For an integer , a -edge colouring of a graph is a mapping such that for any two edges and of that have a common end-vertex. The Edge Colouring problem is to decide if a given graph has a -edge colouring for some given integer . If is fixed, that is, not part of the input, then we denote the problem as:
.99 -Edge Colouring
[2pt] Instance: a graph . Question: does have a -edge colouring?
The chromatic index of a graph is the smallest integer such that has a -edge colouring. The degree of a vertex is the size of its neighbourhood . We let denote the maximum degree of . Vizing showed the following classical result.
Theorem 1.1 ()
The chromatic index of a graph is either or .
Due to Theorem 1.1, we can make the following observation.
Let be a graph. If , then is a yes-instance of -Edge Colouring. If , then is a no-instance of -Edge Colouring.
By Observation 1 we may assume without loss of generality that an input graph of -Edge Colouring has maximum degree . The following well-known hardness result was proven by Holyer for and by Leven and Galil for .
For every , -Edge Colouring is NP-complete even for graphs in which every vertex has degree .
In this note we consider the -Edge Colouring problem from the viewpoint of Parameterized Complexity, where problem inputs are specified by a main part of size and a parameter , which is assumed to be small compared to . The main question is to determine if a problem is fixed-parameter tractabe (FPT), that is, if it can be solved in time , where is a computable (and possibly exponential) function of . The choice of parameter depends on the context. By the above discussion, the most natural choice of parameter for -Edge Colouring is the number of vertices of maximum degree. We prove the following result (note that -Edge Colouring is polynomial-time solvable for ).
For every , -Edge Colouring can be solved in time on graphs with vertices, of which have maximum degree, and edges. Moreover, it is possible to find a -edge colouring of in time (if it exists).
As we assume that is a fixed constant, Theorem 1.3 implies that for every , the -Edge Colouring problem is fixed-parameter tractable when parameterized by the number of vertices of maximum degree. We prove Theorem 1.3 in Section 2 by using the alternative proof of Theorem 1.1 given by Ehrenfeucht, Faber and Kierstead . We first discuss some related work.
Apart from a number of (classical) complexity results (e.g. [1, 2, 11, 12, 15, 21, 22, 23, 26]), most results on edge colouring are related to Theorem 1.1 and are of a more structural nature. That is, they involve the derivation of sufficient or necessary conditions for a graph to be -edge colourable (see, for example [4, 8, 16, 19]), in which case the graph is said to be Class 1 (and Class 2 otherwise). This is also the focus of papers on edge colouring related to the number of maximum-degree vertices (see, for example, [5, 10, 25, 28]). In particular, Fournier  proved that every graph in which the vertices of degree induce a forest is Class 1. As a consequence, a graph with at most two vertices of maximum degree is Class 1. In  and , Chetwynd and Hilton gave necessary and sufficient polynomial-time verifiable conditions for a graph with three, respectively, four vertices of maximum degree to be Class 1. The same authors proved an analogous result for -vertex graphs with vertices of maximum degree in , assuming .
For more on edge colouring we refer to the recent survey of Cao, Chen, Jing, Stiebitz and Toft .
2 The Proof of Theorem 1.3
In this section we prove Theorem 1.3. Let be a graph and let . The graph is the subgraph of induced by . The set is the set of vertices adjacent to at least one vertex in . Let be the set of maximum-degree vertices of . In this context, the graph is said to be the core of , whereas the graph is said to be the semi-core of .
Machado and de Figueiredo  proved the following result.
Theorem 2.1 ()
Let be a graph and be an integer. Then has a -edge colouring if and only if its semi-core has a -edge colouring.
The proof of Theorem 2.1 is based on an application of the recolouring procedure of Vizing  for proving Theorem 1.1 and which was also used by Misra and Gries  for giving a constructive proof of Theorem 1.1. As explained by Zatesko , the proof of Theorem 2.1 immediately yields a polynomial-time algorithm for finding a -edge colouring of a graph given a -edge colouring of its semi-core (if it exists). Below we give a new, short algorithmic proof of Theorem 2.1, with the same time complexity, via a modification of the alternative proof of Theorem 1.1 by Ehrenfeucht et al. , which might be of independent interest. We state this result as a lemma, as we use it to prove our main result.
Let be a graph with a core of size . Given a -edge colouring of its semi-core, it is possible to construct in time a -edge colouring of .
Recall that denotes the set of vertices of maximum degree of . Let be a -edge colouring of the semi-core of and denote by the size of the semi-core of . Note that . If , then is a -edge colouring of . Assume that .
We write . We let and for . Note that .
We define . We now show how to extend vertex by vertex until we obtain a -edge colouring of the whole graph. That is, for , we show how to construct a -edge colouring of given a -edge colouring of
Thus suppose that and suppose that we already have a -edge colouring of (note that at the start of the procedure, when , this is indeed the case, as we have ). We say that a vertex misses colour if none of the edges incident to is coloured by . We denote the set of colours that misses by . Note that . We write . We now iteratively colour the edges incident to in in such a way that at any time, the resulting mapping is a partial -edge colouring of that satisfies the two following properties:
for each uncoloured edge in , we have ;
there is at most one uncoloured edge in with .
Observe that at the start of this process, when no edge in has been coloured, satisfies (1) and (2). This can be seen as follows. Vertex is in only adjacent to vertices of (where we define if ). Every that is adjacent to does not belong to and thus has degree at most . Hence, is adjacent to at most vertices in and thus has . As , this means that each edge in has , implying that (1) and (2) are satisfied.
We will now describe how we can maintain properties (1) and (2) while colouring the edges incident to in one by one. Throughout this process we maintain a set , so at the start consists of all neighbours of in . We distinguish two cases.
Case 1. There exists a colour such that appears in at most one set for some with . Then we assign colour to the edge for which is the smallest over all sets that contain . If all these sets have size at least 3, then afterwards (1) and (2) are satisfied. Otherwise, there is a unique smallest set of size at most 2, which also implies that afterwards (1) and (2) are satisfied.
Case 2. For each colour , there exists at least two distinct vertices such that and both and have size at most 2. This means that
Hence, we deduce that . As has degree at most in , it follows that in , at most edges incident to have already been coloured in this stage. It follows that . Hence, there exists a colour . Consequently, each vertex in must have an incident edge coloured . Now choose a vertex for which is minimum and consider a colour . We swap colours and along the path in that starts in and whose edges are alternatingly coloured and . This yields another partial -edge colouring of . However, now has no incident edge coloured anymore, which means that we can colour the edge with colour . We observe that for all , the set remains unchanged except the end-vertex of if is adjacent to ; in that case is replaced by . However, by minimality of , we have that and property (2) implies that if , then . Thus, . We conclude that both (1) and (2) are satisfied by the newly obtained (partial) -edge colouring of .
After colouring the last uncoloured edge in incident to we have obtained our -edge colouring of . Hence, after doing this for , we have indeed extended to a -edge colouring of .
Finally note that since , the algorithm has at most extension steps and as each takes time, the total running time is . This completes the proof of Lemma 1. ∎
We are now ready to prove Theorem 1.3, which we restate below.
Theorem 1.3. For every , -Edge Colouring can be solved in time on graphs with vertices, of which have maximum degree, and edges. Moreover, it is possible to find a -edge colouring of in time (if it exists).
Let be an instance of -Edge Colouring. We first compute the maximum degree of and the corresponding set of vertices of degree in time (so . Let denote the semi-core of . By Theorem 2.1, it suffices to check if has a -edge colouring. Note that we may assume by Observation 1, hence . This means that
Hence, has at most edges. Checking if a mapping is a -edge colouring takes time. The number of such mappings is at most . Hence, brute force checking if has a -edge colouring takes time . Thus the first statement of the theorem follows. We obtain the second statement by applying Lemma 1. ∎
We proved that -Edge Colouring is fixed-parameter-tractable when parameterized by the number of maximum-degree vertices. We note that our proof does not work to show this for Edge Colouring, as the set may have size . As such, we pose the following open problem: what is the computational complexity of Edge Colouring when parameterized by the number of maximum-degree vertices?
Acknowledgements. We thank Leandro Zatesko for bringing paper  to our attention.
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