© Wuhan University 2024

0 Introduction

Many conclusions about the vertex-distinguishing proper edge coloring ^[1-3] and vertex-distinguishing general edge coloring ^[4-7] of graphs have been made in simple undirected graphs. The vertex-distinguishing total coloring of graphs was proposed by Zhang et al^[8] in 2008. In 2011, Chen et al^[9] introduced vertex-distinguishing E-total colorings of graphs. Liu et al^[10] studied general vertex-distinguishing total coloring of graphs. In 2019, Chen et al^[11] gave a survey on not necessarily proper coloring under which certain pairs of vertices are distinguished by non-multiple color sets. Cao et al^[12] discussed the E-total coloring of wheels and fans, which were vertex-distinguished by multiple sets. Wang et al^[13] discussed the I-total coloring and VI-total coloring of mC₄ vertex-distinguished by multiple sets. In this paper, we study E-total coloring of complete bipartite graphs K_5,n (5≤n≤7 113) which are vertex-distinguished by multiple sets and determine E-total chromatic numbers of K_5,n (5≤n≤7 113) vertex-distinguished by multiple sets.

1 Preliminaries

An E-total coloring of a graph G is an assignment of several colors to the vertices and edges of the graph G, such that any adjacent vertices are assigned different colors, and each edge and its incident vertices are assigned different colors.

Let f be an E-total coloring of a graph G and x be any vertex of G. The (multiple) set of the colors of x and the edges incident with x under f is called the (multiple) chromatic set of x under f, denoted by C̃_f(x) or C̃(x).

Let f be an E-total coloring of a graph G, and for any u, v∈V(G), once u≠v, there exists C̃(u)≠C̃(v), then f is said to be vertex-distinguishable by multiple sets.

A k-E-total coloring of a graph G is an E-total coloring of G using k available colors.

Let ${{\tilde{\chi }}^e}_{vt}(G)$=min{k|there exists k-E-total coloring of G which is vertex-distinguished by multiple sets} and we call ${{\tilde{\chi }}^e}_{vt}(G)$ the E-total chromatic number of G which are vertex-distinguished by multiple sets.

In this study, we use K_m,n to represent complete bipartite graphs, where

V(K_m,n)={u₁,u₂,…,u_m,v₁,v₂,…,v_n}=X∪Y; X={u₁,u₂,…,u_m}; Y={ v₁,v₂,…,v_n}; E(K_m,n)={u_iv_j│i=1,2,…,m; j=1,2,…,n}.

The mapping h: V ∪ E→{1,2,…,k} is defined. Usually, when coloring, the k colors used are represented by {1,2,…,k}.

Lemma 1^[14] The number of repeated combinations of r elements taken from n different elements is $\left(\begin{array}{c}n+r-\mathrm{1}\\ r\end{array}\right)$.

The repeated combination of r elements from n different elements is called r-combination. r-combination is also a multiple subset containing r elements of the set of n different elements; therefore, an r-combination is also called an r-multiple subset or an r-subset.

$\mathbf{L}\mathbf{e}\mathbf{m}\mathbf{m}\mathbf{a}{\mathrm{2}}^{[\mathrm{14}]}\left(\begin{array}{c}n\\ r\end{array}\right)=\left(\begin{array}{c}n-\mathrm{1}\\ r\end{array}\right)+\left(\begin{array}{c}n-\mathrm{1}\\ r-\mathrm{1}\end{array}\right).$

2 Main Results and Their Proofs

Theorem 1   If $\mathrm{5}\le n\le \mathrm{22}$, then ${\tilde{\chi }}_{vt}^e(K_{\mathrm{5},n})=\mathrm{4}$.

Proof   Firstly, we will prove that there is no 3-E-total coloring of K_5,n, which are vertex-distinguished by multiple sets.Suppose that K_5,n has an E-total coloring h in which there are 3 available colors, and the vertices are distinguished by multiple sets. The color set of the vertex in Y under h is a 6-subset of {1,2,3} and "at least one color appears only once in this 6-subset". Such 6-subsets may be {1,2,3,3,3,3}, {1,3,2,2,2,2}, {2,3,1,1,1,1}, {1,2,2,3,3,3},{1,3,3,2,2,2}, {2,3,3,1,1,1}, {2,1,1,3,3,3}, {3,2,2,1,1,1}, {3,1,1,2,2,2}, {1,2,2,2,2,2}, {1,3,3,3,3,3},{2,1,1,1,1,1}, {2,3,3,3,3,3}, {3,1,1,1,1,1}, {3,2,2,2,2,2}, the following 5 cases are discussed.

Case 1: u₁, u₂, u₃, u₄, and u₅ are assigned the same color under h.

Without loss of generality, we assume that h(u₁)=h(u₂)=h(u₃)=h(u₄)=h(u₅)=1. At this time, all edges of K_5,n and all vertices in Y cannot be colored 1. Therefore, the 6-subset of {1,2,3} of the color set of each vertex in Y under h, and "at least one color appears only once in this 6-subset". Such 6-subsets are only {2,3,3,3,3,3}, {3,2,2,2,2,2}. Due to 5≤n≤22, these two 6-subsets cannot distinguish the n vertices in Y, which is a contradiction.

Case 2: u₁, u₂, u₃, u₄, and u₅ are assigned two different colors under h, and four vertices in X have the same color.

Without loss of generality, we assume that h(u₁)=h(u₂)=h(u₃)=h(u₄)=1, h(u₅)=2. At this time, all vertices in Y cannot be colored with colors 1 and 2. Therefore, the 6-subset of {1,2,3} of the color set of each vertex in Y under h, and "at least one color appears only once in this 6-subset". Such a 6-subset is only {3,1,2,2,2,2}. Due to 5≤n≤22, this 6-subset cannot distinguish n vertices in Y, which is a contradiction.

Case 3: u₁, u₂, u₃, u₄, and u₅ are assigned two different colors under h, and three vertices in X have the same color, which is different from the colors of the other two vertices.

Without loss of generality, we assume that h(u₁)=h(u₂)=h(u₃)=1, h(u₄)=h(u₅)=2. At this time, all vertices in Y cannot be colored with colors 1 and 2. Therefore, the 6-subset of {1,2,3} of the color set of each vertex in Y under h, and "at least one color appears only once in this 6-subset". Such a 6-subset is only {3,1,1,2,2,2}. Due to 5≤n≤22, this 6-subset cannot distinguish n vertices in Y, which is a contradiction.

Case 4: u₁, u₂, u₃, u₄, and u₅ are assigned three different colors under h, and the colors of the three vertices in X are the same.

Without loss of generality, we assume that h(u₁)=h(u₂)=h(u₃)=1, h(u₄)=2, h(u₅)=3. At this time, all vertices in Y cannot be colored with colors 1, 2 and 3. Then, the 6-subset of {1,2,3} of the color set of each vertex in Y under h, and "at least one color appears only once in this 6-subset". Since a 6-subset does not exist, therefore, such a 6-subset cannot distinguish n vertices in Y, which is a contradiction.

Case 5: u₁, u₂, u₃, u₄ and u₅ are assigned three different colors under h, and there are no three vertices in X with the same color.

Without loss of generality, we assume that h(u₁)=h(u₂)=1, h(u₃)=h(u₄)=2, h(u₅)=3. At this time, the three colors cannot distinguish the vertices in K_5,n, which is a contradiction.

Secondly, we will give, for each n $\in \left\{\mathrm{5},\mathrm{6},\cdots ,\mathrm{22}\right\}$, there exists a 4-E-total coloring of K_5,n which is vertex-distinguished by multiple sets. It is advisable to color the vertices u₁, u₂, u₃, u₄ and u₅ in X with colors 1, 1, 1, 2, 2, respectively. The following coloring scheme is given using a matrix of 6 rows and 23 columns.

$\left(\begin{array}{ccccccccccccccccccccccc}\mathrm{0}& \mathrm{3}& \mathrm{4}& \mathrm{3}& \mathrm{4}& \mathrm{3}& \mathrm{3}& \mathrm{4}& \mathrm{3}& \mathrm{3}& \mathrm{4}& \mathrm{3}& \mathrm{4}& \mathrm{3}& \mathrm{3}& \mathrm{3}& \mathrm{3}& \mathrm{4}& \mathrm{4}& \mathrm{4}& \mathrm{4}& \mathrm{3}& \mathrm{4}\\ \mathrm{1}& \mathrm{2}& \mathrm{2}& \mathrm{2}& \mathrm{2}& \mathrm{2}& \mathrm{2}& \mathrm{2}& \mathrm{2}& \mathrm{4}& \mathrm{3}& \mathrm{2}& \mathrm{2}& \mathrm{2}& \mathrm{4}& \mathrm{2}& \mathrm{2}& \mathrm{2}& \mathrm{3}& \mathrm{2}& \mathrm{2}& \mathrm{4}& \mathrm{3}\\ \mathrm{1}& \mathrm{2}& \mathrm{2}& \mathrm{4}& \mathrm{3}& \mathrm{2}& \mathrm{4}& \mathrm{3}& \mathrm{2}& \mathrm{4}& \mathrm{3}& \mathrm{4}& \mathrm{3}& \mathrm{2}& \mathrm{4}& \mathrm{2}& \mathrm{2}& \mathrm{2}& \mathrm{3}& \mathrm{2}& \mathrm{2}& \mathrm{4}& \mathrm{3}\\ \mathrm{1}& \mathrm{4}& \mathrm{3}& \mathrm{4}& \mathrm{3}& \mathrm{4}& \mathrm{4}& \mathrm{3}& \mathrm{2}& \mathrm{4}& \mathrm{3}& \mathrm{4}& \mathrm{3}& \mathrm{2}& \mathrm{4}& \mathrm{4}& \mathrm{2}& \mathrm{2}& \mathrm{3}& \mathrm{3}& \mathrm{2}& \mathrm{4}& \mathrm{3}\\ \mathrm{2}& \mathrm{4}& \mathrm{3}& \mathrm{1}& \mathrm{1}& \mathrm{1}& \mathrm{4}& \mathrm{3}& \mathrm{4}& \mathrm{4}& \mathrm{3}& \mathrm{4}& \mathrm{3}& \mathrm{1}& \mathrm{1}& \mathrm{4}& \mathrm{4}& \mathrm{1}& \mathrm{1}& \mathrm{3}& \mathrm{3}& \mathrm{4}& \mathrm{3}\\ \mathrm{2}& \mathrm{1}& \mathrm{1}& \mathrm{1}& \mathrm{1}& \mathrm{1}& \mathrm{1}& \mathrm{1}& \mathrm{1}& \mathrm{1}& \mathrm{1}& \mathrm{4}& \mathrm{3}& \mathrm{1}& \mathrm{1}& \mathrm{4}& \mathrm{4}& \mathrm{1}& \mathrm{1}& \mathrm{3}& \mathrm{3}& \mathrm{4}& \mathrm{3}\end{array}\right)$

The first-row elements (color) of the matrix except 0 are assigned to the vertices v₁, v₂, …, v₂₂ in Y. The first column elements (color) of the matrix except 0 are assigned to the vertices u₁, u₂, u₃, u₄, and u₅ in X. The elements (color) at the intersection of the i+1 row and the j+1 column are assigned to the edges u_iv_j, i=1, 2, 3, 4, 5; j=1, 2, …, 22, then the second row, the third row, the fourth row, the fifth row, and the sixth row are exactly the color sets corresponding to the vertices u₁, u₂, u₃, u₄ and u₅ in X. The second column, the third column, …, and the 23rd column are exactly the color sets corresponding to the vertices v₁, v₂, …, v₂₂ in Y.

It can be obtained from the matrix, when n=5, the color sets of vertices in K_5,5 under f are 6-subsets. The 6-subsets of vertices in X under f are {1,2,2,2,2,2}, {1,2,2,4,3,2}, {1,4,3,4,3,4}, {2,4,3,1,1,1}, {2,1,1,1,1,1}. The 6-subsets of vertices in Y under f are {3,2,2,4,4,1}, {4,2,2,3,3,1}, {3,2,4,4,1,1}, {4,2,3,3,1,1}, {3,2,2,4,1,1}. All 6-subsets in X are different from all 6-subsets in Y under f, so ${{\tilde{\chi }}^e}_{vt}(K_{\mathrm{5,5}})=\mathrm{4}$. When 6≤n≤22, the color sets of vertices in X under f are (n+1)-subsets. The color sets corresponding to the vertices in Y are all 6-subsets. Therefore, the color sets corresponding to the vertices in X under f are different from those corresponding to the vertices in Y under f. According to the specific coloring scheme given by the matrix, the vertices and all edges in K_5,n are colored. It can be obtained that the corresponding color sets of vertices in X under f are different from each other. The corresponding color sets of vertices in Y under f are different from each other, so ${{\tilde{\chi }}^e}_{vt}(K_{\mathrm{5},n})=\mathrm{4}$.

$\mathbf{T}\mathbf{h}\mathbf{e}\mathbf{o}\mathbf{r}\mathbf{e}\mathbf{m}\mathrm{2}\mathrm{I}\mathrm{f}\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}<n\le \left(\begin{array}{c}k+\mathrm{4}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{1}\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{4},\mathrm{5}\le k\le \mathrm{11},\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{n}{\tilde{\chi }}_{vt}^e(K_{\mathrm{5},n})=k.$

Proof   Firstly, we will prove that there is no (k–1)-E-total coloring of K_5,n which are vertex-distinguished by multiple sets.Suppose that K_5,n has an E-total coloring h in which there are k–1 available colors, and the vertices are distinguished by multiple sets; the following 7 cases are discussed.

Case 1: u₁, u₂, u₃, u₄ and u₅ are assigned the same color under h.

Without loss of generality, we assume that h(u₁)=h(u₂)=h(u₃)=h(u₄)=h(u₅)=1. At this time, all edges of K_5,n and all vertices in Y cannot be colored 1. The color set of each vertex in Y under h is a 6-subset of {2,3,…,k–1}, and "at least one color appears only once in this 6-subset". There are

$\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{6}\end{array}\right)+\mathrm{5}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{5}\end{array}\right)+\mathrm{6}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{4}\end{array}\right)+\mathrm{4}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{4}\end{array}\right)+\mathrm{3}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{3}\end{array}\right)+\mathrm{6}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{3}\end{array}\right)+\mathrm{2}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)=\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{1}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{3}\end{array}\right)$

such 6-subsets. When $\mathrm{5}\le k\le \mathrm{11}$,

$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}-\left[\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{1}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{3}\end{array}\right)\right]=\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{1}\\ \mathrm{4}\end{array}\right)-\mathrm{7}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}>\mathrm{0},$

then

$\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{1}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{3}\end{array}\right)<\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}<n.$

Therefore, all such 6-subsets cannot distinguish n vertices in X in Y, which is a contradiction.

Case 2: u₁, u₂, u₃, u₄ and u₅ are assigned two different colors under h, and four vertices have the same color.

Without loss of generality, we assume that h(u₁)= h(u₂)= h(u₃)= h(u₄)=1, h(u₅)=2. At this time, for each s, t, r, p∈{3,4,…,k–1}, {s,t,r,p,1,1}, {s,t,r,2,1,1}, {s,t,r,r,1,1}, {s,t,2,2,1,1}, {s,2,t,t,1,1}, {1,2,s,s,t,t}, {s,t,r,1,1,1}, {s,t,2,1,1,1}, {1,2,s,s,s,s}, {s,t,1,1,1,1}, {s,2,1,1,1,1}, {s,t,t,1,1,1}, {s,2,2,1,1,1}, {1,2,2,s,s,s}, {2,1,1,s,s,s}, {s,1,1,t,t,t}, {s,1,1,2,2,2}, {1,s,s,2,2,2}, {1,s,s,t,t,t}, {2,s,s,1,1,1}, {2,s,s,t,t,t}, {s,1,1,1,1,1}, {s,2,2,2,2,2}, {1,s,s,s,s,s}, {2,s,s,s,s,s}, {1,2,2,2,2,2}, {2,1,1,1,1,1} cannot be the color sets of vertices in Y, and the number of such color sets is $\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\mathrm{2}\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)+\mathrm{7}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)+\mathrm{2}k-\mathrm{4}$. Then there are$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\left[\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\mathrm{2}\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)+\mathrm{7}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)+\mathrm{2}k-\mathrm{4}\right]=\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)-\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{7}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{2}k+\mathrm{4}$ color sets, which can become the color sets of the vertices in Y. When $\mathrm{5}\le k\le \mathrm{11}$,

$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}-\left[\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)-\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{7}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{2}k+\mathrm{4}\right]=\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{3}\end{array}\right)-k+\mathrm{3}>\mathrm{0},$

then$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)-\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{7}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{2}k+\mathrm{4}<\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}<n.$Therefore, all such 6-subsets cannot distinguish n vertices in Y, which is a contradiction.

Case 3: u₁, u₂, u₃, u₄, and u₅ are assigned two different colors under h, and three vertices in X have the same color, which is different from the colors of the other two vertices in X .

Without loss of generality, we assume that h(u₁)= h(u₂)= h(u₃)=1, h(u₄)=h(u₅)=2. At this time, for each s, t, r ∈{3,4,…,k–1}, {1,2,s,s,t,t}, {s,t,r,1,1,1}, {s,t,2,1,1,1}, {1,2,s,s,s,s}, {s,1,2,2,2,2}, {s,t,2,2,2,2}, {s,2,1,1,1,1}, {s,t,1,1,1,1}, {1,s,s,t,t,t}, {2,s,s,t,t,t}, {1,2,2,s,s,s}, {2,1,1,s,s,s}, {1,s,s,2,2,2}, {2,s,s,1,1,1}, {s,t,t,1,1,1}, {s,2,2,1,1,1}, {1,2,2,2,2,2}, {1,s,s,s,s,s}, {2,1,1,1,1,1}, {2,s,s,s,s,s}, {s,1,1,1,1,1}, {s,2,2,2,2,2} cannot be the color sets of vertices in Y, and the number of such color sets is $\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)+\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)+\mathrm{3}k-\mathrm{7}.\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{e}\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\left[\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)+\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)+\mathrm{3}k-\mathrm{7}\right]=\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}$color sets, which can become the color sets of the vertices of Y. When $\mathrm{5}\le k\le \mathrm{11}$,$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}<n.$ Hence, all such 6-subsets cannot distinguish the n vertices in Y, which results in a contradiction.

Case 4: u₁, u₂, u₃, u_4, and u₅ are assigned three different colors under h, and the colors of the three vertices in X are the same.

Without loss of generality, we assume that h(u₁)= h(u₂)= h(u₃)=1, h(u₄)=2, h(u₅)=3. At this time, for each s, t ∈{4,5,…,k–1}, i, j, x ∈{1,2,3}, {i,j,s,s,t,t}, {i,j,x,x,s,s}, {s,3,3,1,1,1}, {s,2,2,1,1,1}, {s,t,t,1,1,1}, {s,t,r,1,1,1}, {s,t,2,1,1,1}, {s,t,3,1,1,1}, {s,2,3,1,1,1}, {i,j,x,s,s,s}, {i,j,s,s,s,s}, {i,j,x,x,x,x}, {s,t,1,1,1,1}, {s,2,1,1,1,1}, {s,3,1,1,1,1}, {i,s,s,t,t,t}, {i,j,j,x,x,x}, {i,j,j,s,s,s}, {i,s,s,j,j,j}, {i,j,j,j,j,j}, {i,s,s,s,s,s}, {s,i,i,i,i,i} cannot be the color sets of vertices in Y, and the number of such color sets is$\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{3}\end{array}\right)+\mathrm{12}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)+\mathrm{17}k-\mathrm{53}$. Then there are$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\left[\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{3}\end{array}\right)+\mathrm{12}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)+\mathrm{17}k-\mathrm{53}\right]=\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{11}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)-\mathrm{16}k+\mathrm{50}$color sets, which can become the color sets of the vertices of Y. When $\mathrm{5}\le k\le \mathrm{11}$,

$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}-\left[\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{11}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)-\mathrm{16}k+\mathrm{50}\right]=\mathrm{3}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)+\mathrm{5}k-\mathrm{19}>\mathrm{0}.$

Then$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{11}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)-\mathrm{16}k+\mathrm{50}<\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}<n.$Therefore, all such 6-subsets cannot distinguish n vertices in Y, which is a contradiction.

Case 5: u₁, u₂, u₃, u₄, and u₅ are assigned three different colors under h, and there are no three vertices in X with the same color.

Without loss of generality, we assume that h(u₁)= h(u₂)=1, h(u₃)=h(u₄)=2, h(u₅)=3. At this time, for each s, t ∈{4,5,…,k–1}, i, j, x ∈{1,2,3}, {i,j,s,s,t,t}, {i,j,x,x,s,s}, {i,j,x,s,s,s}, {i,j,s,s,s,s}, {i,j,x,x,x,x}, {s,t,1,1,1,1}, {s,t,2,2,2,2}, {s,2,1,1,1,1}, {s,1,2,2,2,2}, {s,3,1,1,1,1}, {s,3,2,2,2,2}, {i,j,j,x,x,x}, {i,s,s,t,t,t}, {i,j,j,s,s,s}, {i,s,s,j,j,j}, {i,j,j,j,j,j}, {i,s,s,s,s,s}, {s,i,i,i,i,i} cannot be the color sets of vertices in Y, and the number of such color sets is $\mathrm{11}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)+\mathrm{18}k-\mathrm{57}$. Then there are $\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\left[\mathrm{11}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)+\mathrm{18}k-\mathrm{57}\right]=\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{11}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)-\mathrm{18}k+\mathrm{57}$ color sets, which can become the color sets of the vertices of Y. When $\mathrm{5}\le k\le \mathrm{11}$,

$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}-\left[\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{11}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)-\mathrm{18}k+\mathrm{57}\right]=\mathrm{3}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)-\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)+\mathrm{7}k-\mathrm{26}>\mathrm{0}.$

Then

$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{11}\left(\begin{array}{c}k-\mathrm{3}\\ \mathrm{2}\end{array}\right)-\mathrm{18}k+\mathrm{57}<\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}<n.$

Therefore, all such 6-subsets cannot distinguish n vertices in Y, which is a contradiction.

Case 6: u₁, u₂, u₃, u₄ and u₅ are assigned four different colors under h.

Without loss of generality, we assume that h(u₁)= h(u₂)=1, h(u₃)=2,h(u₄)=3, h(u₅)=4. At this time, for each s, t ∈{5,6,…,k–1}, i, j, x, y ∈{1,2,3,4}, {i,j,x,y,s,s}, {i,j,s,s,t,t}, {i,j,x,x,y,y}, {i,j,x,x,s,s}, {i,j,x,s,s,s}, {i,j,x,y,y,y}, {i,j,x,x,x,x}, {i,j,s,s,s,s}, {s,t,1,1,1,1}, {s,2,1,1,1,1}, {s,3,1,1,1,1}, {s,4,1,1,1,1}, {i,j,j,x,x,x}, {i,j,j,s,s,s}, {i,s,s,j,j,j}, {i,s,s,t,t,t}, {i,j,j,j,j,j}, {i,s,s,s,s,s}, {s,i,i,i,i,i} cannot be the color sets of vertices in Y, and the number of such color sets is $\mathrm{15}\left(\begin{array}{c}k-\mathrm{4}\\ \mathrm{2}\end{array}\right)+\mathrm{43}k-\mathrm{157}$. Then there are

$\begin{array}{l}\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\left[\mathrm{15}\left(\begin{array}{c}k-\mathrm{4}\\ \mathrm{2}\end{array}\right)+\mathrm{43}k-\mathrm{157}\right]\\ =\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{15}\left(\begin{array}{c}k-\mathrm{4}\\ \mathrm{2}\end{array}\right)-\mathrm{43}k+\mathrm{157}\end{array}$

color sets, which can become the color sets of the vertices of Y. When $\mathrm{5}\le k\le \mathrm{11}$,

$\begin{array}{l}\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}-\left[\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{15}\left(\begin{array}{c}k-\mathrm{4}\\ \mathrm{2}\end{array}\right)-\mathrm{43}k+\mathrm{157}\right]\\ =-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)+\mathrm{15}\left(\begin{array}{c}k-\mathrm{4}\\ \mathrm{2}\end{array}\right)+\mathrm{40}k-\mathrm{150}>\mathrm{0}\end{array}$

Then

$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{15}\left(\begin{array}{c}k-\mathrm{4}\\ \mathrm{2}\end{array}\right)-\mathrm{43}k+\mathrm{157}<\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}<n.$

Therefore, all such 6-subsets cannot distinguish n vertices in Y, which is a contradiction.

Case 7: u₁, u₂, u₃, u₄ and u₅ are assigned five different colors under h.

Without loss of generality, we assume that h(u₁)=1, h(u₂)=2, h(u₃)=3, h(u₄)=4, h(u₅)=5. At this time, for each s, t ∈{6,7,…,k–1}, i, j, x, y, w ∈{1,2,3,4,5}, {i,j,x,y,s,s}, {i,j,x,y,w,w}, {i,j,s,s,t,t}, {i,j,x,x,s,s}, {i,j,x,x,y,y}, {i,j,x,s,s,s}, {i,j,x,y,y,y}, {i,j,s,s,s,s}, {i,j,x,x,x,x}, {i,s,s,t,t,t}, {i,j,j,x,x,x}, {i,j,j,s,s,s}, {i,s,s,j,j,j}, {i,j,j,j,j,j}, {i,s,s,s,s,s}, {s,i,i,i,i,i} cannot be the color sets of vertices in Y, and the number of such color sets is $\mathrm{20}\left(\begin{array}{c}k-\mathrm{5}\\ \mathrm{2}\end{array}\right)+\mathrm{85}k-\mathrm{345}$. Then there are $\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\left[\mathrm{20}\left(\begin{array}{c}k-\mathrm{5}\\ \mathrm{2}\end{array}\right)+\mathrm{85}k-\mathrm{345}\right]=\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{20}\left(\begin{array}{c}k-\mathrm{5}\\ \mathrm{2}\end{array}\right)-\mathrm{85}k+\mathrm{345}$color sets, which can become the color sets of the vertices of Y. When $\mathrm{5}\le k\le \mathrm{11}$,

$\begin{array}{l}\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}-\left[\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{20}\left(\begin{array}{c}k-\mathrm{5}\\ \mathrm{2}\end{array}\right)-\mathrm{85}k+\mathrm{345}\right]\\ =-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)+\mathrm{20}\left(\begin{array}{c}k-\mathrm{5}\\ \mathrm{2}\end{array}\right)+\mathrm{82}k-\mathrm{338}>\mathrm{0}\end{array}$

Then

$\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)+\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{3}\end{array}\right)-\mathrm{20}\left(\begin{array}{c}k-\mathrm{5}\\ \mathrm{2}\end{array}\right)-\mathrm{85}k+\mathrm{345}<\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{2}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{2}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{7}<n.$

Therefore, all such 6-subsets cannot distinguish n vertices in Y, which is a contradiction.

Secondly, we will demonstrate that there exists a k-E-total coloring of K_5,n which is vertex-distinguished by multiple sets. It is advisable to color the vertices u₁, u₂, u₃, u₄ and u₅ in X with colors 1, 1, 1, 2, 2, respectively. {a,b,c,d,e,f}, {a,b,c,d,e,e}, {a,b,c,c,d,d}, {a,b,c,d,d,d}, {a,b,c,c,c,c}, {a,b,b,c,c,c}, {a,b,b,b,b,b} can be the color sets of vertices in Y under f. There are $\left(\begin{array}{c}k+\mathrm{4}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{1}\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k+\mathrm{4}$such 6-subsets. We suggest that the vertices v₁, v₂, v₃, v₄ and v₅ in Y correspond to five 6-subsets in turn: {5,2,3,4,4,1}, {5,2,2,3,1,1}, {5,2,2,2,1,1}, {5,2,2,2,3,3}, {5,3,3,3,1,1}, and let the remaining n–5 vertices v_j in Y correspond to any n–5 subsets in the remaining $\left(\begin{array}{c}k+\mathrm{4}\\ \mathrm{6}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{3}\\ \mathrm{5}\end{array}\right)+\left(\begin{array}{c}k+\mathrm{1}\\ \mathrm{4}\end{array}\right)-\mathrm{8}\left(\begin{array}{c}k-\mathrm{1}\\ \mathrm{2}\end{array}\right)-\mathrm{3}k-\mathrm{1}$such 6-subsets, j=6,7,…,n, the specific coloring scheme is as follows. When the 6-subset corresponding to v_j is in the form of {s,t,r,p,q,m}, {s,2,t,r,p,1}, {s,t,r,p,q,1}, {s,2,t,r,p,q}, {s,t,r,p,q,q}, {s,2,2,t,r,p}, {s,t,r,p,1,1}, {s,2,2,t,r,1}, {s,2,t,r,1,1}, {s,2,t,r,r,1}, {s,1,t,r,p,p}, {s,t,r,p,p,2}, {s,t,r,r,p,p}, {s,2,2,t,1,1}, {s,t,r,r,1,1}, {s,2,2,t,r,r}, {s,t,t,r,r,1}, {s,2,t,t,r,r}, {s,2,2,t,t,1}, {s,2,t,t,1,1}, {s,t,r,p,p,p}, {s,2,2,2,t,r}, {s,2,2,2,t,1}, {s,t,r,r,r,1}, {s,2,t,t,t,1}, {s,2,t,r,r,r}, {s,t,r,r,r,r}, {s,t,t,t,t,1}, {s,2,t,t,t,t}, {s,t,t,r,r,r}, {s,2,2,2,1,1}, {s,t,t,t,1,1}, {s,2,2,2,t,t}, {s,2,2,t,t,t} or {s,t,t,t,t,t}, where different colors s, t, r, p, q, m ∈ {3,4,…,k}, v_j and u₁v_j, u₂v_j, u₃v_j, u₄v_j, u₅v_jare colored with the first color, the second color, the third color, the fourth color, fifth color and the sixth color of the color set, respectively.

At this time, the color set of each vertex in Y is exactly the color set corresponding to this vertex in advance. As the sets corresponding to different vertices in Y are different; therefore, no two distinct vertices in Y has the same color set.

The number of elements in the color sets of vertices in X is different from that in the color sets of vertices in Y, so the sets of the colors of vertices in X are different from that in Y.

The following proves that C̃(u₁), C̃(u₂), C̃(u₃), C̃(u₄), C̃(u₅) are not equal to each other. It can be seen from the coloring scheme that both C̃(u₄) and C̃(u₅) contain only one element 2 (because the vertices u₄ and u₅ in X are colored with color 2, the color of the edges associated with these two vertices cannot be 2 ). When 5≤ k ≤11, C̃(u₁), C̃(u₂) and C̃(u₃) contain more than one element 2. Therefore C̃(u₁)≠C̃(u₄), C̃(u₁)≠C̃(u₅), C̃(u₂)≠C̃(u₄), C̃(u₂)≠C̃(u₅), C̃(u₃)≠C̃(u₄), C̃(u₃)≠C̃(u₅). Let C̃(u_i)={f(u_i), f(u_iv₁), …, f(u_iv_n)}, i=1, 2, 3, 4, 5, easy to know f(u₁v₁)=2, f(u₂v₁)=3, f(u₃v₁)=4, f(u₄v₁)=4, f(u₅v₁)=1.

If C̃(u₁), C̃(u₂), C̃(u₃) are all equal, we can exchange colors f(u₁v₁) and f(u₂v₁). At this time, the color sets corresponding to the vertices v₁ and other vertices v_j in Y are unchanged, j = 2,3,…,n. And C̃(u₁) has one more color f(u₂v₁), and one less color f(u₁v₁), there is a color f(u₁v₁) in C̃(u₂), and a color f(u₂v₁) is missing, the color set corresponding to C̃(u₃) is unchanged. Therefore C̃(u₁)≠C̃(u₂)≠C̃(u₃).

If C̃(u₁), C̃(u₂), C̃(u₃) have two values. Without loss of generality, we assume that C̃(u₁)= C̃(u₂)≠C̃(u₃), we can exchange colors f(u₁v₁) and f(u₂v₁). At this time, the color sets corresponding to vertex v₁ and other vertices v_j in Y are unchanged. And C̃(u₁) contains one more a color f(u₂v₁), less a color f(u₁v₁), C̃(u₂) has one more color f(u₁v₁) and one less color f(u₂v₁). The color set corresponding to C̃(u₃) is unchanged. Then if C̃(u₁)=C̃(u₃)≠C̃(u₂), then continue to exchange the second element in C̃(u₁) and C̃(u₃), namely the exchange of colors f(u₂v₁) and f(u₃v₁), so C̃(u₁)≠C̃(u₂)≠C̃(u₃).

If C̃(u₄)=C̃(u₅), we can exchange colors f(u₄v₁) and f(u₅v₁). At this time, the color set corresponding to the vertex v₁ and other vertices v_j in Y is unchanged, j=2,3,…,n. And C̃(u₄) has one more color f(u₅v₁) and one less color f(u₄v₁), C̃(u₅) has one more color f(u₄v₁) and one less color f(u₅v₁). Therefore, C̃(u₄)≠C̃(u₅).

From the above, C̃(u₁), C̃(u₂), C̃(u₃), C̃(u₄), C̃(u₅) are pairwise distinct.

In conclusion, we obtain a k-E-total coloring of K_5,n which are vertex-distinguished by multiple sets.

3 Summary and Prospection

In this study, the E-total chromatic numbers of complete bipartite graphs K_5,n (5≤n≤7 113) have been obtained, which are vertex-distinguished by multiple sets, by using the method of pre-assignment of color sets and the method of constructing specific coloring. The methodologies employed in this paper can encourage people to study further the E-total colorings of complete bipartite graphs K_m,n (6≤m≤n) vertex-distinguished by multiple sets.

References