1 Introduction
In this paper, we only consider finite and simple graphs. Let be a graph, and let be a subset of the vertex set of . We use to denote the subgraph of induced by . We say that is a stable set if has no edges, say that is a clique if is a complete graph, and say that is a cutset if (the subgraph induced by ) has more components than . The clique number of , denote by , is the maximum size of a clique of .
Let be a vertex of . We use to denote the set of neighbors of in (and simply write if ). We say that is complete to if , and say that is anticomplete to if . Let be a subset of . We say that is complete to if each vertex of is complete to , and say that is anticomplete to if each vertex of is anticomplete to .
Let be a positive integer. We use to denote the set .
A coloring of is a partition of into stable sets, of which each is called a color class. We say that is colorable if admits a coloring. The chromatic number, denoted by , of is defined to be the minimum integer such that is colorable. It is certain that .
A hole in is an induced cycle of length at least 4, and an odd (resp. even) hole is a hole of odd (resp. even) length. An antihole is the complement of a hole, and an odd (resp. even) antihole is an antihole with odd (resp. even) number of vertices. Similarly, we define even (resp. odd paths.
For a given graph , we say that induces if has an induced subgraph isomorphic to , and say that is free if it does not induce . Let be a family of graphs, we say that is free if is free for each member .
A graph is said to be perfect if for each of its induced subgraphs . In 2006, Chudnovsky, Robertson, Seymour and Thomas [4] confirmed the Strong Perfect Graph Conjecture of Berge [1, 2], and proved that a graph is perfect if and only if it is (odd hole, odd antihole)free. Confirming a conjecture of Gyárfás [11], Scott and Seymour [14] proved that for every odd hole free graph .
We say that is 2divisible if for each induced subgraph of , is either a stable set, or can be partitioned into two sets and such that (i.e., both and intersect every maximum clique of ). It is easy to see that for each 2divisible graph . Hoáng and McDiarmid [13] introduced this concept, and proved that odd hole, free graphs are 2divisible. They also posed the following conjecture.
Conjecture 1.
A graph is divisible if and only if it is odd hole free.
In [12]), Hoáng introduced and studied another nice property of graphs–the perfect divisibility. A graph is called perfectly divisible if for each induced subgraph , can be partitioned into two subsets and such that is perfect and . There exist perfectly divisible graphs which are not 2divisible (such as odd holes). If is perfectly divisible, then . In [15], Scott and Seymour mentioned a conjecture of Hoáng.
Conjecture 2.
If a graph is odd hole free, then can be partitioned into subsets of which each induces a perfect graph.
Chudnovsky, Robertson, Seymour and Thomas [6] confirmed Conjectures 1 and 2 on free graphs. They proved the following
Theorem 1.1.
Every odd hole, free graph is colorable.
A full house is a graph composed by a vertex adjacent to both ends of an edge in . It is certain that free graph must be full house free. In this paper, we study (odd hole, full house)free graphs, and prove the following
Theorem 1.2.
Let be an odd hole, full housefree graph. Then, , and the equality holds if and only if and induces an odd antihole on seven vertices.
To prove Theorem 1.2, we follow the idea and the structure of proofs of Chudnovsky, Robertson, Seymour and Thomas from [6]. We will prove a structural theorem claiming that if is odd hole, full housefree, then either has a particular cutset, or belongs to two kinds of specific families of graphs.
In Section 2, we present some useful definitions, state our structural theorem, and then prove Theorem 1.2. The proof of the structural theorem will be given in Section 3.
2 Harmonious cutset, heptagramtype graphs, and type graphs
Let and be two subsets of , let and . An path is a path connecting and , and an path is called an path if and .
A cutset is said to be harmonious if can be partitioned into disjoint sets such that

if , then are pairwisely complete to each other, and

for all , if is an induced path, then is even if and odd otherwise.
From the second condition, one sees that each is a stable set. Let be a harmonious cutset of . Suppose that is partitioned into two nonempty sets and , and let for . Following several lemmas, due to Chudnovsky et al [6], are all about harmonious cutsets.
Lemma 2.1.
([6, Lemma 2.1]) If is odd hole free for each , then is odd hole free.
Lemma 2.2.
([6, Lemma 2.3]) Let be an odd hole free graph, and let be a cutset of . Let be a partition of into stable sets, such that if , then the sets are pairwisely complete. Suppose that for all nonadjacent , there exists an induced path , with interior in , such that is even if some contains both and , and odd otherwise. Then has a harmonious cutset.
In [6], Chudnovsky et al also proved that if is colorable for then is colorable. With the same technique as that of Chudnovsky et al, we extend the conclusion to colorability.
Lemma 2.3.
If is colorable for , then is colorable.
Proof. Let be as the definition of a harmonious cutset. By the hypothesis, both and are colourable. The lemma holds trivially when or . Then we may suppose , and so by the definition of a harmonious cutset.
For , let be a colouring of using colours . A vertex is called a compliant if . We will show that
(1) 
We only need to verify the case that by symmetry. Let be a colouring of that maximises the number of compliant vertices. We will show that is a colouring of satisfying the requirement of (1).
Suppose to its contrary, let be a vertex that is not compliant, say and , where . Let be the component containing of the subgraph of induced by the vertices coloured or . We claim that no vertex of is a compliant. For otherwise, let be a compliant, and let be an induced path of . Since is a compliant by our assumption, if and only if for some . From the definition of , if and only if has even length. Note that is a harmonious cutset, has even length if and only if for some . If is even, then as is a harmonious cutset, and so , a contradiction. If is odd, then and as is a harmonious cutset, and so , a contradiction as well. Therefore, no vertex of can be compliant.
Let be the colouring obtained from by swapping the colours and for every vertex of . Then is compliant and it follows that more vertices in are compliants than that of compliants, contrary to the choice of . This proves (1).
Now the colorings and can contribute a colouring of .
The following definitions of type graphs and heptagramtype graphs are totally the same as that of [6]. We present them here for completeness. Let and be two positive integers and . Throughout the paper, by index arithmetic modulo , we always mean that .
Defination 2.1.
A graph is of type if there is a partition of into eleven nonempty stable sets , such that for , is anticomplete to and complete to , with index arithmetic modulo 11.
Let and be two disjoint subsets of . We say that and are linked if every vertex of has a neighbor in , and every vertex of has a neighbor in .
Defination 2.2.
A graph is of heptagramtype if there is a partition of into fourteen stable subsets and , where are nonempty but may be empty, satisfying the following (with index arithmetic modulo 7).

For , is anticomplete to .

For , is complete to , and the subsets and are linked.

For , is complete to ; for , and are linked.

If for , and is complete to , then is adjacent to .

If for , and is anticomplete to , then is nonadjacent to .

For , every vertex in has a neighbor in each of and , and has no neighbor in .

For and each , let be the set of neighbors of in for . Then is complete to , and is anticomplete to , and is anticomplete to , and is complete to .

For , is complete to and anticomplete to .

For , if is not complete to , then is complete to , and is empty.

For , at least one of and is empty.
Now, we can state our structural theorem.
Theorem 2.1.
Let be an odd hole, full housefree graph which has no harmonious cutset and induces an antihole on seven vertices. Then, is either of heptagramtype or of type.
Theorem 2.1 will be proved in latter section. As a direct consequence of Theorem 2.1, we can now prove Theorem 1.2.
Proof of Theorem 1.2. Let be an (odd hole, full house)free graph. By Theorem 1.1, we need only to prove that if . By Lemma 2.3, we may suppose that has no harmonious cutsets. If does not induce an antihole on seven vertices, then is perfect. Otherwise, will be either of heptagramtype or of type by Theorem 2.1, which is colourable with .
3 Proof of Theorem 2.1
This section is devoted to proving Theorem 2.1. The whole proof, similar to that of [6], is divided into 4 subsections, respectively considering type subgraphs, Heptagrams, vertices, and vertices etc.
3.1 type graphs
Firstly, we discuss the type configuration contained in (odd hole, full house)free graphs, and prove that if an odd hole, full housefree graph has no harmonious cutset and contains an induced type subgraph, then the graph itself is a type graph.
Suppose that are disjoint subsets of . A path of form is an induced path , where for all (we denote instead of if is a singleton). We use analogous terminology for holes.
Lemma 3.1.
Let be an odd hole, full housefree graph which has no harmonious cutset. If induces an type subgraph, then itself is type.
Proof. Let be an induced type subgraph of with maximum number of vertices. As in its definition, suppose that is partitioned into stable sets , let , and suppose that .
Let be a subset of such that is connected, and let and be two nonadjacent vertices of of which each has neighbors in . We claim that
(2) 
If it is not the case, we choose to be minimal, and let be a pair of vertices violating (2). By symmetry, we may assume that and . The minimality of implies that there is a positive integer and an induced path with . First we show that
(3) 
For otherwise, suppose that . Let . An induced path is called a path if has both ends in and interior vertices in . As is odd holefree, a path is of length one if it is odd. We suppose that for each , if necessary.
Suppose first that . Since there is no path of the form , we may suppose by symmetry that . If has a neighbor, say , in , then induces a full house, and so
Since there is no path of the form , we have that or . If , then as there is no path of the form , and as there is no path of the form , and as there is no path of the form . Since for each vertex in , cannot be a full house, . By putting into , we get a type subgraph with one more vertex than . This contradiction shows that , and so
which implies that as there is no path of the form .
Since for each vertex in , cannot induce a full house, we have . As there is no path of the form , we may suppose by symmetry that . Consequently, as there is no path of the form , and as cannot induce a full house. By putting into , we get another type subgraph with one more vertex than . This contradiction shows that . Therefore,
and by symmetry, we may suppose that is disjoint from one of and for each . But now, we have that , and a path of the form appears. This contradiction proves (3).
By the minimality of , one of and is anticomplete to . In particular, is anticomplete to for some . Since cannot be an odd hole, we have that is odd.
If , since cannot be an odd hole, we may assume by symmetry that which implies that by the minimality of and by the fact that is odd, then there exists an such that is anticomplete to (since one of and is anticomplete to ), and so an odd hole appears, a contradiction.
Therefore, , and by symmetry, we may assume that for each , one of and is anticomplete to . In particular, is anticomplete to . Now, is an odd hole. This contradiction proves (2).
Now we turn to prove the lemma. Choose such that is connected. For each , let be the set of vertices of which have neighbors in . Let be a maximum set such that for each . By (2), as , and is complete to for all distinct pairs . Let . Then is a cutset of such that for each distinct pair , each induced path with interior in is even if and only if for certain (since or by (2), let , we know that there exists an induced even path as and have a common neighbor in for or , and so the argument holds). By Lemma 2.2, admits an harmonious cutset. This completes the proof of Lemma 3.1.
In the following text, we always suppose that is an (odd hole, full house)free graph which is not a type graph, contains no harmonious cutset, and induces an odd antihole on seven vertices. Since is full house free, it is certain that induces no odd antihole on at least nine vertices. We will show that is a heptagramtype graph.
3.2 Heptagrams
Following the proof technique of [6], we first discuss the local structure around an odd antihole on on seven vertices. Let be an odd antihole on seven vertices. Then, the vertices of can be labelled as in such way that is adjacent to if and only if (here and later all index summations are modulo 7).
Defination 3.1.
A heptagram in is defined as follows (see [6]).

are disjoint, nonempty stable sets,

for , is anticomplete to ,

for , and are pairwise linked,

if , , , and is complete to , then ,

if , , , and is anticomplete to , then , and

if , , , and , then either is adjacent to or is adjacent to .
The following three lemmas are from [6], which are all on the structure of heptagrams.
Lemma 3.2.
([6, Lemma 5.1]) Let be a heptagram in . For , if is complete to , then is complete to and is complete to .
Lemma 3.3.
([6, Lemma 5.2]) Let be a heptagram in . For either is complete to or is complete to .
Lemma 3.4.
([6, Lemma 5.3]) Let be a heptagram in . Then there exists a such that is complete to for all , and is complete to for all . Consequently, for all , if and , then

have common neighbors in , in , and in , and

there is a path of the form .
From now on to the end of this section, we always suppose that is a heptagram in , and still use to denote . As that in [6], we will define and study the properties of vertices associated to .
3.3 vertices
In this subsection, we study the properties of vertices defined below (see [6]).
Defination 3.2.
A vertex is called a Yvertex or a Yvertex of type if the following hold, where for :

are nonempty, and for ,

is complete to , and is anticomplete to , and is anticomplete to ,

is complete to .
Lemma 3.5.
Let , and let . Then one of the following holds:

is a vertex, or

is nonempty for at most two integers , and if there are two such integers, and say, then and is complete to .
Proof. Without loss of generality, we suppose that is not a vertex. For , let and , and let .
(4) 
Suppose to its contrary that by symmetry, and let for . By Lemma 3.3, we have is complete to or is complete to . We may suppose that is complete to and .
Suppose that is a triangle. Then is complete to , as otherwise there exists an such that is a full house by Definition 3.1. For any , if is a triangle, then as otherwise is a full house. But now is a full house. This contradiction shows that
(5) 
By Lemma 3.3, we see that is complete to , and is complete to which implies that is complete to by Definition 3.1. By Lemma 3.2, we see that is complete to . Consequently, as otherwise is a full house. By Lemma 3.2, is complete to . Let . Since is not a path and , we have that .
By the same argument in showing (5) and , we can show that is not a triangle, and consequently prove that . Particularly, and by Definition 3.1. Let . Now, is a 5hole path by Lemma 3.4, which contributes a contradiction. This proves (4).
Next, we show progressively that . We start from proving the following
(6) 
Choose . Since is linked to by Definition 3.1, every vertex of has a neighbor in . If has a neighbor and a neighbor , let , then is a triangle by Definition 3.1 which produces a full house on . So, we may assume by symmetry that is anticomplete to . Now, has a neighbor by Definition 3.1.
Since is not complete to , is complete to and is complete to by Lemma 3.3, and consequently is complete to and is complete to by Lemma 3.2. Choose . If has a neighbor , then a 5hole appears. Therefore, is anticomplete to , and so must have a neighbor, say , in . Since and , by the same argument as above used to , we can show that is anticomplete to . Consequently, we have that has a neighbor as and are linked by Definition 3.1, and that is complete to and is complete to by Lemma 3.3.
By Lemma 3.2, we have is complete to . Let . If is anticomplete to , then has a neighbor , and so a 5hole appears, a contradiction. Let , and .
Since and , we have that by Definition 3.1. Let . Since would produce a 5hole , we see that , and so by Definition 3.1 as .
Consequently, we have by Definition 3.1, and by Definition 3.1, and by Definition 3.1. But now, is a . This proves (6).
After (6), we show
(7) 
If it is not the case, then . By (4), we may assume that . If , a 5hole of the form appears by Lemma 3.4. By symmetry, we assume .
We first consider the case that , i.e., . If is not complete to , let and be two nonadjacent vertices, by Definition 3.1, there exists a vertex , and so there exist a vertex such that by Lemma 3.4, then a 5hole appears, a contradiction. So, we have that
is complete to . 
Let . Note that is anticomplete to by Definition 3.1. Since is complete to , and since the vertices in cannot induce a full house, we know that is anticomplete to . So, has a neighbor by Definition 3.1, and is complete to by Lemma 3.4. Now, a 5hole of form appears. This contradiction shows that . Similarly, .
Now, we may suppose by symmetry. Let . We will deduce a contradiction by showing that is also a heptagram.
It is easy to see that satisfies Definition 3.1. From the symmetry, one sees also that:
To prove that satisfies Definition 3.1, we just need to verify the right side of .
Let , , and . To prove the right side of , we may first suppose that
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