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

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 $k$ colours for some given positive integer $k$ , 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 $k$ cliques. As Colouring and Clique Cover are NP-complete even for $k = 3$ [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

G^{'} \subseteq_{i} G

to denote that the graph

G^{'}

is an induced subgraph of the graph

G

Theorem 1.1 ([22])

Let $H$ be a graph. The Colouring problem on $H$ -free graphs is polynomial-time solvable if $H \subseteq_{i} P_{1} + P_{3}$ or $H \subseteq_{i} P_{4}$ and it is NP-complete otherwise.


$P_{1} + P_{3}$	$P_{4}$

Figure 1: The graphs

H

such that Colouring can be solved in polynomial time on

H

-free graphs.

We study the complexity of Colouring and Clique Cover for $(H_{1}, H_{2})$ -free graphs, that is, when two induced subgraphs $H_{1}$ and $H_{2}$ are both forbidden. Both polynomial-time and NP-completeness results are known for $(H_{1}, H_{2})$ -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 $H_{1}$ and $H_{2}$ of up to four vertices, there are still three open cases left, namely when $(H1,H2)∈{(K1,3,4P1),\allowbreak(K1,3,2P1+P2),\allowbreak(C4,4P1)}$ ; see [26] (the graph $K_{1, 3}$ is the claw and $C_{4}$ is the $4$ -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 $H_{1}$ and $H_{2}$ 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 $(H_{1}, H_{2})$ -free graph classes that are closed under complementation: if a graph $G$ belongs to the class, then so does its complement $¯ ¯¯ ¯ G$ , which is the graph with vertex set $V (G)$ and an edge between two distinct vertices if and only if these two vertices are not adjacent in $G$ . For such graph classes the complexities of Colouring and Clique Cover coincide, so we only need to consider the Colouring problem. A graph $H$ is self-complementary if $H = ¯ ¯¯¯ ¯ H$ . We observe that a class of $(H_{1}, H_{2})$ -free graphs is closed under complementation if either $H_{1}$ and $H_{2}$ are both self-complementary, or $H_{2} =_{1}$ . For the first case we prove the following more general classification in Section 3.

Theorem 1.2

Let $H_{1}, \dots, H_{k}$ be self-complementary graphs. Then Colouring is polynomial-time solvable for $(H_{1}, \dots, H_{k})$ -free graphs if $H_{i} \subseteq_{i} P_{4}$ for some $i$ and it is NP-complete otherwise.

Hence, we may now focus on the case when $H_{2} =_{1}$ and $H_{1}$ is not self-complementary. In this case few results are known. First notice that for any integer $s \geq 1$ , the class of $(s P_{1}, ¯ ¯¯¯¯¯¯ ¯ s P_{1})$ -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 $s P_{1}$ by adding an extra edge, and proved that Colouring is polynomial-time solvable for $(s P_{1} + P_{2}, ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ s P_{1} + P_{2})$ -free graphs. A result of Malyshev [29] implies that Colouring is polynomial-time solvable for $(K_{1, 3},_{1, 3})$ -free graphs. Hoàng and Lazzarato [18] proved the same for $(P_{5},_{5})$ -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 $(P1+P4,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P1+P4)$ -free graphs [4] and $(2P1+P3,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯2P1+P3)$ -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 $s P_{1}$ so that Colouring remains polynomial-time solvable on $(H, ¯ ¯¯¯ ¯ H)$ -free graphs for the resulting graph $H$ .

To narrow down the large number of open cases $(H, ¯ ¯¯¯ ¯ H)$ , we will use the known results that Colouring is NP-complete for graphs of girth at least $p$ for any $p \geq 3$ [12, 22, 25] (the girth of a graph is the length of a shortest induced cycle in it) and for $(K_{1, 3}, K_{4}, ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ 2 P_{1} + P_{2})$ -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 $(P1+2P2,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P1+2P2,\allowbreak2P3,\allowbreak¯¯¯¯¯¯¯¯2P3,\allowbreakP6,¯¯¯¯¯¯P6)$ -free graphs (see Figure 2). We will show that, by combining this result with those above, only the following open cases are left: $H = P_{2} + P_{3}$ (see Figure 3), $H = (s + 1) P_{1} + P_{3}$ or $H = s P_{1} + P_{4}$ for $s \geq 2$ . In Section 4.2 we give a polynomial-time algorithm for Colouring restricted to the class of $(P2+P3,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P2+P3)$ -free graphs (note that this class has unbounded clique-width, as it contains the class of split graphs, which have unbounded clique-width [28]).


$P_{1} + 2 P_{2}$	$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{1} + 2 P_{2}$	$2 P_{3}$	$¯ ¯¯¯¯¯¯ ¯ 2 P_{3}$	$P_{6}$	$_{6}$

Figure 2: The six graphs corresponding to our result that Colouring is NP-complete for

(P1+2P2,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P1+2P2,\allowbreak2P3,\allowbreak¯¯¯¯¯¯¯¯2P3,\allowbreakP6,¯¯¯¯¯¯P6)

-free graphs.


$P_{1} + 2 P_{2}$	$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{1} + 2 P_{2}$	$P_{2} + P_{3}$	$¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{2} + P_{3}$

Figure 3: The graphs

H

on five vertices for which the complexity of Colouring for

(H, ¯ ¯¯¯ ¯ H)

-free graphs was unknown. We show NP-completeness if

H \in {P_{1} + 2 P_{2}, ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{1} + 2 P_{2}}

and polynomial-time solvability if

H \in {P_{2} + P_{3}, ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{2} + P_{3}}

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 $H$ be a graph with $H, ¯ ¯¯¯ ¯ H \notin {(s + 1) P_{1} + P_{3}, s P_{1} + P_{4} | s \geq 2}$ . Then Colouring is polynomial-time solvable for $(H, ¯ ¯¯¯ ¯ H)$ -free graphs if $H$ or $¯ ¯¯¯ ¯ H$ is an induced subgraph of $K_{1, 3}$ , $P_{1} + P_{4}$ , $2 P_{1} + P_{3}$ , $P_{2} + P_{3}$ , $P_{5}$ , or $s P_{1} + P_{2}$ for some $s \geq 0$ and it is NP-complete otherwise.

Theorem 1.3 tells us is that we may only add a few edges to the graph $s P_{1}$ if Colouring stays polynomial-time solvable on $(H, ¯ ¯¯¯ ¯ H)$ -free graphs for the resulting graph $H$ (even in the two missing cases where $H = (s + 1) P_{1} + P_{3}$ or $H = s P_{1} + P_{4}$ for $s \geq 2$ , the graph $H$ is only allowed to contain at most three edges). From a more general perspective, our new hardness result has significantly narrowed the classification for $(H_{1}, H_{2})$ -free graphs.

Our immediate goal is to complete the complexity classification of Colouring for $(H, ¯ ¯¯¯ ¯ H)$ -free graphs, thus to solve the cases when $H = (s + 1) P_{1} + P_{3}$ or $H = s P_{1} + P_{4}$ for $s \geq 2$ . We note that the class of $(3 P_{1} + P_{2}, K_{4})$ -free graphs, and thus the class of $(3P1+P3,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯3P1+P3$ )-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 $k$ -Colouring (the variant of Colouring where the number of colours is fixed) is polynomial-time solvable for $(H, ¯ ¯¯¯ ¯ H)$ -free graphs when $H = (s + 1) P_{1} + P_{3}$ or $H = s P_{1} + P_{4}$ for $s \geq 2$ . This is indeed the case, as Couturier et al. [8] extended the result of [17] on $k$ -Colouring for $P_{5}$ -free graphs by proving that for every pair of integers $k, s \geq 1$ , $k$ -Colouring is polynomial-time solvable even for $(s P_{1} + P_{5})$ -free graphs. However, for other classes of $(H, ¯ ¯¯¯ ¯ H)$ -free graphs, we show that $k$ -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

$4$ -Colouring is NP-complete for $(P_{7},_{8})$ -free graphs, and thus for $(P_{8},_{8})$ -free graphs.

Theorem 1.5

$5$ -Colouring is NP-complete for $(P_{6}, ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{1} + P_{6})$ -free graphs, and thus for $(P_{1} + P_{6}, ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{1} + P_{6})$ -free graphs.

As Colouring is polynomial-time solvable for $(P_{5},_{5})$ -free graphs [18], it would be interesting to solve the following two open problems (see also Table 1):

is there an integer $k$ such that $k$ -Colouring for $(P_{6},_{6})$ -free graphs is NP-complete?
is there an integer $k$ such that $k$ -Colouring for $(P_{7},_{7})$ -free graphs is NP-complete?

As $3$ -Colouring is polynomial-time solvable for $P_{7}$ -free graphs [2], $k$ must be at least $4$ . Similar to summaries for $k$ -Colouring for $P_{t}$ -free graphs (see, for example, [14]) and $(C_{s}, P_{t})$ -free graphs [16], we can survey the known results and the missing cases of $k$ -Colouring for $(P_{t},_{t})$ -free graphs; see Table 1.

$t ∖ k$	$\leq 2$	$3$	$\geq 4$
$\leq 5$	P	P	P
$6$	P	P	?
$7$	P	P	?
$\geq 8$	P	?	NP-c

Table 1: The complexity of

k

-Colouring

(P_{t},_{t})

-free graphs for fixed values of

k

and

t

. 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:

k

-Colouring is polynomial-time solvable for

P_{5}

-free graphs for any

k \geq 1

[17] and

3

-Colouring is polynomial-time solvable for

P_{7}

-free graphs [2].

2 Preliminaries

Throughout our paper we consider only finite, undirected graphs without multiple edges or self-loops. The disjoint union $(V (G) \cup V (H), E (G) \cup E (H))$ of two vertex-disjoint graphs $G$ and $H$ is denoted by $G + H$ and the disjoint union of $r$ copies of a graph $G$ is denoted by $r G$ . For a subset $S \subseteq V (G)$ , we let $G [S]$ denote the subgraph of $G$ induced by $S$ , which has vertex set $S$ and edge set ${u v | u, v \in S, u v \in E (G)}$ . If $S = {s_{1}, \dots, s_{r}}$ then we may write $G [s_{1}, \dots, s_{r}]$ instead of $G [{s_{1}, \dots, s_{r}}]$ . We may also write $G ∖ S$ instead of $G [V (G) ∖ S]$ .

For a set of graphs ${H_{1}, \dots, H_{p}}$ , a graph $G$ is $(H_{1}, \dots, H_{p})$ -free if it has no induced subgraph isomorphic to a graph in ${H_{1}, \dots, H_{p}}$ ; if $p = 1$ , we may write $H_{1}$ -free instead of $(H_{1})$ -free. For a graph $G$ , the set $N (u) = {v \in V | u v \in E}$ denotes the neighbourhood of $u \in V (G)$ .

The graphs $C_{r}$ , $K_{r}$ , $K_{1, r - 1}$ and $P_{r}$ denote the cycle, complete graph, star and path on $r$ vertices, respectively. The graph $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{1} + 2 P_{2}$ is also known as the $5$ -vertex wheel. The graphs $K_{3}$ and $K_{1, 3}$ are also called the triangle and claw, respectively. A graph $G$ is a linear forest if every component of $G$ is a path (on at least one vertex). The graph $S_{h, i, j}$ , for $1 \leq h \leq i \leq j$ , denotes the subdivided claw, that is, the tree that has only one vertex $x$ of degree $3$ and exactly three leaves, which are of distance $h$ , $i$ and $j$ from $x$ , respectively. Observe that $S_{1, 1, 1} = K_{1, 3}$ .

The chromatic number $χ (G)$ of a graph $G$ is the smallest integer $k$ such that $G$ is $k$ -colourable. The clique number $ω (G)$ is the size of a largest clique in $G$ . 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 $(H_{1}, H_{2})$ -free classes of graphs that are closed under complementation, it is sufficient to consider the case where $H_{1}$ and $H_{2}$ are not self-complementary and $H_{2} =_{1}$ . 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 $k \geq 3$ , Colouring is NP-complete for $(C_{3}, C_{4}, \dots, C_{k})$ -free graphs.

Theorem 1.2 (restated). Let $H_{1}, \dots, H_{k}$ be self-complementary graphs. Then Colouring is polynomial-time solvable for $(H_{1}, \dots, H_{k})$ -free graphs if $H_{i} \subseteq_{i} P_{4}$ for some $i$ and it is NP-complete otherwise.

Proof

Let $H$ be a self-complementary graph on $n$ vertices. Then $H$ must have $\frac{1}{2} (\frac{n}{2})$ edges. If $n = 1$ then $H = P_{1}$ . Now $n$ cannot be $2$ or $3$ , since $\frac{1}{2} (\frac{2}{2})$ and $\frac{1}{2} (\frac{3}{2})$ are not integers. If $n = 4$ then $H = P_{4}$ , by inspection. Suppose $n \geq 5$ . Then $\frac{1}{2} (\frac{n}{2}) = \frac{n (n - 1)}{4} \geq n$ , so $H$ must contain a cycle. Thus, if $H$ is a self-complementary graph then it is either an induced subgraph of $P_{4}$ or it contains a cycle.

Let $H_{1}, \dots, H_{k}$ be self-complementary graphs. If $H_{i} \subseteq_{i} P_{4}$ for some $i$ , then the Colouring problem for $(H_{1}, \dots, H_{k})$ -free graphs is polynomial-time solvable by Theorem 1.1. If $H_{i} ⊈_{i} P_{4}$ for all $i$ , then each $H_{i}$ must contain a cycle. In that case Colouring for $(H_{1}, \dots, H_{k})$ -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 $(P1+2P2,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P1+2P2,\allowbreak2P3,\allowbreak¯¯¯¯¯¯¯¯2P3,\allowbreakP6,¯¯¯¯¯¯P6)$ -free graphs in Section 4.1 and polynomial-time solvable for $(P2+P3,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P2+P3)$ -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 $(P1+2P2,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P1+2P2,\allowbreak2P3,\allowbreak¯¯¯¯¯¯¯¯2P3,\allowbreakP6,¯¯¯¯¯¯P6)$ -free graphs.

Figure 4: The graph

G_{W, U}

for the instance

(W, U)

of Exact

3

-Cover where

q = 3

k = 6

W = {1, \dots, 9}

and

U={{1,2,3},\allowbreak{2,3,4},\allowbreak{3,4,7},\allowbreak{4,5,6},\allowbreak{6,7,8},\allowbreak{7,8,9}}

. The edges of the clique

V_{W}

are not shown.

Proof

Let $k$ and $q$ be positive integers with $k \geq q$ . Let $W$ be a set of size $3 q$ . Let $U$ be a collection of $k$ subsets of $W$ each of size $3$ . An exact $3$ -cover for $(W, U)$ is a set $U^{'} \subseteq U$ of size $q$ such that every member of $W$ belongs to one of the subsets in $U^{'}$ . The NP-complete problem Exact $3$ -Cover [13] is that of determining if such a set $U^{'}$ exists. To prove the theorem, we describe a reduction from Exact $3$ -Cover to Colouring for $(P1+2P2,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P1+2P2,\allowbreak2P3,\allowbreak¯¯¯¯¯¯¯¯2P3,\allowbreakP6,¯¯¯¯¯¯P6)$ -free graphs.

Given an instance $(W, U)$ of Exact $3$ -Cover, we construct the graph $G_{W, U}$ as follows (see Figure 4 for an example):

Introduce a set of vertices $V_{W} = {v_{w} ∣ w \in W}$ which forms a clique in $G_{W, U}$ .
Introduce a set of vertices $V_{U} = {v_{u} ∣ u \in U}$ which forms an independent set in $G_{W, U}$ .
Add an edge from $v_{w} \in V_{W}$ to $v_{u} \in V_{U}$ if and only if $w \in u$ .
Introduce a set of vertices $A = {a_{i} ∣ 1 \leq i \leq k - q}$ which forms an independent set in $G_{W, U}$ .
Add an edge from each vertex of $A$ to each vertex of $V_{U}$ .

Claim 1. The vertices of $G_{W, U}$ can be covered by at most $k$ pairwise vertex-disjoint cliques if and only if $U$ contains an exact $3$ -cover $U^{'}$ .
First suppose that $U$ contains an exact $3$ -cover $U^{'}$ . Let $V_{U^{'}} = {v_{u} \in V_{U} ∣ u \in U^{'}}$ . For each $u \in U^{'}$ , let $K^{u}$ be the clique on four vertices containing $v_{u}$ and its three neighbours in $V_{W}$ ; there are $q$ such cliques. Form a perfect matching between the vertices of $V_{U} ∖ V_{U^{'}}$ and the vertices of $A$ , that is, a collection of $k - q$ cliques on two vertices. Together, these $k$ pairwise vertex-disjoint cliques cover $V (G_{W, U})$ .

Now suppose that the vertices of $G_{W, U}$ can be covered by at most $k$ pairwise vertex-disjoint cliques. As $V_{U}$ is independent and $| V_{U} | = k$ , we have exactly $k$ cliques, and each of them contains exactly one vertex of $V_{U}$ . Hence, as $A$ is an independent set and each vertex in $A$ is only adjacent to the vertices of $V_{U}$ , each of the $k - q$ vertices of $A$ must be contained in a clique of size $2$ that consists of a vertex of $A$ and a vertex of $V_{U}$ . There are $q$ other cliques, which, as we deduced, also contain exactly one vertex of $V_{U}$ . Each vertex of $V_{W}$ must be in one of these cliques. As each vertex in $V_{U}$ has exactly three neighbours in $V_{W}$ and there are $3 q$ vertices in $V_{W}$ , this means that each of these cliques must contain four vertices, consisting of one vertex of $V_{U}$ and three vertices of $V_{W}$ . Hence, if we let $u \in U$ belong to $U^{'}$ whenever $v_{u}$ is in a clique of size $4$ , then $U^{'}$ forms an exact $3$ -cover. This completes the proof of Claim 4.1.

We have shown a reduction from Exact $3$ -Cover to the problem of finding $k$ pairwise vertex-disjoint cliques that cover $V (G_{W, U})$ . The latter problem is equivalent to finding a $k$ -colouring of $_{W, U}$ (note that $k$ is part of the input of Exact $3$ -Cover). It remains to show that $_{W, U}$ is $(P1+2P2,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P1+2P2,\allowbreak2P3,\allowbreak¯¯¯¯¯¯¯¯2P3,\allowbreakP6,¯¯¯¯¯¯P6)$ -free.

Since a graph $G$ is $(P1+2P2,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P1+2P2,\allowbreak2P3,\allowbreak¯¯¯¯¯¯¯¯2P3,\allowbreakP6,¯¯¯¯¯¯P6)$ -free if and only if $¯ ¯¯ ¯ G$ is, it is sufficient to show that $G_{W, U}$ is $(P1+2P2,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P1+2P2,\allowbreak2P3,\allowbreak¯¯¯¯¯¯¯¯2P3,\allowbreakP6,¯¯¯¯¯¯P6)$ -free. We do this in Claims 4.1-4.1, but first we prove a useful observation.

Claim 2. Let $J$ be an induced subgraph of $G_{W, U}$ that is the complement of a linear forest on six vertices. Then $J$ contains at least four vertices of $V_{W}$ .
As a linear forest contains no triangle, $J$ contains no independent set on three vertices. So $J$ cannot contain more than two vertices from either of the independent sets $V_{U}$ or $A$ . Thus it contains vertices of $V_{W}$ . But then it cannot contain two vertices from $A$ as, combined with a vertex of $V_{W}$ this would again induce an independent set of size $3$ . Hence there are at least three vertices of $V_{W}$ in $J$ . But this implies that not even one vertex of $A$ belongs to $J$ (as it would have three non-neighbours and every vertex of $J$ is adjacent to all but at most two of the others). This completes the proof of Claim 4.1.

Claim 3. $G_{W, U}$ is $(P_{1} + 2 P_{2})$ -free.
Suppose for contradiction that there is an induced $P_{1} + 2 P_{2}$ in $G_{W, U}$ . Since $V_{U}$ and $A$ are independent sets, every $P_{2}$ in $G_{W, U}$ must either have two vertices in $V_{W}$ or one vertex in $V_{U}$ and the other in $V_{W}$ or $A$ . Since that $V_{W}$ is a clique and every vertex in $A$ is adjacent to every vertex in $V_{U}$ , one of the $P_{2}$ ’s must have both its vertices in $V_{W}$ and the other must have one vertex in $V_{U}$ and the other in $A$ . But every vertex in $G_{W, U}$ has a neighbour in such a $2 P_{2}$ , so no induced $P_{1} + 2 P_{2}$ can exist. This completes the proof of Claim 4.1.

Claim 4. $G_{W, U}$ is $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{1} + 2 P_{2}$ -free.
Suppose for contradiction that there is an induced $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{1} + 2 P_{2}$ in $G_{W, U}$ . Note this subgraph consists of a $C_{4}$ , which we denote $C$ , plus an additional vertex, which we denote $z$ , that is adjacent to every vertex in $C$ . If $z \in A$ , then the vertices of $C$ are all in $V_{U}$ , and if $z \in V_{U}$ , the vertices of the $C$ are either all in $A$ or all in $V_{W}$ . Neither is possible, so we must have $z \in V_{W}$ . Therefore none of the vertices of $C$ is in $A$ . At most two of the vertices of $C$ are in each of $V_{U}$ and $V_{W}$ as $C$ contains neither an independent set nor a clique on three vertices. So $C$ contains exactly two vertices from each of $V_{U}$ and $V_{W}$ , but the pair from $V_{U}$ must be non-adjacent in $C$ and the pair from $V_{W}$ must be adjacent in $C$ . This contradiction completes the proof of Claim 4.1.

Claim 5. $G_{W, U}$ is $2 P_{3}$ -free.
Suppose for contradiction that there is an induced $2 P_{3}$ in $G_{W, U}$ . Denote this graph by $P$ . As $V_{W}$ is a clique, $P$ contains at most two of its vertices. If it contains exactly two, then $P$ must also contain a vertex $v_{u} \in V_{U}$ such that the three vertices together induce a $P_{3}$ . But then $v_{u}$ cannot be adjacent to any other vertex of $P$ so the remaining three vertices must all belong to $V_{U}$ , a contradiction as $V_{U}$ is independent. So $P$ must contain at least five vertices of $A$ and $V_{U}$ . They cannot all belong to one of these two sets as $P$ has no independent set of size $5$ , but if $P$ contains vertices of both sets then there must be a vertex of degree $3$ . This contradiction proves that $P$ does not exist. This completes the proof of Claim 4.1.

Claim 6. $G_{W, U}$ is $¯ ¯¯¯¯¯¯ ¯ 2 P_{3}$ -free.
Suppose for contradiction that there is an induced $¯ ¯¯¯¯¯¯ ¯ 2 P_{3}$ in $G$ . By Claim 4.1, it contains four vertices of $V_{W}$ . We note that $¯ ¯¯¯¯¯¯ ¯ 2 P_{3}$ contains exactly one induced $K_{4}$ and that the other two vertices are adjacent to exactly two vertices in the clique (so must both be in $V_{U}$ ) and to each other (so cannot both be in $V_{U}$ ). This contradiction completes the proof of Claim 4.1.

Claim 7. $G_{W, U}$ is $P_{6}$ -free.
Suppose for contradiction that there is an induced $P_{6}$ in $G_{W, U}$ . Denote this path by $P$ . As $V_{W}$ is a clique, $P$ contains at most two of its vertices. We see that $P$ cannot contain exactly four vertices of either $V_{U}$ or $A$ as they would form an independent set of size $4$ . $P$ cannot contain exactly three vertices of either $V_{U}$ or $A$ as then there would be a vertex in the other of the two sets of degree $3$ . Finally, $P$ cannot contain at least two vertices from each of $V_{U}$ and $A$ as then it would contain an induced $C_{4}$ . These contradictions prove that $P$ does not exist. This completes the proof of Claim 4.1.

Claim 8. $G_{W, U}$ is $_{6}$ -free.
Suppose for contradiction that there is an induced $_{6}$ in $G$ . By Claim 4.1, it contains four vertices of $V_{W}$ . This contradiction – a clique on four vertices is not an induced subgraph of $_{6}$ – completes the proof of Claim 4.1.∎

4.2 A Tractable Case

We show that Colouring is polynomial-time solvable for $(P2+P3,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P2+P3)$ -free graphs. We first introduce some additional terminology. The clique-width $cw (G)$ of a graph $G$ is the minimum number of labels needed to construct $G$ by using the following four operations:

creating a new graph consisting of a single vertex $v$ with label $i$ ;
taking the disjoint union of two labelled graphs $G_{1}$ and $G_{2}$ ;
joining each vertex with label $i$ to each vertex with label $j$ ( $i \neq j$ );
renaming label $i$ to $j$ .

A class of graphs $G$ has bounded clique-width if there is a constant $c$ such that the clique-width of every graph in $G$ is at most $c$ ; otherwise the clique-width is unbounded. For an induced subgraph $G^{'}$ of a graph $G$ , the subgraph complementation operation (acting on $G$ with respect to $G^{'}$ ) replaces every edge present in $G^{'}$ by a non-edge, and vice versa. Similarly, for two disjoint vertex subsets $S$ and $T$ in $G$ , the bipartite complementation operation with respect to $S$ and $T$ acts on $G$ by replacing every edge with one end-vertex in $S$ and the other one in $T$ by a non-edge and vice versa. Let $k \geq 0$ be a constant and let $γ$ be some graph operation. We say that a graph class $G^{'}$ is $(k, γ)$ -obtained from a graph class $G$ if the following two conditions hold:

every graph in $G^{'}$ is obtained from a graph in $G$ by performing $γ$ at most $k$ times, and
for every $G \in G$ there exists at least one graph in $G^{'}$ that is obtained from $G$ by performing $γ$ at most $k$ times.

We say that $γ$ preserves boundedness of clique-width if for any finite constant $k$ and any graph class $G$ , any graph class $G^{'}$ that is $(k, γ)$ -obtained from $G$ has bounded clique-width if and only if $G$ has bounded clique-width.

Vertex deletion preserves boundedness of clique-width [27].
Subgraph complementation preserves boundedness of clique-width [20].
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 $2$ is at most $4$ .

Lemma 3

If a vertex $x$ in a graph $G$ has a false twin then $cw (G) = cw (G ∖ {x})$ .

The following lemma for $(P_{2} + P_{3})$ -free bipartite graphs follows from a result of Lozin [24], who proved that the superclass of $S_{1, 2, 3}$ -free bipartite graphs has bounded clique-width (see [11] for a full classification of the boundedness of clique-width of $H$ -free bipartite graphs).

Lemma 4 ([24])

The class of $(P_{2} + P_{3})$ -free bipartite graphs has bounded clique-width.

A graph $G$ is perfect if $χ (H) = ω (H)$ for every induced subgraph $H$ of $G$ . 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 $C_{r}$ -free and $_{r}$ -free for every odd $r \geq 5$ .

Theorem 4.3 ([15])

Colouring is polynomial-time solvable on perfect graphs.

Theorem 4.4 ([21, 32])

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

If $G$ is a graph, then a (possibly empty) set $C \subseteq V (G)$ is a clique separator if $C$ is a clique and $G ∖ C$ is disconnected. A graph $G$ 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 $G$ , then it is polynomial-time solvable on all graphs in $G$ .

Let $X$ be a set of vertices in a graph $G = (V, E)$ . A vertex $y \in V ∖ X$ is complete (resp. anti-complete) to $X$ if it is adjacent (resp. non-adjacent) to every vertex of $X$ . A set of vertices $Y \subseteq V ∖ X$ is complete (resp. anti-complete) to $X$ if every vertex in $Y$ is complete (resp. anti-complete) to $X$ . If $X$ and $Y$ are disjoint sets of vertices in a graph, we say that the edges between these two sets form a matching if each vertex in $X$ has at most one neighbour in $Y$ 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 $X$ has at most one non-neighbour in $Y$ and vice versa.

In Theorem 4.5 we present an algorithm for Colouring restricted to $(P2+P3,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P2+P3)$ -free graphs. Our algorithm first tries to reduce to perfect graphs, in which case we can use Theorem 4.3. If a $(P2+P3,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P2+P3)$ -free graph is not perfect, then we show that it must contain an induced $C_{5}$ . The clique-width of such graphs is not bounded (we can construct a $(P_{2} + P_{3}, ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{2} + P_{3})$ -free graph of arbitrarily large clique-width by taking a split graph of arbitrarily large clique-width [28] with clique $C$ and independent set $I$ and adding five new vertices that form an induced $C_{5}$ and that are complete to $C$ but anti-complete to $I$ ). 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 $C_{5}$ and then investigate the possible edges both inside these sets and between them. This enables us to use graph operations (see also Facts 1–3) 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 $(P_{2} + P_{3}, ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{2} + P_{3})$ -free graphs.

Proof

Let $G$ be a $(P_{2} + P_{3}, ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{2} + P_{3})$ -free graph. By Lemma 5 we may assume that $G$ 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 $O (| V (G) |^{5})$ time whether $G$ contains an induced $C_{5}$ . If not, then $G$ is $C_{5}$ -free. Since $_{5} = C_{5}$ , $G$ is also $_{5}$ -free. Since $G$ is $(P_{2} + P_{3})$ -free, $G$ is $C_{k}$ -free for all $k \geq 7$ . Since $G$ is $(¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{2} + P_{3})$ -free, $G$ is $_{k}$ -free for all $k \geq 7$ . By Theorem 4.2, this means that $G$ 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 $5$ -vertex cycle $C$ with vertices $v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$ , in that order. For $S \subseteq {1, \dots, 5}$ , let $V_{S}$ be the set of vertices $x \in V (G) ∖ V (C)$ such that $N (x) \cap V (C) = {v_{i} | i \in S}$ . A set $V_{S}$ 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 $5$ and whenever possible we will write $V_{i}$ instead of $V_{{i}}$ and $V_{i, j}$ instead of $V_{{i, j}}$ and so on.

We start with three structural claims.

Claim 1. $V_{\emptyset}$ is an independent set.
If $x, y \in V_{\emptyset}$ are adjacent then $G [x, y, v_{1}, v_{2}, v_{3}]$ is a $P_{2} + P_{3}$ , a contradiction. Therefore $V_{\emptyset}$ is an independent set. This completes the proof of Claim 4.2.

Claim 2. $| V_{i} \cup V_{i, i + 1} | \leq 1$ and $| V_{i + 1} \cup V_{i, i + 1} | \leq 1$ for $i \in {1, 2, 3, 4, 5}$ .
Suppose for contradiction that $x, y \in V_{1} \cup V_{1, 2}$ . If $x$ and $y$ are adjacent then $G [x, y, v_{3}, v_{4}, v_{5}]$ is a $P_{2} + P_{3}$ and if $x$ and $y$ are non-adjacent then $G [v_{3}, v_{4}, x, v_{1}, y]$ is a $P_{2} + P_{3}$ . This contradiction implies that $| V_{1} \cup V_{1, 2} | \leq 1$ . Claim 4.2 follows by symmetry.

Claim 3. $V_{i, i + 2}$ is an independent set for $i \in {1, 2, 3, 4, 5}$ .
Suppose for contradiction that $x, y \in V_{1, 3}$ are adjacent. Then $G [v_{1}, v_{3}, x, v_{2}, y]$ is a $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{2} + P_{3}$ , a contradiction. Therefore $V_{1, 3}$ is an independent set. Claim 4.2 follows by symmetry.

Since $_{5} = C_{5}$ , it follows that $¯ ¯¯ ¯ G$ also contains a $C_{5}$ , namely on the vertices $v_{1}$ , $v_{3}$ , $v_{5}$ , $v_{2}$ and $v_{4}$ in that order. Therefore $¯ ¯¯ ¯ G$ is also a $(P2+P3,\allowbreak¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯P2+P3)$ -free graph containing an induced $C_{5}$ . As a result, we immediately obtain the following claims as corollaries of Claims 4.2–4.2.

Claim 4. $V_{1, 2, 3, 4, 5}$ is a clique.
Claim 5. $| V_{i, i + 1, i + 2, i + 3} \cup V_{i, i + 1, i + 3} | \leq 1$ and $| V_{i, i + 1, i + 2, i + 3} \cup V_{i, i + 2, i + 3} | \leq 1$ for $i \in {1, 2, 3, 4, 5}$ .
Claim 6. $V_{i, i + 1, i + 2}$ is a clique for $i \in {1, 2, 3, 4, 5}$ .

Let $I$ be the set of vertices in $V_{\emptyset}$ that have a non-neighbour in $V_{1, 2, 3, 4, 5}$ .

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

We may assume $V_{\emptyset}$ and $V_{1, 2, 3, 4, 5}$ are non-empty, otherwise $I = \emptyset$ and the claim follows trivially. Note that $V_{\emptyset}$ is an independent set by Claim 4.2 and that $V_{1, 2, 3, 4, 5}$ is a clique by Claim 4.2. By assumption, $V_{1, 2, 3, 4, 5}$ is not a clique separator in $G$ . Since $V_{\emptyset}$ is an independent set, every vertex of $V_{\emptyset}$ must have a neighbour in $V (G) ∖ (V_{\emptyset} \cup V_{1, 2, 3, 4, 5} \cup V (C))$ that is adjacent to some vertex of the cycle $C$ . Let $A$ be the set of vertices outside $V_{1, 2, 3, 4, 5}$ that have a neighbour in $V_{\emptyset}$ . Note that $A$ contains no vertex of $V_{\emptyset}$ , as $V_{\emptyset}$ is an independent set. Hence, every vertex of $A$ is adjacent to least one vertex of $C$ (but not to all vertices of $C$ ).

We claim that $A$ is complete to $V_{1, 2, 3, 4, 5}$ . Indeed, suppose for contradiction that $x \in A$ is non-adjacent to $y \in V_{1, 2, 3, 4, 5}$ . Let $s$ be a vertex in $V_{\emptyset}$ that is a neighbour of $x$ . Without loss of generality, assume that $x$ is adjacent to $v_{1}$ . Then $x$ is adjacent to $v_{3}$ or $v_{4}$ , otherwise $G [v_{3}, v_{4}, s, x, v_{1}]$ would be a $P_{2} + P_{3}$ , a contradiction. Without loss of generality, assume that $x$ is adjacent to $v_{3}$ . Then $x$ is adjacent to $v_{2}$ , otherwise $G [v_{1}, v_{3}, y, x, v_{2}]$ would be a $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{2} + P_{3}$ , a contradiction. Thus $x$ is adjacent to $v_{4}$ or $v_{5}$ , otherwise $G [v_{4}, v_{5}, s, x, v_{2}]$ would be a $P_{2} + P_{3}$ , a contradiction. Without loss of generality, assume that $x$ is adjacent to $v_{4}$ . Now $x$ must be adjacent to $v_{5}$ , otherwise $G [v_{1}, v_{4}, y, x, v_{5}]$ would be a $¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ P_{2} + P_{3}$ , a contradiction. Therefore $x \in V_{1, 2, 3, 4, 5}$ , a contradiction. It follows that $A$ must indeed be complete to $V_{1, 2, 3, 4, 5}$ .

Let $x \in I$ . By definition of $I$ , there is a vertex $y \in V_{1, 2, 3, 4, 5}$ that is non-adjacent to $x$ . We claim that $c (y)$ can also be used to colour $x$ . Indeed, every neighbour of $x$ lies in $A \cup V_{1, 2, 3, 4, 5} ∖ {y}$ . Since $V_{1, 2, 3, 4, 5}$ is a clique and $A$ is complete to $V_{1, 2, 3, 4, 5}$ , no vertex of $A \cup V_{1, 2, 3, 4, 5} ∖ {y}$ is coloured with the colour $c (y)$ . We may therefore colour $x$ with the colour $c (y)$ . Repeating this process, we can extend the $k$ -colouring of $G ∖ I$ to a

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

Graph Isomorphism for (H_1,H_2)-free Graphs: An Almost Complete Dichotomy

Colouring Square-Free Graphs without Long Induced Paths

Coloring graph classes with no induced fork via perfect divisibility

Detecting strong cliques

Clique-width and Well-Quasi-Ordering of Triangle-Free Graph Classes

On Clique Roots of Flat Graphs

Structural domination and coloring of some (P_7, C_7)-free graphs

1 Introduction

Theorem 1.1 ([22])

Theorem 1.2

Theorem 1.3

Theorem 1.4

Theorem 1.5

2 Preliminaries

3 The Proof of Theorem 1.2

Lemma 1 ([12, 22, 25])

Proof

4 The Proof of Theorem 1.3

4.1 The Hardness Result

Theorem 4.1

Proof

4.2 A Tractable Case

Lemma 2

Lemma 3

Lemma 4 ([24])

Theorem 4.2 ([7])

Theorem 4.3 ([15])

Theorem 4.4 ([21, 32])

Lemma 5 ([34])

Theorem 4.5

Proof

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

Related Research

Graph Isomorphism for (H_1,H_2)-free Graphs: An Almost Complete Dichotomy

Colouring Square-Free Graphs without Long Induced Paths

Coloring graph classes with no induced fork via perfect divisibility

Detecting strong cliques

Clique-width and Well-Quasi-Ordering of Triangle-Free Graph Classes

On Clique Roots of Flat Graphs

Structural domination and coloring of some (P_7, C_7)-free graphs

1 Introduction

Theorem 1.1 ([22])

Theorem 1.2

Theorem 1.3

Theorem 1.4

Theorem 1.5

2 Preliminaries

3 The Proof of Theorem 1.2

Lemma 1 ([12, 22, 25])

Proof

4 The Proof of Theorem 1.3

4.1 The Hardness Result

Theorem 4.1

Proof

4.2 A Tractable Case

Lemma 2

Lemma 3

Lemma 4 ([24])

Theorem 4.2 ([7])

Theorem 4.3 ([15])

Theorem 4.4 ([21, 32])

Lemma 5 ([34])

Theorem 4.5

Proof