Classifying k-Edge Colouring for H-free Graphs

10/10/2018 ∙ by Esther Galby, et al. ∙ University of Bergen Durham University University of Fribourg 0

A graph is H-free if it does not contain an induced subgraph isomorphic to H. For every integer k and every graph H, we determine the computational complexity of k-Edge Colouring for H-free graphs.

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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 end-vertex. 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 yes-instance if has maximum degree at most by Theorem 1.1 and that is a no-instance 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 NP-complete if , as shown by Holyer for and by Leven and Galil for .

Theorem 1.2 ([10, 14])

For , -Edge Colouring is NP-complete even for -regular graphs.

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 well-known that they have chromatic index . Hence, Edge Colouring is polynomial-time solvable for bipartite graphs, which are perfect and triangle-free. In contrast, Cai and Ellis [4] proved that for every , -Edge Colouring is NP-complete 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 NP-complete for triangle-free graphs (the graph denotes the cycle on vertices).

Theorem 1.3 ([4])

Let and . Then -Edge Colouring is NP-complete for -regular -free graphs.

Theorem 1.4 ([4])

Let

be an odd integer. Then

-Edge Colouring is NP-complete for -regular line graphs of bipartite graphs.

It is also known that Edge Colouring is polynomial-time solvable for chordless graphs [16], series-parallel graphs [11], split-indifference 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 polynomial-time solvable for -free graphs. Otherwise -Edge Colouring is NP-complete even for -regular -free graphs.

We obtain Theorem 1.5 by combining Theorems 1.3 and 1.4 with two new results. In particular, we will prove a hardness result for -regular claw-free graphs for even integers  (as Theorem 1.4 is only valid when is odd).

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 vertex-disjoint 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 well-known 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 end-vertex 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 claw-free, 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 NP-complete for -regular claw-free graphs.

Proof

Recall that -Edge Colouring for -regular graphs is NP-complete 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 claw-free.

Figure 1: The gadget where is a complete graph of size for . Note that edges inside and are not drawn.

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 constant-time 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 polynomial-time solvable for -free graphs. Otherwise -Edge Colouring is NP-complete 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 constant-time solvable cases and NP-complete 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 polynomial-time 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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