k-Path-Connectivity of Completely Balanced Tripartite Graphs

Abstract

For a graph $G=(V,E)$ and a set $S\subseteq V\left(G\right)$ of a size at least 2, a path in G is said to be an S-path if it connects all vertices of S. Two S-paths $P_1$ and $P_2$ are said to be internally disjoint if $E\left(P_1\right)\cap E\left(P_2\right)=\varnothing$ and $V\left(P_1\right)\cap V\left(P_2\right)=S$; that is, they share no vertices and edges apart from S. Let ${\pi }_G\left(S\right)$ denote the maximum number of internally disjoint S-paths in G. The k-path-connectivity ${\pi }_k\left(G\right)$ of G is then defined as the minimum ${\pi }_G\left(S\right)$, where S ranges over all k-subsets of $V\left(G\right)$. In this paper, we study the k-path-connectivity of the complete balanced tripartite graph $K_{n,n,n}$ and obtain ${\pi }_k\left(K_{n,n,n}\right)=⌊\frac{2n}{k-1}⌋$ for $3\le k\le n.$

1. Introduction

An interconnection network is usually modeled by a connected graph $G=(V,E)$, where vertices represent processors and edges represent communication links between processors. Connectivity is an important parameter to evaluate the reliability and fault tolerance of a network. For a graph G, the $connectivity$  $\kappa \left(G\right)$ is defined as the minimum cardinality of a subset $V^{\prime }$ of vertices of G such that $G-V^{\prime }$ is disconnected or trivial. An equivalent definition of connectivity was given in [1]. For each 2-subset $S=\{u,v\}$ of vertices of G, let $\kappa \left(S\right)$ denote the maximum number of internally disjoint $(u,v)$-paths in G. Then, $\kappa \left(G\right)=$min$\left\{\kappa \right(S\left)\right|S\subseteq V\mathrm{and}\left|S\right|=2\}$.

There exist many generalizations of the classical connectivity, such as conditional connectivity [2], component connectivity [3], tree-connectivity [4,5] and rainbow connectivity [6]. In particular, Hager [7] introduced the concept of path-connectivity, which concerns paths connecting any k vertices in G and not only any two. Given a graph $G=(V,E)$ and a set $S\subseteq V\left(G\right)$ of a size at least 2, a path in G is said to be an S-path if it connects all vertices of S. Two S paths $P_1$ and $P_2$ are said to be internally disjoint if $E\left(P_1\right)\cap E\left(P_2\right)=\varnothing$ and $V\left(P_1\right)\cap V\left(P_2\right)=S$; that is, they share no vertices and edges apart from S. Let ${\pi }_G\left(S\right)$ denote the maximum number of internally disjoint S-paths in G. The k-path connectivity of G, denoted by ${\pi }_k\left(G\right)$, is then defined as ${\pi }_k\left(G\right)=$min$\left\{{\pi }_G\left(S\right)\right|S\subseteq V\left(G\right)and\left|S\right|=k\}$, where $2\le k\le n$. Clearly, ${\pi }_2\left(G\right)$ is exactly the classical connectivity $\kappa \left(G\right)$, and ${\pi }_n\left(G\right)$ is exactly the maximum number of edge-disjoint Hamiltonian paths in G.

In [7], Hager studied the sufficient conditions for ${\pi }_k\left(G\right)$ to be at least ℓ in terms of $\kappa \left(G\right)$. Hager conjectured that if G is a graph with $\kappa \left(G\right)\ge \ell (k-1)$ for $k\ge 2$ and $\ell \ge 1$, then ${\pi }_k\left(G\right)\ge \ell$; moreover, the bound is sharp. He confirmed the conjecture for $2\le k\le 4$. Recently, Li et al. [8] showed that this conjecture also is true for $k=5$. Moreover, they studied the complexity of the path-connectivity. With their conclusions, it is difficult to obtain ${\pi }_k\left(G\right)$ for general G and $k\ge 5$. In [9,10], the path connectivity of lexicographic product graphs was investigated. For special classes of graphs, the exact values of ${\pi }_k\left(G\right)$ were obtained for complete graphs [7] and complete bipartite graphs [7,11].

A complete multipartite graph is $balanced$ if the partite sets all have the same cardinality. In this paper, we study the k-path-connectivity of the complete balanced tripartite graph $K_{n,n,n}$ and obtain ${\pi }_k\left(K_{n,n,n}\right)=⌊\frac{2n}{k-1}⌋$, for $3\le k\le n.$ Moreover, our result implies that Hager’s conjecture is true for $K_{n,n,n}$ and $3\le k\le n$.

2. Main Result

We first introduce some notations and terminology that will be used throughout the paper.

The subgraph of G induced by a vertex set $U\subseteq V\left(G\right)$ is denoted by $G\left[U\right]$. A subset S of V is called an independent set of G if no two vertices of S are adjacent in G. For any two vertices $x,y\in V\left(G\right)$, an $xy$-path is a path starting at x and ending at y. For convenience, let $\left[x_1,x_n\right]=\{x_1,\dots ,x_n\}$. We refer the reader to [12] for the notations and terminology not defined in this paper.

Now we provide our main result.

Theorem 1.

Given any positive integer $n\ge 2$, let $K_{n,n,n}$ denote a complete balanced tripartite graph in which each partite set contains exactly n vertices. Then, we have the following.

$${\pi }_k\left(K_{n,n,n}\right)=⌊\frac{2n}{k-1}⌋,\mathit{for}3\le k\le n.$$

Proof. 

Suppose that X, Y, and Z are the three parts of $K_{n,n,n}$, where $X=[x_1,x_n]$, $Y=[y_1,y_n]$, and $Z=[z_1,z_n]$. Let $G=K_{n,n,n}$ and S be any subset of $V\left(G\right)$ of cardinality k. By the symmetry of $K_{n,n,n}$, we can assume that $S\cap X=A=$ $[x_1,x_a]$, $S\cap Y=B=$ $[y_1,y_b]$, $S\cap X=C=$ $[z_1,z_c]$. Obviously, $a+b+c=k$.

Remember that, when we construct internally disjoint S-paths, each vertex in $V\left(G\right)\hspace{0.17em}\backslash \hspace{0.17em}S$ can appear on one S-path at most. We distinguish three cases as follows.

Case 1: $a=b=0$ and $c=k$.

In this case, $S\subseteq Z$. Therefore, each vertex in S is adjacent to all the vertices in $X\cup Y$, which means that we can use any $k-1$ vertices of $X\cup Y$ to connect all vertices in S into an S-path. On the other hand, since S is an independent set, each S-path needs at least $k-1$ vertices of $X\cup Y$. Thus, $\pi \left(S\right)=⌊\frac{|X\cup Y|}{k-1}⌋=⌊\frac{2n}{k-1}⌋$.

Case 2: $1\le a\le b\le c$.

Note that $3\le k=a+b+c\le n$. We will show $\pi \left(S\right)\ge ⌊\frac{2n}{k-1}⌋$ in this case by constructing $⌊\frac{2(n-c-1)}{k-1}⌋+2$ internally disjoint S-paths and prove that $⌊\frac{2(n-c-1)}{k-1}⌋+2\ge ⌊\frac{2n}{k-1}⌋.$ We divide the construction process into four steps. In Steps 1 and 2, we will construct two S-paths mainly by using some edges in $G\left[S\right]$ and some vertices in $Z\hspace{0.17em}\backslash \hspace{0.17em}C$. In Steps 3 and 4, we will use $n-c-1$ vertices from $X\hspace{0.17em}\backslash \hspace{0.17em}A$ and $n-c-1$ vertices from $Y\hspace{0.17em}\backslash \hspace{0.17em}B$ to construct $⌊\frac{2(n-c-1)}{k-1}⌋$ internally disjoint S-paths. On these S-paths, any two vertices of S are connected by the vertices from $X\hspace{0.17em}\backslash \hspace{0.17em}A$ and $Y\hspace{0.17em}\backslash \hspace{0.17em}B$.

Step 1: Construct the first S-path $P_1$.

Firstly, by using vertices $z_{c+1},\dots ,z_{c+b-1}$ in $Z\hspace{0.17em}\backslash \hspace{0.17em}C$, we can connect all vertices $y_1,\dots ,y_b$ of B into a path, denoted by $P_{11}$, i.e., $P_{11}=y_1{z}_{c+1}{y}_2{z}_{c+2}\dots y_{b-1}{z}_{c+b-1}{y}_b$.

Since $c\ge a$, $A\subseteq \left\{x_1,\dots ,x_c\right\}$. Note that $G\left[\left\{x_1,\dots ,x_c\right\}\cup C\right]=K_{c,c}$. Thus, there must exist a path, denoted by $P_{12}$, connecting all the vertices of $A\cup C$ in $G\left[\left\{x_1,\dots ,x_c\right\}\cup C\right]$. More specifically, let $P_{12}=z_1{x}_1{z}_2\dots x_{c-1}{z}_c{x}_c$.

Finally, using the vertex $x_{c+1}$ to connect $y_b$ and $z_1$, we obtain the first S-path $P_1$, i.e., $P_1=P_{11}\cup \left\{y_b{x}_{c+1}{z}_1\right\}\cup P_{12}$.

Step 2: Construct the second S-path $P_2$.

Firstly, by using the vertices $z_{c+b},\dots ,z_{c+b+a-2}$ in $Z\hspace{0.17em}\backslash \hspace{0.17em}C$, we can connect all the vertices $x_1,\dots ,x_a$ of A into a path, denoted by $P_{21}$, i.e., $P_{21}=x_1{z}_{b+c}\dots x_{a-1}{z}_{a+b+c-2}{x}_a$.

Since $c\ge b$, $B\subseteq \left\{y_1,\dots ,y_c\right\}$. Similarly, in $G\left[\left\{y_1,\dots ,y_c\right\}\cup C\right]$ there must exist a path, denoted by $P_{22}$, connecting all the vertices of $B\cup C$. More specifically, let $P_{22}=z_1{y}_1{z}_2\dots y_{c-1}{z}_c{y}_c$.

Finally, using the vertex $y_{c+1}$ to connect $x_a$ and $z_1$, we obtain the second S-path $P_2$, i.e., $P_2=P_{21}\cup \left\{x_a{y}_{c+1}{z}_1\right\}\cup P_{22}$.

Remark. 

After the first two steps, we have found two S-paths, which are obviously internally disjoint. Moreover, there are $n-c-1$ unused vertices in $X\hspace{0.17em}\backslash \hspace{0.17em}A$ (namely, $x_{c+2},\dots ,x_n$), $n-c-1$ unused vertices in $Y\hspace{0.17em}\backslash \hspace{0.17em}B$ (namely, $y_{c+2},\dots ,y_n$) and $n-k+2$ unused vertices in $Z\hspace{0.17em}\backslash \hspace{0.17em}C$. Set $A^{\prime }=\left[x_{c+2},x_n\right]$ and $B^{\prime }=\left[y_{c+2},y_n\right]$.

Step 3: Construct the next $2l$S-paths, where $l=⌊\frac{n-c-1}{k-1}⌋$.

Note that, if $l=0$, proceed directly to Step 4. Thus, we assume that $l\ge 1$. We now provide a method to construct S-paths in pairs. The outline of the method is as follows.

Firstly, we take $⌈\frac{k-1}{2}⌉$ unused vertices from $A^{{}^{\prime }}$ and $⌊\frac{k-1}{2}⌋$ unused vertices from $B^{{}^{\prime }}$. Then, using the $k-1$ vertices in total, connect all the vertices of S into an S-path. Next, we take $⌊\frac{k-1}{2}⌋$ unused vertices from $A^{{}^{\prime }}$ and $⌈\frac{k-1}{2}⌉$ unused vertices from $B^{{}^{\prime }}$. Using the $k-1$ vertices in total, construct another S-path. Thus, by $⌈\frac{k-1}{2}⌉+⌊\frac{k-1}{2}⌋=k-1$ vertices in $A^{{}^{\prime }}$ and $⌊\frac{k-1}{2}⌋+⌈\frac{k-1}{2}⌉=k-1$ vertices in $B^{{}^{\prime }}$, we can obtain a pair of S-paths. By repeating this process, we can construct $l=⌊\frac{n-c-1}{k-1}⌋$ pairs of S-paths in this step.

Now, we construct the S-paths $P_3$ and $P_4$ to illustrate the specific method. Note that, since $a\ge 1$, $⌊\frac{k-1}{2}⌋=⌊\frac{a+b+c-1}{2}⌋\ge ⌊\frac{b+c}{2}⌋\ge b\ge a$.

The construction of $P_3$.

Firstly, by using $b-1$ vertices $x_{c+2},\dots ,x_{c+b}$ in $A^{\prime }$, connect all vertices $y_1,\dots ,y_b$ of B into a path, denoted by $P_{31}$, i.e., $P_{31}=y_1{x}_{c+2}{y}_2{x}_{c+3}\dots x_{c+b}{y}_b$.

Similarly, by using $a-1$ vertices $y_{c+2},\dots ,y_{c+a}$ in $B^{\prime }$, connect all vertices $x_1,\dots ,x_a$ of A into a path, denoted by $P_{32}$, i.e., $P_{32}=x_1{y}_{c+2}{x}_2{y}_{c+3}\dots y_{c+a}{x}_a$.

Then, join the vertices $y_b$ and $z_1$ by vertex $x_{c+b+1}$. Moreover, join vertices $x_1$ and $z_c$ by vertex $y_{c+a+1}$.

Next, we take $⌈\frac{k-1}{2}⌉-b$ unused vertices $\left[x_{c+b+2},x_{c+1+⌈\frac{k-1}{2}⌉}\right]$ from $A^{{}^{\prime }}$ and take $⌊\frac{k-1}{2}⌋-a$ unused vertices $\left[y_{c+a+2},y_{c+1+⌊\frac{k-1}{2}⌋}\right]$ from $B^{{}^{\prime }}$. Since each vertex in $A^{{}^{\prime }}\bigcup B^{{}^{\prime }}$ is adjacent to all the vertices in C, using the $k-1-a-b=c-1$ vertices in total, we can connect all the vertices of C into a $z_1{z}_c$-path $P_{33}$.

Now, we obtain the third S-path $P_3=P_{31}\cup \left\{y_b{x}_{c+b+1}{z}_1\right\}\cup P_{33}\cup \left\{z_c{y}_{c+a+1}{x}_1\right\}\cup P_{32}$.

The construction of $P_4$ is similar. The only difference is that the subpath $P_{43}$ is constructed by $⌊\frac{k-1}{2}⌋-b$ unused vertices in $A^{{}^{\prime }}$ and $\lceil \frac{k-1}{2}\rceil -a$ unused vertices in $B^{{}^{\prime }}$. It follows that the fourth S-path $P_4$ uses $⌊\frac{k-1}{2}⌋$ unused vertices in $A^{{}^{\prime }}$ and $⌈\frac{k-1}{2}⌉$ unused vertices in $B^{{}^{\prime }}$, respectively.

Step 4: Construct the last path if necessary.

Let $d=n-c-1-l(k-1)$. Thus, there are d unused vertices in $A^{{}^{\prime }}$ and $B^{{}^{\prime }}$, respectively. Since $l=⌊\frac{n-c-1}{k-1}⌋$, $0\le d<k-1$. Now, according to the value of d, we distinguish two cases.

If $0\le d<⌈\frac{k-1}{2}⌉$, then $2l=2⌊\frac{n-c-1}{k-1}⌋=2⌊\frac{l(k-1)+d}{k-1}⌋=⌊\frac{2(n-c-1)}{k-1}⌋$. In this case, we stop constructing any new S-path.

If $⌈\frac{k-1}{2}⌉\le d<k-1$, then $2l=2⌊\frac{n-c-1}{k-1}⌋=⌊\frac{2(n-c-1)}{k-1}⌋-1$. Since $d\ge ⌈\frac{k-1}{2}⌉$, we can take $⌈\frac{k-1}{2}⌉$ and $⌊\frac{k-1}{2}⌋$ remaining vertices from $A^{{}^{\prime }}$ and $B^{{}^{\prime }}$, respectively. Similarly to $P_3$, using the $k-1$ vertices in total, we can obtain a new S-path.

Therefore, by the above four steps, we construct $⌊\frac{2(n-c-1)}{k-1}⌋+2$S-paths, which are obviously internally disjoint.

Moreover, since $1\le a\le b\le c$, $k-1=a+b+c-1\ge c+1$. Hence,

$$⌊\frac{2(n-c-1)}{k-1}⌋+2\ge ⌊\frac{2(n-k+1)}{k-1}⌋+2=⌊\frac{2n}{k-1}⌋.$$

It follows that we can obtain at least $⌊\frac{2n}{k-1}⌋$ internally disjoint S-paths in this case; that is, $\pi \left(S\right)\ge ⌊\frac{2n}{k-1}⌋$.

Case 3:$a=0$ and $1\le b\le c$.

In this case, $S=(B\cup C)$. We will also construct at least $⌊\frac{2n}{k-1}⌋$ internally disjoint S-paths. We divide the construction process into four steps, as follows.

Step 1: Construct the first S-path $P_1$.

By using $b-1$ vertices $z_{c+1},\dots ,z_{c+b-1}$ in $Z\backslash C$, connect all the vertices $y_1,\dots ,y_b$ of B into a path, denoted by $P_{11}$, i.e., $P_{11}=y_1{z}_{c+1}{y}_2{z}_{c+2}\dots y_{b-1}{z}_{c+b-1}{y}_b$.

By using $c-1$ vertices $x_1,\dots ,x_{c-1}$ in X, connect all the vertices $z_1,\dots ,z_c$ of C into a path, denoted by $P_{12}$, i.e., $P_{12}=z_1{x}_1{z}_2{x}_2\dots z_{c-1}{x}_{c-1}{z}_c$.

Finally, using the vertex $x_c$ to connect $y_b$ and $z_1$, we obtain the first S-path $P_1$, i.e., $P_1=P_{11}\cup \left\{y_b{x}_c{z}_1\right\}\cup P_{12}$.

Step 2: Construct the second S-path $P_2$.

Let $P_2=z_1{y}_1{z}_2\dots y_{c-1}{z}_c{y}_c$. Since $c\ge b$, $B\subseteq \left\{y_1,\dots ,y_c\right\}$. Hence, $P_2$ is a path connecting all the vertices of $B\cup C$, and so is an S-path.

Remark. 

After the first two steps, we have found two S-paths, which are obviously internally disjoint. Moreover, there are $n-c$ unused vertices in $X\hspace{0.17em}\backslash \hspace{0.17em}A$ (namely, $x_{c+1},\dots ,x_n$), $n-c$ unused vertices in $Y\hspace{0.17em}\backslash \hspace{0.17em}B$ (namely, $y_{c+1},\dots ,y_n$) and $n-c-\left(b-1\right)=n-k+1$ unused vertices in $Z\hspace{0.17em}\backslash \hspace{0.17em}C$ (namely, $z_{c+b},\dots ,z_n$). Set $A^{\prime }=\left[x_{c+1},x_n\right]$, $B^{\prime }=\left[y_{c+1},y_n\right]$, and $C^{{}^{\prime }}=\left[z_k,z_n\right]$.

Step 3: Construct the next $2l$S-paths, where $l=⌊\frac{n-c}{k-1}⌋$.

The method is similar to case 2. If $l=0$, proceed directly to Step 4. thus, we assume that $l\ge 1$. In general, by $k-1$ vertices in $A^{{}^{\prime }}$ and $k-1$ vertices in $B^{{}^{\prime }}$, we can obtain S-paths in pairs: use $⌈\frac{k-1}{2}⌉$ unused vertices in $A^{{}^{\prime }}$ and $⌊\frac{k-1}{2}⌋$ unused vertices in $B^{{}^{\prime }}$ to construct an S-path; next, use $⌊\frac{k-1}{2}⌋$ unused vertices in $A^{{}^{\prime }}$ and $⌈\frac{k-1}{2}⌉$ unused vertices in $B^{{}^{\prime }}$ to construct another S-path; by repeating this process, we can construct $l=⌊\frac{n-c}{k-1}⌋$ pairs of S-paths.

However, when $b=c$, $⌊\frac{k-1}{2}⌋=⌊\frac{2b-1}{2}⌋=b-1$ and $⌈\frac{k-1}{2}⌉=b$. If we only use $b-1$ vertices in $A^{{}^{\prime }}$ and b vertices in $B^{{}^{\prime }}$ and do not use any other vertex and edge in $E\left(G\right[B\cup C\left]\right)$, we cannot connect all the vertices of $B\cup C$ into a path. Thus, we distinguish two subcases:

Subcase 3.1: $1\le b<c$.

We have $⌊\frac{k-1}{2}⌋\ge b$.

Firstly, by using b vertices in $A^{\prime }$, connect all vertices $y_1,\dots ,y_b$ of B and vertex $z_1$ into a $y_1{z}_1$-path, denoted by $P_{i1}$, where $3\le i\le 2l+2$.

Next, when i is odd (when i is even), take $⌈\frac{k-1}{2}⌉-b$ ($⌊\frac{k-1}{2}⌋-b$) unused vertices from $A^{{}^{\prime }}$, and take $⌊\frac{k-1}{2}⌋$ ($⌈\frac{k-1}{2}⌉$) unused vertices from $B^{{}^{\prime }}$. Using the $k-1-b=c-1$ vertices in total, we can connect all vertices $z_1,\dots ,z_c$ of C into a $z_1{z}_c$-path $P_{i2}$.

Combining these two paths, we obtain an S-path $P_i$, i.e., $P_i=P_{i1}\cup P_{i2}$, where $3\le i\le 2l+2$.

Clearly, when i is odd (when i is even), then the path $P_i$ uses $⌈\frac{k-1}{2}⌉$ ($⌊\frac{k-1}{2}⌋$) vertices in $A^{{}^{\prime }}$ and $⌊\frac{k-1}{2}⌋$ ($⌈\frac{k-1}{2}⌉$) vertices in $B^{{}^{\prime }}$, respectively.

Subcase 3.2: $b=c$.

We have $⌈\frac{k-1}{2}⌉=b$ and $⌊\frac{k-1}{2}⌋=b-1$.

When i is odd $(3\le i\le 2l+2)$, since $⌈\frac{k-1}{2}⌉-b\ge 0$, by the same method as Subcase 3.1, we can construct $P_i$ by $⌈\frac{k-1}{2}⌉$ unused vertices in $A^{{}^{\prime }}$ and $⌊\frac{k-1}{2}⌋$ unused vertices in $B^{{}^{\prime }}$.

However, when i is even, as noted above, $⌊\frac{k-1}{2}⌋$ vertices in $A^{{}^{\prime }}$ and $⌈\frac{k-1}{2}⌉$ vertices in $B^{{}^{\prime }}$ are not enough to obtain an S-path. We will complete the construction with the help of a vertex in $C^{{}^{\prime }}$, as follows.

Firstly, by using $⌊\frac{k-1}{2}⌋=b-1$ vertices in $A^{\prime }$, connect all the vertices $y_1,\dots ,y_b$ of B into a $y_1{y}_b$-path, denoted by $P_{i1}$.

Then, by using $⌈\frac{k-1}{2}⌉-1=b-1$ vertices in $B^{\prime }$, connect all vertices $z_1,\dots ,z_c$ of C into a $z_1{z}_c$-path, denoted by $P_{i2}$.

Finally, by one unused vertex $\widehat{y}$ in $B^{\prime }$ and one unused vertex $\widehat{z}$ in $C^{{}^{\prime }}$, connect vertices $y_b$ and $z_1$. Then, we obtain an S-path $P_i$, i.e., $P_i=P_{i1}\bigcup y_b\widehat{z}\widehat{y}{z}_1\bigcup P_{i2}$, where $3\le i\le 2l+2$ and i is even.

Note that, it remains to show that the vertices in $C^{{}^{\prime }}$ are enough. Therefore, we will prove that $|C^{\prime }|\ge l$.

Since $3\le k\le n$ and $c\ge 2$, we obtain the following.

$$\begin{array}{cc}\hfill \left|C^{\prime }\right|-l& =n-k+1-⌊\frac{n-c}{k-1}⌋\hfill \\ \hfill & \ge n-k+1-\frac{n-c}{k-1}\hfill \\ \hfill & =\frac{n(k-2)-{(k-1)}^2+c}{k-1}\hfill \\ \hfill & \ge \frac{k(k-2)-{(k-1)}^2+c}{k-1}\hfill \\ \hfill & =\frac{c-1}{k-1}\ge 0.\hfill \end{array}$$

Thus, in either case, we can always obtain $2l=2⌊\frac{n-c}{k-1}⌋$S-paths in this step.

Step 4: Construct the last path if necessary.

Let $d=n-c-l(k-1)$. Since $l=⌊\frac{n-c}{k-1}⌋$, $0\le d<k-1$. Similarly to Case 2, according to the value of d, distinguish two cases.

If $0\le d<⌈\frac{k-1}{2}⌉$, then $2l=2⌊\frac{n-c}{k-1}⌋=⌊\frac{2(n-c)}{k-1}⌋$. We stop constructing any new S-path.

If $⌈\frac{k-1}{2}⌉\le d<k-1$, then $2l=2⌊\frac{n-c}{k-1}⌋=⌊\frac{2(n-c)}{k-1}⌋-1$. We can construct one more new S-path by the remaining d vertices in $A^{{}^{\prime }}$ and $B^{{}^{\prime }}$, respectively.

Therefore, by the above four steps, we construct $⌊\frac{2(n-c)}{k-1}⌋+2$S-paths, which are obviously internally disjoint.

Moreover, since $a=0$ and $1\le b\le c$, $k-1=b+c-1\ge c$. Hence,

$$⌊\frac{2(n-c)}{k-1}⌋+2\ge ⌊\frac{2(n-k+1)}{k-1}⌋+2=⌊\frac{2n}{k-1}⌋.$$

Thus, in this case, we can also obtain at least $⌊\frac{2n}{k-1}⌋$ internally disjoint S-paths; that is, $\pi \left(S\right)\ge ⌊\frac{2n}{k-1}⌋$.

From the above discussion, $\pi \left(S\right)\ge ⌊\frac{2n}{k-1}⌋$ in all cases and $\pi \left(S\right)$ is exactly $⌊\frac{2n}{k-1}⌋$ in Case 1. Thus, we can conclude that ${\pi }_k\left(K_{n,n,n}\right)=min\left\{\pi \left(S\right)|S\subseteq V\left(K_{n,n,n}\right),\left|S\right|=k\right\}=⌊\frac{2n}{k-1}⌋.$ □

By Steps 3 and 4 of Case 2 in Theorem 1, we can obtain the following corollary, which may be useful for study on complete tripartite graphs.

Corollary 1.

Let $a,b,c$, and d be positive integers with $1\le a\le b\le c$, and G be a complete tripartite graph with three parts X, Y, and Z, where $\left|X\right|=a+d$, $\left|Y\right|=b+d$, and $\left|Z\right|=c$. For any k-subset S of $V\left(G\right)$, if $|X\cap S|=a$, $|Y\cap S|=b$, and $|Z\cap S|=c$, then there always exist at least $⌊\frac{2d}{k-1}⌋$ internally disjoint S-paths in G, where $k=a+b+c$.

Remark. 

Since $\left(⌊\frac{2n}{k-1}⌋+1\right)\left(k-1\right)>\kappa \left(K_{n,n,n}\right)=2n\ge ⌊\frac{2n}{k-1}⌋\left(k-1\right)$, Theorem 1 implies that Hager’s conjecture is true for $K_{n,n,n}$ and $3\le k\le n$.

3. Conclusions

k-path-connectivity is a natural generalization of the traditional connectivity. In this paper, we showed that the k-path-connectivity of the complete balanced tripartite graph $K_{n,n,n}$ is $⌊\frac{2n}{k-1}⌋$, for $3\le k\le n.$ For future work, we will continue to investigate the k-path-connectivity of $K_{n,n,n}$ for $n+1\le k\le 3n$. It would also be interesting to study the path connectivity of complete r-partite graphs for $r\ge 4$.