{21t^2-39t+18}{2}.$ The edge partitioned of $\Gamma_2$ based on their reverse degrees are shown in Eq (4.1). Reverse Randic index can be calculated by using Eq (4.1). Thus, from Eq (1.1), it follows,

$\mathfrak{R}\mathbb{R}_{\alpha}(\Gamma_2) = (1)^{\alpha}|\mathbf{E}_{1, 1}(\Gamma_2)|+(9)^{\alpha}|\mathbf{E}_{9, 1}(\Gamma_2)| +(81)^{\alpha}|\mathbf{E}_{9, 9}(\Gamma_2)| +(15)^{\alpha}|\mathbf{E}_{15, 1}(\Gamma_2)| +(135)^{\alpha}|\mathbf{E}_{15, 9}(\Gamma_2)|+(225)^{\alpha}|\mathbf{E}_{15, 15}(\Gamma_2)|.$

After simplification, we get

$\mathfrak{R}\mathbb{R}_{\alpha}(\Gamma_2) = \left(\frac{3}{2}+6\, (15)^{\alpha}+3\, (225)^{\alpha} \right) {t}^{2}+ \left(\frac{-21}{2}+6\, (9)^{\alpha}+3\, (81)^{\alpha}-30\, (15)^{\alpha}+\right. \\ \left. 18\, (135)^{\alpha}-6\, (225)^{\alpha} \right) t+18-18\, (9)^{\alpha}-6\, (81)^ {\alpha}+36\, (15)^{\alpha}-30\, (135)^{\alpha}+9\, (225)^{\alpha}.$

Put $\alpha = 1$ and after some calculation, we get reverse second Zagreb index $\mathfrak{R}\mathbb{M}_2(\Gamma_2)$ as:

$\begin{equation} \mathfrak{R}\mathbb{M}_2(\Gamma_2) = {\frac {1533\, {t}^{2}}{2}}+{\frac {1833\, t}{2}}-2115 \end{equation}$

(4.2)

Using the Eq (1.4), we can determine the reverse first Zagreb index $\mathfrak{R}\mathbb{M}_1(\Gamma_2)$ as:

$\mathfrak{R}\mathbb{M}_1(\Gamma_2) = 2\times|\mathbf{E}_{1, 1}(\Gamma_1)|+10\times|\mathbf{E}_{9, 1}(\Gamma_1)| +18\times|\mathbf{E}_{9, 9}(\Gamma_1)|+16\times|\mathbf{E}_{15, 1}(\Gamma_1)| +24\times|\mathbf{E}_{15, 9}(\Gamma_1)|+30\times|\mathbf{E}_{15, 15}(\Gamma_1)|.$

By putting the values of from Eq (4.1) and after simplification, we obtain:

$\begin{equation} \mathfrak{R}\mathbb{M}_1(\Gamma_2) = 189\, {t}^{2}-135\, t-126 \end{equation}$

(4.3)

Using the Eq (1.6), we can determine the reverse hyper Zagreb index $\mathfrak{R}\mathbb{HM}(\Gamma_2)$ as:

$\mathfrak{R}\mathbb{HM}(\Gamma_2) = 4\times|\mathbf{E}_{1, 1}(\Gamma_2)|+100\times|\mathbf{E}_{9, 1}(\Gamma_2)| +324\times|\mathbf{E}_{9, 9}(\Gamma_2)|+256\times|\mathbf{E}_{15, 1}(\Gamma_2)| +576\times|\mathbf{E}_{15, 9}(\Gamma_2)|+900\times|\mathbf{E}_{15, 15}(\Gamma_2)|.$

After simplification, we get

$\begin{equation*} \mathfrak{R}\mathbb{HM}(\Gamma_2) = 4242\, {t}^{2}-1182\, t-3636. \end{equation*}$

Using the Eq (1.7), we can determine the reverse forgotten index $\mathfrak{R}\mathbb{F}(\Gamma_2))$ as:

$\mathfrak{R}\mathbb{F}(\Gamma_2)) = 2\times|\mathbf{E}_{1, 1}(\Gamma_2)|+82\times|\mathbf{E}_{9, 1}(\Gamma_2)| +162\times|\mathbf{E}_{9, 9}(\Gamma_2)|+226\times|\mathbf{E}_{15, 1}(\Gamma_2)| +306\times|\mathbf{E}_{15, 9}(\Gamma_2)|+450\times|\mathbf{E}_{15, 15}(\Gamma_2)|.$

After simplification, we get

$\begin{equation*} \mathfrak{R}\mathbb{F}(\Gamma_2) = 2709\, {t}^{2}-3015\, t+594. \end{equation*}$

In the next theorem, we will calculate the reverse atom bond connectivity index, reverse geometric arithmetic index for $\Gamma_2$ graph.

Theorem 4.2. Let $\Gamma_2$ be the third type of triangular hex-derived network, then

● $\mathfrak{R}\mathbb{ABC}(\Gamma_2) = \left(\frac{2\, \sqrt {210}}{5}+\frac{2\, \sqrt {7}}{5} \right) {t}^{2}+ \left(4\, \sqrt {2}+\frac{4}{3}-2\, \sqrt {210}+\frac{2\, \sqrt {330}}{5}-\frac{4\, \sqrt {7}}{5} \right) t -\frac{8}{3}-12\, \sqrt {2}+{\frac {12\, \sqrt {210}}{5}}-\frac{2\, \sqrt {330}}{3}+\frac{6\, \sqrt {7}}{5}$

● $\mathfrak{R}\mathbb{GA}(\Gamma_2) = \left(\frac{9}{2}+\frac{3\, \sqrt {15}}{4} \right) {t}^{2}+ \left(-{\frac{99}{10}}+\frac{3\, \sqrt {15}}{4} \right) t+{\frac{51}{5}}-3\, \sqrt {15}.$

Proof. The reverse atom bond connectivity ( $\mathfrak{R}\mathbb{ABC}(\Gamma_2)$ ), can be determined by using Eq (1.2) and Eq (4.1), as follows:

$\mathfrak{R}\mathbb{ABC}(\Gamma_2) = 0\times|\mathbf{E}_{1, 1}(\Gamma_2)|+\sqrt{\frac{8}{9}}\times|\mathbf{E}_{9, 1}(\Gamma_2)| +\sqrt{\frac{16}{81}}\times|\mathbf{E}_{9, 9}(\Gamma_2)|+\sqrt{\frac{14}{15}}\times|\mathbf{E}_{15, 1}(\Gamma_2)| +\sqrt{\frac{22}{135}}\times|\mathbf{E}_{15, 9}(\Gamma_2)|+\sqrt{\frac{28}{225}}\times|\mathbf{E}_{15, 15}(\Gamma_2)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ABC}(\Gamma_2) = \left(\frac{2\, \sqrt {210}}{5}+\frac{2\, \sqrt {7}}{5} \right) {t}^{2}+ \left(4\, \sqrt {2}+\frac{4}{3}-2\, \sqrt {210}+\frac{2\, \sqrt {330}}{5}-\frac{4\, \sqrt {7}}{5} \right) t -\frac{8}{3}-12\, \sqrt {2}+{\frac {12\, \sqrt {210}}{5}}-\frac{2\, \sqrt {330}}{3}+\frac{6\, \sqrt {7}}{5}.$

The reverse geometric arithmetic ( $\mathfrak{R}\mathbb{GA}(\Gamma_2)$ ), can be determined by using Eq (1.3) and Eq (4.1), as follows:

$\mathfrak{R}\mathbb{GA}(\Gamma_2) = |\mathbf{E}_{1, 1}(\Gamma_2)|+\frac{2\, \sqrt{9}}{10}\times|\mathbf{E}_{9, 1}(\Gamma_2)| +\frac{2\, \sqrt{81}}{18}\times|\mathbf{E}_{9, 9}(\Gamma_2)|+\frac{2\, \sqrt{15}}{16}\times|\mathbf{E}_{15, 1}(\Gamma_2)| +\frac{2\, \sqrt{135}}{24}\times|\mathbf{E}_{15, 9}(\Gamma_2)|+\frac{2\, \sqrt{225}}{30}\times|\mathbf{E}_{15, 15}(\Gamma_2)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{GA}(\Gamma_2) = \left( \frac{9}{2}+\frac{3\, \sqrt {15}}{4} \right) {t}^{2}+ \left( -{\frac{99}{10}}+\frac{3\, \sqrt {15}}{4} \right) t+{\frac{51}{5}}-3\, \sqrt {15}.$

In the next theorems, we will calculate the reverse redefined Zagreb indices for $\Gamma_2$ graph.

Theorem 4.3. Let $\Gamma_2$ be the third type of triangular hex-derived network, then

● $\mathfrak{R}\mathbb{ReZ}_1(\Gamma_2) = {\frac {49\, {t}^{2}}{5}}-{\frac {649\, t}{15}}+{\frac{734}{15}}$

● $\mathfrak{R}\mathbb{ReZ}_2(\Gamma_2) = {\frac {231\, {t}^{2}}{8}}+{\frac {1671\, t}{40}}-{\frac{1017}{10}}$

● $\mathfrak{R}\mathbb{ReZ}_3(\Gamma_2) = 21693\, {t}^{2}+15513\, t-38142.$

Proof. Reverse redefined Zagreb indices can be calculated by using Eq (4.1), the $\mathfrak{R}\mathbb{ReZ}_1(\Gamma_2)$ by using Eq (1.9) as follows:

$\mathfrak{R}\mathbb{ReZ}_1(\Gamma_2) = 2\times|\mathbf{E}_{1, 1}(\Gamma_2)|+\frac{10}{9}\times|\mathbf{E}_{9, 1}(\Gamma_2)| +\frac{18}{81}\times|\mathbf{E}_{9, 9}(\Gamma_2)|+\frac{16}{15}\times|\mathbf{E}_{15, 1}(\Gamma_2)| +\frac{24}{135}\times|\mathbf{E}_{15, 9}(\Gamma_2)|+\frac{30}{225}\times|\mathbf{E}_{15, 15}(\Gamma_2)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ReZ}_1(\Gamma_2) = {\frac {49\, {t}^{2}}{5}}-{\frac {649\, t}{15}}+{\frac{734}{15}}.$

The $\mathfrak{R}\mathbb{ReZ}_2(\Gamma_2)$ can be determined by using Eq (1.10) as follows:

$\mathfrak{R}\mathbb{ReZ}_2(\Gamma_2) = \frac{1}{2}\times|\mathbf{E}_{1, 1}(\Gamma_2)|+\frac{9}{10}\times|\mathbf{E}_{9, 1}(\Gamma_2)| +\frac{81}{18}\times|\mathbf{E}_{9, 9}(\Gamma_2)|+\frac{15}{16}\times|\mathbf{E}_{15, 1}(\Gamma_2)| +\frac{135}{24}\times|\mathbf{E}_{15, 9}(\Gamma_2)|+\frac{225}{30}\times|\mathbf{E}_{15, 15}(\Gamma_2)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ReZ}_2(\Gamma_2) = {\frac {231\, {t}^{2}}{8}}+{\frac {1671\, t}{40}}-{\frac{1017}{10}}.$

The $\mathfrak{R}\mathbb{ReZ}_3(\Gamma_2)$ can be calculated by using Eq (1.11) as follows:

$\mathfrak{R}\mathbb{ReZ}_3(\Gamma_2) = 2\times|\mathbf{E}_{1, 1}(\Gamma_2)|+90\times|\mathbf{E}_{9, 1}(\Gamma_2)| +1458\times|\mathbf{E}_{9, 9}(\Gamma_2)|+240\times|\mathbf{E}_{15, 1}(\Gamma_2)| +3240\times|\mathbf{E}_{15, 9}(\Gamma_2)|+6750\times|\mathbf{E}_{15, 15}(\Gamma_2)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ReZ}_3(\Gamma_2) = 21693\, {t}^{2}+15513\, t-38142.$

5. Rectangular hex-derived network $RHDN3(p)$

In this section, we calculate certain reverse degree based topological indices of the third type of rectangular hex-derived network, $RHDN_3(t, w)$ of dimension $t = w.$ Now we calculated reverse degree based indices such as: reverse Randic index $\mathfrak{R}\mathbb{R}_{\alpha},$ reverse atom bond connectivity index $\mathfrak{R}\mathbb{ABC},$ reverse geometric arithmetic index $\mathfrak{R}\mathbb{GA},$ first reverse Zagreb index $\mathfrak{R}\mathbb{M}_1,$ second reverse Zagreb index $\mathfrak{R}\mathbb{M}_2$ , reverse forgotten index $\mathfrak{R}\mathbb{F},$ reverse augmented Zagreb index $\mathfrak{R}\mathbb{AZI}$ , reverse hyper Zagreb index $\mathfrak{R}\mathbb{HM}$ and reverse redefined Zagreb indices for $\Gamma_3$ graph.

$\begin{equation} |\mathbf{E}_{\mathfrak{R}(\omega), \mathfrak{R}(\mu)}(\Gamma_3)| = \left\{ \begin{array}{ll} 3t^2-16t+21, & \hbox{for;}\ \mathfrak{R}(\omega) = 1, \mathfrak{R}(\mu) = 1 \\ 8t-20, & \hbox{for;}\ \mathfrak{R}(\omega) = 9, \mathfrak{R}(\mu) = 1 \\ 4t-10, & \hbox{for;}\ \mathfrak{R}(\omega) = 9, \mathfrak{R}(\mu) = 9 \\ 2, & \hbox{for;}\ \mathfrak{R}(\omega) = 12, \mathfrak{R}(\mu) = 1 \\ 4, & \hbox{for;}\ \mathfrak{R}(\omega) = 12, \mathfrak{R}(\mu) = 9 \\ 12t^2-48t+48, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 1 \\ 24t-44, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 9 \\ 8, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 12 \\ 6t^2-12t+10, & \hbox{for;}\ \mathfrak{R}(\omega) = 15, \mathfrak{R}(\mu) = 15 \end{array} \right. \end{equation}$

(5.1)

In the next theorem, we will calculate the reverse Randic index, reverse Zagreb indices, reverse forgotten index for $\Gamma_3$ graph.

Theorem 5.1. Let $\Gamma_3$ be the third type of rectangular hex-derived network, then

● $\mathfrak{R}\mathbb{R}_{\alpha}(\Gamma_3) = \left(3+12\, (15)^{\alpha}+6\, (225)^{\alpha} \right) {t}^{2}+ \left( -16+8\, (9)^{\alpha}+4\, (81)^{\alpha}-48\, (15)^{\alpha}+24\, (135)^{ \alpha}-12\, (225)^{\alpha} \right) t+21-20\, (9)^{\alpha}-10\, (81)^{ \alpha}+2\, (12)^{\alpha}+4\, (108)^{\alpha}+48\, (15)^{\alpha}-44\, (135) ^{\alpha}+8\, (180)^{\alpha}+10\, (225)^{\alpha}$

● $\mathfrak{R}\mathbb{M}_1(\Gamma_3) = 378\, {t}^{2}-432\, t$

● $\mathfrak{R}\mathbb{M}_2(\Gamma_3) = 1533\, {t}^{2}+200\, t-2043$

● $\mathfrak{R}\mathbb{HM}(\Gamma_3) = 8484\, {t}^{2}-7232\, t-1278$

● $\mathfrak{R}\mathbb{F}(\Gamma_3) = 5418\, {t}^{2}-7632\, t+2808.$

Proof. Let $\Gamma_3$ be the third type of rectangular hex-derived network which is shown in . The order of hexagonal hex derived network $\Gamma_3$ is $p = |\Gamma_1| = 7t^2 -12t + 6$ and size is $q = 21t^2-40t+19.$ The edge partitioned of $\Gamma_3$ based on their reverse degrees are shown in Eq (5.1). Reverse Randic index can be calculated by using Eq (5.1). Thus, from Eq (1.1), it follows,

$\mathfrak{R}\mathbb{R}_{\alpha}(\Gamma_3) = \left.(1)^{\alpha}|\mathbf{E}_{1, 1}(\Gamma_3)|+(9)^{\alpha}|\mathbf{E}_{9, 1}(\Gamma_3)| +(81)^{\alpha}|\mathbf{E}_{9, 9}(\Gamma_3)|+(12)^{\alpha}|\mathbf{E}_{12, 1}(\Gamma_3)|\right.\\ +(108)^{\alpha}|\mathbf{E}_{12, 9}(\Gamma_3)| +(15)^{\alpha}|\mathbf{E}_{15, 1}(\Gamma_3)| +(135)^{\alpha}|\mathbf{E}_{15, 9}(\Gamma_3)|+(180)^{\alpha}|\mathbf{E}_{15, 12}(\Gamma_3)| +(225)^{\alpha}|\mathbf{E}_{15, 15}(\Gamma_3)|.$

After Simplification, we get

$\mathfrak{R}\mathbb{R}_{\alpha}(\Gamma_3) = \left(3+12\, (15)^{\alpha}+6\, (225)^{\alpha} \right) {t}^{2}+ \left(-16+8\, (9)^{\alpha}+4\, (81)^{\alpha}-48\, (15)^{\alpha}+24\, (135)^{ \alpha}-12\, (225)^{\alpha} \right) t+21-20\, (9)^{\alpha}-10\, (81)^{ \alpha}+2\, (12)^{\alpha}+4\, (108)^{\alpha}+48\, (15)^{\alpha}-44\, (135) ^{\alpha}+8\, (180)^{\alpha}+10\, (225)^{\alpha}.$

Put $\alpha = 1$ and after some calculation, we get reverse second Zagreb index $\mathfrak{R}\mathbb{M}_2(\Gamma_3)$ as:

$\begin{equation} \mathfrak{R}\mathbb{M}_2(\Gamma_3) = 1533\, {t}^{2}+200\, t-2043. \end{equation}$

(5.2)

Using the Eq (1.4), we can determine the reverse first Zagreb index $\mathfrak{R}\mathbb{M}_1(\Gamma_3)$ as:

$\mathfrak{R}\mathbb{M}_1(\Gamma_3) = 2\times|\mathbf{E}_{1, 1}(\Gamma_3)|+10\times|\mathbf{E}_{9, 1}(\Gamma_3)| +18\times|\mathbf{E}_{9, 9}(\Gamma_3)|+13\times|\mathbf{E}_{12, 1}(\Gamma_3)|+21\times|\mathbf{E}_{12, 9}(\Gamma_3)| +16\times|\mathbf{E}_{15, 1}(\Gamma_3)| +24\times|\mathbf{E}_{15, 9}(\Gamma_3)|+27\times|\mathbf{E}_{15, 12}(\Gamma_3)| +30\times|\mathbf{E}_{15, 15}(\Gamma_3)|.$

By putting the values of from Eq (5.1) and after simplification, we obtain:

$\begin{equation} \mathfrak{R}\mathbb{M}_1(\Gamma_3) = 378\, {t}^{2}-432\, t. \end{equation}$

(5.3)

Using the Eq (1.6), we can determine the reverse hyper Zagreb index $\mathfrak{R}\mathbb{HM}(\Gamma_3)$ as:

$\mathfrak{R}\mathbb{HM}(\Gamma_3) = 4\times|\mathbf{E}_{1, 1}(\Gamma_3)|+100\times|\mathbf{E}_{9, 1}(\Gamma_3)| +324\times|\mathbf{E}_{9, 9}(\Gamma_3)|+169\times|\mathbf{E}_{12, 1}(\Gamma_3)|+441\times|\mathbf{E}_{12, 9}(\Gamma_3)| +256\times|\mathbf{E}_{15, 1}(\Gamma_3)| +576\times|\mathbf{E}_{15, 9}(\Gamma_3)|+729\times|\mathbf{E}_{15, 12}(\Gamma_3)| +900\times|\mathbf{E}_{15, 15}(\Gamma_3)|.$

After simplification, we get

$\begin{equation*} \mathfrak{R}\mathbb{HM}(\Gamma_3) = 8484\, {t}^{2}-7232\, t-1278. \end{equation*}$

Using the Eq (1.7), we can determine the reverse forgotten index $\mathfrak{R}\mathbb{F}(\Gamma_3))$ as:

$\mathfrak{R}\mathbb{F}(\Gamma_3)) = 2\times|\mathbf{E}_{1, 1}(\Gamma_3)|+82\times|\mathbf{E}_{9, 1}(\Gamma_3)| +162\times|\mathbf{E}_{9, 9}(\Gamma_3)|+145\times|\mathbf{E}_{12, 1}(\Gamma_3)|+225\times|\mathbf{E}_{12, 9}(\Gamma_3)| +226\times|\mathbf{E}_{15, 1}(\Gamma_3)| +306\times|\mathbf{E}_{15, 9}(\Gamma_3)|+369\times|\mathbf{E}_{15, 12}(\Gamma_3)| +450\times|\mathbf{E}_{15, 15}(\Gamma_3)|.$

After simplification, we get

$\begin{equation*} \mathfrak{R}\mathbb{F}(\Gamma_3) = 5418\, {t}^{2}-7632\, t+2808. \end{equation*}$

In the next theorem, we will calculate the reverse atom bond connectivity index, reverse geometric arithmetic index for $\Gamma_3$ graph.

Theorem 5.2. Let $\Gamma_3$ be the third type of rectangular hex-derived network, then

● $\mathfrak{R}\mathbb{ABC}(\Gamma_3) = \left(\frac{4\, \sqrt {210}}{5}+\frac{4\, \sqrt {7}}{5} \right) {t}^{2}+ \left(\frac{16\, \sqrt {2}}{3}+{\frac{16}{9}}-{\frac {16\, \sqrt {210}}{5}}+{\frac {8\, \sqrt {330}}{15}}-\frac{8\, \sqrt {7}}{5} \right) t-{\frac{40}{9}}-{\frac {40\, \sqrt {2}}{3}}+\frac{\sqrt {33}}{3}+\frac{2\, \sqrt {57}}{9}+{\frac {16\, \sqrt {210} }{5}}-{\frac {44\, \sqrt {330}}{45}}+\frac{4\, \sqrt {5}}{3}+\frac{4\, \sqrt {7}}{3}.$

● $\mathfrak{R}\mathbb{GA}(\Gamma_3) = \left(9+\frac{3\, \sqrt {15}}{2} \right) {t}^{2}-{\frac {96\, t}{5}}+9+{\frac {264\, \sqrt {3}}{91}}-5\, \sqrt {15}+{\frac {32\, \sqrt {5}}{9}}.$

Proof. The reverse atom bond connectivity ( $\mathfrak{R}\mathbb{ABC}(\Gamma_3)$ ), can be determined by using Eq (1.2) and Eq (5.1), as follows:

$\mathfrak{R}\mathbb{ABC}(\Gamma_3) = 0\times|\mathbf{E}_{1, 1}(\Gamma_3)|+\sqrt{\frac{8}{9}}\times|\mathbf{E}_{9, 1}(\Gamma_3)| +\sqrt{\frac{16}{81}}\times|\mathbf{E}_{9, 9}(\Gamma_3)|+\sqrt{\frac{11}{12}}\times|\mathbf{E}_{12, 1}(\Gamma_3)| +\sqrt{\frac{19}{108}}\times|\mathbf{E}_{12, 9}(\Gamma_3)| +\sqrt{\frac{14}{15}}\times|\mathbf{E}_{15, 1}(\Gamma_3)| +\sqrt{\frac{22}{135}}\times|\mathbf{E}_{15, 9}(\Gamma_3)|+\sqrt{\frac{25}{180}}\times|\mathbf{E}_{15, 12}(\Gamma_3)| +\sqrt{\frac{28}{225}}\times|\mathbf{E}_{15, 15}(\Gamma_3)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ABC}(\Gamma_3) = \left(\frac{4\, \sqrt {210}}{5}+\frac{4\, \sqrt {7}}{5} \right) {t}^{2}+ \left(\frac{16\, \sqrt {2}}{3}+{\frac{16}{9}}-{\frac {16\, \sqrt {210}}{5}}+{\frac {8\, \sqrt {330}}{15}}-\frac{8\, \sqrt {7}}{5} \right) t-{\frac{40}{9}}-{\frac {40\, \sqrt {2}}{3}}+\frac{\sqrt {33}}{3}+\frac{2\, \sqrt {57}}{9}+{\frac {16\, \sqrt {210} }{5}}-{\frac {44\, \sqrt {330}}{45}}+\frac{4\, \sqrt {5}}{3}+\frac{4\, \sqrt {7}}{3}.$

The reverse geometric arithmetic ( $\mathfrak{R}\mathbb{GA}(\Gamma_3)$ ), can be determined by using Eq (1.3) and Eq (5.1), as follows:

$\mathfrak{R}\mathbb{GA}(\Gamma_3) = |\mathbf{E}_{1, 1}(\Gamma_3)|+\frac{2\, \sqrt{9}}{10}\times|\mathbf{E}_{9, 1}(\Gamma_3)| +\frac{2\, \sqrt{81}}{18}\times|\mathbf{E}_{9, 9}(\Gamma_3)|+\frac{2\, \sqrt{12}}{13}\times|\mathbf{E}_{12, 1}(\Gamma_3)| +\frac{2\, \sqrt{108}}{21}\times|\mathbf{E}_{12, 9}(\Gamma_3)| +\frac{2\, \sqrt{15}}{16}\times|\mathbf{E}_{15, 1}(\Gamma_3)| +\frac{2\, \sqrt{135}}{24}\times|\mathbf{E}_{15, 9}(\Gamma_3)|+\frac{2\, \sqrt{180}}{27}\times|\mathbf{E}_{15, 12}(\Gamma_3)| +\frac{2\, \sqrt{225}}{30}\times|\mathbf{E}_{15, 15}(\Gamma_3)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{GA}(\Gamma_3) = \left(9+\frac{3\, \sqrt {15}}{2} \right) {t}^{2}-{\frac {96\, t}{5}}+9+{\frac {264\, \sqrt {3}}{91}}-5\, \sqrt {15}+{\frac {32\, \sqrt {5}}{9}}.$

In the next theorems, we will calculate the reverse redefined Zagreb indices for $\Gamma_3$ graph.

Theorem 5.3. Let $\Gamma_3$ be the third type of rectangular hex-derived network, then

● $\mathfrak{R}\mathbb{ReZ}_1(\Gamma_3) = {\frac {98\, {t}^{2}}{5}}-{\frac {3184\, t}{45}}+{\frac{5977}{90}}$

● $\mathfrak{R}\mathbb{ReZ}_2(\Gamma_3) = {\frac {231\, {t}^{2}}{4}}+{\frac {86\, t}{5}}-{\frac{28460}{273}}$

● $\mathfrak{R}\mathbb{ReZ}_3(\Gamma_3) = 43386\, {t}^{2}-8240\, t-31614.$

Proof. Reverse redefined Zagreb indices can be calculated by using Eq (5.1), the $\mathfrak{R}\mathbb{ReZ}_1(\Gamma_3)$ by using Eq (1.9) as follows:

$\mathfrak{R}\mathbb{ReZ}_1(\Gamma_3) = 2\times|\mathbf{E}_{1, 1}(\Gamma_3)|+\frac{10}{9}\times|\mathbf{E}_{9, 1}(\Gamma_3)| +\frac{18}{81}\times|\mathbf{E}_{9, 9}(\Gamma_3)|+\frac{13}{12}\times|\mathbf{E}_{12, 1}(\Gamma_3)| +\frac{21}{108}\times|\mathbf{E}_{12, 9}(\Gamma_3)| +\frac{16}{15}\times|\mathbf{E}_{15, 1}(\Gamma_3)| +\frac{24}{135}\times|\mathbf{E}_{15, 9}(\Gamma_3)|+\frac{27}{108}\times|\mathbf{E}_{15, 12}(\Gamma_3)| +\frac{30}{225}\times|\mathbf{E}_{15, 15}(\Gamma_3)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ReZ}_1(\Gamma_3) = {\frac {98\, {t}^{2}}{5}}-{\frac {3184\, t}{45}}+{\frac{5977}{90}}.$

The $\mathfrak{R}\mathbb{ReZ}_2(\Gamma_1)$ can be determined by using equation (1.10) as follows:

$\mathfrak{R}\mathbb{ReZ}_2(\Gamma_3) = \frac{1}{2}\times|\mathbf{E}_{1, 1}(\Gamma_3)|+\frac{9}{10}\times|\mathbf{E}_{9, 1}(\Gamma_3)| +\frac{81}{18}\times|\mathbf{E}_{9, 9}(\Gamma_3)|+\frac{12}{13}\times|\mathbf{E}_{12, 1}(\Gamma_3)| +\frac{108}{21}\times|\mathbf{E}_{12, 9}(\Gamma_3)| +\frac{15}{16}\times|\mathbf{E}_{15, 1}(\Gamma_3)| +\frac{135}{24}\times|\mathbf{E}_{15, 9}(\Gamma_3)|+\frac{108}{27}\times|\mathbf{E}_{15, 12}(\Gamma_3)| +\frac{225}{30}\times|\mathbf{E}_{15, 15}(\Gamma_3)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ReZ}_2(\Gamma_3) = {\frac {231\, {t}^{2}}{4}}+{\frac {86\, t}{5}}-{\frac{28460}{273}}.$

The $\mathfrak{R}\mathbb{ReZ}_3(\Gamma_3)$ can be calculated by using Eq (1.11) as follows:

$\mathfrak{R}\mathbb{ReZ}_3(\Gamma_3) = 2\times|\mathbf{E}_{1, 1}(\Gamma_3)|+90\times|\mathbf{E}_{9, 1}(\Gamma_3)| +1458\times|\mathbf{E}_{9, 9}(\Gamma_3)|+156\times|\mathbf{E}_{12, 1}(\Gamma_3)| +2268\times|\mathbf{E}_{12, 9}(\Gamma_3)| +240\times|\mathbf{E}_{15, 1}(\Gamma_3)| +3240\times|\mathbf{E}_{15, 9}(\Gamma_3)|+2916\times|\mathbf{E}_{15, 12}(\Gamma_3)| +6750\times|\mathbf{E}_{15, 15}(\Gamma_3)|.$

After some simplification, we get

$\mathfrak{R}\mathbb{ReZ}_3(\Gamma_3) = 43386\, {t}^{2}-8240\, t-31614.$

6. Conclusions

In this article, we have calculated the exact solutions of reverse degree-based topological descriptors for hex-derived networks of third type. Hex-derived network has a variety of useful applications in pharmacy, electronics, and networking. We obtained the reverse degree-based indices such as reverse Randic index, reverse atom bond connectivity index, reverse geometric arithmetic index, reverse Zagreb indices, reverse redefined Zagreb indices for hex derived networks. These results may be helpful for people working in computer science and chemistry who encounter hex-derived networks.

Conflict of interest

The authors declare that there is no conflict of financial interests regarding the publication of this paper.

Acknowledgments

We thanks Giuseppe Zanotti for reading and discussion of the draft.

^a regarding affiliation the writing work started at Università di Padova and ended at ESRF.

Conflict of interest

The authors declare no conflict of interest.

Author contributions:

AG concepted and designed the work, draft the article. VR edited and revised the manuscript.

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