Defective and Clustered Graph Colouring

Consider the following two ways to colour the vertices of a graph where the requirement that adjacent vertices get distinct colours is relaxed. A colouring has"defect"$d$ if each monochromatic component has maximum degree at most $d$. A colouring has"clustering"$c$ if each monochromatic component has at most $c$ vertices. This paper surveys research on these types of colourings, where the first priority is to minimise the number of colours, with small defect or small clustering as a secondary goal. List colouring variants are also considered. The following graph classes are studied: outerplanar graphs, planar graphs, graphs embeddable in surfaces, graphs with given maximum degree, graphs with given maximum average degree, graphs excluding a given subgraph, graphs with linear crossing number, linklessly or knotlessly embeddable graphs, graphs with given Colin de Verdi\`ere parameter, graphs with given circumference, graphs excluding a fixed graph as an immersion, graphs with given thickness, graphs with given stack- or queue-number, graphs excluding $K_t$ as a minor, graphs excluding $K_{s,t}$ as a minor, and graphs excluding an arbitrary graph $H$ as a minor. Several open problems are discussed.


Introduction
Consider a graph where each vertex is assigned a colour. A monochromatic component is a connected component of the subgraph induced by all the vertices assigned a single colour. A graph G is k-colourable with clustering c if each vertex can be assigned one of k colours such that each monochromatic component has at most c vertices. A graph G is k-colourable with defect d if each vertex of G can be assigned one of k colours such that each vertex is adjacent to at most d neighbours of the same colour; that is, each monochromatic component has maximum degree at most d.
This paper surveys results and open problems regarding clustered and defective graph colouring, where the first priority is to minimise the number of colours, with small defect or small clustering as a secondary goal. We include various proofs that highlight the main methods employed. The emphasis is on general results for broadly defined classes of graphs, rather than more precise results for more specific classes. With this viewpoint the following definitions naturally arise.
The clustered chromatic number of a graph class G, denoted by χ (G), is the minimum integer k for which there exists an integer c such that every graph in G has a k-colouring with clustering c. If there is no such integer k, then G has unbounded clustered chromatic number. A graph class G is defectively k-colourable if there exists an integer d such that every graph in G is k-colourable with defect d. The defective chromatic number of G, denoted by χ ∆ (G), is the minimum integer k such that G is defectively k-colourable. If there is no such integer k, then G has unbounded defective chromatic number. Every colouring of a graph with clustering c has defect c − 1. Thus χ ∆ (G) χ (G) χ (G) for every class G. Tables 1 and 2 summarise the results presented in this survey; see Sections 1.2 and 1.3 for the relevant definitions.

History and Terminology
There is no single origin for the notions of defective and clustered graph colouring, and the terminology used in the literature is inconsistent.
Early papers on defective colouring include [16,17,128,138,166,212], although these did not use the term 'defect'. The definition of "k-colourable with defect d", often written (k, d)- Also for clustered colouring, the literature is inconsistent. One of the early papers is by Kleinberg et al. [151], who defined a colouring to be (k, c)-fragmented if, in our language, it is a k-colouring with clustering c. I prefer "clustering" since as c increases, intuitively the "fragmentation" of the monochromatic components decreases. Edwards and Farr [89, 90] called a monochromatic component a chromon and called the clustered chromatic number of a class the metachromatic number. 8.5 no (k + 1)-path H k log 2 (k + 2) − 1, . . . , 3 log 2 k k+1 2 8.5 g-thickness k T g k 2k + 1 2k + 1 9

Definitions
This section briefly states standard graph theoretic definitions, familiar to most readers.
A clique in a graph is a set of pairwise adjacent vertices.
Let G be a graph. A k-colouring of G is a function that assigns one of k colours to each vertex of G. An edge vw of G is bichromatic if v and w are assigned distinct colours. A vertex v of G is properly coloured if v is assigned a colour distinct from every neighbour of v. A colouring of G is proper if every vertex is properly coloured. The chromatic number of G, denoted χ (G), is the minimum integer k such that there is a proper k-colouring of G.
A graph parameter is a real-valued function f on the class of graphs such that f (G 1 ) = f (G 2 ) whenever graphs G 1 and G 2 are isomorphic. Say f is bounded on a graph class G if there exists c such that f (G) c for every G ∈ G, otherwise f is unbounded on G. For a graph G, let mad(G) be the maximum average degree of a subgraph of G.
A graph is k-degenerate if every non-empty subgraph has a vertex of degree at most k. A greedy colouring algorithm shows that every k-degenerate graph is (k + 1)-colourable.
A graph H is a minor of a graph G if a graph isomorphic to H can be obtained from a subgraph of G by contracting edges. A class of graphs G is minor-closed if for every graph G ∈ G every minor of G is in G, and some graph is not in G. A graph G is H-minor-free if H is not a minor of G. Let M H be the class of H-minor-free graphs.
To subdivide an edge vw in a graph G means to delete vw, add a new vertex x, and add new edges vx and xw. A subdivision of G is any graph obtained from G by repeatedly subdividing edges. The 1-subdivision of G is the graph obtained from G by subdividing each edge of G exactly once. A graph H is a topological minor of a graph G if a graph isomorphic to a subdivision of H is a subgraph of G.
The Euler genus of the orientable surface with h handles is 2h. The Euler genus of the non-orientable surface with c cross-caps is c. The Euler genus of a graph G is the minimum Euler genus of a surface in which G embeds (with no crossings). See [177] for background on embeddings of graphs on surfaces.
A tree decomposition of a graph G is given by a tree T whose nodes index a collection (T x ⊆ V (G) : x ∈ V (T )) of sets of vertices in G called bags, such that (1) for every edge vw of G, some bag T x contains both v and w, and (2) for every vertex v of G, the set {x ∈ V (T ) : v ∈ T x } induces a non-empty (connected) subtree of T . The width of a tree decomposition T is max{|T x | − 1 : x ∈ V (T )}. The treewidth of a graph G, denoted by tw(G), is the minimum width of the tree decompositions of G. See [32,33,118,193,194] for surveys on treewidth.
A layering of a graph G is a partition (V 0 , V 1 , . . . , V ) of V (G) such that for every edge vw ∈ E(G), if v ∈ V i and w ∈ V j , then |i − j| 1. Each set V i is called a layer. If r is a vertex in a connected graph G and V i := {v ∈ V (G) : dist G (v, r) = i} for i 0, then V 0 , V 1 , . . . is a layering called the BFS layering of G starting at r.
The layered treewidth of a graph G is the minimum integer k such that there is a tree decomposition (T x ⊆ V (G) : x ∈ V (T )) of G and a layering (V 0 , V 1 , . . . , V ) of G, such that |V i ∩ T x | k for every i ∈ [0, ] and every x ∈ V (T ). Layered treewidth was introduced by Dujmović et al. [81].
A pair (G 1 , G 2 ) is a separation of a graph G if G 1 and G 2 are induced subgraphs of G such that G = G 1 ∪ G 2 , and V (G 1 ) \ V (G 2 ) = ∅ and V (G 2 ) \ V (G 1 ) = ∅. If, in addition, A balanced separator in a graph G is a set S ⊆ V (G) such that every component of G − S has at most 1 2 |V (G)| vertices. The radius of a connected graph G is the minimum integer r such that for some vertex v of G, every vertex of G is at distance at most r from v.

Choosability
Many defective and clustered colouring results hold in the setting of list colouring. Eaton and Hull [86] first introduced defective list colouring.
A list assignment for a graph G is a function L that assigns a set L(v) of colours to each vertex v ∈ V (G). Define a graph G to be L-colourable if there is a proper colouring of G such that each vertex v ∈ V (G) is assigned a colour in L(v). A list assignment L is a k-list assignment if |L(v)| k for each vertex v ∈ V (G). The choice number of a graph G is the minimum integer k such that G is L-colourable for every k-list-assignment L of G.
For a list-assignment L of a graph G and integer d 0, define G to be L-colourable with defect d if there is a colouring of G with defect d such that each vertex v ∈ V (G) is assigned a colour in L(v). Define G to be k-choosable with defect d if G is L-colourable with defect d for every k-list assignment L of G. Similarly, for an integer c 1, G is L-colourable with clustering c if there is a colouring of G with clustering c such that each vertex v ∈ V (G) is assigned a colour in L(v). Define G to be k-choosable with clustering c if G is L-colourable with clustering c for every k-list assignment L of G.
The defective choice number of a graph class G, denoted by χ ∆ (G), is the minimum integer k for which there exists an integer d 0, such that every graph G ∈ G is k-choosable with defect d. The clustered choice number of a graph class G, denoted by χ (G), is the minimum integer k for which there exists an integer c 1, such that every graph G ∈ G is k-choosable with clustering c.

Standard Examples
The following construction, or variants of it, have been used by several authors [87,119,124,183,185] to provide lower bounds on the defective chromatic number. As illustrated in Figure 1, let S(h, d) be defined recursively as follows. Let S(0, d) be the graph with one vertex and no edges. For h 1, let S(h, d) be the graph obtained from d + 1 disjoint copies of S(h − 1, d) by adding one dominant vertex. Note that S(1, d) = K 1,d+1 , the star with d + 1 leaves.

Proof.
We proceed by induction on h. In the base case, S(1, d) = K 1,d+1 , which obviously has no 1-colouring with defect d. Now assume that h 2 and the claim holds for h − 1. Suppose on the contrary that S(h, d) has an h-colouring with defect d. Let v be the dominant vertex in S(h, d), and say v is blue. Then S(h, d) − v has d + 1 components C 1 , . . . , C d+1 . Since v is dominant and has monochromatic degree at most d, at most d of C 1 , . . . , C d+1 contain a blue vertex. Thus some C i is (h − 1)-coloured with defect d. This is a contradiction since Similarly, as illustrated in Figure 2, let S(h, c) be defined recursively as follows. Let S(1, c) be the path on c + 1 vertices. For h 2, let S(h, c) be the graph obtained from c disjoint copies of S(h − 1, c) by adding one dominant vertex. As illustrated in Figure 3, S(2, d) is planar, S(2, c) is outerplanar, and S(3, c) is planar.

Two Fundamental Observations
The following elementary, but fundamental, result characterises those graph classes with bounded defective or clustered chromatic number.
Proposition 3. The following are equivalent for a graph class G: (1) G has bounded defective chromatic number, (2) G has bounded clustered chromatic number, (3) G has bounded chromatic number.
Proof. We first show that (3) implies (1) and (2). Suppose that G has bounded chromatic number. That is, for some integer k, every graph G in G is properly k-colourable. Thus G is k-colourable with defect 0 and with clustering 1.
We now show that (1) implies (3). Suppose that G has bounded defective chromatic number. That is, for some integers k and d, every graph G in G is k-colourable such that each monochromatic subgraph has maximum degree d. Every graph with maximum degree d is properly (d + 1)-colourable by a greedy algorithm. Apply this result to each monochromatic subgraph of G. Hence G is properly k(d + 1)-colourable, and G has bounded chromatic number.
We finally show that (2) implies (1). Suppose that G has bounded clustered chromatic number. That is, for some integers k and c, every graph G in G is k-colourable such that each monochromatic subgraph has at most c vertices, and therefore has maximum degree at most c − 1. Hence χ ∆ (G) k, and G has bounded defective chromatic number.
We have the following analogous result for defective and clustered choosability.
Proposition 4. The following are equivalent for a graph class G: (1) G has bounded defective choice number, (2) G has bounded clustered choice number, (3) G has bounded choice number.
(4) G has bounded maximum average degree. Proof (1) and (2). As in the proof of Proposition 3, if a graph G is k-choosable with clustering c, then G is k-choosable with defect c − 1. Thus (2) implies (1).

Related Topics
We briefly mention here some related topics not covered in this survey: Proof. Let L be a (k +1)-list assignment for G. We prove by induction on |V (H)|+|E(H)| that every subgraph H of G is L-colourable with defect −k. The base case with |V (H)|+|E(H)| = 0 is trivial. Consider a subgraph H of G. If H has a vertex v of degree at most k, then by induction H − v is L-colourable with defect − k, and there is a colour in L(v) used by no neighbour of v which can be assigned to v. Now assume that H has minimum degree at least k + 1. By assumption, H contains an -light edge xy. By induction, H − xy has an L-colouring c with defect − k. If c(x) = c(y), then c is also an L-colouring of H with defect − k. Now assume that c(x) = c(y). We may further assume that c is not an L-colouring of H with defect − k. Without loss of generality, x has exactly − k + 1 neighbours (including y) coloured by c(x). Since deg H (x) , there are at most k − 1 neighbours not coloured by there is a colour used by no neighbour of x which can be assigned to x.

Islands
Esperet and Ochem [98] introduced the following definition and lemma. A k-island in a graph G is a non-empty set S ⊆ V (G) such that every vertex in S has at most k neighbours in V (G) \ S.
Lemma 6 ([98]). If every non-empty subgraph of a graph G has a k-island of size at most c, then G is (k + 1)-choosable with clustering c.

Proof.
We proceed by induction on |V (G)|. The base case is trivial. Let L be a (k + 1)-list assignment for G. By assumption, G has a k-island S. By induction, G − S is L-colourable with clustering c. Assign each vertex v ∈ S a colour in L(v) not assigned to any neighbour Note that islands generalise the notion of degeneracy, since a graph is k-degenerate if and only if every non-empty subgraph has a k-island of size 1. Thus Lemma 6 with c = 1 is equivalent to the well-known statement that every k-degenerate graph is properly (k + 1)-choosable.
[66] proved the following result for defective colourings of outerplanar graphs.
Moreover, the defect bound in the above results is best possible since the graph shown in Figure 4 is not 2-colourable with defect 1. See [224,226] for more results on defective colouring of outerplanar graphs.

Planar Graphs
Let P be the class of planar graphs. We now discuss defective colourings of P.  Proof. We proceed by induction on |V (G)| with the hypothesis that every planar graph G is 3-colourable such that for each edge v 1 v 2 of G, there is such a 3-colouring of G such that each monochromatic component is a path, and v 1 and v 2 are properly coloured. (Recall that this means that every neighbour of v 1 is assigned a distinct colour from v 1 , and every neighbour of v 2 is assigned a distinct colour from v 2 .) In the base case, if |V (G)| 4, then assign v 1 and v 2 distinct colours, and assign the (at most two) other vertices a third colour. Each monochromatic component is a path. Now assume that |V (G)| 5. By adding edges, we may assume that G is a planar triangulation. Say the faces containing v 1 v 2 are v 1 av 2 and v 1 bv 2 . Let G be obtained from G by deleting the edge v 1 v 2 , and introducing a new vertex x adjacent to v 1 , v 2 , a, b. Then G is a planar triangulation. Let C be a shortest cycle in (G − v 1 ) − v 2 such that axb is a subpath of C. Since G is 3-connected and |V (G)| 5, such a cycle exists. Since C is shortest, C is an induced cycle. Let G 1 and G 2 be the subgraphs of G 'inside' and 'outside' of C including C. That is, By induction, each G i is 3-colourable such that each monochromatic component is a path, and v i and x i are properly coloured. Permute the colours so that x 1 and x 2 get the same colour, and v 1 and v 2 get distinct colours. Colour each vertex in V (C) \ {x} by the colour assigned to x 1 and x 2 . Note that V (C) \ {x} induces a path in G (since x is not a vertex of G). Moreover, V (C) \ {x} is a monochromatic component, since each x i is properly coloured in G i . Every other monochromatic component of G is a monochromatic component of G 1 or G 2 , and is therefore a path in G.

Hex Lemma
Here we consider colourings of planar graphs with bounded degree. The following result is a dual version of the Hex Lemma, which says that the game of Hex cannot end in a draw. The proof is based on the proof of the Hex Lemma by Gale [105]. See [171, Section 6.1] for another proof.
Theorem 9. For every integer k 2 there is planar graph G with maximum degree 6 such that every 2-colouring of G has a monochromatic path of length k.

Proof.
A suitable subgraph of the triangular grid forms an embedded plane graph with maximum degree 6, such that every internal face is a triangle, the outerface is a cycle Colour w and y blue. Colour x and z red. Note that G embeds in the plane, such that every internal face of G is a triangle, and the outerface of G is the 4-cycle (w, x, y, z). The four internal faces of G that share an edge with the outerface are (a, w, z), (b, x, w), (c, x, y) and (d, y, z). Call these faces special. Let H be the graph with one vertex for each internal face of G , where two vertices of H are adjacent if the corresponding faces of H share an edge whose endpoints are coloured differently. H is a subgraph of the dual of G and is therefore planar. Let A, B, C, D be the vertices of H respectively corresponding to the special faces (a, w, z), (b, x, w), (c, x, y), (d, y, z). Since w and z are coloured differently, a has the same colour as exactly one of w and z, implying A has degree 1 in H. Similarly, B, C and D each have degree 1 in H. If some face of G is monochromatic, then the corresponding vertex of H has degree 0. Every non-monochromatic non-special face F has two vertices of one colour and one vertex of the other colour, and F does not share an edge with the outerface. Thus the vertex of H corresponding to F has degree 2 in H. In summary, every vertex of H has degree 0 or 2, except for A, B, C, D, which have degree 1

Defective Colouring of Graphs on Surfaces
For every integer g 0, let E g be the class of graphs with Euler genus at most g. This section considers defective colourings of graphs in E g . Cowen et al. [66] proved that every graph in E g [66] that every graph in E g is defectively 3-colourable. Since S(2, d) is planar, χ ∆ (E g ) = 3 by Lemma 1.
Since d √ 6g + 7, we have 12(g − 2) (2d + 1)(d − 11). That is, 6(g − 2) (d + 1 2 )(d − 11). Since d 12, we have 6(g−2) d−11 − 1 2 d, as claimed. Proof. If B = ∅ then the result is vacuous. Now assume that B = ∅. For each non-triangular face f , add an edge between two non-consecutive vertices in B (which must exist since A is a stable set). We obtain a multigraph triangulation G , in which A is a stable set. Let n i be the number of vertices with degree i in G . Let α be the number of edges in G incident with vertices in A. Let β be the number of edges in G with both endpoints in B. By Euler's formula, Since G is a triangulation, each face of G has at most two edges incident with A, and at least one edge with endpoints in B. It follows that α 2β and Woodall [228] improved Theorem 11 to show that every graph with Euler genus g is 3-choosable with defect max{9, 2 + √ 4g + 6}. Thus Note that the light edge approach also proves 3-choosability, but with a weaker defect bound. In particular, results of Ivančo [130] and Jendro ' l and Tuhársky [134] together imply that every graph in E g with minimum degree at least 3 has an edge vw with deg(v) + deg(w) max{2g + 7, 19}.
(Better results are known for specific surfaces with g 5, and all the bounds are tight.) Thus every graph in E g with minimum degree at least 3 has a max{2g + 4, 16}-light edge. (See Lemma 78 for a more general result with a slightly weaker bound.) Lemma 5 then implies that every graph with Euler genus g is 3-choosable with defect max{2g + 2, 14}. See [57,173,233,234,235,236] for more on defective choosability of graphs embedded on surfaces.

Clustered Colouring of Graphs on Surfaces
This section considers clustered colouring of graphs embeddable on surfaces. Esperet and Ochem [98] proved that every graph of bounded Euler genus has a 4-island of bounded size, and is thus 5-colourable with bounded clustering by Lemma 6. Kawarabayashi and Thomassen [148] also proved that every graph of bounded Euler genus is 5-colourable with bounded clustering. Dvořák and Norin [84] improved 5 to 4 via the following remarkably simple argument.

Lemma 13 ([84]).
Let G be a graph, such that for some constants α, c > 0 and β ∈ (0, 1), and every subgraph of G with n vertices has a balanced separator of size at most cn 1−β . Then G has a k-island of size at most

Proof.
Let := α k+1 . By Lemma 15 below, there exists X ⊆ V (G) of size at most |V (G)| such that if K 1 , . . . , K p are the components of G − X, then each K i has at most 2( c (2 β −1) ) 1/β vertices. Let e(K i ) be the number of edges of G with at least one endpoint in K i . Then Hence e(K i ) < (k+1) |V (K i )| for some i. Repeatedly remove vertices from K i with at least k+1 neighbours outside of K i . Doing so maintains the property that e(K i ) < (k + 1) |V (K i )|. Thus the final set is non-empty. We obtain a k-island of size at most |V . Fix c > 0 and β ∈ (0, 1). Let G be a graph with n vertices such that every subgraph G of G has a balanced separator of size at most c|V (G )| 1−β . Then for all p 1 there exists S ⊆ V (G) of size at most c2 β n (2 β −1)p β such that each component of G − S has at most p vertices. Proof. Run the following algorithm. Initialise S := ∅. While G − S has a component X with more than p vertices, let S X be a balanced separator of X with size at most c|V (X)| 1−β , and add S X to S.
Say a component of G − S at the end of the algorithm has level 0. Say X is a component of G − S at some stage of the algorithm, but X is not a component of G − S at the end of the algorithm. Then X is separated by some set S X , which is then added to S. Define the level of X to 1 plus the maximum level of a component of X − S X . By assumption, level 0 components have at most p vertices. Each level 1 component has more than p vertices. By induction on i, each level i 1 component has more than 2 i−1 p vertices. Let t i be the number of components at level i 1. Say X 1 , . . . , X t i are the components at level i. Since level i components are pairwise disjoint, implying t i < n 2 i−1 p . The number of vertices added to S by separating X 1 , . . . , X t i is at most which is maximised, subject to j |V (X j )| n, when |V (X j )| = n t i . Thus Lemma 14 implies the following result. In the language of Edwards and McDiarmid [91], this lemma provides a sufficient condition for a graph to be 'fragmentable'; this idea is extended by Edwards and Farr [88, 90].

Lemma 15 ([91])
. Fix c > 0 and β ∈ (0, 1). Let G be a graph with n vertices such that every subgraph G of G has a balanced separator of size at most c|V We now reach the main result of this section.

Proof. We proceed by induction on
Let v be any vertex of G. By induction, G − v is L-colourable with clustering 1500(g + 2).

Since S(3, d) is planar, Lemma 2 and Theorem 16 imply
It is still open to determine the best possible clustering function.
Open Problem 17. Does every graph in E g have a 4-colouring with clustering O( √ g)?
The following question also remains open; see Section 8.1 for relevant material.

Open Problem 18. Are graphs with bounded Euler genus and bounded maximum degree 3-choosable with bounded clustering?
The above method extends for embedded graphs with large girth (since |E(G)| < 2(|V (G)| + g) if the girth is at least 4, and |E(G)| < 5 3 (|V (G)| + g) if the girth is at least 5).

Theorem 19 ([84]). Let G be a graph with Euler genus g and girth
See Section 8.2 for more applications of the island method. Also note that Linial et al. [159] use sublinear separators in a slightly different way (compared with Lemma 13) to obtain bounds on the size of monochromatic components in 2-colourings of graphs.

Maximum Degree
The defective chromatic number of any graph class with bounded maximum degree equals 1. Thus defective colourings in the setting of bounded degree graphs are only interesting if one also considers the bound on the defect. Lovász  Proof. Consider a k-colouring of G that maximises the number of bichromatic edges. Suppose that some vertex v is adjacent to at least d + 1 vertices of the same colour. Some other colour is assigned to at most Recolour v this colour. The number of bichromatic edges increases by at least 1. This contradiction shows that every vertex v is adjacent to at most d vertices of the same colour.
We now show that Theorem 20 is best possible. Say G = K n is k-colourable with defect d. Some monochromatic subgraph has at least n k vertices. Thus d n k − 1 and k + 1, which exactly matches the bound in Theorem 20.
Clustered colourings of bounded degree graphs are more challenging than their defective cousins. Let D ∆ be the class of graphs with maximum degree ∆. First note the following straightforward lemma.

Lemma 21.
For ∆ > d 1, By Theorem 20, G is k 1 -colourable with defect d. Each monochromatic subgraph, which has maximum degree d, is k 2 -colourable with clustering c (depending only on d). The product gives We now show a series of improving upper bounds on χ (D ∆ ). Theorem 20 with d = 1 implies every graph with maximum degree ∆ is ( ∆/2 + 1)-colourable with defect 1, and thus with clustering 2. Hence In particular, this shows that every graph with maximum degree 3 is 2-colourable with clustering 2. Alon et al. [11] proved that every graph with maximum degree 4 is 2-colourable with clustering 57. Haxell et al.
[120] improved this bound on the cluster size from 57 to 6. Lemma 21 with d = 4 then implies that Alon et al.
[11] pushed their method further to prove that In fact, Alon et al. [11] showed that for ∈ (0, 3) every graph of maximum degree ∆ is For the sake of brevity, we present slightly weaker results with simpler proofs. The following result was implicitly proved by Alon et al. [11].
Theorem 22. Let G be a graph with maximum degree ∆. If G has a k-colouring with defect 2, then G has a (k + 1)-colouring with clustering 24∆.
Proof. Say an induced cycle or path in G is short if it has at most 8∆ vertices, otherwise it is long. Let X 1 , . . . , X k be a partition of V (G) corresponding to the given k-colouring with defect 2. Thus each G[X i ] is a collection of pairwise disjoint induced cycles and paths. Consider such a cycle or path Y that is long.
Here the last vertex in Y j is adjacent to the first vertex in Y j+1 for j ∈ [1, a], and if Y is a cycle then the last vertex in Y a+1 is adjacent to the first vertex in Y 1 . Let Z 1 , . . . , Z n be the collection of all these induced paths with exactly 8∆ vertices (which might include some Y a+1 ).
Let G be the subgraph of G induced by V (Z 1 ∪ · · · ∪ Z n ). Thus G has maximum degree at most ∆. By Lemma 23 below, G has a stable set We claim that {S, X 1 \ S, . . . , X k \ S} defines a (k + 1)-colouring with clustering 24∆. By The proof of Theorem 22 used the following well-known lemma about 'independent transversals'.
Proof. The proof uses the Lovász Local Lemma [96], which says that if X is a set of events in a probability space, such that each event in X has probability at most p and is mutually independent of all but D other events in X , and 4pD 1, then with positive probability no event in X occurs.
We may assume that Consider an edge vw, where v ∈ V i and w ∈ V j . Let X vw be the event that both v and w are chosen. Thus X vw has probability at most p : By the Lovász Local Lemma, with positive probability, no event X vw occurs. Hence there exist v 1 , . . . , v n such that no event X vw occurs. That is, {v 1 , . . . , v n } is the desired stable set.
The 8∆ term in Lemma 23 was improved to 2∆ by Haxell [121], which means the 24∆ term in Theorem 22 can be improved to 6∆.
Theorem 20 with d = 2 and Theorem 22 imply Haxell et al. [120] proved that every graph with maximum degree 8 is 3-colourable with bounded clustering. Using their result for the ∆ = 5 case and the ∆ = 8 case, Haxell et al. [120] proved that Moreover, Haxell et al.
[120] proved for large ∆ one can do slightly better: for some constants > 0 and for all ∆ ∆ 0 , Note that for both these results by Haxell et al.
[120] the clustering bound is independent of

∆.
It is open whether every graph with maximum degree 9 is 3-colourable with bounded clustering [120]. If this is true, then Lemma 21 with d = 9 would imply which would be the best known upper bound on χ (D ∆ ). Graphs with maximum degree 10 are not 3-colourable with bounded clustering [11], as shown by the following general lower bound, which also implies a 3-colour lower bound for graphs of maximum degree 6.
Theorem 24 ([11, 120]). For every integer ∆ 2, , and the result for ∆ − 1 implies the result for ∆. Thus we may assume that ∆ is even. Let k := ∆+6 4 . Our goal is to show that for every integer c 3 there is a graph G that has no (k − 1)-colouring with clustering c. Erdős and Sachs [94] proved that there is a ( ∆ 2 + 1)-regular graph G 0 with girth greater than c. Say |V (G 0 )| = n. Let G be the line graph of G 0 . Then G is ∆-regular with ( ∆+2 4 )n vertices. Suppose on the contrary that G is (k − 1)-colourable with clustering c. For some colour class Thus X corresponds to a set X of at least n edges in G 0 , which therefore contains a cycle of size greater than c. Thus, X contains a monochromatic component of size greater than c, which is a contradiction.
The following problem remains open for ∆ 9.
Open Problem 25. What is χ (D ∆ )? The best known bounds are and for some > 0 and all ∆ at least some constant ∆ 0 .
Open Problem 26. What is χ (D ∆ )? The best known bounds are where the upper bound follows from known Brooks-type bounds on the choice number [67].
It is interesting that for many of the above results the bound on the clustering is independent of ∆. We now show that from any upper bound on χ (D ∆ ) with clustering linear in ∆, one can obtain a slightly larger upper bound that is independent of ∆. The idea of the proof is by Alon et al. [11].

Maximum Average Degree
Recall that mad(G) is the maximum average degree of a subgraph of G.
We prove Theorem 28 below. The key is the following lemma, which is a slightly weaker version of a result by Havet and Sereni [119], who proved that every graph G with mad(G) < k + kd k+d is k-choosable with defect d.

Lemma 29 (Frédéric Havet).
For all m ∈ R + and k, d ∈ Z + such that 1 Proof. Let G be a counterexample with |V (G)| + |E(G)| minimum (with r, k, d fixed). By Lemma 5, G has minimum degree k and has no d-light edge. Let We may assume that C = ∅, as otherwise G is 1-colourable with defect d. Thus the total charge is greater than m|V (G)|, implying the average degree is greater than m, which is a contradiction.
Proof of Theorem 28.
Maximum average degree is closely related to degeneracy. Recall that a graph G is kdegenerate if every subgraph of G has minimum degree at most k. A greedy algorithm shows that every k-degenerate graph is properly (k + 1)-colourable. Since the standard example S(k, d) is k-degenerate, this bound cannot be improved even for defective colourings. Thus for the class of k-degenerate graphs, the defective chromatic number, defective choice number, clustered chromatic number, clustered choice number, and (proper) chromatic number all equal k + 1.

Excluding a Subgraph
For every graph H, the class of graphs with no H subgraph has bounded chromatic number if and only if H is a forest. The same result holds for defective chromatic number and clustered chromatic number. To see this, observe that if H contains a cycle, then graphs with girth greater than |V (H)| contain no H subgraph, and by the classical result of Erdős [95] there are graphs with arbitrarily large girth and chromatic number. By Proposition 3, the defective and clustered chromatic numbers are also arbitrarily large. Conversely, say F is a forest with n vertices. A well known greedy embedding procedure shows that every graph with minimum degree at least n − 1 contains F as a subgraph. That is, every graph containing no F subgraph is (n − 2)-degenerate, and is thus (n − 1)-colourable. This bound is tight since K n−1 contains no F subgraph and is (n − 1)-chromatic. In short, for the class of graphs containing no F subgraph, the chromatic number equals n − 1. The following result by Ossona de Mendez et al. [185] shows that defective colourings exhibit qualitatively different behaviour.
Theorem 31 ([185]). Let T be a tree with n 2 vertices and radius r 1. Then every graph containing no T subgraph is r-colourable with defect n − 2.
. Thus there are always unmapped vertices in V r−j ∪ · · · ∪ V r to choose as the children of v. Hence T is a subgraph of G. This contradiction shows that G[V r ] has maximum degree at most n − 2, and G is r-colourable with defect n − 2.
The number of colours in Theorem 31 is best possible for the complete binary tree T of radius r. Since S(r − 1, d) contains no T subgraph, Lemma 1 and Theorem 31 imply that the defective chromatic number of the class of graphs containing no T subgraph equals r.

Open Problem 32. For a tree T , what is the clustered chromatic number of the class of graphs with no T subgraph?
The results in Section 4 on χ (D ∆ ) are relevant to this question since a graph has maximum degree at most ∆ if and only if it excludes K 1,∆+1 as a subgraph. Section 8.5 studies colourings of graphs that exclude a given path subgraph.

Excluding a Shallow Minor
As illustrated in Figure 6, for integers s, t 1, let K * s,t be the bipartite graph obtained from K s,t by adding s 2 new vertices, each adjacent to a distinct pair of vertices in the colour class of s vertices in K s,t . Ossona de Mendez et al.
[185] studied defective colourings for graphs excluding K * s,t as a subgraph, where the defect bound depends on the density of shallow topological minors. Let ∇(G) be the maximum average degree of a graph H such that the 1-subdivision of H is a subgraph of G.
Proof. Assume for contradiction that G has minimum degree at least s (thus s δ) and that G contains no -light edge. The case s = 1 is simple: Since G has minimum degree at least 1, G has at least one edge, which is -light since ∆(G) t − 1 and = t − 1. Now assume that s 2.
Let A be the set of vertices in G of degree at most . Let B := V (G) \ A. Let a := |A| and b := |B|. Since G has a vertex of degree at most δ and δ , we deduce that a > 0. Note that no two vertices in A are adjacent.
Since the average degree of G is at most δ, That is, (2) Let G be the graph obtained from G − E(G [B]) by greedily finding a vertex w ∈ A having a pair of non-adjacent neighbours x, y in B and replacing w by an edge joining x and y (by deleting all edges incident with w except xw, yw and contracting xw), until no such vertex w exists.
Proof. If K s,t ∈ G for some s, t 1, then by Theorem 33, every graph in G is s-choosable with defect bounded by a function of s, t, mad(G) and ∇(G). This proves the claimed upper bound.
Conversely, let s := χ ∆ (G). Then for some d, every graph in G is s-choosable with defect d.
By Lemma 35 below, if t = (ds + 1)s s then K s,t is not in G.
Lemma 35. For s 1 and d 0, if t = (ds + 1)s s , then the complete bipartite graph K s,t is not s-choosable with defect d.

Proof. Let
A and B be the colour classes of K s,t with |A| = s and |B| = t. . . , c s }. At least d + 1 of these vertices are assigned the same colour, say c i . Thus v i has monochromatic degree at least d + 1. Hence Note that Theorem 34 generalises several previous results. For example, Theorem 34 says that χ ∆ (O) = 2 since K 2,3 is not outerplanar, but K 1,n is outerplanar for all n. Similarly, χ ∆ (P) = 3 since K 3,3 is not planar, but K 2,n is planar for all n. More generally, Theorem 34 immediately implies: Corollary 36. For every graph H, χ ∆ (M H ) equals the minimum integer s such that H is a minor of K s,t for some integer t.
Theorem 34 also determines the defective choice number for graphs excluding a fixed immersion (see Theorem 52).

Linklessly Embeddable Graphs
A graph is linklessly embeddable if it has an embedding in R 3 with no two linked cycles [199,201]. Let L be the class of linklessly embeddable graphs. Then L is a minor-closed class whose minimal excluded minors are the so-called Petersen family [200], which includes K 6 , Note that Theorem 34 also implies χ ∆ (L) = 4 since K 4,4 is not linkless, but K 3,n is linkless for all n.
We have the following result for clustered colourings of linklessly embeddable graphs.
Theorem 38. Every linklessly embeddable graph is 5-choosable with clustering 62948, and Proof. The upper bound follows from Theorem 58 since every linkless graph contains no K 6 -minor. Since S(3, c) is planar, S(4, c) is apex, and thus linklessly embeddable. The lower bound then follows from Lemma 2.

Knotlessly Embeddable Graphs
A graph is knotlessly embeddable if it has an embedding in R 3 in which every cycle forms a trivial knot; see [191] for a survey. Let K be the class of knotlessly embeddable graphs. Then K is a minor-closed class whose minimal excluded minors include K 7 and K 3,3,  Figure 7. By Lemma 1, χ ∆ (K) 5. Since K 3,3,1,1 is a minor of K * 5,3 , knotlessly embeddable graphs do not contain a K * 5,3 subgraph. Since mad(K) < 10, Theorem 33 implies the following result. We have the following result for clustered colourings of knotlessly embeddable graphs.

Colin de Verdière Parameter
The Theorem 33 with s = k and t = max{k, 3} implies that G is k- Lemma 1 then implies that χ ∆ (V k ) k. Note that the weaker lower bound, χ ∆ (V k ) k, follows from Theorem 34 since K k,max{k,3} ∈ V k .
Theorem 41 generalises Theorem 37 which corresponds to the case k = 4.
Clustered colourings provide a natural approach to the conjecture of Colin de Verdière [61] mentioned above.
Note that Conjecture 42 with k 7 is implied by Theorem 58 below since graphs in V k contain no K k+2 minor.

Crossings
This section considers defective colourings of graphs with linear crossing number. For an integer g 0 and real number k 0, let E k g be the class of graphs G such that every subgraph H of G has a drawing on a surface of Euler genus g with at most k |E(H)| crossings. (In a drawing, we assume that no three edges cross at a common point.) This says that the average number of crossings per edge is at most 2k (for every subgraph). Of course, a graph is planar if and only if it is in E 0 , and a graph has Euler genus at most g if and only if it is in E 0 g . Graphs that can be drawn in the plane with at most k crossings per edge, so called k-planar . It follows that k-planar graphs are O( √ k)-colourable, which is best possible since K n is O(n 2 )-planar. Say a graph is (g, k)-planar if it can be drawn on a surface with Euler genus g with at most k crossings per edge. Such graphs are in E (k/2) g . Also note that even 1-planar graphs do not form a minor-closed class. For example, the n × n × 2 grid graph is 1-planar, as illustrated in Figure 8, but contracting the i-th row in the front grid with the i-column in the back grid (plus the edge joining them) creates K n as a minor. Ossona de Mendez et al. [185] showed that Theorem 33 is applicable for graphs in E k g . In particular, such graphs contain no K 3,3k(2g+3)(2g+2)+2 subgraph and have mad O( (k + 1)(g + 1)) and ∇ O( (k + 1)(g + 1)). The first claim here is proved using a standard technique of counting copies of K 3,3 . The second claim is proved using the crossing lemma. The next theorem follows. It is a substantial generalisation of Theorem 8 (the g = k = 0 case) and Theorem 11 (the k = 0 case), with a worse defect bound.

Theorem 43 ([185]).
For every integer g 0 and real number k 0, In particular, every graph in E k g is 3-choosable with defect O((k + 1) 5/2 (g + 1) 7/2 ). Open Problem 44. What is the clustered chromatic number of k-planar graphs? What is the clustered chromatic number of (g, k)-planar graphs? What is the clustered chromatic number of E k g ? It may be that the answer to all these questions is 4. Here we prove the answer is at most 12 for the first two questions.

Stack and Queue Layouts
A k-stack layout of a graph G consists of a linear ordering v 1 , . . . , v n of V (G) and a partition E 1 , . . . , E k of E(G) such that no two edges in E i cross with respect to v 1 , . . . , v n for each i ∈ [1, k]. Here edges v a v b and v c v d cross if a < c < b < d. A graph is a k-stack graph if it has a k-stack layout. The stack-number of a graph G is the minimum integer k for which G is a k-stack graph. Stack layouts are also called book embeddings, and stack-number is also called book-thickness, fixed outer-thickness and page-number. Let S k be the class of k-stack Proof. The lower bound follows from Lemma 1 since an easy inductive argument shows that S(k, d) is a k-stack graph for all d. For the upper bound, K k+1,k(k+1)+1 is not a k-stack graph [30]; see also [70]. Every k-stack graph G has average degree less than 2k + 2 (see [30,83]) and ∇(G) 20k 2 (see [92,181]). The result follows from Theorem 33 with s = k + 1 and t = k(k + 1) + 1, since (k + 1, k(k + 1) + 1, 2k + 2, 40k 2 ) 2 O(k log k) . Proof. Heath and Rosenberg [122] proved that K 2k+1,2k+1 is not a k-queue graph. Every k-queue graph G has mad(G) < 4k (see [83,122,188]) and ∇(G) < (2k + 2) 2 (see [181]). The result then follows from Theorem 33 with s = 2k + 1 and t = 2k + 1, since (2k + 1, 2k + 1, 4k, 2(2k + 2) 2 ) 2 O(k log k) .
An easy inductive construction shows that S(k, n) has a k-queue layout. Thus k + 1 χ ∆ (Q k ) 2k + 1 by Lemma 1 and Theorem 47. It remains an open problem to determine χ ∆ (Q k ).
Now consider clustered colourings of k-stack and k-queue graphs. The standard example S(2, c) is outerplanar, and thus has a 1-stack layout. An easy inductive construction then shows that S(k, c) has a (k − 1)-stack layout (for k 2) and a k-queue layout. The best known upper bounds come from degeneracy: every k-stack graph is (2k + 1)-degenerate and every k-queue graph is (4k − 1)-degenerate (see [83]). Thus Closing the gap in these bounds is interesting because the existing methods say nothing about clustered colourings of k-stack or k-queue graphs. For example, Lemma 13 is not applicable since 3-stack and 2-queue graphs do not have sublinear balanced separators. Indeed, Dujmović et al. [82] constructed (cubic bipartite) 3-stack expander graphs and (cubic bipartite) 2-queue expander graphs.

Excluded Immersions
This section considers colourings of graphs excluding a fixed immersion. A graph G contains a graph H as an immersion if the vertices of H can be mapped to distinct vertices of G, and the edges of H can be mapped to pairwise edge-disjoint paths in G, such that each edge vw of H is mapped to a path in G whose endpoints are the images of v and w. The image in G of each vertex in H is called a branch vertex. A graph G contains a graph H as a strong immersion if G contains H as an immersion such that for each edge vw of H, no internal vertex of the path in G corresponding to vw is a branch vertex. Let I t be the class of graphs not containing K t as an immersion. Let I t be the class of graphs not containing K t as a strong immersion. [1] independently conjectured that every K t -immersion-free graph is properly (t − 1)-colourable. Often motivated by this question, structural and colouring properties of graphs excluding a fixed immersion have recently been widely studied. The best upper bound, due to Gauthier et al. [106], says that every K timmersion-free graph is properly (3.54t + 3)-colourable

Lescure and Meyniel [157] and Abu-Khzam and Langston
Van den Heuvel and Wood [124] proved that the defective chromatic number of K t -immersionfree graphs equals 2. The proof, presented below, is based on the following structure theorem of DeVos et al. [74]. Almost the same result can be concluded from a structure theorem by Wollan [219]. Since Lemma 48 is not proved explicitly in [74] we include the full proof, which relies on the following definitions. For each edge xy of a tree T , let T (xy) and T (yx) be the components of T − xy, where x is in T (xy) and y is in T (yx). For a tree T and graph G, a Note that a bag may be empty. For each edge e = xy ∈ E(T ), let G(T, xy) = z∈V (T (xy)) T z and G(T, yx) = z∈V (T (yx)) T z , and let E(G, T, e) be the set of edges in G between G(T, xy) and G(T, yx). The adhesion of a T -partition is the maximum, taken over all edges e of T , of |E(G, T, e)|. For each node x of T , the torso of x (with respect to a T -partition) is the graph obtained from G by identifying G(T, yx) into a single vertex for each edge xy incident to x, deleting resulting parallel edges and loops. Note that a tree T for which

Lemma 48 ([74]). For every graph G that does not contain K t as an immersion, there is a tree
T and a T -partition of G with adhesion less than (t − 1) 2 , such that each bag has at most t − 1 vertices. Proof. Gomory and Hu [113] proved that for every graph G there is a tree F with vertex set V (G) such that for all distinct vertices v, w ∈ V (G), the size of the smallest edge-cut in G separating v and w equals where the minimum is taken over all edges e on the vw-path in F . Let S be the set of edges e ∈ E(F ) with ζ(e) < (t − 1) 2 . Suppose that some component X of F − S has at least t vertices. Let x, v 2 , v 3 , . . . , v t be distinct vertices in X. Let G be the multigraph obtained from G by adding a new vertex w and adding t − 1 parallel edges between w and v i for each i ∈ [2, t]. We claim that G contains (t − 1) 2 edge-disjoint xw-paths. Consider a set of vertices R ⊆ V (G) with x ∈ R, such that there are less than (t − 1) 2 edges between R and ∈ [2, t]). Thus in G, there is an edge-cut with less than (t − 1) 2 edges separating x and v i , meaning that ζ(e) < (t − 1) 2 for some edge e on the xv i -path in F . But every such edge e is in X, which implies ζ(e) (t − 1) 2 . This contradiction shows (by Menger's Theorem) that G contains (t − 1) 2 edge-disjoint xw-paths. Hence G contains (t − 1) 2 edge-disjoint paths starting at x, exactly t − 1 of which end at v i for each i ∈ [2, t]. Label these xv i -paths as P i,j , for j ∈ [2, t]. For j = i, the combined path P i,j P j,i is a v i v j -path, while each P i,i is a v i x path. Since all these paths are edge-disjoint, G contains a K t -immersion. This contradiction shows that every component of F − S has at most t − 1 vertices.
Let T be obtained from F by contracting each connected component of F −S into a single vertex. For each node x of T , let T x be the set of vertices contracted into x. Then T x : x ∈ V (T ) is the desired partition. ([124]). If a graph G has a T -partition with V (G) ⊆ V (T ) and adhesion at most k, then G is 2-colourable with defect k.

Proof. We proceed by induction on |V
assign v a colour distinct from its neighbour. Now G is 2-coloured with defect k. Now assume that G has minimum degree at least 2. If two small vertices v and w are adjacent, then by induction, G − vw is 2-colourable with defect k, which is also a 2-colouring of G with defect k. Now assume that the small vertices are a stable set. If G has no large vertices, then every 2-colouring of G has defect k. Now assume that G has some large vertex. Let X be the union of all paths in T whose endpoints are large. Let u be a leaf in X. Thus u is large. Let v be the neighbour of u in X, or any neighbour of u if |V (X)| = 1. Let Y := V (T (uv)) \ {u}. Every vertex in Y is small. Since no two small vertices are adjacent and G has minimum degree at least 2, every vertex in Y has at least one neighbour in T (vu). Also, u has at least Note that Lemma 49 with a O(k 3 ) defect bound can be concluded from Theorem 33 with s = 2.

The following result by Van den Heuvel and Wood [124] is the first main contribution of this section.
Theorem 50 ([124]). Every graph G ∈ I t is 2-colourable with defect (t − 1) 3 , and χ ∆ (I t ) = 2.
Proof. By Lemma 48, there is a tree T and a T -partition of G with adhesion at most (t−1) 2 −1, such that each bag has at most t − 1 vertices. Let Q be the graph with vertex set V (T ), where xy ∈ E(Q) whenever there is an edge of G between T x and T y . Any one edge of Q corresponds to at most t − 1 edges in G. By Lemma 49, the graph Q is 2-colourable with defect (t − 1) 2 − 1. Assign to each vertex v in G the colour assigned to the vertex x in Q with v ∈ T x . Since at most t − 1 vertices of G are in each bag, G is 2-coloured with defect While only two colours suffice for defective colourings of graphs excluding a fixed immersion, significantly more colours are needed for defective list colouring.

Theorem 52. For all t 2,
Proof. DeVos et al. [73] proved that mad(I t ) O(t). If the 1-subdivision of a graph H is a subgraph of G, then G contains H as a strong immersion, implying ∇(I t ) O(t). Theorem 34 then implies that χ ∆ (I t ) equals the minimum integer s such that K s,n contains K t as a strong immersion for some n. It is easily seen that K t−1,( t−1 2 )+1 contains K t as a strong immersion, but K t−2,n does not contain K t as a weak immersion for all n. The result follows.
It is an open problem to determine the clustered chromatic number and clustered choice number of graphs excluding a (strong or weak) K t immersion. We have the following lower bounds. Since every graph with maximum degree at most t − 2 contains no K t immersion, by Theorem 24, By Theorem 52,

Minor-Closed Classes
This section studies defective and clustered colourings of graphs in a minor-closed class.
Hadwiger's Conjecture states that every K t -minor-free graph is (t − 1)-colourable [116]. This is widely considered one of the most important open problems in graph theory; see [203] for a survey. Defective and clustered colourings of K t -minor-free graphs provide an avenue for attacking Hadwiger's Conjecture. Graphs with treewidth k are properly (k + 1)-colourable, which is best possible for K k+1 . Moreover, since the standard example S(k, d) has treewidth k, the defective chromatic number of graphs with treewidth k equals k + 1. On the other hand, Alon et al. [11] proved that every graph with maximum degree ∆ and treewidth k is 2-colourable with clustering 24k∆. The proof was based on a result about tree-partitions by Ding and Oporowski [75], which was improved by Wood [220]. It follows that:

K t -Minor-Free Graphs
First consider defective colourings of K t -minor-free graphs. Edwards et al. [87] proved that every K t -minor-free graph is (t − 1)-colourable with defect O(t 2 log t). Their proof gives the same result for (t − 1)-choosability. This result is implied by Theorem 33 since K * t−1,1 contains a K t minor. Indeed, the proof of Theorem 33 in this case is identical to the proof of Edwards et al. [ Now consider clustered colourings of K t -minor-free graphs. Note that (5) implies In particular, every K t -minor-free graph G is (t − 1)-choosable with clustering Proof. For t 9, the exact extremal function for K t -minor-free graphs is known [77,139,167,168,169,207]. In particular, every K t -minor-free graph on n vertices has less than (t − 2)n edges for t 9. For all t, every such graph has a balanced separator of size t 3/2 n 1/2 [12]. By Lemma 13 with k = t − 2 and α = 1 and β = 1 2 and c = t 3/2 , G has a (t − 2)-island of size at most c t . By Lemma 6, G is (t − 1)-choosable with clustering c t . Since the standard example S(t − 2, d) is K t -minor-free, the clustered chromatic number of K t -minor-free graphs is at least t − 1. Thus for t 9, the clustered chromatic number of K t -minor-free graphs equals The only obstacle for extending Theorem 58 for larger values of t is that the exact extremal function is not precisely known. Moreover, for large t, the maximum average degree of K tminor-free graphs tends to Θ(t √ log t); see [153,154,209,210]. Thus Lemma 13 alone cannot determine the clustered chromatic number of K t -minor-free graphs for large t.  Table 3, to  Table 3 depend on the graph minor structure theorem, so the clustering function is large and not explicit. The exception is the self-contained proof of Van den Heuvel and Wood [124], which we now present. (1) The subgraph H i has maximum degree at most t−2, and can be 2-coloured with clustering 1 2 (t − 2) ;

Lemma 59 ([124]). For every set
(2) For each component C of G − V (H 1 ) ∪ · · · ∪ V (H i ) , at most t − 2 subgraphs in H 1 , . . . , H i are adjacent to C, and these subgraphs are pairwise adjacent. Proof. We may assume that G is connected. We construct H 1 , . . . , H n iteratively, maintaining properties (1) and (2). Let H 1 be the subgraph induced by a single vertex in G. Then (1) and (2) hold for i = 1.
For j ∈ [k], let v j be a vertex in C adjacent to Q j . By Lemma 59 with k t − 2, there is an induced connected subgraph H i+1 of C containing v 1 , . . . , v k that satisfies (1).
and C is not adjacent to H i+1 , implying (2) is maintained for C . Now assume C is contained in C. The subgraphs in H 1 , . . . , H i+1 that are adjacent to C are a subset of Q 1 , . . . , Q k , H i+1 , which are pairwise adjacent. Suppose that k = t − 2 and C is adjacent to all of Q 1 , . . . , Q t−2 , H i+1 . Then C is adjacent to all of Q 1 , . . . , Q t−2 . Contracting each of Q 1 , . . . , Q t−2 , H i+1 , C into a single vertex gives K t as a minor of G, which is a contradiction. Hence C is adjacent to at most t − 2 of Q 1 , . . . , Q t−2 , H i+1 , and property (2) is maintained for C . Since H i has maximum degree t−2, we obtain a (t−1)-colouring of G with defect t − 2. Each H i has a 2-colouring with clustering 1 2 (t − 2) . The product of these colourings is a (2t − 2)-colouring with clustering 1 2 (t − 2) .
Note that the O(t) upper bound on χ (M Kt ) has been extended to the setting of odd minors.
In particular, Kawarabayashi [146] proved that every graph with no odd K t -minor is 496tcolourable with clustering at most some function f (t).

H-Minor-Free Graphs
Hadwiger's Conjecture implies that for every graph H with t vertices, the maximum chromatic number of H-minor-free graphs equals t − 1 (since K t−1 is H-minor-free). However, for clustered and defective colourings, fewer colours often suffice. For example, as discussed in Section 3.4, Archdeacon [21] proved that graphs embeddable on a fixed surface are defectively 3-colourable, whereas the maximum chromatic number for graphs of Euler genus g is Θ( √ g).
The natural question arises: what is the defective or clustered chromatic number of the class of H-minor-free graphs, for an arbitrary graph H? We will see that the answer depends on the structure of H, unlike the chromatic number which only depends on |V (H)|.
Ossona de Mendez et al. [185] observed that the following definition is a key to answering this question. Let T be a rooted tree. The depth of T is the maximum number of vertices on a root-to-leaf path in T . The closure of T is obtained from T by adding an edge between every ancestor and descendent in T , as illustrated in Figure 9. The connected tree-depth 4 of a graph H, denoted by td(H), is the minimum depth of a rooted tree T such that H is a subgraph of the closure of T . Note that the connected tree-depth is closed under taking minors. Figure 9: The closure of a tree of depth 6 contains K 5,9 .
Note that the standard example, S(h, d), is the closure of the complete (d + 1)-ary tree of depth h + 1. By Lemma 1, for every graph H, the defective chromatic number of H-minor-free graphs satisfies Proof. Fix an integer c. We now recursively define graphs G k (depending on c), and show by induction on k that G k has no (2k − 3)-colouring with clustering c, and H k is not a minor of For the base case k = 2, let G 2 be the path on c + 1 vertices. Then G 2 has no H 2 = S(1, 2) = K 1,3 minor, and G 2 has no 1-colouring with clustering c. Assume G k−1 is defined for some k 3, that G k−1 has no (2k − 5)-colouring with clustering c, and H k−1 is not a minor of G k−1 . As illustrated in Figure 10, let G k be obtained from a path (v 1 , . . . , v c+1 ) as follows: for i ∈ {1, . . . , c} add 2c − 1 pairwise disjoint copies of G k−1 complete to {v i , v i+1 }.
Suppose that G k has a (2k − 3)-colouring with clustering c. Then v i and v i+1 receive distinct colours for some i ∈ {1, . . . , c}. Consider the 2c − 1 copies of most c − 1 such copies contain a vertex assigned the same colour as v i , and at most c − 1 such copies contain a vertex assigned the same colour as v i+1 . Thus some copy avoids both colours. Hence G k−1 is (2k − 5)-coloured with clustering c, which is a contradiction. Therefore G k has no (2k − 3)-colouring with clustering c.
It remains to show that H k is not a minor of G k . Suppose that G k contains a model {J x : Note that H k − r consists of three pairwise disjoint copies of H k−1 . The model X of one such copy avoids v i−1 and v j+1 (if these vertices are defined). The next lemma is the heart of the proof of Theorem 63.

Proof.
We proceed by induction on h 1, with w and k fixed. The case h = 1 is trivial since S(0, k − 1) is the 1-vertex graph. Now assume that h 2, the claim holds for h − 1, and G is a S(h − 1, k − 1)-minor-free graph with treewidth at most w. Let V 0 , V 1 , . . . be the BFS layering of G starting at some vertex r.
as a minor, as otherwise contracting V 0 ∪ · · · ∪ V i−1 to a single vertex gives a S(h − 1, k − 1) minor (since every vertex in V i has a neighbour in V i−1 ). Since G has treewidth at most w, so does G[V i ]. By Theorem 66 with . Use disjoint sets of colours for even and odd i, and colour r by one of the colours used for even i. No edge joins V i with V j for j i + 2. Now G is (2 h − 2)-coloured with clustering (k − 1)(w − 1).
To drop the assumption of bounded treewidth, we use the following result of DeVos, Ding, Oporowski, Sanders, Reed, Seymour, and Vertigan [72]. See [223,227] for more about defective choosability in minor-closed classes.

Conjectures
We now present a conjecture of Norin et al.
[183] about the clustered chromatic number of an arbitrary minor-closed class of graphs. Consider the following recursively defined class of graphs, illustrated in Figure 11. Let X 1,c := {P c+1 , K 1,c }. For k 2, let X k,c be the set of graphs obtained by the following three operations. For the first two operations, consider an arbitrary graph G ∈ X k−1,c .
• Let G be the graph obtained from c disjoint copies of G by adding one dominant vertex.
Then G is in X k,c .
• Let G + be the graph obtained from G as follows: for each k-clique D in G, add a stable set of k(c − 1) + 1 vertices complete to D. Then G + is in X k,c .
• If k 3 and G ∈ X k−2,c , then let G ++ be the graph obtained from G as follows: for each (k − 1)-clique D in G, add a path of (c 2 − 1)(k − 1) + (c + 1) vertices complete to D. Then G ++ is in X k,c . A vertex-coloured graph is rainbow if every vertex receives a distinct colour.

Lemma 69 ([183]).
For every c 1 and k 2, for every graph G ∈ X k,c , every colouring of G with clustering c contains a rainbow K k+1 . In particular, no graph in X k,c is k-colourable with clustering c.

Proof.
We proceed by induction on k 1. In the case k = 1, every colouring of P c+1 or K 1,c with clustering c contains a bichromatic edge, and we are done. Now assume the claim for k − 1 and for k − 2 (if k 3). Let G ∈ X k−1,c . Consider a colouring of G with clustering c. Say the dominant vertex v is blue. At most c − 1 copies of G contain a blue vertex. Thus, some copy of G has no blue vertex. By induction, this copy of G contains a rainbow K k . With v we obtain a rainbow K k+1 . Now consider a colouring of G + with clustering c. By induction, the copy of G in G + contains a clique {w 1 , . . . , w k } receiving distinct colours. Let S be the set of k(c − 1) + 1 vertices adjacent to w 1 , . . . , w k in G + . At most c − 1 vertices in S receive the same colour as w i . Thus some vertex in S receives a colour distinct from the colours assigned to w 1 , . . . , w k . Hence G + contains a rainbow K k+1 . Now suppose k 3 and G ∈ X k−2,c . Consider a colouring of G ++ with clustering c. By induction, the copy of G in G ++ contains a clique {w 1 , . . . , w k−1 } receiving distinct colours. Let P be the path of (c 2 − 1)(k − 1) + (c + 1) vertices in G ++ complete to {w 1 , . . . , w k−1 }. Let X i be the set of vertices in P assigned the same colour as w i , and let X := i X i . Thus |X i | c − 1 and |X| (c − 1)(k − 1). Hence P − X has at most (c − 1)(k − 1) + 1 components, and |V (P − X)| (c 2 − 1)(k − 1) + (c + 1) − (c − 1)(k − 1) = c (c − 1)(k − 1) + 1 + 1. Some component of P − X has at least c + 1 vertices, and therefore contains a bichromatic edge xy. Then {w 1 , . . . , w k−1 } ∪ {x, y} induces a rainbow K k+1 in G ++ .
Norin et al. [183] conjectured that every minor-closed class that excludes every graph in X k,c for some c is k-colourable with bounded clustering. More precisely:  For clustered colouring, Lemma 57 implies χ (M Ks,t ) 3s. This bound was improved by Dvořák and Norin [84] who proved that χ (M Ks,t ) 2s + 2, which is the best known upper bound. Van den Heuvel and Wood [124] proved the lower bound, χ (M Ks,t ) s + 1 for t max{s, 3}. Their construction is a special case of the construction shown in Figure 11.

Circumference
The circumference of a graph G is the length of the longest cycle if G contains a cycle, and is 2 if G is a forest. This section studies clustered colourings of graphs of given circumference. Let C k be the class of graphs with circumference at most k. A graph has circumference at most k if and only if it contains no C k+1 minor, where C k+1 is the cycle on k + 1 vertices. Thus C k is a minor-closed class.
Lemma 71. For all k, d 1, the standard example S(k, d) contains no path on 2 k+1 vertices and no cycle of length at least 2 k + 1.

Proof.
We proceed by induction on k 1 with d fixed. In the base case, S(1, d) = K 1,d+1 , which contains no 4-vertex path and no cycle. Now assume the result for S(k − 1, d). Let v be the root vertex of S(k, d).
Suppose that S(k, d) contains a cycle C of length at least 2 k + 1. Since v is a cut vertex, C is contained in one copy of S(k − 1, d) plus v. Thus S(k − 1, d) contains a path on 2 k vertices, which is a contradiction. Thus S(k, d) has no cycle of length at least 2 k + 1.
If S(k, d) contains a path P on 2 k+1 vertices, then P − v contains a path component with least 1 2 (2 k+1 − 1) = 2 k vertices that is contained in a copy of S(k − 1, d), which is a contradiction. Hence S(k, d) contains no path of order 2 k+1 .
By (6), Mohar et al. [176] proved an upper bound within a factor of 3 of this lower bound.
Theorem 72 ([176]). For every integer k 2, every graph G with circumference at most k is (3 log 2 k)-colourable with clustering k. Thus This result is implied by the following lemma with C = ∅.

Lemma 73 ([176]).
For every integer k 2, for every graph G with circumference at most k and for every pre-coloured clique C of size at most 2 in G, there is a 3 log 2 k -colouring of G with clustering k, such that every monochromatic component that intersects C is contained in C.

Proof. We proceed by induction on
First suppose that k = 2. Then G is a forest, which is properly 2-colourable. If |C| ≤ 1 or |C| = 2 and two colours are used on C, we obtain the desired colouring (with 2 < 3 log 2 k colours). Otherwise, |C| = 2 with the same colour on the vertices in C. Contract the edge C and 2-colour the resulting forest by induction, to obtain the desired colouring of G. Now assume that k 3.
Suppose that G is not 3-connected. Then G has a minimal separation (G 1 , G 2 ) with S := V (G 1 ∩ G 2 ) of size at most 2. If |S| = 2, then add the edge on S if the edge is not already present. Consider both G 1 and G 2 to contain this edge. Observe that since the separation is minimal, there is a path in each G j (j = 1, 2) between the two vertices of S. Therefore, adding the edge does not increase the circumference of G. Also note that any valid colouring of the augmented graph will be valid for the original graph. Since C is a clique, we may assume that C ⊆ V (G 1 ). By induction, there is a 3 log 2 k -colouring of G 1 , with C precoloured, such that every monochromatic component of G 1 has order at most k and every monochromatic component of G 1 that intersects C is contained in C. This colours S. By induction, there is a 3 log 2 k -colouring of G 2 , with S precoloured, such that every monochromatic component of G 2 has order at most k and every monochromatic component of G 2 that intersects S is contained in S. By combining the two colourings, every monochromatic component of G has order at most k and every monochromatic component of G that intersects C is contained in C, as required. Now assume that G is 3-connected.
Every 3-connected graph contains a cycle of length at least 4. Thus k 4.
If G contains no cycle of length k, then apply the induction hypothesis for k − 1; thus we may assume that G contains a cycle Q of length k. Let A be the set of cycles in G of length at least 1 2 (k − 5) . Suppose that a cycle A ∈ A is disjoint from Q. Since G is 3-connected, there are three disjoint paths between A and Q. It follows that G contains three cycles with total length at least 2(|A| + |Q| + 3) > 3k. Thus G contains a cycle of length greater than k, which is a contradiction. Hence, every cycle in A intersects Q.
Let S := V (Q) ∪ C. As shown above, G := G − S contains no cycle of length at least 1 2 (k − 5) . Then G has circumference at most 1 2 (k − 7) , which is at most 1 2 k , which is at least 2. By induction (with no precoloured vertices), there is a 3 log 2 1 2 k -colouring of G such that every monochromatic component of G has order at most 1 2 k . Use a set of colours for G disjoint from the (at most two) preassigned colours for C. Use one new colour for S \ C, which has size at most k. In total, there are at most 3 log 2 1 2 k + 3 3 log 2 k colours. Every monochromatic component of G has order at most k, and every monochromatic component of G that intersects C is contained in C.
Similar results are obtained for graph classes excluding a fixed path. Let P k be the path on k vertices. It follows from Lemma 71 that td(P k ) = td(P k ) = log 2 (k + 1) . Let H k be the class of graphs containing no path of order k + 1 (or equivalently, with no P k+1 minor). Thus Conjecture 62, in the case of excluded paths, asserts that Every graph with no P k+1 -minor has circumference at most k. Thus Theorem 72 implies the following upper bound that is which is within a factor of 3 of Conjecture 62 for excluded paths.

Theorem 74 ([176]).
For every integer k 2, every graph G with no path of order k + 1 is (3 log 2 k)-colourable with clustering k. Thus To conclude this section, we show that Theorem 34 determines the defective choosability of C k and P k .
First consider C k . Say s t. Then the circumference of K s,t equals 2s. If s k+1 2 then K s,t contains a (k + 1)-cycle and is not in C k . On the other hand, if s k+1 2 − 1 then 2s k and K s,t ∈ C k . Thus χ ∆ (C k ) = k+1 2 . Now consider P k . Say t > s. Then K s,t contains a path order 2s + 1, and contains no path of order 2s + 2. Thus K s,t ∈ H k if and only if 2s + 1 k. That is, K s,t ∈ H k if and only if

Thickness
The thickness of a graph G is the minimum integer k such that G is the union of k planar subgraphs; see [179] for a survey. Let T k be the class of graphs with thickness k. Graphs with thickness k have maximum average degree less than 6k. Thus χ (T k ) 6k. For k = 2, which corresponds to the so-called earth-moon problem, it is known that χ (T 2 ) ∈ {9, 10, 11, 12}. For k 3, complete graphs provide a lower bound of 6k − 2, implying χ (T k ) ∈ {6k − 2, 6k − 1, 6k}.
It is an open problem to improve these bounds; see [129].

Defective Colouring
This section studies defective colourings of graphs with given thickness. Yancey [231] first proposed studying this topic. The results in this section are due to Ossona de Mendez et al. [185]. Since the maximum average degree is less than 6k, Theorem 28 implies that such graphs are (3k + 1)-choosable with defect O(k 2 ), but gives no result with at most 3k colours.

Proof.
We proceed by induction on k 1. In the base case, S(2, d) is planar, and thus has thickness 1. Let G := S(2k, d) for some k 2. Let r be the vertex of G such that G − r is the disjoint union of d + 1 copies of S (2k − 1, d).
Observe that each component of H is isomorphic to S(2k − 2, d) and by induction, H has thickness at most k − 1. Since G − E(H) consists of d + 1 copies of K 2,d pasted on r for some d , G − E(H) is planar and thus has thickness 1. Hence G has thickness at most k.
Lemmas 1 and 75 imply that χ ∆ (T k ) 2k + 1. Ossona de Mendez et al. [185] proved that equality holds. In fact, the proof works in the following more general setting, implicitly introduced by Jackson and Ringel [131]. For an integer g 0, the g-thickness of a graph G is the minimum integer k such that G is the union of k subgraphs each with Euler genus at most g. Let T g k be the class of graphs with g-thickness k. As an aside, note that the g-thickness of complete graphs is closely related to bi-embeddings of graphs [13,14,15,54].
The lower bound in Theorem 76 follows from Lemma 75. The upper bound follows from Lemma 5 and the next lemma.
Proof. By Euler's Formula, G has at most 3k(n + g − 2) edges, and every spanning bipartite subgraph has at most 2k(n + g − 2) edges. Let X be the set of vertices with degree at most − 1. Since vertices in X have degree at least δ and vertices not in X have degree at least , .
Suppose on the contrary that X is a stable set in G. Let G be the spanning bipartite subgraph of G consisting of all edges between X and V (G) \ X. Since each of the at least δ edges incident with each vertex in X are in G , δ|X| |E(G )| 2k(n + g − 2).
Lemma 78 with k = 1 and = 2g + 13 implies that every graph G with minimum degree at least 3 and Euler genus g has a (2g + 12)-light edge. Note that this bound is within +10 of being tight since K 3,2g+2 has minimum degree 3, embeds in a surface of Euler genus g, and every edge has an endpoint of degree 2g + 2. More precise results, which are typically proved by discharging with respect to an embedding, are known [35,130,134]. Theorem 76 then implies that every graph with Euler genus g is 3-choosable with defect 2g + 10. As mentioned in Section 3.4, this result with a better degree bound was proved by Woodall [228]; also see [59]. The utility of Lemma 78 is that it applies for k > 1.

Clustered Colouring
Consider the clustered chromatic number of T g k . We have the following upper bound.
Proposition 79. For all integers g 0 and k 1, every graph G with g-thickness at most k is (6k + 1)-choosable with clustering max{g, 1}.

Proof.
We proceed by induction on |V (G)|. Let L be a (6k + 1)-list assignment for G. In the base case, the claim is trivial if |V (G)| = 0. Now assume that |V (G)| 1.
Colour v by c. Thus G is L-coloured with clustering g.
Now assume that |V (G)| > 6kg. Every graph with g-thickness at most k and more than 6kg vertices has a vertex v of degree at most 6k, since |E(G)| < 3k(|V (G)| + g) (3k + 1 2 )|V (G)|, implying G has average degree less than 6k + 1. By induction, G − v is (6k + 1)-choosable with clustering max{g, 1}. Since |L(v)| 6k + 1 and deg(v) 6k, some colour c ∈ L(v) is not assigned to any neighbour of v. Colour v by c. Thus v is in a singleton monochromatic component, and G is L-coloured with clustering max{g, 1}.
Since S(3, c) is planar, an analogous proof to that of Lemma 75 shows that S(2k + 1, c) has thickness at most k. By Lemma 2 and Proposition 79, Closing this gap is an interesting problem because the existing methods say nothing for graphs with given thickness. For example, as illustrated in Figure 13, the 1-subdivision of K n has thickness 2. Thus thickness 2 graphs have unbounded ∇. Similarly, Lemma 13 is not applicable since graphs with thickness 2 do not have sublinear balanced separators. Indeed, Dujmović et al. [82] constructed 'expander' graphs with thickness 2, bounded degree, and with no o(n) balanced separators. Returning to the earth-moon problem, it is open whether thickness 2 graphs are 11-colourable with bounded clustering. First consider when f is the treewidth. Recall that DeVos et al. [72] proved that χ tw (G) 2 for every minor-closed class G. Bounded degree classes and χ tw (D ∆ ) look interesting. In particular, what is the maximum integer ∆ such that every graph with maximum degree ∆ is 2-colourable with bounded monochromatic treewidth? The answer is at least 5 since every graph with maximum degree 5 is 2-colourable with bounded clustering [120]. The answer is at most 25 since Berger et al. [25] proved that for every 2-colouring of the n × n × n grid with diagonals (which has maximum degree 26), there is a monochromatic subgraph with unbounded treewidth (as n → ∞). This upper bound is probably easily improved by eliminating some of the diagonals in the 3-dimensional grid. Can the lower bound be improved? In particular, does every graph with maximum degree 6 have a 2-colouring with bounded monochromatic treewidth?
Let η(H) be the maximum integer t such that K t is a minor of H, sometimes called the