THE INCIDENCE CHROMATIC NUMBER OF SOME GRAPH

Throughout the paper, all graphs dealt with are finite, simple, undirected, and loopless. Let G be a graph, and let V(G), E(G), ∆(G), respectively, denote vertex set, edge set, and maximum degree of G. In 1993, Brualdi and Massey [3] introduced the concept of incidence coloring. The order of G is the cardinality |v(G)|. The size of G is the cardinality |E(G)|. Let I(G)= {(v,e) | v ∈V , e ∈ E, v is incident with e} (1.1)


Introduction
Throughout the paper, all graphs dealt with are finite, simple, undirected, and loopless.Let G be a graph, and let V (G), E(G), ∆(G), respectively, denote vertex set, edge set, and maximum degree of G.In 1993, Brualdi and Massey [3] introduced the concept of incidence coloring.The order of G is the cardinality |v(G)|.The size of G is the cardinality be the set of incidences of G.We say that two incidences (v,e) and (w, f ) are adjacent provided one of the following holds: (i) v = w; (ii) e = f ; (iii) the edge vw = e or vw = f .The minimum cardinality of C for which there exists an incidence coloring σ : I(G) → C is called the incidence chromatic number of G, and is denoted by inc(G).A partition {I 1 ,I 2 ,...,I k } of I(G) is called an independence partition of I(G) if each I i is independent in I(G) (i.e., no two incidences of I i are adjacent in I(G)).Clearly, for k ≥ inc(G), G is k -incidence colorable.
We may consider G as a digraph by splitting each edge uv into two opposite arcs (u,v) and (v,u).Let e = uv.We identify (u,e) with the arc (u,v).So I(G) may be identified with the set of all arcs A(G).Two distinct arcs (incidences) (u,v) and (x, y) are adjacent if one of the following holds (see Figure 1.2): ( This concept was first developed by Brualdi and Massey [3] in 1993.They posed the incidence coloring conjecture (ICC), which states that for every graph G, inc(G) ≤ ∆ + 2. In 1997, Guiduli [5] showed that incidence coloring is a special case of directed star arboricity, introduced by Algor and Alon [1].They pointed out that the ICC was solved in the negative following an example in [1].Following the analysis in [1], they showed that inc(G) ≥ ∆ + Ω(log∆), where Ω = 1/8 − • (1).Making use of a tight upper bound for directed star arboricity, they obtained the upper bound inc(G) ≤ ∆ + O(log∆).
Brualdi and Massey determined the incidence chromatic numbers of trees, complete graphs, and complete bipartite graphs [3]; Chen, Liu, and Wang determined the incidence chromatic numbers of paths, cycles, fans, wheels, adding-edge wheels, and complete 3partite graphs [4].
In this paper, we will consider the incidence chromatic number for complete k-partite graphs.We will give the incidence chromatic number of complete k-partite graphs, and also give the incidence chromatic number of three infinite families of graphs.Let k be positive integer, put [k] = 1,2,...,k.We state first the following definitions.Definition 1.1.For a graph G(V ,E) with vertex set V and edge set E, the incidence graph I(G) of G is defined as the graph with vertex set V (I(G)) and edge set E(I(G)).
Definitions not given here may be found in [2].

Some useful lemmas and properties of incidence chromatic number
Lemma 2.1.Let T be a tree of order n ≥ 2 with maximum degree ∆.Then inc(T) = ∆ + 1.

Lemma 2.2. A graph G is k-incidence colorable if and only if its incidence graph
Let M = {(ue, ve) | e = uv ∈ E(G), (ue,ve) ∈ E(I(G))}, then M forms a perfect matching of incidence graph I(G).The following lemmas are obvious.
Lemma 2.3.The incidence graph I(G) of a graph G is a graph with a perfect matching.

an independence-partition of incidence graph I(G), and the induced subgraph G[A v ] of I(G) is a clique graph.
By the definition of incidence graph, it is easy to give the proof.Lemma 2.5.Let ∆ be the maximum degree of graph G, I(G) the incidence graph, then complete graph K ∆+1 is a subgraph of I(G).
Proof.Let d(u) = ∆, p = ∆, and e k = uv 1 , e 2 = uv 2 , ..., e p = uv 1 be the edges of G. p incidences in I u = {(u, e 1 ),(u,e 2 ),...,(u,e p )} are adjacent to each other.For an incidence in is adjacent to all incidences in I u .Since p + 1 incidences (u,e 2 ),...,(u,e p ),(v i ,v i u) are vertices of I(G), by the definition of incidence graph, we can complete the proof.Lemma 2.6.For a simple graph G with order n, inc(G The following corollaries can be easily verified. In fact, for graphs G with order n, we can give each incidence in I(G) proper incidence coloring as follows.Let V (G) = {v 1 ,v 2 ,...,v n } be the vertex set and C = {1, 2,...,n} the color set.For i, j = 1,2,...,n, we let σ(v i ,v i v j ) = j.It is easy to see that the coloring above is an incidence coloring of G only with n colors.That is, inc(G) ≤ ∆ + 2, when Corollary 2.8.Let W n be the wheel graph with order n + 1.
Proof.To prove this lemma, we only need to prove that G 1 ∪ G 2 has an m-incidence coloring.Let {I 1 ,I 2 ,...,I m } be an independence partition of I(G 1 ), and . Hence G has an m-incidence coloring.The proof of the lemma is complete.
Theorem 2.11.Let G be a graph with maximum degree ∆(G)=n−2 and minimum degree 3,...,n).In incidence set I(G), incidences (v i ,v i v j )(i, j = 2,3,...,n, and i = j) are all adjacent to (v i ,v i u) and (v j ,v j u), thus the color n cannot be used to color any incidence in I(G − {u}).Denote by N(v 1 ) = {v i1 ,v i2 ,...,v iδ } the vertices adjacent to v 1 .The incidence coloring σ of graph G may be extended to an incidence coloring σ of graph G.For x, y ∈ V (G) and , thus we can select d(v 1 ) colors to incidence color, thus σ is a proper n-incidence coloring of G.The proof is completed.
For the general case, using the way similar to Theorem 2.11, we can give a stronger result.
Theorem 2.12.For graph G with order n and maximum degree

and uv /
∈ E(G), we say that G is with the property P.
Theorem 2.13.For graph G with order n and maximum degree Proof.By V n = {v 1 ,v 2 ,...,v n } we denote a labeling of the vertices of G and let d(v 1 ) = δ(G).For n = 4,5, the desired result follows from Lemma 2.1.
For the case n ≥ 6, the proof can be divided into two cases.Case 1. G is with the property P. Consider the auxiliary graph By Theorem 2.11, using similar methods as in the proof of Theorem 2.11, we can prove the desired result inc(G , and there exists two vertices We can also construct graph G 3 that is not with the property P. In that way, we can obtain a serial of graphs G,G 1 ,G 2 ,...,G k ,... such that all the graphs are not with the property P and Proof.The proof is by contradiction.Suppose that the graph G has an (∆ + 1)-incidence coloring with color set C = {1, 2,...,∆ + 1}.Let N G (u) = {x 1 ,x 2 ,...,x ∆ } and N G (v) = {y 1 , y 2 ,..., y ∆ }.Then each of the incidences (x i ,x i u) (1 ≤ i ≤ ∆) is colored the same, as are the incidences (y i , y i v).Without loss of generality, suppose k the color that (y i , y i v) has.Because N G (u) = N G (v) and (u,x 1 x 1 ) is adjacent to (y 1 , y 1 v), then (u,ux 1 ) has a color other than k.Because (u,ux 2 ) is adjacent to (y 2 , y 2 v),...,(u,ux ∆ ) which is adjacent to (y ∆ , y ∆ v), then (u,ux 2 ),...,(u,ux ∆ ) also has a color other than k, respectively.Further, the ∆ incidences (u,ux i ) (1 ≤ i ≤ ∆) have different colors, so the color k is different from that of incidences (u,ux i ).On the other hand, (y 1 , y 1 v) and (x 1 ,x 1 u) are neighborly incidences, so the color k is different from that of (x 1 ,x 1 u).Thus k / ∈ C, this gives a contradiction!Hence inc(G) ≥ ∆ + 2.

The incidence chromatic number of complete k-partite graph
(3.1) m=1 n m .The proof can be divided into the following two cases.Case 3.There exists i ∈ {1, 2,...,k} such that n i = 1.We let the vertex set of G be V (G) = {v i ,v 2 ,...,v m }, where m = k i=1 n i .By Lemma 2.6, it easy to draw the conclusion.Case 4. n i ≥ 2 (1 ≤ i ≤ k).To complete the proof, we give an incidence coloring just with ∆ + 2 colors firstly.
For j,t = 1,2,...,k, i = 1,2,...,n j , and s = 1,2,...,n t , we let (n m + s), i = s,t > j or i = s,t < j, To complete the proof, it suffices to prove that G cannot be colored with ∆ + 1 colors.It is obvious that each of the vertices in V 1 is the maximum-degree vertex.For Hence inc(G) ≥ ∆ + 2 follows from Theorem 2.14.Therefore inc(G) = ∆ + 2, and the proof is completed.By Theorem 3.1, it is easy to obtain the theorem in [3,4].In fact, the incidence coloring σ given to determine the incidence chromatic number for complete 3-partite graphs is a special case of the coloring above.Hence, we obtain some corollaries as follows.
The incidence coloring of K 3,4 and K 5 is given in Figure 3.1.

Incidence chromatic number of three families of graphs
The planar graph Q n , which is called triangular prism, is defined by where the vertex set V (G) = u 1 ,u 2 ,...,u n ∪ v 1 ,v 2 ,...,v n , and the edges set E(Q n ) consists  of two n-cycles u 1 ,u 2 ,...,u n and v 1 ,v 2 ,...,v n , and 2n edges ( It is easy to see that the coloring above is a proper 5-incidence coloring of Q n .Thus, we can only consider the case n = 5k.We will first prove that Q n is 6-incidence colorable by explicitly giving a 6-incidence coloring σ of Q n for any integer n ≥ 3.At last, we will give the proof that Q n cannot be incidence coloring just with colors 1,2,3,4,5.The proof can be divided into the following three cases.Case 5. n = 3k (k ≥ 1).Let i = 3s + t (t ≤ 2), i = 1,2,...,n, then Q n has an incidence coloring using 6 colors from the color set C = {1, 2,...,n + r + 1}, as follows: for i = 1,2,...,n, let (4.4) Case 7. n = 3k + 2 (k ≥ 1).Let i = 3s + t (t ≤ 2), for i = 1,2,...,n, and w n+1 = w 1 , w = u,v; w 0 = w n , w = u,v.We let It is easy to show that Q n is 6-incidence colorable.To complete the proof, it remains to be shown that there do not exist an incidence coloring using only 5 colors.Assume, on the contrary, that Q n is 5-incident colorable.For each vertex v ) have the same color, without loss of generality, 1.For i = 1,2,...,n, the case is the same.Because there are 5 colors that can be used in incidence coloring, and the degree of each vertex v )) have the same color.If n = 5k, form the proof above, it is easy to obtain a contradiction.Thus, we have completed the prove.

Figure 1 .Figure 1 . 1 .
Figure 1.1 shows three cases of two incidences being adjacent.An incidence coloring σ of G is a mapping from I(G) to a set C such that no two adjacent incidences of G have the same image.If σ : I(G) → C is an incidence coloring of G and |C| = k, k is a positive integer, then we say that G is k-incidence colorable.

Theorem 4 . 3 .
which is the Cartesian product of path P m and P n , For plane graph C m,n , we have inc(C m,n ) = 5.Proof.∆(C m,n ) = 4, then inc(C m,n ) ≥ 5. We now give a 5-incidence coloring σ of C m,n as follows: (i ∈ [m]; j ∈ [n])

(4. 7 )
It is easy to see that the coloring above is an incidence coloring of C m,n .Thus inc(C m,n ) = 5.