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The Component Connectivity of Leaf-Sort Graphs

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12 November 2023

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14 November 2023

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Abstract
In large-scale parallel computing and communication systems, an interconnect network structure is usually modeled as a graph, in which vertices and edges correspond to processors and communication links, respectively. In a graph, the vertices and edges are likely to fail, so we must think about the fault tolerance of a graph. Connectivity is an important parameter in the study of faulty tolerance for a graph. In this paper, we study a special class of connectivity: $m$-component connectivity, this is a natural generalization of the classical connectivity of graphs defined in terms of the minimum vertex-cut. Let $F$ is a vertex set of $G$ $(i.e, F\subseteq V(G))$, if the following conditions are satisfied, we say $F$ is a $m$-component cut: $(1)$ $G-F$ is disconnected; $(2)$ the number of components in $G-F$ is greater than or equal to $m$. In another word, the $m$-component connectivity $c\kappa_{m}(G)$ is defined as $min\{|F|~|~F\subseteq V(G)$ and $F$ is a $m$-component cut$\}$. Determining $m$-component connectivity is still unsolved in most interconnection networks even for small $m'$s. Leaf-sort graph is a special Cayley graph, it has many special properties that are different from other Cayley graphs. So we need to pay attention to some of it's special properties when we study it. In this paper, we can get the values: $c\kappa_{3}(CF_{n})=3n-6$ $(n$ is odd$)$ and $c\kappa_{3}(CF_{n})=3n-7$ $(n$ is even$)$ for $n\geq3$; $c\kappa_{4}(CF_{n})=\frac{9n-21}{2}$ $(n$ is odd$)$ and $c\kappa_{4}(CF_{n})=\frac{9n-24}{2}$ $(n$ is even$)$ for $n\geq4$; $c\kappa_{5}(CF_{n})=6n-16$ $(n$ is odd$)$ and $c\kappa_{5}(CF_{n})=6n-18$ $(n$ is even$)$ for $n\geq5$.
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Subject: Computer Science and Mathematics  -   Discrete Mathematics and Combinatorics

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 = ( V ( G ) , E ( G ) ) , where V ( G ) is the vertex set and E ( G ) 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 = ( V ( G ) , E ( G ) ) be a simple connected graph, where V ( G ) is the vertex set and E ( G ) is the edge set. Let F ⊆ V ( G ) , we use G − F to denote the subgraph of G with vertex set V ( G ) − F and edge set E ( G ) − { ( u , v ) ∈ E ( G ) | { u , v } ∩ F ≠ ∅ } . 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 , ⋯ , w k , y > , where w 1 , w 2 , ⋯ , w k are distinct ones. The vertices w i , i = 1 , 2 , ⋯ , k are called internal vertices of the path P. For any two vertices { u , v } ⊆ V ( G ) , if there exists a u v -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 κ ( G ) is defined as m i n { | F | | F ⊆ V ( G ) and F is a vertex cut }. Otherwise, we say κ ( G ) = n − 1 , where n is the number of vertices in G. A graph G is k-connected if κ ( G ) ≥ 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 κ m ( G ) was first proposed by Hsu et al. [3]. Let F is a vertex set of G ( i . e , F ⊆ V ( G ) ) , if the following conditions are satisfied, we say F is a m-component cut: ( 1 ) G − F is disconnected; ( 2 ) the number of components in G − F is greater than or equal to m. The m-component connectivity c κ m ( G ) is defined as m i n { | F | | F ⊆ V ( G ) and F ia a m-component cut}. By the above definition, we can easily get that c κ 2 ( G ) = κ ( G ) and c κ m + 1 ( G ) ≥ c κ m ( G ) . By referring to the relevant literature, we can also get there exists a certain relationship between m-component connectivity c κ m ( G ) and m-extra connectivity. The m-extra connectivity, denoted by κ m ( G ) , 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 ∈ V ( G ) , N G ( v ) = { u | ( u , v ) ∈ E ( G ) } is the set of neighbors of v. We let d e g G ( v ) = | N G ( v ) | be the degree of v and δ ( G ) = min { d e g G ( v ) | v ∈ V ( G ) } be the minimum degree of G. If d e g G ( v ) = k for every v ∈ V ( G ) , then G is k-regular. A singleton of G is a vertex v with d e g G ( v ) = 0 . For a vertex set X, N G ( X ) = { ∪ x ∈ X N G ( x ) } − 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 ∩ V 2 = ∅ such that V ( G ) = V 1 ∪ V 2 and for each edge ( u , v ) ∈ E ( G ) , | { u , v } ∩ V 1 | = | { u , v } ∩ 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 κ 3 ( C F n ) = 3 n − 6 ( n is odd) and c κ 3 ( C F n ) = 3 n − 7 ( n is even) for n ≥ 3 ; c κ 4 ( C F n ) = 9 n − 21 2 ( n is odd) and c κ 4 ( C F n ) = 9 n − 24 2 ( n is even) for n ≥ 4 ; c κ 5 ( C F n ) = 6 n − 16 ( n is odd) and c κ 5 ( C F n ) = 6 n − 18 ( n is even) for n ≥ 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 κ 3 ( C F n ) . In Section 4, we prove the value of c κ 4 ( C F n ) . In Section 5, we prove some useful lemmas firstly, and then calculate the value of c κ 5 ( C F n ) . 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 ≤ l 1 ≤ l 2 . Set [ l 1 , l 2 ] = { l | l is an integer with l 1 ≤ l ≤ l 2 } . In the permutation 1 2 ⋯ n p 1 p 2 ⋯ p n , i → p i . For the convenience, we denote the permutation 1 2 ⋯ n p 1 p 2 ⋯ p n by p 1 p 2 ⋯ p n . In [19], every permutation can be denoted by a product of disjoint cycles. For example, 1 2 3 2 3 1 = ( 123 ) . Specially, 1 2 ⋯ n 1 2 ⋯ n = ( 1 ) . The product σ τ of two permutations is the composition function τ followed by σ , e.g., ( 12 ) ( 13 ) = ( 132 ) . 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 , ⋯ , n } , and let S n be the symmetric group on [ 1 , n ] containing all permutations p = p 1 p 2 ⋯ p n of [ 1 , n ] . It is well known that { ( 1 i ) : i = 2 , 3 , ⋯ , n } is a generating set for S n . So { ( 1 , i ) : i = 2 , 3 , ⋯ , n } ∪ { ( j , j + 1 ) : j = 2 , 4 , ⋯ , n − 1 } (n is odd.) is also a generating set for S n and { ( 1 , i ) : i = 2 , 3 , ⋯ , n } ∪ { ( j , j + 1 ) : j = 2 , 4 , ⋯ , 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 ( C F n ) = S n in which two vertices u , v are adjacent if and only if u = v ( 1 , i ) , 2 ≤ i ≤ n , or u = v ( j , j + 1 ) , 2 ≤ j ≤ n − 1 when j is even and n is odd, or u = v ( j , j + 1 ) , 2 ≤ j ≤ n − 2 when j and n are even. By the definition, we can get C F n is a 3 n − 3 2 -regular graph on n ! vertices for odd n, and 3 n − 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 , ⋯ , C F n n where every vertex u = x 1 x 2 ⋯ x n   ∈ C F n i has a fixed integer i in the last position x n for i ∈ [ 1 , n ] . It is obvious that C F n i is isomorphic to C F n − 1 for i ∈ [ 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 ∈ 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 ( C F n ) = E C F n ( V ( C F n i ) , V ( C F n j ) ) for i , j ∈ [ 1 , n ] .
Proposition 2.1.([20]) Let C F n i ( 1 ≤ i ≤ 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 ∈ V ( C F n i ) ( 1 ≤ i ≤ 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 ≠ i ) when n is odd; v ( 1 , n ) belong to C F n j ( j ≠ i ) when n is even.
Proposition 2.3. For any u , v ∈ V ( C F n i ) , N u + ∩ N v + = ∅ when n is odd; u + ≠ v + when n is even.
Proof. Let u = u 1 u 2 ⋯ u n − 1 i and v = v 1 v 2 ⋯ v n − 1 i , where u j ≠ v j for some j ∈ [ 1 , n − 1 ] . Then u + = i u 2 ⋯ u n − 1 u 1 ≠ i v 2 ⋯ v n − 1 v 1 = v + , u − = u 1 u 2 ⋯ i u n − 1 ≠ v 1 v 2 ⋯ i v n − 1 = v − . Moreover, u + ≠ v − and v + ≠ u − . Hence when n is odd, N u + ∩ N v + = ∅ ; when n is even, u + ≠ v + .
Proposition 2.4.([20]) For any integer n ≥ 2 , C F n is bipartite.
Proposition 2.5. For any two vertices x , y ∈ V ( C F n ) ( n ≥ 3 ) , | N C F n ( x ) ∩ N C F n ( y ) | ≤ 3 .
Proof. If d C F n ( x , y ) = 1 or d C F n ( x , y ) ≥ 3 , then | N C F n ( x ) ∩ N C F n ( y ) | = 0 ; Otherwise ( i . e . , | N C F n ( x ) ∩ N C F n ( y ) | ≥ 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 ( x ) ∩ N C F 3 ( y ) | = 3 as C F 3 ≅ K 3 , 3 and d C F 3 ( x , y ) = 2 .
For n = 4 , if x , y ∈ V ( C F 4 i ) , | N C F 4 ( x ) ∩ N C F 4 ( y ) | = 3 as C F 4 i ≅ C F 3 and x + ≠ y + . If x ∈ V ( C F 4 i ) , y ∈ V ( C F 4 j ) ( i ≠ j ) , let x = x 1 x 2 x 3 i and y = y 1 y 2 y 3 j , we know x + ≠ y + . If y + ∈ C F 4 i or x + ∈ C F 4 j , then N C F 4 ( x ) ∩ N C F 4 ( y ) ⊆ { x + , y + } . Thus | N C F 4 ( x ) ∩ N C F 4 ( y ) | ≤ 2 .
Now we assume n ≥ 5 and the result holds for C F n − 1 . If x , y ∈ V ( C F n i ) for i ∈ [ 1 , n ] , then by Proposition 2.3 and the inductive hypothesis, the result holds. So we let x ∈ V ( C F n i ) , y ∈ V ( C F n j ) ( i ≠ j ) , x = x 1 x 2 ⋯ x n − 1 i , y = y 1 y 2 ⋯ y n − 1 j . Then x + = i x 2 ⋯ x n − 1 x 1 , x − = x 1 x 2 ⋯ i x n − 1 , y + = j y 2 ⋯ y n − 1 y 1 , y − = y 1 y 2 ⋯ j y n − 1 . We know x + ≠ x − and y + ≠ y − . Since i ≠ j , x + ≠ y + and x − ≠ y − . As d C F n ( x , y ) = 2 and N C F n ( x ) ∩ N C F n ( y ) ⊆ { x + , x − , y + , y − } , we can assume y + ∈ N C F n ( x ) ∩ N C F n ( y ) .
When n is odd. If y + = x − , then j y 2 y 3 ⋯ y n − 1 y 1 = x 1 x 2 ⋯ i x n − 1 , x 1 = j , y 2 = x 2 , y 3 = x 3 , ⋯ , y n − 2 = x n − 2 , y n − 1 = i , y 1 = x n − 1 . Thus x + = y n − 1 y 2 y 3 ⋯ y n − 2 y 1 j ∈ C F n j and y − = x n − 1 x 2 x 3 ⋯ x n − 2 x 1 i ∈ C F n i , y x + ∈ E ( C F n j ) , x y − ∈ E ( C F n i ) . So | N C F n ( x ) ∩ N C F n ( y ) | = 3 . If y + ∈ C F n i and adjacent to x, then y + = j y 2 ⋯ y n − 1 i , N C F n ( x ) ∩ N C F n ( y ) ⊆ { x + , y + , x − } . Thus | N C F n ( x ) ∩ N C F n ( y ) | ≤ 3 . Furthermore, if and only if y − = x + , there will be | N C F n ( x ) ∩ N C F n ( y ) | = 3 , this is similar to the situation y + = x − . When y − ≠ x + , | N C F n ( x ) ∩ N C F n ( y ) | ≤ 2 . So when n is odd, this result holds.
When n is even. Note that x + ≠ y + . If y + ∈ C F n i or x + ∈ C F n j , then N C F n ( x ) ∩ N C F n ( y ) ⊆ { x + , y + } . Thus | N C F n ( x ) ∩ N C F n ( y ) | ≤ 2 .
In summary, this proposition is proven.
Corollary 2.6. When n ≥ 4 is even, if x and y belong to two different subgraphs in C F n , then | N C F n ( x ) ∩ N C F n ( y ) | ≤ 2 .
Corollary 2.7. When n ≥ 3 is odd, for any two vertices x , y ∈ V ( C F n ) , where x ∈ V ( C F n i ) , y ∈ V ( C F n j ) ( i ≠ j ) . Then | N C F n ( x ) ∩ N C F n ( y ) | = 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 κ ( C F n ) = 3 n − 3 2 when n is odd and κ ( C F n ) = 3 n − 4 2 when n is even.
Lemma 1. ([21]) Let F ⊆ V ( C F n ) with | F | ≤ 3 n − 6 when n is odd ( n ≥ 5 ) and | F | ≤ 3 n − 7 when n is even ( n ≥ 4 ) . If C F n − F is disconnected, then C F n − F satisfies one of the following conditions:
( 1 )   C F n − F has two components, one of which is a singleton;
( 2 )   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 κ 3 ( C F n ) = 3 n − 6 for n ≥ 3 ; When n is even, c κ 3 ( C F n ) = 3 n − 7 for n ≥ 3 . Before proving our main results, we prove some useful lemmas firstly. Let S is a subset of V ( G ) ( i . e . , S ⊆ V ( G ) ) , 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 I n d -set.
Lemma 3.1. When n is odd, if x 1 ∈ V ( C F n i ) , then there exists only ( n − 3 ) vertices in C F n i , which can satisfy that | N C F n i ( x 1 ) ∩ N C F n i ( x 2 ) | = 3 ; When n is even, if x 1 ∈ V ( C F n i ) , then there exists only ( n − 2 ) vertices in C F n i such that | N C F n i ( x 1 ) ∩ N C F n i ( x 2 ) | = 3 . In addition, these vertices are all regular.
Proof. Note that C F n i ≅ C F n − 1 . When n is odd, n − 1 is even, { x 1 , x 2 } ⊆ V ( C F n i ) . Since | N C F n i ( x 1 ) ∩ N C F n i ( x 2 ) | = 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 ( x 1 ) ∩ N C F n i ( x 2 ) | ≤ 2 . So we can assume x 1 = i 1 i 2 i 3 ⋯ i n − 2 j i , x 2 = j 1 j 2 j 3 ⋯ j n − 2 j i . Now we let x 1 ′ = i 1 i 2 i 3 ⋯ i n − 2 , x 2 ′ = j 1 j 2 j 3 ⋯ j n − 2 , then { x 1 ′ , x 2 ′ } ⊆ V ( G 1 ) , G 1 ≅ C F n − 2 , n − 2 is odd. Since | N G 1 ( x 1 ′ ) ∩ N G 1 ( x 2 ′ ) | = 3 , by the proof process of Proposition 2.5, we know x 1 ′ , x 2 ′ have two different situations:
Case 1.  x 1 ′ , x 2 ′ belong to two different subgraphs in G 1 , and ( x 1 ′ ) + = ( x 2 ′ ) − or ( x 1 ′ ) − = ( x 2 ′ ) + .
In this case, we have i n − 2 ≠ j n − 2 . If ( x 1 ′ ) + = i n − 2 i 2 i 3 ⋯ i n − 3 i 1 = ( x 2 ′ ) − = j 1 j 2 j 3 ⋯ j n − 2 j n − 3 , then j 1 = i n − 2 , j 2 = i 2 , … , 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 ⋯ i n − 4 i 1 i n − 3   j i . If ( x 1 ′ ) − = i 1 i 2 i 3 ⋯ i n − 2 i n − 3 = ( x 2 ′ ) + = j n − 2 j 2 j 3 ⋯ j n − 3 j 1 , then j n − 2 = i 1 , j 2 = i 2 , … , 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 ⋯ i n − 4 i n − 2 i 1 j i .
Case 2.  x 1 ′ , x 2 ′ belong to the same subgraph in G 1 .
In this case, we have i n − 2 = j n − 2 , x 1 ′ = i 1 i 2 i 3 ⋯ i n − 2 , x 2 ′ = j 1 j 2 j 3 ⋯ i n − 2 . As G 1 i n − 2 ≅ C F n − 3 , n − 3 is even, | N G 1 i n − 2 ( x 1 ′ ) ∩ N G 1 i n − 2 ( x 2 ′ ) | = 3 , we can get x 1 ′ , x 2 ′ belong to a common subgraph in G 1 i n − 2 . Then i n − 3 = j n − 3 . Let x 1 ′ ′ = i 1 i 2 i 3 ⋯ i n − 4 , x 2 ′ ′ = j 1 j 2 j 3 ⋯ j n − 4 , then { x 1 ′ ′ , x 2 ′ ′ } ⊆ V ( G 2 ) , G 2 ≅ C F n − 4 and | N G 2 ( x 1 ′ ′ ) ∩ N G 2 ( x 2 ′ ′ ) | = 3 . As n − 4 is odd, there are two different situations:
Subcase 2.1.  x 1 ′ ′ , x 2 ′ ′ belong to two different subgraphs in G 2 , and ( x 1 ′ ′ ) + = ( x 2 ′ ′ ) − or ( x 1 ′ ′ ) − = ( x 2 ′ ′ ) + .
If ( x 1 ′ ′ ) + = ( x 2 ′ ′ ) − , we can get j 1 = i n − 4 , j 2 = i 2 , … , 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 ⋯ i n − 6 i 1 i n − 5 i n − 3 i n − 2 j i . If ( x 1 ′ ′ ) − = ( x 2 ′ ′ ) + , then j 1 = i n − 5 , j 2 = i 2 , … , j n − 4 = i 1 , j n − 5 = i n − 4 . Thus x 2 = i n − 5 i 2 i 3 ⋯ i n − 6 i n − 4 i 1 i n − 3 i n − 2 j i .
Subcase 2.2.  x 1 ′ ′ , x 2 ′ ′ 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 ( x 1 ) ∩ N C F n i ( x 2 ) | = 3 , they are i n − 2 i 2 i 3 ⋯ i n − 4 i 1 i n − 3 j i , i n − 3 i 2 i 3 ⋯ i n − 4 i n − 2 i 1 j i , i n − 4 i 2 i 3 ⋯ i n − 6 i 1 i n − 5 i n − 3 i n − 2 j i , i n − 5 i 2 i 3 ⋯ i n − 4 i 1 i n − 3 i n − 2 j i ,…, i 3 i 1 i 2 i 4 ⋯ i n − 2 j i , i 2 i 3 i 1 i 4 ⋯ i n − 2 j i .
When n is even, n − 1 is odd, C F n i ≅ C F n − 1 . Let x 1 = i 1 i 2 i 3 ⋯ i n − 1 i , x 2 = j 1 j 2 j 3 ⋯ j n − 1 i , | N C F n i ( x 1 ) ∩ N C F n i ( x 2 ) | = 3 . Assume x 1 ′ = i 1 i 2 i 3 ⋯ i n − 1 , x 2 ′ = j 1 j 2 j 3 ⋯ j n − 1 , then { x 1 ′ , x 2 ′ } ⊆ V ( G 1 ) , G 1 ≅ C F n − 1 , similarly, we can also divide it into two different situations:
Case 1.  x 1 ′ , x 2 ′ belong to two different subgraphs in G 1 , and ( x 1 ′ ) + = ( x 2 ′ ) − or ( x 1 ′ ) − = ( x 2 ′ ) + .
In this case, we have i n − 1 ≠ j n − 1 . If ( x 1 ′ ) + = ( x 2 ′ ) − , then j 1 = i n − 1 , j 2 = i 2 , … , j n − 1 = i n − 2 , j n − 2 = i 1 . Thus x 2 = i n − 1 i 2 i 3 ⋯ i n − 3 i 1 i n − 2 i . If ( x 1 ′ ) − = ( x 2 ′ ) + , then j 1 = i n − 2 , j 2 = i 2 , … , j n − 2 = i n − 1 , j n − 1 = i 1 . Thus x 2 = i n − 2 i 2 i 3 ⋯ i n − 3 i n − 1 i 1 i .
Case 2.  x 1 ′ , x 2 ′ belong to a same subgraph in G 1 .
In this case, i n − 1 = j n − 1 . Since G 1 i n − 1 ≅ C F n − 2 and | N G 1 i n − 1 ( x 1 ′ ) ∩ N G 1 i n − 1 ( x 2 ′ ) | = 3 , x 1 ′ and x 2 ′ are in a same subgraph in G 1 i n − 1 . So i n − 2 = j n − 2 . Let x 1 ′ ′ = i 1 i 2 i 3 ⋯ i n − 3 , x 2 ′ ′ = j 1 j 2 j 3 ⋯ j n − 3 , then { x 1 ′ ′ , x 2 ′ ′ } ⊆ V ( G 2 ) , G 2 ≅ C F n − 3 , n − 3 is odd. As | N G 2 ( x 1 ′ ′ ) ∩ N G 2 ( x 2 ′ ′ ) | = 3 , we can also consider the following two situations:
Subcase 2.1.  x 1 ′ ′ and x 2 ′ ′ belong to two different subgraphs in G 2 , and ( x 1 ′ ′ ) + = ( x 2 ′ ′ ) − or ( x 1 ′ ′ ) − = ( x 2 ′ ′ ) + .
If ( x 1 ′ ′ ) + = ( x 2 ′ ′ ) − , then j 1 = i n − 3 , j 2 = i 2 , … , j n − 3 = i n − 4 , j n − 4 = i 1 . Thus x 2 = i n − 3 i 2 i 3 ⋯ i n − 5
i 1 i n − 4 i n − 2 i n − 1 i . If ( x 1 ′ ′ ) − = ( x 2 ′ ′ ) + , then j 1 = i n − 4 , j 2 = i 2 , … , j n − 3 = i 1 , j n − 4 = i n − 3 . Thus x 2 = i n − 4 i 2 i 3 ⋯ i n − 5 i n − 3 i 1 i n − 2 i n − 1 i .
Subcase 2.2.  x 1 ′ ′ and x 2 ′ ′ 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 ( x 1 ) ∩ N C F n i ( x 2 ) | = 3 , they are i n − 1 i 2 i 3 ⋯ i n − 3 i 1 i n − 2 i , i n − 2 i 2 i 3 ⋯ i n − 3 i n − 1 i 1 i , i n − 3 i 2 i 3 ⋯ i n − 5 i 1 i n − 4 i n − 2 i n − 1 i , i n − 4 i 2 i 3 ⋯
i n − 5 i n − 3 i 1 i n − 2 i n − 1 i , …, i 3 i 1 i 2 i 4 ⋯ i n − 2 i n − 1 i , i 2 i 3 i 1 i 4 ⋯ 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 } ⊆ V ( C F n i ) . If x 1 = i 1 i 2 i 3 ⋯ i n − 1 i , x 2 = j 1 j 2 j 3 ⋯ j n − 1 i and | N C F n i ( x 1 ) ∩ N C F n i ( x 2 ) | = 3 , then j 1 ≠ i 1 .
Lemma 3.3. For n ≥ 3 , let S is an I n d -set and | S | = 2 , then when n is odd, | N C F n ( S ) | ≥ 3 n − 6 ; when n is even, | N C F n ( S ) | ≥ 3 n − 7 .
Proof. Let S = { x 1 , x 2 } , as S is an I n d -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 ( x 1 ) ∩ N C F n ( x 2 ) | ≤ 3 and C F n is a 3 n − 3 2 -regular graph ( n is odd) or 3 n − 4 2 -regular graph ( n is even). So when n is odd, | N C F n ( S ) | = | N C F n ( x 1 ) | + | N C F n ( x 2 ) | − | N C F n ( x 1 ) ∩ N C F n ( x 2 ) | ≥ 2 × 3 n − 3 2 − 3 = 3 n − 6 . When n is even, | N C F n ( S ) | = | N C F n ( x 1 ) | + | N C F n ( x 2 ) | − | N C F n ( x 1 ) ∩ N C F n ( x 2 ) | ≥ 2 × 3 n − 4 2 − 3 = 3 n − 7 .
Corollary 3.4. For n ≥ 4 , let F is a subset of V ( C F n ) ( i . e . , F ⊆ V ( C F n ) ) and when n is odd, | F | ≤ 3 n − 7 ; when n is even, | F | ≤ 3 n − 8 , then C F n − F contains a big component C, which satisfies | V ( C ) | ≥ n ! − | F | − 1 .
Theorem 1. For n ≥ 3 , when n is odd, c κ 3 ( C F n ) = 3 n − 6 ; when n is even, c κ 3 ( C F n ) = 3 n − 7 .
Proof. For n = 3 , since c κ l + 1 ( C F n ) ≥ c κ l ( C F n ) , we can get c κ 3 ( C F 3 ) ≥ 3 n − 3 2 = 3 = 3 n − 6 by Proposition 2.8. For n ≥ 4 , by Corollary 3.4, we can also get c κ 3 ( C F n ) ≥ 3 n − 6 when n is odd and c κ 3 ( C F n ) ≥ 3 n − 7 when n is even. Next, we will prove that: when n is odd, c κ 3 ( C F n ) ≤ 3 n − 6 and when n is even, c κ 3 ( C F n ) ≤ 3 n − 7 . Since | N C F n ( x 1 ) ∩ N C F n ( x 2 ) | ≤ 3 , we can choose two different vertices x 1 , x 2 ∈ V ( C F n ) , which can satisfy the condition | N C F n ( x 1 ) ∩ N C F n ( x 2 ) | = 3 . From the definition of C F n , we know that C F n is a 3 n − 3 2 -regular ( n is odd) and 3 n − 4 2 -regular graph ( n is even), so when n is odd, | N C F n ( { x 1 , x 2 } ) | = 2 × 3 n − 3 2 − 3 = 3 n − 6 ; when n is even, | N C F n ( { x 1 , x 2 } ) | = 2 × 3 n − 4 2 − 3 = 3 n − 7 . Thus let F = N C F n ( { x 1 , x 2 } ) , we know C F n − F contains three components and two of them only has a singleton. So we can get c κ 3 ( C F n ) ≤ 3 n − 6 ( n is odd) and c κ 3 ( C F n ) ≤ 3 n − 7 ( n is even).

4. The 4-component connectivity of C F n

Lemma 4.1. When n = 4 , let S is an I n d -set and | S | = 3 , | N C F 4 ( S ) | ≥ 6 .
Proof. Let S = { x 1 , x 2 , x 3 } , since S is an I n d -set, x 1 , x 2 , x 3 are nonadjacent with each other. Note that C F 4 i ≅ C F 3 , C F 3 ≅ 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 ≅ K 3 , 3 and S is an I n d -set, | N C F 4 i ( S ) | = 3 . By Proposition 2.3, we know the outgoing neighbors of { x 1 , x 2 , x 3 } are different. Thus, | N C F 4 ( S ) | = 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 ≠ j ) .
In this case, we can let { x 1 , x 2 } ⊆ V ( C F 4 1 ) , x 3 ∈ V ( C F 4 2 ) . Since x 1 and x 2 are nonadjacent, we can get | N C F 4 1 ( { x 1 , x 2 } ) | = 3 , | N C F 4 2 ( x 3 ) | = 3 . By the definition of C F n , we know x i ( i ∈ { 1 , 2 , 3 } ) only has one outgoing neighbor. If the structure shown in Figure 2 (a) exists, | N C F 4 ( S ) | ≥ | N C F 4 1 ( { x 1 , x 2 } ) | + | N C F 4 2 ( x 3 ) | = 6 . Now we can show that this structure does not exist. As | N C F 4 1 ( { x 1 , x 2 } ) | = 3 and { x 1 , x 2 } ⊆ V ( C F 4 1 ) , we assume x 1 = 2341 , then x 2 = 4231 or x 2 = 3421 . Thus x 1 + ∈ V ( C F 4 2 ) , x 2 + ∉ V ( C F 4 2 ) , the structure shown in Figure 2 (a) does not exist. Thus | N C F 4 ( S ) | ≥ 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 ∈ V ( C F 4 i ) . By the definition of C F n , we know | N C F 4 i ( x i ) | = 3 , thus | N C F 4 ( S ) | ≥ 3 × 3 = 9 .
Combining the above three situations, we can get | N C F 4 ( S ) | ≥ 6 .
Lemma 4.2. When n = 5 , let S is an I n d -set and | S | = 3 , | N C F 5 ( S ) | ≥ 12 .
Proof. Let S = { x 1 , x 2 , x 3 } , since S is an I n d -set, x 1 , x 2 , x 3 are nonadjacent with each other. Note that C F 5 i ≅ C F 4 ( 1 ≤ i ≤ 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 ≅ C F 4 , by Lemma 4.1, we can get | N C F 5 i ( S ) | ≥ 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 ( S ) | = | N C F 5 i ( S ) | + 6 ≥ 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 ≠ j ) .
In this case, we can let { x 1 , x 2 } ⊆ V ( C F 5 1 ) , x 3 ∈ V ( C F 5 2 ) . Since C F 5 1 ≅ C F 4 , by Lemma 3.3, we know that | N C F 5 1 ( { x 1 , x 2 } ) | ≥ 3 ( n − 1 ) − 7 = 3 × 5 − 10 = 5 . By Proposition 2.8, we can get | N C F 5 2 ( x 3 ) | = 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 ( S ) | ≥ | N C F 5 1 ( { x 1 , x 2 } ) | + | N C F 5 2 ( x 3 ) | + 2 = 5 + 4 + 2 = 11 . Furthermore, | N C F 5 ( S ) | = 11 if and only if this structure exists. Now, we can prove this structure does not exist.
Since { x 1 , x 2 } ⊆ V ( C F 5 1 ) , C F 5 1 ≅ C F 4 and | N C F 5 1 ( x 1 ) ∩ N C F 5 1 ( x 2 ) | = 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 + = 1 i 1 i 2 i 4 i 3 , x 2 − = i 3 i 1 i 2 1 i 4 . Since x 1 + = 1 i 2 i 3 i 4 i 1 , x 1 − = i 1 i 2 i 3 1 i 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 − } ⊆ V ( C F 5 2 ) . 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 + = 2 i 1 i 3 1 i 2 ∈ V ( C F 5 i 2 ) , x 3 − = i 2 i 1 i 3 21 ∈ V ( C F 5 1 ) , and x 3 − is adjacent to x 2 . Since x 1 + = 1 i 2 i 3 2 i 1 ∈ V ( C F 5 i 1 ) , x 2 + = 1 i 1 i 2 2 i 3 ∈ V ( C F 5 i 3 ) , x 3 + ≠ x 1 + , x 3 + ≠ x 2 + . When x 3 = i 3 i 2 i 1 12 , x 3 + = 2 i 2 i 1 1 i 3 ∈ V ( C F 5 i 3 ) , x 3 − = i 3 i 2 i 1 21 ∈ V ( C F 5 1 ) , and x 3 − is adjacent to x 1 . Since x 1 + = 1 i 2 i 3 2 i 1 , x 2 + = 1 i 1 i 2 2 i 3 and 1 ≠ 2 , x 3 + ≠ x 1 + , x 3 + ≠ x 2 + . When x 3 = i 1 i 3 i 2 12 , x 3 + = 2 i 3 i 2 1 i 1 ∈ V ( C F 5 i 1 ) , x 3 − = i 1 i 3 i 2 21 ∈ V ( C F 5 1 ) , and x 3 − is adjacent to x 2 . Since x 1 + = 1 i 2 i 3 2 i 1 , x 2 + = 1 i 1 i 2 2 i 3 and 1 ≠ 2 , x 3 + ≠ x 1 + , x 3 + ≠ x 2 + . So this structure does not exist.
If x 2 = i 2 i 3 i 1 i 4 1 , then x 2 + = 1 i 3 i 1 i 4 i 2 , x 2 − = i 2 i 3 i 1 1 i 4 . Since x 1 + = 1 i 2 i 3 i 4 i 1 , x 1 − = i 1 i 2 i 3 1 i 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 − } ⊆ V ( C F 5 2 ) . 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 + = 2 i 2 i 1 1 i 3 ∈ V ( C F 5 i 3 ) , x 3 − = i 3 i 2 i 1 21 ∈ V ( C F 5 1 ) , and x 3 − is adjacent to x 2 . Since x 1 + = 1 i 2 i 3 2 i 1 ∈ V ( C F 5 i 1 ) , x 2 + = 1 i 3 i 1 2 i 2 ∈ V ( C F 5 i 2 ) , x 3 + ≠ x 1 + , x 3 + ≠ x 2 + . When x 3 = i 1 i 3 i 2 12 , x 3 + = 2 i 3 i 2 1 i 1 ∈ V ( C F 5 i 1 ) , x 3 − = i 1 i 3 i 2 21 ∈ V ( C F 5 1 ) , and x 3 − is adjacent to x 1 . Since x 1 + = 1 i 2 i 3 2 i 1 , x 2 + = 1 i 3 i 1 2 i 2 and 1 ≠ 2 , x 3 + ≠ x 1 + , x 3 + ≠ x 2 + . When x 3 = i 2 i 1 i 3 12 , x 3 + = 2 i 1 i 3 1 i 2 ∈ V ( C F 5 i 2 ) , x 3 − = i 2 i 1 i 3 21 ∈ V ( C F 5 1 ) , and x 3 − is adjacent to x 2 . Since x 1 + = 1 i 2 i 3 2 i 1 , x 2 + = 1 i 3 i 1 2 i 2 and 1 ≠ 2 , x 3 + ≠ x 1 + , x 3 + ≠ x 2 + .
Thus the structure shown in Figure 2 (b) does not exist, | N C F 5 ( S ) | ≥ 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 ∈ V ( C F 5 1 ) , x 2 ∈ V ( C F 5 2 ) , x 3 ∈ V ( C F 5 3 ) . By the definition of C F n , we can get | N C F 5 i ( x i ) | = 3 ( n − 1 ) − 4 2 = 4 ( i ∈ { 1 , 2 , 3 } ) , thus | N C F 5 ( S ) | ≥ | N C F 5 1 ( x 1 ) | + | N C F 5 2 ( x 2 ) | + | N C F 5 3 ( x 3 ) | = 12 .
Combining the above three situations, we can get | N C F 5 ( S ) | ≥ 12 .
Lemma 4.3. When n is odd, let S = { x 1 , x 2 , x 3 } is an I n d -set, where { x 1 , x 2 } ⊆ V ( C F n i ) , x 3 ∈ V ( C F n j ) ( i ≠ j ) and | N C F n i ( { x 1 , x 2 } ) | = 3 n − 10 . Then | N C F n ( S ) | ≠ 9 n − 23 2 .
Proof. When n is odd, | N C F n ( S ) | = 9 n − 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 } ⊆ V ( C F n 1 ) , then x 1 = i 1 i 2 i 3 ⋯ i n − 2 i n − 1 1 , x 2 = j 1 j 2 j 3 ⋯ j n − 2 j n − 1 1 . As | N C F n i ( { x 1 , x 2 } ) | = 3 n − 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 ≅ 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 ( { x 1 , x 2 } ) | ≤ 2 , a contradiction. So j n − 1 = i n − 1 , x 1 + = 1 i 2 i 3 ⋯ i n − 2 i n − 1 i 1 , x 1 − = i 1 i 2 i 3 ⋯ i n − 2 1 i n − 1 , x 2 + = 1 j 2 j 3 ⋯ j n − 2 i n − 1 j 1 , x 2 − = j 1 j 2 j 3 ⋯ j n − 2 1 i n − 1 . By Corollary 3.2, we know j 1 ≠ 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 − } ⊆ V ( C F n i n − 1 ) and x 3 ∈ V ( C F n i n − 1 ) . Now we assume x 3 = k 1 k 2 k 3 ⋯ 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 ⋯ k n − 2 1 i n − 1 or x 3 = 1 k 2 k 3 ⋯ k n − 2 k n − 1 i n − 1 . When x 3 = k 1 k 2 k 3 ⋯ k n − 2 1 i n − 1 , x 3 + = i n − 1 k 2 k 3 ⋯ k n − 2 1 k 1 , x 3 − = k 1 k 2 k 3 ⋯ k n − 2 i n − 1 1 . In this situation, x 3 − ∈ V ( C F n 1 ) , x 3 + ∈ V ( C F n k 1 ) , since i n − 1 ≠ 1 , we have x 3 + ≠ x 1 + , x 3 + ≠ x 2 + . Thus this structure does not exist. When x 3 = 1 k 2 k 3 ⋯ 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 − ≠ 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 ≠ i 1 , j n − 1 = i n − 1 , { x 1 − , x 2 − } ⊆ V ( C F n i n − 1 ) and x 3 ∈ V ( C F n i n − 1 ) . We let x 3 = k 1 k 2 k 3 ⋯ k n − 2 k n − 1 i n − 1 , then x 3 + = i n − 1 k 2 k 3 ⋯ k n − 2 k n − 1 k 1 , x 3 − = k 1 k 2 k 3 ⋯ k n − 2 i n − 1 k n − 1 . Thus x 3 − ≠ x 1 + and x 3 − ≠ 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 I n d -set, where { x 1 , x 2 } ⊆ V ( C F n i ) , x 3 ∈ V ( C F n j ) ( i ≠ j ) and | N C F n i ( { x 1 , x 2 } ) | = 3 n − 9 . Then | N C F n ( S ) | ≠ 9 n − 24 2 .
Proof. When n is even, | N C F n ( S ) | = 9 n − 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 ≅ C F n − 1 , n − 1 is odd.
Suppose on the contrary, we assume this structure exists, as | N C F n i ( { x 1 , x 2 } ) | = 3 n − 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 ⋯ i n − 2 i n − 1 i , x 2 = j 1 j 2 j 3 ⋯ j n − 2 j n − 1 i . By Corollary 3.2, we know j i ∈ [ i 1 , i n − 1 ] and j 1 ≠ i 1 . Thus x 1 + ∈ V ( C F n i 1 ) , x 2 + ∈ V ( C F n j 1 ) , 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 ( S ) | ≠ 9 n − 24 2 .
Lemma 4.5. When n ≥ 4 , let S is an I n d -set and | S | = 3 , then when n is odd, | N C F n ( S ) | ≥ 9 n − 21 2 ; when n is even, | N C F n ( S ) | ≥ 9 n − 24 2 .
Proof. Let S = { x 1 , x 2 , x 3 } , since S is an I n d -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 ≥ 6 and the result holds for C F n − 1 . Note that C F n i ≅ 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 ( S ) | ≥ 9 ( n − 1 ) − 24 2 = 9 n − 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 ( S ) | = | N C F n i ( S ) | + | N C F n − C F n i ( S ) | ≥ 9 n − 33 2 + 6 = 9 n − 21 2 .
When n is even, by induction hypothesis, we have | N C F n i ( S ) | ≥ 9 ( n − 1 ) − 21 2 = 9 n − 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 ( S ) | = | N C F n i ( S ) | + | N C F n − C F n i ( S ) | ≥ 9 n − 30 2 + 3 = 9 n − 24 2 .
Case 2.  x 1 , x 2 , x 3 belong to two different subgraphs C F n i , C F n j ( i ≠ j ) .
In this case, we can let { x 1 , x 2 } ⊆ V ( C F n i ) , x 3 ∈ V ( C F n j ) . By Lemma 3.3 , we can get: when n is odd, | N C F n i ( { x 1 , x 2 } ) | ≥ 3 ( n − 1 ) − 7 = 3 n − 10 ; when n is even, | N C F n i ( { x 1 , x 2 } ) | ≥ 3 ( n − 1 ) − 6 = 3 n − 9 . By Proposition 2.8, we know when n is odd, | N C F n j ( x 3 ) | = 3 ( n − 1 ) − 4 2 = 3 n − 7 2 ; when n is even, | N C F n j ( x 3 ) | = 3 ( n − 1 ) − 3 2 = 3 n − 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 ( S ) | ≥ | N C F n i ( { x 1 , x 2 } ) | + | N C F n j ( x 3 ) | + 2 = 9 n − 23 2 . By Lemma 4.3, we know | N C F n ( S ) | ≠ 9 n − 23 2 , thus | N C F n ( S ) | ≥ 9 n − 23 2 + 1 = 9 n − 21 2 . When n is even, if the structure of Figure 4 exist, then | N C F n ( S ) | ≥ | N C F n i ( { x 1 , x 2 } ) | + | N C F n j ( x 3 ) | = 3 n − 9 + 3 n − 6 2 = 9 n − 24 2 . By Lemma 4.4, we know this structure does not exist, so | N C F n ( S ) | ≥ 9 n − 24 2 + 1 = 9 n − 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 ∈ V ( C F n 1 ) , x 2 ∈ V ( C F n 2 ) , x 3 ∈ V ( C F n 3 ) . By Proposition 2.8, we have when n is odd, | N C F n i ( x i ) | = 3 n − 7 2 ( i ∈ { 1 , 2 , 3 } ) ; when n is even, | N C F n i ( x i ) | = 3 n − 6 2 ( i ∈ { 1 , 2 , 3 } ) . Thus when n is odd, | N C F n ( S ) | ≥ 3 × 3 n − 7 2 = 9 n − 21 2 ; when n is even, | N C F n ( S ) | ≥ 3 × 3 n − 6 2 = 9 n − 18 2 .
Thus the result holds.
Corollary 4.6. When n is even, let S = { x 1 , x 2 , x 3 } is an I n d -set, if | N C F n ( S ) | = 9 n − 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 | F | ≤ 11 , then C F 5 − F contains a big component C, which satisfies the result | V ( C ) | ≥ n ! − | F | − 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 ∩ V ( C F 5 i ) for i = 1 , 2 , 3 , 4 , 5 with | F i 1 | ≥ | F i 2 | ≥ | F i 3 | ≥ | F i 4 | ≥ | F i 5 | , where i j ∈ { 1 , 2 , 3 , 4 , 5 } . If | F i 3 | = 0 , then | F i 4 | = | F i 5 | = 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 ( r e s p . , 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 | ≥ 1 . Since | F | ≤ 11 , we have | F i 5 | ≤ 2 , | F i 4 | ≤ 2 , 1 ≤ | F i 3 | ≤ 3 , 1 ≤ | F i 2 | ≤ 5 , 1 ≤ | F i 1 | ≤ 9 . Firstly, we proof the following Claim is correct.
Claim 1. If | F i j | ≤ 3 for some i j ∈ [ 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 ∈ [ i j , i 5 ] . On the other hand, as | F i 5 | ≤ 2 , we can get | E p , i 5 ( C F 5 ) | = 12 > 5 ≥ | F p | + | F i 5 | , which implies E p , i 5 ( C F 5 − F ) ≠ ∅ for p ∈ [ i j , i 4 ] . Hence C F 5 [ i j , i 5 ] − F [ i j , i 5 ] is connected.
Since | F i 5 | ≤ | F i 4 | ≤ | F i 3 | ≤ 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 | E i 2 , i 3 ( C F 5 ) | = 12 > 8 ≥ | F i 2 ∪ F i 3 | , C F [ i 2 , i 5 ] − F [ i 2 , i 5 ] is connected. As | E i 1 , i 5 ( C F 5 ) | = 12 > 9 + 2 = 11 ≥ | F i 1 ∪ 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 ∈ { i 1 , i 2 } is disconnected, which leads to the following two cases.
Note that if | F i 3 ∪ F i 4 ∪ F i 5 | ≤ 1 , by Proposition 2.3 , we can get C F 5 i − F i ( i ∈ { i 1 , i 2 } ) has a big component C i and at most two vertices, which has a neighbor in F i 3 ∪ F i 4 ∪ 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 ( 1 ) . 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 ( 2 ) . Hence we only think about this situation: | F i 3 ∪ F i 4 ∪ F i 5 | ≥ 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 | ≥ | F i 2 | ≥ 4 . Since | F i 3 ∪ F i 4 ∪ F i 5 | ≥ 2 , | F i 1 ∪ F i 2 | ≤ | F | − | F i 3 ∪ F i 4 ∪ F i 5 | ≤ 11 − 2 = 9 . Hence | F i 2 | = 4 , 4 ≤ | F i 1 | ≤ 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: ( 1 )   C F 5 i 1 − F i 1 has two components, one of which is a singleton. ( 2 )   C F 5 i 1 − F i 1 has three components, two of which are singletons. For ( 1 ) , let C 1 is the big component and x 1 is the singleton of C F 5 i 1 − F i 1 , since | V ( C 1 ) | = | V ( C 2 ) | = | V ( C F 5 i ) − F i − { x i } | ≥ 4 ! − 5 − 1 = 18   ( i ∈ { i 1 , i 2 } ) and | F i 3 ∪ F i 4 ∪ F i 5 | ≤ 3 , by Proposition 2.3 , we can get C F 5 [ V ( C F 5 [ i 3 , i 5 ] − F [ i 3 , i 5 ] ) ∪ V ( C 1 ) ] is connected. Similarly, we can also get C F 5 [ V ( C F 5 [ i 3 , i 5 ] − F [ i 3 , i 5 ] ) ∪ V ( C 2 ) ] is connected. Thus the result holds. For ( 2 ) , 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 | F | ≥ 12 , this contradicts to the fact | F | ≤ 11 . If x 2 is adjacent to { x 1 , x 3 } , say ( x 1 , x 2 ) ∈ 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 | F | ≥ | N C F 5 ( { x 1 , x 2 } ) ∪ N C F 5 ( x 3 ) | ≥ 5 × 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 ∪ F i 4 ∪ F i 5 | ≥ 2 , | F i 1 ∪ F i 2 | ≤ 9 . As C F 5 i 2 − F i 2 is disconnected, we have | F i 2 | ≥ 4 and then | F i 1 | ≤ 5 . If | F i 2 | = 5 , then | F i 1 | ≥ 5 , | F | ≥ | F i 1 | + | F i 2 | + | F i 3 ∪ F i 4 ∪ F i 5 | ≥ 5 + 5 + 2 = 12 , a contradiction. Thus | F i 2 | = 4 . Since | E i 1 , i 3 ( C F 5 ) | = 12 > 8 ≥ | F i 1 ∪ F i 3 | , C F 5 [ V ( C F 5 [ i 3 , i 5 ] − F [ i 3 , i 5 ] ) ∪ 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 ] ) ∪ V ( C F 5 i 1 − F i 1 ) ∪ V ( S ) ] is connected. Then C F 5 − F must satisfies condition ( 1 ) .
Case 3. Only C F 5 i 1 − F i 1 is disconnected.
In this case, we have | F i 1 | ≥ 4 by Proposition 2.8 and | F i 2 | ≤ 4 since | F | ≤ 11 and | F i 3 ∪ F i 4 ∪ F i 5 | ≥ 2 . As | E i 2 , i 3 ( C F 5 ) | = 12 > 7 ≥ | F i 2 ∪ F i 3 | , we have C F 5 [ i 2 , i 5 ] − F [ i 2 , i 5 ] is connected.
If | F i 1 | ≤ 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 ] ) ∪ V ( S ) ] is connected. Then C F 5 − F must be one of conditions ( 1 ) and ( 2 ) .
Now, we suppose | F i 1 | ≥ 6 . Then | F i 2 ∪ F i 3 ∪ F i 4 ∪ F i 5 | ≤ 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 | W | ≤ | F − F i 1 | ≤ 5 , which implies | W | ≤ 2 . Thus C F 5 − F satisfies ( 1 ) or ( 2 ) .
Combing the above three cases, we know this result holds.
Lemma 4.8. For n = 6 , if | F | ≤ 14 , then C F 6 − F contains a big component C, which satisfies the result | V ( C ) | ≥ n ! − | F | − 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 ∩ V ( C F 6 i ) for i ∈ [ 1 , 6 ] with | F i 1 | ≥ | F i 2 | ≥ | F i 3 | ≥ | F i 4 | ≥ | F i 5 | ≥ | F i 6 | , where i j ∈ { 1 , 2 , 3 , 4 , 5 , 6 } . If | F i 2 | = 0 , then | F i 3 | = | F i 4 | = ⋯ = | F i 6 | = 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 | ≥ 1 . Since | F | ≤ 14 , we have | F i 6 | ≤ 2 , | F i 5 | ≤ 2 , | F i 4 | ≤ 3 , | F i 3 | ≤ 4 , 1 ≤ | F i 2 | ≤ 7 , 1 ≤ | F i 1 | ≤ 13 . Firstly, we proof the following Claim is correct.
Claim 2. If | F i j | ≤ 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 ∈ [ i j , i 6 ] . On the other hand, since | E p , i 6 ( C F 6 ) | = ( 6 − 2 ) ! = 24 > 7 ≥ | F p ∪ F i 6 | for p ∈ [ i j , i 5 ] , we can get E p , i 6 ( C F 6 − F ) ≠ ∅ . 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 | E i 2 , i 3 ( C F 6 ) | = ( 6 − 2 ) ! = 24 > 11 ≥ | F i 2 ∪ F i 3 | , | E i 1 , i 3 ( C F 6 ) | = ( 6 − 2 ) ! = 24 > 17 ≥ | F i 1 ∪ 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 ∈ { 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 ≤ | F i 2 | ≤ | F i 1 | ≤ | F | − | F i 2 | ≤ 8 < 9 . By Corollary 3.4, we know that C F 6 i 1 − F i 1 ( r e s p . , C F 6 i 2 − F i 2 ) has a big component C 1 ( r e s p . , C 2 ) and one singleton x 1 ( r e s p . , x 2 ) . As | E C F 6 − F ( V ( C 1 ) , V ( C F 6 i 3 − F i 3 ) ) | ≥ 24 − 2 − 8 − 1 = 13 > 1 . Thus C F 6 − F [ V ( C F 6 [ i 3 , i 6 ] − F [ i 3 , i 6 ] ) ∪ V ( C 1 ) ] is connected. Similarly, we can get C F 6 − F [ V ( C F 6 [ i 3 , i 6 ] − F [ i 3 , i 6 ] ) ∪ V ( C 2 ) ] 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 | F i 2 | ≥ 6 and then | F i 1 | ≤ 8 . Since | E i 1 , i 3 ( C F 6 ) | = 24 > 10 ≥ | F i 1 ∪ F i 3 | , C F 6 [ V ( C F 6 [ i 3 , i 6 ] − F [ i 3 , i 6 ] ) ∪ V ( C F 6 i 1 − F i 1 ) ] is connected. Since | F i 2 | ≤ | F i 1 | ≤ 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 ( C F 6 ) | = 24 > 11 ≥ | F i 2 ∪ F i 3 | + 1 , we can get C F 6 [ V ( C F 6 [ i 3 , i 6 ] − F [ i 3 , i 6 ] ) ∪ V ( C F 6 i 1 − F i 1 ) ∪ V ( C ) ] is connected. Then C F 6 − F must be one of conditions ( 1 ) .
Case 3. Only C F 6 i 1 − F i 1 is disconnected.
In this case, 6 ≤ | F i 1 | ≤ 13 , | F i 2 | ≤ 8 . As | E i 2 , i 3 ( C F 6 ) | = 24 > 11 > | F i 2 ∪ F i 3 | , we have C F 6 [ i 2 , i 6 ] − F [ i 2 , i 6 ] is connected.
If | F i 1 | ≤ 11 , by Lemma 4.8 , C F 6 i 1 − F i 1 has a big component C with | V ( C ) | ≥ 5 ! − | F i 1 | − 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 ] ) ∪ V ( C ) ] is connected. Thus the result holds.
If | F i 1 | ≥ 12 , then | F i 2 ∪ F i 3 ∪ F i 4 ∪ F i 5 ∪ F i 6 | ≤ 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 | W | ≤ | F − F i 1 | ≤ 2 . Thus the result holds.
Lemma 4.9. Let | F | ≤ 9 n − 23 2 for odd n ( n ≥ 5 ) and | F | ≤ 9 n − 26 2 for even n ( n ≥ 6 ) , then C F n − F contains a big component C, which satisfies that | V ( C ) | ≥ n ! − | F | − 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 ≥ 7 and the result holds for C F n − 1 . Now we suppose C F n − F is disconnected for any F ⊆ V ( C F n ) with | F | ≤ 9 n − 23 2 or | F | ≤ 9 n − 26 2 . Let F i = F ∩ V ( C F n i ) for i ∈ [ 1 , n ] with | F i 1 | ≥ | F i 2 | ≥ ⋯ ≥ | F i n | , where i j ∈ [ 1 , n ] .
When n is odd, if | F i 3 | = 0 , then | F i 4 | = | F i 5 | = ⋯ = | F i n | = 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 | ≥ 1 ; When n is even, if | F i 2 | = 0 , then | F i 3 | = | F i 4 | = ⋯ = | F i n | = 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 | ≥ 1 .
Claim 3. When n is even, if | F i j | ≤ 3 n − 8 2 ( i j ∈ [ 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 | F i j | ≤ 3 n − 9 2 ( i j ∈ [ 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 ∈ [ i j , i n ] . On the other hand, since | E p , i n ( C F n ) | = ( n − 2 ) ! > 3 n − 8 > | F p ∪ F i n | for p ∈ [ i j , i n − 1 ] ( n is even) and | E p , i n ( C F n ) | = 2 ( n − 2 ) ! > 3 n − 9 > | F p ∪ F i n | for p ∈ [ i j , i n − 1 ] ( n is odd), we can get E p , i n ( C F n − F ) ≠ ∅ . Thus C F n [ i j , i n ] − F [ i j , i n ] is connected.
Since | F | ≤ 9 n − 26 2 ( n is even) and | F | ≤ 9 n − 23 2 ( n is odd), we have 3 n − 8 2 > | F i 3 | ≥ | F i 4 | ≥ ⋯ ≥ | F i n | for even n and 3 n − 9 2 ≥ | F i 3 | ≥ | F i 4 | ≥ ⋯ ≥ | F i n | 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 | E i 2 , i 3 ( C F n ) | = ( n − 2 ) ! > 9 n − 26 2 > | F i 2 ∪ F i 3 | ( n is even) and | E i 2 , i 3 ( C F n ) | = 2 ( n − 2 ) ! > 9 n − 23 2 > | F i 2 ∪ 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 ∈ { i 1 , i 2 } ) is disconnected, which leads to the following cases.
Note that, when n is odd, if | F i 3 ∪ F i 4 ∪ F i 5 ∪ ⋯ ∪ F i n | ≤ 1 , by the same argument of Lemma 4.7, we know C F n − F satisfies condition ( 1 ) or ( 2 ) . Hence we assume that | F i 3 ∪ F i 4 ∪ F i 5 ∪ ⋯ ∪ F i n | ≥ 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 3 n − 6 2 ≤ | F i 2 | ≤ | F i 1 | ≤ | F | − | F i 2 | ≤ 9 n − 26 2 − 3 n − 6 2 = 3 n − 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 ( C 1 ) , V ( C F n i 3 − F i 3 ) ) | ≥ ( n − 2 ) ! − 1 − 9 n − 26 2 > 1 , | E C F n − F ( V ( C 2 ) , V ( C F n i 3 − F i 3 ) ) | ≥ ( n − 2 ) ! − 1 − 9 n − 26 2 > 1 , Thus C F n − F [ V ( C F n [ i 3 , i n ] − F [ i 3 , i n ] ) ∪ V ( C 1 ) ∪ V ( C 2 ) ] is connected, the result holds.
When n is odd, we have 3 n − 7 2 ≤ | F i 2 | ≤ | F i 1 | ≤ | F | − | F i 2 | − | F i 3 ∪ F i 4 ∪ ⋯ ∪ F i n | ≤ 9 n − 23 2 − 3 n − 7 2 − 2 = 3 n − 10 .So by Lemma 1, we would consider the following three subcases: ( 1 ) 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; ( 2 ) 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; ( 3 ) 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 | V ( C 1 ) | = | V ( C F n i 1 ) − F i 1 − { x 1 , y 1 } | ≥ ( n − 1 ) ! − ( 3 n − 10 ) − 2 and | F i 3 ∪ F i 4 ∪ ⋯ ∪ F i n | ≤ 3 n − 9 2 , by Proposition 2.3 , we know C F n [ V ( C F n [ i 3 , i n ] − F [ i 3 , i n ] ) ∪ V ( C 1 ) ] is connected. Similarly, we can get C F n [ V ( C F n [ i 3 , i n ] − F [ i 3 , i n ] ) ∪ V ( C 2 ) ] 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 | F | ≥ | N C F n i 1 ( { x 1 , y 1 } ) ∪ N C F n i 2 ( { x 2 , y 2 } ) ∪ ( F i 3 ∪ F i 4 ∪ ⋯ ∪ F i n ) | ≥ [ 3 ( n − 1 ) − 4 2 × 2 − 3 ] × 2 + 2 = 6 n − 18 > 9 n − 23 2 , a contradiction.
If C F n − F has three singletons, then by Lemma 4.5 , we can get | F | ≥ 9 n − 21 2 , this contradicts to the fact | F | ≤ 9 n − 23 2 . So C F n − F has two singletons or only one singleton.
Claim 4. If ( C F n − F ) [ { x 1 , y 1 , x 2 , y 2 } ] has at least one edge, say ( x 1 , x 2 ) ∈ 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 ) ∈ E ( C F n − F ) . If y 1 is a singleton, then | F | ≥ | N C F n ( { x 1 , x 2 } ) ∪ N C F n ( y 1 ) | ≥ ( 3 n − 3 2 − 1 ) × 2 + 3 n − 3 2 − 3 = 9 n − 19 2 > 9 n − 23 2 , a contradiction. If ( y 1 , y 2 ) ∈ E ( C F n − F ) , then | F | ≥ | N C F n ( { x 1 , x 2 } ) ∪ N C F n ( { y 1 , y 2 } ) | ≥ ( 3 n − 3 2 − 1 ) × 4 − 3 × 2 = 6 n − 16 > 9 n − 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 | F i 2 | ≥ 3 n − 6 2 for even n and | F i 2 | ≥ 3 n − 7 2 for odd n, then | F i 1 | ≤ 3 n − 10 < 3 n − 9 ( n is even) and | F i 1 | ≤ 3 n − 10 ( n is odd). Since | E i 1 , i 3 ( C F n ) | = ( n − 2 ) ! > 9 n − 26 2 ≥ | F i 1 ∪ F i 3 | ( n is even) and | E i 1 , i 3 ( C F n ) | = 2 ( n − 2 ) ! > 9 n − 23 2 ≥ | F i 1 ∪ F i 3 | ( n is odd), C F n [ V ( C F n [ i 3 , i n ] − F [ i 3 , i n ] ) ∪ V ( C F n i 1 − F i 1 ) ] is connected. Since | F i 2 | ≤ | F i 1 | ≤ 3 n − 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 ] ) ∪ V ( C F n i 1 − F i 1 ) ∪ V ( C ) ] is connected. Then C F n − F must be one of conditions ( 1 ) and ( 2 ) .
Case 3. Only C F n i 1 − F i 1 is disconnected.
In this case, 3 n − 6 2 ≤ | F i 1 | ≤ 9 n − 28 2 for even n and 3 n − 7 2 ≤ | F i 1 | ≤ 9 n − 29 2 for odd n. As | E i 2 , i 3 ( C F n ) | = ( n − 2 ) ! > 9 n − 26 2 ≥ | F i 2 ∪ F i 3 | ( n is even) and | E i 2 , i 3 ( C F n ) | = 2 ( n − 2 ) ! > 9 n − 23 2 ≥ | F i 2 ∪ 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 | F i 1 | ≤ 9 n − 32 2 , by introduction, we know C F n i 1 − F i 1 has a big component C with | V ( C ) | ≥ ( n − 1 ) ! − | F i 1 | − 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 ] ) ∪ V ( C ) ] is connected. Thus the result holds. If | F i 1 | ≥ 9 n − 30 2 , then | F i 2 ∪ F i 3 ∪ ⋯ ∪ F i n | ≤ 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 | W | ≤ | F − F i 1 | ≤ 2 . Thus the result holds.
When n is odd, if | F i 1 | ≤ 9 n − 35 2 , by introduction, C F n i 1 − F i 1 has a big component C with | V ( C ) | ≥ ( n − 1 ) ! − | F i 1 | − 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 ] ) ∪ V ( C ) ] is connected. Thus the result holds. If | F i 1 | ≥ 9 n − 33 2 , then | F i 2 ∪ F i 3 ∪ ⋯ ∪ F i n | ≤ 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 | W | ≤ | F − F i 1 | ≤ 5 . Then we have | W | ≤ 2 . Thus the result holds.
Theorem 2. For n ≥ 4 , when n is odd, c κ 4 ( C F n ) = 9 n − 21 2 ; when n is even, c κ 4 ( C F n ) = 9 n − 24 2 .
Proof. By Lemma 1, we have c κ 4 ( C F 4 ) ≥ 6 = 9 × 4 − 24 2 . For n ≥ 5 , by Lemma 4.9 , we can get when n is odd, c κ 4 ( C F n ) ≥ 9 n − 21 2 ; when n is even, c κ 4 ( C F n ) ≥ 9 n − 24 2 . Next, we will prove that c κ 4 ( C F n ) ≤ 9 n − 21 2 and c κ 4 ( C F n ) ≤ 9 n − 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 κ 4 ( C F 4 ) ≤ 6 = 9 × 4 − 24 2 . For n ≥ 5 , when n is odd, let S = { x 1 , x 2 , x 3 } , where x 1 = i 1 i 2 i 3 ⋯ i n − 4 3214 , x 2 = 2 i 2 i 3 ⋯ i n − 4 i 1 314 , x 3 = 2 i 2 i 3 ⋯ i n − 4 3 i 1 41 , then | N C F n ( S ) | = 3 n − 10 + 3 n − 7 2 + 3 = 9 n − 21 2 and there are three singletons { x 1 , x 2 , x 3 } in C F n − N C F n ( S ) . Thus when n is odd, c κ 4 ( C F n ) ≤ 9 n − 21 2 . When n is even, let S = { x 1 , x 2 , x 3 } , where x 1 = i 1 i 2 i 3 ⋯ i n − 4 3214 j , x 2 = 2 i 2 i 3 ⋯ i n − 4 i 1 314 j , x 3 = 2 i 2 i 3 ⋯ i n − 4 3 i 1 41 j , then { x 1 , x 2 , x 3 } ⊆ V ( C F n j ) and | N C F n j ( S ) | = 9 ( n − 1 ) − 21 2 = 9 n − 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 ( S ) | = 9 n − 30 2 + 3 = 9 n − 24 2 and x 1 , x 2 , x 3 are three singletons in C F n − N C F n ( S ) . Thus when n is even, c κ 4 ( C F n ) ≤ 9 n − 24 2 .

5. The 5-component connectivity of C F n

Lemma 5.1. For n = 4 , let S is an I n d -set and | S | = 4 , then | N C F 4 ( S ) | ≥ 8 .
Proof. Let S = { x 1 , x 2 , x 3 , x 4 } , since S is an I n d -set, x 1 , x 2 , x 3 , x 4 are nonadjacent to each other. As C F 4 i ≅ C F 3 ( i ∈ { 1 , 2 , 3 , 4 } ) , 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 } ⊆ V ( C F 4 1 ) , { x 3 , x 4 } ⊆ V ( C F 4 2 ) . By the definition of C F n , | N C F 4 1 ( { x 1 , x 2 } ) | = | N C F 4 2 ( { x 3 , x 4 } ) | = 3 . Now we let x 1 = 2341 , then x 2 = 4231 or x 2 = 3421 . Thus x 1 + ∈ V ( C F 4 2 ) , x 2 + = 1234 ∈ V ( C F 4 4 ) or x 2 + = 1423 ∈ V ( C F 4 3 ) . 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 ( S ) | = 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 ( S ) | ≥ 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 } ⊆ V ( C F 4 1 ) , x 4 ∈ V ( C F 4 2 ) . We let x 1 = 2341 , x 2 = 4231 , x 3 = 3421 , then x 1 + = 1342 ∈ V ( C F 4 2 ) , x 2 + = 1234 ∈ V ( C F 4 4 ) , x 3 + = 1423 ∈ V ( C F 4 3 ) . Clearly, | N C F 4 1 ( { x 1 , x 2 , x 3 } ) | = 3 , | N C F 4 2 ( x 4 ) | = 3 . If x 4 + ∈ V ( C F 4 1 ) and is adjacent to { x 1 , x 2 , x 3 } , meanwhile x 1 + is adjacent to x 4 , then | N C F 4 ( S ) | = 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 ( S ) | ≥ 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 } ⊆ V ( C F 4 i ) , x 3 ∈ V ( C F 4 j ) , x 4 ∈ V ( C F 4 k ) . Then | N C F 4 i ( { x 1 , x 2 } ) | = 3 , | N C F 4 j ( x 3 ) | = | N C F 4 k ( x 4 ) | = 3 , | N C F 4 ( S ) | ≥ | N C F 4 i ( { x 1 , x 2 } ) | + | N C F 4 j ( x 3 ) | + | N C F 4 k ( x 4 ) | = 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 ∈ V ( C F n k ) , then | N C F n k ( x k ) | = 3 . Thus | N C F 4 ( S ) | ≥ 4 × | N C F n k ( x k ) | = 4 × 3 = 12 .
Combing the above three cases, we have | N C F 4 ( S ) | ≥ 8 .
Lemma 5.2. When n is odd, let S = { x 1 , x 2 , x 3 , x 4 } is an I n d -set and { x 1 , x 2 } ⊆ V ( C F n i ) , { x 3 , x 4 } ⊆ V ( C F n j ) ( i ≠ j ) . If | N C F n i ( { x 1 , x 2 } ) | = | N C F n j ( { x 3 , x 4 } ) | = 3 n − 10 , then | N C F n − C F n i − C F n j   ( S ) | ≥ 4 .
Proof. Since | N C F n i ( { x 1 , x 2 } ) | = | N C F n j ( { x 3 , x 4 } ) | = 3 n − 10 , by the proof process of Lemma 3.3, we know x 1 , x 2 ( r e s p . , x 3 , x 4 ) have three common neighbors in C F n i ( r e s p . , C F n j ) . So by Lemma 3.1, we can let x 1 = i 1 i 2 ⋯ i n − 2 i n − 1 i , x 2 = k 1 k 2 ⋯ k n − 2 i n − 1 i , where k i ∈ [ i 1 , i n − 2 ] and k 1 ≠ i 1 . Then x 1 + = i i 2 i 3 ⋯ i n − 2 i n − 1 i 1 , x 1 − = i 1 i 2 i 3 ⋯ i n − 2 i i n − 1 , x 2 − = k 1 k 2 ⋯ k n − 2 i i n − 1 , x 2 + = i k 2 k 3 ⋯ k n − 2 i n − 1 k 1 . By Proposition 2.3 and Proposition 2.2, we have 2 ≤ | N C F n − C F n i − C F n j ( { x 1 ,   x 2 } ) | ≤ 4 , 2 ≤ | N C F n − C F n i − C F n j ( { x 3 , x 4 } ) | ≤ 4 . Then we think about the following three cases:
Case 1: | N C F n − C F n i − C F n j ( { x 1 , x 2 } ) | = 4 .
In this case, we can easily get | N C F n − C F n i − C F n j ( S ) | ≥ 4 .
Case 2: | N C F n − C F n i − C F n j ( { x 1 , x 2 } ) | = 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 + ∈ V ( C F n j ) . If | N C F n − C F n i − C F n j ( { x 3 , x 4 } ) | = 4 , then | N C F n − C F n i − C F n j ( S ) | ≥ 4 . If | N C F n − C F n i − C F n j ( { x 3 , x 4 } ) | ≤ 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 ⋯ j n − 2 j n − 1 j or x 3 = j 1 j 2 j 3 ⋯ j n − 2 i j . When x 3 = i j 2 j 3 ⋯ 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 ⋯ l n − 2 j n − 1 j , where l i ∈ ( [ j 2 , j n − 2 ] ∪ { i } ) and l 1 ≠ i . Thus x 3 + = j j 2 j 3 ⋯ j n − 2 j n − 1 i , x 3 − = i j 2 j 3 ⋯ j n − 2 j j n − 1 , x 4 + = j l 2 l 3 ⋯ l n − 2 j n − 1 l 1 , x 4 − = l 1 l 2 l 3 ⋯ l n − 2 j j n − 1 . Then x 3 + ∈ V ( C F n i ) . Since j n − 1 ≠ i , x 4 − ∉ V ( C F n i ) . Thus | N C F n − C F n i − C F n j ( S ) | ≥ 4 as x 4 − ≠ x 1 − , x 4 − ≠ x 2 − , x 4 − ≠ x 2 + . When x 3 = j 1 j 2 j 3 ⋯ j n − 2 i j , then by the proof process of Lemma 3.1, we can let x 4 = l 1 l 2 l 3 ⋯ l n − 2 i j , where l i ∈ [ j 1 , j n − 2 ] and l 1 ≠ j 1 . Then x 3 + = j j 2 j 3 ⋯ j n − 2 i j 1 , x 3 − = j 1 j 2 j 3 ⋯ j n − 2 j i , x 4 + = j l 2 l 3 ⋯ l n − 2 i l 1 , x 4 − = l 1 l 2 l 3 ⋯ l n − 2 j i , { x 3 − , x 4 − } ⊆ V ( C F n i ) . Since j 1 ≠ i , l 1 ≠ i , we know { x 3 + , x 4 + } ⊆ V ( C F n − C F n i − C F n j ) . As i ≠ j and k 1 ≠ j , then x 1 − ≠ x 3 + , x 3 + ≠ x 2 + , x 3 + ≠ x 2 − . Thus | N C F n − C F n i − C F n j ( S ) | ≥ 4 .
Case 3: | N C F n − C F n i − C F n j ( { x 1 , x 2 } ) | = 2 .
In this case, two outgoing neighbors of { x 1 , x 2 } belong to C F n j . Thus { x 1 − , x 2 − } ⊆ V ( C F n j ) and { x 3 , x 4 } ⊆ V ( C F n j ) . Now we let x 3 = j 1 j 2 j 3 ⋯ j n − 2 j n − 1 j , then x 4 = u 1 u 2 u 3 ⋯ u n − 2 j n − 1 j , where u i ∈ [ j 1 , j n − 2 ] and u 1 ≠ j 1 . If | N C F n − C F n i − C F n j ( { x 3 , x 4 } ) | = 4 , clearly | N C F n − C F n i − C F n j ( S ) | ≥ 4 . If | N C F n − C F n i − C F n j ( { x 3 , x 4 } ) | = 3 , the proof process is similar to Case 2, we can get | N C F n − C F n i − C F n j ( S ) | ≥ 4 . So we let | N C F n − C F n i − C F n j ( { x 3 , x 4 } ) | = 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 ⋯ j n − 2 j n − 1 j or x 3 = j 1 j 2 j 3 ⋯ j n − 2 i j . When x 3 = i j 2 j 3 ⋯ j n − 2 j n − 1 j , we can get j n − 1 ≠ i , so x 4 = i u 2 u 3 ⋯ 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 ⋯ j n − 2 i j , x 4 = u 1 u 2 u 3 ⋯ u n − 2 i j . Since x 4 + = j u 2 u 3 ⋯ u n − 2 i u 1 , x 3 + = j j 2 j 3 ⋯ j n − 2 i j 1 and i ≠ j , we have x 3 + ≠ x 1 + , x 3 + ≠ x 2 + , x 4 + ≠ x 1 + , x 4 + ≠ x 2 + . Thus | N C F n − C F n i − C F n j ( S ) | ≥ 4 .
Lemma 5.3. When n is odd, let S = { x 1 , x 2 , x 3 , x 4 } is an I n d -set and { x 1 , x 2 , x 3 } ⊆ V ( C F n i ) , x 4 ∈ V ( C F n j ) ( i ≠ j ) . If | N C F n i ( S ) | = 9 ( n − 1 ) − 24 2 = 9 n − 33 2 , | N C F n j ( S ) | = 3 n − 7 2 , then | N C F n − C F n i − C F n j ( S ) | ≥ 4 .
Proof. Since { x 1 , x 2 , x 3 } ⊆ V ( C F n i ) , C F n i ≅ C F n − 1 , n − 1 is even and | N C F n i ( S ) | = 9 n − 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 ⋯ i n − 2 i n − 1 i , x 2 = j 1 j 2 j 3 ⋯ j n − 2 i n − 1 i , x 3 = k 1 k 2 k 3 ⋯ k n − 2 i n − 1 i . Then x 1 + = i i 2 i 3 ⋯ i n − 2 i n − 1 i 1 , x 1 − = i 1 i 2 i 3 ⋯ i n − 2 i i n − 1 , x 2 + = i j 2 j 3 ⋯ j n − 2 i n − 1 j 1 , x 2 − = j 1 j 2 j 3 ⋯ j n − 2 i i n − 1 , x 3 + = i k 2 k 3 ⋯ k n − 2 i n − 1 k 1 , x 3 − = k 1 k 2 k 3 ⋯ k n − 2 i i n − 1 . If | N C F n − C F n i − C F n j ( { x 1 , x 2 , x 3 } ) | ≥ 4 , then | N C F n − C F n i − C F n j ( S ) | ≥ 4 . Thus we assume | N C F n − C F n i − C F n j ( { x 1 , x 2 , x 3 } ) | = 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 − } ⊆ V ( C F n i n − 1 ) and x 4 ∈ V ( C F n i n − 1 ) .
In this case, we let x 4 = l 1 l 2 l 3 ⋯ l n − 2 l n − 1 i n − 1 , then x 4 + = i n − 1 l 2 l 3 ⋯ l n − 2 l n − 1 l 1 , x 4 − = l 1 l 2 l 3 ⋯ l n − 2 i n − 1 l n − 1 . If x 4 + ∉ V ( C F n i ) and x 4 − ∉ V ( C F n i ) , then | N C F n − C F n i − C F n j ( S ) | ≥ 4 as i n − 1 ≠ i and x 4 + ≠ x 1 + , x 4 + ≠ x 2 + , x 4 + ≠ 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 ⋯ l n − 2 i i n − 1 or x 4 = i l 2 l 3 ⋯ l n − 2 l n − 1 i n − 1 . If x 4 = i l 2 l 3 ⋯ l n − 2 l n − 1 i n − 1 , since | N C F n j ( S ) | = 3 n − 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 ⋯ l n − 2 i i n − 1 , then x 4 + = i n − 1 l 2 l 3 ⋯ l n − 2 i l 1 , x 4 − = l 1 l 2 l 3 ⋯ l n − 2 i n − 1 i . Hence x 4 − ∈ V ( C F n i ) , x 4 + ≠ x 1 + , x 4 + ≠ x 2 + , x 4 + ≠ x 3 + . Thus | N C F n − C F n i − C F n j ( S ) | ≥ 4 .
Case 2: Let i 1 = j 1 = k 1 , { x 1 + , x 2 + , x 3 + } ⊆ V ( C F n i 1 ) and x 4 ∈ V ( C F n i 1 ) .
In this case, x 1 = i 1 i 2 i 3 ⋯ i n − 2 i n − 1 i , x 2 = i 1 j 2 j 3 ⋯ j n − 2 i n − 1 i , x 3 = i 1 k 2 k 3 ⋯ k n − 2 i n − 1 i . Then x 1 + = i i 2 i 3 ⋯ i n − 2 i n − 1 i 1 , x 2 + = i j 2 j 3 ⋯ j n − 2 i n − 1 i 1 , x 3 + = i k 2 k 3 ⋯ k n − 2 i n − 1 i 1 , x 1 − = i 1 i 2 i 3 ⋯ i n − 2 i i n − 1 , x 2 − = i 1 j 2 j 3 ⋯ j n − 2 i i n − 1 , x 3 − = i 1 k 2 k 3 ⋯ k n − 2 i i n − 1 . We let x 4 = l 1 l 2 l 3 ⋯ l n − 2   l n − 1 i 1 , then x 4 + = i 1 l 2 l 3 ⋯ l n − 2 l n − 1 l 1 , x 4 − = l 1 l 2 l 3 ⋯ l n − 2 i 1 l n − 1 . If x 4 + ∉ V ( C F n i ) and x 4 − ∉ V ( C F n i ) , then | N C F n − C F n i − C F n j ( S ) | ≥ 4 as x 4 − ≠ x 1 − , x 4 − ≠ x 2 − , x 4 − ≠ 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 ⋯ i i 1 or x 4 = i l 2 l 3 ⋯ l n − 1 i 1 . When x 4 = l 1 l 2 l 3 ⋯ 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 ⋯ l n − 1 i 1 , x 4 + = i 1 l 2 l 3 ⋯ l n − 1 i , x 4 − = i l 2 l 3 ⋯ i 1 l n − 1 . Then x 4 + ∈ V ( C F n i ) , x 4 − ≠ x 1 − , x 4 − ≠ x 2 − , x 4 − ≠ x 3 − . Thus | N C F n − C F n i − C F n j ( S ) | ≥ 4 .
Lemma 5.4. When n is odd, let S = { x 1 , x 2 , x 3 , x 4 } is an I n d -set and { x 1 , x 2 } ⊆ V ( C F n i ) , { x 3 } ⊆ V ( C F n j ) , { x 3 } ⊆ V ( C F n k ) ( i , j , k are different from each other). If | N C F n i ( { x 1 , x 2 }   ) | = 3 n − 10 , then | N C F n − C F n i − C F n j − C F n k ( S ) | ≥ 1 .
Proof. If | N C F n − C F n i − C F n j − C F n k ( S ) | = 0 , the structure in Figure 5 must exists. Now we can proof this structure does not exist. As | N C F n i ( { x 1 , x 2 } ) | = 3 n − 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 ⋯ i n − 2 j i , then by Lemma 3.1, we can let x 2 = j 1 j 2 j 3 ⋯ j n − 2 j i , where j i ∈ [ i 1 , i n − 2 ] and j 1 ≠ i 1 . Then x 1 + = i i 2 i 3 ⋯ i n − 2 j i 1 , x 2 + = i j 2 j 3 ⋯ j n − 2 j j 1 , x 1 − = i 1 i 2 i 3 ⋯ i n − 2 i j , x 2 − = j 1 j 2 j 3 ⋯ j n − 2 i j . Since i 1 ≠ j and i 1 ≠ 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 ( S ) | ≥ 1 .
Lemma 5.5. For n = 5 , let S is an I n d -set and | S | = 4 , then | N C F 5 ( S ) | ≥ 14 .
Proof. Let S = { x 1 , x 2 , x 3 , x 4 } , since S is an I n d -set, x 1 , x 2 , x 3 , x 4 are nonadjacent to each other. Note that C F 5 i ≅ 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 ( S ) | ≥ 8 . By Proposition 2.3, we know | N C F 5 − C F 5 i ( S ) | = 8 . Thus | N C F 5 ( S ) | = | N C F 5 i ( S ) | + | N C F 5 − C F 5 i ( S ) | ≥ 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 ≠ j ) .
In this case, we need to think about the following two situations:
Subcase 2.1: { x 1 , x 2 } ⊆ V ( C F 5 i ) , { x 3 , x 4 } ⊆ V ( C F 5 j ) .
By Lemma 3.3, we can get | N C F 5 i ( { x 1 , x 2 } ) | ≥ 3 n − 10 = 5 , | N C F 5 j ( { x 3 , x 4 } ) | ≥ 3 n − 10 = 5 . By Lemma 5.2, we have | N C F 5 ( S ) | = | N C F 5 i ( { x 1 , x 2 } ) | + | N C F 5 j ( { x 3 , x 4 } ) | + | N C F 5 − C F 5 i − C F 5 j ( S ) | ≥ 2 × 5 + 4 = 14 .
Subcase 2.2: { x 1 , x 2 , x 3 } ⊆ V ( C F 5 i ) , x 4 ∈ V ( C F 5 j ) .
By Lemma 4.1, we have | N C F 5 i ( { x 1 , x 2 , x 3 } ) | ≥ 6 , | N C F 5 j ( x 4 ) | = 3 × 4 − 4 2 = 4 . By Lemma 5.3, we know | N C F n − C F n i − C F n j ( S ) | ≥ 4 . Thus | N C F n ( S ) | = | N C F 5 i ( { x 1 , x 2 , x 3 } ) | + | N C F 5 j ( x 4 ) | + | N C F n − C F n i − C F n j ( S ) | ≥ 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 } ⊆ V ( C F 5 i ) . Clearly, | N C F 5 i ( { x 1 , x 2 } ) | ≥ 3 n − 10 = 5 , | N C F 5 j ( x 3 ) | = | N C F 5 k ( x 4 ) | = 4 . Thus, by Lemma 5.4, we have | N C F 5 ( S ) | = | N C F 5 i ( { x 1 , x 2 } ) | + | N C F 5 j ( x 3 ) | + | N C F 5 k ( x 4 ) | + | N C F 5 − C F 5 i − C F 5 j − C F 5 k   ( S ) | ≥ 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 ∈ V ( C F 5 k ) . Clearly, | N C F 5 k ( x k ) | = 4 , thus | N C F 5 ( S ) | ≥ 4 × | N C F 5 k ( x k ) | = 4 × 4 = 16 > 14 .
Combing the above four cases, we can get | N C F 5 ( S ) | ≥ 14 .
Lemma 5.6. For n = 6 , let S is an I n d -set and | S | = 4 , then | N C F 6 ( S ) | ≥ 18 .
Proof. Let S = { x 1 , x 2 , x 3 , x 4 } , since S is an I n d -set, x 1 , x 2 , x 3 , x 4 are nonadjacent to each other. Note that C F 6 i ≅ 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 ( S ) | ≥ 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 ( S ) | = | N C F 6 i ( S ) | + 4 ≥ 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 ≠ j ) .
In this case, we also need to think about the following two situations:
Subcase 2.1: { x 1 , x 2 } ⊆ V ( C F 6 i ) , { x 3 , x 4 } ⊆ V ( C F 6 j ) .
By Lemma 3.3, we can get | N C F 6 i ( { x 1 , x 2 } ) | ≥ 3 n − 9 = 9 , | N C F 6 j ( { x 3 , x 4 } ) | ≥ 3 n − 9 = 9 . Thus | N C F 6 ( S ) | ≥ | N C F 6 i ( { x 1 , x 2 } ) | + | N C F 6 j ( { x 3 , x 4 } ) | = 9 + 9 = 18 .
Subcase 2.2: { x 1 , x 2 , x 3 } ⊆ V ( C F 6 i ) , x 4 ∈ V ( C F 6 j ) .
By Lemma 4.2, we have | N C F 6 i ( { x 1 , x 2 , x 3 } ) | ≥ 12 , | N C F 6 j ( x 4 ) | = 3 × 5 − 3 2 = 6 . Thus | N C F 6 ( S ) | ≥ | N C F 6 i ( { x 1 , x 2 , x 3 } ) | + | N C F 6 j ( x 4 ) | = 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 } ⊆ V ( C F 6 i ) . Clearly, | N C F 6 i ( { x 1 , x 2 } ) | ≥ 3 n − 9 = 9 , | N C F 6 j ( x 3 ) | = | N C F 6 k ( x 4 ) | = 6 . Thus, | N C F 6 ( S ) | ≥ | N C F 6 i ( { x 1 , x 2 } ) | + | N C F 6 j ( x 3 ) | + | N C F 6 k ( x 4 ) | = 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 ∈ V ( C F 6 k ) . Clearly, | N C F 6 k ( x k ) | = 6 , thus | N C F 6 ( S ) | ≥ 4 × | N C F 6 k ( x k ) | = 4 × 6 = 24 > 18 .
Combing the above four cases, we can get | N C F 6 ( S ) | ≥ 18 .
Lemma 5.7. For n ≥ 5 , let S is an I n