Acyclic Edge Coloring of 4-Regular Graphs (II)

Abstract

A proper edge coloring is called acyclic if no bichromatic cycles are produced. It was conjectured that every simple graph G with maximum degree $\Delta$ is acyclically edge-$(\Delta + 2)$-colorable. In this paper, combining some known results, we confirm the conjecture for graphs with $\Delta =4$.

1 Introduction

Only simple graphs are considered in this paper. Let G be a graph with vertex set V(G) and edge set E(G). A proper edge-k-coloring is a mapping $c:E(G)\rightarrow \{1, 2, \ldots , k\}$ such that any two adjacent edges receive different colors. The graph G is edge-k-colorable if it has an edge-k-coloring. The chromatic index$\chi '(G)$ of G is the smallest integer k such that G is edge-k-colorable. A proper edge-k-coloring of G is called acyclic if there are no bichromatic cycles in G, that is, the union of any two color classes induces a subgraph of G that is a forest. The acyclic chromatic index of G, denoted $a'(G)$, is the smallest integer k such that G is acyclically edge-k-colorable.

Let $\Delta =\Delta (G)$ denote the maximum degree of a graph G. By Vizing’s theorem [17], $\Delta \le \chi '(G)\le \Delta +1$. So it holds automatically that $a'(G)\ge \chi '(G)\ge \Delta$. Fiamčik [9] and later Alon et al. [2] put forward the following conjecture:

Conjecture 1

For any graph G, $a'(G)\le \Delta + 2$.

Using probabilistic method, Alon et al. [1] proved that $a'(G)\le 64\Delta$ for any graph G. This upper bound was gradually improved to that $a'(G)\le 16\Delta$ in [12], that $a'(G)\le \lceil 9.62(\Delta -1)\rceil$ in [13], that $a'(G)\le 4\Delta$ in [8], and that $a'(G)\le \lceil 3.74(\Delta -1)\rceil + 1$ in [10]. For the class of subcubic graphs, Conjecture 1 was affirmed to be true, see [3, 4]. Other results about this topic can be seen in [6, 7, 11, 14, 16].

In 2009, Basavaraju and Chandran [5] showed that if G is a graph with $\Delta = 4$ and $|E(G)|\le 2|V(G)|- 1$, then $a'(G)\le 6$. Namely, every non-regular graph of $\Delta = 4$ satisfies Conjecture 1. More recently, Shu et al. [15] extended this result by showing that every 4-regular graph G without 3-cycles is acyclically edge-6-colorable. In this paper, we solve the case of 4-regular graphs having at least one 3-cycle. Hence, combining the previously known results, Conjecture 1 is confirmed for all graphs with $\Delta = 4$.

2 Main Results

Assume that c is a partial acyclic edge-k-coloring of a graph G using the color set $C = \{1, 2, \ldots , k\}$. For a vertex $v\in V(G)$, we use C(v) to denote the set of colors assigned to edges incident to v under c. If the edges of a cycle $ux\cdots vu$ are alternately colored with colors i and j, then we call such cycle an $(i, j)_{(u, v)}$-cycle. If the edges of a path $ux\cdots v$ are alternately colored with colors i and j, then we call such path an $(i, j)_{(u, v)}$-path. For simplicity, we use $\{e_{1}, e_2, \ldots , e_m\} \rightarrow a$ to express that all edges $e_{1},e_2,\ldots ,e_m$ are colored or recolored with same color a. In particular, when $m=1$, we write simply $e_1 \rightarrow a$. Moreover, we use $(e_1, e_2, \ldots , e_m)_c= (a_1, a_2, \ldots , a_m)$ to denote that $c(e_i)= a_i$ for $i=1,2,\ldots , m$. Let $(e_1, e_2, \ldots , e_n) \rightarrow (b_1, b_2, \ldots , b_n)$ denote that $e_i$ is colored or recolored with color $b_i$ for $i = 1, 2, \ldots , n$. Note that $b_i$ and $b_j$ may be same for some $i\ne j$.

For a graph G, let $X =\{v_1,v_2, \ldots , v_j\}\subset V(G)$ and $S = \{e_1, e_2,\ldots , e_k\}$ be an edge subset. We use $(G - X)\cup S$ or $G - \{v_1, v_2,\ldots , v_j\} + \{e_1, e_2,\ldots , e_k\}$ to denote the graph obtained by deleting from G the vertices in X together with all the edges incident with some vertex in X and adding the edges in S together with all the new vertices incident with some edge in S. We write $G - v_1 +\{e_1,e_2, \ldots , e_k\}$ if $j = 1$ or $G - \{v_1, v_2,\ldots , v_j\} + e_1$ if $k = 1$.

Several lemmas below will be frequently used in the proof of the main result.

Lemma 1

([15]) Suppose that a graph G has an edge-6-coloring c. Let $P = uv_1v_2 \cdots$$v_kv_{k + 1}$ be a maximal $(a, b)_{(u, v_{k + 1})}$-path in G with $c(uv_1)= a$ and $b\not \in C(u)$. If $w\not \in V(P)$, then there is no $(a, b)_{(u, w)}$-path in G under c.

Lemma 2

([3, 4]) If G is a graph with $\Delta \le 3$, then $a'(G)\le 5$, and $a'(G) = 5$ if and only if $G\in \{K_4, K_{3, 3}\}$.

Lemma 3

([5]) If G is a graph with $\Delta = 4$ that is not 4-regular, then $a'(G)\le 6$.

Lemma 4

([15]) If G is a 4-regular graph without 3-cycles, then $a'(G)\le 6$.

Theorem 1

If G is a 4-regular graph, then $a'(G)\le 6$.

Proof

The proof is proceeded by induction on the number $\sigma (G)=|V(G)|+|E(G)|$. If $\sigma (G)=15$, that is, $|V(G)|=5$, then G is the complete graph $K_5$, and it is easy to show that $a'(G)\le 6$. Let G be a 4-regular graph with $\sigma (G)\ge 16$, so $|V(G)|\ge 6$. Obviously, we may assume that G is 2-connected by Lemma 3. If G contains no 3-cycles, then $a'(G)\le 6$ by Lemma 4. So assume that G contains at least one 3-cycle. For any graph H with $\Delta (H)\le 4$ and $\sigma (H)<\sigma (G)$, by the induction hypothesis or Lemmas 3 and 4, H admits an acyclic edge-6-coloring c using the color set $C=\{1, 2, \ldots , 6\}$. Before constructing an acyclic edge-6-coloring of G, we first prove the following lemma.$\square$

Lemma 5

Let $\emptyset \ne X\subset V(G)$ and put $S=[X, \overline{X}]$ where $\overline{X}=V(G)\setminus X$. If $S= \{x_1y_1, x_2y_2, x_3y_3, x_4y_4\}$ with $x_i\in X$, $y_i\in \overline{X}$, $i= 1, 2, 3, 4$, where $x_1, x_2, x_3, x_4$ are pairwise distinct, but some of $y_i$’s may be identical. Then $(G - X)\cup S$ has an acyclic edge-6-coloring c using the color set $C = \{1, 2, \ldots , 6\}$ such that $c(x_iy_i) = i$ for each $i\in \{1, 2, 3, 4\}$.

Proof

Note that for any graph H with $\Delta (H)\le 4$ and $\sigma (H)<\sigma (G)$, H has an acyclic edge-6-coloring c using $C=\{1, 2, \ldots , 6\}$ by the foregoing discussion. By symmetry, we have to consider the following cases. Let $u\not \in V(G)$ be a new vertex.

• If $y_1, y_2, y_3, y_4$ are identical to a vertex, say v, then we define $H = G - X +uv$ and can assume $c(uv) = 1$ with $2, 3, 4\not \in C(v)$.

• If $y_1, y_2, y_3$ are identical to a vertex v and $y_4\ne v$, then we define $H = G - X + \{uv, uy_4\}$ and assume that $(uv, uy_4)_c = (1, 4)$ with $2, 3\not \in C(v)$.

• If $y_1, y_2$ are identical to a vertex $v_1$, and $y_3, y_4$ are identical to a vertex $v_4$, let $H = G - X + \{ uv_1, uv_4\}$ and we can assume that $(uv_1, uv_4)_c= (1, 4)$ with $2\in C\backslash (C(v_1)\cup \{1, 4\})$, $3\in C\backslash (C(v_4)\cup \{1, 4, 2\})$.

• If $y_1, y_2$ are identical to a vertex v, and $y_3 \ne y_4\ne v$, let $H= G - X + \{uv, uy_3, uy_4\}$ and assume that $(uv, uy_3, uy_4)_c= (1, 3, 4)$ with $2\in C \backslash (C(v)\cup \{3, 4\})$.

• If $y_1, y_2, y_3, y_4$ are all distinct, let $H= G- X + \{uy_1, uy_2, uy_3, uy_4\}$ and assume that $(uy_1, uy_2, uy_3, uy_4)_c= (1, 2, 3, 4)$.

Now we only need to let $x_iy_i\rightarrow i$ for $i= 1, 2, 3, 4$ for all cases above to complete the proof. $\square$

Let $v\in V(G)$ be a vertex adjacent to $v_0, v_1, v_2, v_3$. By Lemma 4, we may assume that v lies in a 3-cycle. To obtain an acyclic edge-6-coloring of G, the proof is divided into the following five cases by symmetry.

Case 1$v_0v_1,v_1v_2, v_2v_3, v_3v_0\in E(G)$.

For $i\in \{0, 1, 2, 3\}$, let $v'_i$ be the neighbor of $v_i$ different from $v,v_{i - 1}, v_{i + 1}$, where all indices are taken modulo 4. Since $|V(G)|\ge 6$, $G\ne K_5$. We only need to consider the following subcases by symmetry:

Case 1.1$v_1v_3\in E(G)$ and $v_2v_0\not \in E(G)$.

Take $X=\{v,v_1,v_2,v_3\}$ and $S=\{v_0v_1, v_2v'_2, v_0v_3, vv_0\}$, and let $H = (G - X)\cup S$. By Lemma 5, H has an acyclic edge-6-coloring c with $(v_0v_1, v_2v'_2, v_0v_3, vv_0)_c = (1, 2, 3, 4)$. To extend c to the whole graph G, we let $(v_1v_3, vv_2, v_1v_2, vv_3, vv_1, v_2v_3)\rightarrow (2, 3, 4, 5,6,6)$.

Case 1.2$v_1v_3, v_2v_0\not \in E(G)$.

Take $X = \{v,v_0, v_1,v_2,v_3\}$ and $S = \{v_1v'_1,v_2v'_2,v_3v'_3,v_0v'_0\}$, and let $H = (G - X)\cup S$. By Lemma 5, H has an acyclic edge-6-coloring c with $(v_1v'_1,v_2v'_2,v_3v'_3,v_0v'_0)_c= (1, 2, 3, 4)$. To extend c to the whole graph G, we let $(v_0v_3, vv_3, v_1v_2, vv_1, vv_2, v_0v_1,$$vv_0, v_2v_3)\rightarrow (1, 2, 3, 4,5,5,6,6).$

Case 2$v_0v_1, v_1v_2, v_2v_3\in E(G)$ and $v_0v_3\not \in E(G)$.

If $v_1v_3, v_0v_2\in E(G)$, then since $v_0v_1, v_0v_2, v_2v_3, v_1v_3\in E(G)$, $v_2$ lies in four 3-cycles and the proof can be reduced to Case 1. Thus, without loss of generality, assume that $v_0v_2\not \in E(G)$. Let $v'_2$ be the neighbor of $v_2$ other than $v,v_1,v_3$.

Case 2.1$v_1v_3\in E(G)$.

Let $v'_3$ be the neighbor of $v_3$ other than $v, v_1, v_2$. Take $X=\{v, v_1,v_2,v_3\}$ and $S=\{v_0v_1, v_2v'_2, v_3v'_3, vv_0\}$, and let $H = (G-X)\cup S$. By Lemma 5, H has an acyclic edge-6-coloring c with $(v_0v_1, v_2v'_2, v_3v'_3, vv_0)_c = (1, 2, 3, 4)$. It suffices to define $(v_2v_3, vv_3, v_1v_2, v_1v_3, vv_2, vv_1)\rightarrow (1, 2, 3, 4, 5, 6)$.

Case 2.2 $v_1v_3\not \in E(G)$

Let $v_5$ be the neighbor of $v_1$ other than $v, v_0, v_2$. There are two possibilities as follows.

Case 2.2.1$v'_2 = v_5$.

By Case 1, we may assume that $v_0v_5,v_3v_5\not \in E(G)$. Let $H= G- \{v, v_1, v_2\} + \{v_0v_5,v_3v_5\}$ and assume that $(v_0v_5,v_3v_5)_c= (1,2)$. First, let $\{v_1v_5, vv_0\}\rightarrow 1$ and $\{v_2v_5, vv_3\}\rightarrow 2$. Next, if $C(v_0)= \{3, 4\}$ and $C(v_3)= \{5, 6\}$, then let $(v_1v_2, v_2v_3, vv_1,$$v_0v_1, vv_2)\rightarrow (3, 4, 5,6,6)$. Otherwise, w.l.o.g., assume that $3\not \in C(v_3)\cup C(v_0)$ and $4\not \in C(v_0)$. Let $(v_0v_1, v_2v_3,v_1v_2, vv_1, vv_2)\rightarrow (3,3,4, 5, 6)$. Obviously, there is neither a $(1, 5)_{(v, v_1)}$-cycle nor a $(2, 6)_{(v, v_2)}$-cycle in G.

Case 2.2.2$v'_2\ne v_5$.

Let $H = G- \{v, v_1, v_2\} +\{uv_5, uv'_2, uv_3, uv_0, v_0v_3\}$ and assume that $(uv_0, uv_3,v_0v_3)_c=(1, 2, 3)$, where u is a new vertex. First, let $v_2v_3\rightarrow 2$ and $(v_1v_5, v_2v'_2)\rightarrow (c(uv_5), c(uv'_2))$.

• Assume that $(uv_5, uv'_2)_c = (3, 4)$.

Let $(v_0v_1, vv_1, vv_0, v_1v_2)\rightarrow (1, 2, 3, 6)$. If $\{5, 6\}\backslash C(v_3)\ne \emptyset$, say $5\not \in C(v_3)$, let $(vv_2, vv_3)\rightarrow (1, 5)$; otherwise, $C(v_3)= \{5, 6\}$, we let $(vv_2, vv_3)\rightarrow (5, 1)$.

• Assume that $(uv_5, uv'_2)_c = (4, 5)$.

If $4\not \in C(v_0)$, then let $(vv_0, v_1v_2,v_0v_1, vv_3,vv_2, vv_1)\rightarrow (1,1,3,3,4, 6)$. Otherwise, $4\in C(v_0)$. Then first let $(v_0v_1, vv_1, vv_0)\rightarrow (1, 2, 3)$. Next, if $4\not \in C(v_3)$, then let $(vv_3, vv_2, v_1v_2)\rightarrow (4, 1, 6)$. Or else, $4\in C(v_3)$, and similarly assume $5\in C(v_3)\cap C(v_0)$. It is enough to let $(vv_3, v_1v_2,vv_2)\rightarrow (1, 3,6)$.

Case 3$v_1v_2, v_2v_3\in E(G)$ and $v_1v_0, v_3v_0\not \in E(G)$.

If $v_1v_3, v_0v_2\in E(G)$, then the proof is reduced to Case 2. Thus, assume that $v_1v_3\not \in E(G)$, or $v_0v_2\not \in E(G)$.

Case 3.1$v_1v_3\in E(G)$ and $v_0v_2\not \in E(G)$.

Let $v_5, v_6, v_7$ be the forth neighbor of $v_1, v_2, v_3$ other than $v, v_0, v_1, v_2, v_3$, respectively. Take $X = \{v, v_1, v_2, v_3\}$ and $S=\{v_1v_5, v_2v_6, v_3v_7, vv_0\}$, and let $H = (G-X)\cup S$. By Lemma 5, H has an acyclic edge-6-coloring c with $(v_1v_5, v_2v_6, v_3v_7, vv_0)_c = (1, 2, 3, 4)$. To extend c to G, we let $(v_2v_3, vv_3, vv_1, vv_2, v_1v_3,v_1v_2)\rightarrow (1, 2, 3, 5,5,6)$.

Case 3.2$v_1v_3\not \in E(G)$ and $v_0v_2\in E(G)$.

Let $H = G- \{v, v_2\} + \{v_0v_1, v_0v_3, v_1v_3\}$ and assume that $(v_0v_1, v_0v_3, v_1v_3)_c = (1, 2, 3)$ with $4\in C\backslash (C(v_1)\cup \{1, 2, 3\})$. First, let $(vv_0,vv_3, v_1v_2, vv_1)\rightarrow (1, 3,3,4)$. If $\{5, 6\}\backslash C(v_3)\ne \emptyset$, say $5\not \in C(v_3)$, let $(v_0v_2, v_2v_3, vv_2)\rightarrow (2, 5, 6)$; otherwise, $C(v_3)= \{5, 6\}$, we let $v_2v_3\rightarrow 1$. Then if $\{5, 6\}\backslash C(v_0)\ne \emptyset$, say $5\not \in C(v_0)$, let $(v_0v_2, vv_2)\rightarrow (2, 5)$; or else, $C(v_0)= \{5, 6\}$, let $( vv_2, v_0v_2)\rightarrow (2,4)$.

Case 3.3$v_1v_3,v_0v_2\not \in E(G)$.

Let $v_5\not \in \{v, v_1, v_3, v_0\}$ be the forth neighbor of $v_2$, and $v_6, v_7\not \in \{v, v_2, v_0, v_3\}$ be the other two neighbors of $v_1$. Note that $v_5, v_6, v_7$ are pairwise distinct by Case 2. Let $H = G - \{v, v_2\} + \{v_0v_1, v_1v_5\}$ and suppose that $(v_1v_5, v_0v_1, v_1v_6, v_1v_7)_c=(1, 2, 3, 4)$. Let $(v_2v_5, vv_0)\rightarrow (1, 2)$.

Case 3.3.1$C(v_3)\cap \{1, 2\}\ne \emptyset$, say $1\in C(v_3)$.

• Assume that $2\not \in C(v_3)$.

Since $\{3, 4\}\backslash C(v_3)\ne \emptyset$ and $\{5, 6\}\backslash C(v_3)\ne \emptyset$, we may assume that 4, 6 $\not \in C(v_3)$. If $6\not \in C(v_5)$, then let $(v_1v_2, vv_3, vv_2, vv_1, v_2v_3)\rightarrow (2, 4, 5, 6, 6)$. If $4\not \in C(v_5)$, then let $(vv_1, v_1v_2, v_2v_3, vv_2, vv_3)\rightarrow (1, 2, 4, 5, 6)$. Otherwise, $4, 6\in C(v_5)$ and $\{3, 5\}\backslash C(v_3)$$\subseteq C(v_5)$. It follows that $2\not \in C(v_5)$, and hence, let $(vv_2, v_2v_5, vv_3, v_1v_2,vv_1, v_2v_3)\rightarrow (1, 2, 4, 5, 6, 6)$.

• Assume that $C(v_3) = \{1, 2\}$.

Assume that $6\not \in C(v_5)$. Let $(v_1v_2, vv_2, v_2v_3)\rightarrow (2, 4, 6)$. If $5\not \in C(v_0)$, then let $(vv_1, vv_3)\rightarrow (1, 5)$. If $3\not \in C(v_0)$, let $(vv_1, vv_3)\rightarrow (5, 3)$. Otherwise, we may assume that $C(v_0)= \{2, 3, 4, 5\}$, and hence, let $(vv_1, vv_3, vv_0)\rightarrow (1, 5, 6)$.

Assume that $\{5, 6\}\subseteq C(v_5)$, and $\{5, 6\}\subseteq C(v_0)$ similarly. If $3, 4\not \in C(v_0)\cup C(v_5)$, then let $(v_1v_2, v_2v_5, vv_3,vv_1, v_2v_3, vv_2)\rightarrow (1, 3, 4, 5, 5, 6)$. Otherwise, we suppose that $C(v_0)= \{2, 4, 5, 6\}$. Let $(vv_1, vv_2, vv_0, vv_3, v_1v_2)$$\rightarrow$ (1, 2, 3, 5, 6) and color $v_2v_3$ with a color in $\{3, 4\}\backslash C(v_5)$.

Case 3.3.2$1, 2\not \in C(v_3)$.

If $\{5, 6\}\backslash C(v_3)\ne \emptyset$, say $6\not \in C(v_3)$, then let $(vv_3, v_1v_2, vv_1, v_2v_3)\rightarrow (1, 2, 6, 6)$, and color $vv_2$ with a color in $\{3, 4, 5\}\backslash C(v_3)$. Otherwise, $C(v_3) = \{5, 6\}$, and it suffices to let $(vv_1, v_2v_3, vv_3, vv_2, v_1v_2)\rightarrow (1, 2, 3, 4, 5)$.

Case 4$v_0v_3,v_1v_2\in E(G)$ and $v_0v_1, v_2v_3, v_1v_3, v_0v_2\not \in E(G)$.

Let $V_i= \{v_{i1}, v_{i2}\}$ be the set of other neighbors of $v_i$ for $i\in \{0, 1, 2, 3\}$. By Cases 1–3, we assume that $V_1\cap V_2= \emptyset$ and $V_0\cap V_3= \emptyset$. Let $H = G - \{v, v_0, v_1, v_2, v_3\} + \{uv_{11}, uv_{12}, uv_{21}, uv_{22}, wv_{31}, wv_{32}, wv_{01}, wv_{02}\}$, where u and w are new vertices added. Assume that $(uv_{11}, uv_{12}, uv_{21}, uv_{22})_c=$ (1, 2, 3, 4), $\{c(wv_{01}), c(wv_{02}), c(wv_{31}), c(wv_{32}) \}$$=\{a, b,c,d\}$. Let $(v_1v_{11}, v_1v_{12}, v_2v_{21}, v_2v_{22}, v_0v_{01}, v_0v_{02}, v_3v_{31}, v_3v_{32})\!\rightarrow (1, 2, 3, 4, a, b, c, d)$. Since $|\{1, 2, 3, 4\}\cap \{a, b, c, d\}|\!\ge 2$, we may assume that $1\in \{a, b\}$. By symmetry, let us handle the following subcases.

Case 4.1$\{a, b\}= \{1, 2\}$ and $\{c, d\}\in \{\{3, 4\}, \{3, 5\}, \{5, 6\}\}$.

If $\{c, d\}\in \{\{3, 5\}, \{5, 6\}\}$, then let $(vv_2, vv_1, vv_3, vv_0, v_1v_2)\rightarrow (1, 3, 4, 5, 6)$ and color $v_0v_3$ with a color in $\{3, 6\}\backslash \{c, d\}$. Otherwise, $\{c, d\} = \{3, 4\}$. If $1\not \in C(v_{21})$, let $(vv_2, vv_0, v_1v_2, vv_3, vv_1, v_0v_3)\rightarrow (1, 3, 5, 5, 6, 6)$. Or else, $1\in C(v_{21})$, and furthermore assume that $1, 2\in C(v_{21})\cap C(v_{22})$. Hence, $\{5, 6\}\backslash C(v_{21})\ne \emptyset$, say $5\not \in C(v_{21})$. Let $(vv_0, vv_1, vv_2, v_0v_3, v_1v_2, vv_3)\rightarrow (3, 4, 5, 5, 6, 6)$. If $6\not \in C(v_{22})$, we are done. Or else, $C(v_{22}) = \{4, 1, 2, 6\}$, we recolor $\{vv_1, v_2v_{22}\}$ with 5, and $vv_2$ with 4.

Case 4.2$\{a, b\} = \{1, 3\}$ and $\{c, d\}\in \{\{2, 4\}, \{2, 5\}, \{4, 5\}, \{5, 6\}\}$.

Note that G contains no $(2, 3)_{(v_0, v_3)}$-path. We first let $(vv_2, vv_0, vv_3, v_1v_2, vv_1)\rightarrow (1, 2, 3, 5, 6)$, then let $v_0v_3\rightarrow 5$ if $\{c, d\} = \{2, 4\}$; $v_0v_3\rightarrow 6$ if $\{c, d\} = \{4, 5\}$; and $v_0v_3 \rightarrow 4$ if $\{c, d\} = \{2, 5\}$ or $\{5, 6\}$.

Case 4.3$\{a, b\}= \{1, 5\}$ and $\{c, d\}\in \{\{2, 6\}, \{3, 6\}\}$.

In this case, it suffices to let $(vv_2, vv_0, v_0v_3, v_1v_2, vv_3, vv_1)\rightarrow (1, 2, 4, 5, 5, 6)$.

Case 5$v_1v_2\in E(G)$ and $v_2v_3, v_0v_3, v_0v_1, v_1v_3, v_0v_2\not \in E(G)$.

Let $V_1,V_2$ be defined similarly as in Case 4. By Cases 1–4, $V_1\cap V_2= \emptyset$. Let $H= G- \{v, v_1, v_2\} + \{uv_{11}, uv_{12}, uv_{21}, uv_{22}, v_0v_3\}$ and assume $(uv_{11}, uv_{12}, uv_{21}, uv_{22})_c= (1, 2, 3, 4)$, where u is a new vertex. Let $(v_1v_{11},v_1v_{12},v_2v_{21},v_2v_{22})\rightarrow (1,2,3,4)$. Without loss of generality, we assume that $c(v_0v_3)\in \{1,5\}$.

Case 5.1$c(v_0v_3) = 1$.

• Assume that $2\not \in C(v_3)$.

If G contains neither a $(2, 3)_{(v_1, v_3)}$-path nor a $(1, 3)_{(v_1, v_0)}$-path, let $(vv_0, vv_3, vv_1,$$v_1v_2, vv_2)\rightarrow (1, 2, 3, 5, 6)$. Otherwise, G contains a $(2, 3)_{(v_1, v_3)}$-path or a $(1, 3)_{(v_1, v_0)}$-path. If $3\in C(v_3)\backslash C(v_0)$, then let $(vv_2, vv_3, vv_0)\rightarrow (1, 2, 3)$ and $vv_1\rightarrow a \in \{4, 5, 6\}\backslash C(v_3)$, $v_1v_2\rightarrow b\in \{5, 6\}\backslash \{a\}$. If $3\in C(v_0)\backslash C(v_3)$, then let $(vv_0, vv_2, vv_3)\rightarrow (1, 2, 3)$ and $vv_1\rightarrow c\in \{4, 5, 6\}\backslash C(v_0)$, $v_1v_2\rightarrow d\in \{5, 6\}\backslash \{c\}$. Otherwise, $3\in C(v_3)\cap C(v_0)$ and $4\in C(v_3)\cap C(v_0)$ similarly. Hence, $\{5, 6\}\backslash C(v_3)\ne \emptyset$, say $5\not \in C(v_3)$. Let $(vv_0, vv_2, v_1v_2, vv_3, vv_1)\rightarrow (1, 2, 5, 5, 6)$. If there is no $(1, 6)_{(v_1, v_0)}$-path in G, we are done. Otherwise, G contains a $(1, 6)_{(v_1, v_0)}$-path, which cannot pass through $v_3$ and $C(v_0) = \{3, 4, 6\}$. It suffices to let $(vv_3, vv_0)\rightarrow (1, 5)$.

• Assume that $2\in C(v_3)\cap C(v_0)$ and $3\not \in C(v_3)$.

If $5\not \in C(v_0)$, let $(vv_3, vv_2, vv_1, vv_0, v_1v_2)\rightarrow (1, 2, 3, 5, 6)$. Otherwise, $5\in C(v_0)$ and $6\in C(v_0)$. Let $(vv_0, vv_3, vv_1, vv_2, v_1v_2)\rightarrow (1, 3, 4, 5, 6)$. If there is no $(3, 5)_{(v_2, v_3)}$-path in G, we are done. Otherwise, it suffices to let $(vv_0, vv_3)\rightarrow (3, 1)$.

• Assume that $C(v_3) = C(v_0) = \{2, 3, 4\}$.

Let $(vv_3, vv_2, v_1v_2, vv_0, vv_1)\rightarrow (1, 2, 5, 5, 6)$. If G contains no $(2, 5)_{(v_1, v_0)}$-path, we are done. Otherwise, it suffices to let $(vv_0, vv_3)\rightarrow (1, 5)$.

Case 5.2$c(v_0v_3) = 5$.

Note that $\{1, 2, 3, 4\}\backslash C(v_3)\ne \emptyset$, say $1\not \in C(v_3)$, by symmetry. If G contains no $(1, i)_{(v_1, v_3)}$-path for some $i\in \{3, 4\}$, let $(vv_3, vv_2, vv_1, vv_0, v_1v_2)\rightarrow (1, 2, i, 5, 6)$. Otherwise, for any $i\in \{3, 4\}$, G contains a $(1, i)_{(v_1, v_3)}$-path, implying that $3, 4\in C(v_{11})\cap C(v_3)$ and G contains no $(1, i)_{(v_2, v_3)}$-path. If $1\not \in C(v_0)$, let $(vv_0, vv_2, vv_1, vv_3, v_1v_2)$$\rightarrow$ (1, 2, 3, 5, 6). Otherwise, $1\in C(v_0)$. If G contains neither a $(2, 5)_{(v_1, v_0)}$-path nor a $(1, 6)_{(v_1, v_3)}$-path, let $(vv_3, vv_2,vv_0, v_1v_2, vv_1)\rightarrow (1, 2, 5, 5, 6)$.

Assume that G contains a $(1, 6)_{(v_1, v_3)}$-path. Thus, $6\in C(v_{11})\cap C(v_3)$, and $C(v_{11}) = \{1, 3, 4, 6\}$ and $C(v_3) = \{3, 4, 6\}$. Let $(vv_3, v_1v_2, vv_2, v_1v_{11}, vv_0)\rightarrow (1, 1, 2, 5, 5)$, and color $vv_1$ with a color in $\{3, 4, 6\}\backslash C(v_0)$.

Assume that G contains a $(2, 5)_{(v_1, v_0)}$-path. Then $2\in C(v_0)$, and $\{3, 4\}\backslash C(v_0)\ne \emptyset$, say $3\not \in C(v_0)$. If $5\not \in C(v_{11})$, then let $(vv_3, vv_2, vv_1, vv_0, v_1v_{11}, v_1v_2)\rightarrow (1, 2, 3, 5, 5, 6)$. Otherwise, $C(v_{11})= \{1, 3, 4, 5\}$, it suffices to let $(v_1v_2, vv_3, vv_0, vv_1,vv_2, v_1v_{11})\rightarrow (1, 1, 3, 4, 5, 6)$. $\square$