An update on lateral flow immunoassay for the rapid detection of SARS-CoV-2 antibodies

1.   Introduction

In mathematics chemistry and biology, a chemical compound can be represented by a molecular graph by converting atoms to vertices and bonds to edges. One of the primary mission of QSAR/QSPR research is to accurately convert molecular graphs into numerical values. Graph theoretic invariants of molecular graphs are called molecular descriptors which can be utilized to simulate the structural information of molecules, in order to make worthwhile physical and chemical properties of these molecules can be acquired by single numerical values. Such kinds of molecular descriptors are also referred to as topological indices.

In the chemical literature, various topological indices relying only on vertex degrees of the molecular graphs can be utilized in QSPR/QSAR investigation on account of them can be obtained directly from the molecular architecture, and can be rapidly calculated for generous molecules (see ^[1,2]), and we call them VDB (vertex–degree–based) topological indices. To be more precise, for designated nonnegative real numbers $\{\psi_{ij}\}$ $(1\leq i\leq j\leq n-1)$, a VDB topological index of a an $n$-order (molecular) graph $G$ is expressed as

where $m_{ij}$ is the amount of edges connecting an $i$-vertex and a $j$-vertex of $G$. A great deal of well–known VDB topological indices can be obtained by different $\psi_{ij}$ in expression (1.1). We list some VDB topological indices in Table 1.

The first Zagreb index ^[3] is the very first VDB topological index, as powerful molecular structure-descriptors ^[2], Zagreb indices can describe the peculiarities of the degree of branching in molecular carbon-atom skeleton. Thereafter, many VDB topological indices have been put forward to simulate physical, chemical, biological, and other attributes of molecules ^[4,5,6,7]. In 2021, Gutman ^[8] introduced a new VDB topological index named as the Sombor index which has a linear correlation with the entropy and the enthalpy of vaporization of octanes ^[9]. Das et al., give sharp bounds for Sombor index of graphs by means of some useful graph parameters and they reveal the relationships between the Sombor index and Zagreb indices of graphs ^[10]. Recently, Steiner Gutman index was introduced by Mao and Das ^[11] which incorporate Steiner distance of a connected graph $G$. Nordhaus-Gaddum-type results for the Steiner Gutman index of graphs were given in ^[12]. In 2022, Shang study the Sombor index and degree-related properties of simplicial networks ^[13]. For more details of VDB topological indices, one can see ^{[3,14,15,16,17,18,19,20,21,22,23,24,25,26]} and the books ^[27,28,29].

Fluoranthene is a eminent conjugated hydrocarbon which abound in coal tar ^[30]. A fluoranthene–type benzenoid system (f-benzenoid for short) is formed from two benzenoid units joined by a pentagon ^[31,32]. The ordinary structure modality of a f-benzenoid $F$ is shown in Figure 1, where segments $X$ and $Y$ are two benzenoid systems. Each f-benzenoid possesses exactly one pentagon ^[32]. More and more attention is paid to f-benzenoids after the flash vacuum pyrolysis experiments of these nonalternant polycyclic aromatic hydrocarbons ^[33].

In the whole article, the terminology and notation are chiefly derived from ^{[34,35,36,37,38,39,40,41]}. A vertex of degree $k$ is called a $k$-vertex, and an edge linking a $k$-vertex and a $j$-vertex is designated as a $(k, j)$-edge. Let $n_{k}$ be the number of $k$-vertices and let $m_{kj}$ be the number of $(k, j)$-edges in the molecular graph $G$. A benzenoid system without internal vertices is said to be catacondensed. Analogously, a f-benzenoid $F$ containing a unique internal vertex is referred to as catacatacondensed. We use $h$-hexagon benzenoid system (or $h$-hexagon f-benzenoid) to represent a benzenoid system (or f-benzenoid) containing $h$ hexagons.

Let $L_{h}$ represent the $h$-hexagon linear chain (as shown in Figure 2(a)). An f-benzenoid $FL_{h}$ $(h\geq3)$ obtaining from pieces $X = L_{2}$ and $Y = L_{h-2}$ is named as f-linear chain (as shown in Figure 2(b)).

A fissure (resp. bay, cove, fjord and lagoon) of a f-benzenoid $F$ is a path of degree sequences $(2, 3, 2)$ (resp. $(2, 3, 3, 2)$, $(2, 3, 3, 3, 2)$, $(2, 3, 3, 3, 3, 2)$ and $(2, 3, 3, 3, 3, 3, 2)$) on the perimeter of $F$ (see Figure 3). Fissures, bays, coves, fjords and lagoons are said to be different kinds of inlets and their number are signified by $f$, $B$, $C$, $F_{j}$ and $L$, respectively ^[32,37]. Inlets determine many electronic and topological properties of f-benzenoids. Then, it can be found that $f+2B+3C+4F_{J}+5L$ is the number of $3$-vertices on the perimeter of $F$. It is noted that lagoons cannot occur in the theory of benzenoid systems. For convenience, let $r = f+B+C+F_{j}+L$ to represent the total number of inlets and $b = B+2C+3F_{j}+4L$ is referred to as the quantity of bay regions, In addition, $b$ is exactly the quantity of $(3, 3)$-edges on the perimeter of $F$. It is obvious that $b\geq2$ for any f-benzenoid $F$.

It is noted that any f-benzenoid $F$ contains merely either $2$-vertex or $3$-vertex. The vertices not on the perimeter are said to be internal, and we use $n_{i}$ to represent their number.

Lemma 1.1. ^[32] Let $F$ be an $n$-order, $h$-hexagon f-benzenoid with $m$ edges and $n_{i}$ internal vertices. Then

${\rm (i)}$ $n = 4h+5-n_{i}$;

${\rm (ii)}$ $m = 5h+5-n_{i}$.

Lemma 1.2. ^[32] Let $F$ be an $n$-order and $h$-hexagon f-benzenoid with $r$ inlets, Then

${\rm (i)}$ $m_{22} = n-2h-r$;

${\rm (ii)}$ $m_{23} = 2r$;

${\rm (iii)}$ $m_{33} = 3h-r$.

From the perspective of mathematics and chemistry, finding the extremal values of some useful $TI$ for significant classes of graphs is very interesting ^{[14,19,23,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56]}.

As a matter of convenience, we use $\Gamma_{m}$ to represent the collection of f-benzenoids containing exactly $m$ edges. In ^[45], we derived extremal values for $TI$ among all f-benzenoids with given order. It is noted that structure of f-benzenoids with given order is different from that of f-benzenoids with given number of edges. And we found that the technique for studying $TI$ among all f-benzenoids with given order can not be used directly to investigate $TI$ for all f-benzenoids with fixed number of edges. For this reason, we concentrate on the research of extremal values for $TI$ among all f-benzenoids with given size.

The main idea of this work is to construct f-benzenoids owning maximal $r$ and minimal $h$ at the same time in $\Gamma_{m}$ depending on the number $m$ is congruent to $0, 1, 2, 3$ or $4$ modulo $5$. By making use of this technique, we obtain the extremum of $TI$ over $\Gamma_{m}$ and characterize their corresponding graphs on the basis of $m$ is congruent to $0, 1, 2, 3$ or $4$ modulo $5$. Afterwards the extremums of some well-known $TI$ over $\Gamma_{m}$ can be got by use of the previous results.

The structure of this paper is as below. We first determine the maximal $r$ in the set $\Gamma_{m}$ in Section 2. By utilizing these results, we find the extremum of several famed $TI$ over $\Gamma_{m}$ in Section 3.

2.   F–benzenoids with maximal $r$ in $\Gamma_{m}$

We will find the f-benzenoids with maximal $r$ in $\Gamma_{m}$ in this section. Figure 4 illustrates three f-benzenoids pertaining to $\Gamma_{42}$.

At first, we try to obtain the maximum and minimum number of hexagons in any $F\in\Gamma_{m}$.

The spiral benzenoid system ^[57] $T_{h}$ is a benzenoid system whose structure is in a "spiral" manner as illustrated in Figure 5. $T_{h}$ has maximal $n_{i}$ in all $h$-hexagon benzenoid systems.

As a matter of convenience, let $SH_{h}$ $(h\geq3)$ represent the collection of f-benzenoids formed by two spiral benzenoids $X$ and $Y$. Particularly, a f-spiral benzenoid is a f-benzenoid $F^{\ast}\in SH_{h}$ in which $X = T_{h-1}$ and $Y = T_{1}$ (as shown in Figure 6). It is easy to see that that

In ^[40], we proved that for every $F'\in SH_{h}$ $(h\geq3)$, the inequality

holds, and the following graph operations were introduced.

Operation 1. For any $h$-hexagon f-benzenoid $F$ having two segments $X$ and $Y$, let $h_{1} = h(X)$ and $h_{2} = h(Y)$. By substituting spiral benzenoid systems $T_{h_{1}}$ and $T_{h_{2}}$ for $X$ and $Y$, severally, another f-benzenoid $F'\in SH_{h}$ can be obtained (as shown in Figure 7).

For any $h$-hexagon f-benzenoid $F$, when $h = 3$, it is easily checked that

When $h\geq4$, let $h_{1} = h(X)$ and $h_{2} = h(Y)$. Another $F'\in SH_{h}$ (as shown in Figure 7) in which $X = T_{h_{1}}$ and $Y = T_{h_{2}}$ can be acquired by applying Operation 1 to $F$. It is apparently that $n_{i}(X)\leq n_{i}(T_{h_{1}})$, $n_{i}(Y)\leq n_{i}(T_{h_{2}})$, therefore

So, the following Lemma can be deduced by Eqs (2.1) and (2.3).

Lemma 2.1. ^[41] Let $F$ be an $h(h\geq3)$-hexagon f-benzenoid. Then

and the equality is established when $F$ is $F^{\ast}$.

For any $F\in \Gamma_{m}$, $h(F)$ over $\Gamma_{m}$ is variable. Sharp bounds for $h(F)$ in $\Gamma_{m}$ is given below.

Theorem 2.1. For any f-benzenoid $F\in \Gamma_{m}$,

where $\lceil x\rceil$ is the smallest integer larger or equal to $x$.

Proof. On one hand, from Lemma 1.1 ${\rm(ii)}$ we know that $m = 5h(F)+5-n_{i}(F)$. Combining the fact that $n_{i}(F)\geq1$ for any $F\in \Gamma_{m}$, we get

On the other hand, by Lemma 2.1 we know that $n_{i}(F)\leq n_{i}(F^{\ast})$. Consequently, from $m = 5h(F)+5-n_{i}(F)$ we have

Hence,

Due to the fact that $3h(F)+(3-m) < 0$, we deduce

i.e., $h(F)\leq m-1-\left\lceil\frac{1}{3} \left(2m+\sqrt{4m-31}\right)\right\rceil.$

Remark 1. Theorem 2.1 implies that f-spiral benzenoid $F^{\ast}$ has the maximal number of hexagons over $\Gamma_{m}$.

For the sake of obtaining the extremum $TI$ among all f-benzenoids in $\Gamma_{m}$, we need to find the f-benzenoids $F\in\Gamma_{m}$ possessing maximal $r$.

Recall that convex benzenoid systems (CBS for brevity) are a particular sort of benzenoid systems lack of bay regions ^[14]. Let $\mathcal{HS}_{h}$ be the collection of benzenoid systems containing $h$ hexagons.

Lemma 2.2. ^[42] Let $H\in \mathcal{HS}_{h}$. Under the below cases, $H$ is definitely not a CBS:

$\rm (i)$ If $h\geq4$ and $n_{i} = 1$;

$\rm (ii)$ If $h\geq5$ and $n_{i} = 2$;

$\rm (iii)$ If $h\geq6$ and $n_{i} = 3$.

Lemma 2.3. ^[52] Let $H\in \mathcal{HS}_{h}$ such that $n_{i}(H) = 4$. Then $H$ is bound to embody a subbenzenoid system given in Figure 8, there does not exist hexagons which are adjacent to fissures.

Lemma 2.4. Let $S\in \mathcal{HS}_{h}$. If $h\geq7$ and $n_{i}(S) = 4$, then $S$ is not a CBS.

Proof. Let $S$ be an $h$ $(h\geq7)$-hexagon benzenoid system, $n_{i}(S) = 4$, then by Lemma 2.3 $S$ must contain one of the benzenoid systems of the form given in Figure 7. The proof is carried out in two cases.

Case 1. If these four internal vertices form a path $P_{4}$ or a $K_{1,3}$, then $S$ contains one of benzenoid systems $(d)$–$(f)$ in Figure 7 as its subbenzenoid systems. It is noted that $h\geq7$, by Lemma 2.2, it must not exist hexagons contiguous to the fissures, so, $S$ has at least one hexagon contiguous to a $(2, 2)$-edge, by means of such hexagons, it is succeeded in converting one of the fissures into a cove, bay or fjord. Hence, $b(S)\geq1$.

Case 2. If these four internal vertices are not adjacent then $S$ has possibility subbenzenoid systems as follows.

1) There exist one type $(a)$ and one type $(c)$ benzenoid systems in $S$;

2) There exist two type $(b)$ benzenoid systems in $S$;

3) There exist two type $(a)$ and one type $(b)$ benzenoid systems in $S$.

4) There exist four type $(a)$ benzenoid systems in $S$

By Lemma 2.2, neither hexagons may be adjacent to the fissures in any of the cases indicated above. Since $h\geq7$, $S$ has at least one hexagon contiguous to a $(2, 2)$-edge, by means of such hexagons, it is succeeded in making one of the fissures become a cove, bay or fjord. Therefore, $b(S)\geq1$.

The proof is completed.

Lemma 2.5. ^[45] Let $F$ be an $h$-hexagon f-benzenoid. Then

$\rm 1)$ If $n_{i} = 1$, then $r(F)\leq r(FL_{h}) = 2h-3\ (h\geq3)$;

$\rm 2)$ If $n_{i} = 2$, then $r(F)\leq r(G_{h}) = 2h-4\ (h\geq4)$;

$\rm 3)$ If $n_{i} = 3$, then $r(F)\leq r(R_{h}) = 2h-5\ (h\geq5)$;

$\rm 4)$ If $n_{i} = 4$, then $r(F)\leq r(Z_{h}) = 2h-6\ (h\geq6)$.

Next we find the f-benzenoids with maximal $r$ in $\Gamma_{m}$ with a fixed $n_{i}$. Recall that $M_{h}$, $N_{h}$ and $Q_{h}$ (see Figure 9) are benzenoid systems, and $G_{h}$ (see Figure 10), $R_{h}$ (see Figure 11), $Z_{h}$ (see Figure 12) are f-benzenoids.

Lemma 2.6. ^[41] Let $F$ be an $h$-hexagon f-benzenoid. Then

Lemma 2.7. ^[32] For any $h$-hexagon f-benzenoid including $n_{i}$ internal vertices and $b$ bay regions, the number of $(2, 2)$-edge and $(2, 3)$-edge are $m_{22} = b+5, m_{23} = 4h-2n_{i}-2b$, respectively.

From Lemmas 1.2 ${\rm(ii)}$ and 8, we get

Furthermore, by Lemma 1.1 ${\rm(ii)}$ and Eq (2.6), we deduce

Theorem 2.2. Let $F$ be an $h$-hexagon f-benzenoid. If $n_{i} = 5$, then $r(F)\leq r(U_{h}) = 2h-7$ $(h\geq7)$.

Proof. Let $h _{1} = h(X)$ and $h_{2} = h(Y)$, $X$ and $Y$ are two segments of $F$. If $n_{i} = 5$, by the structure of f-benzenoid, equality $n_{i}(X)+n_{i}(Y) = 4$ holds, so, we have the following five cases.

Case 1. $n_{i}(X) = 1$, $n_{i}(Y) = 3$, i.e., there exist one internal vertex and three internal vertices in $X$ and $Y$, respectively.

Subcase 1.1. If $h_{1} = 3$, then $X = M_{3}$.

Subcase 1.1.1. If $h_{2} = 5$, i.e., $Y = Q_{5}$, then $F$ is the f-benzenoid $D_{1}$, $D_{2}$ or $D_{3}$ (see Figure 14). It is clear that $r(F) = r(D_{1}) = 8\leq 2h-7$, $r(F) = r(D_{2}) = 7\leq 2h-7$ or $r(F) = r(D_{3}) = 8\leq 2h-7$.

Subcase 1.1.2. If $h_{2}\geq6$, by Lemma 2.2 and the hypothesis that $n_{i}(Y) = 3$, $Y$ is not a CBS, so $b(Y)\geq1$. Furthermore, $b(F)\geq3$, combining Eq (2.6) we obtain $r = 2h-n_{i}-b\leq2h-8 < 2h-7$.

Subcase 1.2. If $h_{1}\geq4$, according to Lemma 2.2, $X$ is definitely not a CBS, i.e., $b(X)\geq1$.

Subcase 1.2.1. If $h_{2} = 5$, i.e., $Y = Q_{5}$. It is clear that $b(F)\geq4$, then Eq (2.6) deduces $r\leq2h-9 < 2h-7$.

Subcase 1.2.2. If $h_{2}\geq6$, $Y$ is definitely not not a CBS according to Lemma 2.2, so, $b(Y)\geq1$. It is clear that $b(F)\geq5$, consequently from Eq (2.6) we obtain $r\leq2h-10 < 2h-7$.

Case 2. $n_{i}(X) = 3$ and $n_{i}(Y) = 1$.

Subcase 2.1. If $h_{1} = 5$, then $X = Q_{5}$.

Subcase 2.1.1. If $h_{2} = 3$, i.e., $Y = M_{3}$, then $F$ is the f-benzenoid $D_{4}$, $D_{5}$, $D_{6}$ (see Figure 14), or $D_{7}$ (as shown in Figure 15). $r(F) = r(D_{4}) = 8\leq 2h-7$, $r(F) = r(D_{5}) = 7\leq 2h-7$, $r(F) = r(D_{6}) = 8\leq 2h-7$, $r(F) = r(D_{7}) = 7\leq 2h-7$.

Subcase 2.1.2. If $h_{2}\geq4$, $Y$ is surely not a CBS in light of Lemma 2.2, i.e., $b(X)\geq1$. Hence, we have $b(F)\geq4$, it follows from Eq (2.6) that $r\leq2h-9 < 2h-7$.

Subcase 2.2. If $h_{1}\geq6$, by Lemma 2.2, $X$ is definitely not a CBS, hence $b(X)\geq1$.

Subcase 2.2.1. If $h_{2} = 3$, i.e., $Y = M_{3}$. We have $b(F)\geq4$, and Eq (2.6) infers that $r\leq2h-9 < 2h-7$.

Subcase 2.2.2. f $h_{2}\geq4$, by Lemma 2.2, $Y$ is certainly not a CBS, i.e., $b(X)\geq1$. Hence we have $b(F)\geq5$, by Eq (2.6), $r\leq2h-10 < 2h-7$.

Case 3. $n_{i}(X) = 2$, $n_{i}(Y) = 2$, i.e., $X$ and $Y$ both have two internal vertices.

Subcase 3.1. If $h_{1} = 4$, we note that $n_{i}(X) = 2$, so $X$ must be the benzenoid system $(b)$ in Figure 9.

Subcase 3.1.1. If $h_{2} = 4$, $Y$ is surely the benzenoid system $(b)$ in Figure 9 according to the hypothesis $n_{i}(Y) = 2$, therefore, $F$ is $D_{8}$ or $D_{9}$ (as shown in Figure 15). We get $r(F) = r(D_{8}) = 8 < 2h-7$ or $r(F) = r(D_{9}) = 7 < 2h-7$.

Subcase 3.1.2. If $h_{2}\geq5$, by Lemma 2.2 and that $n_{i}(Y) = 2$, $Y$ is not a CBS, so we know that $b(X)\geq1$. Then $b(F)\geq4$, by Eq (2.6) and the fact that $n_{i} = 5$, $r\leq2h-9 < 2h-7$.

Subcase 3.2. If $h_{2} = 4$, we note that $n_{i}(Y) = 2$, so $Y$ must be the benzenoid system $(b)$ in Figure 8.

Subcase 3.2.1. If $h_{1} = 4$, $X$ must also be the benzenoid system $(b)$ in Figure 9. Hence, $F$ is $D_{8}$ or $D_{9}$ (as shown in Figure 15). $r(F) = r(D_{8}) = 8\leq 2h-7$ or $r(F) = r(D_{9}) = 7\leq 2h-7$.

Subcase 3.2.2. If $h_{1}\geq5$, by Lemma 2.2 and $n_{i}(X) = 2$, $X$ is definitely not a CBS, i.e., $b(X)\geq1$. Hence, $b(F)\geq4$, by Eq (2.6) and the fact that $n_{i} = 5$, we have $r\leq2h-9 < 2h-7$.

Subcase 3.3. If $h_{1}\geq5$, $h_{2}\geq5$, it is noted that $n_{i}(X) = n_{i}(Y) = 2$, neither $X$ nor $Y$ are definitely CBS according to Lemma 2.2. So, both $b(X)$ and $b(Y)$ are greater than $1$. Hence, $b(F)\geq5$, on the basis of Eq (2.6) we get $r\leq2h-10 < 2h-7$.

Case 4. $n_{i}(X) = 4$ and $n_{i}(Y) = 0$, i.e., $X$ contains four internal vertices, $Y$ is a catacondensed benzenoid system.

Subcase 4.1. If $h_{1} = 6$, then $X$ is the benzenoid system $(d)$, $(e)$ or $(f)$ in Figure 9.

Subcase 4.1.1. If $h_{2} = 1$, $F$ is the f-benzenoid $D_{10}$, $D_{11}$, $D_{12}$ (see Figure 16), $D_{13}$ (see Figure 17) or $U_{7}$ (see Figure 12). $r(F) = r(D_{10}) = 6\leq 2h-7$, $r(F) = r(D_{11}) = 6\leq 2h-7$, $r(F) = r(D_{12}) = 6\leq 2h-7$, $r(F) = r(D_{13}) = 6\leq 2h-7$ or $r(F) = r(U_{7}) = 7 = 2h-7$.

Subcase 4.1.2. If $h_{2}\geq2$, we have $b(F)\geq2$, by Eq (2.6), $r\leq2h-7$.

Subcase 4.2. If $h_{1}\geq7$, in the light of Lemma 2.4, $X$ is definitely not a CBS, hence $b(Y)\geq1$. In this situation $b(F)\geq3$, we get the inequality $r\leq2h-8 < 2h-7$ according to Eq (2.6).

Case 5. $n_{i}(X) = 0$ and $n_{i}(Y) = 4$, i.e., $X$ is a catacondensed benzenoid system, $Y$ has four internal vertices.

Subcase 5.1. If $h_{2} = 6$, then $Y$ is the benzenoid system $(d)$, $(e)$ or $(f)$ in Figure 8.

Subcase 5.1.1. If $h_{1} = 2$, $X$ must be the linear chain $L_{2}$. In this event, $F$ is $D_{14}$, $D_{15}$, $D_{16}$, $D_{17}$, $D_{18}$, $D_{19}$, $D_{20}$ or $D_{21}$ (see Figure 17). By further checking, we gain that $r(F) = r(D_{14}) = 7\leq 2h-7$, $r(F) = r(D_{15}) = 8\leq 2h-7$, $r(F) = r(D_{16}) = 8\leq 2h-7$, $r(F) = r(D_{17}) = 7\leq 2h-7$, $r(F) = r(D_{18}) = 7\leq 2h-7$, $r(F) = r(D_{19}) = 8\leq 2h-7$, $r(F) = r(D_{20}) = 6\leq 2h-7$ or $r(F) = r(D_{21}) = 6\leq 2h-7$.

Subcase 5.1.2. If $h_{1}\geq3$, bearing in mind that $X$ is a catacondensed benzenoid system and $Y$ is the benzenoid system $(d)$, $(e)$ or $(f)$ in Figure 8, then $F$ must have f-benzenoid $D_{14}$, $D_{15}$, $D_{16}$, $D_{17}$, $D_{18}$, $D_{19}$, $D_{20}$ or $D_{21}$ (see Figure 17) as its subgraph.

Subcase 5.1.2.1. If $D_{14}$ is a subgraph in $F$, it is obvious that $D_{14}$ has two coves. Since $X$ is a catacondensed benzenoid system and $h_{1}\geq3$, $F$ has at least one hexagon contiguous to a $(2, 2)$-edge of $X$, and such hexagons can convert one fissure into a bay, or convert one cove into a fjord, or convert one fjord into a lagoon. In this instance $b(F)\geq4$. Consequently, $r\leq2h-9 < 2h-7$ can be got according to Eq (2.6).

Subcase 5.1.2.2. If $D_{15}$, $D_{16}$ or $D_{19}$ is a subpart f-benzenoid in $F$, it is obvious each one of $D_{15}$, $D_{16}$ and $D_{19}$ has a bay and a cove. Since $X$ is a catacondensed benzenoid system and $h_{1}\geq3$, $F$ contains at least one hexagon adjoining a $(2, 2)$-edge of $X$, and such hexagons will make one fissure become a bay, or make one cove become a fjord, or make one fjord become a lagoon. Consequently, $b(F)\geq4$, by Eq (2.6) it follows that $r\leq2h-9 < 2h-7$.

Subcase 5.1.2.3. If $D_{17}$ is a subpart f-benzenoid in $F$, it is obvious that $D_{17}$ has a fjord and a bay. Since $X$ is a catacondensed benzenoid system and $h_{1}\geq3$, $F$ has at least one hexagon adjoining a $(2, 2)$-edge of $X$, and such hexagons will convert one fissure into a bay, or convert one cove into a fjord, or convert one fjord into a lagoon. Consequently, $b(F)\geq4$, by Eq (2.6) it follows that $r\leq2h-9 < 2h-7$.

Subcase 5.1.2.4. If $D_{18}$ is a subpart f-benzenoid in $F$, it is obvious that $D_{18}$ has a fjord and two bays. Since $X$ is a catacondensed benzenoid system and $h_{1}\geq3$, there exists has at least one hexagon adjoining a $(2, 2)$-edge of $X$ in $F$, and these hexagons will convert one of the fissures into a bay, or convert one cove into a fjord, or convert one fjord into a lagoon. Consequently, $b(F)\geq4$, in light of Eq (2.6), $r\leq2h-9 < 2h-7$.

Subcase 5.1.2.5. If $D_{20}$ or $D_{21}$ is a subpart f-benzenoid in $F$, it is obvious that both $D_{20}$ and $D_{21}$ have a bay and two fjords. Since $X$ is a catacondensed benzenoid system and $h_{1}\geq3$, $F$ contains at least one hexagon adjoining a $(2, 2)$-edge of $X$, and such hexagons will make one fissure become a bay, or make one cove become a fjord, or make one fjord become a lagoon. Consequently, $b(F)\geq4$, according to Eq (2.6), $r\leq2h-9 < 2h-7$.

Subcase 5.2. If $h_{2}\geq7$, by Lemma 2.4 and the fact that $n_{i}(Y) = 4$, $Y$ is certainly not a CBS, i.e., $b(Y)\geq1$.

Subcase 5.2.1. If $h_{1} = 2$, i.e., $X = L_{2}$. From the structure of f-benzenoid, $F$ is formed from $X$ and $Y$ joined by a pentagon, it is easily seen that there are at least one bay or one cove arisen in the process of construction of $F$. It is clear that $b(F)\geq2$, by Eq (2.6) we have $r\leq2h-7$.

Subcase 5.2.2. If $h_{1}\geq3$, we know that $F$ is formed by joining from $X$ and $Y$ through a pentagon, in this construction process of $F$, it is easily seen that there are at least one bay or one cove arisen. Then $b(F)\geq2$, by Eq (2.6), $r\leq2h-7$.

The proof is completed.

We recall that $FL_{h}$ is the f-linear chain with $h$ hexagons ^[40]. Extremal f-benzenoids with maximal $r$ in $\Gamma_{m}$ were determined in the following theorem.

Theorem 2.3. Let $F\in\Gamma_{m}$. Then

$\rm 1)$ If $m\equiv0(mod5)$, then $r(F)\leq \frac{2m-35}{5} = r(U_{\frac{m}{5}})$;

$\rm 2)$ If $m\equiv1(mod5)$, then $r(F)\leq \frac{2m-32}{5} = r(Z_{\frac{m-1}{5}})$;

$\rm 3)$ If $m\equiv2(mod5)$, then $r(F)\leq \frac{2m-29}{5} = r(R_{\frac{m-2}{5}})$;

$\rm 4)$ If $m\equiv3(mod5)$, then $r(F)\leq \frac{2m-26}{5} = r(G_{\frac{m-3}{5}})$;

$\rm 5)$ If $m\equiv4(mod5)$, then $r(F)\leq \frac{2m-23}{5} = r(FL_{\frac{m-4}{5}})$.

Proof. We know by Eq (2.5) that

1) If $m\equiv0(mod5)$, then $\left\lceil\frac{1}{5}(m-4)\right\rceil = \frac{m}{5}$. If $h = \frac{m}{5}$, then by Lemma 1.1 ${\rm(ii)}$

it means that $n_{i}(F) = 5$. Furthermore, Theorem 2.2 infers that $r(F)\leq r(U_{\frac{m}{5}})$ and we are done. So assume now that $h(F)\geq\frac{m}{5}+1$, then by equality (2.7) and the fact that $b(F)\geq2$

2) If $m\equiv1(mod5)$, then $\left\lceil\frac{1}{5}(m-4)\right\rceil = \frac{m-1}{5}$. If $h(F) = \frac{m-1}{5}$, then by Lemma 1.1 ${\rm(ii)}$

thus $n_{i}(F) = 4$. Then $r(F)\leq r(Z_{\frac{m-1}{5}})$ by part 4 of Lemma 2.5. Otherwise $h(F)\geq\frac{m-1}{5}+1$, then by equality (2.7) and the obvious fact that $b(F)\geq2$

3) If $m\equiv2(mod5)$, then $\left\lceil\frac{1}{5}(m-4)\right\rceil = \frac{m-2}{5}$. If $h(F) = \frac{m-2}{5}$, then by Lemma 1.1 ${\rm(ii)}$

and so $n_{i}(F) = 3$. Then $r(F)\leq r(R_{\frac{m-2}{5}})$ by part 3 of Lemma 2.5. So assume now that $h(F)\geq\frac{m-2}{5}+1$, then by Eq (2.7) and the fact that $b(F)\geq2$

4) If $m\equiv3(mod5)$, then $\left\lceil\frac{1}{5}(m-4)\right\rceil = \frac{m-3}{5}$. If $h(F) = \frac{m-3}{5}$, then by Lemma 1.1 ${\rm(ii)}$

thus $n_{i}(F) = 2$. By Lemma 2.5, $r(F)\leq r(G_{\frac{m-3}{5}})$ and we are done. If $h(F)\geq\frac{m-3}{5}+1$, then by equality (2.7) and the fact that $b(F)\geq2$

5) If $m\equiv4(mod5)$, then $\left\lceil\frac{1}{5}(m-4)\right\rceil = \frac{m-4}{5}$. Since $h\geq\frac{m-4}{5}$ and $b(F)\geq2$, then by Eq (2.7), we have

3.   Extremal values of $TI$ over $\Gamma_{m}$

In this part, we attempt to find the extremal values of $TI$ over $\Gamma_{m}$.

It is noted that a f-benzenoid $F$ contains only $2$-vertex and $3$-vertex. Hence, equation (1.1) reduces to

In the light of Lemmas 1.1 and 1.2,

If $U, V\in\Gamma_{m}$ then clearly

For convenience, we set $s = \psi_{33}-\psi_{22}$, $q = 2\psi_{23}-\psi_{22}-\psi_{33}$.

Theorem 3.1. For any $F\in\Gamma_{m}$, we have the following results.

${\rm a.}$ If $s\leq0$ and $q\geq0$,

${\rm b.}$ If $s\geq0$ and $q\leq0$,

Proof. Let $F\in\Gamma_{m}$. By Eq (2.5)

i.e., f-benzenoids $U_{\frac{m}{5}}$, $Z_{\frac{m-1}{5}}$, $R_{\frac{m-2}{5}}$, $G_{\frac{m-3}{5}}$ and $FL_{\frac{m-4}{5}}$ have minimal $h$ over the set $\Gamma_{m}$. Meanwhile, by Theorem 2.3, we have

i.e., these five f-benzenoids have maximal number of inlets over $\Gamma_{m}$. Hence, for any f-benzenoids $F\in\Gamma_{m}$ and $V\in\{U_{\frac{m}{5}}, Z_{\frac{m-1}{5}}, R_{\frac{m-2}{5}}, G_{\frac{m-3}{5}}, FL_{\frac{m-4}{5}}\}$, $h(F)-h(V)\geq0$ and $r(F)-r(V)\leq0$ hold simultaneously, from Eq (2.7), we have

If $s\leq0$ and $q\geq0$, then $TI(F)-TI(V)\leq0$, i.e., $V$ reaches the maximum value of $TI$ over $\Gamma_{m}$. If $s\geq0$ and $q\leq0$, then $TI(F)-TI(V)\geq0$, i.e., $V$ reaches the minimum value of $TI$ over $\Gamma_{m}$. Furthermore, which $V\in\{U_{\frac{m}{5}}, Z_{\frac{m-1}{5}}, R_{\frac{m-2}{5}}, G_{\frac{m-3}{5}}, FL_{\frac{m-4}{5}}\}$ is the extremal graph depending on $m$ is congruent to $0, 1, 2, 3$ or $4$ modulo $5$.

Example 1. Values of $s$ and $q$ for several famous $TI$ are listed in Table 2:

Therefore, the minimum extreme value of $TI$ for the second Zagreb index, GA index and the AZI index can be determined in the light of Theorems 2.3 and 3.1, and we can obtain the maximum extreme value of $TI$ for the ABC index.

If f-benzenoid $F\in\Gamma_{m}$, then from the Eqs (2.3) and (2.6) and Lemma 1.1$({\rm ii})$ we have

Consequently, for f-benzenoids $U, V\in\Gamma_{m}$

Set $u = 6(\psi_{33}-\psi_{23})$ and keep in mind that $q = 2\psi_{23}-\psi_{22}-\psi_{33}$. Then

It is noted that Eq (3.6) can be decided only by $h$, $b$ and the signs of $u$ and $q$. For any $F\in\Gamma_{m}$, We know that

and the equality can be achieved precisely when $F$ is the f-spiral benzenoid $F^{\ast}$ ^[41].

In ^[41], we proved that $n_{i}(F^{\ast}) = 2h-\left\lceil\sqrt{12(h-1)-3}\right\rceil$. But, $b(F^{\ast})\neq2$ may occur. It is noticeable if $X$ in $F^{\ast}$ is a CBS, $F^{\ast}$ is a f-benzenoid satisfying that $b(F^{\ast}) = 2$ or $3$. For the sake of simplicity, Let $\mathbb{N}$ be the set of positive integers.

The CBS, $W = H(l_{1}, l_{2}, l_{3}, l_{4}, l_{5}, l_{6})$ (as shown in Figure 18), can be completely determined by the positive integers $l_{1}, l_{2}, l_{3}, l_{4}$ ^[14].

The following lemma gave requirements that there exists CBS with maximal $n_{i}$ ^[53].

Lemma 3.1. ^[53] Let $h\in\mathbb{N}$. The conditions below are isovalent:

(a) There is a CBS $W$ containing $h$ hexagons and $2h+1-\lceil\sqrt{12h-3} \ \rceil$ number of internal vertices.

(b) There exist $l_{1}, l_{2}, l_{3}, l_{4}\in\mathbb{N}$ satisfying the following equation

If for $h\in\mathbb{N}$, Eq (3.7) has a solution $l_{1}, l_{2}, l_{3}, l_{4}\in\mathbb{N}$, then there is a CBS $W$ meeting the conditions that $n_{i}(W) = n_{i}(T_{h})$.

Now, we concentrate on the research for $TI$ of f-benzenoids. For a $h-1\in\mathbb{N}$, supposing that the system below

has a solution $\{l_{1}, l_{2}, l_{3}, l_{4}\}$, then a CBS $W_{h-1}$ containing $n_{i}(W_{h-1}) = 2(h-1)+1-\left\lceil\sqrt{12(h-1)-3}\right\rceil$ number of internal vertices exists. Note that $l_{i} = 2$ in system (3.8), i.e., there exists one fissure on the side of $l_{i}$ of $W_{h-1}$, let $u, w, v$ in Figure 1 represent the three vertices of this fissure. Now, we obtain an f-spiral benzenoid $F^{\ast}_{1}$ in which $X = W_{h-1}$ and $Y = L_{1}$. It is obvious that

and $b(F^{\ast}_{1}) = 2$. (as shown in Figure 19)

Theorem 3.2. Let $h-1\in\mathbb{N}$ such that the Eq (3.8) has a solution, and $m = 3h+5+\left\lceil\sqrt{12(h-1)-3}\right\rceil$. Then for any $F\in\Gamma_{m}$

1) $TI(F^{\ast}_{1})\geq TI(F)$, when $u\geq0$ and $q\geq0$;

2) $TI(F^{\ast}_{1})\leq TI(F)$, when $u\leq0$ and $q\leq0$.

Proof. From Lemma 1.1 $({\rm ii})$ and Eq (3.9), we have

and so

It is obvious that $b(F^{\ast}_{1}) = 2$ and $b(F)\geq2$ for any $F\in\Gamma_{m}$. Hence by Eq (3.6), we have

And by Eq (2.5)

If $u\geq0$ and $q\geq0$ then $TI(F)-TI(F^{\ast}_{1})\leq0$, i.e., $F^{\ast}_{1}$ achieves maximal $TI$ in $\Gamma_{m}$. Similarly, if $u\leq0$ and $q\leq0$ then $TI(F)-TI(F^{\ast}_{1})\geq0$, i.e., $F^{\ast}_{1}$ obtains minimal $TI$ in $\Gamma_{m}$.

Example 2. The values of $u$ and $q$ for some famous $TI$ are listed in the following Table 3:

Hence, by Theorem 3.1 we can deduce the minimal values of the Randć index and the the sum–connectivity index in f-spiral benzenoid $F^{\ast}_{1}$ for those $h$ such that Eq (3.8) holds.

Example 3. Take consideration of the generalized Randć index

where $\alpha\in\mathbb{R}$. Note that

for all $\alpha\in\mathbb{R}$. Moreover, $s = 9^{\alpha}-4^{\alpha}\geq0$ if and only if $\alpha\geq0$ if and only if $u = 6(9^{\alpha}-6^{\alpha})\geq0$. Hence, by Theorem 3.1, the minimal value of $R_{\alpha}(G)$ is obtained for all $\alpha\geq0$, and for any $\alpha\leq0$, the minimal value of $R_{\alpha}(G)$ can be attained by the f-spiral benzenoid $F^{\ast}_{1}$ for those $h$ such that Eq (3.8) holds.

4.   Conclusions

This work investigates extremum $TI$ over the collection of f-benzenoids having same number of edges. In practical terms, there are many other types of very useful topological indices for instance graph energy ^{[58,59,60,61,62]}, Wiener index ^[63], Randić energy ^[64], Wiener polarity index ^[65], incidence energy ^[66], Harary index ^[67], entropy measures ^[68,69] and HOMO-LUMO index ^[70]. So, determining these topological indices for f-benzenoids is going to be extraordinary fascinating.

It is noted that the current framework is for studying topological indices of deterministic networks. But random networks would be a very promising direction. In ^[71,72], the distance Estrada index of random graphs was discussed, and the author went deeply into (Laplacian) Estrada index for random interdependent graphs. So, studying VDB topological indices of random and random interdependent graphs is another interesting problem.

Acknowledgments

This work was supported by Ningbo Natural Science Foundation (No. 2021J234). The authors are very grateful to anonymous referees and editor for their constructive suggestions and insightful comments, which have considerably improved the presentation of this paper.

Conflict of interest

The authors declare there is no conflict of interest.