Classes of graphs with small rank decompositions are bounded
Abstract
A class of graphs is bounded if the chromatic number of graphs in is bounded by a function of the clique number. We show that if a class is bounded, then every class of graphs admitting a decomposition along cuts of small rank to graphs from is bounded. As a corollary, we obtain that every class of graphs with bounded rankwidth (or equivalently, cliquewidth) is bounded.
1 Introduction
For a graph and an integer , a proper coloring of is a function such that for every edge . The chromatic number of is the smallest such that has a proper coloring. The chromatic number is one of the most studied graph parameters, and although determining it precisely is NPcomplete [3], a number of interesting bounds and connections to other graph parameters is known. A natural lower bound for the chromatic number is given by the clique number which is the size of the largest complete subgraph of . At first, it might be natural to believe that there could exist an upper bound on the chromatic number in the terms of the clique number. However, that is far from the truth. Erdős [2] showed that for any there exist graphs of arbitrarily large chromatic number that do not contain cycles of length at most .
Therefore, the graph classes where the chromatic and clique numbers are tied together are of an interest. The most famous example is the class of perfect graphs which are the graphs such that for every induced subgraph of . The Strong Perfect Graph Theorem asserts that this class can be alternatively characterized as the class of graphs such that neither nor its complement contains an induced odd cycle of length at least (Chudnovsky et al. [1]). The class of perfect graphs includes line graphs of bipartite graphs, chordal graphs, comparability graphs and others.
For many graph classes, the connection between and is not straightforward. For example, if is a circlearc graph (an intersection graph of arcs of a circle), then . This leads to the following definition. We say that a class of graphs is bounded if there exists a function such that for every graph . The examples of bounded graph classes include the circlearc graphs, the circle graphs [6], linegraphs [8], and graphs avoiding any fixed tree of radius at most two as an induced subgraph [5]. Also note that any class of graphs with bounded chromatic number is bounded.
Our work is motivated by the following conjecture of Geelen:
Conjecture 1.
For every graph , the class of graphs without a vertexminor isomorphic to is bounded.
Recall that a graph is a vertexminor of a graph if can be obtained by a series of vertex removals and neighborhood complementations. The neighborhood complementation with respect to a vertex of a graph is the following operation: if two neighbors of are joined by an edge, delete this edge from , and if they are not joined, add an edge joining them.
A possible approach to Conjecture 1 could consist of proving that every graph without a vertexminor isomorphic to admits a decomposition to wellbehaved pieces along simplestructured cuts. Such an approach has led to proofs of several deep results in structural graph theory. Examples include the graph minor structure theorem, where the graphs avoiding a minor of some fixed graph are obtained by joining pieces that are (almost) embedded in surfaces of small genus along small vertex cuts [7], and the proof of the Strong Perfect Graph Theorem [1].
In this paper, we introduce a new (somewhat technical) way of decomposing graphs along cuts of small rank. This kind of decomposition will allow us to prove two special cases of Conjecture 1.
Theorem 1.
For any , the class of graphs with rankwidth at most is bounded.
Theorem 2.
The class of graphs without a vertexminor isomorphic to the wheel is bounded.
We derive Theorem 2 from a structural characterization of graphs avoiding by Geelen [4] and another corollary of our main result. Recall that a join of two graphs and with two distinguished vertices and , respectively, is the graph obtained by deleting the vertex from , , and adding an edge between every neighbor of in and every neighbor of in . The corollary we use to derive Theorem 2 is the following result.
Theorem 3.
Let be a bounded class of graphs closed under taking induced subgraphs. If is the class of graphs that can be obtained from graphs of by repeated applications of joins, then the class is also bounded.
2 Decomposition along small cuts
If we want to allow the existence of cliques of arbitrary order, we cannot restrict the size of the cuts along those we split. On the other hand, such cuts should not be completely arbitrary and thus it is natural to restrict their complexity. This leads to the following definition. For a graph and a proper subset of vertices of , the matrix of the cut is the matrix indexed by and such that is if and otherwise. The rank of the cut is the rank of its matrix over . The diversity of the cut is the size of the largest set such that or and the vertices of have mutually distinct neighborhoods in the other side of the cut. In other words, the diversity of the cut is the maximum number of different rows or columns of the its matrix. Note that if the rank of the cut is , then its diversity is at most , and, conversely, if the diversity is , then its rank is at most .
A natural way of decomposing a graph is assigning its vertices to nodes of a tree. Formally, a decomposition of a graph is a tree with a mapping . Each edge of the tree naturally defines a cut in the graph with sides being the preimages of the vertex sets of the two trees obtained from by removing . The rank of a decomposition is the maximum of the ranks of the cuts induced by its edges. Analogously, we define the diversity of a decomposition.
Every graph admits a decomposition of rank one with being a star, so restricting the rank of decompositions is too weak. One way to circumvent this is to restrict the structure of the decomposition. For example, if we require that all inner nodes of have degree three and that the vertices of are injectively mapped by to the leaves of , the rank of the smallest such decomposition is the rankwidth of , a wellstudied parameter in structural graph theory. In this paper, we proceed in a different (more general) way.
If a tree with a mapping is a decomposition of a graph and is a node of , then is the spanning subgraph of such that two vertices and of are adjacent in if they are adjacent in and the node lies on the unique path between and in (possibly, can be or ). In other words, is the spanning subgraph of where we remove edges between vertices and such that and lies in the same component of . For a class of graphs, we say that the decomposition is bounded if the graph belongs to for every .
Our main result is the following theorem.
Theorem 4.
Let be a bounded class of graphs closed under taking induced subgraphs and an integer. The class of graphs admitting a bounded decomposition with rank at most is bounded.
We would like to note that the definition of bounded decompositions can also be extended as follows. Let and be two classes of graphs. We can require that the subgraphs of induced by , , belong to and the subgraphs to . So, the class controls the complexity of subgraphs induced by preimages of individual nodes and the class controls the mutual interaction between different pieces of the decomposition. However, if both and are bounded and closed under taking induced subgraphs, this does not lead to a more general concept. Indeed, let be the class of graphs that admit a vertex partition and such that for . If and are bounded and closed under taking induced subgraphs, so is , and any decomposition with respect to and as defined in this paragraph is bounded.
We now derive Theorems 1–3 from Theorem 4. The decompositions appearing in the definition of the rankwidth of a graph, which is given earlier in this section, are bounded where is the class of colorable graphs. So, an application of Theorem 4 for this class yields Theorem 1.
The proof of Theorem 3 is more complicated.
Proof of Theorem 3..
Let be a graph contained in . By the definition of , there exists a tree with nodes and graphs from such that

for every edge of , the graphs and contain distinguished vertices and , respectively, and these vertices are different fordifferent choices of , and

the graph is obtained from the graphs by joins with respect to vertices and , .
Note that the order of joins does not affect the result of the procedure.
Let be the class of graphs that can be obtained from by adding isolated vertices and blowing up some vertices to independent sets, i.e., replacing a vertex with and independent set and joining each vertex of to the neighbors of . Since is bounded and closed under taking induced subgraphs, so is . Since the tree with mapping that maps a vertex to the node such that is a bounded decomposition of and its rank is at most one, Theorem 3 now follows from Theorem 4. ∎
To derive Theorem 2 from Theorem 3, we need the following result of Geelen [4]. Observe that since the class of circle graphs is bounded [6] and all other basic graphs appearing in Theorem 5 have at most vertices, Theorem 2 directly follows from Theorem 3.
Theorem 5 (Geelen [4], Theorem 5.14).
If is a connected graph without a vertexminor isomorphic to , then one of the following holds:

is a circle graph, or

is can be obtained from a graph isomorphic to , the cube or the graph , that is the cube with one vertex removed, by a sequence of neighborhood complementations, or

there exist connected graphs and without a vertexminor isomorphic to that have fewer vertices than , such that is a join of and .
3 Proof of Main Theorem
In this section, we present the proof of Theorem 4. We start with a lemma.
Lemma 6.
Let and be two integers and a connected graph with at least two vertices. If has a decomposition formed by a tree and a mapping such that the diversity of the decomposition is at most and for every node of , then there exists a (not necessarily proper) coloring of the vertices of by at most colors such that for every color .
Proof.
We can assume (without loss of generality) that has a leaf with and root at this leaf. Fix a sequence of subtrees of such that is the subtree formed by solely, and for every . Let be the vertex of not contained in . For an edge , the origin of is the nearest common ancestor of and in .
For a node of , let be the subtree of rooted at and let be the union . Since the diversity of the decomposition given by and is at most , there exists a partition of such that the vertices of have no neighbors outside and two vertices of belong to the same if and only if they have the same neighbors outside of .
The set of colors used by the constructed coloring will be . We will construct partial (not necessarily proper) colorings , , …, of which we use the colors such that

extends for ,

assigns colors to all vertices incident with edges whose origin belongs to , in particular, to all vertices of ,

the vertices of for and are either all colored with the same color or none of them is colored by , , and

if for an edge of and , then there exists , , a child of , and such that both and are not colored by , they are colored by and they both belong to .
Since is the origin of no edge, we can choose as the empty coloring. Suppose that was already defined and let us describe the construction of from . Note that the vertices of that are not colored by are exactly those of . Furthermore, the restriction of to uses at most colors (one for each of , …, ). Let be the set of these colors.
By the assumption of the lemma, there exists a proper coloring of the graph using colors. Choose such that all twins, i.e., vertices with the same set of neighbors, receive the same color. Let be the vertices of that have a nonzero degree in or are mapped by to . Let be a (not necessarily proper) coloring of the vertices of by colors defined as follows: if , the vertex is assigned the color . For , let be the child of such that . Set to be the index such that (note that ). Finally, set ; is a (not necessarily proper) coloring of the vertices of with colors.
The coloring is an extension of to the vertices of such that uses the (at least) colors of and two vertices of get the same color if and only if they are assigned the same color by .
We now verify that has the properties stated earlier.

This follows directly from the definition of .

If the origin of an edge of belongs to , then it either belongs to or it is . In the former case, and are already colored by . In the latter case, they both belong to and are colored by .

If any vertex of is colored by , then all of them are colored by and they have the same color. So, suppose that none of the vertices of is colored by . Observe that all vertices of are twins in . So, if one of them belongs to , then all of them do. Also, they are assigned the same color by . Let be the child of such that belongs to . Since the vertices of have the same neighbors outside , they all belong to the same set . Consequently, they are assigned the same color by . So, they have the same color assigned by and thus by , too.

Suppose that and is an edge of . If both and are colored by , the property follows from the properties of . By symmetry, we can now assume that is not colored by . Since is not colored, must belong to . If were colored by , it would belong to one of the sets . So, it would hold that but which is impossible. We conclude that neither nor is colored by . Since they are colored by , they both belong to , and, since assigns them the same color, it holds that . The definition of implies that both and belong to the same for a child of (otherwise, the edge would be contained in ). Finally, the definition of implies that they both belong to the same for .
We claim that is the sought coloring. The number of colors it uses does not exceed . Assume that contains a monochromatic maximum clique . The fourth property of the coloring implies that all vertices of belong to for some node of and . In particular, they have a common neighbor which is impossible since is a maximum clique. ∎
The proof of the main theorem now follows.
Proof of Theorem 4..
Let be the function witnessing that is bounded and let . We show that for any graph admitting a bounded decomposition with rank at most by induction on the clique number .
If , then . Assume that . By Lemma 6, the vertices of can be colored with colors in such a way that no clique of of size is monochromatic. Fix such a coloring . Consider the subgraphs of induced by the color classes of . Since each of these subgraphs admits a bounded decomposition with rank at most , it has a proper coloring with colors. Coloring each vertex by a pair consisting of the color assigned by and the color assigned by the coloring of the corresponding color class of yields a proper coloring of with colors. ∎
Acknowledgements
The authors would like to thank DIMACS for hospitality during the DIMACS Workshop on Graph Coloring and Structure held in Princeton, May 2009, where they learnt about Conjecture 1 from Jim Geelen. They are also grateful to Sangil Oum for bringing Theorem 5 to their attention as well as for helpful discussions on the subject.
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