Defective and Clustered Graph Colouring∗

†

David R. Wood Abstract. defect d Consider the following two ways to colour the verticesd of a graph whereclustering the crequirement that adjacent vertices get distinct coloursc is relaxed. A colouring has if each monochromatic component has maximum degree at most . A colouring has if each monochromatic component has at most 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 Verdièreparameter,Kt graphs with givenK circumference,s,t graphs excluding a fixed graph as an immersion,H graphs with given thickness, graphs with given stack- or queue-number, graphs excluding as a minor, graphs excluding as a minor, and graphs excluding an arbitrary graph as a minor. Several open problems are discussed. arXiv:1803.07694v1 [math.CO] 20 Mar 2018

∗ Electronic Journal of Combinatorics c Davidhttp://www.combinatorics.org/DS23 R. Wood. Released under the CC BY license (International 4.0). † This is a preliminary version of a dynamic survey to be published in the [email protected] , #DS23, . School of Mathematical Sciences, Monash University, Melbourne, Australia ( ). Research supported by the Australian Research Council.

1 Contents

1 Introduction3

1.1 History and Terminology ...... 3 1.2 Deﬁnitions ...... 5 1.3 Choosability ...... 6 1.4 Standard Examples ...... 6 1.5 Two Fundamental Observations ...... 8 2 Greedy Approaches 10 1.6 Related Topics ...... 9

2.1 Light Edges ...... 10 3 Graphs on Surfaces 11 2.2 Islands ...... 10

3.1 Outerplanar Graphs ...... 11 3.2 Planar Graphs ...... 11 3.3 Hex Lemma ...... 13 3.4 Defective Colouring of Graphs on Surfaces ...... 14 4 Maximum Degree 19 3.5 Clustered Colouring of Graphs on Surfaces ...... 16 5 Maximum Average Degree 23

6 Excluding a Subgraph 25

7 Excluding a Shallow Minor 26 ∗ Ks,t

7.1 Excluding ...... 26 7.2 Linklessly Embeddable Graphs ...... 28 7.3 Knotlessly Embeddable Graphs ...... 29 7.4 Colin de Verdi`ereParameter ...... 30 7.5 Crossings ...... 31 7.6 Stack and Queue Layouts ...... 32 8 Minor-Closed Classes 36 7.7 Excluded Immersions ...... 33

Kt 8.1 ExcludingH a Minor and Bounded Degree ...... 36 8.2 -Minor-Free Graphs ...... 38 8.3 -Minor-Free Graphs ...... 41 8.4 Conjectures ...... 44 9 Thickness 48 8.5 Circumference ...... 46

9.1 Defective Colouring ...... 49 10 General Setting 52 9.2 Clustered Colouring ...... 51

2 1 Introduction

monochromatic component

ConsiderG a graphk where each vertexclustering is assignedc a colour. A k is a connected component of the subgraph induced by all thec vertices assignedG a singlek colour. A graphdefectisd -colourable withG if each vertexk can be assigned one of colours such that each monochromaticd component has at most vertices. A graph is -colourable with if each vertex ofd can be assigned one of colours such that each vertex is adjacent to at most neighbours of the same colour; that is, each monochromatic component has maximum degree at most . 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 emphasisclustered is on chromatic generalresults number for broadly defined classes of graphs,χ?( ) rather than more precisek G G results for more specific classes.c With this viewpoint the followingk definitions naturally arise. G c k unbounded The of a graphG class , denoted by , is the minimum integer defectively k d for whichG there exists an integer such that every graph in has a -colouring with clusteringG k d defective chromatic number χ∆( ) . If there is no such integer , then has clusteredG chromatic number.G A graph k k k class is -colourableG if there exists an integer such that every graph in is unbounded c G-colourable with defect . The of , denoted by , is the c 1 χ∆( ) χ?( ) χ( ) minimum integer− such thatG 6is defectivelyG 6 G -colourable. If thereG is no such integer , then has defective chromatic number. Every colouring of a graph with clustering has defect . Thus for every class . Tables1 and2 summarise 1.1the results History presented and Terminology in this survey; see Sections 1.2 and 1.3 for the relevant definitions.

There is no single origin for the notions of defective and clustered graph colouring, and the terminology used in the literature is inconsistent.k d (k, d)- Earlycolourable papers on defective colouring include [16, 17, 128, 138, 166, 212], although these did notd-relaxed use the termd-improper ‘defect’. The definition of “ -colourable with defect ”, often written G, was introducedd by Cowen et al. [d66-improper]. This terminology chromatic number is fairly standard,d-chromatic although number or G is sometimes used. The minimumdefective number chromatic of colours number in a colouring of a graph with defect has been called the [144] or [16] of . Cowen et al. [65] introduced the of a graph class (as defined above). (k, c)-fragmented Alsok for clustered colouring, thec literature is inconsistent. Onec of the early papers is by Kleinberg et al. [151], who defined a colouring to be if, in our language, it is a -colouring with clusteringchromon. I prefer “clustering” since as increases, intuitively the “fragmentation”metachromatic of the number monochromatic components decreases. Edwards and Farr [89, 90] called a monochromatic component a and called the clustered chromatic number of a class the .

3 ` χ∆( ) χ ( ) G G ∆ G Table 1: Summary of Results for Defective Colouring O graph class Sect. P outerplanar 6 g g 2 2 3.1 E m m m m + 1 + 1 planar 6 A b 2 c3b 2 c 3 3.2 4 4 Euler genus L 3 3 3.4 5 5 max average degree K 5 k k k k linklessly embeddable6 V 7.2 k k k + 1 k + 1 knotlessly embeddable S 7.3 k k k + 1,..., 2k + 1 2k + 1 Colin de Verdi`ere Q 7.4 Kt t 2 t 1 -stack graphs I − 7.6 -queue graphs6 k k + 1 k + 1 7.6 no immersion6 k 6 ∆ 2 2 7.7 treewidthKt Kt t 1 t 1 8.1 M − td(H)+1 − treewidthH , max degree H td(H) 1,..., 2 4 min s : t H Ks,t 8.1 M − − { ∃ } no Ks,t-minor (s 6 t Ks,t s s 8.2 M k+1 no -minor 6 k k log2 k + 1,..., 3 log2 k 2 8.3 C b c b c d k+1 e no (k +-minor 1) ) k log2(k + 2) 1,..., 3 log2 k 2 8.4 Hg d e − b c b c circumferenceg k 2k + 1 2k + 1 8.5 6 Tk no -path 8.5 -thickness 9

` χ?( ) χ ( ) G G ? G Table 2: SummaryO of Results for Clustered Colouring graph class Sect. P g g outerplanar 6 E 3 3 3.1 planar 6 g 6 ∆ 43 3 4, 4 3.2 ∆+6 ∆+1 ∆+6 Euler genus6 ∆ ∆ 4 ,...,43 4 ,..., 4 ∆ + 1 3.5 D m m Euler genus , max degreem m + 1,..., m + 1 + 1,..., m + 1 8.1 6 A b 2 c b c b 2 c b c 5 5 max degree L 4 6 6 max average degree K 5 k k linklessly embeddable6 V 7.2 k k k + 2,..., 2k + 2 k + 2 ... 2k + 2 knotlessly embeddable S 7.3 Colink de Verdi`ere k k +open 1,..., 4k k + open1,..., 4k 7.4 Q t+4 Kt t t 1 -stack graphs I > b 4 c > − 7.6 -queue graphs6 k k + 1 k + 1 7.6 no immersion6 k 6 ∆ 2 2 7.7 31 Kt K t 1,..., 2t 2 t 1,..., t treewidth M t − − − d 2 e 8.1 treewidthKt , max degree6 ∆ 3 > 3 8.1 td(H)+1 no H -minor H td(H) 1,..., 2 4 8.2 M − − no Ks,t-minor, max(s degreet K s + 1,..., 2s + 2 8.2 6 M s,t k k log k + 1,..., 3 log k no -minor 6 C b 2 c b 2 c open 8.3 no (k +-minor 1) ) k log2(k + 2) 1,..., 3 log2 k open 8.4 Hg d e − b c g k 2k + 2,..., 6k + 1 2k + 2,..., 6k + 1 circumference6 Tk open 8.5 no -path open 8.5 -thickness 9

4 1.2 Deﬁnitions

clique This section brieﬂy states standard graph theoretic deﬁnitions, familiar to most readers. G k-colouring G k A G in a graphvw isG a setbichromatic of pairwise adjacentv w vertices. v LetG properlybe a graph. A v of is a function that assigns one of coloursv to each vertex of G. Anproper edge of is if and are assignedchromatic distinct number colours.G A vertexχ(Gof) is coloured ifk is assigned a colour distinctk from every neighbourG of . A colouring of graphis parameterif every vertex is properly coloured.f The of ,f denoted(G1) = f(G2), is the minimum integerG1 suchG2 that there is a properf -colouringbounded of . G c f(G) c G f unbounded f A is a real-valued6 function∈ G on the class of graphs such thatG f( ) := sup f(G): G wheneverG graphsG and { are isomorphic.∈ G} Say is on a graph class if there f f( ) = max f(G): G exists such that for every , otherwiseG is G on{ . If is bounded∈ G} on , then letG mad(G) . Most graph parameters consideredG in this survey are integer-valued, in which case, if is bounded on , then . k-degenerate k For a graph , let be the maximum averagek degree of a subgraph(k + 1) of . A graphH is minor if everyG non-empty subgraph hasH a vertex of degree at most .A greedyG colouring algorithm shows that every -degenerateminor-closed graph is -colourable.G G ∈ G G G H-minor-free H A graph is a G of a graph if a graphG isomorphic to can be obtained from a subgraph G H H of by contractingM edges. A class of graphs is if for every graph every minorsubdivide of is in , andvw some graph isG not in . A graph vwis if x is not a minor of . Letvx xwbe thesubdivision class of -minor-freeG graphs. G To 1-subdivisionan edge in aG graph means to delete , addG a new vertex , and add new edgesG and .A H oftopologicalis any graph minor obtained fromG by repeatedly subdividing edges. The H of isG the graph obtained from by subdividing each edge of exactlyEuler once. genus A graph is a h of a graph 2hif a graphEuler isomorphic genus to a subdivision of is a subgraphc of . c Euler genus G The of the orientableG surface with handles is . The of the non-orientable surface with cross-caps is . The of a graph is the minimum Eulertree genus decomposition of a surface in which Gembeds (with no crossings).T See [177] for background on (embeddingsTx V (G): ofx graphsV (T on)) surfaces. G bags ⊆ ∈ Avw G Tofx a graph isv given byw a tree whose nodes indexv aG collection x V (T ): v Tx T width { ∈ ∈ } of sets of vertices in called , such that (1) for every edge T max Tx 1 : x V (T ) treewidth G tw(G) of , some bag {|contains| − both∈ and} , and (2) for every vertex of , the set induces a non-empty (connected)G subtree of . The of a tree decomposition is . The of a graph , denoted by , is thelayering minimum width ofG the tree decompositions(V0,V1,...,V of . See`) [32V,(G33), 118, 193, 194] for surveys vwon treewidth.E(G) v Vi w Vj i j 1 Vi layer r ∈ ∈ ∈ | − | 6 G Vi := v V (G) : distG(v, r) = i i 0 V0,V1,... A of a graph is a partition{ ∈ of } such> that for every edge , if BFSand layering, thenG . Eachr set is called a . If is a vertex in a connected graph and for , then is a layering called the of starting at .

5 layered treewidth G k (Tx V (G): x V (T )) G (V0,V1,...,V`) G ⊆ ∈ Vi Tx k i [0, `] x V (T ) The| ∩ | 6 of∈ a graph is the∈ minimum integer such that there is a tree decomposition of and a layering of , such that

(G1,Gfor2) everyseparationand every G G1. LayeredG2 treewidth was introduced byG Dujmovi´cetG al.= [G811]. G2 V (G1) V (G2) = V (G2) V (G1) = ∪ \ 6 ∅ \ 6 ∅ V (G1 G2) k (G1,G2) k-separation (G1,G2) G minimal A| pair ∩ | 6 is a of a graph if and are induced subgraphs of V (G1) V (G2) V (G1) V (G2) V (G2) V (G1) such that ∩ , and and \ . If, in addition,\ balanced separator, then isG a S V.( AG) separation of is G Sif ⊆ − every vertex1 inV (G) has a neighbour in both and . 2 | | A radius in a graphG is a set suchr that every component of v G has at most G vertices. r v The of a connected graph is the minimum integer such that for some vertex of , 1.3every vertexChoosability of is at distance at most from .

Manylist assignmentdefective and clustered colouringG results holdL in the setting of listL(v colouring.) Eaton and Hull [86v] firstV (G introduced) defective listG colouring.L-colourable G ∈ v V (G) L(v) L k-list A for a∈ graph is a function that assigns a set of colours to each assignment L(v) k v V (G) choice number G vertex | . Define| > a graph to be ∈ if there is a proper colouring of such that each vertexk G isL assigned a colour in k . A list assignmentL G is a if L for each vertexG .d The> 0 G ofL a-colourable graph is with the defectminimumd integer such that isG -colourabled for every -list-assignmentv V (ofG) . ∈ For a list-assignmentL(v) Gof a graphk-choosableand integer with defect ,d defineG Lto be d ifk there is a colouringL of G with defect such that eachc > vertex1 G L-colourableis assigned with clustering c G c v V (G) a colour in . Define to be if is -colourable with∈ defect for every -list assignmentL(v) of G. Similarly,k-choosable for an integer with clustering, cis G L if therec is a colouringk of withL clusteringG such that each vertex is assigneddefective a colour choice in number. Define to be χ` ( ) if is -colourable G ∆ G withk clustering for every -list assignmentd 0 of . G k > ∈ G d clustered choice number χ`( ) The of a graph class , denotedG by , is the? G minimum integer k c 1 G k for which there exists an integer , such> that every graph is∈ G-choosable with defect . The c of a graph class , denoted by , is the minimum integer for which there exists an integer , such that every graph is -choosable with1.4 clustering Standard. Examples

TheS following(h, d) construction, or variants of it, haveS been(0, d) used by several authors [87, 119, 124, h 1 S(h, d) d + 1 S(h 1, d) 183, 185] to provide> lower bounds on the defective chromatic number. As illustrated in Figure− 1, let be deﬁned recursively as follows.S(1 Let, d) = K1,dbe+1 the graph withd one+ 1 vertex and no edges. For , let be the graph obtained from disjoint copies of by adding one dominant vertex. Note that , the star with leaves.

6 S1,d

Sh 1,d Sh 1,d Sh 1,d Sh 1,d − − − −

d +1 d +1

S(h, d)

Figure 1: The standard example . Lemma 1. For integers h > 1 and d > 0, the graph S(h, d) has no h-colouring with defect d.

Proof. h S(1, d) = K1,d+1 d h 2 h 1 > − We proceed byS( inductionh, d) on h. In the base case, d v , which obviously has S(h, d) v S(h, d) v d + 1 C1,...,Cd+1 v no 1-colouring with defect . Now assume− that and the claim holds for . Suppose on the contrary that has an -colouring withd defect d. LetC1,...,Cbe thed dominant+1 vertex in Ci (h 1) d Ci , and say is blue. Then− has components . Since is S(h 1, d) S(h, d) h d dominant and has− monochromatic degree at most , at most of contain a blue vertex. Thus some is -coloured with defect . This is a contradiction since is isomorphic to . Thus hasS(h, no c) -colouring with defect . S(1, c) c + 1 h > 2 S(h, c) c S(h 1, c) Similarly, as illustrated− in Figure2, let be deﬁned recursively as follows. Let be the path on vertices. For , let be the graph obtained from disjoint copies of by adding one dominant vertex.

S1,c

Sh 1,c Sh 1,c Sh 1,c Sh 1,c − − − −

c +1 c

S(h, c)

Lemma 2. For integers h,Figure d > 1, 2:the The graph standardS(h, c) examplehas no h-colouring. with clustering c.

Proof. h S(1, c) c + 1 c h > 2 We proceed by induction on . In the base case, is the path on vertices, which obviously has no 1-colouring with clustering . Now assume that and the claim

7 h 1 S(h, c) h c − v S(h, c) v S(h, c) v c − holdsC1,...,C for c . Supposev on the contrary that has an -colouring withv defect . Letc c 1 C1,...,Cc Ci (h 1) be the dominant− vertex in , and say is blue. Then has −components c Ci S(h 1, c) S(h, d) . Since is dominant and the monochromatic component containing− has at most vertices,h at most of c contain a blue vertex. Thus some is -coloured with clustering . This is a contradiction since is isomorphic to . Thus has no -colouring with clusteringS(2, d) . S(2, c) S(3, c)

As illustrated in Figure3, is planar, is outerplanar, and is planar. b b b

b b

bbb b b b b b b

b b b

bbb bbbbbb b b b b b b b b b

b b b b b b

S(2, d) S(2,c) S(3,c)

Figure 3: Planar standard examples. 1.5 Two Fundamental Observations

The following elementary, but fundamental, result characterises those graph classes with Proposition 3. The following are equivalent for a graph class : bounded defective or clustered chromatic number. G (1) has bounded defective chromatic number, G (2) has bounded clustered chromatic number, G (3) has bounded chromatic number. G Moreover, χ∆( ) χ?( ) χ( ). G 6 G 6 G Proof. G k G k G G k χ∆( ) χ?( ) k We ﬁrst show that (3) implies (1) and (2). SupposeG that6 hasG 6 bounded chromatic number. That is, for some integer , every graph in is properly -colourable. Thus is G -colourable with defect 0 andk withd clustering 1. HenceG k . G We now show that (1) implies (3). Suppose thatd has bounded defective chromatic number.d That is, for(d + some 1) integers and , every graph in is -colourable such that each G G k(d + 1) monochromatic subgraph has maximum degree . Every graphG with maximum degree is properly -colourable by a greedy algorithm. Apply this result to each monochromatic G subgraph of . Hence is properlyk c -colourable,G and hask bounded chromatic number. G We ﬁnally show that (2) implies (1). Suppose that has bounded clustered chromatic number. That is, for some integers and , every graph in is -colourable such that each

8 c c 1 χ∆( ) k − G 6 G monochromatic subgraph has at most vertices, and therefore has maximum degree at most . Hence , and has bounded defective chromatic number. Proposition 4. The following are equivalent for a graph class : We have the following analogous result for defective and clusteredG choosability. (1) has bounded defective choice number, G (2) has bounded clustered choice number, G (3) has bounded choice number. G (4) has bounded maximum average degree. G Proof.

A greedy algorithm shows that (4) impliesG (3).k Alon [10] proved that (3)c impliesG (4). k c 1 More generally, Kang [141−] proved that (1) implies (4). It is immediate that (3) implies (1) and (2). As in the proof of Proposition3, if a graph is -choosable with clustering , then is 1.6-choosable Related with Topics defect . Thus (2) implies (1).

• We brieﬂy mention here some related topics not covered in this survey: • defective colourings of random graphs [142, 143, 144], • defective versions of Ohba’s list colouring conjecture [215, 230], • defective Nordhaus-Gaddum type results [2,3], • acyclic defective colourings [5, 34, 49, 100, 101], • uniquely defectively colourable graphs [104] • weighted defective colouring [20, 22, 123]. • defective colourings of triangle-free graphs [4, 204, 205], • defective colouring of directed graphs [165], • defective edge colouring [125, 126, 224], • defective circular and fractional colouring [99, 111, 152, 174], • equitable defective colouring [218], defective co-colouringG [6], G Also note that defective colouring is computationally hard. In particular, it is NP-complete to decide if a given graph is 3-colourable with defect 1, even when has maximum degree 6 or is planar with maximum degree 7 [18]. See [24, 28, 52, 55, 64, 115, 155, 178] for more computational aspects of defective and clustered clustered colouring.

9 2 Greedy Approaches

2.1 Light Edges

e `-light e ` Light edges provide a generic method for proving results about defective colourings. An edge in a graph is if both endpoints of have degree at most . There is a large literature on light edges in graphs; see [35, 48, 50, 133, 135, 136] for example. Several results on defective Lemma 5. For integers ` k 1, if every subgraph H of a graph G has a vertex of degree at colouring use, sometimes> implicitly,> the following lemma [87, 119, 158, 185, 206]. most k or an `-light edge, then G is (k + 1)-choosable with defect ` k. − Proof. L (k+1) G V (H) + E(H) | | | | H G L ` k V (H) + E(H) = − | | | | 0 Let be a -list assignmentH G for .H We prove by inductionv on k that H v L ` k L(v) every subgraph− of is -colourable with defect− . The base case with is trivial. Considerv a subgraph of . If v has a vertex ofH degree at most , then by k + 1 H ` xy H xy induction is -colourable with defect , and there is a colour in − used by L c ` k c(x) = c(y) c L H no neighbour of which can− be assigned6 to . Now assume that has minimum degree at ` k c(x) = c(y) c L least− . By assumption, contains an -light edge . By induction, has an H ` k x ` k + 1 -colouring with− defect . If , then is also an− -colouring of with defect y c(x) deg (x) ` k 1 . Now assume that H .6 We may further assume− that is not an -colouring of c(xwith) defectL(v) . Withoutk loss of generality, chas(x) exactly neighbours (including )x coloured by . Since x , there are at most neighbours not coloured by . Since contains colours diﬀerent from , there is a colour used by no neighbour 2.2of which Islands can be assigned to .

k-island G S V (G) S k ⊆ V (G) S Esperet\ and Ochem [98] introduced the following deﬁnition and lemma. A in a graph is a non-empty set such that every vertex in has at most neighbours in Lemma 6 . 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. ([98]) Proof. V (G) L (k + 1) | | G G k S G S L − c v S L(v) We proceed by induction on ∈ . The base case is trivial. Let be a -list v V (G) S S assignment for\ . By assumption, has a -island . By induction, is -colourable G S S c G L c G k with clustering −. Assign each| | 6 vertex a colour in not assigned to any neighbour of in c . Each monochromatic component is contained in or is a monochromatic component of . Since , is -coloured with clustering . Thus is -choosable with clustering . k k c = 1 Note that islands generalise the notion ofk degeneracy, since a graph is -degenerate(k + 1) if and only if every non-empty subgraph has a -island of size 1. Thus Lemma6 with is equivalent to the well-known statement that every -degenerate graph is properly -choosable.

10 3 Graphs on Surfaces

3.1 Outerplanar Graphs

O S(2, c)

Let be the class of outerplanar graphs. Every` outerplanar graph is 2-degenerate, and thus χ?( ) = χ ( ) = 3. is properly 3-colourable and 3-choosable.O Since? O is outerplanar, by Lemma2,

Theorem 7 . Every outerplanar graph G is 2-colourable such that each monochromatic Cowen et al. [66] proved the following result for defective colourings of outerplanar graphs. component is a path (and thus with defect 2). ([66]) Proof. G V0,V1,... r V0 = r G[Vi] i 1 { } > V0 Vi−1 K2,3 We may∪· assume · ·∪ is connected. Let be the BFS layering starting from G[Vi] G[Vi] V0 Vi−1 some vertex . Thus . If has maximum degree at least 3, for some ∪ · · · ∪, then contracting intoK a4 single vertex gives a -minor, which is not outerplanar. Thus G[Vi] has maximum degree at most 2. If Vi containsi mod 2 a cycle,G then contracting into a single vertex gives a -minor, which is not outerplanar. Thus each component of is a path. Colour each vertex in by . Then is 2-coloured such that each monochromatic component is a path. 2 K1,d+1 Eatond and Hull [86] generalised Theorem7 by showing that every outerplanar graph is - ` χ∆( ) = χ ( ) = 2. choosable with defect 2. By Lemma1, theO outerplanar∆ O graph is not 1-colourable with defect . Thus

Moreover, the defect bound in the above results is best possible since the graph shown in Figure4 is not 2-colourable with defect 1.

Figure 4: An outerplanar graph that is not 2-colourable with defect 1.

3.2See [224 Planar, 226] Graphs for more results on defective colouring of outerplanar graphs.

P P

Let be the class of planar graphs. We now discuss defective colourings of . Firstvw note that many results about light edges in a planar graphs are known [136, 137]. Borodin [35] proved that every planar graph with minimum degree at least 3 contains an edge with

11 deg(v) + deg(w) 6 13 3 8 (which is best possible for the graph obtained from the icosahedron by stellating each face). This edge is 10-light. By Lemma5, every planar graph is -choosable with defect . Cowen et al. [66] improved the defect bound here to 2, and Poh [189] proved an analogous result in which each monochromatic component is a path (see [124] for an alternative Theorem 8 . Every planar graph is 3-colourable such that each monochromatic component proof). is a path (and thus with defect 2). ([189]) Proof. V (G) G | | v1v2 G G We proceed by induction on v1with thev2 hypothesis that every planar graph is 3-colourable such that for each edgev1 of , there is such a 3-colouringv1 of such that each v2 v2 V (G) 4 monochromatic component is a path, and and are properly coloured.| | 6 (Recall that this meansv1 thatv2 every neighbour of is assigned a distinct colour from , and every neighbour V (G) 5 of is assigned a distinct colour from .) In the base| case,| > if , then assign and distinctG colours, and assign the (at most two) other verticesv1v2 a thirdv1av colour.2 v Each1bv2 0 monochromaticG componentG is a path. Now assumev1 thatv2 . By adding edges,x we may 0 0 assumev1, v2 that, a, b is aG planar triangulation. Say the facesC containing are (G andv1) v2. 0 − − Let beaxb obtained from byC deleting theG edge , and introducingV (G) > a new5 vertex adjacent 0 0 | | 0 to C . Then Cis a planar triangulation. LetG1 beG a2 shortest cycle in G 0 0 0 0 0 such that is aC subpath of C. Since isG 3-connected= G1 G2 and V (G1) V,( suchG2) a = cycleV (C exists.) 0 0 ∪ 0 ∩ E(G ) E(G ) = E(C) vi V (G ) i 1, 2 Since1 ∩ is shortest,2 is an induced cycle. Let and∈ bei the subgraphs∈ { } of ‘inside’ and ‘outside’ of including . That is, and and V (G) = V (G0) 1 = V (G0 ) + V (G0 ) V (C) 1. | |. Without| | loss − of generality,| 1 | | 2 | − | for | − . Note that 00 0 00 G G C xi vixi E(G ) i i ∈ i V (G00) = V (G0 ) V (C) + 1 | i | | i | − | | Let be obtained from by contracting into vertex . Then and V (G) = ( V (G00) + V (C) 1) + ( V (G00) + V (C) 1) V (C) 1 | | | 1 | | . Thus| − | 2 | | | − − | | − = V (G00) + V (G00) + V (C) 3. | 1 | | 2 | | | − V (G00) 2 V (C) 3 V (G) V (G00) + 2 V (G00) < | 2 | > | | > | | > | 1 | | 1 | V (G) V (G00) < V (G) G00 | | | 2 | | | i Since and , wevi have xi , implying .x Similarly,1 x2 . By induction,v1 v2 each is 3-colourable such that each V (C) x x1 x2 V (C) x G monochromatic\{ } component is a path, and and are properly coloured.\{ } Permute the colours so thatx and get theG same colour, andV (C) andx get distinct colours. Colour each vertex 00 \{ } inxi by the colourGi assigned to and . Note that Ginduces a path in 00 00 (since is notG a1 vertexG2 of ). Moreover, Gis a monochromatic component, since each is properly coloured in . Every other monochromatic component of is a monochromatic component of or , and is therefore a path in . 3

Eaton and Hull [86] strengthenedS(2, d) Theorem8 by showing that every planar graph is -choosable with defect 2. (See Theorem 33 or an alternative` proof of a more general result with a weaker χ∆( ) = χ ( ) = 3. defect bound.) Since is planar, byP Lemma∆ 1P,

12 See [36, 38, 60, 110, 158, 206, 214, 224, 229] for more on defective colourings of planar graphs. See [216, 225, 232] for more on defective4 choosability of planar graphs. Now consider clustered colourings4 of planar graphs. The 4-colour theorem [19, 196] says that every planar graph is properly -colourable. Cowen et al. [66] proved the weaker result that every planar graph is -colourable4 with defect 1 and thus with clustering 2 (with a computer-freeS(3, c) elementary proof). Cushing and Kierstead [68] strengthened this result by ` χ?( ) = χ ( ) = 4. proving that every planar graph is -choosableP with? P defect 1 and thus with clustering 2. Since is planar, by Lemma2, 5 χ`( ) = 5 P Thomassen [211] proved that every planar graph is properly -choosable. Voigt [213] and later 3.3Mirzakhani Hex Lemma [175] constructed planar graphs that are not 4-choosable. Thus .

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 Theorem 9. For every integer k 2 there is planar graph G with maximum degree 6 such proof. > that every 2-colouring of G has a monochromatic path of length k.

Proof.

(a, . . . , b, . . . , c, . . . , d, . . . ) a, . . . , b c, . . . , d k A suitable subgraph of the triangular grid{ forms} an embedded{ } plane graph with b, . . . , c d, . . . , a k maximum degree 6, such{ that} every{ internal} face is a triangle, the outerface is a cycle G a, . . . , b c, . . . , d , the distance between { and } { is at} least , and b, . . . , c d, . . . , a k {the distance} between{ } and is at least . By Lemma 10 below, every 2-colouring of contains a monochromatic path between and or between Lemma 10.andLet G be an, whichembedded has length plane atgraph least with. outerface (a, . . . , b, . . . , c, . . . , d, . . . ), such that every internal face is a triangle, and a, b, c, d are distinct. Then for every 2-colouring of G there is a monochromatic path between a, . . . , b and c, . . . , d or between b, . . . , c { } { } { } and d, . . . , a . { } Proof. G0 G w, x, y, z N(w) = a, . . . , b z, x N(x) = { } ∪ { } b, . . . ,Say c thew, ycoloursN are(y) blue = c, and . . . , red. d Asx, shownz inN( Figurez) = 5d,, . let . . , a bey, obtained w fromw { } ∪ { } { } ∪ { 0 } { } ∪ { } byy adding four newx verticesz , whereG and 0 0 G and G and (w, x, y, z) . Colour 0 and blue.G Colour and red. Note that embeds(a, in w, the z) plane,(b, x, w such) (c, that x, y every) internal(d, y, z) 0 face of is a triangle,special and theH outerface of is the 4-cycle . The four internalG faces of that shareH an edge with the outerface are , H , and . 0 Call these faces . Let beH the graph with one vertex forG each internal face of , whereA, B,two C, vertices D of are adjacentH if the corresponding faces of share an edge(a, whose w, z) endpoints are coloured diﬀerently. is a subgraph of the dual of and is therefore planar. Let be the vertices of respectively corresponding to the special faces ,

13 (b, x, w) (c, x, y) (d, y, z) w z a w z A H B C D 0 , H , . SinceG and are coloured diﬀerently, has the same colourH as exactly one of and , implying has degree 1 in F. Similarly, , and each have degree 1 in . If some face of isF monochromatic, then the corresponding vertex of has degree 0.H Every non-monochromaticF non-specialH face has two vertices of oneH colour and one vertex of the otherA, B, C,colour, D and does not share an edge with the outerface.H Thus the vertex of corresponding to has degree 2 in A,. B, In C, summary, D every vertex of hasA, B, degree C, D 0 or 2, except for , which have degree 1. Thus each componentH of is eitherP an isolated vertex,A aB cycle, or a path joining two of . The two paths joiningP d, . . . , a b, . . . , c are disjoint{ and} do{ not cross.} Thus, without loss of generality, contains a path with endpoints and . The red vertices on the faces corresponding to vertices in form a walk from to , which contains the desired red path.

3.4See [25 Defective, 145, 170 Colouring, 172] for multi-dimensional of Graphs on Surfaces generalisations of the Hex Lemma.

g 0 g g > E g g E E For every integer , let be the class of graphs with Euler genus at most . This section considers defective colourings of graphs in .y Cowen et al. [66] proved that every graph in

d c

z x b b b b b b b b b b b

b b b

b b

b b a b A b b B

Figure 5: Proof of the Hex Lemma

14 g S(2, d) χ∆( g) = 3 E E is defectively 4-colourable. Archdeacon [21] proved the conjecture of Cowen et al. [66] that 2 every graph in is defectivelyKn 3-colourable.3 Θ( Sincen ) is planar, by Lemma1. The following proof of Cowen et al. [65] provides a defect bound that is within a constant Theorem 11 . Every graph G with Euler genus g is 3-colourable with defect d := factor of optimal (since has Euler genus ). max 12, √6g + 7 . { d e } ([65]) Proof. V (G) + E(G) v | | | | G v d v − We proceed by induction on v G . If some vertex has degreed at most 2, thenG by induction, is 3-colourable withA defect . Assign a colour diﬀerent fromd the v, w A vw G vw d colours∈ assigned to the neighbours of . Then −is 3-coloured with defect . Now assume that has minimumG degree at leastd 3. Let be the setA of vertices with degreeB at most . If max{12(g−2),0} for some edge ,d then+ 1 by induction,d 12 is 3-colourable withB defect , which > | | 6 d−11 is a 3-colouring of withA defect . Now|B| assume that is aB stable set. Let be the set of d 2 e verticesB with degree at least . Since and by Lemma 12 below, . Colour the vertices in blue. Colour of the vertices in red. Colour the12(g other−2) vertices g 6 2 B = 0 6 d B 6 d−11 in green. ∅ | | |B| 1 g 3, 4 d = 12 B 24 d 2 e − ∈ { } | | 6 11If then and theg defect5 is , as desired. Otherwise|B|−1 6(g−2) 1 , and the > 2 6 d−11 − 2 defect is at most d . If g > 5 then 12g + 36and6 12g + 7√and6g the defect is at most , as desired. Otherwise, , and thep defect is at mostp p . We now show 12(g 2) 12g + (15 8) 6g 60 = (2 6g + 15)( 6g 4). this bound is at most− .6 Since ,− we have − , implying− d √6g + 7 12(g 2) (2d + 1)(d 11) 6(g 2) (d + 1 )(d 11) > − 6 − − 6 2 − d 12 6(g−2) 1 d > d−11 − 2 6 Since , we have . That is, . Lemma 12 . Let G be a graph with Euler genus g and minimum degree at least 3. Fix an Since , we have , as claimed. integer d. Let A be the set of vertices with degree at most d. Assume that A is a stable set. Let B be the set of vertices with degree at least d + 1. Then (d 11) B max 12(g 2), 0 . ([65]) − | | 6 { − } Proof. B = B = ∅ 6 ∅ f B A 0 If then the result is vacuous. Now assumeG that .A For each non-triangularni 0 0 face , add an edge between two non-consecutivei G α vertices in (which must existG since 0 is a stable set).A We obtainβ a multigraph triangulationG , in which is a stableB set. Let be the number of vertices with degree in . Let be the number of edges in incident with vertices in . LetX be the numberX of edges in Xwith both endpoints in . By Euler’s 6(g 2) = (i 6)ni = (i 6)ni + (i 6)ni formula, − − − − i>3 36i6d i>d+1 X X i 1 X = (i 6)ni + ( 6)ni + ini − 2 − 2 36i6d i>d+1 i>d+1

max 12, √6g + 6 { } 3 Cowen et al. [65] actually claim an upper bound on the defect of . We could not replicate this calculation.

15 X X i 1 = (i 6)ni + ( 6)ni + (α + 2β). − 2 − 2 36i6d i>d+1 G0 G0 A B α 6 2β Since is a triangulation, each face of has at most two edges incident with , and at X X i 6(g 2) (i 6)ni + ( 6)ni + α least one edge with endpoints− in> . It follows− that 2 −and 36i6d i>d+1 X X i = (2i 6)ni + ( 6)ni − 2 − 36i6d i>d+1 0 + d+1 6 B . > 2 − | |

The result follows. g max 9, 2 + √4g + 6 { } Woodall [228] improved Theorem 11 to show that` every graph with Euler genus is 3-choosable χ∆( g) = χ ( g) = 3. with defect . ThusE ∆ E

See Theorems 43 and 76 for generalisations of this result, and see [58, 59, 117, 190, 224, 235] for further results on defective colourings of graphs embeddableG on surfaces.(d1, . . . , d Onek)-colourable direction of interest is the followingV1,...,V deﬁnition,k V which(G) allows for results that bridge the gapG[Vi between] proper and defective colouringsdi [39, 40, 56, 58, 59, 69, 178]: a graph is if g > 0 (0, 0, 0, 9g 4) there is a partition of such that each induced subgraph− has maximum degree at most . For example, Choi and Esperet [59] proved the following analogue of the 4-colour theorem: every graph with Euler genus is -colourable.

g vw deg(v)+deg(w) max 2g +7, 19 NoteE that the light edge approach also proves 3-choosability, but with a weaker6 defect{ bound. In} particular, results of Ivanˇco [130] and Jendrol’ and Tuh´arskyg 6[1345 ] together imply that every graph g max 2g + 4, 16 in with minimum degreeE at least 3 has an edge with { } . (Better results are known for speciﬁc surfaces with , and all the bounds are tight.) g 3 max 2g + 2, 14 Thus every graph in with minimum degree at least 3 has a {-light edge.} (See Lemma 78 for a more general result with a slightly weaker bound.) Lemma5 then implies that every graph with Euler genus is -choosable with defect . See [57, 173, 233, 234, 235, 236] for more on defective choosability of graphs embedded on 3.5surfaces. 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 Lemma6. Kawarabayashi and Thomassen [148] also proved that every graph of bounded Euler genus is 5-colourable with bounded clustering. Dvoˇr´akand Norin [84] improved 5 to 4 via the following remarkably simple argument.

16 Lemma 13 . Let G be a graph, such that for some constants α, c > 0 and β (0, 1), ∈ E(G) < (k + 1 α) V (G) , ([84]) | | − | | and every subgraph of G with n vertices has a balanced separator of size at most cn1−β. Then G has a k-island of size at most & ' c(k + 1) 1/β 2 . α(2β 1) − α Proof. := k+1 X V (G) V (G) ⊆ | c | 1/β K1,...,Kp G X Ki 2( ) − d (2β −1) e Let e(Ki) . By Lemma 15 below, thereG exists of size at mostKi such thatX if are the components of , then each has at most e(Ki) E(G) < (k + 1 α) V (G) = (1 )(k + 1) V (G) (k + 1) V (G) X vertices. Let6 | be| the number− of| edges| of with− at least| one endpoint| 6 in | . Then\ | i X = (k + 1) V (Ki) . | | i

e(Ki) < (k+1) V (Ki) i Ki k+1 | | Ki e(Ki) < (k + 1) V (Ki) c(k|+1) 1/β| Hence for somek. Repeatedly remove verticesV (Ki) from 2( withβ at) least | | 6 d α(2 −1) e neighbours outside of . Doing so maintains the property that . Thus the ﬁnal set is non-empty. We obtain a -island of size at most .

The above proof depends on the following result by Edwards and McDiarmid [91]. Lipton and Lemma 14 . Fix c > 0 and β (0, 1). Let G be a graph with n vertices such that every Tarjan [160, 0161] implicitly proved∈ an analogous result for planar graphs.0 1−β subgraph G of G has a balanced separator of size at most c V (G ) . Then for all p > 1 c2β n | | there exists([91S ]) V (G) of size at most β β such that each component of G S has at ⊆ (2 −1)p − most p vertices.

Proof. S := G S X ∅ − 1−β p SX X c V (X) | | SXRunS the following algorithm. Initialise . While has a component with more than vertices,G let S be a balanced separator of withlevel size at mostX , and − Gadd S to . X G S − − Say a componentX of at the end of theS algorithmX has 0. SayS is a componentlevel of X X SX at some stage of the algorithm, but is not a− component of at the end of the algorithm. Then is separated by some set , whichp is then added to . Deﬁne the of i−1 top 1 plus the maximum level ofi a componenti of> 1 . 2 p By assumption,ti level 0 components have at mosti > 1vertices.X1 Each,...,X levelti 1 component has more than i vertices. Byi induction on , each level component has more than vertices. Let be the number of components at level . Say are the components at ti level . Since level components arei− pairwise1 X disjoint, ti2 p < V (Xj) n, | | 6 j=1

17 n ti < 2i−1p S X1,...,Xti

ti implying . The number of verticesX added1 to−β by separating is at most c V (Xj) , | | j=1

P n j V (Xj) 6 n V (Xj) = | | | | ti

ti 1−β β 1−i β which isX maximised,1 subject−β to n β 1−,β when n 1.−β Thus 2 c V (Xj) cti = ct n < c n = cn . | | 6 t i 2i−1p p j=1 i

1−i β β X 2 cn X 1−i β c 2 n S 6 c n = (2 ) = . Hence | | p pβ (2β 1)pβ i>1 i>1 −

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 Lemma 15 . Fix c > 0 and β (0, 1). Let G be a graph with n vertices such that every Edwards and Farr [88, 90]. ∈ subgraph G0 of G has a balanced separator of size at most c V (G0) 1−β. Then for > 0 there | | exists S V (G) of size at most V (G) such that each component of G S has at most ⊆ ([91]) | | − 2( c )1/β vertices. d (2β −1) e

Theorem 16 . Every graph G with Euler genus g is 4-choosable with clustering 1500(g+2). We now reach the main result of this section. Proof. V (G) L 4 G ([84]) | | V (G) = 0 V (G) 1 V (G) 6000g | | | | > | | 6 v G G v L 1500(g + 2) We proceed by induction on .− Let be a -list assignment for . The claim L(v) = 4 V (G v) < 6000g c L(v) 1500g is trivial| if | | . Now− assume| that . First∈ suppose that . G v v c G L 1500(g + 2) Let be any− vertex of . By induction, is -colourable with clustering . V (G) > 6000g α := 1999 k := 3 Since | and| , some2000 colour is assigned to at most E(G) < 3( V (G) + g) (k + 1 α) V (G) vertices| in | |. Colour| 6by . Thus− | is -coloured| with clustering . Now assumen that . Deﬁne gand . It follows from Euler’sO formula(√gn) p that . Various2 authors(2g + 3) [9n, 78, 93, 109] proved that 1 everyβ = 2 -vertex graphG with Euler3 genus at most has a balanced separator of size . Dujmovi´cet al. [81] proved a concrete upper bound of . Thus Lemma 13 with  !2 implies that4 2√2ghas+ 3 a -island of size at most 2 ·  6 747(2g + 3) < 747(2g + 3) + 1 < 1500(g + 2).  1999 (√2 1)  d e  2000 −  G S L 1500(g + 2) − G L 1500(g + 2) By induction, is -colourable with clustering . By the argument in Lemma6, is -colourable with clustering .

18 S(3, d)

` χ?( g) = χ ( g) = 4. Since is planar, Lemma2 and TheoremE 16? E imply

Open Problem 17. Does every graph in g have a 4-colouring with clustering O(√g)? It is still open to determine the best possibleE clustering function.

Open Problem 18. Are graphs with bounded Euler genus and bounded maximum degree The3-choosable following with question bounded also clustering? remains open; see Section 8.1 for relevant material.

E(G) < 2( V (G) + g) | | | | E(G) < 5 ( V (G) + g) | | 3 | | The above method extends for embedded graphs with large girth (since Theorem 19 . Let G be a graph with Euler genus g and girth k. If k > 4, then G is if the girth is at least 4, and if the girth is at least 5). 3-choosable with clustering O(g). If k > 5, then G is 2-choosable with clustering O(g). ([84])

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. 4 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. Lov´asz [164] proved the following result for defective Theorem 20 . For d > 0, every graph with maximum degree ∆ is k-colourable with defectcolouringsd, where of boundedk := degree∆ + 1graphs;. see [29, 43, 107, 156] for related results and extensions. b d+1 c ([164]) Proof. k G v d + 1 (deg(v) d 1)/(k 1) d v v Consider a -colouringb of− −that maximises− c 6 the number of bichromatic edges. Suppose that some vertex is adjacent to at least vertices of the same colour. Some other colour is assignedv to at most d neighbours of . Recolour this colour. The number of bichromatic edges increases by at least 1. This contradiction shows that every vertex is adjacent to at most vertices of the same colour.G = Kn k d n n n k d > k 1 k > d+1 = ∆(G)+1 d e ∆(G)+1 d e −∆(G)+1 Wed+1 now show that Theoremk 20 is bestn possible.k > Sayd+1 is -colourablek > d+1 with+ defect1 . Some monochromatic subgraph has at least vertices. Thus b andc . Moreover, if does not divide , then , implying , which exactly matches the bound in Theorem 20. ∆ ∆ D Clustered colourings of bounded degree graphs are more challenging than their defective cousins. Let be the class of graphs with maximum degree . First note the following straightforward lemma.

19 Lemma 21. For ∆ > d > 1, ∆ χ?( ∆) + 1 χ?( d). D 6 d + 1 D

∆ Proof. k1 := + 1 k2 := χ?( d) G ∆ b d+1 c D G k1 d Let d k2 and . Letc be a graph withd maximum degree . k1k2 G c χ?( ∆) k1k2 By Theorem 20, is -colourable with defect .D Each6 monochromatic subgraph, which has maximum degree , is -colourable with clustering (depending only on ). The product gives χ?( ∆) d = 1 a -colouring of with clustering . Thus D . ∆ ( ∆/2 + 1) b c We now show a series of improving upper bounds on . Theorem 20 with implies every graph with maximum degree is ∆ -colourable with defect 1, and thus with χ?( ∆) 6 + 1. clustering 2. Hence D 2 3 2

In particular, this shows that every graph with maximum degree is -colourable with clustering d = 4 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 χ?( ∆) 6 2 + 1 . then implies that D 5

∆ + 2 Alon et al. [11] pushed their methodχ further?( ∆) to prove that. D 6 3

(0, 3) ∆ ∈ (∆ + 2)/(3 ) c() ∆ d − e In fact, Alon et al. [11] showed that for every graph of maximum degree is -colourable with clustering (independent of ). For the sake of brevity, we present slightly weaker results with simpler proofs. The following Theorem 22. Let G be a graph with maximum degree ∆. If G has a k-colouring with defect 2, result was implicitly proved by Alon et al. [11]. then G has a (k + 1)-colouring with clustering 24∆.

Proof. G short 8∆ long X1,...,Xk V (G) k 2 Say anG induced[Xi] cycle or path in is if it has at most vertices, otherwise it is Y Y = a(8∆) + b a 1 b [1, 8∆] . Let be a partition of| | corresponding to the> given -colouring∈ with defect Y Y1,...,Ya,Ya+1 V (Yj) = 8∆ j [1, a] V (Ya+1) = b . Thus each is a collection of pairwise disjoint| | induced cycles∈ and paths.| Consider| such Yj Yj+1 j [1, a] Y a cycle or path that is long. Then for some and∈ . Partition into a set of paths Ya+1 , where for Y1 , andZ1,...,Zn . Here the last vertex in is adjacent to the first8∆ vertex in for , and if Yais+1 a 0 0 cycleG then the last vertex inG is adjacentV (Z1 to the firstZn) vertex inG . Let be the ∪ · · · ∪ collection∆ of all these inducedG paths with exactlyS = verticesv1, . . . , (which vn mightvi includeZi somei [n).] { } ∈ ∈ S, X1 S, . . . , Xk S (k + 1) 24∆ Let be the subgraph{ \ of induced\ } by . Thus has maximum degree at most . By Lemma 23 below, has a stable set with for each . We claim that defines a -colouring with clustering . By

20 S G Q G[Xi S] Q \ Y G[Xi Y Q 8∆ | | 6 construction,Y (isY1 a,...,Y stablea,Y seta+1 in) . Say is a componentY of . ThenY1,...,Yis containeda S Q Yj Yj+1 Yj+2 j [1, a + 1] Ya+2 Y1 in some component of ]. If∪ is∪ short, then ∈ as desired. Now assume that Ya+3 Y2 Q 24∆ is long. Let | | 6 be the above partition of . Since each of has a vertex in , is contained in for some , where means and means . Thus . Lemma 23. Let G be a graph with maximum degree at most ∆. Let V ,...,V be a partition The proof of Theorem 22 used the following well-known lemma about ‘independent1 n transversals’. of V (G), with Vi 8∆ for each i [n]. Then G has a stable set v1, . . . , vn with vi Vi | | > ∈ { } ∈ for each i [n]. ∈ Proof. X p X D 4pD 1 The proof uses the Lov´aszLocal LemmaX [96], which6 says that if is a set of events in a probabilityX space, such that each event in has probability at most and is mutually independent of all but Vi other= 8∆ eventsi in[n] , and i ,[ thenn] with positive probability no | | ∈ ∈ 1 event in occurs.vi Vi Vi ∈ 8∆ We mayvw assume thatv Vi wfor Vj .X Forvw each , independentlyv andw randomly ∈ ∈ 1 chooseX onevw vertex . Each vertexp := in64∆2is chosen with probabilityXvw at most . Consider an edge ,X wherexy x andVi Vj .y Let Vi Vbej the eventXvw that both and are chosen. 6∈ ∪ 2 6∈ ∪ 1 2 D := ∆( Vi + Vj ) = 16∆ 4pD = 4( 2 )(16∆ ) 1 Thus has probability| | at| most| . Observe that is mutually64∆ independent6 of every event where and . Thus Xisvw mutually independent of all v1, . . . , vn Xvw v1, . . . , vn but at most other events.{ Thus } . By the Lov´aszLocal Lemma, with positive probability, no event occurs. Hence there exist 8∆ such that no event occurs. That2∆ is, is the desired stable24∆ set. 6∆ The term in Lemmad = 2 23 was improved to by Haxell [121], which means the term in Theorem 22 can be improved to . ∆ Theorem 20 with and Theoremχ 22?( imply∆) + 2. D 6 3

(1)

χ?( ∆) = 2 Answering a question of Alon et al. [11], Haxell et al. [120] provedD that every graph with ∆ 2,..., 5 d = 5 maximum∈ { degree} 5 is 2-colourable with clustering less than 20000. For two colours, maximum degree 5 is best possible, by the Hex Lemma (Theorem9). Thus if and only if ∆ . Lemma 21 with χ?( then∆) implies2 + 1 . D 6 6

∆ = 5 ∆ = 8 Haxell et al. [120] proved that every graph with maximum degree 8 is 3-colourable with bounded ∆ + 1 clustering. Using their result for the χ?( ∆)case6 and the . case, Haxell et al. [120] proved that D 3

21 ∆ > 0 ∆ > ∆0 1 Moreover, Haxell et al. [120] provedχ for?( large∆) 6 one can ∆ do. slightly better: for some constants and for all , D 3 −

∆ Note that for both these results by Haxell et al. [120] the clustering bound is independent of . d = 9 It is open whether every graph with maximum degree 9 is 3-colourable with bounded cluster- ∆ ing [120]. If this is true, then Lemmaχ?( 21∆ )with3 would+ 1 imply, D 6 10

χ?( ∆) D which would be the best known upper bound on . Graphs with maximum degree 10 are not 3-colourable with bounded clustering [11], as shown by the following general lower bound, Theorem 24 . For every integer ∆ 2, which also implies a 3-colour lower bound for> graphs of maximum degree 6. ∆ + 6 ([11, 120]) χ?( ∆) . D > 4

Proof. ∆ ∆+6 = (∆−1)+6 ∆ 1 b 4 c b 4 c − ∆ ∆ k := ∆+6 b 4 c Ifc > is3 odd, then G ,( andk the1) result for implies thec result for ∆ − . Thus we may assume that is( even.2 + 1) Let .G Our0 goal is to show that forc every ∆+2 V (G0) = n G G0 G ∆ ( )n |integer| there is a graph that has no -colouring with clustering4 . Erd˝osand G (k 1) c Sachs [94] proved that there is a − -regular graph with girth greater than . Say X G . Let be the line graph of . Then is -regular with vertices. V (G) (∆ + 2)n Suppose on the contrary that Xis > | -colourable| = with> clusteringn. . For some colour class of , | | k 1 4k 4 0 − − X X n G0 c X c Thus corresponds to a set of at least edges in , which therefore contains a cycle of size greater than . Thus, contains a monochromatic component of size greater than , which is a contradiction. ∆ > 9

Open Problem 25. What is χ?( ∆)? The best known bounds are The following problem remains openD for . ∆ + 6 ∆ + 1 χ?( ∆) , 4 6 D 6 3 and for some > 0 and all ∆ at least some constant ∆0. 1 χ?( ∆) ∆. D 6 3 −

22 ` Open Problem 26. What is χ ( ∆)? The best known bounds are ? D ∆ + 6 ` χ ( ∆) ∆, 4 6 ? D 6 where the upper bound follows from known Brooks-type bounds on the choice number [67].

∆ χ?( ∆) ∆ D 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 with clustering linear in , one can obtain a slightly larger upper bound that is independent of . The∆ idea of the proof is by Theorem 27. Suppose that every graph with maximum degree ∆ is ( x + y)-colourable with Alonclustering et al.α [11∆]., for some constants α, x, y > 0. Then for every > 0 there is a number ∆ c = c(, α, x, y) such that every graph with maximum degree ∆ is ((1 + ) x + y)-colourable with clustering c.

2 2αd Proof. d := (xy 1) 1 c := max αd, α∆ 6 c d − e − 2αd{ } ∆ α∆ > c > d 6 2 G k ∆ Let d k :=andd+1 + 1 . If thenk the result holds d b c 0 by assumption.( x Now+ y) assume that αdimplyingG k . By Theorem 20, is - colourableαd 6 c with defect , where . By assumption, each of these monochromatic subgraphs is -colourable ∆ with clusteringd ∆. Thus∆(xy is 1)-colourabled with clustering k0 := + 1 + y = + − + + y. , where d + 1 x x x(d + 1) x

xy 1 (d + 1) d ∆ − 6 2 6 2 ∆ ∆ ∆ (1 + )∆ Since andk0 + , + + y = + y. 6 x 2x 2x x

See [26, 27, 89, 159] for more results about clustered colourings of graphs with given maximum degree. 5 Maximum Average Degree

+ mad(G) G m R m ∈ A G mad(G) m χ( m) m + 1 6 A 6 b c χ∆( m) Recall that is the maximum averageA degree of a subgraph of . For , let be the class of graphs with + . A greedy algorithm shows that . Theorem 28 . For m R , Havet and Sereni [119] determined∈ as follows. ` jmk χ ( m) = χ∆( m) = + 1. ([119]) ∆ A A 2

kd G mad(G) < k + k+d Wek prove Theorem 28 below.d 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 with is -choosable with defect .

23 + + 1 1 2 Lemma 29 . For all m R and k, d Z such that + , every graph ∈ ∈ k d 6 m G with mad(G) 6 m is k-choosable with defect d. (FrédéricHavet) Proof. G V (G) + E(G) r, k, d | | | | G k d Let be a counterexample with minimum (with fixed). By A := v V (G) : deg(v) m , Lemma5, has minimum degree{ ∈and has no -light6 edge.} Let B := v V (G): m < deg(v) d { ∈ 6 } C := v V (G): d < deg(v) . { ∈ } and 1 1 2 m m m m + 6 1 6 1 γ 1 6 γ 6 1 k d m k − − d k − P − d v deg(v) v deg(v) = 2 E(G) vw v A w γ Since| | we have . Let be a real number with ∈ . v w C G d v A Associate with each vertex∈ an initial charge of . The total charge is ∈ (1 + γ) deg(v) (1 + γ)k m v B . Redistribute> the charge> as follows: for each edge with∈ , let send v C (1 γ) deg(v) > (1 γ)d m C = charge to∈ . Note that since− has no -light− edge.> Now, each vertex 6 has∅ G d m V (G) charge . The charge for each vertex is unchanged.| Each| vertex has charge at least m . We may assume that , as otherwise is 1-colourable with defect . Thus the total charge is greater2 than , Proof of Theorem 28. k := m + 1 k > m d := m + m implying the average degree is greaterb 2 c than , which is2 a contradiction.d 4k−2m 2 e 4k > 2m Let . Thus m2 . Let , which is well-defined since . Then 4d 2m , − > k m − 2

m m2 m(2d m) + m2 md 1 k > + = − = = . implying 2 4d 2m 4d 2m 2d m 2 1 − − − m − d 2d > m 1 2 1 6 Since , k m − d 1 1 2 m k + d 6 m G mad(G) 6 m ( 2 + 1) ` m b c d χ∆( m) χ ( m) + 1 A 6 ∆ A 6 2 and . By Lemma h29:=, everym graph with is S(h, d)-choosable with b 2 c defect . Thus 2h 6 m h. d χ∆( m) h + 1 ForA the lower> bound, let . Then the standard example has maximum average degree less than and is not -colourable with defect by Lemma1. Thus , as required.

See [37, 39, 40, 41, 42, 44, 45, 46, 47, 79, 149, 150] for more results about defective colourings of graphs with given maximum average degree. m Open Problem 30. What is χ?( m)? The best known bounds are + 1 χ?( m) Little is known about clustered colouringsA of graphs with given maximumb 2 c average6 degree.A 6 m + 1. b c

24 G k G k Maximum averagek degree is closely related(k to+ 1) degeneracy. Recall that a graph is - degenerateS(k, d) k if every subgraph of has minimum degree at most . A greedy algorithm shows that every -degeneratek graph is properly -colourable. Since the standard example is -degenerate, this bound cannot be improved even for defective colourings. Thus fork + the 1 class of -degenerate graphs, the defective chromatic number, defective choice number, clustered chromatic number, clustered choice number, and (proper) chromatic number all equal . 6 Excluding a Subgraph

H H H For every graph , the class of graphs with no Hsubgraph has bounded chromatic number if V (H) H and only if | is a forest.| The same result holds for defective chromatic number and clustered chromatic number. To see this, observe that if contains a cycle, then graphs with girth greater than contain no subgraph, and by the classical result ofF Erd˝os [95] there aren graphs with arbitrarily large girth and chromatic number. By Proposition3, the defective n 1 F F and clustered chromatic− numbers are also arbitrarily large. Conversely, say is a forest with (n 2) (n 1) Kn−1 vertices.− A well known greedy embedding− procedure shows that every graph with minimum F (n 1) F degree at least contains− as a subgraph. That is, every graph containing no subgraph n 1 is -degenerate, and is thus -colourable.− This bound is tight since contains no subgraph and is -chromatic. In short, for the class of graphs containing no subgraph, the chromatic number equals . The following result by Ossona de Mendez Theorem 31 . Let T be a tree with n 2 vertices and radius r 1. Then every graph et al. [185] shows that defective colourings exhibit> qualitatively diﬀerent> behaviour. containing no T subgraph is r-colourable with defect n 2. − ([185]) Proof. i = 1, 2, . . . , r 1 Vi v V (G) (V1 Vi−1) − ∈ \ ∪ · · · ∪ n 2 V (G) (V1 Vi−1) Vr := V (G) (V1 Vr−1) − \ ∪ · · · ∪ \ ∪ · · · ∪ V1 Vr V (G) i [1, r 1] G[Vi] For∪ · · · ∪ , let be the set∈ of vertices− that have at most n neighbours2 in G[Vr] . Let n 1 . − − 0 Then T is a partition of G . For , byv construction,T has maximumv degreeG x at most , as desired.T Suppose that T has maximumG degree at least . We 0 now showx T that isx a subgraph of , wherex each vertex of is mappedn 1 toG a[Vr vertex] of − 0 . Let be the centre of . Mapv the verticesj of[1, r]to verticesx inT in order of their distancev ∈ 0 0 fromVr−j+1 in , whereVr j =is 0 mappedv = tox a vertex with degreev = atx least inn 1 . The key ∪ · · · ∪ 0 − invariantVr j is that[1, r each1] vertex at distance v from n in1 is mapped toV ar− vertexj Vinr ∈ 0 − − ∪ · · · ∪ v Vr−j Vr−j Vr . If then and by assumption, has at least neighbours∪ · · · ∪ in . If then byv construction,T has at least G neighbours in G[Vr] n 2 G r n 2 (otherwise would be in ). Thus− there are always unmapped vertices in − to choose as the children of . Hence is a subgraph of . This contradiction shows that has maximum degree at most , and is -colourable with defect T. r S(r 1, d) T − The number of colours in Theorem 31 is best possible forT the complete binaryr tree of radius . Since contains no subgraph, Lemma1 and Theorem 31 imply that the defective chromatic number of the class of graphs containing no subgraph equals .

25 ∗ K7,13

Open Problem 32. For a tree T ,Figure what is 6: the The clustered graph chromatic. number of the class of graphs with no T subgraph?

χ?( ∆) D ∆ K1,∆+1 The results in Section4 on are relevant to this question since a graph has maximum degree at most if and only if it excludes as a subgraph. Section 8.5 studies colourings of graphs that exclude a given path subgraph. 7 Excluding a Shallow Minor

∗ 7.1 Excluding Ks,t

∗ s, t > 1 Ks,t s Ks,t 2 Ass illustrated inK Figures,t 6, for integers , let be the bipartite graph obtained from ∗ by addingKs,t new vertices, each adjacent to a distinct pair of vertices in the colour class (G) H of vertices in . Ossona∇ de Mendez et al. [185] studied defective colourings for graphs excluding asH a subgraph, whereG the defect bound depends on the density of shallow Theoremtopological 33 minors.. LetEvery graphbeG thewith maximum no K∗ subgraph average degree is s-choosable of a graph with defectsuch` thats + the 1, s,t − 1-subdivisionwhere δ = mad( of G) isand a subgraph= (G of) and. ∇ ∇ ([185])  b∇c 1  (δ s) (t 1) + + δ if s > 2, b − s−1 − 2 ∇ c ` := `(s, t, δ, ) := 1 (δ 2) t + δ if s = 2, ∇ b 2 − ∇ c t 1 if s = 1. − Proof. G s s 6 δ G ` s = 1 G Assume for contradiction that has minimum degree at least (thus ) and that contains no -light edge. The case is simple: Since has minimum degree at least 1,

26 G ` ∆(G) t 1 ` = t 1 6 − − s > 2 hasA at least one edge, whichG is -light since ` Band:= V (G) A. Nowa assume:= A that \ | | b := .B G δ δ ` a > 0 | | 6 Let be the set of verticesA in of degree at most . Let . Let and . Since has a vertexG of degreeδ at most and , we deduce that . Note that no two vertices in are adjacent. (` + 1)b + sa 2 E(G) δ(a + b). Since the average degree of is at most 6, | | 6

(` + 1 δ)b (δ s)a. − 6 − That is, G0 G E(G[B]) w A − ∈ (2) x y B w x y Let be the graph obtained fromw xw ywby greedily ﬁndingxw a vertex havingw a pair of non-adjacent neighbours , in and replacing by an edge joining and (by deletingA0 := allV ( edgesG0) B incidenta0 with:= A0 except , 1 and contractingG0[B]), until no such vertexG \ | | exists. G0[B] G0[B] a a0 ∇ − Let and . Clearly the -subdivision of is a subgraph of . So a a0 1 b. every subgraph of has average degree− at6 most2 ∇ . Since contains at least edges, M s G0[B] G0[B] b∇c (3) Let be the number of cliques of sizeM in b∇c. Sinceb is -degenerate, 6 s 1 − s = 2

M 1 b. (See [180, p. 25] or [222]). If , then the6 following2 ∇ better inequality holds: v A0 v G0 ∈ v B s a0 > M(t 1) t 0 − For each vertexA , since s was not contracted in theB creationt of , the set of neighbourss B A A0 K∗ G of in is a clique of size at least .− Thus if s,t , then there are at least vertices in sharing at least common neighbours in . These vertices and their common a0 M(t 1). neighbours in with the vertices in 6 form− a subgraph of , contradicting our assumption. Thus, M 1 (4) ` + 1 6 (δ s) (t 1) + 2 + δ, By (2), (3) and (4), − b − ∇ ` G s 1 G − ` contradicting the deﬁnition of . Thus has minimum degree at most or contains an -light edge. The theorem now follows from Lemma5.

Theorem 33, in conjunction with the standard example, determines the defective chromatic number for several graph classes of interest; see Sections 7.2 to 7.6. Moreover, Theorem 33 determines the defective choice number for a very broad class of graphs—complete bipartite subgraphs are the key.

27 Theorem 34 . Let be a subgraph-closed class of graphs with mad( ) and ( ) G G ∇ G bounded (which holds if is minor-closed). Then χ` ( ) equals the minimum integer s such G ∆ G that Ks,t for some integer t. 6∈ G([84, 185])

Proof. Ks,t s, t 1 s 6∈ G > G s t mad( ) ( ) ` G ∇ G If s :=forχ some∆( ) , then byd Theorem 33, every graphs in is -choosable withd G s G t = (ds + 1)s Ks,t defect bounded by a function of , , and . ThisG proves the claimed upper bound. Conversely, let . Then for some , every graph in is -choosable with defect . Lemma 35. For s 1 and d 0, if t = (ds + 1)ss, then the complete bipartite graph K is By Lemma 35 below,> if > then is not in . s,t not s-choosable with defect d.

Proof. A B Ks,t A = s B = t A = | | | | v1, . . . , vs X1,...,Xs s L { } s Ks,t L(vi) := Xi vi A Let and be the colour classes of with and∈ . Say (c1, . . . , cs) . Let ci Xi be pairwisei [s] disjointL(x) sets := ofc1 colours,, . . . , cs eachds of+ size 1 . Let xbe theB ∈ ∈s { } B = (ds + 1)s L Ks,t vi following -list assignment| | for . Let for each vertex . For each vector ci L(vi) Xi Xj = ci = cj i, j [s] ∈with for each∩ ∅ , let 6 for ∈ vertices in . ds + 1 x B L(x) = c1, . . . , cs d + 1 This is possible since ∈ . Consider{ an -colouring} of . Say each vertex is coloured . Since ci , wevi have for distinct . Byd construction,+ 1 thereKs,t are L vertices withd Ks,t . Ats least of these verticesd are assigned the same colour, say . Thus has monochromatic degree at least . Hence is not -colourable with defect . Therefore is not -choosable with defect . ` χ∆( ) = 2 K2,3 K1,n n ` O χ ( ) = 3 K3,3 K2,n n Note∆ P that Theorem 34 generalises several previous results. For example, Theorem 34 says that since is not outerplanar, but is outerplanar for all . Similarly, since is not planar, but` is planar for all . More generally, Theorem 34 Corollary 36. For every graph H, χ ( H ) equals the minimum integer s such that H is a immediately implies: ∆ M minor of Ks,t for some integer t.

Theorem 34 also determines the defective choice number for graphs excluding a ﬁxed immersion (see7.2 Theorem Linklessly 52). Embeddable Graphs

3 linklessly embeddable R L L A graph is if it has an embedding in with no two linked cycles [199K6, 201K4,4]. Let be the class of linklessly embeddable graphs. Then is a minor-closed classK6 whose minimal excluded5 minors are the so-called8 Petersen family [200], whichK includes6 , minus an edge, and the Petersen graph.6 Since linklessly embeddable graphs exclude minors, they are -colourable [198] and -choosableχ∆( ) [23]. It is open whetherapex -minor-free graphs or linklessly embeddable graphs are -choosableL [23]. OssonaS(2, d) de MendezS(3 et, d al.) [185] determined as follows. A graph is ifχ deleting∆( ) > at4 ` L χ ( ) 4 K4,4 most one vertex makes it planar. Every∆ L apex> graph is linklessly embeddable [199]. Since6∈ L is planar, is apex, and thus linklessly embeddable. By Lemma1, . Note that the weaker lower bound, , follows from Theorem 34 since .

28 K6

Mader’s theoremK4 [168,4 ] for -minor-free graphs implies that linklessly embeddable graphs Theoremhave average 37 degree. Every less than linklessly 8 and minimum embeddable degree graph at mostis 4-choosable 7. Since linklessly with defect embeddable440, and graphs exclude minors, Theorem 33 implies the upper bound in the following theorem. ` χ∆( ) = χ ( ) = 4. ([185]) L ∆ L ` χ ( ) = 4 K4,4 K3,n ∆ L n Note that Theorem 34 also implies since is not linkless, but is linkless for all . Theorem 38. Every linklessly embeddable graph is 5-choosable with clustering 62948, and We have the following result for clustered colourings of linklessly embeddable graphs. ` χ?( ) = χ ( ) = 5. L ? L Proof. K6 S(3, c) S(4, c) The upper bound follows from Theorem 58 since every linkless graph contains no -minor. Since is planar, is apex, and thus linklessly embeddable. The lower 7.3bound Knotlessly then follows Embeddable from Lemma2. Graphs

3 knotlessly embeddable R K K7 K3,3,1,1 AK graph is if it has an embedding in in which every cycle forms a trivial knot; see [191] for a survey. Let be the class of knotlessly embeddable graphs. Then is a minor-closed class whose minimal excluded minors include K7and [63, 1028]. More than 260 minimal excluded minors are knownK7 [112], but the full list of minimal excluded minors is unknown. Since knotlessly9 embeddable graphs10 exclude minors, they are - colourableK7 [7, 132]. Mader [168] proved that -minor-free graphs6 have average7 degree less than 10, which implies they are -degenerate and thus -choosable. It is open whether -minor-free graphs or knotlessly embeddable2-apex graphs are -colourable or -choosable [23]. Ossona de Mendez et al. [185] determined the defective chromatic number2 of knotlessly embeddable graphs as follows. A graph is S(4, d) if2 deletingS(4 at, d most) two vertices makes it ∗ χ∆( ) 5 K3,3,1,1 K planar. Blain et al. [31] and Ozawa and TsutsumiK > [186] proved that every -apex5,3 graph is K∗ mad( ) < 10 knotlessly embeddable. Since every block5,3 of is -apex, K is knotlessly embeddable, as illustrated in Figure7. By Lemma1, . Since is a minor of , knotlessly Theoremembeddable 39 graphs. doEvery not contain knotlessly a embeddablesubgraph. graph Since is 5-choosable, with Theorem defect 33660 implies, and the following result. ` χ∆( ) = χ ( ) = 5. ([185]) K ∆ K

Theorem 40. Every knotlessly embeddable graph is 6-choosable with clustering 99958, and We have the following result for clustered colourings of knotlessly embeddable graphs. ` χ?( ) = χ ( ) = 6. K ? K

29 b b b

S2,d S2,d S2,d S2,d planar planar planar planar

S(4, d)

Proof. Figure 7: is knotlessly embeddable. K7 S(5, d) 2 S(5, d) The upper bound follows from Theorem 58 since every knotless graph contains no -minor. Since every block of is -apex, is knotlessly embeddable. The lower 7.4bound Colin then follows de Verdi`ereParameter from Lemma2.

µ(G) µ(G) 6 1 G The Colin de Verdièreparameterµ(G) 6 2 is an importantG graph invariantµ(G introduced) 6 3 by Colin deG Verdière [61, 62µ](;G see) 6 [1274 , 202] for surveys.G It is known that if and only if is a disjoint union of paths, χif(G and) 6 onlyµ(G if) + 1is outerplanar, if and only if is planar, and if and only if is linklessly embeddable. A famous conjecture of k := G : µ(G) k Colin de Verdière [61] states that V { (which6 implies} the 4-colour theorem, and is implied by Hadwiger’s Conjecture). Ossona de Mendez et al. [185] showed that for defective Theorem 41 . For k 1, colourings one fewer colour> suffices. Let . ` χ∆( k) = χ ( k) = k. ([185]) V ∆ V

Proof. k µ(Ks,t) = s+1 V t max s, 3 µ(G) k G K mad(G) > { } 6 k,max(k,3) 6 2 (G) 6 isO a(k minor-closed√log k) class [61, 62]. vans der= k Holst ett = al. max[127] provedk, 3 that G k ∇ O(k log log k) ` { } for . Thus,2 if thenχ∆( kcontains) 6 χ∆( nok) 6 k minor, and V V 0 0 . Theoremµ(G) 33 with and µ(G ) implies that is G- G G v µ(G) = µ(G v) + 1 choosable with defect . Thus − . For the lower bound, van der Holst et al.S(k[1271],proved d) thatk d >equals2 the maximum of , takenχ∆( overk) > thek components − ` V V χ ( k) k K k of , and if has a dominant∆ V > vertex , then . Itk, followsmax{k,3} that6∈ V the standard example is in for . Lemma1 then implies that . Note that the weaker lower bound, , follows from Theorem 34 since k = 4 .

Theorem 41 generalises Theorem 37 which corresponds to the case . Clustered colourings provide a natural approach to the conjecture of Colin de Verdi`ere [61] mentioned above.

30 Conjecture 42. χ?( k) = k + 1. V

k 7 k 6 V Kk+2 Note that Conjecture 42 with is implied by Theorem 58 below since graphs in contain no7.5 Crossingsminor.

g 0 k 0 k G > > Eg H G g k E(H) This section considers defective colourings of graphs with linear crossing| | number. For an integer and real number , let be the class of graphs such that every subgraph of has a drawing on a surface of Euler2k genus with at most crossings. (In a 0 0 g drawing, we assume thatE no three edges cross at a common point.) This says that the averageEg number of crossings per edge is at most (for everyk subgraph). Of course, a graph isk-planar planar (k/2) graphsif and only if it is in , and a graph has Euler genus at most if and only ifk it is in . E0 Graphs that can be drawnO(√k in) the plane withk at most crossings perO(√ edge,k) so called 2 , are examplesKn of graphsO(n ) in . Pach and T´oth(g,[187 k)-planar] proved that -planar graphs (k/2) have average degree g . It follows thatk -planar graphs are -colourable,g which is E best possible since is -planar. Say a graph is if it can be drawnn n on a2 × × surface with Euler genus with at most crossings per edge. Such graphsi are in . Also note thati even 1-planar graphs do not form a minor-closed class. For example,Kn the grid graph is 1-planar, as illustrated in Figure8, but contracting the -th row in the front grid with the -column in the back grid (plus the edge joining them) creates as a minor.

n n 2 × × k Figure 8: The grid graph is 1-planar. g p E K3,3k(2g+3)(2g+2)+2 mad 6 O( (k + 1)(g + 1)) p Ossona deO Mendez( (k + et 1)( al.g[+185 1))] showed that Theorem 33 is applicable for graphs in . In particu- ∇ 6 lar, such graphs containK3,3 no subgraph and have and . The ﬁrst claim here is proved using a standard technique of counting copies of . The second claim is proved using the crossing lemma. The next

31 g = k = 0 k = 0 theorem follows. It is a substantial generalisation of Theorem8 (the case) and Theorem 43 . For every integer g > 0 and real number k > 0, Theorem 11 (the case), with a worse defect bound. k ` k χ∆( ) = χ ( ) = 3. ([185]) Eg ∆ Eg In particular, every graph in k is 3-choosable with defect O((k + 1)5/2(g + 1)7/2). Eg 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 k? Eg

PropositionIt may be that 45. theEvery answer(g, kto)-planar all these graph questionsG is 12-colourable is 4. Here we with prove clustering the answerO(( isk + at 1) most7/2(g 12+ 1)for9/ the2). ﬁrst two questions.

Proof. G O((k + 1)5/2(g + 1)7/2) G G1 G2 G3 G 5/2 7/2 S O((k +By 1) Theorem(g + 1) 43,) is 3-colourableV (G) = withi V (G defecti) . That is, (containsg, k) three induced subgraphs , andO((g + 1)(of k +each 1)) with maximum degree at mostGi ,V where1,...,Vn Gi .G Dujmovi´ceti[Vj] al. [80] Oproved((g + that1)(k + every 1)) 7/2 9/2 -planar graphGi[V hasj] layered treewidth O((k + 1). Apply(g + this 1) result) to eachGi . Thus, for some layeringj of , each layer j has treewidth . Vi 3 2 2 = 12 By Theorem 55, is 2-colourable with clustering × × . Within , use two colours for odd and two distinct colours for even . Each monochromatic component is 7.6contained Stack in some and Queue. The Layouts total number of colours is .

k-stack layout G v1, . . . , vn V (G) E1,...,Ek E(G) Ei v1, . . . , vn i [1, k] vavb vcvd cross a < c < b < d k-stack graph A∈ of a graph consists of a linear ordering of and a partition k of suchstack-number that no two edges in Gcross with respect to k for eachG k . Here edges and if book embeddings. A graph is a if it book-thickness fixed outer-thickness page-number k k has a -stack layout. The of a graph is the minimumS integer for which is χ( k) 2k, 2k + 1, 2k + 2 a -stack graph. Stack layouts are also called S ∈ { , and stack-number} is also called , and k + 1 . Let be the class of -stack Theoremgraphs. Dujmovićand 46 . The Wood class[83 of ]kshowed-stack graphs that has defective chromatic number. For and defective defective choicecolourings, number Ossona equal de to Mendezk + 1. et In al. particular, [185] showed every thatk-stack graphcolours is ( suffice.k + 1)-choosable with O(k log k) defect 2 ([185. ])

Proof. S(k, d) k d Kk+1,k(k+1)+1 k The lower bound follows fromk Lemma1 sinceG an easy inductive argument shows2k that+ 2 2 is a -stack( graphG) 6 for20k all . For the upper bound, is not a -stack ∇ 2 O(k log k) graphs = k + [30 1]; seet = alsok(k [70+]. 1) + Every 1 -stack`(k + graph 1, k(k + 1)has + 1average, 2k + 2, degree40k ) 6 less2 than (see [30, 83]) and (see [92, 181]). The result follows from Theorem 33 with and , since .

32 k-queue layout G v1, . . . , vn V (G) E1,...,Ek E(G) Ei v1, . . . , vn i [1, k] vavb vcvd nested a < c < d < b queue-number A ∈ of a graph consists of a linear ordering of and a partition G of such that nok two edges inG arek nested with respect to k-queuefor graph k k k each . Here edges and Qare if . The of χ( k) a graph is the minimum integer forQ which has a -queue layout. A graph is a if it2 hask + 1 a -queue4k layout. Let be the class of -queue graphs. Dujmovi´cand Wood [83] state that determining is an open problem, and showed lower and upper bounds of and . Ossona de Mendez et al. [185] proved the following partial answer Theorem 47 . Every k-queue graph is (2k + 1)-choosable with defect 2O(k log k). to this question.

Proof. ([185]) K2k+1,2k+1 k k G mad(G) < 4k (G) < (2k + 2)2 ∇ Heath and Rosenberg [122] proved thats = 2k + 1 ist = not 2k a+ 1-queue` graph.(2k + 1 Every, 2k + 2 O(k log k) 1,-queue4k, 2(2k graph+ 2) ) 6has2 (see [83, 122, 188]) and (see [181]). The result then follows from Theorem 33 with and , since . S(k, n) k k + 1 6 χ∆( k) 2k + 1 Q 6 χ∆( k) An easyQ inductive construction shows that has a -queue layout. Thus by Lemma1 and Theoremk 47. Itk remains an open problem to determine S(2, c) . S(k, c) (k 1) k 2 k Now consider clustered colourings− of -stack and -queue> graphs. The standard example is outerplanar, and thus has a 1-stack layout.k An easy inductive(2k + 1) construction then k (4k 1) shows that has a − -stack layout (for ) and a -queue layout. The best known upper bounds come from degeneracy:` every `-stack graph is -degenerate and k + 2 χ?( k) χ ( k) χ ( k) 2k + 2 every -queue graph is 6 -degenerateS 6 ? S (see6 [83]).S Thus6 ` ` k + 1 6 χ?( k) 6 χ?( k) 6 χ ( k) 6 4k. Q Q Q and

k k Closing the gap in these bounds is interesting because the existing methods say nothing about clustered colourings of -stack or -queue graphs. For example, Lemma 13 is not applicable since 3-stack and 2-queue graphs do not have sublinear balanced separators. Indeed, Dujmovi´c et al. [82] constructed (cubic bipartite) 3-stack expander graphs and (cubic bipartite) 2-queue 7.7expander Excluded graphs. Immersions

G H immersion H G This sectionH considers colourings of graphs excluding a ﬁxed immersion.G A graph containsvw H a graph as an G if the vertices of can be mappedv to distinctw vertices of G, and the edges of H can be mappedbranch to pairwise vertex edge-disjointG paths in , suchH that eachstrong edge immersionof isG mapped toH a path in whose endpoints are the imagesvw of Hand . The image in of each vertexG in is called a vw . A graph containst a graph as a Kift 0 I Kt contains as an immersionIt such that for each edge of , no internal vertex of the path in corresponding to is a branch vertex. Let be the class of graphs not containing as an immersion. Let be the class of graphs not containing as a strong immersion.

33 Kt (t 1) − Lescure and Meyniel [157] and Abu-Khzam and Langston [1] independently conjectured that every -immersion-free graph is properly -colourable. Often motivated by this question,Kt structural and colouring properties(3 of.54 graphst + 3) excluding a fixed immersion have recently been widely studied. The best upper bound, due to Gauthier et al. [106], says thatKt every - immersion-free graph is properly -colourable Van den Heuvel and Wood [124] proved that the defective chromatic number of -immersion- free graphs equals 2. The proof, presented below, is based on the following structure theorem of DeVos et al. [74]. Almost the same result canxy be concludedT fromT a(xy structure) T ( theoremyx) by T xy x T (xy) y T (yx) T G Wollan [219]. Since− Lemma 48 is not proved explicitly in [74] we include the full proof, which T -partition G (Tx V (G): x V (T )) V (G) T relies on the following definitions. For⊆ each edge∈ of a tree , let and be the componentsTx of , wherebag is in and is in . For a tree ande = xy graphE(T, a) S S ∈ G(T, xy)of = isz∈V a(T partition(xy)) Tz G(T, yx) = z∈V (T (yxof)) Tz indexedE(G, by T, the e) nodes of . Each set Gis called aG(T,. xy Note) thatG(T, a bag yx) may beadhesion empty. ForT each edge , e T E(G, T, e) x T torso of x let |and | , and let be the set of edgesT in between and G. The Gof(T, a yx-partition) is the maximum, taken over allxy edges of x, of . For each node of , the (with respectT V (G) V (T ) T G Tx = x V (G) to a -partition)⊆ is the graph obtained from by identifying into{ } a ∩ single vertex for x V (T ) each edge∈ incident to , deleting resulting parallel edges and loops. Note that a tree for which implicitly defines a -partition of with for each Lemma 48 . For every graph G that does not contain K as an immersion, there is a tree node . t T and a T -partition of G with adhesion less than (t 1)2, such that each bag has at most − t 1 vertices. − ([74]) Proof. G F V (G) v, w V (G) G ∈ Gomoryv andw Hu [113] proved that for every graph there is a tree with vertex set such that for all distinct verticesζ(e) := min E(G, F,, the e) , size of the smallest edge-cut in e | | separating and equals e vw F S e E(F ) ζ(e) < (t 1)2 ∈ − where the minimum is taken overX all edgesF S on the -patht in . Let bex, thev2, v set3, . .of . , edges vt − with X G0 . G w 0 Supposet that1 some component of w vhasi at leasti [2vertices., t] Let G be 2 − ∈ (distinctt 1) vertices in . Letxw be the multigraph obtained fromR byV adding(G) a newx vertexR and − 2 0 ⊆ 0 ∈ 0 adding parallel(t edges1) between and R for eachV (G ) R . WeG claim thatw V (containsG ) R 2 − − 0 ∈ − edge-disjoint(t 1) -paths. Considervi aw set of verticesV (G ) R with i [2,, such t] that − 2 − ∈ thereG are less than edges between(t 1)and in .x Sincevi 2 − ζ(e) < (t 1) e xvi F e X has degree− , some neighbour of is also in (where ). Thus ζ(e) (t 1)2 G0 in , there is> an edge-cut− with less than edges separating and , meaning that (t 1)2 xw G (t 1)2 x − for some edge on the -path in .− But every such edge is in , which t 1 vi i [2, t] xvi Pi,j j [2, t] implies − . This contradiction∈ shows (by Menger’s Theorem) that contains∈ j = i Pi,jPj,i vivj Pi,i vix 6 edge-disjoint -paths. Hence contains edge-disjoint paths starting at , exactly of which end at for each . Label these -paths as , for . For , the combined path is a -path, while each is a path. Since all

34 G Kt F S t 1 − − theseT paths are edge-disjoint,F contains a -immersion. This contradictionF S shows that every − component of x Thas atTx most vertices. x Tx : x V (T ) ∈ Let be obtained from by contracting each connected component of into a single vertex. For each node of , let be the set of vertices contracted into . Then is Lemma 49 . If a graph G has a T -partition with V (G) V (T ) and adhesion at most k, the desired partition. ⊆ then G is 2-colourable with defect k. ([124]) Proof. V (G) + E(G) V (G) 2 | | | | | | 6 V (G) 3 v G large deg (v) k + 1 v small | | > G > deg (v) 1 v G v k G We6 proceed by induction on .− The result is trivial if . Now assumev . Call a vertex of Gif ; otherwisek is . If G for some vertex , then by2 induction, is 2-colourablev w with defect ; assign G vw k G a colour distinct− from its neighbour. Now is 2-coloured with defect . Now assume thatk has minimum degree at least . If two small verticesG and are adjacent, then by induction, G is 2-colourablek with defect ,G which is also a 2-colouring ofX with defect . Now assumeT that the small vertices are a stableu set. If has noX large vertices,u then everyv u X u V (X) = 1 Y := V (T (uv)) u 2-colouring of has defect . Now assume that has| some| large vertex. Let be the\{ union} of all paths in Ywhose endpoints are large. Let be a leaf in . Thus Gis large. Let be the neighbour of in , or anyY neighbour of if .T Let(vu) u . deg (u) Y T (vu) E(G, T, uv) Y + deg (u) Y k + 1 EveryG vertex− | | in is small. Since no two| small vertices| > | are| adjacentG and− | | >has minimum degree at least 2, every vertex in has at least one neighbour in . Also, has at least neighbours in . Thus , which 3 is a contradiction. O(k ) s = 2

Note that Lemma 49 with a defect bound can be concluded from Theorem 33 with .

The following result by Van den Heuvel and Wood [124] is the ﬁrst main contribution3 of this Theorem 50 . Every graph G t is 2-colourable with defect (t 1) , and section. ∈ I −

χ∆( t) = 2. ([124]) I Proof. T T G (t 1)2 1 − − t 1 Q V (T ) − xy E(Q) G Tx Ty Q By∈ Lemma 48, there is a tree and a -partition of with adhesion at most , t 1 G Q 2 such that each bag has at− most vertices. Let be the graph with vertex set , (t 1)2 1 v G x Q where − − whenever there is an edge of between and . Any one edge of v Tx t 1 G G 2 corresponds∈ to at most −edges in . By Lemma 49, the graph is -colourable with (t 1) (t 1)2 1 + (t 2) < (t 1)3 defect− · − −. Assign− to each vertex− in the colour assigned to the vertex in with . Since at most vertices of are in each bag, is -coloured with defect .

Van den Heuvel and Wood [124] proved the following result for excluded strong immersions. Theorem 51 . Every graph G 0 is 2-colourable with defect at most some function The omitted proof is based on a more∈ involved It structure theorem of Dvoˇr´akand Wollan [85]. d(t), and thus 0 χ∆( ) = 2. ([124]) It

35 While only two colours suffice for defective colourings of graphs excluding a fixed immersion, Theorem 52. For all t > 2, significantly more colours are needed` for defective` 0 list colouring. χ ( t) = χ ( ) = t 1. ∆ I ∆ It − 0 Proof. mad( t) 6 O(t) H I 0 G G H ( t) 6 O(t) ` 0 ∇ I χ ( ) s Ks,n Kt DeVos et al.∆[73It] proved that . If the 1-subdivision of a graph is a n K t−1 Kt subgraph of , then contains as a strong immersion,t−1,( 2 )+1 implying . Theorem 34 thenK impliest−2,n that equalsKt the minimum integer such thatn contains as a strong immersion for some . It is easily seen that contains as a strong immersion, but does not contain as a weak immersion for all . The result follows. Kt t 2 Kt It is an open problem to determine the clustered chromatic− number and clustered choice number of graphs excluding a (strong or weak) immersion. We have the following lower 0 t + 4 bounds. Since every graphχ with?( t) maximum> χ?( t) degree> χ?( att− most2) > contains. no immersion, by Theorem 24, I I D 4

` 0 ` ` χ ( ) χ ( t) χ ( t) = t 1. ? It > ? I > ∆ I − By Theorem 52, 8 Minor-Closed Classes

Kt (t 1) − This section studies defective and clustered colourings of graphs in a minor-closed class. Hadwiger’s Conjecture states that every -minor-freeKt graph is -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 -minor-free graphs provide an avenue for 8.1attacking Excluding Hadwiger’s a Minor Conjecture. and Bounded Degree

All the known examples of planar graphs that are not 3-colourable3 with bounded clustering have unbounded maximum degree. This observation motivated several authors [11, 151, 159] to ask whether planar graphs with bounded maximum degree are -colourable with bounded clustering. Esperet and Joret [97] solved this question in the aﬃrmative and extended the Theorem 53 . Every graph with maximum degree ∆ and Euler genus g is 3-colourable result to graphs of bounded Euler genus. with clustering f(∆, g), for some function f. ([97])

Open Problem 54 . Are triangle-free planar graphs with bounded maximum degree Esperet and Joret [97] posed the following open problem 2-colourable with bounded clustering? ([97])

36 k (k + 1) Kk+1 S(k, d) k Graphs with treewidth kare properlyk + 1 -colourable, which is best possible for . Moreover, since the standard example∆ hask treewidth , the defective chromatic24k∆ number of graphs with treewidth equals . On the other hand, Alon et al. [11] proved that every graph with maximum degree and treewidth is 2-colourable with clustering . The proof was based on a result about tree-partitions by Ding and Oporowski [75], which was Theorem 55. Every graph with maximum degree ∆ and treewidth k is 2-colourable with improved by Wood [220]. It follows that: clustering 5 (k + 1)( 7 ∆ 1). 2 2 −

∆ k f(k, ∆) f Liu [162] proved a list colouring analogue of Theorem 55: every graph with maximum degree and treewidth is 2-choosable with clustering , for some function . DeVos et al. [72] proved the conjecture of Robin Thomas that graphs excluding a ﬁxed minor can be 2-coloured so that each monochromatic4 subgraph has bounded treewidth. Alon et al. [11] observed that this result and Theorem 55 together imply that graphs excluding a ﬁxed minor and with bounded maximum degree are -colourable with bounded clustering. Answering a question of Esperet and Joret [97], this result was improved by Liu and Oum [163] (using the Theorem 56 . Every graph containing no H-minor and with maximum degree ∆ is graph minor structure theorem): 3-colourable with clustering f(∆,H), for some function f. ([163])

3 This theorem generalises Theorem 53 above. It is open whether graphs with bounded maximum degree and excluding a ﬁxed minor are -choosable with bounded clustering (Open Problem 18 is a special case). The above results lead to the following connection between clustered and defective colourings, which was implicitly observed by Edwards et al. [87]. The second observation was made by Lemma 57 . For every minor-closed class , Norin et al. [183]. G

χ?( ) 3χ∆( ). ([87, 183]) G 6 G Moreover, if some planar graph is not in then G

χ?( ) 2χ∆( ). G 6 G Proof. d c = c( , d) d 3 G G c k := χ∆( ) d G k As mentioned above,G Liu and Oum [163] proved that for every integerG there is an integer d such that every graph in with maximum degree has a -colouring with clusteringG . Let . That is, ford some integerG 3,k every graph in is -colourablec χ?( ) 3k withG defect6 . Apply the result of Liu and Oum [163] to each monochromatic component of , which has maximum degree at most . Then is -colourable with clustering , and . For the second claim, Robertson and Seymour [197] proved that a minor-closed class not containing all planar graphs has bounded treewidth. The result follows from the method used above, with Theorem 55 in place of the result of Liu and Oum [163].

37 8.2 Kt-Minor-Free Graphs

Kt 2 Kt (t 1) O(t log t) − (t 1) K∗ First consider defective− colourings of -minor-free graphs. Edwards et al. [87t]−proved1,1 that everyKt -minor-free graph is -colourable with defect . Their proof gives the same result for -choosability. This result is implied by Theorem 33 since contains t 2 a minor. Indeed, the proof of Theorem 33 in this case− is identical to the proof of Edwards S(t 2, d) Kt et al. [87] (which predated [185]). Van den Heuvel and− Wood [124] improved the upper bound on the defect in the result of Edwards et al. [87] to ; see Theorem 61 below. Edwards et al. [87] also showed that the standard example is -minor-free. Thus, Lemma1 ` implies the following defectiveχ version∆( K of)= Hadwiger’sχ ( K ) Conjecture: = t 1. M t ∆ M t −

Kt (5)

χ?( K ) t 1. Now consider clustered colourings of -minor-freeM t > − graphs. Note that (5) implies

χ?( K ) M t t 6 9 We now show that the island-based method of Dvoˇr´akand Norin [84] determines for Theorem 58 . For t 9, . 6 χ?( K ) = t 1. M t − ([84]) In particular, every Kt-minor-free graph G is (t 1)-choosable with clustering −  !2 5t3/2 ct := 2  .  √2 1   − 

Proof. t 6 9 Kt Kt n (t 2)n 3/2 1/2 − Fort 6 9 , the exactt extremal function for -minor-free graphs is knownt [n77, 139, 167, k = t 2 α = 1 β = 1 c = t3/2 G (t 2) 168, 169, 207]. In particular,− every -minor-free2 graph on vertices has− less than ct G (t 1) ct edges for . For all , every− such graph has a balanced separator of size [12]. By S(t 2, d) Kt Kt Lemma− 13 with and and and , has a -island of size at t 1 t 9 Kt most −. By Lemma6, 6is -choosable with clustering . Since the standard example t 1 − is -minor-free, the clustered chromatic number of -minor-free graphs is at least . Thus for , the clustered chromatic number of -minor-free graphs equals . t t Kt The only obstacle for extendingΘ(t√log Theoremt) 58 for larger values of is that the exact extremal function is not precisely known. Moreover, forKt large , the maximum averaget degree of - minor-free graphs tends to ; see [153, 154, 209, 210]. Thus Lemma 13 alone cannot O(t) χ?( Kt ) determine the clustered chromatic number of -minor-free graphs for largeM . Kawarabayashi and Mohar [147] ﬁrst proved a upper bound on . The constants χ?( K ) 2t 2. in this result have been successively improved,M t 6 as shown− in Table3, to

38 Kt t > 3 χ Table 3: Upper bounds on the clustered chromatic?( K numbert ) 6 of -minor-free graphs ( ). 31 M 2 t c(t) d 7t−e3 clusteringc(t) choosability d 2 e 4t 4 c(t) Kawarabayashi1 and Mohar [147] − yes 3t 3 c(t) Wood [221] − yes 2t 2 c(t) Edwards et al. [87] − 2t 2 t−2 Liu and Oum2 [163] − d 2 e 2t 2 c(t) Norin [182] − Van den Heuvel and Wood [124] Dvoˇr´akand Norin [84] ` χ?( K ) = χ ( K ) = t 1 M t ? M t −

Dvoˇr´akand Norin [84] have announced a proof that . Most of the results shown in Table3 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 Lemma 59 . For every set A of k 1 vertices in a connected graph G, and for every den Heuvel and Wood [124], which we now> present. minimal induced connected subgraph H of G containing A, (1) H has([124 maximum]) degree at most k, and 1 (2) H can be 2-coloured with clustering 2 k .

Proof. v V (H) T H H ∈ v H T A T degSayH (v) . Let k be a spanningdegH (v tree) 6 k of that includes∆(H) 6 eachk edge of incident with . By the minimality of V, every(H) leaf of is in . Thus,V (H the) = numberA = k of leaves in is at | | | | | | least and at mostV (H. Hence) > k V (H ,A implying) = ∅ . H | | − 6 H A H L H L To prove (2)− by induction on . In the base case, and the result is L A V (L v) 1 k trivial. Now assume that . Thus , and by the minimality| − of |,6 every2 L v H L H0 = H V (L v) vertex in is a cut-vertex of . Consider a leaf-block of . Every vertex− in −, except A0 = (A V (L)) v H0 G A0 the one cut-vertex\ ∪ { in} , is in . There are at least two leaf-blocks. Thus A0 k H0 2 1 k for some| | 6 leaf block , where is the one cut-vertex of in . Let2 and L v v H0 H 2 1 k \{ } . Then is a minimal induced connected subgraph of containing2 , and . By induction, has a -colouring with clustering . Colour every vertex in by the colourH1 not assignedH2 to in G . Nowadjacentis -coloured with clusteringG . H1 H2 Disjoint subgraphs and in a graph are if some edge of has one endpoint Lemma 60 . For t 4, for every Kt-minor-free graph G, there are induced subgraphs in and one endpoint in> . H1,...,Hn of G that partition V (G), and for i [n]: ∈ (1) The subgraph([124]) Hi has maximum degree at most t 2, and can be 2-coloured with clustering − 1 (t 2); 2 −

1 2 This result depended on a result announced by Norine and Thomas [184, 208] which has not yet been written. See [203] for some of the details.

39 (2) For each component C of G V (H1) V (Hi) , at most t 2 subgraphs in − ∪ · · · ∪ − H1,...,Hi are adjacent to C, and these subgraphs are pairwise adjacent.

Proof. G H1,...,Hn H1 G We may assumei = 1 that is connected. We construct iteratively, maintaining properties (1)H and1,...,H (2).i Let be the subgraph inducedi > by1 a singleV (H vertex1),...,V in (H.i) Then (1) and (2) holdV (G for) C. G V (H1) V (Hi) Q1,...,Qk − ∪ · · · ∪ H1,...,Hi C Q1,...,Qk Assume that{ }satisfy (1) and (2) for some , but do not k t 2 G k 1 partition6 − . Let be a component of> . Let be the elementsj [k of] vj that areC adjacent to Qj. By (2), arek pairwiset 2 adjacent ∈ 6 − and . Since is connected,Hi+1 C . v1, . . . , vk 0 0 0 For , let be aC vertexG in V (adjacentH1) toV (H.i By+1) Lemma 59C with , thereC isC an 0 − ∪· · ·∪0 induced connectedC C subgraph ofC containingC thatG satisﬁesV (H (1).1) V (Hi) 0 0 − ∪ · · · ∪ ConsiderC a component ofHi+1 . EitherC is disjoint from , or is 0 0 contained in C. If is disjointC from , then is a componentH1,...,Hi of+1 C 0 and is notQ1 adjacent,...,Qk,H to i+1 , implying (2) is maintained for . k = t 2 C − Now assume isQ contained1,...,Qt− in2,H.i+1 The subgraphsC in thatQ1,...,Q are adjacentt−2 to are 0 a subsetQ of1,...,Qt−2,Hi+1,C, which are pairwise adjacent. SupposeKt that G and is 0 adjacent to all of C . Then tis adjacent2 Q1,...,Q to allt of−2,Hi+1 . Contracting 0 − each of C into a single vertex gives as a minor of , which is a contradiction. Hence is adjacent to at most of , and property (2) is maintained for .

Theorem 61 . For t 4, every Kt-minor-free graph G is (t 1)-colourable with defect The following is the main result> of Van den Heuvel and Wood [124−]. t 2 and is (2t 2)-colourable with clustering 1 (t 2) . − − d 2 − e ([124]) Proof. H1,...,Hn Hi t 2 H1,...,Hi−1 i = 1, 2, . . . , n Hi − t 1 t 2 Let− be the subgraphs from Lemma 60. By property (2) in− Lemma 60, each subgraphH1,...,Hi−1 is adjacentH toi at mostHi of . Fort 2 (tcolour1) with − 1 − G t 2 Hi 2 (t 2) one of colours− diﬀerent from the colours assigned to the at2 most− subgraphs in (2t 2) 1 (t 2) adjacent to − . Since has maximum degreed 2 − , wee obtain a -colouring of with defect . Each has a -colouring with clustering . The product of O(t) χ?( K ) these colourings is a -colouringM witht clustering . Kt 496t Note that the upper bound on has beenf(t) extended to the496 settingt of odd minors. 10t 13 In particular,− Kawarabayashi [146] proved that every graph with no odd -minor is - colourable with clustering at most some function . The bound of was improved to Kby3 Kang and Oum [140]. By Proposition4, these results do not generalise to the setting of choosability (even for defective colourings), since complete bipartite graphs contain no odd minor and have unbounded maximum average degree.

40 8.3 H-Minor-Free Graphs

H t H t 1 Kt−1 H − Hadwiger’s Conjecture implies that for every graph with vertices, the maximum chromatic number of -minor-free graphs equals (since is -minor-free). However, for clustered and defective colourings, fewer colours often suffice. For example, as discussedg Θ(√g in) Section 3.4, Archdeacon [21] proved that graphs embeddable on a fixed surface are defectively 3-colourable,H whereas the maximum chromatic numberH for graphs of Euler genus is . H V (H) The natural question arises: what is the defective or clustered chromatic| number| of the class of -minor-free graphs, for an arbitrary graph ? We will see that the answer depends on the structure of T, unlike the chromatic numberdepth whichT only depends on . Ossona de MendezT et al. [185closure] observedT that the followingT definition is a key to answering this question. Let be a rootedT tree. The of is the maximumconnected number tree-depth of vertices on a root–to–leafH pathtd( inH). The of is obtained from byT adding an edgeH between4 every ancestor and descendentT in , as illustrated in Figure9. The of a graph , denoted by , is the minimum depth of a rooted tree such that is a subgraph of the closure of . Note that the connected tree-depth is closed under taking minors.

K5,9

Figure 9: The closureS(h, d) of a tree of depth 6 contains (d. + 1) h + 1 H H Note that the standard example, , is the closure of the complete -ary tree of χ∆( H ) td(H) 1, depth . By Lemma1, for every graphM ,> the defective− chromatic number of -minor-free graphs satisﬁes (6) Conjecture 62 . For every graph H, as observed by Ossona de Mendez et al. [185], who conjectured that equality holds.

χ∆( H ) = td(H) 1. ([185]) M − tree-depth H td(H) 4 H This deﬁnition isH a variant of the moretd( commonlyH) = td( studiedH) notiontd( ofH the) = td(H) of H, denoted by , which equalsH the1 maximumH2 connectedtd(H1) = tree-depth td(H2) = of td( theH) connected componentstd(H) =of td(.H See) + [180 1 ] for background on tree-depth. If is connected, then . In fact, unless has two connected components and with , in which case . Norin et al. [183] introduced connected tree-depth to avoid this distinction.

41 Ks,t min s, t + 1 { } H = Ks,t Sincetd(H) = 3 has connected tree-depth , by Theorem 33, Conjecture 62 is true if χ?( H ) td(H) χ∆( H ) M , as proved by Ossona de Mendez et al. [185], who also proved Conjecture M62 if χ?( H ) td(H) M. Norin et al. [183] provided further evidence for the conjecture by showing that is bounded from above by some function of . This implies that both Theorem 63 . For every graph H, and are tied to . td(H)+1 χ?( H ) 2 4. ([183]) M 6 −

χ∆( H ) = td(H) 1 M − χ?( H ) M While Ossona de Mendez et al. [185] conjectured that , the following lower bound by Norin et al. [183] shows that might be larger, thus providing some Theorem 64 . For each k 2, if Hk is the the standard example S(k 1, 2) then Hk distinction between defective and> clustered colourings. − has connected tree-depth k and ([183]) χ?( H ) 2k 2. M k > −

Proof. c Gk c k Gk (2k 3) c Hk − Gk Fix an integer . We now recursively deﬁne graphs (depending on ), and show by induction on thatk = 2has noG2 -colouringc + with1 clustering G, and2 isH not2 = aS minor(1, 2) =of K1,.3 G2 c

For the baseGk−1 case , let bek the3 path onGk−1 vertices.(2k Then5) has no > − c minor,Hk−1 and has no 1-colouringGk−1 with clustering . Gk (v1, . . . , vc+1) i 1, . . . , c 2c 1 Gk−1 Assume is deﬁned for some ∈ {, that } has no− -colouring with clustering vi, vi+1 , and {is not a} minor of . As illustrated in Figure 10, let be obtained from a path Gk as follows:(2k 3) for add c pairwisevi disjointvi+1 copies of − complete to i 1. , . . . , c 2c 1 Gk−1 vi, vi+1 ∈ { } − { } Suppose that has a -colouring with clustering . Then and receive distinct colours for some . Consider the copies of complete to . At 2c 1 2c 1 2c 1 2c 1 2c 1 2c 1 − − − − − −

b b b b b b b b b b b b b b b b b b

Gk 1 Gk 1 Gk 1 Gk 1 Gk 1 Gk 1 Gk 1 Gk 1 Gk 1 Gk 1 Gk 1 Gk 1 − − − − − − − − − − − −

v1 v2 v3 v4 v5 v6 b b b vc+1

Figure 10: Construction of in Theorem 64.

42 c 1 vi c 1 − − vi+1 Gk−1 (2k 5) c Gk most such copies− contain a vertex assigned the same colour as , and at most such (2k 3) c copies contain− a vertex assigned the same colour as . Thus some copy avoids both colours. Hence is -colouredHk with clusteringGk , which is a contradiction.Gk Therefore Jhasx : { nox V (Hk)-colouringHk withr clustering . Hk Hk P∈ } V (Jx) Jr (v1, . . . , vc+1) Jr = (vi, . . . , vj) It remainsx∈V (H) | to show| that is not a minor of . Suppose that contains a model Hk r Hk−1 X of− . Let be the root vertex in . Choose the -model to minimise .vi Thus−1 visj+1 a connected subgraph of H.k Say−1 X . Gk vi−1, . . . , vj+1 (vi, . . . , vj) Note that consists of three− { pairwise disjoint} copies of . The model of one such copy avoids andGk−1 (if theseHk−1 vertices are deﬁned).Gk−1 Since is connected, is containedHk−1 in a componentGk of and is adjacent to . Each such component is a copy of . Thus is a minor of , which is a contradiction. Thus is not a minor of . Conjecture 65 . For every graph H, Norin et al. [183] conjectured an analogous upper bound:

χ?( H ) 2 td(H) 2. ([183]) M 6 − H td(H) = 3 χ∆( H ) = 2 H M χ?( H ) 2χ∆( H ) = 4 = 2 td(H) 2 Norin et al. [183] observed thatM Conjecture6 M65 holds for every graph− with . In this case, as mentioned above, Ossona de Mendez et al. [185] proved that . Since is planar, by Lemma 57, , asH claimed. p > 1 Inp H the remainder of this sectionp we proveH Theorem 63. The proof depends on the following Erd˝os-P´osaTheorem by Robertson and Seymour [197]. For a graph and integer , let Theorem 66 . For every graph H with c connected components be the disjoint union of copies of . and for all integers p, w > 1, for every graph G with treewidth at most w and with no p H minor, there is a set X V (G) of size at most (p 1)(wc 1) such that G X has no H ([197]; see [⊆192, Lemma 3.10]) − − − minor.

Lemma 67 . For all integers h, k, w 1, every S(h 1, k 1)-minor-free graph G of The next lemma is the heart of the proof of> Theorem 63. − − treewidth at most w is (2h 2)-colourable with clustering (k 1)(w 1). − − − ([183]) Proof. h > 1 w k h = 1 S(0, k 1) h 2 h 1 G − > − S(h 1, k 1) w V0,V1,... We− proceed− by induction on , with and ﬁxed. The case is trivial since Gis the 1-vertex graph. Nowr assume that , the claim holds for , and is a i 1 G[-minor-freeVi] graphk Swith(h treewidth2, k 1) at most . Let be the BFS > − − Vlayering0 ofVi−1starting at some vertex . S(h 1, k 1) Vi ∪ · · · ∪ − − Fix . ThenVi−1 containsG no was a minor,G[ asVi] otherwise contracting H = S(h 2, k to1) a singlec = vertex 1 gives a Xi Vi minor (since(k every1)(w vertex1) in has − − ⊆ h−1− − G[Vi Xi] S(h 2, k 1) G[Vi Xi] (2 2) a neighbour\ in ).− Since− has treewidth at most , so\ does . By− Theorem 66 with and , there is a set of size at most such that has no minor. By induction, is -colourable with

43 h−1 (k 1)(w 1) Xi G[Vi] (2 1) − − − (k 1)(w 1) i − − clusteringr . Use one newi colour for . ThusVi Vj is j > i + 2-colourableG (2h 2) (k 1)(w 1) with− clustering . Use− disjoint− sets of colours for even and odd , and colour by one of the colours used for even . No edge joins with for . Now is -coloured with clustering .

To drop the assumption of bounded treewidth, we use the following result of DeVos, Ding, Theorem 68 . For every graph H there is an integer w such that for every graph G with Oporowski, Sanders, Reed, Seymour, and Vertigan [72]. no H-minor, there is a partition V1,V2 of V (G) such that G[Vi] has treewidth at most w, for i 1, 2 . ∈ { } ([72]) Proof of Theorem 63. k := V (H) h := td(H) H | | S(h 1, k 1) H G C(h, k) − − G (2h+1 4) Let and− . By deﬁnition, is a subgraph of g(h, k) . Thus every -minor-free graph contains no -minor. Lemma 67 and Theorem 68 implies that is -colourable with clustering at most some function td(H) . The claim follows. χ?( H ) 4 M 6

Note that Norin et al. [183] proved a weaker bound, roughly , with a self- contained proof avoiding Theorem 68 and thus avoiding the graph minor structure theorem. 8.4See [223 Conjectures, 227] for more about defective choosability in minor-closed classes.

1,c := Pc+1,K1,c k 2 k,c We now present a conjecture of NorinX et al. {[183] about} the clustered> chromaticX number of an arbitrary minor-closed class of graphs. Consider the following recursively deﬁned class of G k−1,c graphs, illustrated∈ in X Figure 11. Let . For , let be the set of graphs• obtainedG0 by the following three operations.c For theG ﬁrst two operations, consider an 0 arbitrary graphG k,c . X • Let G+be the graph obtained from Gdisjoint copies of byk addingD one dominantG vertex. + Then k(cis in1) + 1. D G k,c − X ++ • Letk 3be theG graphk obtained−2,c from G as follows: for each -clique inG , add a stable > ∈ X set of(k 1) verticesD G complete to .(c Then2 1)(k is1) in + (c +. 1) − ++ − − D G k,c If and X , then let be the graph obtained from as follows: for each -clique rainbowin , add a path of vertices complete to . Then is in . Lemma 69 . For every c 1 and k 2, for every graph G k,c, every colouring of G A vertex-coloured graph is > if every> vertex receives a distinct∈ X colour. with clustering c contains a rainbow Kk+1. In particular, no graph in k,c is k-colourable with X clustering c([. 183])

Proof. k > 1 k = 1 Pc+1 K1,c c k 1 k 2 k 3 − We proceed− by induction> on . In the case , every colouring of or with clustering contains a bichromatic edge, and we are done. Now assume the claim for and for (if ).

44 G k 1,c G k 1,c G k 2,c ∈ X − ∈ X − ∈ X − k(c 1)+1 (c 1)2(k 1)+ c +1 if then if − then if − − then G′ :=

G+ := G++ :=

Gb b b G G G k (k 1) ∀ ∀ −

c -clique -clique Xk,c Xk,c Xk,c

is in is in is in k,c X

Figure 11: Construction0 of . G k−1,c G c v ∈ X c 1 G G − Let . ConsiderG a colouring of withK clusteringk v . Say the dominantK vertexk+1 is blue. At most copies of G+contain a blue vertex.c Thus, some copy of hasG noG blue+ vertex. By induction,w1, this. . . , wcopyk of contains a rainbow . WithS we obtain ak( rainbowc 1) + 1 . { } + − Now considerw1 a, . colouring . . , wk G of with clusteringc 1 . By induction,S the copy of in wcontainsi − + a clique S receiving distinct colours. Let be the set ofw1, . . . , wk verticesG adjacent to Kk+1in . At most vertices in receive the same colour as . Thus ++ some vertex in k >receives3 aG colourk distinct−2,c from the colours assignedG to . Hencec ∈++ X contains a rainbow G . G w1, . . . , wk−1 2 { ++ } NowP suppose and(c 1)(k 1) +. (Considerc + 1) a colouringG of with clusteringw1, . . . , w.k− By1 − − { S } induction,Xi the copy of in Pcontains a clique wi receivingX distinct:= i X colours.i Xi c 1 X (c 1)(k 1) P X (c 1)(k 1) + 1 |Let| 6be− the path| of| 6 − − −vertices in complete− − to . V (P X) (c2 1)(k 1) + (c + 1) (c 1)(k 1) = c(c 1)(k 1) + 1 + 1 Let | be− the set| > of vertices− in− assigned− the− same colour− as ,− and let− . Thus andP X c +. 1 Hence has at most components,xy − ++ w1, . . . , wk−1 x, y Kk+1 G and { } ∪ { } . Some component of has at least vertices, and therefore contains a bichromatic edge . k,c Then induces a rainbow in . X c k Norin et al. [183] conjectured that every minor-closed class that excludes every graph in Conjecture 70 . For every minor-closed class of graphs, χ?( ) equals the minimum for some is -colourable with bounded clustering.M More precisely: M integer k such that k,c = for some integer c. M ∩ X ∅ ([183])

k = 1 k = 2 Note that the lower bound in Conjecture 70 follows from Lemma 69. Conjecture 70 is trivial when , and Norin et al. [183] proved it when . It is easily seenKs,t that Conjecture 70 simplies6 t Conjecture 65; see [183]. Now consider the class of graphs excluding the complete` bipartite graph as a minor, where χ∆( K ) = χ ( K ) = s. .Theorem 33 and (6) imply thatM s,t ∆ M s,t

45 n S(k, 1) k = log (n + 1) d 2 e

Figure 12: The path on vertices is a subgraph of , where . χ?( K ) 3s M s,t 6 χ?( K ) 2s + 2 M s,t 6 χ?( K ) s + 1 For clustered colouring, Lemma 57 implies . This boundM wass,t improved> by t max s, 3 Dvoˇr´akand> { Norin} [84] who proved that , which is the best known upper χ?( K ) = s + 1 bound. Van den Heuvel andM Woods,t [124] proved the lower bound, for . Their construction is a special case of the construction shown in Figure 11. 8.5Conjecture Circumference 70 says that .

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

Proof. k > 1 d S(1, d) = K1,d+1 S(k 1, d) v − We proceedS( byk, d induction) on with ﬁxed. In the base case, , which containsS no(k, 4-vertex d) path and noC cycle. Now assume the2k + result 1 for v . Let beC the root vertex of . S(k 1, d) v S(k 1, d) 2k − − k Suppose that containsS a( cyclek, d) of length at least . Since2 + 1is a cut vertex, isS contained(k, d) in one copy ofP 2k+1 plus . ThusP v contains a path on vertices, which1 k+1 is a contradiction.k Thus has no cycle of− length at least . 2 (2 1) = 2 S(k 1, d) d − e k+1 − If S(k,contains d) a path on vertices,2 then contains a path component with least vertices that is contained in a copy of , which is a contradiction. Hence contains no path of order .

46 td(Ck+1) = 1 + log (k + 1) = 2 + log k . It follows from Lemma 71 that d 2 e b 2 c

χ?( k) χ∆( k) td(Ck+1) 1 = 1 + log k . C > C > − b 2 c By (6),

Theorem 72 . For every integer k > 2, every graph G with circumference at most k is Mohar et al. [176] proved an upper bound within a factor of 3 of this lower bound. (3 log2 k)-colourable with clustering k. Thus ([176]) χ∆( k) χ?( k) 3 log k. C 6 C 6 2 C = ∅ Lemma 73 . For every integer k > 2, for every graph G with circumference at most k This result is implied by the following lemma with . and for every pre-coloured clique C of size at most 2 in G, there is a 3 log k -colouring of G b 2 c with clustering([176k]), such that every monochromatic component that intersects C is contained in C.

Proof. k + V (G) V (G) 2 | | | | 6 V (G) 3 | | > We proceed byk = induction 2 onG . The result is trivial if . NowC assume1 | | ≤ C = 2 . C 2 < 3 log k | | b 2 c C = 2 C First suppose that | | . Then is a forest, which is properly 2-colourable. If or C and two colours are used on , we obtain the desired colouring (with G colours). Otherwise,k > 3 with the same colour on the vertices in . Contract the edge

and 2-colourG the resulting3 forest by induction,G to obtain the desired colouring(G1,G2) of .S Now:= Vassume(G1 thatG2) . S = 2 S ∩ | | Suppose that is not G-connected.1 G2 Then has a minimal separation with of size at most 2. IfGj j = 1,, then2 add the edge on if theS edge is not already present. Consider both and to contain thisG edge. Observe that since the separation is minimal, there is a path in each ( ) between the two verticesC of . Therefore, adding C V (G1) 3 log k G1 C the edge⊆ does not increase the circumferenceb of2 c. Also note that any valid colouring of the augmented graph will be valid for the originalG1 graph. Since kis a clique, we may assume that G1. By induction, thereC is a -colouringC of ,S with precoloured, such 3 log k G2 S thatb every2 c monochromatic component of has order at most and every monochromatic componentG2 of that intersectsk is contained in . This colours G. By2 induction, thereS is a -colouringS of , with precoloured, such that every monochromatic componentG of has orderk at most and every monochromatic componentG of Cthat intersects Cis contained in . By combiningG the two colourings, every monochromatic component of has order at most and every monochromatic component of that intersectsk > 4 is contained in , as required. Now assume that is 3-connected. G k k 1 − Every 3-connectedG graph containsQ a cycle of lengthk at least 4. Thus . G A 1 (k 5) A Q G If containsd 2 − noe cycle of length , then apply∈ theA induction hypothesis for ; thus we may assume that contains a cycle of length . Let be the set of cycles in of length at least . Suppose that a cycle is disjoint from . Since is 3-connected,

47 A Q G 2( A + Q + 3) > 3k G k | | | | Q there are three disjoint paths between and A. It follows that contains three cycles with totalS length:= V ( atQ) leastC G.0 Thus:= G containsS a cycle of length greater than , which1 is a contradiction.∪ 0 Hence, every cycle in 1intersects− . 1 2 (k 5) G 2 (k 7) 2 k d − e d − e 1 b c Let . As shown above, contains no3 cycle log2 of2 k length at least 0 0 b b1 cc G . Then has circumference at most G , which is at most2 k , which is 0 b c at least 2.G By induction (with no precoloured vertices), there is a C -colouring of S C k 3 log 1 k + 3 3 log k such\ that every monochromatic component of has orderb at most2b 2 cc . Use6 b a set2 ofc colours for disjoint from the (at most two)G preassigned coloursk for . Use one new colour for , whichG has size at mostC . In total, thereC are at most colours. Every monochromatic component of has order at most , and every monochromatic component of that intersects is contained in . Pk k

Similar results are obtained for graph classes excluding a ﬁxed path. Let be the path on td(Pk) = td(Pk) = log2(k + 1) . vertices. It follows from Lemma 71 that d e k k + 1 Pk+1 H

Let be the class of graphs containing no path of order (or equivalently, with no χ∆( k) = td(Pk+1) 1 = log2(k + 2) 1. minor). Thus Conjecture 62,H in the case of excluded− d paths, assertse − that Pk+1 k

TheoremEvery graph 74 with no. For every-minor integer has circumferencek > 2, every at graph mostG .with Thus no Theorem path of 72 order impliesk + 1 theis (3following log2 k)-colourable upper bound with that clustering is which isk. within Thus a factor of 3 of Conjecture 62 for excluded paths.

([176]) χ∆( k) χ?( k) 3 log k , H 6 H 6 b 2 c

k C k P To conclude this section, we show that Theorem 34 determines the defective choosabilityk+1 of k s 6 t Ks,t 2s s > 2 Ks,t and . C k+1 d e (k + 1) k s 6 2 1 2s 6 k ` k+1 C d e − FirstK considers,t k . Sayχ ( k). = Then the circumference of equals . If then ∈ C ∆ C d 2 e contains a k -cyclet >and s is notK ins,t . On the other hand, if2s + 1 then P and 2s +. 2 Thus Ks,t k . 2s + 1 6 k Ks,t k k−1 ∈` H k+1 6∈ H s > χ ( k) = Now consider2 . Say ∆.H Then b 2 containsc a path order , and contains no path of order . Thus if and only if . That is, if and only if . By Theorem 34, . 9 Thickness

thickness G k G k k k T k 6k χ( k) 6k k = 2 The of a graph is the minimum integer such that T is6 the union of planar χ( 2) 9, 10, 11, 12 subgraphs; see [179] for a survey. Let be the class of graphs withT thickness∈ { . Graphs} with k 3 6k 2 χ( k) 6k 2, 6k 1, 6k thickness> have maximum average degree less than− . Thus T ∈ {. For− −, which} corresponds to the so-called earth-moon problem, it is known that . For , complete graphs provide a lower bound of , implying . It is an open problem to improve these bounds; see [129].

48 9.1 Defective Colouring

This section studies defective colourings of graphs with given6k thickness. Yancey [231] ﬁrst 2 proposed studying(3k + 1) this topic. The resultsO( ink this) section are due to Ossona de3k Mendez Lemmaet al. [185 75]. Since. theThe maximum standard average example degreeS(2k, is d) lesshas than thickness, Theorem at most 28k. implies that such graphs are -choosable with defect , but gives no result with at most colours. Proof. ([185]) k > 1 S(2, d) G := S(2k, d) k 2 r G G r > − d + 1 S(2k 1, d) i [d + 1] vi i We proceed by induction on − . In the base∈ case, is planar, and thus has Ci G r Ci vi d+1 S(2k 2, d) thickness 1. Let − for some− . Let be the vertex of such that − is the H := G r, v1, v2, . . . , vd+1 H S(2k 2, d) disjoint−{ union of copies} of . For , let be the vertex of the− -th component of H such that is thek disjoint1 unionG ofE(H) copies of d + 1 . Let 0 − − K2,d0 r d G E(H) G . Observe− that each component of is isomorphic to and by induction, k has thickness at most . Since consists of copies of pasted on for some , is planar and thus has thickness 1. Hence has χ∆( k) 2k + 1 thickness at most . T >

Lemmas1 and 75 imply that . Ossonag > de0 Mendezg-thickness et al. [185] provedG that equality holds. In fact,k the proofG works in the followingk more general setting, implicitly g g g k g introducedTk by Jackson and Ringel [131]. For an integer , the of a graph is the minimum integer such that is the union of subgraphs each with Euler genus at most . Let be the class of graphs with -thickness . As an aside, note that the -thickness of Theorem 76 . For integers g > 0 and k > 1, complete graphs is closely related to bi-embeddings of graphs [13, 14, 15, 54]. g ` g χ∆( ) = χ ( ) = 2k + 1. ([185]) Tk ∆ Tk In particular, every graph with g-thickness at most k is (2k + 1)-choosable with defect 2kg + 8k2 + 2k.

The lower bound in Theorem 76 follows from Lemma 75. The upper bound follows from Lemma5 Lemma 77 . For integers g > 0 and k > 1, every graph with minimum degree at least 2andk + the 1 and nextg-thickness lemma. at most k has an (` 1)-light edge, where ` := 2kg + 8k2 + 2k + 1. − ([185]) Proof. δ = 2k + 1 β := (4k 1)(2k + 1) + 2k(g 1) γ := 4k(2k + 1)(g 1) 2 − 2 − 1 p 2 − We claim` that Lemmaβ` γ >780 below with ` impliesβ` γ the result.2 (β + Equationsβ + 4γ) (7) and (8) γ − −2α − − γ are immediatelyβ + β satisﬁed.β + β Let> 0 and ` > β + β . (9) requires that . The larger root of is , which is at most since . Elementary manipulations show that . Thus (9) is Lemma 78. Let G be a graph with n vertices, g-thickness at most k, and minimum degree at satisﬁed. least δ, where

6k > δ > 2k + 1,

(7)

49 (δ 2k)` > 4kδ, and − (δ 2k)`2 (4k 1)δ + 2k(g 1)` 4k(g 1)δ > 0. − − − − − − (8) Then G has an (` 1)-light edge. − (9) Proof. G 3k(n + g 2) − 2k(n + g 2) X − ` 1 X δ X ` − By Euler’s Formula, has at most edges, and every spanning bipartite subgraph has at most edges.X Let be the set of vertices with degree at most δ X + (n X )` deg(v) = 2 E(G) 6k(n + g 2). . Since vertices| | in −have | | degree6 at least and vertices| | 6 not in have− degree at least , v∈V (G)

(` 6k)n 6k(g 2) (` δ) X . − − − 6 − | | Thus X G G0 G X V (G) X δ 0 \ Suppose on the contrary thatX is a stableG set in . Let be the spanning bipartite subgraph of consisting of all edges between and . Since each of the at least edges δ X E(G0) 2k(n + g 2). incident with each vertex in |are| 6 in | , | 6 − ` > 4k δ > δ ` δ > 0 δ 0 δ−2k − > δ(` 6k)n 6k(g 2)δ δ(` δ) X (` δ)(2k)(n + g 2) Since (hence− − )− and 6 , − | | 6 − − = δ(` 6k) 2k(` δ)n (` δ)2k(g 2) + 6k(g 2)δ ⇒ − − − 6 − − − = (δ 2k)` 4kδn 2k(g 2)` + 4k(g 2)δ. ⇒ − − 6 − − n ` (` 1) n `+1 (δ 2k)` 4kδ > 0 6 − > − − (δ 2k)` 4kδ(` + 1) 2k(g 2)` + 4k(g 2)δ. If then every edge− is −-light. Now assume6 that− .− Since ,

(δ 2k)`2 (4k 1)δ + 2k(g 1)` 4k(g 1)δ 0, − − − − − − 6 Thus X G (` 1) − which is a contradiction. Thus is not a stable set. Hence contains an -light edge. k = 1 ` = 2g + 13 G 3 g (2g + 12) +10 Lemma 78 with Kand3,2g+2 implies that every graph with minimum degree atg least and Euler genus has a 2-lightg + 2 edge. Note that this bound is within of being tight since has minimum degree 3, embeds in a surface of Euler genus , and every edge has an endpoint of degree .g More3 precise results, which are2g + typically 10 proved by discharging with respect to an embedding, are known [35, 130, 134]. Theorem 76 then implies that every graph with Euler genus is -choosablek > 1 with defect . As mentionedg in= Section 0 k 3.4=, 2 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 . 2 The case and relates to the famous earth–moon problem [8, 108, 129, 131, 195], which asks for the maximum chromatic number of graphs with thickness . The answer is known

50 9, 10, 11, 12 G { } 12 G k d (k, d) (7, 18), (8, 9), (9, 5), (10, 3), (11, 2) to be in . Since the maximum∈ {average degree of every graph with} thickness 2 is less than , the result of Havet and Sereni [119] mentioned in Section5 implies that is -choosable with defect , for . This result gives5 no bound with at most 6 colours. On the other hand,G Theorem 76 saysk that the class of d (k, d) (5, 36), (6, 19), (7, 12), (8, 9), (9, 6), (10, 4), (11, 2) graphs with thickness∈ { 2 has defective chromatic number and defective choice} number equal to . In particular, the method shows that every graph with thickness 2 is -choosable with defect , for . This 11-colouring result, which is also implied by the result of Havet and Sereni [119], is close to the conjecture that graphs with thickness 2 are 11-colourable. Improving these degree bounds provides an 9.2approach Clustered for attacking Colouring the earth–moon problem.

g Tk Proposition 79. For all integers g 0 and k 1, every graph G with g-thickness at most k Consider the clustered chromatic number> of >. We have the following upper bound. is (6k + 1)-choosable with clustering max g, 1 . { } Proof. V (G) L (6k + 1) G | | V (G) = 0 V (G) 1 | | | | > We proceedV by(G induction) 6kg on g .1 Let vbe a -listG assignment for G. Inv the | | 6 > − Lbase case, the claim is trivialg if L(v) . Now6g assumeV (G thatv) < 6kg . c L(v) | | > | − | ∈ g 1 G v v c G L First suppose that − . Thus −. Let be any vertex of . By induction, is -colourableg with clustering . Since and , some colour is assigned to at mostV (G) > 6kgvertices in . Colourg by . Thus k is -coloured with6kg | | clustering . v 6k E(G) < 3k( V (G) + g) (3k + 1 ) V (G) | | | | 6 2 | | G 6k + 1 G v (6k + 1) Now assume that . Every graph with -thickness at most− and more than max g, 1 L(v) 6k + 1 deg(v) 6k c L(v) vertices has a vertex { of} degree at| most| >, since 6 ∈ , implying has average degree lessv than v .c By induction,v is -choosable G L max g, 1 with clustering . Since and{ } , some colour is not assigned to any neighbour of . Colour by . Thus is in a singleton monochromatic component,S(3, c and) is -coloured with clustering . S(2k + 1, c) k

Since is planar, an analogous proof to thatg of Lemma 75 shows that has 2k + 2 χ?( ) 6k + 1. thickness at most . By Lemma2 and Proposition6 Tk 796,

Kn 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 has thickness 2. Thus thickness 2 graphs have unbounded . Similarly, Lemma 13 is not applicableo(n) since graphs with thickness 2 do not have sublinear balanced separators. Indeed, Dujmovi´c et al. [82] constructed ‘expander’ graphs with thickness 2, bounded degree, and with no balanced separators. Returning to the earth-moon problem, it is open whether thickness 2 graphs are 11-colourable with bounded clustering.

51 Kn

10 General SettingFigure 13: The 1-subdivision of has thickness 2.

f f

Consider a graph parameter . Several authors have studied colourings of graphs such that χf( ) k c G Gis bounded forG each monochromatic subgraph, or equivalently each monochromatic∈ subgraph G satisfiesk a particular property;f(H) 6 seec [51, 53, 71, 103, 138, 217] for example.H G For a graph class , define to be the minimum integer such that for some , every graph f(H) has a H χf( ) = χ?( ) -colouring such that for each monochromatic subgraph ofG . ThisG definition f(H) = ∆(H) χf( ) = χ∆( ) f incorporates the clusteredG and defectiveG chromatic numbers. In particular,G if is the maximum number off vertices in a connected component of , then χtw(, and) if2 G 6 then . There are many choices forχtw(and∆) . G D First consider when is the treewidth. Recall∆ that DeVos et al. [72] proved that ∆ for every minor-closed class . Bounded degree classes and look interesting. In particular, what is the maximum integer such that every graph with maximum degree is n n n 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 n most 25 since Berger→ ∞ et al. [25] proved that for every 2-colouring of the grid with diagonals (which has maximum degree 26), there is a monochromatic subgraph with unbounded treewidth (as ). 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, doesη( everyH) graph with maximum degreet 6 have aK 2-colouringt withH bounded monochromatic Hadwigertreewidth? number H η(H) 6 1 H η(H) 6 2 Let be theH maximum integer such that is a minor of , sometimesG called the k [1, η(G)] G k η(H) η(G) k + 1 ∈ of . For example, 6 if− and only if is edgeless, and ifH andG only if isk a= forest.η(G) Ding et al. [76] conjectured that for every graph and integer k = 2 k = 3 χη( ∆) , is -colourable with for each monochromatic subgraphD of . The case is Hadwiger’s Conjecture. Alternately 2-colouring the layers in a BFS layering proves this conjecture for . Gon¸calves [114] proved it for . also looks interesting.

52 Acknowledgements

Many thanks to Jan van den Heuvel, Sergey Norin, Alex Scott, Paul Seymour, Patrice Ossona de Mendez, Bojan Mohar, Sang-il Oum and Bruce Reed with whom I have collaborated on defective and clustered colourings. This survey has also greatly benefited from insightful comments from ZdenˇekDvoˇrák,Louis Esperet, Jacob Fox, FrédéricHavet, GwenaëlJoret, Chun-Hung Liu and Tibor Szabó,for which I am extremely grateful. I am particularly thankful to FrédéricHavet for allowing his unpublished proof of Lemma 29 to be included. Many thanks to the referees for detailed and insightful comments that have greatly improved the paper. References

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