On the maximum average degree and the incidence chromatic number of a graph

We prove that the incidence chromatic number of every 3 -degenerated graph G is at most ∆( G ) + 4 . It is known that the incidence chromatic number of every graph G with maximum average degree mad ( G ) < 3 is at most ∆( G ) +3 . We show that when ∆( G ) ≥ 5 , this bound may be decreased to ∆( G ) + 2 . Moreover, we show that for every graph G with mad ( G ) < 22 / 9 (resp. with mad ( G ) < 16 / 7 and ∆( G ) ≥ 4 ), this bound may be decreased to ∆( G ) + 2 (resp. to ∆( G ) + 1 ).


Introduction
The concept of incidence coloring was introduced by Brualdi and Massey (3) in 1993.
Let G = (V (G), E(G)) be a graph.An incidence in G is a pair (v, e) with v ∈ V (G), e ∈ E(G), such that v and e are incident.We denote by I(G) the set of all incidences in G.For every vertex v, we denote by I v the set of incidences of the form (v, vw) and by A v the set of incidences of the form (w, wv).Two incidences (v, e) and (w, f ) are adjacent if one of the following holds: (i) v = w, (ii) e = f or (iii) the edge vw equals e or f .
A k-incidence coloring of a graph G is a mapping σ of I(G) to a set C of k colors such that adjacent incidences are assigned distinct colors.The incidence chromatic number χ i (G) of G is the smallest k such that G admits a k-incidence coloring.
For a graph G, let ∆(G), δ(G) denote the maximum and minimum degree of G respectively.It is easy to observe that for every graph G we have χ i (G) ≥ ∆(G) + 1 (for a vertex v of degree ∆(G) we must use ∆(G) colors for coloring I v and at least one additional color for coloring A v ).Brualdi and Massey proved the following upper bound: Guiduli (4) showed that the concept of incidence coloring is a particular case of directed star arboricity, introduced by Algor and Alon (1).Following an example from (1), Guiduli proved that there exist graphs G with χ i (G) ≥ ∆(G) + Ω(log ∆(G)).He also proved that For every graph G, χ i (G) ≤ ∆(G) + O(log ∆(G)).
• For every cubic graph G, χ i (G) ≤ 5 and this bound is tight (6).
The maximum average degree of a graph G, denoted by mad(G), is defined as the maximum of the average degrees ad(H) = 2 • |E(H)|/|V (H)| taken over all the subgraphs H of G.
In this paper we consider the class of 3-degenerated graphs (recall that a graph G is k-degenerated if δ(H) ≤ k for every subgraph H of G), which includes for instance the class of triangle-free planar graphs and the class of graphs with maximum average degree at most 3.More precisely, we shall prove the following: In fact we shall prove something stronger, namely that one can construct for these classes of graphs incidence colorings such that for every vertex v, the number of colors that are used on the incidences of the form (w, wv) is bounded by some fixed constant not depending on the maximum degree of the graph.
More precisely, we define a (k, ℓ)-incidence coloring of a graph G as a k-incidence coloring σ of G such that for every vertex We end this section by introducing some notation that we shall use in the rest of the paper.Let G be a graph.If v is a vertex in G and vw is an edge in G, we denote by Let G be a graph and σ ′ a partial incidence coloring of G, that is an incidence coloring only defined on some subset I of I(G).For every uncolored incidence (v, vw) ∈ I(G) \ I, we denote by F σ ′ G (v, vw) the set of forbidden colors of (v, vw), that is: ( We shall often say that we extend such a partial incidence coloring σ ′ to some incidence coloring σ of G.In that case, it should be understood that we set σ(v, vw) = σ ′ (v, vw) for every incidence (v, vw) ∈ I.
We shall make extensive use of the fact that every (k, ℓ)-incidence coloring may be viewed as a (k ′ , ℓ)incidence coloring for any k ′ > k.Drawing convention.In a figure representing a forbidden configuration, all the neighbors of "black" or "grey" vertices are drawn, whereas "white" vertices may have other neighbors in the graph.
Therefore, we assume ∆(G) ≥ 4 and we prove the theorem by induction on the number of vertices of G.If G has at most 5 vertices then G ⊆ K 5 .Since for every k > 0, χ i (K n ) = n, we obtain χ i (G) ≤ χ i (K 5 ) = ∆(K 5 ) + 1 = 5, and every 5-incidence coloring of G is obviously a (∆(G) + 4, 3)incidence coloring.We assume now that G has n + 1 vertices, n ≥ 5, and that the theorem is true for all 3-degenerated graphs with at most n vertices.
Let v be a vertex of G with minimum degree.Since G is 3-degenerated, we have d G (v) ≤ 3. We consider three cases according to d G (v).
Let w denote the unique neighbor of v in G (see Figure 1.(1)).Due to the induction hypothesis, the graph there is a color a such that a / ∈ F σ ′ G (w, wv).We then set σ(w, wv) = a and σ(v, vw) = b, for any color b in σ ′ (A w ).
Let u, w be the two neighbors of v in G (see Figure 1.( 2)).Due to the induction hypothesis, the graph G ′ = G \ v admits a (∆(G) + 4, 3)-incidence coloring σ ′ .We extend σ ′ to a (∆(G) + 4, 3)incidence coloring σ of G as follows.We first set σ(v, vu) = a for a color We still have to color the two incidences (u, uv) and (w, wv).Since a ∈ σ ′ (A u ), we have Let u 1 , u 2 and u 3 be the three neighbors of v in G (see Figure 1.( 3)).Due to the induction hypothesis, the graph Observe first that for every i, Moreover, if |A ui | < 3 then any of these five colors may be assigned to the incidence (v, vu i ) whereas we have only three possible choices (among these five) if |A ui | = 3.In the following, we shall see that having only three available colors is enough, and therefore assume that |σ ′ (A ui )| = 3 for every i, 1 ≤ i ≤ 3.
We define the sets B and B i,j as follows: We consider now four subcases according to the degrees of u 1 , u 2 and u 3 : In this case, since we have 3 colors for the incidence (v, vu i ) for every i, We still have to color the three incidences

Only one of the vertices u i is of degree ∆(G).
We can suppose without loss of generality that , there exist two distinct colors a 2 and a 3 distinct from a 1 such that a 2 ∈ σ ′ (A u2 ) and a 3 ∈ σ ′ (A u3 ).We set σ(v, vu i ) = a i for every i, 1 ≤ i ≤ 3.
We still have to color the three incidences of form (u i , u i v).Since a 1 ∈ B 3,1 and a 3 ∈ σ ′ (A u3 ) we have: and since a i ∈ σ ′ (A ui ) for every i = 1, 2 we have: Therefore, there exist three colors b

Only one vertex among the u i 's is of degree less than ∆(G).
We can suppose without loss of generality that Similarly to the previous case, we have B 2,1 = ∅ and B 3,2 = ∅.We consider two cases: We still have to color the three incidences and since a i ∈ B i+1,i for i = 1, 2 and a j ∈ σ ′ (A uj ) for j = 2, 3, we have: We still have to color the three incidences and since a 1 ∈ B 2,1 = B 3,2 and a j ∈ σ ′ (A uj ) for j = 2, 3, we have: Therefore, there exist three colors b Similarly to the case (b) we have B i,j = ∅ for every i and j, 1 ≤ i, j ≤ 3 and thus |B| ≥ 1.
Let a 1 and a 2 be two distinct colors in B. We can suppose without loss of generality that a 1 ∈ B 2,1 and a 2 ∈ B 3,2 .
We consider the two following subcases: for every i, 1 ≤ i ≤ 3, we have: Therefore, there exist three colors b and since a 2 ∈ σ ′ (I u1 ) ∪ σ ′ (A u1 ) and a 1 ∈ σ ′ (A − u 1 ) we have: Therefore, there exist three colors b It is easy to check that in all cases we obtain a (∆(G) + 4, 3)-incidence coloring of G and the theorem is proved. ✷ Since every triangle free planar graph is 3-degenerated, we have: Corollary 3 For every triangle free planar graph G, χ i (G) ≤ ∆(G) + 4.

Graphs with bounded maximum average degree
In this section we study the incidence chromatic number of graphs with bounded maximum average degree.The following result has been proved in (5).
Proof: By Observation 6 we have mad(G) < 2g/(g − 2) ≤ (2 × 6)/(6 − 2) = 3 and we get the result from Corollary 5. ✷ If the graph has maximum degree at least 5, the previous result can be improved: Proof: Suppose that the theorem is false and let G be a minimal counter-example (with respect to the number of vertices).We first show that G must avoid all the configurations depicted in Fig. 2.
, there is a color a such that a / ∈ F σ G (w, wv).We set σ(w, wv) = a and σ(v, vw) = b, for any color b in σ ′ (A w ).( 2) Let w 1 , w 2 denote the two neighbors of v in G. Due to the minimality of G, the graph denote the five neighbors of v and w i denote the other neighbor of u i in G (see Figure 2.( 3)).Due to the minimality of G, the graph G ′ = G \ v admits a (∆(G) + 2, 2)-incidence coloring σ ′ .We extend σ ′ to a (∆(G) + 2, 2)-incidence coloring σ of G. Let Since we have ∆(G) + 2 ≥ 7 colors, there is a color x distinct from a i for every i, 1 ≤ i ≤ 5.
we have two possible colors for the incidence (u i , u i w i ) for every i, 1 ≤ i ≤ 5. So, we can suppose that σ ′ (u i , u i w i ) = x for every i, , and since we have at least 7 colors, there is 5 distinct colors c 1 , c 2 ,. .., c 5 such that c i / ∈ {x, σ ′ (u i , u i w i )}, 1 ≤ i ≤ 5, and we set σ(v, vu i ) = c i for every i, 1 ≤ i ≤ 5.
It is easy to check that in every case we have obtained a (∆(G) + 2, 2)-incidence coloring of G, which contradicts our assumption.
We now associate with each vertex v of G an initial charge d(v) = d G (v), and we use the following discharging procedure: each vertex of degree at least 5 gives 1/2 to each of its 2-neighbors.
We shall prove that the modernized degree d * of each vertex of G is at least 3 which contradicts the assumption mad(G) < 3 (since u∈G d * (u) = u∈G d(u)).Let v be a vertex of G; we consider the possible cases for old degree d G (v) of v (since G does not contain the configuration 2(1), we have Since G does not contain the configuration 2(2) the two neighbors of v are of degree at least 5. Therefore, v receives 1/2 from each of its neighbors so that In this case we have Since G does not contain the configuration 2(3) at least one of the neighbors of v is of degree at least 3 and v gives at most 4 × 1/2 = 2.We obtain Therefore, every vertex in G gets a modernized degree of at least 3 and the theorem is proved.✷

Remark 9
The previous result also holds for graphs with maximum degree 2 and for graphs with maximum degree 3 (by the result from ( 6)) but the question remains open for graphs with maximal degree 4.
For graphs with maximum average degree less than 22/9, we have: Theorem 11 Every graph G with mad(G) < 22/9 admits a (∆(G)+2, 2)-incidence coloring.Therefore, Proof: It is enough to consider the case of graphs with maximum degree at most 4, since for graphs with maximum degree at least 5 the theorem follows from Theorem 8. Suppose that the theorem is false and let G be a minimal counter-example (with respect to the number of vertices and edges).Observe first that we have ∆(G) ≥ 3 since otherwise we obtain by Theorem 1 that We first show that G cannot contain any of the configurations depicted in Figure 3.
(1) This case is similar to case 1 of Theorem 8. (3) Let u 1 , u 2 and u 3 denote the three neighbors of v and w i denotes the other neighbor of u i , 1 Suppose that a i = σ ′ (w i , w i u i ), 1 ≤ i ≤ 3. Since we have ∆(G) + 2 ≥ 5 colors, there is a color x distinct from a i for every i, 1 ≤ i ≤ 3.
we have at least two colors for the incidence (u i , u i w i ) for every i, 1 ≤ i ≤ 3. Thus, we can suppose σ ′ (u i , u i w i ) = x for every i, 1 ≤ i ≤ 3. We then set σ(u i , u i v) = x for every i, 1 ≤ i ≤ 3.
Since F σ G (v, vu i ) = {x, σ ′ (u i , u i w i )} for every i, 1 ≤ i ≤ 3, and since we have at least 5 colors, there are Therefore, in all cases we obtain a (∆(G) + 2, 2)-incidence coloring of G, which contradicts our assumption.
We now associate with each vertex v of G an initial charge d(v) = d G (v), and we use the following discharging procedure: each vertex of degree at least 3 gives 2/9 to each of its 2-neighbors.
We shall prove that the modernized degree d * of each vertex of G is at least 22/9 which contradicts the assumption mad(G) < 22/9.Let v be a vertex of G; we consider the possible cases for old degree d G (v) of v (since G does not contain the configuration 3(1), we have d G (v) ≥ 2): Since G does not contain the configuration 3(2) the two neighbors of v are of degree at least 3. Therefore, v receives then 2/9 from each of its neighbors so that d * (v) = 2 + 2/9 + 2/9 = 22/9.Therefore, every vertex in G gets a modernized degree of at least 3 and the theorem is proved.✷ By considering cycles of length ℓ ≡ 0 (mod 3), we get that the upper bound of Theorem 11 is tight.previously, for planar graphs we obtain: Corollary 12 Every planar graph G of girth g ≥ 11 admits a (∆(G) + 2, 2)-incidence coloring.Therefore, χ i (G) ≤ ∆(G) + 2.
Proof: Since for every graph G, χ i (G) ≥ ∆(G) + 1, it is enough to prove that G admits a (∆(G) + 1, 1)incidence coloring.Suppose that the theorem is false and let G be a minimal counter-example (with respect to the number of vertices).We first show that G cannot contain any of the configurations depicted in Figure 4.
(1) This case is similar to case 1 of Theorem 8.
(2) Let u i , i = 1, 2, be the two neighbors of v and w i denote the other neighbor of u i in G. Due to the minimality of G, the graph , we can suppose that a = c.We then set σ(v, vu 1 ) = a and σ(v, vu 2 ) = c.

2.
All the u i 's are of degree 2. Let x i be the other neighbor of u i in G, 1 ≤ i ≤ 3.
(a) One of the x i 's is of degree at least 3, say x 1 .In this case v gives 1/7 to u 1 , at most 2/7 to u 2 and at most 2/7 to u 3 .We then have d * (v) ≥ 3 − 1/7 − 2/7 − 2/7 = 16/7.(b) All the x i 's are of degree 2. Let w i be the other neighbor of x i in G, 1 ≤ i ≤ 3. Since G does not contain the configuration 4(2) we have d G (w i ) ≥ 3 for every i, 1 ≤ i ≤ 3, and since G does not contain the configuration 4(3), at most one of the w i 's, 1 ≤ i ≤ 3, can be of degree 3. Thus, we can suppose without loss of generality that d G (w 1 ) and d G (w 2 ) ≥ 4. In this case, v gives 1/7 to w 1 , 1/7 to w 2 and at most 2/7 to w 3 .We then have d * (v) ≥ 3 − 1/7 − 1/7 − 2/7 = 17/7 ≥ 16/7.
Therefore, every vertex in G gets a modernized degree of at least 16/7 and the theorem is proved.✷ Considering the lower bound discussed in Section 1, we get that the upper bound of Theorem 13 is tight.
Remark 14 For every graph G, the square of G, denoted by G 2 , is the graph obtained from G by linking any two vertices at distance at most 2. It is easy to observe that providing a (k, 1)-incidence coloring of G is the same as providing a proper k-vertex-colouring of G 2 , for every k (by identifying for every vertex v the color of A v in G with the color of v in G 2 ).By considering the cycle C 4 on 4 four vertices (note that C 2 4 = K 4 ) we get that the previous result cannot be extended to the case ∆ = 2. Consider now the graph H obtained from the cycle C 5 on five vertices by adding one pending edge with a new vertex.Since H 2 contains a subgraph isomorphic to K 5 , we similarly get that the previous result cannot be extended to the case ∆ = 3.
the graph obtained from G by deleting the vertex v and by G \ vw the graph obtained from G by deleting the edge vw.