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Article

Neighbor Distinguishing Colorings of Graphs with the Restriction for Maximum Average Degree

School of Mathematics and Physics, Hebei University of Engineering, Handan 056038, China
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Author to whom correspondence should be addressed.
Axioms 2023, 12(12), 1132; https://doi.org/10.3390/axioms12121132
Submission received: 27 October 2023 / Revised: 9 December 2023 / Accepted: 13 December 2023 / Published: 18 December 2023
(This article belongs to the Special Issue Recent Advances in Graph Theory with Applications)

Abstract

:
Neighbor distinguishing colorings of graphs represent powerful tools for solving the channel assignment problem in wireless communication networks. They consist of two forms of coloring: neighbor distinguishing edge coloring, and neighbor distinguishing total coloring. The neighbor distinguishing edge (total) coloring of a graph G is an edge (total) coloring with the requirement that each pair of adjacent vertices contains different color sets. The neighbor distinguishing edge (total) chromatic number of G is the smallest integer k in cases where a neighbor distinguishing edge (total) coloring exists through the use of k colors in G. The maximum average degree of G is the maximum of the average degree of its non-empty subgraphs. In this paper, we characterize the neighbor distinguishing edge (total) chromatic numbers of graphs with a maximum average degree less than four by means of the discharging method.

1. Introduction

Many real-world situations can conveniently be described by means of a graph composed of vertices and edges joining certain pairs of these vertices. For example, in a social network, each vertex represents a person and each edge represents a relationship where two people are friends. In a protein network, each vertex represents a protein and each edge represents the interaction of different proteins. Graph colorings began with the famous Four Color Problem in 1852, and have grown to be one of the most interesting branches in graph theory. They deal with the fundamental problem of partitioning a set of objects into classes, according to some prescribed rules.
Neighbor distinguishing colorings of graphs originate from the channel assignment problem, which are caused by the rapid development of mobile communication and sharp increase in number of users, resulting in the contradiction between the growing number of users and the limited expansion of communication network resources. In a wireless communication network, each vertex represents a communication equipment and each edge represents a relationship where two communication devices transmit data to each other. If two edges are assigned different channels, they can transmit data independently. In the first place, adjacent edges incident to a vertex are required synchronously to transmit or receive without interference. In the second place, the edge between any two vertices is considered to be composed of two directed edges in opposite directions. These bidirectional edges are required to transmit and receive data simultaneously without interference. Driven by the channel assignment with the above two properties in a wireless communication network, neighbor distinguishing colorings of graphs have attracted the attention of researchers.
Neighbor distinguishing colorings of graphs consist of two forms of coloring, one is the neighbor distinguishing edge coloring, and the other is the neighbor distinguishing total coloring. Neighbor distinguishing colorings of graphs expand the concept and category of edge coloring and total coloring in the traditional sense, and consider the theory of edge coloring and total coloring under stronger conditions, and are also a manifestation of the further development and promotion of the edge coloring and the total coloring.
The maximum average degree of graphs is an important measure to determine the sparseness of an arbitary graph (which need not be planar). It is known that the maximum average degree of a graph can be computed in polynomial time through the use of the Matroid Partitioning Algorithm. In this paper, we conduct research on the neighbor distinguishing edge chromatic number and the neighbor distinguishing total chromatic number of graphs with the restriction for maximum average degree. All the graphs under consideration are finite, undirected and simple. For the statement that a graph G has the vertex set V ( G ) and the edge set E ( G ) , we usually write G = ( V ( G ) , E ( G ) ) .
If a vertex v is in V ( G ) , then the degree d G ( v ) of v is the number of edges in G that are incident with v. Clearly, d G ( v ) can also be interpreted as the number of vertices in G that are adjacent to v. If d G ( v ) = k ( d G ( v ) k , d G ( v ) k ) , then v is said to be a k-vertex ( k + -vertex, k -vertex). The maximum degree Δ ( G ) of G is the largest degree among the vertices in V ( G ) , that is, Δ ( G ) = m a x { d G ( v ) : v V ( G ) } , and abbreviated to Δ . The average degree of G is defined as v V ( G ) d G ( v ) | V ( G ) | = 2 | E ( G ) | | V ( G ) | and denoted by a d ( G ) . The maximum average degree of G, denoted by m a d ( G ) , is the maximum of the average degree among the non-empty subgraphs of G. So we can immediately say that m a d ( G ) = m a x { a d ( H ) : H G } = m a x { 2 | E ( H ) | | V ( H ) | : H G } .
The length of a cycle is the number of edges encountered in the cycle. A cycle of length k (at least k) is said to be a k-cycle ( k + -cycle) in G. The girth of G is the length of the smallest cycle in G and denoted by g ( G ) . For three positive integers p , q , r with the condition that p is less than or equal to q, the set of positive integers { p , p + 1 , , q } is denoted by [ p , q ] . If r [ p , q ] , then we use { p q , r } to denote the set of positive integers { p , p + 1 , , q , r } = [ p , q ] { r } .
An edge k-coloring of G is considered as a function ϕ from E ( G ) to [ 1 , k ] for which no pair of adjacent edges is given the same color. The edge chromatic number (or chromatic index) χ ( G ) of G is the minimum positive integer k in which case there is an edge k-coloring of G. For an edge k-coloring ϕ of G and a vertex v V ( G ) , C ϕ ( v ) = { ϕ ( x v ) | x v E ( G ) } is the set of colors assigned to the edges that are incident with v in G. The edge k-coloring ϕ of G is said to be neighbor distinguishing (or ϕ is an NDE-k-coloring of G) if C ϕ ( u ) C ϕ ( v ) for each two adjacent vertices u and v in G. The neighbor distinguishing edge chromatic number (or neighbor distinguishing chromatic index) χ a ( G ) of G is the minimum positive integer k in which case there is an NDE-k-coloring of G.
If there are no isolated edges in a graph G, then G is considered normal. Clearly, there exists an NDE-coloring of G if and only if G is normal. Therefore, when we are dealing with the NDE-coloring of G, we are assuming that G is normal. For any graph G, we can easily observe that χ a ( G ) χ ( G ) Δ . If there is a pair of adjacent Δ -vertices in G, then Δ + 1 is a rather obvious lower bound for χ a ( G ) . If G 1 , G 2 , , G k are the connected components of a graph G, then χ a ( G ) = m a x { χ a ( G i ) | i [ 1 , k ] } .
In 2002, to address problems associated with the optimal connectivity of switches in LANs, the NDE-coloring of graphs was first introduced by Zhang et al. [1]. Subsequently, they determined the neighbor distinguishing edge chromatic numbers of specific graphs, such as complete graphs, complete bipartite graphs, cycles, paths and trees, and proposed the challenging conjecture in this area as follows.
Conjecture 1.
If G is a connected graph with | V ( G ) | 3 and G C 5 , then χ a ( G ) Δ + 2 .
Note that χ a ( C 5 ) = 5 and Δ ( C 5 ) = 2 . It follows that χ a ( C 5 ) = Δ ( C 5 ) + 3 . To date, it is unknown whether C 5 is the unique exception graph. Balister et al. [2] verified that the conjecture above is true for graphs with Δ = 3 and bipartite graphs. Akbari et al. [3] provided an upper bound 3 Δ for the neighbor distinguishing edge chromatic number χ a ( G ) of an arbitrary graph G in terms of the maximum degree Δ of G. By means of the edge-partitions of graphs, this upper bound was improved gradually to that of 2.5 Δ + 5 due to Zhang et al. [4], to that of 2.5 Δ due to Wang et al. [5] and to that of 2 Δ + 2 due to Vučković [6]. Hatami [7] used the probabilistic method to establish that χ a ( G ) Δ + 300 for every graph G having a large enough Δ . Later, this upper bound was reduced by Joret and Lochet [8] to Δ + 19 .
If a graph G is drawn in a plane where no two of its edges cross, then G is said to be planar. For planar graphs with Δ 10 , Conjecture 1 is true. This was demonstrated in [9] for Δ 12 , in [10] for Δ = 11 and in [11] for Δ = 10 . Furthermore, for planar graphs with Δ 16 , Wang and Huang [12] characterized their neighbor distinguishing edge chromatic numbers. Afterwards, this condition is improved to the case Δ 14 in [13] and further to the case Δ 13 in [14].
Since every graph with maximum average degree less than three is 2-degenerate, Wang et al. [15] showed that if m a d ( G ) < 3 and Δ 6 , then Δ χ a ( G ) Δ + 1 ; and χ a ( G ) = Δ + 1 if and only if there is a pair of adjacent Δ -vertices in G. In this paper, our first main purpose is to prove the following theorem.
Theorem 1.
Let G be a graph with m a d ( G ) < 4 .
(a)
If Δ 8 , then Δ χ a ( G ) Δ + 1 .
(b)
If Δ 9 , then χ a ( G ) = Δ + 1 if and only if there is a pair of adjacent Δ -vertices in G.
In the case of planar graphs, the following relationship is due to Montassier and Raspaud [16] and established between the maximum average degree and the girth.
Proposition 1.
Let G be a planar graph. Then m a d ( G ) < 2 g ( G ) g ( G ) 2 .
If G is a planar graph with g ( G ) 4 , by Proposition 1, then m a d ( G ) < 2 g ( G ) g ( G ) 2 = 2 + 4 g ( G ) 2 4 . With the aid of Theorem 1, we have the following.
Corollary 1.
Let G be a planar graph with g ( G ) 4 .
(a)
If Δ 8 , then Δ χ a ( G ) Δ + 1 .
(b)
If Δ 9 , then χ a ( G ) = Δ + 1 if and only if there is a pair of adjacent Δ -vertices in G.
If G is a planar graph without 3-cycles, then g ( G ) 4 . The results in Corollary 1 have improved the results in [17] on planar graphs without 3-cycles.
An element of a graph G is a member of V ( G ) E ( G ) . If two elements in G are adjacent vertices, or adjacent edges, or incident vertex and edge, then they are said to be adjacent. A total k-coloring of G is considered as an assignment of k colors to V ( G ) E ( G ) in the case each pair of adjacent elements is given the distinct colors. The total chromatic number χ ( G ) of G is the smallest positive integer k in the case there is a total k-coloring of G. For a total k-coloring φ of G and a vertex v in V ( G ) , C φ ( v ) = { φ ( x v ) | x v E ( G ) } { φ ( v ) } is the set of colors assigned to v and its incident edges. The total k-coloring φ of G is said to be neighbor distinguishing (or φ is an NDT-k-coloring of G) if C φ ( u ) C φ ( v ) for each two adjacent vertices u and v in G. The neighbor distinguishing total chromatic number χ a ( G ) of G is the smallest positive integer k in which case there is an NDT-k-coloring of G. For any graph G, it is easy to observe that χ a ( G ) χ ( G ) Δ + 1 . If there is a pair of adjacent Δ -vertices in G, then Δ + 2 is a lower bound for χ a ( G ) . If G 1 , G 2 , , G k are the connected components of a graph G, then χ a ( G ) = m a x { χ a ( G i ) | i [ 1 , k ] } .
In 2005, the NDT-coloring of graphs was first investigated by Zhang et al. [18]. At the same time, the neighbor distinguishing total chromatic numbers were determined for a few special graphs, such as paths, trees, cycles, wheels, stars, complete graphs and complete bipartite graphs. With the aid of these results, they put forward the conjecture about the NDT-coloring below.
Conjecture 2.
If G is a connected graph with | V ( G ) | 2 , then χ a ( G ) Δ + 3 .
Conjecture 2 was verified for graphs with Δ = 3 in [19,20,21], and graphs with Δ = 4 in [22,23]. Huang et al. [24] demonstrated that 2 Δ is an upper bound for χ a ( G ) when G is a graph with Δ 3 . This result was improved by Vučković [25], who showed that any graph G has χ a ( G ) 5 Δ + 8 3 . The probabilistic result, due to Coker and Johannson [26], stated that a sufficiently large constant C exists in which case any graph G has χ a ( G ) Δ + C .
If we restrict ourselves to consider planar graphs, then the truth of Conjecture 2 is establiashed when Δ { 5 , 6 , 7 } . This was demonstrated in [27] for Δ 11 , in [28] for Δ = 10 , in [29,30] for Δ = 9 and in [31] for Δ = 8 . In addition, for planar graphs with Δ 14 , Wang and Huang [32] characterized their neighbor distinguishing total chromatic numbers. Later, this result was enhanced to the case Δ 13 in [33] and further to the case Δ 12 in [34].
In 2008, Wang and Wang [35] demonstrated that if m a d ( G ) < 3 and Δ 5 , then Δ + 1 χ a ( G ) Δ + 2 ; and χ a ( G ) = Δ + 2 if and only if there is a pair of adjacent Δ -vertices in G. In this paper, our second main purpose is to prove the following theorem.
Theorem 2.
Let G be a graph with m a d ( G ) < 4 .
(a)
If Δ 6 , then Δ + 1 χ a ( G ) Δ + 2 .
(b)
If Δ 7 , then χ a ( G ) = Δ + 2 if and only if there is a pair of adjacent Δ -vertices in G.
By Proposition 1 and Theorem 2, we deduce the following corollary easily.
Corollary 2.
Let G be a planar graph with g ( G ) 4 .
(a)
If Δ 6 , then Δ + 1 χ a ( G ) Δ + 2 .
(b)
If Δ 7 , then χ a ( G ) = Δ + 2 if and only if there is a pair of adjacent Δ -vertices in G.
The results given in Corollary 2 have completed the characterization for the neighbor distinguishing total chromatic numbers of planar graphs without three-cycles.

2. Preliminaries

Suppose that V ( H ) V ( G ) and E ( H ) E ( G ) , which means that H is a subgraph of a graph G. For a vertex v V ( H ) , a neighbor of v is a vertex that is adjacent to v in H. Let N H ( v ) be the set of vertices that are adjacent to v in H. N H ( v ) can also be understood as the set of the neighbors of v in H. Let N k H ( v ) be the set of k-vertices that are adjacent to v in H. In general, d H ( v ) is the number of vertices in N H ( v ) , and d k H ( v ) is the number of k-vertices in N k H ( v ) . In a similar way, the expressions N k + H ( v ) , N k H ( v ) , d k + H ( v ) and d k H ( v ) can be defined. To avoid confusing the context, the letter G is dropped from the expressions d G ( v ) , d k G ( v ) , d k + G ( v ) and d k G ( v ) when referring to the graph G.
Given a graph G, the number of i-vertices in V ( G ) is denoted by n i ( G ) for i [ 1 , Δ ( G ) ] . A graph H is said to be smaller than a graph G if ( | E ( H ) | , n t ( H ) , n t 1 ( H ) , , n 1 ( H ) ) precedes ( | E ( G ) | , n t ( G ) , n t 1 ( G ) , , n 1 ( G ) ) in terms of the standard lexicographic order, where t = m a x { Δ ( H ) , Δ ( G ) } . A graph is considered minimal in relation to a given property P if a smaller graph satisfying the property P does not exist.
Lemma 1
(Wang and Wang, [35]). Let G be a graph. If e is an edge in E ( G ) , then m a d ( G e ) m a d ( G ) . If v is an 1-vertex in V ( G ) , then m a d ( G v ) m a d ( G ) .

3. Proof of Theorem 1a

In this section, our proof proceeds by means of reductio ad absurdum. Let k = m a x { 9 , Δ + 1 } . Assume that G with m a d ( G ) < 4 is a minimal counterexample of Theorem 1a, to put it another way, there exists no NDE-k-coloring of G with m a d ( G ) < 4 , whereas there exists an NDE-k-coloring of any other graph H with m a d ( H ) < 4 and smaller than G. Clearly, G is a connected graph. Let C = [ 1 , k ] be the color set in the following discussion. Then | C | = k 9 .
Suppose that an NDE-k-coloring ϕ of G exists and an edge u v is in E ( G ) . If d G ( u ) is not equal to d G ( v ) , then C ϕ ( u ) is not equal to C ϕ ( v ) . If the adjacent vertices u and v meet the requirement that C ϕ ( u ) is equal to C ϕ ( v ) , then they are regarded as conflict under the coloring ϕ . The edge u v is considered legally colored if its color is not the same as that of its adjacent edges in E ( G ) and neither u nor v conflicts with their neighbors.

3.1. Structural Properties in G

A series of auxiliary claims are used to give the structural properties of the minimal counterexample G with m a d ( G ) < 4 in this subsection.
Claim 1.
There exists no edge u v E ( G ) with d ( v ) 5 and d ( u ) = 1 .
Proof.
Suppose that the statement is false. There is then an edge u v with d ( v ) 5 and d ( u ) = 1 in G. The graph H = G u v satisfies the inequality E ( H ) < E ( G ) . It follows that m a d ( H ) m a d ( G ) < 4 according to Lemma 1. With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists using the color set C. Because d ( v ) 5 , we use a color in C C ϕ ( v ) to color u v for which v and its neighbors do not conflict. □
Claim 2.
Let v be a p-vertex in V ( G ) .
(1)
If p = 2 , then d 2 ( v ) = 0 .
(2)
If p [ 3 , 4 ] , then d p ( v ) 1 .
(3)
If p [ 2 , 4 ] , then d i ( v ) = 0 for i [ 1 , p 1 ] .
Proof.
(1) Suppose that the statement is false. Then d 2 ( v ) 1 . Let N ( v ) = { u , v 1 } and N ( u ) = { v , u 1 } , where d ( u ) = 2 . Below, we break it down into two cases.
Case 1.  v 1 = u 1 = x . Then the graph H = G u v satisfies the inequality E ( H ) < E ( G ) . We have m a d ( H ) m a d ( G ) < 4 according to Lemma 1. With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C. Since ϕ ( u x ) ϕ ( v x ) , we use a color in C { ϕ ( u x ) , ϕ ( v x ) } to color u v properly. It follows that an NDE-k-coloring of G is obtained. This is a contradiction.
Case 2.  v 1 u 1 . We delete the vertices u and v from G and add one 2-vertex x in G { u , v } for which x is adjacent to v 1 and u 1 . This produces a smaller graph H satisfying the inequalities E ( H ) < E ( G ) and n 2 ( H ) < n 2 ( G ) .
Now we will show that m a d ( H ) < 4 . Let H * be the surgraph of H. If x V ( H * ) , then H * is the surgraph of G and a d ( H * ) m a d ( G ) < 4 . Otherwise, x V ( H * ) . Then H * * = H * x is the surgraph of G and a d ( H * * ) m a d ( G ) < 4 . As 2 | E ( H * * ) | | V ( H * * ) | < 4 , we have a d ( H * ) = 2 | E ( H * ) | | V ( H * ) | 2 ( | E ( H * * ) | + 2 ) | V ( H * * ) | + 1 < 4 | V ( H * * ) | + 4 | V ( H * * ) | + 1 = 4 . It follows from the arbitrariness of H * that m a d ( H ) < 4 .
With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C. Observe that ϕ ( v 1 x ) ϕ ( u 1 x ) . We color u u 1 with ϕ ( u 1 x ) and v v 1 with ϕ ( v 1 x ) . Then we use a color in C { ϕ ( u 1 x ) , ϕ ( v 1 x ) } to color u v for which u and v do not conflict with their neighbors.
(2) Suppose that the statement is false. Then d p ( v ) 2 . Let N ( v ) = { v 1 , v 2 , , v p } , N ( v 1 ) = { v , u 1 , u 2 , , u p 1 } and N ( v 2 ) = { v , w 1 , w 2 , , w p 1 } , where d ( v 1 ) = d ( v 2 ) = p . The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 . With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C. Again, let ϕ ( v v i ) = i 1 for i [ 2 , p ] and ϕ ( v 1 u j ) = a j for j [ 1 , p 1 ] . Based on the values of p, we split the proof into these two cases.
Case 1.  p = 3 . If { a 1 , a 2 } { 1 , 2 } , then we use a color in C { C ϕ ( v ) C ϕ ( v 1 ) } to color v v 1 for which v and v 1 do not conflict with their neighbors. Otherwise, { a 1 , a 2 } = { 1 , 2 } . Then we use a color in C ( C ϕ ( v 2 ) { 2 } ) to recolor v v 2 for which v 2 and its neighbors other than v do not conflict. This brings us back to the above discussion.
Case 2.  p = 4 . Let t = | { a 1 , a 2 , a 3 } { 1 , 2 , 3 } | . We now consider these four subcases on the basis of the value of t.
Subcase 2.1.  t = 0 . Let a i = i + 3 for i [ 1 , 3 ] . We suppose that v v 1 are not legally colored, and need to analyse these two possibilities by symmetry.
  • C ϕ ( v i ) = { 1 3 , i + 5 } for i [ 2 , 4 ] . When { 4 , 5 , 6 , 7 } { C ϕ ( u 1 ) , C ϕ ( u 2 ) , C ϕ ( u 3 ) } , then we color v v 1 with 7 and recolor v v 3 with a color in { 4 , 5 , 6 , 9 } for which v 3 and its neighbors do not conflict. When { 4 , 5 , 6 , α } { C ϕ ( u 1 ) , C ϕ ( u 2 ) , C ϕ ( u 3 ) } , where α [ 8 , 9 ] , then we use α to color v v 1 and a color in { 4 , 5 , 6 , [ 8 , 9 ] { α } } to recolor v v 2 for which v 2 and its neighbors do not conflict. Suppose that { C ϕ ( u 1 ) , C ϕ ( u 2 ) , C ϕ ( u 3 ) } = { { 4 , 5 , 6 , 7 } , { 4 , 5 , 6 , 8 } , { 4 , 5 , 6 , 9 } } . Let C ϕ ( u i ) = { 4 , 5 , 6 , i + 6 } for i [ 1 , 3 ] . We use 9 to color v v 1 and a color in { 4 , 5 , 6 , 8 } to recolor v v 2 for which v 2 and its neighbors do not conflict, and recolor v 1 u 1 with a color in { 1 , 2 , 3 , 8 } for which u 1 and its neighbors do not conflict.
  • C ϕ ( v i ) = { 1 3 , i + 5 } for i [ 2 , 3 ] and C ϕ ( u 1 ) = { 4 , 5 , 6 , 9 } . When { C ϕ ( u 2 ) , C ϕ ( u 3 ) } = { { 4 , 5 , 6 , 7 } , { 4 , 5 , 6 , 8 } } , the proof is reduced to the above possibility. When { C ϕ ( u 2 ) , C ϕ ( u 3 ) } { { 4 , 5 , 6 , 7 } , { 4 , 5 , 6 , 8 } } , we can see that there exists an element A { { 4 , 5 , 6 , 7 } , { 4 , 5 , 6 , 8 } } that satisfies A { C ϕ ( u 2 ) , C ϕ ( u 3 ) } . Suppose that { 4 , 5 , 6 , 7 } { C ϕ ( u 2 ) , C ϕ ( u 3 ) } . Then we color v v 1 with 7 and recolor v v 3 with a color β { 4 , 5 , 6 , 9 } for which v 3 and its neighbors do not conflict. If C ϕ ( v 4 ) { 1 , 3 , 7 , β } , then we are done. Otherwise, C ϕ ( v 4 ) = { 1 , 3 , 7 , β } . We use a color in { 4 , 5 , 6 , 8 , 9 } { β } to recolor v v 2 for which v 2 and its neighbors do not conflict.
Subcase 2.2.  t = 1 . Let a 1 [ 1 , 3 ] , a 2 = 4 and a 3 = 5 . We suppose that v v 1 are not legally colored, and need to analyse these two possibilities by symmetry.
  • C ϕ ( v i ) = { 1 3 , i + 4 } for i [ 2 , 4 ] and C ϕ ( u 1 ) = { a 1 , 4 , 5 , 9 } . Then v v 1 is colored with a color α [ 6 , 8 ] for which v 1 and its neighbors other than v do not conflict. If α = 6 , then v v 3 is recolored with a color in { 4 , 5 , 8 , 9 } for which v 3 and its neighbors do not conflict. If α [ 7 , 8 ] , then we use a color in { 4 , 5 , [ 7 , 8 ] { α } , 9 } to recolor v v 2 for which v 2 and its neighbors do not conflict.
  • C ϕ ( v i ) = { 1 3 , i + 4 } for i [ 2 , 3 ] and C ϕ ( u j ) = { a 1 , 4 , 5 , j + 7 } for j [ 1 , 2 ] . When C ϕ ( u 3 ) { { a 1 , 4 , 5 , 6 } , { a 1 , 4 , 5 , 7 } } , the proof goes back to the above possibility. Suppose that C ϕ ( u 3 ) { { a 1 , 4 , 5 , 6 } , { a 1 , 4 , 5 , 7 } } . We then color v v 1 with 6 and recolor v v 3 with a color β { 4 , 5 , 8 , 9 } for which v 3 and its neighbors do not conflict. If C ϕ ( v 4 ) { 1 , 3 , 6 , β } , then we are finished. Otherwise, C ϕ ( v 4 ) = { 1 , 3 , 6 , β } . We color v v 1 with 7, recolor v v 3 with 2, and v v 2 with a color in { 4 , 5 , 8 , 9 } for which v 2 and its neighbors do not conflict.
Subcase 2.3.  t = 2 . Let a 1 , a 2 [ 1 , 3 ] and a 3 = 4 . Suppose that v v 1 are not legally colored. Then we can obtain that C ϕ ( v i ) = { 1 3 , i + 3 } for i [ 2 , 4 ] and C ϕ ( u j ) = { a 1 , a 2 , 4 , j + 7 } for j [ 1 , 2 ] . We use a color in { 5 , 6 , 7 , 9 } to recolor v 1 u 1 for which u 1 and its neighbors other than v 1 do not conflict. The proof goes back to Subcase 2.2.
Subcase 2.4.  t = 3 . Let a 1 = 1 , a 2 = 2 and a 3 = 3 . If we may use a color in [ 4 , 9 ] to recolor v v 2 legally, then the proof returns to Subcase 2.3. Otherwise, our assumption is that v v 2 can not be recolored, which implies that ϕ ( v 2 w i ) = i + 3 and C ϕ ( w i ) = { 4 , 5 , 6 , i + 6 } for i [ 1 , 3 ] . Then v 2 w 1 is recolored with a color γ { 2 , 3 , 8 , 9 } for which w 1 and its neighbors other than v 2 do not conflict. We use a color in { 8 , 9 } { γ } to recolor v v 2 . At this time, the proof returns to Subcase 2.3.
(3) Suppose that the statement is false. Then d i ( v ) 1 . Let d ( u ) = i and u v E ( G ) . Then H = G u v satisfies the inequality E ( H ) < E ( G ) . It follows that m a d ( H ) m a d ( G ) < 4 according to Lemma 1. With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C. Since i < p 4 , by Claims 2(1) and 2(2), we have d p ( v ) 1 and d i ( u ) 1 , meaning that there is at most one conflict vertex for each of u and v. We use a color in C ( C ϕ ( u ) C ϕ ( v ) ) to color u v for which v and u do not conflict with their neighbors. This gives rise to an NDE-k-coloring of G, drawing a contradiction. □
Claim 3.
Let v be a vertex in V ( G ) .
(1)
If d 1 ( v ) p [ 1 , 2 ] , then d s ( v ) = 0 for s [ 2 , p + 1 ] .
(2)
d 2 ( v ) 1 .
(3)
If d ( v ) 6 , then d 1 ( v ) d ( v ) 4 .
Proof.
(1) Suppose that the statement is false. Then d s ( v ) 1 for s [ 2 , p + 1 ] [ 2 , 3 ] . Let v 1 , v 2 , , v p + 1 be the vertices adjacent to v in which case d ( v i ) = 1 for i [ 1 , p ] and d ( v p + 1 ) = s . Let N ( v p + 1 ) = { u 1 , u 2 , , u s 1 , v } . By Claims 2(1) and 2(2), we have d s ( v p + 1 ) 1 . Now there are these two cases, based on whether d s ( v p + 1 ) = 0 or 1.
Case 1.  d s ( v p + 1 ) = 0 . We delete the edge v v p + 1 from G and add one 1-vertex x 1 in G v v p + 1 such that x 1 is adjacent to v. This produces a smaller graph H satisfying the expressions E ( H ) = E ( G ) , n s ( H ) < n s ( G ) , n s 1 ( H ) > n s 1 ( G ) and n 1 ( H ) > n 1 ( G ) .
Now we will show that m a d ( H ) < 4 . Let H * be the surgraph of H. If x 1 V ( H * ) , then H * is the surgraph of G and a d ( H * ) m a d ( G ) < 4 . Otherwise, x 1 V ( H * ) . Then H * * = H * x 1 is the surgraph of G and a d ( H * * ) m a d ( G ) < 4 . As 2 | E ( H * * ) | | V ( H * * ) | < 4 , we have a d ( H * ) = 2 | E ( H * ) | | V ( H * ) | = 2 ( | E ( H * * ) | + 1 ) | V ( H * * ) | + 1 < 4 | V ( H * * ) | + 2 | V ( H * * ) | + 1 < 4 | V ( H * * ) | + 4 | V ( H * * ) | + 1 = 4 . It follows from the arbitrariness of H * that m a d ( H ) < 4 .
With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C in which case ϕ ( v v i ) = i for i [ 1 , p ] . Observe that ϕ ( v x 1 ) [ 1 , p ] . In order to restore the original graph G, the vertices v p + 1 and x 1 are stuck together. If ϕ ( v x 1 ) C ϕ ( v p + 1 ) , then we can easily see that ϕ is an NDE-k-coloring of G. Otherwise, the colors of v x 1 and v v j are exchanged for j [ 1 , p ] ( C ϕ ( v p + 1 ) { ϕ ( v x 1 ) } ) . Whichever possibility arises, an NDE-k-coloring of G is always obtained. This is a contradiction.
Case 2.  d s ( v p + 1 ) = 1 . Let d ( u s 1 ) = d ( v p + 1 ) = s . The graph H = G v p + 1 u s 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 . With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C in which case ϕ ( v v i ) = i for i [ 1 , p ] . Observe that ϕ ( v v p + 1 ) [ 1 , p ] . If C ϕ ( u s 1 ) C ϕ ( v p + 1 ) , then we color v p + 1 u s 1 properly. Otherwise, the colors of v v t and v v p + 1 are exchanged for t [ 1 , p ] ( C ϕ ( v p + 1 ) { ϕ ( v v p + 1 ) } ) and v p + 1 u s 1 is properly colored. Thus, an NDE-k-coloring of G is always obtained. This is a contradiction.
(2) Suppose that the statement is false. Then d 2 ( v ) 2 . Let v 1 , v 2 be the vertices adjacent to v satisfying the equality d ( v 1 ) = d ( v 2 ) = 2 . By Claim 2(1), v 1 is not adjacent to v 2 in G. These two cases are then considered.
Case 1. Assume that v, v 1 and v 2 are in a 4-cycle [ v v 1 u v 2 ] , that is, N ( v 1 ) N ( v 2 ) = { v , u } . We delete the edges v v 1 and v v 2 from G and add two 1-vertices x 1 and x 2 in G { v v 1 , v v 2 } for which x 1 and x 2 are adjacent to v, respectively. This produces a smaller graph H satisfying the expressions E ( H ) = E ( G ) , n 2 ( H ) < n 2 ( G ) and n 1 ( H ) > n 1 ( G ) .
Now we will show that m a d ( H ) < 4 . Let H * be the surgraph of H. Then these four possibilities need to be handled.
  • x 1 , x 2 V ( H * ) . Then H * is the surgraph of G and a d ( H * ) m a d ( G ) < 4 .
  • x 1 V ( H * ) and x 2 V ( H * ) . Then H * * = H * x 1 is the surgraph of G and a d ( H * * ) m a d ( G ) < 4 . As 2 | E ( H * * ) | | V ( H * * ) | < 4 , we have a d ( H * ) = 2 | E ( H * ) | | V ( H * ) | = 2 ( | E ( H * * ) | + 1 ) | V ( H * * ) | + 1 < 4 | V ( H * * ) | + 2 | V ( H * * ) | + 1 < 4 | V ( H * * ) | + 4 | V ( H * * ) | + 1 = 4 .
  • x 1 V ( H * ) and x 2 V ( H * ) . Then H * * = H * x 2 is the surgraph of G and a d ( H * * ) m a d ( G ) < 4 . As 2 | E ( H * * ) | | V ( H * * ) | < 4 , we have a d ( H * ) = 2 | E ( H * ) | | V ( H * ) | = 2 ( | E ( H * * ) | + 1 ) | V ( H * * ) | + 1 < 4 | V ( H * * ) | + 2 | V ( H * * ) | + 1 < 4 | V ( H * * ) | + 4 | V ( H * * ) | + 1 = 4 .
  • x 1 , x 2 V ( H * ) . Then H * * = H * { x 1 , x 2 } is the surgraph of G and a d ( H * * ) m a d ( G ) < 4 . As 2 | E ( H * * ) | | V ( H * * ) | < 4 , we have a d ( H * ) = 2 | E ( H * ) | | V ( H * ) | = 2 ( | E ( H * * ) | + 2 ) | V ( H * * ) | + 2 < 4 | V ( H * * ) | + 4 | V ( H * * ) | + 2 < 4 | V ( H * * ) | + 8 | V ( H * * ) | + 2 = 4 .
According to the analysis of the above four possibilities, we can always get a d ( H * ) < 4 . It follows from the arbitrariness of H * that m a d ( H ) < 4 .
With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C. The vertices x i and v i are stuck together for i [ 1 , 2 ] , and the colors of v x 1 and v x 2 are exchanged if necessary. It follows that an NDE-k-coloring of G is given. This is a contradiction.
Case 2. Assume that v, v 1 and v 2 are in a 5 + -cycle. Let N ( v 1 ) = { v , u 1 } and N ( v 1 ) = { v , u 2 } , where u 1 u 2 . We delete the edges v 1 u 1 and v 2 u 2 from G and add one 2-vertex x in G { v 1 u 1 , v 2 u 2 } for which x is adjacent to u 1 and u 2 . This produces a smaller graph H satisfying the expressions E ( H ) = E ( G ) , n 2 ( H ) < n 2 ( G ) and n 1 ( H ) > n 1 ( G ) .
Now we will show that m a d ( H ) < 4 . Let H * be the surgraph of H. If x V ( H * ) , then H * is the surgraph of G and a d ( H * ) m a d ( G ) < 4 . If x V ( H * ) , then H * * = H * x is the surgraph of G and a d ( H * * ) m a d ( G ) < 4 . As 2 | E ( H * * ) | | V ( H * * ) | < 4 , We may derive that a d ( H * ) = 2 | E ( H * ) | | V ( H * ) | 2 ( | E ( H * * ) | + 2 ) | V ( H * * ) | + 1 < 4 | V ( H * * ) | + 4 | V ( H * * ) | + 1 = 4 . It follows from the arbitrariness of H * that m a d ( H ) < 4 .
With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C. First, we split x into x 1 and x 2 in which case x i is adjacent to u i for i [ 1 , 2 ] . Second, the vertices x j and v j are stuck together for j [ 1 , 2 ] , and the colors of v v 1 and v v 2 are exchanged if necessary. Thus, an NDE-k-coloring of G is given. This is a contradiction.
(3) Let d ( v ) = r 6 and N ( v ) = { v 1 , v 2 , , v r } with d ( v 1 ) d ( v 2 ) d ( v r ) . Suppose that the statement is false. Then d 1 ( v ) d ( v ) 3 = r 3 3 . Observe that d ( v i ) = 1 for i [ 1 , 3 ] and d r ( v ) d ( v ) d 1 ( v ) 3 . The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 .
With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C in which case ϕ ( v v p ) = p 1 for p [ 2 , r ] . In the first place, we use any color in { r , r + 1 } to color v v 1 . In the second place, for each i [ 2 , 3 ] , we use a color b i { r , r + 1 } to recolor v v i and { r , r + 1 } { b i } to color v v 1 . As a consequence, at least 2 + ( 3 1 ) = 4 different ways to recolor or color some edges incident with v exists. On account of at most three vertices conflict with v, it follows that ϕ is extended to an NDE-k-coloring of the entire graph G. This is a contradiction. □
Claim 4.
If v is a 5-vertex of G, then d 3 ( v ) 1 .
Proof.
Let N ( v ) = { v 1 , v 2 , , v 5 } with d ( v 1 ) d ( v 2 ) d ( v 5 ) . Suppose that the statement is false. Then d 3 ( v ) 2 . Let d ( v 1 ) = s 3 , d ( v 2 ) 3 and N ( v 1 ) = { v , u 1 , , u s 1 } . The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 . With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C in which case ϕ ( v v p ) = p 1 for p [ 2 , 5 ] . On the basis of Claim 1, we have s 2 . Now these two cases are considered, depending on whether s = 2 or 3.
Case 1.  s = 2 . By Claim 2(1), we have d ( u 1 ) 2 . We assume, without loss of generality, that ϕ ( v 1 u 1 ) [ 1 , 5 ] . Then v v 1 is colored with a color in [ 6 , 9 ] for which there is no conflict between v and v 3 , v 4 , v 5 . Therefore, an NDE-k-coloring of G is given, producing a contradiction.
Case 2.  s = 3 . By Claim 2(2), it follows that d 3 ( v 1 ) 1 . When d 3 ( v 1 ) = 0 , we set ϕ ( v 1 u 1 ) = a 1 and ϕ ( v 1 u 2 ) = a 2 . When d 3 ( v 1 ) = 1 , i.e., d ( u 1 ) = 3 , we use a color in [ 1 , 4 ] ( { C ϕ ( u 1 ) { ϕ ( v 1 u 1 ) } } { ϕ ( v 1 u 2 ) } ) to recolor v 1 u 1 , and set a 1 C ϕ ( u 1 ) C ϕ ( v 1 ) and ϕ ( v 1 u 2 ) = a 2 . We assume, without loss of generality, that a 1 , a 2 [ 1 , 6 ] . In the first place, we use any color in [ 7 , 9 ] to color v v 1 . In the second place, we use a color γ [ 5 , 9 ] { C ϕ ( v 2 ) } to recolor v v 2 for which v 2 and its neighbors do not conflict, and a color in [ 7 , 9 ] { γ } to color v v 1 . As a consequence, at least 3 + 2 = 5 different ways to recolor or color some edges incident with v exists. On account of at most three vertices conflict with v, it follows that ϕ is extended to an NDE-k-coloring of the entire graph G. This is a contradiction. □
Claim 5.
If v is a 6-vertex of G, then d 3 ( v ) 2 .
Proof.
Let N ( v ) = { v 1 , v 2 , , v 6 } with d ( v 1 ) d ( v 2 ) d ( v 6 ) . Suppose that the statement is false. Then d 3 ( v ) 3 . Let d ( v 1 ) = s 3 , and d ( v i ) 3 for i [ 2 , 3 ] , and N ( v 1 ) = { v , u 1 , , u s 1 } . The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 . With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C in which case ϕ ( v v p ) = p 1 for p [ 2 , 6 ] . We now consider these three cases, in accordance with the values of s.
Case 1.  s = 1 . Then d ( v 1 ) = 1 . We use a color in [ 6 , 9 ] to color the edge v v 1 in which case there is no conflict between v and v 4 , v 5 , v 6 . Hence, an NDE-k-coloring of G is given, producing a contradiction.
Case 2.  s = 2 . By Claim 2(1), we have d ( u 1 ) 2 . We assume, without loss of generality, that ϕ ( v 1 u 1 ) [ 1 , 6 ] . In the first place, we use any color in [ 7 , 9 ] to color v v 1 . In the second place, we use a color α [ 6 , 9 ] { C ϕ ( v 2 ) } to recolor v v 2 for which v 2 and its neighbors do not conflict, and a color in [ 7 , 9 ] { α } to color v v 1 . As a consequence, at least 3 + 2 = 5 different ways to recolor or color some edges incident with v exists. On account of at most three vertices conflict with v, it follows that ϕ is extended to an NDE-k-coloring of the entire graph G. This is a contradiction.
Case 3.  s = 3 . The inequality d 3 ( v 1 ) 1 is guaranteed by Claim 2(2). When d 3 ( v 1 ) = 0 , we set ϕ ( v 1 u 1 ) = a 1 and ϕ ( v 1 u 2 ) = a 2 . When d 3 ( v 1 ) = 1 , i.e., d ( u 1 ) = 3 , we use a color in [ 1 , 5 ] ( C ϕ ( u 1 ) { ϕ ( v 1 u 2 ) } ) to recolor v 1 u 1 , and set a 1 C ϕ ( u 1 ) C ϕ ( v 1 ) and ϕ ( v 1 u 2 ) = a 2 . We assume, without loss of generality, that a 1 , a 2 [ 1 , 7 ] . In the first place, we use any color in [ 8 , 9 ] to color v v 1 . In the second place, for each i [ 2 , 3 ] , we use a color β i [ 6 , 9 ] { C ϕ ( v i ) } to recolor v v i for which v i and its neighbors do not conflict, and a color in [ 8 , 9 ] { β i } to color v v 1 . As a consequence, at least 2 + 1 + 1 = 4 different ways to recolor or color some edges incident with v exists. On account of at most three vertices conflict with v, it follows that ϕ is extended to an NDE-k-coloring of the entire graph G. This is a contradiction. □
Claim 6.
Let v be a 7-vertex of G.
(1)
If d 2 ( v ) 1 , then d 3 ( v ) 3 .
(2)
If d 2 ( v ) = 0 , then d 3 ( v ) 6 .
Proof.
Let N ( v ) = { v 1 , v 2 , , v 7 } with d ( v 1 ) d ( v 2 ) d ( v 7 ) .
(1) Suppose that the statement is false. Then d 3 ( v ) 4 if d 2 ( v ) 1 . Let d ( v 1 ) = s 2 and d ( v i ) 3 for i [ 2 , 4 ] . The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 . With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C in which case ϕ ( v v p ) = p 1 for p [ 2 , 7 ] .
For each i [ 2 , 4 ] , by Claims 1 and 2, we have d 3 ( v i ) 1 . When d ( v i ) 2 or d ( v i ) = 3 with d 3 ( v i ) = 0 , we use a color α i [ 7 , 9 ] { C ϕ ( v i ) } to recolor v v i for which v i and its neighbors do not conflict. When d ( v i ) = 3 with d 3 ( v i ) = 1 , we assume that w i , 1 N ( v i ) and d ( w i , 1 ) = 3 . Then we use a color in [ 1 , 6 ] { C ϕ ( w i , 1 ) C ϕ ( v i ) } to recolor v i w i , 1 and a color α i [ 7 , 9 ] { C ϕ ( v i ) { ϕ ( v i w i , 1 ) } } to recolor v v i for which v i and its neighbors do not conflict. So we can always find a color α i satisfying that v i and its neighbors do not conflict. We now consider these two cases, according to whether s = 1 or 2.
Case 1.  s = 1 . Then d ( v 1 ) = 1 . In the first place, we use any color in [ 7 , 9 ] to color v v 1 . In the second place, as stated above, we use a color α 2 to recolor v v 2 for which v 2 and its neighbors do not conflict, and a color in [ 7 , 9 ] { α 2 } to color v v 1 . As a consequence, at least 3 + 2 = 5 different ways to recolor or color some edges incident with v exists. On account of at most three vertices conflict with v, it follows that ϕ is extended to an NDE-k-coloring of the entire graph G. This is a contradiction.
Case 2.  s = 2 . By Claim 2(1), we have d 2 ( v 1 ) = 0 . Let u 1 N ( v 1 ) and u 1 v . We assume, without loss of generality, that ϕ ( v 1 u 1 ) [ 1 , 7 ] . In the first place, we use any color in [ 8 , 9 ] to color v v 1 . In the second place, for each i [ 2 , 4 ] , as stated above, we use a color α i to recolor v v i for which v i and its neighbors do not conflict, and a color in [ 8 , 9 ] { α i } to color v v 1 . As a consequence, at least 2 + 1 + 1 + 1 = 5 different ways to recolor or color some edges incident with v exists. Because there are at most three vertices conflict with v, it follows that ϕ is extended to an NDE-k-coloring of the entire graph G. This is a contradiction.
(2) Suppose that the statement is false. Then d 3 ( v ) = 7 if d 2 ( v ) = 0 . Let d ( v i ) = 3 for i [ 1 , 7 ] and N ( v 1 ) = { v , u 1 , u 2 } . The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 . With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C in which case ϕ ( v v p ) = p 1 for p [ 2 , 7 ] . The inequality d 3 ( v 1 ) 1 is guaranteed by Claim 2(2). When d 3 ( v 1 ) = 0 , we set ϕ ( v 1 u 1 ) = a 1 and ϕ ( v 1 u 2 ) = a 2 . When d 3 ( v 1 ) = 1 , i.e., d ( u 1 ) = 3 , we use a color in [ 1 , 6 ] ( C ϕ ( u 1 ) { ϕ ( v 1 u 2 ) } ) to recolor v 1 u 1 , and set a 1 C ϕ ( u 1 ) C ϕ ( v 1 ) and ϕ ( v 1 u 2 ) = a 2 . We assume, without loss of generality, that a 1 , a 2 [ 1 , 8 ] . Then we color v v 1 with 9. So an NDE-k-coloring of G is obtained. This is a contradiction. □
Claim 7.
Let v be a 8-vertex of G. If d 2 ( v ) 1 , then d 3 ( v ) 7 .
Proof.
Let N ( v ) = { v 1 , v 2 , , v 8 } with d ( v 1 ) d ( v 2 ) d ( v 8 ) . Suppose that the statement is false. Then d 3 ( v ) = 8 if d 2 ( v ) 1 . Let d ( v 1 ) = s 2 and d ( v i ) 3 for i [ 2 , 8 ] . The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 . With the aid of the minimality of G, an NDE-k-coloring ϕ of H exists through the use of the color set C in which case ϕ ( v v p ) = p 1 for p [ 2 , 8 ] . When s = 1 , we color v v 1 with 8. When s = 2 , by Claim 2(1), we have d 2 ( v 1 ) = 0 . Let u 1 N ( v 1 ) and u 1 v . We assume without loss of generality, that ϕ ( v 1 u 1 ) [ 1 , 8 ] . Then we color v v 1 with 9. So an NDE-k-coloring of G is obtained. This is a contradiction. □

3.2. Discharging Analysis in G

In this subsection, we employ the discharging method [36] for the completion of the proof. Firstly, an initial charge function w ( v ) = d ( v ) 4 is given for each vertex v in V ( G ) . Secondly, the discharging rules are designed and the charges on the vertices are redistributed accordingly. After the discharging process is complete, we have access to the new charge function w on the vertices. While the discharging is occurring, the sum of all charges remains constant, that is to say, v V ( G ) w ( v ) = v V ( G ) w ( v ) .
Lemma 2.
Assume that V 1 , V 2 is the vertex partition of G, where the members of the set V 1 are 3 + -vertices and the members of the set V 2 are 2 -vertices. If v V 1 w ( v ) 0 and v V 2 w ( v ) v V 2 ( w ( v ) + d V 1 ( v ) ) , where d V 1 ( v ) is the number of the vertices in V 1 that are adjacent to v, then m a d ( G ) 4 .
Proof.
From the definition of the vertex partition of G, we have V 1 V 2 = V ( G ) and V 1 V 2 = . Likewise, d V 2 ( v ) is used to denote the number of the vertices in V 2 that are adjacent to v. It follows that d ( v ) = d V 1 ( v ) + d V 2 ( v ) and v V 2 d V 1 ( v ) = | E ( V 1 , V 2 ) | = v V 1 d V 2 ( v ) , where the members of the set E ( V 1 , V 2 ) are the edges joining a vertex of V 1 and a vertex of V 2 in G. Since
v V 1 ( d V 1 ( v ) 4 ) v V 1 ( d V 1 ( v ) 4 ) + v V 2 ( w ( v ) + d V 1 ( v ) ) v V 2 w ( v ) = v V 1 ( d V 1 ( v ) 4 ) + v V 2 ( d ( v ) 4 + d V 1 ( v ) ) v V 2 w ( v ) = v V 1 ( d V 1 ( v ) 4 ) + v V 2 d V 1 ( v ) + v V 2 ( d ( v ) 4 ) v V 2 w ( v ) = v V 1 ( d V 1 ( v ) 4 ) + | E ( V 1 , V 2 ) | + v V 2 ( d ( v ) 4 ) v V 2 w ( v ) = v V 1 ( d V 1 ( v ) 4 ) + v V 1 d V 2 ( v ) + v V 2 ( d ( v ) 4 ) v V 2 w ( v ) = v V 1 ( d V 1 ( v ) 4 + d V 2 ( v ) ) + v V 2 ( d ( v ) 4 ) v V 2 w ( v ) = v V 1 ( d ( v ) 4 ) + v V 2 ( d ( v ) 4 ) v V 2 w ( v ) = v V ( G ) w ( v ) v V 2 w ( v ) = v V ( G ) w ( v ) v V 2 w ( v ) = v V 1 w ( v ) 0 ,
we obtain 0 v V 1 ( d V 1 ( v ) 4 ) | V 1 | ( m a d ( G [ V 1 ] ) 4 ) , where the subgraph G [ V 1 ] of G is induced by V 1 . Therefore we can infer a conclusion that m a d ( G ) m a d ( G [ V 1 ] ) 4 . □
According to Lemma 2, we need to demonstrate that for a vertex v in V ( G ) , w ( v ) 1 4 + 1 = 2 if d ( v ) = 1 , w ( v ) 2 4 + 2 = 0 if d ( v ) = 2 , w ( v ) 0 if d ( v ) 3 , contradicting the condition that m a d ( G ) < 4 . Hence, it is proved that there is no such minimal counterexample. Assume that an edge u v satisfying the inequalities d ( u ) 3 and d ( v ) 5 is in E ( G ) . The following is our definition of the discharge rules.
(R1) If d ( u ) 2 , then v gives 1 to u.
(R2) If d ( u ) = 3 , then v gives 1 2 to u.
Suppose that a vertex v with d ( v ) = p is in V ( G ) . We will analyze the new charge w ( v ) of the vertex v based on the values of p.
Case 1.  p = 1 . The equality w ( v ) = 1 4 = 3 is established. It follows from Claim 1 that the neighbor of v is a 6 + -vertex and d V 1 ( v ) = 1 . According to the rule (R1), we have w ( v ) = 3 + 1 = 2 .
Case 2.  p = 2 . The equality w ( v ) = 2 4 = 2 is established. It follows from Claims 1 and 2 that the neighbors of v are two 5 + -vertices and d V 1 ( v ) = 2 . According to the rule (R1), we have w ( v ) = 2 + 1 × 2 = 0 .
Case 3.  p = 3 . The equality w ( v ) = 3 4 = 1 is established. It follows from Claims 1 and 2 that there are at least two 5 + -vertices in the neighbors of v. According to the rule (R2), we have w ( v ) 1 + 1 2 × 2 = 0 .
Case 4.  p = 4 . The equality w ( v ) = 4 4 = 0 is established. To comply with the rules (R1) and (R2), we have w ( v ) = 0 .
Case 5.  p = 5 . The equality w ( v ) = 5 4 = 1 is established. It follows from Claim 4 that d 3 ( v ) 1 . To comply with the rules (R1) and (R2), we have w ( v ) 1 1 = 0 .
Case 6.  p = 6 . The equality w ( v ) = 6 4 = 2 is established. It follows from Claim 5 that d 3 ( v ) 2 . To comply with the rules (R1) and (R2), we have w ( v ) 2 1 × 2 = 0 .
Case 7.  p = 7 . The equality w ( v ) = 7 4 = 3 is established. If d 2 ( v ) 1 , then it follows from Claim 6(1) that d 3 ( v ) 3 . To comply with the rules (R1) and (R2), we have w ( v ) 3 1 × 3 = 0 . Otherwise, d 2 ( v ) = 0 . By Claim 6(2), it follows that d 3 ( v ) 6 . Hence, by (R2), we have w ( v ) 3 1 2 × 6 = 0 .
Case 8.  p = 8 . The equality w ( v ) = 8 4 = 4 is established. It follows from Claim 3(3) that d 1 ( v ) 4 . If d 1 ( v ) 2 , then d 2 ( v ) = d 3 ( v ) = 0 , which is guaranteed by Claim 3(1). According to the rule (R1), we have w ( v ) = 4 1 × d 1 ( v ) 4 1 × 4 = 0 . If d 1 ( v ) = 1 , by Claims 3(1) and 7, then d 2 ( v ) = 0 and d 3 ( v ) 6 . To comply with the rules (R1) and (R2), we have w ( v ) 4 1 1 2 × 6 = 0 . If d 1 ( v ) = 0 and d 2 ( v ) 1 , by Claims 3(2) and 7, then d 2 ( v ) = 1 and d 3 ( v ) 6 . To comply with the rules (R1) and (R2), we have w ( v ) 4 1 1 2 × 6 = 0 . If d 1 ( v ) = d 2 ( v ) = 0 , applying (R2), then we have w ( v ) 4 1 2 × 8 = 0 .
Case 9.  p 9 . The equality w ( v ) = p 4 is established. It follows from Claim 3(3) that d 1 ( v ) p 4 . If d 1 ( v ) 2 , then d 2 ( v ) = d 3 ( v ) = 0 , which is guaranteed by Claim 3(1). According to the rule (R1), we have w ( v ) = p 4 1 × d 1 ( v ) p 4 1 × ( p 4 ) = 0 . If d 1 ( v ) = 1 , then d 2 ( v ) = 0 , which is guaranteed by Claim 3(1). To comply with the rules (R1) and (R2), we have w ( v ) p 4 1 1 2 × ( p 1 ) = 1 2 × ( p 9 ) 0 . If d 1 ( v ) = 0 , by Claim 3(2), then d 2 ( v ) 1 . To comply with the rules (R1) and (R2), we have w ( v ) p 4 1 1 2 × ( p 1 ) = 1 2 × ( p 9 ) 0 .
From the discussion so far, we obtain that for v V ( G ) , w ( v ) = 2 if d ( v ) = 1 , w ( v ) 0 if d ( v ) 2 . Therefore, we have completed the proof of Theorem 1a. □

4. Proof of Theorem 1b

Firstly, we consider the necessity of Theorem 1b. It is a well-known fact that Δ + 1 is a lower bound for χ a ( G ) if a pair of adjacent Δ -vertices is in G. From Theorem 1a, it follows that Δ + 1 is an upper bound for χ a ( G ) if m a d ( G ) < 4 and Δ 8 . As a consequence, the necessity of Theorem 1b is clearly true. Secondly, we consider the sufficiency of Theorem 1b. Every graph G possesses the property of Δ being a lower bound for χ a ( G ) . We only need to demonstrate the following theorem.
Theorem 3.
Let G be a graph with m a d ( G ) < 4 and Δ 9 . If there are no pairs of adjacent Δ -vertices in G, then χ a ( G ) = Δ .
Proof.
We use reductio ad absurdum to prove this theorem. Let k = m a x { 9 , Δ } . Assume that G with m a d ( G ) < 4 is a minimal counterexample of Theorem 3, in other words, no NDE-k-coloring of G with m a d ( G ) < 4 exists, while an NDE-k-coloring of any other graph H with m a d ( H ) < 4 and smaller than G exists. Apparently, G is a connected graph. Then the color set C = [ 1 , k ] satisfies the expression | C | = k 9 . As there is not any pair of adjacent Δ -vertices in G, we can easily obtain that no Δ -vertex adjacent to a 1-vertex exists. Similar to the discussion of Claims 1–7, we can deduce the following structural properties of G.
Claim 8.
Assume that v is a vertex in V ( G ) .
(1)
If d ( v ) = 1 , then d 5 ( v ) = 0 .
(2)
If d ( v ) = 2 , then d 2 ( v ) = 0 .
(3)
If d ( v ) [ 3 , 4 ] , then d k ( v ) 1 .
(4)
If d ( v ) [ 2 , 4 ] , then d i ( v ) = 0 for i [ 1 , k 1 ] .
(5)
If d 1 ( v ) p [ 1 , 2 ] , then d s ( v ) = 0 for s [ 2 , p + 1 ] .
(6)
d 2 ( v ) 1 .
(7)
If d ( v ) [ 6 , Δ 1 ] , then d 1 ( v ) d ( v ) 4 .
(8)
If d ( v ) = 5 , then d 3 ( v ) 1 .
(9)
If d ( v ) = 6 , then d 3 ( v ) 2 .
(10)
Let d ( v ) = 7 . If d 2 ( v ) 1 , then d 3 ( v ) 3 . If d 2 ( v ) = 0 , then d 3 ( v ) 6 .
(11)
Let d ( v ) = 8 . If d 2 ( v ) 1 , then d 3 ( v ) 7 .
(12)
If d ( v ) = Δ , then d 1 ( v ) = 0 .
For the derivation of a contradiction, at the very beginning, the same charge function w ( v ) = d ( v ) 4 is used for each vertex v in V ( G ) . Afterwards, the same discharging rules (R1) and (R2) are defined. By Lemma 2, it is sufficient to demonstrate that the new charge function w ( v ) complies with the requirements that v V 1 w ( v ) 0 and v V 2 w ( v ) v V 2 ( w ( v ) + d V 1 ( v ) ) . Actually, this can be justified in a similar way for v V ( G ) meeting the condition d ( v ) Δ 1 . Hence, our assumption is that d ( v ) = Δ 9 . By Claims 8(6) and 8(12), our inference is that d 1 ( v ) = 0 and d 2 ( v ) 1 . To comply with the rules (R1) and (R2), we calculate w ( v ) = d ( v ) 4 d 2 ( v ) 1 2 × d 3 ( v ) Δ 4 d 2 ( v ) 1 2 × ( Δ d 2 ( v ) ) = 1 2 × ( Δ d 2 ( v ) ) 4 1 2 × ( Δ 1 ) 4 = 1 2 × ( Δ 9 ) 0 . Therefore, this completes the proof of Theorem 1b. □

5. Proof of Theorem 2a

Let k = m a x { 8 , Δ + 2 } . Assume that G with m a d ( G ) < 4 is a minimal counterexample of Theorem 2a, to put it another way, there exists no NDT-k-coloring of G with m a d ( G ) < 4 , whereas an NDT-k-coloring of any other graph H with m a d ( H ) < 4 and smaller than G exists. The graph G is a connected graph, as can be observed. We then use the color set C = [ 1 , k ] with k 8 .
Suppose that an NDT-k-coloring φ of a graph G exists and an edge u v is in E ( G ) . If d G ( u ) is different from d G ( v ) , then C φ ( u ) is different from C φ ( v ) . If the adjacent vertices u and v meet the requirement C φ ( u ) is equal to C φ ( v ) , then they are regarded as conflict under the coloring φ . The edge u v is considered legally colored if its color is not the same as that of its adjacent elements in V ( G ) E ( G ) and neither u nor v conflicts with their neighbors.
Remark 1.
Assume that v is a r-vertex in G with the condition that N G ( v ) = { v 1 , v 2 , , v r } for r [ 1 , 3 ] . A partial NDT-k-coloring φ of G satisfies the conditions that φ ( v v i ) = i for i [ 1 , r ] and only v is not colored. Let | { φ ( v 1 ) , φ ( v 2 ) , , φ ( v r ) } [ 1 , r ] | = p , and φ ( v 1 ) , φ ( v 2 ) , , φ ( v p ) [ 1 , r ] . Since k r ( r p ) = ( k 2 r ) + p p + 2 , we use a color in [ r + 1 , k ] { φ ( v p + 1 ) , φ ( v p + 2 ) , , φ ( v r ) } to color the vertex v for which there is no conflict between v and v 1 , v 2 , , v p . As a result, φ is extended to an NDT-k-coloring of the entire graph G.
According to Remark 1, for the purpose that an NDT-k-coloring of G is given, the colors of 3 -vertices may firstly be removed, and secondly these 3 -vertices are recolored once the other edges and vertices have been legally colored. A partial NDT-k-coloring of G is described as nice if only the 3 -vertices in V ( G ) are not colored. It follows that any partial nice NDT-k-coloring of G can be extended to an NDT-k-coloring of G.

5.1. Structural Properties in G

Claim 9.
There exists no edge u v E ( G ) with d ( v ) 4 and d ( u ) 3 .
Proof.
Suppose that the statement is false. Then there is an edge u v with d ( v ) 4 and d ( u ) 3 in G. The graph H = G u v satisfies the inequality E ( H ) < E ( G ) . We have m a d ( H ) m a d ( G ) < 4 according to Lemma 1. With the aid of the minimality of G, an NDT-k-coloring φ of H exists through the use of the color set C. Now our proof is divided into these two cases in the light of whether d ( v ) 3 or d ( v ) = 4 .
Case 1.  d ( v ) 3 . The colors of the vertices u and v are removed firstly. Since the edge u v has at most four forbidden colors, we use a color in C ( ( C φ ( u ) C φ ( v ) ) { φ ( u ) , φ ( v ) } ) to color the edge u v properly. Afterwards, the vertices u and v are colored according to Remark 1.
Case 2.  d ( v ) = 4 . Let N ( u ) = { v , u 1 , u 2 , , u p 1 } and N ( v ) = { u , v 1 , v 2 , v 3 } , where p = d ( u ) 3 . The color of the vertex u is removed. Suppose that φ ( v ) = 1 , φ ( v v i ) = i + 1 for i [ 1 , 3 ] , and φ ( u u j ) = a j for j [ 1 , p 1 ] . Then d ( u j ) p for j [ 1 , p 1 ] based on the previous case. We assume, without loss of generality, that a j [ 1 , p + 3 ] [ 1 , 6 ] for j [ 1 , p 1 ] . If we can color u v legally, then a nice NDT-k-coloring of G is obtained. Otherwise, our assumptions are that C φ ( v 1 ) = { 1 4 , 7 } and C φ ( v 2 ) = { 1 4 , 8 } . Similar to the discussion in Remark 1, we use 7 to color the edge u v and a color in { 5 , 6 } { φ ( v 3 ) } to recolor the vertex v for which there is no conflict between v and v 3 . Finally, we color u according to Remark 1. □
Claim 10.
Let v be a vertex in V ( G ) .
(1)
If d 1 ( v ) p [ 1 , 2 ] , then d s ( v ) = 0 for s [ 2 , p + 1 ] .
(2)
d 2 ( v ) 1 .
(3)
If d ( v ) 6 , then d 1 ( v ) d ( v ) 4 .
Proof.
(1) Suppose that the statement is false. Then d s ( v ) 1 for s [ 2 , p + 1 ] [ 2 , 3 ] . Let v 1 , v 2 , , v p + 1 be the vertices adjacent to v in which case d ( v i ) = 1 for i [ 1 , p ] and d ( v p + 1 ) = s . Let N ( v p + 1 ) = { u 1 , u 2 , , u s 1 , v } . By Claim 9, we obtain d s ( v p + 1 ) = 0 . We delete the edge v v p + 1 from G and add one 1-vertex x 1 in G v v p + 1 such that x 1 is adjacent to v. This produces a smaller graph H satisfying the expressions E ( H ) = E ( G ) , n s ( H ) < n s ( G ) , n s 1 ( H ) > n s 1 ( G ) and n 1 ( H ) > n 1 ( G ) . As with the discussion of Case 1 in Claim 3(1), it is easy to show that m a d ( H ) < 4 .
With the aid of the minimality of G, an NDT-k-coloring φ of H exists through the use of the color set C in which case φ ( v v i ) = i for i [ 1 , p ] . We can easily observe that φ ( v x 1 ) [ 1 , p ] . The colors of v 1 , v 2 , , v p + 1 and x 1 are removed. In order to restore the original graph G, the vertices v p + 1 and x 1 are stuck together. If φ ( v x 1 ) C φ ( v p + 1 ) , then we can easily see that φ is a nice NDT-k-coloring of G. Otherwise, the colors of v x 1 and v v j are exchanged for j [ 1 , p ] ( C φ ( v p + 1 ) { φ ( v x 1 ) } ) . Whichever of these two possibilities happens, a nice NDT-k-coloring of G is obtained. This is a contradiction.
(2) Suppose that the statement is false. Then d 2 ( v ) 2 . Let v 1 , v 2 be the vertices adjacent to v meeting the equality d ( v 1 ) = d ( v 2 ) = 2 . By Claim 9, v 1 is not adjacent to v 2 in G. Now these two cases are considered.
Case 1. Assume that v, v 1 and v 2 are in a 4-cycle [ v v 1 u v 2 ] , that is, N ( v 1 ) N ( v 2 ) = { v , u } . We delete the edges v v 1 and v v 2 from G and add two 1-vertices x 1 and x 2 in G { v v 1 , v v 2 } such that x 1 and x 2 are adjacent to v, respectively. This produces a smaller graph H satisfying the expressions E ( H ) = E ( G ) , n 2 ( H ) < n 2 ( G ) and n 1 ( H ) > n 1 ( G ) . Analogous to Case 1 in Claim 3(2), it is easy for us to show that m a d ( H ) < 4 .
With the aid of the minimality of G, an NDT-k-coloring φ of H exists through the use of the color set C. Remove the colors of x 1 , x 2 , v 1 , v 2 . The vertices x i and v i are stuck together for i = 1 , 2 , and the colors of v x 1 and v x 2 are exchanged if necessary. Thus, a nice NDT-k-coloring of G is given, producing a contradiction.
Case 2. Assume that v, v 1 and v 2 are in a 5 + -cycle. Let N ( v 1 ) = { v , u 1 } and N ( v 1 ) = { v , u 2 } , where u 1 u 2 . We delete the edges v 1 u 1 and v 2 u 2 from G and add one 2-vertex x in G { v 1 u 1 , v 2 u 2 } for which x is adjacent to u 1 and u 2 . This produces a smaller graph H meeting the expressions E ( H ) = E ( G ) , n 2 ( H ) < n 2 ( G ) and n 1 ( H ) > n 1 ( G ) . Analogous to Case 2 in Claim 3(2), it is easy for us to show that m a d ( H ) < 4 .
With the aid of the minimality of G, an NDT-k-coloring φ of H exists through the use of the color set C. Remove the colors of x , v 1 , v 2 . In the first place, we split x into x 1 and x 2 such that x 1 is the neighbor of u 1 and x 2 is the neighbor of u 2 . In the second place, the vertices x i and v i are stuck together for i = 1 , 2 , and the colors of v v 1 and v v 2 are exchanged if necessary. Hence a nice NDT-k-coloring of G is given. This leads to a contradiction.
(3) Let d ( v ) = r 6 and N ( v ) = { v 1 , v 2 , , v r } with d ( v 1 ) d ( v 2 ) d ( v r ) . Suppose that the statement is false. Then d 1 ( v ) d ( v ) 3 = r 3 3 . Let d ( v i ) = 1 for i [ 1 , 3 ] and d r ( v ) d ( v ) d 1 ( v ) 3 . This means that there are at most three vertices conflict with v. The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 .
With the aid of the minimality of G, an NDT-k-coloring φ of H exists through the use of the color set C. In this case, φ ( v ) = 1 and φ ( v v p ) = p for p [ 2 , r ] . The colors of the vertices v 1 , v 2 , v 3 are removed. In the first place, we use any color in { r + 1 , r + 2 } to color v v 1 . In the second place, for each i [ 2 , 3 ] , we use a color c i { r + 1 , r + 2 } to recolor v v i , and { r + 1 , r + 2 } { c i } to color v v 1 . As a consequence, at least 2 + 1 + 1 = 4 different ways to recolor or color some edges incident with v exists. Because there are at most three vertices that conflict with v, it follows that φ is extended into a nice NDT-k-coloring of the entire graph G, which is a contradiction. □
Claim 11.
Let v be a 5-vertex of G.
(1)
If d 2 ( v ) 1 , then d 3 ( v ) 1 .
(2)
If d 2 ( v ) = 0 , then d 3 ( v ) 2 .
Proof.
Let N ( v ) = { v 1 , v 2 , , v 5 } with d ( v 1 ) d ( v 2 ) d ( v 5 ) .
(1) Suppose that the statement is false. Then d 3 ( v ) 2 if d 2 ( v ) 1 . Let d ( v 1 ) = s 2 , d ( v 2 ) = t 3 , N ( v 1 ) = { v , u 1 , , u s 1 } and N ( v 2 ) = { v , w 1 , , w t 1 } . Applying Claim 9, we obtain the expressions d ( u i ) s and d ( w j ) t for i [ 1 , s 1 ] and j [ 1 , t 1 ] . The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . According to Lemma 1, we have m a d ( H ) m a d ( G ) < 4 .
With the aid of the minimality of G, an NDT-k-coloring φ of H exists through the use of the color set C. In this case, φ ( v ) = 1 and φ ( v v p ) = p for p [ 2 , 5 ] . The colors of the vertices v 1 , v 2 are removed. We assume, without loss of generality, that φ ( v 1 u i ) [ 1 , 4 + s ] [ 1 , 6 ] for i [ 1 , s 1 ] . If we can color v v 1 legally, then a nice NDT-k-coloring of G is obtained. Otherwise, our assumptions are that C φ ( v 3 ) = { 1 5 , 7 } and C φ ( v 4 ) = { 1 5 , 8 } . After we use a color c [ 6 , 8 ] { φ ( v 2 w 1 ) , , φ ( v 2 w t 1 ) } to recolor the edge v v 2 , these two possibilities are considered according to whether c = 6 or c { 7 , 8 } .
  • c = 6 . Then we use a color in { 7 , 8 } to color the edge v v 1 in which case there is no conflict between v and v 5 . Thus, a nice NDT-k-coloring of G is given, producing a contradiction.
  • c { 7 , 8 } . If C φ ( v 5 ) { 1 , 3 5 , 7 , 8 } , then we color v v 1 with { 7 , 8 } { c } . If C φ ( v 5 ) = { 1 , 3 5 , 7 , 8 } , then we use 6 to recolor the vertex v and { 7 , 8 } { c } to color the edge v v 1 . It follows that a nice NDT-k-coloring of G is obtained, producing a contradiction.
(2) Suppose that the statement is false. Then d 3 ( v ) 3 if d 2 ( v ) = 0 . Let d ( v i ) = 3 for i [ 1 , 3 ] and N ( v 1 ) = { v , u 1 , u 2 } . It follows that d ( u i ) 3 for i [ 1 , 2 ] by Claim 9. The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . In accordance with Lemma 1, we have m a d ( H ) m a d ( G ) < 4 .
With the minimality of G, an NDT-k-coloring φ of H exists through the use of the color set C. In this case, φ ( v ) = 1 and φ ( v v p ) = p for p [ 2 , 5 ] . The colors of the vertices v 1 , v 2 , v 3 are removed. We assume, without loss of generality, that φ ( v 1 u i ) [ 1 , 7 ] for i [ 1 , 2 ] . If v v 1 can be legally colored, then a nice NDT-k-coloring of G is obtained. Otherwise, our assumption is that C φ ( v 4 ) = { 1 5 , 8 } . As with the discussion in Remark 1, we use 8 to color the edge v v 1 and a color in { 6 , 7 } { φ ( v 5 ) } to recolor the vertex v for which there is no conflict between v and v 5 . Hence a nice NDT-k-coloring of G is given, which is a contradiction. □
Claim 12.
Let v be a 6-vertex of G. If d 2 ( v ) 1 , then d 3 ( v ) 4 .
Proof.
Let N ( v ) = { v 1 , v 2 , , v 6 } with d ( v 1 ) d ( v 2 ) d ( v 6 ) . Suppose that the statement is false. Then d 3 ( v ) 5 if d 2 ( v ) 1 . Set d ( v 1 ) = s 2 , d ( v i ) 3 for i [ 2 , 5 ] and N ( v 1 ) = { v , u 1 , , u s 1 } . It follows that d ( u j ) s for j [ 1 , s 1 ] from Claim 9. The graph H = G v v 1 satisfies the inequality E ( H ) < E ( G ) . In accordance with Lemma 1, we have m a d ( H ) m a d ( G ) < 4 .
With the aid of the minimality of G, an NDT-k-coloring φ of H exists through the use of the color set C. In this case, φ ( v ) = 1 and φ ( v v p ) = p for p [ 2 , 6 ] . The colors of the vertices v 1 , v 2 , v 3 , v 4 , v 5 are removed. We assume, without loss of generality, that φ ( v 1 u j ) [ 1 , 5 + s ] [ 1 , 7 ] for j [ 1 , s 1 ] . If we can color v v 1 legally, then a nice NDT-k-coloring of G is obtained. Otherwise, we assume that C φ ( v 6 ) = { 1 6 , 8 } . Then we use 7 to recolor the vertex v and 8 to color the edge v v 1 . Hence a nice NDT-k-coloring of G is given, which is a contradiction. □

5.2. Discharging Analysis in G

In the first place, an initial charge function w ( v ) = d ( v ) 4 is defined for each vertex v in V ( G ) . In the second place, the discharging rules are designed and the charges on the vertices are redistributed accordingly. After the discharging process is finished, this produces the new charge function w on the vertices. While the discharging is in process, this equation v V ( G ) w ( v ) = v V ( G ) w ( v ) holds.
According to Lemma 2, we need to demonstrate that for a vertex v in V ( G ) , w ( v ) 1 4 + 1 = 2 if d ( v ) = 1 , w ( v ) 2 4 + 2 = 0 if d ( v ) = 2 , w ( v ) 0 if d ( v ) 3 , contradicting the condition that m a d ( G ) < 4 . Hence, it is proved that there is no such minimal counterexample. Assume that an edge u v satisfying the inequalities d ( u ) 3 and d ( v ) 5 is in E ( G ) . The following is our definition of the discharging rules.
(R1) If d ( u ) 2 , then v gives 1 to u.
(R2) If d ( u ) = 3 , then v gives 1 3 to u.
Suppose that v is a q-vertex in V ( G ) . Based on the value of q, we will calculate the new charge w ( v ) of the vertex v.
Case 1.  q = 1 . The equality w ( v ) = 1 4 = 3 is established. It follows from Claim 9 that the neighbor of v is a 5 + -vertex and d V 1 ( v ) = 1 . According to the rule (R1), we have w ( v ) = 3 + 1 = 2 .
Case 2.  q = 2 . The equality w ( v ) = 2 4 = 2 is established. It follows from Claim 9 that the neighbors of v are two 5 + -vertices and d V 1 ( v ) = 2 . According to the rule (R1), we have w ( v ) = 2 + 1 × 2 = 0 .
Case 3.  q = 3 . The equality w ( v ) = 3 4 = 1 is established. It follows from Claim 9 that the neighbors of v are three 5 + -vertices. According to the rule (R2), we have w ( v ) = 1 + 1 3 × 3 = 0 .
Case 4.  q = 4 . The equality w ( v ) = 4 4 = 0 is established. To comply with the rules (R1) and (R2), we have w ( v ) = 0 .
Case 5.  q = 5 . The equality w ( v ) = 5 4 = 1 is established. If d 2 ( v ) 1 , then it follows from Claim 11(1) that d 3 ( v ) 1 . According to the rule (R1), we have w ( v ) 1 1 = 0 . Otherwise, d 2 ( v ) = 0 . It follows from Claim 11(2) that d 3 ( v ) 2 . According to the rule (R2), we have w ( v ) 1 1 3 × 2 = 1 3 .
Case 6.  q = 6 . The equality w ( v ) = 6 4 = 2 is established. It follows from Claim 10(3) that d 1 ( v ) 6 4 = 2 . If d 1 ( v ) = 2 , then d 2 ( v ) = d 3 ( v ) = 0 , which is guaranteed by Claim 10(1). According to the rule (R1), we have w ( v ) 2 1 × 2 = 0 . If d 1 ( v ) = 1 , then d 2 ( v ) = 0 and d 3 ( v ) 4 , which are guaranteed by Claims 10(1) and 12. According to the rules (R1) and (R2), we have w ( v ) 2 1 1 3 × 3 = 0 . If d 1 ( v ) = 0 and d 2 ( v ) 1 , then d 2 ( v ) = 1 and d 3 ( v ) 4 , which are guaranteed by Claims 10(2) and 12. According to the rules (R1) and (R2), we have w ( v ) 2 1 1 3 × 3 = 0 . If d 1 ( v ) = d 2 ( v ) = 0 , applying the rule (R2), then w ( v ) 2 1 3 × 6 = 0 .
Case 7.  q 7 . The equality w ( v ) = q 4 is established. It follows from Claim 10(3) that d 1 ( v ) q 4 . If d 1 ( v ) 2 , then d 2 ( v ) = d 3 ( v ) = 0 , which is guaranteed by Claim 10(1). According to the rule (R1), we have w ( v ) = q 4 1 × d 1 ( v ) q 4 1 × ( q 4 ) = 0 . If d 1 ( v ) = 1 , then d 2 ( v ) = 0 , which is guaranteed by Claim 10(1). To comply with the rules (R1) and (R2), we have w ( v ) q 4 1 1 3 × ( q 1 ) = 2 3 × ( q 7 ) 0 . If d 1 ( v ) = 0 , then d 2 ( v ) 1 , which is guaranteed by Claim 10(2). To comply with the rules (R1) and (R2), we have w ( v ) q 4 1 1 3 × ( q 1 ) = 2 3 × ( q 7 ) 0 .
Thus, we obtain that for v V ( G ) , w ( v ) = 2 if d ( v ) = 1 , w ( v ) 0 if d ( v ) 2 . The proof of Theorem 2a is accomplished. □

6. Proof of Theorem 2b

To prove the necessary and sufficient conditions in Theorem 2b, by Theorem 2a, we only need to demonstrate the following theorem.
Theorem 4.
Let G be a graph with m a d ( G ) < 4 and Δ 7 . If there is not any pair of adjacent Δ -vertices in G, then χ a ( G ) = Δ + 1 .
Proof.
Reductio ad absurdum is used to demonstrate this theorem. Let k = m a x { 8 , Δ + 1 } . Assume that G with m a d ( G ) < 4 is a minimal counterexample of Theorem 4, to put it another way, no NDT-k-coloring of G with m a d ( G ) < 4 exists, whereas an NDT-k-coloring of any other graph H with m a d ( H ) < 4 and smaller than G exists. The graph G is connected, which we can easily observe. The color set C = [ 1 , k ] satisfies the expression | C | = k 8 . Since there are no pairs of adjacent Δ -vertices in G, we can obtain easily that no Δ -vertex adjacent to a 1-vertex exists. Combining the similar discussion of Claims 9–12, we deduce the following structural properties of G.
Claim 13.
Assume that v is a vertex in V ( G ) .
(1)
If d ( v ) 4 , then d 3 ( v ) = 0 .
(2)
If d 1 ( v ) p [ 1 , 2 ] , then d s ( v ) = 0 for s [ 2 , p + 1 ] .
(3)
d 2 ( v ) 1 .
(4)
If d ( v ) [ 6 , Δ 1 ] , then d 1 ( v ) d ( v ) 4 .
(5)
Let d ( v ) = 5 . If d 2 ( v ) 1 , then d 3 ( v ) 1 . If d 2 ( v ) = 0 , then d 3 ( v ) 2 .
(6)
Let d ( v ) = 6 . If d 2 ( v ) 1 , then d 3 ( v ) 4 .
(7)
If d ( v ) = Δ , then d 1 ( v ) = 0 .
For the purpose of deriving contradiction, at the very beginning, the same charge function w ( v ) = d ( v ) 4 is used for each vertex v in V ( G ) . Afterwards, the same discharging rules (R1) and (R2) are defined. By Lemma 2, it is sufficient to demonstrate that the new charge function w ( v ) complies with the requirements that v V 1 w ( v ) 0 and v V 2 w ( v ) v V 2 ( w ( v ) + d V 1 ( v ) ) . For v V ( G ) meeting the condition d ( v ) Δ 1 , this has been justified in Section 5.2. Hence, our assumption is that d ( v ) = Δ 7 . In accordance with Claims 13(3) and 13(7), we have d 1 ( v ) = 0 and d 2 ( v ) 1 . To comply with the rules (R1) and (R2), we calculate w ( v ) = d ( v ) 4 d 2 ( v ) 1 3 × d 3 ( v ) Δ 4 d 2 ( v ) 1 3 × ( Δ d 2 ( v ) ) = 2 3 ( Δ d 2 ( v ) ) 4 2 3 ( Δ 1 ) 4 = 2 3 ( Δ 7 ) 0 . Therefore, the proof of Theorem 2b is completed. □

7. Conclusions and Future Work

Neighbor distinguishing colorings of graphs represent mathematical problems based on the actual background of channel assignment in wireless communication networks. In this paper, characterizations are achieved for the neighbor distinguishing edge chromatic numbers of the graph G with m a d ( G ) < 4 and Δ 9 and the neighbor distinguishing total chromatic numbers of the graph G with m a d ( G ) < 4 and Δ 7 through the use of the discharging method.
These results will provide us the exact numbers of channels to be assigned in the wireless networks of topological structures with a maximum average degree of less than four. It realizes the synchronous transmission or reception of data without interference between the adjacent edges incident to the same vertex. At the same time, it also enables bidirectional edges between adjacent vertices to transmit and receive data simultaneously without interfering with each other. Thus, the utilization efficiency of channel resources is improved in wireless communication networks.
For the graph G with m a d ( G ) < 4 , it is our opinion that the condition Δ 9 in the neighbor distinguishing edge coloring of G and the condition Δ 7 in the neighbor distinguishing total coloring of G can probably be reduced slightly. For this reason, we raise two problems as follows.
Problem 1.
Give the minimum positive integer C 1 ( 8 ) , satisfying that if G is a graph with m a d ( G ) < 4 and Δ C 1 , then Δ χ a ( G ) Δ + 1 , and further χ a ( G ) = Δ + 1 if and only if there is a pair of adjacent Δ -vertices in G.
Problem 2.
Give the minimum positive integer C 2 ( 6 ) , satisfying that if G is a graph with m a d ( G ) < 4 and Δ C 2 , then Δ + 1 χ a ( G ) Δ + 2 , and further χ a ( G ) = Δ + 2 if and only if there is a pair of adjacent Δ -vertices in G.

Author Contributions

Writing—original draft preparation, J.H.; formal analysis, S.W. and Y.C.; writing—review and editing, M.L. All authors have read and agreed to the published version of the manuscript.

Funding

This research was supported by the Natural Science Foundation of Hebei Province (A2020402006) and the National Natural Science Foundation of China (11701136).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

Conflicts of Interest

The authors declare no conflict of interest.

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Huo, J.; Wen, S.; Chen, Y.; Li, M. Neighbor Distinguishing Colorings of Graphs with the Restriction for Maximum Average Degree. Axioms 2023, 12, 1132. https://doi.org/10.3390/axioms12121132

AMA Style

Huo J, Wen S, Chen Y, Li M. Neighbor Distinguishing Colorings of Graphs with the Restriction for Maximum Average Degree. Axioms. 2023; 12(12):1132. https://doi.org/10.3390/axioms12121132

Chicago/Turabian Style

Huo, Jingjing, Sensen Wen, Yulong Chen, and Mingchao Li. 2023. "Neighbor Distinguishing Colorings of Graphs with the Restriction for Maximum Average Degree" Axioms 12, no. 12: 1132. https://doi.org/10.3390/axioms12121132

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