Narrowing the Complexity Gap for
Colouring Free Graphs^{†}^{†}thanks: An extended abstract of this paper appeared in the proceedings of AAIM 2014 [17].
Abstract
For a positive integer and graph , a colouring of is a mapping such that whenever . The Colouring problem is to decide, for a given , whether a colouring of exists. The Precolouring Extension problem is to decide, for a given , whether a colouring of a subset of can be extended to a colouring of . A list assignment of a graph is an allocation of a list — a subset of — to each vertex, and the List Colouring problem is to decide, for a given , whether has a colouring in which each vertex is coloured with a colour from its list. We continued the study of the computational complexity of these three decision problems when restricted to graphs that do not contain a cycle on vertices or a path on vertices as induced subgraphs (for fixed positive integers and ).
1 Introduction
Let be a graph. A colouring of is a mapping such that whenever . We call the colour of . A colouring of is a colouring with for all . We study the following decision problem:
Colouring
Instance : A graph .
Question : Is colourable?
It is wellknown that Colouring is NPcomplete even if [22], and so the problem has been studied for special graph classes; see the surveys of Randerath and Schiermeyer [25] and Tuza [27], and the very recent survey of Golovach, Johnson, Paulusma and Song [9]. In this paper, we consider graph classes defined in terms of forbidden induced subgraphs, and study the computational complexity of Colouring and some related problems that we introduce now before stating our results.
A precolouring of is a mapping for some subset . A colouring is an extension of if for each .
Precolouring Extension
Instance : A graph and a precolouring of .
Question : Can be extended to a colouring of ?
A list assignment of a graph is a function that assigns a list of admissible colours to each . If for each , then is also called a list assignment. A colouring respects if for all . Here is our next decision problem:
List Colouring
Instance : A graph and a list assignment for .
Question : Is there a colouring of that respects ?
Note that Colouring can be viewed as a special case of Precolouring Extension which is, in turn, a special case of List Colouring.
A graph is free if it has no subgraph isomorphic to either , the cycle on vertices, or , the path on vertices. Several papers [4, 10, 14] have considered the computational complexity of our three decision problems when restricted to free graphs. In this paper, we continue this investigation. Our first contribution is to state the following theorem that provides a complete summary of our current knowledge. The cases marked with an asterisk are new results presented in this paper. We use ptime to mean polynomialtime throughout the paper.
Theorem 1.1
Let be three positive integers. The following statements hold for free graphs.

List Colouring is NPcomplete if

, and

, and .


List Colouring is ptime solvable if

, and

, and

, and

, and

, and

, and

, and .


Precolouring Extension is NPcomplete if

, and

, and

, and

, and

, and

, and

, and .


Precolouring Extension is ptime solvable if

, and

, and

, and

, and

, and

, and

, and .


Colouring is NPcomplete if

, and

, and

, and

, and

, and

, and where is a constant that only depends on

, and

, and .


Colouring is ptime solvable if

, and

, and

, and

, and

, and

, and

, and

, and

, and

, and

, and .

The new results on List Colouring, Precolouring Extension and Colouring are in Sections 2, 3 and 4 respectively. In these three sections we often prove stronger statements, for example, on (chordal) bipartite graphs, to strengthen existing results in the literature as much as we can. In Section 5, we prove Theorem 1.1 by combining a number of previously known results with our new results, and in Section 6 we summarize the open cases and pose a number of related open problems.
We introduce some more terminology that we will need. Let be a graph. The chromatic number of is the smallest integer for which has a colouring. Let be a set of graphs. We say that is free if has no induced subgraph isomorphic to a graph in ; if , we write free instead of free. The complement of , denoted by , has vertex set and an edge between two distinct vertices if and only if these vertices are not adjacent in . The disjoint union of two graphs and is denoted , and the disjoint union of copies of is denoted . The girth of is the number of vertices of a shortest cycle in or infinite if has no cycle. Note that a graph has girth at least for some if and only if it is free. To add a pendant vertex to a vertex , means to obtain a new graph from by adding one more vertex and making it adjacent only to . We denote the complete graph on vertices by . A graph is chordal bipartite if it is bipartite and every induced cycle has exactly four vertices.
We complete this section by providing some context for our work on free graphs. We comment that it can be seen as a natural continuation of investigations into the complexity of Colouring and List Colouring for free graphs (see [9]). The sharpest results are the following. Hoàng et al. [15] proved that, for all , List Colouring is ptime solvable on free graphs. Huang [16] proved that Colouring is NPcomplete for free graphs and that Colouring is NPcomplete for free graphs. Recently, Chudnovsky, Maceli and Zhong [5, 6] announced a ptime algorithm for solving 3Colouring on free graphs. Broersma et al. [3] proved that List 3Colouring is ptime solvable for free graphs. Golovach, Paulusma and Song [11] proved that List 4Colouring is NPcomplete for free graphs. These results lead to the following table (in which the open cases are denoted by “?”).
Colouring  Precolouring Extension  List Colouring  
P  P  P  P  P  P  P  P  P  P  P  P  
P  ?  NPc  NPc  P  ?  NPc  NPc  P  NPc  NPc  NPc  
P  NPc  NPc  NPc  ?  NPc  NPc  NPc  ?  NPc  NPc  NPc  
?  NPc  NPc  NPc  ?  NPc  NPc  NPc  ?  NPc  NPc  NPc 
2 New Results for List Colouring
In this section we give two results on List Colouring.
We first prove that List Colouring is NPcomplete for the class of free graphs. (We observe that is also known as the 5vertex wheel and is sometimes called the gem or the 5vertex fan.) This result strengthens an analogous result on List Colouring of free graphs [11], and is obtained by a closer analysis of the hardness reduction used in the proof of that result. The reduction is from the problem NotAllEqual 3Sat with positive literals only which was shown to be NPcomplete by Schaefer [26] and is defined as follows. The input consists of a set of variables, and a set of 3literal clauses over in which all literals are positive. The question is whether there exists a truth assignment for such that each contains at least one true literal and at least one false literal. We describe a graph and 4list assignment that are defined using the instance :

type and type vertices: for each clause , contains two clause components and each isomorphic to . Considered along the paths the vertices in are with lists of admissible colours , respectively, and the vertices in are with lists of admissible colours , , , , , respectively.

type vertices: for each variable , contains a vertex with list of admissible colours .

For every clause , its variables are ordered in an arbitrary (but fixed) way, and in there are edges and for .

There is an edge in from every type vertex to every type vertex.
See Figure 1 for an example of the graph . In this figure, is a clause with ordered variables . The thick edges indicate the connection between these vertices and the type vertices of the two copies of the clause gadget. Indices from the labels of the clause gadget vertices have been omitted to aid clarity.
The following two lemmas are known.
Lemma 1 ([11])
The graph has a colouring that respects if and only if has a satisfying truth assignment in which each clause contains at least one true and at least one false literal.
Lemma 2 ([11])
The graph is free.
We are now ready to prove the main result of this section.
Theorem 2.1
The List Colouring problem is NPcomplete for the class of free graphs.
Proof
Lemma 1 shows that the List 4Colouring problem is NPhard for the class of graphs , where is an instance of NotAllEqual 3Sat with positive literals only, in which every clause contains either two or three literals and in which each literal occurs in at most three different clauses. Lemma 2 shows that each is free. As the List 4Colouring problem is readily seen to be in NP, it remains to prove that each contains no induced subgraph in the set . For contradiction, we assume that some has an induced subgraph isomorphic to a graph in .
First suppose that . The total number of type and type vertices can be at most , as otherwise contains an induced or a vertex of degree at least 3. Because and the subgraph of induced by its type and type vertices is connected, must contain at least two adjacent type vertices. This is not possible.
Now suppose that . Because the type and type vertices induce a bipartite graph, must contain an type vertex. Every type vertex has degree at most 3. If it has degree 3, then it has two nonadjacent neighbours (which are of type); a contradiction.
Finally suppose that . Let be the vertex that has degree 4 in . Then cannot be of type, because no type vertex has more than three neighbours in . If is of type, then every other vertex of is either of type or of type, and because vertices of the same type are not adjacent, must contain two type vertices and two type vertices. Then an type vertex is adjacent to two type vertices; a contradiction. Thus must be of type, and so every other vertex of is either of type or of type. Because vertices of the same type are nonadjacent, must contain two type vertices and two type vertices. But then is adjacent to two type vertices in the same clausecomponent. This is not possible. ∎
Our second result modifies the same reduction. From and , we obtain a new graph and list assignment by subdividing every edge between an type vertex and an type vertex and giving each new vertex the list . We say that these new vertices are of type.
Lemma 3
The graph is free and chordal bipartite.
Proof
We first prove that is free (but not free). Let be an induced path in . If contains no type vertex, then contains vertices of at most one clausecomponent together with at most two type vertices. This means that . If contains no type vertex, then can contain at most one type vertex (as any two type vertices can only be connected by a path that uses at least one type vertex). Consequently, can have at most two type vertices and at most two type vertices. Hence, in this case. From now on assume that contains at least one type vertex and at least one type vertex. Also note that can contain in total at most three vertices of type and type.
First suppose that contains exactly three vertices of type and type. Then these vertices form a 3vertex subpath in of types or . In both cases we can extend both ends of the subpath only by an type vertex and an adjacent type vertex, which means that . Now suppose that contains exactly two vertices of type and type. Because these vertices are of different type, they are adjacent and we can extend both ends of the corresponding 2vertex subpath of only by an type vertex and an adjacent type vertex. This means that . We conclude that is free.
We now prove that is chordal bipartite. The graph is readily seen to be bipartite. Because is free, it contains no induced cycle on ten or more vertices. Hence, in order to prove that is chordal bipartite, it remains to show that is free.
For contradiction, let be an induced subgraph of that is isomorphic to or . Then must contain at least one vertex of type and at least one vertex of type. The subgraph of induced by its vertices of type and type is connected and has size at most 3. This means that the subgraph of induced by its vertices of type and type is also connected and has size at least 3. This is not possible. We conclude that is chordal bipartite.∎
The following lemma can be proven by exactly the same arguments that were used to prove Lemma 1.
Lemma 4
The graph has a colouring that respects if and only if has a satisfying truth assignment in which each clause contains at least one true and at least one false literal.
Theorem 2.2
List Colouring is NPcomplete for free chordal bipartite graphs.
3 New Results for Precolouring Extension
In this section we give three results on Precolouring Extension.
Let . Consider the bipartite graph with its list assignment from Section 2 that was defined immediately before Lemma 3. The list of admissible colours of each vertex is a subset of . We add pendant vertices to and precolour these vertices with different colours from . This results in a graph with a precolouring , where is the set of all the new pendant vertices in .
Lemma 5
For all , the graph is free and chordal bipartite.
Proof
Because is free and chordal bipartite by Lemma 3, and we only added pendant vertices, is free and chordal bipartite.∎
Lemma 6
For all , the graph has a colouring that is an extension of if and only if has a satisfying truth assignment in which each clause contains at least one true and at least one false literal.
Lemmas 5 and 6 imply the first result of this section which extends a result of Kratochvíl [21] who showed that 5Precolouring Extension is NPcomplete for free bipartite graphs.
Theorem 3.1
For all , Precolouring Extension is NPcomplete for the class of free chordal bipartite graphs.
Here is our second result.
Theorem 3.2
The Precolouring Extension problem is NPcomplete for the class of free graphs.
Proof
Let be the instance with list assignment as constructed at the start of Section 2. Instead of considering lists, we introduce new vertices, which we precolour (we do not precolour any of the original vertices). For each clause component we add five new vertices, , , , , . We add edges , and for . We precolour , , , , with colours , , , , , respectively. For each clause component we add five new vertices, , , , , . We add edges , and for . We precolour , , , , with colours , , , , , respectively. Finally, we add two new vertices , which we make adjacent to all type vertices, and two new vertices , which we make adjacent to all type vertices. We colour , , , with colours , , , , respectively. This results in a new graph . Because can be viewed as type vertices and as type vertices, because every other new vertex is a pendant vertex and because is free (by Theorem 2.1), we find that is free. Moreover, our precolouring forces the list upon every vertex of . Hence, has a 4colouring extending this precolouring if and only if has a colouring that respects . By Lemma 1 the latter is true if and only if has a satisfying truth assignment in which each clause contains at least one true and at least one false literal.∎
Broersma et al. [3] showed that Precolouring Extension for free graphs is NPcomplete. It can be shown that the gadget constructed in their NPhardness reduction is free. By adding dominating vertices, precoloured with colours , to each vertex in their gadget, we can extend their result from to . This leads to the following theorem.
Theorem 3.3
For all , Precolouring Extension is NPcomplete for free graphs.
4 New Results for Colouring
The first result that we show in this section is that Colouring is NPcomplete for free graphs. We will prove this by modifying the graph from Section 3. It improves the result of Golovach et al. [10] who showed that 4Colouring is NPcomplete for free graphs.
First we review a wellknown piece of graph theory. The Mycielski construction of a graph is created by adding, for each vertex of , a new vertex that is adjacent to each vertex , and then adding a further vertex that is adjacent to each of the other new vertices. For example, the Mycielski construction of is a 5cycle, and the Mycielski construction of a 5cycle is the wellknown Grötzsch graph. These examples are the first in an infinite sequence of graphs where and , , is the Mycielski construction of . The graph is displayed in Figure 2. As we will make considerable use of this graph, let us explain its construction carefully. Let us suppose that we start with where and (note that, for clarity, we denote an edge between two vertices and by instead of ). Then is made by adding each vertex , , and making it adjacent to the neighbours of vertex and to a further vertex 11. Finally is obtained by adding a vertex , , with the same neighbours as vertex and a further last new vertex, 23. Mycielski [23] showed that each is free and has chromatic number . Moreover, any proper subgraph of is colourable (see for example [1]).
Let be the graph obtained from by removing the edge from 17 to 23. We need the following lemma.
Lemma 7
Let be a colouring of . Then , and moreover, .
Proof
If , then is a 4colouring of , which is not possible. As , we must have else there is a 4colouring which disagrees with only on 17 and so, again, is a 4colouring of . Thus we also have .∎
Let . We present three lemmas about induced paths in with endvertices in . Proving each of these three lemmas is straightforward but tedious. We therefore omitted their proofs.
Lemma 8
Every induced path in with both endvertices in contains at most vertices.
Lemma 9
Every induced path in with an endvertex in contains at most vertices.
Lemma 10
contains no induced subgraph isomorphic to such that each of the two paths has an endvertex in . Also, contains no induced subgraph isomorphic to such that each of the two paths has an endvertex in .
For a graph and subset , we let denote the graph obtained from after removing all vertices in .
To prove our result we also use the graphs and defined in Sections 2 and 3, respectively. Let , , , denote the sets of of type, type, type and type vertices in , respectively. Note these sets also exist in . We need an upper bound on the length of an induced path in some specific induced subgraphs of and .
Lemma 11
Every induced path in that starts with an type or type vertex has at most vertices.
Proof
Let be a maximal induced path in . Suppose that is of type. Then, as contains no type vertices, we find that . Consequently, or . In both cases we must have and . Afterward we cannot extend any further. Suppose that is of type. Because there is no type vertex in , we find that . Hence, or . In both cases, cannot be extended further. We conclude that . ∎
Lemma 12
Every induced path in has

at most vertices;

at most vertices if it contains only one pendant vertex of ;

at most vertices if it contains no pendant vertex of .
Proof
We observe that is a forest in which each tree can be constructed as follows. Take a star. Subdivide each of its edges exactly once. Afterward add one or more pendant vertices to each vertex. The tree obtained is free and every induced contains two pendant vertices as its endvertices. Moreover, every induced has at least one pendant vertex. ∎
We will now modify the (bipartite) graph with the precolouring in the following way.

Remove all vertices that are pendant to any vertex in .

Add a copy of . Write , , and .

Let be the set of vertices pendant to any vertex in . For each do as follows. If then add the edge for all .

Add an edge between every vertex in and for .

Add an edge between every vertex in and for .
We call the resulting graph . Note that is not free in general. In order to see this we say that the pendant vertices of are of type and that the vertices of are of type and then take an induced 21vertex path with vertices of type
Note that such a path only uses two vertices of , namely and (as and are adjacent to all vertices of type). As such, trying to optimize Lemmas 8–10 (which we believe is possible) does not help us with improving our result. In the following lemma we show that this length is maximum.
Lemma 13
The graph is free.
Proof
Because and are both independent sets and the graphs and are bipartite, is free. Below we show that is free. Let be an induced path in . Let be the number of vertices of in .
Case 1. .
Then either , and so by Lemma 5, or , and so .
Case 2. .
Then for some , where the subpaths and are each fully contained in either or . If both of them are contained in , then and so .
If one of them is contained in and the other one is contained in ,
then by Lemmas 5 and 9. Otherwise both of them are
contained in and so by Lemma 5.
Case 3. .
Then for some , where each of , and is fully contained
in either or . We need the following claim.
Claim 1. and .
We prove Claim 1 as follows. By symmetry it suffices to show it for only. First assume that is contained in . We distinguish two cases.
Case a. .
Let be the right endvertex of .
First assume that .
Then must be in by the construction of .
Because is adjacent to every vertex in , we have .
Hence, .
We also have , as is induced and each vertex in is adjacent to three vertices of .
Then, by Lemma 12, we obtain .
Now assume .
Then for the same reason as before.
Hence contains only nonpendant vertices of .
Then, by Lemma 12, we obtain .
Case b. .
Let be the right endvertex of .
If then , and so is an induced path of .
By Lemma 12 we find that .
Now assume that . Since is adjacent to all vertices of , we have .
Hence, by construction, we find that . In fact we have ,
as is induced and each vertex in is adjacent to three vertices of .
Thus, is contained in .
As , its neighbour on must be in (if this neighbour exists).
Then, by Lemma 11, we find that .
Now suppose that is contained in . Using Lemma 10 with respect to the paths and we conclude that . This completes the proof of Claim 1.
By Claim 1 we have that and . If this means that and we are done. So, it suffices to prove that , as we will do below.
If is contained in , then by Lemma 8 we have that . Assume that . First suppose that . If contains a type vertex, then must be a single type vertex and thus . Otherwise we have , and hence by Lemma 12.
Now suppose that . If contains a type vertex, then must be a single type vertex and thus . Otherwise both endvertices of are in and all internal vertices of are contained in . Because every neighbour of a vertex in in belongs to , we find that by Lemma 11.
We observe that the pairs and are symmetric in terms of the edges between them to and . Moreover, and can be seen as symmetric, and the same holds for and . This has the following consequence. Suppose that . Then we may assume without loss of generality that , say .
If the left endvertex of is not in , then and so by Lemma 12. Similarly, if the right endvertex of is not in then all vertices except the left endvertex of are in . In particular, the right endvertex of belongs to in that case, and as such has its neighbour on in (if this neighbour exists). Then by Lemma 11. Therefore we may assume that starts with a vertex in , ends at a vertex in . As is induced and each vertex in is adjacent to three vertices of , we find that contains no vertices of . Moreover, no internal vertex of belongs to , as these vertices would be adjacent to one of . Hence all internal vertices of are in . Then only has one internal vertex, which is either in or in . Hence, . This completes the proof of Case 3.
Case 4. .
Then for some , where the subpaths , , and are each fully contained in either or .
As is induced and each vertex in is adjacent to three vertices of .
we find that if () contains a vertex of .
We observe that Claim 1 is also valid here (as exactly the same arguments can be used).
Hence, and .
By the aforementioned symmetry relations between vertices and vertex pairs of , we may assume without loss of generality that . According to the relative positions
among these three vertices on the path we have the following two subcases.
Case 4.1 .
We first prove that either is contained in or that . Suppose that is not contained in .
The right endvertex of cannot be in due to the presence of . Hence, it must be in