1. Introduction
Treechromatic number is a hybrid of the graph parameters treewidth and chromatic number, recently introduced by Seymour [17]. Here is the definition.
A treedecomposition of a graph is a pair where is a tree and is a collection of subsets of vertices of , called bags, satisfying:

for each , there exists such that , and

for each , the set of all such that induces a nonempty subtree of .
A graph is colourable if each vertex of can be assigned one of colours, such that adjacent vertices are assigned distinct colours. The chromatic number of a graph is the minimum integer such that is colourable.
For a treedecomposition of , the chromatic number of is . The treechromatic number of , denoted , is the minimum chromatic number taken over all treedecompositions of . The pathchromatic number of , denoted , is defined analogously, where we insist that is a path instead of an arbitrary tree. Henceforth, for a subset , we will abbreviate by . For , let be the set of neighbours of and .
The purpose of this paper is to survey the known results on tree and pathchromatic number, and to present some new results and conjectures.
Clearly, and are monotone under the subgraph relation, but unlike treewidth, they are not monotone under the minor relation. For example, , but the graph obtained by subdividing each edge of is bipartite and so .
By definition, for every graph ,
Section 2 reviews results that show that each of these inequalities can be strict and in fact, both of the pairs and can be arbitrarily far apart.
We present our new results and conjectures in Sections 35. In Section 3, we propose a version of Hadwiger’s Conjecture for treechromatic number and show how it is related to a ‘local’ version of Hadwiger’s Conjecture. In Section 4, we prove that minorfree graphs have treechromatic number at most , without using the Four Colour Theorem. We finish in Section 5, by presenting some hardness results and conjectures for computing and .
2. Separating , and
Complete graphs are a class of graphs with unbounded treechromatic number. Are there more interesting examples? The following lemma of Seymour [17] leads to an answer. A separation of a graph is a pair of edgedisjoint subgraphs whose union is .
Lemma .
For every graph , there is a separation of such that and
Seymour [17] noted that Lemma 2 shows that the random construction of Erdős [6]
of graphs with large girth and large chromatic number also have large treechromatic number with high probability.
Interestingly, it is unclear if the known explicit constructions of large girth, large chromatic graphs also have large treechromatic number. For example, shift graphs are one of the classic constructions of trianglefree graphs with unbounded chromatic number, as first noted in [7]. The vertices of the th shift graph are all intervals of the form , where and are integers satisfying . Two intervals and are adjacent if and only if or . The following lemma (first noted in [17]) shows that the gap between and is unbounded on the class of shift graphs.
Lemma .
For all , and .
Proof.
The fact that is wellknown; we include the proof for completeness. Let and be a proper colouring of . For each let . We claim that for all , . By definition, . If , then for some . But this is a contradiction, since and are adjacent in . Since there are subsets of , , as required.
We now show that . For each , let . Let be the path with vertex set (labelled in the obvious way). We claim that is a pathdecomposition of . First observe that if and only if . Next, for each edge , . Finally, observe that for all , and is a bipartition of . Therefore, has pathchromatic number , as required. ∎
Given that shift graphs contain large complete bipartite subgraphs, the following question naturally arises.
Open Problem .
Does there exist a function such that for all and all free graphs , ?
It is not obvious that the parameters and are actually different. Indeed, Seymour [17] asked if for all graphs ? Huynh and Kim [10] answered the question in the negative by exhibiting for each , an infinite family of connected graphs for which . They also prove that the Mycielski graphs [14] have unbounded pathchromatic number.
However, can and be arbitrarily far apart? Seymour [17] suggested the following family as a potential candidate. Let be the complete binary rooted tree with leaves. A path in is called a if the vertex of closest to the root (which we call the low point of the ) is an internal vertex of . Let be the graph whose vertices are the s of , where two s are adjacent if the low point of one is an endpoint of the other.
Lemma ([17]).
For all , and .
Proof.
For each , let be the set of s in which contain . We claim that is a treedecomposition of with chromatic number . First observe that if is a , then , which induces a nonempty subtree of . Next, if and are adjacent s with , then . Finally, for each , let be the elements of whose low point is and let . Then is a bipartition of , implying that .
For the second claim, it is easy to see that contains a subgraph isomorphic to the th shift graph . Thus, , by Lemma 2. ∎
BarreraCruz, Felsner, Mészáros, Micek, Smith, Taylor, and Trotter [1] subsequently proved that for all . However, with a slight modification of the definition of , they were able to construct a family of graphs with treechromatic number 2 and unbounded pathchromatic number.
Theorem 2.1 ([1]).
For each integer , there exists a graph with and .
The definition of is as follows. A subtree of the complete binary tree is called a if it has three leaves and the vertex of the closest to the root of is one of its three leaves. The vertices of are the s and s of . Two s are adjacent if the low point of one is an endpoint of the other. Two s are adjacent if the lowest leaf of one is an upper leaf of the other. A is adjacent to a if the low point of the is an upper leaf of the . The proof that uses Ramsey theoretical methods for trees developed by Milliken [13].
3. Hadwiger’s Conjecture for and
One could hope that difficult conjectures involving might become tractable for or , thereby providing insightful intermediate results. Indeed, the original motivation for introducing was a conjecture of Gyárfás [8] from 1985, on boundedness of trianglefree graphs without long holes ^{1}^{1}1A hole in a graph is an induced cycle of length at least ..
Conjecture (Gyárfás’s Conjecture [8]).
For every integer , there exists such that every trianglefree graph with no hole of length greater than has chromatic number at most .
Theorem 3.1 ([17]).
For all integers and , if is a graph with no hole of length greater than and for all , then .
Note that Theorem 3.1 with implies that for every trianglefree graph with no hole of length greater than . A proof of Gyárfás’s Conjecture [8] (among other results) was subsequently given by Chudnovsky, Scott, and Seymour [3].
The following is another famous conjectured upper bound on , due to Hadwiger [9]; see [16] for a survey.
Conjecture ([9]).
If is a graph without a minor, then .
We propose the following weakenings of Hadwiger’s Conjecture.
Conjecture .
If is a graph without a minor, then .
Conjecture .
If is a graph without a minor, then .
By Theorem 2.1, and can be arbitrarily far apart, so Conjecture 3 may be easier to prove than Conjecture 3. By Theorem 2, and can be arbitrarily far apart, so Conjecture 3 may be easier to prove than Hadwiger’s Conjecture. We give further evidence of this in the next section, by proving Conjecture 3 for , without using the Four Colour Theorem.
Robertson, Seymour, and Thomas [15] proved that every minorfree graph is colourable. Their proof uses the Four Colour Theorem and is pages long. Thus, even if we are allowed to use the Four Colour Theorem, it would be interesting to find a short proof that every minorfree graph has treechromatic number at most .
Conjectures 3 and 3 are also related to a ‘local’ version of Hadwiger’s Conjecture via the following lemma.
Lemma .
Let be a optimal treedecomposition of , with minimal. Then there are vertices and such that .
Proof.
Let be a leaf of and be the unique neighbour of in . If , then contradicts the minimality of . Therefore, there is a vertex such that for all . It follows that , as required. ∎
Lemma 3 immediately implies that the following ‘local version’ of Hadwiger’s Conjecture follows from Conjecture 3.
Conjecture .
If is a graph without a minor, then there exists such that .
It is even open whether Conjectures 3, 3, or 3 hold with an upper bound of instead of . Finally, the following apparent weakening of Hadwiger’s Conjecture (and strengthening of Conjecture 3) is actually equivalent to Hadwiger’s Conjecture.
Conjecture .
If is a graph without a minor, then for all .
Proof of equivalence to Hadwiger’s Conjecture.
4. minorfree graphs
As evidence that Conjecture 3 may be more tractable than Hadwiger’s Conjecture, we now prove it for minorfree graphs without using the Four Colour Theorem. We begin with the planar case.
Theorem 4.1.
For every planar graph , .
Proof.
We use the same treedecomposition previously used by Eppstein [5] and Dujmović, Morin, and Wood [4].
Say has vertices. We may assume that and that is a plane triangulation. Let be the set of faces of . By Euler’s formula, and . Let be a vertex of . Let be the bfs layering of starting from . Let be a bfs tree of rooted at . Let be the subgraph of the dual with vertex set , where two vertices are adjacent if the corresponding faces share an edge not in . Thus
By the Jordan Curve Theorem, is connected. Thus is a tree.
For each vertex of , if corresponds to the face of , let , where is the vertex set of the path in , for each . See [5, 4] for a proof that is a treedecomposition of .
We now prove that is 4colourable. Let be the largest index such that . For each , let . Note that . We prove by induction on that is 4colourable. This clearly holds fof , since .
For the inductive step, let . For each , let . Since contains at most one vertex from each of , and , . If , then for all . Since all edges of are between consecutive layers or within a layer, clearly implies that is colourable. So, we may assume . By induction, let be a colouring of . Let .
If , then we can extend to a 4colouring of by using to 3colour .
Suppose . By induction, has a 4colouring . If is a stable set, then we can extend to a colouring of such that all vertices of are the same colour. Thus, , and we are done by the previous case. Let such that . Let be the other vertex of (if it exists). By relabeling, we may assume that , and . Let be the set of neighbours of in and be the set of neighbours of in . Observe that extends to a colouring of unless . However, if, , then we obtain a minor in by using to contract onto and onto (if exists). This contradicts planarity.
The remaining case is . In this case, extends to a colouring of , unless there exist distinct vertices such that and are both adjacent to all vertices of . Again we obtain a minor in by using to contract onto and contracting all but one edge of the – path in . ∎
We finish the proof by using Wagner’s characterization of minorfree graphs [19], which we now describe. Let and be two graphs with , where is a clique of size in both and . The sum of and (along ) is the graph obtained by gluing and together along (and keeping all edges of ). The Wagner graph is the graph obtained from an cycle by adding an edge between each pair of antipodal vertices.
Theorem 4.2 (Wagner’s Theorem [19]).
Every edgemaximal minorfree graph can be obtained from , , and sums of planar graphs and .
Theorem 4.3.
For every minorfree graph , .
Proof.
Let be a minorfree graph. We proceed by induction on . We may assume that is edgemaximal. First note that if , then . Next, if is planar, then by Theorem 4.1 (whose proof avoids the Four Colour Theorem). By Theorem 4.2, we may assume that is a sum of two graphs and , for some . Let be the clique in along which the sum is performed. Since and are both minorfree graphs, and by induction. For , let be a treedecomposition of with chromatic number at most . Since is a clique in , for some and . Let be the tree obtained from the disjoint union of and by adding an edge between and . Then is a treedecomposition of with chromatic number at most . ∎
5. Computing and
We finish by showing some hardness results for computing and . We need some preliminary results. For a graph , let be the graph consisting of disjoint copies of and all edges between distinct copies of .
Lemma .
For all and all graphs without isolated vertices,
Proof.
Let be a optimal treedecomposition of , with minimal. By Lemma 3, there exists and such that . Since has no isolated vertices, has a neighbour in the same copy of in which it belongs. Therefore,
For the other inequalities, . ∎
We also require the following hardness result of Lund and Yannakakis [12].
Theorem 5.1 ([12]).
There exists , such that it is NPhard to correctly determine within a multiplicative factor of for every vertex graph .
Our first theorem is a hardness result for approximating and .
Theorem 5.2.
There exists , such that it is NPhard to correctly determine within a multiplicative factor of for every vertex graph . The same hardness result holds for with the same .
Proof.
We show the proof for . The proof for is identical. Let , where is the constant from Theorem 5.1. Let be an vertex graph.
Note that has vertices, and . If , then by Lemma 5. Therefore, if we can approximate within a factor of , then we can approximate within a factor of . ∎
For the decision problem, we use the following hardness result of Khanna, Linial, and Safra [11].
Theorem 5.3 ([11]).
Given an input graph with , it is NPcomplete to decide if or .
As a corollary of Theorem 5.3, we obtain the following.
Theorem 5.4.
It is NPcomplete to decide if . It is also NPcomplete to decide if .
Proof.
Let be a graph without isolated vertices and . By Lemma 5, if , then and if , then . Same for . Finally, a tree or pathdecomposition and a colouring of each bag is a certificate that or . ∎
Combining the standard time dynamic programming for computing pathwidth exactly (see Section 3 of [18]) and the time algorithm of Björklund, Husfeldt, and Koivisto [2] for deciding if , yields a time algorithm to decide to . As far as we know, there is no faster algorithm for deciding (except for small values of , where faster algorithms for deciding colourability can be used instead of [2]).
Finally, unlike for , we conjecture that it is still NPcomplete to decide if .
Conjecture .
It is NPcomplete to decide if . It is also NPcomplete to decide if .
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