Equitable Coloring of IC-Planar Graphs with Girth g ≥ 7

: An equitable k -coloring of a graph G is a proper vertex coloring such that the size of any two color classes differ at most 1. If there is an equitable k -coloring of G , then the graph G is said to be equitably k -colorable. A 1-planar graph is a graph that can be embedded in the Euclidean plane such that each edge can be crossed by other edges at most once. An IC-planar graph is a 1-planar graph with distinct end vertices of any two crossings. In this paper, we will prove that every IC-planar graph with girth g ≥ 7 is equitably ∆ ( G ) -colorable, where ∆ ( G ) is the maximum degree of G .


Introduction
Chromatic graph theory originates from the Four Color Conjecture [1], is a focal problem in graph theory.As a special case of chromatic graph theory, equitable coloring has been widely used in industrial production, enterprise management and biology [2].Especially, it plays a crucial role in the study of schedule [3], partition [4][5][6][7] and load balancing [8].In this paper, we mainly discuss the equitable coloring of IC-planar graphs with girth g ≥ 7. Some relevant definitions are as follows.
Only undirected, finite and simple graphs are considered in this paper.In a graph G, V(G), E(G), |G|, δ(G) and ∆(G) (or ∆ in short) are used to denote the vertex set, edge set, order, minimum degree and maximum degree of G, respectively.The girth of a graph G is the length of shortest cycles of G, denoted by g(G).For v ∈ V(G), we use d G (v) to denote the degree of a vertex v in G.The set of all neighbors of v is denoted by N G (v).
A graph that can be drawn in the Euclidean plane such that any two edges intersect only at their ends is called the planar graph.A planar graph with such a particular drawing is called a plane graph.The face set of a plane graph G is denoted by F(G).For f ∈ F(G), we use d G ( f ) to denote the degree of a face f in G.That is, the number of edges on the boundary of f .We denote are the vertices on the boundary of f and arranged in clockwise order.In a graph G, a vertex is called as a k-vertex, a k − -vertex or a k + -vertex if its degree is equal to k, at most k or at least k, respectively.Similarly, we can define a k-face, a k − -face or a k + -face.For a k-face f ∈ F(G), we denote A 1-planar graph is a graph that can be embedded in the Euclidean plane such that each edge can be crossed by other edges at most once.An IC-planar graph is a 1-planar graph with distinct end vertices of any two crossings, which is introduced by Alberson [9] in 2008.
A proper k-coloring of a graph G is a mapping φ: V(G) → {1, 2, • • • , k} such that φ(u) = φ(v) for any two adjacent vertices u and v.The minimum positive integer k such that G has a proper k-coloring is called the chromatic number of G, denoted by χ(G).And we use V i (1 ≤ i ≤ k) to denote the set of vertices colored with i.Notice that φ is an equitable k-coloring of G, or G is equitably k-colorable.The minimum positive integer k such that G has an equitable k-coloring is called the equitable chromatic number of G, denoted by χ e (G).
Obviously, χ e (G) ≥ χ(G), and the inequality can be strictly held.The rest of this article is organized as follows: In Section 2, we will introduce the history and recent progress on equitable coloring, and we also present our motivation and contribution of this paper.In Section 3, we provide some lemmas which are helpful to prove our main theorem.In Section 4, we discuss the structures of IC-planar graphs with girth g ≥ 7 and get an important property, which is presented in Lemma 5.In Section 5, we use discharging method to prove Lemma 5, which is used to prove Lemma 6 and Theorem 1.In Section 6, to show Theorem 1, we first give proof of Lemma 6 and then get a corollary.Finally, we summarize our result and get the main theorem.

Related Work
In 1973, Meyer [10] introduced the concept of equitable coloring.At the same time, he proposed the following conjecture.
In 1964, Erdős [11] conjectured that every graph is equitably k-colorable for any k ≥ ∆ + 1, which was confirmed by Hajnal and Szemerédi [12] in 1970.In 2010, by applying algorithm analysis, Kierstead et al. [13] gave a new and short proof of Erdős's conjecture.So we are interested in the sufficient conditions for graphs to be equitably ∆-colorable.In 1994, Chen, Lih and Wu [14] put forth the following conjecture.
Similar with planar graphs, we also want to see whether Conjecture 2 holds for ICplanar graphs with girth limitation.In this paper, we will concern on the equitable coloring of IC-planar graphs with large girth g and will prove the result in the following: Theorem 1.Every IC-planar graph with girth g ≥ 7 is equitably ∆-colorable.

Properties of IC-Planar Graphs with Girth g ≥ 7
Every IC-planar graph with girth g ≥ 7 in the following is assumed to be embedded in the Euclidean plane with the number of crossings as small as possible.The set of all crossing vertices and the edges with no crossings are denoted by C(G) and E 0 (G), respectively.Let E 1 (G) = {xz, zy| xy ∈ E(G) \ E 0 (G), and z is the crossing vertex on edge xy}.The associated plane graph of G, denoted by G × , is a plane graph such that V(G × ) = V(G) ∪ C(G), and with at least one false vertex; otherwise, f is true.Let f be a false i-face and v be a true vertex, where i ∈ {3, 4}, f is called the pendant false i-face of v if v is not incident with f but is adjacent to the false vertex incident with f .Finally, we use f i (v), n i (G) and n G i (v) to denote the number of i-faces incident with the vertex v in the associated plane graph G × , the number of i-vertices in the graph G and the number of i-vertices adjacent to the vertex v in the graph G, respectively.
It is easy to check that the following two properties hold by the fact that G is the IC-planar graph with girth g ≥ 7.
Fact 1 (1) Any two false vertices are not adjacent.
(2) A true vertex is adjacent to at most one false vertex.Fact 2 A true vertex is incident with at most one false 3-face.
Next, let's introduce some facts, which is also involved in the proofs of the next lemma.Since the IC-planar graph has been embedded with minimal number of crossings, the following fact holds.
Fact 3 A false 3-face is not incident with any 2-vertex.Suppose that v is a false vertex.Let v i (1 ≤ i ≤ 4) be the vertices adjacent to v in G × and arranged in clockwise order.Let f i = [v i+1 vv i • • • ] be the false face with vv i and vv i+1 as boundary edges, where 1 ≤ i ≤ 4 and , where i ∈ {2, 3, 4}.Hence, v is incident with at most one false 3-face.
Next we will show that d G × ( f 3 ) ≥ 6. Assume to the contrary that Hence, v is incident with at most one false 4-face.
Next we will show that Lemma 5. Let G be a connected IC-planar graph with g ≥ 7 and |G| ≥ 6, then G contains one of the following configurations H .
Remark 1.Each configuration depicted above is such that: (1) the vertices which are labelled as x k , x k−1 , x k−2 in every configuration are different while the other vertices may be not distinct; (2) the degree of solid vertices are exactly shown; (3) the degree of hollow vertices may be larger than or equal to the degree shown in the figure, except for specially pointed.

Proof of Lemma 5
On the contrary, assume that Lemma 5 is not true.Let G be a counterexample.Then G is a connected IC-planar graph with g ≥ 7 and |G| ≥ 6, but without configurations H 1 ∼H 9 .Now we consider G × , the associated plane graph of G.
In the following, we use Euler's formula and discharging analysis on G × to derive a contradiction.We define the initial weight function w such that w Next, we will design the appropriate discharging rules to redistribute the weights on V(G × ) ∪ F(G × ) depending on the value of δ(G × ).Once the discharging process is finished, a new weight function w (x) is produced while the total sum of weights is kept fixed.Then we will show that w(x) to derive the contradiction.For x, y ∈ V(G × ) ∪ F(G × ), let τ(x → y) be the weight transferred from x to y.By Lemma 4, we have δ(G) ≤ 3.So δ(G × ) ≤ 3. Thus, the proof will be divided into the following cases depending on the value of δ(G × ).

Case 1. δ(G
Our discharging rules are defined as follows.R1.Suppose that v is a false 4-vertex and f is the face incident with v.
2 .Next, we are going to show that for each element be the vertices adjacent to v in G × and arranged in clockwise order.Let f i be the face with vv i and vv i+1 as boundary edges, where 1 then by Fact 1, Fact 4 and Fact 5, v has at most one pendant false i-face, i ∈ {3, 4}.If v has one pendant false 3-face f , say v 1 is the false vertex incident with f , then min{d Suppose that k = 3.Then w( f ) = −3, and τ(u Suppose that k = 4. Then w( f ) = −2, and τ(u w (x) ≥ 0, which is a contradiction.
Hence, the counterexample graph G does not exist and Lemma 5 holds.

Case 2. δ(G
In Case 2, the discharging rules are the same as those in Case 1.With the same arguments as Case 1, we can prove that w (x) Proof.By g ≥ 7, we have f is a false face.By Fact 3, we have 4 , which is a contradiction.Hence, the counterexample graph G does not exist and Lemma 5 holds.

Case 3. δ(G
Since G does not contain H 2 and H 3 as subgraphs, G satisfies the following properties.Claim 3.1 2-vertex is not adjacent to any 2-vertex.Claim 3.2 Any 3 + -vertex is adjacent to at most one 2-vertex.Proof.Suppose that v is a special 4-vertex and v i (1 ≤ i ≤ 4) are the neighbors of v in G × and arranged in clockwise order.Let f i be the face with vv i and vv i+1 as boundary edges, where 1 ≤ i ≤ 4 and v 5 = v 1 .
Without loss of generality, assume that where t ≥ 0. If t = 0, then y 2 v 4 ∈ E(G × ).Since y 2 is a false vertex, we have y 1 v 4 ∈ E(G).Thus, G contains a 4-cycle vv 2 y 1 v 4 v, a contradiction.If t = 1, then y 2 z 1 ∈ E(G × ), and z 1 is a true vertex.Since y 2 is a false vertex, we have y 1 z 1 ∈ E(G).Thus, G contains a 5-cycle Suppose that x is a bad 4-vertex.By Claim Corollary 1.If f is a false 4-face incident with a poor 4-vertex, then f is incident with at least two true 4 + -vertices.

Proof. Suppose that
In Subcase 4.2, the discharging rules are the same as those in Case 3. By Claim 3.3, there is at most one special 2-vertex in G × .With the same arguments as Case 3, we can prove that w (x) ≥ 0 for any x ∈ V(G × ) ∪ F(G × ) − V , where V is the set of 1-vertex and special 2-vertex.Note that w (u) = w(u) = −4 for the 1-vertex u and w (v) ≥ −2 for the special 2-vertex v. Thus, −12 Since G does not contain H 9 as a subgraph, G contains exactly two 1-vertices and no 2-vertex.In Subcase 4.3, the discharging rules are the same as those in Case 1.With the same arguments as Case 1, we can prove that w (x) ≥ 0 for any x ∈ V(G × ) ∪ F(G × ) − V , where V is the set of 1-vertices.Note that w (u) = w(u) = −4 for each 1-vertex u.

Proof of Theorem 1
Before we prove Theorem 1, it is necessary to present the following result.
Proof.Assume to the contrary that G is a counterexample with fewest vertices of Lemma 6.If the order of each component of G is at most five, then ∆ ≤ 4. Thus, k ≥ max{∆, 5} ≥ ∆ + 1, and G is equitably k-colorable by Lemma 2. Otherwise, there is a component of G with order at least six.By Lemma 5, G contains one of the configurations H 1 ∼H 9 in Lemma 6.By Lemma 1, we need to choose the subset S, and define the subset S as follows.If G contains configurations H i , i ∈ {1, 4, 5, 7, 8}, then S = {x k , x k−1 , x k−2 , x k−3 , x 1 }.If G contains configurations H i , i ∈ {3, 6}, then S = {x k , x k−1 , x k−2 , x 2 , x 1 }.If G contains configurations H i , i ∈ {2, 9}, then S = {x k , x k−1 , x k−2 , x 1 }.By Lemma 4, G is 3-degenerate, thus the remaining unspecified vertices in S can be remarked by choosing the minimum degree vertex in a graph obtained from G by deleting the remarked vertices at each step from highest to lowest indices.It is easy to confirm that for every 1 ≤ i ≤ k, the subset S defined above satisfies that | N G (x i ) − S | ≤ k − i.
Let H = G − S ⊆ G. Then ∆(H) ≤ ∆.If ∆(H) < ∆, then H is equitably k-colorable by Lemma 2. Otherwise, H is equitably k-colorable since G is a counterexample with fewest vertices.Therefore, G is equitably k-colorable by Lemma 1.
By Lemma 2, we have the following corollary: Corollary 3. Let G be an IC-planar graph with g ≥ 7. If ∆ ≥ 5, then G is equitably ∆-colorable.
Combining Corollary 3, and the conclusions in [14,15], we complete the proof of Theorem 1.
With this conclusion, we can solve the problem of equitable coloring of IC-planar graphs with large girth g (g ≥ 7) limitation.

3 . 4
H 4 as a subgraph, the following claim holds.Claim 3.3 G × has at most one special 2-vertex.A 4-vertex v ∈ V(G) is called a special 4-vertex if v is adjacent to a 2-vertex in G; a bad 4-vertex if v is special and adjacent to a 3-vertex in G; a poor 4-vertex if v is special and not adjacent to any 3-vertex.Since G does not contain H 3 , H 5 and H 6 as subgraphs, the following three claims hold.Claim Suppose that v is a poor 4-vertex.Thenn G 2 (v) = 1 and n G 4 + (v) = 3. Claim 3.5 Suppose that v is a bad 4-vertex.Then n G 2 (v) = 1, n G 3 (v) = 1 and n G 5 + (v) = 2. Claim 3.6 A special 4-vertex is not adjacent to any special 4-vertex in G. Claim 3.7 In G × , if v is a special 4-vertex incident with a 4-face, then f 6 + (v) ≥ 1.

Corollary 2 .
is a false 4-face, where v 1 is a false vertex.Then v 2 , v 3 and v 4 are true vertices by Fact 1.If v 2 or v 4 is a poor 4-vertex, say v 2 is a poor 4-vertex, then max{d G (v 3 ), d G (v 4 )} ≥ 4 by Claim 3.8.Therefore, f is incident with at least two true 4 + -vertices.If v 3 is a poor 4-vertex, then max{d G (v 2 ), d G (v 4 )} ≥ 4 by Claim 3.4.Therefore, f is incident with at least two true 4 + -vertices.If f is a false 5-face incident with a special 4-vertex, then f is incident with at least two true 4 + -vertices.

1 -
vertex u and w , which is a contradiction.Hence, the counterexample graph G does not exist and Lemma 5 holds.
, which is a contradiction.Hence, the counterexample graph G does not exist and Lemma 5 holds.