Planar Graphs of Girth at least Five are Square (Δ+2) -Choosable
aa r X i v : . [ m a t h . C O ] J un Planar Graphs of Girth at least Five are Square (∆ + 2) -Choosable
Marthe Bonamy CNRS, LaBRI, Universit´e de BordeauxBordeaux, Franceand
Daniel W. Cranston Department of Mathematics and Applied MathematicsVirgina Commonwealth UniversityRichmond, Virginia, USA, 23284and
Luke Postle Department of Combinatorics and OptimizationUniversity of WaterlooWaterloo, ON, Canada, N2l 3G1
ABSTRACT
We prove a conjecture of Dvoˇr´ak, Kr´al, Nejedl´y, and ˇSkrekovski that planar graphs of girthat least five with maximum degree ∆ are square (∆ + 2)-colorable for large enough ∆. Infact, we prove the stronger statement that such graphs are square (∆ + 2)-choosable andeven square (∆ + 2)-paintable. [email protected] . This author is supported by the ANR Grant EGOS (2012-2015)12 JS02 002 01. [email protected] . This author is partially supported by NSA Grant H98230-15-1-0013. [email protected] . This author is partially supported by NSERC Discovery Grant No. 2014-06162.c (cid:13) h i . This manuscript version is made available under the CC-BY-NC-ND 4.0 license http://creativecommons.org/licenses/by-nc-nd/4.0/ Introduction
Graph coloring is a central area of research in discrete mathematics. Historically, muchwork has focused on coloring planar graphs, particularly in an effort to prove the 4 ColorTheorem. Since its proof in 1976, research has expanded to numerous related problems.One that has received significant attention is coloring the square G of a planar graph G ,where V ( G ) = V ( G ) and uv ∈ E ( G ) if dist G ( u, v ) ≤
2. Wegner [12] conjectured thatevery planar graph G with maximum degree ∆ ≥ χ ( G ) ≤ (cid:4) (cid:5) + 1. He alsoconstructed graphs showing that this number of colors may be needed (his construction is aminor variation on that shown in Figure 2, which requires (cid:4) (cid:5) colors). The girth of a graph G , denoted g ( G ), is the length of its shortest cycle. Since Wegner’s construction containsmany 4-cycles, it is natural to ask about coloring the square of a planar graph G with girthat least 5. First, we need a few more definitions.A list assignment L for a graph G assigns to each vertex v ∈ V ( G ) a list of allowablecolors L ( v ). A proper L -coloring ϕ is a proper vertex coloring of G such that ϕ ( v ) ∈ L ( v )for each v ∈ V ( G ). A graph G is k -choosable if G has a proper L -coloring from each listassignment L with | L ( v ) | = k for each v ∈ V ( G ). The list chromatic number χ ℓ ( G ) is theminimum k such that G is k -choosable. Finally, a graph G is square k -choosable if G is k -choosable. Conjecture 1.1 (Wang and Lih [11]) . For every k ≥ there exists ∆ k such that if G is aplanar graph with girth at least k and ∆ ≥ ∆ k , then χ ( G ) = ∆ + 1 . Borodin et al. [4] proved the Wang–Lih Conjecture for k ≥
7. Specifically, they showedthat χ ( G ) = ∆ + 1 whenever G is a planar graph with girth at least 7 and ∆ ≥
30. Incontrast, for each integer D at least 2, they constructed a planar graph G D with girth 6 and∆ = D such that χ ( G D ) ≥ ∆ + 2.In 2008, Dvoˇr´ak et al. [9] showed that for k = 6 the Wang–Lih Conjecture fails only by 1.More precisely, let G be a planar graph with girth at least 6. They showed that if ∆ ≥ χ ( G ) ≤ ∆ + 2. Borodin and Ivanova strengthened this result: in 2009 they showed [5]that ∆ ≥
18 implies χ ( G ) ≤ ∆ + 2 (and also [6] that ∆ ≥
36 implies χ ℓ ( G ) ≤ ∆ + 2).Dvorak et al. conjectured that a similar result holds for girth 5. Conjecture 1.2 (Dvoˇr´ak, Kr´al, Nejedl´y, and ˇSkrekovski [9]) . There exists ∆ such that if G is a planar graph with girth at least and ∆( G ) ≥ ∆ then χ ( G ) ≤ ∆( G ) + 2 . Our main result verifies Conjecture 1.2, even for list-coloring. Further, in Section 4,we extend the result to paintability, also called online list-coloring. Using a version ofLemma 3.21 in [8], we can also extend the result to correspondence coloring.2 heorem 1.3.
There exists ∆ such that if G is a planar graph with girth at least fiveand ∆( G ) ≥ ∆ , then G is square ( ∆( G ) + 2 )-choosable. In particular, we can let ∆ =1 , + 1 = 2 , , . The number of colors in Theorem 1.3 is optimal, as shown by the family of graphsintroduced in [4] and depicted in Figure 1. The vertex u and its p neighbors together require p + 1 colors; since v is at distance 2 from each of them, in total we need p + 2 distinct colors. u v p Figure 1: A graph G p with girth 5, ∆( G p ) = p , and χ ( G p ) = ∆( G p ) + 2.The girth assumption is tight as well, due to a construction directly inspired from Shan-non’s triangle (see Figure 2). When coloring the square, all 3 p degree 2 vertices need distinctcolors, since each pair has a common neighbor. u vw u u p v v p w w p Figure 2: A graph G p with girth 4, ∆( G p ) = 2 p , and χ ( G p ) = 3 p .Theorem 1.3 is also optimal in another sense. But before we can explain it, we mustintroduce a more refined measure of a graph’s sparsity: its maximum average degree. Theaverage degree of a graph G , denoted ad( G ), is P v ∈ V d ( v ) | V | = | E || V | . The maximum averagedegree of G , denoted mad( G ), is the maximum of ad( H ) over every subgraph H of G . Forplanar graphs, Euler’s formula links girth and maximum average degree. Lemma 1.4 (Folklore) . For every planar graph G , (mad( G ) − g ( G ) − < . Note that every planar graph G with g ( G ) ≥ G ) < . It was proved [3]that Conjecture 1.1 is true not only for planar graphs with g ≥
7, but also for all graphswith mad < (in fact even for all graphs with mad < − ǫ , for any fixed ǫ > G with3ad( G ) < G ) ≥
17 satisfies χ ℓ ( G ) ≤ ∆( G ) + 2. These results suggested thatperhaps sparsity was the single decisive characteristic when square list coloring planar graphsof high girth. However, as we show below, Theorem 1.3 cannot be strengthened to requireonly mad( G ) < (rather than planar, with girth 5).Charpentier [7] generalized the family of graphs shown in Figure 1 to obtain for each C ∈ Z + a family of graphs with maximum average degree less than C +2 C +1 , with unboundedmaximum degree, and whose squares have chromatic number ∆ + C + 1. For C = 2 theconstruction, shown in Figure 3, yields a family of graphs with arbitrarily large maximumdegree, maximum average degree less than , and whose squares are not (∆ + 2)-colorable.In the square, all p + 4 vertices u, v , . . . , v p , w , w , x must receive distinct colors, since theyare pairwise adjacent. The maximum average degree of G p is reached on the graph G p itself.We can argue by induction that mad( G p ) < : note that mad( G ) = < and that G p +1 is built from G p by adding precisely 3 vertices and 5 edges (if a +5) b +3 ≥ then necessarily ab ≥ ). uv w x v p w v Figure 3: A graph G p with ∆( G p ) = p + 1, mad( G p ) = p +63 p +4 < and χ ( G p ) = p + 4. Most of our definitions and notation are standard; for reference, though, we collect thembelow. Let G be a multigraph with no loops. The neighborhood of a vertex v in G , denoted N ( v ), is the set of neighbors of v , i.e., N ( v ) = { u : uv ∈ E ( G ) } . The closed neighborhood ,denoted N [ v ], is defined by N [ v ] = N ( v ) ∪ { v } . For a vertex set S , let N ( S ) = ∪ v ∈ S N ( v )and N [ S ] = ∪ v ∈ S N [ v ]. For a multigraph G or digraph D and a subset S of its vertices, G [ S ]or D [ S ] is the subgraph induced by S . The degree of a vertex v in a multigraph G is thenumber of incident edges. The degree of a vertex v is denoted d G ( v ), or d ( v ) for short. So,in particular, we may have d G ( v ) > | N G ( v ) | . If H is a subgraph of G , then d H ( v ) is thenumber of edges of H incident to v . For vertices u and v with u ∈ N ( v ), the multiplicity of4he edge uv is the number of edges with u and v as their two endpoints. The maximum andminimum degrees of G are ∆( G ) and δ ( G ), respectively; when G is clear from context, wemay write simply ∆ or δ . The girth of G is the length of its shortest cycle.A multigraph G is planar if it can be drawn in the plane with no crossings. A planemap is a planar embedding of a planar multigraph such that each face has length at least3. An ℓ -face (resp. ℓ + -face ) is a face with boundary walk of length equal to (resp. at least) ℓ . The underlying map G ′ of a plane embedding of a planar multigraph G is formed fromthe embedding of G by suppressing all of its 2-faces (that is, repeatedly identifying the twoboundary edges of some remaining 2-face, until no 2-face remains). Suppose that d G ( v ) = 2and that N ( v ) = { u, w } . To suppress v , delete v and add an edge uw . Let ∆ = 1 , + 1 = 2 , ,
901 and let k ≥ ∆ . To prove Theorem 1.3, it suffices to provethat every plane graph G of girth at least five with ∆( G ) ≤ k is square ( k + 2)-choosable.Assume, for a contradiction, that this does not hold, and consider a counterexample G =( V, E ) that minimizes | V | + | E | . We fix an embedding of G . Let L be a list assignment of( k + 2) colors to each vertex of G such that G has no L -coloring. We reach a contradiction,by showing that G must contain some subgraph H such that every L -coloring of ( G \ E ( H )) can be extended to an L -coloring of G ; such an H is reducible . An unusual feature of ourproof is that we do not use the discharging method. Instead, we use only the fact that everyplanar map has a vertex of degree at most 5.A vertex v of G is big if deg( v ) ≥ √ k , and otherwise v is small . The sets of big and smallvertices of G are denoted, respectively, by B and S . Further, let S i = { v ∈ S : | N G ( v ) ∩ B | = i } , i.e., small vertices with exactly i big neighbors. By Lemma 1.4, mad( G ) < , so | E ( G ) | < | V ( G ) | /
3. Thus, only a tiny fraction of V ( G ) can be big vertices. Likewise, byplanarity S i ≥ S i has size linear in the number of big vertices, again a very small fractionof | V ( G ) | . Hence, the vast majority of V ( G ) is the subset S i =0 S i . We show that S , theset of small vertices with only small neighbors, induces an independent set. Thus, we candecompose the planar embedding into regions, each defined by a pair of big vertices. Weprove that any region with many vertices is reducible. To complete the proof, we show thatsome big vertex v is adjacent to few regions (here we use that every planar map has a vertexof degree at most 5), so v must be adjacent to a region with many vertices.We begin with a few simple observations about G . Lemma 3.1.
Graph G is connected and δ ( G ) ≥ .Proof. If G is not connected, then one of its components is a smaller counterexample, con-tradicting the minimality of G . 5f G contains a vertex u of degree 1, color G \{ u } by minimality, and extend the coloring to G as follows. Vertex u has exactly one neighbor v , whose degree is at most k , by assumptionon G . So, | N G ( u ) | = | N ( v ) \ { u }| + |{ v }| ≤ k . Since | L ( u ) | ≥ k + 2, some color c in L ( u ) isavailable to use on u . Coloring u with c gives an L -coloring of G , a contradiction.A key observation is the next lemma, which shows that in a minimal counterexample atleast one endpoint of every edge is either big or adjacent to a big vertex. Lemma 3.2.
For every edge uv of G , either u ∈ N [ B ] or v ∈ N [ B ] .Proof. Assume, for a contradiction, that there is an edge uv ∈ E ( G ) such that u, v N [ B ].In other words, u and v and all their neighbors have degree at most √ k . By the minimalityof G , there exists an L -coloring ϕ of ( G − uv ) . Now we recolor both u and v to obtain an L -coloring of G . Since u has at most √ k neighbors in G , each of which has degree at most √ k , vertex u has at most k neighbors in G . Thus, at most k colors appear on N G ( u ). Since | L ( u ) | = k + 2, at least two colors remain available for u , counting its own color. Similarly, v has at least two available colors. Thus, we may extend ϕ to u and v to obtain a proper L -coloring of G , a contradiction.The next lemma extends Lemma 3.2, by showing that if both endpoints of an edge havedegree two then both endpoints are adjacent to big vertices. Lemma 3.3. If u and v are adjacent vertices of degree two, then u ∈ N ( B ) and v ∈ N ( B ) .Proof. Suppose not. Let u and v be adjacent vertices of degree 2 such that v N ( B ). Let w be the neighbor of v distinct from u . By the minimality of G , there exists an L -coloring ϕ of ( G \ { u, v } ) . Since u has at most k + 1 neighbors in G that are already colored, wecan extend ϕ to u . Now | N G ( v ) | = | N ( w ) \ { v }| + |{ w, u }| + | N ( u ) \ { v }| ≤ d ( w ) + 2.Since w B , d ( w ) < √ k , so we can extend ϕ to v , which yields an L -coloring of G , acontradiction.The intuition behind much of the proof is that small vertices with only small neighborscan always be colored last. A key ingredient in formalizing this intuition is a new planemultigraph. Let G ′ denote the plane multigraph obtained from G by first suppressing verticesof degree 2 in S \ N ( B ) (defined at the end of Section 2) and then contracting each edgewith one endpoint in each of S and B . Note that there is a natural bijection between thefaces of G ′ and those of G . We will use Lemmas 3.2 and 3.3 to prove structural propertiesof G ′ .Since big vertices of G are not identified with each other in the construction of G ′ , wealso let B denote the vertices of G ′ that contain a big vertex of G . Let S ′ = V ( G ′ ) \ B .Note that neither suppressing nor contracting decreases the degree of a vertex in B ; thus,we conclude the following. 6 bservation 3.4. For every vertex v in B , we have d G ′ ( v ) ≥ d G ( v ) . Let G ′′ denote the underlying map of G ′ . We will next show that there is a big vertex(in G and G ′ ) whose degree in G ′′ is small; in other words, v has many edges in G ′ to thesame neighbor. But first we need the following general lemma about plane maps with certainproperties, the hypotheses of which (as we will show) are satisfied by G ′′ . Lemma 3.5.
Let H be a plane map and A , C , and D be disjoint vertex sets such that V ( H ) = A ∪ C ∪ D , every v ∈ C satisfies | N ( v ) ∩ D | ≥ , and for all v , v ∈ C such that | N ( v ) ∩ D | = | N ( v ) ∩ D | = 2 , it holds that N ( v ) ∩ D = N ( v ) ∩ D . If A is an independentset and d ( v ) ≥ for all v ∈ A , then there exists u ∈ D with d H [ D ] ( u ) ≤ and d H ( u ) ≤ . Before proving this lemma, we show how we apply it.
Lemma 3.6.
There exists v ∈ B with d G ′′ [ B ] ( v ) ≤ and d G ′′ ( v ) ≤ . Further, there exists u ∈ V ( G ′ ) (recall that V ( G ′′ ) = V ( G ′ ) ) such that at least √ k − consecutive faces of lengthtwo in G ′ have boundary ( u, v ) .Proof. To prove the first statement, we apply Lemma 3.5 to G ′′ with D = B , C = S i ≥ S i ,and A = { v ∈ S : d ( v ) ≥ } . So we must show that this application satisfies the necessaryhypotheses. (Recall that when forming G ′ , we suppressed all w ∈ S with d ( w ) = 2 andwe contracted into B all w ∈ S .) Note that A is an independent set, by Lemma 3.2.Now suppose there exist v , v ∈ C such that | N G ′′ ( v ) ∩ B | = | N G ′′ ( v ) ∩ B | = 2 and N G ′′ ( v ) ∩ B = N G ′′ ( v ) ∩ B ; say N G ′′ ( v ) ∩ B = { b , b } . When G ′′ was formed from G ,vertices v and v may have gained neighbors in B , but they did not lose neighbors. Since v , v ∈ S i ≥ S i , each already had 2 neighbors in B in G ; these must be b and b . Hence, G contains the 4-cycle b v b v , contradicting the assumption that G has girth at least 5.Thus, we can apply Lemma 3.5, as desired; the guaranteed vertex u is our desired vertex v ,which proves the first statement of the lemma.Now consider the second statement. Since d G ( v ) ≥ √ k , also d G ′ ( v ) ≥ √ k . Since d G ′′ ( v ) ≤
40, by Pigeonhole some edge uv in G ′ has multiplicity at least √ k ; furthermore, these copiesof the edge uv are embedded in G ′ to create at least √ k − uv , say t copies, are embedded in G ′ , and thus in G ′′ , such that theydo not create a 2-face. However, now the t copies of uv contribute t to d G ′′ ( v ), so they donot impede this Pigeonhole argument.)Now we prove a slight strengthening of Lemma 3.5. Lemma 3.7.
Let H be a multigraph embedded in the plane so that no face is a 2-face,except for possibly the outer face. Let w be some specified vertex on the outer face. Let A, C, D be disjoint vertex sets such that w ∈ D , V ( H ) = A ∪ C ∪ D , every v ∈ C satisfies | N ( v ) ∩ D | ≥ , and for all v , v ∈ C such that | N ( v ) ∩ D | = | N ( v ) ∩ D | = 2 we have ( v ) ∩ D = N ( v ) ∩ D . If A is an independent set and d ( v ) ≥ for all v ∈ A , then thereexists u ∈ D \ { w } such that d H [ D ] ( u ) ≤ and d H ( u ) ≤ . Before proving the lemma, we note that it implies Lemma 3.5, as follows. Suppose that H = ( V, E, F ) and V ( H ) = A ∪ C ∪ D , satisfy the hypothesis of Lemma 3.5. Choose anarbitrary vertex w on the outer face; if it is not in D , then move it to D . Now the vertex u guaranteed by Lemma 3.7 also satisfies the conclusion of Lemma 3.5. Proof of Lemma 3.7.
Suppose, to the contrary, that the lemma is false. Among all coun-terexamples, choose one, call it H = ( V, E, F ) with V = A ∪ C ∪ D such that the followingproperties hold:(1) | D | is minimized; and, subject to that,(2) the number of cut-vertices in C ∪ D is minimized; and, subject to that,(3) if possible, the outer face is a 2-face with a vertex of A on its boundary; and, subjectto that,(4) the number of parallel edges incident with vertices in A is minimized; and, subject tothat,(5) the number of edges incident to D is maximized; and, subject to that,(6) the number of edges incident to A ∪ C is maximized.We prove the lemma via a series of claims. Claim 1. H is connected, and no vertex of C ∪ D is a cut-vertex.Proof. If H is disconnected, then each of its components contains a vertex of D , so thecomponent containing w contradicts (1). Thus, H is connected. Now suppose, to thecontrary, that there exists v ∈ C ∪ D that is a cut-vertex. As a subclaim, we show thatthere is some component H i of H − v , with w / ∈ V ( H i ), such that H [ V ( H i ) ∪ { v } ] has anembedding in which no face is a 2-face except for possibly the outer face. To see this, weconsider two cases: (i) every component of H − v has a vertex on the outer face of H and(ii) some component does not. In (i), we simply take some H i such that w / ∈ V ( H i ). Now H [ V ( H i ) ∪ { v } ] has no 2-faces, except for possibly the outer face. So we are done. In (ii), wetake some component H i such that no other component H j lies inside a face of H [ V ( H i ) ∪{ v } ]in H . That such a component exists follows from the fact that every rooted tree has at leastone leaf that is not the root. (We construct a rooted forest where every component of H − v with a vertex on the outer face of H is a root of its own tree, and a component H j of H − v is a child of a component H k if H j lies inside a face of H [ V ( H k ) ∪ { v } ].) This proves thesubclaim. 8et H ′ = H [ V ( H i ) ∪ { v } ]. Let D ′ = ( D ∩ V ( H ′ )) ∪ { v } , C ′ = ( C ∩ V ( H ′ )) \ { v } , A ′ = A ∩ V ( H ′ ), and w ′ = v . Note that | D ′ | ≤ | D | , since D ′ \ { v } ⊆ D and w ∈ D \ D ′ .(Of course, if v = w , then w / ∈ D \ D ′ . However, then we do not move v from C ′ to D ′ ,so again | D ′ | ≤ | D | .) Further, v is a cut-vertex of H , but not of H ′ . So, by (1) and (2) inour choice of H , the lemma holds for H ′ , with w ′ = v ; that is, there exists u ∈ D ′ \ { v } with the desired properties. This proves the lemma for H , since d H [ D ] ( u ) = d H ′ [ D ] ( u ) and d H ( u ) = d H ′ ( u ). Claim 2.
Every + -face f contains a vertex of A on its boundary.Proof. Suppose, to the contrary, there exists a 3 + -face f with no vertex of A on its boundary.Now we can add a new vertex, v , to A and make v adjacent to every vertex on f . Thiscontradicts (5) or (6) in our choice of H . Claim 3.
Every + -face f is a 3-face, and the boundary of every -face contains exactlyone vertex of A . For every vertex u ∈ V ( H ) and every pair of vertices v , v such that uv and uv appear consecutively in the cyclic order of edges incident to u , there exists an edge v v such that edges v u , uv , v v induce a face of length (possibly with additional paralleledges).Proof. Since v u and uv are consecutive around u , they lie on a 3 + -face f . Suppose, to thecontrary, that some face f has length at least 4. By Claim 2, f contains a vertex v ∈ A onits boundary. Since d ( v ) ≥
3, set A is independent, and D ∪ C contains no cut-vertex, thevertices that immediately precede and succeed v on the boundary of f are distinct and arenot in A ; call them v and v . Since f has length at least 4, we can add the edge v v in theinterior of f , while maintaining planarity (and without creating a 2-face). This contradicts(5) or (6) in our choice of H . Thus, f has length 3. By Claim 2, V ( f ) contains at least onevertex in A . Since A is an independent set, V ( f ) contains exactly one vertex of A . Finally,by definition v , u, v appear along the boundary of some face, f . That f is a 3-face followsfrom the first statement. Claim 4.
The outer face is a 2-face with a vertex of A on its boundary.Proof. Suppose, to the contrary, this is false. Let f denote the outer face. By Claim 3,either f is a 2-face with no vertex of A on its boundary, or f is a 3-face with a vertex of A on its boundary. Suppose the former, and let w, v be the boundary vertices of f (where w is specified in the hypotheses of the lemma). Add a new vertex a ∈ A , add edge av and twocopies of edge aw , so that the outer 2-face is bounded by a and w . This contradicts (3) inour choice of H . Instead, assume the latter, and let a, v, w be the boundary vertices of f .Now add a second copy of aw , so that the new outer face is a 2-face, bounded by a and w .Again, this contradicts (3) in our choice of H . This proves the claim.9 laim 5. Every vertex in C has at most 3 neighbors in D .Proof. Suppose, to the contrary, that some vertex u ∈ C has neighbors v , v , . . . , v p ( p ≥ D . We add an edge v v and replace u by two new vertices u and u in C , where u isadjacent to v , v , v and u to v , . . . , v p , v . Further, u inherits all neighbors of u in A ∪ C that appear between v and v or between v and v in the cyclic order of neighbors of u .Likewise, u inherits all other neighbors of u in A ∪ C . This contradicts (4) or (5) in ourchoice of H . Claim 6.
No vertex in A has incident parallel edges, with a single exception: on the outer2-face, with boundary vertices a ∗ ∈ A and w ∈ D , vertex a ∗ has exactly two edges to w .Further, ≤ d ( a ∗ ) ≤ , and every vertex in A \ { a ∗ } has degree and at least one neighborin C . Finally, if d ( a ∗ ) = 4 , then a ∗ has a neighbor in C .Proof. Let A ∗ = A \ { a ∗ } . Assume there is u ∈ A ∗ and v ∈ C ∪ D such that H has paralleledges between u and v ; choose such a pair u, v and two such edges that bound a face f in H [ { u, v } ] (other than the outer face), so as to minimize the number of vertices embeddedinside f . By Claim 1, vertex v is not a cut-vertex, so u must have a neighbor inside f . Let V ′ be the set of vertices that lie inside f (including u and v ). Let H ′ = H [ V ′ ], let A ′ = A ∩ V ′ ,let C ′ = ( C ∩ V ′ ) \ { v } , let D ′ = ( D ∩ V ′ ) ∪ { v } , and let w ′ = v . Now H ′ contradicts (4)in our choice of H . If a ∗ has more than 2 edges to w or has parallel edges to a vertex otherthan w , then nearly the same argument gives a contradiction, using a ∗ in place of u . Thisproves the first statement.Suppose, to the contrary, there is a vertex u in A ∗ of degree at least 4, and denote itsconsecutive neighbors in a cyclic order by v , . . . , v p ( p ≥ v v and replace u by two new vertices u and u in A , where u is adjacent to v , v , v and u to v , . . . , v p , v .This contradicts (5) or (6) in our choice of H . If d ( a ∗ ) ≥
5, then we do something similar,as follows. Let v , . . . , v p denote the neighbors of a ∗ in cyclic order, with v = w . Remove a ∗ and add three new vertices a ∗ , a ∗ , a ∗ . Let a ∗ inherit from a ∗ edges to v , v , v . Let a ∗ inheritfrom a ∗ edges to v , . . . , v p , v . Lastly, let a ∗ have two edges to v = w (bounding the outerface), and also an edge to v . This contradicts (5) or (6) in our choice of H .Finally, suppose that u ∈ A ∗ and all 3 of its neighbors are in D . Now we can move u to C , and add a new vertex u ′ to A in one of the triangular faces f incident to u , making u ′ adjacent to every vertex on f . This contradicts (5) in our choice of H . Thus, each vertex in A ∗ has a neighbor in C . Nearly the same modification works if d ( a ∗ ) = 4 and all neighborsof a ∗ are in D . However, now we must ensure that (3) holds for the modified graph, since(3) holds for H . So we put u ′ into the outer 2-face of H and join it with a ∗ and (by twoedges) with w . This creates a new outer 2-face bounded by u ′ ∈ A and w . Claim 7.
Every face contains a vertex of D , and C is an independent set. roof. Assume, for a contradiction, that there are faces containing only vertices of A ∪ C ,and take f to be one adjacent to a face f ′ containing a vertex of D . By hypothesis, w ∈ D and w is on the outer face. So any face sharing an edge with the outer face has a vertex of D on its boundary. Since every non-outer face in H is a 3-face, this yields a 4-cycle ( u, v , v , v )where u ∈ D , v , v , v ∈ A ∪ C and each of ( v , u, v ) and ( v , v , v ) induces a face. ByClaim 2, either v ∈ A or v ∈ A . By symmetry, assume that v ∈ A .Since A induces an independent set, v , v ∈ C . Consider the other face f ′′ = ( v , v , v )incident to v v . Note that v = v , since v ∈ A , so d ( v ) ≤
4. Also, v = u , since v ∈ A and u ∈ D . Let u ′ denote the third neighbor of v . Now we delete the three edges v v , v v , and v v , and add the two edges uv and u ′ v . This contradicts (5) in our choice of H ,unless u ′ = u , so in the modified graph u is a cut-vertex. However, in this case we proceedas in the proof of Claim 1. Since v and v each have a neighbor in D \ { u } , the graph H ′ satisfies | D ′ | < | D | , which contradicts (1) in our choice of H . So, we can assume that u = u ′ .One other possible exception is that v = a ∗ and u ′ = w , and one of the edges from a ∗ to w stops us from adding the edge u ′ v . Now we simply delete v a ∗ and add v w , contradicting(5) in our choice of H . Consequently, every face contains a vertex of D .If there is an edge v v between two vertices in C , then the two incident 3-faces musteach contain a vertex in D , respectively u and u . Now we delete edge v v and add anedge u u , which contradicts (5) in our choice of H . Thus, C is an independent set. Claim 8.
Let H ′ denote H [ C ∪ D ] , the plane multigraph induced by the vertices of C and D , with the plane embedding inherited from H . Every face of H ′ is a 3-face, except possiblythe outer face, and d H ′ ( u ) = d H ( u )2 for every u ∈ ( C ∪ D ) \ { w } .Proof. Let a ∗ denote the vertex of A on the outer 2-face, and let A ∗ = A \ { a ∗ } . Let f bean arbitrary non-outer face of H ′ . Since every non-outer face of H contains a vertex of A ,face f does not exist in H ; so f was formed by deleting one or more vertices of A . Since A is an independent set in H and every non-outer face of H is a 3-face, f contained exactlyone vertex of A . Since d H ( a ) = 3 for all a ∈ A ∗ , face f has length 3, and the vertices of A and C ∪ D alternate in the neighborhood of every vertex u ∈ ( C ∪ D ) \ w in H . This impliesthat d H ′ ( u ) = d H ( u )2 . Claim 9.
Let H ′′ denote H [ D ] , the plane multigraph induced in H by the vertices of D . Form H ′′′ from H ′′ by repeatedly deleting one edge in a 2-face until the resulting graph contains no2-faces. For every vertex u ∈ D that is not on the outer 2-face of H ′ (if the outer face of H ′ is a 2-face), we have d H ′′ ( u ) = d H ′ ( u )2 and d H ′′′ ( u ) ≥ d H ′′ ( u )2 . Further, there exists vertex u ∈ D \ { w } (also not on the outer face of H ′ if it is a 2-face) with d H ′′′ ( u ) ≤ . Thus, d D ( u ) = d H ′′ ( u ) ≤ and d H ( u ) ≤ d H ′′′ ( u ) ≤ . roof. By Claim 8, every face of H ′ is a 3-face, except possibly the outer face. Further, eachsuch 3-face has a vertex of C on its boundary, since no vertex in A has all three neighborsin D , by Claim 6. Since C is an independent set (by Claim 7), there is in fact a bijectionbetween C and the faces of H ′′ (excluding the outer face of H ′′ if the outer face of H ′ is a2-face bounded by two vertices in D ). So, for every vertex u ∈ D not on an outer 2-face of H ′ , we have d H ′′ ( u ) = d H ′ ( u )2 . By hypothesis on H , no two vertices of C , each with exactlytwo neighbors in D , have the same two neighbors in D . Thus, no two faces of length 2 in H ′′ share an edge (except that one can possibly share an edge with the outer 2-face, if itexists). Therefore, by deleting exactly one edge in every face of length 2, we obtain a planemultigraph H ′′′ with no 2-face except possibly the outer face, and such that for every vertex u ∈ D (not on an outer 2-face of H ′ ) we have d H ′′′ ( u ) ≥ d H ′′ ( u )2 . Since every non-outerface of H ′′′ is a 3 + -face, Euler’s formula implies that | E ( H ′′′ ) | ≤ | D | −
5. Thus, there issome u ∈ D that is not on an outer 2-face of H ′ such that d H ′′′ ( u ) ≤
5. As a consequence, d D ( u ) = d H ′′ ( u ) ≤
10 and d H ( u ) ≤ d H ′′′ ( u ) ≤ c b b b c c H c b b b c c H ′ b b b H ′′ b b b H ′′′ Figure 4: The evolution, in the proof of Lemma 3.7, from H to H ′′′ of a subgraphof H . Here b , b , b ∈ D , c , c , c ∈ C . Each represents a vertex in A adjacentto all three vertices incident to the face. (Black vertices have all incident edgesdrawn, but white vertices may have more incident edges.)Recall that S ′ = V ( G ′ ) \ B . Lemma 3.8.
No vertex in S ′ is incident to or more consecutive faces of length in G ′ .Proof. Assume, for a contradiction, that there is an edge uv in G ′ , with u ∈ S ′ , such that atleast 3 consecutive faces have boundary ( u, v ). First consider the case where v ∈ S ′ . In theconstruction of G ′ from G , an edge is added between u and v only when there is a vertexof degree 2 adjacent to both u and v that is suppressed. Hence, regardless of whether uv belongs to E ( G ) or is formed from the suppression of a vertex of degree 2 adjacent to u and v , there exists a cycle in G of length at most 4, contradicting that G has girth at least five.So we may assume that v ∈ B . Let T = S ∩ N ( v ), that is the set of small neighborsof v with exactly one big neighbor (which must be v ). In the construction of G ′ , an edge is12dded between u and v only if either there is a neighbor of u in T , or if there is a vertexof degree 2 adjacent to both u and to a vertex in T . Let U denote the set of vertices ofdegree 2 that are adjacent to u and also to a vertex in T . Note that each copy of uv in G ′ corresponds to a path of length at most 3 in G with all vertices in { u, v } ∪ U ∪ T . Thus, uv / ∈ E ( G ), since this would create a 3-cycle or 4-cycle in G , contradicting that G has girthat least 5.Since uv has multiplicity at least four, and the copies of uv form 3 consecutive faces oflength 2 in G ′ , we know 4 ≤ | U | + | N ( u ) ∩ T | ; further, there exist four vertices, w , . . . , w ,in T ∪ U that are consecutive in the cyclic neighborhood of u in G . Since G has girth atleast five, | N ( u ) ∩ N ( v ) | ≤
1, so | N ( u ) ∩ T | ≤
1. Thus, at least one of w and w is in U ;by symmetry, assume w ∈ U . Let x be the neighbor of w in G distinct from u . Notethat x ∈ T . Since G has girth at least five, x is neither adjacent to a neighbor of v nor toa neighbor of a neighbor of v . Regardless of whether w belongs to U or T , it follows byplanarity that w and v are the only neighbors of x in G . Therefore d ( x ) = 2, which is acontradiction to Lemma 3.3, since w N ( B ).Now we use Lemma 3.8 to strengthen the final conclusion of Lemma 3.6. Recall that G ′′ is formed from G ′ by suppressing 2-faces. Corollary 3.9.
There exist u, v ∈ B ∩ V ( G ′ ) such that there are at least √ k − consecutivefaces of length two with boundary ( u, v ) in G ′ .Proof. Let v be as in Lemma 3.6. By Lemma 3.8, each edge from v to a small vertex in G ′′ accounts for at most 3 consecutive edges in G ′ incident to v . Thus, the | N G ′′ ( v ) ∩ B | big neighbors of v account for the remaining at least d G ( v ) − v is big, d G ′ ( v ) ≥ d G ( v ) ≥ √ k . By Pigeonhole, some edge from v to a big vertex, say u , accountsfor at least ( √ k − /d G ′′ [ B ] ( v ) consecutive edges incident to v in G ′ ; and the number ofconsecutive faces with boundary ( u, v ) is one less. Since d G ′′ [ B ] ( v ) ≤
10, some big neighbor u shares with v at least √ k −
13 consecutive faces of length 2.If b , b ∈ B ∩ V ( G ′ ) are such that at least r consecutive faces f ′ , . . . , f ′ r of G ′ haveboundary ( b , b ), then these faces are an r -region R ′ of G ′ ; see Figure 5. Analogously, an r -region R of G is a set of faces which contract to an r -region R ′ in G ′ . We define V ( R ) as( S ri =1 V ( f i )) \ { b , b } . Observation 3.10. If R is an r -region of G , then V ( R ) = B ∪ B ∪ D , where B , B ,and D are disjoint vertex sets such that B ⊆ N ( b ) and B ⊆ N ( b ) for some b , b ∈ B ;further, D is an independent set of degree two vertices, each of which has one neighbor in B and the other neighbor in B .Proof. Let R be an r -region of G . By definition, there exist b , b ∈ B ∩ V ( G ′ ) such that the r -region R ′ consists of at least r consecutive faces in G ′ , each with boundary ( b , b ). Recall13 k + b √ k + b Gf f f f f B D B √ k + b √ k + b G ′ f f f f f Figure 5: A 5-region in G and the corresponding 5-region in G ′ .that G ′ is formed from G by suppressing the degree 2 vertices in S \ N ( B ) and contractingeach edge joining S and B . By Lemma 3.3, these suppressed degree 2 vertices form anindependent set. Thus, each copy of b b in R ′ in G ′ corresponds to a path of length 1, 3, or4 joining b and b in G . Each such path P (of length at least 3) must contain a vertex fromeach of B and B . If P contains another vertex w , then w must be suppressed in forming G ′ , so w must be a degree 2 vertex with a neighbor in each of B and B . This proves theobservation.Hereafter, we use B , B , D , b , b , and V ( R ) as defined in the previous observation. Lemma 3.11. If R is an r -region of G , then B and B are independent sets, and each v ∈ B ∪ B satisfies | N ( v ) ∩ V ( R ) | ≤ .Proof. The fact that B and B are independent sets follows from the assumption that G has girth at least five. Now choose v ∈ B ∪ B and suppose, for a contradiction, that | N ( v ) ∩ V ( R ) | ≥
4. Without loss of generality, we may assume that v ∈ B ; see Figure 6.Recall that V ( R ) ⊆ B ∪ B ∪ D . Since B is independent, | N ( v ) ∩ B | = 0; thus | N ( v ) ∩ ( B ∪ D ) | ≥
4. Since G has girth at least five, | N ( v ) ∩ B | ≤
1, so | N ( v ) ∩ D | ≥
3. Hence,by planarity, there exists u ∈ N ( v ) ∩ D such that if w is the other neighbor of u , then vw ∈ E ( G ′ ) (actually v gets contracted into b and w gets contracted into b when forming G ′ ) and vw is incident with two faces, each of length two, in region R ′ . Since G has girth atleast five, it follows that w has degree two in G . But now u and w are adjacent vertices ofdegree two, yet u N ( B ), which contradicts Lemma 3.3.To complete the proof of Theorem 1.3, we need one more reducible configuration; inLemma 3.14, we show that an r -region is reducible, if r ≥ χ ′ ℓ ( G ) = ∆( G ) for every bipartite graph G (here χ ′ ℓ denotes the edge list chromatic number).14 k + b vuw √ k + b G D B √ k + b √ k + b G ′ Figure 6: An illustration of the proof of Lemma 3.11.A kernel in a digraph D is an independent set F of vertices such that each vertex in V ( D ) \ F has an out-neighbor in F . A digraph D is kernel-perfect if for every A ⊆ V ( D ),the digraph D [ A ] has a kernel. To prove our next result, we will need the following lemma ofBondy, Boppana, and Siegel (see [1, p. 129] and [10, p. 155]). For completeness, we includean easy proof. Lemma 3.12.
Let D be a kernel-perfect digraph with underlying graph G . If L is a list-assignment of V ( G ) such that for all v ∈ V ( G ) , | L ( v ) | ≥ d + ( v ) + 1 , then G is L -colorable.Proof. We use induction on | V ( G ) | . Choose some color c ∈ ∪ v ∈ V ( G ) L ( v ). Let A c be the setof vertices with color c in their lists. By assumption, D [ A c ] contains a kernel, F c . Use color c on each vertex of F c . Now let D ′ = D \ F c and L ′ ( v ) = L ( v ) − c for each v ∈ V ( D ′ ).By induction, the remaining uncolored digraph D ′ can be colored from its lists L ′ ; we mustonly check that D ′ and L ′ satisfy the hypothesis of the lemma. Since D is kernel-perfect,so is D ′ . Further, each vertex of D ′ lost at most one color from its list (namely, c ). Moreprecisely, each vertex of A c \ F c lost one color from its list and each other vertex lost no colors.Fortunately, since F c is a kernel for A c , we get d + D ′ ( v ) ≤ d + D ( v ) − v ∈ A c \ F c . Thus, | L ′ ( v ) | ≥ d + D ′ ( v ) + 1 for every v ∈ V ( D ′ ), as desired.We now use Lemma 3.12 to prove the following lemma, which we will use to show thatlarge regions are reducible for square (∆ + 2)-choosability. Lemma 3.13.
Let H be a graph covered by two disjoint cliques B and B , L be a list-assignment for V ( H ) , and S ⊆ B and S ⊆ B be such that • if v ∈ B i , then | N ( v ) ∩ V ( B − i ) | ≤ , if v ∈ B i \ S i , then | L ( v ) | ≥ | B i | , • if v ∈ S i , then | L ( v ) | ≥ | B i | − .Now if | B | ≥ , | B | ≥ , | S | ≤ , and | S | ≤ , then H is L -colorable.Proof. We construct a kernel-perfect orientation D of H satisfying Lemma 3.12 as follows.Let x , x , . . . , x | B | be an ordering of the vertices of B and y , y , . . . , y | B | be an orderingof the vertices of B such that • x i ∈ S iff 1 ≤ i ≤ | S | , and • y i ∈ S iff 1 ≤ i ≤ | S | , and • N B ( x a ) ∩ N B ( N B ( y b )) = ∅ for each a and b such that | B | − ≤ a ≤ | B | and | B | − ≤ b ≤ | B | .It is helpful to restate the third condition in words: there is no path of length 1 or 3that starts at one of the final 3 vertices in B , ends at one of the final 3 vertices in B , andalternates between B and B . We claim that such an ordering exists. To see this, let thevertices of S be x , . . . , x | S | in any order and similarly for S . Now it suffices to ensure thethird condition holds. Note that | B | − | S | ≥
10. Suppose there exists u ∈ B \ N ( S )with d B ( u ) = 3. Choose N B ( u ) to be the three final vertices of B , and call this set Z . x . . . . . . . . . x p x k x r x | B | B y . . . . . . . . . y q y j y s y | B | B S S WZ Figure 7: The proof of Lemma 3.13, constructing the orientation D of H , whichshows that this situation cannot occur, due to our choices of W and Z .Now | N B ( Z ) \{ u }| ≤
6, so | N B ( N B ( Z )) \ Z | ≤ | N B ( N B ( N B ( Z )) \ Z ) | ≤ | B | − | S | − − |{ u }| ≥
3, we can choose the desired 3 final vertices of B ; call this set W . If no such u exists, then there exist 3 vertices v , v , v ∈ B \ ( S ∪ N ( S ))such that d B ( v i ) ≤ i ∈ { , , } . Now swap the roles of B and B and let Z = { v , v , v } . The analysis is essentially the same, except that now we have no vertex u .This proves the claim that such an ordering exists.16et D be obtained from H by directing the edges of H as follows. For each edge withboth endpoints in B or both endpoints in B , direct the edge from the vertex with higherindex to the vertex with lower index. For each edge between B and B , direct the edge inboth directions, unless one endpoint is among the final three vertices of B or B ; in thatcase, only direct the edge into the vertex among the final three (recall that no edge has oneendpoint among the final three vertices of B and the other endpoint among the final threevertices of B ).We claim that D is a kernel-perfect orientation. Let A ⊆ V ( H ). Let p = min { i : x i ∈ A } and q = min { j : y j ∈ A } . If A ∩ V ( B ) = ∅ , then { x p } is a kernel of A as desired. Similarly if A ∩ V ( B ) = ∅ , then { y q } is a kernel of A as desired. So we may assume that A ∩ V ( B ) = ∅ and A ∩ V ( B ) = ∅ . If x p y q E ( H ), then { x p , y q } is a kernel of A as desired. So we assumethat x p y q ∈ E ( H ).Let r = min { k : x k ∈ A, x k N H ( y q ) } and s = min { ℓ : y ℓ ∈ A, y ℓ N H ( x p ) } . Now { x p , y s } is a kernel of A , unless there exists j with q ≤ j < s such that y j ∈ A and x p y j iseither not an edge of H or is only directed from x p to y j . Given the choice of s , it must bethat x p y j is only directed from x p to y j . (If { ℓ : y ℓ ∈ A, y ℓ / ∈ N ( x p ) } = ∅ , then the sameargument works with { x p } in place of { x p , y s } .) Thus, we conclude that y j is among thefinal 3 vertices of B . Now, we instead take as our kernel { y q , x r } . This is a kernel unlessthere exists k with p ≤ k < r such that x k ∈ A and either x k y q is not an edge or it isonly directed from y q to x k . Given our choice of r , we know that x k y q is an edge. But if x k y q is only directed from y q to x k , then x k is among the final 3 vertices of B . (Similar toabove, if { k : x k ∈ A, x k / ∈ N ( y q ) } = ∅ , then we use { y q } in place of { y q , x r } .) However,this is impossible, since now the path x k y q x p y j contradicts the third condition. Thus, D iskernel-perfect, as desired.Finally, we claim that | L ( v ) | ≥ d + D ( v )+1 for all v ∈ V ( H ). First suppose that v ∈ S ∪ S .Now v has at most 11 out-neighbors within its clique and at most 3 out-neighbors in theother clique, so d + D ( v ) ≤
14. Since | B | ≥
18 and | B | ≥
18, we have | L ( v ) | ≥ | B i | − ≥ ≥ d + D ( v ) + 1. Next, suppose that v ∈ ( B ∪ B ) \ ( S ∪ S ), but v is not among the final 3vertices of either B i . By symmetry, we can assume that v ∈ B . Since v has no out-neighborsamong the final 3 vertices of B , it has at most | B | − B . Since v has atmost 3 out-neighbors in B , we have | L ( v ) | ≥ | B | = ( | B | −
4) + 3 + 1 ≥ d + D ( v ) + 1. Nowsuppose that v is among the final 3 vertices of B or B ; by symmetry, assume that v ∈ B .Since all out-neighbors of v are in B , we get d + D ( v ) ≤ | B | −
1; thus, | L ( v ) | ≥ d + D ( v ) + 1. Lemma 3.14.
For every r ≥ , graph G does not have an r -region.Proof. Suppose, to the contrary, that G has such an r -region R , with r ≥ B , B , D , b , and b be as in Observation 3.10. Let v and v be adjacent vertices of B ∪ B ∪ D such that every vertex within distance 2 in G of v or v is in { b , b } ∪ N ( b ) ∪ ( b ) ∪ V ( R ). To see that such vertices exist, pick v ∈ B such that each face containing v is in R , and let v be a neighbor of v in B ∪ D . By the minimality of G , we can L -color( G − v v ) ; call this coloring ϕ . Now we uncolor many of the vertices in V ( R ) and extendthe coloring to G using Lemma 3.13, as well as greedily coloring vertices of D last. Thedetails forthwith.Note that | N ( b ) ∩ N ( b ) | ≤
1, since G has girth at least 5. In fact, if v ∈ N ( b ) ∩ N ( b ),then v ∈ V ( G ′ ), since v is neither suppressed nor contracted into a big vertex. Thus, v / ∈ V ( R ). So V ( R ) ∩ N ( b ) ∩ N ( b ) = ∅ .Let S be the set of vertices in B ∪ B that are incident with a face of G not in R . Let B ′ = B \ N [ S ] and B ′ = B \ N [ S ]. Note that B ′ and B ′ are independent sets in G butare cliques in G . Let H = G [ B ′ ∪ B ′ ]. For each v ∈ V ( H ), let L ′ ( v ) = L ( v ) \ { c : ϕ ( w ) = c for some w ∈ N G ( v ) \ ( V ( H ) ∪ D ) } . Let S i = B ′ i ∩ N ( N [ N [ S ]] ∩ D ) for each i ∈ { , } .Note that S i consists of vertices of B ′ i that are adjacent in G to (colored) vertices in N [ N [ S ]]via vertices in D .To color H by Lemma 3.13, we first verify that each v ∈ B ′ i has at most 3 neighborsin B ′ − i in H . By symmetry, assume v ∈ B ′ . Now each neighbor of v in B ′ in H is eitheradjacent to v in G or has a common neighbor with v in D . Further, each neighbor in G in V ( R ) yields at most one such neighbor in B ′ , since B ′ is independent in G and each vertexin D has degree 2. So we are done by Lemma 3.11.We must also verify that S and S are small enough and that B ′ , B ′ , and all of the lists L ′ are big enough. Note that | B ∩ S | = 2 and | B ∩ S | = 2, since each vertex of S mustbe on the first or last edge of the r -region, in G ′ , and each of these edges has exactly onevertex in each of B and B . Each v ∈ N ( S ) ∩ B has at most three neighbors, in G , in B by Lemma 3.11. Further, one of these three is in S . So | S | = | B ′ ∩ N ( N [ N [ S ]] ∩ D ) | ≤ | S ∩ B | + 2 | S ∩ B | ≤ | S | ≤ v ∈ B ′ ∪ B ′ ; by symmetry, assume v ∈ B ′ . Note that | L ′ ( v ) | ≥ | B ′ | whenever v ∈ B ′ \ S , since each v ∈ B ′ \ S loses at most one color for each vertex in ( { b , b }∪ N ( b )) \ ( V ( H ) ∪ D ), and D ∩ ( { b , b } ∪ N ( b )) = ∅ . Each v ∈ B has at most three neighbors in B . Thus, each vertex v ∈ S has at most three colored neighbors, in G , in B \ B ′ . So, v loses at most three more colors than in the analysis for vertices in B ′ \ S . Hence, each v ∈ S has | L ′ ( v ) | ≥ | B ′ | −
3. Similarly, for each v ∈ S we get | L ′ ( v ) | ≥ | B ′ | − B ′ and B ′ are big enough. The number of edges of G ′ incident withthe region R ′ is | R ′ | + 1. By Lemma 3.11, every vertex of B or B is in at most three ofthose edges, so | B | ≥ ( | R | + 1) / | B | ≥ ( | R | + 1) /
3; we can actually get better boundsusing planarity, but we omit that argument to keep the proof simpler. Now | S ∩ B | = 2 and | N ( S ) ∩ B | ≤ | S ∩ B | ≤
6, so | N [ S ] ∩ B | ≤
8. Hence | B ′ | ≥ ( | R | + 1) / − | N [ S ] ∩ B | ≥ (161 + 1) / − | B ′ | ≥ V ( H ). After coloring V ( H ),18or each vertex x ∈ D , we can color it arbitrarily from its list, since | L ( x ) | ≥ k + 2 and d G ( x ) ≤ √ k . Hence, G has an L -coloring, a contradiction. Proof of Theorem 1.3.
Recall, from the start of Section 3, that G is a minimal counterex-ample to Theorem 1.3. By Corollary 3.9, G contains some r -region with √ k − ≤ r . ByLemma 3.14, G contains no r -region with r > √ k − ≤ k ≤ , = 2 , , ≥ , + 1 we reach a contradiction, whichproves the theorem (our main result).By relying more heavily on planarity, we can reduce the value of ∆ . However, thatapproach adds numerous complications, which we prefer to avoid. In this section, we explain how the proof of Theorem 1.3 yields a polynomial algorithmto color G from its lists. In fact, we give an algorithm for the more general context ofpaintability (which we define below). Essentially, we construct a vertex order σ such thatwe can consider the vertices of G in order σ and color them greedily from their lists, butthere is a wrinkle. If vertices appear together in an r -region, for r ≥ Theorem 4.1.
Let G be a planar graph with girth at least 5 and fix k ≥ max { ∆( G ) , +1 } . As in Section 3, let B denote the set of vertices w with d ( w ) ≥ √ k . Now G contains atleast one of the following:(a) only a single vertex,(b) 2 or more components,(c) a vertex of degree at most 1,(d) an edge uv such that u / ∈ N [ B ] and v / ∈ N [ B ] ,(e) adjacent 2-vertices u and v such that u / ∈ N ( B ) or v / ∈ N ( B ) , and(f ) an r -region with r ≥ .Proof. The proof is essentially the same as the proof of Corollary 3.9. The key observationis that in that proof, and the results upon which it depends, we do not explicitly use that G is a minimal counterexample to Theorem 1.3; we only need that G has no instance of(a), (b), (c), (d), or (e). So, as in Corollary 3.9, we conclude that G has an r -region with r ≥ √ k −
13. Since k > , this gives r > b -paintability (also called online b -list-coloring ) is played between two play-ers, Lister and Painter. On round i , Lister presents a set J i of uncolored vertices. Painterresponds by choosing some independent set I i ⊆ J i to receive color i . If Painter eventuallycolors every vertex of the graph, then Painter wins. If instead Lister presents some uncol-ored vertex on b rounds, but Painter never colors it, then Lister wins. The paint number χ p ( G ) is the minimum b such that Painter can win regardless of how Lister plays. Let G be a planar graph with girth at least five. We show that if G has maximum degree∆ ≥ + 1 = 2 , , χ p ( G ) ≤ ∆( G ) + 2. Theorem 4.2.
Let G be a planar graph with girth at least 5. Let k = max { ∆( G ) , + 1 } .Now χ p ( G ) ≤ k + 2 .Proof. A weak order of a vertex set V ( G ) is a generalization of a total order, where wepartition V ( G ) into subsets and then form a total order on these subsets. In many cases,the subsets will be singletons, though not always. Further, each non-singleton subset gets alabel that is all vertices in the subset. (Later, we may remove vertices from a subset, butwe never change its initial label.) For convenience, we simply list the subsets to reflect thetotal order (from least to greatest).Let G satisfy the hypothesis. We first construct a weak order σ of V ( G ), by induction on | V ( G ) | + | E ( G ) | , using the six cases in Theorem 4.1, applying the first case that is applicable.(a) G = v and σ = v .(b) Suppose G has 2 or more components G , . . . , G t , with t ≥
2. By hypothesis, constructweak orders σ , . . . , σ t , and form σ by concatenating these, in any order.(c) If G has a vertex v of degree at most 1, then let G ′ = G − v . Let σ ′ be the weak orderfor G ′ and form σ by appending v to σ ′ .(d) If G has an edge uv such that u / ∈ N [ B ] and v / ∈ N [ B ], then let G ′ = G − uv . Let σ ′ be the order for G ′ . Form σ from σ ′ by removing u and v from their places in theweak order and appending u, v .(e) If G has adjacent 2-vertices u and v such that u / ∈ N ( B ) or v / ∈ N ( B ), then bysymmetry, assume u / ∈ N ( B ). Let G ′ = G \ { u, v } and let σ ′ be the order for G ′ . Form σ from σ ′ by appending v, u .(f) If G contains an r -region with r ≥ B ′ , B ′ , and D as in the proof ofLemma 3.14. Let v and v be vertices of B ′ ∪ B ′ ∪ D such that every vertex withindistance 2 in G of v or v is in { b , b } ∪ N ( b ) ∪ N ( b ) ∪ V ( R ). G ′ = G − v v andlet σ ′ be the order for G ′ . Form σ from σ ′ by removing the vertices of B ′ , B ′ , and D from wherever they appear, possibly in labeled (non-singleton) subsets, appending20he subset B ′ ∪ B ′ (with label B ′ ∪ B ′ ), followed by each of the vertices of D (assingletons) in arbitrary order.This completes the construction of the weak order σ of V ( G ). Note that any labeledsubset in σ must arise from some B ′ ∪ B ′ in (f). Now we use σ to describe a strategy forPainter to win the ( k + 2)-painting game on G .For a given round i , suppose Lister lists the set J i . Let σ i be the restriction to J i of σ . Painter greedily constructs an independent set I i as follows. If the least element, v ,in σ i is unlabeled, add it to I i and modify σ i by deleting all vertices adjacent to v in G .Suppose instead the least element is a labeled subset, call it T j , which arose in (f) from some r -region, R . Let D be the digraph formed in the proof of Lemma 3.13 corresponding to R (the vertices of D are encoded in the label of T j ). Since D is kernel-perfect, D [ T j ] has akernel, T ′ j . Now add the vertices of T ′ j to I i , and delete from σ i every vertex adjacent in G to one or more vertices of T ′ j . This completes the description of Painter’s strategy. It canclearly be implemented in polynomial time. Determining if an arbitrary graph has a kernelis NP-hard. However, the proof of Lemma 3.13 is constructive and gives rise to a simplealgorithm to find a kernel.Finally, we show that Painter’s strategy described above always wins the ( k + 2)-paintinggame on G . We need to consider vertices that were put into σ by each of (a) and (c)–(f).(In the process of recursively building σ , a vertex v may possibly be removed from a weakorder for a smaller graph, and reinserted at the end, as in (d) or (f). In this case, we classify v according to the final step that placed it in σ .)(a) Suppose v was put into σ by (a). Now v has no earlier neighbors (in G ) in σ , so v iscolored on the first round on which it appears.(c) Suppose vertex v was put into σ by (c). This means that v has at most k vertices thatappear earlier in σ and are adjacent to v in G . Thus, v can appear in J i \ I i on atmost k rounds. So, when the game ends, v is colored.(d) Suppose vertex v was put into σ by (d). Since v / ∈ N [ B ], in G vertex v has at most( √ k ) = k neighbors. So, when the game ends, v is colored.(e) Suppose vertex v was put into σ by (e). If the other 2-vertex u put into σ by (e) follows v , then at most k + 1 vertices w that are adjacent in G to v precede v in σ , so whenthe game ends v will be colored. Otherwise, v / ∈ N [ B ], so in G , vertex v has at most k neighbors, among vertices earlier in σ . So, when the game ends, v is colored.(f) Finally, suppose v was put into σ by (f). If v was in D for some r -region, then at most k neighbors in G of v precede v in σ , so v will be colored when the game ends. Thus,we assume v ∈ B ′ ∪ B ′ for some r -region (with r ≥ ∈ B ′ . Let D be the digraph formed in the proof of Lemma 3.13 by orienting edgesof G [ B ′ ∪ B ′ ]. Recall that d + D ( v ) ≤ | B ′ | − v ∈ B ′ \ S . Also, d + D ( v ) ≤ | B ′ | − v ∈ S . First, suppose v ∈ B ′ \ S . By construction, v has at most k +2 −| B ′ | neighborsin G \ ( B ′ ∪ B ′ ∪ D ). So, v appears in J i \ T j , due to neighbors in G \ ( B ′ ∪ B ′ ∪ D ),at most k + 2 − | B ′ | times. The number of times v appears in T j \ T ′ j is at most d + D ( v ) ≤ | B ′ | −
1. Thus, v appears in J i \ T ′ j at most ( k + 2 − | B ′ | ) + | B ′ | − k + 1times. So v is colored when the game ends. When v ∈ S , a similar analysis shows v appears in J i \ T j at most k + 2 − | B ′ | + 4 times and in T j \ T ′ j at most 14 times. Thus, v appears in J i \ T ′ j at most k + 1 times. So when the game ends, v is colored.This completes the proof that Painter wins the ( k + 2)-painting game on G . Acknowledgment
Thanks to an anonymous referee, whose careful reading of earlier versions of the paper caughtnumerous inaccuracies, as well as more serious mistakes in a previous proof of Lemma 3.7.
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