Characterizing the fullerene graphs with the minimum forcing number 3

TL;DR

This paper characterizes all fullerene graphs with the minimum forcing number 3, resolving an open problem and revealing unique properties of certain fullerene types.

Contribution

It provides a complete characterization of fullerenes with minimum forcing number 3 using a construction approach, and identifies the unique exception with minimum forcing number 2.

Findings

Exactly one fullerene $F_{24}$ has minimum forcing number 2.

All other fullerenes with anti-forcing number 4 have minimum forcing number 3.

Nanotube fullerenes of type (4, 2) also have minimum forcing number 3.

Abstract

The minimum forcing number of a graph $G$ is the smallest number of edges simultaneously contained in a unique perfect matching of $G$. Zhang, Ye and Shiu \cite{HDW} showed that the minimum forcing number of any fullerene graph was bounded below by $3$. However, we find that there exists exactly one excepted fullerene $F_{24}$ with the minimum forcing number $2$. In this paper, we characterize all fullerenes with the minimum forcing number $3$ by a construction approach. This also solves an open problem proposed by Zhang et al. We also find that except for $F_{24}$, all fullerenes with anti-forcing number $4$ have the minimum forcing number $3$. In particular, the nanotube fullerenes of type $(4, 2)$ are such fullerenes.

Full text

Keywords: Fullerene graph; Perfect matching; Minimum forcing number; Generalized patch

1 Introduction

A fullerene graph (simply fullerene) is a cubic $3$-connected plane graph with only pentagonal and hexagonal faces. By Euler’s formula, a fullerene graph has exactly twelve pentagonal faces. Such graphs are suitable models for carbon fullerene molecules: carbon atoms are represented by vertices, whereas edges represent chemical bonds between two atoms (see [2, 11]). Grünbaum and Motzkin [3] showed that a fullerene graph with $n$ vertices exists for $n=20$ and for all even $n\geq 24$.

A perfect matching $M$ of a graph $G$ is an edge set such that each vertex of $G$ is incident with exactly one edge in $M$. Let $G$ be a graph with a perfect matching $M$. A set $S\subseteq M$ is called a forcing set of $M$ if $S$ is not contained in any other perfect matchings of $G$. The forcing number of $M$, first proposed in organic chemistry by Randić and Klein [8, 12, 13] under name innate degree of freedom in correlation with resonance structure, is defined as the minimum size of all forcing sets of $M$ by Harary et al. [4], denoted by $f(G,M)$. The minimum forcing number of $G$, denoted by $f(G)$, is the minimum value of the forcing numbers of all perfect matchings of $G$.

D. Vukičević and N. Trinajstić [15, 16] recently introduced the anti-forcing number of a graph $G$ as the smallest number of edges whose removal results in a subgraph with a single perfect matching, denoted by $af(G)$. For a fullerene graph $F$, Yang et al. [17] showed that $af(F)\geq 4$, and further gave a procedure to construct all fullerenes with the anti-forcing number $4$. For a $(3,6)$-fullerene graph $H$, two of the present authors proved [14]that $af(H)\geq 2$ and equality holds if and only if $H$ either has connectivity $2$ or is isomorphic to $K_{4}$, and determined all the $(3,6)$-fullerenes with the anti-forcing number $3$. Jiang and Zhang [5] characterized all the $(4,6)$-fullerenes with the minimum forcing number $2$.

In Ref. [19], Zhang, Ye and Shiu gave a main result that the forcing number of of every perfect matching of any fullerene graph was bounded below by $3$ (See Theorem 2.7). We find that in the last paragraph of its proof, they neglected the trivial case of cyclic 5-edge cut $S$, that is, the claim “Clearly, $S$ is non-trivial” is not right. In fact, when $S$ is trivial, we obtain a unique fullerene $F_{24}$ with the minimum forcing number $2$ (see Fig. 1). So the main result (Theorem 1.1 or 2.7) in [19] can be corrected as

Theorem 1.1**.**

Let $F$ be a fullerene graph. Then $f(F)\geq 3$ except for $F_{24}$.

We also should point out the fact by Yang et al. [17] that the anti-forcing number of any fullerene graphs are at least 4 still holds, although they applied the wrong lower bound $3$ of the minimum forcing number. This is because $F_{24}$ has the anti-forcing number $4$, and all the discussions in that fact are not affected if we exclude the fullerene $F_{24}$.

In this paper, we focus on studying properties of fullerenes with the minimum forcing number $3$. By applying these properties we obtain a procedure to generate all fullerenes with the minimum forcing number $3$. Hence we give a solution to an open Problem 4.1 proposed by Zhang et al. [19].

2 Preliminaries

For a graph $G$ with $\emptyset\not=X\subset V(G)$, let $\partial X$ be the set of edges with only one end in $X$. Then $\partial X$ is an edge-cut of $G$. For a subgraph $H$ of $G$ with $V(H)\neq V(G)$, we denote by $\triangledown_{G}(H)$ the edge set of $G$ with only one end in $V(H)$. An edge-cut $C$ of $G$ is trivial if its edges are all incident with the same vertex, and cyclic if $G-C$ has at least two components, each containing a cycle. A cyclic $k$-edge-cut (with $k$ edges) of $G$ is trivial if at least one of the two components is a single cycle of length $k$. The cyclic edge-connectivity of $G$ is the minimum size of cyclic edge-cuts of $G$

Lemma 2.1** ([1, 17]).**

For a fullerene graph, every $3$-edge-cut is trivial, every $4$-edge-cut isolates an edge, and the cyclic edge-connectivity is $5$.

We denote by $G^{k}$ the tubular fullerene graph comprised of two pentacaps and $k$ layers of hexagons between them. For example, see Fig. 2 for $k=2$.

Theorem 2.2** ([7, 10]).**

A fullerene graph has a non-trivial cyclic $5$-edge-cut if and only if it is isomorphic to the graph $G^{k}$ for some integer $k\geq 1$.

From the proof of the theorem or Ref. [11] we easily have the following lemma.

Lemma 2.3**.**

Let $S$ be a non-trivial cyclic $5$-edge-cut of fullerene $G^{k}$. Then $G^{k}-S$ has two components $H_{1}$ and $H_{2}$, and $H_{i}$ consists of a pentacap and $l$ ($0\leq l\leq k-1$)* layers of hexagons around it. Moreover, for any $2$-degree vertex $x$ in $H_{i}$, $H_{i}-x$ is $2$-connected.*

A cyclic edge-cut $C$ of a fullerene graph $F$ is non-degenerate if both components of $G-C$ contain precisely six pentagons, and degenerate otherwise. So the trivial cyclic edge-cuts of $F$ are degenerate and the non-trivial cyclic $5$-edge-cuts of $F$ are non-degenerate.

A patch of a fullerene graph $F$ is a $2$-connected subgraph of $F$ whose all interior faces are faces of $F$ and all vertices not on the outer face have degree $3$. A generalized patch of $F$ is a connected plane subgraph where all interior faces (if exists) are faces of $F$, and vertices not on the outer face have degree $3$ and vertices on the outer face have degree $1$, $2$ or $3$. Clearly, a patch is also a generalized patch.

By Theorem $3$ in [7], we have the following lemma.

Lemma 2.4**.**

Let $H$ be a patch of a fullerene $F$ and $\triangledown_{F}(H)$ a degenerate cyclic $6$-edge-cut of $F$ such that $H$ has at most five pentagons. If $H$ has eaxctly five pentagons or $H$ has at least $14$ vertices, then $H\cong P^{1}$ or $P^{2}$ as shown in Fig. 3.

From Ref. [6], we have the following.

Lemma 2.5**.**

Let $H$ be a generalized patch of a fullerne $F$ with at most five pentagons and $\triangledown_{F}(H)$ be a degenerate cyclic $7$-edge-cut of $F$. Then $H$ is isomorphic to one of the patches $D_{01}$, …, $D_{57}$ as shown in Fig. 7 in Ref. [6]. Moreover, if $H$ is $2$-connected and has a $2$-degree vertex $x$ such that $H-x$ has a unique perfect matching, then $H$ is one of the patches as depicted in Fig. 4, and such $2$-degree vertices are the white vertices as shown in Fig. 4.

In the following, we set $\mathcal{D}:=\{D_{05},D_{08},D_{09},D_{12},D_{13},D_{18},D_{19},D_{28},D_{29},D_{39},D_{43}\}$.

Lemma 2.6** ([6]).**

Let $C$ be an edge-cut in a fullerene graph $F$ and $H$ a component of $F-C$. Let $n_{1}$ and $n_{2}$ be the numbers of vertices of degree one and two, $f_{5}$ the number of pentagons, and $l$ the size of the outer face of $H$. Then, $6-f_{5}=4n_{1}+2n_{2}-l$.

Recall that a bridge of a graph $G$ is an edge $e$ such that $G-e$ has more connected components than $G$.

Theorem 2.7** ([9]).**

Let $G$ be a graph with a unique perfect matching. Then $G$ has a bridge belonging to the perfect matching.

Proposition 2.8**.**

*Suppose that $H$ is an induced subgraph of a fullerene $F$ and $H$ has a unique perfect matching $M$. If $\triangledown_{F}(H)$ is a cyclic $6$-edge-cut, then $H\cong J_{1}$ *(*see Fig. 3**). If $\triangledown_{F}(H)$ is an $8$-edge-cut and $H$ has not $1$-degree vertex, then $H\cong J_{2}$ *(see Fig. 3**).

Proof.

Since $H$ has a unique perfect matching $M$, $H$ has a bridge $e$ with $e\in M$ by Theorem 2.7. Clearly, deleting $e$ from $H$ makes two new connected components. We denote one of them by $H_{1}$, and set $H_{2}:=H-H_{1}$. $|\triangledown_{F}(H_{i})|$ is odd and $|\triangledown_{F}(H_{i})|\geq 3$, $i=1,2$ since $H_{i}$ has odd number of vertices and $F$ is a $3$-connected cubic graph. Without loss of generality, we suppose that $|\triangledown_{F}(H_{1})|\leq|\triangledown_{F}(H_{2})|$.

If $\triangledown_{F}(H)$ is a cyclic $6$-edge-cut of $F$, then $|\triangledown_{F}(H_{1})|+|\triangledown_{F}(H_{2})|=8$. So $\triangledown_{F}(H_{1})$ is a $3$-edge-cut and $\triangledown_{F}(H_{2})$ is a $5$-edge-cut. By Lemma 2.1, $H_{1}$ is an isolated vertex. So $H_{2}$ has a cycle, and further $\triangledown_{F}(H_{2})$ is a cyclic $5$-edge-cut. By Lemma 2.3, $\triangledown_{F}(H_{2})$ is a trivial cyclic $5$-edge-cut, otherwise, $H$ has at least two perfect matchings, a contradiction. Since $\triangledown_{F}(H)$ is a cyclic $6$-edge-cut, $F-H$ is connected and has cycles. So $F-H$ has at least six vertices. It follows that $H_{2}$ is a $5$-cycle. Then $H\cong J_{1}$.

If $\triangledown_{F}(H)$ is an $8$-edge-cut of $F$, then $|\triangledown_{F}(H_{1})|+|\triangledown_{F}(H_{2})|=10$. Since $H$ has not $1$-degree vertex, both $H_{1}$ and $H_{2}$ have cycles. So $\triangledown_{F}(H_{i})$ is a cyclic edge-cut of $F$, $i=1,2$. By Lemma 2.1, $|\triangledown_{F}(H_{i})|\geq 5$. So $\triangledown_{F}(H_{i})$ is a cyclic $5$-edge-cut of $F$. Since $H$ has a unique perfect matching, $H_{i}$ is a $5$-cycle. So $H\cong J_{2}$. ∎

3 Properties of fullerenes with $f(F)=3$

We start with some notations concerning a generalized patch $P$ of a fullerene introduced in [17]. Label clockwise (counterclockwise) the half-edges of $P$ by $t_{1},t_{2},\ldots,t_{k}$, and set $a_{i}$ as the number of vertices from $t_{i}$ to $t_{i+1}$ in a clockwise (counterclockwise) scan of the boundary of $P$. Then the cyclic sequence $[a_{1}a_{2}\ldots a_{k}]$ is called a distance-array of $P$. Since a fullerene graph has only pentagonal and hexagonal faces, $1\leq a_{i}\leq 6$. For instance, a distance-array to describe the boundary of $J_{1}$ (see Fig. 3) could be $[132223]$. Since we might start reading the boundary from different position and in clockwise or counterclockwise direction, the boundary of a generalized patch may has more than one distance-arrays to describe it. However, we easily see that for the same boundary, the distance-arrays by rotations and reversions are regarded as equivalent.

We note that if $P$ has a distance-array $[a_{1}a_{2}\ldots a_{k}]$, it has at most $2k$ distinct distance-arrays describing it. Clearly, the smallest one in the numerical is uniquely determined. In the following, we call such distance-array the min-distance-array of $P$. For example, for $P$ with a distance-array $[3516]$, there are eight distinct distance-arrays $[3516],[5163],[1635],[6351],[3615],[5361],[1536],[6153]$ to describe it, where $[1536]$ is the min-distance-array of $P$.

For nine caps in Fig. 5, $A_{1},A_{3}$ and $A_{6}$ have the min-distance-array $[234234]$, $A_{2},A_{5}$ and $A_{8}$ have $[233424]$, and $A_{4},A_{7}$ and $A_{9}$ have $[233343]$. A nanotube fullerene of type $(4,2)$ is comprised of two caps in Fig. 5 with the same min-distance-array, and some layers of hexagons between them. The graph shown in Fig. 6 is a nanotube fullerene of type $(4,2)$ with caps $A_{1}$ and $A_{3}$.

By Theorem $4$ and Corollary $5$ in [7], the following lemma holds.

Lemma 3.1**.**

Let $F$ be a fullerene graph with a non-degenerate cyclic $6$-edge-cut $C$. If the configurations of the six pentagons in one component of $F-C$ are as depicted in Fig. 3 $P^{3}$, then $F$ is a nanotube of type $(4,2)$.

Lemma 3.2**.**

*Let $F$ be a nanotube fullerene of type $(4,2)$. Then $f(F)=3$ and $F$ has a forcing set $S$ of size $3$ such that $F[V(S)]$ is isomorphic to $P_{6}$ with the incident edge set as depicted in $L_{3}$ *(see Fig. 12**).

Proof.

By the definition of a nanotube fullerene of type $(4,2)$, $F$ has a cap $A_{i}$, $i\in\{1,\ldots,9\}$ (see Fig. 5). Let $S$ be the set of the three dark edges in $A_{i}$ (see Fig. 5). Then we can check that $S$ is a forcing set of $F$. For example, $S:=\{e_{1},e_{2},e_{3}\}$ is a forcing set of the fullerene $F$ as shown in Fig. 6. ∎

Let $G$ be a connected graph and $e$ a cut edge of $G$. The edge $e$ has one end $x$ in one component $G_{1}$ of $G-e$. If $G_{1}-x$ is empty or has a unique perfect matching, then we call $e$ a pendent edge and $G_{1}$ a pendent blossom, and say that $G_{1}$ is incident with $e$, and vice versa. The number of the adjacent vertices of $v$ in $G$ is called the degree of $v$, denoted by $d_{G}(v)$.

Let $F$ be a fullerene and $S=\{e_{1},e_{2},e_{3}\}$ a forcing set of $F$. We define the following notations:

$F_{0}^{\prime}:=F[V(S)]$, $F_{0}^{\prime\prime}:=F-F_{0}^{\prime}$;

If $F_{i}^{\prime\prime}$ has a $1$-degree vertex, then $F_{i+1}^{\prime\prime}$ is obtained from $F_{i}^{\prime\prime}$ by deleting this $1$-degree vertex and its adjacent vertex;

$F_{i+1}^{\prime}:=F-F_{i+1}^{\prime\prime}$;

$X_{i}$ denotes the set of edges in $F$ from $F_{i}^{\prime}$ to $F_{i}^{\prime\prime}$.

Suppose that $F_{k}^{\prime\prime}$ ($k\geq 0$) is the first such subgraph that has no $1$-degree vertices. In the following, we let $F^{\prime\prime}:=F_{k}^{\prime\prime}$ and $F^{\prime}:=F-F^{\prime\prime}$.

Lemma 3.3**.**

For integer $k\geq 1$, $|X_{i+1}|\leq|X_{i}|\leq 12$, $i=0,1,\ldots,k-1$.

Proof.

Since $F_{0}^{\prime}=F[V(S)]$ and $S$ is a set of three independent edges, $|X_{0}|\leq 12$.

If $F_{i+1}^{\prime\prime}$ is obtained from $F_{i}^{\prime\prime}$ by deleting a $1$-degree vertex and its adjacent vertex, then $|X_{i+1}|=|X_{i}|-2+\delta_{e}$. Since $F$ is a cubic graph and $|F_{i}^{\prime\prime}|$ and $|F_{i+1}^{\prime\prime}|$ are even, $\delta_{e}=-2,0$, or $2$. So $|X_{i+1}|\leq|X_{i}|$. ∎

For two subgraphs $A$ and $B$ of a graph $G$, we denote by $E(A,B)$ the set of edges of $G$ with one end in $A$ and the other end in $B$, and set $e(A,B)=|E(A,B)|$.

Theorem 3.4**.**

*If $F^{\prime\prime}=\emptyset$, then $F=F^{\prime}$. If $F^{\prime\prime}\neq\emptyset$, then $|\triangledown_{F}(F^{\prime\prime})|=8,10,$ or $12$. Moreover, $F^{\prime\prime}\cong J_{2}$ *(*see Fig. 3**) if $|\triangledown_{F}(F^{\prime\prime})|=8$. Suppose that $F$ is not a nanotube fullerene of type $(4,2)$. Then $F^{\prime\prime}\cong PP$ *(see Fig. 3) or consists of a pentagon and a patch $H\in\mathcal{D}$ connecting by an edge if $|\triangledown_{F}(F^{\prime\prime})|=10$, and $F^{\prime\prime}$ consists of two patches $H_{1}\in\mathcal{D}$ and $H_{2}\in\mathcal{D}$ connecting by an edge if $F^{\prime\prime}$ has no pendent pentagons and $|\triangledown_{F}(F^{\prime\prime})|=12$.

Proof.

Recall that $S=\{e_{1},e_{2},e_{3}\}$ is a forcing set of $F$. We denote by $M$ the unique perfect matching of $F$ with $S\subseteq M$. We suppose that $F^{\prime\prime}\neq\emptyset$. Clearly, $F^{\prime\prime}$ is an induced plane subgraph of $F$ and it can be extended to $F$ on the plane. By Lemma 3.3, $|\triangledown_{F}(F^{\prime\prime})|\leq 12$. Since $F$ is $3$-regular and the order of $F^{\prime\prime}$ is even, $|\triangledown_{F}(F^{\prime\prime})|$ is even. Since $F^{\prime\prime}$ has no $1$-degree vertices and has a unique perfect matching, $F^{\prime\prime}$ is connected by Lemma 2.1 and Proposition 2.8. Since $F^{\prime\prime}$ has a unique perfect matching $M\cap E(F^{\prime\prime})$, by Theorem 2.7, $F^{\prime\prime}$ has a cut-edge $e^{\prime\prime}=v_{1}v_{2}\in M\cap E(F^{\prime\prime})$ which separates $F^{\prime\prime}$ into two components $H_{1}$ and $H_{2}$ with $v_{i}\in V(H_{i})$, $i=1,2$. Clearly, the order of $H_{i}$ is odd. So $|\triangledown_{F}(H_{i})|$ is odd. Since $F^{\prime\prime}$ has not $1$-degree vertices, both $\triangledown_{F}(H_{1})=E(H_{1},F^{\prime})\cup\{e^{\prime\prime}\}$ and $\triangledown_{F}(H_{2})=E(H_{2},F^{\prime})\cup\{e^{\prime\prime}\}$ are cyclic edge-cut of $F$. Since $c\lambda(F)=5$, we suppose that $|\triangledown_{F}(H_{1})|\geq|\triangledown_{F}(H_{2})|\geq 5$. So $|\triangledown_{F}(F^{\prime\prime})|=|\triangledown_{F}(H_{1})|+|\triangledown_{F}(H_{2})|-2\geq 8$. Hence $|\triangledown_{F}(F^{\prime\prime})|=8,10,$ or $12$. If $|\triangledown_{F}(F^{\prime\prime})|=8$, then $F^{\prime\prime}\cong J_{2}$ by Proposition 2.8. Next, we suppose that $|\triangledown_{F}(F^{\prime\prime})|=10$ or $12$.

Case $1$. $|\triangledown_{F}(F^{\prime\prime})|=10$.

Since $|\triangledown_{F}(H_{1})|+|\triangledown_{F}(H_{2})|=|\triangledown_{F}(F^{\prime\prime})|+2=12$, $|\triangledown_{F}(H_{2})|=5$ and $|\triangledown_{F}(H_{1})|=7$. By Lemma 2.3, $\triangledown_{F}(H_{2})$ is a trivial cyclic 5-edge-cut of $F$ since $H_{2}-v_{2}$ has a unique perfect matching. So $H_{2}$ is isomorphic to a $5$-cycle. Clearly, $H_{1}$ has not $1$-degree vertex or has a unique $1$-degree vertex $v_{1}$. For case $d_{H_{1}}(v_{1})=1$, $\triangledown_{F}(H_{1}-v_{1})$ is a cyclic $6$-edge-cut of $F$. Since $H_{1}-v_{1}$ has a unique perfect matching, $H_{1}-v_{1}\cong J_{1}$ by Proposition 2.8. So $F^{\prime\prime}\cong PP$. If $d_{H_{1}}(v_{1})=2$, then $H_{1}$ is $2$-connected, otherwise, $F$ has a cyclic edge-cut of size at most four, a contradiction. Moreover, any inner vertex of $H_{1}$ has degree $3$. So $H_{1}$ is a patch of $F$, and the boundary of $H_{1}$ is a cycle, denoted by $C$. We claim that $H_{1}$ has at most five pentagons (then $H_{1}\in\mathcal{D}$ by Lemma 2.5). By the contrary, we suppose that $H_{1}$ has at least six pentagons.

Since $|\triangledown_{F}(H_{1})|=7$, by Lemma 2.6, the length of $C$ is at least $14$. Let $C=v_{1}x_{1}x_{2}\cdots$

$x_{k-1}x_{k}v_{1}$, $k\geq 13$. We note that there are exactly seven $2$-degree vertices in $H_{1}$, all of which belong to $C$. Clearly, $\triangledown_{F}(H_{1}-v_{1})$ is an $8$-edge-cut of $F$. If $H_{1}-v_{1}$ has not $1$-degree vertex, then $H_{1}-v_{1}\cong J_{2}$ by Proposition 2.8 since $H_{1}-v_{1}$ has a unique perfect matching. This contradicts that $H_{1}-v_{1}$ has at least five pentagons. So $H_{1}-v_{1}$ has a $1$-degree vertex. Since $d_{H_{1}}(v_{1})=2$ and $H_{1}$ is $2$-connected, any $1$-degree vertex of $H_{1}-v_{1}$ is adjacent to $v_{1}$ in $H_{1}$. So only $x_{1}$ and $x_{k}$ may be $1$-degree vertex in $H_{1}-v_{1}$. We consider the following two cases.

Subcase $1.1.$ $d_{H_{1}-v_{1}}(x_{1})=1$ and $d_{H_{1}-v_{1}}(x_{k})\neq 1$.

In this case, $d_{H_{1}}(x_{1})=2$ and $d_{H_{1}}(x_{k})=3$. So $x_{1}x_{2}\in M$. Clearly, $H_{1}^{1}:=H_{1}-\{v_{1},x_{1},x_{2}\}$ has a unique perfect matching, and has at least 5 pentagons if $d_{H_{1}}(x_{2})=2$, at least 4 pentagons if $d_{H_{1}}(x_{2})=3$. If $d_{H_{1}}(x_{2})=2$, then $\triangledown_{F}(H_{1}^{1})$ is a cyclic $6$-edge-cut of $F$. By Proposition 2.8, $H_{1}^{1}\cong J_{1}$, a contradiction. Hence $d_{H_{1}}(x_{2})=3$ and $\triangledown_{F}(H_{1}^{1})$ is an $8$-edge-cut of $F$. By Proposition 2.8, $H_{1}^{1}$ has a $1$-degree vertex. We can check that any inner vertex of $H_{1}$ can not be $1$-degree vertex in $H_{1}^{1}$, and $d_{H_{1}^{1}}(x_{k})\neq 1$. So $d_{H_{1}^{1}}(x_{3})=1$. That is, $d_{H_{1}}(x_{3})=2$ and $x_{3}x_{4}\in M$. For $H_{1}^{2}:=H_{1}-\{v_{1},x_{1},x_{2},x_{3},x_{4}\}$, we can similarly show that $d_{H_{1}}(x_{4})=3$, $d_{H_{1}}(x_{5})=2$ and $x_{5}x_{6}\in M$. Set $H_{1}^{3}:=H_{1}-\{v_{1},x_{1},\ldots,x_{6}\}$. $H_{1}^{3}$ has a unique perfect matching, and has at least 3 pentagons if $d_{H_{1}}(x_{6})=2$, at least 2 pentagons if $d_{H_{1}}(x_{6})=3$. As the above discussion, $d_{H_{1}}(x_{6})=3$. If $H_{1}^{3}$ has no 1-degree vertex, then $H_{1}^{3}\cong J_{2}$ by Proposition 2.8. So $H_{1}\cong G_{1}$ (see Fig. 7 ($a$)).

Since the length of any facial-cycle of $F$ is five or six, $v_{2}$ is adjacent to two vertices in $H_{1}$, this contradicts that there is exactly one edge connecting $H_{1}$ and $H_{2}$. So $H_{1}^{3}$ has a $1$-degree vertex. We can check that only $x_{7}$ can be a $1$-degree vertex in $H_{1}^{3}$. Hence $d_{H_{1}}(x_{7})=2$ and $x_{7}x_{8}\in M$. For $H_{1}^{4}:=H_{1}-\{v_{1},x_{1},\ldots,x_{8}\}$, as the discussion of $H_{1}^{3}$, we have $d_{H_{1}}(x_{8})=3,d_{H_{1}}(x_{9})=2$ and $x_{9}x_{10}\in M$. Since $|\triangledown_{F}(H_{1})|=7$, $e(H_{1},H_{2})=1$ and any facial cycle of $F$ is of length $5$ or $6$, $d_{H_{1}}(x_{10})=3$. Set $H_{1}^{5}:=H_{1}-\{v_{1},x_{1},\ldots,x_{10}\}$. So $\triangledown_{F}(H_{1}^{5})$ is an $8$-edge-cut of $F$ and $H_{1}^{5}$ has a $1$-degree vertex. We can check that $d_{H_{1}^{5}}(x_{11})=1$. Hence the incident vertex set of $\triangledown_{F}(H_{1})$ in $H_{1}$ is $\{v_{1},x_{1},x_{3},x_{5},x_{7},x_{9},x_{11}\}$ (see Fig. 7 (b)). So $d_{H_{1}}(x_{i})=3$, $i=12,\ldots,k$. Since $H_{2}$ is a pentagon and the length of any facial cycle of $F$ is five or six, $k=13$ and $x_{11}$ has an adjacent vertex in $H_{2}$. So $e(H_{1},H_{2})\geq 2$. This contradicts that there is only one edge between $H_{1}$ and $H_{2}$. So Subcase $1.1$ can not happen.

Subcase $1.2.$ $d_{H_{1}-v_{1}}(x_{1})=1$ and $d_{H_{1}-v_{1}}(x_{k})=1$.

Then $d_{H_{1}}(x_{1})=2$ and $d_{H_{1}}(x_{k})=2$, $x_{1}x_{2}\in M$ and $x_{k}x_{k-1}\in M$. Let $H_{1,1}=H_{1}-\{v_{1},x_{1},x_{k}\}$ and $\bar{H}_{1,1}=F-H_{1,1}$. We note that $\triangledown_{F}(H_{1,1})$ is a cyclic $6$-edge-cut of $F$ and $|\bar{H}_{1,1}|\geq|\{v_{1},x_{1},x_{k}\}\cup V(H_{2})\cup V(S)|=3+5+2\times 3=14$.

Claim 1: $\triangledown_{F}(H_{1,1})$ is a non-degenerate cyclic $6$-edge-cut of $F$.

By the contrary, we suppose that $\triangledown_{F}(H_{1,1})$ is degenerate. So $H_{1,1}$ or $\bar{H}_{1,1}$ has at most five pentagons. If $H_{1,1}$ has at most five pentagons, then it has exactly five pentagons since $H_{1}$ has at least six pentagons. So $x_{2}$ and $x_{k-1}$ are adjacent in $H_{1}$. By Lemma 2.4, $H_{1,1}\cong P^{1}$ or $P^{2}$. We notice that vertices $x_{2}$ and $x_{k-1}$ are two $2$-degree adjacent vertices in $H_{1,1}$. It is easy to check that $H_{1,1}-\{x_{2},x_{k-1}\}$ is $2$-connected and has at least two perfect matchings. So $H_{1}-v_{1}$ has at least two perfect matchings, a contradiction. If $\bar{H}_{1,1}$ has at most five pentagons, by Lemma 2.4, $\bar{H}_{1,1}\cong P^{1}$ or $P^{2}$ since $|\bar{H}_{1,1}|\geq 14$. We notice that both $x_{1}$ and $x_{k}$ are $2$-degree vertices in $\bar{H}_{1,1}$ which are connected by a $2$-path $x_{1}v_{1}x_{k}$ on the boundary of $\bar{H}_{1,1}$. If $\bar{H}_{1,1}\cong P^{2}$, then for any case of $\{x_{1},x_{k}\}$, $\bar{H}_{1,1}-(\{x_{1},v_{1},x_{k}\}\cup V(H_{2}))$ is a path $P_{8}$ of order $8$. Clearly, $S\subset E(P_{8})$ can not force a perfect matching of $P_{8}$ in the initiative graph $F$, a contradiction. For the case $\bar{H}_{1,1}\cong P^{1}$, we can show that $\bar{H}_{1,1}-(\{x_{1},v_{1},x_{k}\}\cup V(H_{2}))$ has not perfect matching, a contradiction. Hence $\triangledown_{F}(H_{1,1})$ is a non-degenerate cyclic $6$-edge-cut of $F$.

By Lemma 2.6, the boundary $C$ of $H_{1}$ is a $14$-cycle, $k=13$. By Claim 1, $x_{2}x_{12}\notin E(F)$. Further, $d_{H_{1}}(x_{2})=3$ and $d_{H_{1}}(x_{12})=3$, otherwise, $\triangledown_{F}(H_{1}-\{v_{1},x_{1},x_{2},x_{13}\})$ or $\triangledown_{F}(H_{1}-\{v_{1},x_{1},x_{13},x_{12}\})$ is a non-degenerate cyclic $5$-edge-cut of $F$, then $H_{1}-v_{1}$ has at least two perfect matchings, a contradiction. So there is a vertex $y_{1}\in V(H_{1,1})$ lying in the interior area of $C$ such that $x_{2}y_{1}$, $x_{12}y_{1}\in E(H_{1,1})$. Since the cyclic edge-connectivity of $F$ is five and each $5$-cycle (resp. $6$-cycle) is a facial cycle, $y_{1}$ is not adjacent to $x_{3},x_{4},\ldots,x_{11}$. So there is $y_{2}\in V(H_{1})\setminus\{v_{1},x_{1},\ldots,x_{13}\}$ such that $y_{1}y_{2}\in M$. We note that $H_{1,1}$ is a patch of $F$ with $\triangledown_{F}(H_{1,1})$ being a non-degenerate cyclic $6$-edge-cut, and $x_{2}y_{1}x_{12}$ is a $2$-path on the boundary of $H_{1,1}$, $H_{1,1}-\{x_{2},x_{12}\}$ has a unique perfect matching $M\cap E(H_{1,1}-\{x_{2},x_{12}\})$.

Claim 2: Exactly one of $x_{3}$ and $x_{11}$ is $2$-degree vertex in $H_{1}$. Suppose that $d_{H_{1}}(x_{3})=2$, then $d_{H_{1}}(x_{4})=3$ and $y_{1}$, $x_{2},x_{3},x_{4}$ belong to a hexagonal face $f_{1}$ of $H_{1}$ (see Fig. 8 (a)).

Clearly, $H_{1,2}:=H_{1}-\{v_{1},x_{1},x_{2},x_{12},x_{13},y_{1},y_{2}\}$ has a unique perfect matching. Since $\triangledown_{F}(H_{1,2})$ is an $8$-edge-cut of $F$ and $H_{1,2}$ has at least three pentagons, $H_{1,2}$ has a $1$-degree vertex by Proposition 2.8. We can check that only $x_{3}$ and $x_{11}$ may be $1$-degree in $H_{1,2}$.

If both $x_{3}$ and $x_{11}$ are $1$-degree in $H_{1,2}$, then $x_{3}x_{4},x_{11}x_{10}\in M$. Let $H_{1,1}^{\prime}:=H_{1}-\{v_{1},x_{1},x_{2},x_{3},x_{11},x_{12},x_{13},y_{1}\}$. Since $\triangledown_{F}(H_{1,1}^{\prime})$ is a non-trivial cyclic $5$-edge-cut of $F$, by Lemma 2.3, $H_{1,1}^{\prime}$ is a patch that consists of a pentacap and $i\geq 0$ layers of hexagons around it. We notice that $x_{4},y_{2},x_{10}$ are three consecutive $2$-degree vertices on the boundary of $H_{1,1}^{\prime}$. It is easy to check that $H_{1,1}^{\prime}-\{x_{4},y_{2},x_{10}\}$ has at least two perfect matchings. So $H_{1}-v_{1}$ has at least two perfect matchings, a contradiction. So exactly one of $x_{3}$ and $x_{11}$ is $2$-degree in $H_{1}$.

We suppose that $d_{H_{1}}(x_{11})=3$, $d_{H_{1}}(x_{3})=2$. Then $x_{3}x_{4}\in M$. Let $H_{1,1}^{\prime\prime}:=H_{1}-\{v_{1},x_{1},x_{2},x_{3},x_{4},x_{12},x_{13},y_{1}\}$. If $d_{H_{1}}(x_{4})=2$, then $\triangledown_{F}(H_{1,1}^{\prime\prime})$ is a non-degenerate cyclic $5$-edge-cut. So $H_{1,1}^{\prime\prime}-y_{2}$ has at least two perfect matchings. It follows that $H_{1}-v_{1}$ has at least two perfect matchings, a contradiction. So $d_{H_{1}}(x_{4})=3$.

Clearly $H_{1,2}^{\prime}:=H_{1}-\{v_{1},x_{1},x_{2},x_{3},x_{4},x_{12},x_{13},y_{1},y_{2}\}$ has a unique perfect matching $M\cap E(H_{1,2}^{\prime})$. The two vertices $y_{2}$ and $x_{4}$ are not adjacent in $F$, otherwise, $\triangledown_{F}(H_{1,2}^{\prime})$ is a cyclic $6$-edge-cut of $F$, then $H_{1,2}\cong J_{1}$ by Proposition 2.8, this contradicts that $H_{1,2}^{\prime}$ has at least three pentagons. So $f_{1}$ is a hexagon (see Fig. 8($a$)). Then the Claim 2 is done.

Let $H_{1,3}:=H_{1}-\{v_{1},x_{1},x_{2},x_{3},x_{12},x_{13},y_{1}\}$. We notice that $\triangledown_{F}(H_{1,3})$ is a cyclic $6$-edge-cut of $F$ and $M\cap E(H_{1,3})$ is not a perfect matching of $H_{1,3}$. We can check that $H_{1,3}$ has not $1$-degree vertex. By Lemma 2.1, $H_{1,3}$ is $2$-connected. So $H_{1,3}$ is a patch of $F$.

If $f_{6}$ (see Fig. 8($a$)) is a pentagon, then $\triangledown_{F}(H_{1,3})$ is degenerate and $H_{1,3}$ has exactly five pentagons. So $H_{1,3}\cong P^{1}$ or $P^{2}$ by Lemma 2.4. We notice that $x_{4},y_{2},x_{11}$ are three $2$-degree vertices in $H_{1,3}$, and $x_{4},y_{2}$ are connected by a $2$-path, $y_{2},x_{11}$ are connected by a $2$-path on the boundary of $H_{1,3}$. Since $H_{1,3}-\{x_{4},y_{2}\}$ has a unique perfect matching $M\cap E(H_{1,3}-\{x_{4},y_{2}\})$, $H_{1}\cong R_{1}$ or $R_{2}$ (see Fig. 9).

If $f_{6}$ (see Fig. 8($a$)) is a hexagon, then $\triangledown_{F}(H_{1,3})$ is non-degenerate. We can check that $H_{1,3}$ is a patch of $F$, $x_{4}y_{3}y_{2}$ is a $2$-path on the boundary of $H_{1,3}$ and $H_{1,3}-\{x_{4},y_{2}\}$ has a unique perfect matching $M\cap E(H_{1,3}-\{x_{4},y_{2}\})$. As the proof of Claim 2, we obtain that $d_{H_{1}}(x_{5})=2$ and $d_{H_{1}}(x_{6})=3$, $f_{2}$ is a hexagon (see Fig. 8($b$)). Let $H_{1,4}:=H_{1}-\{v_{1},x_{1},x_{2},x_{3},x_{4},x_{5},x_{12},x_{13},y_{1},y_{2},y_{3}\}$. Clearly, $\triangledown_{F}(H_{1,4})$ is a cyclic $6$-edge-cut of $F$. As the proof of $H_{1,3}$, we know that $H_{1,4}$ is a patch of $F$. If $g_{2}$ is a pentagon, then $\triangledown_{F}(H_{1,4})$ is degenerate and $H_{1,4}$ has five pentagons. So $H_{1,4}\cong P^{1}$ or $P^{2}$. We note that the distance-array of $H_{1,4}$ has three successive $3$ and the first $3$ corresponds to $x_{6}x_{5}$ and $y_{4}y_{3}$. Since $H_{1,4}-\{x_{6},y_{4}\}$ has a unique perfect matching, $H_{1,4}\cong P^{2}$. So $H_{1}\cong R_{3}$. If $g_{2}$ is a hexagon, then $\triangledown_{F}(H_{1,4})$ is non-degenerate. We note that $x_{6}y_{5}y_{4}$ is a $2$-path on the boundary of $H_{1,4}$ and $H_{1,4}-\{x_{6},y_{4}\}$ has a unique perfect matching $M\cap E(H_{1,4}-\{x_{6},y_{4}\})$. As the proof of Claim 2, we have $d_{H_{1}}(x_{7})=2$ and $d_{H_{1}}(x_{8})=3$, $f_{3}$ is a hexagon (see Fig. 8($c$)). We can check that $H_{1,5}:=H_{1}-\{v_{1},x_{1},x_{2},\ldots,x_{7},x_{12},x_{13},y_{1},y_{2},\ldots,y_{5}\}$ is a patch of $F$ and $\triangledown_{F}(H_{1,5})$ is a cyclic $6$-edge-cut. If $g_{3}$ is a pentagon, then $\triangledown_{F}(H_{1,5})$ is degenerate and $H_{1,5}$ has five pentagons. By Lemma 2.4, $H_{1,5}\cong P^{2}$ since the distance-array of $H_{1,5}$ has four consecutive $3$. So $H_{1}\cong R_{4}$ since $H_{1,5}-\{x_{8},y_{6}\}$ has a unique perfect matching. If $g_{3}$ is a hexagon, then $\triangledown_{F}(H_{1,5})$ is non-degenerate, as the above discussion, we know that $d_{H_{1}}(x_{9})=2$, $d_{H_{1}}(x_{10})=3$ and $f_{4}$ is a hexagon (see Fig. 8(d)). Then the incident vertex set of the seven edges in $\triangledown_{F}(H_{1})$ is $\{v_{1},x_{1},x_{3},x_{5},x_{7},x_{9},x_{13}\}$(see Fig. 8($d$)). Set $H_{1,6}:=H_{1}-\{y_{1}\}\cup V(C)$. If $f_{5}$ is a pentagon, then $\triangledown_{F}(H_{1,6})$ is a degenerate cyclic $6$-edge-cut of $F$ such that $H_{1,6}$ has five pentagons. So $H_{1,6}\cong P^{2}$ since the distance array of $H_{1,6}$ has five consecutive $3$. We notice that $y_{2}$ and $y_{3}$ are two adjacent $2$-degree vertices on the boundary of $H_{1,6}$. So $H_{1}\cong R_{5}$ (see Fig. 9). If $f_{5}$ is a hexagon, then $\triangledown_{F}(H_{1,6})$ is a non-degenerate cyclic $6$-edge-cut of $F$ and has the same distance-array as the patch $H_{1,1}$. In an inductive way, we can show that the configurations of the six pentagons in the component $H_{1,1}$ are as depicted in Fig. 3 $P^{3}$. By Lemma 3.1, $F$ is a nanotube fullerene of type $(4,2)$, a contradiction. So Subcase 1.2 can not happen.

Case $2$. $|\triangledown_{F}(F^{\prime\prime})|=12$ and $F^{\prime\prime}$ has no pendent pentagons.

Since $F^{\prime\prime}$ has no pendent pentagons, both $\triangledown_{F}(H_{1})$ and $\triangledown_{F}(H_{2})$ are not cyclic $5$-edge-cut of $F$, otherwise, $H_{1}$ or $H_{2}$ is a pendent pentagon of $F^{\prime\prime}$ since $F^{\prime\prime}$ has a unique perfect matching. So $|\triangledown_{F}(H_{i})|\geq 7$, $i=1,2$. This implies that $e(H_{1},F^{\prime})\geq 6$ and $e(H_{2},F^{\prime})\geq 6$. Hence $e(H_{1},F^{\prime})+e(H_{2},F^{\prime})\geq 12$. Since $|\triangledown_{F}(F^{\prime\prime})|=e(H_{1},F^{\prime})+e(H_{2},F^{\prime})=12$, both $\triangledown_{F}(H_{1})$ and $\triangledown_{F}(H_{2})$ are cyclic $7$-edge-cuts of $F$.

Since $F^{\prime\prime}$ is connected, $H_{1}$ is connected. We claim that $H_{1}$ has not $1$-degree vertex. If not, $H_{1}$ has a $1$-degree vertex. We note that only $v_{1}$ can be a $1$-degree vertex in $H_{1}$ since $F^{\prime\prime}$ has not $1$-degree vertices. So $d_{H_{1}}(v_{1})=1$ and $\triangledown_{F}(H_{1}-v_{1})$ is a cyclic $6$-edge-cut of $F$. Since $H_{1}-v_{1}$ has a unique perfect matching $M\cap E(H_{1}-v_{1})$, $H_{1}-v_{1}\cong J_{1}$ by Proposition 2.8. So $F^{\prime\prime}$ has a pendant pentagon, a contradiction. It means that $H_{1}$ has not $1$-degree vertex. Since $c\lambda(F)=5$ and $|\triangledown_{F}(H_{1})|=7$, $H_{1}$ is $2$-connected and any inner vertex of $H_{1}$ has degree $3$. So $H_{1}$ is a patch of $F$. Similarly, $H_{2}$ is also a patch of $F$.

Obviously, $F$ has exactly twelve pentagons. We claim that $H_{i}$ has at most five pentagons, $i=1,2$. If not, we suppose that $H_{1}$ has at least six pentagons. As the proof of Subcases 1.1 and 1.2, we can also obtain a contradiction. So $H_{1}$ has at most five pentagons. Similarly, $H_{2}$ has at most five pentagons. Since both $H_{1}-v_{1}$ and $H_{2}-v_{2}$ have a unique perfect matching, $H_{1},H_{2}\in\mathcal{D}$ by Lemma 2.5. ∎

Corollary 3.5**.**

In the condition of Theorem 3.4, $F^{\prime\prime}\cong PP$ $($see Fig. 3$)$ or $F^{\prime\prime}\cong T_{1},T_{2},\ldots T_{16}$ $($see Fig. 28$)$ if $|\triangledown_{F}(F^{\prime\prime})|=10$. $F^{\prime\prime}\cong T_{17},T_{18},\ldots T_{180}$ $($see Fig. 29, 30 and 31$)$ if $|\triangledown_{F}(F^{\prime\prime})|=12$ and $F^{\prime\prime}$ has no pendent pentagons.

Lemma 3.6**.**

Let $H$ be an induced subgraph of fullerene $F$ with a unique perfect matching $M$ and having no $1$-degree vertices. Then for any isolated edge $e_{3}$ in $F-H$, we have

(i) If $H$ consists of two patches $H_{x}$ and $H_{y}$ connecting by an edge $xy\in M$ with $x\in H_{x},y\in H_{y}$ and $|\triangledown_{F}(H_{x})|=|\triangledown_{F}(H_{y})|=7$, then $e(e_{3},H_{x})=e(e_{3},H_{y})=2$. Moreover, if $T:=F-H-V(e_{3})$ is a generalized patch of $F$, then $F[V(H)\cup V(e_{3})]$ is a patch as shown in Fig. 10, and $\triangledown_{F}(T)$ and $\triangledown_{F}(F[V(H)\cup V(e_{3})])$ merge successively along the boundaries of $T$ and $F[V(H)\cup V(e_{3})]$, respectively. If we label the edges in $\triangledown_{F}(F[V(H)\cup V(e_{3})])$ as shown in Fig. 10, then $T$ has the distance-array $[a_{1}a_{2}\cdots a_{8}]$ with $a_{1}=1$ and $a_{5}\in\{1,2\}$.

(ii) If $H$ has a pendent pentagon $P_{x}$ connecting by an edge $xy$ to $R:=H-P_{x}$ with $x\in P_{x}$ and $y\in R$, and $|\triangledown_{F}(R)|\leq 9$, then $e(e_{3},P_{x})=1$. Moreover, if $T:=F-H-V(e_{3})$ is a generalized patch of $F$ and $T$ has a perfect matching, then the three edges in $E(P_{x},T)$ are incident with three successive vertices on $P_{x}$ respectively, and there are two elements $a_{i}$ and $a_{i+1}$ in the distance-array $[a_{1}a_{2}\cdots a_{8}]$ of $T$ such that $a_{i},a_{i+1}\in\{3,4\}$ (here $a_{9}:=a_{1}$).

Proof.

From the assumption in (i), both $H_{x}$ and $H_{y}$ have cycles, and $|H_{x}|\geq 7$, $|H_{y}|\geq 7$. Since the cyclic edge-connectivity of $F$ is five, $e(e_{3},H_{x})\geq 1$ and $e(e_{3},H_{y})\geq 1$, otherwise, $\triangledown_{F}(H_{y}\cup e_{3})$ or $\triangledown_{F}(H_{x}\cup e_{3})$ is a cyclic $3$-edge-cut of $F$, a contradiction. If $e(e_{3},H_{x})=1$, then $\triangledown_{F}(H_{y}\cup e_{3})$ is a cyclic $5$-edge-cut of $F$ that separates $H_{x}$ and $H_{y}$. So $\triangledown_{F}(H_{y}\cup e_{3})$ is a nontrivial cyclic $5$-edge-cut of $F$. Let $A^{i}$ be a patch that consists of a pentacap and $i$ layers of hexagons (see Fig. 11 for $A^{1}$). By Lemma 2.3, $F[V(H_{y})\cup V(e_{3})]\cong A^{i}$ for some integer $i\geq 0$, and $e_{3}$ is a boundary edge of $F[V(H_{y})\cup V(e_{3})]$. Let $y^{\prime}$ be a $2$-degree vertex in $A^{i}$. For any boundary edge $e^{\prime}$ of $A^{i}$ that is not incident with $y^{\prime}$, $A^{i}-(\{y^{\prime}\}\cup V(e^{\prime}))$ has at least two perfect matchings or $A^{i}-V(e^{\prime})$ has a $1$-degree vertex different from $y^{\prime}$. So $H_{y}-y$ has at least two perfect matchings or $H_{y}$ has a $1$-degree vertex different from $y$. This contradicts that $H_{y}-y$ has a unique perfect matching and $H$ has not $1$-degree vertex. Hence $e(e_{3},H_{x})\geq 2$. Similarly, $e(e_{3},H_{y})\geq 2$. Since $e(e_{3},H_{x})+e(e_{3},H_{y})=4$, $e(e_{3},H_{x})=e(e_{3},H_{y})=2$. Since any facial cycle of $F$ is a pentagon or hexagon and $E(H_{x},H_{y})=\{xy\}$, the two edges in $E(e_{3},H_{x})$ are incident with two ends of $e_{3}$. Since $T$ is connected and $F$ is a plane graph, the two edges in $E(H_{x},e_{3})$ are incident with two successive $2$-degree vertices on the boundary of $H_{x}$ and the four edges in $E(H_{x},T)$ are incident to four successive $2$-degree vertices on the boundary of $H_{x}$, respectively. For the six edges in $\triangledown_{F}(H_{y})\setminus\{xy\}$, we have the same property. So $F[V(H)\cup V(e_{3})]$ is a patch of $F$ as depicted in Fig. 10. Further, since $F$ is a plane graph, the edges in $\triangledown_{F}(T)$ and the edges in $\triangledown_{F}(H\cup V(e_{3}))$ merge successively along the boundaries of $T$ and $F[V(H)\cup V(e_{3})]$. Since $E(H_{x},H_{y})=1$ and any facial cycle of $F$ is a pentagon or hexagon, $b_{1}=5$, $b_{5}\in\{4,5\}$ in the distance-array $[b_{1}\cdots b_{8}]$ of $F[V(H)\cup V(e_{3})]$. So we have $a_{1}=1$, $a_{5}\in\{1,2\}$ in the distance-array $[a_{1}a_{2}\cdots a_{8}]$ of $T$.

For (ii), since $F^{\prime\prime}$ has no $1$-degree vertices, $R:=F^{\prime\prime}-P_{x}$ has a cycle. Since $P_{x}$ is a pentagon, $e(e_{3},P_{x})\leq 1$. If $e(e_{3},P_{x})=0$, then $e(e_{3},R)=4$ and $\triangledown_{F}(R\cup e_{3})$ is a cyclic edge-cut of $F$ of size at most five since $|\triangledown_{F}(R)|\leq 9$. So $\triangledown_{F}(R\cup e_{3})$ is a non-trivial cyclic $5$-edge-cut of $F$ with one component being $F[V(R)\cup V(e_{3})]$. By Lemma 2.3, $F[V(R)\cup V(e_{3})]\cong A^{i}$ for some integer $i\geq 0$ ($A^{i}$ is as depicted in the above case). If $i=0$, then $F\cong G^{1}$ (see Fig. 11) since $E(R\cup e_{3},P_{x})=\{xy\}$ and $P_{x}$ is isomorphic to a $5$-cycle. This implies that $T\cong J_{3}$ as depicted in Fig. 11. We note that $J_{3}$ has not perfect matching, a contradiction. So $i\geq 1$. Since $e(e_{3},R)=4$, the degrees of the two ends of $e_{3}$ both are three in $F[V(R)\cup V(e_{3})]$. So $e_{3}$ is not on the boundary of $F[V(R)\cup V(e_{3})]$. Let $y^{\prime}$ be a $2$-degree vertex of $A^{i}$. For each edge $e^{\prime}$ not on the boundary of $A^{i}$, it is easy to check that $A^{i}-V(e^{\prime})\cup\{y^{\prime}\}$ has at least two perfect matchings. So $R-y$ has at least two perfect matchings. This contradicts that $H$ has a unique perfect matching. Hence $e(e_{3},P_{x})=1$. As the discussion of case $(i)$, if $T$ is a generalized patch of $F$, then the three edges in $E(P_{x},T)$ are incident with three successive vertices on $P_{x}$, respectively. We suppose that the distance-array of $T$ is $[a_{1}\cdots a_{8}]$. Clearly, the three edges in $E(P_{x},T)$ are three consecutive edges $t_{i},t_{i+1},t_{i+2}$ in $\triangledown_{F}(T)$. Since any facial cycle of $F$ is a pentagon or hexagon, $a_{i},a_{i+1}\in\{3,4\}$. ∎

4 Generalized patches of fullerenes with $f(F)=3$

A fullerene graph $F$ is called to be $1$-resonant if for each hexagon $f$ of $F$, the subgraph $F-V(f)$ has a perfect matching.

Theorem 4.1** ([18, 11]).**

Fullerene graph is $1$-resonant.

Theorem 4.2**.**

*Let $F$ be a fullerene that is not nanotube of type $(4,2)$. If $S=\{e_{1},e_{2},e_{3}\}$ forces a perfect matching $M$ of $F$ and any proper subset of $S$ is not a forcing set of $F$, then $F\cong W_{i}$, $i\in\{1,\ldots,81\}$ *(*see Figs. 24 and 27**), or $F$ has a generalized patch $L\in\mathcal{L}$ such that $S\subseteq E(L)$ forces a perfect matching $M\cap E(L)$ of $L$ in the sense of $F$, where $\mathcal{L}:=\{L_{1},L_{2},\ldots,L_{95}\}$ *(see Figs. 12–25**).

Proof.

By the assumption, $M$ is a unique perfect matching of $F$ with $S\subseteq M$. Set $F_{0}^{\prime}:=F[V(S)]$ and $F_{0}^{\prime\prime}:=F-F_{0}^{\prime}$. Obviously, $F_{0}^{\prime}$ has three cases as follows.

Case 1. $F_{0}^{\prime}$ is connected.

Since the order of $F_{0}^{\prime}$ is six, $F_{0}^{\prime}\cong P_{6}$, or $C_{6}$, or $J_{1}$ (see Fig. 3), or $T^{\prime}$ (see Fig. 12). Since any proper subset of $S$ is not a forcing set of $F$, $F_{0}^{\prime}\not\cong J_{1}$, $T^{\prime}$. Clearly, each $6$-cycle of $F$ is a facial cycle. By Theorem 4.1, $F$ is $1$-resonant. Since $F-V(C_{6})$ is $2$-connected, $F-V(C_{6})$ has at least two perfect matchings. So $F_{0}^{\prime}\cong P_{6}$. It follows that $F$ has a generalized patch $L_{i}$, $i\in\{1,\ldots,5\}$ (see Fig. 12).

Case 2. $F_{0}^{\prime}$ has exactly two connected components, say $B_{1}$ and $B_{2}$.

Without loss of generality, we suppose that $|B_{1}|\geq|B_{2}|$. Then $B_{1}$ is a path with four vertices and $B_{2}$ is an edge. Clearly, $|\triangledown_{F}(F_{0}^{\prime})|=10$. We notice that $F_{0}^{\prime\prime}:=F-F_{0}^{\prime}$ has a unique perfect matching $M\setminus S$.

We claim that $F_{0}^{\prime\prime}$ has a $1$-degree vertex. If not, then $F_{0}^{\prime\prime}$ has a pendent pentagon $P_{x}$ by Theorem 3.4. Obviously, $\triangledown_{F}(B_{1})$ has exactly two cases as shown in Fig. 13 and 14. So the distance-array of $B_{1}$ is $[132132]$ or $[122214]$. This is impossible by Lemma 3.6 $(ii)$. So $F_{0}^{\prime\prime}$ has a $1$-degree vertex.

For $F_{0}^{\prime}$ in Fig. 13, the $1$-degree vertex of $F_{0}^{\prime\prime}$ has one adjacent vertex in $B_{1}$ and one in $B_{2}$ since any $5$-cycle of $F$ is a facial cycle and the length of the shortest cycle in $F$ is five. So $F$ has a generalized patch $L_{i}$, $i\in\{6,\ldots,15\}$.

For $F_{0}^{\prime}$ in Fig. 14, the $1$-degree vertex of $F_{0}^{\prime\prime}$ has one adjacent vertex in $B_{1}$ and one in $B_{2}$, or has two adjacent vertices in $B_{1}$. For the first case, $F$ has a generalized patch $L_{i}$, $i\in\{16,\ldots,25\}$. For the second case, $F$ has a subgraph $L_{a1}$ or a generalized patch $L_{i}$, $i=26$ or $27$. Since $|\triangledown_{F}(L_{a1})|=10$ and the distance-array of $L_{a1}-V(e_{3})$ is $[132223]$, by Lemma 3.6 (ii), $F-V(L_{a1})$ has a $1$-degree vertex. Clearly, this $1$-degree vertex connects the two components of $L_{a1}$. So $F$ has a generalized patch $L_{i}$, $i\in\{28,29,30\}$ or a generalized patch $L_{j}$, $j\in\{18,\ldots,25\}$.

Case 3. $F_{0}^{\prime}$ consists of three independent edges $e_{1}$, $e_{2}$ and $e_{3}$ (see Fig. 16).

$F_{0}^{\prime\prime}:=F-F_{0}^{\prime}$ has a unique perfect matching $M\setminus S$ and $|\triangledown_{F}(F_{0}^{\prime\prime})|=12$. If $F_{0}^{\prime\prime}$ has not $1$-degree vertices, then we claim that $F_{0}^{\prime\prime}$ has not pendent pentagons. Suppose that $F_{0}^{\prime\prime}$ has a pendent pentagon $U$. Let edge $e\in M\setminus S$ be the pendent edge that is incident with $U$. By Lemma 3.6 (ii), $e(U,e_{i})=1$ for $i=1,2,3$. So $e(U,F_{0}^{\prime\prime}-U)=2$, a contradiction. Hence $F_{0}^{\prime\prime}$ has not pendent pentagons. By Theorem 3.4, $F_{0}^{\prime\prime}$ consists of two patches $U_{0},V_{0}\in\mathcal{D}$ connecting by an edge $x_{0}\in M\setminus S$. By Lemma 3.6 $(i)$, $e(U_{0},e_{i})=2=e(V_{0},e_{i})$ for $i=1,2,3$. Since $F$ is a plane graph, the two edges in $E(U_{0},e_{i})$ are two edges in $\triangledown_{F}(U_{0})$ (and $\triangledown_{F}(e_{i})$) that are consecutive along the boundary of $U_{0}$ (and $e_{i}$) for $i=1,2,3$.

Similarly, the two edges in $E(V_{0},e_{i})$ are two edges in $\triangledown_{F}(V_{0})$ (and $\triangledown_{F}(e_{i})$) that are consecutive along the boundary of $V_{0}$ (and $e_{i}$) for $i=1,2,3$. Since $E(U_{0},V_{0})=\{x_{0}\}$ and any facial cycle of $F$ is $5$-cycle or $6$-cycle, the two edges in $E(U_{0},e_{i})$ are incident with two ends of $e_{i}$ respectively, $i=1,2,3$. So both $F[V(U_{0})\cup V(S)]$ and $F[V(V_{0})\cup V(S)]$ have the structures as depicted in Fig. 15 $(a)$. It implies that both $U_{0}$ and $V_{0}$ have the distance-array $[a_{1}\wedge a_{2}a_{3}2a_{5}2a_{7}]$, where $a_{1},a_{2}=2$ or $3$, $a_{3},a_{5},a_{7}=3$ or $4$, and $\wedge$ marks the position from which the edge $x_{0}$ issues out. We can check that the distance-arrays $[2\wedge 232323]$ of $D_{08}$ and $[2\wedge 243234]$ of $D_{29}$ satisfy the condition of $U_{0}$. So $U_{0}=D_{08}$ or $D_{29}$. Similarly, $V_{0}=D_{08}$ or $D_{29}$. Hence, $F\cong W_{1},W_{2},$ or $W_{3}$ (see Fig. 24).

If $F_{0}^{\prime\prime}$ has a $1$-degree vertex, then this vertex has one adjacent vertex in $V(e_{1})$ and one in $V(e_{2})$. So $F$ has a subgraph that is isomorphic to $L_{b1}$, $L_{b2}$, or $L_{b3}$, or has a generalized patch $L_{i}$, $i\in\{31,\ldots,36\}$ (see Fig. 16). Now we consider the following three cases.

Subcase 3.1. $F$ has a subgraph $L_{b1}$.

Let $F_{1}^{\prime}:=L_{b1}$ and $F_{1}^{\prime\prime}:=F-F_{1}^{\prime}$. Clearly, $F_{1}^{\prime\prime}$ has a $1$-degree vertex that is adjacent to one end of $e_{1}$ and one end of $e_{2}$. Then $F$ has a subgraph $F_{2}^{\prime}$ as depicted in Fig. 17, or has a generalized patch $L_{31}$ or $L_{32}$ (see Fig. 16). Let $F_{2}^{\prime\prime}:=F-F_{2}^{\prime}$. If $F_{2}^{\prime\prime}$ has not $1$-degree vertices, then it also has not pendent pentagons by Lemma 3.6 $(ii)$ since the distance-array of $F_{2}^{\prime}-V(e_{3})$ is $[13231323]$. By Theorem 3.4, $F_{2}^{\prime\prime}$ consists of two patches $U_{2}^{1},V_{2}^{1}\in\mathcal{D}$ connecting by only an edge $x_{2}^{1}$.

By Lemma 3.6 $(i)$, $F[F_{2}^{\prime\prime}\cup V(e_{3})]$ is a patch that has the structure as shown in Fig. 10, and the edges in $\triangledown_{F}(L_{b1}-V(e_{3}))$ and $\triangledown_{F}(F[F_{2}^{\prime\prime}\cup V(e_{3})])$ merge successively along the boundaries of $L_{b1}-V(e_{3})$ and $F[F_{2}^{\prime\prime}\cup V(e_{3})]$, respectively. So the structures of $F[V(U_{2}^{1})\cup V(F_{2}^{\prime})]$ and $F[V(V_{2}^{1})\cup V(F_{2}^{\prime})]$ are as depicted in Fig. 15 $(b)$. The distance-arrays of $U_{2}^{1}$ and $V_{2}^{1}$ are both $[a_{1}\wedge a_{2}a_{3}2a_{5}a_{6}a_{7}]$, where $a_{1},a_{2},a_{5},a_{7}=2$ or $3$ and $a_{3},a_{6}=3$ or $4$. It is easy to check that the distance-arrays $[2\wedge 232232]$ of $D_{05}$, $[3\wedge 232232]$ of $D_{08}$, $[2\wedge 242233]$ of $D_{09}$, $[2\wedge 332233]$ of $D_{12}$, $[3\wedge 242242]$ of $D_{13}$, $[2\wedge 242333]$ of $D_{18}$, $[2\wedge 342243]$ of $D_{19}$ satisfy the above condition of $U_{2}^{1}$. Similarly, we have the same conclusion for $V_{2}^{1}$. So $U_{2}^{1},V_{2}^{1}\in\{D_{05},D_{08},D_{09},D_{12},D_{13},D_{18},D_{19}\}$. Hence $F\cong W_{i}$, $i\in\{4,\ldots,26\}$ (see Fig. 24). If $F_{2}^{\prime\prime}$ has a $1$-degree vertex, then this vertex connects the two components of $F_{2}^{\prime}$. So $F$ has a patch $L_{i}$, $i\in\{37,\ldots,42\}$.

Subcase 3.2. $F$ has a subgraph $L_{b2}$.

By the above discussions, we suppose that any two edges in $S$ can not be as $\{e_{1},e_{2}\}$ in $L_{b1}-V(e_{3})$. Let $F_{1}^{\prime}:=L_{b2}$ and $F_{1}^{\prime\prime}:=F-F_{1}^{\prime}$. If $F_{1}^{\prime\prime}$ has not $1$-degree vertices, then it also has no pendent pentagons by Lemma 3.6 (ii) since the distance-array of $L_{b2}-V(e_{3})$ is $[12321414]$. By Theorem 3.4, $F_{1}^{\prime\prime}$ consists of two patches $U_{1}^{2},V_{1}^{2}\in\mathcal{D}$ connecting by only one edge $x_{1}^{2}$. By Lemma 3.6 (i), the structures of $F[V(U_{1}^{2})\cup V(F_{1}^{\prime})]$ and $F[V(V_{1}^{2})\cup V(F_{1}^{\prime})]$ are as depicted in Fig. 15 $(c),(d)$ respectively. So the distance-array of $U_{1}^{2}$ is $[a_{1}\wedge a_{2}a_{3}22a_{6}2]$, of $V_{1}^{2}$ is $[a_{1}^{\prime}\wedge a_{2}^{\prime}a_{3}^{\prime}2a_{5}^{\prime}a_{6}^{\prime}a_{7}^{\prime}]$, where $a_{1},a_{2}=2$ or $3$, $a_{3}=3$ or $4$, $a_{6}=4$ or $5$, $a_{1}^{\prime},a_{2}^{\prime},a_{6}^{\prime}=2$ or $3$, $a_{3}^{\prime},a_{5}^{\prime},a_{7}^{\prime}=3$ or $4$. We can check that the distance-array $[3\wedge 242242]$ of $D_{13}$ satisfies the condition of $U_{1}^{2}$, and the distance-arrays $[2\wedge 232323]$ of $D_{08}$, $[2\wedge 242333]$ of $D_{18}$, $[2\wedge 242424]$ of $D_{29}$ satisfy the condition of $V_{1}^{2}$. So $F\cong W_{27},W_{28},$ or $W_{29}$ (see Fig. 24).

If $F_{1}^{\prime\prime}$ has a $1$-degree vertex, then this vertex connects the two components of $F_{1}^{\prime}$ or has two adjacent vertices in $L_{b2}-V(e_{3})$. For the first case, $F$ has a generalized patch $L_{i}$, $i\in\{43,\ldots,51\}$. For the second case, $F$ has a subgraph $L_{c1}$ or a generalized patch $L_{i}$, $i\in\{52,53,54,55\}$.

Now, we suppose that $F$ has a subgraph $L_{c1}$. Let $F_{2}^{\prime}:=L_{c1}$ and $F_{2}^{\prime\prime}:=F-F_{2}^{\prime}$. We notice that $F_{2}^{\prime\prime}$ has a unique perfect matching $M\cap E(F_{2}^{\prime\prime})$.

If $F_{2}^{\prime\prime}$ has not $1$-degree vertices, then $F_{2}^{\prime\prime}$ is connected by Lemma 2.1 and Proposition 2.8. If $F_{2}^{\prime\prime}$ has a pendent pentagon, say $P_{x}$, then by Lemma 3.6 (ii) the three edges in $E(P_{x},L_{c1}-V(e_{3}))$ are three consecutive edges in $\triangledown_{F}(P_{x})$ along the boundary of $P_{x}$. So the three edges $r_{1},r_{2},r_{3}$ in $L_{c1}$ (see Fig. 18) are those three edges in $E(P_{x},L_{c1}-V(e_{3}))$. Hence $F$ has a generalized patch $L_{56}$ or $L_{57}$.

If $F_{2}^{\prime\prime}$ has no pendent pentagons, then $F_{2}^{\prime\prime}$ consists of two patches $U_{2}^{2},V_{2}^{2}\in\mathcal{D}$ connecting by only an edge $x_{2}^{2}$ by Theorem 3.4. By Lemma 3.6 (i), the structures of $F[V(U_{2}^{2})\cup V(F_{2}^{\prime})]$ and $F[V(V_{2}^{2})\cup V(F_{2}^{\prime})]$ are as depicted in Fig. 20. So the distance-array of $V_{2}^{2}$ is $[2\wedge a_{2}^{\prime}a_{3}^{\prime}2a_{5}^{\prime}2a_{7}^{\prime}]$, where $a_{2}^{\prime},a_{5}^{\prime}=2$ or $3$, $a_{3}^{\prime}=3$ or $4$, $a_{7}^{\prime}=4$ or $5$. We can check that the distance-array of each patch in $\mathcal{D}$ does not satisfy the condition of $V_{2}^{2}$. So $V_{2}^{2}\notin\mathcal{D}$, a contradiction.

If $F_{2}^{\prime\prime}$ has a $1$-degree vertex, then this vertex connects the two components of $F_{2}^{\prime}$ or not. For the first case, $F$ has a generalized patch $L_{i}$, $i\in\{58,\ldots,63\}$, or has a generalized patch $L_{j}$, $j\in\{43,45,46,\ldots,51\}$. For the second case, $F$ has a generalized patch $L_{i}$, $i\in\{54,55\}$, or has a subgraph $L_{d1}$ or $L_{d2}$ (see Fig. 19). If $F$ has a subgraph $L_{d1}$, then by Theorem 3.4 and Lemma 3.6 (ii) $F-V(L_{d1})$ has a $1$-degree vertex since $|\triangledown_{F}(L_{d1})|=10$ and the distance-array of $L_{d1}-V(e_{3})$ is $[232323]$. Clearly, this $1$-degree vertex connects the two components of $L_{d1}$. So $F$ has a generalized patch $L_{i}$, $i\in\{48,49,50,51,54,55\}$.

If $F$ has a subgraph $L_{d2}$, then $F-V(L_{d2})$ has a $1$-degree vertex that has two adjacent vertices in $L_{d2}-V(e_{3})$. So $F$ has a subgraph $L_{e1}$ (see Fig. 19), or has a generalized patch $L_{64}$ or $L_{65}$. For the subgraph $L_{e1}$, $F-V(L_{e1})$ has a $1$-degree vertex or a pendent pentagon by Theorem 3.4 and Lemma 3.6 (i) since $|\triangledown_{F}(L_{e1})|=12$ and the distance-array of $L_{e1}-V(e_{3})$ is $[13323233]$. So we can check that $F$ has a generalized patch $L_{i}$, $i\in\{66,67,68,69\}$, or has a generalized patch $L_{i}$, $i\in\{48,49,50,51,61,62\}$.

Subcase 3.3. $F$ has a subgraph $L_{b3}$.

By the above discussions, we suppose that any two edges in $S$ can not be as $\{e_{1},e_{2}\}$ in $L_{b1}-V(e_{3})$ or $L_{b2}-V(e_{3})$. Let $F_{1}^{\prime}:=L_{b3}$ and $F_{1}^{\prime\prime}:=F-F_{1}^{\prime}$.

If $F_{1}^{\prime\prime}$ has not $1$-degree vertices, then by Lemma 3.6 (ii) $F_{1}^{\prime\prime}$ has no pendent pentagons since the distance-array of $L_{b3}-V(e_{3})$ is $[12314124]$. By Theorem 3.4, $F_{1}^{\prime\prime}$ consists of two patches $U_{1}^{3},V_{1}^{3}\in\mathcal{D}$ connecting by only an edge $x_{1}^{3}$. By Lemma 3.6 (i), the structures of $U_{1}^{3}$ and $V_{1}^{3}$ are as depicted in Fig. 22 $(a),(b)$. So the distance-array of $U_{1}^{3}$ is $[2\wedge a_{2}a_{3}22a_{6}a_{7}]$ and of $V_{1}^{3}$ is $[2\wedge a_{2}^{\prime}a_{3}^{\prime}2a_{5}^{\prime}2a_{7}^{\prime}]$, where $a_{2},a_{7}=2$ or $3$, $a_{3}=3$ or $4$, $a_{6}=4$ or $5$, and $a_{2}^{\prime}=2$ or $3$, $a_{3}^{\prime},a_{5}^{\prime}=3$ or $4$, $a_{7}^{\prime}=4$ or $5$. We can check that the distance-array $[2\wedge 342243]$ of $D_{19}$ satisfies the condition of $U_{1}^{3}$, and the distance-array $[2\wedge 242424]$ of $D_{29}$ satisfies the condition of $V_{1}^{3}$. So $U_{1}^{3}\cong D_{19}$ and $V_{1}^{3}\cong D_{29}$. Clearly, $F[V(U_{1}^{3})\cup\{z\}]\cong P^{3}$ (see Fig. 3) and $\triangledown_{F}(F[V(U_{1}^{3})\cup\{z\}])$ is a non-degenerate cyclic $6$-edge-cut of $F$. By Lemma 3.1, $F$ is a nanotube fullerene of type $(4,2)$, a contradiction.

If $F_{1}^{\prime\prime}$ has a $1$-degree vertex, then this vertex connects the two components of $F_{1}^{\prime}$ or not. For the first case, $F$ has a generalized patch $L_{i}$, $i\in\{70,\ldots,78\}$ or a generalized patch $L_{33}$ or $L_{35}$. For the second case, $F$ has a subgraph $L_{h1}$ or $L_{h2}$ (see Fig. 21), or has a generalized patch $L_{j}$, $j\in\{79,80,81,82\}$, or a generalized patch $L_{r}$, $r=33,34,$ or $35$.

If $F$ has a subgraph $L_{h1}$, then let $F_{2}^{\prime}:=L_{h1}$ and $F_{2}^{\prime\prime}:=F-F_{2}^{\prime}$. We note that $F_{2}^{\prime\prime}$ has a unique perfect matching. If $F_{2}^{\prime\prime}$ has not $1$-degree vertices, then $F_{2}^{\prime\prime}$ has a pendent pentagon or not since the distance-array of $L_{h1}-V(e_{3})$ is $[13241233]$. For the first case, as the discussion of $L_{c1}$ (see Fig. 18), we can show that $F$ has a generalized patch $L_{83}$. For the second case, $F_{2}^{\prime\prime}$ consists of two patches $U_{2}^{3},V_{2}^{3}\in\mathcal{D}$ connecting by only an edge $x_{2}^{3}$ by Theorem 3.4. By Lemma 3.6 (i), the structures of $U_{2}^{3}$ and $V_{2}^{3}$ are as depicted in Fig. 22 $(c)(d)$, or as Fig. 22 $(e)(f)$. If the structures of $U_{2}^{3}$ and $V_{2}^{3}$ are as depicted in Fig. 22 $(c)(d)$, the distance-array of $U_{2}^{3}$ is $[a_{1}\wedge a_{2}a_{3}2a_{5}a_{6}a_{7}]$ and of $V_{2}^{3}$ is $[a_{1}^{\prime}\wedge a_{2}^{\prime}a_{3}^{\prime}2a_{5}^{\prime}a_{6}^{\prime}2]$, where $a_{1},a_{2},a_{5},a_{6}=2$ or $3$, $a_{3},a_{7}=3$ or $4$, and $a_{1}^{\prime},a_{2}^{\prime},a_{5}^{\prime}=2$ or $3$, $a_{3}^{\prime},a_{6}^{\prime}=3$ or $4$. We can check that the distance-arrays $[2\wedge 232223]$ of $D_{05}$, $[2\wedge 232323]$ of $D_{08}$, $[2\wedge 242233]$ of $D_{09}$, $[2\wedge 332233]$ of $D_{12}$, $[2\wedge 242333]$ of $D_{18}$ satisfy the condition of $U_{2}^{3}$,

and the distance-arrays $[2\wedge 232232]$ of $D_{05}$, $[3\wedge 232232]$ of $D_{08}$, $[3\wedge 242242]$ of $D_{13}$ satisfy the condition of $V_{2}^{3}$. Then $F\cong W_{i}$, $i\in\{30,\ldots,44\}$ (see Fig. 24). If the structures of $U_{2}^{3}$ and $V_{2}^{3}$ are as depicted in Fig. 22 $(e)(f)$, we can similarly show that $U_{2}^{3}=D_{08}$ or $D_{18}$, and $V_{2}^{3}=D_{05},D_{08},D_{09},D_{12},D_{13}$ or $D_{19}$. So $F\cong W_{i}$, $i\in\{45,\ldots,56\}$ (see Fig. 24, 27).

If $F_{2}^{\prime\prime}$ has a $1$-degree vertex, then this vertex connects the two components of $F_{2}^{\prime}$ or not. For the first case, $F$ has a generalized patch $L\in\{L_{84},L_{85},L_{86},L_{87}\}$ or a generalized patch $L_{i}$, $i\in\{33,35,72,76,77,78\}$. For the second case, $F$ has a subgraph $L_{k1}$ (see Fig. 23), or a generalized patch $L_{j}$, $j\in\{33,34,35\}$.

If $F$ has a subgraph $L_{k1}$, then let $F_{3}^{\prime}:=L_{k1}$ and $F_{3}^{\prime\prime}:=F-F_{3}^{\prime}$. If $F_{3}^{\prime\prime}$ has not $1$-degree vertices, then we claim that it has a pendent pentagon by Theorem 3.4 and Lemma 3.6 $(i)$ since the distance-array of $L_{k1}-V(e_{3})$ is $[13323143]$. By Lemma 3.6 (ii), $F$ has a generalized patch $L_{83}$, or has a generalized patch $L_{88}$ or $L_{89}$ (see Fig. 25). If $F_{3}^{\prime\prime}$ has a $1$-degree vertex, then this vertex connects the two components of $F_{3}^{\prime}$ or not. For the first case, $F$ has a generalized patch $L_{i}$, $i\in\{70,71,72,73,77,78,84,85,86,87\}$. For the second case, $F$ has a generalized patch $L_{81}$ or $L_{82}$, or has a subgraph $L_{k2}$ or $L_{k3}$ (see Fig. 25). If $F$ has a subgraph $L_{k3}$, then it has a generalized patch $L_{j}$, $j\in\{72,78,83,86,87\}$. If $F$ has a subgraph $L_{k2}$, then $F-L_{k2}$ has a $1$-degree vertex that is adjacent to two vertices in $L_{k2}-V(e_{3})$. So $F$ has a subgraph $L_{r1}$, or a generalized patch $L_{90}$, $L_{91}$. For the subgraph $L_{r1}$, if $F_{4}^{\prime\prime}:=F-L_{r1}$ has a $1$-degree vertex, then this vertex connects the two components of $L_{r1}$. So $F$ has a generalized patch $L_{i}$, $i\in\{72,78,87,92,93,94\}$. Next, we suppose that $F_{4}^{\prime\prime}$ has no $1$-degree vertices. Then $F_{4}^{\prime\prime}$ has a pendent pentagon or not. If $F_{4}^{\prime\prime}$ has a pendent pentagon, then we can check that $F$ has a generalized patch $L_{95}$. If $F_{4}^{\prime\prime}$ has no pendent pentagons, then by Theorem 3.4 and Lemma 3.6 (i), $F_{4}^{\prime\prime}$ consists of two patches $U_{3}^{3}$, $V_{3}^{3}\in\mathcal{D}$, and the structures of $U_{3}^{3}$ and $V_{3}^{3}$ are as depicted in Fig. 26 $(a)$. We can check that the distance-arrays $[2\wedge 232223]$ and $[2\wedge 232232]$ of $D_{05}$, $[2\wedge 232323]$ of $D_{08}$, $[2\wedge 242233]$ of $D_{09}$, $[2\wedge 332233]$ of $D_{12}$, $[2\wedge 242333]$ of $D_{18}$ satisfy the condition of $U_{3}^{3}$ (resp. $V_{3}^{3}$). So $F\cong W_{i}$, $i\in\{57,\ldots,76\}$ (see Fig. 27).

Now, we suppose that $F$ has a subgraph $L_{h2}$ (see Fig. 21). If $F-L_{h2}$ has no $1$-degree vertices, then $F-L_{h2}$ has no pendent pentagons by Lemma 3.6 (ii) since the distance-array of $L_{h2}-V(e_{3})$ is $[14132314]$. By Theorem 3.4, $F-L_{h2}$ consists of two patches $U_{4}^{3},V_{4}^{3}\in\mathcal{D}$ connecting by only one edge. Since the distance-array of $L_{h2}-V(e_{3})$ is $[14132314]$, by Lemma 3.6 (i), the structures of $U_{4}^{3}$ and $V_{4}^{3}$

are as depicted in Fig. 26 (b) or as depicted in Fig. 26 (c) (d). If the structures of $U_{4}^{3}$ and $V_{4}^{3}$ are both as depicted in Fig. 26 $(b)$, then we can check that only the distance-array $[2\wedge 342243]$ of $D_{19}$ satisfies the condition of $U_{4}^{3}$ (resp. $V_{4}^{3}$). But then we will obtain a $7$-cycle, a contradiction. If the structures of $U_{4}^{3}$ and $V_{4}^{3}$ are as depicted in Fig. 26 $(c),(d)$, then we can check that only the distance-array $[3242242]$ of $D_{13}$ satisfies the condition of $V_{4}^{3}$. It implies that the distance-array of $U_{4}^{3}$ is $[2\wedge a_{2}a_{3}2a_{5}a_{6}a_{7}]$, where $a_{2},a_{5},a_{7}=2$ or $3$, $a_{3},a_{6}=3$ or $4$. We can check that the distance-arrays $[2\wedge 232232]$ of $D_{05}$, $[2\wedge 242233]$ of $D_{09}$, $[2\wedge 332233]$ of $D_{12}$, $[2\wedge 242333]$ of $D_{18}$, $[2\wedge 342243]$ of $D_{19}$ satisfy the condition of $U_{4}^{3}$. So $U_{4}^{3}=D_{05},D_{09},D_{12},D_{18}$ or $D_{19}$. Hence $F\cong W_{i}$, $i\in\{77,\ldots,81\}$. If $F-L_{h2}$ has a $1$-degree vertex, then this vertex connects the two components of $L_{h2}$ or not. For the first case, we can show that $F$ has a generalized patch $L_{j}$, $j\in\{70,71,72,73,74,78\}$. For the second case, we can show that $F$ has a subgraph $L_{k1}$ (see Fig. 23), or has a generalized patch $L_{i}$, $i\in\{79,80,81,82\}$. Note that we have considered $L_{k1}$ in the above paragraph.

Hence $F\cong W_{i}$, $i\in\{1,\ldots,81\}$ (see Fig. 24, 27), or has a generalized patch $L\in\mathcal{L}$. It is easy to check that $S\subset E(L)$ forces a perfect matching of $L$ in the sense of $F$. ∎

5 Constructing all fullerenes with $f(F)=3$

Up to now, we know that the nanotube fullerenes of type $(4,2)$ and fullerenes $W_{1},\ldots,W_{81}$ (see Fig. 24, 27) have the minimum forcing number three. In the sequel, we want to know all the other fullerene graphs with the minimum forcing number three. Starting from a generalized patch $L\in\mathcal{L}$, our idea is to expand $L$ to a larger generalized patch $L^{\prime}$ such that $S=\{e_{1},e_{2},e_{3}\}$ forces a perfect matching of $L^{\prime}$ in the sense of $F$. By a serious of such expansions, we will finally obtain a fullerene with $S$ being a forcing set.

Let $L$ be a generalized patch of fullerene $F$ and $[a_{1}\cdots a_{k}]$ a distance-array of $L$. As the operations defined in Ref. [17], we define the following operations on $L$:

$(O_{1})$ If $k\geq 4$, $a_{i}\in\{4,5\}$ and $a_{i-1},a_{i+1}\in\{1,2,3,4\}$, then let $t_{i}$ and $t_{i+1}$ be incident with a new vertex $u$. Add another new vertex $v$, an edge $uv$, and attach two half edges to $v$. A distance-array for the resulting generalized patch $L^{\prime}$ would be $[a_{1},\ldots,a_{i-2},a_{i-1}+2,1,a_{i+1}+2,a_{i+2},\ldots,a_{k}]$.

$(O_{2})$ If $k\geq 4$, $a_{i}\in\{5,6\}$ and $a_{i-1}+a_{i+1}\leq 6$, then let the half edges $t_{i}$ and $t_{i+1}$ merge into one edge. A distance-array for the resulting generalized patch $L^{\prime}$ would be $[a_{1},\ldots,a_{i-2},a_{i-1}+a_{i+1},a_{i+2},\ldots,a_{k}]$.

$(O_{3})$ If $k=2$ and $a_{1},a_{2}\in\{5,6\}$, then we merge the half edges $t_{1}$ and $t_{2}$ into one edge.

The distance-array of the resulting graph $L^{\prime}$ is the empty distance-array $[]$.

$(O_{4})$ If $k=8$, $a_{i},a_{i+4}\in\{1,2\}$ for some $i$, and all other $a_{j}$ are $3$ or $4$, then we connect each half-edge of $L$ to a $2$-degree vertex in $J_{2}$ (see Fig. 3) in the only admissible way. The distance-array of the resulting graph $L^{\prime}$ is the empty distance-array $[]$.

$(O_{5})$ If $k=10$, $a_{i}\in\{1,2,3,4\}$ for $i\in\{1,2,\ldots,10\}$, then we check that whether the ten half edges of $L$ and the ten half edges of some generalized patch $PP$ or $T_{r}$, $r\in\{1,2,\ldots,16\}$ (see Fig. 28) can merge such that the resulting ten new faces are pentagons or hexagons, that is, we check that whether the sum $[a_{i_{1}}+b_{1},\ldots,a_{i_{10}+b_{10}}]$ of some distance-array $[a_{i_{1}},\ldots,a_{i_{10}}]$ of $L$ and the min-distance-array $[b_{1},\ldots,b_{10}]$ of $PP$ or $T_{r}$, $r\in\{1,2,\ldots,16\}$ has $a_{i_{s}}+b_{s}=5$ or $6$ for each $s=1,\ldots,10$. We record all efficient cases and merge the ten edges according to the various efficient cases, respectively. Here, we may obtain different fullerenes. However, the distance-array corresponding to the resulting graph $L^{\prime}$ is the empty distance-array $[]$.

$(O_{6})$ If $k=12$, $a_{i}\in\{1,2,3,4\}$ for $i\in\{1,2,\ldots,12\}$, and $a_{j},a_{j+1},a_{j+2}\in\{3,4\}$, $a_{j-1},a_{j+3}\in\{1,2,3\}$, then let the four half edges $t_{j},t_{j+1},t_{j+2},t_{j+3}$ being incident with the four vertices $v_{1},v_{2},v_{3},$ $v_{4}$ of a pentagon $P:=v_{1}\cdots v_{5}$. Add another new vertex $v$, a new edge $v_{5}v$, and attach two half edges to $v$. A distance-array for the resulting generalized patch $L^{\prime}$ would be $[a_{1},\ldots,a_{j-2},a_{j-1}+3,1,a_{j+3}+3,a_{j+4},\ldots,a_{k}]$.

$(O_{7})$ If $k=12$, $a_{i}\in\{1,2,3,4\}$ for $i\in\{1,2,\ldots,12\}$, then we check that whether the twelve half edges of $L$ and the twelve half edges of some terminal generalized patch $T_{r}$, $r\in\{17,18,\ldots,180\}$ (see Fig. 29, 30 and 31) can merge such that the resulting twelve new faces are pentagons or hexagons, that is,

we check that whether the sum $[a_{i_{1}}+b_{1},\ldots,a_{i_{12}+b_{12}}]$ of some distance-array $[a_{i_{1}},\ldots,a_{i_{12}}]$ of $L$ and the min-distance-array $[b_{1},\ldots,b_{12}]$ of $T_{r}$, $r\in\{17,\ldots,180\}$ has $a_{i_{s}}+b_{s}=5$ or $6$ for each $s=1,\ldots,12$. We record all efficient cases and merge the twelve edges according to the various efficient cases, respectively. Here, we may obtain different fullerenes. However, the distance-array of the resulting graph $L^{\prime}$ is the empty distance-array $[]$.

We note that operation $O_{1}$ preserves the length of the distance-array. Operations $O_{2}$ and $O_{6}$ reduce the length of the distance-array by $2$. Clearly, the application of the operations $O_{3},O_{4},O_{5}$ and $O_{7}$ produces a fullerene graph, and no further operation can be performed. So we call such operations the terminal operations. Set $F_{t}^{\prime}$ be a subgraph obtained from some $L\in\mathcal{L}$ by performing $t\geq 0$ times operation $O_{1}$. Clearly, $F_{t}^{\prime}$ is connected and $|\triangledown_{F}(F_{t}^{\prime})|\leq 12$. Let $F_{t}^{\prime\prime}:=F-F_{t}^{\prime}$.

Lemma 5.1**.**

Suppose that $F_{t}^{\prime\prime}$ has a $1$-degree vertex and any $1$-degree vertex is not incident with two consecutive edges in $\triangledown_{F}(F_{t}^{\prime})$ along the boundary of $F_{t}^{\prime}$, then $|\triangledown_{F}(F_{t}^{\prime})|=12$. Moreover, $F_{t}^{\prime\prime}$ is two copies of $J_{1}$, and $\triangledown_{F}(F_{t}^{\prime})$ has four consecutive edges along the boundary of $F_{t}^{\prime}$ such that they are incident with four vertices of a pendent pentagon, respectively.

Proof.

We note that $F_{t}^{\prime}$ is connected, $|\triangledown_{F}(F_{t}^{\prime})|$ is an even number and $|\triangledown_{F}(F_{t}^{\prime})|\leq 12$. Since $F_{t}^{\prime\prime}$ has not a $1$-degree vertex such that it is incident with two consecutive edges in $\triangledown_{F}(F_{t}^{\prime})$ along the boundary of $F_{t}^{\prime}$, $F_{t}^{\prime\prime}$ is not connected and has not an isolated edge. By Lemma 2.1, $F_{t}^{\prime\prime}$ has exactly two connected components (say $B_{1}$ and $B_{2}$) and $|\triangledown_{F}(B_{i})|=6$, $i=1,2$ since $F_{t}^{\prime\prime}$ has a perfect matching and $|\triangledown_{F}(F_{t}^{\prime})|\leq 12$. Hence $|\triangledown_{F}(F_{t}^{\prime})|=12$. We claim that $B_{i}$ contains a cycle, $i=1,2$. If not, $B_{i}$ is a path with $4$ vertices, then $F_{t}^{\prime\prime}$ has a $1$-degree vertex that is incident with two consecutive edges in $\triangledown_{F}(F_{t}^{\prime})$ along the boundary of $F_{t}^{\prime}$, a contradiction. So $\triangledown_{F}(B_{i})$ is a cyclic $6$-edge-cut. Since $B_{i}$ has a unique perfect matching, by Proposition 2.8, $B_{i}\cong J_{1}$, $i=1,2$. This lemma holds since $F$ is a plane graph. ∎

We denote the min-distance-array of each $L_{i}\in\mathcal{L}$ by $L_{i}$, $i=1,\ldots,95$. Then we have

$L_{1}=[12223125],L_{2}=[12232134],L_{3}=[12241224],L_{4}=[12331233],\ldots$, $L_{95}=[12433335]$, Starting from these $95$ initial min-distance-arrays, we describe the following procedure to generate a directed graph $D$ which is called the distance-array digraph.

Algorithm 5.2**.**

(**Generating the Distance-Array Digraph $D$)**

$(S1)$

Set $V=\{L_{1},L_{2},\ldots,L_{95}\}$ and $A=\emptyset$.

$(S2)$

Select $L\in V$ on which the operations in $(S3)$ have not been made.

$(S3)$

Implying all possible operations from $O_{1}$ to $O_{7}$ for the distance-array $L$, we obtain a set $R^{\prime}$ of some distance-arrays. Replacing each distance-array in $R^{\prime}$ with its min-distance-array, we obtain a set $R$. Set $V:=V\cup R$ and $A:=A\cup\{(L,L^{\prime})|L^{\prime}\in R\}$, where $(L,L^{\prime})$ is an arc of $D$. Particularly, if $[]\in R$ and there are $t\geq 1$ different ways to obtain $[]$ from $L$ (note that we may have many generalized patches $T_{i}$ such that the edges in $\triangledown_{F}(L)$ and $\triangledown_{F}(T_{i})$ can merge suitably, and for a generalized patch $T$, the edges in $\triangledown_{F}(L)$ and $\triangledown_{F}(T)$ may have different ways to merge suitably), then we use $t$ multiple arcs $(L,[])$ to represent those various ways to merge.

$(S4)$

If all min-distance-arrays in $V$ have been selected to make operations in $(S3)$, then go to $(S5)$. Otherwise, go to $(S2)$.

$(S5)$

If $[]\notin V$, then $D:=\emptyset$. Otherwise, for every $L\in V$, delete $L$ if there is not directed path from $L$ to the empty distance-array $[]$. Then we obtain the final directed graph $D$ with vertex-set $V$ and arc-set $A$.

Using MATLAB we obtain that $D$ has $7802$ vertices and $28379$ arcs, and $\mathcal{L}^{*}\subset V(D)$, where $\mathcal{L}^{*}:=\mathcal{L}\setminus\{L_{1},L_{2},L_{4},L_{14},L_{44},L_{45},L_{70},L_{75}\}$. In the sequel, we call $L_{i}\in\mathcal{L}^{*}$ the valid initial generalized patch of $F$. Starting from each valid initial generalized patch we can construct a fullerene with a forcing set of size $3$ by implying the seven operations $O_{1}$ to $O_{7}$. We note that $D$ has exactly four distance-arrays $[135151351515]$, $[1335135135]$, $[1333333335]$ and $[13333335]$, each of which is incident with a loop.

Since $D$ is too big, we take a subgraph of $D$ to demonstrate how to generate $D$. In the first step $(S1)$, we let $V=\{L_{40}\}$. By Algorithm 5.2, we finally obtain the directed graph $D^{\prime}$ (see Fig. 32). Let $F^{\prime}$ be a generalized patch of $F$. If $F^{\prime}$ has distance-array $[133233133233]$, then we can check that the twelve edges in $\triangledown_{F}(F^{\prime})$ only can merge suitably with the twelve edges in $\triangledown_{F}(T_{i})$, $i\in\{22,27,60,67,75,78,92,97,168,170,176\}$, and $\triangledown_{F}(F^{\prime})$ and $\triangledown_{F}(T_{i})$ has only one way to merge suitably. So the dark edge $b$ in Fig. 32 depicts eleven arcs from $[133233133233]$ to $[]$. Similarly, if $F^{\prime}$ has distance-array $[1332331334]$, then the ten edges in $\triangledown_{F}(F^{\prime})$ only can merge suitably with the ten edges in $\triangledown_{F}(T_{i})$, $i\in\{2,5,7,13,15\}$, and for each such $T_{i}$, $\triangledown_{F}(T_{i})$ and $\triangledown_{F}(F^{\prime})$ have two ways to merge suitably. So the dark edge $c$ in Fig. 32 depicts ten arcs from $[1332331334]$ to $[]$. For $F^{\prime}$ with distance-array $[1333233333]$, the ten edges in $\triangledown_{F}(F^{\prime})$ only can merge suitably with the ten edges in $\triangledown_{F}(T_{i})$, $i\in\{1,2,3,4,5,8,9,13,15\}$. Checking each such $T_{i}$, we know that the dark edge $d$ in Fig. 32 depicts $13$ arcs from $[1333233333]$ to $[]$.

For example, along the directed walk $[123151323135]\rightarrow[1323142315]\rightarrow[1323314315]$

$\rightarrow[1323331515]\rightarrow[13332325]\rightarrow[13333234]\rightarrow[13333325]\rightarrow[13333334]\rightarrow[13333335]\rightarrow\cdots\rightarrow[13333335]\rightarrow[333334]\rightarrow[153335]\rightarrow[135335]\rightarrow[3435]\rightarrow[1555]\rightarrow[56]\rightarrow[]$ in $D^{\prime}$ (there are $7k$ (integer $k\geq 1$) times $O_{1}$ operations from $[13333335]$ to $[13333335]$), we can construct a nanotube fullerene with $k$-layers hexagons of type $(6,1)$ (see Fig. 33 (a)) and capped on the two ends by the two caps $A$ and $B$ (see Fig. 33 (b), (c)), respectively. We can check that $\{e_{1},e_{2},e_{3}\}$ (see Fig. 33 (b)) is a forcing set of this nanotube fullerene.

We note that although $F_{24}$ (see Fig. 1) has the minimum forcing number $2$, it also has a minimal forcing set of size $3$, where minimal means that its any proper subset is not a forcing set any more. So $F_{24}$ can also be generated from a forcing set of size $3$.

In the following, we will prove that $D$ represents all possible ways to obtain fullerene graphs from the initial generalized patches in $\mathcal{L}$ by implying the seven operations $O_{1}$ to $O_{7}$ (in any order).

Theorem 5.3**.**

*A fullerene graph has the minimum forcing number $3$ if and only if it is isomorphic to some $W_{i}$, $i\in\{1,\ldots,81\}$ *(see Fig. 24, 27), or its order is not $24$ and can be constructed from some valid initial generalized patches by implementing operations $O_{1}$ to $O_{7}$ along a directed walk in $D$ from its min-distance-array to $[]$.

Proof.

By Lemma 3.2, a nanotube fullerene of type $(4,2)$ has a forcing set $S$ of size $3$ such that $F[V(S)]\cong L_{3}$ (see Fig. 12). We note that $L_{3}\in D$. Further, we can check that a nanotube fullerene of type $(4,2)$ can be constructed from $L_{3}$ by applying several times operations $O_{1},O_{2}$ and $O_{3}$.

Next, we suppose that $F$ is a fullerene with the minimum forcing number $3$ and is not a nanotube fullerene of type $(4,2)$. Then $F$ has a forcing set $S:=\{e_{1},e_{2},e_{3}\}$. By Theorem 4.2 and the above discussions, $F$ is isomorphic to some $W_{i}$, $i\in\{1,\ldots,81\}$ or $F$ has a generalized patch $L\in\mathcal{L}$ such that $S\subseteq E(L)$ forces a perfect matching of $L$ in the sense of $F$. Since $S$ is a forcing set of $F$, $F-L$ has a unique perfect matching. Clearly, $|\triangledown_{F}(L)|\leq 12$. If $F-L=\emptyset$, then $L$ is not an induced subgraph of $F$, that is, there exits $u,v$ on the boundary of $L$ such that $uv\in E(F)\setminus E(L)$. We can show that operation $O_{2}$ or $O_{3}$ can be done. Next, we suppose that $F-L\neq\emptyset$. It implies that $|\triangledown_{F}(L)|\geq 4$.

Case 1. $L$ is not an induced subgraph of $F$.

Then there exits two vertices on the boundary of $L$ such that they are adjacent in $F$ and are not adjacent in $L$. So $|\triangledown_{F}(L)|\leq 10$. If there are two such vertices $u,v\in V(L)$ on the boundary of $L$ such that the half edge incident with $u$ and the half edge incident with $v$ are consecutive along the boundary of $L$, then operation $O_{2}$ can be done. Otherwise, $F-L$ is not connected. Since $F-L$ has a perfect matching and $|\triangledown_{F}(L)|\leq 10$, $F-L$ has exactly two components, denoted by $H_{1}$ and $H_{2}$. Clearly, $|\triangledown_{F}(H_{i})|$ is even and $|\triangledown_{F}(H_{i})|\geq 4$. Since $|\triangledown_{F}(H_{1})|+|\triangledown_{F}(H_{2})|=|\triangledown_{F}(L)|\leq 10$, there is some $i\in\{1,2\}$ with $|\triangledown_{F}(H_{i})|=4$. Without loss of generality, we suppose that $|\triangledown_{F}(H_{1})|=4$. Hence $H_{1}$ is an edge by Lemma 2.1, that is, $F-L$ has a $1$-degree vertex that is incident with two consecutive edges in $\triangledown_{F}(L)$ along the boundary of $L$. So operation $O_{1}$ can be done.

Case 2. $L$ is an induced subgraph of $F$.

We consider the following two cases.

Subcase 2.1. $F-L$ has a $1$-degree vertex.

If $F-L$ has a $1$-degree vertex that is incident with two consecutive edges in $\triangledown_{F}(L)$ along the boundary of $L$, then operation $O_{1}$ can be done. If $F-L$ has a $1$-degree vertex and any such vertex is not incident with two consecutive edges in $\triangledown_{F}(L)$ along the boundary of $L$, then $|\triangledown_{F}(L)|=12$ and $F-L\cong 2J_{1}$ by Lemma 5.1. So operation $O_{6}$ may be done.

Subcase 2.2. $F-L$ has no $1$-degree vertices.

By Lemma 2.1 and Proposition 2.8, $F-L$ is connected. Moreover, by Theorem 3.4 and Corollary 3.5, $F-L$ is isomorphic to $J_{2}$ if $|\triangledown_{F}(L)|=8$, and is isomorphic to $PP$ (see Fig. 3) or some $T_{i}$, $i\in\{1,\ldots,16\}$ (see Fig. 28) if $|\triangledown_{F}(L)|=10$, and is isomorphic to some $T_{i}$, $i\in\{17,\ldots,180\}$ (see Fig. 29, 30 and 31) if $F-L$ has not pendent pentagons and $|\triangledown_{F}(L)|=12$. So operations $O_{4},O_{5},O_{7}$ can be done. If $F-L$ has a pendent pentagons and $|\triangledown_{F}(L)|=12$, then operation $O_{6}$ can be done.

Let $L^{t}$ ($t\geq 1$) be the subgraph of $F$ obtained from $L$ by implementing $t$ times operations $O_{1}$, $O_{2}$, $\ldots$, or $O_{7}$. If the distance-array of $L^{t}$ is $[]$, then we are done. Next, we suppose that the distance-array of $L^{t}$ is not $[]$. It means that $L^{t}$ is obtained from $L$ by implementing $t$ times operations $O_{1},O_{2}$, or $O_{6}$. By the definition of the operations $O_{1},O_{2}$, and $O_{6}$, $F-L^{t}$ has a unique perfect matching and $|\triangledown_{F}(L^{t})|\leq 12$. We note that if operation $O_{6}$ is implemented at least once in the process of obtaining $L^{t}$ from $L$, then $|\triangledown_{F}(L^{t})|\leq|\triangledown_{F}(L)|-2$, and the conclusions of Theorem 3.4 and Corollary 3.5 also hold for $|\triangledown_{F}(L^{t})|=8$ or $10$ with $F-L^{t}$ having no $1$-degree vertices. As the discussion of $L$, we can show that at least one of the operations $O_{1}$, $O_{2}$, $\ldots$, $O_{7}$ can be done for $L^{t}$.

Since $F$ is finite, there is an integer $t_{0}\geq 1$ such that the distance-array of $L^{t_{0}}$ is $[]$. Hence $L\in D$ and $F$ can be constructed from valid initial generalized patch $L$ by implementing operations $O_{1}$ to $O_{7}$ along a directed walk in $D$ from $L$ to $[]$.

Conversely, each fullerene $W_{i}$, $i\in\{1,\ldots,81\}$ has the minimum forcing number three. It is sufficient to consider a fullerene $F$ constructed from some valid initial generalized patch $L_{i}$ by implementing operations $O_{1}$ to $O_{7}$ along a directed walk in $D$ from $L_{i}$ to $[]$. Clearly, $L_{i}$ has an edge set $S:=\{e_{1},e_{2},e_{3}\}$ such that $S$ forces a perfect matching of $L_{i}$ in the sense of $F$. By the description of the operations $O_{1}$ to $O_{7}$, $S$ is a forcing set of $F$. Since the order of $F$ is not $24$, $F\ncong F_{24}$. So $f(F)=3$. ∎

We notice that there is only one fullerene $F_{24}$ (see Fig. 1) of order $24$.

Corollary 5.4**.**

Except for fullerene $F_{24}$, each fullerene with anti-forcing number $4$ has the minimum forcing number $3$.

Proof.

Let $F$ be a fullerene with anti-forcing number $4$ and $F\ncong F_{24}$. Then the order of $F$ is not $24$. From the Theorem $4.3$ in Ref. [17], $F$ can be constructed from $L_{5},L_{26}$, or $L_{27}$ by implying operations $O_{1}$ to $O_{4}$. So $F$ has the minimum forcing number $3$ by Theorem 5.3. ∎

We note that a nanotube fullerene of type $(4,2)$ has anti-forcing number $4$ and minimum forcing number $3$. By Theorem $4.4$ in Ref [17] and Corollary 5.4, the following corollary holds.

Corollary 5.5**.**

For any even $n\geq 20$ $(n\neq 22,24)$, there is a fullerene $F$ of order $n$ such that $F$ has the minimum forcing number $3$.