The Component Connectivity of Leaf-Sort Graphs

1. Introduction

With the rapid development of technologies, interconnection networks play an important role in large multiprocessor system. An interconnection network is usually modeled as an undirected graph $G=\left(V\right(G),E(G\left)\right)$, where $V\left(G\right)$ is the vertex set and $E\left(G\right)$ is the edge set. In this graph, the vertices and edges correspond to the process and communication links respectively. In the large-scale interconnected network, the failure of vertices or edges is inevitable. Therefore, in order to have an unimpeded interconnection network, we must think about the fault tolerance of a graph. Connectivity is an important parameter to measure the fault tolerance of an interconnection network, so the research of connectivity is very important. Connectivity can be divided into many kinds: maximal connectivity, local connectivity, maximal local connectivity, generalized connectivity and so on. In this paper, we study a special class of connectivity: m-component connectivity. Next, we firstly introduce the traditional connectivity of a graph G.

Let $G=\left(V\right(G),E(G\left)\right)$ be a simple connected graph, where $V\left(G\right)$ is the vertex set and $E\left(G\right)$ is the edge set. Let $F\subseteq V\left(G\right)$, we use $G-F$ to denote the subgraph of G with vertex set $V\left(G\right)-F$ and edge set $E\left(G\right)-\left\{\right(u,v)\in E(G\left)\right|\{u,v\}\cap F\ne \varnothing \}$. Let x and y be any two distinct vertices, a path P between them is a sequence of adjacent vertices $<x,w_1,w_2,\cdots ,w_k,y>$, where $w_1,w_2,\cdots ,w_k$ are distinct ones. The vertices $w_i$, $i=1,2,\cdots ,k$ are called internal vertices of the path P. For any two vertices $\{u,v\}\subseteq V\left(G\right)$, if there exists a $uv$-path, we say G is connected. Furthermore, if $G-F$ is disconnected or has only one vertex, we called F is a vertex cut. Meanwhile, we call the biggest component in $G-F$ a big component. It’s well know that, when G is not a complete graph, the traditional connectivity $\kappa \left(G\right)$ is defined as $min\left\{\right|F\left|\right|F\subseteq V\left(G\right)$ and F is a vertex cut }. Otherwise, we say $\kappa \left(G\right)=n-1$, where n is the number of vertices in G. A graph G is k-connected if $\kappa \left(G\right)\ge k$.

As an extension of traditional connectivity, let’s look at the m-component connectivity of a graph G. By referring to the relevant literature, we can find that the notion concerning the number of components in $G-F$ was first introduced by Chartrand et al. [1] and Sampathkumar [2]. Furthermore, the definition of $c{\kappa }_m\left(G\right)$ was first proposed by Hsu et al. [3]. Let F is a vertex set of G$(i.e,F\subseteq V(G\left)\right)$, if the following conditions are satisfied, we say F is a m-component cut: $\left(1\right)$$G-F$ is disconnected; $\left(2\right)$ the number of components in $G-F$ is greater than or equal to m. The m-component connectivity $c{\kappa }_m\left(G\right)$ is defined as $min\left\{\right|F\left|\right|F\subseteq V\left(G\right)$ and F ia a m-component cut}. By the above definition, we can easily get that $c{\kappa }_2\left(G\right)=\kappa \left(G\right)$ and $c{\kappa }_{m+1}\left(G\right)\ge c{\kappa }_m\left(G\right)$. By referring to the relevant literature, we can also get there exists a certain relationship between m-component connectivity $c{\kappa }_m\left(G\right)$ and m-extra connectivity. The m-extra connectivity, denoted by ${\kappa }_m\left(G\right)$, is defined as the minimum number of vertices whose removal from G results in every component in $G-F$ has at least $(m+1)$ vertices [4]. Li et al. [5], Hao et al. [6] and Guo et al. [7] have studied the relationship between extra (edge) connectivity and component (edge) connectivity in networks. So far, the m-component (edge) connectivity of many graphs has been studied [8,9,10,11,12,13,14,15,16,17]. However, these results most are about small m. If we want to get a result about a bigger m, we must expend greater effort. Next, we give some definitions that will be used in the following sections.

For a vertex $v\in V\left(G\right)$, $N_G\left(v\right)=\left\{u\right|(u,v)\in E\left(G\right)\}$ is the set of neighbors of v. We let $de{g}_G\left(v\right)=\left|N_G\left(v\right)\right|$ be the degree of v and $\delta \left(G\right)=min\left\{de{g}_G\left(v\right)\right|v\in V\left(G\right)\}$ be the minimum degree of G. If $de{g}_G\left(v\right)=k$ for every $v\in V\left(G\right)$, then G is k-regular. A singleton of G is a vertex v with $de{g}_G\left(v\right)=0$. For a vertex set X, $N_G\left(X\right)=\left\{{\cup }_{x\in X}{N}_G\left(x\right)\right\}-X$ is the neighbor of X. The distance between any two vertices u and v, denoted by $d_G(u,v)$, is the length of the shortest path from u to v. G is bipartite if there exist two vertex subsets $V_1,V_2$ with $V_1\cap V_2=\varnothing$ such that $V\left(G\right)=V_1\cup V_2$ and for each edge $(u,v)\in E\left(G\right)$, $|\{u,v\}\cap V_1|=|\{u,v\}\cap V_2|=1$. It is well known that bipartite graphs contain no odd cycles. We use Bondy and Murty [18] for terminology and notation not defined here.

In this paper, we study the m-component connectivity of $C{F}_n$, prove that $c{\kappa }_3\left(C{F}_n\right)=3n-6$$(n$ is odd) and $c{\kappa }_3\left(C{F}_n\right)=3n-7$$(n$ is even) for $n\ge 3$; $c{\kappa }_4\left(C{F}_n\right)=\frac{9n-21}{2}$$(n$ is odd) and $c{\kappa }_4\left(C{F}_n\right)=\frac{9n-24}{2}$$(n$ is even) for $n\ge 4$; $c{\kappa }_5\left(C{F}_n\right)=6n-16$$(n$ is odd) and $c{\kappa }_5\left(C{F}_n\right)=6n-18$$(n$ is even) for $n\ge 5$.

The detailed arrangement of the paper ia as follows: Section 2 introduces the definition of $C{F}_n$ and gives some properties of $C{F}_n$. In Section 3, we discuss $c{\kappa }_3\left(C{F}_n\right)$. In Section 4, we prove the value of $c{\kappa }_4\left(C{F}_n\right)$. In Section 5, we prove some useful lemmas firstly, and then calculate the value of $c{\kappa }_5\left(C{F}_n\right)$. Section 6 concludes this paper. Next, let’s first introduce the Leaf-sort graph.

2. Preliminaries

Let $l_1$, $l_2$ be two integers with $1\le l_1\le l_2$. Set $[l_1,l_2]=\left\{l\right|l$ is an integer with $l_1\le l\le l_2\}$. In the permutation $\left(\begin{array}{cccc}1& 2& \cdots & n\\ p_1& p_2& \cdots & p_n\end{array}\right)$, $i\to p_i$. For the convenience, we denote the permutation $\left(\begin{array}{cccc}1& 2& \cdots & n\\ p_1& p_2& \cdots & p_n\end{array}\right)$ by $p_1{p}_2\cdots p_n$. In [19], every permutation can be denoted by a product of disjoint cycles. For example, $\left(\begin{array}{c}123\\ 231\end{array}\right)=\left(123\right)$. Specially, $\left(\begin{array}{c}12\cdots n\\ 12\cdots n\end{array}\right)=\left(1\right)$. The product $\sigma \tau$ of two permutations is the composition function $\tau$ followed by $\sigma$, e.g., $\left(12\right)\left(13\right)=\left(132\right)$. For terminology and notation not defined here, we follow [19]. Now we give the definition of n-dimensional leaf-sort graphs $C{F}_n$.

Let $[1,n]=\{1,2,\cdots ,n\}$, and let $S_n$ be the symmetric group on $[1,n]$ containing all permutations $p=p_1{p}_2\cdots p_n$ of $[1,n]$. It is well known that $\left\{\right(1i):i=2,3,\cdots ,n\}$ is a generating set for $S_n$. So $\left\{\right(1,i):i=2,3,\cdots ,n\}\cup \left\{\right(j,j+1):j=2,4,\cdots ,n-1\}$ (n is odd.) is also a generating set for $S_n$ and $\left\{\right(1,i):i=2,3,\cdots ,n\}\cup \left\{\right(j,j+1):j=2,4,\cdots ,n-2\}$ (n is even.) is also a generating set for $S_n$. The n-dimensional leaf-sort graph $C{F}_n$ is the graph with vertex set $V\left(C{F}_n\right)=S_n$ in which two vertices $u,v$ are adjacent if and only if $u=v(1,i),2\le i\le n$, or $u=v(j,j+1),2\le j\le n-1$ when j is even and n is odd, or $u=v(j,j+1),2\le j\le n-2$ when j and n are even. By the definition, we can get $C{F}_n$ is a $\frac{3n-3}{2}$-regular graph on $n!$ vertices for odd n, and $\frac{3n-4}{2}$-regular graph on $n!$ vertices for even n. The graphs $C{F}_2$, $C{F}_3$ and $C{F}_4$ are depicted in Figure 1.

We can partition $C{F}_n$ into n subgraphs $C{F}_n^1,C{F}_n^2,\cdots ,C{F}_n^n$ where every vertex $u=x_1{x}_2\cdots x_n$ $\in C{F}_n^i$ has a fixed integer i in the last position $x_n$ for $i\in [1,n]$. It is obvious that $C{F}_n^i$ is isomorphic to $C{F}_{n-1}$ for $i\in [1,n]$ [20]. The edges whose end vertices in different $C{F}_n^i$ are called cross-edges. Two edges are said to be independent if they are not adjacent. For any $u\in C{F}_n^i$, we denote $u^{+}=u(1,n)$, $u^{-}=u(n-1,n)$, and $N_u^{+}=\{u^{+},u^{-}\}$, which are called outgoing neighbors of u. Denote $E_{i,j}\left(C{F}_n\right)=E_{C{F}_n}(V\left(C{F}_n^i\right),V\left(C{F}_n^j\right))$ for $i,j\in [1,n]$.

Proposition 2.1.([20]) Let $C{F}_n^i$$(1\le i\le n)$ be defined as above. Then there are $2(n-2)!$ independent cross-edges between two different $C{F}_n^i$ when n is odd; there are $(n-2)!$ independent cross-edges between two different $C{F}_n^i$ when n is even.

Proposition 2.2. ([20]) Let $v\in V\left(C{F}_n^i\right)$$(1\le i\le n)$ which be defined as above. Then $v(1,n)$ and $v(n-1,n)$ belong to two different $C{F}_n^j$’s$(j\ne i)$ when n is odd; $v(1,n)$ belong to $C{F}_n^j$$(j\ne i)$ when n is even.

Proposition 2.3. For any $u,v\in V\left(C{F}_n^i\right)$, $N_u^{+}\cap N_v^{+}=\varnothing$ when n is odd; $u^{+}\ne v^{+}$ when n is even.

Proof. Let $u=u_1{u}_2\cdots u_{n-1}i$ and $v=v_1{v}_2\cdots v_{n-1}i$, where $u_j\ne v_j$ for some $j\in [1,n-1]$. Then $u^{+}=i{u}_2\cdots u_{n-1}{u}_1\ne i{v}_2\cdots v_{n-1}{v}_1=v^{+}$, $u^{-}=u_1{u}_2\cdots i{u}_{n-1}\ne v_1{v}_2\cdots i{v}_{n-1}=v^{-}$. Moreover, $u^{+}\ne v^{-}$ and $v^{+}\ne u^{-}$. Hence when n is odd, $N_u^{+}\cap N_v^{+}=\varnothing$; when n is even, $u^{+}\ne v^{+}$.

Proposition 2.4.([20]) For any integer $n\ge 2$, $C{F}_n$ is bipartite.

Proposition 2.5. For any two vertices $x,y\in V\left(C{F}_n\right)$$(n\ge 3)$, $|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|\le 3$.

Proof. If $d_{C{F}_n}(x,y)=1$ or $d_{C{F}_n}(x,y)\ge 3$, then $|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|=0$; Otherwise $(i.e.,|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|\ge 1$), there will be a 3-circle or $d_{C{F}_n}(x,y)=2$, a contradiction. So we consider $d_{C{F}_n}(x,y)=2$. Next, we proof this result by induction on n.

For $n=3$, $|N_{C{F}_3}\left(x\right)\cap N_{C{F}_3}\left(y\right)|=3$ as $C{F}_3\cong K_{3,3}$ and $d_{C{F}_3}(x,y)=2$.

For $n=4$, if $x,y\in V\left(C{F}_4^i\right)$, $|N_{C{F}_4}\left(x\right)\cap N_{C{F}_4}\left(y\right)|=3$ as $C{F}_4^i\cong C{F}_3$ and $x^{+}\ne y^{+}$. If $x\in V\left(C{F}_4^i\right)$, $y\in V\left(C{F}_4^j\right)$$(i\ne j)$, let $x=x_1{x}_2{x}_{3}i$ and $y=y_1{y}_2{y}_{3}j$, we know $x^{+}\ne y^{+}$. If $y^{+}\in C{F}_4^i$ or $x^{+}\in C{F}_4^j$, then $N_{C{F}_4}\left(x\right)\cap N_{C{F}_4}\left(y\right)\subseteq \{x^{+},y^{+}\}$. Thus $|N_{C{F}_4}\left(x\right)\cap N_{C{F}_4}\left(y\right)|\le 2$.

Now we assume $n\ge 5$ and the result holds for $C{F}_{n-1}$. If $x,y\in V\left(C{F}_n^i\right)$ for $i\in [1,n]$, then by Proposition $2.3$ and the inductive hypothesis, the result holds. So we let $x\in V\left(C{F}_n^i\right)$, $y\in V\left(C{F}_n^j\right)$$(i\ne j)$, $x=x_1{x}_2\cdots x_{n-1}i$, $y=y_1{y}_2\cdots y_{n-1}j$. Then $x^{+}=i{x}_2\cdots x_{n-1}{x}_1$, $x^{-}=x_1{x}_2\cdots i{x}_{n-1}$, $y^{+}=j{y}_2\cdots y_{n-1}{y}_1$, $y^{-}=y_1{y}_2\cdots j{y}_{n-1}$. We know $x^{+}\ne x^{-}$ and $y^{+}\ne y^{-}$. Since $i\ne j$, $x^{+}\ne y^{+}$ and $x^{-}\ne y^{-}$. As $d_{C{F}_n}(x,y)=2$ and $N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)\subseteq \{x^{+},x^{-},y^{+},y^{-}\}$, we can assume $y^{+}\in N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)$.

When n is odd. If $y^{+}=x^{-}$, then $j{y}_2{y}_3\cdots y_{n-1}{y}_1=x_1{x}_2\cdots i{x}_{n-1}$, $x_1=j,y_2=x_2,y_3=x_3,\cdots ,y_{n-2}=x_{n-2},y_{n-1}=i,y_1=x_{n-1}$. Thus $x^{+}=y_{n-1}{y}_2{y}_3\cdots y_{n-2}{y}_{1}j\in C{F}_n^j$ and $y^{-}=x_{n-1}{x}_2{x}_3\cdots x_{n-2}{x}_{1}i\in C{F}_n^i$, $y{x}^{+}\in E\left(C{F}_n^j\right)$, $x{y}^{-}\in E\left(C{F}_n^i\right)$. So $|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|=3$. If $y^{+}\in C{F}_n^i$ and adjacent to x, then $y^{+}=j{y}_2\cdots y_{n-1}i$, $N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)\subseteq \{x^{+},y^{+},x^{-}\}$. Thus $|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|\le 3$. Furthermore, if and only if $y^{-}=x^{+}$, there will be $|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|=3$, this is similar to the situation $y^{+}=x^{-}$. When $y^{-}\ne x^{+}$, $|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|\le 2$. So when n is odd, this result holds.

When n is even. Note that $x^{+}\ne y^{+}$. If $y^{+}\in C{F}_n^i$ or $x^{+}\in C{F}_n^j$, then $N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)\subseteq \{x^{+},y^{+}\}$. Thus $|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|\le 2$.

In summary, this proposition is proven.

Corollary 2.6. When $n\ge 4$ is even, if x and y belong to two different subgraphs in $C{F}_n$, then $|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|\le 2$.

Corollary 2.7. When $n\ge 3$ is odd, for any two vertices $x,y\in V\left(C{F}_n\right)$, where $x\in V\left(C{F}_n^i\right)$, $y\in V\left(C{F}_n^j\right)$$(i\ne j)$. Then $|N_{C{F}_n}\left(x\right)\cap N_{C{F}_n}\left(y\right)|=3$ if and only if $y^{+}=x^{-}$ or $y^{-}=x^{+}$.

Proposition 2.8.([20]) Let $C{F}_n$ be the leaf-sort graph. Then the connectivity $\kappa \left(C{F}_n\right)=\frac{3n-3}{2}$ when n is odd and $\kappa \left(C{F}_n\right)=\frac{3n-4}{2}$ when n is even.

Lemma 1. ([21]) Let $F\subseteq V\left(C{F}_n\right)$ with $\left|F\right|\le 3n-6$ when n is odd $(n\ge 5)$ and $\left|F\right|\le 3n-7$ when n is even $(n\ge 4)$. If $C{F}_n-F$ is disconnected, then $C{F}_n-F$ satisfies one of the following conditions:

$\left(1\right)$ $C{F}_n-F$ has two components, one of which is a singleton;

$\left(2\right)$ $C{F}_n-F$ has three components, two of which are singletons.

The conclusion of Lemma 1 is closely linked to the proof of m-component connectivity of $C{F}_n$, that is why we listed it first. Next, we will discuss the component connectivity of $C{F}_n$.

3. The 3-component connectivity of $C{F}_n$

In this section, we will discuss the 3-component connectivity of $C{F}_n$, and will prove that: when n is odd, $c{\kappa }_3\left(C{F}_n\right)=3n-6$ for $n\ge 3$; When n is even, $c{\kappa }_3\left(C{F}_n\right)=3n-7$ for $n\ge 3$. Before proving our main results, we prove some useful lemmas firstly. Let S is a subset of $V\left(G\right)$$(i.e.,S\subseteq V(G\left)\right)$, if any two vertices $x_1$ and $x_2$ in S are nonadjacent, we call S an independent set. For convenience, we can simply write the independent set as $Ind$-set.

Lemma 3.1. When n is odd, if $x_1\in V\left(C{F}_n^i\right)$, then there exists only $(n-3)$ vertices in $C{F}_n^i$, which can satisfy that $|N_{C{F}_n^i}\left(x_1\right)\cap N_{C{F}_n^i}\left(x_2\right)|=3$; When n is even, if $x_1\in V\left(C{F}_n^i\right)$, then there exists only $(n-2)$ vertices in $C{F}_n^i$ such that $|N_{C{F}_n^i}\left(x_1\right)\cap N_{C{F}_n^i}\left(x_2\right)|=3$. In addition, these vertices are all regular.

Proof. Note that $C{F}_n^i\cong C{F}_{n-1}$. When n is odd, $n-1$ is even, $\{x_1,x_2\}\subseteq V\left(C{F}_n^i\right)$. Since $|N_{C{F}_n^i}\left(x_1\right)\cap N_{C{F}_n^i}\left(x_2\right)|=3$, by Corollary 2.6, we know that $x_1,x_2$ must belong to a common subgraph in $C{F}_n^i$; Otherwise, if $x_1,x_2$ are in different subgraphs in $C{F}_n^i$, then $|N_{C{F}_n^i}\left(x_1\right)\cap N_{C{F}_n^i}\left(x_2\right)|\le 2$. So we can assume $x_1=i_1{i}_2{i}_3\cdots i_{n-2}ji$, $x_2=j_1{j}_2{j}_3\cdots j_{n-2}ji$. Now we let $x_1^{\prime }=i_1{i}_2{i}_3\cdots i_{n-2}$, $x_2^{\prime }=j_1{j}_2{j}_3\cdots j_{n-2}$, then $\{x_1^{\prime },x_2^{\prime }\}\subseteq V\left(G_1\right)$, $G_1\cong C{F}_{n-2}$, $n-2$ is odd. Since $|N_{G_1}\left(x_1^{\prime }\right)\cap N_{G_1}\left(x_2^{\prime }\right)|=3$, by the proof process of Proposition 2.5, we know $x_1^{\prime },x_2^{\prime }$ have two different situations:

Case 1. $x_1^{\prime },x_2^{\prime }$ belong to two different subgraphs in $G_1$, and ${\left(x_1^{\prime }\right)}^{+}={\left(x_2^{\prime }\right)}^{-}$ or ${\left(x_1^{\prime }\right)}^{-}={\left(x_2^{\prime }\right)}^{+}$.

In this case, we have $i_{n-2}\ne j_{n-2}$. If ${\left(x_1^{\prime }\right)}^{+}=i_{n-2}{i}_2{i}_3\cdots i_{n-3}{i}_1={\left(x_2^{\prime }\right)}^{-}=j_1{j}_2{j}_3\cdots j_{n-2}{j}_{n-3}$, then $j_1=i_{n-2},j_2=i_2,\dots ,j_{n-4}=i_{n-4},j_{n-3}=i_1,j_{n-2}=i_{n-3}$. Thus $x_2=i_{n-2}{i}_2{i}_3\cdots i_{n-4}{i}_1{i}_{n-3}$ $ji$. If ${\left(x_1^{\prime }\right)}^{-}=i_1{i}_2{i}_3\cdots i_{n-2}{i}_{n-3}={\left(x_2^{\prime }\right)}^{+}=j_{n-2}{j}_2{j}_3\cdots j_{n-3}{j}_1$, then $j_{n-2}=i_1,j_2=i_2,\dots ,j_{n-4}=i_{n-4},j_{n-3}=i_{n-2},j_1=i_{n-3}$. Thus $x_2=i_{n-3}{i}_2{i}_3\cdots i_{n-4}{i}_{n-2}{i}_{1}ji$.

Case 2. $x_1^{\prime },x_2^{\prime }$ belong to the same subgraph in $G_1$.

In this case, we have $i_{n-2}=j_{n-2}$, $x_1^{\prime }=i_1{i}_2{i}_3\cdots i_{n-2}$, $x_2^{\prime }=j_1{j}_2{j}_3\cdots i_{n-2}$. As $G_1^{i_{n-2}}\cong C{F}_{n-3}$, $n-3$ is even, $|N_{G_1^{i_{n-2}}}\left(x_1^{\prime }\right)\cap N_{G_1^{i_{n-2}}}\left(x_2^{\prime }\right)|=3$, we can get $x_1^{\prime },x_2^{\prime }$ belong to a common subgraph in $G_1^{i_{n-2}}$. Then $i_{n-3}=j_{n-3}$. Let $x_1^{\prime \prime }=i_1{i}_2{i}_3\cdots i_{n-4}$, $x_2^{\prime \prime }=j_1{j}_2{j}_3\cdots j_{n-4}$, then $\{x_1^{\prime \prime },x_2^{\prime \prime }\}\subseteq V\left(G_2\right)$, $G_2\cong C{F}_{n-4}$ and $|N_{G_2}\left(x_1^{\prime \prime }\right)\cap N_{G_2}\left(x_2^{\prime \prime }\right)|=3$. As $n-4$ is odd, there are two different situations:

Subcase 2.1. $x_1^{\prime \prime },x_2^{\prime \prime }$ belong to two different subgraphs in $G_2$, and ${\left(x_1^{\prime \prime }\right)}^{+}={\left(x_2^{\prime \prime }\right)}^{-}$ or ${\left(x_1^{\prime \prime }\right)}^{-}={\left(x_2^{\prime \prime }\right)}^{+}$.

If ${\left(x_1^{\prime \prime }\right)}^{+}={\left(x_2^{\prime \prime }\right)}^{-}$, we can get $j_1=i_{n-4},j_2=i_2,\dots ,j_{n-6}=i_{n-6},j_{n-4}=i_{n-5},j_{n-5}=i_1$. Thus $x_2=i_{n-4}{i}_2{i}_3\cdots i_{n-6}{i}_1{i}_{n-5}{i}_{n-3}{i}_{n-2}ji$. If ${\left(x_1^{\prime \prime }\right)}^{-}={\left(x_2^{\prime \prime }\right)}^{+}$, then $j_1=i_{n-5},j_2=i_2,\dots ,j_{n-4}=i_1,j_{n-5}=i_{n-4}$. Thus $x_2=i_{n-5}{i}_2{i}_3\cdots i_{n-6}{i}_{n-4}{i}_1{i}_{n-3}{i}_{n-2}ji$.

Subcase 2.2. $x_1^{\prime \prime },x_2^{\prime \prime }$ belong to a same subgraph in $G_2$.

This case is similar to case 2, this is a finite cycle process.

Finally, when n is odd, we can get $(n-3)$ vertices such that $|N_{C{F}_n^i}\left(x_1\right)\cap N_{C{F}_n^i}\left(x_2\right)|=3$, they are $i_{n-2}{i}_2{i}_3\cdots i_{n-4}{i}_1{i}_{n-3}ji$, $i_{n-3}{i}_2{i}_3\cdots i_{n-4}{i}_{n-2}{i}_{1}ji$, $i_{n-4}{i}_2{i}_3\cdots i_{n-6}{i}_1{i}_{n-5}{i}_{n-3}{i}_{n-2}ji$, $i_{n-5}{i}_2{i}_3\cdots i_{n-4}{i}_1{i}_{n-3}{i}_{n-2}ji$,…,$i_3{i}_1{i}_2{i}_4\cdots i_{n-2}ji$, $i_2{i}_3{i}_1{i}_4\cdots i_{n-2}ji$.

When n is even, $n-1$ is odd, $C{F}_n^i\cong C{F}_{n-1}$. Let $x_1=i_1{i}_2{i}_3\cdots i_{n-1}i$, $x_2=j_1{j}_2{j}_3\cdots j_{n-1}i$, $|N_{C{F}_n^i}\left(x_1\right)\cap N_{C{F}_n^i}\left(x_2\right)|=3$. Assume $x_1^{\prime }=i_1{i}_2{i}_3\cdots i_{n-1}$, $x_2^{\prime }=j_1{j}_2{j}_3\cdots j_{n-1}$, then $\{x_1^{\prime },x_2^{\prime }\}\subseteq V\left(G_1\right)$, $G_1\cong C{F}_{n-1}$, similarly, we can also divide it into two different situations:

Case 1. $x_1^{\prime },x_2^{\prime }$ belong to two different subgraphs in $G_1$, and ${\left(x_1^{\prime }\right)}^{+}={\left(x_2^{\prime }\right)}^{-}$ or ${\left(x_1^{\prime }\right)}^{-}={\left(x_2^{\prime }\right)}^{+}$.

In this case, we have $i_{n-1}\ne j_{n-1}$. If ${\left(x_1^{\prime }\right)}^{+}={\left(x_2^{\prime }\right)}^{-}$, then $j_1=i_{n-1},j_2=i_2,\dots ,j_{n-1}=i_{n-2},j_{n-2}=i_1$. Thus $x_2=i_{n-1}{i}_2{i}_3\cdots i_{n-3}{i}_1{i}_{n-2}i$. If ${\left(x_1^{\prime }\right)}^{-}={\left(x_2^{\prime }\right)}^{+}$, then $j_1=i_{n-2},j_2=i_2,\dots ,j_{n-2}=i_{n-1},j_{n-1}=i_1$. Thus $x_2=i_{n-2}{i}_2{i}_3\cdots i_{n-3}{i}_{n-1}{i}_{1}i$.

Case 2. $x_1^{\prime },x_2^{\prime }$ belong to a same subgraph in $G_1$.

In this case, $i_{n-1}=j_{n-1}$. Since $G_1^{i_{n-1}}\cong C{F}_{n-2}$ and $|N_{G_1^{i_{n-1}}}\left(x_1^{\prime }\right)\cap N_{G_1^{i_{n-1}}}\left(x_2^{\prime }\right)|=3$, $x_1^{\prime }$ and $x_2^{\prime }$ are in a same subgraph in $G_1^{i_{n-1}}$. So $i_{n-2}=j_{n-2}$. Let $x_1^{\prime \prime }=i_1{i}_2{i}_3\cdots i_{n-3}$, $x_2^{\prime \prime }=j_1{j}_2{j}_3\cdots j_{n-3}$, then $\{x_1^{\prime \prime },x_2^{\prime \prime }\}\subseteq V\left(G_2\right)$, $G_2\cong C{F}_{n-3}$, $n-3$ is odd. As $|N_{G_2}\left(x_1^{\prime \prime }\right)\cap N_{G_2}\left(x_2^{\prime \prime }\right)|=3$, we can also consider the following two situations:

Subcase 2.1. $x_1^{\prime \prime }$ and $x_2^{\prime \prime }$ belong to two different subgraphs in $G_2$, and ${\left(x_1^{\prime \prime }\right)}^{+}={\left(x_2^{\prime \prime }\right)}^{-}$ or ${\left(x_1^{\prime \prime }\right)}^{-}={\left(x_2^{\prime \prime }\right)}^{+}$.

If ${\left(x_1^{\prime \prime }\right)}^{+}={\left(x_2^{\prime \prime }\right)}^{-}$, then $j_1=i_{n-3},j_2=i_2,\dots ,j_{n-3}=i_{n-4},j_{n-4}=i_1$. Thus $x_2=i_{n-3}{i}_2{i}_3\cdots i_{n-5}$

$i_1{i}_{n-4}{i}_{n-2}{i}_{n-1}i$. If ${\left(x_1^{\prime \prime }\right)}^{-}={\left(x_2^{\prime \prime }\right)}^{+}$, then $j_1=i_{n-4},j_2=i_2,\dots ,j_{n-3}=i_1,j_{n-4}=i_{n-3}$. Thus $x_2=i_{n-4}{i}_2{i}_3\cdots i_{n-5}{i}_{n-3}{i}_1{i}_{n-2}{i}_{n-1}i$.

Subcase 2.2. $x_1^{\prime \prime }$ and $x_2^{\prime \prime }$ belong to a same subgraph in $G_2$.

This case is similar to case 2, this is also a finite cycle process.

Finally, when n is even, we can get $(n-2)$ vertices such that $|N_{C{F}_n^i}\left(x_1\right)\cap N_{C{F}_n^i}\left(x_2\right)|=3$, they are $i_{n-1}{i}_2{i}_3\cdots i_{n-3}{i}_1{i}_{n-2}i$, $i_{n-2}{i}_2{i}_3\cdots i_{n-3}{i}_{n-1}{i}_{1}i$, $i_{n-3}{i}_2{i}_3\cdots i_{n-5}{i}_1{i}_{n-4}{i}_{n-2}{i}_{n-1}i$, $i_{n-4}{i}_2{i}_3\cdots$

$i_{n-5}{i}_{n-3}{i}_1{i}_{n-2}{i}_{n-1}i$, …, $i_3{i}_1{i}_2{i}_4\cdots i_{n-2}{i}_{n-1}i$, $i_2{i}_3{i}_1{i}_4\cdots i_{n-2}{i}_{n-1}i$.

Corollary 3.2. Let $C{F}_n$ is an n-dimension leaf-sort graph, $\{x_1,x_2\}\subseteq V\left(C{F}_n^i\right)$. If $x_1=i_1{i}_2{i}_3\cdots i_{n-1}i$, $x_2=j_1{j}_2{j}_3\cdots j_{n-1}i$ and $|N_{C{F}_n^i}\left(x_1\right)\cap N_{C{F}_n^i}\left(x_2\right)|=3$, then $j_1\ne i_1$.

Lemma 3.3. For $n\ge 3$, let S is an $Ind$-set and $\left|S\right|=2$, then when n is odd, $|N_{C{F}_n}\left(S\right)|\ge 3n-6$; when n is even, $|N_{C{F}_n}\left(S\right)|\ge 3n-7$.

Proof. Let $S=\{x_1,x_2\}$, as S is an $Ind$-set, so $x_1$ and $x_2$ are nonadjacent. By Proposition 2.5 and the definition of $C{F}_n$, we know that $|N_{C{F}_n}\left(x_1\right)\cap N_{C{F}_n}\left(x_2\right)|\le 3$ and $C{F}_n$ is a $\frac{3n-3}{2}$-regular graph $(n$ is odd) or $\frac{3n-4}{2}$-regular graph $(n$ is even). So when n is odd, $|N_{C{F}_n}\left(S\right)|=|N_{C{F}_n}\left(x_1\right)|+|N_{C{F}_n}\left(x_2\right)|-|N_{C{F}_n}\left(x_1\right)\cap N_{C{F}_n}\left(x_2\right)|\ge 2\times \frac{3n-3}{2}-3=3n-6$. When n is even, $|N_{C{F}_n}\left(S\right)|=|N_{C{F}_n}\left(x_1\right)|+|N_{C{F}_n}\left(x_2\right)|-|N_{C{F}_n}\left(x_1\right)\cap N_{C{F}_n}\left(x_2\right)|\ge 2\times \frac{3n-4}{2}-3=3n-7$.

Corollary 3.4. For $n\ge 4$, let F is a subset of $V\left(C{F}_n\right)$$(i.e.,F\subseteq V\left(C{F}_n\right)$) and when n is odd, $\left|F\right|\le 3n-7$; when n is even, $\left|F\right|\le 3n-8$, then $C{F}_n-F$ contains a big component C, which satisfies $\left|V\right(C\left)\right|\ge n!-\left|F\right|-1$.

Theorem 1. For $n\ge 3$, when n is odd, $c{\kappa }_3\left(C{F}_n\right)=3n-6$; when n is even, $c{\kappa }_3\left(C{F}_n\right)=3n-7$.

Proof. For $n=3$, since $c{\kappa }_{l+1}\left(C{F}_n\right)\ge c{\kappa }_l\left(C{F}_n\right)$, we can get $c{\kappa }_3\left(C{F}_3\right)\ge \frac{3n-3}{2}=3=3n-6$ by Proposition 2.8. For $n\ge 4$, by Corollary 3.4, we can also get $c{\kappa }_3\left(C{F}_n\right)\ge 3n-6$ when n is odd and $c{\kappa }_3\left(C{F}_n\right)\ge 3n-7$ when n is even. Next, we will prove that: when n is odd, $c{\kappa }_3\left(C{F}_n\right)\le 3n-6$ and when n is even, $c{\kappa }_3\left(C{F}_n\right)\le 3n-7$. Since $|N_{C{F}_n}\left(x_1\right)\cap N_{C{F}_n}\left(x_2\right)|\le 3$, we can choose two different vertices $x_1,x_2\in V\left(C{F}_n\right)$, which can satisfy the condition $|N_{C{F}_n}\left(x_1\right)\cap N_{C{F}_n}\left(x_2\right)|=3$. From the definition of $C{F}_n$, we know that $C{F}_n$ is a $\frac{3n-3}{2}$-regular $(n$ is odd) and $\frac{3n-4}{2}$-regular graph $(n$ is even), so when n is odd, $|N_{C{F}_n}\left(\{x_1,x_2\}\right)|=2\times \frac{3n-3}{2}-3=3n-6$; when n is even, $|N_{C{F}_n}\left(\{x_1,x_2\}\right)|=2\times \frac{3n-4}{2}-3=3n-7$. Thus let $F=N_{C{F}_n}\left(\{x_1,x_2\}\right)$, we know $C{F}_n-F$ contains three components and two of them only has a singleton. So we can get $c{\kappa }_3\left(C{F}_n\right)\le 3n-6$$(n$ is odd) and $c{\kappa }_3\left(C{F}_n\right)\le 3n-7$$(n$ is even).

4. The 4-component connectivity of $C{F}_n$

Lemma 4.1. When $n=4$, let S is an $Ind$-set and $\left|S\right|=3$, $|N_{C{F}_4}\left(S\right)|\ge 6$.

Proof. Let $S=\{x_1,x_2,x_3\}$, since S is an $Ind$-set, $x_1,x_2,x_3$ are nonadjacent with each other. Note that $C{F}_4^i\cong C{F}_3$, $C{F}_3\cong K_{3,3}$. Next, we will think about the following three cases.

Case 1. $x_1,x_2,x_3$ belong to the same subgraph $C{F}_4^i$.

In this case, since $C{F}_4^i\cong K_{3,3}$ and S is an $Ind$-set, $|N_{C{F}_4^i}\left(S\right)|=3$. By Proposition 2.3, we know the outgoing neighbors of $\{x_1,x_2,x_3\}$ are different. Thus, $|N_{C{F}_4}\left(S\right)|=3+3=6$.

Case 2. $x_1,x_2,x_3$ belong to two different subgraphs $C{F}_4^i$, $C{F}_4^j$$(i\ne j)$.

In this case, we can let $\{x_1,x_2\}\subseteq V\left(C{F}_4^1\right)$, $x_3\in V\left(C{F}_4^2\right)$. Since $x_1$ and $x_2$ are nonadjacent, we can get $|N_{C{F}_4^1}\left(\{x_1,x_2\}\right)|=3$, $|N_{C{F}_4^2}\left(x_3\right)|=3$. By the definition of $C{F}_n$, we know $x_i$$(i\in \{1,2,3\left\}\right)$ only has one outgoing neighbor. If the structure shown in Figure 2 (a) exists, $|N_{C{F}_4}\left(S\right)|\ge |N_{C{F}_4^1}\left(\{x_1,x_2\}\right)|+|N_{C{F}_4^2}\left(x_3\right)|=6$. Now we can show that this structure does not exist. As $|N_{C{F}_4^1}\left(\{x_1,x_2\}\right)|=3$ and $\{x_1,x_2\}\subseteq V\left(C{F}_4^1\right)$, we assume $x_1=2341$, then $x_2=4231$ or $x_2=3421$. Thus $x_1^{+}\in V\left(C{F}_4^2\right)$, $x_2^{+}\notin V\left(C{F}_4^2\right)$, the structure shown in Figure 2 (a) does not exist. Thus $|N_{C{F}_4}\left(S\right)|\ge 7$.

Case 3. $x_1,x_2,x_3$ belong to three different subgraphs $C{F}_4^i$, $C{F}_4^j$, $C{F}_4^k$$(i,j,k$ are different from each other).

In this case, we can let $x_i\in V\left(C{F}_4^i\right)$. By the definition of $C{F}_n$, we know $|N_{C{F}_4^i}\left(x_i\right)|=3$, thus $|N_{C{F}_4}\left(S\right)|\ge 3\times 3=9$.

Combining the above three situations, we can get $|N_{C{F}_4}\left(S\right)|\ge 6$.

Lemma 4.2. When $n=5$, let S is an $Ind$-set and $\left|S\right|=3$, $|N_{C{F}_5}\left(S\right)|\ge 12$.

Proof. Let $S=\{x_1,x_2,x_3\}$, since S is an $Ind$-set, $x_1,x_2,x_3$ are nonadjacent with each other. Note that $C{F}_5^i\cong C{F}_4$$(1\le i\le 5)$. We think about the following three cases.

Case 1. $x_1,x_2,x_3$ belong to the same subgraph $C{F}_5^i$.

Since $C{F}_5^i\cong C{F}_4$, by Lemma 4.1, we can get $|N_{C{F}_5^i}\left(S\right)|\ge 6$. By the definition of $C{F}_n$ and Proposition 2.3, we know the outgoing neighbors of $\{x_1,x_2,x_3\}$ are different and every vertex has two different outgoing neighbors. Thus $|N_{C{F}_5}\left(S\right)|=|N_{C{F}_5^i}\left(S\right)|+6\ge 6+6=12$.

Case 2. $x_1,x_2,x_3$ belong to two different subgraphs $C{F}_5^i$, $C{F}_5^j$$(i\ne j)$.

In this case, we can let $\{x_1,x_2\}\subseteq V\left(C{F}_5^1\right)$, $x_3\in V\left(C{F}_5^2\right)$. Since $C{F}_5^1\cong C{F}_4$, by Lemma 3.3, we know that $|N_{C{F}_5^1}\left(\{x_1,x_2\}\right)|\ge 3(n-1)-7=3\times 5-10=5$. By Proposition 2.8, we can get $|N_{C{F}_5^2}\left(x_3\right)|=\frac{3(n-1)-4}{2}=4$. By Proposition 2.2, we can get $x_1$ and $x_2$ have at most two neighbors which belong to $C{F}_5^2$. In other words, there are at least two neighbors of $x_1$ and $x_2$ belong to $C{F}_5-C{F}_5^1-C{F}_5^2$. If the structure shown in Figure 2 (b) exists, then $|N_{C{F}_5}\left(S\right)|\ge |N_{C{F}_5^1}\left(\{x_1,x_2\}\right)|+|N_{C{F}_5^2}\left(x_3\right)|+2=5+4+2=11$. Furthermore, $|N_{C{F}_5}\left(S\right)|=11$ if and only if this structure exists. Now, we can prove this structure does not exist.

Since $\{x_1,x_2\}\subseteq V\left(C{F}_5^1\right)$, $C{F}_5^1\cong C{F}_4$ and $|N_{C{F}_5^1}\left(x_1\right)\cap N_{C{F}_5^1}\left(x_2\right)|=3$, by Corollary 2.6, we can know that $x_1,x_2$ must belong to a common subgraph in $C{F}_5^1$. So we can assume $x_1=i_1{i}_2{i}_3{i}_{4}1$, $x_2=j_1{j}_2{j}_3{i}_{4}1$. As the subgraph of $C{F}_5^1$ is isomorphic to $C{F}_3$, thus $x_2=i_3{i}_1{i}_2{i}_{4}1$ or $x_2=i_2{i}_3{i}_1{i}_{4}1$$(i_1,i_2,i_3,i_4$ are different from each other).

If $x_2=i_3{i}_1{i}_2{i}_{4}1$, then $x_2^{+}=1i_1{i}_2{i}_4{i}_3$, $x_2^{-}=i_3{i}_1{i}_{2}1i_4$. Since $x_1^{+}=1i_2{i}_3{i}_4{i}_1$, $x_1^{-}=i_1{i}_2{i}_{3}1i_4$ and one of the two outgoing neighbors of $x_1,x_2$ belong to $C{F}_5^2$, we can get $i_4=2$. Thus $\{x_1^{-},x_2^{-}\}\subseteq V\left(C{F}_5^2\right)$. As $x_1^{-},x_2^{-}$ are adjacent to $x_3$, so $x_3=i_2{i}_1{i}_{3}12$ or $x_3=i_3{i}_2{i}_{1}12$ or $x_3=i_1{i}_3{i}_{2}12$. When $x_3=i_2{i}_1{i}_{3}12$, $x_3^{+}=2i_1{i}_{3}1i_2\in V\left(C{F}_5^{i_2}\right)$, $x_3^{-}=i_2{i}_1{i}_{3}21\in V\left(C{F}_5^1\right)$, and $x_3^{-}$ is adjacent to $x_2$. Since $x_1^{+}=1i_2{i}_{3}2i_1\in V\left(C{F}_5^{i_1}\right)$, $x_2^{+}=1i_1{i}_{2}2i_3\in V\left(C{F}_5^{i_3}\right)$, $x_3^{+}\ne x_1^{+}$, $x_3^{+}\ne x_2^{+}$. When $x_3=i_3{i}_2{i}_{1}12$, $x_3^{+}=2i_2{i}_{1}1i_3\in V\left(C{F}_5^{i_3}\right)$, $x_3^{-}=i_3{i}_2{i}_{1}21\in V\left(C{F}_5^1\right)$, and $x_3^{-}$ is adjacent to $x_1$. Since $x_1^{+}=1i_2{i}_{3}2i_1$, $x_2^{+}=1i_1{i}_{2}2i_3$ and $1\ne 2$, $x_3^{+}\ne x_1^{+}$, $x_3^{+}\ne x_2^{+}$. When $x_3=i_1{i}_3{i}_{2}12$, $x_3^{+}=2i_3{i}_{2}1i_1\in V\left(C{F}_5^{i_1}\right)$, $x_3^{-}=i_1{i}_3{i}_{2}21\in V\left(C{F}_5^1\right)$, and $x_3^{-}$ is adjacent to $x_2$. Since $x_1^{+}=1i_2{i}_{3}2i_1$, $x_2^{+}=1i_1{i}_{2}2i_3$ and $1\ne 2$, $x_3^{+}\ne x_1^{+}$, $x_3^{+}\ne x_2^{+}$. So this structure does not exist.

If $x_2=i_2{i}_3{i}_1{i}_{4}1$, then $x_2^{+}=1i_3{i}_1{i}_4{i}_2$, $x_2^{-}=i_2{i}_3{i}_{1}1i_4$. Since $x_1^{+}=1i_2{i}_3{i}_4{i}_1$, $x_1^{-}=i_1{i}_2{i}_{3}1i_4$ and one of the two outgoing neighbors of $x_1,x_2$ belong to $C{F}_5^2$, we can get $i_4=2$. Thus $\{x_1^{-},x_2^{-}\}\subseteq V\left(C{F}_5^2\right)$. As $x_1^{-},x_2^{-}$ are adjacent to $x_3$, so $x_3=i_3{i}_2{i}_{1}12$ or $x_3=i_1{i}_3{i}_{2}12$ or $x_3=i_2{i}_1{i}_{3}12$. When $x_3=i_3{i}_2{i}_{1}12$, $x_3^{+}=2i_2{i}_{1}1i_3\in V\left(C{F}_5^{i_3}\right)$, $x_3^{-}=i_3{i}_2{i}_{1}21\in V\left(C{F}_5^1\right)$, and $x_3^{-}$ is adjacent to $x_2$. Since $x_1^{+}=1i_2{i}_{3}2i_1\in V\left(C{F}_5^{i_1}\right)$, $x_2^{+}=1i_3{i}_{1}2i_2\in V\left(C{F}_5^{i_2}\right)$, $x_3^{+}\ne x_1^{+}$, $x_3^{+}\ne x_2^{+}$. When $x_3=i_1{i}_3{i}_{2}12$, $x_3^{+}=2i_3{i}_{2}1i_1\in V\left(C{F}_5^{i_1}\right)$, $x_3^{-}=i_1{i}_3{i}_{2}21\in V\left(C{F}_5^1\right)$, and $x_3^{-}$ is adjacent to $x_1$. Since $x_1^{+}=1i_2{i}_{3}2i_1$, $x_2^{+}=1i_3{i}_{1}2i_2$ and $1\ne 2$, $x_3^{+}\ne x_1^{+}$, $x_3^{+}\ne x_2^{+}$. When $x_3=i_2{i}_1{i}_{3}12$, $x_3^{+}=2i_1{i}_{3}1i_2\in V\left(C{F}_5^{i_2}\right)$, $x_3^{-}=i_2{i}_1{i}_{3}21\in V\left(C{F}_5^1\right)$, and $x_3^{-}$ is adjacent to $x_2$. Since $x_1^{+}=1i_2{i}_{3}2i_1$, $x_2^{+}=1i_3{i}_{1}2i_2$ and $1\ne 2$, $x_3^{+}\ne x_1^{+}$, $x_3^{+}\ne x_2^{+}$.

Thus the structure shown in Figure 2 (b) does not exist, $|N_{C{F}_5}\left(S\right)|\ge 12$.

Case 3. $x_1,x_2,x_3$ belong to three different subgraphs $C{F}_5^i$, $C{F}_5^j$, $C{F}_5^k$$(i,j,k$ are different from each other).

Without loss of generality, we can let $x_1\in V\left(C{F}_5^1\right)$, $x_2\in V\left(C{F}_5^2\right)$, $x_3\in V\left(C{F}_5^3\right)$. By the definition of $C{F}_n$, we can get $|N_{C{F}_5^i}\left(x_i\right)|=\frac{3(n-1)-4}{2}=4$$(i\in \{1,2,3\left\}\right)$, thus $|N_{C{F}_5}\left(S\right)|\ge |N_{C{F}_5^1}\left(x_1\right)|+|N_{C{F}_5^2}\left(x_2\right)|+|N_{C{F}_5^3}\left(x_3\right)|=12$.

Combining the above three situations, we can get $|N_{C{F}_5}\left(S\right)|\ge 12$.

Lemma 4.3. When n is odd, let $S=\{x_1,x_2,x_3\}$ is an $Ind$-set, where $\{x_1,x_2\}\subseteq V\left(C{F}_n^i\right)$, $x_3\in V\left(C{F}_n^j\right)$$(i\ne j)$ and $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|=3n-10$. Then $|N_{C{F}_n}\left(S\right)|\ne \frac{9n-23}{2}$.

Proof. When n is odd, $|N_{C{F}_n}\left(S\right)|=\frac{9n-23}{2}$ occurs if and only if the structure in Figure 3 appears. Next, we will prove that these structures can not appear.

Firstly, we prove that the structure shown in Figure 3 (a) does not exist. Suppose on the contrary, we assume this structure exists and $\{x_1,x_2\}\subseteq V\left(C{F}_n^1\right)$, then $x_1=i_1{i}_2{i}_3\cdots i_{n-2}{i}_{n-1}1$, $x_2=j_1{j}_2{j}_3\cdots j_{n-2}{j}_{n-1}1$. As $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|=3n-10$, by the proof process of Lemma 3.3, we can know that $x_1,x_2$ must have three common neighbors in $C{F}_n^1$. Note that n is odd, then $n-1$ is even, and $C{F}_n^i\cong C{F}_{n-1}$. By Corollary 2.6, we can get $x_1,x_2$ must belong to a common subgraph in $C{F}_n^1$; Otherwise, if $x_1,x_2$ belong to two different subgraphs in $C{F}_n^1$, then $|N_{C{F}_n^1}\left(\{x_1,x_2\}\right)|\le 2$, a contradiction. So $j_{n-1}=i_{n-1}$, $x_1^{+}=1i_2{i}_3\cdots i_{n-2}{i}_{n-1}{i}_1$, $x_1^{-}=i_1{i}_2{i}_3\cdots i_{n-2}1i_{n-1}$, $x_2^{+}=1j_2{j}_3\cdots j_{n-2}{i}_{n-1}{j}_1$, $x_2^{-}=j_1{j}_2{j}_3\cdots j_{n-2}1i_{n-1}$. By Corollary 3.2, we know $j_1\ne i_1$. As one of the two outgoing neighbors of $x_1$ and $x_2$ belong to a common subgraph with $x_3$, so $\{x_1^{-},x_2^{-}\}\subseteq V\left(C{F}_n^{i_{n-1}}\right)$ and $x_3\in V\left(C{F}_n^{i_{n-1}}\right)$. Now we assume $x_3=k_1{k}_2{k}_3\cdots k_{n-2}{k}_{n-1}{i}_{n-1}$. As one of the two outgoing neighbors of $x_3$ belongs to $C{F}_n^1$, so $x_3=k_1{k}_2{k}_3\cdots k_{n-2}1i_{n-1}$ or $x_3=1k_2{k}_3\cdots k_{n-2}{k}_{n-1}{i}_{n-1}$. When $x_3=k_1{k}_2{k}_3\cdots k_{n-2}1i_{n-1}$, $x_3^{+}=i_{n-1}{k}_2{k}_3\cdots k_{n-2}1k_1$, $x_3^{-}=k_1{k}_2{k}_3\cdots k_{n-2}{i}_{n-1}1$. In this situation, $x_3^{-}\in V\left(C{F}_n^1\right)$, $x_3^{+}\in V\left(C{F}_n^{k_1}\right)$, since $i_{n-1}\ne 1$, we have $x_3^{+}\ne x_1^{+}$, $x_3^{+}\ne x_2^{+}$. Thus this structure does not exist. When $x_3=1k_2{k}_3\cdots k_{n-2}{k}_{n-1}{i}_{n-1}$, as $x_1^{-},x_2^{-}$ are adjacent to $x_3$, we can get $x_1^{-}=x_3(1,n-1)$, $x_2^{-}=x_3(1,n-1)$, this contradicts to the fact $x_1^{-}\ne x_2^{-}$, thus this structure does not exist.

Next we will prove that the structure in Figure 3 (b) does not exist. Similarly, we know that $j_1\ne i_1$, $j_{n-1}=i_{n-1}$, $\{x_1^{-},x_2^{-}\}\subseteq V\left(C{F}_n^{i_{n-1}}\right)$ and $x_3\in V\left(C{F}_n^{i_{n-1}}\right)$. We let $x_3=k_1{k}_2{k}_3\cdots k_{n-2}{k}_{n-1}{i}_{n-1}$, then $x_3^{+}=i_{n-1}{k}_2{k}_3\cdots k_{n-2}{k}_{n-1}{k}_1$, $x_3^{-}=k_1{k}_2{k}_3\cdots k_{n-2}{i}_{n-1}{k}_{n-1}$. Thus $x_3^{-}\ne x_1^{+}$ and $x_3^{-}\ne x_2^{+}$, the structure in Figure 3 (b) does not exist.

Lemma 4.4. When n is even, let $S=\{x_1,x_2,x_3\}$ is an $Ind$-set, where $\{x_1,x_2\}\subseteq V\left(C{F}_n^i\right)$, $x_3\in V\left(C{F}_n^j\right)$$(i\ne j)$ and $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|=3n-9$. Then $|N_{C{F}_n}\left(S\right)|\ne \frac{9n-24}{2}$.

Proof. When n is even, $|N_{C{F}_n}\left(S\right)|=\frac{9n-24}{2}$ occurs if and only if the structure in Figure 4 appears. Next, we will prove that this structure does not exist. Note that $C{F}_n^i\cong C{F}_{n-1}$, $n-1$ is odd.

Suppose on the contrary, we assume this structure exists, as $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|=3n-9$, we know $x_1,x_2$ must have three common neighbors in $C{F}_n^i$ by Lemma 3.3. Now, we let $x_1=i_1{i}_2{i}_3\cdots i_{n-2}{i}_{n-1}i$, $x_2=j_1{j}_2{j}_3\cdots j_{n-2}{j}_{n-1}i$. By Corollary 3.2, we know $j_i\in [i_1,i_{n-1}]$ and $j_1\ne i_1$. Thus $x_1^{+}\in V\left(C{F}_n^{i_1}\right)$, $x_2^{+}\in V\left(C{F}_n^{j_1}\right)$, $x_1^{+}$ and $x_2^{+}$ can not belong to a common subgraph in $C{F}_n$, this contradicts to this structure. Thus $|N_{C{F}_n}\left(S\right)|\ne \frac{9n-24}{2}$.

Lemma 4.5. When $n\ge 4$, let S is an $Ind$-set and $\left|S\right|=3$, then when n is odd, $|N_{C{F}_n}\left(S\right)|\ge \frac{9n-21}{2}$; when n is even, $|N_{C{F}_n}\left(S\right)|\ge \frac{9n-24}{2}$.

Proof. Let $S=\{x_1,x_2,x_3\}$, since S is an $Ind$-set, $x_1,x_2,x_3$ are nonadjacent with each other. We proof this result by induction on n. By Lemma $4.1$ and Lemma $4.2$, we know when $n=4,5$, this result holds. Now we assume that $n\ge 6$ and the result holds for $C{F}_{n-1}$. Note that $C{F}_n^i\cong C{F}_{n-1}$. Next, we think about the following three cases:

Case 1. $x_1,x_2,x_3$ belong to a same subgraph $C{F}_n^i$.

When n is odd, by induction hypothesis, we have $|N_{C{F}_n^i}\left(S\right)|\ge \frac{9(n-1)-24}{2}=\frac{9n-33}{2}$. By Proposition 2.3 and the definition of $C{F}_n$, we know the neighbors of $x_1,x_2,x_3$ in $C{F}_n-C{F}_n^i$ are different and every vertex has two outgoing neighbors. Thus $|N_{C{F}_n}\left(S\right)|=|N_{C{F}_n^i}\left(S\right)|+|N_{C{F}_n-C{F}_n^i}\left(S\right)|\ge \frac{9n-33}{2}+6=\frac{9n-21}{2}$.

When n is even, by induction hypothesis, we have $|N_{C{F}_n^i}\left(S\right)|\ge \frac{9(n-1)-21}{2}=\frac{9n-30}{2}$. By Proposition 2.3 and the definition of $C{F}_n$, we know the neighbors of $x_1,x_2,x_3$ in $C{F}_n-C{F}_n^i$ are different and every vertex has only one outgoing neighbor. Thus $|N_{C{F}_n}\left(S\right)|=|N_{C{F}_n^i}\left(S\right)|+|N_{C{F}_n-C{F}_n^i}\left(S\right)|\ge \frac{9n-30}{2}+3=\frac{9n-24}{2}$.

Case 2. $x_1,x_2,x_3$ belong to two different subgraphs $C{F}_n^i$, $C{F}_n^j$$(i\ne j)$.

In this case, we can let $\{x_1,x_2\}\subseteq V\left(C{F}_n^i\right)$, $x_3\in V\left(C{F}_n^j\right)$. By Lemma $3.3$, we can get: when n is odd, $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|\ge 3(n-1)-7=3n-10$; when n is even, $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|\ge 3(n-1)-6=3n-9$. By Proposition 2.8, we know when n is odd, $|N_{C{F}_n^j}\left(x_3\right)|=\frac{3(n-1)-4}{2}=\frac{3n-7}{2}$; when n is even, $|N_{C{F}_n^j}\left(x_3\right)|=\frac{3(n-1)-3}{2}=\frac{3n-6}{2}$. When n is odd, by Proposition 2.2, we know $x_1,x_2$ have at most two outgoing neighbors can belong to $C{F}_n^j$, in another word, there are at least two outgoing neighbors of $\{x_1,x_2\}$ can belong to $C{F}_n-C{F}_n^i-C{F}_n^j$. So $|N_{C{F}_n}\left(S\right)|\ge |N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_n^j}\left(x_3\right)|+2=\frac{9n-23}{2}$. By Lemma 4.3, we know $|N_{C{F}_n}\left(S\right)|\ne \frac{9n-23}{2}$, thus $|N_{C{F}_n}\left(S\right)|\ge \frac{9n-23}{2}+1=\frac{9n-21}{2}$. When n is even, if the structure of Figure 4 exist, then $|N_{C{F}_n}\left(S\right)|\ge |N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_n^j}\left(x_3\right)|=3n-9+\frac{3n-6}{2}=\frac{9n-24}{2}$. By Lemma 4.4, we know this structure does not exist, so $|N_{C{F}_n}\left(S\right)|\ge \frac{9n-24}{2}+1=\frac{9n-22}{2}$.

Case 3. $x_1,x_2,x_3$ belong to three different subgraphs $C{F}_n^i$, $C{F}_n^j$, $C{F}_n^k$$(i,j,k$ are different from each other).

Without loss of generality, we can let $x_1\in V\left(C{F}_n^1\right)$, $x_2\in V\left(C{F}_n^2\right)$, $x_3\in V\left(C{F}_n^3\right)$. By Proposition 2.8, we have when n is odd, $|N_{C{F}_n^i}\left(x_i\right)|=\frac{3n-7}{2}$$(i\in \{1,2,3\left\}\right)$; when n is even, $|N_{C{F}_n^i}\left(x_i\right)|=\frac{3n-6}{2}$$(i\in \{1,2,3\left\}\right)$. Thus when n is odd, $|N_{C{F}_n}\left(S\right)|\ge 3\times \frac{3n-7}{2}=\frac{9n-21}{2}$; when n is even, $|N_{C{F}_n}\left(S\right)|\ge 3\times \frac{3n-6}{2}=\frac{9n-18}{2}$.

Thus the result holds.

Corollary 4.6. When n is even, let $S=\{x_1,x_2,x_3\}$ is an $Ind$-set, if $|N_{C{F}_n}\left(S\right)|=\frac{9n-24}{2}$, then $x_1,x_2,x_3$ belong to a same subgraph in $C{F}_n$.

Lemma 4.7. For $n=5$, if $\left|F\right|\le 11$, then $C{F}_5-F$ contains a big component C, which satisfies the result $\left|V\right(C\left)\right|\ge n!-\left|F\right|-2$.

Proof. We are not going to think about $C{F}_5-F$ is connected for the moment, so we assume that $C{F}_5-F$ is disconnected. Let $F_i=F\cap V\left(C{F}_5^i\right)$ for $i=1,2,3,4,5$ with $\left|F_{i_1}\right|\ge \left|F_{i_2}\right|\ge \left|F_{i_3}\right|\ge \left|F_{i_4}\right|\ge \left|F_{i_5}\right|$, where $i_j\in \{1,2,3,4,5\}$. If $\left|F_{i_3}\right|=0$, then $\left|F_{i_4}\right|=\left|F_{i_5}\right|=0$ and $C{F}_5^{[i_3,i_5]}-F^{[i_3,i_5]}$ is connected. By Proposition $2.2$, we know there exists a vertex in $C{F}_5^{i_1}-F_{i_1}$$(resp.,C{F}_5^{i_2}-F_{i_2})$, which has neighbor in $C{F}_5^{[i_3,i_5]}-F^{[i_3,i_5]}$. So $C{F}_5-F$ is connected, a contradiction. Hence we consider $|F_{i_3}|\ge 1$. Since $\left|F\right|\le 11$, we have $\left|F_{i_5}\right|\le 2$, $\left|F_{i_4}\right|\le 2$, $1\le \left|F_{i_3}\right|\le 3$, $1\le \left|F_{i_2}\right|\le 5$, $1\le |F_{i_1}|\le 9$. Firstly, we proof the following Claim is correct.

Claim 1. If $|F_{i_j}|\le 3$ for some $i_j\in [i_1,i_4]$, then $C{F}_5^{[i_j,i_5]}-F^{[i_j,i_5]}$ is connected.

Proof of Claim 1. By Proposition $2.8$, we can get $C{F}_5^j-F_j$ is connected for each $j\in [i_j,i_5]$. On the other hand, as $|F_{i_5}|\le 2$, we can get $|E_{p,i_5}\left(C{F}_5\right)|=12>5\ge |F_p|+|F_{i_5}|$, which implies $E_{p,i_5}(C{F}_5-F)\ne \varnothing$ for $p\in [i_j,i_4]$. Hence $C{F}_5^{[i_j,i_5]}-F^{[i_j,i_5]}$ is connected.

Since $|F_{i_5}|\le |F_{i_4}|\le |F_{i_3}|\le 3$, by Claim 1, we can get $C{F}_5^{[i_3,i_5]}-F^{[i_3,i_5]}$ is connected. If $C{F}_5^{i_1}-F_{i_1}$ and $C{F}_5^{i_2}-F_{i_2}$ are all connected, we know $C{F}_5-F$ is connected. As $\left|E_{i_2,i_3}\left(C{F}_5\right)\right|=12>8\ge |F_{i_2}\cup F_{i_3}|$, $C{F}^{[i_2,i_5]}-F^{[i_2,i_5]}$ is connected. As $\left|E_{i_1,i_5}\left(C{F}_5\right)\right|=12>9+2=11\ge |F_{i_1}\cup F_{i_5}|$, we have $C{F}_5-F$ is connected. Since $C{F}_5-F$ is disconnected, at least one of $C{F}_5^i-F_i,i\in \{i_1,i_2\}$ is disconnected, which leads to the following two cases.

Note that if $|F_{i_3}\cup F_{i_4}\cup F_{i_5}|\le 1$, by Proposition $2.3$, we can get $C{F}_5^i-F_i$$(i\in \{i_1,i_2\})$ has a big component $C_i$ and at most two vertices, which has a neighbor in $F_{i_3}\cup F_{i_4}\cup F_{i_5}$. Thus, if $C{F}_5^{i_1}-F_{i_1}$ or $C{F}_5^{i_2}-F_{i_2}$ is connected, then $C{F}_5-F$ satisfies the condition $\left(1\right)$. If $C{F}_5^{i_1}-F_{i_1}$ and $C{F}_5^{i_2}-F_{i_2}$ are all disconnected, then $C{F}_5-F$ satisfies the condition $\left(2\right)$. Hence we only think about this situation: $|F_{i_3}\cup F_{i_4}\cup F_{i_5}|\ge 2$.

Case 1. Both $C{F}_5^{i_1}-F_{i_1}$ and $C{F}_5^{i_2}-F_{i_2}$ are disconnected.

In this case, we know $|F_{i_1}|\ge |F_{i_2}|\ge 4$. Since $|F_{i_3}\cup F_{i_4}\cup F_{i_5}|\ge 2$, $|F_{i_1}\cup F_{i_2}|\le |F|-|F_{i_3}\cup F_{i_4}\cup F_{i_5}|\le 11-2=9$. Hence $|F_{i_2}|=4$, $4\le |F_{i_1}|\le 5$. By Corollary 3.4, we know $C{F}_5^{i_2}-F_{i_2}$ has a big component $C_2$ and a singleton $x_2$. By Lemma 1, $C{F}_5^{i_1}-F_{i_1}$ should consider the following two situations: $\left(1\right)$ $C{F}_5^{i_1}-F_{i_1}$ has two components, one of which is a singleton. $\left(2\right)$ $C{F}_5^{i_1}-F_{i_1}$ has three components, two of which are singletons. For $\left(1\right)$, let $C_1$ is the big component and $x_1$ is the singleton of $C{F}_5^{i_1}-F_{i_1}$, since $|V\left(C_1\right)|=|V\left(C_2\right)|=|V\left(C{F}_5^i\right)-F_i-\left\{x_i\right\}|\ge 4!-5-1=18$ $(i\in \{i_1,i_2\})$ and $|F_{i_3}\cup F_{i_4}\cup F_{i_5}|\le 3$, by Proposition $2.3$, we can get $C{F}_5[V(C{F}_5^{[i_3,i_5]}-F^{[i_3,i_5]})\cup V\left(C_1\right)]$ is connected. Similarly, we can also get $C{F}_5[V(C{F}_5^{[i_3,i_5]}-F^{[i_3,i_5]})\cup V\left(C_2\right)]$ is connected. Thus the result holds. For $\left(2\right)$, let $C_1$ is the big component and $x_1$, $x_3$ are singletons of $C{F}_5^{i_1}-F_{i_1}$. If $x_2$ is nonadjacent to $\{x_1,x_3\}$, then $\{x_1,x_2,x_3\}$ are three singletons in $C{F}_5-F$. By Lemma $4.2$, we can get $\left|F\right|\ge 12$, this contradicts to the fact $\left|F\right|\le 11$. If $x_2$ is adjacent to $\{x_1,x_3\}$, say $(x_1,x_2)\in E(C{F}_5-F)$, then $C{F}_5-F$ only has two components; Otherwise, we let $C{F}_5-F$ has three components, then $x_3$ is a singleton in $C{F}_5-F$. By Proposition 2.5, we have $\left|F\right|\ge |N_{C{F}_5}\left(\{x_1,x_2\}\right)\cup N_{C{F}_5}\left(x_3\right)|\ge 5\times 2+6-3=13>11$, a contradiction. Thus the result holds.

Case 2. Only $C{F}_5^{i_2}-F_{i_2}$ is disconnected.

Since $|F_{i_3}\cup F_{i_4}\cup F_{i_5}|\ge 2$, $|F_{i_1}\cup F_{i_2}|\le 9$. As $C{F}_5^{i_2}-F_{i_2}$ is disconnected, we have $|F_{i_2}|\ge 4$ and then $|F_{i_1}|\le 5$. If $|F_{i_2}|=5$, then $|F_{i_1}|\ge 5$, $\left|F\right|\ge |F_{i_1}|+|F_{i_2}|+|F_{i_3}\cup F_{i_4}\cup F_{i_5}|\ge 5+5+2=12$, a contradiction. Thus $|F_{i_2}|=4$. Since $|E_{i_1,i_3}\left(C{F}_5\right)|=12>8\ge |F_{i_1}\cup F_{i_3}|$, $C{F}_5[V(C{F}_5^{[i_3,i_5]}-F^{[i_3,i_5]})\cup V(C{F}_5^{i_1}-F_{i_1})]$ is connected. As $|F_{i_2}|=4$, by Corollary $3.4$, we know $C{F}_5^{i_2}-F_{i_2}$ has a big component S and at most one singleton. By the same argument as that of Case 1, we can get $C{F}_5[V(C{F}_5^{[i_3,i_5]}-F^{[i_3,i_5]})\cup V(C{F}_5^{i_1}-F_{i_1})\cup V\left(S\right)]$ is connected. Then $C{F}_5-F$ must satisfies condition $\left(1\right)$.

Case 3. Only $C{F}_5^{i_1}-F_{i_1}$ is disconnected.

In this case, we have $|F_{i_1}|\ge 4$ by Proposition $2.8$ and $|F_{i_2}|\le 4$ since $\left|F\right|\le 11$ and $|F_{i_3}\cup F_{i_4}\cup F_{i_5}|\ge 2$. As $|E_{i_2,i_3}\left(C{F}_5\right)|=12>7\ge |F_{i_2}\cup F_{i_3}|$, we have $C{F}_5^{[i_2,i_5]}-F^{[i_2,i_5]}$ is connected.

If $|F_{i_1}|\le 5$, by Lemma 1, $C{F}_5^{i_1}-F_{i_1}$ has a big component S and one single and two singletons. By the same argument as that of Case 1, we can get $C{F}_5[V(C{F}_5^{[i_2,i_5]}-F^{[i_2,i_5]})\cup V\left(S\right)]$ is connected. Then $C{F}_5-F$ must be one of conditions $\left(1\right)$ and $\left(2\right)$.

Now, we suppose $|F_{i_1}|\ge 6$. Then $|F_{i_2}\cup F_{i_3}\cup F_{i_4}\cup F_{i_5}|\le 5$. Let W be the union of components of $C{F}_5-F$, whose vertices, which are totally contained in $C{F}_5^{i_1}-F_{i_1}$, are not connected with $C{F}_5^{[i_2,i_5]}-F^{[i_2,i_5]}$. By Proposition $2.2$ and Proposition $2.3$, we know $2\left|W\right|\le |F-F_{i_1}|\le 5$, which implies $\left|W\right|\le 2$. Thus $C{F}_5-F$ satisfies $\left(1\right)$ or $\left(2\right)$.

Combing the above three cases, we know this result holds.

Lemma 4.8. For $n=6$, if $\left|F\right|\le 14$, then $C{F}_6-F$ contains a big component C, which satisfies the result $\left|V\right(C\left)\right|\ge n!-\left|F\right|-2$.

Proof. Similarly, we do not think about the situation $C{F}_6-F$ is connected, so we let $C{F}_6-F$ is disconnected. Let $F_i=F\cap V\left(C{F}_6^i\right)$ for $i\in [1,6]$ with $\left|F_{i_1}\right|\ge \left|F_{i_2}\right|\ge \left|F_{i_3}\right|\ge \left|F_{i_4}\right|\ge \left|F_{i_5}\right|\ge \left|F_{i_6}\right|$, where $i_j\in \{1,2,3,4,5,6\}$. If $\left|F_{i_2}\right|=0$, then $\left|F_{i_3}\right|=\left|F_{i_4}\right|=\cdots =\left|F_{i_6}\right|=0$ and $C{F}^{[i_2,i_6]}-F^{[i_2,i_6]}$ is connected. By Proposition $2.2$, we can get $C{F}_6-F$ is connected. So we assume $|F_{i_2}|\ge 1$. Since $\left|F\right|\le 14$, we have $\left|F_{i_6}\right|\le 2$, $\left|F_{i_5}\right|\le 2$, $\left|F_{i_4}\right|\le 3$, $\left|F_{i_3}\right|\le 4$, $1\le \left|F_{i_2}\right|\le 7$, $1\le \left|F_{i_1}\right|\le 13$. Firstly, we proof the following Claim is correct.

Claim 2. If $\left|F_{i_j}\right|\le 5$, then $C{F}_6^{[i_j,i_6]}-F^{[i_j,i_6]}$ is connected.

Proof of Claim 2. By Proposition $2.8$, we know $C{F}_6^j-F_j$ is connected for each $j\in [i_j,i_6]$. On the other hand, since $\left|E_{p,i_6}\left(C{F}_6\right)\right|=(6-2)!=24>7\ge |F_p\cup F_{i_6}|$ for $p\in [i_j,i_5]$, we can get $E_{p,i_6}(C{F}_6-F)\ne \varnothing$. Thus $C{F}_6^{[i_j,i_6]}-F^{[i_j,i_6]}$ is connected.

By Claim 2, we know $C{F}_6^{[i_3,i_6]}-F^{[i_3,i_6]}$ is connected. If both $C{F}_6^{i_1}-F_{i_1}$ and $C{F}_6^{i_2}-F_{i_2}$ are connected, we can get $C{F}_6-F$ is connected. As $\left|E_{i_2,i_3}\left(C{F}_6\right)\right|=(6-2)!=24>11\ge |F_{i_2}\cup F_{i_3}|$, $\left|E_{i_1,i_3}\left(C{F}_6\right)\right|=(6-2)!=24>17\ge |F_{i_1}\cup F_{i_3}|$ and $C{F}_6^{[i_3,i_6]}-F^{[i_3,i_6]}$ is connected, thus $C{F}_6-F$ is connected. Since $C{F}_6-F$ is disconnected, at least one of $C{F}_6^i-F_i$$(i\in \{i_1,i_2\})$ is disconnected, which leads to the following cases.

Case 1. Both $C{F}_6^{i_1}-F_{i_1}$ and $C{F}_6^{i_2}-F_{i_2}$ are disconnected.

In this case, we know $6\le \left|F_{i_2}\right|\le \left|F_{i_1}\right|\le \left|F\right|-|F_{i_2}|\le 8<9$. By Corollary 3.4, we know that $C{F}_6^{i_1}-F_{i_1}$$(resp.,C{F}_6^{i_2}-F_{i_2})$ has a big component $C_1$$(resp.,C_2)$ and one singleton $x_1$$(resp.,x_2)$. As $|E_{C{F}_6-F}(V\left(C_1\right),V(C{F}_6^{i_3}-F_{i_3}))|\ge 24-2-8-1=13>1$. Thus $C{F}_6-F[V(C{F}_6^{[i_3,i_6]}-F^{[i_3,i_6]})\cup V\left(C_1\right)]$ is connected. Similarly, we can get $C{F}_6-F[V(C{F}_6^{[i_3,i_6]}-F^{[i_3,i_6]})\cup V\left(C_2\right)]$ is also connected. Thus the result holds.

Case 2. Only $C{F}_6^{i_2}-F_{i_2}$ is disconnected.

As $C{F}_6^{i_2}-F_{i_2}$ is disconnected, we have $\left|F_{i_2}\right|\ge 6$ and then $\left|F_{i_1}\right|\le 8$. Since $\left|E_{i_1,i_3}\left(C{F}_6\right)\right|=24>10\ge |F_{i_1}\cup F_{i_3}|$, $C{F}_6[V(C{F}_6^{[i_3,i_6]}-F^{[i_3,i_6]})\cup V(C{F}_6^{i_1}-F_{i_1})]$ is connected. Since $\left|F_{i_2}\right|\le \left|F_{i_1}\right|\le 8$, by Corollary 3.4, $C{F}_6^{i_2}-F_{i_2}$ has a big component C and one singleton. Since $|E_{i_2,i_3}\left(C{F}_6\right)|=24>11\ge |F_{i_2}\cup F_{i_3}|+1$, we can get $C{F}_6[V(C{F}_6^{[i_3,i_6]}-F^{[i_3,i_6]})\cup V(C{F}_6^{i_1}-F_{i_1})\cup V\left(C\right)]$ is connected. Then $C{F}_6-F$ must be one of conditions $\left(1\right)$.

Case 3. Only $C{F}_6^{i_1}-F_{i_1}$ is disconnected.

In this case, $6\le \left|F_{i_1}\right|\le 13$, $|F_{i_2}|\le 8$. As $\left|E_{i_2,i_3}\left(C{F}_6\right)\right|=24>11>|F_{i_2}\cup F_{i_3}|$, we have $C{F}_6^{[i_2,i_6]}-F^{[i_2,i_6]}$ is connected.

If $\left|F_{i_1}\right|\le 11$, by Lemma $4.8$, $C{F}_6^{i_1}-F_{i_1}$ has a big component C with $\left|V\left(C\right)\right|\ge 5!-\left|F_{i_1}\right|-2$. By the same argument as that of Case 2, we can get $C{F}_6[V(C{F}_6^{[i_2,i_6]}-F^{[i_2,i_6]})\cup V\left(C\right)]$ is connected. Thus the result holds.

If $\left|F_{i_1}\right|\ge 12$, then $|F_{i_2}\cup F_{i_3}\cup F_{i_4}\cup F_{i_5}\cup F_{i_6}|\le 2$. Let W be the union of components of $C{F}_6-F$, whose vertices, which are totally contained in $C{F}_6^{i_1}-F_{i_1}$, and are not connected with $C{F}_6^{[i_2,i_6]}-F^{[i_2,i_6]}$. By Proposition $2.2$ and Proposition $2.3$, we have $\left|W\right|\le |F-F_{i_1}|\le 2$. Thus the result holds.

Lemma 4.9. Let $\left|F\right|\le \frac{9n-23}{2}$ for odd n$(n\ge 5)$ and $\left|F\right|\le \frac{9n-26}{2}$ for even n$(n\ge 6)$, then $C{F}_n-F$ contains a big component C, which satisfies that $\left|V\right(C\left)\right|\ge n!-\left|F\right|-2$.

Proof. By Lemma $4.7$ and Lemma $4.8$, the result holds for $n=5,6$. We proof this result by induction on n. Assume $n\ge 7$ and the result holds for $C{F}_{n-1}$. Now we suppose $C{F}_n-F$ is disconnected for any $F\subseteq V\left(C{F}_n\right)$ with $\left|F\right|\le \frac{9n-23}{2}$ or $\left|F\right|\le \frac{9n-26}{2}$. Let $F_i=F\cap V\left(C{F}_n^i\right)$ for $i\in [1,n]$ with $\left|F_{i_1}\right|\ge \left|F_{i_2}\right|\ge \cdots \ge \left|F_{i_n}\right|$, where $i_j\in [1,n]$.

When n is odd, if $\left|F_{i_3}\right|=0$, then $\left|F_{i_4}\right|=\left|F_{i_5}\right|=\cdots =\left|F_{i_n}\right|=0$ and $C{F}^{[i_3,i_n]}-F^{[i_3,i_n]}$ is connected, and thus, by Proposition $2.2$, we can get $C{F}_n-F$ is connected. Now we assume $|F_{i_3}|\ge 1$; When n is even, if $\left|F_{i_2}\right|=0$, then $\left|F_{i_3}\right|=\left|F_{i_4}\right|=\cdots =\left|F_{i_n}\right|=0$ and $C{F}^{[i_2,i_n]}-F^{[i_2,i_n]}$ is connected, and thus, by Proposition $2.2$, we can get $C{F}_n-F$ is connected. Now we assume $|F_{i_2}|\ge 1$.

Claim 3. When n is even, if $\left|F_{i_j}\right|\le \frac{3n-8}{2}$$(i_j\in [i_1,i_{n-1}])$, then $C{F}_n^{[i_j,i_n]}-F^{[i_j,i_n]}$ is connected; When n is odd, if $\left|F_{i_j}\right|\le \frac{3n-9}{2}$$(i_j\in [i_1,i_{n-1}])$, then $C{F}_n^{[i_j,i_n]}-F^{[i_j,i_n]}$ is connected;

Proof of Claim 3. By Proposition $2.8$, we know $C{F}_n^j-F_j$ is connected for each $j\in [i_j,i_n]$. On the other hand, since $\left|E_{p,i_n}\left(C{F}_n\right)\right|=(n-2)!>3n-8>|F_p\cup F_{i_n}|$ for $p\in [i_j,i_{n-1}]$$(n$ is even) and $\left|E_{p,i_n}\left(C{F}_n\right)\right|=2(n-2)!>3n-9>|F_p\cup F_{i_n}|$ for $p\in [i_j,i_{n-1}]$$(n$ is odd), we can get $E_{p,i_n}(C{F}_n-F)\ne \varnothing$. Thus $C{F}_n^{[i_j,i_n]}-F^{[i_j,i_n]}$ is connected.

Since $\left|F\right|\le \frac{9n-26}{2}$$(n$ is even) and $\left|F\right|\le \frac{9n-23}{2}$$(n$ is odd), we have $\frac{3n-8}{2}>\left|F_{i_3}\right|\ge \left|F_{i_4}\right|\ge \cdots \ge \left|F_{i_n}\right|$ for even n and $\frac{3n-9}{2}\ge \left|F_{i_3}\right|\ge \left|F_{i_4}\right|\ge \cdots \ge \left|F_{i_n}\right|$ for odd n. By Claim 3, we can get $C{F}_n^{[i_3,i_n]}-F^{[i_3,i_n]}$ is connected. If $C{F}_n^{i_1}-F_{i_1}$ and $C{F}_n^{i_2}-F_{i_2}$ are all connected, as $\left|E_{i_2,i_3}\left(C{F}_n\right)\right|=(n-2)!>\frac{9n-26}{2}>|F_{i_2}\cup F_{i_3}|$$(n$ is even) and $\left|E_{i_2,i_3}\left(C{F}_n\right)\right|=2(n-2)!>\frac{9n-23}{2}>|F_{i_2}\cup F_{i_3}|$$(n$ is odd), then $C{F}_n^{[i_2,i_n]}-F^{[i_2,i_n]}$ is connected. Similarly, we can also get $C{F}_n-F$ is connected. So at least one of $C{F}_n^i-F_i$$(i\in \{i_1,i_2\})$ is disconnected, which leads to the following cases.

Note that, when n is odd, if $|F_{i_3}\cup F_{i_4}\cup F_{i_5}\cup \cdots \cup F_{i_n}|\le 1$, by the same argument of Lemma 4.7, we know $C{F}_n-F$ satisfies condition $\left(1\right)$ or $\left(2\right)$. Hence we assume that $|F_{i_3}\cup F_{i_4}\cup F_{i_5}\cup \cdots \cup F_{i_n}|\ge 2$.

Case 1. Both $C{F}_n^{i_1}-F_{i_1}$ and $C{F}_n^{i_2}-F_{i_2}$ are disconnected.

When n is even, we have $\frac{3n-6}{2}\le \left|F_{i_2}\right|\le \left|F_{i_1}\right|\le \left|F\right|-|F_{i_2}|\le \frac{9n-26}{2}-\frac{3n-6}{2}=3n-10$. By Corollary 3.4, we know $C{F}_n^{i_1}-F_{i_1}$ and $C{F}_n^{i_2}-F_{i_2}$ all have a big component $C_1,C_2$ and one singleton. As $|E_{C{F}_n-F}(V\left(C_1\right),V(C{F}_n^{i_3}-F_{i_3}))|\ge (n-2)!-1-\frac{9n-26}{2}>1$, $|E_{C{F}_n-F}(V\left(C_2\right),V(C{F}_n^{i_3}-F_{i_3}))|\ge (n-2)!-1-\frac{9n-26}{2}>1$, Thus $C{F}_n-F[V(C{F}_n^{[i_3,i_n]}-F^{[i_3,i_n]})\cup V\left(C_1\right)\cup V\left(C_2\right)]$ is connected, the result holds.

When n is odd, we have $\frac{3n-7}{2}\le \left|F_{i_2}\right|\le \left|F_{i_1}\right|\le \left|F\right|-|F_{i_2}|-|F_{i_3}\cup F_{i_4}\cup \cdots \cup F_{i_n}|\le \frac{9n-23}{2}-\frac{3n-7}{2}-2=3n-10$.So by Lemma 1, we would consider the following three subcases: $\left(1\right)$ Both $C{F}_n^{i_1}-F_{i_1}$ and $C{F}_n^{i_2}-F_{i_2}$ have three components, two of which are singletons; $\left(2\right)$ Only one of $C{F}_n^{i_1}-F_{i_1}$ and $C{F}_n^{i_2}-F_{i_2}$ has three components, two of which are singletons; $\left(3\right)$ Both $C{F}_n^{i_1}-F_{i_1}$ and $C{F}_n^{i_2}-F_{i_2}$ have two components, one of which is singleton. Now, we just proof the first subcase and the other two subcases could be proved by the same argument. Let $x_1,y_1,C_1$ (resp., $x_2,y_2,C_2$) be the two singletons and the other big component of $C{F}_n^{i_1}-F_{i_1}$ (resp., $C{F}_n^{i_2}-F_{i_2}$). Since $\left|V\left(C_1\right)\right|=|V\left(C{F}_n^{i_1}\right)-F_{i_1}-\{x_1,y_1\}|\ge (n-1)!-(3n-10)-2$ and $|F_{i_3}\cup F_{i_4}\cup \cdots \cup F_{i_n}|\le \frac{3n-9}{2}$, by Proposition $2.3$, we know $C{F}_n[V(C{F}_n^{[i_3,i_n]}-F^{[i_3,i_n]})\cup V\left(C_1\right)]$ is connected. Similarly, we can get $C{F}_n[V(C{F}_n^{[i_3,i_n]}-F^{[i_3,i_n]})\cup V\left(C_2\right)]$ is connected.

If $x_1,x_2,y_1,y_2$ are four singletons in $C{F}_n-F$, then by Proposition $2.5$, we know $\left|F\right|\ge |N_{C{F}_n^{i_1}}\left(\{x_1,y_1\}\right)\cup N_{C{F}_n^{i_2}}\left(\{x_2,y_2\}\right)\cup (F_{i_3}\cup F_{i_4}\cup \cdots \cup F_{i_n})|\ge [\frac{3(n-1)-4}{2}\times 2-3]\times 2+2=6n-18>\frac{9n-23}{2}$, a contradiction.

If $C{F}_n-F$ has three singletons, then by Lemma $4.5$, we can get $\left|F\right|\ge \frac{9n-21}{2}$, this contradicts to the fact $\left|F\right|\le \frac{9n-23}{2}$. So $C{F}_n-F$ has two singletons or only one singleton.

Claim 4. If $(C{F}_n-F)\left[\{x_1,y_1,x_2,y_2\}\right]$ has at least one edge, say $(x_1,x_2)\in E(C{F}_n-F)$, then $C{F}_n-F$ only has two components.

Proof of Claim 4. Suppose $C{F}_n-F$ has at least three components. Then $y_1$ is a singleton or $(y_1,y_2)\in E(C{F}_n-F)$. If $y_1$ is a singleton, then $\left|F\right|\ge |N_{C{F}_n}\left(\{x_1,x_2\}\right)\cup N_{C{F}_n}\left(y_1\right)|\ge (\frac{3n-3}{2}-1)\times 2+\frac{3n-3}{2}-3=\frac{9n-19}{2}>\frac{9n-23}{2}$, a contradiction. If $(y_1,y_2)\in E(C{F}_n-F)$, then $\left|F\right|\ge |N_{C{F}_n}\left(\{x_1,x_2\}\right)\cup N_{C{F}_n}\left(\{y_1,y_2\}\right)|\ge (\frac{3n-3}{2}-1)\times 4-3\times 2=6n-16>\frac{9n-23}{2}$, a contradiction.

Thus, by Claim 4, the result holds.

Case 2. Only $C{F}_n^{i_2}-F_{i_2}$ is disconnected.

As $C{F}_n^{i_2}-F_{i_2}$ is disconnected, we have $\left|F_{i_2}\right|\ge \frac{3n-6}{2}$ for even n and $\left|F_{i_2}\right|\ge \frac{3n-7}{2}$ for odd n, then $\left|F_{i_1}\right|\le 3n-10<3n-9$$(n$ is even) and $\left|F_{i_1}\right|\le 3n-10$$(n$ is odd). Since $\left|E_{i_1,i_3}\left(C{F}_n\right)\right|=(n-2)!>\frac{9n-26}{2}\ge |F_{i_1}\cup F_{i_3}|$$(n$ is even) and $\left|E_{i_1,i_3}\left(C{F}_n\right)\right|=2(n-2)!>\frac{9n-23}{2}\ge |F_{i_1}\cup F_{i_3}|$$(n$ is odd), $C{F}_n[V(C{F}_n^{[i_3,i_n]}-F^{[i_3,i_n]})\cup V(C{F}_n^{i_1}-F_{i_1})]$ is connected. Since $\left|F_{i_2}\right|\le \left|F_{i_1}\right|\le 3n-10$, when n is even, by Corollary 3.4, we know $C{F}_n^{i_2}-F_{i_2}$ has a big component C and one singleton; When n is odd, by Lemma 1, we know $C{F}_n^{i_2}-F_{i_2}$ has a big component C and at most two singletons. By the same argument as that of Case 1, we can get $C{F}_n[V(C{F}_n^{[i_3,i_n]}-F^{[i_3,i_n]})\cup V(C{F}_n^{i_1}-F_{i_1})\cup V\left(C\right)]$ is connected. Then $C{F}_n-F$ must be one of conditions $\left(1\right)$ and $\left(2\right)$.

Case 3. Only $C{F}_n^{i_1}-F_{i_1}$ is disconnected.

In this case, $\frac{3n-6}{2}\le \left|F_{i_1}\right|\le \frac{9n-28}{2}$ for even n and $\frac{3n-7}{2}\le \left|F_{i_1}\right|\le \frac{9n-29}{2}$ for odd n. As $\left|E_{i_2,i_3}\left(C{F}_n\right)\right|=(n-2)!>\frac{9n-26}{2}\ge |F_{i_2}\cup F_{i_3}|$$(n$ is even) and $\left|E_{i_2,i_3}\left(C{F}_n\right)\right|=2(n-2)!>\frac{9n-23}{2}\ge |F_{i_2}\cup F_{i_3}|$$(n$ is odd), we have $C{F}_n^{[i_2,i_n]}-F^{[i_2,i_n]}$ is connected.

When n is even, if $\left|F_{i_1}\right|\le \frac{9n-32}{2}$, by introduction, we know $C{F}_n^{i_1}-F_{i_1}$ has a big component C with $\left|V\left(C\right)\right|\ge (n-1)!-\left|F_{i_1}\right|-2$. By the same argument as that of Case 1, we can get $C{F}_n[V(C{F}_n^{[i_2,i_n]}-F^{[i_2,i_n]})\cup V\left(C\right)]$ is connected. Thus the result holds. If $\left|F_{i_1}\right|\ge \frac{9n-30}{2}$, then $|F_{i_2}\cup F_{i_3}\cup \cdots \cup F_{i_n}|\le 2$. Let W be the union of components of $C{F}_n-F$, whose vertices, which are totally contained in $C{F}_n^{i_1}-F_{i_1}$, and are not connected with $C{F}_n^{[i_2,i_n]}-F^{[i_2,i_n]}$. By Proposition $2.2$ and Proposition $2.3$, we have $\left|W\right|\le |F-F_{i_1}|\le 2$. Thus the result holds.

When n is odd, if $\left|F_{i_1}\right|\le \frac{9n-35}{2}$, by introduction, $C{F}_n^{i_1}-F_{i_1}$ has a big component C with $\left|V\left(C\right)\right|\ge (n-1)!-\left|F_{i_1}\right|-2$. By the same argument as that of Case 1, we can get $C{F}_n[V(C{F}_n^{[i_2,i_n]}-F^{[i_2,i_n]})\cup V\left(C\right)]$ is connected. Thus the result holds. If $\left|F_{i_1}\right|\ge \frac{9n-33}{2}$, then $|F_{i_2}\cup F_{i_3}\cup \cdots \cup F_{i_n}|\le 5$. Let W be the union of components of $C{F}_n-F$, whose vertices, which are totally contained in $C{F}_n^{i_1}-F_{i_1}$, and are not connected with $C{F}_n^{[i_2,i_n]}-F^{[i_2,i_n]}$. By Proposition $2.3$, $2\left|W\right|\le |F-F_{i_1}|\le 5$. Then we have $\left|W\right|\le 2$. Thus the result holds.

Theorem 2. For $n\ge 4$, when n is odd, $c{\kappa }_4\left(C{F}_n\right)=\frac{9n-21}{2}$; when n is even, $c{\kappa }_4\left(C{F}_n\right)=\frac{9n-24}{2}$.

Proof. By Lemma 1, we have $c{\kappa }_4\left(C{F}_4\right)\ge 6=\frac{9\times 4-24}{2}$. For $n\ge 5$, by Lemma $4.9$, we can get when n is odd, $c{\kappa }_4\left(C{F}_n\right)\ge \frac{9n-21}{2}$; when n is even, $c{\kappa }_4\left(C{F}_n\right)\ge \frac{9n-24}{2}$. Next, we will prove that $c{\kappa }_4\left(C{F}_n\right)\le \frac{9n-21}{2}$ and $c{\kappa }_4\left(C{F}_n\right)\le \frac{9n-24}{2}$. For $n=4$, if we let $F=\{2314,3124,1234,4213,4132,4321\}$, then $C{F}_n-F$ has three singletons: $x_1=3214,x_2=2134,x_3=1324$. Thus $c{\kappa }_4\left(C{F}_4\right)\le 6=\frac{9\times 4-24}{2}$. For $n\ge 5$, when n is odd, let $S=\{x_1,x_2,x_3\}$, where $x_1=i_1{i}_2{i}_3\cdots i_{n-4}3214$, $x_2=2i_2{i}_3\cdots i_{n-4}{i}_{1}314$, $x_3=2i_2{i}_3\cdots i_{n-4}3i_{1}41$, then $|N_{C{F}_n}\left(S\right)|=3n-10+\frac{3n-7}{2}+3=\frac{9n-21}{2}$ and there are three singletons $\{x_1,x_2,x_3\}$ in $C{F}_n-N_{C{F}_n}\left(S\right)$. Thus when n is odd, $c{\kappa }_4\left(C{F}_n\right)\le \frac{9n-21}{2}$. When n is even, let $S=\{x_1,x_2,x_3\}$, where $x_1=i_1{i}_2{i}_3\cdots i_{n-4}3214j$, $x_2=2i_2{i}_3\cdots i_{n-4}{i}_{1}314j$, $x_3=2i_2{i}_3\cdots i_{n-4}3i_{1}41j$, then $\{x_1,x_2,x_3\}\subseteq V\left(C{F}_n^j\right)$ and $|N_{C{F}_n^j}\left(S\right)|=\frac{9(n-1)-21}{2}=\frac{9n-30}{2}$. As $x_1,x_2,x_3$ belong to a common subgraph, by Proposition 2.3, we know $x_1,x_2,x_3$ have different outgoing neighbors. So $|N_{C{F}_n}\left(S\right)|=\frac{9n-30}{2}+3=\frac{9n-24}{2}$ and $x_1,x_2,x_3$ are three singletons in $C{F}_n-N_{C{F}_n}\left(S\right)$. Thus when n is even, $c{\kappa }_4\left(C{F}_n\right)\le \frac{9n-24}{2}$.

5. The 5-component connectivity of $C{F}_n$

Lemma 5.1. For $n=4$, let S is an $Ind$-set and $\left|S\right|=4$, then $|N_{C{F}_4}\left(S\right)|\ge 8$.

Proof. Let $S=\{x_1,x_2,x_3,x_4\}$, since S is an $Ind$-set, $x_1,x_2,x_3,x_4$ are nonadjacent to each other. As $C{F}_4^i\cong C{F}_3$$(i\in \{1,2,3,4\left\}\right)$, we know $x_1,x_2,x_3,x_4$ can not belong to a same subgraph of $C{F}_4$. So we need think about the following cases:

Case 1: $x_1,x_2,x_3,x_4$ belong to two different subgraphs of $C{F}_4$.

In this case, we can divide it into two subcases:

Subcase 1.1: There are two subgraphs of $C{F}_4$ which contain only two vertices of S. Without loss of generality, we can let $\{x_1,x_2\}\subseteq V\left(C{F}_4^1\right)$, $\{x_3,x_4\}\subseteq V\left(C{F}_4^2\right)$. By the definition of $C{F}_n$, $|N_{C{F}_4^1}\left(\{x_1,x_2\}\right)|=|N_{C{F}_4^2}\left(\{x_3,x_4\}\right)|=3$. Now we let $x_1=2341$, then $x_2=4231$ or $x_2=3421$. Thus $x_1^{+}\in V\left(C{F}_4^2\right)$, $x_2^{+}=1234\in V\left(C{F}_4^4\right)$ or $x_2^{+}=1423\in V\left(C{F}_4^3\right)$. Hence $x_1^{+}$ and $x_2^{+}$ can not belong to a common subgraph of $C{F}_4$. Similarly, we know $x_3^{+}$ and $x_4^{+}$ can not belong to a common subgraph of $C{F}_4$. If $x_1^{+}$ or $x_2^{+}$ belong to $C{F}_4^2$ and adjacent to $\{x_3,x_4\}$, meanwhile $x_3^{+}$ or $x_4^{+}$ belong to $C{F}_4^1$ and adjacent to $\{x_1,x_2\}$, then $|N_{C{F}_4}\left(S\right)|=3+3+2=8$. Now, we illustrate this structure exists. Let $x_3=3142$, $x_4=1432$, then $x_1^{+}$ is adjacent to $x_3$, $x_4^{+}$ is adjacent to $x_1$. Thus $|N_{C{F}_4}\left(S\right)|\ge 8$.

Subcase 1.2: There is a subgraph of $C{F}_4$ which contains three vertices of S. In this subcase, we can let $\{x_1,x_2,x_3\}\subseteq V\left(C{F}_4^1\right)$, $x_4\in V\left(C{F}_4^2\right)$. We let $x_1=2341$, $x_2=4231$, $x_3=3421$, then $x_1^{+}=1342\in V\left(C{F}_4^2\right)$, $x_2^{+}=1234\in V\left(C{F}_4^4\right)$, $x_3^{+}=1423\in V\left(C{F}_4^3\right)$. Clearly, $|N_{C{F}_4^1}\left(\{x_1,x_2,x_3\}\right)|=3$, $|N_{C{F}_4^2}\left(x_4\right)|=3$. If $x_4^{+}\in V\left(C{F}_4^1\right)$ and is adjacent to $\{x_1,x_2,x_3\}$, meanwhile $x_1^{+}$ is adjacent to $x_4$, then $|N_{C{F}_4}\left(S\right)|=3+3+2=8$. Let $x_4=1432$, then $x_4^{+}=2431$. Thus $x_1^{+}$ is adjacent to $x_4$ and $x_4^{+}$ is adjacent to $x_1$. Thus $|N_{C{F}_4}\left(S\right)|\ge 8$.

Case 2: $x_1,x_2,x_3,x_4$ belong to three different subgraphs of $C{F}_4$.

In this case, there exists a subgraph $C{F}_4^i$ which must contains two vertices of S. Now we can let $\{x_1,x_2\}\subseteq V\left(C{F}_4^i\right)$, $x_3\in V\left(C{F}_4^j\right)$, $x_4\in V\left(C{F}_4^k\right)$. Then $|N_{C{F}_4^i}\left(\{x_1,x_2\}\right)|=3$, $|N_{C{F}_4^j}\left(x_3\right)|=|N_{C{F}_4^k}\left(x_4\right)|=3$, $|N_{C{F}_4}\left(S\right)|\ge |N_{C{F}_4^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_4^j}\left(x_3\right)|+|N_{C{F}_4^k}\left(x_4\right)|=9$.

Case 3: $x_1,x_2,x_3,x_4$ belong to four different subgraphs of $C{F}_4$.

In this case, we can let $x_k\in V\left(C{F}_n^k\right)$, then $|N_{C{F}_n^k}\left(x_k\right)|=3$. Thus $|N_{C{F}_4}\left(S\right)|\ge 4\times |N_{C{F}_n^k}\left(x_k\right)|=4\times 3=12$.

Combing the above three cases, we have $|N_{C{F}_4}\left(S\right)|\ge 8$.

Lemma 5.2. When n is odd, let $S=\{x_1,x_2,x_3,x_4\}$ is an $Ind$-set and $\{x_1,x_2\}\subseteq V\left(C{F}_n^i\right)$, $\{x_3,x_4\}\subseteq V\left(C{F}_n^j\right)$$(i\ne j)$. If $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|=|N_{C{F}_n^j}\left(\{x_3,x_4\}\right)|=3n-10$, then $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}$ $\left(S\right)|\ge 4$.

Proof. Since $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|=|N_{C{F}_n^j}\left(\{x_3,x_4\}\right)|=3n-10$, by the proof process of Lemma 3.3, we know $x_1,x_2$$(resp.,x_3,x_4)$ have three common neighbors in $C{F}_n^i$$(resp.,C{F}_n^j)$. So by Lemma 3.1, we can let $x_1=i_1{i}_2\cdots i_{n-2}{i}_{n-1}i$, $x_2=k_1{k}_2\cdots k_{n-2}{i}_{n-1}i$, where $k_i\in [i_1,i_{n-2}]$ and $k_1\ne i_1$. Then $x_1^{+}=i{i}_2{i}_3\cdots i_{n-2}{i}_{n-1}{i}_1$, $x_1^{-}=i_1{i}_2{i}_3\cdots i_{n-2}i{i}_{n-1}$, $x_2^{-}=k_1{k}_2\cdots k_{n-2}i{i}_{n-1}$, $x_2^{+}=i{k}_2{k}_3\cdots k_{n-2}{i}_{n-1}{k}_1$. By Proposition 2.3 and Proposition 2.2, we have $2\le |N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\right\{x_1,$ $x_2\left\}\right)|\le 4$, $2\le |N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_3,x_4\}\right)|\le 4$. Then we think about the following three cases:

Case 1: $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_1,x_2\}\right)|=4$.

In this case, we can easily get $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$.

Case 2: $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_1,x_2\}\right)|=3$.

In this case, only one of the outgoing neighbors of $\{x_1,x_2\}$ belong to $C{F}_n^j$. So $i_1=j$ or $k_1=j$. We assume $i_1=j$, then $x_1^{+}\in V\left(C{F}_n^j\right)$. If $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_3,x_4\}\right)|=4$, then $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$. If $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_3,x_4\}\right)|\le 3$, one of the outgoing neighbors of $\{x_3,x_4\}$ must belong to $C{F}_n^i$, we assume $x_3^{+}$ or $x_3^{-}$ belong to $C{F}_n^i$, then $x_3=i{j}_2{j}_3\cdots j_{n-2}{j}_{n-1}j$ or $x_3=j_1{j}_2{j}_3\cdots j_{n-2}ij$. When $x_3=i{j}_2{j}_3\cdots j_{n-2}{j}_{n-1}j$, then by the proof process of Lemma 3.1, we can let $x_4=l_1{l}_2{l}_3\cdots l_{n-2}{j}_{n-1}j$, where $l_i\in ([j_2,j_{n-2}]\cup \left\{i\right\})$ and $l_1\ne i$. Thus $x_3^{+}=j{j}_2{j}_3\cdots j_{n-2}{j}_{n-1}i$, $x_3^{-}=i{j}_2{j}_3\cdots j_{n-2}j{j}_{n-1}$, $x_4^{+}=j{l}_2{l}_3\cdots l_{n-2}{j}_{n-1}{l}_1$, $x_4^{-}=l_1{l}_2{l}_3\cdots l_{n-2}j{j}_{n-1}$. Then $x_3^{+}\in V\left(C{F}_n^i\right)$. Since $j_{n-1}\ne i$, $x_4^{-}\notin V\left(C{F}_n^i\right)$. Thus $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$ as $x_4^{-}\ne x_1^{-}$, $x_4^{-}\ne x_2^{-}$, $x_4^{-}\ne x_2^{+}$. When $x_3=j_1{j}_2{j}_3\cdots j_{n-2}ij$, then by the proof process of Lemma 3.1, we can let $x_4=l_1{l}_2{l}_3\cdots l_{n-2}ij$, where $l_i\in [j_1,j_{n-2}]$ and $l_1\ne j_1$. Then $x_3^{+}=j{j}_2{j}_3\cdots j_{n-2}i{j}_1$, $x_3^{-}=j_1{j}_2{j}_3\cdots j_{n-2}ji$, $x_4^{+}=j{l}_2{l}_3\cdots l_{n-2}i{l}_1$, $x_4^{-}=l_1{l}_2{l}_3\cdots l_{n-2}ji$, $\{x_3^{-},x_4^{-}\}\subseteq V\left(C{F}_n^i\right)$. Since $j_1\ne i$, $l_1\ne i$, we know $\{x_3^{+},x_4^{+}\}\subseteq V(C{F}_n-C{F}_n^i-C{F}_n^j)$. As $i\ne j$ and $k_1\ne j$, then $x_1^{-}\ne x_3^{+}$, $x_3^{+}\ne x_2^{+}$, $x_3^{+}\ne x_2^{-}$. Thus $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$.

Case 3: $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_1,x_2\}\right)|=2$.

In this case, two outgoing neighbors of $\{x_1,x_2\}$ belong to $C{F}_n^j$. Thus $\{x_1^{-},x_2^{-}\}\subseteq V\left(C{F}_n^j\right)$ and $\{x_3,x_4\}\subseteq V\left(C{F}_n^j\right)$. Now we let $x_3=j_1{j}_2{j}_3\cdots j_{n-2}{j}_{n-1}j$, then $x_4=u_1{u}_2{u}_3\cdots$$u_{n-2}{j}_{n-1}j$, where $u_i\in [j_1,j_{n-2}]$ and $u_1\ne j_1$. If $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_3,x_4\}\right)|=4$, clearly $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$. If $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_3,x_4\}\right)|=3$, the proof process is similar to Case 2, we can get $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$. So we let $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_3,x_4\}\right)|=2$, then one of the outgoing neighbors of $x_3$ and $x_4$ belong to $C{F}_n^i$, we assume $x_3=i{j}_2{j}_3\cdots j_{n-2}{j}_{n-1}j$ or $x_3=j_1{j}_2{j}_3\cdots j_{n-2}ij$. When $x_3=i{j}_2{j}_3\cdots j_{n-2}{j}_{n-1}j$, we can get $j_{n-1}\ne i$, so $x_4=i{u}_2{u}_3\cdots u_{n-2}{j}_{n-1}j$. By Corollary 3.2, we know $x_3$ and $x_4$ can not have three common neighbors in $C{F}_n^j$, so $x_3=j_1{j}_2{j}_3\cdots j_{n-2}ij$, $x_4=u_1{u}_2{u}_3\cdots u_{n-2}ij$. Since $x_4^{+}=j{u}_2{u}_3\cdots u_{n-2}i{u}_1$, $x_3^{+}=j{j}_2{j}_3\cdots j_{n-2}i{j}_1$ and $i\ne j$, we have $x_3^{+}\ne x_1^{+}$, $x_3^{+}\ne x_2^{+}$, $x_4^{+}\ne x_1^{+}$, $x_4^{+}\ne x_2^{+}$. Thus $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$.

Lemma 5.3. When n is odd, let $S=\{x_1,x_2,x_3,x_4\}$ is an $Ind$-set and $\{x_1,x_2,x_3\}\subseteq V\left(C{F}_n^i\right)$, $x_4\in V\left(C{F}_n^j\right)$$(i\ne j)$. If $|N_{C{F}_n^i}\left(S\right)|=\frac{9(n-1)-24}{2}=\frac{9n-33}{2}$, $|N_{C{F}_n^j}\left(S\right)|=\frac{3n-7}{2}$, then $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$.

Proof. Since $\{x_1,x_2,x_3\}\subseteq V\left(C{F}_n^i\right)$, $C{F}_n^i\cong C{F}_{n-1}$, $n-1$ is even and $|N_{C{F}_n^i}\left(S\right)|=\frac{9n-33}{2}$, by Corollary 4.6, we know $x_1,x_2,x_3$ must belong to a common subgraph in $C{F}_n^i$. So we let $x_1=i_1{i}_2{i}_3\cdots i_{n-2}{i}_{n-1}i$, $x_2=j_1{j}_2{j}_3\cdots j_{n-2}{i}_{n-1}i$, $x_3=k_1{k}_2{k}_3\cdots k_{n-2}{i}_{n-1}i$. Then $x_1^{+}=i{i}_2{i}_3\cdots i_{n-2}{i}_{n-1}{i}_1$, $x_1^{-}=i_1{i}_2{i}_3\cdots i_{n-2}i{i}_{n-1}$, $x_2^{+}=i{j}_2{j}_3\cdots j_{n-2}{i}_{n-1}{j}_1$, $x_2^{-}=j_1{j}_2{j}_3\cdots j_{n-2}i{i}_{n-1}$, $x_3^{+}=i{k}_2{k}_3\cdots k_{n-2}{i}_{n-1}{k}_1$, $x_3^{-}=k_1{k}_2{k}_3\cdots k_{n-2}i{i}_{n-1}$. If $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_1,x_2,x_3\}\right)|\ge 4$, then $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$. Thus we assume $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(\{x_1,x_2,x_3\}\right)|=3$, then one of the two outgoing neighbors of $x_1,x_2,x_3$ must belong to a common subgraph of $C{F}_n$, we need to think about the following two situations:

Case 1: $\{x_1^{-},x_2^{-},x_3^{-}\}\subseteq V\left(C{F}_n^{i_{n-1}}\right)$ and $x_4\in V\left(C{F}_n^{i_{n-1}}\right)$.

In this case, we let $x_4=l_1{l}_2{l}_3\cdots l_{n-2}{l}_{n-1}{i}_{n-1}$, then $x_4^{+}=i_{n-1}{l}_2{l}_3\cdots l_{n-2}{l}_{n-1}{l}_1$, $x_4^{-}=l_1{l}_2{l}_3\cdots l_{n-2}{i}_{n-1}{l}_{n-1}$. If $x_4^{+}\notin V\left(C{F}_n^i\right)$ and $x_4^{-}\notin V\left(C{F}_n^i\right)$, then $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$ as $i_{n-1}\ne i$ and $x_4^{+}\ne x_1^{+}$, $x_4^{+}\ne x_2^{+}$, $x_4^{+}\ne x_3^{+}$. If one of the two outgoing neighbors of $x_4$ belong to $C{F}_n^i$, we can assume $x_4=l_1{l}_2{l}_3\cdots l_{n-2}i{i}_{n-1}$ or $x_4=i{l}_2{l}_3\cdots l_{n-2}{l}_{n-1}{i}_{n-1}$. If $x_4=i{l}_2{l}_3\cdots l_{n-2}{l}_{n-1}{i}_{n-1}$, since $|N_{C{F}_n^j}\left(S\right)|=\frac{3n-7}{2}$, we know $x_1^{-},x_2^{-},x_3^{-}$ are adjacent to $x_4$, so $x_1^{-}=x_4(1,n-1)$, $x_2^{-}=x_4(1,n-1)$, $x_3^{-}=x_4(1,n-1)$, this contradicts to the fact that $x_1^{-},x_2^{-},x_3^{-}$ are different from each other. So $x_4=l_1{l}_2{l}_3\cdots l_{n-2}i{i}_{n-1}$, then $x_4^{+}=i_{n-1}{l}_2{l}_3\cdots l_{n-2}i{l}_1$, $x_4^{-}=l_1{l}_2{l}_3\cdots l_{n-2}{i}_{n-1}i$. Hence $x_4^{-}\in V\left(C{F}_n^i\right)$, $x_4^{+}\ne x_1^{+}$, $x_4^{+}\ne x_2^{+}$, $x_4^{+}\ne x_3^{+}$. Thus $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$.

Case 2: Let $i_1=j_1=k_1$, $\{x_1^{+},x_2^{+},x_3^{+}\}\subseteq V\left(C{F}_n^{i_1}\right)$ and $x_4\in V\left(C{F}_n^{i_1}\right)$.

In this case, $x_1=i_1{i}_2{i}_3\cdots i_{n-2}{i}_{n-1}i$, $x_2=i_1{j}_2{j}_3\cdots j_{n-2}{i}_{n-1}i$, $x_3=i_1{k}_2{k}_3\cdots k_{n-2}{i}_{n-1}i$. Then $x_1^{+}=i{i}_2{i}_3\cdots i_{n-2}{i}_{n-1}{i}_1$, $x_2^{+}=i{j}_2{j}_3\cdots j_{n-2}{i}_{n-1}{i}_1$, $x_3^{+}=i{k}_2{k}_3\cdots k_{n-2}{i}_{n-1}{i}_1$, $x_1^{-}=i_1{i}_2{i}_3\cdots i_{n-2}i{i}_{n-1}$, $x_2^{-}=i_1{j}_2{j}_3\cdots j_{n-2}i{i}_{n-1}$, $x_3^{-}=i_1{k}_2{k}_3\cdots k_{n-2}i{i}_{n-1}$. We let $x_4=l_1{l}_2{l}_3\cdots l_{n-2}$ $l_{n-1}{i}_1$, then $x_4^{+}=i_1{l}_2{l}_3\cdots l_{n-2}{l}_{n-1}{l}_1$, $x_4^{-}=l_1{l}_2{l}_3\cdots l_{n-2}{i}_1{l}_{n-1}$. If $x_4^{+}\notin V\left(C{F}_n^i\right)$ and $x_4^{-}\notin V\left(C{F}_n^i\right)$, then $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$ as $x_4^{-}\ne x_1^{-}$, $x_4^{-}\ne x_2^{-}$, $x_4^{-}\ne x_3^{-}$. If one of the two outgoing neighbors of $x_4$ belong to $C{F}_n^i$, we can assume $x_4=l_1{l}_2{l}_3\cdots i{i}_1$ or $x_4=i{l}_2{l}_3\cdots l_{n-1}{i}_1$. When $x_4=l_1{l}_2{l}_3\cdots i{i}_1$, as $x_1^{+},x_2^{+},x_3^{+}$ are adjacent to $x_4$, so $x_1^{+}=x_4(1,n-1)$, $x_2^{+}=x_4(1,n-1)$, $x_3^{+}=x_4(1,n-1)$, this contracts to the fact $x_1^{+},x_2^{+},x_3^{+}$ are different from each other. So $x_4=i{l}_2{l}_3\cdots l_{n-1}{i}_1$, $x_4^{+}=i_1{l}_2{l}_3\cdots l_{n-1}i$, $x_4^{-}=i{l}_2{l}_3\cdots i_1{l}_{n-1}$. Then $x_4^{+}\in V\left(C{F}_n^i\right)$, $x_4^{-}\ne x_1^{-}$, $x_4^{-}\ne x_2^{-}$, $x_4^{-}\ne x_3^{-}$. Thus $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$.

Lemma 5.4. When n is odd, let $S=\{x_1,x_2,x_3,x_4\}$ is an $Ind$-set and $\{x_1,x_2\}\subseteq V\left(C{F}_n^i\right)$, $\left\{x_3\right\}\subseteq V\left(C{F}_n^j\right)$, $\left\{x_3\right\}\subseteq V\left(C{F}_n^k\right)$$(i,j,k$ are different from each other). If $|N_{C{F}_n^i}(\{x_1,x_2\}$ $\left)\right|=3n-10$, then $|N_{C{F}_n-C{F}_n^i-C{F}_n^j-C{F}_n^k}\left(S\right)|\ge 1$.

Proof. If $|N_{C{F}_n-C{F}_n^i-C{F}_n^j-C{F}_n^k}\left(S\right)|=0$, the structure in Figure 5 must exists. Now we can proof this structure does not exist. As $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|=3n-10$, we can get $x_1,x_2$ must have three common neighbors in $C{F}_n^i$. So we can assume $x_1=i_1{i}_2{i}_3\cdots i_{n-2}ji$, then by Lemma 3.1, we can let $x_2=j_1{j}_2{j}_3\cdots j_{n-2}ji$, where $j_i\in [i_1,i_{n-2}]$ and $j_1\ne i_1$. Then $x_1^{+}=i{i}_2{i}_3\cdots i_{n-2}j{i}_1$, $x_2^{+}=i{j}_2{j}_3\cdots j_{n-2}j{j}_1$, $x_1^{-}=i_1{i}_2{i}_3\cdots i_{n-2}ij$, $x_2^{-}=j_1{j}_2{j}_3\cdots j_{n-2}ij$. Since $i_1\ne j$ and $i_1\ne j_1$, $x_1^{+}$ can not belong to a common subgraph with $x_2^{+}$ or $x_2^{-}$. Thus the structure in Figure 5 does not exist, $|N_{C{F}_n-C{F}_n^i-C{F}_n^j-C{F}_n^k}\left(S\right)|\ge 1$.

Lemma 5.5. For $n=5$, let S is an $Ind$-set and $\left|S\right|=4$, then $|N_{C{F}_5}\left(S\right)|\ge 14$.

Proof. Let $S=\{x_1,x_2,x_3,x_4\}$, since S is an $Ind$-set, $x_1,x_2,x_3,x_4$ are nonadjacent to each other. Note that $C{F}_5^i\cong C{F}_4$. Now we think about the following four cases:

Case 1: $S=\{x_1,x_2,x_3,x_4\}$ belong to a same subgraph $C{F}_5^i$.

By Lemma 5.1, we have $|N_{C{F}_5^i}\left(S\right)|\ge 8$. By Proposition 2.3, we know $|N_{C{F}_5-C{F}_5^i}\left(S\right)|=8$. Thus $|N_{C{F}_5}\left(S\right)|=|N_{C{F}_5^i}\left(S\right)|+|N_{C{F}_5-C{F}_5^i}\left(S\right)|\ge 8+8=16$.

Case 2: $S=\{x_1,x_2,x_3,x_4\}$ belong to two different subgraphs $C{F}_5^i$, $C{F}_5^j$$(i\ne j)$.

In this case, we need to think about the following two situations:

Subcase 2.1: $\{x_1,x_2\}\subseteq V\left(C{F}_5^i\right)$, $\{x_3,x_4\}\subseteq V\left(C{F}_5^j\right)$.

By Lemma 3.3, we can get $|N_{C{F}_5^i}\left(\{x_1,x_2\}\right)|\ge 3n-10=5$, $|N_{C{F}_5^j}\left(\{x_3,x_4\}\right)|\ge 3n-10=5$. By Lemma 5.2, we have $|N_{C{F}_5}\left(S\right)|=|N_{C{F}_5^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_5^j}\left(\{x_3,x_4\}\right)|+|N_{C{F}_5-C{F}_5^i-C{F}_5^j}\left(S\right)|\ge 2\times 5+4=14$.

Subcase 2.2: $\{x_1,x_2,x_3\}\subseteq V\left(C{F}_5^i\right)$, $x_4\in V\left(C{F}_5^j\right)$.

By Lemma 4.1, we have $|N_{C{F}_5^i}\left(\{x_1,x_2,x_3\}\right)|\ge 6$, $|N_{C{F}_5^j}\left(x_4\right)|=\frac{3\times 4-4}{2}=4$. By Lemma 5.3, we know $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$. Thus $|N_{C{F}_n}\left(S\right)|=|N_{C{F}_5^i}\left(\{x_1,x_2,x_3\}\right)|+|N_{C{F}_5^j}\left(x_4\right)|+|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 6+4+4=14$.

Case 3: $S=\{x_1,x_2,x_3,x_4\}$ belong to three different subgraphs $C{F}_5^i$, $C{F}_5^j$, $C{F}_5^k$$(i,j,k$ are different from each other).

In this case, there exists a subgraph $C{F}_5^i$, which contains two vertices of S, we let $\{x_1,x_2\}\subseteq V\left(C{F}_5^i\right)$. Clearly, $|N_{C{F}_5^i}\left(\{x_1,x_2\}\right)|\ge 3n-10=5$, $|N_{C{F}_5^j}\left(x_3\right)|=|N_{C{F}_5^k}\left(x_4\right)|=4$. Thus, by Lemma 5.4, we have $|N_{C{F}_5}\left(S\right)|=|N_{C{F}_5^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_5^j}\left(x_3\right)|+|N_{C{F}_5^k}\left(x_4\right)|+|N_{C{F}_5-C{F}_5^i-C{F}_5^j-C{F}_5^k}$ $\left(S\right)|\ge 5+4+4+1=14$.

Case 4: $S=\{x_1,x_2,x_3,x_4\}$ belong to four different subgraphs.

In this case, we can let $x_k\in V\left(C{F}_5^k\right)$. Clearly, $|N_{C{F}_5^k}\left(x_k\right)|=4$, thus $|N_{C{F}_5}\left(S\right)|\ge 4\times |N_{C{F}_5^k}\left(x_k\right)|=4\times 4=16>14$.

Combing the above four cases, we can get $|N_{C{F}_5}\left(S\right)|\ge 14$.

Lemma 5.6. For $n=6$, let S is an $Ind$-set and $\left|S\right|=4$, then $|N_{C{F}_6}\left(S\right)|\ge 18$.

Proof. Let $S=\{x_1,x_2,x_3,x_4\}$, since S is an $Ind$-set, $x_1,x_2,x_3,x_4$ are nonadjacent to each other. Note that $C{F}_6^i\cong C{F}_5$. Now we think about the following four cases:

Case 1: $S=\{x_1,x_2,x_3,x_4\}$ belong to a same subgraph $C{F}_6^i$.

By Lemma $5.5$, we can get $|N_{C{F}_6^i}\left(S\right)|\ge 14$. By the definition of $C{F}_n$, we know every vertex in $C{F}_6$ has only one outgoing neighbor. Thus $|N_{C{F}_6}\left(S\right)|=|N_{C{F}_6^i}\left(S\right)|+4\ge 18$.

Case 2: $S=\{x_1,x_2,x_3,x_4\}$ belong to two different subgraphs $C{F}_6^i$, $C{F}_6^j$$(i\ne j)$.

In this case, we also need to think about the following two situations:

Subcase 2.1: $\{x_1,x_2\}\subseteq V\left(C{F}_6^i\right)$, $\{x_3,x_4\}\subseteq V\left(C{F}_6^j\right)$.

By Lemma 3.3, we can get $|N_{C{F}_6^i}\left(\{x_1,x_2\}\right)|\ge 3n-9=9$, $|N_{C{F}_6^j}\left(\{x_3,x_4\}\right)|\ge 3n-9=9$. Thus $|N_{C{F}_6}\left(S\right)|\ge |N_{C{F}_6^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_6^j}\left(\{x_3,x_4\}\right)|=9+9=18$.

Subcase 2.2: $\{x_1,x_2,x_3\}\subseteq V\left(C{F}_6^i\right)$, $x_4\in V\left(C{F}_6^j\right)$.

By Lemma 4.2, we have $|N_{C{F}_6^i}\left(\{x_1,x_2,x_3\}\right)|\ge 12$, $|N_{C{F}_6^j}\left(x_4\right)|=\frac{3\times 5-3}{2}=6$. Thus $|N_{C{F}_6}\left(S\right)|\ge |N_{C{F}_6^i}\left(\{x_1,x_2,x_3\}\right)|+|N_{C{F}_6^j}\left(x_4\right)|=12+6=18$.

Case 3: $S=\{x_1,x_2,x_3,x_4\}$ belong to three different subgraphs $C{F}_6^i$, $C{F}_6^j$, $C{F}_6^k$$(i,j,k$ are different from each other).

In this case, there exists a subgraph $C{F}_6^i$, which contains two vertices of S, we let $\{x_1,x_2\}\subseteq V\left(C{F}_6^i\right)$. Clearly, $|N_{C{F}_6^i}\left(\{x_1,x_2\}\right)|\ge 3n-9=9$, $|N_{C{F}_6^j}\left(x_3\right)|=|N_{C{F}_6^k}\left(x_4\right)|=6$. Thus, $|N_{C{F}_6}\left(S\right)|\ge |N_{C{F}_6^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_6^j}\left(x_3\right)|+|N_{C{F}_6^k}\left(x_4\right)|=9+6+6=21>18$.

Case 4: $S=\{x_1,x_2,x_3,x_4\}$ belong to four different subgraphs.

In this case, we can let $x_k\in V\left(C{F}_6^k\right)$. Clearly, $|N_{C{F}_6^k}\left(x_k\right)|=6$, thus $|N_{C{F}_6}\left(S\right)|\ge 4\times |N_{C{F}_6^k}\left(x_k\right)|=4\times 6=24>18$.

Combing the above four cases, we can get $|N_{C{F}_6}\left(S\right)|\ge 18$.

Lemma 5.7. For $n\ge 5$, let S is an $Ind$-set and $\left|S\right|=4$, then when n is odd, $|N_{C{F}_n}\left(S\right)|\ge 6n-16$; when n is even, $|N_{C{F}_n}\left(S\right)|\ge 6n-18$.

Proof. We proof this result by induction on n. By Lemma 5.5 and Lemma 5.6, we know when $n=5,6$, this result holds. Now we assume $n\ge 7$ and the result holds for $C{F}_{n-1}$. Let $S=\{x_1,x_2,x_3,x_4\}$, since S is an $Ind$-set, $x_1,x_2,x_3,x_4$ are nonadjacent to each other. Note that $C{F}_n^i\cong C{F}_{n-1}$. Now we think about the following four cases:

Case 1: $S=\{x_1,x_2,x_3,x_4\}$ belong to a same subgraph $C{F}_n^i$.

By induction hypothesis, we know when n is odd, $|N_{C{F}_n^i}\left(S\right)|\ge 6(n-1)-18=6n-24$; when n is even, $|N_{C{F}_n^i}\left(S\right)|\ge 6(n-1)-16=6n-22$. By Proposition 2.3, we know when n is odd, $|N_{C{F}_n-C{F}_n^i}\left(S\right)|=8$; when n is even, $|N_{C{F}_n-C{F}_n^i}\left(S\right)|=4$. Thus $|N_{C{F}_n}\left(S\right)|=|N_{C{F}_n^i}\left(S\right)|+|N_{C{F}_n-C{F}_n^i}\left(S\right)|\ge 6n-24+8=6n-16$(n is odd) and $|N_{C{F}_n}\left(S\right)|\ge 6n-22+4=6n-18$(n is even).

Case 2: $S=\{x_1,x_2,x_3,x_4\}$ belong to two different subgraphs $C{F}_n^i$, $C{F}_n^j$$(i\ne j)$.

In this case, we need to think about two situations:

Subcase 2.1: $\{x_1,x_2\}\subseteq V\left(C{F}_n^i\right)$, $\{x_3,x_4\}\subseteq V\left(C{F}_n^j\right)$.

By Lemma 3.3, we can get when n is odd, $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|$$\ge 3n-10$, $|N_{C{F}_n^j}\left(\{x_3,x_4\}\right)|\ge 3n-10$; when n is even, $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|\ge 3n-9$, $|N_{C{F}_n^j}\left(\{x_3,x_4\}\right)|\ge 3n-9$. When n is odd, by Lemma 5.2, we have $|N_{C{F}_n}\left(S\right)|=|N_{C{F}_n^i}$$\left(\{x_1,x_2\}\right)|+|N_{C{F}_n^j}\left(\{x_3,x_4\}\right)|+|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 2\times (3n-10)+4=6n-16$. When n is even, $|N_{C{F}_n}\left(S\right)|\ge |N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_n^j}\left(\{x_3,x_4\}\right)|=2\times (3n-9)=6n-18$.

Subcase 2.2: $\{x_1,x_2,x_3\}\subseteq V\left(C{F}_n^i\right)$, $x_4\in V\left(C{F}_n^j\right)$.

When n is odd, by Lemma 4.5, we have $|N_{C{F}_n^i}\left(\{x_1,x_2,x_3\}\right)|\ge \frac{9(n-1)-24}{2}=\frac{9n-33}{2}$. By Lemma 5.3, we know $|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge 4$. As $|N_{C{F}_n^j}\left(x_4\right)|=\frac{3\times (n-1)-4}{2}=\frac{3n-7}{2}$, so $|N_{C{F}_n}\left(S\right)|=|N_{C{F}_n^i}\left(\{x_1,x_2,x_3\}\right)|+|N_{C{F}_n^j}\left(x_4\right)|+|N_{C{F}_n-C{F}_n^i-C{F}_n^j}\left(S\right)|\ge \frac{9n-33}{2}+\frac{3n-7}{2}+4=6n-16$. When n is even, by Lemma 4.5, we have $|N_{C{F}_n^i}\left(\{x_1,x_2,x_3\}\right)|\ge \frac{9(n-1)-21}{2}=\frac{9n-30}{2}$. As $|N_{C{F}_n^j}\left(x_4\right)|=\frac{3\times (n-1)-3}{2}=\frac{3n-6}{2}$, so $|N_{C{F}_n}\left(S\right)|\ge |N_{C{F}_n^i}\left(\{x_1,x_2,x_3\}\right)|+|N_{C{F}_n^j}\left(x_4\right)|=\frac{9n-30}{2}+\frac{3n-6}{2}=6n-18$.

Case 3: $S=\{x_1,x_2,x_3,x_4\}$ belong to three different subgraphs $C{F}_n^i$, $C{F}_n^j$, $C{F}_n^k$$(i,j,k$ are different from each other).

In this case, there exists a subgraph $C{F}_n^i$, which contains two vertices of S, we let $\{x_1,x_2\}\subseteq V\left(C{F}_n^i\right)$. When n is odd, by Lemma 3.3, we have $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|\ge 3n-10$. Clearly, $|N_{C{F}_n^j}\left(x_3\right)|=|N_{C{F}_n^k}\left(x_4\right)|=\frac{3n-7}{2}$. Thus, by Lemma 5.4, we have $|N_{C{F}_n}\left(S\right)|=|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_n^j}\left(x_3\right)|+|N_{C{F}_n^k}\left(x_4\right)|+|N_{C{F}_n-C{F}_n^i-C{F}_n^j-C{F}_n^k}\left(S\right)|\ge 3n-10+2\times \frac{3n-7}{2}+1=6n-16$. When n is even, by Lemma 3.3, we have $|N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|\ge 3n-9$. Clearly, $|N_{C{F}_n^j}\left(x_3\right)|=|N_{C{F}_n^k}\left(x_4\right)|=\frac{3n-6}{2}$. Thus, $|N_{C{F}_n}\left(S\right)|\ge |N_{C{F}_n^i}\left(\{x_1,x_2\}\right)|+|N_{C{F}_n^j}\left(x_3\right)|+|N_{C{F}_n^k}\left(x_4\right)|=3n-9+2\times \frac{3n-6}{2}=6n-15>6n-18$.

Case 4: $S=\{x_1,x_2,x_3,x_4\}$ belong to four different subgraphs.

In this case, we can let $x_i\in V\left(C{F}_n^i\right)$. Clearly, when n is odd, $|N_{C{F}_n^i}\left(x_i\right)|=\frac{3n-7}{2}$; when n is even, $|N_{C{F}_n^i}\left(x_i\right)|=\frac{3n-6}{2}$. Thus when n is odd, $|N_{C{F}_n}\left(S\right)|\ge 4\times |N_{C{F}_n^i}\left(x_i\right)|=4\times \frac{3n-7}{2}=6n-14>6n-16$; when n is even, $|N_{C{F}_n}\left(S\right)|\ge 4\times |N_{C{F}_n^i}\left(x_i\right)|=4\times \frac{3n-6}{2}=6n-12>6n-18$.

Combing the above four cases, we know the result holds.

Lemma 5.8. For $n=5$, if F satisfies the condition $\left|F\right|\le 13$, then $C{F}_5-F$ contains a big component C with $\left|V\right(C\left)\right|\ge 5!-\left|F\right|-3$.

Proof. In this Lemma, we do not think about the situation $C{F}_5-F$ is connected, so we let $C{F}_5-F$ is disconnected. Let $F_i=F\cap C{F}_n^i$$(i\in [1,5\left]\right)$. By Proposition 2.8, we know $\kappa \left(C{F}_5^i\right)=\frac{3(n-1)-4}{2}=\frac{3\times 4-4}{2}=4$. Since $\left|F\right|\le 14$, we can get there exists at most three vertex set $F_i$, which can satisfies the condition $|F_i|\ge 4$. Now we think about the following situations:

Case 1: $|F_i|\le 3$ for every $i\in [1,5]$.

By Proposition 2.8, we know $C{F}_5^i-F_i$ is connected for $i\in [1,5]$. Since $E_{i,j}\left(C{F}_5\right)=2(n-2)!=12>6\ge |F_i|+|F_j|$, $C{F}_5^i$ and $C{F}_5^j$ are connected. Thus we can get $C{F}_5-F$ is connected, this contradicts to the assumption $C{F}_5-F$ is disconnected.

Case 2: There exists only one $F_i$, which can satisfies that $|F_i|\ge 4$.

In this case, we can let $|F_1|\ge 4$, then $|F_i|\le 3$ for $i\in [2,5]$. Hence by Proposition 2.8, we can get $C{F}_5^i-F_i$ is connected $(i\in [2,5\left]\right)$. Let $M=C{F}_5-F-C{F}_5^1$, similarly to the discussion of Case 1, we know M is connected. Now we think about the following two situations:

Subcase 2.1: $|F-F_1|\le 7$.

By the definition of $C{F}_n$ and Proposition 2.3, we know every vertex has two outgoing neighbors, and these outgoing neighbors are different from each other. Thus, if $C{F}_5^1-F_1$ is connected and $|C{F}_5^1-F_1|\ge 4$, there must exists a vertex $x_1$ in $C{F}_5^1-F_1$ such that it has a good neighbor in M. Thus $C{F}_5-F$ is connected, this contradicts to the assumption $C{F}_5-F$ is disconnected. If $C{F}_5^1-F_1$ is connected and $|C{F}_5^1-F_1|\le 3$, the result is certainly true. If $C{F}_5^1-F_1$ is disconnected, we can assume $C_1$ is the vertex set in $C{F}_5^1-F_1$, which has no good neighbors in M. As $|F-F_1|\le 7$, we can get $\left|V\right(C_1\left)\right|\le 3$, so the result holds.

Subcase 2.2: $|F-F_1|\ge 8$.

In this case, if $C{F}_5^1-F_1$ is connected, similar to the case 1, we can get $C{F}_5-F$ is connected. Now we assume $C{F}_5^1-F_1$ is disconnected. Since $|F-F_1|\ge 8$, we have $|F_1|\le |F|-8\le 13-8=5$, by Lemma 1, we know $C{F}_5^1-F_1$ has a big component $C_1$ and one singleton or two singletons. Since $|V\left(C_1\right)|\ge |V(C{F}_5^1-F_1)|-2\ge 4!-5-2=17>13>|F_2\cup F_3\cup F_4\cup F_5|$, $C_1$ is connected to M. Thus $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-2$.

Case 3: There exists two vertex set $F_i$, $F_j$, which satisfy that $|F_i|\ge 4$, $|F_j|\ge 4$.

In this case, we can let $|F_1|\ge |F_2|\ge 4$. Then $|F_i|\le 3$$(i\in [3,5\left]\right)$, $|F_1|\le |F|-|F_2|\le 13-4=9$. Let $M=C{F}_5-F-C{F}_5^1-C{F}_5^2$, similarly, we can get M is connected. If $C{F}_5^i-F_i$$(i\in \{1,2\left\}\right)$ is connected, by the same argument with Subcase 2.2, we know it connected to M. Thus, if $C{F}_5^1-F_1$ and $C{F}_5^2-F_2$ are all connected, then $C{F}_5-F$ is connected, this contradicts to the assumption that $C{F}_5-F$ is disconnected. Hence at least one of $C{F}_5^i-F_i$$(i\in \{1,2\left\}\right)$ is disconnected. Now we think about the following three cases:

Subcase 3.1: $|F_2|=|F_1|=4$.

By Corollary 3.4, we know if $C{F}_5^i-F_i$$(i\in \{1,2\left\}\right)$ is disconnected, then it has a big component $C_i$ and one singleton. Similarly, we can get $C_i$ is connected to M. Thus $C{F}_5-F$ has a component C with $\left|V\right(C\left)\right|\ge 5!-\left|F\right|-2$.

Subcase 3.2: $|F_1|=5$.

In this case, we know $4\le |F_2|\le |F_1|=5$. By Lemma 1, we can get if $C{F}_5^1-F_1$ is disconnected, then it has a big component $C_1$ and one singleton or two singletons. If $|F_2|=4$, by Corollary 3.4, we can get $C{F}_5^2-F_2$ has a big component $C_2$ and at most one singleton. Similarly, we can also get $C_i$ is connected to M. Thus $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$, the result holds. If $|F_2|=5$, then $|F-F_1-F_2|\le 13-5-5=3$. By Lemma 1, we know, if $C{F}_5^i-F_i$$(i\in \{1,2\left\}\right)$ is disconnected, it has a big component $C_i$ and one singleton or two singletons. If one of $C{F}_5^i-F_i$ is connected or has only one singleton, then the result holds. Now we consider $C{F}_5^1-F_1$ and $C{F}_5^2-F_2$ are all disconnected and they all have two singletons, we let $\{x_1,x_2\}\subseteq V(C{F}_5^1-F_1)$, $\{y_1,y_2\}\subseteq V(C{F}_5^2-F_2)$. Similarly, we can know that $C_i$ must connected to M, then $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge 5!-\left|F\right|-4$. Since $|F_1|=|F_2|=5$ and $x_1,x_2$ (resp.,$y_1,y_2$) are two singletons, $x_1,x_2$ must have three common neighbors in $C{F}_5^1$ (resp.,$y_1,y_2$ must have three common neighbors in $C{F}_5^2)$. If $x_1,x_2,y_1,y_2$ are singletons in $C{F}_5-F$, then by Lemma 5.2, we can get $|N_{C{F}_5-C{F}_5^1-C{F}_5^2}\left(\{x_1,x_2,y_1,y_2\}\right)|\ge 4$. Since $|F-F_1-F_2|\le 3$, we know at least one vertex in $\{x_1,x_2,y_1,y_2\}$ must connected to M. Thus $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge 5!-\left|F\right|-3$. If $(C{F}_5-F)\left[\{x_1,x_2,y_1,y_2\}\right]$ has at least one edge, we assume $(x_2,y_1)\in E(C{F}_5-F)$, then $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge 5!-\left|F\right|-3$. Suppose $\left|V\right(C\left)\right|=5!-\left|F\right|-4$, then $x_1,y_2$ are singletons in $C{F}_5-F$ or $(x_1,y_2)\in E(C{F}_5-F)$. If $x_1,y_2$ are singletons in $C{F}_5-F$, then $\left|F\right|\ge |N_{C{F}_5}\left(\{x_2,y_1\}\right)\cup N_{C{F}_5}\left(x_1\right)\cup N_{C{F}_5}\left(y_2\right)|\ge 5\times 2+6\times 2-3\times 2=16>13$, a contradiction. If $(x_1,y_2)\in E(C{F}_5-F)$, then $\left|F\right|\ge |N_{C{F}_5}\left(\{x_2,y_1\}\right)\cup N_{C{F}_5}\left(\{x_1,y_2\}\right)\ge 5\times 4-3\times 2=14>13$, a contradiction. Thus $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$.

Subcase 3.3: $6\le |F_1|\le 9$.

In this case, we can get $4\le |F_2|\le |F|-|F_1|\le 13-6=7$. If $|F_2|=7$, then $|F_1|\ge 7$, $\left|F\right|\ge |F_1|+|F_2|\ge 14$, a contradiction. Thus $4\le |F_2|\le 6$.

If $|F_2|=4$ and $C{F}_5^2-F_2$ is disconnected, then $C{F}_5^2-F_2$ has a big component $C_2$ and one singleton $x_4$. Furthermore, we have $\left|F\right|-|F_1|-|F_2|\le 13-6-4=3$. When $\left|F\right|-|F_1|-|F_2|\le 2$, since every vertex in $C{F}_5$ has two outgoing neighbors, there are at most two vertices in $C{F}_5^1-F_1$, which can satisfy that one of the two outgoing neighbors belong to $F_2$ and the other belongs to $F-F_1-F_2$. Thus the result holds. When $\left|F\right|-|F_1|-|F_2|=3$, $C{F}_5^1-F_1$ has at most three vertices, which can satisfy that one of their outgoing neighbors belongs to $F_2$ and the other belongs to $F-F_1-F_2$. We let they are $x_1,x_2,x_3$. If $C{F}_5^1-F_1$ is connected or has at most two vertices, then the result holds. Now we consider there are three vertices in $C{F}_5^1-F_1$, we can get $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge 5!-\left|F\right|-4$. If $\left|V\right(C\left)\right|=5!-\left|F\right|-4$, the structure in Figure 6 must exists. If there exists one edge between $\{x_1,x_2,x_3\}$, then there will be a 5-circle in $C{F}_5$, a contradiction. So $x_1,x_2,x_3$ are three singletons in $C{F}_5^1-F_1$. If $x_4$ is adjacent to $x_1,x_2,x_3$, then there exists a 3-circle in $C{F}_5$. Thus $x_1,x_2,x_3,x_4$ are four singletons in $C{F}_5-F$. By Lemma 5.3, we know $|N_{C{F}_5-C{F}_5^1-C{F}_5^2}\left(S\right)|\ge 4$, this contradicts to the fact $\left|F\right|-|F_1|-|F_2|=3$, thus $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$.

If $5\le |F_2|\le 6$, then $\left|F\right|-|F_1|-|F_2|\le 13-5-6\le 2$. When $\left|F\right|-|F_1|-|F_2|\le 1$, $C{F}_5-F$ has a big component and at most two vertices $x_1,x_2$, where $x_i\in V(C{F}_5^i-F_i)$ and they have common outgoing neighbor vertex in $F\setminus (F_1\cup F_2)$ and the other outgoing neighbor vertex belong to $C{F}_5^2$ or $C{F}_5^1$, so the result holds. When $\left|F\right|-|F_1|-|F_2|=2$, we can get $|F_1|=6$, $|F_2|=5$, and if $C{F}_5^i-F_i$$(i\in \{1,2\left\}\right)$ is disconnected, then it has at most two vertices which has neighbors in $F-F_1-F_2$. Thus $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge 5!-\left|F\right|-4$. Now, we proof $\left|V\right(C\left)\right|=5!-\left|F\right|-4$ can not exist. Suppose on the contrary, the structure in Figure 7 exists. Since $|F_2|=5$, we can get $x_1,x_2$ are two singletons and have three common neighbors in $C{F}_5^2$; Otherwise, if $x_1$ is adjacent to $x_2$, then $|F_2|=6$, a contradiction. So by Lemma 3.1, we let $x_1=i_1{i}_2{i}_3{i}_{4}2$ and $x_2=j_1{j}_2{j}_3{i}_{4}2$, where $j_i\in [i_1,i_3]$ and $j_1\ne i_1$. Then $x_1^{+}=2i_2{i}_3{i}_4{i}_1$, $x_1^{-}=i_1{i}_2{i}_{3}2i_4$, $x_2^{+}=2j_2{j}_3{i}_4{j}_1$, $x_2^{-}=j_1{j}_2{j}_{3}2i_4$. Since one of the out neighbor vertices of $x_1,x_2$ belong to $C{F}_5^1$ and $j_1\ne i_1$, $i_1\ne i_4$, $j_1\ne i_4$, we have $i_4=1$. Thus $x_1^{+}=2i_2{i}_{3}1i_1$, $x_2^{+}=2j_2{j}_{3}1j_1$. From the structure of Figure 7, we have ${\left(x_1^{+}\right)}^{-}=2i_2{i}_3{i}_{1}1=x_3$, ${\left(x_2^{+}\right)}^{-}=2j_2{j}_3{j}_{1}1=x_4$. Let $x_3^{\prime }=2i_2{i}_3{i}_1$, $x_4^{\prime }=2j_2{j}_3{j}_1$, then $\{x_3^{\prime },x_4^{\prime }\}\subseteq V\left(G_1\right)$, $G_1\cong C{F}_4$. Since $j_1\ne i_1$, $x_4^{\prime }$ and $x_3^{\prime }$ belong to different subgraph in $G_1$. As ${\left(x_3^{\prime }\right)}^{+}=i_1{i}_2{i}_{3}2$, ${\left(x_4^{\prime }\right)}^{+}=j_1{j}_2{j}_{3}2$, we know ${\left(x_3^{\prime }\right)}^{+}\ne {\left(x_4^{\prime }\right)}^{+}$. Thus $x_3$ and $x_4$ have no common neighbors in $C{F}_5^1$. Clearly, we have $x_3$ is nonadjacent to $x_4$; Otherwise, there is a 7-circle in $C{F}_5$(as shown in Figure 7 by read line). Thus, $|F_1|\ge 8$, this contradicts to the fact $|F_1|=6$, this structure does not exist. So $C{F}_5-F$ has a large component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$.

Case 4: There exists three vertex set $F_i$, $F_j$, $F_k$, which can satisfy that $|F_i|\ge 4$, $|F_j|\ge 4$ and $|F_k|\ge 4$$(i,j,k$ are different from each other).

In this case, we have $|F_i|\le |F|-|F_j|-|F_k|\le 13-4-4=5$. Similarly, we can get $|F_j|\le 5$, $|F_k|\le 5$. Let $M=C{F}_5-C{F}_5^i-C{F}_5^j-C{F}_5^k$, we can get M is connected. If $|F_i|\le 4$, $|F_j|\le 4$ and $|F_k|\le 4$, by Corollary 3.4, we know this result holds. If $|F_i|=5$, $|F_j|\le 4$ and $|F_k|\le 4$, then by Lemma 1, there are at most two singletons in $C{F}_5^i-F_i$. Thus $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge 5!-\left|F\right|-4$. If $\left|V\right(C\left)\right|=n!-\left|F\right|-4$, then $C{F}_5^i-F_i$ has two singletons $\{x_1,x_2\}$ and $C{F}_5^j-F_j$, $C{F}_5^k-F_k$ only has one singleton. Since $|F_i|=5$, $x_1$ and $x_2$ have three common neighbors in $C{F}_5^i$. Since $\left|F\right|-|F_i|-|F_j|-|F_k|\le 13-5-4-4=0$, we know $x_1^{+}$ belong to a common subgraph with $x_2^{+}$ or $x_2^{-}$. By the proof process of Lemma 5.4, we know this situation will not exist. So $\left|V\right(C\left)\right|\ne n!-\left|F\right|-4$, $C{F}_5-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$. If $|F_i|=5$, $|F_j|=5$, $|F_k|\le 4$, then $\left|F\right|\ge |F_i|-|F_j|-|F_k|\ge 5+5+4=14$, a contradiction. If there are three $F_i$, $F_j$, $F_k$, such that $|F_i|=5$, $|F_j|=5$, $|F_k|=5$, then $\left|F\right|\ge |F_i|-|F_j|-|F_k|\ge 5+5+5=15$, a contradiction.

Lemma 5.9. For $n=6$, if F satisfies the condition $\left|F\right|\le 17$, then $C{F}_6-F$ contains a big component C with $\left|V\right(C\left)\right|\ge 6!-\left|F\right|-3$.

Proof. In this Lemma, we do not think about $C{F}_6-F$ is connected, so we assume that $C{F}_6-F$ is disconnected. Let $F_i=F\cap C{F}_6^i$, note that $C{F}_6^i\cong C{F}_5$. By Proposition 2.8, we have $\kappa \left(C{F}_6^i\right)=\frac{3n-6}{2}=6$. Since $\left|F\right|\le 17$, we can get there exists at most two vertex set $F_i$, which can satisfy $|F_i|\ge 6$. Next we think about the following three cases:

Case 1: $|F_i|\le 5$ for every $i\in [1,6]$.

In this case, we know $C{F}_6^i-F_i$ is connected. Since there are $(n-2)!=4!=24$ cross-edges in different $C{F}_6^i$ and $24>5+5=10$, $C{F}_6-F$ is connected, a contradiction.

Case 2: There exists only one vertex set $F_i$, which can satisfies the condition $|F_i|\ge 6$.

In this case, we know $|F_j|\le 5$$(j\in [6]\setminus \{i\left\}\right)$. So $C{F}_6^j-F_j$ is connected. Let $M=C{F}_6-F-C{F}_6^i$, similarly, we can get M is connected. When $C{F}_6^i-F_i$ is connected, since $24>17$, $C{F}_6-F$ is connected, this contradicts to the assumption $C{F}_6-F$ is disconnected. When $C{F}_6^i-F_i$ is disconnected, let S be the set of vertices in $C{F}_6^i-F_i$, which have no good neighbors in M. If $\left|F\right|-|F_i|\le 3$, at most three vertices in $C{F}_6^i-F_i$ such that their out neighbor vertex belong to $F-F_i$. Thus $\left|V\right(S\left)\right|\le 3$. If $\left|F\right|-|F_i|\ge 4$, then $|F_i|\le |F|-4=17-4=13$. By Lemma 5.8, we know the result holds.

Case 3: There exists two vertex set $F_i$, $F_j$, which can satisfy that $|F_i|\ge 6$ and $|F_j|\ge 6$.

In this case, we have $|F_i|\le |F|-|F_j|\le 17-6=11$. Similarly, we can get $|F_j|\le 11$. By Lemma 4.7, we know $C{F}_6^i-F_i$ and $C{F}_6^j-F_j$ contain a big component $C_k$$(k\in \{i,j\left\}\right)$ with $|V\left(C_k\right)|\ge 5!-|F_k|-2$. If $6\le |F_i|\le 8$, by Corollary 3.4, we have $C{F}_6^i-F_i$ has a big component $C_i$ with $|V\left(C_i\right)|\ge 5!-|F_i|-1$. Thus $C{F}_6-F$ has a big component C with $\left|V\right(C\left)\right|\ge 6!-\left|F\right|-3$. If $9\le |F_i|\le 11$, then $|F_j|\le |F|-|F_i|\le 17-9=8$. By Corollary 3.4, we can also get $C{F}_6^j-F_j$ has a big component $C_j$ with $|V\left(C_j\right)|\ge 5!-|F_j|-1$. Thus $C{F}_6-F$ has a big component C with $\left|V\right(C\left)\right|\ge 6!-\left|F\right|-3$.

Lemma 5.10. For $n\ge 5$, if F satisfies that $\left|F\right|\le 6n-17$$(n$ is odd) and $\left|F\right|\le 6n-19$$(n$ is even), then $C{F}_n-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$.

Proof. In this Lemma, we only think about the case $C{F}_n-F$ is disconnected. Let $F_i=F\cap C{F}_n^i$, we proof this result by induction on n. By Lemma 5.8 and Lemma 5.9, we know this result holds for $n=5,6$. Now we assume $n\ge 7$ and the result holds for $C{F}_{n-1}$. Note that $C{F}_n^i\cong C{F}_{n-1}$. By Proposition 2.8, we know when n is odd, $\kappa \left(C{F}_n^i\right)=\frac{3n-7}{2}$; when n is even, $\kappa \left(C{F}_n^i\right)=\frac{3n-6}{2}$. Since $\left|F\right|\le 6n-17$(n is odd) and $\left|F\right|\le 6n-19$(n is even), we can get there exists at most three vertex set $F_i$, which can satisfy that $|F_i|\ge \frac{3n-7}{2}$(n is odd) and $|F_i|\ge \frac{3n-6}{2}$(n is even); Otherwise, for $n\ge 7$, when n is odd, $\left|F\right|\ge 4\times \frac{3n-7}{2}=6n-14>6n-17$; when n is even, $\left|F\right|\ge 4\times \frac{3n-6}{2}=6n-12>6n-19$. Next, we will think about the following situations:

Case 1. When n is odd, $|F_i|\le \frac{3n-7}{2}-1=\frac{3n-9}{2}$ for every $i\in [1,n]$; when n is even, $|F_i|\le \frac{3n-6}{2}-1=\frac{3n-8}{2}$ for every $i\in [1,n]$.

By Proposition 2.8, we can get $C{F}_n^i-F_i$ is connected for $i\in [1,n]$. By Proposition 2.1, we know when n is odd, there are $2(n-2)!$ cross-edges between $C{F}_n^i$ and $C{F}_n^j$$(i\ne j)$; when n is even, there are $(n-2)!$ cross-edges between $C{F}_n^i$ and $C{F}_n^j$$(i\ne j)$. Since $n\ge 7$, $2(n-2)!\ge 6n-17$ and $(n-2)!\ge 6n-19$, there are at least one cross-edge between $C{F}_n^i-F_i$ and $C{F}_n^j-F_j$. Thus $C{F}_n-F$ is connected, a contradiction.

Case 2. When n is odd, there exists only one vertex set $F_i$, which satisfies that $|F_i|\ge \frac{3n-7}{2}$; when n is even, there exists only one vertex set $F_i$, which satisfies that $|F_i|\ge \frac{3n-6}{2}$$(i\in [1,n\left]\right)$.

In this case, we can assume $|F_1|\ge \frac{3n-7}{2}$(n is odd) and $|F_1|\ge \frac{3n-6}{2}$(n is even). Then we can get $|F_j|\le \frac{3n-9}{2}$(n is odd) and $|F_j|\le \frac{3n-8}{2}$(n is even) for every $j\in \left[n\right]\setminus \left\{1\right\}$. By Proposition 2.8, we know $C{F}_n^j-F_j$ is connected. Let $M=C{F}_n-(F\cup C{F}_n^1)$, by the same argument with Case 1, we know M is connected. Now we think about the following two subcases:

Subcase 2.1: When n is odd, $|F-F_1|\le 7$; when n is even, $|F-F_1|\le 3$.

By the definition of $C{F}_n$, we know when n is odd, every vertex in $C{F}_n$ has two outgoing neighbors; when n is even, every vertex in $C{F}_n$ has only one outgoing neighbor. If $C{F}_n^1-F_1$ is connected with $|C{F}_n^1-F_1|\ge 4$, since $|F-F_1|\le 7$(n is odd) and $|F-F_1|\le 3$(n is even), we can get $C{F}_n^1-F_1$ is connected to M. Thus we can get $C{F}_n-F$ is connected, this contradicts to the assumption $C{F}_n-F$ is disconnected. If $C{F}_n^1-F_1$ is connected with $|C{F}_n^1-F_1|\le 3$, the conclusion is certainly true. If $C{F}_n^1-F_1$ is disconnected, we can let S is the set of vertices in $C{F}_n^1-F_1$ which has no good neighbors in M. Since $|F-F_1|\le 7$(n is odd) and $|F-F_1|\le 3$(n is even), we can get $\left|V\right(S\left)\right|\le 3$, the result holds.

Subcase 2.2: When n is odd, $|F-F_1|\ge 8$; when n is even, $|F-F_1|\ge 4$.

In this case, since $|V\left(C{F}_n^1\right)-F_1|-|F-F_1|\ge (n-1)!-(6n-17)>1$(n is odd) and $|V\left(C{F}_n^1\right)-F_1|-|F-F_1|\ge (n-1)!-(6n-19)>1$(n is even) for $n\ge 7$, we know if $C{F}_n^1-F_1$ is connected, then $C{F}_n-F$ is also connected, this contradicts to the assumption $C{F}_n-F$ is disconnected. Now we assume $C{F}_n^1-F_1$ is disconnected. Since $|F-F_1|\ge 8$(n is odd) and $|F-F_1|\ge 4$(n is even), we can get when n is odd, $|F_1|\le |F|-8\le 6n-17-8=6n-25=6(n-1)-19$; when n is even, $|F_1|\le |F|-4\le 6n-19-4=6n-23=6(n-1)-17$. Thus by inductive hypothesis, we know $C{F}_n^1-F_1$ has a big component $C_1$ with $|V\left(C_1\right)|\ge (n-1)!-\left|F_1\right|-3$. Next, we will prove that $C_1$ is connected to M. Clearly, we can get $|F-F_1|\le 6n-17-8=6n-25$(n is odd) and $|F-F_1|\le 6n-19-4=6n-23$(n is even). As $2|V\left(C_1\right)|-(6n-25)\ge 2[(n-1)!-|F_1|-3]-(6n-25)\ge 1$(n is odd) and $|V\left(C_1\right)|-(6n-13)\ge (n-1)!-\left|F_1\right|-3-(6n-23)\ge 1$(n is even) for $n\ge 7$, we know $C_1$ must has one vertex which must has a good neighbor in M. So the result holds.

Case 3. When n is odd, there exists two vertex set $F_i$, $F_j$, which can satisfies that $|F_i|\ge \frac{3n-7}{2}$ and $|F_j|\ge \frac{3n-7}{2}$; when n is even, there exists two vertex set $F_i$, $F_j$, which can satisfies that $|F_i|\ge \frac{3n-6}{2}$ and $|F_j|\ge \frac{3n-6}{2}$$(i\ne j)$.

In this case, we know $|F_k|\le \frac{3n-9}{2}$$(n$ is odd) and $|F_k|\le \frac{3n-8}{2}$$(n$ is even), where $k\in \left[n\right]\setminus \{i,j\}$. By Proposition 2.8, we know $C{F}_n^k-F_k$ is connected. Let $M=C{F}_n-(F\cup V\left(C{F}_n^i\right)\cup V\left(C{F}_n^j\right))$, by the same argument with Case 1, we know M is connected.

Firstly, we think about n is odd. Now we can let $|F_1|\ge |F_2|\ge \frac{3n-7}{2}$, then as $\left|F\right|\ge |F_1|+|F_2|$, we can get $|F_1|\le |F|-|F_2|\le 6n-17-\frac{3n-7}{2}=\frac{9n-27}{2}$. Now, we think about the following three cases:

Case 3.1: $\frac{3n-7}{2}\le |F_2|\le |F_1|\le 3n-11$.

In this case, since $\left|F\right|-|F_1|-|F_2|\le 6n-17-(3n-7)=3n-10$, $|V\left(C{F}_n^1\right)-F_1|\ge (n-1)!-(3n-11)$, $|V\left(C{F}_n^2\right)-F_2|\ge (n-1)!-(3n-11)$ and $(n-1)!-(3n-11)-(3n-10)>1$, we can get if $C{F}_n^1-F_1$ or $C{F}_n^2-F_2$ is connected, then they are connected to M. Hence, if $C{F}_n^1-F_1$ and $C{F}_n^2-F_2$ are all connected, then $C{F}_n-F$ is connected, this contradicts to the assumption $C{F}_n-F$ is disconnected. Now we assume $C{F}_n^i-F_i$$(i\in \{1,2\left\}\right)$ is disconnected. Since $C{F}_n^i\cong C{F}_{n-1}$ and $\frac{3n-7}{2}\le |F_2|\le |F_1|\le 3n-11=3(n-1)-8$, by Corollary 3.4, we know $C{F}_n^i-F_i$ has a big component $C_i$ with $|V\left(C_i\right)|\ge (n-1)!-\left|F_i\right|-1$. Since $(n-1)!-(3n-11)-1-(3n-10)>1$ for $n\ge 7$, we know $C_i$ must connected to M. Thus $C{F}_n-F$ contains a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-2$.

Case 3.2: $3n-10\le |F_1|\le \frac{9n-35}{2}$.

In this case, we can get $\frac{3n-7}{2}\le |F_2|\le |F-F_1|\le 6n-17-(3n-10)=3n-7$. If $C{F}_n^1-F_1$ and $C{F}_n^2-F_2$ are all connected, by the similarly discussion with Case 3.1, we can get $C{F}_n-F$ is connected, this contradicts to the assumption $C{F}_n-F$ is disconnected. Thus at least one of $C{F}_n^1-F_1$ and $C{F}_n^2-F_2$ is disconnected. Without loss of generality, we let $C{F}_n^1-F_1$ is disconnected. Since $|F_1|\le \frac{9n-35}{2}=\frac{9(n-1)-26}{2}$, by Lemma 4.9, we can get $C{F}_n^1-F_1$ has a big component $C_1$ with $|V\left(C_1\right)|\ge (n-1)!-\left|F_1\right|-2$. Similarly, we can get $C_1$ is connected to M. If $C{F}_n^2-F_2$ is connected, we know $C{F}_n^2-F_2$ is connected to M, thus $C{F}_n-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-2$, the result holds. Now we consider $C{F}_n^2-F_2$ is disconnected. If $|F_2|\le 3n-11=3(n-1)-8$, by Corollary 3.4, we know $C{F}_n^2-F_2$ has a big component and a singleton, the conclusion is certainly true. Now we think about $3n-10\le |F_2|\le 3n-7$, then $|F-F_1-F_2|\le 6n-17-2(3n-10)=3$. If $|F-F_1-F_2|=0$, we know the vertices in $C{F}_n^1-F_1$ and $C{F}_n^2-F_2$ are all connected to M, thus $C{F}_n-F$ is connected, a contradiction. If $|F-F_1-F_2|=1$, then $C{F}_n-F$ has a big component and at most two vertices $u,v$, where $u\in V\left(C{F}_n^1\right)$, $v\in V\left(C{F}_n^2\right)$ and they have a common outgoing neighbor in $F\setminus (F_1\cup F_2)$ and the other outgoing neighbor of u (resp., v) belong to $F_2$$(resp.,F_1)$.

If $|F-F_1-F_2|=3$, then $|F_1|=|F_2|=3n-10$. Since $|F_2|\le |F_1|\le \frac{9n-35}{2}=\frac{9(n-1)-26}{2}$, by Lemma 4.9, we can get $C{F}_n^1-F_1$ and $C{F}_n^2-F_2$ have a big component $C_i$ with $|V\left(C_i\right)|\ge (n-1)!-\left|F_i\right|-2$$(i\in \{1,2\left\}\right)$. If one of $C_i$ satisfies $|V\left(C_i\right)|\ge (n-1)!-\left|F_i\right|-1$, the conclusion is certainly true. Now we consider $|V\left(C_i\right)|=(n-1)!-\left|F_i\right|-2$ for all $i\in \{1,2\}$, in another word, $C{F}_n^i-F_i$ has a big component and two vertices. Similarly, we can get $C_i$ is connected to M. Since $|F_1|=|F_2|=3n-10$, we know $C{F}_n^i-F_i$ contains two singletons $u_i,v_i$ and these two singletons have three common neighbors in $C{F}_n^i$; Otherwise, if $u_i$ is adjacent to $v_i$, then $|F_i|\ge (\frac{3n-7}{2}-1)\times 2=3n-9>3n-10$, a contradiction. Thus $C{F}_n-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-4$. Let $S=\{u_1,u_2,v_1,v_2\}$. If S is an $Ind$-set, by Lemma 5.2, we have $|N_{C{F}_n-C{F}_n^1-C{F}_n^2}\left(S\right)|\ge 4$. Since $|F-F_1-F_2|=3$, $C{F}_n-F$ can not has a big component and four vertices. Thus $C{F}_n-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$. If there exists at least one edge in S, say $(v_1,u_2)\in E(C{F}_n-F)$, then $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$. Suppose on the contrary, we let $\left|V\right(C\left)\right|=n!-\left|F\right|-4$. If there exists only one edge between S and $u_1,v_2$ are singletons in $C{F}_n-F$, then $\left|F\right|\ge |N_{C{F}_n}\left(\{v_1,u_2\}\right)|+|N_{C{F}_n}\left(u_1\right)|+|N_{C{F}_n}\left(v_2\right)|-3\times 2=(\frac{3n-3}{2}-1)\times 2+\frac{3n-3}{2}\times 2-3\times 2=6n-14>6n-17$, a contradiction. If there exists two edges between S in $C{F}_n-F$, then $(v_1,u_2)\in E(C{F}_n-F)$, $(v_2,u_1)\in E(C{F}_n-F)$. Thus $\left|F\right|\ge |N_{C{F}_n}\left(\{v_1,u_2\}\right)|+|N_{C{F}_n}\left(\{v_2,u_1\}\right)|-3\times 2=(\frac{3n-3}{2}-1)\times 4-3\times 2=6n-16>6n-17$, a contradiction.

If $|F-F_1-F_2|=2$, then $|F_1|=3n-10$ or $|F_2|=3n-10$. We assume $|F_1|=3n-10$, then $|F_2|=3n-9$. By the same argument with the above discussion, we know if one of $C_i$ satisfies $|V\left(C_i\right)|\ge (n-1)!-\left|F_i\right|-1$, the result holds, so we consider $|V\left(C_i\right)|=(n-1)!-\left|F_i\right|-2$ for all $i\in \{1,2\}$. Hence $C{F}_n-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-4$. If we want $\left|V\right(C\left)\right|=n!-\left|F\right|-4$, the structure in Figure 8 must exists. Now we proof this structure can not exist. Since $|F_1|=3n-10$, $C{F}_n^1-F_1$ contains two singletons $u_1,v_1$ and this two singletons have three common neighbors in $C{F}_n^1$. Let $C{F}_n^2-F_2$ has a big component and two vertices $u_2,v_2$, then $u_2,v_2$ are nonadjacent; Otherwise, there exists a 7-circle (as shown by the red line in Figure 8). By Lemma 3.1, we can let $u_1=i_1{i}_2{i}_3\cdots i_{n-2}{i}_{n-1}1$, then $v_1=j_1{j}_2{j}_3\cdots j_{n-2}{i}_{n-1}1$, where $j_i\in [i_1,i_{n-2}]$ and $j_1\ne i_1$. So $u_1^{+}=1i_2{i}_3\cdots i_{n-2}{i}_{n-1}{i}_1$, $u_1^{-}=i_1{i}_2{i}_3\cdots i_{n-2}1i_{n-1}$, $v_1^{+}=1j_2{j}_3\cdots j_{n-2}{i}_{n-1}{j}_1$, $v_1^{-}=j_1{j}_2{j}_3\cdots j_{n-2}1i_{n-1}$. Since $j_1\ne i_1$, $i_1\ne i_{n-1}$ and $i_{n-1}\ne j_1$, we can get $i_{n-1}=2$, then $u_1^{-},v_1^{-}\in V\left(C{F}_n^2\right)$ and $u_2,v_2\in V\left(C{F}_n^2\right)$. By the structure in Figure 8, we know $u_2=1i_2{i}_3\cdots i_{n-2}{i}_{1}2$, $v_2=1j_2{j}_3\cdots j_{n-2}{j}_{1}2$. Let $u_2^{\prime }=1i_2{i}_3\cdots i_{n-2}{i}_1$, $v_2^{\prime }=1j_2{j}_3\cdots j_{n-2}{j}_1$, then $\{u_2^{\prime },v_2^{\prime }\}\subseteq V\left(G_1\right)$, $G_1\cong C{F}_{n-1}$ and $u_2^{\prime },v_2^{\prime }$ belong to different subgraphs in $G_1$. Since $n-1$ is even and ${\left(u_2^{\prime }\right)}^{+}=i_1{i}_2{i}_3\cdots i_{n-2}1$, ${\left(v_2^{\prime }\right)}^{+}=j_1{j}_2{j}_3\cdots j_{n-2}1$, we know ${\left(u_2^{\prime }\right)}^{+}\ne {\left(v_2^{\prime }\right)}^{+}$. Thus $u_2^{\prime }$ and $v_2^{\prime }$ have no common neighbors in $G_1$. In other words, $u_2$ and $v_2$ have no common neighbors in $C{F}_n^2$. So $|F_2|\ge 3n-7$, this contradicts to the fact that $|F_2|=3n-9$. Thus this structure does not exist, $C{F}_n-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$.

Case 3.3: $\frac{9n-33}{2}\le \left|F_1\right|\le \frac{9n-27}{2}$.

By the same argument with Case 3.2, we know if $C{F}_n^1-F_1$ and $C{F}_n^2-F_2$ are all connected, then $C{F}_n-F$ is connected, this contradicts to the assumption $C{F}_n-F$ is disconnected. Thus we assume $C{F}_n^2-F_2$ is disconnected. In this case, we can get $|F-F_1-F_2|\le 6n-17-\frac{9n-33}{2}-\frac{3n-7}{2}=3$, $\frac{3n-7}{2}\le |F_2|\le |F|-|F_1|\le 6n-17-\frac{9n-33}{2}=\frac{3n-1}{2}$. Thus if $C{F}_n^2-F_2$ is disconnected, then it has a big component and only one singleton; Otherwise, we assume $C{F}_n^2-F_2$ has two vertices $u_2,v_2$. If $u_2$ is adjacent to $v_2$, then $|F_2|=3n-9>\frac{3n-1}{2}$, a contradiction. If $u_2$ is nonadjacent to $v_2$, then $|F_2|\ge 3n-10>\frac{3n-1}{2}$, a contradiction. If $C{F}_n^1-F_1$ is connected, the result is certainly true. Now, we assume $C{F}_n^1-F_1$ is disconnected. If $|F-F_1-F_2|\le 1$, be the similarly discussion with Case 3.2, the conclusion is certainly true. If $|F-F_1-F_2|=2$, then $C{F}_n^1-F_1$ contains a big component $C_1$ and at most two vertices, which have no neighbors in $F-F_1-F_2$, the result holds.

When $|F-F_1-F_2|=3$, we have $|F_1|=\frac{9n-33}{2}$ and $|F_2|=\frac{3n-7}{2}$. Thus we can let H is the vertex set in $C{F}_n^1-F_1$, which are not adjacent to M, as $|F-F_1-F_2|=3$, we can get $\left|V\right(H\left)\right|\le 3$. Hence $C{F}_n-F$ contains a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-4$. Furthermore, $\left|V\right(C\left)\right|=n!-\left|F\right|-4$ if and only if the structure in Figure 9 exists. Now we can proof this structure does not exist. Let $C{F}_n^1-F_1$ has a big component and three vertices $x_2,x_3,x_4$ and $C{F}_n^2-F_2$ has a big component and one singleton $x_1$. Firstly, we can get $x_2,x_3,x_4$ are three singletons in $C{F}_n^1-F_1$; Otherwise, there exist a 5-circle in $C{F}_n$(as shown by the red line in Figure 9). Furthermore, we can get $x_1$ is nonadjacent to $\{x_2,x_3,x_4\}$; Otherwise, there exists a 3-circle in $C{F}_n$. Thus $\{x_1,x_2,x_3,x_4\}$ is an $Ind$-set in $C{F}_n-F$. We let $x_1=i_1{i}_2{i}_3\cdots i_{n-2}{i}_{n-1}2$, then $x_1^{+}\in V\left(C{F}_n^{i_1}\right)$, $x_1^{-}\in V\left(C{F}_n^{i_{n-1}}\right)$. Since $|F-F_1-F_2|=3$, $x_1^{+}$ or $x_1^{-}$ must equal to one of the outgoing neighbors of $\{x_2,x_3,x_4\}$ in $F\setminus (F_1\cup F_2)$. We assume $x_1^{+}=x_3^{-}$, then $x_3=2i_2{i}_3\cdots i_{n-2}{i}_1{i}_{n-1}$. Thus $i_{n-1}=1$, $x_1^{-}\in V\left(C{F}_n^1\right)$. The neighbors of $x_1$ in $C{F}_n^2$ are follows: $a_1=i_2{i}_1{i}_3\cdots i_{n-2}12$, …, $a_{n-2}=1i_2{i}_3{i}_4\cdots i_{n-2}{i}_{1}2$, $b_{n-1}=i_1{i}_3{i}_3\cdots i_{n-2}12$, …, $b_{\frac{3n-7}{2}}=i_1{i}_2{i}_3\cdots i_{n-2}{i}_{n-3}12$. In these vertices, we need to choose three vertices such that one of their outgoing neighbor belong to $C{F}_n^1$. Then $x_2,x_3,x_4$ must belong to A, where $A=\{a_1^{-},\dots ,a_{n-3}^{-},a_{n-2}^{+},b_{n-1}^{-},\dots ,b_{\frac{3n-7}{2}}^{-}\}$. Now, we choose a vertex in A such that one of it’s outgoing neighbors equal to $x_1^{+}$, we assume this vertex is $x_3$, $x_3=2i_2{i}_3{i}_4\cdots i_{n-2}{i}_{1}1$. As $C{F}_n^1\cong C{F}_{n-1}$, $x_2,x_3,x_4$ are three singletons, and $|F_1|=\frac{9n-33}{2}=\frac{9(n-1)-24}{2}$, by Corollary 4.6, we can get $x_2,x_3,x_4$ must belong to a common subgraph of $C{F}_n^1$. From the choose of $x_2,x_3,x_4$, we know there have not one vertex which belongs to a common subgraph with $x_3$ in $C{F}_n^1$. Thus this structure does not exist, $C{F}_n-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$.

When n is even, we can assume $|F_1|\ge |F_2|\ge \frac{3n-6}{2}$. Then $|F_1|\le |F|-|F_2|\le 6n-19-\frac{3n-6}{2}=\frac{9n-32}{2}$, we think about the following cases:

Case 3.1: $\frac{3n-6}{2}\le |F_2|\le |F_1|\le 3n-10$.

In this case, since $\left|F\right|-|F_1|-|F_2|\le 6n-17-(3n-6)=3n-11$, $|V\left(C{F}_n^1\right)-F_1|\ge (n-1)!-(3n-10)$, $|V\left(C{F}_n^2\right)-F_2|\ge (n-1)!-(3n-10)$ and $(n-1)!-(3n-10)-(3n-11)>1$, we can get if $C{F}_n^1-F_1$ or $C{F}_n^2-F_2$ is connected, then it is connected to M. If $C{F}_n^1-F_1$ and $C{F}_n^2-F_2$ are all connected, then $C{F}_n-F$ is connected, this contradicts to the assumption $C{F}_n-F$ is disconnected. Thus we assume $C{F}_n^i-F_i$$(i\in \{1,2\left\}\right)$ is disconnected. Since $C{F}_n^i\cong C{F}_{n-1}$ and $\frac{3n-6}{2}\le |F_2|\le |F_1|\le 3n-10=3(n-1)-7$, by Corollary 3.4, we know $C{F}_n^i-F_i$$(i\in \{1,2\left\}\right)$ contains a big component $C_i$, which satisfies that $|V\left(C_i\right)|\ge (n-1)!-\left|F_i\right|-1$. Similarly to the above discussion, we know the big component of $C{F}_n^i-F_i$ is connected to M. Thus $C{F}_n-F$ contains a big component C, which can satisfies that $\left|V\right(C\left)\right|\ge n!-\left|F\right|-2$.

Case 3.2: $3n-9\le |F_1|\le \frac{9n-32}{2}$.

In this case, we know $|F_2|\le |F|-|F_1|\le 6n-19-(3n-9)=3n-10=3(n-1)-7$. By the same argument with Case 3.1, we know if $C{F}_n^1-F_1$ and $C{F}_n^2-F_2$ are all connected, then $C{F}_n-F$ is connected, a contradiction. Thus we assume $C{F}_n^2-F_2$ is disconnected. By Corollary 3.4, we know $C{F}_n^2-F_2$ contains a big component $C_2$, which can satisfies that $|V\left(C_2\right)|\ge (n-1)!-\left|F_2\right|-1$. If $C{F}_n^1-F_1$ is connected, the conclusion is certainly true. Now we think about $C{F}_n^1-F_1$ is disconnected. Since $\le |F_1|\le \frac{9n-32}{2}=\frac{9(n-1)-23}{2}$, $C{F}_n^1-F_1$ has a big component $C_1$ with $|V\left(C_1\right)|\ge (n-1)!-\left|F_1\right|-2$. Thus the result holds.

Case 4. When n is odd, there exists three vertex set $F_i$, $F_j$, $F_k$, which can satisfy that $|F_i|\ge \frac{3n-7}{2}$, $|F_j|\ge \frac{3n-7}{2}$, $|F_k|\ge \frac{3n-7}{2}$; when n is even, there exists three vertex set $F_i$, $F_j$, $F_k$, which can satisfy that $|F_i|\ge \frac{3n-6}{2}$, $|F_j|\ge \frac{3n-6}{2}$, $|F_k|\ge \frac{3n-6}{2}$$(i,j,k$ are different from each other).

In this case, we let $M=C{F}_n-(F\cup C{F}_n^i\cup C{F}_n^j\cup C{F}_n^k)$.

When n is even, we have $|F_i|\le |F|-|F_j|-|F_k|\le 6n-19-2\times \frac{3n-6}{2}=3n-13<3n-10$, by Corollary 3.4, we know this result holds.

When n is odd, we have $|F_i|\le |F|-|F_j|-|F_k|\le 6n-17-2\times \frac{3n-7}{2}=3n-10$. If $|F_i|\le 3n-11$, $|F_j|\le 3n-11$, $|F_k|\le 3n-11$, by Corollary 3.4, we know the result holds. If $|F_i|=3n-10$, $|F_j|\le 3n-11$ and $|F_k|\le 3n-11$. By Lemma 1, we know there exists one singleton or two singletons in $C{F}_n^i-F_i$. Thus $C{F}_n-F$ contains a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-4$. If $\left|V\right(C\left)\right|=n!-\left|F\right|-4$, then $C{F}_n^i-F_i$ has two singletons $x_1,x_2$ and $C{F}_n^j-F_j$, $C{F}_n^k-F_5$ only has one singleton, meanwhile, these singletons are not connected to M, then $\left|F\right|-|F_i|-|F_j|-|F_k|\le 6n-17-(3n-10)-2\times \frac{3n-7}{2}=0$. Since $|F_i|=3n-10$, by the proof process of Lemma 5.4, we know $x_1^{+}$ can not belong to a common subgraph with $x_2^{+}$ and $x_2^{-}$. Thus $\left|F\right|-|F_i|-|F_j|-|F_k|\ge 1$, this contradicts to the fact $\left|F\right|-|F_i|-|F_j|-|F_k|\le 0$. Thus $\left|V\right(C\left)\right|\ne n!-\left|F\right|-4$, $C{F}_n-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$. If $|F_i|=|F_j|=3n-10$, then $\left|F\right|\ge |F_i|+|F_j|+|F_k|=2\times (3n-10)+\frac{3n-7}{2}=\frac{15n-47}{2}>6n-17$, a contradiction. If $|F_i|=|F_j|=|F_k|=3n-10$, then $\left|F\right|\ge |F_i|+|F_j|+|F_k|=3\times (3n-10)=9n-30>6n-17$, a contradiction.

Thus, for $n\ge 5$, if F satisfies the condition $\left|F\right|\le 6n-17$$(n$ is odd) and $\left|F\right|\le 6n-19$$(n$ is even), then $C{F}_n-F$ has a big component C with $\left|V\right(C\left)\right|\ge n!-\left|F\right|-3$.

Theorem 3. For $n\ge 5$, when n is odd, $c{\kappa }_5\left(C{F}_n\right)=6n-16$; when n is even, $c{\kappa }_5\left(C{F}_n\right)=6n-18$.

Proof. By Lemma 5.10, we can get when n is odd, $c{\kappa }_5\left(C{F}_n\right)\ge 6n-16$; when n is even, $c{\kappa }_5\left(C{F}_n\right)\ge 6n-18$. Next, we will proof $c{\kappa }_5\left(C{F}_n\right)\le 6n-16$(n is odd) and $c{\kappa }_5\left(C{F}_n\right)\le 6n-18$(n is even). When n is odd, let $x_1=i_1{i}_2{i}_3\cdots i_{n-2}ji$, $x_2=i_{n-2}{i}_2{i}_3\cdots i_1{i}_{n-3}ji$, $x_3=i_3{i}_2{i}_1{i}_4\cdots i_{n-2}ij$, $x_4=i_{n-2}{i}_2{i}_1{i}_4\cdots i_3{i}_{n-3}ij$. Then $|N_{C{F}_n}\left(\{x_1,x_2,x_3,x_4\}\right)|=6n-16$. Let $F=N_{C{F}_n}\left(\{x_1,x_2,x_3,x_4\}\right)$, then $C{F}_n-F$ has a big component and four singletons $x_1,x_2,x_3,x_4$. Thus $c{\kappa }_5\left(C{F}_n\right)\le 6n-16$. When n is even, we let $x_1=i_1{i}_2{i}_3\cdots i_{n-2}jik$, $x_2=i_{n-2}{i}_2{i}_3\cdots i_1{i}_{n-3}jik$, $x_3=i_3{i}_2{i}_1{i}_4\cdots i_{n-2}ijk$, $x_4=i_{n-2}{i}_2{i}_1{i}_4\cdots i_3{i}_{n-3}ijk$. Then $\{x_1,x_2,x_3,x_4\}$ $\subseteq V\left(C{F}_n^k\right)$ and $|N_{C{F}_n^k}\left(\{x_1,x_2,x_3,x_4\}\right)|=6(n-1)-16=6n-22$. In $C{F}_n^k$, if we remove $N_{C{F}_n^k}\left(\{x_1,x_2,x_3,x_4\}\right)$, we can get four singletons. From the definition of $C{F}_n$, we know any two vertices in $C{F}_n^k$ have different outgoing neighbors. So, if we remove $N_{C{F}_n^k}\left(\{x_1,x_2,x_3,x_4\}\right)\cup \{x_1^{+},x_2^{+},x_3^{+},x_4^{+}\}$ in $C{F}_n$, we can get a big component and four singletons $x_1,x_2,x_3,x_4$. Thus $c{\kappa }_5\left(C{F}_n\right)\le 6n-22+4=6n-18$.

6. Conclusions

It is very useful to study the connectivity of a graph. In this paper, we study the m-component connectivity of $C{F}_n$. The Leaf-sort graph is a special Cayley graph, it has many special properties. We have shown that for $n\ge 3$, $c{\kappa }_3\left(C{F}_n\right)=3n-6$$(n$ is odd) and $c{\kappa }_3\left(C{F}_n\right)=3n-7$$(n$ is even); for $n\ge 4$, $c{\kappa }_4\left(C{F}_n\right)=\frac{9n-21}{2}$$(n$ is odd) and $c{\kappa }_4\left(C{F}_n\right)=\frac{9n-24}{2}$$(n$ is even); for $n\ge 5$, $c{\kappa }_5\left(C{F}_n\right)=6n-16$$(n$ is odd) and $c{\kappa }_5\left(C{F}_n\right)=6n-18$$(n$ is even). So far, we only get the value of $c{\kappa }_3\left(C{F}_n\right)$, $c{\kappa }_4\left(C{F}_n\right)$ and $c{\kappa }_5\left(C{F}_n\right)$, for $m\ge 6$, this problem is still unsolved. So in the future, this problem is well worth studying. Furthermore, by referring to the references, we can find that: there is a regularity between $c{\kappa }_m\left(B{S}_n\right)$ and ${\kappa }^{\left(m\right)}\left(B{S}_n\right)$. Similarly, $c{\kappa }_m\left(B{P}_n\right)$ and ${\kappa }^{\left(m\right)}\left(B{P}_n\right)$ also have regularity. So we can think about, is there some sort of relationship between $c{\kappa }_m\left(C{F}_n\right)$ and ${\kappa }^{\left(m\right)}\left(C{F}_n\right)$?