Rainbow $C_4$’s and Directed $C_4$’s: The Bipartite Case Study

Abstract

In this paper we obtain a new sufficient condition for the existence of directed cycles of length 4 in oriented bipartite graphs. As a corollary, a conjecture of Li is confirmed. As an application, a sufficient condition for the existence of rainbow cycles of length 4 in bipartite edge-colored graphs is obtained.

1 Introduction

For terminology and notation not defined here, we refer to [2]. Let $G=(V,E)$ be a simple graph. An edge-coloring of $G$ is a mapping $C: E\rightarrow \mathbb N^{+}$, where $\mathbb N^{+}$ is the set of positive integers. We call $G^c$ an edge-colored graph (or briefly, a colored graph) if $G$ is assigned an edge-coloring $C$. Let $v$ be a vertex of $G^c$. The color degree of $v$ in $G^c$, denoted by $d^{c}_{G}(v)$ (or briefly, $d^{c}(v)$), is the number of colors of the edges incident to $v$. A color neighborhood set $N^c (v)$ of $v$ is a subset of $N(v)$, the neighborhood of $v$, such that the colors of edges between $v$ and $N^c (v)$ are pairwise distinct. Let $H$ be a subgraph of $G$, then $C(H)=\{C(e):e\in E(H)\}$ is called the color set of $H$.

A subgraph of a colored graph is called rainbow (sometimes called heterochromatic or colorful) if all edges of it have distinct colors. The existence of rainbow subgraphs has been studied for a long time. A problem of rainbow Hamilton cycles in colored complete graphs was mentioned by Erdös et al. [6], and later studied by Hahn and Thomassen [8], Frieze and Reed [7] and Albert et al. [1], respectively. Rainbow matchings were studied by Wang and Li [16], Lesaulnier et al. [11], and Kostochka and Yancey [10]. Chen and Li [4, 5] studied the existence of long rainbow paths. A recent article on strong rainbow connection can be found in [15]. For a survey on the study of rainbow subgraphs in colored graphs, we refer to [9].

In particular, rainbow short cycles have received much attention. Broersma et al. [3] studied the existence of rainbow $C_3$’s and $C_4$’s under color neighborhood union condition. Later, Li and Wang [14] obtained two results on the existence of rainbow $C_3$’s and $C_4$’s under colored degree condition.

Theorem 1

(Li and Wang [14]) Let $G^c$ be a colored graph of order $n\ge 3$. If $d^{c}(v)\ge (4\sqrt{7}/7-1)n+3-4\sqrt{7}/7$ for each $v\in V(G)$, then $G^c$ has either a rainbow $C_3$ or a rainbow $C_4$.

Theorem 2

(Li and Wang [14]) Let $G^c$ be a colored graph of order $n\ge 3$. If $d^{c}(v)\ge (\sqrt{7}+1)n/6$ for each $v\in V(G)$, then $G^c$ has a rainbow $C_3$.

Li and Wang [14] conjectured that every colored graph $G^c$ of order $n\ge 3$ has a rainbow $C_3$ if $d^{c}(v)\ge (n+1)/2$ for each $v\in V(G)$. This conjecture was proved by Li [13] and stronger results were proved by Li et al. [12] with different methods as follows.

Theorem 3

(Li [13]) Let $G^c$ be a colored graph of order $n\ge 3$. If $d^{c}(v)\ge (n+1)/2$ for each $v\in V(G)$, then $G^c$ has a rainbow $C_3$.

Theorem 4

(Li et al. [12]) Let $G^c$ be a colored graph of order $n\ge 3$. If $\sum \limits _{v\in V(G)}d^c(v)\ge n(n+1)/2$, then $G^c$ has a rainbow $C_3$.

Theorem 5

(Li et al. [12]) Let $G^c$ be a colored graph of order $n\ge 3$. If $d^{c}(v)\ge n/2$ for each $v\in V(G)$, then $G^c$ has a rainbow $C_3$ or $G\in \{K_{n/2,n/2}, K_4-e, K_4 \}$.

The existence of rainbow $C_4$’s in special colored graphs has also been studied. Wang et al. [17] obtained a result on the existence of rainbow $C_4$’s in triangle-free colored graphs. Recently, Li [13] got a result on the existence of rainbow $C_4$’s in balanced bipartite colored graphs.

Theorem 6

(Wang et al. [17]) Let $G^c$ be a triangle-free colored graph of order $n\ge 9$. If $d^{c}(v)\ge (3-\sqrt{5})n/2+1$ for each $v\in V(G)$, then $G^c$ has a rainbow $C_4$.

Theorem 7

(Li [13]) Let $G^c$ be a balanced bipartite colored graph of order $2n$ with bipartition $(A,B)$. If $d^c(v)> 3n/5+1$ for each $v\in A\cup B$, then $G^c$ has a rainbow $C_4$.

While in [13], Li made a tiny error in the proof of Theorem 7. Notice that $K_{3,3}$ is 3-edge-colorable, and a proper 3-edge-coloring of $K_{3,3}$ satisfies the condition of Theorem 7, but it has no rainbow $C_4$ since there are only 3 colors. We point out that, in order to correct it, the condition $d^c(u)>3n/5+1$ should be changed into $d^c(u)>(3n+8)/5$.

Now we turn to finite simple oriented graphs, i.e., finite graphs without multiple edges and loops in which each edge is replaced by exactly one arc. Let $D[A,B]$ be an oriented bipartite graph with bipartition $(A,B)$. When there is no ambiguity, we use $D$ instead of $D[A,B]$. For $A_1\subseteq A$ and $B_1\subseteq B$, we denote by $A_{D}(A_1,B_1)$ the set of arcs from $A_1$ to $B_1$ in $D[A,B]$.

The study of rainbow cycles in colored graphs is largely related to the study of oriented cycles in digraphs. For a wonderful example, see the introduction of [13]. In particular, motivated by the study of short rainbow cycles in colored graphs, Li [13] proposed the following nice conjecture and proved for balanced oriented bipartite graphs.

Conjecture 1

(Li [13]) Let $D$ be an oriented bipartite graph with bipartition $(A,B)$. If $d^{+}(u)>|B|/3$ for each $u\in A$ and $d^{+}(v)>|A|/3$ for each $v\in B$, then $D$ has a directed $C_4$.

Theorem 8

(Li [13]) Let $D$ be a balanced oriented bipartite graph with bipartition $(A,B)$, where $|A|=|B|=n$. If $d^{+}(v)>n/3$ for each $v\in A\cup B$, then $D$ has a directed $C_4$.

We state a construction from [17] to show that if Conjecture 1 holds, then it would be almost the best possible. Let $m$ and $n$ be two positive integers divisible by 3. Let $|M_0|=|M_1|=|M_2|=m/3$ and $|N_0|=|N_1|=|N_2|=n/3$. We construct an oriented bipartite graph with bipartition $(M, N)$, where $M=M_0\cup M_1\cup M_2$ and $N=N_0\cup N_1\cup N_2$, by creating all possible arcs from $M_i$ to $N_i$, and from $N_i$ to $M_{i+1}$, $i=0,1,2$ (modulo 3). In the rest parts, we use $D^*(m,n)$ to denote the construction above.

The first purpose of this paper is to confirm Conjecture 1. In fact, we prove a stronger result as follows.

Theorem 9

Let $D$ be an oriented bipartite graph with bipartition $(A,B)$, where $|A|=m\ge 2$ and $|B|=n\ge 2$. If $d^{+}(u)\ge n/3$ for each $u\in A$ and $d^{+}(v)\ge m/3$ for each $v\in B$, then either $D$ has a directed $C_4$ or $D=D^*(m,n)$.

By using Theorem 9, we can extend Theorem 7 as follows.

Theorem 10

Let $G^c$ be a bipartite colored graph with bipartition $(A,B)$. If $d^c(u)\ge (3|B|+8)/5$ for each $u\in A$ and $d^c(v)\ge (3|A|+8)/5$ for each $v\in B$, then $G^c$ has a rainbow $C_4$.

2 Proofs

2.1 Proof of Theorem 9

Let $\mathcal {D}(m,n)$ be the family of digraphs consisting of those oriented bipartite graphs with bipartition $(A,B)$ which satisfy the condition of Theorem 9, where $m=|A|$ and $n=|B|$.

First, we claim that it is sufficient to prove for those $m$ and $n$ which are both multiples of 3. Suppose Theorem 9 holds for $\mathcal {D}(m,n)$ with $m\equiv n\equiv 0~(\mathrm{mod}~3)$. For any $D\in \mathcal {D}(m',n')$, where $m'$ and $n'$ are not both multiples of 3, let $s_1=3\lceil \frac{m'}{3}\rceil -m'$, $A^{*}=\{u_1,\ldots ,u_{s_1}\}$, and $A'=A\cup A^{*}$. Let $s_2=3\lceil \frac{n'}{3}\rceil -n'$, $B^{*}=\{v_1,\ldots ,v_{s_2}\}$ and $B'=B\cup B^{*}$. Now we construct a new oriented bipartite graph $D'$ with bipartition $(A',B')$, where $A(D')=A(D)\cup \{u'v,v'u: u\in A,u'\in A^{*},v\in B, v'\in B^{*}\}$. Notice that $d^{+}_{D'}(u)\ge \lceil \frac{|B|}{3}\rceil =\frac{|B'|}{3}$ for each $u\in A$ and $d^{+}_{D'}(u')=n'>\frac{|B'|}{3}$ for each $u'\in A^{*}$. Similarly, $d^{+}_{D'}(v)\ge \lceil \frac{|A|}{3}\rceil =\frac{|A'|}{3}$ for each $v\in B$ and $d^{+}_{D'}(v')=m'>\frac{|A'|}{3}$ for each $v'\in B^{*}$. It follows that $D'\in \mathcal {D}(3\lceil \frac{m'}{3}\rceil , 3\lceil \frac{n'}{3}\rceil )$. Hence $D'$ has a directed $C_4$ or $D'=D^*(3\lceil \frac{m'}{3}\rceil , 3\lceil \frac{n'}{3}\rceil )$. Since $A^{*}$ and $B^{*}$ are not both empty sets and the vertices in $A^{*}$ and $B^{*}$ only have outdegrees, $D'\ne D^*(3\lceil \frac{m'}{3}\rceil , 3\lceil \frac{n'}{3}\rceil )$, and moreover, the directed $C_4$ in $D'$ is also in $D$. The proof of our claim is complete.

Now assume $D\in \mathcal {D}(m, n)$, where $m=3m_1$, $n=3n_1$, and $m_1,n_1$ are two positive integers. Let $D_1$ be a spanning subdigraph of $D$ satisfying $d_{D_1}^{+}(u)=n_1$ for each $u\in A$ and $d_{D_1}^{+}(v)=m_1$ for each $v\in B$. Suppose $D$ has no directed $C_4$, obviously, $D_1$ also has no directed $C_4$. Let $u_0$ be a vertex with maximum indegree $k_1$ among $A$, and $v_0$ be a vertex with maximum indegree $k_2$ among $B$. Let $B_1=N_{D_1}^{-}(u_0)$, $B_2=N_{D_1}^{+}(u_0)$, $A_3=N_{D_1}^{+}(B_2)$, and $B_3=N_{D_1}^{+}(A_3)-B_2$, where $|B_1|=k_1$, $|B_2|=n_1$. Since $D_1$ has no directed $C_4$, we have $N_{D_1}^{+}(A_3)\cap B_1=\emptyset$. Since $k_2$ is the maximum indegree of all vertices in $B$, we get

$$\begin{aligned} |B_3|k_2\ge |N_{D_1}^{-}(B_3)|=|A_{D_1}(A_3,B_3)|. \end{aligned}$$

(1)

Since $D_1$ has no directed $C_4$, there is no arc from $A_3$ to $B_1$, which implies all arcs starting from $A_3$ have heads in $B_2\cup B_3$. Hence $|A_{D_1}(A_3,B_3)|=|A_3|n_1-|A_{D_1}(A_3,B_2)|$. Since

$$\begin{aligned} |A_{D_1}(A_3,B_2)|\le |A_3||B_2|-|A_{D_1}(B_2,A_3)|, \end{aligned}$$

(2)

we obtain

$$\begin{aligned} |A_{D_1}(A_3,B_3)|\ge |A_3|n_1-(|A_3||B_2|-|A_{D_1}(B_2,A_3)|) =|A_{D_1}(B_2,A_3)|=m_1n_1. \end{aligned}$$

(3)

Together with (1) and (3), we obtain $|B_3|\ge \frac{m_1n_1}{k_2}$. Therefore,

$$\begin{aligned} 3n_1=|B|\ge |B_1|+|B_2|+|B_3|\ge k_1+n_1+\frac{m_1n_1}{k_2}\ge 3\root 3 \of {\frac{k_1m_1n_1^2}{k_2}}. \end{aligned}$$

(4)

It follows that $k_2n_1\ge k_1m_1$. By symmetry, we also have $k_1m_1\ge k_2n_1$. Thus, $k_1m_1=k_2n_1$ and all the inequalities (1)–(4) are actually equalities. These facts imply $|B_1|=|B_2|=|B_3|=n_1=k_1$, $m_1=k_2$, $|A_{D_1}(B_2,A_3)|=|A_{D_1}(A_3,B_3)|=m_1n_1$, and all vertices in $A$ have indegree $n_1$ in $D_1$, all vertices in $B$ have indegree $m_1$ in $D_1$.

Next we show $|A_3|=m_1$. First, choose a vertex $v'\in B_2$, $d^+(v')=m_1$, so $|A_3|\ge m_1$ by the definition of $A_3$. Assume that $|A_3|>m_1$. Since the inequality (2) becomes equality, the underlying graph of $D_1[A_3,B_2]$ is a complete bipartite graph. Hence there exists a vertex, say $u'\in A_3$, such that $u'v'\in D_1$. Since $u'\in A_3$, there exist a vertex $v''\in B_2$, such that $v''u'\in D_1$ by the choice of $A_3$. Note that $d^+_{D[B_2,A_3]}(v')=d^+_{D[B_2,A_3]}(v'')=m_1$ and $u'\notin N^+(v')$. It follows that there exists a vertex, say $u''\in A_3$, such that $v'u''\in D_1$ and $v''u''\notin D_1$. Since the underlying graph of $D_1[B_2,A_3]$ is a complete bipartite graph, $u''v''\in D_1$. Now $C=u'v'u''v''u'$ is a directed $C_4$ in $D_1$, a contradiction. Hence $|A_3|=m_1$.

Now let $A_1=N_{D_1}^{+}(B_3)$ and $A_2=N_{D_1}^{-}(B_2)$. Note that $|\{vu: v\in B_2, u\in A_3\}|=m_1n_1=|A_{D_1}(B_2,A_3)|$. It follows that $A_{D_1}(B_2, A_3)=\{vu: v\in B_2, u\in A_3\}$. Similarly, $A_{D_1}(A_3, B_3)=\{uv: u\in A_3, v\in B_3\}$. So there is no arc with tail in $A_3$ and head in $B_2$, or arc with tail in $B_3$ and head in $A_3$, follows $A_1\cap A_3=A_2\cap A_3=\emptyset$. Since $D_1$ has no directed $C_4$, $A_1\cap A_2=\emptyset$.

Since $d_{D_1}^+(u)=d_{D_1}^-(u)=n_1$ for each $u\in A$ and $d_{D_1}^+(v)=d_{D_1}^-(v)=m_1$ for each $v\in B$, we obtain $|A_1|\ge m_1$ by its definition, and $n_1|A_2|\ge |A_{D_1}(A_2, B_2)|=\sum _{v\in B_2} d_{D_1}^- (v)=m_1n_1$, follows that $|A_2|\ge m_1$. Since $|A_1|+|A_2|+|A_3|=3m_1$ and $A_1, A_2, A_3$ are pairwise disjoint, $|A_1|=|A_2|=|A_3|=m_1$. Now apparently, $A_{D_1}(A_2,B_2)=\{uv:u\in A_2,v\in B_2\}$ and $A_{D_1}(B_3,A_1)=\{vu:v\in B_3,u\in A_1\}$. Since $A_1$ and $A_2$ are disjoint, $A_{D_1}(A_1,B_2)=\emptyset$. Furthermore, $A_{D_1}(A_1,B_3)=\emptyset$, hence $N^+(A_1)\subset B_1$. $\sum _{v\in A_1}d^+(v)=m_1n_1$ implies $A_{D_1}(A_1,B_1)=\{uv:u\in A_1,v\in B_1\}$. Similarly, $A_{D_1}(B_1,A_2)= \{vu: v\in B_1,u\in A_2\}$. Therefore $D_1=D^{*}(m,n)$. If there is any arc in $D$ but not in $D_1$, then obviously, there would be a directed $C_4$ in $D$, a contradiction. Thus, $D=D_1=D^{*}(m,n)$.

The proof is complete. $\square$

2.2 Proof of Theorem 10

First note that the color degree condition implies that $|A|\ge 4$ and $|B|\ge 4$.

Suppose not. Let $G^c$ be a colored graph which satisfies the condition of Theorem 10 but has no rainbow $C_4$. Set $|A|=n_1$ and $|B|=n_2$.

Choose an edge $e=xy\in E(G^c)$ such that $C(xy)=c_0$. Let $N^c(x)=\{y,y_1,y_2,\ldots ,y_{r-1}\}$ and $N^c(y)=\{x,x_1,x_2,\ldots ,x_{s-1}\}$. Since $d^c(x)\ge \frac{3n_2+8}{5}$ and $d^c(y)\ge \frac{3n_1+8}{5}$, we can set $r=\lceil \frac{3n_2+8}{5}\rceil$ and $s=\lceil \frac{3n_1+8}{5}\rceil$. Let $A_1=\{x_1,x_2,\ldots ,x_{s-1}\}$ and $B_1=\{y_1,y_2,\ldots ,y_{r-1}\}$. Note that $G^c[A_1,B_1]$ is also a bipartite colored graph.

The following claim can be deduced immediately from the definition of color neighborhood set and the assumption that $G^c$ has no rainbow $C_4$.

Claim 1

For any edge $x_iy_j$ and $C(xy_j)\ne C(yx_i)$, where $1\le i\le s-1$ and $1\le j\le r-1$, we have $C(x_iy_j)\in \{C(xy),C(xy_j),C(yx_i)\}$.

Now we construct an oriented bipartite graph $D=D[A_1,B_1]$ as follows. For any edge $x_iy_j\in E(G[A_1,B_1])$, such that $C(x_iy_j)\ne C(xy)$ and $C(xy_j)\ne C(yx_i)$, then $C(x_iy_j)=C(xy_j)$ or $C(x_iy_j)=C(yx_i)$ by Claim 1. If $C(x_iy_j)=C(xy_j)$, we define an arc $x_i y_j$ in $D$, and if $C(x_iy_j)= C(yx_i)$, we define an arc $y_j x_i$ in $D$. Let $G'=G[D]$ be the underlying graph of $D$.

In the following, for convenience, when we mention the color of an arc in $D$, we mean the color of the corresponding edge in $E(G')$.

Claim 2

There is no directed $C_4$ in $D$.

Proof

Suppose $Q=x_iy_jx_py_qx_i$ is a directed $C_4$ in $D$. By the definition of $D$, we have $C(x_iy_j)=C(xy_j)$, $C(y_jx_p)=C(yx_p)$, $C(x_py_q)=C(xy_q)$, and $C(y_qx_i)=C(yx_i)$. The existence of arcs $x_iy_j$ and $y_jx_p$ implies $C(xy_j)\ne C(yx_i)$ and $C(xy_j)\ne C(yx_p)$, hence $C(x_iy_j)\ne C(y_qx_i)$ and $C(x_iy_j)\ne C(y_jx_p)$. We have $C(xy_j)\ne C(xy_q)$ from the definition of $B_1$, hence $C(x_iy_j)\ne C(x_py_q)$. So $C(x_iy_j)$ is different from the colors of all other three edges in the cycle $Q$ in $G^c$. Similarly, we can prove that all edges in $Q$ receive distinct colors in $G^c$, and therefore, $Q$ is a rainbow $C_4$ in $G^c$, a contradiction. $\square$

Claim 3

$D\ne D^*(|A_1|,|B_1|)$.

Proof

Assume that $D=D^*(|A_1|,|B_1|)$. Without loss of generality, set $A_1=X_1\cup X_2\cup X_3$ and $B_1=Y_1\cup Y_2\cup Y_3$, where $|X_1|=|X_2|=|X_3|$ and $|Y_1|=|Y_2|=|Y_3|$. Since $s-1\equiv 0~(\mathrm{mod}~3)$ and $r-1\equiv 0~(\mathrm{mod}~3)$, we have $\lceil \frac{3n_i+3}{5}\rceil \equiv 0~(\mathrm{mod}~3)$, $i=1, 2$. It follows that $n_i\equiv 3\ \text {or} \ 4~(\mathrm{mod}~5)$, $i=1, 2$.

First, we claim that $D\ne D^*(3,3)$. Suppose not. Then $\lceil \frac{3n_1+3}{5}\rceil =\lceil \frac{3n_2+3}{5}\rceil =3$. Hence $n_1=n_2=4$. In this case, we may suppose that $X_i=\{x_i\}$ and $Y_i=\{y_i\}$ for $i=1,2,3$. The existence of arc $x_1y_1$ in $D$ implies that $C(x_1y_1)=C(xy_1)$. Hence $d^c(y_1)\le 3< \frac{3|A|+8}{5}$, a contradiction.

Let $v_0$ be an arbitrary vertex of $D$. Without loss of generality, assume $v_0\in Y_1$. If $n_1=5k+\lambda$, $\lambda =3\ \text {or}\ 4$, then $|X_1|=|X_2|=|X_3|=k+1$. From the definition of $D$, we know that each edge in $\{uv_0:u\in X_1\}$ has the same color $C(v_0x)$, and there are $k+1$ different colors in $\{v_0u: u\in X_2\}$. Since $|A\backslash (A_1\cup \{x\})|=5k+\lambda -(3k+3)-1=2k+\lambda -4$, there are at most $2k+\lambda -4$ different colors in the edge set $\{v_0u:u\in A\backslash (A_1\cup \{x\})\}$. These facts mean that there are at most $3k+\lambda -2$ different colors in the edge set $\{v_0u:u\in A\backslash X_3\}$. Since $d^c(v)\ge 3k+\lambda$, there exists an edge $v_0u_0\in \{v_0u:u\in X_3\}$, such that $v_0u_0$ has a new color and $v_0u_0$ is not in $E(G')$.

Next we show that any directed path of length 3 in $D=D^*(|A_1|,|B_1|)$ is rainbow. Without loss of generality, we choose a directed path $uv'u'v$, where $u'\in X_1$, $v\in Y_1$, $u\in X_3$, and $v'\in Y_3$. By the construction of $D$, $C(u'v)=C(xv)$, $C(v'u')=C(yu')$, and $C(uv')=C(xv')$. Since $u'v$ exists in $D$, $C(xv)\ne C(yu')$. It follows that $C(u'v)\ne C(v'u')$. Similarly, we have $C(v'u')\ne C(uv')$. Since $C(u'v)=C(xv)$ and $C(uv')=C(xv')$ and $C(xv)\ne C(xv')$ by the choice of $N^c(x)$, we have $C(u'v)\ne C(uv')$.

Now we fix a vertex $v_0\in Y_1$. By the analysis above, there exists an edge $v_0u_0\in \{v_0u\in E(G): u\in X_3\}$, which is not in $E(G')$ and satisfies $C(v_0u_0)\notin \{C(uv_0):u\in X_1\}$. Since $D=D^*(|A_1|,|B_1|)\ne D^*(3,3)$, there are at least two arcs in $A_D(Y_3,X_1)$ with distinct colors, and we can choose one of them, say $v_0'u_0'\in A_D(Y_3,X_1)$, such that $C(v_0'u_0')\ne C(v_0u_0)$. By the analysis before, there is an edge $u_0v_0''\in E(G^c)$, where $v_0''\in Y_1$, such that it is not in $E(G')$ and satisfies $C(u_0v_0'')\ne C(u_0v_0')$. Now we will show that $v_0''=v_0$. First, the deletion of $u_0v_0$ means $C(u_0v_0)=C(xy)$ or $C(xv_0)=C(yu_0)$. If $C(u_0v_0)=C(xy)$, then the existence of arcs $u_0'v_0$, $u_0'v_0'$, and $u_0v_0'$ in $D$ implies $C(xy)\ne C(u_0'v_0)$, $C(xy)\ne C(u_0'v_0')$, and $C(xy)\ne C(u_0v_0')$. Since colors of arcs in $\{u_0'v_0,v_0'u_0',u_0v_0'\}$ are pairwise distinct, $u_0'v_0'u_0v_0u_0'$ is a rainbow $C_4$ in $G^c$, a contradiction. Hence $C(xv_0)=C(yu_0)$. Similarly, we may obtain $C(xv_0'')=C(yu_0)$ from the deletion of $u_0v_0''$ when constructing $D$. It follows that $C(xv_0)=C(xv_0'')$. Now we can get $v_0=v_0''$ directly from the definition of $N^c(x)$. From the analysis above, colors of arcs in $\{u_0'v_0,v_0'u_0',u_0v_0'\}$ are pairwise distinct, and $C(v_0u_0)$ is different from all of the three. Therefore, $u_0'v_0u_0v_0'u_0'$ is also a rainbow $C_4$ in $G^c$, a contradiction.

$\square$

By Claims 2 and 3, $D$ has no directed $C_4$ and $D\ne D^*(|A_1|,|B_1|)$. By Theorem 9, there exists a vertex, without loss of generality, say $y_j\in B_1$, such that $d^{+}_{D}(y_j)<\frac{|A_1|}{3}$. By the construction of $D$, we know there are less than $\frac{|A_1|}{3}+1$ different colors in $C(E_{G'}(y_j, A_1))$. For any edge $e$ adjacent to $y_j$ which is in $E(G[A_1, B_1])\backslash E(G')$, $C(e)=C(xy)$ or there exists an edge $yx_i$ such that $C(yx_i)=C(xy_j)$ in $G^c$, and in this case, $e=x_iy_j$ and $C(e)$ may be missed in $C(E_{G'}(y_j, A_1))$. This implies that there are at most three colors in $C(xy_j)\cup C(E_{G}(y_j, A_1))\backslash C(E_{G'}(y_j, A_1))$. However, $C(xy_j)$ is also in $C(E_{G'}(y_j,A_1))$. Hence there are less than $\frac{|A_1|}{3}+3$ different colors in $C(E_{G}(y_j,\{x\}\cup A_1))$. It follows that there are more than $d^c(y_j)-3-\frac{|A_1|}{3}\ge s-3-\frac{s-1}{3}$ different colors in the color set $\{y_jx':x'\in A\backslash (A_1\cup \{x\})\}$. Hence $y_j$ has more than $s-3-\frac{s-1}{3}$ different neighbors in $A\backslash (A_1\cup \{x\})$. Now we have

$$\begin{aligned} n_1= & {} |A|\ge |A_1|+|\{x\}|+|N(y_j)\backslash (A_1\cup \{x\})|\\> & {} (s-1)+1+s-3-\frac{s-1}{3}=\frac{5s-8}{3}\ge n_1, \end{aligned}$$

a contradiction.

The proof is complete. $\square$