The Steiner k-Wiener Index of Cacti

Abstract

Let G be a connected graph. The Steiner k-Wiener index $S{W}_k\left(G\right)$ of graph G is defined as $S{W}_k\left(G\right)={\displaystyle \sum _{S\subseteq V\left(G\right),\left|S\right|=k}}{d}_G\left(S\right)$, where $d_G\left(S\right)$ represents the minimum size of a connected subgraph of G that connects S. Using some graph operations, we obtain the minimum value and the second minimum value of the Steiner k-Wiener index for cacti with order n and t cycles, and we characterize the corresponding extremal graphs by exploiting structural symmetries.

1. Introduction

In this paper, all graphs are undirected, finite, simple, and connected. The terminology and concepts used herein are based on the reference [1]. Let G be a simple connected graph, where $V\left(G\right)$ and $E\left(G\right)$ are its vertex set and edge set, respectively. For $u\in V\left(G\right)$, $d_G\left(u\right)$ denotes the degree of u in G, which is the number of edges incident to u. A vertex of degree 1 is called a pendant vertex, and an edge incident to a pendant vertex is called a pendant edge. For $u,v\in V\left(G\right)$, $d_G(u,v)$ represents the distance between u and v, specifically the length of the shortest path connecting u and v. Let $G-uv$ be the graph that arises from G by deleting the edge $uv\in E\left(G\right)$. Similarly, $G+uv$ is a graph that arises from G by adding an edge $uv\in E\left(G\right)$, where $u,v\in V\left(G\right)$. Let $V_c=\{v\in V\left(G\right):v\mathrm{is}\mathrm{a}\mathrm{cut}\mathrm{vertex}\mathrm{of}\mathrm{G}\}$.

The Wiener index was introduced by the chemist Wiener in 1947 during his study of paraffin boiling points [2], and the Wiener index of a graph G is defined as the sum of distances between all pairs of vertices in G, expressed as

$${\displaystyle W\left(G\right)=\sum _{\{u,v\}\subseteq V\left(G\right)}{d}_G(u,v).}$$

For more details on the Wiener index, we refer to [3,4,5,6,7,8,9].

In 1994, Chartrand et al. [10] put forward the notion of Steiner distance. For a subset $S\subseteq V\left(G\right)$, the Steiner distance $d_G\left(S\right)$ is defined as the minimum size of a connected subgraph of G connecting S. In particular, if $S=\{u,v\}$, then $d_G\left(S\right)=d_G(u,v)$. In 2014, Li et al. [11] formulated the Steiner k-Wiener index, which generalizes the Wiener index by incorporating the definition of Wiener index and Steiner distance. For a positive integer k, $2\le k\le n-1$, the Steiner k-Wiener index $S{W}_k\left(G\right)$ of a graph G is defined as

$$S{W}_k\left(G\right)=\sum _{S\subseteq V\left(G\right),\left|S\right|=k}{d}_G\left(S\right).$$

If $k=2$, the Steiner 2-Wiener index coincides with the Wiener index. Subsequently, some studies on the Steiner k-Wiener index were reported. In 2020, Li et al. [12] calculated the Steiner Szeged index and the Steiner k-Wiener index of cycles and wheels. In 2021, Lai et al. [13] determined upper and lower bounds of the Steiner $(n-1)$-Wiener index among all unicyclic graphs. Based on this, in 2022, Fan et al. [14] determined upper and lower bounds of the Steiner k-Wiener index among all unicyclic graphs. For results on the Steiner k-Wiener index, one can refer to [15,16,17]. And other topological indices have begun to receive extensive attention, such as the Trinajstić index [18].

A connected graph whose every block is either an edge or a cycle is called a cactus graph, or simply a cactus. Denote by $C(n,t)$ the set of all cacti with n vertices and t cycles, where $0\le t\le ⌊{\displaystyle \frac{n-1}{2}}⌋$. Clearly, $C(n,0)$ represents the set of all trees of order n, and $C(n,1)$ represents the set of all unicyclic graphs of order n. A cycle is called an end-block if it contains exactly one vertex whose degree is not equal to 2. A bundle is a cactus in which all cycles share exactly one common vertex. Let $G_n(l_1,l_2,\dots ,l_t)$ be a bundle consisting of t cycles of lengths $l_1,l_2,\dots ,l_t$, where $n+k-{\sum }_{i=1}^t(l_i-1)$ pendant edges are attached to the unique common vertex of all cycles. If $l_1=l_2=\cdots =l_t=3$, define $G^0(n,t)\cong G_n\left(\underset{t}{{\displaystyle \underbrace{3,3,\dots ,3}}}\right)$ (see Figure 1).

Figure 1. Graph $G^0(n,t)$.

Cacti hold significant importance in graph theory and network analysis, as they are widely employed to describe and analyze closed structures and cyclic relationships in various real-world systems. Consequently, investigating the Steiner k-Wiener index of cacti is of great value. In 2007, Liu et al. [8] gave the minimum Wiener index among graphs in $C(n,t)$. In 2021, Deng et al. [9] determined the extremal graphs with second smallest Wiener index among cacti with order n and t cycles. However, the lower bound on the Steiner k-Wiener index of n-vertex cacti with exactly t cycles remains unknown for $3\le k\le n-1$. Therefore, in this paper, using graph transformation methods and the symmetric structure of the graph, we obtain the minimum and second minimum Steiner k-Wiener indices for cacti with order n and t cycles, and we characterize the corresponding extremal graphs.

Theorem 1. 

Let $G\in C(n,t)$, $n\ge 6$.

(a) If $3\le k\le n-2$, then

$$S{W}_k\left(G\right)\ge \left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)(n-1),$$

with equality if and only if $G\cong G^0(n,t)$.

(b) If $k=n-1$, then

$S{W}_k\left(G\right)\ge \left\{\begin{array}{cc}{(n-1)}^2\hfill & t=0,\mathit{with}\mathit{equality}\mathit{if}\mathit{and}\mathit{only}\mathit{if}G\cong S_n,\hfill \\ n(n-2)\hfill & t=1,\mathit{with}\mathit{equality}\mathit{if}\mathit{and}\mathit{only}\mathit{if}G\cong C_n,\hfill \\ {(n-1)}^2\hfill & t\ge 2,\mathit{with}\mathit{equality}\mathit{if}\mathit{and}\mathit{only}\mathit{if}G\cong G_n(l_1,l_2,\dots ,l_t).\hfill \end{array}\right.$

The double star $S_{n_1,n_2}$ (assume $n_1\ge n_2>1$) is defined as a tree in which the end u of an edge $e=uv$ is connected to $n_1-1$ pendant edges, and the other end v is connected to $n_2-1$ pendant edges(see Figure 2). Clearly, $n_1+n_2=n$.

Figure 2. Graphs $S_{n_1,n_2}$ and $H_n(t_1,s_1;t_2,s_2)$.

The graph $H_n(t_1,s_1;t_2,s_2)$ is defined as a graph in which one end u of an edge $e=uv$ is connected to $t_1$ cycles and $s_1$ pendant edges, and the other end v is connected to $t_2$ cycles and $s_2$ pendant edges(see Figure 2). Here, $d_G\left(u\right)\ge 2$, $d_G\left(v\right)\ge 2$ and $t_1+t_2=t$.

Theorem 2. 

Let $G\in C(n,t)$, $n\ge 6$.

(a) If $3\le k\le n-2$ and $G\in C(n,t)\setminus \left\{G^0(n,t)\right\}$, then

if $t=0$, then

$$S{W}_k\left(G\right)\ge \left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)(n-1)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k-1}}\right),$$

with equality if and only if $G\cong S_{n-2,2}$;

if $t\ge 1$, then

$$S{W}_k\left(G\right)\ge \left\{\begin{array}{cc}\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{2}}\right)(n-1)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)-1\hfill & k=3,\hfill \\ \left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)(n-1)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right)\hfill & 4\le k\le n-2,\hfill \end{array}\right.$$

with equality if and only if $G\cong G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}})$ (see Figure 3).

Figure 3. Graph $G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}})$.

(b) If $k=n-1$, then

if $t=0$ and $G\in C(n,0)\setminus \left\{S_n\right\}$, $S{W}_k\left(G\right)\ge n^2-2n+2$, with equality if and only if $G\cong S_{n_1,n_2}$;

if $t=1$ and $G\in C(n,1)\setminus \left\{C_n\right\}$, $S{W}_k\left(G\right)\ge {(n-1)}^2$, with equality if and only if $G\cong G_n\left(l_1\right)(l_1\ne n)$;

if $t\ge 2$ and $G\in C(n,t)\setminus \left\{G_n(l_1,l_2,\dots ,l_t)\right\}$,$S{W}_k\left(G\right)\ge n^2-2n+2$, with equality if and only if $G\cong H_n(t_1,s_1;t_2,s_2)$.

2. Preliminaries

To establish the main results, we present some crucial lemmas.

Lemma 1. 

Let H, X, and Y be three pairwise disjoint connected graphs. Suppose that $u,v\in V\left(H\right)$, $v^{\prime }\in V\left(X\right)$, and $u^{\prime }\in V\left(Y\right)$. Let G be the graph obtained from $H,X,Y$ by identifying v with $v^{\prime }$ and u with $u^{\prime }$. Let $G^{\prime }$ be the graph obtained from $H,X,Y$ by identifying v, $v^{\prime }$, and $u^{\prime }$, and let $G^{″}$ be the graph obtained from $H,X,Y$ by identifying u, $v^{\prime }$, and $u^{\prime }$ (see Figure 4). If $2\le k\le n-1$, then

$$S{W}_k\left(G\right)>S{W}_k\left(G^{\prime }\right)\mathrm{or}S{W}_k\left(G\right)>S{W}_k\left(G^{″}\right).$$

Figure 4. The transformation used in Lemma 1.

Proof. 

Let $V\left(G\right)=V\left(H\right)\cup (V\left(X\right)\setminus \left\{v^{\prime }\right\})\cup (V\left(Y\right)\setminus \left\{u^{\prime }\right\})$. Now we will examine the change in Steiner distance from G to $G^{\prime }$.

• If $S\subseteq V\left(H\right)$ or $S\subseteq V\left(X\right)\setminus \left\{v^{\prime }\right\}$ or $S\subseteq V\left(Y\right)\setminus \left\{u^{\prime }\right\}$, then $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$.

• If $S\cap V\left(H\right)=\varnothing$, $S\cap V\left(X\right)\setminus \left\{v^{\prime }\right\}\ne \varnothing$, and $S\cap V\left(Y\right)\setminus \left\{u^{\prime }\right\}\ne \varnothing$, then $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)>0$.

• If $S\cap V\left(H\right)\ne \varnothing$, $S\cap V\left(X\right)\setminus \left\{v^{\prime }\right\}=\varnothing$, and $S\cap V\left(Y\right)\setminus \left\{u^{\prime }\right\}\ne \varnothing$, then $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=d_H(S^{\prime }\cup \left\{u\right\})-d_H(S^{\prime }\cup \left\{v\right\})$, where $S^{\prime }=S\cap V\left(H\right)$.

• If $S\cap V\left(H\right)\ne \varnothing$, $S\cap V\left(X\right)\setminus \left\{v^{\prime }\right\}\ne \varnothing$, and $S\cap V\left(Y\right)\setminus \left\{u^{\prime }\right\}=\varnothing$, then $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$.

• If $S\cap V\left(H\right)\ne \varnothing$, $S\cap V\left(X\right)\setminus \left\{v^{\prime }\right\}\ne \varnothing$, and $S\cap V\left(Y\right)\setminus \left\{u^{\prime }\right\}\ne \varnothing$, then $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)\ge 0$.

Thus, we have

$$S{W}_k\left(G\right)-S{W}_k\left(G^{\prime }\right)>\sum _{S^{\prime }=S\cap V\left(H\right)}[d_H(S^{\prime }\cup \left\{u\right\})-d_H(S^{\prime }\cup \left\{v\right\})].$$

Similarly, we also have

$$S{W}_k\left(G\right)-S{W}_k\left(G^{″}\right)>\sum _{S^{\prime }=S\cap V\left(H\right)}[d_H(S^{\prime }\cup \left\{v\right\})-d_H(S^{\prime }\cup \left\{u\right\})].$$

If ${\sum }_{S^{\prime }=S\cap V\left(H\right)}[d_H(S^{\prime }\cup \left\{u\right\})-d_H(S^{\prime }\cup \left\{v\right\})]>0$, then $S{W}_k\left(G\right)-S{W}_k\left(G^{\prime }\right)>0$; if ${\sum }_{S^{\prime }=S\cap V\left(H\right)}[d_H(S^{\prime }\cup \left\{u\right\})-d_H(S^{\prime }\cup \left\{v\right\})]<0$, then $S{W}_k\left(G\right)-S{W}_k\left(G^{″}\right)>0$. □

Let $C_n$ be a cycle of order n, with vertex set $V\left(C_n\right)=\{u_1,u_2,\dots ,u_n\}$. For each k-subset $S\subseteq V\left(C_n\right)$, let $S=\{a_1,a_2,\dots ,a_k\}$, where the vertices are selected in a clockwise sequence from $V\left(C_n\right)$. We can associate this subset with a corresponding vector $(x_1,x_2,\dots ,x_k)$, where $x_i$ is the length of the path from $a_i$ to $a_{i+1}$ in the clockwise direction for $1\le i\le k-1$, and $x_k$ is the length of the path from $a_k$ to $a_1$ in the clockwise direction. Obviously, ${\sum }_{i=1}^k{x}_i=n$. Thus, $d_G\left(S\right)=n-max\{x_1,x_2,\dots ,x_k\}$ (see Figure 5). For example, consider the cycle $C_{10}$ and the subset $S=\{u_2,u_4,u_9\}$; it is obvious that $(x_1,x_2,x_3)=(2,5,3)$ and $d_G\left(S\right)=5$.

Figure 5. Graph $C_n$.

Lemma 2. 

Let G be a graph of order n, which is obtained from a connected graph H and a cycle $C_g=u_1{u}_2\dots u_g{u}_1$ ($g\ge 4$) by identifying the vertex $u_1$ with a vertex u of the graph H (see Figure 6). Let $G^{\prime }=G-u_2{u}_3+u_1{u}_3$.

Figure 6. The transformation used in Lemma 2.

(a) If $3\le k\le n-2$, then $S{W}_k\left(G\right)\ge S{W}_k\left(G^{\prime }\right)$;

(b) If $k=n-1$, then $S{W}_{n-1}\left(G\right)=S{W}_{n-1}\left(G^{\prime }\right)$.

Proof. 

Let $V\left(G\right)=\left\{u_2\right\}\cup X\cup Y$, where $X=V\left(H\right)\setminus \left\{u_1\right\}$ and $Y=V\left(C_g\right)\setminus \left\{u_2\right\}$.

(a) For any subset $S\subseteq V\left(G\right)=V\left(G^{\prime }\right)$ with $\left|S\right|=k$, where $3\le k\le n-2$, the following cases arise:

Case 1. $u_2\in S$.

Subcase 1.1. If $S\setminus \left\{u_2\right\}\subseteq X$, it is clear that $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$.

Subcase 1.2. If $S\setminus \left\{u_2\right\}\subseteq Y$, let $S=\{u_2,a_1,a_2,\dots ,a_{k-1}\}$, where $a_1\ne u_1$. Then $d_G\left(S\right)=d_{C_g}\left(\{u_2,a_1,a_2,\dots ,a_{k-1}\}\right)$, $d_{G^{\prime }}\left(S\right)=d_{C_{g-1}}\left(\{u_1,a_1,a_2,\dots ,a_{k-1}\}\right)+1$. Let the corresponding vector of $\{u_2,a_1,a_2,\dots ,a_{k-1}\}$ in G be $(x_1,x_2,\dots ,x_{k-1},x_k)$; thus, the corresponding vector of $\{u_1,a_1,a_2,\dots ,a_{k-1}\}$ in $G^{\prime }$ is $(x_1,x_2,\dots ,x_{k-1},x_k-1)$.

• If $max\{x_1,x_2,\dots ,x_{k-1},x_k\}=x_i$ in G, where $x_i\ge x_k$ for $1\le i\le k-1$, then $max\{x_1,x_2,\dots ,$$x_{k-1},x_k-1\}=x_i$ in $G^{\prime }$. In this case, we obtain $d_G\left(S\right)=g-x_i,d_{G^{\prime }}\left(S\right)=(g-1)-x_i+1=g-x_i$; thus, $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$.

• If $max\{x_1,x_2,\dots ,x_{k-1},x_k\}=x_k$ in G, where $x_k>x_i$ for all $1\le i\le k-1$, then $max\{x_1,x_2,\dots ,x_{k-1},x_k-1\}=x_k-1\ge x_i$ in $G^{\prime }$. In this case, we obtain $d_G\left(S\right)=g-x_k$, $d_{G^{\prime }}\left(S\right)=(g-1)-(x_k-1)+1=g-x_k+1$; thus, $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=-1$.

Subcase 1.3. If $S\setminus \left\{u_2\right\}\cap X\ne \varnothing$ and $S\setminus \left\{u_2\right\}\cap Y\ne \varnothing$, let $S\setminus \left\{u_2\right\}\cap Y=\{a_1,a_2,\dots ,a_s\}$, where $1\le s\le k-2$ and $a_1\ne u_1$. Then, we have $|S\setminus \{u_2\}\cap Y|=s$ and $|S\setminus \{u_2\}\cap X|=k-s-1$. Let $d_H(S^{\prime }\cup \left\{u_1\right\})=m$, where $S^{\prime }=S\cap X$; then, $d_G\left(S\right)=d_{C_g}\left(\{u_1,u_2,a_1,a_2,\dots ,a_s\}\right)+m$, $d_{G^{\prime }}\left(S\right)=d_{C_{g-1}}\left(\{u_1,a_1,a_2,\dots ,a_s\}\right)+m+1$. Let the corresponding vector of $\{u_1,u_2,a_1,a_2,\dots ,a_s\}$ in G be $(x_1=1,x_2,\dots ,x_{s+1},x_{s+2})$; thus, the corresponding vector of $\{u_1,a_1,a_2,\dots ,a_s\}$ in $G^{\prime }$ is $(x_2,x_3,\dots ,x_{s+1},x_{s+2})$. It is noted that $x_i\ge 1$ for $2\le i\le s+1$, and $x_{s+2}\ge 0$.

• If $max\{1,x_2,\dots ,x_{s+1},x_{s+2}\}=x_i\ge x_1$ in G, then $max\{x_2,x_3,\dots ,x_{s+1},$$x_{s+2}\}=x_i$ in $G^{\prime }$. In this case, we obtain $d_G\left(S\right)=g-x_i+m$, $d_{G^{\prime }}\left(S\right)=(g-1)-x_i+m+1=g-x_i+m$; thus, $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$.

Case 2. $u_2\notin S$.

Subcase 2.1. If $S\subseteq X$, obviously, $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$.

Subcase 2.2. If $S\subseteq Y$, let $S=\{a_1,a_2,\dots ,a_k\}$, where $a_1\ne u_1$. Then, $d_G\left(S\right)=d_{C_g}\left(\{a_1,a_2,\dots ,a_k\}\right)$, $d_{G^{\prime }}\left(S\right)=d_{C_{g-1}}\left(\{a_1,a_2,\dots ,a_k\}\right)$. Let the corresponding vector of $\{a_1,a_2,\dots ,a_k\}$ in G be $(x_1,x_2,\dots ,x_{k-1},x_k)$; thus, its corresponding vector in $G^{\prime }$ is $(x_1,x_2,\dots ,x_{k-1},x_k-1)$.

• If $max\{x_1,x_2,\dots ,x_{k-1},x_k\}=x_i$ in G, where $x_i>x_k$ for $i\ne k$, then $max\{x_1,x_2,$$\dots ,x_{k-1},x_k-1\}=x_i$ in $G^{\prime }$. In this case, we obtain $d_G\left(S\right)=g-x_i$, $d_{G^{\prime }}\left(S\right)=g-1-x_i$; thus, $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=1$.

• If $max\{x_1,x_2,\dots ,x_{k-1},x_k\}=x_k$ in G, where $x_k>x_i$ for all $1\le i\le k-1$, then $max\{x_1,x_2,\dots ,x_{k-1},x_k-1\}=x_k-1\ge x_i$ in $G^{\prime }$. In this case, we obtain $d_G\left(S\right)=g-x_k$, $d_{G^{\prime }}\left(S\right)=(g-1)-(x_k-1)=g-x_k$; thus, $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$.

• If $max\{x_1,x_2,\dots ,x_{k-1},x_k\}=x_k$ in G and there exists $x_i$ for $1\le i\le k-1$ such that $x_k=x_i$, then $max\{x_1,x_2,\dots ,x_{k-1},x_k-1\}=x_i$ in $G^{\prime }$. In this case, we obtain $d_G\left(S\right)=g-x_k$, $d_{G^{\prime }}\left(S\right)=(g-1)-x_i=g-1-x_k$; thus, $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=1$.

Subcase 2.3. If $S\cap X\ne \varnothing$ and $S\cap Y\ne \varnothing$, let $S\cap Y=\{a_1,a_2,\dots ,a_s\}$ with $1\le s\le k-1$, where $a_1\ne u_1$. Then, $|S\cap Y|=s$ and $|S\cap X|=k-s$. Let $d_H(S^{\prime }\cup \left\{u_1\right\})=m$, where $S^{\prime }=S\cap X$; then, $d_G\left(S\right)=d_{C_g}\left(\{u_1,a_1,a_2,\dots ,a_s\}\right)+m$, $d_{G^{\prime }}\left(S\right)=d_{C_{g-1}}\left(\{u_1,a_1,a_2,\dots ,a_s\}\right)+m$. Let the vector corresponding to $\{u_1,a_1,a_2,\dots ,a_s\}$ in G be $(x_1,x_2,\dots ,x_s,x_{s+1})$; thus, its corresponding vector in $G^{\prime }$ is $(x_1-1,x_2,\dots ,x_s,x_{s+1})$.

• If $max\{x_1,x_2,\dots ,x_s,x_{s+1}\}=x_1$ in G, where $x_1>x_i$ for all $2\le i\le s+1$, then $max\{x_1-1,x_2,\dots ,x_s,x_{s+1}\}=x_1-1\ge x_i$ in $G^{\prime }$. In this case, we obtain $d_G\left(S\right)=g-x_1+m$ and $d_{G^{\prime }}\left(S\right)=(g-1)-(x_1-1)+m=g-x_1+m$; thus, $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$.

• If $max\{x_1,x_2,\dots ,x_s,x_{s+1}\}=x_i$ in G, where $x_i\ge x_1$ for $2\le i\le s+1$, then $max\{x_1-1,x_2,\dots ,x_s,x_{s+1}\}=x_i$ in $G^{\prime }$. In this case, we obtain $d_G\left(S\right)=g-x_i+m$, $d_{G^{\prime }}\left(S\right)=(g-1)-x_i+m=g-1-x_i+m$; thus, $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=1$.

In all cases outlined above, except for Subcases 1.2, 2.2, and 2.3, we have $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$. Define ${\rho }^{-}$ as the set of all k-subsets S in Subcase 1.2, where $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=-1$. Similarly, let ${\rho }^{+}$ be the set of all k-subsets S in Subcases 2.2 and 2.3, where $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=1$. To prove $S{W}_k\left(G\right)\ge S{W}_k\left(G^{\prime }\right)$, now we need to demonstrate that $|{\rho }^{+}| \ge |{\rho }^{-}|$.

If $k\ge g$, then $|{\rho }^{-}| = 0$, implying $|{\rho }^{+}| \ge |{\rho }^{-}|$; now we consider $k < g$. Define ${\rho }^{-}={\rho }_1+{\rho }_2+\dots +{\rho }_k$, where ${\rho }_i=\{S\mid |S|=k,u_2,u_3,\dots ,u_{i+1}\in S,u_{i+2}\notin S,S\setminus \{u_2,u_3,\dots ,u_{i+1}\}\subseteq \{u_{i+3},\dots ,u_g,u_1\}\}$ for $1\le i\le k$. Each k-subset S in ${\rho }_i$ satisfies $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=-1$; let $|{\rho }_i|=s_i$ for $1\le i\le k$. Obviously, $|{\rho }_k| =s_k=1$.

For each $S_{ij}\in {\rho }_i$, where $1\le j\le s_i$, let

$$S_{ij}=\{u_2,u_3,\dots ,u_i,u_{i+1},a_1,a_2,\dots ,a_{k-i}\}.$$

By rotating each element in $S_{ij}$ in a counterclockwise manner by i units on the cycle, we can obtain

$$S_{ij}^{\prime }=\{a_1^{\prime },a_2^{\prime },\dots ,a_{k-i}^{\prime },u_{g+2-i},u_{g+3-i},\dots ,u_g,u_1\}.$$

Let the vector corresponding to $S_{ij}$ be $(x_1,\dots ,x_i,x_{i+1},\dots ,x_k)$; thus, the vector for $S_{ij}^{\prime }$ is $(y_1,\dots ,y_{k-i},$$y_{k-i+1},\dots ,y_k)$, where $y_1=x_{i+1},\dots ,y_{k-i}=x_k,y_{k-i+1}=x_1,\dots ,y_k=x_i$. Since $S_{ij}\in {\rho }_i$, and following from Subcase 1.2, we have $max\{x_1,x_2,\dots ,x_{k-1},x_k\}=x_k>x_i$ for all $1\le i\le k-1$ in G, and $max\{x_1,x_2,\dots ,x_{k-1},x_k-1\}=x_k-1\ge x_i$ in $G^{\prime }$.

When $1\le i\le k-1$, it follows that $S_{ij}^{\prime }$ satisfies $max\{y_1,\dots ,y_{k-i},y_{k-i+1},\dots ,y_k\}=y_{k-i}$ in G, where $y_{k-i}>y_k$ for $k-i\ne k$, and $max\{y_1,\dots ,y_{k-i},y_{k-i+1},\dots ,y_k-1\}=y_{k-i}$ in $G^{\prime }$. And because $u_{i+2}\notin S_{ij}$, then $u_2\notin S_{ij}^{\prime }$. According to Subcase 2.2, this implies $d_G\left(S_{ij}^{\prime }\right)-d_{G^{\prime }}\left(S_{ij}^{\prime }\right)=1$, and hence $S_{ij}^{\prime }\in {\rho }^{+}$. Define ${\rho }_i^{\prime }=\{S_{ij}^{\prime }\mid 1\le j\le s_i\}$. If $S_{ij},S_{il}\in {\rho }_i$ and $S_{ij}\ne S_{il}$, by the definition of $S_{ij}^{\prime }$ and $S_{il}^{\prime }$, then $S_{ij}^{\prime }\ne S_{il}^{\prime }$. Let $S_{ij}=\{u_2,u_3,\dots ,u_i,u_{i+1},a_1,a_2,\dots ,a_{k-i}\}\in {\rho }_i$, $S_{mj} = \{u_2,u_3,\dots ,u_i,u_{i+1},\dots ,u_m,u_{m+1},$$\dots ,b_1,b_2,\dots ,b_{k-m}\} \in {\rho }_m$, assuming $1\le i<m\le k-1$. Hence, we can get

$$S_{ij}^{\prime }=\{a_1^{\prime },a_2^{\prime },\dots ,a_{k-i}^{\prime },u_{g+2-i},u_{g+3-i},\dots ,u_g,u_1\},$$

$$S_{mj}^{\prime }=\{b_1^{\prime },b_2^{\prime },\dots ,b_{k-m}^{\prime },u_{g+2-m},u_{g+3-m},\dots ,u_{g+1-i},u_{g+2-i},u_{g+3-i},\dots ,u_g,u_1\}.$$

If $S_{ij}^{\prime }=S_{mj}^{\prime }$, it would follow that $a_{k-i}^{\prime }=u_{g+1-i}\in S_{ij}^{\prime }$, implying $x_k=y_{k-i}=1$, which is contradictory with $max\{x_1,x_2,\dots ,x_{k-1},x_k\}=x_k>x_i$ for all $1\le i\le k-1$ in G; then, $S_{ij}^{\prime }\ne S_{mj}^{\prime }$. Therefore, we know that ${\sum }_{i=1}^{k-1}|{\rho }_i| ={\sum }_{i=1}^{k-1}\left|{\rho }_i^{\prime }\right|$.

Now we consider case $i=k$. Clearly, ${\rho }_k=\left\{S_{k1}\right\}$, where $S_{k1}=\{u_2,u_3,\dots ,u_k,u_{k+1}\}$. When $g<n$, since $g>k$, then $g-k+1\ge 2$; consider the k-subset $S_{k1}^{\prime }=\{v,u_3,u_4,\dots ,$$u_k,u_1\}$, where $v\in V\left(H\right)\cap N_G\left(u_1\right)$. Obviously, $S_{k1}^{\prime }\cap X\ne \varnothing$ and $S_{k1}^{\prime }\cap Y\ne \varnothing$; then, the corresponding vector of $\{u_1,u_3,u_4,\dots ,u_k,u_1\}$ is $(2,1,1,\dots ,1,g-k+1,0)$. Because $u_2\notin S_{k1}^{\prime }$, $max\{2,1,1,\dots ,1,g-k+1,0\}=g-k+1=x_{k-1}\ge x_1$ in G, and $max\{1,1,1,\dots ,1,g-k+1,0\}=x_{k-1}$ in $G^{\prime }$, then according to Subcase 2.3, we can get $d_G\left(S_{k1}^{\prime }\right)-d_{G^{\prime }}\left(S_{k1}^{\prime }\right)=1$, so $S_{k1}^{\prime }\in {\rho }^{+}$.

When $g=n$, since $k\le n-2=g-2$, then $g-k\ge 2$; consider the situation of the parity of $g-k$. If $g-k$ is odd, there exists a k-subset $S_{k1}^{\prime }=\{u_3,u_4,\dots ,u_k,u_{k+2},u_{{\displaystyle \frac{g+k+5}{2}}\left(modg\right)}\}$. Evidently, $S_{k1}^{\prime }\subseteq Y$, and the corresponding vector of $S_{k1}^{\prime }$ is $(1,1,\dots ,1,2,{\displaystyle \frac{g-k+1}{2}},{\displaystyle \frac{g-k+1}{2}})$, where ${\displaystyle \frac{g-k+1}{2}}\ge 2$. Because $u_2\notin S_{k1}^{\prime }$, $max\{1,1,\dots ,1,$$2,{\displaystyle \frac{g-k+1}{2}},{\displaystyle \frac{g-k+1}{2}}\}={\displaystyle \frac{g-k+1}{2}}=x_k=x_{k-1}$ in G, and $max\{1,1,\dots ,1,2,$${\displaystyle \frac{g-k+1}{2}},{\displaystyle \frac{g-k+1}{2}}-1\}=x_{k-1}$ in $G^{\prime }$, then according to Subcase 2.2, we can get $d_G\left(S_{k1}^{\prime }\right)-d_{G^{\prime }}\left(S_{k1}^{\prime }\right)=1$, so $S_{k1}^{\prime }\in {\rho }^{+}$; if $g-k$ is even, there exists a k-subset $S_{k1}^{\prime }=\{u_3,u_4,\dots ,u_{k-1},u_k,u_{k+1},u_{{\displaystyle \frac{g+k+4}{2}}\left(modg\right)}\}$. Evidently, $S_{k1}^{\prime }\subseteq Y$, and the corresponding vector of $S_{k1}^{\prime }$ is $(1,1,\dots ,1,1,{\displaystyle \frac{g-k+2}{2}},{\displaystyle \frac{g-k+2}{2}})$, where ${\displaystyle \frac{g-k+2}{2}}\ge 2$. Because $u_2\notin S_{k1}^{\prime }$, $max\{1,1,\dots ,1,1,$${\displaystyle \frac{g-k+2}{2}},{\displaystyle \frac{g-k+2}{2}}\}={\displaystyle \frac{g-k+2}{2}}=x_k=x_{k-1}$ in G, and $max\{1,1,\dots ,1,1,$${\displaystyle \frac{g-k+2}{2}},{\displaystyle \frac{g-k+2}{2}}-1\}=x_{k-1}$ in $G^{\prime }$, then according to Subcase 2.2, we can get $d_G\left(S_{k1}^{\prime }\right)-d_{G^{\prime }}\left(S_{k1}^{\prime }\right)=1$, so $S_{k1}\in {\rho }^{+}$.

Thus, $|{\rho }^{-}|={\sum }_{i=1}^{k-1}|{\rho }_i|+1={\sum }_{i=1}^{k-1}|{\rho }_i^{\prime }|+1\le |{\rho }^{+}|$. Hence, we conclude that $S{W}_k\left(G\right)\ge S{W}_k\left(G^{\prime }\right)$.

(b) For any subset $S\subseteq V\left(G\right)$ with $\left|S\right|=n-1$, it is evident that $n-2\le d\left(S\right)\le n-1$. Let $\left\{v\right\}=V\left(G\right)\setminus S$. If $v\in X\cup Y$, then $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$. Similarly, if $v=u_1$, then $d_G\left(S\right)-d_{G^{\prime }}\left(S\right)=0$. Thus, we conclude that $S{W}_{n-1}\left(G\right)=S{W}_{n-1}\left(G^{\prime }\right)$. □

Lemma 3. 

Let G be a connected graph with order n, then

$$S{W}_{n-1}\left(G\right)=n(n-2)+p,$$

where p is the number of cut vertices in G.

Proof. 

For any subset $S\subseteq V\left(G\right)$ with $\left|S\right|=n-1$, it is obvious that $n-2\le d\left(S\right)\le n-1$.

Let $\left\{v\right\}=V\left(G\right)\setminus S$, $|V_c|=p$, where $p\ge 0$. If $v\in V_c$, then $d_G\left(S\right)=n-1$, and its contribution to $S{W}_{n-1}\left(G\right)$ is $p(n-1)$; if $v\notin V_c$, then $d_G\left(S\right)=n-2$, and its contribution to $S{W}_{n-1}\left(G\right)$ is $(n-p)(n-2)$. Thus, we can get $S{W}_{n-1}\left(G\right)=p(n-1)+(n-p)(n-2)=n(n-2)+p$. □

3. Minimum Values

In this section, we attain the graph with the minimum Steiner k-Wiener index in $C(n,t)$.

Proof of Theorem 1. 

According to Lemma 3, (b) is obviously true; now we prove (a). We first prove two Claims.

Claim 1. 

Let $G\in C(n,t)$ with the minimum Steiner k-Wiener index; then, $|V_c|=1$.

Assume that $|V_c|>1$. Let $u,v\in V_c$ and H be a component of G containing both u and v, such that $N_G\left(u\right)\setminus N_H\left(u\right)\ne \varnothing$ and $N_G\left(v\right)\setminus N_H\left(v\right)\ne \varnothing$. Let $N_G\left(u\right)\setminus N_H\left(u\right)=\{u_1,u_2,\dots ,u_s\}$ and $N_G\left(v\right)\setminus N_H\left(v\right)=\{v_1,v_2,\dots ,v_t\}$, where $s,t\ge 1$. Let

$$\begin{array}{cc}\hfill G_1^{\prime }& =G-\{u{u}_1,\dots ,u{u}_s\}+\{v{u}_1,\dots ,v{u}_s\},\hfill \\ \hfill G_2^{\prime }& =G-\{v{v}_1,\dots ,v{v}_t\}+\{u{v}_1,\dots ,u{v}_t\}.\hfill \end{array}$$

Apparently, $G_1^{\prime },G_2^{\prime }\in C(n,t)$, and by Lemma 1, either $S{W}_k\left(G\right)>S{W}_k\left(G_1^{\prime }\right)$ or $S{W}_k\left(G\right)>S{W}_k\left(G_2^{\prime }\right)$, a contradiction. Therefore, $|V_c|=1$.

Let u be the unique cut vertex of G. By Claim 1 and Lemma 2, we can get $S{W}_k\left(G\right)\ge S{W}_k\left(G_n(l_1,l_2,\dots ,l_t)\right)\ge S{W}_k\left(G^0(n,t)\right)$. Now we need to prove that $G^0(n,t)$ is the unique graph with the minimum Steiner k-Wiener index.

Claim 2. 

If $n\ge 6$, then $S{W}_k(G_n(4,\underset{t-1}{\underbrace{l_2,\dots ,l_t}}))>S{W}_k(G_n(3,\underset{t-1}{\underbrace{l_2,\dots ,l_t}}))$.

For convenience, let $G^{\prime }=G_n(4,\underset{t-1}{\underbrace{l_2,\dots ,l_t}})$, $G^{″}=G_n(3,\underset{t-1}{\underbrace{l_2,\dots ,l_t}})$. And let the cycle $C_4=u{w}_1{w}_2{w}_{3}u$ in $G^{\prime }$; then, $G^{″}=G^{\prime }-w_1{w}_2+u{w}_2$. Therefore, for any subset $S\subseteq V\left(G^{\prime }\right)=V\left(G^{″}\right)$ with $\left|S\right|=k$, there are the following cases: (1) $w_2\notin S$; (2) $w_1\in S$, $w_2\in S$,$w_3\in S$; (3) $w_1\notin S$, $w_2\in S$, $w_3\in S$; (4) $w_1\in S$, $w_2\in S$, $w_3\notin S$; (5) $w_1\notin S$, $w_2\in S$, $w_3\notin S$.

For $k=3$, in case (2), $d_{G^{\prime }}\left(S\right)-d_{G^{″}}\left(S\right)=-1$; in case (5), $d_{G^{\prime }}\left(S\right)-d_{G^{″}}\left(S\right)=1$; for all other cases, $d_{G^{\prime }}\left(S\right)-d_{G^{″}}\left(S\right)=0$. Therefore, it follows that

$$\begin{array}{cc}\hfill S{W}_3\left(G^{\prime }\right)-S{W}_3\left(G^{″}\right)=& \sum _{S\subseteq V\left(G^{\prime }\right)}{d}_{G^{\prime }}\left(S\right)-\sum _{S\subseteq V\left(G^{\prime \prime }\right)}{d}_{G^{″}}\left(S\right)\hfill \\ \hfill =& \left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)-1.\hfill \end{array}$$

Since $n\ge 6$, then $S{W}_3\left(G^{\prime }\right)-S{W}_3\left(G^{″}\right)>0$.

For $4\le k\le n-2$, in case (5), $d_{G^{\prime }}\left(S\right)-d_{G^{″}}\left(S\right)=1$; for all other cases, $d_{G^{\prime }}\left(S\right)-d_{G^{″}}\left(S\right)=0$. Therefore, it follows that

$$\begin{array}{cc}\hfill S{W}_k\left(G^{\prime }\right)-S{W}_k\left(G^{″}\right)=& \sum _{S\subseteq V\left(G^{\prime }\right)}{d}_{G^{\prime }}\left(S\right)-\sum _{S\subseteq V\left(G^{\prime \prime }\right)}{d}_{G^{″}}\left(S\right)\hfill \\ \hfill =& \left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right).\hfill \end{array}$$

Since $k-1\le n-3$, then $S{W}_k\left(G^{\prime }\right)-S{W}_k\left(G^{″}\right)>0$.

By Claim 2, we know that $G^0(n,t)$ is the unique graph with the minimum Steiner k-Wiener index. Now we compute the Steiner k-Wiener index of $G^0(n,t)$. For any subset $S\subseteq V\left(G^0(n,t)\right)$ with $\left|S\right|=k$, if $u\in S$, then $d_{G^0(n,t)}\left(S\right)=k-1$, and there are $\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)$ such subsets, contributing $\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)(k-1)$ to $S{W}_k\left(G^0(n,t)\right)$. If $u\notin S$, then $d_{G^0(n,t)}\left(S\right)=k$, and there are $\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k}}\right)$ such subsets, contributing $\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k}}\right)k$ to $S{W}_k\left(G^0(n,t)\right)$. Thus, we obtain

$$S{W}_k\left(G^0(n,t)\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)(k-1)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k}}\right)k=\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)(n-1).$$

□

4. Second Minimum Values

In this section, we determine the graph with the smallest Steiner k-Wiener index in $G\in C(n,t)\setminus \left\{G^0(n,t)\right\}$. If $G\in C(n,t)\setminus \left\{G^0(n,t)\right\}$, there are three possibilities:

(1) G contains a cycle that is not a triangle.

(2) G contains a cut edge that is not a pendant edge.

(3) G contains a cycle that is not an end-block.

Lemma 4. 

Let $G\in C(n,t)\setminus \left\{G^0(n,t)\right\}$, where $n\ge 6$. If G contains a cycle that is not triangle, then

$$S{W}_k(G)\ge \left\{\begin{array}{cc}\left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{2}}\right)(n-1)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)-1,\hfill & k=3,\hfill \\ \left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)(n-1)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right),\hfill & 4\le k\le n-2,\hfill \end{array}\right.$$

with equality if and only if $G\cong G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}})$.

Proof. 

We first prove that for $n\ge 6$, then $S{W}_k(G_n(5,\underset{t-1}{\underbrace{l_2,\dots ,l_t}}))>S{W}_k(G_n(4,\underset{t-1}{\underbrace{l_2,\dots ,l_t}}))$.

Let $G^{\prime }=G_n(5,\underset{t-1}{\underbrace{l_2,\dots ,l_t}})$, $G^{″}=G_n(4,\underset{t-1}{\underbrace{l_2,\dots ,l_t}})$, and let $C_5=u{w}_1{w}_2{w}_3{w}_{4}u$ in $G^{\prime }$; then, $G^{″}=G^{\prime }-w_1{w}_2+u{w}_2$. For any subset $S\subseteq V(G^{\prime })=V(G^{″})$ with $\left|S\right|=k$, the following cases are considered: (1) $w_2\notin S$; (2) $w_2\in S$, $w_1\notin S$, $w_3\notin S$, $w_4\notin S$; (3) $w_2\in S$, $w_1\in S$, $w_3\notin S$, $w_4\notin S$; (4) $w_2\in S$, $w_1\notin S$, $w_3\in S$, $w_4\notin S$; (5) $w_2\in S$, $w_1\notin S$, $w_3\notin S$, $w_4\in S$; (6) $w_2\in S$, $w_1\notin S$, $w_3\in S$, $w_4\in S$; (7) $w_2\in S$, $w_1\in S$, $w_3\notin S$, $w_4\in S$; (8) $w_2\in S$, $w_1\in S$, $w_3\in S$, $w_4\notin S$; (9) $w_2\in S$, $w_1\in S$, $w_3\in S$, $w_4\in S$.

For $k=3$, it is clear that case (9) does not exist. In cases (2), (4), and (5), $d_{G^{\prime }}(S)-d_{G^{″}}(S)=1$; in case (8), $d_{G^{\prime }}(S)-d_{G^{″}}(S)=-1$; in all other cases, $d_{G^{\prime }}(S)-d_{G^{″}}(S)=0$. Since case (8) contains only one subset, it follows that $S{W}_3(G^{\prime })>S{W}_3(G^{″})$.

For $k=4$, in cases (2), (4), and (5), $d_{G^{\prime }}(S)-d_{G^{″}}(S)=1$; in case (9), $d_{G^{\prime }}(S)-d_{G^{″}}(S)=-1$; in all other cases, $d_{G^{\prime }}(S)-d_{G^{″}}(S)=0$. Since case (9) contains only one subset, it follows that $S{W}_4(G^{\prime })>S{W}_4(G^{″})$.

For $5\le k\le n-2$, in cases (2), (4), and (5), $d_{G^{\prime }}(S)-d_{G^{″}}(S)=1$; in all other cases, $d_{G^{\prime }}(S)-d_{G^{″}}(S)=0$. It follows that $S{W}_k(G^{\prime })>S{W}_k(G^{″})$. According to the above cases, we claim that $S{W}_k(G^{\prime })>S{W}_k(G^{″})$ for $3\le k\le n-2$.

If G contains no cycle of length 4, then by Lemmas 1 and 2, we know that $S{W}_k(G)\ge S{W}_k(G_n(l_1,l_2,\dots ,l_t))\ge S{W}_k(G_n(5,\underset{t-1}{\underbrace{3,\dots ,3}}))>S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))$.

If G contains a cycle of length 4, then by Lemmas 1, 2 and Claim 2 in Theorem 1, we know that $S{W}_k(G)\ge S{W}_k(G_n(l_1,l_2,\dots ,l_t))\ge S{W}_k(\begin{array}{c}\hfill G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}})\end{array})$, with equality if and only if $G\cong G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}})$.

Now we compute the Steiner k-Wiener index of $G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}})$. Based on Claim 2 in Theorem 1, we know that

$$S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))-S{W}_k(G^0(n,t))=\left\{\begin{array}{cc}\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)-1,\hfill & k=3,\hfill \\ \left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right),\hfill & 4\le k\le n-2.\hfill \end{array}\right.$$

□

Lemma 5. 

Let $G\in C(n,t)\setminus \left\{G^0(n,t)\right\}$, where $n\ge 6$. If G contains a cut edge that is not a pendant edge, and $3\le k\le n-2$, then

$$S{W}_k(G)\ge \left({\displaystyle \genfrac{}{}{0pt}{}{n-1}{k-1}}\right)(n-1)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k-1}}\right),$$

with equality if and only if $G\cong G_2$ (see Figure 7).

Figure 7. $G^{\prime }$, $G_2$, and $G_3$ used in Lemma 5.

Proof. 

Let $G^{\prime }$ be the graph in which all cut edges except for $e=uv$ are pendant edges, and all cycles are end-blocks and triangles (see Figure 7). By Lemmas 1 and 2, it is clear that $S{W}_k(G)\ge S{W}_k(G^{\prime })$. Suppose that vertex u is attached with $t_1$ triangles and $s_1$ pendant edges, and vertex v is attached with $t_2$ triangles and $s_2$ pendant edges in $G^{\prime }$, where $t_1+t_2=t$ and $s_1+s_2=n-2t-2$.

Let $G_2$ be the graph obtained by moving the $t_2$ triangles and $s_2-1$ pendant edges from vertex v to vertex u in graph $G^{\prime }$, and $G_3$ be the graph obtained by moving the $t_2-1$ triangles and $s_2$ pendant edges from vertex v to vertex u in graph $G^{\prime }$ (see Figure 7). According to Lemma 1, it is evident that $S{W}_k(G^{\prime })\ge min\{S{W}_k(G_2),S{W}_k(G_3)\}$, with equality if and only if $G^{\prime }\cong G_2$ or $G^{\prime }\cong G_3$. Next, we calculate the Steiner k-Wiener indices for $G_2$ and $G_3$.

For any set $S\subseteq V(G^0(n,t))=V(G_2)$ with $\left|S\right|=k$, several cases are as follows: (1) $w_1\notin S$; (2) $v\in S$, $w_1\in S$; (3) $v\notin S$, $w_1\in S$. In case (3), $d_{G_2}(S)-d_{G^0(n,t)}(S)=1$; for other cases, $d_{G_2}(S)-d_{G^0(n,t)}(S)=0$. Therefore, we can derive

$$\begin{array}{cc}\hfill S{W}_k(G_2)-S{W}_k(G^0(n,t))& =\sum _{S\subseteq V(G_2)}{d}_{G_2}(S)-\sum _{S\subseteq V(G^0(n,t))}{d}_{G^0(n,t)}(S)\hfill \\ & =\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k-1}}\right).\hfill \end{array}$$

Thus, $S{W}_k(G_2)=S{W}_k(G^0(n,t))+\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k-1}}\right)$.

For any set $S\subseteq V(G^0(n,t))=V(G_3)$ with $\left|S\right|=k$, there are the following cases: (1) $w_2\notin S$, $w_3\notin S$; (2) $w_2\in S$, $w_3\notin S$, $v\notin S$; (3) $w_2\in S$, $w_3\notin S$, $v\in S$; (4) $w_2\notin S$, $w_3\in S$, $v\notin S$; (5) $w_2\notin S$, $w_3\in S$, $v\in S$; (6) $w_2\in S$, $w_3\in S$, $v\notin S$; (7) $w_2\in S$, $w_3\in S$, $v\in S$.

For $k=3$, in cases (2), (4), and (6), $d_{G_3}(S)-d_{G^0(n,t)}(S)=1$; in case (7), $d_{G_3}(S)-d_{G^0(n,t)}(S)=-1$; for other cases, $d_{G_3}(S)-d_{G^0(n,t)}(S)=0$. Therefore, we can derive

$$\begin{array}{cc}\hfill S{W}_3(G_3)-S{W}_3(G^0(n,t))=& \sum _{S\subseteq V(G_3)}{d}_{G_3}(S)-\sum _{S\subseteq V(G^0(n,t))}{d}_{G^0(n,t)}(S)\hfill \\ \hfill =& 2\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{1}}\right)-1\hfill \\ \hfill =& \left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)-1.\hfill \end{array}$$

For $4\le k\le n-2$, in cases (2), (4), and (6), $d_{G_3}(S)-d_{G^0(n,t)}(S)=1$; for other cases, $d_{G_3}(S)-d_{G^0(n,t)}(S)=0$. Therefore, we can derive

$$\begin{array}{cc}\hfill S{W}_k(G_3)-S{W}_k(G^0(n,t))=& \sum _{S\subseteq V(G_3)}\begin{array}{c}\hfill d_{G_3}(S)\end{array}-\sum _{S\subseteq V(G^0(n,t))}{d}_{G^0(n,t)}(S)\hfill \\ \hfill =& \left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right)\hfill \\ \hfill =& \left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right).\hfill \end{array}$$

Thus, we can get

$$S{W}_k(G_3)-S{W}_k(G^0(n,t))=\left\{\begin{array}{cc}\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)-1,\hfill & k=3,\hfill \\ \left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right),\hfill & 4\le k\le n-2.\hfill \end{array}\right.$$

If $k=3$, since $n\ge 6$, then $S{W}_3(G_3)-S{W}_3(G_2)=\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)-1>0$; if $4\le k\le n-2$, since $k-1\le n-3$, then $S{W}_k(G_3)-S{W}_k(G_2)=\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right)>0$. □

Lemma 6. 

Let $G\in C(n,t)\setminus \left\{G^0(n,t)\right\}$, where $n\ge 6$. If G contains a cycle that is not an end-block, and $3\le k\le n-2$, then

$$S{W}_k(G)>S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}})).$$

Proof. 

If G contains a cycle that is not a triangle, since G has a cycle that is not an end-block, it follows from Lemma 4 that $S{W}_k(G)>S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))$.

If all cycles in G are triangles, by Lemmas 1 and 2, we know that $S{W}_k(G)\ge S{W}_k(G^{″})$. Here, $G^{″}$ has exactly one triangle that is not an end-block, and the remaining cycles are triangles and end-blocks. Moreover, all cut edges in $G^{″}$ are pendant edges (see Figure 8). Let $C_3=u_1{u}_2{u}_3{u}_1$ be the non-end-block triangle in $G^{″}$. Suppose that $u_i$ is attached with $t_i$ triangles and $s_i$ pendant edges, where $1\le i\le 3$. Clearly, ${\sum }_{i=1}^3{t}_i=t$ and ${\sum }_{i=1}^3{s}_i=n-2t-3$. By moving the $t_3$ triangles and $s_3$ pendant edges from vertex $u_3$ to vertex $u_1$ in $G^{″}$, we obtain the graph $G^{‴}$ (see Figure 8). According to Lemma 1, we have $S{W}_k(G^{″})\ge S{W}_k(G^{‴})$, with equality if and only if $d_{G^{″}}(u_3)=2$.

Figure 8. $G^{″}$ and $G^{‴}$ used in Lemma 6.

Now, let $G_4$ be the graph obtained by moving the $t_2$ triangles and $s_2-1$ pendant edges from $u_2$ to $u_1$ in $G^{‴}$, and $G_5$ be the graph obtained by moving the $t_2-1$ triangles and $s_2$ pendant edges from $u_2$ to $u_1$ in $G^{‴}$ (see Figure 9). By Lemma 1, we can conclude $S{W}_k(G^{‴})\ge min\{S{W}_k(G_4),S{W}_k(G_5)\}$. Next we calculate the Steiner k-Wiener indices of $G_4$ and $G_5$, respectively.

Figure 9. $G_4$ and $G_5$ used in Lemma 6.

For any set $S\subseteq V(G^0(n,t))=V(G_4)$ with $\left|S\right|=k$, there are the following cases: (1) $w_1\notin S$; (2) $w_1\in S$, $u_2\notin S$, $u_3\notin S$; (3) $w_1\in S$, $u_2\notin S$, $u_3\in S$; (4) $w_1\in S$, $u_2\in S$, $u_3\notin S$; (5) $w_1\in S$, $u_2\in S$, $u_3\in S$.

For $k=3$, in cases (2) and (3), $d_{G_4}(S)-d_{G^0(n,t)}(S)=1$; in case (5), $d_{G_4}(S)-d_{G^0(n,t)}(S)$$=-1$; for other cases, $d_{G_4}(S)-d_{G^0(n,t)}(S)=0$. Accordingly, we can get

$$\begin{array}{cc}\hfill S{W}_3(G_4)-S{W}_3(G^0(n,t))=& \sum _{S\subseteq V(G_4)}{d}_{G_4}(S)-\sum _{S\subseteq V(G^0(n,t))}{d}_{G^0(n,t)}(S)\hfill \\ \hfill =& \left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{1}}\right)-1.\hfill \end{array}$$

For $4\le k\le n-2$, in cases (2) and (3), $d_{G_4}(S)-d_{G^0(n,t)}(S)=1$; for other cases, $d_{G_4}(S)$$-d_{G^0(n,t)}(S)=0$. Accordingly, we can get

$$\begin{array}{cc}\hfill S{W}_k(G_4)-S{W}_k(G^0(n,t))=& \sum _{S\subseteq V(G_4)}{d}_{G_4}(S)-\sum _{S\subseteq V(G^0(n,t))}{d}_{G^0(n,t)}(S)\hfill \\ \hfill =& \left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right).\hfill \end{array}$$

Thus, we know that

$$S{W}_k(G_4)-S{W}_k(G^0(n,t))=\left\{\begin{array}{cc}\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{1}}\right)-1,\hfill & k=3,\hfill \\ \left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right),\hfill & 4\le k\le n-2.\hfill \end{array}\right.$$

If $k=3$, because $n\ge 6$, then $S{W}_3(G_4)-S{W}_3(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))=n-3>0$; if $4\le k\le n-2$, because $k-2\le n-4$, then $S{W}_k(G_4)-S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))=\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-2}}\right)>0$.

For any set $S\subseteq V(G^0(n,t))=V(G_5)$ with $\left|S\right|=k$, there are the following cases: (1) $w_2\notin S$, $w_3\notin S$; (2) $w_2\in S$, $w_3\notin S$, $u_2\notin S$, $u_3\notin S$; (3) $w_2\in S$, $w_3\notin S$, $u_2\notin S$, $u_3\in S$; (4) $w_2\in S$, $w_3\notin S$, $u_2\in S$, $u_3\notin S$; (5) $w_2\in S$, $w_3\notin S$, $u_2\in S$, $u_3\in S$; (6) $w_2\notin S$, $w_3\in S$, $u_2\notin S$, $u_3\notin S$; (7) $w_2\notin S$, $w_3\in S$, $u_2\notin S$, $u_3\in S$; (8) $w_2\notin S$, $w_3\in S$, $u_2\in S$, $u_3\notin S$; (9) $w_2\notin S$, $w_3\in S$, $u_2\in S$, $u_3\in S$; (10) $w_2\in S$, $w_3\in S$, $u_2\notin S$, $u_3\notin S$; (11) $w_2\in S$, $w_3\in S$, $u_2\notin S$, $u_3\in S$; (12) $w_2\in S$, $w_3\in S$, $u_2\in S$, $u_3\notin S$; (13) $w_2\in S$, $w_3\in S$, $u_2\in S$, $u_3\in S$.

For $k=3$, it is clear that case (13) does not exist. In cases (2), (3), (6), (7), and (10), $d_{G_5}(S)-d_{G^0(n,t)}(S)=1$; in cases (5), (9), and (12), $d_{G_5}(S)-d_{G^0(n,t)}(S)=-1$; for other cases, $d_{G_5}(S)-d_{G^0(n,t)}(S)=0$. Accordingly, we can get

$$\begin{array}{cc}\hfill S{W}_3(G_5)-S{W}_3(G^0(n,t))=& \sum _{S\subseteq V(G_5)}{d}_{G_5}(S)-\sum _{S\subseteq V(G^0(n,t))}{d}_{G^0(n,t)}(S)\hfill \\ \hfill =& 2\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{2}}\right)+2\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{1}}\right)-3\hfill \\ \hfill =& 2\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)+n-7.\hfill \end{array}$$

For $k=4$, in cases (2), (3), (6), (7), (10), and (11), $d_{G_5}(S)-d_{G^0(n,t)}(S)=1$; in case (13), $d_{G_5}(S)-d_{G^0(n,t)}(S)=-1$; for other cases, $d_{G_5}(S)-d_{G^0(n,t)}(S)=0$. Accordingly, we can get

$$\begin{array}{cc}\hfill S{W}_4(G_5)-S{W}_4(G^0(n,t))=& \sum _{S\subseteq V(G_5)}{d}_{G_5}(S)-\sum _{S\subseteq V(G^0(n,t))}{d}_{G^0(n,t)}(S)\hfill \\ \hfill =& 2\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{3}}\right)+2\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{1}}\right)-1\hfill \\ \hfill =& 2\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{3}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)-1\hfill \\ \hfill =& \left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{3}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{3}}\right)-1.\hfill \end{array}$$

For $5\le k\le n-2$, in cases (2), (3), (6), (7), (10), and (11), $d_{G_5}(S)-d_{G^0(n,t)}(S)=1$; for other cases, $d_{G_5}(S)-d_{G^0(n,t)}(S)=0$. Accordingly, we can get

$$\begin{array}{cc}\hfill S{W}_k(G_5)-S{W}_k(G^0(n,t))=& \sum _{S\subseteq V(G_5)}{d}_{G_5}(S)-\sum _{S\subseteq V(G^0(n,t))}{d}_{G^0(n,t)}(S)\hfill \\ \hfill =& 2\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{k-1}}\right)+2\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{k-2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{k-2}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-4}{k-3}}\right)\hfill \\ \hfill =& 2\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-2}}\right)\hfill \\ \hfill =& \left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right).\hfill \end{array}$$

Thus, we know that

$$S{W}_k(G_5)-S{W}_k(G^0(n,t))=\left\{\begin{array}{cc}2\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)+n-7,\hfill & k=3,\hfill \\ \left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{3}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{3}}\right)-1,\hfill & k=4,\hfill \\ \left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k-1}}\right)+\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-1}}\right),\hfill & 5\le k\le n-2.\hfill \end{array}\right.$$

If $k=3$, since $n\ge 6$, then $S{W}_3(G_5)-S{W}_3(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))=\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{2}}\right)+n-6>0$; if $k=4$, then $S{W}_4(G_5)-S{W}_4(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))=\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{3}}\right)-1>0$; if $5\le k\le n-2$, since $k-1\le n-3$, then $S{W}_k(G_5)-S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))=\left({\displaystyle \genfrac{}{}{0pt}{}{n-2}{k-1}}\right)>0$. In summary, it can be concluded that $S{W}_k(G)>S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}})).$ □

Proof of Theorem 2. 

By Lemma 3, (b) is straightforward. We now turn to the proof of (a).

When $t=0$, since the cases described in Lemma 4 and Lemma 6 do not arise, the conclusion follows directly from Lemma 5; when $t\ge 1$, if $k = 3$, since $n\ge 6$, we have $S{W}_3(G_2)-S{W}_3(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))=n-2>0$. If $4\le k\le n-2$, since $k-2\le n-4$, we have $S{W}_k(G_2)-S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))=\left({\displaystyle \genfrac{}{}{0pt}{}{n-3}{k-2}}\right)>0$. Thus, $S{W}_k(G_2)>S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))$. By Lemmas 4 to 6, we derive $S{W}_k(G)\ge S{W}_k(G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}}))$, with equality if and only if $G\cong G_n(4,\underset{t-1}{\underbrace{3,\dots ,3}})$. □

5. Conclusions

In this paper, we first determine the minimum Steiner k-Wiener index of cacti and the corresponding extremal graphs by employing edge-contraction and cycle-contraction transformations. Next, we divide cacti except the extremal graph attaining the minimum value into three types and derive the minimum Steiner k-Wiener index for each type; by comparing these minimum values, we establish the second-minimum Steiner k-Wiener index of cacti together and the corresponding extremal graphs. Based on the cacti, we can consider studying the largest Steiner k-Wiener index among all cacti with order n and t cycles.