1 Introduction
The vertexarboricity of a graph is the minimum such that the vertices of can be partitioned into subsets each of which induces a forest. We contrast this with the chromatic number of , which is the minimum number such that the vertices of can be partitioned into subsets each of which is independent. Like the chromatic number, determining the vertex arboricity of graphs is NPhard in general garey1979 . We focus our attention on the class of cographs, where both problems are polynomialtime solvable. We define a common generalization as follows. We say that a graph is partitionable if the vertex set of can be partitioned into forests and independent sets. This problem is NPhard in general as well, as long as (and is polynomialtime solvable otherwise). On the other hand, it follows from damaschkeJGT14 that this problem also has a polynomialtime algorithm for any , when restricted to cographs. Moreover, it follows from damaschkeJGT14 that the number of minimal obstructions for cograph partitionability is finite, for any . We investigate such minimal obstructions for partitionability of cographs, i.e., for arboricity . We give a complete answer only for arboricity , and give some useful information for general . We also give a concrete dynamic programming algorithm to decide whether a cograph is partitionable after the deletion of at most vertices. This last problem, allowing the deletion of vertices, is natural for the dynamic programming algorithm, but it is an interesting problem which can be formulated as follows.
Let and be nonnegative integers and let be a graph. A partition of is a partition of its vertex set such that the subgraph induced on has vertexarboricity , the subgraph induced on is colourable, and has at most vertices. We say that is partitionable if it admits a partition, and we say that is a minimal obstruction if it is not partitionable but every induced subgraph is. (When , we simplify to in all the notation.) Note that finding the minimum such that is partitionable is the wellknown problem of finding the maximum colourable subgraph; finding the minimum such that is partitionable is the problem of finding the maximum subgraph of arboricity ; finding the minimum such that is partitionable is the minimum vertex feedback set problem; finding the minimum such that is partitionable is the problem of finding the smallest such that has a colourable vertex feedback set.
If and are graphs, then we denote the disjoint union of and by , and so, if is a positive integer, the disjoint union of different copies of will be denoted by . The join of and will be denoted by .
A cograph is a graph than can be obtained recursively from the following rules

is a cograph.

If is a cograph, then is a cograph.

If and are cographs, then is a cograph.
There are many interesting characterizations of the family of cographs corneilDAM3 , but there are two that are particularly useful when dealing with minimal obstructions for a hereditary property. A graph is a cograph if and only if it is free, if and only if the complement of any of its nontrivial connected subgraphs is disconnected. Notice also that the complement operation can be replaced by the join of two graphs ().
The rest of the article is organized as follows. In Section 2, cographs that are partitionable are characterized in terms of 7 minimal obstructions; some families of cograph minimal obstructions for partitions are studied. In Section 3 we consider minimal obstructions for partitions, and notice how these partitions are related to the independent feedback vertex set problem. Although finite, the cograph minimal obstructions for partition can be very large, both in size and in number, Section 4 is devoted to present lower and upper bounds for these parameters, as well as an upper bound on the height of the cotree of a minimal obstruction. A polynomial algorithm to determine the arboricity of a cograph is presented in Section 5. In Section 6 we present conclusions and related open problems.
2 All Minimal Cograph Obstructions for Arboricity 2
Note that a graph has arboricity one if and only if it has no cycles. Thus there are precisely two cograph minimal obstructions for arboricity one, the cycles and .
We now introduce a family of cographs consisting of
These graphs are depicted in Figure 1.
Lemma 1.
Each graph in is a cograph minimal obstruction to arboricity .
Proof.
It is clear from the descriptions that each graph in is a cograph. We claim that each of them is not partitionable into two forests, but whenever a vertex is deleted, it becomes so partitionable.
We prove the first, fourth and last cases (, , ); the rest of the cases can be handled similarly. Consider first . It is clear that in a complete graph each (acyclic) colour class has at most vertices, therefore is a obstruction. To check the minimality, remove any vertex of , the remaining graph is a which is easily colourable. Therefore, is actually a minimal obstruction.
Let us assume that . First notice that to colour by using only colours, at least two vertices receive the same colour, and then we can use that colour on at most one other vertex outside . Since we cannot use this colour anywhere else, without loss of generality we can assume that all vertices of have the same colour and we are using this colour on one other vertex as well. So we have coloured (at most) vertices using one colour. There are two disjoint copies of (minus one vertex) still uncoloured. On each of the copies we can use one colour for at most vertices. Therefore, at most vertices can be coloured, and thus at most vertices can be coloured with colours. Since has vertices, we conclude that is a obstruction. To verify minimality, first consider the case where . Then we can colour one vertex of along with one copy of in each of the two parts to colour . Now consider the case where belongs to . Let be the duplicate of in the other copy of and colour along with all vertices of using one colour. The remaining vertices induce an acyclic graph and can be coloured with one colour. This shows that is (acyclic) colourable. Therefore is a minimal obstruction.
For our final case, let . Notice that has vertices and each (acyclic) colour class has size at most 3 (two vertices of a and another vertex). Therefore is a obstruction. By removing any vertex of we will have vertices and at least copies of . Considering each of these ’s along with some other vertex (maybe in another ) will give us an acyclic colouring where each colour class has size exactly 3. Thus, is actually a minimal obstruction. ∎
Theorem 2.
A cograph has vertex arboricity if and only if it is free.
Proof.
Let be a cograph. If the vertex arboricity of is at most , then it is clearly free. So, suppose that is free. We may assume without loss of generality that is connected.
Since is a connected cograph, there exist cographs and such that . If and are forests, we are done. So, at least one of them must contain an induced cycle. Without loss of generality suppose that it is . Since is a cograph, this cycle should be a triangle or a cycle. Suppose that is triangle free, then it must contain a cycle. Since is free, then is trianglefree, and, since is free, then is free, i.e., is a disjoint union of ’s and ’s; but is free, then is free, and thus, it is either a , or an empty graph. As a trianglefree cograph, is bipartite, with bipartition . If , colour red one vertex in , together with all the vertices in , and colour blue the other vertex in (possibly none), together with the vertices in , this colouring of realizes the vertex arboricity . If , as is free, we have that every component of , different from the one containing the induced cycle, is a star. The component of containing the induced cycle is a bipartite connected cograph, and thus, it is a complete bipartite graph; moreover, since is free, one of the two parts of this component has less than three vertices. colour red one of the vertices in this small part, together with all the vertices in , and colour blue all the remaining vertices of . Clearly, the red vertices induce a star, and the blue vertices induce a disjoint union of stars.
Now, suppose that contains an induced triangle. Using again that is free, we conclude that is an empty graph. Let the set induce a triangle in , and let be the component of containing it. Now, is a connected cograph, and there are cographs and such that . Recall that is free, and thus, both and are trianglefree, so, we assume without loss of generality that , and ; moreover, must be an independent set.
We will consider two cases; suppose first that has at least two vertices. Then, since is and free, we have that has precisely one component, namely , which is not acyclic. Again, we have two cases. First, suppose that has at least two vertices. From the fact that is free, we obtain that is connected, and, since is free, then is a path on two vertices, and . So, we can colour , together will all the vertices in red, and the rest of the vertices in blue; it is not hard to verify that each colour class induces a forest. So, we may now suppose that has only one vertex. Since is trianglefree, it is bipartite with bipartition . Again, as is free, we have that is acyclic, and we can colour the only vertex in together with all the vertices in red, and the rest of the vertices in blue; again, each colour class induces a forest.
As a second case, suppose that has only one vertex . Since is free, and is a universal vertex in , we have that is free. It follows from Theorem 4 with that contains an independent feedback vertex set, . Thus, colouring together with the vertices in red, and the rest of the vertices of blue, clearly yields acyclic colour classes. ∎
The family has a natural generalization for higher arboricity. Let be an integer, , and denote by the following family of cographs.
Lemma 3.
Let be an integer, . Each graph in the family is a cograph minimal obstruction for arboricity .
Proof.
The proof is analogous to that of Lemma 1. ∎
3 Minimal Cograph Obstructions for Colourable Vertex Feedback Set
By analogy with the independent vertex feedback set problem, we say that a cograph has a colourable vertex feedback set if it has a partition. It turns out there are exactly two minimal cograph obstructions for partition, namely, the complete graph , and the complete partite graph with two vertices in each part, .
Theorem 4.
Let be a nonnegative integer. A cograph has a colourable vertex feedback set if and only if it is free.
Proof.
Clearly, a partitionable cograph is free. We prove the converse by induction on . The base case follows from the simple fact that a cograph is a forest is and only if it is free. Suppose that the claim holds for all , and let be a free cograph. Without loss of generality, we may assume is connected. The fact that is free implies , and the claim holds by taking an independent set as the forest if , so we may assume .
Since is a connected cograph, there exists a family of cographs such that . Notice that in any colouring of , each colour class is contained in for some . If there is a colouring of with a colour class with a single vertex , then together with any other colour class induces a forest. By taking this forest and the remaining colour classes we obtain a partition of , so we may assume that every colour class has at least two vertices.
Since , if has an induced for every , then has an induced copy of , thus is free for some . By induction hypothesis, has a partition (notice that if , the fact that each colour class has at least two vertices would imply the existence of an induced , and so we must have that ). Since has chromatic number , a proper colouring of , together with the partition of gives us a partition of . ∎
We can use the Theorem to derive a minmax relationship. For the purposes of its statement, we shall call a thin clique and a thick clique. The strength of a thin clique is defined to be , and the strength of a thick clique is defined as . We let denote the maximum strength of a (thin or thick) clique in . We also let denote the minimum number of colours such that admits a colourable vertex feedback set.
Corollary 5.
If is a cograph, then .
We note that the maximum strength of a clique in a cograph can be computed by a cotree bottomup procedure analogous to the wellknown algorithm for computing the size of a maximum complete subgraph of a cograph corneilDAM3 .
4 Bounds on Minimal Cograph Obstructions for Arboricity
As we mentioned at the end of Section 2, we do not have a complete description for all cograph minimal obstructions for arboricity , . In this section we illustrate the fact that there are exponentially many. We will construct this obstructions as joins of star forests.
Proposition 6.
Let be an integer, . There are at least minimal obstructions for arboricity .
Proof.
Let us first observe that if we add some edges to a minimal obstruction, the resulting graph is still an obstruction but might not be minimal. Consider the complete multipartite graph (which is a minimal obstruction for arboricity ), and add edges to of the parts to make them nonempty forests.
Let be the set of all nonempty forests on vertices which are cographs. Notice that a tree cograph on a fixed number of vertices is unique (it is a star). Therefore, , where is the is the partition number of (the number of possible partitions of ) . In marotiEJCNT3 the following lower bound is proved for
Let be an integer, , and define the graph as , where for . Notice that when , does not receive arguments, and we obtain , and when we have that is isomorphic to for every , and thus, the only possible graph that can be obtained is . We denote the set of all graphs for all selections of by .
Claim 0.
For every , each is not colourable.
Let be a member of . The number of vertices of is . On the other hand, notice that we can use each colour class in at most two parts of (otherwise we will get a monochromatic cycle). Also if we want to use a colour in two parts, in one of them we are using it at most once (or we will get a monochromatic ). Since each is nonempty, if we colour it with just one colour, then we cannot use that colour for any other vertices (or we will get a monochromatic triangle). Therefore, in each colour class we can have at most vertices. Hence, using colours we can colour at most vertices of . Therefore is not colourable for .
Claim 0.
For every , every is a minimal obstruction for arboricity .
We have proved that is an obstruction, now we need to show that it is minimal. If we remove a vertex from , then we can use each colour for one vertex of and one of the parts of , which is an independent set (using colours). Also we can use one colour for each of the remaining (using colours).
If we remove a vertex from a part of , then we can colour the remaining graph by using only colours; use one colour for each vertex of this part together with all the vertices in another of the parts, i.e., use each vertex of this part as the center of a star having all the vertices in some other part as leaves (using colours) and use one colour for each (using colours).
Now we need to calculate . For a fixed value of , we have to consider the effect of permutation (switching and ) by dividing each term by . It is easy to see that for two different sets of forests we will get different obstructions. So we have
Notice that if , then members of and have different number of vertices and therefore they are different. So the total number of different minimal obstruction that we will get from this structure is
∎
Next we focus on upperbound for minimal cograph obstructions to arboricity . First, we consider the cotree height.
Theorem 7.
Let be an integer, . If is a minimal cograph obstruction for arboricity with cotree , then the height of is at most .
Proof.
We may assume that is connected and therefore the root vertex of the cotree is a join vertex, . Notice that to have a unique cotree all children of a join vertex should be union or single vertices and all children of union vertices should be join or single vertices. Recall that , whose cotree has height two, is a minimal obstruction and therefore no other minimal obstruction can contain it. For simplicity, in the following we will use instead of to denote the clique number of the subgraph of induced by the vertex set .
Let be a join vertex of the cotree with degree whose children are . It is easy to see that . In particular, implies that for any that is a child of . This implies that any path from to a leaf of contains at most join vertices, hence the height of the cotree is at most . ∎
Corollary 8.
Let be an integer, . Let be a minimal cograph obstruction for arboricity and let be its cotree.
If , then every join vertex in has at most children.
Theorem 9.
Let and be minimal cograph obstructions for vertex arboricity such that for . Let denote the height of the cotree of a cograph . If is an independent set of size , then the cograph satisfies:

.

is a cograph minimal obstruction for arboricity.

Proof.
It is easy to see that , and . If , then there exists a partition of into induced forests, and hence, at least one forest in contains two distinct vertices of . This implies that there exists such that , and so the restriction of to is a partition of into forests, which is a contradiction, and so holds.
To show , let . If , let be , and take a colouring, , of for . Notice that for every , the set induces a forest in , which shows . Suppose now that , and let and take and . Let be a partition of into forests for . (Such partitions exist due to the minimality of and .) Let be given by
It is easy to see that induces a partition of into forests, which shows .
Part follows directly from the construction of . ∎
We did not succeed to obtain an analog to Corollary 8 in terms of union vertices, which would yield an upper bound on the size of a cograph minimal obstruction for arboricity . Instead, we derive, from the algorithm in the next section, the following result.
Theorem 10.
Each minimal cograph obstruction for arboricity has at most vertices.
5 A polynomial algorithm
The following simple observation describes the recursive structure of partitions in cographs. The only thing to remember is that for the second statement (join of two cographs), a forest cannot intersect both sides in more than one vertex.
Proposition 11.

Let be a cograph with nonempty subgraphs and . Then has a partition if and only if there exist integers such that and and have a partition and a partition, respectively.

Let be a cograph with nonempty subgraphs and . Then has a partition if and only if there exist integers such that , , and and have a partition and partition, respectively.
Notice that the integers in Proposition 11 are not necessarily unique. If is a cograph, then there exists cographs and such that either or . Suppose and have an partition and an partition, respectively. Then, the triples such that has a partition as the one described in Proposition 11 are said to be derived from and .
This structure can be used to obtain an efficient algorithm to solve the partition problem in cographs. To this end, define the weight of a triple to be .
Proposition 12.
Given two triples and with weights at most , the set of all triples derived from and can be generated in time.
Proof.
If then according to the first item in Proposition 11, we have that as the only option. But if then we may have more options for . In this case, by setting , any triple can be produced, given that all the components of the triple are nonnegative. ∎
Theorem 13.
Given a cograph with vertices, there exists an algorithm that computes in all the triples with weight at most such that admits a partition.
Proof.
We build an algorithm recursively as follows. The algorithm is trivial when is a clique or an independent set. So suppose either or . Then apply to and to obtain the lists and . Then for each pair , add all the triples derived from and to some list , which is our final answer. Let be the run time of this algorithm on a cograph with vertices. Suppose and have , and vertices, respectively. Considering Observation 12 and the fact that lists and each have at most members, we get the following recursion:
which implies . ∎
Finally, we use a similar approach, based on triples, to prove Theorem 10. Let be a set of triples of nonnegative integers. We say a graph is a minimal cograph obstruction for the set if the following conditions hold:

does not admit a partition for any .

For any vertex , there exists a triple such that admits a partition.
Given a triple , by the weight of now we mean (and not ). The following observation is immediate.
Observation 14.
Suppose a triple is derived from triples and . Then . Furthermore, if , is partitionable, and and are partitionable and partitionable, respectively, then .
For integers , let be the smallest integer with the following property: any minimal obstruction with respect to a set of triples with weight at most and at most triples with weight exactly has size at most
. Now we embark on estimating
using recursion. Note that . For the sake of convenience, we assume .Theorem 15.
Proof.
Let be a set of triples with weight at most and at most triples having weight . Let a minimal obstruction with respect to . Ignoring trivial cases, we may assume that either or , where both and are nonempty. For denote by the set of triples with weight at most such that admits an partition. We say a triple is dangerous for (, respectively) if there exists a triple (, respectively) such that from and we can derive a triple in . Let (, respectively) be the set of all triples dangerous for . Note that is nonempty if has at least two vertices. To see this, let be an arbitrary vertex. Then has a partition for some . Applying Observation 11, there must be triples and such that is derived from and and and admit an partition and an partition, respectively. This means and . A similar argument shows that must be a minimal obstruction with respect to if it has at least two vertices. Note that since is nonempty so any triple in has non zero weight. This implies the weight of triples in is at most according to Observation 14. In fact, if , the weight of the triples in (and ) are at most . So in this case, we have , and thus:
(1) 
Now suppose . Suppose contains a triple with weight . This implies admits an partition. So , which means (otherwise would admit an partition). This means that for some integer , we have
(2) 
∎
6 Concluding remarks
We have already observed that the partition problem can be considered as a general framework that includes interesting problems, e.g., colouring, arboricity , or independent feedback vertex set. In these cases, the value of is . Notice that can be used as an additional input value to state some classic decision problems arising from optimization problems. We discuss two examples.
Recall that the vertex cover optimization problem asks, given a graph , to find the size of a minimum vertex cover of . There is a decision problem associated to this optimization problem. The problem Vertex Cover takes as input a graph and a nonnegative integer , and asks whether contains a vertex cover with at most vertices. Now, notice that a partition of a graph is a partition into an independent set, and a set with at most vertices , this is, all the edges of must have at least one end in the set . From here, it is easy to conclude that has a vertex cover with at most vertices if and only if admits a partition.
The odd cycle transversal problem asks to find the minimum set of vertices having a nonempty intersection with every odd cycle in a graph
. Again, this optimization problem has an associated decision problem. Consider the Bipartization problem, with input , where is a graph and is a nonnegative integer, and where we have to decide whether or not there is a subset of at most vertices of such that is a bipartite graph. Notice that alternatively, we could ask whether admits a partition. It was proved in lewisJCSS20 that Bipartization is NPcomplete, even when restricted to planar graphs.In this setting, it is easy to notice that partition corresponds to the Feedback Vertex Set problem. We think that partitions represent a nice framework where many seemingly unrelated problems converge.
Before proceeding, we make a simple observation about the structure of minimal obstructions to the partition problem.
Theorem 16.
Let and be nonnegative integers. Every disconnected cograph minimal obstruction is of the form where is a cograph minimal obstruction for and .
Proof.
Let be a minimal obstruction and the set of components of . Since is a minimal obstruction, we know that for every there exists such that , , and is colourable.
Claim 1.
For every , every choice of and , we have for each .
Suppose otherwise and let , and be such that . Since is a minimal obstruction for partition, then for every . This means that satisfies , and so is not colourable. Since for every , , it follows that , but , contradicting the choice of .
Claim 2.
Let and take . For every , the cograph is a minimal obstruction for partition.
Due to the choice of and Claim 1, is an obstruction to partition. To see that it is minimal, let and . Since is colourable and , then is a minimal obstruction for partition.
Since , we get that , completing the proof of the theorem. ∎
We propose a question that might turn out to be interesting. Observe that for the two cases and the number of cograph minimal obstructions is independent of . (There is precisely one cograph minimal obstruction for partition, because cographs are perfect, and it follows from Theorem 4 that there are exactly two cograph minimal obstructions for partition.) We wonder whether it is always the case that the number of cograph minimal obstructions is independent of .
Problem 17.
Let be a fixed nonnegative integer. Is it true that there is an integer such that for every nonnegative integer , the number of cograph minimal obstructions for partition is ?
If the answer to Problem 17 is negative, then it could be interesting to determine for which values of the number of cograph minimal obstructions for partition is independent of .
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