Maximum Colored Cuts in Edge-Colored Complete k-Partite Graphs and Complete Graphs

Abstract

The Maximum Colored Cut problem aims to seek a bipartition of the vertex set of a graph, maximizing the number of colors in the crossing edges. It is a classical Max-Cut problem if the host graph is rainbow. Let $mcc\left(G\right)$ denote the maximum number of colors in a cut of an edge-colored graph G. Let $C_k$ be a cycle of length k; we say G is PC-$C_k$-free if G contains no properly colored $C_k$. We say G is a p-edge-colored graph if there exist p colors in G. In this paper, we first show that if G is a PC-$C_3$-free p-edge-colored complete 4-partite graph, then $mcc\left(G\right)=p$. Let $k\ge 3$ be an integer. Then, we show that if G is a PC-$C_4$-free p-edge-colored complete k-partite graph, then $mcc\left(G\right)\ge min\{p-1,15p/16\}$. Finally, for a p-edge-colored complete graph G, we prove that $mcc\left(G\right)\ge p-1$ if G is PC-$C_4$-free, and $mcc\left(G\right)\ge min\{p-6,7p/8\}$ if G is PC-$C_5$-free and $p\ge 7$.

1. Introduction

The Max-Cut problem is a classical partition problem: for a given graph $G(V,E)$, we aim to seek a bipartition $(V_1,V_2)$ of $V\left(G\right)$ that maximizes the number of edges joining $V_1$ and $V_2$. It is one of the earliest proven NP-hard problems, and has a wide range of applications in different disciplines and industries, including statistical physics, VLSI design, and machine learning. It has become one of core topics in graph theory and combinatorics, and has gained a lot of attention in the past few decades. For a graph G with m edges, let $mc\left(G\right)$ be the size of maximum bipartite subgraph of G. A basic result is that $mc\left(G\right)\ge m/2$, which can be seen by considering the expected value of crossing edges in a uniformly random bipartition. In 1973, Edwards [1,2] showed that

$$mc\left(G\right)\ge {\displaystyle \frac{m}{2}}+{\displaystyle \frac{\sqrt{8m+1}-1}{8}},$$

and the bound is tight as evidenced by complete graphs with odd orders. For more references on maximum cuts in graphs, we refer readers to [3,4,5,6,7,8,9].

Compared with the Max-Cut problem, a more difficult and general issue is the Maximum Colored Cut problem: given a simple edge-colored graph $G(V,E)$, the objective is to find a bipartition $(V_1,V_2)$ of $V\left(G\right)$ which maximizes $\left|col\right(V_1,V_2\left)\right|$, the number of colors of edges joining $V_1$ and $V_2$. It is a classical Max-Cut problem if the given edge-colored graph G is rainbow. Faria, Klein, Sau, Souza and Sucupira [10] mentioned this problem but did not explore it in depth. Like the maximum cut problem, this problem can also be applied to many practical problems. For example, in VLSI design, the chip routing problem can be modeled as an edge-colored graph, and different colors represent different constraints (timing, crosstalk, power consumption); the Maximum Colored Cut can identify the boundary with the most complex constraints in the routing region, and guide the allocation of routing resources. For convenience, we now introduce some concepts. We say an edge-colored graph G is $monochromatic$ if the number of colors appearing in G equals 1; G is $rainbow$ if the number of colors appearing in G equals $\left|E\right(G\left)\right|$; and G is $properly$ $colored$ if any two edges with the same color in G are not adjacent. Let H be a graph; we say G is PC-H-free if G contains no properly colored H as a subgraph. A cycle of length k is denoted by $C_k$, and a properly colored $C_3$ is always called a rainbow triangle.

Given an edge-colored graph, the Maximum Colored Cuts are close to the maximum cuts if the number of colors is close to the number of edges. Hence, we are more interested in the cases where there is a significant difference in the number of edges and colors. Naturally, this can be ensured when the given edge-colored graph does not contain certain given edge-colored subgraphs, whose edges are colored by many different colors. In [11], using the method of seeking maximum cuts in graphs, we provided several lower bounds of Maximum Colored Cuts in rainbow-triangle-free or PC-$C_4$-free edge-colored complete graphs, respectively. In this paper, we focus our attention on edge-colored complete k-partite graphs and complete graphs. We call G a p-edge-colored graph if there exist p colors in G. Let $mcc\left(G\right)$ denote the maximum number of colors in a cut of G. We give the following theorems.

Theorem 1.

Let G be a rainbow-triangle-free p-edge-colored complete 4-partite graph. Then, $mcc\left(G\right)=p$.

Theorem 2.

Let $k\ge 3$ be an integer and G be a PC-$C_4$-free p-edge-colored complete k-partite graph. Then, $mcc\left(G\right)\ge min\{p-1,15p/16\}$.

By Theorem 2, we obtain that $mcc\left(G\right)\ge min\{p-1,15p/16\}$ if G is a PC-$C_4$-free p-edge-colored complete graph. In fact, for such a graph G, we prove the following result.

Theorem 3.

Let G be a PC-$C_4$-free p-edge-colored complete graph. Then, $mcc\left(G\right)\ge p-1$, and the bound is tight.

For PC-$C_5$-free edge-colored complete graphs, combining graph structure analysis and probabilistic methods, we give the following result.

Theorem 4.

Let G be a PC-$C_5$-free p-edge-colored complete graph, where $p\ge 7$. Then, $mcc\left(G\right)\ge min\{p-6,7p/8\}$.

The rest of the article is organized as follows. Some notations and two important lemmas are given in Section 2. The proofs of Theorems 1 and 2 are given in Section 3. In Section 4, we provide the proofs of Theorems 3 and 4. Finally, we sum up several concluding remarks and propose an open problem.

2. Preliminaries

In this section, we introduce some notations and two lemmas which would be used in our proofs. We list the required notations one by one through the following

Table 1. Some notations.

p	a positive integer
$\left[p\right]$	$\{1,2,\dots ,p\}$
G	a p-edge-colored graph
$col\left(G\right)$	the color set which contains all colors appearing in G
$i\in \left[p\right]$, $G^i$	the edge-induced subgraph of G by the edges with color i
X,Y	two disjoint subsets of $V\left(G\right)$
$col(X,Y)$	the set containing colors appearing on the edges between X and Y
$col(x,Y)$	$col\left(\right\{x\},Y)$
$G\left[X\right]$	the induced subgraph of G by X
$col\left(X\right)$	$col\left(G\right[X\left]\right)$
$c\left(e\right)$	the color of e

.

A vertex is called a $dominating$ $vertex$ of G if it is adjacent to all other vertices of G. Properly colored $C_4$ is an important structure in edge-colored graphs. The first lemma is a good result for PC-$C_4$-free edge-colored complete graphs.

Lemma 1

(Martin, Magnant, Salehi Nowbandegani [12]). Let G be a PC-$C_4$-free edge-colored complete graph. Then, $G^i$ has a dominating vertex for each color i.

The following fundamental result (see [11]), proved by a simple randomization method, plays a key role in our proofs.

Lemma 2

(Ma [11]). Let G be a p-edge-colored graph and $\Delta ={min}_{i\in \left[p\right]}\{\Delta \left(G^i\right)\}$. If $\Delta \ge k$, then $mcc\left(G\right)\ge (1-1/2^k)p.$

3. Complete k-Partite Graphs

For the convenience of readers, we restate each theorem before its proof.

3.1. Proof of Theorem 1

Theorem 5.

Let G be a rainbow-triangle-free p-edge-colored complete 4-partite graph. Then, $mcc\left(G\right)=p$.

Proof. 

Let G be a rainbow-triangle-free p-edge-colored complete 4-partite graph with vertex classes $A_1, A_2, A_3, A_4$. By symmetry, assume that

$$|col(A_1,A_2)|=\underset{1\le i<j\le 4}{max}\left\{\right|col(A_i,A_j)\left|\right\}.$$

If $\left|col\right(A_1,A_2\left)\right|=p$, then $(A_1,V\left(G\right)\setminus A_1)$ is a desired partition, we are finished.

So, we suppose that $\left|col\right(A_1,A_2\left)\right|<p$. Let $R^{\left(0\right)}=\left[p\right]\setminus col(A_1,A_2)$, $A^{\left(0\right)}=A_1\cup A_2$, and $B^{\left(0\right)}=V\left(G\right)\setminus A^{\left(0\right)}$. Note that G contains no rainbow triangles. We claim that for each $v\in B^{\left(0\right)}$,

$$\begin{array}{c}\hfill col(v,A_1)\cap R^{\left(0\right)}\subseteq col(v,A_2)\cap R^{\left(0\right)}\end{array}$$

(1)

or

$$\begin{array}{c}\hfill col(v,A_2)\cap R^{\left(0\right)}\subseteq col(v,A_1)\cap R^{\left(0\right)}.\end{array}$$

(2)

Since otherwise, this would yield a rainbow triangle.

If $R^{\left(0\right)}\subseteq col(A^{\left(0\right)},B^{\left(0\right)})$, then for each $v\in B^{\left(0\right)}$, we place v into $A_1$ if $\left(1\right)$ holds, and place v into $A_2$ if $\left(2\right)$ holds. By placing each vertex in $B^{\left(0\right)}$ into $A_1$ or $A_2$ in this way, the final partition is a desired partition, and we are finished.

So, we suppose that $R^{\left(0\right)}⊈col(A^{\left(0\right)},B^{\left(0\right)})$. Let $R^{{\left(0\right)}^{\prime }}=R^{\left(0\right)}\setminus col(A^{\left(0\right)},B^{\left(0\right)})$. Then, $R^{{\left(0\right)}^{\prime }}=\left[p\right]\setminus (col(A_1,A_2)\cup col(A^{\left(0\right)},B^{\left(0\right)}))$. Assume that $1\in R^{{\left(0\right)}^{\prime }}$. Then, there exists an edge $e_1=u_1{v}_1$ in $G\left[B^{\left(0\right)}\right]$ colored 1, where $u_1\in A_3$ and $v_1\in A_4$. Recall that G contains no rainbow triangles. We claim that for each $x\in A^{\left(0\right)}$, $c\left(u_{1}x\right)=c\left(v_{1}x\right)$. Otherwise, it would yield a rainbow triangle. Now, we place $u_1$ into $A_1$ and $v_1$ into $A_2$. Let $A_1^{\left(1\right)}=A_1\cup \left\{u_1\right\}$ and $A_2^{\left(1\right)}=A_2\cup \left\{v_1\right\}$. Then, $1\in col(A_1^{\left(1\right)},A_2^{\left(1\right)})$ and $col(\{u_1,v_1\},A^{\left(0\right)})\subseteq col(A_1^{\left(1\right)},A_2^{\left(1\right)})$.

Let $A^{\left(1\right)}=A_1^{\left(1\right)}\cup A_2^{\left(1\right)}$, $B^{\left(1\right)}=B^{\left(0\right)}\setminus \{u_1,v_1\}$ and $R^{\left(1\right)}=R^{\left(0\right)}\setminus [col(\{u_1,v_1\},A^{\left(0\right)})\cup \left\{1\right\}]$. If $R^{\left(1\right)}\subseteq col(A^{\left(1\right)},B^{\left(1\right)})$, then by the above analysis, for each $v\in B^{\left(1\right)}$,

$$\begin{array}{c}\hfill col(v,A_1^{\left(1\right)})\cap R^{\left(1\right)}\subseteq col(v,A_2^{\left(1\right)})\cap R^{\left(1\right)}\end{array}$$

(3)

or

$$\begin{array}{c}\hfill col(v,A_2^{\left(1\right)})\cap R^{\left(1\right)}\subseteq col(v,A_1^{\left(1\right)})\cap R^{\left(1\right)}.\end{array}$$

(4)

We place v into $A_1^{\left(1\right)}$ if $\left(3\right)$ holds, and place v into $A_2^{\left(1\right)}$ if $\left(4\right)$ holds. By placing each vertex in $B^{\left(1\right)}$ into $A_1^{\left(1\right)}$ or $A_2^{\left(1\right)}$ in this way, the final partition is a desired partition, and we are finished. If $R^{\left(1\right)}⊈col(A^{\left(1\right)},B^{\left(1\right)})$, then let $R^{{\left(1\right)}^{\prime }}=R^{\left(1\right)}\setminus col(A^{\left(1\right)},B^{\left(1\right)})$ and let $2\in R^{{\left(1\right)}^{\prime }}$. Then, there exists an edge $e_2=u_2{v}_2$ in $G\left[B^{\left(1\right)}\right]$ colored 2, where $u_2\in A_3$ and $v_2\in A_4$. Then, for each $x\in A^{\left(0\right)}$, $c\left(u_{2}x\right)=c\left(v_{2}x\right)$, and $u_1{u}_2\notin E\left(G\right)$, $v_1{v}_2\notin E\left(G\right)$. We place $u_2$ into $A_1^{\left(1\right)}$ and $v_2$ into $A_2^{\left(1\right)}$. Let $A_1^{\left(2\right)}=A_1^{\left(1\right)}\cup \left\{u_2\right\}$ and $A_2^{\left(2\right)}=A_2^{\left(1\right)}\cup \left\{v_2\right\}$. Then, $2\in col(A_1^{\left(2\right)},A_2^{\left(2\right)})$ and $col(\{u_2,v_2\},A^{\left(1\right)})\subseteq col(A_1^{\left(2\right)},A_2^{\left(2\right)})$.

We continue to operate the above process until $R^{\left(s\right)}\subseteq col(A^{\left(s\right)},B^{\left(s\right)})$. Then, place the remaining vertices in $B^{\left(s\right)}$ into $A_1^{\left(s\right)}$ or $A_2^{\left(s\right)}$ one by one. The final partition is a desired partition, and we complete the proof of the theorem. □

Remark 1.

By the same method, we can show that for a rainbow-triangle-free p-edge-colored complete graph G, $mcc\left(G\right)=p$.

Remark 2.

For rainbow-triangle-free complete 3-partite graphs, we have the same result. Meanwhile, for a complete k-partite graph ($k\ge 5$), this method will no longer be ineffective. Also, we wonder if there is the same result for $k\ge 5$.

3.2. Proof of Theorem 2

Theorem 6.

Let $k\ge 3$ be an integer and G be a PC-$C_4$-free p-edge-colored complete k-partite graph. Then, $mcc\left(G\right)\ge min\{p-1,15p/16\}$.

Proof. 

Let $k\ge 3$ be an integer and G be a PC-$C_4$-free p-edge-colored complete k-partite graph. We may suppose that $p\ge 4$ as the result is trivial if $p\le 3$. Let $A_1, A_2,\dots , A_k$ be the k vertex classes of $V\left(G\right)$ and $\Delta ={min}_{1\le i\le p}\{\Delta \left(G^i\right)\}$. If $\Delta \ge 4$, then by Lemma 2, $f^c\left(G\right)\ge 15p/16$. So, we suppose that $\Delta \le 3$. We split the discussion into the following three cases according to the value of $\Delta$.

Case 1. $\Delta =1$.

Without loss of generality, assume that $\Delta \left(G^1\right)=1$. Clearly, $G^1$ is a match. Note that G is PC-$C_4$-free, We claim that $G^1$ contains only one edge. Otherwise, it would yield a PC-$C_4$. Let $e=uv$ be the only edge-colored 1 in G. We can obtained the following fact: For each $i\in \left[p\right]\setminus \left\{1\right\}$, there exists an edge-colored i which is adjacent to e. Otherwise, it would yield a PC-$C_4$. By the above fact, let $V_1=\{u,v\}$, $V_2=V\left(G\right)\setminus V_1$, we have $mcc\left(G\right)\ge |col(V_1,V_2)|=p-1$. The desired result follows.

Case 2. $\Delta =2$.

Assume that $\Delta \left(G^1\right)=2$ and let $c\left(uv\right)=c\left(uw\right)=1$. Let $V_1=\{u,v,w\}$, $V_2=V\left(G\right)\setminus V_1$. Then, $V_2\ne \varnothing$ as $p\ge 4$. If for each $i\in \left[p\right]\setminus \left\{1\right\}$, there always exists an edge-colored i which is incident with exactly one vertex in $V_1$, then $mcc\left(G\right)\ge |col(V_1,V_2)| \ge p-1$. Therefore, we may suppose that $T=\left[p\right]\setminus (\left\{1\right\}\cup col(V_1,V_2))\ne \varnothing$. Assume that $2\in T$ and $c\left(xy\right)=2$. Note that if $vw\in E\left(G\right)$, then $c\left(vw\right)\notin T$. Otherwise, it would yield a PC-$C_4$ by the fact that $\Delta =2$. Hence, $x,y\notin V_1$. Recall that x and y are not in the same vertex class. Then, there must exist $vx\in E\left(G\right)$ or $vy\in E\left(G\right)$. Now, we give the following claim.

Claim 1.

If the edge $vx$ (resp. $vy$) exists, then $c\left(vx\right)=1$ (resp. $c\left(vy\right)=1$).

Proof. 

Recall that x and y are not in the same vertex class. We may first suppose that there is exactly one vertex in $\{x,y\}$ such that it is in the same vertex class with v. By symmetry, assume that x and v are in the same vertex class. Then, $vy\in E\left(G\right)$ as v and y are in different vertex classes. Recall that $\Delta \left(G^1\right)=2$ and $x,y\notin V_1$. Then, $c\left(ux\right)\ne 1$. Also $c\left(ux\right)\ne 2$ as $2\in T$, which implies that $c\left(vy\right)=1$. Otherwise, $vuxyv$ would be a PC-$C_4$. Now, we suppose that x, y and v are in distinct vertex classes. Then, $vx, vy\in E\left(G\right)$. By symmetry, assume that u and x are in different vertex classes. Similar to the analysis above, we have $c\left(ux\right)\notin \{1,2\}$, which also implies that $c\left(vy\right)=1$. □

We may suppose that $\left|T\right|\ge 2$. Since if $T=\left\{2\right\}$, then by Claim 1, $1\in col(V_1,V_2)$, and by the fact that if $vw\in E\left(G\right)$, then $c\left(vw\right)\in \left\{1\right\}\cup col(V_1,V_2)$, we have $mcc\left(G\right)\ge |col(V_1,V_2)| \ge p-1$. Let $G^{\prime }$ be a subgraph of G by the edges whose colors belong to T. Clearly, $G^{\prime }$ has no isolated vertices and $u,v,w\notin V\left(G^{\prime }\right)$. Let $\nu \left(G^{\prime }\right)$ denote the matching number of $G^{\prime }$. We give the following claim.

Claim 2.

$\nu \left(G^{\prime }\right)=1$ and x or y is a dominating vertex of $G^{\prime }$.

Proof. 

By contradiction, assume that M is a maximum matching of $G^{\prime }$ and $\left|M\right|\ge 2$. Let $x_1{y}_1,x_2{y}_2\in M$. By Claim 1, $c\left(v{x}_1\right)=1$ or $c\left(v{y}_1\right)=1$ and $c\left(v{x}_2\right)=1$ or $c\left(v{y}_2\right)=1$, which contradicts with the fact that $\Delta \left(G^1\right)=2$. Thus, we have proven $\nu \left(G^{\prime }\right)=1$. Also, we assume that neither x nor y is a dominating vertex of $G^{\prime }$. Recall that $\left|T\right|\ge 2$. Then, there must exist another vertex $z\in V\left(G^{\prime }\right)$ such that $xz,yz\notin E\left(G^{\prime }\right)$. Note that z is not an isolated vertex of $G^{\prime }$, which contradicts the fact that $\nu \left(G^{\prime }\right)=1$. □

By symmetry, assume that x is a dominating vertex of $G^{\prime }$. We show $col(x,V\left(G^{\prime }\right)\setminus \left\{x\right\})=T$. Since if $T-col(x,V\left(G^{\prime }\right)\setminus \left\{x\right\})\ne \varnothing$, then $\nu \left(G^{\prime }\right)\ge 2$ as $\Delta =2$, a contradiction. Let $W_1=\{u,v,w,x\}$ and $W_2=V\left(G\right)\setminus W_1$. Now, we consider the colors in $col\left(W_1\right)$ and complete this case by the following 6 subcases.

Subcase 1. $v,w$ are in the same vertex class, and $u,x$ are in the same vertex class.

Note that u and y are in different vertex class. Then, $c\left(uy\right)\ne 1$ as $\Delta \left(G^1\right)=2$ and $c\left(uy\right)\ne 2$ as $2\in T$, which implies that $c\left(vx\right)=c\left(wx\right)=1$, and $col\left(W_1\right)=\left\{1\right\}$. Thus, $mcc\left(G\right)\ge |col(W_1,W_2)| \ge p-1$.

Subcase 2. $x,v,w$ are in the same vertex class.

Since $c\left(ux\right)\notin \{1,2\}$, $c\left(vy\right)=c\left(wy\right)=1$. That is to say, $1\in col(W_1,W_2)$. Then, $mcc\left(G\right)\ge |col(W_1,W_2)| \ge p-1$.

Subcase 3. $v,w$ are in the same vertex class, and $u,v,x$ are in distinct vertex classes.

If v and y are in the same vertex class, then $c\left(vx\right)=c\left(wx\right)=1$. Then, $ux$ is the only edge-colored $c\left(ux\right)$ in $G\left[W_1\right]$. Recall that $\Delta =2$. Then, $c\left(ux\right)\in col(W_1,W_2)$. Otherwise, it would yield a PC-$C_4$. Thus, $mcc\left(G\right)\ge |col(W_1,W_2)| \ge p-1$. Now, we suppose that v and y are in different vertex classes. Since $u,x$ are in distinct vertex classes, $c\left(vy\right)=c\left(wy\right)=1$. Then, $1\in col(W_1,W_2)$. For color set $C=\left\{c\right(ux),c(vx),c(wx\left)\right\}$, if $\left|C\right|=1$, then $mcc\left(G\right)\ge |col(W_1,W_2)| \ge p-1$. If $\left|C\right|=3$, then $ux$ is the only edge-colored $c\left(ux\right)$ in $G\left[W_1\right]$. Then, $c\left(ux\right)\in col(W_1,W_2)$. So are $c\left(vx\right)$ and $c\left(wx\right)$. Thus, $mcc\left(G\right)=|col(W_1,W_2)| =p$. We may suppose that $\left|C\right|=2$. By symmetry, assume that $c\left(ux\right)\ne c\left(vx\right)=c\left(wx\right)$. Then, $ux$ is the only edge-colored $c\left(ux\right)$ in $G\left[W_1\right]$. We have $c\left(ux\right)\in col(W_1,W_2)$ as $\Delta =2$. Thus, $mcc\left(G\right)\ge |col(W_1,W_2)| \ge p-1$.

Subcase 4. $u,v,w$ are in distinct vertex classes, and $u,x$ are in the same vertex class.

Note that u and y are in different vertex class. Then, $c\left(uy\right)\ne 1$ as $\Delta \left(G^1\right)=2$ and $c\left(uy\right)\ne 2$ as $2\in T$, which implies that $c\left(vx\right)=c\left(wx\right)=1$. If $c\left(vw\right)=1$, then we are finished. So, we suppose that $c\left(vw\right)\ne 1$. Then, $vw$ is the only edge-colored $c\left(vw\right)$ in $G\left[W_1\right]$, which implies that $c\left(vw\right)\in col(W_1,W_2)$ as $\Delta =2$. Thus, $mcc\left(G\right)\ge |col(W_1,W_2)| \ge p-1$.

Subcase 5. $u,v,w$ are in distinct vertex classes, and $v,x$ are in the same vertex class.

Note that u and x are not in the same vertex class, which implies that $vy$ or $wy\in E\left(G\right)$, and $c\left(vy\right)=1$ or $c\left(wy\right)=1$ if exists. Thus, $1\in col(W_1,W_2)$. For color set $C^{\prime }=\{c\left(ux\right),c\left(wx\right),c\left(vw\right)\}$, we complete the proof by the same method as in Subcase 3.

Subcase 6. $u,v,w,x$ are in distinct vertex classes.

If u and y are not in the same vertex class, then $c\left(vx\right)=c\left(wx\right)=1$, which implies that $c\left(vw\right)=1$. Otherwise, $uxwvu$ would be a PC-$C_4$. Thus, $ux$ is the only edge-colored $c\left(ux\right)$ in $G\left[W_1\right]$, which implies that $c\left(ux\right)\in col(W_1,W_2)$. We have $mcc\left(G\right)\ge |col(W_1,W_2)| \ge p-1$. So, we suppose that u and y are in the same vertex class. Then, $c\left(vy\right)=c\left(wy\right)=1$ as u and x are not in the same vertex class. Then, $1\in col(W_1,W_2)$. Recall that $c\left(vx\right),c\left(wx\right)\ne 2$ as $2\in T$. Then, $c\left(vw\right)=1$ or $c\left(vw\right)=c\left(vx\right)$. Otherwise, $wvxyw$ would be a PC-$C_4$. Assume that $c\left(vw\right)=1$. For color set $\left\{c\right(ux),c(vx),c(wx\left)\right\}$, we complete the proof by the same method as in Subcase 3. So, we may suppose that $c\left(vw\right)=c\left(vx\right)$. Then, for color set $\left\{c\right(ux),c(vx),c(wx\left)\right\}$, we can also complete the proof by the same method as in Subcase 3.

Case 3. $\Delta =3$.

Assume that $\Delta \left(G^1\right)=3$ and let $c\left(uv\right)=c\left(uw\right)=c\left(ux\right)=1$. Let $U_1=\{u,v,w,x\}$, $U_2=V\left(G\right)\setminus U_1$. We may suppose that $U_2\ne \varnothing$. Otherwise, $G=G\left[U_1\right]$ and $\left|col\right(G\left)\right|\le 4$. It is easy to show that $mcc\left(G\right)\ge p-1$. If for each $i\in \left[p\right]\setminus \left\{1\right\}$, there always exists an edge-colored i which is incident with exactly one vertex in $U_1$, then $mcc\left(G\right)\ge |col(U_1,U_2)| \ge p-1$. Therefore, we may suppose that $T^{\prime }=\left[p\right]\setminus (\left\{1\right\}\cup col(U_1,U_2))\ne \varnothing$. Assume that $2\in T^{\prime }$ and $c\left(yz\right)=2$. Clearly, $2\notin col\left(U_1\right)$ (since otherwise, it would yield a PC-$C_4$), which implies that $y,z\notin U_1$. Now, we give the following claim.

Claim 3.

$1\in col(U_1,U_2)$.

Proof. 

Note that y and z are in different vertex classes. We may first suppose that there is exactly one vertex in $\{y,z\}$ such that it is in the same vertex class as v. By symmetry, assume that y and v are in the same vertex class. Then, v and z are in different vertex classes and $vz\in E\left(G\right)$. $c\left(uy\right)\ne 1$ as $\Delta \left(G^1\right)=3$. $c\left(uy\right)\ne 2$ as $2\in T^{\prime }$. Then, $c\left(vz\right)=1$. Otherwise, $zvuyz$ would be a PC-$C_4$. Hence, $1\in col(U_1,U_2)$. Now, we suppose that $v,y,z$ are in distinct vertex classes. By symmetry, assume that u and y are in different vertex classes. Then, $c\left(uy\right)\notin \{1,2\}$. By the analysis above, we also have $c\left(vz\right)=1$, as claimed. □

By Claim 3, there is $u^{\prime }\in U_2$, such that $1\in col(u^{\prime },\{v,w,x\})$. By symmetry, assume that $c\left(u^{\prime }v\right)=1$. Let $G_1$ be the edge-induced subgraph of G by $uv,uw,ux$ and $u^{\prime }v$. It is not hard to verify that for any 2-partition of $V\left(G_1\right)$, the number of crossing edges is at least 1. Now, we give a random 2-partition of $V\left(G\right)$. For each $v\in V\left(G\right)$, we place v into A or B, independently, with probability 1/2. Let $X_i$ be the indicator random variable which is 1 if $i\in col(A,B)$; otherwise, it is 0. From the above analysis, we can deduce that for each i with $\Delta \left(G^i\right)=3$,

$$E\left(X_i\right)=Pr(i\in col(A,B))\ge 1-2\times {\displaystyle \frac{1}{32}}={\displaystyle \frac{15}{16}}.$$

And for each j with $\Delta \left(G^j\right)\ge 4$,

$$E\left(X_j\right)=Pr(j\in col(A,B))\ge 1-2\times {\displaystyle \frac{1}{32}}={\displaystyle \frac{15}{16}}.$$

It follows that

$$E\left(\right|col(A,B)\left|\right)=\sum _{i\in \left[p\right]}E\left(X_i\right)\ge {\displaystyle \frac{15}{16}}p.$$

Therefore, there exists a partition $V\left(G\right)=A\cup B$ such that $\left|col\right(A,B\left)\right|\ge 15p/16$. Based on the above analyses, we have $mcc\left(G\right)\ge min\{p-1,15p/16\}$.

4. Complete Graphs

4.1. Proof of Theorem 3

Theorem 7.

Let G be a PC-$C_4$-free p-edge-colored complete graph. Then, $mcc\left(G\right)\ge p-1$, and the bound is tight.

Proof. 

Let G be a PC-$C_4$-free edge-colored complete graph with p colors and let ${min}_{1\le i\le p}\{\Delta \left(G^i\right)\}=k$. Without loss of generality, assume that $\Delta \left(G^1\right)=k$. By Lemma 1, $G^1$ contains a dominating vertex. Assume $A=V\left(G^1\right)=\{v_0,v_1,v_2,\dots ,v_k\}$ and $v_0$ is a dominating vertex of $G^1$. We claim that for each $i\in \left[p\right]\setminus \left\{1\right\}$, there exists an edge $e_i$ colored i such that $e_i$ is incident with the vertices in A. By contradiction, assume that all edges colored i are not incident with the vertices in A. Let $e_i=uv$ be an edge-colored i in $E\left(G\right)$. Then, $v_0{v}_{1}vu{v}_0$ would be a PC-$C_4$.

Now we claim that for each $i\in \left[p\right]\setminus \left\{1\right\}$, $i\in col(A,V(G)\setminus A)$. Let x be a dominating vertex of $G^i$. There exist the following three cases: (1) $x=v_0$; (2) $x\in A\setminus \left\{v_0\right\}$; (3) $x\in V\left(G\right)\setminus A$. If (1) occurs, then $i\in col(v_0,V\left(G\right)\setminus A)\subseteq col(A,V\left(G\right)\setminus A).$ If (2) occurs, then $i\in col(x,V(G)\setminus A)\subseteq col(A,V(G)\setminus A)$ by the fact that $\Delta \left(G^1\right)={min}_{1\le i\le p}\{\Delta \left(G^i\right)\}$. If (3) occurs, then $i\in col(x,A)\subseteq col(A,V(G)\setminus A)$ by the above claim. Hence, $mcc\left(G\right)\ge \left|col\right(A,V\left(G\right)\setminus A\left)\right|=p-1$.

In the following, we show the bound is tight. Let G be a PC-$C_4$-free edge-colored complete graph with exactly one rainbow triangle. Clearly, such a graph G exists. For example, let $G\cong K_4$ with $V\left(G\right)=\{u,v,x,y\}$ and $c\left(uv\right)=1$, $c\left(vx\right)=2$, $c\left(xu\right)=3$, $c\left(uy\right)=c\left(vy\right)=c\left(xy\right)=4$. It is not hard to calculate that $mcc\left(G\right)=p-1$. □

4.2. Proof of Theorem 4

Theorem 8.

Let G be a PC-$C_5$-free p-edge-colored complete graph, where $p\ge 7$. Then, $mcc\left(G\right)\ge min\{p-6,7p/8\}$.

Proof. 

Let G be a PC-$C_5$-free p-edge-colored complete graph, where $p\ge 7$. If $\Delta ={min}_{1\le i\le p}\{\Delta \left(G^i\right)\}\ge 3$, then by Lemma 2, $mcc\left(G\right)\ge 7p/8$. So we suppose that $\Delta \le 2$. We finish this proof by considering the following two cases.

Case 1. $\Delta =1$.

Suppose that $\Delta \left(G^1\right)=1$. Then, $E\left(G^1\right)$ is a match and assume that $e=uv\in E\left(G^1\right)$. If for each $i\in \left[p\right]\setminus \left\{1\right\}$, there exists an edge-colored i which is adjacent to e, then $mcc\left(G\right)\ge |col(V_1,V_2)| \ge p-1$, where $V_1=\{u,v\}$, $V_2=V\left(G\right)\setminus V_1$. So we suppose that $2\in \left[p\right]$ that all edges colored 2 are not adjacent to e. Assume that $e^{\prime }=xy$ is colored 2 and not adjacent to e. Let $A=\{u,v,x,y\}$. Then, $G\left[A\right]$ is a $K_4$ and $\left|col\right(A\left)\right|\le 6$. If for each $i\in \left[p\right]\setminus col\left(A\right)$, there exists an edge-colored i which is incident with one vertex in A, then $mcc\left(G\right)\ge |col(V_1,V_2)| \ge p-6$, where $V_1=A$, $V_2=V\left(G\right)\setminus A$. So we suppose that $I\ne \varnothing$, where

$$I=\{i\in [p]\setminus col(A):\mathrm{all}\mathrm{edges}\mathrm{colored}i\mathrm{are}\mathrm{not}\mathrm{incident}\mathrm{with}\mathrm{the}\mathrm{vertices}\mathrm{in}A\}.$$

For each $i\in I$, choose one edge-colored i in $E\left(G\right)$ and put it into T. Let $e^{″}=wz\in T$ and $col\left(e^{″}\right)=3$. Then, $w,z\notin A$ as $3\in I$. Now, we claim that $c\left(vx\right)=c\left(uy\right)$.

Claim 4.

$c\left(vx\right)=c\left(uy\right)$.

Proof. 

By contradiction, assume that $c\left(vx\right)\ne c\left(uy\right)$. Recall that $\Delta \left(G^1\right)=1$; all edges colored 2 are not adjacent to $e=uv$; and $3\in I$. Then, $c\left(uy\right),c\left(vx\right),c\left(uw\right)\notin \{1,2,3\}$. Assume that $c\left(vx\right)=4$ and $c\left(uy\right)=5$. Note that $uvxzwu$ is not a PC-$C_5$, which implies that $c\left(xz\right)=4$. Then, $c\left(uw\right)=5$ as $uwzxyu$ is not a PC-$C_5$, and $c\left(vz\right)=4$ as $vuyxzv$ is not a PC-$C_5$. Recall that $c\left(vw\right)\notin \{1,3\}$ and $uvwzyu$ is not a PC-$C_5$. Then, $c\left(zy\right)=5$. $c\left(vw\right)=4$ as $vwzyxv$ is not a PC-$C_5$. $c\left(ux\right)=4$ as $vuxzwv$ is not a PC-$C_5$. Thus, $uxyzwu$ would be a PC-$C_5$, a contradiction. □

Let $c\left(vx\right)=c\left(uy\right)=4$. We claim that for any one edge $e^{″}=wz\in T$, $col\left(\right\{w,z\},A)=\left\{4\right\}$.

Claim 5.

$col\left(\right\{w,z\},A)=\left\{4\right\}$, $c\left(ux\right)=c\left(vy\right)=4$.

Proof. 

Note that $c\left(uw\right)\notin \{1,3\}$. Then, $c\left(zx\right)=4$ as $xzwuvx$ is not a PC-$C_5$. $c\left(zv\right)=4$ as $vzxyuv$ is not a PC-$C_5$; and $c\left(wu\right)=4$ as $wuyxzw$ is not a PC-$C_5$. Note that $c\left(wv\right)\notin \{1,3\}$. Then, $c\left(zy\right)=4$ as $zyuvwz$ is not a PC-$C_5$. $c\left(zu\right)=4$ as $zuvxyz$ is not a PC-$C_5$. Thus, $col(z,A)=\left\{4\right\}$. $c\left(wv\right)=4$ as $wvxyzw$ is not a PC-$C_5$; $c\left(wx\right)=4$ as $wxvuzw$ is not a PC-$C_5$; $c\left(wy\right)=4$ as $wyuvzw$ is not a PC-$C_5$. Thus, $col(w,A)=\left\{4\right\}$.

Note that $c\left(ux\right)\ne 1$. Then, $c\left(ux\right)=4$ as $uxwzvu$ is not a PC-$C_5$. $c\left(vy\right)=4$ as $vyxwzv$ is not a PC-$C_5$. □

Claim 6.

T is a match.

Proof. 

Due to the contradiction that T is not a match, let $e_1=u^{\prime }{v}^{\prime }\in T$ and $e_2=v^{\prime }{w}^{\prime }\in T$. By the definition of T, $c\left(e_1\right)\ne c\left(e_2\right)$. By Claim 5, $col(\{u^{\prime },v^{\prime },w^{\prime }\},A)=\left\{4\right\}$. Thus, $uv{w}^{\prime }{v}^{\prime }{u}^{\prime }u$ would be a PC-$C_5$, a contradiction. □

Let $\left|I\right|=k$ and $T=\{u_1{v}_1,u_2{v}_2,\dots ,u_k{v}_k\}$. The result clearly holds if $k=1$. So we suppose that $k\ge 2$. Let $A^{\prime }=A\cup \{u_1,u_2,\dots ,u_k\}$. Then, for each $1\le i<j\le k$, $col(u_i,A)=\left\{4\right\}$ and $c\left(u_i{u}_j\right)\notin I$. Hence, $mcc\left(G\right)\ge \left|col\right(A^{\prime },V\left(G\right)\setminus A^{\prime }\left)\right|\ge p-2$. We complete the proof of this case.

Case 2. $\Delta =2$.

Assume that $\Delta \left(G^1\right)=2$. Let $c\left(uv\right)=c\left(uw\right)=1$ and $B=\{u,v,w\}$. Let

$$I^{\prime }=\{i\in \left[p\right]\setminus col\left(B\right):\mathrm{all}\mathrm{edges}\mathrm{colored}i\mathrm{are}\mathrm{not}\mathrm{incident}\mathrm{with}\mathrm{the}\mathrm{vertices}\mathrm{in}B\}.$$

We may suppose that $|I^{\prime }|\ge 5$. Otherwise, $mcc\left(G\right)\ge \left|col\right(B,V\left(G\right)\setminus B\left)\right|\ge p-6$, and we are finished. Let $2,3\in I^{\prime }$ and $c\left(x_1{y}_1\right)=2$ and $c\left(x_2{y}_2\right)=3$. Then, $x_1,x_2,y_1,y_2\notin B$. We give the following claim.

Claim 7.

$1\in col(B,V(G)\setminus B)$.

Proof. 

Suppose for a contradiction that $1\notin col(B,V(G)\setminus B)$. Then, we have $1,2,3\notin col(B,\{x_1,x_2,y_1,y_2\})$. We first show that $x_1{y}_1$ and $x_2{y}_2$ are not adjacent. Otherwise, assume that $y_1=y_2$. Then, $uv{x}_1{y}_1{x}_{2}u$ would be a PC-$C_5$. From this, we see that $2,3\notin col(\{x_1,y_1\},\{x_2,y_2\})$, which implies that $c\left(u{x}_1\right)=c\left(u{y}_2\right)$, $c\left(v{x}_1\right)=c\left(v{y}_2\right)$ and $c\left(w{x}_1\right)=c\left(w{y}_2\right)$. Note that $uv{x}_1{x}_2{y}_{2}u$ is not a PC-$C_5$. Then, $c\left(x_1{x}_2\right)=c\left(v{x}_1\right)=c\left(v{y}_2\right)$.

For each $i\in I^{\prime }$, choose exactly one edge-colored i in $E\left(G\right)$ and put it into $T^{\prime }$. Then, $T^{\prime }$ is a match. Let $T^{\prime }=\{w_1{z}_1,w_2{z}_2,\dots ,w_{|I^{\prime }|}{z}_{|I^{\prime }|}\}$, $B^{\prime }=B\cup \{w_1,w_2,\dots ,w_{|I^{\prime }|}\}$ and $B^{″}=V\left(G\right)\setminus B^{\prime }$. By the analysis above, we know that for any $1\le i<j\le |I^{\prime }|$, $c\left(u{w}_i\right),c\left(v{w}_i\right),c\left(w{w}_i\right),c\left(w_i{w}_j\right)\in col(B^{\prime },B^{″})$. Thus, $mcc\left(G\right)\ge |col(B^{\prime },B^{″})| \ge p-2$. Therefore, we may suppose that $1\in col(B,V(G)\setminus B)$. □

Recall that $\Delta \left(G^1\right)=2$. Then, by Claim 7, $G^1$ contains a path of length 3. For each $v\in V\left(G\right)$, we place v into S or T, independently, with probability 1/2. Let $Y_i$ be the indicator random variable which is 1 if $i\in col(S,T)$; otherwise, it is 0. From the above analysis, we can deduce that for each i with $\Delta \left(G^i\right)=2$, $G^i$ contains a path of length 3, and we have

$$E\left(Y_i\right)=Pr(i\in col(S,T))\ge 1-2\times {\displaystyle \frac{1}{16}}={\displaystyle \frac{7}{8}}.$$

And for each j with $\Delta \left(G^j\right)\ge 3$,

$$E\left(Y_j\right)=Pr(j\in col(S,T))\ge 1-2\times {\displaystyle \frac{1}{16}}={\displaystyle \frac{7}{8}}.$$

It follows that

$$E\left(\right|col(S,T)\left|\right)=\sum _{i\in \left[p\right]}E\left(Y_i\right)\ge {\displaystyle \frac{7}{8}}p.$$

Therefore, there exists a partition $V\left(G\right)=S\cup T$ such that $\left|col\right(S,T\left)\right|\ge 7p/8$. Based on the analyses above, we have $mcc\left(G\right)\ge min\{p-6,7p/8\}$.

5. Concluding Remarks

We conclude several remarks and propose an open problem in this section.

• For Theorem 2 and Theorem 4, we provide the corresponding lower bounds of $mcc\left(G\right)$ by using graph structure analysis and probabilistic methods. We remark that it is difficult to strengthen them significantly by the same method. Maybe the bound $p-6$ of Theorem 4 can be improved slightly by more detailed analyses.
• It is an interesting direction to replace PC-$C_k$ with rainbow-$C_k$ for forbidden edge-colored subgraphs and ask the same question to determine the bound of $mcc\left(G\right)$, but it could be more challenging.
• Motivated by the results we obtained, we pose the following problem.

Problem 1.

For any integer $k\ge 5$, let G be a PC-$C_k$-free p-edge-colored complete graph. It is unclear whether there exists a constant $C\left(k\right)$ such that $mcc\left(G\right)\ge p-C\left(k\right)$.