Hereditary Graph Classes: When the Complexities of Colouring and Clique Cover Coincide

07/22/2016 ∙ by Alexandre Blanché, et al. ∙ Durham University 0

A graph is (H_1,H_2)-free for a pair of graphs H_1,H_2 if it contains no induced subgraph isomorphic to H_1 or H_2. In 2001, Král', Kratochvíl, Tuza, and Woeginger initiated a study into the complexity of Colouring for (H_1,H_2)-free graphs. Since then, others have tried to complete their study, but many cases remain open. We focus on those (H_1,H_2)-free graphs where H_2 is H_1, the complement of H_1. As these classes are closed under complementation, the computational complexities of Colouring and Clique Cover coincide. By combining new and known results, we are able to classify the complexity of Colouring and Clique Cover for (H,H)-free graphs for all cases except when H=sP_1+ P_3 for s≥ 3 or H=sP_1+P_4 for s≥ 2. We also classify the complexity of Colouring on graph classes characterized by forbidding a finite number of self-complementary induced subgraphs, and we initiate a study of k-Colouring for (P_r,P_r)-free graphs.

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1 Introduction

A colouring of a graph is an assignment of labels, called colours, to its vertices in such a way that no two adjacent vertices have the same label. The corresponding decision problem, Colouring, which is that of deciding whether a given graph can be coloured with at most  colours for some given positive integer , is a central problem in discrete optimization. Its complementary problem Clique Cover is that of deciding whether the vertices of a graph can be covered with at most  cliques. As Colouring and Clique Cover are NP-complete even for  [23], it is natural to restrict their input to some special graph class. A classic result in this area is due to Grötschel, Lovász, and Schrijver [15], who showed that Colouring is polynomial-time solvable for perfect graphs. However, for both problems, finding the exact borderline between tractable and computationally hard graph classes is still open.

A graph class is hereditary

if it can be characterized by some set of forbidden induced subgraphs, or equivalently, if it is closed under vertex deletion. The aforementioned class of perfect graphs is an example of such a class, as a graph is perfect if and only if it contains no induced odd holes and no induced odd antiholes 

[7]. For the case where exactly one induced subgraph is forbidden, Král’, Kratochvíl, Tuza, and Woeginger [22] were able to prove a complete dichotomy; see also Figure 1 (we write to denote that the graph  is an induced subgraph of the graph ).

Theorem 1.1 ([22])

Let  be a graph. The Colouring problem on -free graphs is polynomial-time solvable if or and it is NP-complete otherwise.

Figure 1: The graphs  such that Colouring can be solved in polynomial time on -free graphs.

We study the complexity of Colouring and Clique Cover for -free graphs, that is, when two induced subgraphs  and  are both forbidden. Both polynomial-time and NP-completeness results are known for -free graphs, as shown by many teams of researchers (see, for example, [5, 6, 9, 10, 16, 18, 22, 26, 29, 30, 31, 33]), but the complexity classification is far from complete: even if we forbid two graphs  and  of up to four vertices, there are still three open cases left, namely when ; see [26] (the graph  is the claw and  is the -vertex cycle). We refer to [14, Theorem 21] for a summary and to the recent paper [31], in which the number of missing cases when  and  are both connected graphs on at most five vertices was reduced from ten to eight.

To narrow the complexity gap between hard and easy cases and to increase our understanding of it, we initiate a systematic study into -free graph classes that are closed under complementation: if a graph  belongs to the class, then so does its complement , which is the graph with vertex set  and an edge between two distinct vertices if and only if these two vertices are not adjacent in . For such graph classes the complexities of Colouring and Clique Cover coincide, so we only need to consider the Colouring problem. A graph  is self-complementary if . We observe that a class of -free graphs is closed under complementation if either  and  are both self-complementary, or . For the first case we prove the following more general classification in Section 3.

Theorem 1.2

Let be self-complementary graphs. Then Colouring is polynomial-time solvable for -free graphs if for some  and it is NP-complete otherwise.

Hence, we may now focus on the case when and  is not self-complementary. In this case few results are known. First notice that for any integer , the class of -free graphs consists of graphs with no large independent set and no large clique. The number of vertices in such graphs is bounded by a constant by Ramsey’s Theorem. Dabrowski et al. [10] researched the effect on the complexity of changing the forbidden subgraph  by adding an extra edge, and proved that Colouring is polynomial-time solvable for -free graphs. A result of Malyshev [29] implies that Colouring is polynomial-time solvable for -free graphs. Hoàng and Lazzarato [18] proved the same for -free graphs. By combining results of [21] and [32], Colouring is polynomial-time solvable on graph classes of bounded clique-width. It is known that -free graphs [4] and -free graphs [1] have bounded clique-width, so Colouring is polynomial-time solvable for these two graph classes. In light of these results, it is a natural research question to find out to what extent we may add edges to  so that Colouring remains polynomial-time solvable on -free graphs for the resulting graph .

To narrow down the large number of open cases , we will use the known results that Colouring is NP-complete for graphs of girth at least  for any [12, 22, 25] (the girth of a graph is the length of a shortest induced cycle in it) and for -free graphs [22]. Even after using these results there are still many open cases left, and to get a handle on them we need another NP-hardness result for a very restricted graph class. In Section 4.1 we will prove such a result, namely, we show that Colouring is NP-complete even for -free graphs (see Figure 2). We will show that, by combining this result with those above, only the following open cases are left: (see Figure 3), or for . In Section 4.2 we give a polynomial-time algorithm for Colouring restricted to the class of -free graphs (note that this class has unbounded clique-width, as it contains the class of split graphs, which have unbounded clique-width [28]).

Figure 2: The six graphs corresponding to our result that Colouring is NP-complete for -free graphs.
Figure 3: The graphs  on five vertices for which the complexity of Colouring for -free graphs was unknown. We show NP-completeness if and polynomial-time solvability if .

In Section 4.3 we show that our new hardness result and new polynomial-time algorithm combine together with the aforementioned known results to give us the main theorem of our paper.

Theorem 1.3

Let  be a graph with . Then Colouring is polynomial-time solvable for -free graphs if or is an induced subgraph of , , , , , or for some and it is NP-complete otherwise.

Theorem 1.3 tells us is that we may only add a few edges to the graph if Colouring stays polynomial-time solvable on -free graphs for the resulting graph  (even in the two missing cases where or for , the graph  is only allowed to contain at most three edges). From a more general perspective, our new hardness result has significantly narrowed the classification for -free graphs.

Our immediate goal is to complete the complexity classification of Colouring for -free graphs, thus to solve the cases when or for . We note that the class of -free graphs, and thus the class of )-free graphs, has unbounded clique-width [10]. We emphasize that our long-term goal is to increase our understanding of the computational complexity of Colouring for hereditary graph classes. Another natural question is whether -Colouring (the variant of Colouring where the number of colours is fixed) is polynomial-time solvable for -free graphs when or for . This is indeed the case, as Couturier et al. [8] extended the result of [17] on -Colouring for -free graphs by proving that for every pair of integers , -Colouring is polynomial-time solvable even for -free graphs. However, for other classes of -free graphs, we show that -Colouring turns out to be NP-hard by using a construction of Huang [19], that is, we show the following results in Section 5.

Theorem 1.4

-Colouring is NP-complete for -free graphs, and thus for -free graphs.

Theorem 1.5

-Colouring is NP-complete for -free graphs, and thus for -free graphs.

As Colouring is polynomial-time solvable for -free graphs [18], it would be interesting to solve the following two open problems (see also Table 1):

  • is there an integer such that -Colouring for -free graphs is NP-complete?

  • is there an integer  such that -Colouring for -free graphs is NP-complete?

As -Colouring is polynomial-time solvable for -free graphs [2], must be at least . Similar to summaries for -Colouring for -free graphs (see, for example, [14]) and -free graphs [16], we can survey the known results and the missing cases of -Colouring for -free graphs; see Table 1.

P P P
P P ?
P P ?
P ? NP-c
Table 1: The complexity of -Colouring -free graphs for fixed values of  and . Here, P means polynomial-time solvable and NP-c means NP-complete. The entries in this table originate from Theorem 1.4 and the following two results: -Colouring is polynomial-time solvable for -free graphs for any  [17] and -Colouring is polynomial-time solvable for -free graphs [2].

2 Preliminaries

Throughout our paper we consider only finite, undirected graphs without multiple edges or self-loops. The disjoint union of two vertex-disjoint graphs  and  is denoted by  and the disjoint union of  copies of a graph  is denoted by . For a subset , we let  denote the subgraph of  induced by , which has vertex set  and edge set . If then we may write instead of . We may also write instead of .

For a set of graphs , a graph  is -free if it has no induced subgraph isomorphic to a graph in ; if , we may write -free instead of -free. For a graph , the set denotes the neighbourhood of .

The graphs , , and  denote the cycle, complete graph, star and path on  vertices, respectively. The graph is also known as the -vertex wheel. The graphs  and  are also called the triangle and claw, respectively. A graph  is a linear forest if every component of  is a path (on at least one vertex). The graph , for , denotes the subdivided claw, that is, the tree that has only one vertex  of degree  and exactly three leaves, which are of distance  and  from , respectively. Observe that .

The chromatic number  of a graph  is the smallest integer  such that  is -colourable. The clique number  is the size of a largest clique in . A graph is bipartite if its vertex set can be partitioned into two (possibly empty) independent sets.

3 The Proof of Theorem 1.2

As mentioned in Section 1, if we study the complexity of Colouring for -free classes of graphs that are closed under complementation, it is sufficient to consider the case where  and  are not self-complementary and . We will give a short proof for this claim. To do so, we will need the following lemma.

Lemma 1 ([12, 22, 25])

For any integer , Colouring is NP-complete for -free graphs.

Theorem 1.2 (restated). Let be self-complementary graphs. Then Colouring is polynomial-time solvable for -free graphs if for some  and it is NP-complete otherwise.

Proof

Let  be a self-complementary graph on  vertices. Then  must have edges. If then . Now  cannot be  or , since and are not integers. If then , by inspection. Suppose . Then , so  must contain a cycle. Thus, if  is a self-complementary graph then it is either an induced subgraph of  or it contains a cycle.

Let be self-complementary graphs. If for some , then the Colouring problem for -free graphs is polynomial-time solvable by Theorem 1.1. If for all , then each  must contain a cycle. In that case Colouring for -free graphs is NP-complete by Lemma 1.∎

4 The Proof of Theorem 1.3

In this section we prove Theorem 1.3. As part of the proof we first show that Colouring is NP-complete for -free graphs in Section 4.1 and polynomial-time solvable for -free graphs in Section 4.2.

4.1 The Hardness Result

In this section we prove the following result.

Theorem 4.1

Colouring is NP-complete for -free graphs.

Figure 4: The graph  for the instance  of Exact -Cover where , ,  and . The edges of the clique  are not shown.
Proof

Let  and  be positive integers with . Let  be a set of size . Let  be a collection of  subsets of  each of size . An exact -cover for is a set of size  such that every member of  belongs to one of the subsets in . The NP-complete problem Exact -Cover [13] is that of determining if such a set  exists. To prove the theorem, we describe a reduction from Exact -Cover to Colouring for -free graphs.

Given an instance of Exact -Cover, we construct the graph  as follows (see Figure 4 for an example):

  • Introduce a set of vertices which forms a clique in .

  • Introduce a set of vertices which forms an independent set in .

  • Add an edge from to if and only if .

  • Introduce a set of vertices which forms an independent set in .

  • Add an edge from each vertex of  to each vertex of .

Claim 1. The vertices of can be covered by at most  pairwise vertex-disjoint cliques if and only if  contains an exact -cover .
First suppose that  contains an exact -cover . Let . For each , let  be the clique on four vertices containing  and its three neighbours in ; there are  such cliques. Form a perfect matching between the vertices of and the vertices of , that is, a collection of cliques on two vertices. Together, these  pairwise vertex-disjoint cliques cover .

Now suppose that the vertices of  can be covered by at most  pairwise vertex-disjoint cliques. As  is independent and , we have exactly  cliques, and each of them contains exactly one vertex of . Hence, as  is an independent set and each vertex in  is only adjacent to the vertices of , each of the vertices of  must be contained in a clique of size  that consists of a vertex of  and a vertex of . There are  other cliques, which, as we deduced, also contain exactly one vertex of . Each vertex of  must be in one of these cliques. As each vertex in  has exactly three neighbours in  and there are  vertices in , this means that each of these cliques must contain four vertices, consisting of one vertex of  and three vertices of . Hence, if we let belong to  whenever  is in a clique of size , then  forms an exact -cover. This completes the proof of Claim 4.1.

We have shown a reduction from Exact -Cover to the problem of finding  pairwise vertex-disjoint cliques that cover . The latter problem is equivalent to finding a -colouring of  (note that  is part of the input of Exact -Cover). It remains to show that  is -free.

Since a graph  is -free if and only if  is, it is sufficient to show that  is -free. We do this in Claims 4.1-4.1, but first we prove a useful observation.

Claim 2. Let  be an induced subgraph of  that is the complement of a linear forest on six vertices. Then  contains at least four vertices of .
As a linear forest contains no triangle, contains no independent set on three vertices. So  cannot contain more than two vertices from either of the independent sets  or . Thus it contains vertices of . But then it cannot contain two vertices from  as, combined with a vertex of  this would again induce an independent set of size . Hence there are at least three vertices of  in . But this implies that not even one vertex of  belongs to  (as it would have three non-neighbours and every vertex of  is adjacent to all but at most two of the others). This completes the proof of Claim 4.1.

Claim 3. is -free.
Suppose for contradiction that there is an induced in . Since  and  are independent sets, every  in  must either have two vertices in  or one vertex in  and the other in  or . Since that  is a clique and every vertex in  is adjacent to every vertex in , one of the ’s must have both its vertices in  and the other must have one vertex in  and the other in . But every vertex in has a neighbour in such a , so no induced can exist. This completes the proof of Claim 4.1.

Claim 4. is -free.
Suppose for contradiction that there is an induced in . Note this subgraph consists of a , which we denote , plus an additional vertex, which we denote , that is adjacent to every vertex in . If , then the vertices of  are all in , and if , the vertices of the  are either all in  or all in . Neither is possible, so we must have . Therefore none of the vertices of  is in . At most two of the vertices of  are in each of  and  as  contains neither an independent set nor a clique on three vertices. So  contains exactly two vertices from each of  and , but the pair from  must be non-adjacent in  and the pair from  must be adjacent in . This contradiction completes the proof of Claim 4.1.

Claim 5. is -free.
Suppose for contradiction that there is an induced  in . Denote this graph by . As  is a clique,  contains at most two of its vertices. If it contains exactly two, then  must also contain a vertex such that the three vertices together induce a . But then  cannot be adjacent to any other vertex of  so the remaining three vertices must all belong to , a contradiction as  is independent. So  must contain at least five vertices of  and . They cannot all belong to one of these two sets as  has no independent set of size , but if  contains vertices of both sets then there must be a vertex of degree . This contradiction proves that  does not exist. This completes the proof of Claim 4.1.

Claim 6. is -free.
Suppose for contradiction that there is an induced  in . By Claim 4.1, it contains four vertices of . We note that  contains exactly one induced  and that the other two vertices are adjacent to exactly two vertices in the clique (so must both be in ) and to each other (so cannot both be in ). This contradiction completes the proof of Claim 4.1.

Claim 7. is -free.
Suppose for contradiction that there is an induced  in . Denote this path by . As  is a clique,  contains at most two of its vertices. We see that  cannot contain exactly four vertices of either  or  as they would form an independent set of size . cannot contain exactly three vertices of either  or  as then there would be a vertex in the other of the two sets of degree . Finally, cannot contain at least two vertices from each of  and  as then it would contain an induced . These contradictions prove that  does not exist. This completes the proof of Claim 4.1.

Claim 8. is -free.
Suppose for contradiction that there is an induced  in . By Claim 4.1, it contains four vertices of . This contradiction – a clique on four vertices is not an induced subgraph of  – completes the proof of Claim 4.1.∎

4.2 A Tractable Case

We show that Colouring is polynomial-time solvable for -free graphs. We first introduce some additional terminology. The clique-width  of a graph  is the minimum number of labels needed to construct  by using the following four operations:

  1. creating a new graph consisting of a single vertex  with label ;

  2. taking the disjoint union of two labelled graphs  and ;

  3. joining each vertex with label  to each vertex with label  ();

  4. renaming label  to .

A class of graphs  has bounded clique-width if there is a constant  such that the clique-width of every graph in  is at most ; otherwise the clique-width is unbounded. For an induced subgraph  of a graph , the subgraph complementation operation (acting on  with respect to ) replaces every edge present in  by a non-edge, and vice versa. Similarly, for two disjoint vertex subsets  and  in , the bipartite complementation operation with respect to  and  acts on  by replacing every edge with one end-vertex in  and the other one in  by a non-edge and vice versa. Let be a constant and let  be some graph operation. We say that a graph class  is -obtained from a graph class  if the following two conditions hold:

  1. every graph in  is obtained from a graph in  by performing  at most  times, and

  2. for every there exists at least one graph in  that is obtained from  by performing  at most  times.

We say that  preserves boundedness of clique-width if for any finite constant  and any graph class , any graph class  that is -obtained from  has bounded clique-width if and only if  has bounded clique-width.

  1. Vertex deletion preserves boundedness of clique-width [27].

  2. Subgraph complementation preserves boundedness of clique-width [20].

  3. Bipartite complementation preserves boundedness of clique-width [20].

Two (non-adjacent) vertices are false twins if they have the same neighbourhood. We will use the following two lemmas, which are readily seen.

Lemma 2

The clique-width of a graph of maximum degree at most  is at most .

Lemma 3

If a vertex  in a graph  has a false twin then .

The following lemma for -free bipartite graphs follows from a result of Lozin [24], who proved that the superclass of -free bipartite graphs has bounded clique-width (see [11] for a full classification of the boundedness of clique-width of -free bipartite graphs).

Lemma 4 ([24])

The class of -free bipartite graphs has bounded clique-width.

A graph  is perfect if for every induced subgraph  of . Besides the above lemmas we will also apply the following three well-known theorems.

Theorem 4.2 ([7])

A graph is perfect if and only if it is -free and -free for every odd .

Theorem 4.3 ([15])

Colouring is polynomial-time solvable on perfect graphs.

Theorem 4.4 ([21, 32])

For any constant , Colouring is polynomial-time solvable on graphs of clique-width at most .

If  is a graph, then a (possibly empty) set is a clique separator if  is a clique and is disconnected. A graph  is an atom if it has no clique separator. Due to the following result of Tarjan, when considering the Colouring problem in a hereditary class, we only need to consider atoms in the class.

Lemma 5 ([34])

If Colouring is polynomial-time solvable on atoms in an hereditary class , then it is polynomial-time solvable on all graphs in .

Let  be a set of vertices in a graph . A vertex is complete (resp. anti-complete) to  if it is adjacent (resp. non-adjacent) to every vertex of . A set of vertices is complete (resp. anti-complete) to  if every vertex in  is complete (resp. anti-complete) to . If  and  are disjoint sets of vertices in a graph, we say that the edges between these two sets form a matching if each vertex in  has at most one neighbour in  and vice versa (if each vertex has exactly one such neighbour, we say that the matching is perfect). Similarly, the edges between these sets form a co-matching if each vertex in  has at most one non-neighbour in  and vice versa.

In Theorem 4.5 we present an algorithm for Colouring restricted to -free graphs. Our algorithm first tries to reduce to perfect graphs, in which case we can use Theorem 4.3. If a -free graph is not perfect, then we show that it must contain an induced . The clique-width of such graphs is not bounded (we can construct a -free graph of arbitrarily large clique-width by taking a split graph of arbitrarily large clique-width [28] with clique  and independent set  and adding five new vertices that form an induced  and that are complete to  but anti-complete to ). Therefore, we do some pre-processing to simplify the graph. This enables us to bound the clique-width so that we may then apply Theorem 4.4. Our general scheme for bounding the clique-width is to partition the remaining vertices into sets according to their neighbourhood in the  and then investigate the possible edges both inside these sets and between them. This enables us to use graph operations (see also Facts 13) that do not change the clique-width by “too much” to partition the input graph into disjoint pieces known to have bounded clique-width.

Theorem 4.5

Colouring is polynomial-time solvable for -free graphs.

Proof

Let  be a -free graph. By Lemma 5 we may assume that  is an atom and so has no clique separator (we will use this assumption in the proof of Claim 4.2). We first test in time whether  contains an induced . If not, then  is -free. Since , is also -free. Since  is -free, is -free for all . Since  is -free, is -free for all . By Theorem 4.2, this means that  is a perfect graph. Therefore, by Theorem 4.3, we can solve Colouring in polynomial time in this case.

Now assume our algorithm found an induced -vertex cycle  with vertices , in that order. For , let  be the set of vertices such that . A set  is large if it contains at least three vertices, otherwise it is small.

To ease notation, in the remainder of the proof, subscripts on vertex sets should be interpreted modulo  and whenever possible we will write  instead of  and  instead of  and so on.

We start with three structural claims.

Claim 1. is an independent set.
If are adjacent then is a , a contradiction. Therefore is an independent set. This completes the proof of Claim 4.2.

Claim 2. and for .
Suppose for contradiction that . If  and  are adjacent then is a and if  and  are non-adjacent then is a . This contradiction implies that . Claim 4.2 follows by symmetry.

Claim 3. is an independent set for .
Suppose for contradiction that are adjacent. Then is a , a contradiction. Therefore is an independent set. Claim 4.2 follows by symmetry.

Since , it follows that  also contains a , namely on the vertices , , , and  in that order. Therefore  is also a -free graph containing an induced . As a result, we immediately obtain the following claims as corollaries of Claims 4.24.2.

Claim 4. is a clique.
Claim 5. and for .
Claim 6. is a clique for .

Let  be the set of vertices in  that have a non-neighbour in .

Claim 7. is -colourable if and only if is -colourable.
The “only if” direction is trivial. Suppose  is -colourable. Fix a -colouring  of . We will show how to extend to all vertices of .

We may assume  and  are non-empty, otherwise and the claim follows trivially. Note that is an independent set by Claim 4.2 and that  is a clique by Claim 4.2. By assumption, is not a clique separator in . Since  is an independent set, every vertex of  must have a neighbour in that is adjacent to some vertex of the cycle . Let  be the set of vertices outside  that have a neighbour in . Note that  contains no vertex of , as  is an independent set. Hence, every vertex of  is adjacent to least one vertex of (but not to all vertices of ).

We claim that  is complete to . Indeed, suppose for contradiction that  is non-adjacent to . Let  be a vertex in  that is a neighbour of . Without loss of generality, assume that  is adjacent to . Then  is adjacent to  or , otherwise would be a , a contradiction. Without loss of generality, assume that  is adjacent to . Then  is adjacent to , otherwise would be a , a contradiction. Thus  is adjacent to  or , otherwise would be a , a contradiction. Without loss of generality, assume that  is adjacent to . Now  must be adjacent to , otherwise would be a , a contradiction. Therefore , a contradiction. It follows that  must indeed be complete to .

Let . By definition of , there is a vertex that is non-adjacent to . We claim that  can also be used to colour . Indeed, every neighbour of  lies in . Since  is a clique and  is complete to , no vertex of is coloured with the colour . We may therefore colour  with the colour