General Vertex-Distinguishing Total Colorings of Complete Bipartite Graphs

Abstract

Let G be a simple graph. A general total coloring f of G refers to a coloring of the vertices and edges of G. Let $C\left(x\right)$ be the set of colors of vertex x and edges incident with x under f. For a general total coloring f of G in which k colors are available, if $C\left(u\right)\ne C\left(v\right)$ for any two different vertices u and v in $V\left(G\right)$, then f is called a k-general vertex-distinguishing total coloring of G, or a k-GVDTC of G for short. The minimum number of colors required for a GVDTC of G is denoted by ${\chi }_{gvt}\left(G\right)$ and is called the general vertex-distinguishing total chromatic number, or the GVDT chromatic number of G for short. GVDTCs of complete bipartite graphs are studied in this paper.

1. Introduction and Terminologies

For an edge coloring (proper or not) of a graph G and a vertex x of G, denote by $S\left(x\right)$ the set of colors used to color the edges incident with x.

A proper edge coloring of a graph G is said to be vertex-distinguishing if for $\forall u,v\in V\left(G\right),u\ne v$, we have $S\left(u\right)\ne S\left(v\right)$. In other words, $S\left(u\right)\ne S\left(v\right)$ whenever $u\ne v$. A graph G has a vertex-distinguishing proper edge coloring if and only if it has no more than one isolated vertex and no isolated edges. Such a graph will be referred to as a $vdec$-graph. The minimum number of colors required for a vertex-distinguishing proper edge coloring of a $vdec$-graph G is denoted by ${\chi }_s^{\prime }\left(G\right)$. Vertex-distinguishing proper edge coloring has been considered in several papers [1,2,3,4,5,6,7].

A general edge coloring (not necessarily proper) of a graph G is said to be point-distinguishing if $S\left(u\right)\ne S\left(v\right)$ is required for any two distinct vertices $u,v$. The point-distinguishing chromatic index of a $vdec$-graph G, denoted by ${\chi }_0\left(G\right)$, refers to the minimum number of colors required for a point-distinguishing general edge coloring of G. This parameter was introduced by Harary and Plantholt in [8]. Although the structure of complete bipartite graphs is simple, it seems that the problem of determining ${\chi }_0\left(K_{m,n}\right)$ is not easy, especially in the case $m=n$, as documented by papers of M. Horňák and R. Soták [9,10], N. Zagaglia Salvi [11,12] and M. Horňák and N. Zagaglia Salvi [13].

A proper total coloring of a graph G is an assignment of several colors to the vertices and edges of G such that the following three conditions are satisfied:

Condition (V): No two adjacent vertices receive the same color;

Condition (E): No two adjacent edges receive the same color;

Condition (I): No edge receives the same color as any one of its incident vertices.

A general total coloring of a graph G is an assignment of several colors to the vertices and edges of G. Note that three conditions (V), (E) and (I) are not required; i.e., under a general total coloring of a graph G, two adjacent or incident elements of G may receive the same color.

For a total coloring (proper or general) f of G and a vertex x of G, denote by $C_f\left(x\right)$ (or simply $C\left(x\right)$ if no confusions arise) the set of colors used to color the vertex x or the edges incident with x. Note that $C_f\left(x\right)$ is not a multiple set. And $C_f\left(x\right)$ is called the color set of vertex x under f. Let $\overline{C\left(x\right)}$ be the complementary set of $C\left(x\right)$ in the set of all available colors. Obviously $\left|C\left(x\right)\right|\le d_G\left(x\right)+1$, and the equality holds if the total coloring is proper.

For a proper total coloring, if $C\left(u\right)\ne C\left(v\right)$ for any two distinct vertices u and v, then the coloring is called vertex-distinguishing proper total coloring, and the minimum number of colors required for a vertex-distinguishing proper total coloring is denoted by ${\chi }_{vt}\left(G\right)$. This concept was considered in [14,15]. In [15], the following conjecture was given.

Conjecture 1 

([15]). Suppose G is a simple graph and $n_d$ is the number of vertices of degree d, $\delta \le d\le \mathrm{\Delta }$. Let k be the minimum positive integer such that $\left(\genfrac{}{}{0pt}{}{k}{d+1}\right)\ge n_d$ for all d such that $\delta \le d\le \mathrm{\Delta }$; then ${\chi }_{vt}\left(G\right)=k$ or $k+1$.

From [15] we know that the above conjecture is valid for complete graphs, complete bipartite graphs, wheels, fans, regular double stars, paths, cycles, $P_n\vee P_n$, $P_n\vee C_n$, $C_n\vee C_n$, and 2-order n-separate trees. An upper bound $\left|V\right(G\left)\right|+2$ for ${\chi }_{vt}\left(G\right)$ is also given in [15].

General vertex-distinguishing total coloring of a graph is introduced in [16]. The relationship between this coloring and vertex-distinguishing proper total coloring is similar to the relationship between vertex-distinguishing general edge coloring and vertex-distinguishing proper edge coloring.

After vertex-distinguishing proper edge coloring was investigated by Burris and Schelp et al. in [4], Zhongfu Zhang et al. in [15] began by studying vertex-distinguishing proper total coloring, obtained several results and proposed the Vizing-like conjecture (Conjecture 1) of this coloring. Similarly, Chanjuan Liu and Enqiang Zhu in [16] researched general vertex-distinguishing total coloring and obtained many results, after vertex-distinguishing general edge coloring was studied by Harary et al. in [8]. General vertex-distinguishing total coloring (in [16]) is nothing but vertex-distinguishing general total coloring. In this paper we introduce an index $\xi \left(G\right)$ of a graph G and propose a Vizing-like conjecture of the coloring. For reasons such as these, general vertex-distinguishing total coloring turns out to be rather important.

A general total coloring of graph G in which k colors are available is also called a k-general total coloring of graph G. If f is a general total coloring of graph G and $C\left(u\right)\ne C\left(v\right)$ for any two different vertices $u,v$ in $V\left(G\right)$, then f is called a vertex-distinguishing general total coloring of G or a general vertex-distinguishing total coloring of G and denoted by GVDTC in brief. A general vertex-distinguishing total coloring of G in which k colors are available is also called a k-general vertex-distinguishing total coloring of G and denoted by k-GVDTC in brief.

The positive integer $min\left\{k\right|Ghasak-GVDTC\}$ is called the general vertex-distinguishing total chromatic number of graph G (or GVDT chromatic number of G) and is denoted by ${\chi }_{gvt}\left(G\right)$.

The following proposition is obviously true (since a vertex-distinguishing proper total coloring of a graph G is also a vertex-distinguishing general total coloring of G).

Proposition 1.

${\chi }_{gvt}\left(G\right)\le {\chi }_{vt}\left(G\right)$.

For a graph G, let $n_i$ denote the number of vertices of degree i, $\delta \le i\le \mathrm{\Delta }$. Let

$$\xi \left(G\right)=min\left\{k\right|\left(\genfrac{}{}{0pt}{}{k}{1}\right)+\left(\genfrac{}{}{0pt}{}{k}{2}\right)+\left(\genfrac{}{}{0pt}{}{k}{3}\right)+\dots +\left(\genfrac{}{}{0pt}{}{k}{s+1}\right)\ge n_{\delta }+n_{\delta +1}+\dots +n_s,\delta \le s\le \mathrm{\Delta }\}.$$

Obviously, we have the following proposition.

Proposition 2.

${\chi }_{gvt}\left(G\right)\ge \xi \left(G\right)$.

The GVDTCs of a path, cycle, star, double star, tristar, fan, wheel and complete graph are investigated and the GVDT chromatic numbers of these graphs are obtained in [16]. The conclusions on the GVDT chromatic numbers of star $S_n$ and complete graph $K_n$ are listed in the following. A conjecture (the following Conjecture 2) is proposed in [16].

Proposition 3

([16]). Suppose $S_n(n\ge 1)$ is a star of order $n+1$; i.e., $S_n=K_{1,n}$. We have

$${\chi }_{gvt}\left(S_n\right)=\left\{\begin{array}{c}2,n=1,2;\hfill \\ 3,n=3;\hfill \\ \lceil \frac{\sqrt{8n+1}-1}{2}\rceil ,n\ge 4.\hfill \end{array}\right.$$

Proposition 4

([16]). If $K_n(n\ge 1)$ is a complete graph of order n, then we have ${\chi }_{gvt}\left(K_n\right)=1+\lceil {log}_{2}n\rceil .$

Conjecture 2

([16]). If G is a graph, then ${\chi }_{gvt}\left(G\right)\le 1+\lceil {log}_{2}n\rceil .$

In this paper we will consider the GVDTC of complete bipartite graphs.

For a set S, let $2^S$ denote the collection of all subsets of S. Of course $2^S$ is also a set and $2^S$ has $2^{\left|S\right|}$ members.

An i-subset of S is a subset of S that has i elements.

For complete bipartite graph $K_{m,n}$ ($m\le n$), we suppose that the vertex set of $K_{m,n}$ is $X\cup Y$, where $X=\{x_1,x_2,\dots ,x_m\}$, $Y=\{y_1,y_2,\dots ,y_n\}$ and the edge set of $K_{m,n}$ is $\left\{x_i{y}_j\right|i=1,2,\dots ,m;j=1,2,\dots ,n\}$.

For a GVDTC of complete bipartite graph $K_{m,n}$ ($m\le n$), we use $\mathcal{C}\left(X\right)$ to denote the collection of the color sets of all vertices in X under given GVDTC; i.e., $\mathcal{C}\left(X\right)=\{C\left(x_1\right),C\left(x_2\right),\dots ,C\left(x_m\right)\}$. We use $\mathcal{C}\left(Y\right)$ to denote the collection of the color sets of all vertices in Y under given GVDTC and $\mathcal{C}(X\cup Y)$ to denote the collection of the color sets of all vertices in $X\cup Y$ under given GVDTC; i.e., $\mathcal{C}(X\cup Y)$=$\mathcal{C}\left(X\right)\cup \mathcal{C}\left(Y\right)$.

Generally, if we mention a k-GVDTC, then the available colors in this coloring are $1,2,\dots ,k$. If A is a subset of $\{1,2,\dots ,k\}$, then we denote $\{1,2,\dots ,k\}\setminus A$ by $\overline{A}$.

2. Preliminaries

For complete bipartite graphs $K_{m,n}(1\le m<n)$, $\xi \left(K_{m,n}\right)$ is the minimum positive integer l such that

$$\left(\genfrac{}{}{0pt}{}{l}{1}\right)+\left(\genfrac{}{}{0pt}{}{l}{2}\right)+\left(\genfrac{}{}{0pt}{}{l}{3}\right)+\dots +\left(\genfrac{}{}{0pt}{}{l}{m+1}\right)\ge n,$$

(1)

$$\left(\genfrac{}{}{0pt}{}{l}{1}\right)+\left(\genfrac{}{}{0pt}{}{l}{2}\right)+\left(\genfrac{}{}{0pt}{}{l}{3}\right)+\dots +\left(\genfrac{}{}{0pt}{}{l}{n+1}\right)\ge n+m.$$

(2)

The following Lemma 1 is obviously true.

Lemma 1. 

(i) If a is a positive real number, then function $f\left(x\right)=2^x+2^{a-x}$ is decreasing when $x\in [0,\frac{a}{2}]$ and is increasing when $x\in [\frac{a}{2},a]$.

(ii) If $a,x$ are positive integers, $1\le x\le a-1$, $a\ge 2$, then $2^x+2^{a-x}\ge 2^{\frac{a}{2}+1}$ when a is an even number and $2^x+2^{a-x}\ge 2^{\frac{a-1}{2}}+2^{\frac{a+1}{2}}$ when a is an odd number.

Proof. 

By simple calculation we have $f^{\prime }\left(x\right)=(2^x-2^{a-x})ln2$. So $f^{\prime }\left(x\right)<0$ when $0<x<\frac{a}{2}$ and $f^{\prime }\left(x\right)>0$ when $\frac{a}{2}<x<a$. Thus (i) is true. But (ii) is the direct consequence of (i). □

Lemma 2.

Let $a_m=2\lceil {log}_2(m+1)\rceil$, $b_m=2\lceil {log}_2\frac{2m+1}{3}\rceil +1$, where m is a positive integer. Then $a_m-b_m=1$ or $b_m-a_m=1$.

Proof. 

Suppose q is a non-negative integer such that $2^q\le m<2^{q+1}$. Then $a_m=2(q+1)$. If $m=1$, then $a_1=2,b_1=1$, $a_1-b_1=1$. Thus we may assume that $m\ge 2,q\ge 1$.

(i) When $2^q\le m\le 3·2^{q-1}-1$, we have $2^{q-1}<\frac{1}{3}(2^{q+1}+1)\le \frac{1}{3}(2m+1)$$\le \frac{1}{3}(3·2^q-1)<2^q$. Thus $b_m=2q+1$ and $a_m-b_m=1$.

(ii) When $3·2^{q-1}\le m\le 2^{q+1}-1$, we have $2^q<\frac{1}{3}(3·2^q+1)\le \frac{1}{3}(2m+1)$$\le \frac{1}{3}(2^{q+2}-1)<2^{q+1}$. Thus $b_m=2q+3$ and $b_m-a_m=1$. □

Lemma 3.

Let s be the minimum positive integer such that $2^s\ge 3m$. If (i) $s\le r\le m+1$, $2^r-2m<n\le 2^{r+1}-2m$ or (ii) $r=s-1$, $m\le n\le 2^{r+1}-2m$, then $K_{m,n}$ has an $(r+1)$-GVDTC.

Proof. 

We will give an $(r+1)$-GVDTC of $K_{m,n}$ in which the available colors are $1,2,\dots ,r,r+1$ in the following. There are four cases to be considered.

Case 1.

 $r=m+1$.

In this case, only condition (i) is satisfied. Let $D\left(x_i\right)=\{1,2,\dots ,r+1\}\setminus \left\{i\right\}$, $i=1,2,\dots ,r-2$, and $D\left(x_{r-1}\right)=\{1,2,\dots ,r+1\}$. Let $D\left(y_j\right)=\{j,r+1\}$, $j=1,2,\dots ,r-2$; $D\left(y_{r-1}\right)=\{r-1\}$; $D\left(y_r\right)=\left\{r\right\}$; $D\left(y_{r+1}\right)=\{r+1\}$; $D\left(y_{r+2}\right)=\{1,2,\dots ,r+1\}\setminus \{r-1\}$; $D\left(y_{r+3}\right)=\{1,2,\dots ,r+1\}\setminus \left\{r\right\}$; and $D\left(y_{r+4}\right)=\{1,2,\dots ,r+1\}\setminus \{r+1\}$. Let $D\left(y_{r+5}\right)$, $D\left(y_{r+6}\right)$, …, $D\left(y_n\right)$ be different subsets (of $\{1,2,\dots ,r+1\}$ ) containing at least 2 and at most $r-1$ colors except for all $\{j,r+1\}$ ($j=1,2,3,\dots ,r-2$). This is well established since $n-6\le \left(\genfrac{}{}{0pt}{}{r+1}{2}\right)+\left(\genfrac{}{}{0pt}{}{r+1}{3}\right)+\dots +\left(\genfrac{}{}{0pt}{}{r+1}{r-1}\right)$.

Construct a general total coloring as follows. Let $x_i$ receive color $r+1$, $i=1,2,\dots ,m$. Let $x_j{y}_j$ and $y_j$ receive color $r+1$ and $x_i{y}_j$ receive color j ($i\in \{1,2,\dots ,m\}\setminus \left\{j\right\}$), $j=1,2,\dots ,r-2$. Let $x_i{y}_j$ ($i\in \{1,2,\dots ,m\}$) and $y_j$ receive color j, $j=r-1,r,r+1$. For each $j\in \{r+2,r+3,\dots ,n\}$, let $D\left(y_j\right)=\{a_1,a_2,\dots ,a_l\}$, $a_1<a_2<\dots <a_l$, then $2\le l\le r$. Note that $a_1,a_2,\dots ,a_l$ depend on j. Let $y_j$ receive color $a_1$; let edge $x_{a_p-1}{y}_j$ receive color $a_p$ ($p=2,3,\dots ,l$); and let edge $x_i{y}_j$ ($i\ne a_2-1,a_3-1,\dots ,a_l-1$) receive the color $min[D\left(x_i\right)\cap D\left(y_j\right)]$.

Obviously we have $r\le n$ since $n>2^r-2m\ge m=r-1$. From the colors of the vertices in X and the edges incident with $y_j$, $j=1,2,\dots ,r$, we know that $C\left(x_i\right)=D\left(x_i\right)$, $i=1,2,\dots ,m$. From the coloring process we know that $C\left(y_j\right)=D\left(y_j\right),j=1,2,\dots ,n$. Thus the resulting coloring is an $(r+1)$-GVDTC of $K_{m,n}$.

Case 2.

 $r=m$.

If $m\ge 4$, then condition (i) is satisfied and $n>2^r-2m\ge m=r$. If $1\le m\le 3$, then condition (ii) is satisfied and $n\ge m=r$.

Let $D\left(x_i\right)=\{1,2,\dots ,r+1\}\setminus \left\{i\right\}$, $i=1,2,\dots ,r-1$, and $D\left(x_r\right)=\{1,2,\dots ,r+1\}$. Let $D\left(y_j\right)=\{j,r+1\}$, $j=1,2,\dots ,r-1$; $D\left(y_r\right)=\left\{r\right\}$; $D\left(y_{r+1}\right)=\{r+1\}$; $D\left(y_{r+2}\right)=\{1,2,\dots ,r+1\}\setminus \left\{r\right\}$; and $D\left(y_{r+3}\right)=\{1,2,\dots ,r\}$. Let $D\left(y_{r+4}\right)$, $D\left(y_{r+5}\right)$, …, $D\left(y_n\right)$ be different subsets (of $\{1,2,\dots ,r+1\}$ ) containing at least 2 and at most $r-1$ colors except for all $\{j,r+1\}$ ($j=1,2,3,\dots ,r-1$). This is well established since $n-4\le \left(\genfrac{}{}{0pt}{}{r+1}{2}\right)+\left(\genfrac{}{}{0pt}{}{r+1}{3}\right)+\dots +\left(\genfrac{}{}{0pt}{}{r+1}{r-1}\right)$.

Note that $r\le n$; we can give an $(r+1)$-GVDTC of $K_{m,n}$ similar to the case $r=m+1$ such that $C\left(x_i\right)=D\left(x_i\right)$, $i=1,2,\dots ,m$, $C\left(y_j\right)=D\left(y_j\right),j=1,2,\dots ,n$.

Case 3.

 $r\le m-1$ and when condition (ii) is satisfied we suppose $m\ne 5$.

In this case if condition (i) is satisfied then $m\ge 5$; if condition (ii) is satisfied then $m\ge 4$ and $m\ne 5$.

Fact 1. 

 $2^r\ge m+r+1$.

Proof of Fact 1. 

If condit ion (i) is satisfied, we have $2^r\ge 2^s\ge 3m>2m\ge m+r+1.$ If condition (ii) is satisfied, we will prove that $2^{s-1}\ge m+s$ by induction on s.

When $m=4$, $s=4$, we have $2^{s-1}\ge m+s$. Since $6\le m\le 10$ when $s=5$, $11\le m\le 21$ when $s=6$ and $22\le m\le 42$ when $s=7$, we have $2^{s-1}\ge m+s$ for $s=5,6,7$. Let $s_0\ge 7$ and $m_0$ be the maximum positive integer such that $2^{s_0}\ge 3m_0$. Then the maximum positive integer $m_1$ is such that $2^{s_0+1}\ge 3m_1$ does not exceed $2m_0+1$ (otherwise $2^{s_0+1}\ge 3(2m_0+2)$. So $2^{s_0}\ge 3(m_0+1)$. This is a contradiction to the choice of $m_0$). Suppose $2^{s_0-1}\ge m_0+s_0$ (induction assumption) and $s_1=s_0+1$. Then $2^{s_1-1}=2^{s_0}\ge 2m_0+2s_0\ge 2m_0+s_1-2+s_1\ge m_1+s_1$. The proof of Fact 1 is completed. □

Let $D\left(x_i\right)=\{1,2,\dots ,r+1\}\setminus \left\{i\right\}$, $i=1,2,\dots ,r$, and $D\left(x_{r+1}\right)=\{1,2,\dots ,r+1\}$. Let $D\left(x_{r+2}\right)$, $D\left(x_{r+3}\right)$, …, $D\left(x_m\right)$ be different subsets (of $\{1,2,\dots ,r+1\}$) containing color $r+1$ such that $3\le \left|D\right(x_i\left)\right|\le r-1$ ($j=r+2,r+3,\dots ,m$). Let $D\left(y_j\right)=\{j,r+1\}$, $j=1,2,\dots ,r$; $D\left(y_{r+1}\right)=\{r+1\}$; and $D\left(y_{r+2}\right)=\{1,2,\dots ,r\}$. Let $D\left(y_{r+3}\right)$, $D\left(y_{r+4}\right)$, …, $D\left(y_n\right)$ be different subsets (of $\{1,2,\dots ,r+1\}$ ) containing at least 2 and at most $r-1$ colors except for all $D\left(x_i\right)$, $\overline{D\left(x_i\right)}$ ($i=r+2,r+3,\dots ,m$) and $\{j,r+1\}$ ($j=1,2,\dots ,r$). This is well established since $\left(\genfrac{}{}{0pt}{}{r}{2}\right)+\left(\genfrac{}{}{0pt}{}{r}{3}\right)+\dots +\left(\genfrac{}{}{0pt}{}{r}{r-2}\right)\ge m-r-1$ (obtained by Fact 1) and $n-2\le \left(\genfrac{}{}{0pt}{}{r+1}{2}\right)+\left(\genfrac{}{}{0pt}{}{r+1}{3}\right)+\dots +\left(\genfrac{}{}{0pt}{}{r+1}{r-1}\right)-2(m-r-1)$.

Construct a general total coloring as follows. Let $x_i$ receive color $r+1$, $i=1,2,\dots ,m$. Let $x_j{y}_j$ and $y_j$ receive color $r+1$ and $x_i{y}_j$ receive color j ($i\in \{1,2,\dots ,m\}\setminus \left\{j\right\}$), $j=1,2,\dots ,r$. Let $x_i{y}_{r+1}$ ($i\in \{1,2,\dots ,m\}$) and $y_{r+1}$ receive color $r+1$. For each $j\in \{r+2,r+3,\dots ,n\}$, let $D\left(y_j\right)=\{a_1,a_2,\dots ,a_l\}$, then $2\le l\le r$. Let $y_j$ receive color $a_1$; let edge $x_{a_p-1}{y}_j$ receive the color $a_p$ ($p=2,3,\dots ,l$); and let edge $x_i{y}_j$ ($i\ne a_2-1,a_3-1,\dots ,a_l-1$) receive the color $min\{D\left(x_i\right)\cap D\left(y_j\right)\}$.

Obviously we have $r+1\le m\le n$. From the color of $x_i$ and colors of the edges incident with $y_j$, $j=1,2,\dots ,r+1$, we know that $C\left(x_i\right)=D\left(x_i\right)$, $i=1,2,\dots ,m$. From the coloring process we know that $C\left(y_j\right)=D\left(y_j\right),j=1,2,\dots ,n$. Thus the resulting coloring is an $(r+1)$-GVDTC of $K_{m,n}$.

Case 4. 

$r\le m-1$, $m=5$ and condition (ii) is satisfied.

In this case we have that $n=5$ or $n=6$. The coloring given in Table 1 is a 4-GVDTC of $K_{5,6}$. The restriction of this 4-GVDTC of $K_{5,6}$ on the subgraph induced by $X\cup \{y_1,y_2,\dots ,y_5\}$ is a 4-GVDTC of $K_{5,5}$ (see Figure 1).

Table 1. A 4-GVDTC of $K_{5,6}$.

	${\mathit{y}}_1$; 4	${\mathit{y}}_2$; 1	${\mathit{y}}_3$; 2	${\mathit{y}}_4$; 4	${\mathit{y}}_5$; 4	${\mathit{y}}_6$; 1
$x_1$; 4	4	3	3	4	2	2
$x_2$; 4	4	3	3	1	4	3
$x_3$; 4	4	1	2	1	2	2
$x_4$; 4	4	1	3	1	2	2
$x_5$; 4	4	3	3	4	4	3

Figure 1. A 4-GVDTC of $K_{5,5}$.

3. Main Results

Theorem 1.

Let $m\ge 1$, $n\ge 2^{m+2}-m$ and $k\ge m+3$. If ${\displaystyle \sum _{i=1}^{m+1}}\left(\genfrac{}{}{0pt}{}{k-1}{i}\right)-m+1<n\le {\displaystyle \sum _{i=1}^{m+1}}\left(\genfrac{}{}{0pt}{}{k}{i}\right)-m+1$, then ${\chi }_{gvt}\left(K_{m,n}\right)=k$.

Proof. 

Firstly, we show that $K_{m,n}$ does not have a $(k-1)$-GVDTC.

Otherwise, we assume that g is a $(k-1)$-GVDTC of $K_{m,n}$ in which the available colors are $1,2,\dots ,k-1$.

When $k=m+3$, from $\mathcal{C}(X\cup Y)\subseteq 2^{\{1,2,\dots ,k-1\}}\setminus \{\varnothing \}$ we know that $m+n\le {\displaystyle \sum _{i=1}^{m+2}}\left(\genfrac{}{}{0pt}{}{m+2}{i}\right)$; i.e., $n\le {\displaystyle \sum _{i=1}^{m+1}}\left(\genfrac{}{}{0pt}{}{k-1}{i}\right)-m+1$. This is a contradiction. So we let $k\ge m+4$. Suppose that $\mathcal{A}$ is the collection of all 1-subsets, 2-subsets, …, $(m+1)$-subsets of $\{1,2,\dots ,k-1\}$.

If $|{\cup }_{i=1}^m\overline{C\left(x_i\right)}|\ge m-1$, say $\{1,2,\dots ,m-1\}\subseteq {\cup }_{i=1}^m\overline{C\left(x_i\right)}$, then $\left\{1\right\}$, $\left\{2\right\}$, …, $\{m-1\}$ are not in $\mathcal{C}\left(Y\right)$ and $\mathcal{C}\left(Y\right)\subseteq \mathcal{A}\setminus \left\{\right\{1\}$, $\left\{2\right\}$, …, $\{m-1\}\}$. Thus $n\le {\displaystyle \sum _{i=1}^{m+1}}\left(\genfrac{}{}{0pt}{}{k-1}{i}\right)-m+1$. This is a contradiction.

If $|{\cup }_{i=1}^m\overline{C\left(x_i\right)}|\le m-2$, then $\overline{C\left(x_i\right)}\in \mathcal{A}\cup \{\varnothing \}$ and $\overline{C\left(x_i\right)}\notin \mathcal{C}\left(Y\right)$, $i=1,2,\dots ,m$. Thus $\mathcal{C}\left(Y\right)\subseteq (\mathcal{A}\cup \{\varnothing \})\setminus \{\overline{C\left(x_1\right)}$, $\overline{C\left(x_2\right)}$, …, $\overline{C\left(x_m\right)}\}$ and $n\le {\displaystyle \sum _{i=1}^{m+1}}\left(\genfrac{}{}{0pt}{}{k-1}{i}\right)-m+1$. This is a contradiction.

Secondly, we will give a k-GVDTC of $K_{m,n}$ in which the available colors are $1,2,\dots ,k$.

Put $D\left(x_i\right)=\{1,2,\dots ,k\}\setminus \left\{i\right\},i=1,2,\dots ,m-1;$ $D\left(x_m\right)=\{1,2,\dots ,k\};$

$D\left(y_j\right)=\{j,k\},j=1,2,\dots ,m-1;$$D\left(y_j\right)=\left\{j\right\},j=m,m+1,\dots ,k.$

Now distribute other subsets of $\{1,2,\dots ,k\}$ with cardinality between 2 and $m+1$ to vertices $y_{k+1},y_{k+2},\dots ,y_n$ such that different vertices correspond different subsets. The subset that corresponds to $y_j$ is denoted by $D\left(y_j\right),j=k+1,k+2,\dots ,n$. This is well established since $n-k\le {\displaystyle \sum _{i=2}^{m+1}}\left(\genfrac{}{}{0pt}{}{k}{i}\right)-(m-1)$.

Construct a general total coloring as follows. Let $x_i$ receive color k, $i=1,2,\dots ,m$; let $x_j{y}_j$ and $y_j$ receive color k and $x_i{y}_j$ receive color j ($i\in \{1,2,\dots ,m\}\setminus \left\{j\right\}$), $j=1,2,\dots ,m-1$; and let $x_i{y}_j$ ($i\in \{1,2,\dots ,m\}$) and $y_j$ receive color j, $j=m,m+1,\dots ,k$.

Let $j\in \{k+1,k+2,\dots ,n\}$.

If $D\left(y_j\right)\cap \{1,2,\dots ,m-1\}$ contains at least two different colors, then let $D\left(y_j\right)\cap \{1,2,\dots ,m-1\}=\{a_1,a_2,\dots ,a_l\}$, where $2\le l\le m-1$. Let $y_j$ receive color $a_1$. Let edge $x_{a_p-1}{y}_j$ receive color $a_p$ ($p=2,3,\dots ,l$). The number of colors in $D\left(y_j\right)\cap \{m,m+1,\dots ,k\}$ is not larger than the number of the uncolored edges incident with $y_j$, so we can assign the colors in $D\left(y_j\right)\cap \{m,m+1,\dots ,k\}$ to the edges in $\left\{x_i{y}_j\right|i\ne a_2-1,a_3-1,\dots ,a_l-1\}$ such that each color in $D\left(y_j\right)\cap \{m,m+1,\dots ,k\}$ is used to color at least one edge incident with $y_j$.

If $D\left(y_j\right)\cap \{1,2,\dots ,m-1\}$ contains just one color $a_1$, then let $y_j$ receive color $a_1$. The number of colors in $D\left(y_j\right)\cap \{m,m+1,\dots ,k\}$ is not larger than the number of the all (uncolored) edges incident with $y_j$, so we can assign the colors in $D\left(y_j\right)\cap \{m,m+1,\dots ,k\}$ to the edges incident with $y_j$ such that each color in $D\left(y_j\right)\cap \{m,m+1,\dots ,k\}$ is used to color at least one edge incident with $y_j$.

If $D\left(y_j\right)\cap \{1,2,\dots ,m-1\}$ is empty, then the number of colors in $D\left(y_j\right)\cap \{m,m+1,\dots ,k\}$ is not larger than $d_{K_{m,n}}\left(y_j\right)$, so we can assign the colors in $D\left(y_j\right)\cap \{m,m+1,\dots ,k\}$ to the vertex $y_j$ and the edges incident with $y_j$ such that each color in $D\left(y_j\right)\cap \{m,m+1,\dots ,k\}$ is used to color the vertex $y_j$ or one edge incident with $y_j$.

By simple calculation we see that $k\le n$. From the color of each $x_i$ and the coloring of the incident edges of vertices $y_j$, $j=1,2,\dots ,k$, we know that $C\left(x_i\right)=D\left(x_i\right)$, $i=1,2,\dots ,m$. From the coloring process we know that $C_f\left(y_j\right)=D\left(y_j\right),j=1,2,\dots ,n$. Thus the resulting coloring is a k-GVDTC of $K_{m,n}$. □

Example 1.

According to the proof of Theorem 1, a 5-GVDTC of $K_{2,24}$ is shown in Figure 2.

Figure 2. A 5-GVDTC of $K_{2,24}$.

Remark 1.

Note that $\left(\genfrac{}{}{0pt}{}{m+2}{i}\right)\ge m+2$ when $i\in \{1,2,\dots ,m+1\}$, $m\ge 2$. It is easy to see that ${\displaystyle \sum _{i=1}^{m+1}}\left(\genfrac{}{}{0pt}{}{m+2}{i}\right)-2m+3\ge m+3$. Let $n_0={\displaystyle \sum _{i=1}^{m+1}}\left(\genfrac{}{}{0pt}{}{m+2}{i}\right)-m+2$. So the restriction of the $(m+3)$-GVDTC of $K_{m,n_0}$ obtained in the proof of Theorem 1 (take $k=m+3$) on the subgraph induced by $X\cup \{y_1,y_2,\dots ,y_n\}$ is an $(m+3)$-GVDTC of $K_{m,n}$ when ${\displaystyle \sum _{i=1}^{m+1}}\left(\genfrac{}{}{0pt}{}{m+2}{i}\right)-2m+3\le n\le {\displaystyle \sum _{i=1}^{m+1}}\left(\genfrac{}{}{0pt}{}{m+2}{i}\right)-m+1$.

Theorem 2.

Let $m\ge 2,\left(\genfrac{}{}{0pt}{}{m+2}{1}\right)+\left(\genfrac{}{}{0pt}{}{m+2}{2}\right)+\dots +\left(\genfrac{}{}{0pt}{}{m+2}{m+1}\right)-2m+2<n\le \left(\genfrac{}{}{0pt}{}{m+2}{1}\right)+\left(\genfrac{}{}{0pt}{}{m+2}{2}\right)+\dots +\left(\genfrac{}{}{0pt}{}{m+2}{m+1}\right)-m+1$; i.e., $2^{m+2}-2m<n\le 2^{m+2}-m-1$. Then ${\chi }_{gvt}\left(K_{m,n}\right)=m+3$.

Proof. 

By Remark 1, we have ${\chi }_{gvt}\left(K_{m,n}\right)\le m+3$. So we need only to prove that $K_{m,n}$ does not have an $(m+2)-$GVDTC. Otherwise, suppose g is an $(m+2)-$GVDTC of $K_{m,n}$ in which the available colors are $1,2,\dots ,m+2$.

If there are two members $C\left(x_{i_1}\right)$ and $C\left(x_{i_2}\right)$ in $\mathcal{C}\left(X\right)$ such that $C\left(x_{i_1}\right)$$=\{1,2,\dots ,m+2\}\setminus C\left(x_{i_2}\right)$ (i.e., the two members are complementary), then each nonempty subset in $2^{C\left(x_{i_1}\right)}\cup 2^{C\left(x_{i_2}\right)}$ is not in $\mathcal{C}\left(Y\right)$. Since $2^{m+2}-m+1\le n+m\le 2^{m+2}-1$, there exist at most $m-2$ nonempty subsets of $\{1,2,\dots ,m+2\}$, which are not the color set of any vertices in $X\cup Y$. Therefore at least $2^t-1+2^{m+2-t}-1-(m-2)$ members in $2^{C\left(x_{i_1}\right)}\cup 2^{C\left(x_{i_2}\right)}$ are contained in $\mathcal{C}\left(X\right)$. We have $2^t-1+2^{m+2-t}-1-(m-2)\le m$. By Lemma 1, we have $2^{\frac{m+2}{2}+1}\le 2m$; i.e., $2^{\frac{m}{2}}\le \frac{m}{2}$. This is a contradiction.

If any subset in $\mathcal{C}\left(X\right)$ is not the complement of any subset in $\mathcal{C}\left(X\right)$, then each $\overline{C\left(x_i\right)}$ is not the color set of any vertices. Thus $\mathcal{C}(X\cup Y)\subseteq 2^{\{1,2,\dots ,m+2\}}\setminus \{\overline{C\left(x_1\right)},\overline{C\left(x_2\right)},\dots ,\overline{C\left(x_m\right)}\}$ and $m+n\le 2^{m+2}-m$. This is a contradiction. □

Theorem 3.

Let $r_m=min\{2\lceil {log}_2(m+1)\rceil ,2\lceil {log}_2\frac{2m+1}{3}\rceil +1\}$ and $r_m\le r\le m+1$. If $2^r-2m<n\le 2^{r+1}-2m$, then ${\chi }_{gvt}\left(K_{m,n}\right)=r+1$.

Proof. 

By Lemma 3, we know that ${\chi }_{gvt}\left(K_{m,n}\right)\le r+1$ since $r_m\ge s$. So we need only to prove that $K_{m,n}$ does not have an r-GVDTC under the given conditions.

Otherwise, let g be an r-GVDTC of $K_{m,n}$ in which the available colors are $1,2,\dots ,r$. We will give six claims as follows.

Claim 1. 

$C\left(x_i\right)\cap C\left(y_j\right)\ne \varnothing$, $i=1,2,\dots ,m,j=1,2,\dots ,n$.

This is obvious because the color of edge $x_i{y}_j$ is in both $C\left(x_i\right)$ and $C\left(y_j\right)$.

Claim 2. 

$C\left(x_1\right)\cap C\left(x_2\right)\cap \dots \cap C\left(x_m\right)=\varnothing$.

Otherwise, say $1\in C\left(x_1\right)\cap C\left(x_2\right)\cap \dots \cap C\left(x_m\right)$. Then, by Claim 1, each $\overline{C\left(x_i\right)}$ is not the color set of any vertex and we have that $m+n\le 2^r-m$. This is a contradiction.

Claim 3. 

$\left|C\right(y_j\left)\right|\ge 2,j=1,2,\dots ,n$.

From Claim 2, we can deduce Claim 3 easily.

Claim 4. 

$C\left(y_1\right)\cap C\left(y_2\right)\cap \dots \cap C\left(y_n\right)=\varnothing$.

Otherwise, say $1\in C\left(y_1\right)\cap C\left(y_2\right)\cap \dots \cap C\left(y_n\right)$. Then, by Claim 1, each $\overline{C\left(y_j\right)}$ is not the color set of any vertex and we have that $m+n\le 2^r-n\le 2^r-m$. This is a contradiction.

Claim 5. 

$\left|C\right(x_i\left)\right|\ge 2,i=1,2,\dots ,m$.

This is deduced from Claim 4 directly.

Claim 6. 

If D is a subset of $\{1,2,\dots ,r\}$ with $2\le \left|D\right|\le r-2$, then either D or $\overline{D}$ is not the color set of any vertex in X, where $\overline{D}=\{1,2,\dots ,r\}\setminus D$.

Otherwise, D and $\overline{D}$ are the color sets of vertices in X; then each nonempty subset in $2^D\cup 2^{\overline{D}}$ is not in $\mathcal{C}\left(Y\right)$. Since $2^r-m+1\le n+m\le 2^{r+1}-m$, there exist at most $m-2$ nonempty subsets of $\{1,2,\dots ,r\}$, which are not the color set of any vertices in $X\cup Y$. Therefore at least $2^t-1+2^{r-t}-1-(m-2)$ members in $2^D\cup 2^{\overline{D}}$ are contained in $\mathcal{C}\left(X\right)$, where $\left|D\right|=t$. We have $2^t-1+2^{r-t}-1-(m-2)\le m$.

If r is even, by Lemma 1, we have $2^{\frac{r}{2}+1}\le 2m$; i.e., $2^{\frac{r}{2}}\le m$. In addition, by Lemma 2 from $r\ge r_m$ we know that $r\ge 2\lceil {log}_2(m+1)\rceil$. Thus $\frac{r}{2}\ge \lceil {log}_2(m+1)\rceil$; i.e., $\frac{r}{2}\ge {log}_2(m+1)$ and we have $2^{\frac{r}{2}}\ge m+1$. This is a contradiction.

If r is odd, by Lemma 1, we have $2^{\frac{r-1}{2}}+2^{\frac{r+1}{2}}\le 2m$; i.e., $3·2^{\frac{r-1}{2}}\le 2m$. In addition, by Lemma 2 from $r\ge r_m$ we know that $r\ge 2\lceil {log}_2\frac{2m+1}{3}\rceil +1$. Thus $\frac{r-1}{2}\ge {log}_2\frac{2m+1}{3}$; i.e., $3·2^{\frac{r-1}{2}}\ge 2m+1$. This is a contradiction, too.

We consider the following two cases.

Case 1.

$\overline{\left\{1\right\}}$, $\overline{\left\{2\right\}}$, …, $\overline{\left\{r\right\}}$, $\overline{\varnothing }$ are in $\mathcal{C}\left(X\right)$.

Of course $r+1\le m$. If $2\le \left|C\right(x_i\left)\right|\le r-2$, then $\overline{C\left(x_i\right)}$ is not the color set of any vertex in $X\cup Y$ by Claim 1 and Claim 6. Thus, by Claim 3 and Claim 5, we have that $\mathcal{C}\left(Y\right)\subseteq \mathcal{A}$, where $\mathcal{A}$ is the collection of all 2-subsets, 3-subsets, …, $(r-2)$-subsets, except for all $C\left(x_i\right)$ and $\overline{C\left(x_i\right)}$ with $2\le \left|C\right(x_i\left)\right|\le r-2$. Therefore $n\le \left(\genfrac{}{}{0pt}{}{r}{2}\right)+\left(\genfrac{}{}{0pt}{}{r}{3}\right)+\dots +\left(\genfrac{}{}{0pt}{}{r}{r-2}\right)-2(m-r-1)$; i.e., $2^r-2m\ge n$. This is a contradiction.

Case 2.

At least one subset among all $(r-1)$-subsets and r-subsets is not the color set of any vertex in X.

If $r\le m$, then the complementary sets of the color sets of at least $m-r$ vertices in X (whose color set is not $(r-1)$-subset or r-subset) are not the color set of any vertex in $X\cup Y$. Thus $m+n\le 2^r-1-r-(m-r)=2^r-m-1$. This is a contradiction.

If $r=m+1$, then $\mathcal{C}(X\cup Y)\subseteq \mathcal{B}$, where $\mathcal{B}$ is the collection of all 2-subsets, 3-subsets, …, r-subsets. Thus $m+n\le 2^r-1-r=2^r-m-2$. This is a contradiction, too.

Thus $K_{m,n}$ does not have an r-GVDTC under the given conditions. The proof is completed. □

Theorem 4.

Let $r_m=min\{2\lceil {log}_2(m+1)\rceil ,2\lceil {log}_2\frac{2m+1}{3}\rceil +1\}$ and s be the minimum positive integer such that $2^s\ge 3m$. Suppose

$$a=\left\{\begin{array}{cc}{2}^{\frac{r}{2}+1}-m-2,\hfill & riseven;\hfill \\ 3·2^{\frac{r-1}{2}}-m-2,\hfill & risodd.\hfill \end{array}\right.$$

If (i) $s\le r<r_m$, $2^r-m\le n\le 2^{r+1}-2m$, or (ii) $r=s-1$, $max\{2^r-m,m\}\le n\le 2^{r+1}-2m$ or (iii) $a>0$, $s\le r<r_m$, $2^r-a-m\le n<2^r-m$, then ${\chi }_{gvt}\left(K_{m,n}\right)=r+1$.

Proof. 

By Lemma 3, we know that $K_{m,n}$ has an $(r+1)$-GVDTC when one of three conditions (i), (ii) or (iii) holds. When conditions (i) and (ii) are valid, we have that ${\chi }_{gvt}\left(K_{m,n}\right)\ge r+1$, and the results are right since all the nonempty subsets of the set composed by r colors cannot distinguish n vertices. So we suppose that condition (iii) holds in the following. By use of Lemma 2 and the definitions of a and $r_m$, we can deduce that $a<m$ from $r<r_m$. By Lemma 3, we need only to prove that $K_{m,n}$ does not have an r-GVDTC.

Assume that $K_{m,n}$ has an r-GVDTC g in which the available colors are $1,2,\dots ,r$. Firstly, we will give six claims.

Claim 1. 

$C\left(x_i\right)\cap C\left(y_j\right)\ne \varnothing$, $i=1,2,\dots ,m,j=1,2,\dots ,n$.

This is obvious because the color of edge $x_i{y}_j$ is in both $C\left(x_i\right)$ and $C\left(y_j\right)$.

Claim 2. 

$C\left(x_1\right)\cap C\left(x_2\right)\cap \dots \cap C\left(x_m\right)=\varnothing$.

Otherwise, say $1\in C\left(x_1\right)\cap C\left(x_2\right)\cap \dots \cap C\left(x_m\right)$. Then each $\overline{C\left(x_i\right)}$ is not the color set of any vertex and we have that $m+n\le 2^r-m<2^r-a$. This is a contradiction.

Claim 3. 

$\left|C\right(y_j\left)\right|\ge 2,j=1,2,\dots ,n$.

From Claim 2, we can deduce Claim 3 easily.

Claim 4. 

$C\left(y_1\right)\cap C\left(y_2\right)\cap \dots \cap C\left(y_n\right)=\varnothing$.

Otherwise, say $1\in C\left(y_1\right)\cap C\left(y_2\right)\cap \dots \cap C\left(y_n\right)$. Then each $\overline{C\left(y_j\right)}$ is not the color set of any vertex and we have that $m+n\le 2^r-n\le 2^r-m<2^r-a$. This is a contradiction.

Claim 5. 

$\left|C\right(x_i\left)\right|\ge 2,i=1,2,\dots ,m$.

This is deduced from Claim 4 directly.

Claim 6. 

If D is a subset of $\{1,2,\dots ,r\}$ with $2\le \left|D\right|\le r-2$, then either D or $\overline{D}$ is not the color set of vertex in X, where $\overline{D}=\{1,2,\dots ,r\}\setminus D$.

Otherwise, D and $\overline{D}$ are the color sets of vertices in X; then each nonempty subset in $2^D\cup 2^{\overline{D}}$ is not in $\mathcal{C}\left(Y\right)$. Since $2^r-a\le n+m<2^r$, there exist at most $a-1$ nonempty subsets of $\{1,2,\dots ,r\}$, which are not the color set of any vertex in $X\cup Y$. Therefore at least $2^t-1+2^{r-t}-1-(a-1)$ members in $2^D\cup 2^{\overline{D}}$ are contained in $\mathcal{C}\left(X\right)$, where $\left|D\right|=t$. We have $2^t-1+2^{r-t}-1-(a-1)\le m$.

If r is even, by Lemma 1, we have $2·2^{\frac{r}{2}}-1\le m+a$; i.e., $2·2^{\frac{r}{2}}-1\le m+2^{\frac{r}{2}+1}-m-2$. This is a contradiction.

If r is odd, by Lemma 1, we have $2^{\frac{r-1}{2}}+2^{\frac{r+1}{2}}-1\le m+a$; i.e., $2^{\frac{r-1}{2}}+2^{\frac{r+1}{2}}-1\le m+3·2^{\frac{r-1}{2}}-m-2$. This is a contradiction, too.

The proof of Claim 6 is completed.

Finally we consider the following two cases.

Case 1.

$\overline{\left\{1\right\}}$, $\overline{\left\{2\right\}}$, …, $\overline{\left\{r\right\}}$, $\overline{\varnothing }$ are in $\mathcal{C}\left(X\right)$.

If $2\le \left|C\right(x_i\left)\right|\le r-2$, then $\overline{C\left(x_i\right)}$ is not the color set of any vertex in $X\cup Y$ by Claim 6. Let $\mathcal{C}\left(X\right)\setminus \{$$\overline{\left\{1\right\}}$, $\overline{\left\{2\right\}}$, …, $\overline{\left\{r\right\}}$, $\overline{\varnothing }$$\}=\{A_1,A_2,\dots ,A_{m-r-1}\}$. Then, by Claim 3 and Claim 5, we have

$$\mathcal{C}(X\cup Y)\subseteq 2^{\{1,2,\dots ,r\}}\setminus \{\left\{1\right\},\left\{2\right\},\dots ,\left\{r\right\},\varnothing ,\overline{A_1},\overline{A_2},\dots ,\overline{A_{m-r-1}}\}.$$

Thus $m+n\le 2^r-r-1-(m-r-1)$; i.e., $m+n\le 2^r-m<2^r-a$. This is a contradiction.

Case 2.

At least one subset among all $(r-1)$-subsets and r-subsets is not the color set of vertices in X.

Let $\mathcal{C}\left(X\right)\setminus \{$$\overline{\left\{1\right\}}$, $\overline{\left\{2\right\}}$, …, $\overline{\left\{r\right\}}$, $\overline{\varnothing }$$\}=\{A_1,A_2,\dots ,A_{\alpha }\}$ with $\alpha \ge m-r$. Then we have

$$\mathcal{C}(X\cup Y)\subseteq 2^{\{1,2,\dots ,r\}}\setminus \{\left\{1\right\},\left\{2\right\},\dots ,\left\{r\right\},\varnothing ,\overline{A_1},\overline{A_2},\dots ,\overline{A_{\alpha }}\}.$$

Thus $m+n\le 2^r-r-1-\alpha \le 2^r-r-1-(m-r)$; i.e., $m+n\le 2^r-m-1<2^r-a-1$. This is a contradiction.

Thus $K_{m,n}$ does not have an r-GVDTC under the given conditions. The proof is completed. □

4. An Application to the Case $\mathbf{m}=\mathbf{10}$

By Theorems 1–4, we can determine the ${\chi }_{gvt}\left(K_{10,n}\right)$ when $n\ge 4086$, $4077\le n\le 4085$, $109\le n\le 4076$ and $50\le n\le 108$, $22\le n\le 44$, respectively. From Theorem 4 (when condition (iii) holds) we can know that ${\chi }_{gvt}\left(K_{10,n}\right)=5$ when $n=10,11$ and 12. We will discuss other cases in the following.

Theorem 5.

${\chi }_{gvt}\left(K_{10,13}\right)=5$.

Proof. 

Obviously, we have ${\chi }_{gvt}\left(K_{10,13}\right)\ge 5$. So we need only to give a 5-GVDTC of $K_{10,13}$ in which the available colors are $1,2,3,4,5$. Let

$$D\left(x_1\right)=\{1,2,3\},D\left(x_2\right)=\{4,5\},D\left(x_3\right)=\{2,4,5\},D\left(x_4\right)=\{3,4,5\},$$

$$D\left(x_5\right)=\{2,3,4,5\},D\left(x_6\right)=\{1,3,4,5\},D\left(x_7\right)=\{1,2,4,5\},$$

$$D\left(x_8\right)=\{1,2,3,5\},D\left(x_9\right)=\{1,2,3,4\},D\left(x_{10}\right)=\{1,2,3,4,5\};$$

$$D\left(y_1\right)=\{1,4\},D\left(y_2\right)=\{1,5\},D\left(y_3\right)=\{2,4\},D\left(y_4\right)=\{2,5\},$$

$$D\left(y_5\right)=\{3,4\},D\left(y_6\right)=\{3,5\},D\left(y_7\right)=\{2,3,5\},$$

$$D\left(y_8\right)=\{2,3,4\},D\left(y_9\right)=\{1,3,5\},D\left(y_{10}\right)=\{1,3,4\},$$

$$D\left(y_{11}\right)=\{1,2,5\},D\left(y_{12}\right)=\{1,2,4\},D\left(y_{13}\right)=\{1,4,5\}.$$

We can give a 5-GVDTC of $K_{10,13}$ easily such that $C\left(x_i\right)=D\left(x_i\right)$, $i=1,2,\dots ,10;$$C\left(y_j\right)=D\left(y_j\right)$, $j=1,2,\dots ,13$ (see Table 2). □

Table 2. A 5-GVDTC of $K_{10,13}$.

	${\mathit{y}}_1$; 4	${\mathit{y}}_2$; 5	${\mathit{y}}_3$; 4	${\mathit{y}}_4$; 5	${\mathit{y}}_5$; 4	${\mathit{y}}_6$; 5	${\mathit{y}}_7$; 5	${\mathit{y}}_8$; 4	${\mathit{y}}_9$;5	${\mathit{y}}_{10}$; 4	${\mathit{y}}_{11}$; 5	${\mathit{y}}_{12}$; 4	${\mathit{y}}_{13}$; 5
$x_1$; 3	1	1	2	2	3	3	2	2	1	1	1	1	1
$x_2$; 5	4	5	4	5	4	5	5	4	5	4	5	4	5
$x_3$; 5	4	5	2	2	4	5	2	2	5	4	2	2	5
$x_4$; 5	4	5	4	5	3	3	3	3	5	4	5	4	5
$x_5$; 5	4	5	2	2	3	3	3	3	5	4	5	4	5
$x_6$; 5	1	1	4	5	3	3	3	3	5	4	5	4	5
$x_7$; 5	1	1	2	2	4	5	2	2	5	4	5	4	5
$x_8$; 5	1	1	2	2	3	5	5	3	5	3	5	2	5
$x_9$; 4	1	1	4	2	4	3	2	2	3	3	2	2	1
$x_{10}$; 5	1	1	4	2	4	3	2	2	3	3	2	2	1

Theorem 6.

If $14\le n\le 21$, then ${\chi }_{gvt}\left(K_{10,n}\right)=6$.

Proof. 

By Lemma 3, we know that $K_{10,n}$ has 6-GVDTC when $14\le n\le 21$. So we need only to show that $K_{10,n}$ does not have any 5-GVDTC when $14\le n\le 21$.

Assume that $K_{10,n}$ has a 5-GVDTC g when $14\le n\le 21$.

Claim 1. 

$\left|C\right(x_i\left)\right|\ge 2$, $i=1,2,\dots ,10$.

Otherwise, without loss of generality, we suppose that $C\left(x_1\right)=\left\{1\right\}$. Then each $\overline{C\left(y_j\right)}$ is not the color set of any vertex and $10+n\le 2^5-n$. This is a contradiction.

Claim 2. 

$\left|C\right(y_j\left)\right|\ge 2$, $j=1,2,\dots ,n$.

Otherwise, without loss of generality, we suppose that $C\left(y_1\right)=\left\{1\right\}$. Then each $\overline{C\left(x_i\right)}$ is not the color set of any vertex and $10+n\le 2^5-10$. This is a contradiction.

From Claim 1 and Claim 2, we have the following claim.

Claim 3. 

$\mathcal{C}(X\cup Y)\subseteq \mathcal{A}$ and $14\le n\le 16$, where $\mathcal{A}$ is the collection of all subsets (of $\{1,2,3,4,5\}$) that contains at least 2 elements.

All 2-subsets of $\{1,2,3,4,5\}$ except for at most two 2-subsets $A,B$ are in $\mathcal{C}(X\cup Y)$.

We consider two cases.

Case 1.

$A\cap B\ne \varnothing$. We may suppose $A=\{1,2\},B=\{1,3\}$. Since the color set of each vertex in X is not disjoint with the color set of any vertex in Y, we have that $\{1,4\}$, $\{1,5\}$, $\{2,3\}$, $\{2,4\}$, $\{2,5\}$, $\{3,4\}$, $\{3,5\}$, $\{4,5\}$ are the color sets of vertices in one part. Thus the color set of each vertex in another part is one of the following sets:

$$\{2,4,5\},\{3,4,5\},\overline{\left\{1\right\}},\overline{\left\{2\right\}},\overline{\left\{3\right\}},\overline{\left\{4\right\}},\overline{\left\{5\right\}},\{1,2,3,4,5\}.$$

This is a contradiction.

Case 2.

$A\cap B=\varnothing$. We may suppose $A=\{1,2\},B=\{3,4\}$. Since the color set of each vertex in X is not disjoint with the color set of any vertex in Y, we have that $\{1,3\}$, $\{1,4\}$, $\{1,5\}$, $\{2,3\}$, $\{2,4\}$, $\{2,5\}$, $\{3,5\}$, $\{4,5\}$ are the color sets of vertices in one part. Thus the color set of each vertex in another part is one of the following sets:

$$\{3,4,5\},\{1,2,5\},\overline{\left\{1\right\}},\overline{\left\{2\right\}},\overline{\left\{3\right\}},\overline{\left\{4\right\}},\overline{\left\{5\right\}},\{1,2,3,4,5\}.$$

This is a contradiction. □

Theorem 7.

If $45\le n\le 49$, then ${\chi }_{gvt}\left(K_{10,n}\right)=7$.

Proof. 

By Lemma 3, we know that $K_{10,n}$ has a 7-GVDTC when $45\le n\le 49$. So we need only to show that $K_{10,n}$ does not have any 6-GVDTC when $45\le n\le 49$.

Assume that $K_{10,n}$ has a 6-GVDTC g when $45\le n\le 49$.

Claim 1. 

$\left|C\right(x_i\left)\right|\ge 2$, $i=1,2,\dots ,10$.

Otherwise, without loss of generality, we suppose that $C\left(x_1\right)=\left\{1\right\}$. Then each $C\left(y_j\right)$ contains color 1. But $\{1,2,3,4,5,6\}$ has at most 31 subsets that contain 1 and have at least two elements. This is a contradiction.

Claim 2. 

$\left|C\right(y_j\left)\right|\ge 2$, $j=1,2,\dots ,n$.

Otherwise, without loss of generality, we suppose that $C\left(y_1\right)=\left\{1\right\}$. Then each $\overline{C\left(x_i\right)}$ is not the color set of any vertex and $10+n\le 2^6-10$. This is a contradiction.

From Claim 1 and Claim 2, we have the following claim.

Claim 3. 

$\mathcal{C}(X\cup Y)\subseteq \mathcal{A}$ and $45\le n\le 47$, where $\mathcal{A}$ is the collection of all subsets ( of $\{1,2,3,4,5,6\}$ ) that contains at least two elements.

All 2-subsets of $\{1,2,3,4,5,6\}$ except for at most two 2-subsets $A,B$ are in $\mathcal{C}(X\cup Y)$.

We consider two cases.

Case 1.

$A\cap B\ne \varnothing$. We may suppose $A=\{1,2\},B=\{1,3\}$. We can deduce that other 2-subsets are the color sets of vertices in one part. Thus the color set of each vertex in another part is one of the following sets:

$$\{2,4,5,6\},\{3,4,5,6\},\overline{\left\{1\right\}},\overline{\left\{2\right\}},\overline{\left\{3\right\}},\overline{\left\{4\right\}},\overline{\left\{5\right\}},\overline{\left\{6\right\}},\{1,2,3,4,5,6\}.$$

This is a contradiction.

Case 2.

$A\cap B=\varnothing$. We may suppose $A=\{1,2\},B=\{3,4\}$. We can deduce that other 2-subsets are the color sets of vertices in one part. Thus the color set of each vertex in another part is one of the following sets:

$$\{3,4,5,6\},\{1,2,5,6\},\overline{\left\{1\right\}},\overline{\left\{2\right\}},\overline{\left\{3\right\}},\overline{\left\{4\right\}},\overline{\left\{5\right\}},\overline{\left\{6\right\}},\{1,2,3,4,5,6\}.$$

This is a contradiction. □

5. A Conjecture

It is easy to know that Conjecture 2 is right for the bipartite graphs mentioned in Theorems 1–4 and $K_{10,n}(n\ge 10)$.

By simple calculation we can obtain that ${\chi }_{gvt}=\xi$ or ${\chi }_{gvt}=\xi +1$ for the above graphs. For example, for the graph $K_{m,n}$ in Theorem 4, we have ${\chi }_{gvt}\left(K_{m,n}\right)=r+1.$ If conditions (i) and (ii) hold, then $\xi \left(K_{m,n}\right)=r+1={\chi }_{gvt}\left(K_{m,n}\right)$; i.e., ${\chi }_{gvt}\left(K_{m,n}\right)=\xi \left(K_{m,n}\right)$. When condition (iii) holds, we get $\xi \left(K_{m,n}\right)=r={\chi }_{gvt}\left(K_{m,n}\right)-1$; i.e., ${\chi }_{gvt}\left(K_{m,n}\right)=\xi \left(K_{m,n}\right)+1$.

So we propose the following Vizing-like conjecture.

Conjecture 3. 

If G is a simple graph, then ${\chi }_{gvt}\left(G\right)=\xi \left(G\right)$ or ${\chi }_{gvt}\left(G\right)=\xi \left(G\right)+1$.

6. Conclusions

In this paper we have investigated the general vertex-distinguishing total colorings of complete bipartite graphs and determined the general vertex-distinguishing total chromatic numbers for many complete bipartite graphs (Theorems 1–4). Furthermore, as an application, we have given the general vertex-distinguishing total chromatic number of complete bipartite graphs $K_{10,n}$. What is more, our results show that Conjecture 2 holds for the complete bipartite graphs considered in this paper, and we have proposed a Vizing-like conjecture (Conjecture 3).