1 Introduction
There has been recently keen interest in finding polynomialtime algorithms to optimally color graphs that do not contain any graph in a list as an induced subgraphs. Particular attention is focused on graphs whose forbidden list contains graphs with four vertices, and recent papers of Lozin and Malyshev [15], and of Fraser, Hamel, Hoàng, Holmes and LaMantia [10] discuss the state of the art on this problem, identifying three outstanding classes: , claw), , claw, codiamond), and ). As a resolution of these cases is likely challenging, a productive approach would be to consider increasing the number of graphs in . But it makes sense to ask, which is an interesting one to consider? We are particularly interested in cases that are slightly larger than the class ), which is one of the unresolved cases. Before we introduce the problems we need some definitions:
A hole is an induced cycle with at least four vertices. A hole is even if its number of vertices is even. The problem of coloring evenholefree graphs has been much studied. A theorem of AddarioBerry, Chudnovsky, Havet, Reed, and Seymour [1] shows that for an evenholefree graph , the chromatic number of is at most two times its clique number (the number of vertices in a largest clique of ). It is currently not known whether evenholefree graphs can be colored in polynomial time.
We offer the following four problems for consideration:
Problem 1.1
What is the complexity of coloring ()free graphs?
This is the original problem, and is likely the most challenging.
Problem 1.2
What is the complexity of coloring evenholefree graphs?
Problem 1.2 might even be NPcomplete. Combining Problems 1.1 and 1.2, we get the following problem:
Problem 1.3
What is the complexity of coloring (, even hole)free graphs?
This problem appears to be more tractable than the previous two. In this paper, we study Problem 1.3. Since a free graph does not contain a hole of length at least 8, Problem 1.3 is equivalent to the following:
Problem 1.4
What is the complexity of coloring ()free graphs?
Even though we have not been able to solve Problem 1.4, we have succeeded, in some sense, in solving “half” of it, as follows. Consider a ()free graph . We may assume is not perfect (there are known algorithms to color perfect graphs). Thus has to contain a or . If contains a , then our result shows that can be colored in polynomial time. The case where contains a but not a is open. Investigation into this problem led us to a proof that there is a polynomial time algorithm to color a (, twin)free graph.
In Section 2, we discuss the background of the problem and state the main results. In Section 3, we study ()free graphs that contain a and show that such graphs can be colored in polynomial time. In Section 4, we study ()free graphs that contain a , but no . In Section 5, we give a polynomial time algorithm to color a (, twin)free graph. Finally, in Section 6, we dicuss open problems related to our work.
2 Background and results
Before discussing our results in more detail, we need introduce a few definitions. Let be a graph. A colouring of a graph is a mapping for some nonnegative integer such that whenever . The chromatic number, denoted , is the minimum number of colors needed to colour a graph . VERTEX COLORING is the problem of determining the chromatic number of a graph.
Consider the following operations to build a graph.
 (i)

Create a vertex labeled by integer .
 (ii)

Disjoint union (i.e., cojoin)
 (iii)

Join between all vertices with label and all vertices with label for , denoted by (that is, add all edges between vertices of label and label ).
 (iv)

Relabeling all vertices of label by label , denoted by
The clique width of a graph , denoted by , is the minimum number of labels needed to build the graph with the above four operations. It is wellknown [8] that if the clique width of a graph is bounded then so is that of its complement. Clique widths have been intensively studied. In Rao [17], the following result is established.
Theorem 2.1
VERTEX COLORING is polynomial time solvable for graphs with bounded clique width.
We will need the folowing well known observation that is easy to establish (for example, see [6]).
Observation 2.2
Let be a graph and be the graph obtained from by removing a constant number of vertices. Then has bounded clique width if and only if does.
The symbol denotes the number of vertices in a largest clique of . A graph is perfect if for each induced subgraph of , we have . A hole is an induced cycle of length at least , i.e. for . A hole is even or odd depending on the parity of the vertices in the hole. An antihole is the complement of a hole.
Two important results are known about perfect graphs. The Perfect Graph Theorem, proved by Lovász [14], states that a graph is perfect if and only if its complement is. The Strong Perfect Graph Theorem, proved by Chudnovsky, Robertson, Seymour, and Thomas [7], states that a graph is perfect if and only if it is oddholefree and oddantiholefree. Both of the above results were long standing open problems proposed by Berge [3]. Grötschel, Lovász and Schrijver [12] designed a polynomialtime algorithm for finding a largest clique and a minimum coloring of a perfect graph.
Suppose we want to color a ()free graph . Note that contains no for because is free. By the result of Grötschel et al, we may assume is not perfect. The result of Chudnovsky et al implies contains a or as an induced subgraph (note that the antihole of length at least six contains a and the is selfcomplementary.) If contains a , then we will show that has bounded cliquewidth.
Theorem 2.3
Let be ()free graph that contains a . Then has bounded clique width.
We will use Theorem 2.3 to prove the following theorem that is the main result of this paper.
Theorem 2.4
VERTEX COLORING can be solved in polynomial time for the class of ()free graphs that contain a .
Two adjacent vertices of a graph are twins if for any vertex different from and , is an edge if and only if is an edge. A holetwin is the graph obtained from a hole by adding a vertex that form twins with some vertex of the hole. Figure 1 shows the twin. Holetwins play an interesting role in graph theory. They are among the forbidden induced subgraphs for linegraphs (Beineke [2]).
By adding the twin to the list of forbidden induced subgraphs for our graph class, we obtain the following theorem.
Theorem 2.5
VERTEX COLORING can be solved in polynomial time for the class of (, twin)free graphs.
We will rely on a theorem of Fraser et al [10]. To explain this theorem, we will need to introduce a few definitions. Given sets of vertices , we write to mean there is no edge between any vertex in and any vertex in (also called a cojoin). Given sets of vertices , we write to mean there are all edges between and (also called a join).
Consider a partition of the vertices of into sets such that each induces a clique. A set is uniform to a set if or . The set is uniform in if every set , is uniform to it. The partition is uniform if every set is uniform in . A set is nearuniform if there is at most one set that is not uniform to . The partition is nearuniform if every set is nearuniform. Thus, a uniform partition is nearuniform. A nearuniform partition of is a partition of the vertices of into nearuniform sets. In such a partition, the pair of sets , such that each set is not uniform to the other is call a uniformpair. The following is proved in Fraser et al [10].
Theorem 2.6
We will establish the following theorem.
Theorem 2.7
Let be ()free graph that contains a . Then admits a uniform partition, for some constant .
3 When the graphs contain a
Assume that the graph is ()free and contains a . In this section, we examine the structure of the neighborhood of the . Then we will prove Theorem 2.7.
We will need first to establish a number of preliminary results. Given a hole, , and a vertex not in , we say is a vertex (for ) if has exactly neighbours in .
For all the claims below, we shall now assume that contains an induced with vertices . The vertex numbers of the are taken modulo 7.
Let

denote the set of vertices adjacent to in the ,

denote the set of vertices on in the ,

denote the set of vertices on in the , and

denote the set of vertices in the .
We will show that the sets form a partition of .
Observation 3.1
The has no vertex in for .
Proof. Suppose that there is a vertex for . This creates an induced , which is forbidden. Therefore, there is no vertex. Now suppose that there is a vertex that is adjacent to . This creates an induced , which is forbidden. Therefore, there is no vertex. Next, suppose that there is a vertex for . It is easy to see that and some three vertices in the form a , a contradiction. Consequently, there is no vertex. Next, suppose that there is a vertex for . Then contains a or , both of which are forbidden. Therefore, there is no vertex. Finally, suppose that there exists a vertex for . Then, contains an induced , which is forbidden. Therefore, there is no vertex for .
It is a routine matter to verify the two observations below.
Observation 3.2
Let be a 3vertex for the . Then for some .
Observation 3.3
Let be a 5vertex for the . Then for some .
The above three observations imply the following observation.
Observation 3.4
Let be the graph obtained from by removing the . Then the sets form a partition of .
Our aim is to show that has bounded clique width. In fact, we will show that is an uniform partition of .
We now examine the adjacencies between the sets of the partition .
Observation 3.5
Each of the sets of the partition is a clique.
Proof. It is easy to see that if a set of is not a clique then there is a .
The next sequence of observations will imply that is nearuniform in .
Observation 3.6
Proof. Suppose there are vertices and such that . Then there exists a , which is forbidden. Therefore, . By symmetry, we have .
Observation 3.7
.
Proof. Consider a vertex . Suppose there is a vertex such that . This creates a , which is forbidden. Therefore, we have , and by symmetry . Now suppose that there is a vertex such that . This creates a , which is forbidden. Therefore, we have , and by symmetry, .
Observation 3.8
.
Proof. Consider a vertex . Suppose there is a vertex such that . This creates a , which is forbidden. Therefore, . Now suppose there is a vertex with . This creates a , which is forbidden. Therefore . Finally, suppose there is a vertex with . This creates a , which is forbidden. Therefore . Consequently, .
Observation 3.9
Proof. Consider a vertex . Suppose there is a vertex with . This creates a , which is forbidden. Therefore, we have , and by symmetry, . Now suppose that there is a vertex with . This creates a , which is forbidden. Therefore, , and by symmetry, .
Observation 3.10
Proof. Consider a vertex . Suppose there is a vertex such that . This creates a , which is forbidden. Therefore, we have , and by symmetry, . Now suppose there is a vertex such that . This creates a , which is forbidden. Therefore, we have , and by symmetry, . Finally, suppose that there is a vertex such that . This creates a , which is forbidden. Therefore, we have , and we are done.
Observation 3.11
Proof. Consider a vertex . Suppose there is a vertex such that . This creates a , which is forbidden. Therefore, . Next, suppose that there is a vertex such that . This also creates a , which is forbidden. Therefore, .
Observation 3.12
Proof. Suppose that there are vertices and such that . This creates a , which is forbidden. Therefore we have .
Lemma 3.13
For every , the set is uniform in the partition .
Next, we examine the sets .
Observation 3.14
At most two of the sets can be nonempty. In particular, if , then , and we have or , but not both.
Proof. Suppose . Let be a vertex in . Suppose the set contains a vertex . If , then there exists a , which is forbidden. If , then there is a (), which is forbidden. So we have , and by symmetry, . Now, suppose is nonempty and contains a vertex . If , then there is a (), a contradiction. But if , then there exists a , a contradiction. So, we have , and by symmetry, .
The first part of this proof shows that if , then . So only one of the two sets can be nonempty.
Observation 3.15
Proof. Consider a vertex . Suppose there is a vertex such that . There is a , a contradiction. So, we have , and by symmetry, .
Observation 3.16
If , then
Proof. Assume that and . Consiser vertices and . If , then there exists a , which is forbidden. However, if , then there exists a , which is also forbidden. Therefore if , then , and by symmetry, .
Observation 3.17
Proof. Consider a vertex . Let . If , then there is a , which is forbidden. Therefore, .
Consider a vertex . Then , and is adjacent to a vertex . If , there there is a , which is forbidden. Therefore, .
Observation 3.18
Proof.Consider vertices and such that . This creates a , which is forbidden. Therefore .
Lemma 3.19
For every , the set is uniform in the partition .
Next, we examine the sets .
Observation 3.20
If , then and .
Proof. Suppose that . Also, suppose . Consider vertices and . If , then there is a (), a contradiction. If , then there is a (), a contradiction. So is empty, and by symmetry, is empty.
Observation 3.21
There can exist at most distinct sets of vertices for .
Proof. Follows from Observation 3.20.
Observation 3.22
Proof. Consider a vertex and a vertex with . If is adjacent to two nonadjacent vertices, say and , of the set , then , for otherwise there is a (). Observe that any vertex in is adjacent to two nonadjacent vertices of . The Observation follows.
Lemma 3.23
For every , the set is uniform in the partition .
We can now prove our main results.
Proof of Theorem 2.7. Let be a )free graph with a . Define the sets as above. Let . Observations 3.4 implies that the sets form a partition of . Lemmas 3.13, 3.19, and 3.23 show that each of the sets is uniform. Now the set is also uniform because every other set is uniform to . Thus, the partition is uniform.
Proof of Theorem 2.3. Let be a )free graph with a . The graph has bounded clique width by Theorem 2.7 and Theorem 2.6, has bounded clique width. Thus, has bounded clique width by Observation 2.2.
Proof of Theorem 2.4. Let be a , even hole)free graph with a . By Theorem 2.3, has bounded clique width. By Theorem 2.1, can be optimally colored in polynomial time.
So, if our graphs contain a , we know how to color them. If they do not contain a , then we know they must contain a , for otherwise they are perfect and we would know how to color them. In the next section, we discuss the case of the .
4 When the graphs contains a
In this section, we assume is (, , )free. For the all the claims below, we will also assume that contains an induced with vertices . Let denote the set of vertices for this , let be the set of vertices adjacent to , let be the set of vertices with neighbors , let be the set of vertices with neighbors and let denote the set of vertices.
The following observation is immediate.
Observation 4.1
The sets form a partition of the vertex set of
Observation 4.2
Each of form a clique.
Proof. Consider two nonadjacent vertices of . If both belong to , then and and some two nonadjacent vertices of the form a , a contradiction. Similarly, we can see that cannot both belong to , or to . If both belong to or to , then and some two nonadjacent vertices of form a .
Observation 4.3
.
Proof. Consider a vertex and a vertex . If , then and some two nonadjacent vertices in the form a .
Observation 4.4
If then for all .
Proof. Consider vertices , , with . We must have , for otherwise and some two vertices in the form a . If , then there is a (). So we have , and by symmetry, . If , then there is a (). So we have , and by symmetry, .
Observation 4.5
.
Proof. Let , and let . If , then and some two nonadjacent vertices of the form a . So . By symmetry (with the case ), we have .
Observation 4.6
.
Proof. Let , and let . If , then there is a (). So we have , and by symmetry, .
Observation 4.7
.
Proof. Let , and let . If , then there is a (). So we have .
We note that vertices of may have neighbors and nonneighbors in .
Observation 4.8
for all .
Proof. Consider vertices . If , then there is a (). So we have , and by symmetry, . Now consider a vertex . If , then there is a (). So we have , and by symmetry, .
Observation 4.9
.
Proof. Consider vertices . If , then there is a ().
Observation 4.10
.
Proof. Consider vertices . If , then there is a ().
In the next section, we will use the results of this section to prove Theorem 2.5.
5 Clique cutset decomposition
In this section, we present a proof of Theorem 2.5. We will need to introduce definitions and background for the problem.
Consider a graph . A clique cutset of is a set of vertices such that is a clique and is disconnected. Consider the following procedure to decompose . If has a clique cutset , then is decomposed into subgraphs and where and ( denotes the subgraph of induced by for a subset of vertices of ). Given optimal colourings of , we can obtain an optimal colouring of by identifying the colouring of in with that of in . In particular, we have . If () has a clique cutset, then we can recursively decompose in the same way. This decomposition can be represented by a binary tree whose root is and the two children of are and , which are in turn the roots of subtrees representing the decompositions of and . Each leaf of corresponds to an induced subgraph of G that contains no clique cutset; we will call such graph an atom. Algorithmic aspects of the clique cutset decomposition are studied in Tarjan [18] and Whiteside [19]. In particular, the decomposition tree can be constructed in time such that the total number atoms is at most [18] (Here, as usual, , resp., , denotes the number of vertices, resp., edges, of the graph ). This discussion can be summarized by the theorem below.
Theorem 5.1 ([18], [19])
Let be a graph. If every atom of can be colored in polynomial time, then can be colored in polynomial time.
We will need the following theorem that illustrates the structure of (twin)free graphs.
Theorem 5.2
Let be a (twin)free graph. If contains a , then one of the following holds:
 (i)

contains a clique cutset.
 (ii)

contains a .
 (iii)

has bounded clique width.
 (iv)

is the join of a (possibly empty) clique and a . In this case, (iii) is also satisfied.
Proof of Theorem 5.2. Let be a ()free graph and suppose contains a . Assume that contains no clique cutset and no , for otherwise we are done. Define the sets as above. Note that for all because contains no twin. By Observation 4.4, at most one set can be nonempty. We will assume that this one set, it it exists, is . Define .
We will show that
.  (1) 
Suppose . Note that is a cutset, separating from the . Let be a minimal ()separator of that is contained in . By assumption, is not a clique. Consider two nonadjacent vertices in . By the minimality of , there is a chordless path with endpoints being , and interior vertices belonging to . Since is a clique, has at most three edges.
Suppose first that has three edges. Enumerate the vertices of as with . Since is not adjacent to , by Observation 4.3, we have . Similarily, we have . But by Observation 4.2, is an edge, a contradiction.
So has two edges. Enumerate the vertices of as with . Note that both have neighbors in the . No vertex can be adjacent to both , for otherwise, there is a (). So we have , in particular, . Since is not an edge, by Observation 4.2, either or , or both, belongs to some . We may assume , that is, is a 2vertex.
Let be a vertex in the that is adjacent to . Let be a the vertex in that is adjacent to and is closest to in the . Then ) is an induced path. If has length at least four, the induces a chordless cycle of length at least six, a contradiction. So we know . Since and is an edge, we know . It follows that . The vertex may or may not be adjacent to . If , let be the chordless path ; otherwise, let
Comments
There are no comments yet.