1 Introduction
Coloring a graph consists in finding an assignment of colors such that any pair of adjacent vertices receive different colors. The minimum integer such that a coloring exists is called the chromatic number of , denoted by . We denote by the maximum degree of the graph . Using a greedy algorithm, it is easy to show that every graph can be colored with colors. A seminal result from Brooks characterizes the cases when this bound is tight:
Theorem 1 (brooks1941colouring ).
Every connected graph satisfies excepted when or and
is an odd cycle.
Given an integer and a graph , the th power of is the graph obtained from by adding an edge between vertices at distance at most in . In this paper, we are interested in bounding the chromatic number by a function of the integer and the maximum degree .
Up to replacing any graph by each of its connected components, we may assume that all the graphs we consider are connected. Moreover, coloring powers of paths and cycles is already settled by the following result.
Proposition 2 (prowse2003choosability ).
Let and be two integers. Then


If , then where and .

otherwise.
Therefore, we may only consider graphs with maximum degree at least 3, so that in particular, none of the graphs we consider in this paper is a cycle.
For the case of squares of graphs (i.e. ), we have (and this can be tight). Therefore, applying Brooks’ theorem on states that colors are sufficient excepted when is a clique on vertices. Such graphs are called Moore graphs, and Hoffman and Singleton hoffman2003moore proved that only finitely many of them exist. For all the other graphs, the bound can actually be improved, as shown by the following result.
Theorem 3 (cranston2016painting ).
If is not a Moore graph, then .
These results have been generalized for higher powers of graphs. Assume that . Then, the maximum possible value of is , where
is the number of nodes of a ary tree of height , without its root (see Figure 1).
Brooks’ theorem gives that colors are sufficient to color every graph with maximum degree , as soon as it is not a generalized Moore graph, i.e. is not a clique on vertices. However, no such graph exists when (see hoffman2003moore ). Therefore, the bound always holds. Moreover, this bound is improved by the following result.
Theorem 4 (bonamy2014brooks ).
For and every graph , we have .
When , note that . Hence, together with Theorem 3, this result settles a conjecture of miao2014distance , stating that two colors can be spared from the naive upper bound , excepted when and is a Moore graph.
Bonamy and Bousquet conjecture in bonamy2014brooks that we can improve this result, by sparing colors for higher values of , excepted for a finite number of graphs.
Conjecture 5 (bonamy2014brooks ).
For every , only finitely many graphs satisfy .
To study how far can be from , we introduce the following parameter.
For every integer , the gap of a graph , denoted by , is the difference .
The aforementioned results can then be rephrased in terms of gap: for every integer and every graph , we have if and only if and is a Moore graph. Moreover, Conjecture 5 actually expresses finiteness of the class of graphs satisfying . In Section 2, we prove the following generic result towards this conjecture.
Theorem 6.
Let and be two integers with . Then, there are only finitely many graphs of maximum degree such that .
Observe that Brooks’ theorem gives an infinite list of graphs such that , but for every , this list contains only one graph of given maximum degree . From that perspective, Theorem 6 can be seen as a generalization of Brooks’ theorem for powers of graphs. However, note that the bound we obtain for is worse than the one given by Brooks’ theorem.
Theorem 7.
There exists a function when such that for all integers and , only finitely many graphs of maximum degree satisfy .
Observe that when and are large enough, this improved version implies that there are finitely many graphs with maximum degree and satisfying . However, this does not imply Conjecture 5, since the finiteness result is only valid when is fixed.
2 Proof of Theorem 6
In this section, we give a proof of Theorem 6. First note that the cases are consequences of Brooks’ theorem and Theorem 3. Thus, we only consider the case . In the following, we fix two integers and greater or equal to 3, and denote by a graph of maximum degree . The proof of Theorem 6 relies on the following proposition.
Proposition 8.
If , then the diameter of is at most .
Indeed, this ensures that if , then has at most vertices. In particular, there are only finitely many choices for , which concludes the proof of Theorem 6.
It now remains to prove Proposition 8. The structure of the proof is as follows: we assume by contrapositive that contains two vertices and at distance , and we design an ordering of the vertices such that the associated greedy procedure gives a valid coloring of with colors. As we will see, this ordering roughly consists in considering the vertices by decreasing distance to the middle of a fixed shortest path between and . This shows that , and proves Proposition 8.
This section is organized as follows. In a first subsection, we explain how we count the number of available colors for a given vertex when it is considered in the greedy procedure. Then, in the second subsection, we present the ordering of the vertices and show that each time a vertex is considered, it has at least an available color. This ensures that we obtain a valid coloring of when the procedure terminates.
2.1 A framework for counting available colors
To check whether the coloring procedure we will describe later yields a valid coloring of with colors, we need to count the number of available colors for every vertex
at the moment it is considered by the procedure. Let
be the closed neighborhood of at distance in , i.e. .Observe that when induces a ary tree of height in , the number of uncolored vertices in minus is a lower bound on the number of available colors for . However, does not necessarily have this nice structure. In particular, there may be cycles or vertices of degree less than in , and the resulting lower bound is not strong enough to conclude. We now explain how to deal with this issue.
First, let be a tree of height , where each node at height less than has children. We define as the graph obtained from by adding an edge between each vertex and the root of copies of . In particular, observe that for every , the neighborhood contains only vertices of degree .
Remark 9.
In many proofs about graph colorings, we usually remove elements of , and use some minimality argument to obtain a coloring to extend. In this case, we instead add some vertices. This is not related to some inductive argument (we do not even color all of these new vertices). The goal of this modification is to obtain a better lower bound on the number of available colors for . Indeed, we can refine the previous bound by considering the number of uncolored vertices in minus , which is always an improvement since is an induced subgraph of .
Observe that the number of available colors for also depends on whether some colors appear several times in , or similarly if this neighborhood contains cycles. To capture these cases, instead of counting uncolored vertices, we will consider (in any fixed order) all the nonempty nonbacktracking walks of length at most in starting from (meaning that we allow the same edge to be used twice, but not in a row). In the following, each walk is implicitly assumed to be nonbacktracking. The number of available colors for thus depends on the number of such walks ending on an uncolored vertex, or ending with vertices with the same color (possibly the same vertices).
We say that such a walk is nice if its endpoint (possibly ) either is in , or is uncolored, or is the endpoint of an already considered walk. Each time we find a nice walk starting at , we say that saves a color. The number of forbidden colors is the number of nonnice walks. Therefore, the number of available colors for is the number of nice walks starting from , minus .
Remark 10.
When is a ary tree of height , then each nonempty walk corresponds to a unique vertex in . In particular, the nice walks are the ones ending at uncolored vertices or at vertices whose color appears at the endpoint of an already considered walk. Therefore, counting nice walks is actually a generalization of counting uncolored vertices.
2.2 The coloring procedure
We may now give the coloring procedure we consider, and show that it yields a valid coloring of . Since has diameter at least , there is a shortest path of length between two vertices and . First, we set the color of each (resp. ) as . Note that this is a proper partial coloring of : if , there is a path from to of length , contradicting the hypothesis.
We fix the following ordering of the vertices of : . Let be a vertex of . We define the root of as the largest vertex in on a shortest path from to .
Let . We first color the vertices of by decreasing distance to , then color the vertices of , also by decreasing distance to (so will be the last vertex to be colored).
Let be an uncolored vertex of outside of . Let and . Denote by a shortest path from to . We now consider several types of nice walks in , ending on vertices shown in Figure 2:

The subpaths of ending at some internal vertex of . By construction, we have
hence is uncolored when we consider , and the corresponding walk is nice. Note that for , there is such a walk of length .

For each internal vertex of with , the walk , where is a neighbor of different from and (which exists since has degree in ). Indeed, we have
hence is uncolored when we consider . For every , there is at least one such walk of length .

If , the walks , where and is a neighbor of in different from and . Indeed, we have
hence is uncolored when we consider . For , there is such a walk of length .

If (hence ), the walks obtained by concatenating with the subpath of between and , and ending at vertices in (which are uncolored by construction). There is at least one such walk of length for every .

If and , the two walks and end with vertices sharing the same color, hence one of them is nice.

If and , then is a nice walk since is the last vertex to be colored.
We now have to make sure that at least such walks have length at most . First consider the case where . Among the nice walks of length at most starting at , the previous counting ensures that there are:

walks of type 1.

walks of type 2 if , and 0 otherwise.

walks of type 3.

walks of type 4.
If , there are at least walks of type 1 and 2, hence we may assume that . We separate four cases depending on whether , and whether .

If and , then there are at least walks of type 3.

If and , then there are at least walks of type 3 and 4.

If and , then there are at least
walks of type 1, 2, or 3.

If and , then there are at least
walks of type 1, 2, 3 or 4.
It now remains to consider the case when . We consider several cases depending on the value of .

If , there are nice walks of type 1 and 2.

If , there are walks of type 1, 2 and 6.

If , observe that there are at least nice walks of type 5 and length at most (ending either at or for each ). Therefore, there are
nice walks of type 1, 2, 5 and 6.
3 Improving the gap bound
This section is devoted to the proof of Theorem 7. We thus take two integers and greater or equal to three, and a graph of maximum degree . We again show that if has a small gap, then it should have a small diameter, which ensures that can take only finitely many values. More precisely, Theorem 7 relies on the following statement.
Proposition 11.
For every integer , if , then has diameter at most .
To obtain Theorem 7, we apply the proposition with , and observe that .
Remark 12.
When , we can take in Proposition 11, an obtain that graphs of maximum degree satisfying have diameter at most . In particular, when , this strengthens Proposition 8 by giving more insight on the structure of graphs satisfying . Moreover, when , this also implies a weak version of Conjecture 5: for every , there are only finitely many graphs of maximum degree such that . This is actually valid when , using some slight optimizations in the upcoming proof.
The end of this section is devoted to the proof of Proposition 11. We use similar ideas as for Proposition 8. Indeed, we assume again that contains two vertices and at distance for some , and describe a greedylike procedure allowing to color with colors. The ordering we choose is very similar: we again color vertices by decreasing distance to the middle of a shortest path, excepted for the vertices whose root is in the middle, that we color last. The difference comes from the fact that we do not assign colors to the vertices of the path beforehand, but only to the endpoints of the path and their neighbors in .
We also reuse part of the framework introduced in the previous section. More precisely, we construct a graph containing such that the neighborhood of every vertex contains only vertices of degree . We also count the number of nice walks in order to show that the greedy procedure does not use more than colors.
Let and be the internal vertices of a shortest path between and (with if is odd).
We first precolor all vertices in with colors in . To this end, let be the walks of length at most starting at , and color each vertex in with the smallest index such that ends in . We apply the same procedure to color . The resulting coloring is a proper partial coloring of since otherwise there would be a path of length at most between and , which is impossible by hypothesis.
First note that every vertex satisfying saves colors. Indeed, for each color , we have two cases:

Either the color appears in both and , hence there are two walks starting at and ending on vertices colored with (one going through and the other through ). This ensures that at least one of these walks is nice.

Or, by symmetry, the color does not appear in . Then the endpoint of is also the endpoint of some for . In particular, is a nice walk starting at .
Therefore, there are at least nice walks starting at , i.e. saves colors.
Observe that every vertex in is at distance at most from every vertex in . In particular, the vertices in always save colors even if they are the last ones to be colored. We may thus focus on the vertices of outside of . We color them greedily by decreasing distance to .
Let be a vertex of not in . We may assume that for some since the case is similar. We consider two cases depending on whether the distance is small or large.

Assume that . Let be a shortest path between and , and let . Then
hence is uncolored when we consider , unless .
In that case, there is a path of length at most between and (going through and or ). This is impossible since
Therefore, all vertices in are uncolored when we consider . In particular, there are nice walks of length at most starting at .

Assume that . Let and . If , the vertex is uncolored by construction. We may thus assume that . In that case, since , we have
hence again is uncolored when we consider , unless .
In that case, observe that there is a path of length at most from (or ) to . However, by hypothesis, we have and , a contradiction.
Therefore, all vertices in are uncolored when we consider . In particular, there are nice walks of length at most
starting at .
4 Conclusion
Theorem 6 is a first step towards Conjecture 5. A very similar proof allows us to replace by when . However, even for large , the current proof cannot yield any better bound on (say, ). Indeed, the bottleneck is reached by vertices in whose neighbors in (excepted ) are all colored and only the colors appear twice in their neighborhood at distance . Therefore, these vertices can save at most colors in the worst case regardless of the value of .
Observe also that while bounding the diameter of graphs with is sufficient to obtain Theorem 6, we would need more insight on their structure to prove Conjecture 5. It is not hard to extend the previous methods to show that such graphs are regular, and that their girth is at least .
Regarding Theorem 7, there is no reason to believe that is the largest possible value of such that for every integers and , there are only finitely many graphs of maximum degree such that . Indeed, it may be possible to improve this value by considering a more involved coloring procedure. For example, trying to merge the procedures we presented in Propositions 8 and 11 may lead to a better bound. However, no generic upper bound seem to be known. This leads to the following question.
For all integers and , what is the smallest such that infinitely many graphs of maximum degree satisfy ?
References
 [1] M. Bonamy and N. Bousquet. Brooks’ theorem on powers of graphs. Discrete Mathematics, 325:12–16, 2014.
 [2] R.L. Brooks. On colouring the nodes of a network. In Mathematical Proceedings of the Cambridge Philosophical Society, volume 37, pages 194–197. Cambridge University Press, 1941.
 [3] D.W. Cranston and L. Rabern. Painting squares in shades. The Electronic Journal of Combinatorics, 23(2):P2–50, 2016.
 [4] A.J. Hoffman and R.R. Singleton. On Moore graphs with diameters 2 and 3. volume 4, pages 497–504. IBM Corp., 1960.
 [5] L.Y. Miao and Y.Z. Fan. The distance coloring of graphs. Acta Mathematica Sinica, English Series, 30(9):1579–1587, 2014.
 [6] A. Prowse and D.R. Woodall. Choosability of powers of circuits. Graphs and Combinatorics, 19(1):137–144, 2003.
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