Decomposing edge-colored graphs under color degree constraints

# Decomposing edge-colored graphs under color degree constraints

## Abstract

For an edge-colored graph , the minimum color degree of means the minimum number of colors on edges which are adjacent to each vertex of . We prove that if is an edge-colored graph with minimum color degree at least then can be partitioned into two parts such that each part induces a subgraph with minimum color degree at least . We show this theorem by proving a much stronger form. Moreover, we point out an important relationship between our theorem and Bermond-Thomassen’s conjecture in digraphs.

Keywords: Bermond-Thomassen’s conjecture; edge-colored graph; vertex partition

## 1 Introduction

When we try to solve a problem in dense graphs, decomposing a graph into two dense parts sometimes plays an important role in the proof argument. This is because one can apply an induction hypothesis to one of the parts so as to obtain a partial configuration, and then use the other part to obtain a desired configuration. Motivated by this natural strategy, many work has been done along this line, and now we have a variety of results in this partition problem. To name a few, Stiebitz [[8]] showed a nice theorem, which states that every graph with minimum degree at least can be decomposed into two parts and such that has minimum degree at least and has minimum degree at least . We see that the bound is best possible by considering the complete graph of order . By the same example, Thomassen [[12], [13]] conjectured that every -connected graph can be decomposed into two parts and in such a way that is -connected and is -connected. It was shown by Thomassen himself [[10]] that if , then the conjecture is true. However, rather surprisingly, even for the case this conjecture is widely open until now. Likewise, there are some other partition problems to find the partition so that both and have a certain property, respectively. The digraph version of this problem was proposed at the Prague Midsummer Combinatorial Workshop in 1995: For a digraph , let be the minimum out degree of . For integers and , does there exists a smallest value such that each digraph with admits a vertex partition () satisfying and ? In [[1], [2]] Alon posed the problem: Is there a constant such that ?We only know that holds by a result of Thomassen [[11]]. No much progress has been made for this problem. Recently Stiebitz [[9]] propose this problem again when he deals with the coloring number of graphs. As observed from the above known results, it seems that these partition problems are very difficult even if we restrict our consideration to a very specific case.

In this paper, we would like to consider a similar problem in edge-colored graphs. To state our results, we introduce some notation and definitions. Throughout this paper, all graphs are finite and simple. Let be an edge-colored graph. For an edge , we use to denote the color of . For a vertex , let be the color degree of in , that is, the number of colors on edges which are adjacent to . The minimum color degree of is denoted by . For a subgraph of with , let be the set of colors appeared in . Also, for a pair of vertex-disjoint subgraphs in , let be the set of colors on edges between and in . For a vertex of , let . By definition, note that . When there is no ambiguity, we often write for , for , for and for . A graph is called a properly colored graph (briefly, PC graph) if no two adjacent edges have the same color. Let and be integers with . A pair is called (a,b)-feasible if and are disjoint, non-empty subsets of such that and ; in particular, if contains an -feasible pair with then we say that has an -feasible partition.

Again, motivated by the same complete graph having mutually distinct colored edges (that is, the rainbow ), we propose the following conjecture.

###### Conjecture 1.

Let be integers with , and be an edge-colored graph with . Then has an -feasible partition.

The main purpose of our paper is to give the solution of this conjecture for the case .

###### Theorem 1.1.

Conjecture 1 is true for .

To consider our problem, utilizing the structure of minimal subgraphs with will be very important. An edge-colored graph is -colored if . Specifically, we say a graph is minimally -colored if holds but any proper subgraph of has minimum color degree less than in . By definition, note that, every PC cycle is a minimally -colored graph. An edge-colored graph obtained from two disjoint cycles by joining a path is a generalized bowtie (more briefly, call it g-bowtie). We allow the case where the path joining two cycles is empty. In that case, the g-bowtie becomes a graph obtained from two disjoint cycles by identifying one vertex in each cycle. Note also that (that is, a graph obtained from two disjoint triangles by identifying one vertex of each triangle) is a g-bowtie with minimum order.

We have the following characterization of minimally -colored graphs, which will be used to prove our main result.

###### Theorem 1.2.

If an edge-colored graph is minimally -colored, then is either a PC cycle or a 2-colored g-bowtie without containing PC cycles.

In fact Theorem 1.1 will be given by proving a much stronger result. We generalize the concept of -feasible partitions as follows. For if can be partitioned into parts such that holds for each then we say that has an -feasible partition. In this paper, we will mainly focus on the case where . For simplicity, let us call -feasible partition in this special case (thus, -feasible partitions are equivalent to -feasible partitions). To state our result, we shall introduce the following theorem, which is on the existence of vertex-disjoint directed cycles in digraphs.

###### Theorem 1.3 (Thomassen [[11]]).

For each natural number there exists a (smallest) number such that every digraph with contains vertex-disjoint directed cycles.

Bermond and Thomassen [[3]] conjectured that and Alon [[1]] showed that .

As above, for let be a function such that every directed graph satisfying contains disjoint directed cycles. Define a function as follows.

 g(k)={2,k=1;max{f(k)+1,g(k−1)+3},k≥2.%

Our main result is following.

###### Theorem 1.4.

Let be an edge-colored graph with . Then has a -feasible partition.

We then focus on the case in Conjecture 1. We obtained the following partial result.

###### Theorem 1.5.

Let be an integer with , and let be an edge-colored complete graph of order with . Then has an -feasible partition.

Also, in [[4]], it is shown that any edge-colored complete bipartite graph with contains a PC . This yields the following.

###### Theorem 1.6.

If an edge-colored complete bipartite graph satisfies , then admits an -feasible partition.

Regarding Conjecture 1 in the general case, by using the probabilistic method, we get the following result.

###### Theorem 1.7.

Let be integers with . If is an edge-colored graph with and , then has an -feasible partition.

Although our results might look a bit modest, proving Conjecture 1 even for the case seems quite hard. This is because we could give a big improvement on the Alon’s bound if it is true.

###### Theorem 1.8.

If Conjecture 1 is true for , then .

In view of Theorem 1.8, it tells us that solving Conjecture 1 completely seems a very difficult problem.

This paper is organized as follows. In Sections 2, 3 and 4, we give the proofs of Theorems 1.2, 1.4 and 1.7, respectively. In Section 5, we prove Theorems 1.5 and 1.8. In particular, Theorem 1.8 is obtained by a much stronger result (see Proposition 4 in Section 5).

## 2 Proof of Theorem 1.2

In order to prove this theorem, we first introduce a structural theorem characterizing edge-colored graphs without containing PC cycles.

###### Theorem 2.1 (Grossman and Häggkist [[6]], Yeo [[14]]).

Let be an edge-colored graph containing no PC cycles. Then there is a vertex such that no component of is joint to with edges of more than one color.

Proof of Theorem 1.2.

Let be a minimally 2-colored graph. If contains a subgraph which is a PC cycle or a 2-colored g-bowtie without containing PC cycles, then (otherwise, by deleting vertices in or edges in , we obtain a smaller 2-colored graph). Hence, it is sufficient to prove that if contains no PC cycle, then contains a 2-colored g-bowtie. Apply Theorem 2.1 to . Since is minimally -colored, we may assume that is connected and there is a vertex such that consists of two components and with all the edges between and has color for .

Let and , respectively, be longest PC paths in and starting from . Set and . Since and for arbitrary vertices and , we have and there exist vertices and for some with and such that and . Since contains no PC cycle, we have and . Together, the path and cycles and form a 2-colored g-bowtie.

The proof is complete.

## 3 Proof of Theorem 1.4

First we prove the following proposition.

###### Proposition 1.

Let be an edge-colored graph with . If contains an -feasible pair, then there exists an -feasible partition of .

###### Proof.

Let be an -feasible pair such that is maximal. If is not an -feasible partition, then with . Since is maximal, is not a feasible pair. Hence there exists a vertex in such that . Recall that . So . Thus is a feasible pair, which is a contradiction with the maximality of . This proves that is an -feasible partition of . ∎

It is easy to check that the following proposition is also true.

###### Proposition 2.

Let be an edge-colored graph with . If contains disjoint subgraphs such that for , then admits an -feasible partition.

In what follows, we will keep the above propositions in mind and use these facts as a matter of course.

Proof of Theorem 1.4.

We prove the theorem by contradiction. Let be a counterexample such that is chosen according to the following order of preferences.

(i) is minimum; (ii) is minimum; (iii) is minimum; (iv) is maximum.

By the choice of , we know that , and contains no rainbow triangles. Let . Then the following two claims obviously hold:

for all .

###### Claim 2.

For each edge , either or .

Now we prove the following claims.

###### Claim 3.

For each color , the subgraph induced by edges colored by is a star.

###### Proof.

By the choice of , we know that contains no monochromatic triangles or monochromatic ’s. Thus for every color , each component of is a star. If contains more than one component, then color one of the components with a color not in . Thus, we get a counterexample with more colors than , which contradicts to the choice of . ∎

###### Claim 4.

For , if and , then .

###### Proof.

Suppose to the contrary that there exist vertices satisfying , and . Then appears only once at and more than once at . By Claim 3, the color can only appear at , particularly, not at . Now we construct a colored graph by deleting the vertex and adding edges to with all of them colored by (since , this is possible without resulting multi-edges). For each vertex , we have . For each vertex , we have . Since the color does not appear at , we have . This implies that . Note that . By the assumption of , we know that must admit a -feasible partition. By Theorem 1.2, contains disjoint subgraphs such that is either a PC cycle or a minimally 2-colored g-bowtie without containing PC cycles for . If , then we can find a -partition of as desired, a contradiction. If , then all the edges in form a monochromatic star with the vertex as a center. Thus, without loss of generality, assume that .

Since is either a PC cycle or a minimally 2-colored g-bowtie without containing PC cycles, for each vertex and each color , the color appears at most times at in . Thus we have .

If , then let be the unique edge in . Replace in with the path (see Figure 1(a)). We obtain a colored graph in with . Thus implies a -feasible partition of , a contradiction.

If , then let . Since , we know that is a minimally 2-colored g-bowtie with being an end vertex of the connecting path in . Delete the edges and add vertex and edges in (see Figure 1(b)). We obtain a g-bowtie in with . Thus implies a -feasible partition of , a contradiction. ∎

###### Claim 5.

There exists an edge such that and .

###### Proof.

Suppose not. Then by Claim 2 , we can construct an oriented graph by orienting each edge from to if and only if . Then for each vertex . Let .

###### Subclaim 1.

For each vertex and colors with , if and , then the following statements hold:
and .
contains at least one edge.

###### Proof.

By the definition, we know that . Since and , we know that . Let and . Then colors and appears only once at and , respectively. If , then is a rainbow triangle, a contradiction. So we have .

Suppose that is empty for some color with . Then choose . We have and . Apply Claim 4 to and , we obtain . For each color with , by Subclaim 1 and the assumption that is empty, we have . Note that

 N+D(v)=⋃|Ti′(v)|≥2,i′∈col(G)Ti′(v).

We have . Recall that and . There must exist a vertex . It is easy to check that is a rainbow triangle in , a contradiction. ∎

###### Subclaim 2.

For each vertex , there is exactly one color such that .

###### Proof.

Given a vertex , by Claim 1, we can find a vertex . By the assumption of , we have . Let . Then . This implies that for each vertex , there is at least one color such that . Now, suppose to the contrary that there exists a vertex and colors with satisfying and . By Subclaim 1, we can choose edges from and from . Let . Then . Now we will discuss on the minimum color degree of .

If , then by the assumption of , has a -feasible partition. Together with , we obtain a -feasible partition of , a contradiction. So we have . Let be a vertex satisfying . Since and , we have

 4≤|col(x,F)|≤5.

For vertices and , if , then it is easy to check that either or is a rainbow triangle, a contradiction. So we have . Note that . This forces that and . Thus is a rainbow cycle of length . Suppose that there exists a vertex such that . Then . Note that . Thus either or is a rainbow triangle, a contradiction. Hence we have . By the assumption of , the graph has a -feasible partition. Together with , we get a -feasible partition of , a contradiction. ∎

Subclaim 2 implies that there are at least colors appear only once at for each vertex . Thus, we have . So contains disjoint directed cycles, which correspond to disjoint PC cycles in , a contradiction. ∎

###### Claim 6.

For each edge satisfying and , we have
, and
, where and for with .

###### Proof.

Since contains no rainbow triangles and appears only once at and , respectively. we have for all . Now let . Then is well defined and for all vertices in . Let be the new vertex resulted by contracting the edge .

Suppose that , then . Thus we have . By the choice of , we know that must admit a -feasible partition. By Theorem 1.2, contains disjoint subgraphs such that () is either a PC cycle or a minimally 2-colored g-bowtie without containing PC cycles.

If , then are -disjoint subgraphs of . This implies a -feasible partition of , a contradiction. So we can assume that . Apparently, .

If , then let (see Figure 2). If , then replace with . If and , then replace the path with . In all cases, we can transform into a graph such that and for . Thus imply the existence of a -feasible partition of , a contradiction.

If , then must be a minimally 2-colored g-bowtie with being an end-vertex of the connecting path. Let with on a same cycle in (see Figure 3). If , then replace with . If and , then replace with . If and , then replace with . Constructions of the remaining cases are similar. Finally, in all cases, we can transform into a graph such that and for . Thus implies a -feasible partition of , a contradiction.

If , then is a minimally 2-colored g-bowtie with two cycles overlapped on the vertex . Let with on one cycle and on the other cycle (see Figure 4). If , then replace with . If and , then replace the path with . If and , then split into the edge such that the resulting graph is still a g-bowtie. If and , then split into the edge in an orthogonal direction such that the resulting graph is a cycle with one chord . Constructions of the remaining cases are similar. Finally, in all cases, we can transform into a graph such that and for . Thus implies a -feasible partition of , a contradiction.

By Claim 6 and the fact that , we have and . For each color and , since is a star and the color appears at and , we know that must be leaf vertices of . Let be the center of . The proof is complete. ∎

Now let be the set of vertices described in Claim 6. Without loss of generality, let for . Let be the subgraph of induced by and .

For , .

###### Proof.

Suppose to the contrary that there exists a vertex such that . If for some with and , then (since is not a rainbow triangle). Since the color appears at least 2 times at , we know that , a contradiction. Now the vertex must belong to . Since each is a star and , we have . If , then by applying Claim 6 to the edge , we have . Since , we have , namely, , a contradiction. So we have . Applying Claim 4 to , we obtain a vertex . Note that and contains no rainbow triangle, we have . Let . It is easy to check that .

We will show that for each vertex , . For , the assertion holds since has no neighbor to or . Thus we may assume that for some with and . If or , then we have the desired conclusion. So we may assume that is adjacent to and (otherwise, is a rainbow triangle). Since there is no rainbow triangle and is a star, we can easily check that . So satisfies the desired property.

Now, . So admits a -feasible partition. Together with , we obtain a -feasible partition of , a contradiction. ∎

###### Claim 8.

There exists a vertex with such that .

###### Proof.

Suppose not. Then there exists a vertex for all with . By Claim 7, for . Let . Then . By the choice of , the graph must admit a -feasible partition, which implies that has a -feasible partition, a contradiction. ∎

We are now in a position to prove the theorem. Let be the vertex in Claim 8. Since and , there is a vertex . Note that . By Claim 2, we have . Now apply Claim 4 to the edge , we have . This implies that either or , a contradiction.

This completes the proof of Theorem 1.4.

## 4 Proof of Theorem 1.7

###### Lemma 1.

Let be positive integers and a non-negative integer with . Let be a set of vertices such that each vertex is colored by . Divide these vertices into two sets and , randomly and independently, with . Let be the probability of the event that there are at most () differently colored vertices in . Then

 PS(x0,x1,…,xk)≤x0∑j=0(kj)(12)k. (4.1)
###### Proof.

For convenience, we say a vector is good if are positive integers and is a non-negative integer with . Proving Lemma 1 is equivalent to verify Inequation (4.1) for all good vectors. For good vectors