In this section, our proof proceeds by means of reductio ad absurdum. Let . Assume that G with is a minimal counterexample of Theorem 1a, to put it another way, there exists no NDE-k-coloring of G with , whereas there exists an NDE-k-coloring of any other graph H with and smaller than G. Clearly, G is a connected graph. Let be the color set in the following discussion. Then .
3.1. Structural Properties in G
A series of auxiliary claims are used to give the structural properties of the minimal counterexample G with in this subsection.
Claim 1. There exists no edge with and .
Proof.
Suppose that the statement is false. There is then an edge with and in G. The graph satisfies the inequality . It follows that according to Lemma 1. With the aid of the minimality of G, an NDE-k-coloring of H exists using the color set C. Because , we use a color in to color for which v and its neighbors do not conflict. □
Claim 2. Let v be a p-vertex in .
- (1)
If , then .
- (2)
If , then .
- (3)
If , then for .
Proof.
(1) Suppose that the statement is false. Then . Let and , where . Below, we break it down into two cases.
Case 1. . Then the graph satisfies the inequality . We have according to Lemma 1. With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C. Since , we use a color in to color properly. It follows that an NDE-k-coloring of G is obtained. This is a contradiction.
Case 2. . We delete the vertices u and v from G and add one 2-vertex x in for which x is adjacent to and . This produces a smaller graph H satisfying the inequalities and .
Now we will show that . Let be the surgraph of H. If , then is the surgraph of G and . Otherwise, . Then is the surgraph of G and . As , we have . It follows from the arbitrariness of that .
With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C. Observe that . We color with and with . Then we use a color in to color for which u and v do not conflict with their neighbors.
(2) Suppose that the statement is false. Then . Let , and , where . The graph satisfies the inequality . According to Lemma 1, we have . With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C. Again, let for and for . Based on the values of p, we split the proof into these two cases.
Case 1. . If , then we use a color in to color for which v and do not conflict with their neighbors. Otherwise, . Then we use a color in to recolor for which and its neighbors other than v do not conflict. This brings us back to the above discussion.
Case 2. . Let . We now consider these four subcases on the basis of the value of t.
Subcase 2.1. . Let for . We suppose that are not legally colored, and need to analyse these two possibilities by symmetry.
for . When , then we color with 7 and recolor with a color in for which and its neighbors do not conflict. When , where , then we use to color and a color in to recolor for which and its neighbors do not conflict. Suppose that . Let for . We use 9 to color and a color in to recolor for which and its neighbors do not conflict, and recolor with a color in for which and its neighbors do not conflict.
for and . When , the proof is reduced to the above possibility. When , , we can see that there exists an element , that satisfies . Suppose that . Then we color with 7 and recolor with a color for which and its neighbors do not conflict. If , then we are done. Otherwise, . We use a color in to recolor for which and its neighbors do not conflict.
Subcase 2.2. . Let , and . We suppose that are not legally colored, and need to analyse these two possibilities by symmetry.
for and . Then is colored with a color for which and its neighbors other than v do not conflict. If , then is recolored with a color in for which and its neighbors do not conflict. If , then we use a color in to recolor for which and its neighbors do not conflict.
for and for . When , the proof goes back to the above possibility. Suppose that . We then color with 6 and recolor with a color for which and its neighbors do not conflict. If , then we are finished. Otherwise, . We color with 7, recolor with 2, and with a color in for which and its neighbors do not conflict.
Subcase 2.3. . Let and . Suppose that are not legally colored. Then we can obtain that for and , for . We use a color in to recolor for which and its neighbors other than do not conflict. The proof goes back to Subcase 2.2.
Subcase 2.4. . Let , and . If we may use a color in to recolor legally, then the proof returns to Subcase 2.3. Otherwise, our assumption is that can not be recolored, which implies that and for . Then is recolored with a color for which and its neighbors other than do not conflict. We use a color in to recolor . At this time, the proof returns to Subcase 2.3.
(3) Suppose that the statement is false. Then . Let and . Then satisfies the inequality . It follows that according to Lemma 1. With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C. Since , by Claims 2(1) and 2(2), we have and , meaning that there is at most one conflict vertex for each of u and v. We use a color in to color 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 .
- (1)
If , then for .
- (2)
.
- (3)
If , then .
Proof.
(1) Suppose that the statement is false. Then for . Let , be the vertices adjacent to v in which case for and . Let . By Claims 2(1) and 2(2), we have . Now there are these two cases, based on whether or 1.
Case 1. . We delete the edge from G and add one 1-vertex in such that is adjacent to v. This produces a smaller graph H satisfying the expressions , , and .
Now we will show that . Let be the surgraph of H. If , then is the surgraph of G and . Otherwise, . Then is the surgraph of G and . As , we have . It follows from the arbitrariness of that .
With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C in which case for . Observe that . In order to restore the original graph G, the vertices and are stuck together. If , then we can easily see that is an NDE-k-coloring of G. Otherwise, the colors of and are exchanged for . Whichever possibility arises, an NDE-k-coloring of G is always obtained. This is a contradiction.
Case 2. . Let . The graph satisfies the inequality . According to Lemma 1, we have . With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C in which case for . Observe that . If , then we color properly. Otherwise, the colors of and are exchanged for and 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 . Let be the vertices adjacent to v satisfying the equality . By Claim 2(1), is not adjacent to in G. These two cases are then considered.
Case 1. Assume that v, and are in a 4-cycle , that is, . We delete the edges and from G and add two 1-vertices and in for which and are adjacent to v, respectively. This produces a smaller graph H satisfying the expressions , and .
Now we will show that . Let be the surgraph of H. Then these four possibilities need to be handled.
. Then is the surgraph of G and .
and . Then is the surgraph of G and . As , we have .
and . Then is the surgraph of G and . As , we have .
. Then is the surgraph of G and . As , we have .
According to the analysis of the above four possibilities, we can always get . It follows from the arbitrariness of that .
With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C. The vertices and are stuck together for , and the colors of and are exchanged if necessary. It follows that an NDE-k-coloring of G is given. This is a contradiction.
Case 2. Assume that v, and are in a -cycle. Let and , where . We delete the edges and from G and add one 2-vertex x in for which x is adjacent to and . This produces a smaller graph H satisfying the expressions , and .
Now we will show that . Let be the surgraph of H. If , then is the surgraph of G and . If , then is the surgraph of G and . As , We may derive that . It follows from the arbitrariness of that .
With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C. First, we split x into and in which case is adjacent to for . Second, the vertices and are stuck together for , and the colors of and are exchanged if necessary. Thus, an NDE-k-coloring of G is given. This is a contradiction.
(3) Let and with . Suppose that the statement is false. Then . Observe that for and . The graph satisfies the inequality . According to Lemma 1, we have .
With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C in which case for . In the first place, we use any color in to color . In the second place, for each , we use a color to recolor and to color . As a consequence, at least 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 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 .
Proof.
Let with . Suppose that the statement is false. Then . Let , and . The graph satisfies the inequality . According to Lemma 1, we have . With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C in which case for . On the basis of Claim 1, we have . Now these two cases are considered, depending on whether or 3.
Case 1. . By Claim 2(1), we have . We assume, without loss of generality, that . Then is colored with a color in for which there is no conflict between v and . Therefore, an NDE-k-coloring of G is given, producing a contradiction.
Case 2. . By Claim 2(2), it follows that . When , we set and . When , i.e., , we use a color in to recolor , and set and . We assume, without loss of generality, that . In the first place, we use any color in to color . In the second place, we use a color to recolor for which and its neighbors do not conflict, and a color in to color . 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 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 .
Proof.
Let with . Suppose that the statement is false. Then . Let , and for , and . The graph satisfies the inequality . According to Lemma 1, we have . With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C in which case for . We now consider these three cases, in accordance with the values of s.
Case 1. . Then . We use a color in to color the edge in which case there is no conflict between v and . Hence, an NDE-k-coloring of G is given, producing a contradiction.
Case 2. . By Claim 2(1), we have . We assume, without loss of generality, that . In the first place, we use any color in to color . In the second place, we use a color to recolor for which and its neighbors do not conflict, and a color in to color . 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 is extended to an NDE-k-coloring of the entire graph G. This is a contradiction.
Case 3. . The inequality is guaranteed by Claim 2(2). When , we set and . When , i.e., , we use a color in to recolor , and set and . We assume, without loss of generality, that . In the first place, we use any color in to color . In the second place, for each , we use a color to recolor for which and its neighbors do not conflict, and a color in to color . 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 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 , then .
- (2)
If , then .
Proof.
Let with .
(1) Suppose that the statement is false. Then if . Let and for . The graph satisfies the inequality . According to Lemma 1, we have . With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C in which case for .
For each , by Claims 1 and 2, we have . When or with , we use a color to recolor for which and its neighbors do not conflict. When with , we assume that and . Then we use a color in to recolor and a color to recolor for which and its neighbors do not conflict. So we can always find a color satisfying that and its neighbors do not conflict. We now consider these two cases, according to whether or 2.
Case 1. . Then . In the first place, we use any color in to color . In the second place, as stated above, we use a color to recolor for which and its neighbors do not conflict, and a color in to color . 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 is extended to an NDE-k-coloring of the entire graph G. This is a contradiction.
Case 2. . By Claim 2(1), we have . Let and . We assume, without loss of generality, that . In the first place, we use any color in to color . In the second place, for each , as stated above, we use a color to recolor for which and its neighbors do not conflict, and a color in to color . 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 is extended to an NDE-k-coloring of the entire graph G. This is a contradiction.
(2) Suppose that the statement is false. Then if . Let for and . The graph satisfies the inequality . According to Lemma 1, we have . With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C in which case for . The inequality is guaranteed by Claim 2(2). When , we set and . When , i.e., , we use a color in to recolor , and set and . We assume, without loss of generality, that . Then we color 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 , then .
Proof.
Let with . Suppose that the statement is false. Then if . Let and for . The graph satisfies the inequality . According to Lemma 1, we have . With the aid of the minimality of G, an NDE-k-coloring of H exists through the use of the color set C in which case for . When , we color with 8. When , by Claim 2(1), we have . Let and . We assume without loss of generality, that . Then we color 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
is given for each vertex
v in
. 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
on the vertices. While the discharging is occurring, the sum of all charges remains constant, that is to say,
.
Lemma 2. Assume that is the vertex partition of G, where the members of the set are -vertices and the members of the set are -vertices. If and , where is the number of the vertices in that are adjacent to v, then .
Proof.
From the definition of the vertex partition of
G, we have
and
. Likewise,
is used to denote the number of the vertices in
that are adjacent to
v. It follows that
and
, where the members of the set
are the edges joining a vertex of
and a vertex of
in
G. Since
we obtain
, where the subgraph
of
G is induced by
. Therefore we can infer a conclusion that
. □
According to Lemma 2, we need to demonstrate that for a vertex v in , if , if , if , contradicting the condition that . Hence, it is proved that there is no such minimal counterexample. Assume that an edge satisfying the inequalities and is in . The following is our definition of the discharge rules.
(R1) If , then v gives 1 to u.
(R2) If , then v gives to u.
Suppose that a vertex v with is in . We will analyze the new charge of the vertex v based on the values of p.
Case 1. . The equality is established. It follows from Claim 1 that the neighbor of v is a -vertex and . According to the rule (R1), we have .
Case 2. . The equality is established. It follows from Claims 1 and 2 that the neighbors of v are two -vertices and . According to the rule (R1), we have .
Case 3. . The equality is established. It follows from Claims 1 and 2 that there are at least two -vertices in the neighbors of v. According to the rule (R2), we have .
Case 4. . The equality is established. To comply with the rules (R1) and (R2), we have .
Case 5. . The equality is established. It follows from Claim 4 that . To comply with the rules (R1) and (R2), we have .
Case 6. . The equality is established. It follows from Claim 5 that . To comply with the rules (R1) and (R2), we have .
Case 7. . The equality is established. If , then it follows from Claim 6(1) that . To comply with the rules (R1) and (R2), we have . Otherwise, . By Claim 6(2), it follows that . Hence, by (R2), we have .
Case 8. . The equality is established. It follows from Claim 3(3) that . If , then , which is guaranteed by Claim 3(1). According to the rule (R1), we have . If , by Claims 3(1) and 7, then and . To comply with the rules (R1) and (R2), we have . If and , by Claims 3(2) and 7, then and . To comply with the rules (R1) and (R2), we have . If , applying (R2), then we have .
Case 9. . The equality is established. It follows from Claim 3(3) that . If , then , which is guaranteed by Claim 3(1). According to the rule (R1), we have . If , then , which is guaranteed by Claim 3(1). To comply with the rules (R1) and (R2), we have . If , by Claim 3(2), then . To comply with the rules (R1) and (R2), we have .
From the discussion so far, we obtain that for , if , if . Therefore, we have completed the proof of Theorem 1a. □