1 Introduction
A graph is edge colourable for some integer if there exists a mapping such that for any two edges and of that have a common endvertex. The chromatic index of is the smallest integer such that is edge colourable. Vizing proved the following classical result.
Theorem 1.1 ([20])
The chromatic index of a graph with maximum degree is either or .
The Edge Colouring problem is to decide if a given graph is edge colourable for some given integer . Note that is a yesinstance if has maximum degree at most by Theorem 1.1 and that is a noinstance if has maximum degree at least . If is fixed, that is, is not part of the input, then we denote the problem by Edge Colouring. It is trivial to solve this problem for . However, the problem is NPcomplete if , as shown by Holyer for and by Leven and Galil for .
Due to the above hardness results we may wish to restrict the input to some special graph class. A natural property of a graph class is to be closed under vertex deletion. Such graph classes are called hereditary and they form the focus of our paper. To give an example, bipartite graphs form a hereditary graph class, and it is wellknown that they have chromatic index . Hence, Edge Colouring is polynomialtime solvable for bipartite graphs, which are perfect and trianglefree. In contrast, Cai and Ellis [4] proved that for every , Edge Colouring is NPcomplete for regular comparability graphs, which are also perfect. They also proved the following two results, the first one of which shows that Edge Colouring is NPcomplete for trianglefree graphs (the graph denotes the cycle on vertices).
Theorem 1.3 ([4])
Let and . Then Edge Colouring is NPcomplete for regular free graphs.
Theorem 1.4 ([4])
Let
be an odd integer. Then
Edge Colouring is NPcomplete for regular line graphs of bipartite graphs.It is also known that Edge Colouring is polynomialtime solvable for chordless graphs [16], seriesparallel graphs [11], splitindifference graphs [19] and for graphs of treewidth at most for any constant [1].
It is not difficult to see that a graph class is hereditary if and only if it can be characterized by a set of forbidden induced subgraphs (see, for example, [12]). Malyshev determined the complexity of Edge Colouring for every hereditary graph class , for which consists of graphs that each have at most five vertices, except perhaps two graphs that may contain six vertices [17]. Malyshev performed a similar complexity study for Edge Colouring for graph classes defined by a family of forbidden (but not necessarily induced) graphs with at most seven vertices and at most six edges [18].
We focus on the case where consists of a single graph . A graph is free if does not contain an induced subgraph isomorphic to . We obtain the following dichotomy for free graphs.
Theorem 1.5
Let be an integer and be a graph. If is a linear forest, then Edge Colouring is polynomialtime solvable for free graphs. Otherwise Edge Colouring is NPcomplete even for regular free graphs.
2 Preliminaries
The graphs , and denote the path, cycle and complete graph on vertices, respectively. A set is an independent set of a graph if all vertices of are pairwise nonadjacent in . A graph is bipartite if its vertex set can be partitioned into two independent sets and . If there exists an edge between every vertex of and every vertex of , then is complete bipartite. The claw is the complete bipartite graph with and .
Let and be two vertexdisjoint graphs. The join operation adds an edge between every vertex of and every vertex of . The disjoint union operation merges and into one graph without adding any new edges, that is, . We write to denote the disjoint union of copies of a graph .
A forest is a graph with no cycles. A linear forest is a forest of maximum degree at most 2, or equivalently, a disjoint union of one or more paths. A graph is a cograph if can be generated from by a sequence of join and disjoint union operations. A graph is a cograph if and only if it is free (see, for example, [3]). The following wellknown lemma follows from this equivalence and the definition of a cograph.
Lemma 1
Every connected free graph on at least two vertices has a spanning complete bipartite subgraph.
Let be a graph. For a subset , the graph denotes the subgraph of induced by . We say that is connected if is connected. Recall that a graph is free for some graph if does not contain as an induced subgraph. A subset is dominating if every vertex of is adjacent to least one vertex of . We will need the following result of Camby and Schaudt.
Theorem 2.1 ([5])
Let and be a connected free graph. Let be any minimum connected dominating set of . Then is either free or isomorphic to .
Let be some graph. The degree of a vertex is equal to the size of its neighbourhood . The graph is regular if every vertex of has degree . The line graph of is the graph , which has vertex set and an edge between two distinct vertices and if and only if and have a common endvertex in .
3 The Proof of Theorem 1.5
To prove our dichtomy, we first consider the case where the forbidden induced subgraph is a claw. As line graphs are clawfree, we only need to deal with the case where the number of colours is even due to Theorem 1.4. For proving this case we need another result of Cai and Ellis, which we will use as a lemma. Let be a edge colouring of a graph . Then a vertex misses colour if none of the edges incident to is coloured .
Lemma 2 ([4])
For even , the complete graph has a edge colouring with the property that can be partitioned into sets , such that for , vertices and miss the same colour, which is not missed by any of the other vertices.
We use Lemma 2 to prove the following result, which solves the case where is even and .
Lemma 3
Let be an even integer. Then Edge Colouring is NPcomplete for regular clawfree graphs.
Proof
Recall that Edge Colouring for regular graphs is NPcomplete for every integer due to Theorem 1.2. Consider an instance of Edge Colouring, where is regular for some even integer . From we construct a graph as follows. First we replace every vertex in by the gadget shown in Figure 1. Next we connect the different gadgets in the following way. Every gadget has exactly pendant edges, which are incident with vertices , respectively. As is regular, every vertex has neighbours in . Hence, we can identify each edge of with a unique edge in , which is a pendant edge of both and . It is readily seen that is regular and clawfree.
First suppose that is edge colourable. Let be a edge colouring of . Consider a vertex . For every neighbour of in , we colour the pendant edge in corresponding to the edge with colour . As assigned different colours to the edges incident to , the pendant edges of will receive pairwise distinct colours, which we denote by . By Lemma 2, we can colour the edges of in such a way that for , and miss colour . For , we can therefore assign colour to edge . Similarly, we may assume that for , and miss colour . For , we can therefore assign colour to edge . Recall that the colours , are all different. Hence, doing this procedure for each vertex of yields a edge colouring of .
Now suppose that is edge colourable. Let be a edge colouring of . Consider some . Denote the pendant edges of by for , where is incident to (and to some vertex in a gadget for each neighbour of in ). Suppose that gave colour to an edge for some , say to , but not to any edge for . Note that cannot be coloured . As every vertex of has degree , every with and every with is incident to some edge coloured . As is neither the colour of nor the colour of , the complete graph contains a perfect matching all of whose edges have colour . However, has odd size . Hence, this is not possible. We conclude that each of the (pairwise distinct) colours of , which we denote by , is the colour of an edge for some .
Let be the (pairwise distinct) colours of , respectively. By the same arguments as above, we find that each of those colours is also the colour of a pendant edge of that is incident to a vertex for some . Note that , are pairwise distinct colours, as they are colours of edges incident to the same vertex, namely vertex . Hence, we can define a colouring of by setting for every edge with corresponding edge .∎
We note that the graph in the proof of Lemma 3 is not a line graph, as the gadget is not a line graph: the vertices form a diamond and by adding the pendant edge incident to and the edge we obtain an induced subgraph of that is not a line graph.
To handle the case where the forbidden induced subgraph is a path, we make the following observation.
Observation 1
If a graph of maximum degree has a dominating set of size at most , then has at most vertices.
We use Observation 1 in the proof of the following lemma.
Lemma 4
Let and . Then Edge Colouring is constanttime solvable for free graphs.
Proof
Let be a free graph with maximum degree . If , then is edge colourable by Theorem 1.1. If , then is not edge colourable. Hence, we may assume that . We may assume without loss of generality that is connected. We claim that has at most vertices for some function that only depends on and . As we assume that and are constants, this means that we can check in constant time if is edge colourable.
We prove the above claim by induction on . First suppose (and observe that if the claim holds for , it also holds for ). As is connected, has a dominating set of size due to Lemma 1. Hence, by Observation 1, has at most vertices. Now suppose . Let be an arbitrary minimum connected dominating set of . By Theorem 2.1, is either free or isomorphic to . In the first case we use the induction hypothesis to conclude that has at most vertices. Hence, has at most vertices by Observation 1. In the second case, we find that has at most vertices. We set . ∎
We are now ready to prove Theorem 1.5, which we restate below.
Theorem 1.5. (restated) Let be an integer and be a graph. If is a linear forest, then Edge Colouring is polynomialtime solvable for free graphs. Otherwise Edge Colouring is NPcomplete even for regular free graphs.
Proof
First suppose that contains a cycle for some . Then the class of free graphs is a superclass of the class of free graphs. This means that we can apply Theorem 1.3. From now on assume that contains no cycle, so is a forest. Suppose that contains a vertex of degree at least . Then the class of free graphs is a superclass of the class of free graphs, which in turn forms a superclass of the class of line graphs. Hence, if is odd, then we apply Theorem 1.4, and if is even, then we apply Lemma 3. From now on assume that contains no cycle and no vertex of degree at least . Then is a linear forest, say with connected components. Let . Then the class of free graphs is contained in the class of free graphs. Hence we may apply Lemma 4. This completes the proof of Theorem 1.5. ∎
4 Conclusions
We gave a complete complexity classification of Edge Colouring for free graphs, showing a dichotomy between constanttime solvable cases and NPcomplete cases. We saw that this depends on being a linear forest or not. It would be interesting to prove a dichotomy result for Edge Colouring restricted to free graphs. Note that due to Theorem 1.5 we only need to consider the case where is a linear forest. However, even determining the complexity for small linear forests , such as the cases where and , turns out to be a difficult problem. In fact, the computational complexity of Edge Colouring for split graphs, or equivalently, free graphs [8] and for free graphs has yet to be settled, despite the efforts towards solving the problem for these graph classes [6, 7, 15].
On a side note, a graph is edge colourable if and only if its line graph is vertex colourable. In contrast to the situation for Edge Colouring, the computational complexity of Vertex Colouring has been fully classified for free graphs [13]. However, the complexity status of Vertex Colouring is still open for free graphs if and after Bonomo et al. [2] proved polynomialtime solvability for and ; see the survey [9] for further background information.
Acknowledgements
The present work was done when the second author was visiting the University of Fribourg funded by a scholarship of the University of Fribourg.
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