Some Results on the Erdős–Faber–Lovász Conjecture

Abstract

Erdős–Faber–Lovász conjecture states that if a graph G is a union of the n edge-disjoint copies of complete graph $K_n$, that is, each pair of complete graphs has at most one shared vertex, then the chromatic number of graph G is n. In fact, we only need to consider the graphs where each pair of complete graphs has exactly one shared vertex. However, each shared vertex may be shared by more than two complete graphs. Therefore, this paper first considers the graphs where each shared vertex happens to be shared by two complete graphs, and then discusses the graphs with only one shared vertex shared by more than two complete graphs. The conjecture is correct for these two kinds of graphs in this work. Finally, the graph where each shared vertex happens to be shared by three complete graphs has been studied, and the conjecture also holds for such graphs when $n=13$. The graphs discussed in this paper have certain symmetric properties. The symmetry of graphs plays an important role in coloring. This work is an attempt to combine the symmetry of graphs with the coloring of graphs.

1. Introduction

The Erdős–Faber–Lovász (EFL for short) conjecture is a graph coloring problem proposed by Erdős, Faber and Lovász in 1972. The conjecture is described in the form of vertex coloring as follows:

(EFL conjecture) If a graph G is the union of n complete graphs of order n, and each pair of complete graphs has at most one shared vertex, then the vertices of graph G can be properly colored with n colors. In fact, there are many other equivalent forms of this conjecture, such as set coloring and hypergraph coloring. These equivalent forms are introduced below.

Set form: There are n sets, each set has n elements and any two sets have at most one element in common. Is it possible to color the elements in the union of these n sets with n colors so that each set has no two elements in the same color? This is the initial formulation when Erdős, Faber and Lovász met at a party in Boulder Colorado in September 1972 [1].

Conference seating arrangement: In 2004, Haddad and Tardif [2] introduced a story about seating arrangements in committees: Assume that there are n committees in a university, and each committee has exactly n members, and every two committees have at most one common member. Each committee meets in a meeting room with exactly n chairs. Is it possible to assign the committee members to chairs in such a way that each member sits in the same chair for all the different committees to which he or she belongs? In this model of committee seating arrangements, committees are equivalent to complete graphs, committee members are equivalent to the vertices of the graph, and chairs are equivalent to vertex colors.

Hypergraph language: The n complete graphs of order n in EFL conjecture can be regarded as n hyperedges of a n-uniform linear hypergraph. Therefore, the EFL conjecture is described in hypergraph language as follows: For any n-uniform linear hypergraph with n hyperedges, the vertices of the hypergraph can be colored with n colors so that no two vertices in the same edge receive the same color. Since it is trivial to color the vertices of degree 1 in the uniform linear hypergraph, the vertices of degree 1 in the linear hypergraph can be deleted, so that the following simplified equivalent form of the conjecture can be obtained: For any n-uniform linear hypergraph with n hyperedges, in which every vertex has degree at least two, then the vertices of the hypergraph can be colored with n colors so that no two vertices in the same edge receive the same color. In addition, it is easy to verify that the dual graph of a linear hypergraph is also a linear hypergraph. And the degree of the vertices in the hypergraph is at least two which is equivalent to that each edge in the dual graph of the hypergraph contains at least two vertices, that is, acyclic, and coloring the vertices of a hypergraph is equivalent to coloring the edges of the dual graph of the hypergraph. Hence, the following equivalent hyperedge coloring version is a matter of course: For any simple hypergraph with n vertices, its edge chromatic number does not exceed n. Please refer to [3] for more details.

Erdős initially offered $50 for a proof or disproof of this conjecture, but soon realized that the problem was very difficult, so the bonus for proof or disproof of the conjecture was increased to $500. It has been almost 50 years since the conjecture was proposed. A lot of work has been done during the past decades. In 1981, Hindman [4] proved that the conjecture was correct when $n\le 10$. In [5], Romero and Alonso-Pecina adopted a heuristic algorithm, combining simulated annealing algorithm, genetic algorithm and tabu search to advance this result to $n\le 12$. At the same time, the conjecture was verified for some special graphs. For example, it was proved in [6] that the conjecture holds for Cyclic Steiner 2-designs, and Füredi [7] proved that the conjecture was correct when any two hyperedges intersect. In addition, the upper bounds of the chromatic number in the conjecture have been established. In 1988, Chang and Lawler’s simple proof in [8] showed that the upper bound of the edge chromatic number in the conjecture was $\frac{3}{2}n-2$; Two years later, the upper bound of the vertex chromatic number of the conjecture has been improved to be $\frac{3}{2}n$ by Horák in [9], which is the best asymptotically; In 1992, Kahn proved in [10] that it always exists proper $n+o(n)$ edge coloring for the conjecture, which is an approximate version of the conjecture. In the same year, Kahn and Seymour proved in the literature [11] that the fractional version of the EFL conjecture was valid. In recent years, research results on this conjecture have also appeared in various journals. In 2008, Sánchez-Arroyo proved that the conjecture is valid for dense hypergraphs, see literature [12]. In [13], Paul and Germina generalized the results of Sánchez-Arroyo. Please refer to [14,15,16,17,18,19,20] for more information about this conjecture.

The graph-related terminologies used in this article refer to [21]. Let $G={\cup }_{i=1}^n{G}_i$ represent the graph in EFL conjecture, where each $G_i(1\le i\le n)$ is a copy of a complete graph of order n, and each pair of complete graphs has at most 1 shared vertex. The shared degree $\tilde{d}(v)$ of a vertex $v\in G$ is given by $\tilde{d}(v)=|G_i:v\in G_i,1\le i\le n|$. If $\tilde{d}(v)=1$, then v is called a free vertex; else, v is called a shared vertex. Color the shared vertices of the graph G first, and then color the free vertices. If the shared vertices can be (properly) n colored, then there is no difficulty in coloring free vertices within each complete graph, thus the graph G can be n-colored. Therefore, it only needs to color the shared vertices of the graph G in the conjecture. In addition, if there exist two complete graphs $G_i$ and $G_j$ in G which have no shared vertex, then $G_i$ and $G_j$ have free vertices x and y respectively. Let x and y be merged into a single vertex z, then z is the shared vertex of $G_i$ and $G_j$. Do it along this way until any pair of complete graphs has shared vertex. If the ultimate graph can be properly n-colored, then the initial graph can also be properly n-colored. Therefore, it only needs to consider the graph in which any pair of complete graphs has shared vertex. Especially, denote by ${\mathcal{S}}_n$ the set of graphs that each shared vertex has shared degree 2, and ${\mathcal{T}}_n$ by the set of graphs that not every shared vertex has shared degree 2. Furthermore, let ${\mathcal{T}}_{n,j}$ be the set of graphs obtained by splitting the shared vertices that shared by each pair of some $j(3\le j\le n)$ complete graphs in $S_n\in {\mathcal{S}}_n$, and then merged them into a single vertex of shared degree j. In addition, we use ${\mathcal{T}}_n(k)$ to represent the set of graphs with shared degree k for each shared vertex. In terms of graph structure, all graphs considered above have certain symmetry characteristics. For example, ${\mathcal{S}}_3$ is the well-known pyramid graph. Please see Figure 1 below. These special symmetry structure characteristics of graphs enable us to quickly find a specific vertex coloring method for these graphs. The next section witness the correctness of the EFL conjecture for ${\mathcal{S}}_n$ and ${\mathcal{T}}_{n,j}$. Last but not least, the EFL conjecture is also valid for ${\mathcal{T}}_{13}(3)$. We end up this work with some concluding remarks.

Figure 1. ${\mathcal{S}}_3$ or pyramid.

2. Main Results

For each $S_n\in {\mathcal{S}}_n$, let $S_n={\cup }_{i=1}^n{G}_i$, where each complete graph $G_i$ has $(n-1)$ shared vertices, and each shared vertex is shared by 2 complete graphs. Therefore, $S_n$ has $\frac{n(n-1)}{2}$ shared vertices. In addition, every complete graph has exactly 1 free vertex, so $S_n$ has n free vertices. Therefore, $S_n$ has a total of $\frac{n(n+1)}{2}$ vertices. We only need to color the shared vertices of $S_n$.

Theorem 1.

Let n be any positive integer, for each $S_n\in {\mathcal{S}}_n$, $\chi (S_n)=n$.

Proof. 

Let $s_{i,j}(1\le i<j\le n)$ be the shared vertex of $G_i$ and $G_j$. The colors be used are $0,1,2,\dots ,n-1$. For $1\le i<j\le n$, color $s_{i,j}$ with the color $(i+j)modn$. □

For each $T_{n,j}\in {\mathcal{T}}_{n,j}$, it can be concluded from the aforementioned that $T_{n,j}$ has $\frac{n(n-1)}{2}-\frac{j(j-1)}{2}$ vertices of shared degree 2, and 1 single vertex of shared degree j.

Theorem 2.

For each $T_{n,j}\in {\mathcal{T}}_{n,j}$, where $n\ge 3$ and $3\le j\le n$, $\chi (T_{n,j})=n$.

Proof. 

Let $\frac{n(n-1)}{2}-\frac{j(j-1)}{2}$ vertices of shared degree 2 of $T_{n,j}$ are as follows: $t_{1,2}$, $t_{1,3}$, …, $t_{1,n}$; $t_{2,3}$, $t_{2,4}$, …, $t_{2,n}$; …; $t_{n-j,n-j+1}$, $t_{n-j,n-j+2}$, …, $t_{n-j,n}$. Let $t_j$ be the single vertex of shared degree j. The colors be used are $0,1,2,\dots ,n-1$. For $1\le i\le n-j$ and $i<k\le n$, color $t_{i,k}$ with the color $(i+k)modn$. Color $t_j$ with the color $(n-j+1)modn$. It can be verified that adjacent shared vertices receive different colors under such coloring. □

For each $T_n(k)\in {\mathcal{T}}_n(k)$, since the shared degree of each shared vertex in $T_n(k)$ is k, we call it the k-regular shared graph, where k is the regular shared degree of $T_n(k)$; otherwise, we call them non-regular shared graphs. In fact, let $k=2$, then $T_n(2)=S_n$, therefore, $S_n$ is called a 2-regular shared graph. As mentioned above, in $S_n$, each complete graph has $(n-1)$ shared vertices, and each shared vertex is shared by just two complete graphs, so $S_n$ has $\frac{(n-1)\times n}{1\times 2}$ shared vertices. Similarly, for $k\ge 3$, in $T_n(k)$, each complete graph has exactly $\frac{n-1}{k-1}$ shared vertices, and each shared vertex is shared by k complete graphs, so $T_n(k)$ has $\frac{\frac{n-1}{k-1}\times n}{k}=\frac{(n-1)\times n}{(k-1)\times k}$ shared vertices for $(k-1)\times k\mid (n-1)\times n$ and $(k-1)\mid (n-1)$. The following two theorems are the results of a little thought.

Theorem 3.

Let $k\ge 2$ and $n=k^2$, then $\chi (T_n(k))=n$.

Proof. 

Denote by $\Gamma$ the vertex induced graph of the shared vertices of $T_n(k)$. It can be seen from the above analysis that $\Gamma$ is $k(\frac{n-1}{k-1}-1)$ -regular with $\frac{(n-1)\times n}{(k-1)\times k}$ vertices. If $n=k^2$, then $\Gamma$ is a n-regular graph with $n+\sqrt{n}$ vertices, which can not be a complete graph since $n+\sqrt{n}>n+1$. Thus $\chi (T_n(k))=n$ follows from the well-known Brooks’ theorem. □

Theorem 4.

Let $n=m^2-m+1(m\in Z^{+})$, then $\chi (T_n(m))=n$.

Proof. 

There are $\frac{(n-1)\times n}{(m-1)\times m}$ shared vertices in $T_n(m)$. If $n=m^2-m+1(m\in Z^{+})$, then $\chi (T_n(m))=n$ follows immediately since there are just n shared vertices in $T_n(m)$. □

A few conclusions may be obtained by Theorem 4, such as $\chi (T_7(3))=7$ and $\chi (T_{13}(4))=13$ etc. Generally, if the number of shared vertices in $T_n(k)$ is less than or equal to n, that is $\frac{(n-1)\times n}{(k-1)\times k}\le n$, i.e., $k\ge \frac{1+\sqrt{4n-3}}{2}$, then $T_n(k)=n$. Therefore, it only needs to consider the case where the number of shared vertices is greater than n, that is $\frac{(n-1)\times n}{(k-1)\times k}>n$, i.e., $k<\frac{1+\sqrt{4n-3}}{2}$. The following concluding result is revealed with a bit of work.

Theorem 5.

For each $T_{13}(3)\in {\mathcal{T}}_{13}(3)$, $\chi (T_{13}(3))=13$.

Proof. 

It can be concluded from the above statements that $T_{13}(3)$ has 26 shared vertices, and the shared degree of each shared vertex is 3. To be more precise, each complete graph has six shared vertices, and each shared vertex must be the common vertex of some three complete graphs $G_i$, $G_j$ and $G_k$$(1\le i<j<k\le 13)$, denoted by $v_{i,j,k}$. What are these 26 vertices? To this end, for $1\le i\le 13$, let $X_i=(x_{i,1},x_{i,2},\cdots ,x_{i,26})$ be the unknown 0–1 vector, and $X={(X_1,X_2,\cdots ,X_{13})}^T$ be the unknown 0–1 matrix. In addition, let $I_{13}$ and $I_{26}$ be the all 1s column vectors. The solutions of the following equations implies what are these 26 vertices.

$$\left\{\begin{array}{c}X·I_{26}=6I_{13},\hfill \\ X^T·I_{13}=3I_{26},\hfill \\ X_i·X_j^T=1,i\ne j,i,j\in \{1,2,\cdots ,13\}.\hfill \end{array}\right.$$

(1)

With the help of MATLAB and GUROBI, one solution of the above equations implies the following 26 shared vertices, presented in Table 1, as well as one of their 13-colorings.

Table 1. One 26 shared vertices in $T_{13}(3)$ and one 13-coloring.

Vertices	$v_{1,5,11}$, $v_{8,10,12}$	$v_{4,5,8}$, $v_{7,10,13}$	$v_{5,9,10}$, $v_{6,11,13}$	$v_{3,8,13}$, $v_{4,10,11}$
Colors	1	2	3	4
Vertices	$v_{3,5,6}$, $v_{7,8,11}$	$v_{2,8,9}$, $v_{3,7,12}$	$v_{1,7,9}$, $v_{2,11,12}$	$v_{2,3,4}$, $v_{5,12,13}$
Colors	5	6	7	8
Vertices	$v_{1,2,13}$, $v_{4,6,7}$	$v_{1,3,10}$, $v_{6,9,12}$	$v_{1,4,12}$, $v_{2,6,10}$	$v_{1,6,8}$, $v_{4,9,13}$
Colors	9	10	11	12
Vertices	$v_{2,5,7},v_{3,9,11}$
Colors	13

The other solution of the Equations (1) implies the following other 26 shared vertices, presented in Table 2, as well as one of their 13-colorings.

Table 2. The other 26 shared vertices in $T_{13}(3)$ and one 13-coloring.

Vertices	$v_{1,4,13}$, $v_{3,9,12}$	$v_{4,8,9}$, $v_{1,5,10}$	$v_{1,6,9}$, $v_{4,5,7}$	$v_{2,4,6}$, $v_{5,8,13}$
Colors	1	2	3	4
Vertices	$v_{3,8,11}$, $v_{7,9,13}$	$v_{3,5,6}$, $v_{10,11,13}$	$v_{1,3,7}$, $v_{6,12,13}$	$v_{2,5,12}$, $v_{6,8,10}$
Colors	5	6	7	8
Vertices	$v_{5,9,11}$, $v_{7,10,12}$	$v_{1,8,12}$, $v_{2,9,10}$	$v_{1,2,11}$, $v_{3,4,10}$	$v_{2,7,8}$, $v_{4,11,12}$
Colors	9	10	11	12
Vertices	$v_{2,3,13},v_{6,7,11}$
Colors	13

In general, except for $v_{i,j,k}$$(1\le i<j<k\le 13)$, there are five remaining shared vertices within each of the three complete graphs $G_i$, $G_j$ and $G_k$. The vertex induced subgraph of these 15 vertices is an 8-regular graph, which can be 8-colored owing to the Brooks’ theorem. The vertex induced subgraph of the remaining ten vertices, denoted by H, is a 6-regular graph. Thus, $\overline{H}$ is a cubic graph.

Case 1. If $\overline{H}$ has a bridge, then $\overline{H}$ is isomorphic to the graph in Figure 2, which has a perfect matching. By the way, in the sense of isomorphism, there is only one 3-regular graph of order 10 with bridge. The symmetric drawing in Figure 2 has clearly shown the symmetry structural characteristics of this graph.

Figure 2. A 10 order cubic graph with a bridge and a perfect matching (red edges).

Case 2. If $\overline{H}$ is bridgeless, then $\overline{H}$ has a perfect matching.

In a word, $\overline{H}$ has a perfect matching, which means H can be five colored. Therefore, no matter what are these 26 vertices, they can be 13-colored as long as $v_{i,j,k}$ be colored with any color from the color set of the vertices in H. □

3. Concluding Remarks

Actually, it never needs 13 colors to color the shared vertices in Table 1. For example, color $v_{2,5,7}$ with the color 4, and $v_{3,9,11}$ with the color 9, then color 13 is released. Add a little more thought, 8 colors are enough for these shared vertices, please turn to the following Table 3. But two less color is not enough. In other words, 6 colors are not enough, because the independent number of the induced subgraph of these 26 shared vertices is 4. What about 7 colors? This is a challenge job for interested readers.

Table 3. A 8-coloring for the 26 vertices in Table 1.

Vertices	$v_{1,2,13}$, $v_{4,6,7}$, $v_{3,9,11}$, $v_{8,10,12}$	$v_{2,5,7}$, $v_{3,8,13}$, $v_{4,10,11}$, $v_{6,9,12}$
Colors	1	2
Vertices	$v_{1,5,11}$, $v_{2,6,10}$, $v_{4,9,13}$, $v_{3,7,12}$	$v_{1,4,12}$, $v_{2,8,9}$, $v_{3,5,6}$, $v_{7,10,13}$
Colors	3	4
Vertices	$v_{4,5,8}$, $v_{1,7,9}$, $v_{2,11,12}$	$v_{2,3,4}$, $v_{7,8,11}$, $v_{5,9,10}$
Colors	5	6
Vertices	$v_{1,3,10}$, $v_{6,11,13}$	$v_{1,6,8}$, $v_{5,12,13}$
Colors	7	8

In addition, let the vertex induced subgraph of the 26 shared vertices in Table 1 be ${\Gamma }_1$, and the vertex induced subgraph of the 26 shared vertices in Table 2 be ${\Gamma }_2$. We think ${\Gamma }_1\cong {\Gamma }_2$, or even more boldly, the $T_{13}(3)$ is unique in the sense of graph isomorphism. In other words, these 26 vertices are indistinguishable, indicating the symmetry structure of the graph. By the way, $\chi (T_9(3))=9$ by Theorem 3. Ingenious coloring methods require more in-depth work as n increases as long as $6\mid (n-1)n$ and $2\mid (n-1)$, such as $T_{15}(3)$, $T_{19}(3)$ etc.

At last, from the foregoing, it can be seen that $\chi (T_{13}(2))=\chi (S_{13})=13$ by Theorem 1 and $\chi (T_{13}(4))=13$ due to Theorem 4. In addition, $T_{13}(k)(5\le k\le 12)$ does not exist, while $T_{13}(13)$ has just one shared vertex. The elegant theoretic coloring schemes are not available right now for the other non-regular shared graphs in ${\mathcal{T}}_{13}$, except $\chi (T_{13,j})=13$ for $3\le j\le n$ by Theorem 2. The culprit is the existence of various shared vertices with different shared degree in such non-regular shared graphs. These are all problems for further study.