# 2-Distance vertex-distinguishing index of subcubic graphs

• Victor Loumngam Kamga
• Weifan Wang
• Ying Wang
• Min Chen
Article

## Abstract

A 2-distance vertex-distinguishing edge coloring of a graph G is a proper edge coloring of G such that any pair of vertices at distance 2 have distinct sets of colors. The 2-distance vertex-distinguishing index $$\chi ^{\prime }_{\mathrm{d2}}(G)$$ of G is the minimum number of colors needed for a 2-distance vertex-distinguishing edge coloring of G. Some network problems can be converted to the 2-distance vertex-distinguishing edge coloring of graphs. It is proved in this paper that if G is a subcubic graph, then $$\chi ^{\prime }_{\mathrm{d2}}(G)\le 6$$. Since the Peterson graph P satisfies $$\chi ^{\prime }_{\mathrm{d2}}(P)=5$$, our solution is within one color from optimal.

## Keywords

Subcubic graph Edge coloring 2-Distance vertex-distinguishing index Star-chromatic index

## 1 Introduction

Graph coloring is an important branch of graph theory and combinatorial optimization, because it can be widely applied to various practical problems such as frequency channel assignment, timelabeling and scheduling, planning of experiments, coding and cryptology, etc. The problem in which we are interested is a particular case of the great variety of different ways of labeling a graph. The original motivation of studying this problem came from irregular networks. One of the important tasks in network studies is to be able to identify and distinguish their elements, e.g., vertices, using local substructures and small labels. Among the multiple ways to achieve this, one of the most natural ones is to distinguish vertices by the set of labels in their neighborhoods. Related definitions and results about the problem of distinguishing any two vertices in a graph refer to the literatures (Aigner and Triesch 1990; Balister et al. 2002; Bazgan et al. 1999; Burris 1993; Burris and Schelp 1997). As a special case, a number of authors considered the problem of only distinguishing any pair of adjacent vertices in a graph, see Balister et al. (2007); Hatami (2005); Zhang et al. (2002). The aim of this paper is to deal with another special case in which any two vertices at distance 2 in a graph are required to be distinguish. In fact, this problem also corresponds to a strong application background in networks.

Only simple graphs are considered in this paper. Formally, for a graph G, we use V(G), E(G),  and $$\Delta (G)$$ ($$\Delta$$, for short) to denote its vertex set, edge set, and maximum degree, respectively. A cubic graph is a 3-regular graph, and a subcubic graph is a graph of maximum degree at most 3. The distance, denoted by d(uv), between two vertices u and v is the length of a shortest path connecting them.

A proper edge k-coloring of a graph G is a mapping $$\phi : E(G)\rightarrow \{1,2,\ldots ,k\}$$ such that $$\phi (e)\ne \phi (e')$$ for any two adjacent edges e and $$e'$$. The chromatic index, denoted $$\chi ^{\prime }(G)$$, of a graph G is the smallest integer k such that G has a proper edge k-coloring. Let C(v) denote the set of colors assigned to those edges incident to a vertex v, i.e., $$C(v)=\{\phi (uv)| uv\in E(G)\}$$. The coloring $$\phi$$ is called 2-distance vertex-distinguishing if $$C(u)\ne C(v)$$ for any pair of vertices u and v with $$d (u,v)=2$$. The 2-distance vertex-distinguishing index $$\chi ^{\prime }_{\mathrm{d2}}(G)$$ of a graph G is the smallest integer k such that G has a 2-distance vertex-distinguishing edge coloring using k colors.

The 2-distance vertex-distinguishing edge coloring of graphs is a special case of the r-strong edge coloring of graphs, which was introduced by Akbari et al. (2006), and independently by Zhang et al. (2006). Let $$r\ge 1$$ be an integer. The r-strong chromatic indxe $$\chi ^{\prime }_{\mathrm{s}}(G,r)$$ of a graph G is the minimum number of colors required for a proper edge coloring of G such that any two vertices u and v with $$d(u,v)\le r$$ have $$C (x)\ne C (y)$$.

When $$r=1$$, $$\chi ^{\prime }_s(G,1)=\chi ^{\prime }_a(G)$$, which is called the neighbor-distinguishing chromatic index of G. Note that $$\chi ^{\prime }_{\mathrm{a}}(G)$$ can be well defined if and only if G does not contain isolated edges. Zhang et al. (2002) first investigated the neighbor-distinguishing edge coloring of graphs and proposed the following conjecture:

## Conjecture 1

If G is a connected graph with at least six vertices, then $$\chi ^{\prime }_{\mathrm{a}}(G)\le \Delta +2$$.

Balister et al. (2007) comfirmed Conjecture 1 for bipartite graphs and subcubic graphs. Using probabilistic method, Hatami (2005) showed that every graph G with $$\Delta >10^{20}$$ has $$\chi ^{\prime }_{\mathrm{a}}(G)\le \Delta +300$$. Akbari et al. (2006) proved that every graph G satisfies $$\chi ^{\prime }_{\mathrm{a}}(G)\le 3\Delta$$. Wang et al. (2015) improved this result by showing that every graph G has $$\chi ^{\prime }_{\mathrm{a}}(G)\le 2.5\Delta$$ if $$\Delta \ge 7$$, and $$\chi ^{\prime }_{\mathrm{a}}(G)\le 2\Delta$$ if $$\Delta \le 6$$. The currently best known upper bound that $$\chi ^{\prime }_{\mathrm{a}}(G)\le 2\Delta +2$$ for any graph G was obtained by Vučković (2017).

Wang et al. (2016) first introduced the 2-distance vertex-distinguishing edge coloring of graphs and raised the following challenging conjecture:

## Conjecture 2

For a graph G, $$\chi ^{\prime }_{\mathrm{d2}}(G)\le \Delta +2$$.

It follows from the definition that $$\chi ^{\prime }_{\mathrm{d2}}(G)\ge \chi ^{\prime }(G)\ge \Delta$$, and moreover $$\chi ^{\prime }_{\mathrm{d2}}(G)\ge \Delta +1$$ if G contains two vertices of maximum degree at distance 2. In Wang et al. (2016), Conjecture 2 was confirmed for some basic graphs such as cycles, paths, trees, complete bipartite graphs, unicycle graphs, and Halin graphs. A graph G is called outerplanar if it can be embedded in the plane such that all vertices lie in the boundary of the unbounded face. Using an algorithmic analysis, Wang et al. (2016) proved that every outerplanar graph G satisfies $$\chi ^{\prime }_{\mathrm{d2}}(G)\le \Delta +8$$. In addition, it was shown in Huang et al. (2015) that if G is a bipartite outerplanar graph, then $$\chi ^{\prime }_{\mathrm{d2}}(G)\le \Delta +2$$.

A proper edge coloring of a graph G is called star-edge coloring if there do not exist bichromatic paths and cycles of length four. That is, at least three colors occur in a path and cycle of length four. The star-chromatic index, denoted $$\chi ^{\prime }_{\mathrm{st}}(G)$$, of G is the smallest integer k such that G has a star-edge k-coloring.

The following relationship is an easy observation, which provides a natural upper bound for $$\chi ^{\prime }_{\mathrm{d2}}(G)$$:

## Proposition 1

For any graph G, $$\chi ^{\prime }_{\mathrm{d2}}(G)\le \chi ^{\prime }_{\mathrm{st}}(G)$$.

More recently, Bezegová et al. (2016) considered the star-edge coloring of trees and outerplanar graphs by showing the following results: (i) every tree T has $$\chi ^{\prime }_{\mathrm{st}}(T)\le \lfloor 1.5\Delta \rfloor$$, and the upper bound is tight; (ii) every outerplanar graph G has $$\chi ^{\prime }_{\mathrm{st}}(G)\le \lfloor 1.5\Delta \rfloor +12$$. Dvořák et al. (2013) proved that every subcubic graph G has $$\chi ^{\prime }_{\mathrm{st}}(G)\le 7$$ and conjectured that 7 can be replaced by 6.

Using Proposition 1, we get immediately that $$\chi ^{\prime }_{\mathrm{d2}}(G)\le 7$$ for any subcubic graph G. In this paper, we will improve this result by reducing 7 to 6.

## 2 Main results

Before showing our main results, we introduce a few of concepts and notation. Let G be a graph. A k-vertex $$(k^-$$-vertex, $$k^+$$-vertex, respectively) of G is a vertex of degree k (at most k, at least k, respectively). For a vertex $$v\in V(G)$$, let N(v) denote the set of neighbors of v in G. Let $$\chi (G)$$ denote the chromatic number of G, which is the least integer k for which G has a vertex coloring using k colors so that any two adjacent vertices get distinct colors.

In what follows, a 2-distance vertex-distinguishing edge k-coloring of G is shortly written as a 2DVDE-k-coloring. Suppose that $$\phi$$ is a partial 2DVDE-coloring of G using the color set C. We say that two vertices u and v with $$d(u,v)=2$$ are conflict under $$\phi$$ (or simply conflict) if $$C (u)=C (v)$$. An edge uv is said to be legally colored if its color is distinct from the colors of all its neighbors and no pair of conflict vertices at distance 2 are produced.

The following Lemmas 2 and 3 display two famous consequences in graph theory, which will play an important role in the proof of the main result in this paper.

## Lemma 2

(Petersen 1891) Every connected cubic graph G without cut edges can be edge-partitioned into a perfect matching and a class of cycles.

## Lemma 3

(Brooks 1941) If G is a connected graph and is neither an odd cycle nor a complete graph, then $$\chi (G)\le \Delta$$.

## Lemma 4

Let G denote the Peterson graph. Then $$\chi ^{\prime }_{\mathrm{d2}}(G)=5$$.

## Proof

Firstly, we give a 2DVDE-5-coloring of G, as shown in Fig. 1b. Thus, $$\chi ^{\prime }_{\mathrm{d2}}(G)\le 5$$. To show that $$\chi ^{\prime }_\mathrm{d2}(G)\ge 5$$, we assume that G admits a 2DVDE-4-coloring $$\phi$$ using the color set $$C=\{1,2,3,4\}$$. Let $${\mathcal {S}}$$ denote the set of 3-sets of C, i.e.,
\begin{aligned} {\mathcal {S}}=\{\{1,2,3\}, \{1,2,4\},\{1,3,4\},\{2,3,4\}\}. \end{aligned}
Since $$|C|=4$$, we have $$C(v)\in {\mathcal {S}}$$ for any $$v\in V(G)$$. By symmetry, we may assume that $$C(u_1)=\{1,2,3\}$$. Since $$d(u_1,x)=2$$ for each $$x\in \{v_2,v_3,v_4,v_5,u_3,u_4\}$$, we derive that $$C(x)\ne \{1,2,3\}$$. Since any two vertices in $$\{v_1,u_2,u_5\}$$ have distance 2, there exists at most one in $$\{v_1, u_2, u_5\}$$ with $$\{1, 2, 3\}$$ as the color set. Hence at most two in V(G) have $$\{1,2,3\}$$ as the color set. It follows that at most $$2| {\mathcal {S}|}=8$$ vertices can be legally colored, contradicting the fact that $$|V(G)|=10$$. $$\square$$

## Theorem 5

If G is a subcubic graph, then $$\chi ^{\prime }_{\mathrm{d{2}}}(G)\le 6$$.

## Proof

Assume that the theorem is not true. Let G be a counterexample with the minimum number of |E(G)|. Then G is connected, and every proper subgraph H of G satisfies $$\chi ^{\prime }_{\mathrm{d{2}}}(H)\le 6$$. Suppose that $$\phi$$ is a 2DVDE-6-coloring of a proper subgraph H of G using the color set $$C=\{1,2,\ldots ,6\}$$. By showing a series of claims below, we will derive a contradiction.

## Claim 1

G does not contain 1-vertices.

## Proof

Suppose to the contrary that G contains a 1-vertex v. Let u be the neighbor of v. Consider the subgraph $$H=G-v$$. Then H admits a 2DVDE-6-coloring $$\phi$$ using the color set C by the minimality of G. If $$d(u)=2$$, then let w denote the neighbor of u other than v. It suffices to color uv with a color in $$C{\setminus } C(w)$$ to extend $$\phi$$ to the whole graph G. Otherwise, $$d(u)=3$$. Let $$u_1,u_2$$ denote the neighbors of u other than v. If at least one of $$u_1$$ and $$u_2$$ is a $$2^-$$-vertex, then it suffices to color uv with a color in $$C{\setminus } (C(u_1)\cup C(u_2))$$. Otherwise, assume that $$d(u_1)=d(u_2)=3$$. Let $$N(u_1)=\{u,w_1,w_2\}$$ and $$N(u_2)=\{u,w_3,w_4\}$$. Furthermore, we assume that $$\phi (uu_1)=1$$ and $$\phi (uu_2)=2$$. If uv can not be legally colored, without loss of generality, suppose that $$C(w_i)=\{1,2,i+2\}$$ for $$i=1,2,3,4$$. If $$C(u_1)=\{1, 3, 4\}$$, then we recolor $$uu_1$$ with 6 and color uv with 5. Otherwise, by symmetry, we assume $$C(u_1)=\{1, 2, 4\}$$. Let $$w'_2$$ be the neighbor of $$w_2$$ such that $$\phi (w_2w'_2)=2$$. If $$4\notin C(w'_2)$$, then we recolor $$uu_1$$ with 6 and color uv with 5. Otherwise, $$C(w'_2)=\{2, 4, a\}$$. Then we recolor $$uu_1$$ with a color $$b\in \{5,6\}{\setminus } \{a\}$$ and color uv with a color in $$\{5, 6\}{\setminus }\{b\}$$. Thus, we always get a 2DVDE-6-coloring of G, a contradiction. $$\square$$

## Claim 2

G does not contain adjacent 2-vertices.

## Proof

Suppose to the contrary that G contains two adjacent 2-vertices u and v with $$N(u)=\{x,v\}$$ and $$N(v)=\{u, y\}$$. Then $$d(x),d(y)\ge 2$$ by Claim 1. If G is a cycle, then G is 2DVDE-6-colorable by the result in Wang et al. (2016). Now we assume that $$d(x)=3$$, for otherwise we consider the neighbor of x until such 3-vertex is encountered. By the minimality of G, $$G-uv$$ admits a 2DVDE-6-coloring $$\phi$$ using the color set C. If $$x=y$$, then we color uv with a color in $$C{\setminus } C(x)$$. Otherwise, $$x\ne y$$. If $$d(y)=2$$, then we color uv with a color in $$C{\setminus } (C(x)\cup C(y))$$. Otherwise, we assume that $$d(y)=3$$. Let $$N(x)=\{u,x_1,x_2\}$$ and $$N(y)=\{v,y_1,y_2\}$$. If $$|C{\setminus } (C(x)\cup C(y))|\ge 1$$, then we color uv with a color in $$C{\setminus } (C(x)\cup C(y))$$. Otherwise, we may assume that $$\phi (ux)=1$$, $$\phi (xx_1)=2$$, $$\phi (xx_2)=3$$, $$\phi (vy)=4$$, $$\phi (yy_1)=5$$, and $$\phi (yy_2)=6$$. If there exists at least one 3-vertex in $$\{x_1, x_2, y_1, y_2\}$$, say $$d(x_1)=3$$, then we color uv with 2. Or else, $$d(x_i)=d(y_i)=2$$ for $$i=1,2$$. In this case, it suffices to recolor ux with 6 and then color uv with 1. $$\square$$

A 3-vertex u is called a $$3_i$$-vertex if the number of 2-vertices adjacent to u is exactly i.

## Claim 3

G does not contain a $$3_{3}$$-vertex.

## Proof

Suppose to the contrary that G contains a $$3_3$$-vertex u. Let $$N(u)=\{u_1, u_2, u_3\}$$. Then $$d(u_i)=2$$, let $$N(u_{i})=\{u, v_{i}\}$$ for $$i=1,2,3$$. By Claims 1 and 2, $$v_1,v_2,v_3$$ are 3-vertices, let $$N(v_1)=\{u_1,v'_1,v''_1\}$$, and $$N(v_2)=\{u_2,v'_2,v''_2\}$$. By the minimality of G, $$G-uu_3$$ admits a 2DVDE-6-coloring $$\phi$$ using the color set C. Without loss of generality, assume that $$C(v_3)=\{1,2,3\}$$ with $$\phi (u_3v_3)=1$$, $$\phi (uu_1)=c_1$$, and $$\phi (uu_2)=c_2$$.

The proof is split into the following two cases.

Case 1. There is a color $$a\in \{4,5,6\}{\setminus } (C(u_1)\cup C(u_2))$$, say $$a=4$$.

If $$C(v_1)\ne \{4,c_1,c_2\}$$ and $$C(v_2)\ne \{4,c_1,c_2\}$$, then we color $$uu_3$$ with 4. Otherwise, by symmetry, we may assume that $$C(v_1)=\{4,c_1,c_2\}$$. It is easy to see that $$\phi (u_1v_1)=c_2$$. We color $$uu_3$$ with 4 and then recolor $$uu_1$$ with a color $$b\in C{\setminus } \{4,c_1,c_2,\phi (u_2v_2)\}$$ such that u does not conflict with $$v_2$$.

Case 2. $$4,5,6\in C(u_1)\cup C(u_2)$$.

Without loss of generality, assume that $$\phi (uu_1)=4$$, $$\phi (u_1v_1)=5$$, and $$6\in C(u_2)$$. There are two possibilities to be handled.
• $$c_2=6$$. If $$\phi (u_2v_2)\notin \{4,5\}$$, then we recolor $$uu_1$$ with a color in $$\{1,2,3\}$$ such that $$u_1$$ does not conflict with $$v'_1$$ and $$v''_1$$. Then the proof is reduced to Case 1. So assume that $$\phi (u_2v_2)\in \{4,5\}$$. It suffices to recolor $$uu_2$$ with a color in $$\{1,2,3\}$$ such that $$u_2$$ does not conflict with $$v'_2$$ and $$v''_2$$, and then the proof is reduced to Case 1.

• $$c_2\ne 6$$. Thus, $$\phi (u_2v_2)=6$$. If $$c_2=5$$, then we recolor $$uu_1$$ with a color in $$\{1,2,3\}$$ such that $$u_1$$ does not conflict with $$v'_1$$ and $$v''_1$$ and then the proof is reduced to Case 1. Otherwise, $$c_2\in \{1,2,3\}$$. If $$C(v_1)\ne \{4,5,c_2\}$$, then we color $$uu_3$$ with 5. Otherwise, $$C(v_1)=\{4,5,c_2\}$$, say $$\phi (v_1v'_1)=4$$ and $$\phi (v_1v''_1)=c_2$$. In this case, we color $$uu_3$$ with 4, and recolor $$uu_1$$ with a color in $$\{1,2,3\}{\setminus } \{c_2\}$$.

$$\square$$

## Claim 4

G does not contain a path $$P=v_0v_1\ldots v_7$$ such that $$d(v_1)=d(v_3)=d(v_5)=3$$ and $$d(v_2)=d(v_4)=d(v_6)=2$$.

## Proof

Suppose to the contrary that G contains such a path P, as shown in Fig. 2. For $$i=1,3,5$$, we denote by $$u_i$$ the neighbor of $$v_i$$ other than $$v_{i-1}$$ and $$v_{i+1}$$. By Claim 3, we see that $$d(u_3)=d(u_5)=3$$, and there exists at least one 3-vertex in $$\{v_0, u_1\}$$, say $$d(u_1)=3$$. Let $$N(u_3)=\{v_3,x_3,y_3\}$$. Let $$H=G-v_2v_3-v_3v_4$$. By the minimality of G, H has a 2DVDE-6-coloring $$\phi$$ with the color set C such that $$\phi (u_3v_3)=1$$, $$\phi (u_3x_3)=2$$, and $$\phi (u_3y_3)=3$$. Let $$S_1=\{4,5,6\}{\setminus } C(v_1)$$ and $$S_5=\{4,5,6\}{\setminus } C(v_5)$$. To extend $$\phi$$ from H to G, we consider the following cases.

Case 1. $$|S_1|\ge 1$$ and $$|S_5|\ge 1$$.

Let $$a\in S_1$$ and $$b\in S_5$$. Then the proof splits into two subcases.

Case 1.1. $$a\ne b$$, say $$a=4$$ and $$b=5$$.

Assign 4 to $$v_2v_3$$ and 5 to $$v_3v_4$$. If $$v_2$$ and $$v_4$$ are not conflict under the extended coloring, we are done. Otherwise, we derive that $$\phi (v_4v_5)=4$$ and $$\phi (v_1v_2)=5$$. In this case, if $$6\in S_1$$, or $$6\in S_5$$, we recolor $$v_2v_3$$ or $$v_3v_4$$ with 6, and the proof is complete. Otherwise, $$6\notin S_1\cap S_5$$. If $$v_3v_4$$ can be legally recolored with some color $$c\in \{2,3,6\}$$, we are done. Otherwise, $$\phi (v_5v_6)\in \{1, 6\}$$, $$C(x_3)= \{1,2,4\}$$, and $$C(y_3)=\{1,3,4\}$$. We only need recolor $$v_2v_3$$ with a color $$c\in \{2,3\}$$ such that $$v_2$$ does not conflict with $$v_0$$.

Case 1.2. $$a=b$$.

Suppose that $$a=b=4$$. Furthermore, we may assume that $$5, 6\in C(v_1)\cap C(v_5)$$ by the proof of Case 1.1. Assign 4 to $$v_2v_3$$. Afterward, we consider the following subcases by the symmetry of colors used.
• $$\phi (v_4v_5)=1$$. We color $$v_3v_4$$ with a color in $$\{2,3,5,6\}$$ such that $$v_3$$ does not conflict with $$x_3,y_3$$, and $$v_4$$ does not conflict with $$v_6$$.

• $$\phi (v_4v_5)=2$$. We color $$v_3v_4$$ with a color in $$\{5,6\}$$ such that $$v_4$$ does not conflict with $$v_6$$.

• $$\phi (v_4v_5)=5$$. If $$v_3v_4$$ can be legally colored with some color in $$\{2,3,6\}$$, we are done. Otherwise, it is immediate to conclude that $$\phi (v_5v_6)=6$$, $$\phi (v_6v_7)=5$$, $$C(x_3)=\{1,2,4\}$$, and $$C(y_3)=\{1,3,4\}$$. If $$\phi (v_1v_2)=1$$, then we recolor $$v_2v_3$$ with 2 and color $$v_3v_4$$ with 3. If $$\phi (v_1v_2)\in \{2,3\}$$, say $$\phi (v_1v_2)=2$$, then we recolor $$v_2v_3$$ with 3 and color $$v_3v_4$$ with 2. Otherwise, $$\phi (v_1v_2)\in \{5,6\}$$, we recolor $$v_2v_3$$ with a color $$c\in \{2,3\}$$ such that $$v_2$$ does not conflict with $$v_0$$, and then color $$v_3v_4$$ with a color in $$\{2,3\}{\setminus } \{c\}$$.

Case 2. $$|S_1|=|S_5|=0$$.

It follows that $$C(v_1)=C(v_5)=\{4,5,6\}$$, say $$\phi (v_1v_2)=4$$. If neither $$C(x_3)$$ nor $$C(y_3)$$ is $$\{1,2,3\}$$, then we color $$v_2v_3$$ with 2 and $$v_3v_4$$ with 3. Otherwise, we may assume that $$C(x_3)=\{1,2,3\}$$ by symmetry. Assign 3 to $$v_2v_3$$. If $$\phi (v_4v_5)\ne \phi (v_6v_7)$$, then we color $$v_3v_4$$ with a color in $$\{4,5,6\}{\setminus } \{\phi (v_4v_5)\}$$ such that $$v_3$$ does not conflict with $$y_3$$. Otherwise, $$\phi (v_4v_5)=\phi (v_6v_7)$$.

By symmetry, we may first assume that $$\phi (v_4v_5)=\phi (v_6v_7)=5$$. If $$v_3v_4$$ can not be legally colored with any color in $$\{4,6\}$$, then $$C(y_3)=\{1,3,4\}$$ and $$\phi (v_5v_6)=6$$; or $$C(y_3)=\{1,3,6\}$$ and $$\phi (v_5v_6)=4$$. In this case, we recolor $$v_2v_3$$ with 2 and color $$v_3v_4$$ with 4 or 6 such that $$v_4$$ does not conflict with $$v_6$$.

Next assume that $$\phi (v_4v_5)=\phi (v_6v_7)=4$$. If $$v_3v_4$$ can not be legally colored with any color in $$\{5,6\}$$, then $$C(y_3)=\{1,3,5\}$$ and $$\phi (v_5v_6)=6$$; or $$C(y_3)=\{1,3,6\}$$ and $$\phi (v_5v_6)=5$$. It suffices to recolor $$v_2v_3$$ with 2 and color $$v_3v_4$$ with 5 or 6 such that $$v_4$$ does not conflict with $$v_6$$.

Case 3. $$|S_1|\ge 1$$ and $$|S_5|=0$$.

This means that $$C(v_5)=\{4,5,6\}$$. Let $$4\in S_1$$. Assign 4 to $$v_2v_3$$. If $$v_3v_4$$ can not be legally colored with 5 and 6, then $$\phi (v_4v_5)=4$$ and $$\phi (v_1v_2)\in \{5, 6\}$$, or $$\phi (v_4v_5)\in \{5, 6\}$$. Therefore, if $$v_3v_4$$ can not be legally colored with 2 and 3, we have $$C(x_3)=\{1, 2, 4\}$$ and $$C(y_3)=\{1, 3, 4\}$$. If there is $$c\in \{5, 6\}\cap S_1$$, then we recolor $$v_2v_3$$ with c, and color $$v_3v_4$$ with 2 or 3 such that $$v_2$$ does not conflict with $$v_4$$. Otherwise, $$5,6\in C(v_1)$$. If $$\phi (v_1v_2)\in \{2, 3\}$$, then we recolor $$v_2v_3$$ with $$c\in \{2, 3\}{\setminus }\{\phi (v_1v_2)\}$$, and color $$v_3v_4$$ with a color in $$\{2,3\}{\setminus } \{c\}$$. If $$\phi (v_1v_2)\notin \{2, 3\}$$, then we recolor $$v_2v_3$$ with $$c\in \{2, 3\}$$ such that $$v_2$$ does not conflict with $$v_0$$, and color $$v_3v_4$$ with a color in $$\{2,3\}{\setminus } \{c\}$$.

Case 4. $$|S_1|=0$$ and $$|S_5|\ge 1$$.

Note that $$C(v_1)=\{4,5,6\}$$. Let $$4\in S_5$$, and assign 4 to $$v_3v_4$$.

First, assume that $$\phi (v_1v_2)=4$$. If $$v_2v_3$$ can not be legally colored with 2, 3, 5 or 6, then we can assume that $$C(x_3)=\{1,2,4\}$$, $$C(y_3)=\{1,3,4\}$$, $$C(v_0)=\{4,5\}$$, and $$C(v_4)=\{4,6\}$$. Recolor $$v_3v_4$$ with a color $$c\in \{2,3\}$$ such that $$v_4$$ does not conflict with $$v_6$$, and then color $$v_2v_3$$ with a color in $$\{2,3\}{\setminus } \{c\}$$.

Next, assume that $$\phi (v_1v_2)\in \{5,6\}$$, say $$\phi (v_1v_2)=5$$. If $$v_2v_3$$ can not be legally colored with 2, 3, or 6, then we can assume that $$C(x_3)=\{1,2,4\}$$, $$C(y_3)=\{1,3,4\}$$, and $$C(v_0)=\{5,6\}$$. If there is a color $$c\in \{5, 6\}{\setminus } C(v_5)$$, then we recolor $$v_3v_4$$ with c, and then color $$v_2v_3$$ with a color in $$\{2, 3\}$$ such that $$v_2$$ does not conflict with $$v_4$$. Or else, $$5, 6\in C(v_5)$$. If $$\phi (v_4v_5)\in \{2, 3\}$$, then we recolor $$v_3v_4$$ with a color in $$\{2, 3\}{\setminus } \{\phi (v_4v_5)\}$$, and then color $$v_2v_3$$ with $$\phi (v_4v_5)$$. If $$\phi (v_4v_5)\notin \{2, 3\}$$, then we recolor $$v_3v_4$$ with a color $$c\in \{2,3\}$$ such that $$v_4$$ does not conflict with $$v_6$$, and then color $$v_2v_3$$ with a color in $$\{2,3\}{\setminus } \{c\}$$. $$\square$$

## Claim 5

G does not contain a $$3_{2}$$-vertex.

## Proof

Suppose to the contrary that G contains a 3-vertex u adjacent to two 2-vertices $$u_1,u_2$$ and a vertex $$u_3$$, as shown in Fig. 3. By Claims 1 and 3, $$d(u_3)=3$$. Let $$N(u_3)=\{u,x_3,y_3\}$$, $$N(u_1)=\{u,v_1\}$$, and $$N(u_2)=\{u,v_2\}$$. By Claims 1 and 2, $$d(v_i)=3$$ for $$i=1,2$$. Let $$H=G-uu_1-uu_2$$. By the minimality of G, H has a 2DVDE-6-coloring $$\phi$$ with the color set C such that $$\phi (uu_3)=1$$, $$\phi (u_3x_3)=2$$, and $$\phi (u_3y_3)=3$$. To extend $$\phi$$ to G, we have to consider the following two cases.

Case 1. $$v_1=v_2$$.

Let $$N(v_1)=\{u_1, u_2, x_1\}$$. By Claims 1 and 3, $$x_1$$ is a 3-vertex. If $$C(v_1)=\{4,5,6\}$$ with $$\phi (u_1v_1)=4$$ and $$\phi (u_2v_1)=5$$, then we color $$uu_1$$ with 6 and $$uu_2$$ with 4. Otherwise, assume that $$4\notin C(v_1)$$, and assign 4 to $$uu_1$$. If there is a color $$c\in \{5, 6\}{\setminus } C(v_1)$$, then we color $$uu_2$$ with c. Otherwise, $$5, 6\in C(v_1)$$. When $$\phi (u_2v_1)\in \{2, 3\}$$, we derive that $$\phi (u_1v_1)\in \{5, 6\}$$, and so color $$uu_2$$ with $$\phi (u_1v_1)$$. When $$\phi (u_2v_1)\notin \{2, 3\}$$, we color $$uu_2$$ with a color in $$\{5,6\}{\setminus } \{\phi (u_2v_1)\}$$.

Case 2. $$v_1\ne v_2$$.

Note that, for $$i=1,2$$, each of the other neighbors of $$v_i$$, distinct from $$u_i$$, are of degree 3 by Claim 4. Let $$T_1=\{4,5,6\}{\setminus } C(v_1)$$ and $$T_2=\{4,5,6\}{\setminus } C(v_2)$$. By symmetry, we consider three possibilities as follows.

Case 2.1. $$|T_1|\ge 1$$ and $$|T_2|\ge 1$$.

Let $$a\in T_1$$ and $$b\in T_2$$. First assume that $$a\ne b$$, say $$a=4$$ and $$b=5$$. Assign 4 to $$uu_1$$ and 5 to $$uu_2$$. If $$C(u_1)\ne C(u_2)$$, we are done. Otherwise, we deduce that $$\phi (u_1v_1)=5$$ and $$\phi (u_2v_2)=4$$. If $$6\in T_1$$, then we recolor $$uu_1$$ with 6. If $$6\in T_2$$, then we recolor $$uu_2$$ with 6. So suppose that $$6\in C(v_1)\cap C(v_2)$$. If $$C(x_3)\ne \{1,2,4\}$$ or $$C(y_3)\ne \{1,3,4\}$$, then we recolor $$uu_2$$ with 2 or 3. Otherwise, we recolor $$uu_1$$ with 2.

Next assume that $$T_1=T_2=\{4\}$$, implying that $$5,6\in C(v_1)\cap C(v_2)$$. Assign 4 to $$uu_1$$. If $$\phi (u_2v_2)\in \{1, 2, 3\}$$, then we color $$uu_2$$ with 5 or 6. Otherwise, $$\phi (u_2v_2)\in \{5, 6\}$$, say $$\phi (u_2v_2)=5$$, we color $$uu_2$$ with 6.

Case 2.2. $$|T_1|=|T_2|=0$$.

It follows that $$C(v_1)=C(v_2)=\{4,5,6\}$$. Let $$c\in \{4,5,6\} {\setminus } \{\phi (u_1v_1),\phi (u_2v_2)\}$$, say $$c=6$$. Thus, we can always color $$\{uu_1,uu_2\}$$ with two of $$\{2,3,6\}$$ such that u does not conflict with $$x_3$$ and $$y_3$$.

Case 2.3. $$|T_1|=0$$ and $$|T_2|\ge 1$$.

Then $$C(v_1)=\{4,5,6\}$$. Let $$4\in T_2$$. Assign 4 to $$uu_2$$. If $$\phi (u_1v_1)\ne 4$$, by symmetry, we can assume that $$\phi (u_1v_1)=5$$. Color $$uu_1$$ with a color in $$\{2,3,6\}$$ such that u does not conflict with $$x_3$$ and $$y_3$$. Otherwise, $$\phi (u_1v_1)=4$$. Then we color $$uu_1$$ with some color in $$\{2,3,5,6\}$$ such that u does not conflict with $$x_3,y_3$$, and $$u_1$$ does not conflict with $$u_2$$. $$\square$$

## Claim 6

No 2-vertex lies in a 3-cycle.

## Proof

Suppose to the contrary that G contains a 2-vertex x which lies on the boundary of a 3-cycle xyzx. By Claim 2, both y and z are 3-vertices. Let $$N(y)=\{x,z,u\}$$ and $$N(z)=\{x,y,v\}$$. By Claim 5, $$d(u)=d(v)=3$$. Let $$N(u)=\{y,u_1,u_2\}$$. By the minimality of G, $$G-xy$$ admits a 2DVDE-6-coloring $$\phi$$ with the color set C such that $$\phi (xz)=1$$, $$\phi (yz)=2$$, and $$\phi (vz)=3$$.

If $$\phi (uy)=1$$, then we color xy with a color in $$\{4,5,6\}$$ such that y does not conflict with $$u_1$$ and $$u_2$$.

If $$\phi (uy)=3$$ and xy can not be legally colored with 4, 5, or 6, then we may assume that $$\phi (uy)\in \{4, 5, 6\}$$, say $$\phi (uy)=4$$. If xy can not be legally colored with 3, 5 or 6, then it is easy to derive, without loss of generality, that $$C(v)=\{2,3,4\}$$, $$C(u_1)=\{2,4,5\}$$, and $$C(u_2)=\{2,4,6\}$$. Switch the colors of xz and yz and then color xy with 3. $$\square$$

## Claim 7

G does not contain a 2-vertex.

## Proof

Using Claims 16, we suppose to the contrary that G contains a 2-vertex $$u_1$$ adjacent to two 3-vertices u and $$v_1$$. Let $$N(v_1)=\{u_1,x_1,w_1\}$$ and $$N(u)=\{u_1,u_2,u_3\}$$. By Claim 5, $$d(u_2)=d(u_3)=d(x_1)=d(w_1)=3$$. Let $$N(u_2)=\{u,v_2,v_3\}$$ and $$N(u_3)=\{u,v_4,v_5\}$$. Let $$H=G-u_1+uv_1$$. By Claim 6, H is a simple subcubic graph with $$|E(H)|<|E(G)|$$. By the minimality of G, H admits a 2DVDE-6-coloring $$\phi$$ with the color set C such that $$\phi (uv_1)=1$$, $$\phi (uu_2)=2$$, $$\phi (uu_3)=3$$, $$\phi (x_1v_1)=a$$, and $$\phi (w_1v_1)=b$$.

To extend $$\phi$$ to the whole graph G, we first assign 1 to $$u_1v_1$$. If $$uu_1$$ can not be legally colored with 4, 5, or 6, then we may assume that $$C(v_i)=\{2,3,i+2\}$$ for $$i\in \{2, 3, 4\}$$. This implies that $$d(v_i)=3$$ for $$i=2,3,4$$. Let $$N(v_i)=\{u_2, x_i, w_i\}$$ for $$i\in \{2, 3\}$$, $$N(v_j)=\{u_3, x_j, w_j\}$$ for $$i\in \{4, 5\}$$.

If $$C(u_2)=\{2, 4, 5\}$$, then we recolor $$uu_2$$ with 6 and color $$uu_1$$ with 4 or 5 such that u does not conflict with $$v_5$$. Otherwise, $$C(u_2)\ne \{2, 4, 5\}$$, by symmetry, we assume that $$\phi (u_2v_2)=4$$, $$\phi (u_2v_3)=\phi (v_2w_2)=3$$, and $$\phi (v_3w_3)=5$$. If $$C(w_2)\ne \{1,3,4\}$$, then we recolor $$uu_2$$ with 1 and color $$uu_1$$ with 4, 5, or 6 such that u does not conflict with $$v_1$$ and $$v_5$$. Now suppose that $$C(w_2)=\{1,3,4\}$$. If $$C(w_3)\ne \{3,4,5\}$$, then we recolor $$uu_2$$ with 5 and color $$uu_1$$ with 4 or 6 such that u does not conflict with $$v_5$$. Now we furthermore suppose that $$C(w_3)=\{3,4,5\}$$. If $$C(u_3)\ne \{3,4,6\}$$, then we recolor $$uu_2$$ with 6 and color $$uu_1$$ with 4 or 5 such that u does not conflict with $$v_5$$. Otherwise, $$C(u_3)=\{3,4,6\}$$. This implies that $$\phi (u_3v_4)=6$$ and $$\phi (u_3v_5)=4$$. So assume that $$\phi (v_4x_4)=2$$ and $$\phi (v_4w_4)=3$$.

Since $$2\le d(v_5)\le 3$$, we have to deal with two subcases, depending on the size of $$d(v_5)$$.

Case 1. $$d(v_5)=2$$.

Let $$N(v_5)=\{u_3,y_5\}$$. If $$C(y_5)\ne \{4,5,6\}$$, then we recolor $$uu_3$$ with 5 and color $$uu_1$$ with 4. If $$C(y_5)= \{4,5,6\}$$, then we recolor $$uu_3$$ with 1 and color $$uu_1$$ with a color in $$\{4,5,6\}$$ such that u does not conflict with $$v_1$$.

Case 2. $$d(v_5)=3$$.

If neither $$C(x_5)$$ nor $$C(w_5)$$ is $$\{4,5,6\}$$, then we recolor $$uu_3$$ with 5 and color $$uu_1$$ with 6. Otherwise, assume that $$C(w_5)=\{4,5,6\}$$ by symmetry. If $$C(x_5)\ne \{1,4,6\}$$, then we recolor $$uu_3$$ with 1 and color $$uu_1$$ with 5 or 6 such that u does not conflict with $$v_1$$. Otherwise, assume that $$C(x_5)= \{1,4,6\}$$. If $$C(x_4)\ne \{2, 4, 6\}$$, then we recolor $$uu_3$$ with 2 and $$uu_2$$ with 6, and then color $$uu_1$$ with 5. So assume that $$C(x_4)=\{2, 4, 6\}$$.

First assume that $$C(v_5)=\{1,4,5\}$$. If $$C(w_4)\ne \{2,3,6\}$$, then we recolor $$u_3v_5$$ with 2 and $$uu_2$$ with 6, and color $$uu_1$$ with 4. If $$C(w_4)=\{2,3,6\}$$, then we recolor $$u_3v_5$$ with 2 and $$uu_3$$ with 5, and color $$uu_1$$ with 4.

Now assume that $$C(v_5)\ne \{1,4,5\}$$, say $$c(v_5x_5)=1$$ and $$c(v_5w_5)=6$$. Let $$z_5$$ denote the neighbor of $$x_5$$ with $$x_5z_5$$ being colored with 6. If $$C(w_4)\ne \{2,3,6\}$$ and $$C(z_5)\ne \{1,2,6\}$$, then we recolor $$u_3v_5$$ with 2 and $$uu_2$$ with 6 and color $$uu_1$$ with 4. Otherwise, we consider two possibilities as follows.
• Assume that $$C(w_4)=\{2,3,6\}$$. If $$C(z_5)\ne \{1,3,6\}$$, then we recolor $$u_3v_5$$ with 3 and $$uu_3$$ with 5, and color $$uu_1$$ with 4. Otherwise, $$C(z_5)=\{1,3,6\}$$, we can recolor $$u_3v_5$$ with 2 and $$uu_3$$ with 5, and color $$uu_1$$ with 4.

• Assume that $$C(z_5)=\{1,2,6\}$$. If $$C(w_4)\ne \{1,3,6\}$$, then we recolor $$u_3v_5$$ with 3 and $$uu_3$$ with 1, and color $$uu_1$$ with 4 or 5 such that u does not conflict with $$v_1$$. Otherwise, $$C(w_4)=\{1,3,6\}$$, we recolor $$u_3v_5$$ with 3, $$uu_3$$ with 2, $$uu_2$$ with 6, and color $$uu_1$$ with 5.

## Claim 8

G does not contain a cut edge.

## Proof

Assume to the contrary that G contains a cut edge uv. Then $$G-uv$$ consists of exactly two components $$H_1$$ and $$H_2$$ with $$u\in V(H_1)$$ and $$v\in V(H_2)$$. Let $$G_1=G[V(H_1)\cup \{v\}]$$ and $$G_2=G[V(H_2)\cup \{u\}]$$. Then $$G_1$$ and $$G_2$$ are proper subgraphs of G. By the minimality of G, $$G_1$$ has a 2DVDE-6-coloring $$\phi _1$$ using the color set C such that $$C(u)=\{1,2,3\}$$ and $$\phi _1(uv)=1$$, and $$G_2$$ has a 2DVDE-6-coloring $$\phi _2$$ using the same color set C such that $$C(v)=\{1,4,5\}$$ and $$\phi _2(uv)=1$$. (This can be accomplished by exchanging reasonably the colors of $$G_2$$ under $$\phi _2$$, if necessarily.) It is easy to inspect that combining $$\phi _1$$ and $$\phi _2$$ yields a 2DVDE-6-coloring of G. This is a contradiction. $$\square$$

Up to now, we have shown that G is a 3-regular simple graph without cut edges. By Lemma 2, G can be edge-partitioned into a perfect matching M and a class $${\mathcal {A}}$$ of cycles. To establish a 2DVDE-6-coloring of G, we first color legally the edges of M using the fewest colors in C. Equivalently, we need to determine the smallest integer l for which M can be decomposed into subsets $$M_1,M_2,\ldots ,M_l$$ such that each $$M_i$$ is an induced matching in G. To do this, let $$G^*$$ be the graph obtained from G by contracting each edge of M. Since G is 3-regular, we see that $$\Delta (G^*)\le 4$$. By Lemma 3, $$\chi (G^*)\le 4$$ unless $$G^*$$ is isomorphic to the complete graph $$K_5$$. If $$G\cong K_5$$, then it is easy to observe that G is isomorphic to the Petersen graph. By Lemma 4, $$\chi ^{\prime }_\mathrm{d2}(G)=5$$ and hence the theorem holds, a contradiction. So, assume that $$\chi (G^*)\le 4$$. Let c be a 4-coloring of $$V(G^*)$$ using the colors 3, 4, 5, 6. For $$i=3,4,5,6$$, we define the subset of edges in M as follows.
\begin{aligned} M_i=\{e\in M\ |\ e \hbox { corresponds to a vertex } v^*\hbox { in }G^*\hbox { with } c(v^*)=i\}. \end{aligned}
Then $$M=M_3\cup M_4\cup M_5\cup M_6$$. Construct a partial edge coloring $$\pi$$ of G by assigning the color i to $$M_i$$ for $$i=3,4,5,6$$. Obviously, each $$M_i$$ forms an induced matching of G. Afterward, we are going to extend $$\pi$$ into a 2DVDE-6-coloring $$\phi$$ of G using $$C=\{1,2,\ldots ,6\}$$.
Let $$A=v_0v_1\cdots v_{m-1}v_0$$ be an arbitrary cycle in $${\mathcal {A}}$$, where $$m\ge 3$$. For each vertex $$v_i\in V(A)$$, we use $$u_i$$ to denote the neighbor of $$v_i$$ other than $$v_{i-1}$$ and $$v_{i+1}$$. We say that $$v_iu_i$$ is a pendant edge of A at the vertex $$v_i$$. However, it should be noticed that $$v_iu_i$$ might be a chord of A. Here and in the following discussion, all indices are taken modulo m. For convenience, for each edge $$e_i:=v_iv_{i+1}$$ in E(A), we define a list $$L(e_i)$$ as follows:
\begin{aligned} L(e_i)=\{3,4,5,6\}{\setminus } \{\pi (v_{s}u_{s})\hbox { for }s=i-1,i,i+1,i+2\}. \end{aligned}
The definition of $$\pi$$ implies that $$0\le |L(e_i)|\le 2$$ for each $$i=0,1,\ldots ,m-1$$. The edge $$e_i$$ is called good if $$|L(e_i)|\ge 1$$, and bad if $$|L(e_i)|=0$$. The cycle A is said to be good if $$|L(e_i)|\ge 1$$ for all $$e_i\in E(A)$$. Otherwise, A is called bad. The proof is split into two cases as follows.

Case 1. A is good.

It follows that $$|L(e_i)|\ge 1$$ for all $$i=0,1,\ldots , m-1$$. Let us discuss two subcases below.

Case 1.1. m is even.

We first color $$e_i$$ with 1 for $$i=0,2,\ldots ,m-2$$, and $$e_i$$ with 2 for $$i=1,3,\ldots ,m-1$$. If $$m\equiv 0$$(mod 4), then we recolor $$e_i$$ with a color in $$L(e_i)$$ for $$i=0,4,8,\ldots ,m-4$$. If $$m\equiv 2$$(mod 4), then we recolor $$e_i$$ with a color in $$L(e_i)$$ for $$i=0,4,8,\ldots ,m-6$$ and moreover recolor $$e_{m-3}$$ with a color in $$L(e_{m-3})$$.

Case 1.2. m is odd.

Since m is odd, there must exist a pair of pendant edges $$v_ku_k$$ and $$v_{k+2}v_{k+2}$$ whose colors are different under $$\pi$$, say $$k=m-3$$. So, we first color $$e_0$$ with a color $$a\in L(e_0)$$, then color $$e_i$$ with 1 for $$i=1,3,\ldots ,m-2$$, and $$e_i$$ with 2 for $$i=2,4,\ldots ,m-1$$. If $$m\equiv 1$$(mod 4), then we recolor $$e_i$$ with a color in $$L(e_i)$$ for $$i=4,8,12, \ldots , m-5$$. If $$m\equiv 3$$(mod 4), then we recolor $$e_i$$ with a color in $$L(e_i)$$ for $$i=4,8,12, \ldots , m-3$$.

Note that there is at least one bad edge in E(A). If each of the edges of A is bad, then it is easy to verify that $$m\equiv 0$$(mod 4). It suffices to alternately color the edges of A with 1 and 2 in clockwise order. It is easy to inspect that the resultant coloring is a 2DVDE-6-coloring of G. Otherwise, assume that A has at least one good edge. Then A can be edge-partitioned into 2r ($$r\ge 1$$) sections ($$P_1,Q_1, P_2, Q_2,\ldots , P_r, Q_r$$) in clockwise order, as shown in Fig. 4, where $$P_i$$ is a maximal sub-path of A with all edges being bad, and $$Q_i$$ is a maximal sub-path of A with all edges being good.
We first give an initial coloring for E(A) as follows:
• If |E(A)| is even, then we alternately color the edges of A with 1 and 2 in clockwise order.

• If |E(A)| is odd, then we choose an edge $$e^*\in E(Q_1)$$, which is called special edge, and then alternately color the edges of $$E(A){\setminus } \{e^*\}$$ with 1 and 2 in clockwise order.

It is an easy observation that, in the current coloring, any two edges in $$P_j$$ for $$1\le j\le r$$ are legal, and furthermore any two edges in $$E(P_1)\cup E(P_2)\cup \cdots \cup E(P_r)$$ are legal. Hence it is enough to recolor certain edges in $$Q_i$$’s to get a 2DVDE-6-coloring of G.

Without loss of generality, assume that that $$Q_i=v_1v_2\cdots v_{k+1}$$, i.e., $$Q_i=(e_1,e_2,\ldots ,e_k)$$. By the maximality of $$Q_i$$, both $$e_0$$ and $$e_{k+1}$$ are bad. If $$k = 1$$, then $$Q_i = \{e_1\}$$ and we assign some color in $$L(e_1)$$ to $$e_1$$. Otherwise, $$k \ge 2$$, we define Recoloring Procedures I and II, where (I) runs if $$Q_i$$ contains no special edge, and (II) applies only in the case when $$Q_i$$ contains a special edge, i.e., A is odd and $$i = 1$$.

Recoloring Procedure I.

If $$k\equiv 2$$(mod 4), then we recolor $$e_2$$ with a color in $$L(e_2)$$, and then recolor $$e_i$$ with a color in $$L(e_i)$$ for $$i=5,9,13,\cdots ,k-1$$.

If $$k\equiv 3$$(mod 4), then we recolor $$e_2$$ with a color in $$L(e_2)$$, and then recolor $$e_i$$ with a color in $$L(e_i)$$ for $$i=6,10,14,\cdots ,k-1$$.

If $$k\equiv 0$$(mod 4), then we recolor $$e_2$$ with a color in $$L(e_2)$$, and then recolor $$e_i$$ with a color in $$L(e_i)$$ for $$i=6,10,14,\cdots ,k-2$$.

If $$k\equiv 1$$(mod 4), then we recolor $$e_3$$ with a color in $$L(e_3)$$, and then recolor $$e_i$$ with a color in $$L(e_i)$$ for $$i=7,11,15,\cdots ,k-2$$.

Recoloring Procedure II.

If $$k\not \equiv 1$$(mod 4), then we choose $$e_2$$ as special edge and color $$e_2$$ with a color in $$L(e_2)$$. If $$k\equiv 1$$(mod 4), then we choose $$e_3$$ as special edge and color $$e_3$$ with a color in $$L(e_3)$$. Finally, we recolor other edges (if necessarily) according to Recoloring Procedure I.

Now we come to complete the recoloring for the whole cycle A. If A is an even cycle, then we apply (I) for every path $$Q_i$$ for $$i=1,2,\ldots ,r$$. If A is an odd cycle, then we apply (I) for every path $$Q_i$$ for $$i=2,3,\ldots ,r$$ and (II) for $$Q_1$$.

It is easy to inspect that the coloring defined above is a 2DVDE-6-coloring of G. This contradicts the choice of G. $$\square$$

## 3 Concluding remarks

In this paper, we show that every subcubic graph G satisfies $$\chi ^{\prime }_{\mathrm{d2}}(G)\le 6$$. The upper bound value 6 seems not to be best possible. Hence we like to put forward the following probelm:

## Conjecture 3

For a subcubic graph G, $$\chi ^{\prime }_{\mathrm{d2}}(G)\le 5$$.

Lemma 4 asserts that if Conjecture 3 were true, then the upper bound 5 is tight.

Recall that the proof of Theorem 5 consists of basically two parts: (i) reducing the vertices of degree at most 2 and cut edges; and (ii) dealing with 2-connected 3-regular graphs. For (ii), we first decompose E(G) into a 1-factor M and a 2-factor H. Then we establish a strong edge coloring for M, which is restricted in the original graph G, and then extend the coloring of M to the uncolored subgraph H to form a 2DVDE-6-coloring of the whole graph G. We feel that this coloring scheme can be probably applied to other coloring problems of graphs.

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## Authors and Affiliations

• Victor Loumngam Kamga
• 1
• Weifan Wang
• 1
• Ying Wang
• 1
• Min Chen
• 1
1. 1.Department of MathematicsZhejiang Normal UniversityJinhuaChina