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Article

On the Total Neighbor Sum Distinguishing Index of IC-Planar Graphs

School of Mathematics and Computer Application, Shangluo University, Shangluo 726000, China
*
Author to whom correspondence should be addressed.
Symmetry 2021, 13(10), 1787; https://doi.org/10.3390/sym13101787
Submission received: 26 August 2021 / Revised: 17 September 2021 / Accepted: 22 September 2021 / Published: 26 September 2021
(This article belongs to the Special Issue Research on Symmetry Applied in Graph Theory)

Abstract

:
A proper total k-coloring ϕ of G with z E G ( u ) { u } ϕ ( z ) z E G ( v ) { v } ϕ ( z ) for each u v E ( G ) is called a total neighbor sum distinguishing k-coloring, where E G ( u ) = { u v | u v E ( G ) } . Pilśniak and Woźniak conjectured that every graph with maximum degree Δ exists a total neighbor sum distinguishing ( Δ + 3 ) -coloring. In this paper, we proved that any IC-planar graph with Δ 12 satisfies this conjecture, which improves the result of Song and Xu.

1. Introduction

Let G = ( V ( G ) , E ( G ) ) be a simple graph. Set T ( G ) = V ( G ) E ( G ) and E G ( u ) = { u v | u v E ( G ) } . A total neighbor sum distinguishing k-coloring (for short, tnsd) is a mapping ϕ : T ( G ) { 1 , 2 , , k } satisfying the following conditions:
(i)
ϕ ( z 1 ) ϕ ( z 2 ) for any two adjacent or incident elements z 1 , z 2 in T ( G ) ;
(ii)
z E G ( u ) { u } ϕ ( z ) z E G ( v ) { v } ϕ ( z ) for each u v E ( G ) .
The total neighbor sum distinguishing index of G, denoted by χ Σ ( G ) , is the smallest integer k such that G has a tnsd k-coloring.
In 2015, Pilśniak and Woźniak [1] first studied the coloring and gave the conjecture about the tnsd index as follows:
Conjecture 1
([1]). For every graph G, χ Σ ( G ) Δ ( G ) + 3 .
Up to now, many special classes of graphs had been proved to satisfy Conjecture 1, including complete graphs, bipartite graphs, subcubic graphs [1], 2-degenerate graphs with maximum degree Δ 6 [2], K 4 -minor-free graphs [3], planar graphs with maximum degree Δ 10 [4], and planar graphs with maximum degree Δ 7 but without triangles [5].
A graph is 1-planar if it can be embedded on the plane so that each edge is crossed at most once. In 1965, Ringel introduced the notion of 1-planar graphs to color a planar graph and its dual simultaneously. In particularly, a 1-planar graph such that each vertex is incident with, at most, a crossing is called an IC-planar (independent crossing planar) graph, which was introduced by Albertson [6] while considering the relationship between the crossing number and the chromatic number.
Recently, the tnsd index on IC-planar graphs has been extensively studied. There are some recent results in the following.
Theorem 2.
Each of the following classes of graphs satisfies Conjecture 1.
(1)
IC-planar graph with Δ 7 but without triangles ([7]).
(2)
IC-planar graph with Δ 14 but without 2-vertex incident with crossed edge ([8]).
(3)
IC-planar graph with Δ 13 ([9]).
There is also a result of list version about the tnsd index of IC-planar graphs, see [10].
In this paper, we discuss any IC-planar graph with maximum degree Δ 12 and obtain the following result, which extends the result of (3) in Theorem 2.
Theorem 3.
Let G be an IC-planar graph. Then, χ Σ ( G ) max { Δ ( G ) + 3 , 15 } .

2. Preliminaries

Let G be a simple graph and u v be an edge of G. For a vertex u V ( G ) , we use d G ( u ) and N G ( u ) to denote the degree and neighbors of u in G, respectively. If d G ( u ) = (resp., d G ( u ) , d G ( u ) ), we call u an -vertex (resp., + -vertex, -vertex) and an -neighbor (resp., + -neighbor, -neighbor) of v. We use n G ( u ) (resp., n G + ( u ) , n G ( u ) ) to represent the number of -vertices (resp., + -vertices, -vertices) adjacent to u in G. For two subsets A and B of V ( G ) , we denote by [ A , B ] the set of edges with one end in A and the other in B. In particular, for u V ( G ) , we write [ u , B ] for [ { u } , B ] .
Theorem 4
([11]). Assume that F is an arbitrary field and P F [ x 1 , , x n ] with degree d e g ( P ) = k = 1 n i k , where i k 0 is an integer. If the coefficient c P ( x 1 i 1 , , x n i n ) of the monomial x 1 i 1 x n i n in P is nonzero, and if S 1 , , S n are subsets of F with | S k | > i k , there are s 1 S 1 , , s n S n such that P ( s 1 , , s n ) 0 .
The planar graph G × obtained from a 1-planar graph G by turning all crossings of G into new 4-vertices on the plane is called the associated planar graph of G. Let u v E ( G × ) . Then, u is called a false vertex and a false neighbor of v if u V ( G × ) \ V ( G ) , and a real vertex and a real neighbor of v otherwise. We use F ( G × ) to denote the set of faces in G × . For each face f F ( G × ) , we call f a false face if it is incident with one false vertex, and a real face otherwise. A face with degree is called an -face. Unless stated otherwise, an -face means a real or false -face throughout each section in the following.
Lemma 1
([12]). Assume that G is a 1-planar graph and G × is its associated planar graph. If a real 3-vertex u is adjacent to a false vertex v in G × , then u v is not incident with two 3-faces in G × .

3. Proof of Theorem 3

Let V i ( G ) ( V i ( G ) , V i + ( G ) , respectively) represent the set of vertices in G of degree i (at most i, at least i, respectively). For the graph G, a graph G is smaller than it if one of the below conditions is satisfied.
(i)
| E ( G ) | < | E ( G ) | ;
(ii)
In the lexicographic order, ( | V t ( G ) | , | V t 1 ( G ) | , , | V 1 ( G ) | ) precedes ( | V t ( G ) | , | V t 1 ( G ) | , , | V 1 ( G ) | ) , where t = max { Δ ( G ) , Δ ( G ) } .
We say a graph is minimal for some property if no smaller graph admits the property. To obtain a smaller graph, G , than G, we will frequently delete some edges or split some vertices of G in the following.
From now on, let G be a minimal graph not satisfying Theorem 3 and k = max { Δ ( G ) + 3 , 15 } . Then, any graph G that is smaller than G has a tnsd k-coloring ϕ . By splitting a vertex v into two vertices v 1 and v 2 , we obtain a smaller graph G than G, then G has a tnsd k-coloring ϕ . If we can obtain a tnsd k-coloring of G from the ϕ by sticking v 1 and v 2 into v, we say that v 1 and v 2 can be properly stunk. In the following, we will often extend a tnsd k-coloring ϕ of G to a tnsd k-coloring ϕ of G to obtain a contradiction. Set m ϕ ( u ) = z E G ( u ) { u } ϕ ( z ) . Assume that X is a subset of T ( G ) and ψ : X { 1 , , k } is a mapping. For each z T ( G ) , set
S ( z ) = { 1 , , k } \ { ψ ( x ) | x is   adjacent   or   incident   to   z   in   G } .
Assume that u is a 4 -vertex in G and ϕ : T ( G ) \ { u } { 1 , , k } is a mapping satisfying the conditions as follows:
(i)
ϕ ( z 1 ) ϕ ( z 2 ) for any adjacent or incident elements z 1 , z 2 T ( G ) \ { u } ;
(ii)
z E G ( z 1 ) { z 1 } ϕ ( z ) z E G ( z 2 ) { z 2 } ϕ ( z ) for any two adjacent vertices z 1 , z 2 V ( G ) \ { u } .
Since S ( u ) = { 1 , , k } \ { ϕ ( x ) | x is   adjacent   or   incident   to   u   in   G } , | S ( u ) | k 2 d G ( u ) > d G ( u ) . Thus, we can extend the ϕ to a tnsd k-coloring ϕ of G, which is a contradiction. For simplicity, we will omit the colors of all 4 -vertices in the following discussion.
In Figure 1 and Figure 2, a black vertex represents that its neighbors are all depicted and a white vertex means that its neighbors may not be all depicted.
Claim 1.
The graph G contains no configuration in Figure 1.
Proof. 
Suppose that G contains the configuration in Figure 1. By splitting the two 3-vertices v 2 , v 4 of the configuration in Figure 1 into v 21 , v 22 and v 41 , v 42 , respectively, we obtain a smaller graph G than G. Then, G has a tnsd k-coloring ϕ . Without loss of generality, set ϕ ( v v 1 ) = c 1 , ϕ ( v v 22 ) = c 2 , ϕ ( v v 3 ) = c 3 , ϕ ( v v 42 ) = c 4 , ϕ ( v v 5 ) = c 5 , ϕ ( v 1 v 21 ) = c 6 , ϕ ( v 21 v 3 ) = c 7 , ϕ ( v 3 v 41 ) = c 8 , and ϕ ( v 41 v 5 ) = c 9 (see Figure 2).
If we can properly stick v 21 , v 41 with v 22 , v 42 , respectively, we can obtain a tnsd k-coloring ϕ of G, which is a contradiction. Thus, assume that at least one of v 21 and v 41 cannot be stuck properly below. Let Y = T ( G ) \ { v 21 , v 22 , v 41 , v 42 } .
Case 1. Suppose that one of v 21 and v 41 cannot be stuck properly. If v 21 can not be stuck properly, c 2 { c 6 , c 7 } . Without loss of generosity, set c 2 = c 6 .
Set c 8 c 6 , c 9 c 6 , and c 4 c 7 . Then, we assign ϕ ( v v 22 ) = c 4 , ϕ ( v v 42 ) = c 2 , and ϕ ( z ) = ϕ ( z ) for every z Y \ { v v 22 , v v 42 } .
Set c 8 c 6 , c 9 c 6 , and c 4 = c 7 . Then, we assign ϕ ( v v 22 ) = c 4 , ϕ ( v v 42 ) = c 2 , ϕ ( v 21 v 3 ) = c 8 , ϕ ( v 3 v 41 ) = c 7 , and ϕ ( z ) = ϕ ( z ) for every z Y \ { v v 22 , v v 42 , v 21 v 3 , v 3 v 41 } .
Set c 8 = c 6 . If c 9 c 3 , we assign ϕ ( v v 22 ) = c 3 , ϕ ( v v 3 ) = c 2 , ϕ ( v 3 v 41 ) = c 3 , and ϕ ( z ) = ϕ ( z ) for every z Y \ { v v 22 , v v 3 , v 3 v 41 } . If c 9 = c 3 and c 5 c 7 , we assign ϕ ( v v 22 ) = c 5 , ϕ ( v v 3 ) = c 8 , ϕ ( v 3 v 41 ) = c 3 , ϕ ( v v 5 ) = c 9 , ϕ ( v 41 v 5 ) = c 5 , and ϕ ( z ) = ϕ ( z ) for every z Y \ { v v 22 , v v 3 , v 3 v 41 , v v 5 , v 41 v 5 } . If c 9 = c 3 and c 5 = c 7 , we assign ϕ ( v v 1 ) = c 6 , ϕ ( v 1 v 21 ) = c 1 , ϕ ( v v 22 ) = c 4 , ϕ ( v v 42 ) = c 1 , and ϕ ( z ) = ϕ ( z ) for every z Y \ { v v 1 , v 1 v 21 , v v 22 , v v 42 } .
Set c 9 = c 6 . If c 8 c 5 and c 5 c 7 , we assign ϕ ( v v 22 ) = c 5 , ϕ ( v v 5 ) = c 2 , ϕ ( v 41 v 5 ) = c 5 , and ϕ ( z ) = ϕ ( z ) for every z Y \ { v v 22 , v v 5 , v 41 v 5 } . If c 8 c 5 and c 5 = c 7 , we assign ϕ ( v v 22 ) = c 4 , ϕ ( v v 5 ) = c 9 , ϕ ( v 41 v 5 ) = c 5 , ϕ ( v v 3 ) = c 7 , ϕ ( v 21 v 3 ) = c 3 , ϕ ( v v 42 ) = c 3 , and ϕ ( z ) = ϕ ( z ) for every z Y \ { v v 22 , v 21 v 3 , v v 3 , v v 42 , v v 5 , v 41 v 5 } . If c 8 = c 5 , we assign ϕ ( v v 22 ) = c 3 , ϕ ( v v 5 ) = c 9 , ϕ ( v 41 v 5 ) = c 5 , ϕ ( v v 3 ) = c 8 , ϕ ( v 3 v 41 ) = c 3 , and ϕ ( z ) = ϕ ( z ) for every z Y \ { v v 22 , v v 3 , v 3 v 41 , v v 5 , v 41 v 5 } .
In each case above, by sticking v i 1 and v i 2 into v i for i = 2 , 4 , we obtain the graph G. Since both v 2 and v 4 are 3-vertices in G, we can obtain a tnsd k-coloring ϕ of G from ϕ . By symmetry, if v 41 cannot be stuck properly, we can also obtain a tnsd k-coloring ϕ of G. It is a contradiction.
Case 2. Assume that both v 21 and v 41 cannot be stuck properly. Then, c 2 = c 6 = c 8 and c 4 = c 7 = c 9 . Let ϕ ( v v 1 ) = c 6 , ϕ ( v 1 v 21 ) = c 1 , ϕ ( v v 5 ) = c 9 , ϕ ( v 41 v 5 ) = c 5 , ϕ ( v v 22 ) = c 5 , ϕ ( v v 42 ) = c 1 , and ϕ ( z ) = ϕ ( z ) for every z Y \ { v v 1 , v 1 v 21 , v v 5 , v 41 v 5 , v v 22 , v v 42 } . By sticking v i 1 and v i 2 into v i for i = 2 , 4 , we obtain the graph G. Since both v 2 and v 4 are 3-vertices in G, we can obtain a tnsd k-coloring ϕ of G from ϕ , which is a contradiction. □
The below Claim 2 follows from (3) in Theorem 2.
Claim 2.
Δ ( G ) 12 .
Claim 3.
If u V 6 ( G ) , then | [ u , V 5 ( G ) ] | = 0 .
Proof. 
By contradiction, suppose that there is a vertex u V 6 ( G ) and a vertex v V 5 ( G ) such that u v [ u , V 5 ( G ) ] . Let G = G u v . Then, G has a tnsd k-coloring ϕ . We first erase the colors on u and v from ϕ . Without loss of generality, set d G ( u ) = 6 and d G ( v ) = 5 . Then,
| S ( u ) | k 2 × ( 6 1 ) 5 , | S ( u v ) | k ( 6 1 ) ( 5 1 ) 6   and | S ( v ) | k 2 × ( 5 1 ) 7 .
Let ϕ be a map on T ( G ) obtained from ϕ by assigning ϕ ( u ) = x 1 , ϕ ( v ) = x 2 , ϕ ( u v ) = y 1 , and ϕ ( z ) = ϕ ( z ) for every z T ( G ) \ { u , v , u v } . Set
P ( x 1 , x 2 , y 1 ) = z N G ( u ) \ { v } m ϕ ( u ) m ϕ ( z ) z N G ( v ) \ { u } m ϕ ( v ) m ϕ ( z ) · m ϕ ( u ) m ϕ ( v ) ( x 1 x 2 ) ( x 1 y 1 ) ( x 2 y 1 ) .
With the help of MATHEMATICA, we have c P ( x 1 4 x 2 4 y 1 5 ) = 25 . By Theorem 4, there is a color c 1 S ( u ) , a color c 2 S ( v ) , and a color c 3 S ( u v ) such that P ( c 1 , c 2 , c 3 ) 0 . Therefore, we can extend the ϕ to a tnsd k-coloring ϕ of G by recoloring u , v with colors c 1 , c 2 and coloring u v with color c 3 , respectively. It is a contradiction. □
Claim 4.
If u V 7 ( G ) , | [ u , V 4 ( G ) ] | = 0 and | [ u , V 5 ( G ) ] | 1 .
Proof. 
We first prove | [ u , V 4 ( G ) ] | = 0 when u V 7 ( G ) . By contradiction, suppose that there is a vertex u V 7 ( G ) and a vertex v V 4 ( G ) such that u v [ u , V 4 ( G ) ] . Let G = G u v . Then, G has a tnsd k-coloring ϕ . We first erase the colors on u and v from ϕ . Note that v is a 4 -vertex. Then,
| S ( u ) | k 2 × ( 7 1 ) 3 and | S ( u v ) | k ( 7 1 ) ( 4 1 ) 6 .
Let ϕ be a map on T ( G ) \ { v } obtained from ϕ by assigning ϕ ( u ) = x 1 , ϕ ( u v ) = y 1 , and ϕ ( z ) = ϕ ( z ) for every z T ( G ) \ { u , v , u v } . Set
P ( x 1 , y 1 ) = z N G ( u ) \ { v } m ϕ ( u ) m ϕ ( z ) ( x 1 y 1 ) .
By applying MATHEMATICA, we have c P ( x 1 2 y 1 5 ) = 9 . By Theorem 4, there is a color c 1 S ( u ) and a color c 2 S ( u v ) such that P ( c 1 , c 2 ) 0 .
Since v is a 4 -vertex, we can obtain a tnsd k-coloring of G from the ϕ by recoloring u with color c 1 and coloring u v with color c 2 , respectively. This is a contradiction.
Next, we show | [ u , V 5 ( G ) ] | 1 when u V 7 ( G ) . Suppose, to the contrary, that there is a vertex u V 7 ( G ) and two vertices v 1 , v 2 V 5 ( G ) such that u v 1 , u v 2 [ u , V 5 ( G ) ] . Let G = G u v 1 u v 2 . Then, G has a tnsd k-coloring ϕ . We first erase the colors on u, v 1 , and v 2 from ϕ . Then,
| S ( u ) | k 2 × ( 7 2 ) 5 , | S ( u v i ) | k ( 7 2 ) ( 5 1 ) 6 and | S ( v i ) | k 2 × ( 5 1 ) 7 for i = 1 , 2 .
Let ϕ be a map on T ( G ) obtained from ϕ by assigning ϕ ( u ) = x 1 , ϕ ( v i ) = x i + 1 , and ϕ ( u v i ) = y i for i = 1 , 2 , and ϕ ( z ) = ϕ ( z ) for every z T ( G ) \ { u , v 1 , v 2 , u v 1 , u v 2 } . Set
P ( x 1 , x 2 , x 3 , y 1 , y 2 ) = z N G ( u ) \ { v 1 , v 2 } m ϕ ( u ) m ϕ ( z ) i = 1 2 m ϕ ( u ) m ϕ ( v i ) · i = 1 2 z N G ( v i ) \ { u } m ϕ ( v i ) m ϕ ( z ) i = 1 2 ( x 1 x i + 1 ) · ( y 1 y 2 ) i = 1 2 ( x 1 y i ) ( x i + 1 y i ) .
With the help of MATHEMATICA, we have c P ( x 1 4 x 2 4 x 3 5 y 1 4 y 2 5 ) = 120 . By Theorem 4, there is a color c 1 S ( u ) , a color c i + 1 S ( v i ) , and a color c i + 3 S ( u v i ) for i = 1 , 2 such that P ( c 1 , , c 5 ) 0 . Therefore, we can extend the ϕ to a tnsd k-coloring ϕ of G by recoloring u , v i with colors c 1 , c i + 1 and coloring u v i with color c i + 3 for i = 1 , 2 , respectively. This is a contradiction. □
Claim 5.
If u V ( G ) with 8 9 , then,
(1)
| [ u , V 4 ( G ) ] | 7 ,
(2)
| [ u , V 5 ( G ) ] | 7 when | [ u , V 4 ( G ) ] | 1 .
Proof. 
Suppose to be contrary that there is a vertex u V ( G ) , a vertex v 1 V 4 ( G ) , and ( 7 ) vertices v 2 , , v 6 V 5 ( G ) with 8 9 such that u v 1 [ u , V 4 ( G ) ] and u v i [ u , V 5 ( G ) ] for i = 2 , , 6 . Let G = G { u v i | i = 1 , , 6 } . Then, G has a tnsd k-coloring ϕ . We first erase the colors on u and v i with 1 i 6 from ϕ . Without loss of generality, set d G ( v i ) = 5 with 2 i 6 . Note that v 1 is a 4 -vertex. Then,
| S ( u ) | k 2 × ( ( 6 ) ) 3 , | S ( u v 1 ) | k ( ( 6 ) ) ( 4 1 ) 6 , | S ( u v i ) | k ( ( 6 ) ) ( 5 1 ) 5 and | S ( v i ) | k 2 × ( 5 1 ) 7 for i = 2 , , 6 .
Let ϕ be a map on T ( G ) \ { v 1 } obtained from ϕ by assigning
ϕ ( u ) = x 1 , ϕ ( v 2 ) = x 2 , , ϕ ( v 6 ) = x 6 , ϕ ( u v 1 ) = y 1 , , ϕ ( u v 6 ) = y 6
and ϕ ( z ) = ϕ ( z ) for every z T ( G ) \ { u , v 1 , , v 6 , u v 1 , , u v 6 } . Set
P ( x 1 , , x 6 , y 1 , , y 6 ) = z N G ( u ) \ { v 1 , , v 6 } m ϕ ( u ) m ϕ ( z ) i = 2 6 m ϕ ( u ) m ϕ ( v i ) · i = 2 6 z N G ( v i ) \ { u } m ϕ ( v i ) m ϕ ( z ) i = 2 6 ( x 1 x i ) ( x i y i ) · i = 1 6 ( x 1 y i ) 1 i < j 6 ( y i y j ) .
When = 8 , 6 = 2 , and we have c P ( x 1 2 x 2 6 y 1 4 y 2 4 ) = 20 by applying MATHEMATICA. When = 9 , 6 = 3 , and we have c P ( x 1 2 x 2 6 x 3 5 y 1 5 y 2 4 y 2 4 ) = 59 by applying MATHEMATICA. By Theorem 4, there is a color c 1 S ( u ) , a color c i S ( v i ) with 2 i 6 , and a color c i + 6 S ( u v i ) with 1 i 6 such that P ( c 1 , , c 2 12 ) 0 when 8 9 . Since v 1 is a 4 -vertex, we can extend the ϕ to a tnsd k-coloring of G by recoloring u , v 2 , , v 6 with colors c 1 , , c 6 and coloring u v 1 , , u v 6 with colors c 5 , , c 2 12 , respectively. This is a contradiction. □
Claim 6.
If u V ( G ) with 10 11 , then
(1)
| [ u , V 3 ( G ) ] | 7 ,
(2)
| [ u , V 4 ( G ) ] | 7 when | [ u , V 3 ( G ) ] | 1 ,
(3)
| [ u , V 5 ( G ) ] | = 0 when | [ u , V 3 ( G ) ] | 1 and | [ u , V 4 ( G ) ] | = 7 .
Proof. 
By contradiction, suppose that there is a vertex u V ( G ) , a vertex v 1 V 5 ( G ) , a vertex v 2 V 3 ( G ) , ( 8 ) vertices v 3 , , v 6 V 4 ( G ) with 10 11 such that u v 1 [ u , V 5 ( G ) ] , u v 2 [ u , V 3 ( G ) ] , and u v i [ u , V 4 ( G ) ] for i = 3 , , 6 . Let G = G { u v i | i = 1 , , 6 } . Then, G has a tnsd k-coloring ϕ . We first erase the colors on u and v i with 1 i 6 from ϕ . Without loss of generality, set d G ( v 1 ) = 5 . Note that v 2 is a 3 -vertex and v i is a 4 -vertex with 3 i 6 . Then,
| S ( u ) | k 2 × ( ( 6 ) ) 3 , | S ( v 1 ) | k 2 × ( 5 1 ) 7 , | S ( u v 1 ) | k ( ( 6 ) ) ( 5 1 ) 5 , | S ( u v 2 ) | k ( ( 6 ) ) ( 3 1 ) 7 and | S ( u v i ) | k ( ( 6 ) ) ( 4 1 ) 6 for i = 3 , , 6 .
Let ϕ be a map on T ( G ) \ { v 2 , , v 6 } obtained from ϕ by assigning
ϕ ( u ) = x 1 , ϕ ( v 1 ) = x 2 , ϕ ( u v 1 ) = y 1 , , ϕ ( u v 6 ) = y 6
and ϕ ( z ) = ϕ ( z ) for every z T ( G ) \ { u , v 1 , , v 6 , u v 1 , , u v 6 } . Set
P ( x 1 , x 2 , y 1 , , y 6 ) = z N G ( u ) \ { v 1 , , v 6 } m ϕ ( u ) m ϕ ( z ) z N G ( v 1 ) \ { u } m ϕ ( v 1 ) m ϕ ( z ) · m ϕ ( u ) m ϕ ( v 1 ) ( x 1 x 2 ) ( x 2 y 1 ) · i = 1 6 ( x 1 y i ) 1 i < j 6 ( y i y j ) .
When = 10 , 6 = 4 , and we have c P ( x 1 2 x 2 6 y 1 4 y 2 6 y 3 4 y 4 ) = 20 with the help of MATHEMATICA. When = 11 , 6 = 5 , and we have c P ( x 1 2 x 2 6 y 1 4 y 2 6 y 3 5 y 4 4 y 5 ) = 4 with the help of MATHEMATICA. By Theorem 4, there is a color c 1 S ( u ) , a color c 2 S ( v 1 ) , and a color c i + 2 S ( u v i ) with 1 i 6 such that P ( c 1 , , c 4 ) 0 when 10 11 . Since v i is a 4 -vertex with 2 i 6 , we can obtain a tnsd k-coloring of G from the ϕ by recoloring u , v 1 with colors c 1 , c 2 and coloring u v 1 , , u v 6 with colors c 3 , , c 4 , respectively. This is a contradiction. □
Claim 7.
If u V 12 ( G ) , then
(1) | [ u , V 2 ( G ) ] | 5 ,
(2) | [ u , V 3 ( G ) ] | 5 when | [ u , V 2 ( G ) ] | 1 ,
(3) | [ u , V 4 ( G ) ] | 5 when | [ u , V 2 ( G ) ] | 1 and | [ u , V 3 ( G ) ] | 2 ,
(4) | [ u , V 5 ( G ) ] | = 0 when | [ V 2 ( G ) , u ] | 1 , | [ V 3 ( G ) , u ] | 2 and | [ u , V 4 ( G ) ] | = 5 .
Proof. 
Suppose to the contrary that there is a vertex u V 12 ( G ) , a vertex v 1 V 5 ( G ) , a vertex v 2 V 2 ( G ) , a vertex v 3 V 3 ( G ) , and three vertices v 4 , v 5 , v 6 V 4 ( G ) such that u v 1 [ u , V 5 ( G ) ] , u v 2 [ u , V 2 ( G ) ] , u v 3 [ u , V 3 ( G ) ] , and u v i [ u , V 4 ( G ) ] for i = 4 , 5 , 6 . Let G = G { u v i | i = 1 , , 6 } . Then, G has a tnsd k-coloring ϕ . We first erase the colors on u and v i with 1 i 6 from ϕ . Without loss of generality, set d G ( v 1 ) = 5 . Note that v 2 is a 2 -vertex, v 3 is a 3 -vertex, and v i is a 4 -vertex with 4 i 6 . Then,
| S ( u ) | k 2 × ( 12 6 ) 3 , | S ( v 1 ) | k 2 × ( 5 1 ) 7 , | S ( u v 1 ) | k ( 12 6 ) ( 5 1 ) 5 , | S ( u v 2 ) | k ( 12 6 ) ( 2 1 ) 8 , | S ( u v 3 ) | k ( 12 6 ) ( 3 1 ) 7 and | S ( u v i ) | k ( 12 6 ) ( 4 1 ) 6 for i = 4 , 5 , 6 .
Let ϕ be a map on T ( G ) \ { v 2 , , v 6 } obtained from ϕ by assigning
ϕ ( u ) = x 1 , ϕ ( v 1 ) = x 2 , ϕ ( u v 1 ) = y 1 , , ϕ ( u v 6 ) = y 6
and ϕ ( z ) = ϕ ( z ) for every z T ( G ) \ { u , v 1 , , v 6 , u v 1 , , u v 6 } . Set
P ( x 1 , x 2 , y 1 , , y 6 ) = z N G ( u ) \ { v 1 , , v 6 } m ϕ ( u ) m ϕ ( z ) z N G ( v 1 ) \ { u } m ϕ ( v 1 ) m ϕ ( z ) · m ϕ ( u ) m ϕ ( v 1 ) ( x 1 x 2 ) ( x 2 y 1 ) · i = 1 6 ( x 1 y i ) 1 i < j 6 ( y i y j ) .
With the help of MATHEMATICA, we have c P ( x 1 2 x 2 6 y 1 4 y 2 7 y 3 6 y 4 5 y 5 4 ) = 4 . By Theorem 4, there is a color c 1 S ( u ) , a color c 2 S ( v 1 ) , and a color c i + 2 S ( u v i ) with 1 i 6 such that P ( c 1 , , c 8 ) 0 . Since v i is a 4 -vertex with 2 i 6 , we can obtain a tnsd k-coloring of G from the ϕ by recoloring u , v 1 with colors c 1 , c 2 and coloring u v 1 , , u v 6 with colors c 3 , , c 8 , respectively. This is a contradiction. □
Claim 8.
Let C 3 = v 1 v 2 v 3 v 1 be a 3-cycle in G. Then, ( d G ( v 1 ) , d G ( v 2 ) , d G ( v 3 ) ) ( 6 , 6 , 7 ) .
The following Equation (1) follows from Claims 3∼7.
n G 5 ( u ) = 0 if d G ( u ) 6 ; n G 5 ( u ) 1 if d G ( u ) = 7 ; n G 4 ( u ) d G ( u ) 7 if 8 d G ( u ) 9 ; n G 3 ( u ) d G ( u ) 7 if 10 d G ( u ) 11 ; n G 2 ( u ) 5 if d G ( u ) = 12 .
Let H = G V 2 ( G ) . Then, d H ( u ) = d G ( u ) n G 2 ( u ) for each u V ( H ) .
Claim 9.
If u V ( H ) , then,
(1)
d H ( u ) 3 ,
(2)
d H ( u ) = d G ( u ) if 3 d G ( u ) 7 ,
(3)
d H ( u ) 7 if 8 d G ( u ) 12 ,
(4)
n H 5 ( u ) = 0 if d H ( u ) = 7 and 8 d G ( u ) 12 ,
(5)
n H 5 ( u ) = 0 if d H ( u ) 6 .
Proof. 
Note that 3 d G ( u ) 12 by the definition of H and Claim 2 for every u V ( H ) . By the definition of H, (1), (2), (3), and (4) follow from Equation (1). Below, we prove (5).
(5) Suppose that n H 5 ( u ) 1 when d H ( u ) 6 . Without loss of generosity, set d H ( u ) = 6 . Then, d G ( u ) = 6 by (2). Thus, n G 5 ( u ) = 0 by Claim 3, and u has at least one 7 + -neighbor v in G with d H ( v ) 5 by (2). Hence, d H ( v ) = d G ( v ) n G 2 ( v ) 7 by (3) and the definition of H, which is a contradiction. Therefore, the statement (5) holds. □
By a similar discussion to that of Claim 9, we can obtain the following Claims 10∼12.
Claim 10.
If d H ( u ) = 7 , then n H 5 ( u ) 1 .
Claim 11.
If d H ( u ) = with 8 9 , then,
(1)
n H 4 ( u ) 7 ,
(2)
n H 5 ( u ) 7 when n H 4 ( u ) 1 .
Claim 12.
If d H ( u ) = with 10 11 , then,
(1)
n H 3 ( u ) 7 ,
(2)
n H 4 ( u ) 7 when n H 3 ( u ) 1 ,
(3)
n H 5 ( u ) = 0 when n H 3 ( u ) 1 and n H 4 ( u ) = 7 .
By Claims 8 and 9, the below claim is immediate.
Claim 13.
Let C 3 = v 1 v 2 v 3 v 1 be a 3-cycle in H. Then,
(1)
( d H ( v 1 ) , d H ( v 2 ) , d H ( v 3 ) ) ( 6 , 6 , 6 ) ,
(2)
n H 5 ( v 3 ) = 0 when ( d H ( v 1 ) , d H ( v 2 ) , d H ( v 3 ) ) = ( 6 , 6 , 7 ) .
From now on, we always assume that H has been embedded on the plane such that the number of crossings is as small as possible. Let H × be the associated planar graph of H and u v E ( H × ) . If u v is incident with two 3-faces and u is a real -vertex in H × , we call u a bad-neighbor of v. If u v is incident with exactly one 3-face and u is a real 3-vertex in H × , we call u a special 3-neighbor of v. Let n H × b ( v ) ( n H × 3 s ( v ) ) denote the number of bad -neighbors (special 3-neighbors) of v in H × . Unless stated otherwise, an -vertex represents a real -vertex in the following. A 4-face is called a good 4-face if it is incident at most a real 5 -vertex and a bad 4-face otherwise. By Claims 1 and 9, the below Claims 14 and 15 are immediate.
Claim 14.
If d H ( u ) = 12 , then,
(1)
n H × 3 b ( u ) 4 , n H × 3 s ( u ) 6 , n H × 4 b ( u ) 6 and n H × 5 b ( u ) 6 ,
(2)
n H × 3 s ( u ) + n H × 4 b ( u ) + n H × 5 b ( u ) = 0 when n H × 3 b ( u ) = 4 .
Claim 15.
Every k-face is incident with at most k 2 real 5 -vertices in H × .
Let γ ( H × ) denote the number of connected components of H × . Note that each plane graph has exactly one unbounded face. Then, we have
f F ( H × ) d H × ( f ) = 2 | E ( H × ) |
and Euler’s formula
| V ( H × ) | | E ( H × ) | + | F ( H × ) | = 1 + γ ( H × ) .
Below, we will apply the discharging method [13] on the planar graph H × to complete the proof of Theorem 3. Set ω ( v ) = d H × ( v ) 6 for any v V ( H × ) and ω ( f ) = 2 d H × ( f ) 6 for any f F ( H × ) . By Euler’s formula,
| V ( H × ) | | E ( H × ) | + | F ( H × ) | = 1 + γ ( H × )
and
v V ( H × ) d H × ( v ) = f F ( H × ) d H × ( f ) = 2 | E ( H × ) | .
One can obtain
v V ( H × ) ( d H × ( v ) 6 ) + f F ( H × ) ( 2 d H × ( f ) 6 ) = 6 ( 1 + γ ( H × ) ) .
In order to redistribute charges on T ( H × ) and keep the total charges unchanged, we make some discharging rules as follows.
(R1) 
Let u v be an edge of H × .
(R1.1) 
If u is a bad 3-neighbor of v, u receives 1 from v.
(R1.2) 
If u is a special 3-neighbor of v, u receives 1 2 from v.
(R1.3) 
If u is a bad 4-neighbor of v, u receives 1 2 from v.
(R1.4) 
If u is a bad 5-neighbor of v, u receives 1 5 from v.
(R2) 
Each real -vertex with 3 5 receives 1 from every 4 + -face incident with it.
(R3) 
Let u v be an edge and u be a false 4-vertex of H × .
(R3.1) 
If d H × ( v ) = 7 , u receives 1 when v is not adjacent to any real 5 -vertex and 4 5 otherwise from v.
(R3.2) 
If d H × ( v ) = with 8 12 , u receives 2 when n H × 3 ( v ) < n H 3 ( v ) and 1 otherwise from v.
(R3.3) 
If f is a good 4-face or 5 + -face incident with u, u receives 1 from f.
For each z V ( H × ) F ( H × ) , we use ω ( z ) to represent the new charge after applying the rules. Then, one can obtain
z V ( H × ) F ( H × ) ω ( z ) = v V ( H × ) ( d H × ( v ) 6 ) + f F ( H × ) ( 2 d H × ( f ) 6 ) = 6 ( 1 + γ ( H × ) ) < 0 .
Thus, there exists a vertex or a face whose charge is negative.
Firstly, it is easy to verify that the new charge of every face is non-negative by Claim 15 and the discharging rules.
Next, we show that the new charge of every real vertex is non-negative. Pick arbitrarily a real vertex v from V ( H ) . Note that v is adjacent to at most a false 4-vertex as G (and thus, H) is an IC-planar graph. Note also that 3 d H ( v ) = d H × ( v ) 12 and n H × 3 ( v ) n H 3 ( v ) by the definition of H × and Claim 9.
( 1 ) For v V 3 ( H ) . If v is incident with three 4 + -faces, then ω ( v ) = 3 6 + 3 × 1 = 0 by (R2). If v is incident with two 4 + -faces, it is a common special 3-neighbor of two neighbors, and so ω ( v ) = 3 6 + 2 × 1 + 2 × 1 2 = 0 by (R1.2) and (R2). If v is incident with one 4 + -face, then it is a special 3-neighbor of one neighbor and a common bad 3-neighbor of the other two neighbors, respectively, and so ω ( v ) = 3 6 + 1 × 1 + 1 × 1 + 2 × 1 2 = 0 by (R1.1), (R1.2), and (R2). If v is not incident with any 4 + -faces, it is a common bad 3-neighbor of three neighbors; so, ω ( v ) = 3 6 + 3 × 1 = 0 by (R1.1).
( 2 ) For v V 4 ( H ) . If v is incident with at least two 4 + -faces, ω ( v ) 4 6 + 2 × 1 = 0 by (R2). If v is incident with one 4 + -face, it is a common bad 4-neighbor of two neighbors; so, ω ( v ) = 4 6 + 1 × 1 + 2 × 1 2 = 0 by (R1.3) and (R2). If v is not incident with any 4 + -faces, it is a common bad 4-neighbor of four neighbors; so, ω ( v ) = 4 6 + 4 × 1 2 = 0 by (R1.3).
( 3 ) For v V 5 ( H ) . If v is incident with at least one 4 + -faces, ω ( v ) 5 6 + 1 × 1 = 0 by (R2). If v is not incident with any 4 + -faces, it is a common bad 5-neighbor of five neighbors; so, ω ( v ) = 5 6 + 5 × 1 5 = 0 by (R1.4).
( 4 ) For v V 6 ( H ) . ω ( v ) = 6 6 = 0 as no rule is applied to it.
( 5 ) For v V 7 ( H ) . Note that v is adjacent to at most a real 5-vertex in H (and so, in H × ) by Claim 10. If v is not adjacent to any real 5-vertex in H × , ω ( v ) 7 6 1 × 1 = 0 by (R3.1). If v is adjacent to a real 5-vertex in H × , ω ( v ) 7 6 1 × 4 5 1 × 1 5 = 0 by (R1.4) and (R3.1).
( 6 ) For v V ( H ) with 8 9 . If n H × 3 ( v ) = n H 3 ( v ) , ω ( v ) 6 1 × 1 max { ( 7 ) × 1 , 2 × 1 5 } = 0 by (R1), (R3.2), and Claim 11. If n H × 3 ( v ) < n H 3 ( v ) , n H 3 ( v ) 1 ; so, v is adjacent to at most ( 8 ) real 5 -vertices in H × by Claim 11 and the definition of H × . Thus, ω ( v ) 6 1 × 2 ( 8 ) × 1 = 0 by (R1) and (R3.2).
( 7 ) For v V ( H ) with 10 11 . If n H × 3 ( v ) = n H 3 ( v ) , ω ( v ) 6 1 × 1 max { ( 7 ) × 1 , 2 × 1 2 } = 0 by (R1), (R3.2), and Claim 12. If n H × 3 ( v ) < n H 3 ( v ) , n H 3 ( v ) 1 ; so, v is adjacent to at most ( 8 ) real 4 -vertices in H × by Claim 12 and the definition of H × . Thus, ω ( v ) 6 1 × 2 max { ( 8 ) × 1 , 2 × 1 5 } = 0 by (R1) and (R3.2).
( 8 ) For v V 12 ( H ) . By Claim 14, ω ( v ) 12 6 1 × 2 max { 4 × 1 , 2 × 1 2 } = 0 by (R1) and (R3.3).
Hence, the new charge of every real vertex is non-negative.
Finally, we prove the new charge of every false 4-vertex is non-negative. Choose arbitrarily a false 4-vertex u from V ( H × ) \ V ( H ) . Note that u is adjacent to at most two 5 -vertices in H × by the definition of H × and Claim 9. Assume that the edges v 1 v 2 and v 3 v 4 intersect at u in H. Then, N H × ( u ) = { v 1 , v 2 , v 3 , v 4 } . The subgraph of H × induced by N H × ( u ) { u } is denoted by H × [ N H × ( u ) { u } ] . Note that if v i ( v i + 2 ) is a 5 -vertex with i = 1 , 2 , v i + 2 ( v i ) is a 7 + -vertex and v i + 2 ( v i ) is not adjacent to any real 5 -vertex when v i + 2 ( v i ) is a 7-vertex in H × by the definition of H × and Claim 9.
( 9 ) Let H × [ N H × ( u ) { u } ] contain no 3-face (see Figure 3). Then, u is incident with at least three faces, each of which either is a good 4-face or a 5 + -face since n H × 5 ( u ) 2 by Claim 9; so, ω ( u ) 4 6 + 3 × 1 = 1 by (R3.3).
( 10 ) Let H × [ N H × ( u ) { u } ] contain exactly one 3-face (see Figure 4). Note that n H × 5 ( u ) 2 and at most one of d H × ( v 1 ) and d H × ( v 2 ) is equal to 3 by Claim 9. If d H × ( v 1 ) 3 and d H × ( v 2 ) 3 , ω ( u ) 4 6 + 2 × 1 = 0 by (R3.3) since u is incident with at least two faces, each of which either is a good 4-face or a 5 + -face. If d H × ( v 1 ) is equal to 3, ω ( u ) 4 6 + 2 × 1 + 1 × 1 1 × 1 2 = 1 2 by (R1.2) and (R3) since u is incident with at least two faces, each of which either is a good 4-face or a 5 + -face. By symmetry, ω ( u ) 1 2 when d H × ( v 2 ) is equal to 3.
( 11 ) Let H × [ N H × ( u ) { u } ] contain two 3-faces (see Figure 5). Note that n H × 5 ( u ) 2 by Claim 9 and v 1 of (b) in Figure 5 is not a 3-vertex by Lemma 1. If n H × 5 ( u ) = 0 , ω ( u ) 4 6 + 2 × 1 = 0 by (R3.3) since u is incident with two faces, each of which either is a good 4-face or a 5 + -face. If n H × 5 ( u ) = 1 , ω ( u ) 4 6 + 2 × 1 + 1 × 1 1 × 1 2 = 1 2 by (R1.2) and (R3.3), since u is incident with two faces, each of which either is a good 4-face or a 5 + -face. If n H × 5 ( u ) = 2 , ω ( u ) 4 6 + 1 × 1 + 2 × 1 2 × 1 2 = 0 by (R1.2) and (R3.3) since u is incident with one face, which either is a good 4-face or a 5 + -face.
( 12 ) Let H × [ N H × ( u ) { u } ] contain three 3-faces (see Figure 6). Note that n H × 5 ( u ) 2 by Claim 9 and neither v 1 nor v 2 is a 3-vertex by Lemma 1 in Figure 6. If n H × 5 ( u ) = 0 , at least one neighbor of u is a 7 + -vertex, and when one neighbor of u is 7-vertex, it is not adjacent to any real 5 -vertex by Claim 13; so, ω ( u ) 4 6 + 1 × 1 + 1 × 1 = 0 by (R1) and (R3.3) since u is incident with one face, which either is a good 4-face or a 5 + -face. If n H × 5 ( u ) = 1 and n H × 3 ( u ) = 0 , u is adjacent to at least two 7 + -vertices; so, ω ( u ) 4 6 + 1 × 1 + 2 × 4 5 1 × 1 2 = 1 10 by (R1.3), (R1.4), and (R3.3) since u is incident with one face, which either is a good 4-face or a 5 + -face. If n H × 5 ( u ) = 1 and n H × 3 ( u ) = 1 , d H × ( v 3 ) = 3 or d H × ( v 4 ) = 3 . Since v 3 and v 4 are symmetrical in Figure 6, we can set d H × ( v 3 ) = 3 . Then, d H × ( v 1 ) 8 , d H × ( v 2 ) 8 , and n H × 3 ( v 1 ) < n H 3 ( v 1 ) ; so, ω ( u ) 4 6 + 1 × 1 + 1 × 2 1 × 1 2 = 1 2 by (R1.2), (R3.2), and (R3.3) since u is incident with one face, which either is a good 4-face or a 5 + -face. If n H × 5 ( u ) = 2 , d H × ( v 3 ) 5 , d H × ( v 4 ) 5 , d H × ( v 1 ) 8 , and d H × ( v 2 ) 8 by Claims 9 and 10. If n H × 3 ( u ) = 0 , ω ( u ) 4 6 + 2 × 1 = 0 by (R3.2). If n H × 3 ( u ) = 1 , then we may as well set d H × ( v 3 ) = 3 since v 3 and v 4 are symmetrical in Figure 6. Thus, n H × 3 ( v 1 ) < n H 3 ( v 1 ) ; so, ω ( u ) 4 6 + 1 × 1 + 1 × 2 1 × 1 2 = 1 2 by (R1.2) and (R3.2). If n H × 3 ( u ) = 2 , d H × ( v 3 ) = d H × ( v 4 ) = 3 . Thus, n H × 3 ( v 1 ) < n H 3 ( v 1 ) and n H × 3 ( v 2 ) < n H 3 ( v 2 ) . So, ω ( u ) 4 6 + 2 × 2 2 × 1 2 = 1 by (R1.2) and (R3.2).
( 13 ) Let H × [ N H × ( u ) { u } ] contain four 3-faces (see Figure 7). Then, d H × ( v i ) 3 for i = 1 , 2 , 3 , 4 by Lemma 1. Note that n H × 5 ( u ) 1 by Claim 9. If n H × 5 ( u ) = 0 and n H × 8 + ( u ) = 0 , then u is adjacent to at least two 7-vertices by Claim 13. When u is adjacent to two 7-vertices, each of the two 7-vertices is not adjacent to any real 5 -vertex by Claim 13, and so ω ( u ) 4 6 + 2 × 1 = 0 by (R3.1). When u is adjacent to at least three 7-vertices, ω ( u ) 4 6 + 3 × 4 5 = 2 5 by (R3.1). If n H × 5 ( u ) = 0 and n H × 8 + ( u ) = 1 , we may as well set d H × ( v 1 ) 8 by symmetry. Then, at least one of v 2 , v 3 and v 4 is a 7-vertex by Claim 13. When one of v 2 , v 3 and v 4 is a 7-vertex, this 7-vertex is not adjacent to any real 5 -vertex by Claim 13, and so ω ( u ) 4 6 + 1 × 1 + 1 × 1 = 0 by (R3.1) and (R3.2). When at least two of v 2 , v 3 and v 4 are 7-vertices, ω ( u ) 4 6 + 2 × 4 5 + 1 × 1 = 3 5 by (R3.1) and (R3.2). If n H × 5 ( u ) = 0 and n H × 8 + ( u ) 2 , ω ( u ) 4 6 + 2 × 1 = 0 by (R3.2). If n H × 5 ( u ) = 1 and n H × 4 ( u ) = 1 , u is adjacent to three 8 + -vertices by Claim 9; so, ω ( u ) 4 6 + 3 × 1 1 2 = 1 2 by (R1.3) and (R3.2). If n H × 5 ( u ) = 1 and n H × 4 ( u ) = 0 , u is adjacent to three 7 + -vertices by Claim 9; so, ω ( u ) 4 6 + 3 × 4 5 1 5 = 1 5 by (R1.4), (R3.1), and (R3.2).
Hence, the new charge of every false 4-vertex is non-negative.
By the above analysis, we obtain that the new charge of every member in V ( H × ) F ( H × ) is non-negative. This is a contradiction. The proof of Theorem 3 is completed.

4. Conclusions

In this paper, we proved that any IC-planar graph with Δ 12 satisfies the tnsd index conjecture by the discharging method and Combinatorial Nullstellensatz. Whether the bound Δ 12 can be improved is still open for future work. The difficulty in improving this bound is how to reduce the number of bad 3-neighbors of one vertex. Therefore, finding suitable technology to reduce the number of bad 3-neighbors of one vertex is our main future work in the study of the tnsd index of IC-planar graph.

Author Contributions

Methodology, D.Z.; software, F.C.; formal analysis, D.Z.; writing—original draft preparation, D.Z.; writing—review and editing, C.L. All authors have read and agreed to the published version of the manuscript.

Funding

This research was funded by Shangluo Science and Technology Plan Project (No. SK2017-40).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

The authors are grateful to all anonymous reviewers and the editor for their inspiring and constructive comments and suggestions.

Conflicts of Interest

The authors declare no conflict of interest.

References

  1. Pilśniak, M.; Woźniak, M. On the total-neighbor-distinguishing index by sums. Graphs Comb. 2015, 31, 771–782. [Google Scholar] [CrossRef] [Green Version]
  2. Xu, C.; Ge, S.; Li, J. Neighbor sum distinguishing total chromatic number of 2-degenerate graphs. Discret. Appl. Math. 2018, 251, 349–352. [Google Scholar] [CrossRef]
  3. Li, H.; Liu, B.; Wang, G. Neighbor sum distinguishing total coloring of K4-minor-free graphs. Front. Math. China 2013, 8, 1351–1366. [Google Scholar] [CrossRef]
  4. Yang, D.; Sun, L.; Yu, X.; Wu, J.; Zhou, S. Neighbor sum distinguishing total chromatic number of planar graphs with maximum degree 10. Appl. Math. Comput. 2017, 314, 456–468. [Google Scholar] [CrossRef]
  5. Wang, J.; Ma, Q.; Han, X. Neighbor sum distinguishing total colorings of triangle free planar graphs. Acta Math. Sin. (Engl. Ser.) 2015, 2, 216–224. [Google Scholar] [CrossRef]
  6. Albertson, M.O. Chromatic number, independent ratio, and crossing number. Ars Math. Contemp. 2008, 1, 1–6. [Google Scholar] [CrossRef] [Green Version]
  7. Song, W.; Duan, Y.; Miao, L. Neighbor Sum Distinguishing Total Coloring of Triangle Free IC-planar Graphs. Acta Math. Sin. (Engl. Ser.) 2020, 36, 292–304. [Google Scholar] [CrossRef]
  8. Song, W.; Duan, Y.; Miao, L. Neighbor sum distinguishing total coloring of IC-planar graphs. Discret. Math. 2020, 343, 111918. [Google Scholar] [CrossRef]
  9. Song, C.; Xu, C. Neighbor sum distinguishing total colorings of IC-planar graphs with maximum degree 13. J. Comb. Optim. 2020, 39, 293–303. [Google Scholar] [CrossRef]
  10. Zhang, D. Neighbor Sum Distinguishing Total Choosability of IC-Planar Graphs without Theta Graphs Θ2,1,2. Mathematics 2021, 9, 708. [Google Scholar] [CrossRef]
  11. Alon, N. Combinatorial nullstellensatz. Combin. Probab. Comput. 1999, 8, 7–29. [Google Scholar] [CrossRef] [Green Version]
  12. Sun, L.; Yu, G.; Li, X. Neighbor sum distinguishing total choosability of 1-planar graphs with maximum degree at least 24. Discrete Math. 2021, 344, 112190. [Google Scholar] [CrossRef]
  13. Daniel, W.; Douglas, B. An introduction to the discharging method via graph coloring. Discrete Math. 2017, 340, 766–793. [Google Scholar]
Figure 1. No configuration in G.
Figure 1. No configuration in G.
Symmetry 13 01787 g001
Figure 2. No configuration with colors in G.
Figure 2. No configuration with colors in G.
Symmetry 13 01787 g002
Figure 3. H × [ { u } N H × ( u ) ] contains no 3-face.
Figure 3. H × [ { u } N H × ( u ) ] contains no 3-face.
Symmetry 13 01787 g003
Figure 4. H × [ { u } N H × ( u ) ] contains one 3-face.
Figure 4. H × [ { u } N H × ( u ) ] contains one 3-face.
Symmetry 13 01787 g004
Figure 5. H × [ { u } N H × ( u ) ] contains two 3-faces; (a) two nonadjacent 3-faces; (b) two adjacent 3-faces.
Figure 5. H × [ { u } N H × ( u ) ] contains two 3-faces; (a) two nonadjacent 3-faces; (b) two adjacent 3-faces.
Symmetry 13 01787 g005
Figure 6. H × [ { u } N H × ( u ) ] contains three 3-faces.
Figure 6. H × [ { u } N H × ( u ) ] contains three 3-faces.
Symmetry 13 01787 g006
Figure 7. H × [ { u } N H × ( u ) ] contains four 3-faces.
Figure 7. H × [ { u } N H × ( u ) ] contains four 3-faces.
Symmetry 13 01787 g007
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Zhang, D.; Li, C.; Chao, F. On the Total Neighbor Sum Distinguishing Index of IC-Planar Graphs. Symmetry 2021, 13, 1787. https://doi.org/10.3390/sym13101787

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Zhang D, Li C, Chao F. On the Total Neighbor Sum Distinguishing Index of IC-Planar Graphs. Symmetry. 2021; 13(10):1787. https://doi.org/10.3390/sym13101787

Chicago/Turabian Style

Zhang, Donghan, Chao Li, and Fugang Chao. 2021. "On the Total Neighbor Sum Distinguishing Index of IC-Planar Graphs" Symmetry 13, no. 10: 1787. https://doi.org/10.3390/sym13101787

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