Upper Bounds of the Generalized Competition Indices of Symmetric Primitive Digraphs with d Loops

Abstract

A digraph (D) is symmetric if $(u,v)$ is an arc of D and if $(v,u)$ is also an arc of D. If a symmetric digraph is primitive and contains d loops, then it is said to be a symmetric primitive digraph with d loops. The m-competition index (generalized competition index) of a digraph is an extension of the exponent and the scrambling index. The m-competition index has been applied to memoryless communication systems in recent years. In this article, we assume that $S_n\left(d\right)$ represents the set of all symmetric primitive digraphs of n vertices with d loops, where $1\le d\le n.$ We study the m-competition indices of $S_n\left(d\right)$ and give their upper bounds, where $1\le m\le n.$ Furthermore, for any integer m satisfying $1\le m\le n,$ we find that the upper bounds of the m-competition indices of $S_n\left(d\right)$ can be reached.

1. Introduction

In this paper, we follow the terminology and notation used in [1,2,3]. Let $D=(V,E)$ represent a digraph (directed graph) with n vertices, where the vertex set is $V=V\left(D\right),$ the arc set is $E=E\left(D\right).$ Multiple arcs are not allowed, but loops are allowed. A walk from u to v in a digraph (D) is a sequence of vertices ($u,v_1,\cdots ,v_t,v\in V\left(D\right)$) and a sequence of arcs ($(u,v_1),(v_1,v_2),\cdots ,(v_t,v)\in E\left(D\right)$), where the vertices and arcs are not necessarily distinct. If $u=v,$ then the walk is said to be a closed walk from u to $v.$ A cycle refers to a closed walk from u to v with different vertices except for $u=v.$ The length of a walk is the number of arcs in the walk. Let $u\stackrel{t}{\to }v$ denote a walk of length t from u to $v.$ Let $d_D(u,v)$ (for short, $d(u,v)$) represent the distance from vertex u to vertex v in $D,$ that is, the length of the shortest walk from u to $v.$

Let $u,v$ be any pair of vertices in a digraph (D). If there exists a positive integer (k) such that there exists a walk of length k from u to v in $D,$ we call such a digraph (D) primitive. The smallest such k is denoted as $exp\left(D\right),$ and it is called the exponent of $D.$ Let the greatest common factor of the length of all cycles in D be $c\left(D\right).$ As is well known, D is primitive if and only if D is strongly connected and $c\left(D\right)=1.$

As a generalization of the competition index in [3], the author of [4] introduced the m-competition index. For a primitive digraph (D) with n vertices and an integer (m) satisfying $1\le m\le n,$ the m-competition index (generalized competition index) of D is denoted as $k_m\left(D\right),$ which is the smallest integer (k) such that for every pair of vertices (u and v), there are m different vertices ($v_1,v_2,\cdots ,v_m$) in D that satisfy $u\stackrel{k}{\to }{v}_i$ and $v\stackrel{k}{\to }{v}_i,$ for each $i=1,2,\cdots ,m.$

Several studies [5,6,7] have investigated exponents and their generalization. The scrambling index of a primitive digraph (D) is denoted as $k\left(D\right)$, as introduced in [2,8]. For a primitive digraph (D), we have $k\left(D\right)=k_1\left(D\right),$ where $k_1\left(D\right)$ is the 1-competition index of D. The scrambling index of primitive digraphs has been studied by some researchers [9,10,11].

For a primitive digraph (D), according to [12], we define the following notation:

$$N^{+}(D^k:u)=\{w\in V\left(D\right)|u\stackrel{k}{\to }w\},$$

(1)

$$N^{+}(D^k:u,v)=N^{+}(D^k:u)\cap N^{+}(D^k:v),$$

(2)

$$k_m(D:u,v)=min\{k:|N^{+}(D^t:u,v)|\ge m,\mathrm{where}t\ge k\},$$

(3)

$$k_m(D:u)=\underset{v\in V\left(D\right)}{max}\left\{k_m(D:u,v)\right\}.$$

(4)

Then, we have

$$k_m\left(D\right)=\underset{u\in V\left(D\right)}{max}{k}_m(D:u)=\underset{u,v\in V\left(D\right)}{max}{k}_m(D:u,v).$$

(5)

Based on the definitions above, it is easy to derive that $k_m(D:u,v)\le k_m(D:u)\le k_m\left(D\right).$ In addition, we have $k\left(D\right)=k_1\left(D\right)\le k_2\left(D\right)\le \cdots \le k_n\left(D\right)=exp\left(D\right).$ Thus, the m-competition index of a digraph is an extension of the exponent and the scrambling index (the 1-competition index).

There has been interest recently in a generalized competition index [12,13,14,15,16,17]. Suppose a memoryless communication system is represented by a primitive digraph of n vertices; then, the m-competition indices represent the longest first time for m vertices to know all 2 bits of the information (see [5,18]). For literature on the generalized $\mu$-scrambling indices, see [18,19,20]. Sombor index was studied in [21,22].

For any pair of vertices ($u,v$) of a digraph (D), if $(u,v)\in E\left(D\right)$ and $(v,u)\in E\left(D\right),$ then we call D a symmetric digraph. If a symmetric digraph (D) is primitive, then we call D a symmetric primitive digraph. If a symmetric primitive digraph (D) contains exactly d loops, then we call D a symmetric primitive digraph with d loops.Let $S_n\left(d\right)$ denote the set of all symmetric primitive digraphs of order n with d loops, where $1\le d\le n.$ We can treat a symmetric digraph (possibly with loops) as an undirected graph. For some studies on graphs, see [23,24,25].

In the following, for convenience, we use undirected graph terms such as edge, edge set, etc., to describe a symmetric digraph. For a symmetric digraph (a graph, G), let $E\left(G\right)$ represent the set of all edges in $G.$ The notation $[u,v]\in E\left(G\right)$ indicates that there is an edge in G with end vertices u and $v.$

In this paper, let $⌊s⌋$ represent the largest integer not greater than $s,$ and let $⌈t⌉$ represent the smallest integer not less than $t.$ For any symmetric primitive digraph (D) with n vertices, we know that its exponent ($exp\left(D\right)$) satisfies $exp\left(D\right)\le 2n-2$ [6]. In Theorem 3.2 of [9], let the minimum odd cycle length of the symmetric primitive digraph be 1 and $n\ge 3$. Combined with Theorem 4.1 in [9], Proposition 1 is obtained, which gives the upper bounds of the 1-competition index of $S_n\left(d\right).$

Proposition 1.

Let $G\in S_n\left(d\right).$ Let n and d be integers such that $n\ge 3$ and $1\le d\le n.$ Then,

(1)

If $1\le d\le n-⌈\frac{n-1}{2}⌉,$ then $k\left(G\right)=k_1\left(G\right)\le n-d.$

(2)

If $n-⌈\frac{n-1}{2}⌉\le d\le n,$ then $k\left(G\right)=k_1\left(G\right)\le ⌈\frac{n-1}{2}⌉.$

Symmetry is widely used in the field of graph theory. Clearly, a symmetric primitive digraph with d loops is a special kind of digraph. Using a graph-theoretic approach, for any integer m satisfying $1\le m\le n,$ we get the upper bounds of the m-competition indices of such symmetric digraphs.

2. Preliminary

In this section, for the convenience of the following research, we provide some notations and definitions, as well as Observations 1 and 2.

Let $G=(V,E)\in S_n\left(d\right).$ If graph $G^{\prime }$ satisfies $V\left(G^{\prime }\right)\subseteq V\left(G\right)$ and $E\left(G^{\prime }\right)\subseteq E\left(G\right),$ then $G^{\prime }$ is called a subgraph of $G.$ Let $\varnothing \ne H\subseteq V\left(G\right).$ The subgraph is composed of H as the vertex set and all the edges with both end vertices in H as the edge set, which is called the induced subgraph of G from $H,$ and recorded as $G\left[H\right].$

Let $G\in S_n\left(d\right).$ For any pair of vertices $x,y$ in $G,$ the notation $x\stackrel{t}{\to }y$ is used to indicate that there exists a walk of length t from x to $y.$ If $t=0,$ then $x=y.$ Let $W_G(x,y)$ (abbreviated as $W(x,y)$) represent a walk from x to y in G, $\left|W\right(x,y\left)\right|$ represent the length of $W(x,y)$, and $V\left(W\right(x,y\left)\right)$ represent the set of all vertices on the walk ($W(x,y)$) (including $x,y$). Let $\left|U\right|$ represent the number of all elements in the set $U.$$V\left(L\right(G\left)\right)$ is used to denote the set of d loop vertices in $G,$ and $V\left(U\right(G\left)\right)$ is used to denote the set of all non-loop vertices in $G.$ Then, $\left|V\right(L\left(G\right)\left)\right|=d,$$\left|V\right(U\left(G\right)\left)\right|=n-d.$ If the walk $W(x,y)$ passes through a loop vertex, we say that $V\left(W\right(x,y\left)\right)\cap V\left(L\right(G\left)\right)\ne \varnothing .$ Otherwise, we say that $V\left(W\right(x,y\left)\right)\cap V\left(L\right(G\left)\right)=\varnothing .$ We define $d(x,X)=min\left\{d\right(x,v):v\in X\},$ where the vertex $x\in V\left(G\right)$ and the set $X\subseteq V\left(G\right).$ For $x\in X,$ obviously, $d(x,X)=0.$

Let $G\in S_n\left(d\right).$ After deleting any edge $[u,v]$ in G that satisfies $u\ne v,$ if the obtained graph ($G^{*}$) satisfies $G^{*}\notin S_n\left(d\right)$ (that is to say, $G^{*}$ is not connected), then we call $G\in S_n^{\prime }\left(d\right).$ In fact, if $G\in S_n^{\prime }\left(d\right),$ then G is a tree with loops. If $G\in S_n^{\prime }\left(d\right),$ let $x,y$ be any pair of vertices in G such that $x\ne y$; then, there is only one unique path from x to $y.$ Furthermore, $S_n^{\prime }\left(d\right)\subseteq S_n\left(d\right).$

Let $G_1\in S_n^{\prime }\left(d\right)$ and $G_2\in S_n\left(d\right).$ Suppose $G_1,$$G_2$ satisfies $V\left(G_1\right)=V\left(G_2\right)$ and $E\left(G_1\right)\subseteq E\left(G_2\right).$ For any m that satisfies $1\le m\le n,$ then $k_m\left(G_2\right)\le k_m\left(G_1\right).$ Thus, assuming $G\in S_n\left(d\right),$ we study the upper bounds on $k_m\left(G\right),$ we need only to study the graphs in $S_n^{\prime }\left(d\right).$ Thus, we provide the following definitions (1–3).

Definition 1.

Let $G\in S_n^{\prime }\left(d\right)$ and $1\le d\le n.$ Let $x,y$ be any pair of vertices in G. We define the following notation:

(1)

Let $P_G(x,y)$(for short, $P(x,y)$) denote the path from x to y in $G,$ and let $\left|P\right(x,y\left)\right|$ represent the length of $P(x,y).$ Let $V\left(P\right(x,y\left)\right)$ denote the set of all vertices on the path $P(x,y)$ (including $x,y$). If $x=y,$ let us record $V\left(P\right(x,y\left)\right)=\left\{x\right\}=\left\{y\right\}.$

(2)

If the path $P(x,y)$ from x to y passes through a loop vertex (including x or y as a loop vertex), we say that $V\left(P\right(x,y\left)\right)\cap V\left(L\right(G\left)\right)\ne \varnothing .$ Otherwise, we say that $V\left(P\right(x,y\left)\right)\cap V\left(L\right(G\left)\right)=\varnothing .$

Definition 2.

Let $G\in S_n^{\prime }\left(d\right)$ and $1\le d\le n$. We define the following notation and label all vertices of G as follows:

(1)

Suppose that the longest path in G is denoted by $P.$ The number of vertices in the path (P) is denoted by $l.$ Let us assume that the path (P) passes through the vertices $v_1,v_2,\cdots ,v_l$ sequentially. Then, the diameter of G is $l-1,$ that is, $l-1=max\{d(x,y):x,y\in V\left(G\right)\}=d(v_1,v_l).$ We record $c=⌈\frac{l+1}{2}⌉,$ then $c-1=⌈\frac{l-1}{2}⌉$ and $c+1=⌈\frac{l+3}{2}⌉.$ Let $V\left(P\right)$ represent the set of all vertices in the path (P), that is, $V\left(P\right)=\{v_1,v_2,\cdots ,v_l\}.$

(2)

For any vertex, $v\in V\left(G\right)\setminus V\left(P\right),$ since $G\in S_n^{\prime }\left(d\right)$; then, there exists a unique integer (h) such that $d(v,V\left(P\right))=d(v,v_h),$ where $v_h\in V\left(P\right)$ and $2\le h\le l-1.$ Then, we assume $v\in V\left(P_h\right).$ For any vertex $v_k\in V\left(P\right)$, we assume $v_k\in V\left(P_k\right)$ and $v_k=v_{k,0},$ where $1\le k\le l.$ That is, $v_{k,0}$ and $v_k$ are the same vertex. Let $J_{k,i}=\{v:v\in V\left(P_k\right)\setminus V\left(P\right)\mathrm{and}d(v,v_k)=i\},$ where $2\le k\le l-1$ and $i\ge 1.$ Let $|J_{k,i}|=g.$ If $g=1,$ we record the unique vertex in $J_{k,i}$ as $v_{k,i}.$ If $g\ge 2,$ we record the vertices in $J_{k,i}$ as $v_{k,i,1},v_{k,i,2},\cdots ,v_{k,i,g}.$ In the following, for convenience, if $g\ge 2,$ we take any vertex in $J_{k,i}$ that satisfies the given conditions and record the vertex as $v_{k,i},$ which does not affect the conclusion of this paper.

(3)

Let $V\left(T_1\right)={\bigcup }_{k=1}^{c-1}V\left(P_k\right),$$V\left(T_2\right)={\bigcup }_{k=c}^{l}V\left(P_k\right).$

Definition 3.

We define the following notation:

(1)

Let $S_{n,1}^{\prime }\left(d\right)=\{G:V\left(T_1\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(T_2\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ where $G\in S_n^{\prime }\left(d\right)$ and $d\ge 2\}.$

(2)

Let $S_{n,2}^{\prime }\left(d\right)=\{G:V\left(T_1\right)\cap V\left(L\left(G\right)\right)=\varnothing$ or $V\left(T_2\right)\cap V\left(L\left(G\right)\right)=\varnothing ,$ where $G\in S_n^{\prime }\left(d\right)$ and $d\ge 2\}.$

Remark 1.

According to Definition 1, if $G\in S_n^{\prime }\left(d\right),$ for any vertex $x,y$ in $G,$ then $x\stackrel{d(x,y)}{\to }y$ denotes the unique path $P(x,y)$ of length $d(x,y)$ from x to $y.$ According to Definition 2, P is the longest path in $G,$ and l is the number of vertices on the path (P). Therefore, for any vertex $v\in V\left(G\right)$, we have $d(v,v_k)\le max\{k-1,l-k\}$, where $v_k\in V\left(P\right)$ and $1\le k\le l.$ Moreover, we have $\{v_1,v_2,\cdots ,v_{c-1}\}\subseteq V\left(T_1\right),\{v_c,v_{c+1},\cdots ,v_l\}\subseteq V\left(T_2\right).$ According to Definition 3, for $d\ge 2,$ then $S_n^{\prime }\left(d\right)=S_{n,1}^{\prime }\left(d\right)\cup S_{n,2}^{\prime }\left(d\right).$

In the next following, we assume that $n,d$ and m are any integers satisfying $n\ge 5,$$1\le d\le n,1\le m\le n.$ Let $x,y$ be any pair of vertices of $G.$ Let a be any integer satisfying $a\ge 0.$

Observation 1.

Let $G\in S_n\left(d\right).$ If there exists a walk $W(x,y)$ from x to y satisfying $V\left(W\right(x,y\left)\right)\cap V\left(L\right(G\left)\right)\ne \varnothing ,$ then there exists a walk of length s from x to y for any integer $s\ge \left|W\right(x,y\left)\right|.$

Proof. 

If $x\in V\left(U\right(G\left)\right)$ and $y\in V\left(U\right(G\left)\right)$, suppose a walk ($W(x,y)$) from x to y through a loop vertex denoted by $x\stackrel{h}{\to }u\stackrel{\left|W\right(x,y\left)\right|-h}{\to }y,$ where u is the loop vertex and $1\le h\le \left|W\right(x,y\left)\right|-1.$ Since u is a loop vertex, then the length of the walk ($u\stackrel{s+1-\left|W\right(x,y\left)\right|}{\to }u$) from u to u is $s+1-\left|W\right(x,y\left)\right|.$ Therefore, the length of the walk ($x\stackrel{h}{\to }u\stackrel{s+1-\left|W\right(x,y\left)\right|}{\to }u\stackrel{\left|W\right(x,y\left)\right|-h}{\to }y$) is $s+1.$ Moreover, $s+1$ is any integer satisfying $s+1\ge \left|W\right(x,y\left)\right|+1.$

If $x\in V\left(L\right(G\left)\right)$ or $y\in V\left(L\right(G\left)\right)$, assume that $x\in V\left(L\right(G\left)\right).$ Then, the length of the walk ($x\stackrel{s+1-\left|W\right(x,y\left)\right|}{\to }x\stackrel{\left|W\right(x,y\left)\right|}{\to }y$) is $s+1.$

Therefore, the conclusion holds. □

Observation 2. 

Let $G\in S_n\left(d\right).$ Let $f,g$ be integers satisfying $f\ge 0,g\ge 0.$ Suppose that vertex v in G satisfies $\left\{v\right\}\subseteq N^{+}(G^{f+a}:x,y).$ For any vertex z in G that satisfies $d(z,v)\le g,$ then $\left\{z\right\}\subseteq N^{+}(G^{f+g}:x,y).$

Proof. 

Since $\left\{v\right\}\subseteq N^{+}(G^{f+a}:x,y),$ then $\left\{v\right\}\subseteq N^{+}(G^{f+g-d(z,v)}:x,y),$ where a is any integer satisfying $a\ge 0.$ The length of the walk ($x\stackrel{f+g-d(z,v)}{\to }v\stackrel{d(z,v)}{\to }z$) is $f+g.$ Then, $\left\{z\right\}\subseteq N^{+}(G^{f+g}:x).$ Similarly, $\left\{z\right\}\subseteq N^{+}(G^{f+g}:y).$ Therefore, we have $\left\{z\right\}\subseteq N^{+}(G^{f+g}:x,y).$ □

3. Some Results of the $\mathbf{m}$-Competition Indices of ${\mathbf{S}}_{\mathbf{n}}^{\prime }\left(\mathbf{d}\right)$

In this section, for $d\ge 2,$ we divide $S_n^{\prime }\left(d\right)$ into $S_{n,1}^{\prime }\left(d\right)$ and $S_{n,2}^{\prime }\left(d\right).$ For $G\in S_n^{\prime }\left(d\right),$ in order to obtain the upper bounds of $k_m\left(G\right),$ we need to discuss $k_m(G:x,y),$ where $x,y$ are any pair of vertices in $G.$ For any vertex $x,y$ in $G,$ we consider cases in which $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$$V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing ,$$V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing$ and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing .$ To facilitate the proof presented below, we propose the following propositions (2,3) and Lemma 1.

Proposition 2.

Let $G\in S_n^{\prime }\left(d\right).$ Suppose that $\left\{v\right\}\subseteq N^{+}(G^{h+a}:x,y)$ and $\{v_{c-1},v_{c+1},v_c\}\subseteq N^{+}(G^{h+1+a}:x,y),$ where v is a vertex of G and h is an integer satisfying $h\ge 0.$ Then, $k_m(G:x,y)\le h+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Proof. 

Since $\left\{v\right\}\subseteq N^{+}(G^{h+a}:x,y),$ we have $k_1(G:x,y)\le h.$ If $1\le r\le \lfloor \frac{l-1}{2}\rfloor ,$ then $d(z,v_{c-1})\le r-1,$ where any vertex $z\in \{v_{c-r},v_{c-r+1},\cdots ,v_{c-1}\}.$ If $1\le r\le \lfloor \frac{l-1}{2}\rfloor ,$ then $d(z,v_{c+1})\le r-1,$ where any vertex $z\in \{v_{c+1},v_{c+2},\cdots ,v_{c+r}\}.$ In addition, we have $\{v_{c-1},v_{c+1}\}\subseteq N^{+}(G^{h+1+a}:x,y).$ For $1\le r\le \lfloor \frac{l-1}{2}\rfloor ,$ according to Observation 2, then $\{v_{c-r},v_{c-r+1},\cdots ,v_{c+r}\}\subseteq N^{+}(G^{h+r}:x,y).$ Then, we have $k_m(G:x,y)\le h+⌊\frac{m}{2}⌋,$ where $2\le m\le 2\lfloor \frac{l-1}{2}\rfloor +1.$ Next, we consider $2\lfloor \frac{l-1}{2}\rfloor +1\le m\le n.$

If l is odd, for any vertex $v\in V\left(G\right)$, we have $d(v,v_c)\le \lfloor \frac{l-1}{2}\rfloor .$ In addition, we have $\left\{v_c\right\}\subseteq N^{+}(G^{h+1+a}:x,y).$ According to Observation 2, $N^{+}(G^{h+\lfloor \frac{l+1}{2}\rfloor }:x,y)=V\left(G\right).$

We consider that l is even. For any vertex $v\in V\left(P_k\right)$ such that $1\le k\le c-1,$ we have $d(v,v_{c-1})\le d(v_1,v_{c-1})=⌈\frac{l-3}{2}⌉=\lfloor \frac{l-1}{2}\rfloor .$ For any vertex $v\in V\left(P_k\right)$ such that $c\le k\le l,$ we have $d(v,v_c)\le d(v_c,v_l)=\lfloor \frac{l-1}{2}\rfloor .$ Moreover, $\{v_{c-1},v_c\}\subseteq N^{+}(G^{h+1+a}:x,y).$ Furthermore, according to Observation 2, we have $N^{+}(G^{h+\lfloor \frac{l+1}{2}\rfloor }:x,y)=V\left(G\right).$ Then, whether l is odd or even, we have $k_n(G:x,y)\le h+\lfloor \frac{l+1}{2}\rfloor .$ Therefore, for $2\lfloor \frac{l-1}{2}\rfloor +1\le m\le n,$ we have $k_m(G:x,y)\le k_n(G:x,y)\le h+\lfloor \frac{l+1}{2}\rfloor .$

Therefore, the conclusion holds. □

Proposition 3.

Let $G\in S_n^{\prime }\left(d\right).$ Suppose that u is a vertex that satisfies $u\in V\left(G\right)\setminus V\left(P\right)$ and $d(u,v_c)=1.$ Suppose $\left\{v\right\}\subseteq N^{+}(G^{h+a}:x,y)$ and $\{v_{c-1},v_c,u\}\subseteq N^{+}(G^{h+1+a}:x,y),$ where v is a vertex of G and h is an integer satisfying $h\ge 0.$ Then, $k_m(G:x,y)\le h+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Proof. 

Since $\left\{v\right\}\subseteq N^{+}(G^{h+a}:x,y),$ we have $k_1(G:x,y)\le h.$ If $1\le r\le ⌈\frac{l-1}{2}⌉,$ then $d(z,v_{c-1})\le r-1,$ where any vertex $z\in \{v_{c-r},v_{c-r+1},\cdots ,v_{c-1}\}.$ If $1\le r\le ⌈\frac{l-1}{2}⌉,$ then $d(z,v_c)\le r-1,$ where any vertex $z\in \{v_c,v_{c+1},\cdots ,v_{c-1+r}\}.$ In addition, we have $\{v_{c-1},v_c\}\subseteq N^{+}(G^{h+1+a}:x,y).$ According to Observation 2, $\{v_{c-r},v_{c-r+1},\cdots ,v_{c-1+r}\}\cup \left\{u\right\}\subseteq N^{+}(G^{h+r}:x,y),$ where $1\le r\le ⌈\frac{l-1}{2}⌉.$ Then, we have $k_m(G:x,y)\le h+⌊\frac{m}{2}⌋,$ where $2\le m\le 2⌈\frac{l-1}{2}⌉+1.$ Next, we consider $2⌈\frac{l-1}{2}⌉+1\le m\le n.$

For any vertex $z\in V\left(G\right)$, we have $d(z,v_c)\le d(v_1,v_c)=⌈\frac{l-1}{2}⌉.$ Moreover, $\left\{v_c\right\}\subseteq N^{+}(G^{h+1+a}:x,y).$ Furthermore, according to Observation 2, we have $N^{+}(G^{h+⌈\frac{l+1}{2}⌉}:x,y)=V\left(G\right).$ Then, we have $k_n(G:x,y)\le h+⌈\frac{l+1}{2}⌉.$ Therefore, for $2⌈\frac{l-1}{2}⌉+1\le m\le n,$ we have $k_m(G:x,y)\le k_n(G:x,y)\le h+⌈\frac{l+1}{2}⌉.$

Therefore, this conclusion is clearly established. □

Lemma 1.

Let $G\in S_n^{\prime }\left(d\right).$ If the walks $W(x,v_c),$$W(y,v_c)$ such that $V\left(W(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(W(y,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ then $k_m(G:x,y)\le max\left\{\right|W(x,v_c)|,|W(y,v_c)\left|\right\}+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Proof. 

Let $M=max\left\{\right|W(x,v_c)|,|W(y,v_c)\left|\right\}.$ According to Observation 1, there exists a walk of length s from y to $v_c$ for any integer $s\ge \left|W\right(y,v_c\left)\right|$. Similarly, there exists a walk of length t from x to $v_c$ for any integer $t\ge \left|W\right(x,v_c\left)\right|$. Therefore, we have $\left\{v_c\right\}\subseteq N^{+}(G^{M+a}:x,y).$ Then, $k_1(G:x,y)\le M.$ For any vertex $v\in \{v_{c-r},v_{c-r+1},\cdots ,v_{c+r}\}$, $d(v,v_c)\le r,$ where $1\le r\le \lfloor \frac{l-1}{2}\rfloor .$ Moreover, $\left\{v_c\right\}\subseteq N^{+}(G^{M+a}:x,y).$ According to Observation 2, $\{v_{c-r},v_{c-r+1},\cdots ,v_{c+r}\}\subseteq N^{+}(G^{M+r}:x,y),$ where $1\le r\le \lfloor \frac{l-1}{2}\rfloor .$ Then, we have $k_m(G:x,y)\le M+⌊\frac{m}{2}⌋,$ where $2\le m\le 2\lfloor \frac{l-1}{2}\rfloor +1.$

For any vertex $v\in V\left(G\right)$, we have $d(v,v_c)\le d(v_1,v_c)=⌈\frac{l-1}{2}⌉\le \lfloor \frac{l+1}{2}\rfloor .$ Moreover, $\left\{v_c\right\}\subseteq N^{+}(G^{M+a}:x,y).$ Furthermore, according to Observation 2, we have $N^{+}(G^{M+\lfloor \frac{l+1}{2}\rfloor }:x,y)=V\left(G\right).$ Then, $k_m(G:x,y)\le M+\lfloor \frac{l+1}{2}\rfloor ,$ where $1\le m\le n.$

Therefore, the conclusion holds. □

3.1. Results of $S_{n,1}^{\prime }\left(d\right)$

Corollary 1.

Let $G\in S_{n,1}^{\prime }\left(d\right).$ If $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ then $k_m(G:x,y)\le ⌈\frac{n-1}{2}⌉+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Proof. 

We have $d(x,v_c)\le ⌈\frac{l-1}{2}⌉\le ⌈\frac{n-1}{2}⌉.$ Similarly, $d(y,v_c)\le ⌈\frac{n-1}{2}⌉.$ In addition, $S_{n,1}^{\prime }\left(d\right)\subseteq S_n^{\prime }\left(d\right).$ According to Lemma 1, we have $k_m(G:x,y)\le max\{d(x,v_c),d(y,v_c)\}+⌊\frac{m}{2}⌋\le ⌈\frac{n-1}{2}⌉+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$ Therefore, the conclusion holds. □

Let $G\in S_{n,1}^{\prime }\left(d\right).$ If $v_c\in V\left(L\left(G\right)\right),$ then $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$, and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing .$ Now, let us consider the case in which $v_c\in V\left(U\left(G\right)\right).$ For the sake of research, we propose Definition 4.

Definition 4.

Let $G\in S_{n,1}^{\prime }\left(d\right)$, and let $v_c\in V\left(U\left(G\right)\right)$. We define the following notation:

(1)

Let $i,j$ be non-negative integers satisfying $i\ge 0$ and $j\ge 0.$ We assume $v_{s,i},v_{t,j}$ are two different loop vertices closest to $v_c,$ where $s\le c-1,t\ge c.$ Then, $v_{s,i}\in V\left(T_1\right)$, and $v_{t,j}\in V\left(T_2\right).$ Moreover, we know that $d(v_{s,i},v_s)=i$ and $d(v_{t,j},v_t)=j.$ If $t=c,$ let $v_{c,1}$ be the vertex such that $v_{c,1}\in V\left(P(v_c,v_{c,j})\right)$ and $d(v_{c,1},v_c)=1.$

(2)

Let z be any vertex such that $V\left(P(z,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing .$ Let $d(z,V\left(P(v_{s,i},v_{t,j})\right))=d(z,v_{f,e}),$ where $v_{f,e}\in V\left(P(v_{s,i},v_{t,j})\right).$ For any vertex $v\in V\left(P(v_{s,i},v_{t,j})\right)$ and $t\ge c$, the walk from z to v through $v_{s,i}$ is denoted by $W_1(z,v),$ that is, $z\stackrel{d(z,v_{s,i})}{\to }{v}_{s,i}\stackrel{d(v_{s,i},v)}{\to }v.$ The walk from z to v through $v_{t,j}$ is denoted by $W_2(z,v),$ that is, $z\stackrel{d(z,v_{t,j})}{\to }{v}_{t,j}\stackrel{d(v_{t,j},v)}{\to }v.$ The shorter of the two walks ($W_1(z,v)$ and $W_2(z,v)$) is denoted as $W_m(z,v).$

Remark 2.

According to Definition 4, if $t\ge c+1,$ then $\{v_{c-1},v_{c+1},v_c\}\subseteq V\left(P(v_{s,i},v_{t,j})\right).$ If $t=c,$ then $\{v_{c-1},v_{c,1},v_c\}\subseteq V\left(P(v_{s,i},v_{t,j})\right).$ In Definition 4(2), suppose $v_{f,e}\in V\left(P(v_{s,i},v_{c-1})\right)$ and $t\ge c+1.$ Then, the walk from z to $v_c$ through $v_{s,i}$ is denoted by $W_1(z,v_c),$ that is, $z\stackrel{d(z,v_{f,e})}{\to }{v}_{f,e}\stackrel{d(v_{f,e},v_{s,i})}{\to }{v}_{s,i}\stackrel{d(v_{s,i},v_{f,e})}{\to }{v}_{f,e}\stackrel{d(v_{f,e},v_{c-1})}{\to }{v}_{c-1}\stackrel{d(v_{c-1},v_c)}{\to }{v}_c.$ The walk from z to $v_c$ through $v_{t,j}$ is denoted by $W_2(z,v_c),$ that is, $z\stackrel{d(z,v_{f,e})}{\to }{v}_{f,e}\stackrel{d(v_{f,e},v_c)}{\to }{v}_c\stackrel{d(v_c,v_{t,j})}{\to }{v}_{t,j}\stackrel{d(v_{t,j},v_{c+1})}{\to }{v}_{c+1}\stackrel{d(v_{c+1},v_c)}{\to }{v}_c.$

Proposition 4.

Let $G\in S_{n,1}^{\prime }\left(d\right)$ and $v_c\in V\left(U\left(G\right)\right).$ Let z be any vertex such that $V\left(P(z,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing .$ Then, for any vertex $v\in V\left(P(v_{s,i},v_{t,j})\right)$, there exists a walk ($W_m(z,v)$) such that $V\left(W_m(z,v)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $|W_m(z,v)|\le n-d+1.$ If the equation holds, then $V\left(U\left(G\right)\right)\subseteq V\left(W_1(z,v)\right)\cup V\left(W_2(z,v)\right).$

Proof. 

Note that $v_c\in V\left(P(v_{s,i},v_{t,j})\right),$ and we consider $v=v_c.$$W_1(z,v_c),$$W_2(z,v_c),$$v_{f,e}$, which are defined in Definition 4. Let us assume that $v_{f,e}\in V\left(P(v_{s,i},v_{c-1})\right).$ We have $(V\left(W_1(z,v_c)\right)\cup V\left(W_2(z,v_c)\right))\cap V\left(L\left(G\right)\right)=\{v_{s,i},v_{t,j}\}.$ Then,

$$\begin{array}{cc}\hfill & |W_1(z,v_c)|+|W_2(z,v_c)|\hfill \\ \hfill =& 2d(z,v_{f,e})+2d(v_{f,e},v_{s,i})+2d(v_{f,e},v_c)+2d(v_c,v_{t,j})\hfill \\ \hfill =& 2d(z,v_{f,e})+2d(v_{s,i},v_{t,j})\hfill \\ \hfill =& 2\left(\right|V\left(W_1(z,v_c)\right)\cup V\left(W_2(z,v_c)\right)|-1)\le 2(n-d+1).\hfill \end{array}$$

For $v_{f,e}\in V\left(P(v_c,v_{t,j})\right),$ similar to $v_{f,e}\in V\left(P(v_{s,i},v_{c-1})\right),$ there are two different walks from z to $v_c$, each passing through a loop vertex, and the sum of the lengths of the two walks is not greater than $2(n-d+1).$

For any vertex v satisfying $v\in V\left(P(v_{s,i},v_{t,j})\right),$ we have $V\left(W_1(z,v)\right)\cup V\left(W_2(z,v)\right)=V\left(W_1(z,v_c)\right)\cup V\left(W_2(z,v_c)\right).$ Similar to $v=v_c,$ for any vertex v satisfying $v\in V\left(P(v_{s,i},v_{t,j})\right),$ we can conclude that $|W_1(z,v)|+|W_2(z,v)|=2d(z,v_{f,e})+2d(v_{s,i},v_{t,j})=2\left(\right|V\left(W_1(z,v)\right)\cup V\left(W_2(z,v)\right)|-1)\le 2(n-d+1).$ Moreover, we have $V\left(W_1(z,v)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(W_2(z,v)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ and $(V\left(W_1(z,v)\right)\cup V\left(W_2(z,v)\right))\cap V\left(L\left(G\right)\right)=\{v_{s,i},v_{t,j}\}.$ Therefore, $|W_m(z,v)|=min\{|W_1(z,v)|,|W_2(z,v)\left|\right\}\le n-d+1.$

If $|W_m(z,v)|=n-d+1.$ Since $|W_1(z,v)|+|W_2(z,v)|\le 2(n-d+1),$ we have $|W_1(z,v)|=|W_2(z,v)|=n-d+1.$ Then, $|W_1(z,v)|+|W_2(z,v)|=2(n-d+1).$ Therefore, we have $|V\left(W_1(z,v)\right)\cup V\left(W_2(z,v)\right)|=n-d+2.$ Furthermore, $V\left(U\left(G\right)\right)\subseteq V\left(W_1(z,v)\right)\cup V\left(W_2(z,v)\right).$ □

Corollary 2.

Let $G\in S_{n,1}^{\prime }\left(d\right)$, and let $v_c\in V\left(U\left(G\right)\right).$ Let z be any vertex such that $V\left(P(z,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing$. Then,

(1)

If $t\ge c+1,$ then there exist two different walks ($W(z,v_{c-1}),$$W(z,v_{c+1})$) such that $V\left(W(z,v_{c-1})\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(W(z,v_{c+1})\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ and $|W(z,v_{c-1})|=|W_1(z,v_c)|-1$ and $|W(z,v_{c+1})|=|W_2(z,v_c)|-1.$

(2)

If $t=c,$ then there exist two different walks ($W(z,v_{c-1}),$$W(z,v_{c,1})$) such that $V\left(W(z,v_{c-1})\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(W(z,v_{c,1})\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ and $|W(z,v_{c-1})|=|W_1(z,v_c)|-1$ and $|W(z,v_{c,1})|=|W_2(z,v_c)|-1.$

Proof. 

(1) Let us assume that $v_{f,e}\in V\left(P(v_{s,i},v_{c-1})\right).$ There exists a walk ($W(z,v_{c-1})$) from z to $v_{c-1},$ that is $z\stackrel{d(z,v_{f,e})}{\to }{v}_{f,e}\stackrel{d(v_{f,e},v_{s,i})}{\to }{v}_{s,i}\stackrel{d(v_{s,i},v_{f,e})}{\to }{v}_{f,e}\stackrel{d(v_{f,e},v_{c-1})}{\to }{v}_{c-1}.$ There exists another walk ($W(z,v_{c+1})$) from z to $v_{c+1},$ that is $z\stackrel{d(z,v_{f,e})}{\to }{v}_{f,e}\stackrel{d(v_{f,e},v_c)}{\to }{v}_c\stackrel{d(v_c,v_{t,j})}{\to }{v}_{t,j}\stackrel{d(v_{t,j},v_{c+1})}{\to }{v}_{c+1}.$ Obviously, $|W(z,v_{c-1})|=|W_1(z,v_c)|-1$, and $|W(z,v_{c+1})|=|W_2(z,v_c)|-1.$

(2)

The proof is similar to (1), so we can omit it.

□

Lemma 2.

Let $G\in S_{n,1}^{\prime }\left(d\right)$ and $2\le d\le n-⌈\frac{n-1}{2}⌉-1.$ Suppose that $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing .$ Then, $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Proof. 

Since the vertex y satisfies $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing ,$ then $v_c\in V\left(U\left(G\right)\right).$ According to Definition 4, $v_{s,i},v_{t,j}$ are two different loop vertices closest to $v_c,$ where $s\le c-1,t\ge c.$ According to Proposition 4, the walk $W_m(y,v_c)$ from y to $v_c$ satisfies $V\left(W_m(y,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing .$ Moreover, $|W_m(y,v_c)|\le n-d+1.$ Since $d\le n-⌈\frac{n-1}{2}⌉-1,$ we conclude that $⌈\frac{n-1}{2}⌉\le n-d-1.$ Then, $d(x,v_c)\le ⌈\frac{n-1}{2}⌉\le n-d-1.$ If $|W_m(y,v_c)|\le n-d,$ we have $\left\{v_c\right\}\subseteq N^{+}(G^{n-d}:x,y).$ According to Lemma 1, the conclusion holds. Next, we consider $|W_m(y,v_c)|=n-d+1.$ Then, $|W_1(y,v_c)|=|W_2(y,v_c)|=n-d+1.$

Case 1: $t\ge c+1.$

According to Corollary 2(1), there exists a walk ($W(y,v_{c+1})$) from y to $v_{c+1},$ and $V\left(W(y,v_{c+1})\right)\cap V\left(L\left(G\right)\right)\ne \varnothing .$ There exists a walk ($W(y,v_{c-1})$) from y to $v_{c-1},$ and $V\left(W(y,v_{c-1})\right)\cap V(L\left(G\right)\ne \varnothing .$ Note that $|W(y,v_{c-1})|=|W_1(y,v_c)|-1=n-d$ and $|W(y,v_{c+1})|=|W_2(y,v_c)|-1=n-d.$ There exists a walk ($W(x,v_{c-1})$) from x to $v_{c-1},$ that is $x\stackrel{d(x,v_c)}{\to }{v}_c\stackrel{d(v_c,v_{c-1})}{\to }{v}_{c-1}.$ Similarly, there exists a walk ($W(x,v_{c+1})$) from x to $v_{c+1}$ that is $x\stackrel{d(x,v_c)}{\to }{v}_c\stackrel{d(v_c,v_{c+1})}{\to }{v}_{c+1}.$ Note that $\left|W\right(x,v_{c-1}\left)\right|\le n-d$ and $\left|W\right(x,v_{c+1}\left)\right|\le n-d.$

Then, $\left\{v_{c-1}\right\}\subseteq N^{+}(G^{n-d+a}:x,y)$, and $\{v_{c-1},v_{c+1},v_c\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ According to Proposition 2, this conclusion holds.

Case 2: $t=c.$

According to Corollary 2(2), there are two different walks ($W(y,v_{c,1}),W(y,v_{c-1})$) through a loop vertex. We have $\left|W\right(y,v_{c,1}\left)\right|=n-d$ and $\left|W\right(y,v_{c-1}\left)\right|=n-d.$ Similar to $t\ge c+1,$ there exists a walk ($W(x,v_{c-1})$) through a loop vertex, and $W(x,v_{c-1})$ satisfies $\left|W\right(x,v_{c-1}\left)\right|\le n-d.$ Similarly, there exists a walk ($W(x,v_{c,1})$) through a loop vertex, and $W(x,v_{c,1})$ satisfies $\left|W\right(x,v_{c,1}\left)\right|\le n-d.$

Note that $\left\{v_{c-1}\right\}\subseteq N^{+}(G^{n-d+a}:x,y)$, and $\{v_{c-1},v_{c,1},v_c\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ According to Proposition 3, this conclusion holds. □

Lemma 3.

Let $G\in S_{n,1}^{\prime }\left(d\right)$ and $d=n-⌈\frac{n-1}{2}⌉.$ Suppose that $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$, and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing .$ Then, $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Proof. 

Because $n\ge 5,$ then $d\ge 2.$ Since $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing ,$ then $v_c$ is not a loop vertex. According to Definition 4, $v_{s,i},v_{t,j}$ are two different loop vertices closest to $v_c,$ where $s\le c-1,t\ge c.$ According to Proposition 4, there exists a walk ($W_m(y,v_c)$) from y to $v_c$ that satisfies $V\left(W_m(y,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing .$ Moreover, $|W_m(y,v_c)|\le n-d+1.$ Since $d=n-⌈\frac{n-1}{2}⌉,$ then $d(x,v_c)\le ⌈\frac{n-1}{2}⌉=n-d.$

If $|W_m(y,v_c)|\le n-d,$ we have $\left\{v_c\right\}\subseteq N^{+}(G^{n-d}:x,y).$ According to Lemma 1, the conclusion holds. If $|W_m(y,v_c)|=n-d+1$, and $d(x,v_c)\le ⌈\frac{n-1}{2}⌉-1=n-d-1.$ The following proof is similar to the proof of Lemma 2. According to Corollary 2, we have $\left\{v_{c-1}\right\}\subseteq N^{+}(G^{n-d+a}:y).$ Furthermore, $\left\{v_{c-1}\right\}\subseteq N^{+}(G^{n-d+a}:x,y).$ If $t\ge c+1,$ we conclude that $\{v_{c-1},v_{c+1},v_c\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ If $t=c,$ we conclude that $\{v_{c-1},v_{c,1},v_c\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ Therefore, this conclusion holds.

Next, we consider $d(x,v_c)=n-d=⌈\frac{n-1}{2}⌉$, and $|W_m(y,v_c)|=n-d+1.$ Then, $d(x,v_c)=⌈\frac{n-1}{2}⌉\le ⌈\frac{l-1}{2}⌉.$ We can draw the following two conclusions: if n is odd, then $l=n$ or $l=n-1;$ if n is even, then $l=n.$

Case 1: n is odd, and $l=n.$

Then, $d=\frac{n+1}{2}$, and $c=\frac{n+1}{2}.$ Since $d(x,v_c)=\frac{n-1}{2},$ we have $x=v_1$ or $v_n.$ Let us assume that $x=v_1.$ Since $s\le c-1,$ then $d(x,v_{c-1})=n-d-1.$ Therefore, we have $\left\{v_{c-1}\right\}\subseteq N^{+}(G^{n-d}:x).$ Note that $|W_m(y,v_c)|=n-d+1.$ According to Corollary 2(1), we have $\{v_{c-1},v_{c+1}\}\subseteq N^{+}(G^{n-d}:y).$ Furthermore, we have $\left\{v_{c-1}\right\}\subseteq N^{+}(G^{n-d+a}:x,y)$, and $\{v_{c-1},v_c,v_{c+1}\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ According to Proposition 2, this conclusion holds. For $x=v_n,$ it is similar to $x=v_1,$ so we can omit it.

Case 2: n is odd, and $l=n-1.$

Then, $d=\frac{n+1}{2}$, and $c=\frac{n+1}{2}.$ We have $d(x,v_c)=\frac{n-1}{2}.$ Moreover, we know that $v_c=v_{\frac{n+1}{2}}$, and $v_{c-1}=v_{\frac{n-1}{2}}.$ If $v_{2,1}\notin V\left(G\right),$ then $x=v_1.$ If $v_{2,1}\in V\left(G\right),$ then $x=v_1$ or $x=v_{2,1}.$ Since $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ then the path from x to $v_{\frac{n-1}{2}}$ goes through a loop vertex. In addition, we have $d(x,v_{\frac{n-1}{2}})=n-d-1.$ Then, $\left\{v_{\frac{n-1}{2}}\right\}\subseteq N^{+}(G^{n-d}:x).$ Note that $|W_m(y,v_{\frac{n+1}{2}})|=n-d+1.$ According to Corollary 2, $\left\{v_{\frac{n-1}{2}}\right\}\subseteq N^{+}(G^{n-d}:y).$ Furthermore, we have $\left\{v_{\frac{n-1}{2}}\right\}\subseteq N^{+}(G^{n-d+a}:x,y).$ For $1\le r\le \frac{n-3}{2},$ we have $\{v_{\frac{n-1}{2}-r},v_{\frac{n-1}{2}-r+1},\cdots ,v_{\frac{n-1}{2}+r}\}\subseteq N^{+}(G^{n-d+r}:x,y).$ For any vertex $v\in V\left(G\right)$, we have $d(v,v_{\frac{n-1}{2}})\le \frac{n-1}{2}.$ Therefore, $N^{+}(G^{n-1}:x,y)=V\left(G\right).$

Case 3: n is even.

Then, $l=n$, and $c=\frac{n}{2}+1.$ Moreover, we have $d=\frac{n}{2}$ and $x=v_1.$ Then, $d(x,v_{c-1})=n-d-1.$ Since $s\le c-1,$ we have $\left\{v_{c-1}\right\}\subseteq N^{+}(G^{n-d}:x).$ Note that $|W_m(y,v_c)|=n-d+1.$ According to Corollary 2(1), we have $\{v_{c-1},v_{c+1}\}\subseteq N^{+}(G^{n-d}:y).$ Furthermore, we have $\left\{v_{c-1}\right\}\subseteq N^{+}(G^{n-d+a}:x,y)$, and $\{v_{c-1},v_c,v_{c+1}\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ According to Proposition 2, this conclusion holds. □

Lemma 4.

Let $G\in S_{n,1}^{\prime }\left(d\right).$ Suppose that $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing$, and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing$. Then, $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Proof. 

According to Definition 4, $v_{s,i},v_{t,j}$ are two different loop vertices closest to $v_c,$ where $s\le c-1,t\ge c.$ It is known that $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing .$ Let $d(x,V\left(P(v_{s,i},v_{t,j})\right))=d(x,v_{f,e}),$ where $v_{f,e}\in V\left(P(v_{s,i},v_{t,j})\right).$ Then, the length of the path $x\stackrel{d(x,v_{f,e})}{\to }{v}_{f,e}\stackrel{d(v_{f,e},v_{s,i})}{\to }{v}_{s,i}$ is $d(x,v_{f,e})+d(v_{f,e},v_{s,i}).$ Then, $d(x,v_{f,e})+d(v_{f,e},v_{s,i})\le n-d.$ Then, $\left\{v_{s,i}\right\}\subseteq N^{+}(G^{n-d+a}:x).$ Similarly, $\left\{v_{t,j}\right\}\subseteq N^{+}(G^{n-d+a}:x).$ Furthermore, we have $\{v_{s,i},v_{t,j}\}\subseteq N^{+}(G^{n-d+a}:x,y).$

Case 1: $t\ge c+1.$

Then, $\{v_{c-1},v_{c+1},v_c\}\subseteq V\left(P(v_{s,i},v_{t,j})\right).$ Let v be any vertex satisfying $v\in \{v_{c-1},v_c,v_{c+1}\}.$ According to Proposition 4, there exists a walk $W_m(x,v)$ such that $V\left(W_m(x,v)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing .$ Moreover, $|W_m(x,v)|\le n-d+1.$ Similarly, we have $|W_m(y,v)|\le n-d+1.$ Then, $\{v_{s,i},v_{t,j}\}\subseteq N^{+}(G^{n-d+a}:x,y)$, and $\{v_{c-1},v_{c+1},v_c\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ According to Proposition 2, this conclusion holds.

Case 2: $t=c.$

Then, $\{v_{c-1},v_c,v_{c,1}\}\subseteq V\left(P(v_{s,i},v_{t,j})\right).$ Let any vertex $v\in \{v_{c-1},v_c,v_{c,1}\}$. According to Proposition 4, there exists a walk ($W_m(x,v)$) such that $V\left(W_m(x,v)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing .$ Moreover, $|W_m(x,v)|\le n-d+1.$ Similarly, we have $|W_m(y,v)|\le n-d+1.$ Then, $\{v_{s,i},v_{t,j}\}\subseteq N^{+}(G^{n-d+a}:x,y)$, and $\{v_{c-1},v_c,v_{c,1}\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ According to Proposition 3, this conclusion holds. □

Lemma 5.

Let $G\in S_{n,1}^{\prime }\left(d\right),$ where n is even. If $l=n$ and $\frac{n}{2}+1\le d\le n,$ then $k_m\left(G\right)\le \frac{n}{2}+⌊\frac{m-1}{2}⌋,$ where $1\le m\le n.$

Proof. 

Let $v_s,$$v_t$ be the nearest loop vertices to $v_{\frac{n}{2}+1},$ where $1\le s\le \frac{n}{2}$ and $\frac{n}{2}+1\le t\le n.$ If $V\left(P(x,v_{\frac{n}{2}+1})\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ we have $d(x,v_{\frac{n}{2}+1})\le \frac{n}{2}.$ If $V\left(P(x,v_{\frac{n}{2}+1})\right)\cap V\left(L\left(G\right)\right)=\varnothing$, then there exists a walk ($W_m(x,v_{\frac{n}{2}+1})$) such that $V\left(W_m(x,v_{\frac{n}{2}+1})\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ and $|W_m(x,v_{\frac{n}{2}+1})|\le n-d+1\le \frac{n}{2}.$ Therefore, $\left\{v_{\frac{n}{2}+1}\right\}\subseteq N^{+}(G^{\frac{n}{2}}:x).$ Similarly, we have $\left\{v_{\frac{n}{2}}\right\}\subseteq N^{+}(G^{\frac{n}{2}}:x).$ Furthermore, we have $\{v_{\frac{n}{2}},v_{\frac{n}{2}+1}\}\subseteq N^{+}(G^{\frac{n}{2}+a}:x,y).$ For any integer r such that $1\le r\le \frac{n}{2}-1,$ we have $\{v_{\frac{n}{2}-r},v_{\frac{n}{2}-r+1},\cdots ,v_{\frac{n}{2}+r+1}\}\subseteq N^{+}(G^{\frac{n}{2}+r}:x,y).$ Therefore, the conclusion holds. □

3.2. Results of $S_{n,2}^{\prime }\left(d\right)$

In order to obtaining results of $S_{n,2}^{\prime }\left(d\right),$ we need to provide the following proposition, its corollary, and Definition 5.

Proposition 5.

Let $G\in S_n\left(d\right)$ and $x\in V\left(G\right).$ Let $G^{\prime }$ be a subgraph of G such that $G^{\prime }$ is connected. Suppose that the set $X_1$ satisfies $\varnothing \ne X_1\subseteq V\left(G^{\prime }\right),$$X_1\subseteq N^{+}(G^{L+a}:x)$, and suppose that $X_2=V\left(G^{\prime }\right)\setminus X_1\ne \varnothing .$ Then, $|N^{+}(G^{L+r}:x)\cap V\left(G^{\prime }\right)|\ge |X_1|+r,$ where r is any integer such that $1\le r\le |X_2|.$

Proof. 

Since $X_1\subseteq N^{+}(G^{L+a}:x)\cap V\left(G^{\prime }\right),$$X_1\subseteq N^{+}(G^{L+r}:x)\cap V\left(G^{\prime }\right),$ where r is any integer such that $1\le r\le |X_2|.$

Let $h=max\{d_{G^{\prime }}(v,X_1):v\in X_2\}.$ Then, there exists a path ($P_{G^{\prime }}(z,w)$) from z to w in $G^{\prime }$ that satisfies $|P_{G^{\prime }}(z,w)|=h,$ and the path $P_{G^{\prime }}(z,w)$ does not pass through other vertices in $X_1$, except $w,$ where $z\in X_2$ and $w\in X_1.$ Let us denote z as $z_h.$ We denote the vertex sequence of path $P_{G^{\prime }}(z,w)$ as $z_h,z_{h-1},\cdots ,z_1,w.$ It is easy to obtain $z_i\in X_2$ and $d_{G^{\prime }}(z_i,X_1)=i,$ where i is any integer such that $1\le i\le h.$ Therefore, for any integer g satisfying $1\le g\le h,$ there exists $u\in X_2$ such that $d_{G^{\prime }}(u,X_1)=g.$

Next, we label the vertices in $X_2$ according to their distance from $X_1.$ Let $X_{2,i}=\{v:d_{G^{\prime }}(v,X_1)=i,\mathrm{where}v\in X_2\mathrm{and}1\le i\le h\},$$|X_{2,i}|=s_i.$ Let $f_0=0$, and $f_i=s_1+s_2+\cdots +s_i,$ where $1\le i\le h.$ We label the vertices in $X_2$ according to $i=1,2,\cdots ,h$ in order and label the vertices in $X_{2,i}$ as $u_{f_{i-1}+1},u_{f_{i-1}+2},\cdots ,u_{f_{i-1}+s_i}.$ Moreover, we have $f_{h-1}+s_h=s_1+s_2+\cdots +s_h=\left|X_2\right|.$ For any vertex $u_r\in X_2$, we have $d(u_r,X_1)\le r,$ where $1\le r\le |X_2|.$ Moreover, $X_1\subseteq N^{+}(G^{L+a}:x).$ According to Observation 2, $\left\{u_r\right\}\subseteq N^{+}(G^{L+r}:x),$ where $1\le r\le |X_2|.$ Furthermore, we have $\{u_1,u_2,\cdots ,u_r\}\subseteq N^{+}(G^{L+r}:x).$ Therefore, $X_1\cup \{u_1,u_2,\cdots ,u_r\}\subseteq N^{+}(G^{L+r}:x)\cap V\left(G^{\prime }\right).$ Therefore, the conclusion holds. □

Corollary 3.

Let $G\in S_n\left(d\right)$ and $x\in V\left(G\right).$ Let $\varnothing \ne V_1\subseteq V\left(G\right),$$\varnothing \ne V_2\subseteq V\left(G\right)$, and $V_1,V_2$ denote that $G\left[V_1\right]$ and $G\left[V_2\right]$ are connected. Let $\varnothing \ne Z_1\subseteq V_1$ and $\varnothing \ne Z_2\subseteq V_2.$ Let $Z_1^{\prime }=V_1\setminus Z_1,$$Z_2^{\prime }=V_2\setminus Z_2,$ and $Z_1^{\prime },Z_2^{\prime }$ satisfy $Z_1^{\prime }\cap Z_2^{\prime }=\varnothing$ and $Z_i^{\prime }\cap (Z_1\cup Z_2)=\varnothing ,$ where $i=1,2.$ If $Z_1\cup Z_2\subseteq N^{+}(G^{L+a}:x),$ then $|N^{+}(G^{L+r}:x)|\ge |Z_1\cup Z_2|+2r,$ where r is any integer such that $0\le r\le min\left\{\right|Z_1^{\prime }|,|Z_2^{\prime }\left|\right\}.$

Proof. 

We know that $G\left[V_1\right],G\left[V_2\right]$ are the induced subgraphs from $V_1,V_2,$ respectively. Since $Z_1\cup Z_2\subseteq N^{+}(G^{L+a}:x),$ then $Z_1\cup Z_2\subseteq N^{+}(G^{L+r}:x),$ where r is any integer such that $0\le r\le min\left\{\right|Z_1^{\prime }|,|Z_2^{\prime }\left|\right\}.$

If $Z_1^{\prime }=\varnothing$ or $Z_2^{\prime }=\varnothing ,$ then $r=min\left\{\right|Z_1^{\prime }|,|Z_2^{\prime }\left|\right\}=0.$ We have $|N^{+}(G^L:x)|\ge |Z_1\cup Z_2|.$ Therefore, the conclusion holds.

Next, we consider $Z_1^{\prime }\ne \varnothing$ and $Z_2^{\prime }\ne \varnothing .$ It is known that $G\left[V_1\right]$ is a subgraph of G, and $G\left[V_1\right]$ is connected. Furthermore, it is known that $\varnothing \ne Z_1\subseteq V_1$, and $Z_1\subseteq N^{+}(G^{L+a}:x).$ According to Proposition 5, we have $|N^{+}(G^{L+r}:x)\cap V\left(G\left[V_1\right]\right)|\ge \left|Z_1\right|+r,$ where $1\le r\le |Z_1^{\prime }|.$ Similarly, we have $|N^{+}(G^{L+r}:x)\cap V\left(G\left[V_2\right]\right)|\ge \left|Z_2\right|+r,$ where $1\le r\le |Z_2^{\prime }|.$ In addition, $Z_1^{\prime }\cap Z_2^{\prime }=\varnothing$, and $Z_i^{\prime }\cap (Z_1\cup Z_2)=\varnothing ,$ where $i=1,2.$ Therefore, we have $|N^{+}(G^{L+r}:x)|\ge |Z_1\cup Z_2|+2r,$ where r is any integer such that $0\le r\le min\left\{\right|Z_1^{\prime }|,|Z_2^{\prime }\left|\right\}.$ □

Definition 5.

Let $G\in S_{n,2}^{\prime }\left(d\right)$, and let z be any vertex of G. If $V\left(L\left(G\right)\right)\subseteq V\left(T_2\right),$ in Proposition 6 and Lemma 6, we define the following notation:

(1)

Suppose that $k=min\{i:V\left(P_i\right)\cap V\left(L\left(G\right)\right)\ne \varnothing \}$; then, $⌈\frac{l+1}{2}⌉\le k\le l.$

(2)

If $v_k\notin V\left(L\left(G\right)\right),$$v_k$ is not a loop vertex. Let $v_{k,b}$ be the nearest loop vertex to $v_k$, where $d(v_{k,b},v_k)=b\ge 1$ and $v_k\in V\left(P\right).$ If there are at least two loop vertices nearest to $v_k$ on $V\left(P_k\right)$, we take any of their vertices as $v_{k,b},$ which does not affect the conclusion of Proposition 6 and Lemma 6. The walk from z to $v_k$ is recorded as $W_1(z,v_k),$ that is, $z\stackrel{d\left(z,v_k\right)}{\to }{v}_k\stackrel{d\left(v_k,v_{k,b}\right)}{\to }{v}_{k,b}\stackrel{d\left(v_{k,b},v_k\right)}{\to }{v}_k.$

(3)

If $v_k\notin V\left(L\left(G\right)\right)$ and $V\left(L\left(G\right)\right)⊈V\left(P_k\right)$, we assume that $h=min\{i:V\left(P_i\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,\mathrm{where}i\ge k+1\};$ then, $k+1\le h\le l.$ Let $v_{h,g}$ be the nearest loop to $v_h$, where $d(v_{h,g},v_h)=g$ and $g\ge 0.$ If $g=0,$ we know that $v_{h,g}=v_h.$ The walk from z to $v_k$ is recorded as $W_2(z,v_k),$ that is, $z\stackrel{d\left(z,v_k\right)}{\to }{v}_k\stackrel{d\left(v_k,v_h\right)}{\to }{v}_h\stackrel{d\left(v_h,v_{h,g}\right)}{\to }{v}_{h,g}\stackrel{d\left(v_{h,g},v_h\right)}{\to }{v}_h\stackrel{d\left(v_h,v_k\right)}{\to }{v}_k,$ where $g\ne 0.$ If $g=0,$ the walk ($W_2(z,v_k)$) from z to $v_k$ is $z\stackrel{d\left(z,v_k\right)}{\to }{v}_k\stackrel{d\left(v_k,v_h\right)}{\to }{v}_h\stackrel{d\left(v_h,v_k\right)}{\to }{v}_k.$

(4)

Let $i=1,2,3$ represent the cases of $v_k\in V\left(L\left(G\right)\right),$$v_k\notin V\left(L\left(G\right)\right)$ and $V\left(L\left(G\right)\right)\subseteq V\left(P_k\right),$$v_k\notin V\left(L\left(G\right)\right)$ and $V\left(L\left(G\right)\right)⊈V\left(P_k\right)$ respectively. When $i=1,2,3,$ let $V_{1,i}=V\left(P(v_1,v_k)\right),V_{2,i}={\bigcup }_{i=k}^{l}V\left(P_i\right).$ Then, $G\left[V_{1,i}\right]$ and $G\left[V_{2,i}\right]$ are connected. When $i=1,2,3,$ let $Z_{1,i}=\left\{v_k\right\}.$ When $i=1,2,$ let $Z_{2,i}=\left\{v_k\right\}.$ Let $Z_{2,3}=\{v_k,v_{k,b},v_{h,g}\}.$ When $i=1,2,3,$ let $Z_{1,i}^{\prime }=V_{1,i}\setminus Z_{1,i},Z_{2,i}^{\prime }=V_{2,i}\setminus Z_{2,i}.$ Then, $Z_{1,i}^{\prime }\cap Z_{2,i}^{\prime }=\varnothing ,$$Z_{1,i}^{\prime }\cap (Z_{1,i}\cup Z_{2,i})=\varnothing$, and $Z_{2,i}^{\prime }\cap (Z_{1,i}\cup Z_{2,i})=\varnothing ,$ where $i=1,2,3.$ Moreover, for $i=1,2,3,$ we have $|Z_{1,i}^{\prime }|=d(v_1,v_k)=k-1.$

Proposition 6.

Let $G\in S_{n,2}^{\prime }\left(d\right).$ Suppose that $V\left(L\left(G\right)\right)\subseteq V\left(T_2\right).$ Then,

(1)

If $i=1,2,$ then $\left\{v_k\right\}\subseteq N^{+}(G^{n-d+a}:x,y).$

(2)

If $i=3,$ then $\{v_{k,b},v_{h,g}\}\subseteq N^{+}(G^{n-d+a}:x,y)$, and $\{v_{k,b},v_k,v_{h,g}\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$

Proof. 

(1) We consider $i=1$; then, $v_k\in V\left(L\left(G\right)\right).$ We have $d\left(x,v_k\right)\le d\left(v_1,v_k\right)\le n-d.$ Similarly, $d\left(y,v_k\right)\le n-d.$ Then, $\left\{v_k\right\}\subseteq N^{+}(G^{n-d}:x,y)$, and $\{v_{k-1},v_k,v_{k+1}\}\subseteq N^{+}(G^{n-d+1}:x,y).$ Furthermore, we have $\left\{v_k\right\}\subseteq N^{+}(G^{n-d+a}:x,y).$

• We consider $i=2$; then, $v_k$ is not a loop vertex, and $V\left(L\left(G\right)\right)\subseteq V\left(P_k\right).$ According to Definition 5(2), $v_{k,b}$ is the nearest loop vertex to $v_k.$ Then, the path from $v_1$ to $v_{k,b}$ does not pass through any loop vertex other than $v_{k,b}.$ Therefore, we have $\left|W_1(v_1,v_k)\right|=d(v_1,v_{k,b})+d(v_{k,b},v_k)\le n-d-d(v_k,v_l)+d(v_{k,b},v_k)\le n-d.$ Because $d\left(x,v_k\right)\le d\left(v_1,v_k\right),$ we have $\left|W_1(x,v_k)\right|\le \left|W_1(v_1,v_k)\right|\le n-d.$ Similarly, $\left|W_1(y,v_k)\right|\le n-d.$ Then, $\left\{v_k\right\}\subseteq N^{+}(G^{n-d}:x,y)$, and $\{v_{k-1},v_k,v_{k+1}\}\subseteq N^{+}(G^{n-d+1}:x,y).$ Further, we have $\left\{v_k\right\}\subseteq N^{+}(G^{n-d+a}:x,y).$

(2)

We consider $i=3$; then, $v_k$ is not a loop vertex, and $V\left(L\left(G\right)\right)⊈V\left(P_k\right).$ According to Definition 5, we have $(V\left(W_1(v_1,v_k)\right)\cup V\left(W_2(v_1,v_k)\right))\cap V\left(L\left(G\right)\right)=\{v_{k,b},v_{h,g}\},$ and we have

$$\begin{array}{cc}\hfill \left|W_1(v_1,v_k)\right|+\left|W_2(v_1,v_k)\right|& =2d\left(v_1,v_k\right)+2d\left(v_k,v_{k,b}\right)+2d\left(v_k,v_h\right)+2d(v_h,v_{h,g})\hfill \\ \hfill & =2\left(\right|V\left(W_1(v_1,v_k)\right)\cup V\left(W_2(v_1,v_k)\right)|-1)\le 2(n-d+1).\hfill \end{array}$$

Therefore, $min\{\left|W_1(v_1,v_k)\right|,\left|W_2(v_1,v_k)\right|\}\le n-d+1.$ Note that $\left|W_1(x,v_k)\right|\le \left|W_1(v_1,v_k)\right|$, and $\left|W_2(x,v_k)\right|\le \left|W_2(v_1,v_k)\right|.$ Then, $min\{\left|W_1(x,v_k)\right|,\left|W_2(x,v_k)\right|\}\le n-d+1.$ Similarly, we have $min\{\left|W_1(y,v_k)\right|,\left|W_2(y,v_k)\right|\}\le n-d+1.$

In addition, the length of the walk $x\stackrel{d\left(x,v_k\right)}{\to }{v}_k\stackrel{d\left(v_k,v_{k,b}\right)}{\to }{v}_{k,b}$ is $d(x,v_k)+d(v_k,v_{k,b}).$ We have $d(x,v_k)+d(v_k,v_{k,b})\le d(v_1,v_{k,b})\le n-d.$ The length of the walk $x\stackrel{d\left(x,v_h\right)}{\to }{v}_h\stackrel{d\left(v_h,v_{h,g}\right)}{\to }{v}_{h,g}$ is $d(x,v_h)+d(v_h,v_{h,g}).$ Therefore, $d(x,v_h)+d(v_h,v_{h,g})\le d(v_1,v_{h,g})\le n-d.$ Then, we have $\{v_{k,b},v_{h,g}\}\subseteq N^{+}(G^{n-d}:x,y)$, and $\{v_{k,b},v_k,v_{h,g}\}\subseteq N^{+}(G^{n-d+1}:x,y).$ Furthermore, we have $\{v_{k,b},v_{h,g}\}\subseteq N^{+}(G^{n-d+a}:x,y)$, and $\{v_{k,b},v_k,v_{h,g}\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$

□

Lemma 6.

Let $G\in S_{n,2}^{\prime }\left(d\right).$ Then,

(1)

If $2\le d\le n-⌈\frac{n-1}{2}⌉-1,$ then

$$k_m\left(G\right)\le \left\{\begin{array}{cc}n-d+⌊\frac{m}{2}⌋,\hfill & \mathrm{where}1\le m\le 2d-1,\hfill \\ n-2d+m,\hfill & \mathrm{where}2d\le m\le n.\hfill \end{array}\right.$$

(2)

If $d\ge n-⌈\frac{n-1}{2}⌉,$ then $k_m\left(G\right)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Proof. 

Since $V\left(T_1\right)\cap V\left(L\left(G\right)\right)=\varnothing$ or $V\left(T_2\right)\cap V\left(L\left(G\right)\right)=\varnothing ,$ let us assume that $V\left(T_1\right)\cap V\left(L\left(G\right)\right)=\varnothing .$ Since $V\left(T_1\right)\cap V\left(L\left(G\right)\right)=\varnothing ,$ then $V\left(L\left(G\right)\right)\subseteq V\left(T_2\right).$ The case of $V\left(T_2\right)\cap V\left(L\left(G\right)\right)=\varnothing$ is similar to that of $V\left(T_1\right)\cap V\left(L\left(G\right)\right)=\varnothing ,$ so we omit it. When $i=1,2,3,$ we know that $G\left[V_{1,i}\right]$ and $G\left[V_{2,i}\right]$ are connected. Moreover, $Z_{1,i}^{\prime }\cap Z_{2,i}^{\prime }=\varnothing ,$$Z_{1,i}^{\prime }\cap (Z_{1,i}\cup Z_{2,i})=\varnothing$, and $Z_{2,i}^{\prime }\cap (Z_{1,i}\cup Z_{2,i})=\varnothing .$ According to Proposition 6, if $i=1,2,$ then $\left\{v_k\right\}\subseteq N^{+}(G^{n-d+a}:x,y).$ If $i=3,$ then $\{v_{k,b},v_{h,g}\}\subseteq N^{+}(G^{n-d+a}:x,y)$, and $\{v_{k,b},v_k,v_{h,g}\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$

(1)

$2\le d\le n-⌈\frac{n-1}{2}⌉-1.$

We know that $\left|Z_{2,1}^{\prime }\right|\ge d-1,$$\left|Z_{2,2}^{\prime }\right|\ge d$, and $\left|Z_{2,3}^{\prime }\right|\ge d-2.$ For $i=1,2,3,$ next, we can conclude that Case a and Case b hold.

Case a: If $k-1\ge \left|Z_{2,i}^{\prime }\right|+1,$

We consider $i=1,2.$ According to Corollary 3, $|N^{+}(G^{n-d+r}:x,y)|\ge 2r+1,$ where $0\le r\le \left|Z_{2,i}^{\prime }\right|.$ Hence, we have $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le 2\left|Z_{2,i}^{\prime }\right|+1.$

Now, let us consider the case in which $i=1,2$ and $2\left|Z_{2,i}^{\prime }\right|+2\le m\le n.$ In Proposition 5, let $G=G^{\prime },$$X_1=V_{2,i}\cup V\left(P(v_{k-|Z_{2,i}^{\prime }|},v_k)\right)$ and $X_2=V\left(G\right)\setminus X_1.$ Then, $\left|X_1\right|=2\left|Z_{2,i}^{\prime }\right|+1.$ For any vertex $v\in X_1$, we have $d(v,v_k)\le \left|Z_{2,i}^{\prime }\right|.$ Moreover, $\left\{v_k\right\}\subseteq N^{+}(G^{n-d+a}:x,y).$ Therefore, $X_1\subseteq N^{+}(G^{n-d+|Z_{2,i}^{\prime }|+a}:x,y).$ According to Proposition 5, we have $|N^{+}(G^{n-d+|Z_{2,i}^{\prime }|+r}:x,y)|\ge |X_1|+r,$ where r is any integer such that $1\le r\le |X_2|.$ Therefore, for $2\left|Z_{2,i}^{\prime }\right|+2\le m\le n,$ we have $k_m(G:x,y)\le n-d-1+m-\left|Z_{2,i}^{\prime }\right|.$

In particular, let us consider the case in which $k-1=\left|Z_{2,i}^{\prime }\right|+1.$ If $k-1=\left|Z_{2,i}^{\prime }\right|+1,$ for any vertex $v\in V\left(G\right)$, we have $d(v,v_k)\le d(v_1,v_k)=k-1.$ Moreover, $\left\{v_k\right\}\subseteq N^{+}(G^{n-d+a}:x,y).$ Therefore, $|N^{+}(G^{n-d+k-1}:x,y)|=V\left(G\right).$ If $k-1=\left|Z_{2,i}^{\prime }\right|+1,$ we have $k_m(G:x,y)\le n-d+k-1=n-d+\left|Z_{2,i}^{\prime }\right|+1,$ where $2\left|Z_{2,i}^{\prime }\right|+2\le m\le n.$

Since for $i=1,2,$ we have $\left|Z_{2,i}^{\prime }\right|\ge d-1,$ then $2\left|Z_{2,i}^{\prime }\right|+1\ge 2d-1.$ Therefore, the conclusion holds.

We consider $i=3.$ According to Corollary 3, $|N^{+}(G^{n-d+1+r}:x,y)|\ge 2r+3,$ where $0\le r\le \left|Z_{2,i}^{\prime }\right|.$ Hence, we have $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le 2\left|Z_{2,i}^{\prime }\right|+3.$

Now, let us consider the case in which $i=3$ and $2\left|Z_{2,i}^{\prime }\right|+4\le m\le n.$ In Proposition 5, let $G=G^{\prime },$$X_1^{\prime }=V_{2,i}\cup V\left(P(v_{k-|Z_{2,i}^{\prime }|},v_k)\right)$ and $X_2^{\prime }=V\left(G\right)\setminus X_1^{\prime }.$ It is known that $V_{2,i}=Z_{2,i}\cup Z_{2,i}^{\prime }.$ Then, $\left|X_1^{\prime }\right|=2\left|Z_{2,i}^{\prime }\right|+3.$ For any vertex $v\in Z_{2,i}^{\prime }$, we have $d(v,Z_{2,i})\le \left|Z_{2,i}^{\prime }\right|.$ Moreover, $Z_{2,i}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ Thus, $Z_{2,i}^{\prime }\subseteq N^{+}(G^{n-d+1+|Z_{2,i}^{\prime }|+a}:x,y).$ Therefore, $X_1^{\prime }\subseteq N^{+}(G^{n-d+1+|Z_{2,i}^{\prime }|+a}:x,y).$ According to Proposition 5, we have $|N^{+}(G^{n-d+1+|Z_{2,i}^{\prime }|+r}:x,y)|\ge |X_1^{\prime }|+r,$ where r is any integer such that $1\le r\le |X_2^{\prime }|.$ Therefore, for $2\left|Z_{2,i}^{\prime }\right|+4\le m\le n,$ we have $k_m(G:x,y)\le n-d-2+m-\left|Z_{2,i}^{\prime }\right|.$

In particular, let us consider the case in which $k-1=\left|Z_{2,i}^{\prime }\right|+1.$ If $k-1=\left|Z_{2,i}^{\prime }\right|+1,$ for any vertex $v\in V\left(G\right)$, we have $d(v,v_k)\le d(v_1,v_k)=k-1.$ Moreover, $\left\{v_k\right\}\subseteq N^{+}(G^{n-d+1+a}:x,y).$ Therefore, $|N^{+}(G^{n-d+k}:x,y)|=V\left(G\right).$ If $k-1=\left|Z_{2,i}^{\prime }\right|+1,$ we have $k_m(G:x,y)\le n-d+k=n-d+\left|Z_{2,i}^{\prime }\right|+2,$ where $2\left|Z_{2,i}^{\prime }\right|+4\le m\le n.$

Since $\left|Z_{2,i}^{\prime }\right|\ge d-2,$ then $2\left|Z_{2,i}^{\prime }\right|+3\ge 2d-1.$ Therefore, the conclusion holds. As demonstrated above, for $i=1,2,3,$ if $k-1=\left|Z_{2,i}^{\prime }\right|+1,$ we have $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Case b: If $k-1\le \left|Z_{2,i}^{\prime }\right|$,

We consider $i=1,2.$ According to Corollary 3, $|N^{+}(G^{n-d+r}:x,y)|\ge 2r+1,$ where $0\le r\le k-1.$ Moreover, $N^{+}(G^{n-d+k-1}:x,y)=V\left(G\right).$

We consider $i=3.$ According to Corollary 3, $|N^{+}(G^{n-d+1+r}:x,y)|\ge 2r+3,$ where $0\le r\le k-1.$ Moreover, $N^{+}(G^{n-d+k}:x,y)=V\left(G\right).$

Hence, for $i=1,2,3,$ we have $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

(2)

$d\ge n-⌈\frac{n-1}{2}⌉.$

We consider $i=1,2.$ If $i=1,$ then $d(v_1,v_k)\le n-d.$ If $i=2,$ then $d(v_1,v_k)\le \left|W_1(v_1,v_k)\right|\le n-d.$ Therefore, for $i=1,2,$ we have $d(v_1,v_k)\le n-d\le ⌈\frac{n-1}{2}⌉,$$\left|Z_{2,i}^{\prime }\right|\ge d-1\ge n-1-⌈\frac{n-1}{2}⌉.$ If n is odd, then $k-1\le \left|Z_{2,i}^{\prime }\right|.$ If n is even, then $k-1\le \frac{n}{2}$, and $\left|Z_{2,i}^{\prime }\right|\ge \frac{n}{2}-1.$ Whether n is odd or even, we have $k-1\le \left|Z_{2,i}^{\prime }\right|$ or $k-1=\left|Z_{2,i}^{\prime }\right|+1.$ Combining $k-1=\left|Z_{2,i}^{\prime }\right|+1$ in Case a and Case b, we have $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

We consider $i=3.$ Note that $k-1\le n-d-1\le ⌈\frac{n-1}{2}⌉-1,\left|Z_{2,i}^{\prime }\right|\ge d-2\ge n-2-⌈\frac{n-1}{2}⌉.$ If n is odd, then $k-1\le \left|Z_{2,i}^{\prime }\right|.$ If n is even, then $k-1\le \frac{n}{2}-1$, and $\left|Z_{2,i}^{\prime }\right|\ge \frac{n}{2}-2.$ Whether n is odd or even, we have $k-1\le \left|Z_{2,i}^{\prime }\right|$ or $k-1=\left|Z_{2,i}^{\prime }\right|+1.$ Combining $k-1=\left|Z_{2,i}^{\prime }\right|+1$ in Case a and Case b, we have $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

□

4. Upper Bounds of the $\mathbf{m}$-Competition Indices

In this section, combined with some conclusions presented in Section 3, we derive the upper bounds of the m-competition indices of $S_n\left(d\right),$ where $1\le m\le n.$ See Theorems 1–3 below.

Theorem 1.

Let $G\in S_n\left(d\right)$ and $1\le d\le n-⌈\frac{n-1}{2}⌉-1.$ Then,

$$k_m\left(G\right)\le \left\{\begin{array}{cc}n-d+⌊\frac{m}{2}⌋,\hfill & \mathrm{if}1\le m\le 2d-1,\hfill \\ n-2d+m,\hfill & \mathrm{if}2d\le m\le n.\hfill \end{array}\right.$$

Proof. 

We only need to consider $G\in S_n^{\prime }\left(d\right).$ Let $d=1.$ Let us assume that u is the loop vertex. In Proposition 5, let $G=G^{\prime },$$X_1=\left\{u\right\}$ and $X_2=V\left(G\right)\setminus X_1.$ Then, $d(x,u)\le n-1$, and $d(y,u)\le n-1.$ We have $X_1\subseteq N^{+}(G^{n-1+a}:x,y).$ Then, $k_1\left(G\right)\le n-1.$ According to Proposition 5, we have $|N^{+}(G^{n-1+r}:x,y)|\ge |X_1|+r=r+1,$ where r is any integer such that $1\le r\le n-1.$ Therefore, $k_m(G:x,y)\le n+m-2,$ where $2\le m\le n.$ Therefore, the conclusion is clearly established. Next, we consider $2\le d\le n-⌈\frac{n-1}{2}⌉-1.$

Since $d\le n-⌈\frac{n-1}{2}⌉-1,$ we have $⌈\frac{n-1}{2}⌉\le n-d-1.$ According to Corollary 1, Lemma 2, Lemma 4, and Lemma 6, we conclude that this conclusion holds. □

Theorem 2.

Let $G\in S_n\left(d\right)$ and n be odd. If $\frac{n+1}{2}\le d\le n,$ then $k_m\left(G\right)\le \frac{n-1}{2}+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Proof. 

We only need to consider $G\in S_n^{\prime }\left(d\right).$ Suppose that $G\in S_{n,1}^{\prime }\left(d\right).$ We consider $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$ and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing .$ If $d\ge \frac{n+3}{2}$, we have $d(x,v_c)\le \frac{n-1}{2}.$ Moreover, according to Proposition 4, there exists a walk ($W_m(y,v_c)$) such that $V\left(W_m(y,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing ,$ and $|W_m(y,v_c)|\le n-d+1.$ Therefore, we have $|W_m(y,v_c)|\le n-d+1\le \frac{n-1}{2}.$ According to Lemma 1, we have $k_m(G:x,y)\le \frac{n-1}{2}+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$ If $d=\frac{n+1}{2},$ the conclusion holds according to Lemma 3. Combining Corollary 1, Lemma 4, and Lemma 6, the conclusion holds. □

Theorem 3.

Let $G\in S_n\left(d\right)$ and n be even. Then,

(1)

If $d=\frac{n}{2},$ then $k_m\left(G\right)\le \frac{n}{2}+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

(2)

If $\frac{n}{2}+1\le d\le n,$ then $k_m\left(G\right)\le \frac{n}{2}+⌊\frac{m-1}{2}⌋,$ where $1\le m\le n.$

Proof. 

We only need to consider $G\in S_n^{\prime }\left(d\right).$

(1)

According to Corollary 1, Lemmas 3–4, and Lemma 6, this conclusion holds.

(2)

Case 1: $G\in S_{n,1}^{\prime }\left(d\right).$

Case 1.1: $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$, and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing .$

If $l=n,$ according to Lemma 5, the conclusion holds. If $l\le n-1,$ we have $d(x,v_c)\le ⌈\frac{l-1}{2}⌉\le \frac{n-2}{2}$, and $d(y,v_c)\le ⌈\frac{l-1}{2}⌉\le \frac{n-2}{2}.$ Then, $k_m(G:x,y)\le \frac{n-2}{2}+⌊\frac{m}{2}⌋=\frac{n}{2}-1+⌊\frac{m}{2}⌋\le \frac{n}{2}+⌊\frac{m-1}{2}⌋,$ where $1\le m\le n.$

Case 1.2: $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)\ne \varnothing$, and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing .$

Then, $|W_m(y,v_c)|\le n-d+1\le \frac{n}{2}$, and $d(x,v_c)\le ⌈\frac{l-1}{2}⌉\le \frac{n}{2}.$ If $d(x,v_c)\le \frac{n}{2}-2,$ the proof is similar to that of Lemma 2, and we have $k_m(G:x,y)\le \frac{n}{2}-1+⌊\frac{m}{2}⌋\le \frac{n}{2}+⌊\frac{m-1}{2}⌋,$ where $1\le m\le n.$ If $d(x,v_c)\ge \frac{n}{2}-1,$ then $⌈\frac{l-1}{2}⌉\ge \frac{n}{2}-1.$ We can conclude that $l=n,$$l=n-1$, or $l=n-2.$

If $l=n,$ according to Lemma 5, the conclusion holds. Next, we consider $l=n-1$ or $l=n-2.$ Whether $l=n-1$ or $l=n-2,$ we have $v_c=v_{\frac{n}{2}}.$ Then, $d(x,v_c)=d(x,v_{\frac{n}{2}})=\frac{n}{2}-1.$ Furthermore, $|W_m(y,v_c)|\le n-d+1\le \frac{n}{2}.$ According to Corollary 2, we can obtain $\left\{v_{\frac{n}{2}-1}\right\}\subseteq N^{+}(G^{\frac{n}{2}}:y).$ Furthermore, we have $\{v_{\frac{n}{2}-1},v_{\frac{n}{2}}\}\subseteq N^{+}(G^{\frac{n}{2}+a}:x,y).$ For $1\le r\le \frac{n}{2}-2,$$\{v_{\frac{n}{2}-1-r},v_{\frac{n}{2}-r},\cdots ,v_{\frac{n}{2}+r}\}\subseteq N^{+}(G^{\frac{n}{2}+r}:x,y).$ For any vertex $v\in V\left(G\right)$, we have $d(v,v_{\frac{n}{2}})\le \frac{n}{2}-1.$ Therefore, $N^{+}(G^{n-1}:x,y)=V\left(G\right).$

Case 1.3: $V\left(P(x,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing$, and $V\left(P(y,v_c)\right)\cap V\left(L\left(G\right)\right)=\varnothing .$

According to Lemma 4, we have $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋\le \frac{n}{2}-1+⌊\frac{m}{2}⌋\le \frac{n}{2}+⌊\frac{m-1}{2}⌋,$ where $1\le m\le n.$

Case 2: $G\in S_{n,2}^{\prime }\left(d\right).$

According to Lemma 6, we have $k_m(G:x,y)\le n-d+⌊\frac{m}{2}⌋\le \frac{n}{2}-1+⌊\frac{m}{2}⌋\le \frac{n}{2}+⌊\frac{m-1}{2}⌋,$ where $1\le m\le n.$ □

5. The $\mathbf{m}$-Competition Indices of ${\mathbf{L}}_{\mathbf{n},\mathbf{d}}$

In this section, the m-competition indices of $L_{n,d}$ are given, where $L_{n,d}$ is defined as follows:

Definition 6.

Let $L_{n,d}\in S_n^{\prime }\left(d\right).$ Let $V\left(L_{n,d}\right)=\{v_1,v_2,\cdots ,v_n\}$, and let $E\left(L_{n,d}\right)=\left\{[v_i,v_{i+1}]\right|1\le i\le n-1\}\cup \left\{[v_i,v_i]\right|n-d+1\le i\le n\},$ where $1\le d\le n.$

Remark 3.

According to the above definition, the conclusions in $L_{n,d}$ hold:

(1) If $1\le d\le n-⌈\frac{n-1}{2}⌉-1,$ then $v_1$ and $v_2$ are not loop vertices.

(2) If n is odd and $\frac{n+1}{2}\le d\le n$, then $v_{\frac{n+1}{2}}$ is a loop vertex.

(3) If n is even and $d=\frac{n}{2},$ then $v_1$ and $v_2$ are not loop vertices.

(4) If n is even and $\frac{n}{2}+1\le d\le n,$ then $v_{\frac{n}{2}},v_{\frac{n}{2}+1}$ are two loop vertices.

In Figure 1, since $1\le d\le n-⌈\frac{n-1}{2}⌉-1,$ then $n-d+1\ge ⌈\frac{n+3}{2}⌉.$ In Figure 2, since n is even and $d=\frac{n}{2},$ then $n-d+1=\frac{n}{2}+1.$

The authors of [26] studied the generalized competition indices of several classes of primitive digraphs and obtained the m-competition indices of $L_{n,d}$ or isomorphic to $L_{n,d},$ where $1\le m\le n.$ According to the conclusions reported in [26] about the m-competition indices of $L_{n,d}$, we propose Theorems 4–6. To facilitate the understanding of Theorems 4–6, we provide remarks below each theorem.

Theorem 4.

If $1\le d\le n-⌈\frac{n-1}{2}⌉-1,$ then

$$k_m\left(L_{n,d}\right)=\left\{\begin{array}{cc}n-d+⌊\frac{m}{2}⌋,\hfill & \mathrm{if}1\le m\le 2d-1,\hfill \\ n-2d+m,\hfill & \mathrm{if}2d\le m\le n.\hfill \end{array}\right.$$

Remark 4.

If $1\le d\le n-⌈\frac{n-1}{2}⌉-1,$ then $L_{n,d}$ is shown in Figure 1. For $0\le r\le d-1,$ we have $N^{+}(L_{n,d}^{n-d+r}:v_1,v_2)=\{v_{n-d+1-r},v_{n-d+2-r},\cdots ,v_{n-d+1+r}\}.$ For $1\le m\le 2d-1,$ we have $k_m\left(L_{n,d}\right)=k_m(L_{n,d}:v_1,v_2)=n-d+⌊\frac{m}{2}⌋.$ For $2d-1\le m\le n-1,$ then $k_{m+1}\left(L_{n,d}\right)=k_{m+1}(L_{n,d}:v_1,v_2)=k_m(L_{n,d}:v_1,v_2)+1.$ Therefore, Theorem 4 can be obtained.

Theorem 5.

If n is odd and $\frac{n+1}{2}\le d\le n$, then $k_m\left(L_{n,d}\right)=\frac{n-1}{2}+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

Remark 5.

Suppose that n is odd and $\frac{n+1}{2}\le d\le n.$ For $0\le r\le \frac{n-1}{2},$ we have $N^{+}(L_{n,d}^{\frac{n-1}{2}+r}:v_1,v_n)=\{v_{\frac{n+1}{2}-r},v_{\frac{n+3}{2}-r},\cdots ,v_{\frac{n+1}{2}+r}\}.$ Therefore, Theorem 5 can be obtained.

Theorem 6.

Let n be even; then,

(1)

If $d=\frac{n}{2},$ then $k_m\left(L_{n,d}\right)=\frac{n}{2}+⌊\frac{m}{2}⌋,$ where $1\le m\le n.$

(2)

If $\frac{n}{2}+1\le d\le n,$ then $k_m\left(L_{n,d}\right)=\frac{n}{2}+⌊\frac{m-1}{2}⌋,$ where $1\le m\le n.$

Remark 6.

Suppose that n is even. If $d=\frac{n}{2},$ then $L_{n,d}$ is shown in Figure 2. For $0\le r\le \frac{n}{2}-1,$ we have $N^{+}(L_{n,d}^{\frac{n}{2}+r}:v_1,v_2)=\{v_{\frac{n}{2}+1-r},v_{\frac{n}{2}+2-r},\cdots ,v_{\frac{n}{2}+1+r}\}.$ Furthermore, $N^{+}(L_{n,d}^n:v_1,v_2)=V\left(L_{n,d}\right).$ If $\frac{n}{2}+1\le d\le n,$ for $0\le r\le \frac{n}{2}-1,$ we have $N^{+}(L_{n,d}^{\frac{n}{2}+r}:v_1,v_n)=\{v_{\frac{n}{2}-r},v_{\frac{n}{2}+1-r},\cdots ,v_{\frac{n}{2}+1+r}\}.$ Therefore, Theorem 6 can be obtained.

According to Theorems 4–6, for any m satisfying $1\le m\le n,$ the upper bounds of the m-competition indices of $S_n\left(d\right)$ can be reached.

6. Conclusions and Discussion

In this paper, for any m satisfying $1\le m\le n,$ we studied the upper bounds of the m-competition indices of $S_n\left(d\right)$ and obtained the upper bounds. Furthermore, we found that the upper bounds can be reached. In Theorems 1–3, let $m=1,$ the upper bounds of the 1-competition indices obtained are consistent with the case in $n\ge 5$ in Proposition 1. Furthermore, for $G\in S_n\left(d\right),$ according to Theorem 1, if $1\le d\le n-⌈\frac{n-1}{2}⌉-1,$ then the upper bound of $k_m\left(G\right)$ is related to $n,d,$ and $m.$ According to Theorems 2 and 3, whether n is odd or even, if $⌈\frac{n}{2}⌉\le d\le n,$ then the upper bound of $k_m\left(G\right)$ depends on $n,m,$ but not on $d.$

In future work, for some special digraphs, in addition to upper bounds of the generalized competition indices, the set of generalized competition indices deserves to be studied. It would be meaningful to solve these problems in future studies.