Strengthening ( a , b )-Choosability Results to ( a , b )-Paintability

Let a, b ∈ N. A graph G is (a, b)-choosable if for any list assignment L such that |L(v)| > a, there exists a coloring in which each vertex v receives a set C(v) of b colors such that C(v) ⊆ L(v) and C(u)∩C(w) = ∅ for any uw ∈ E(G). In the online version of this problem, on each round, a set of vertices allowed to receive a particular color is marked, and the coloring algorithm chooses an independent subset of these vertices to receive that color. We say G is (a, b)-paintable if when each vertex v is allowed to be marked a times, there is an algorithm to produce a coloring in which each vertex v receives b colors such that adjacent vertices receive disjoint sets of colors. We show that every odd cycle C2k+1 is (a, b)-paintable exactly when it is (a, b)chosable, which is when a > 2b + db/ke. In 2009, Zhu conjectured that if G is (a, 1)-paintable, then G is (am,m)-paintable for any m ∈ N. The following results make partial progress towards this conjecture. Strengthening results of Tuza and Voigt, and of Schauz, we prove for any m ∈ N that G is (5m,m)-paintable when G is planar. Strengthening work of Tuza and Voigt, and of Hladky, Kral, and Schauz, we prove that for any connected graph G other than an odd cycle or complete graph and any m ∈ N, G is (∆(G)m,m)-paintable.


Introduction
Let G be a graph, and let g : V (G) → N. A g-fold coloring of G assigns to each vertex v a set of g(v) distinct colors such that adjacent vertices have disjoint sets of colors.When g(v) = m for all v, we call this an m-fold coloring.When all colors come from {1, . . ., k}, we call this a g-fold k-coloring and say that G is (k, g)-colorable.When G is (k, g)-colorable and g(v) = m for all v, we say that G is (k, m)-colorable.An ordinary proper k-coloring is also a 1-fold k-coloring.
More generally, let L be a list assignment specifying for each vertex v a set L(v) of available colors.A g-fold L-coloring of G is a g-fold coloring φ of G such that φ(v) ⊆ L(v) for each vertex v.A graph G is (f, g)-choosable if there is a g-fold L-coloring for any list assignment L such that |L(v)| f (v) for all v ∈ V (G).Introduced by Erdős, Rubin, and Taylor [2], when f (v) = k and g(v) = m for all v ∈ V (G) and G is (f, g)-choosable, we say that G is (k, m)-choosable.When g(v) = 1 for every v ∈ V (G), we shorten (f, g)-choosable to f -choosable.Erdős, Rubin, and Taylor [2] conjectured that if G is k-choosable, then G is (km, m)-choosable for all m ∈ N. Schauz [5] introduced an online version of choosability, and Zhu [9] generalized (f, g)choosability.Vertices are colored in rounds, where the coloring algorithm must decide on round i which vertices will receive color i, without knowing which colors will appear on any vertices later in the lists.This concept is formalized in the following game.Definition 1.1.Let G be a graph where each vertex v is assigned a nonnegative number f (v) of tokens and a nonnegative number g(v) specifying how many times v must be colored.The (f, g)-paintability game is played by two players: Lister and Painter.Each round, Lister marks a nonempty subset M of vertices that have been colored fewer than g(v) times; every vertex in M loses one token.Painter responds by coloring an independent subset D of M ; every vertex of D gains a color distinct from those used on earlier rounds.Lister wins the game by marking a vertex that has no tokens remaining, and Painter wins by coloring each vertex v on g(v) distinct rounds.
We say G is (f, g)-paintable when Painter has a winning strategy on G in the (f, g)paintability game.If f (v) = a and g(v) = b for every v ∈ V (G) and Painter has a winning strategy, then we say G is (a, b)-paintable.We say Mahoney, Meng, and Zhu [4] proved that Conjecture 1.2 is true for all 2-paintable graphs.
In Section 2, we prove that planar graphs are (5m, m)-paintable for all m 1, which strengthens the previous results and makes partial progress towards Conjecture 1.2.
Let G be a connected graph other than an odd cycle or a complete graph, and let ∆(G) denote the maximum degree of G. Brooks' Theorem [1] states that G is ∆(G)colorable.Stronger versions of Brooks' Theorem are proved by Tuza and Voigt [7] and by Hladky, Kral, and Schauz [3].In Section 3, we prove that G is (∆(G)m, m)-paintable, strengthing both results and making partial progress towards Conjecture 1.2.
In [8], Voigt proved that if C 2k+1 is (a, b)-choosable, then a 2b + b/k .We conclude the introduction by strengthening this result, characterizing the (a, b)-paintability and (a, b)-choosability of odd cycles.If |M | = 2k + 1, then we keep track of how many times moves of this type have occurred in the game.If a move of this type has been played i times before (mod 2k + 1), then Painter colors {v i , v i+2 , . . ., v i+2k−2 }.There are exactly 2k + 1 distinct independent sets of size k for C 2k+1 .In this strategy, Painter balances which of these independent sets is colored by cycling through all possible choices.
Suppose Lister can win against this particular Painter strategy when each vertex has 2b + b/k tokens.Let Lister's marked sets be M 1 , . . ., M t , where Lister wins on round t, and let Painter's responses be D 1 , . . ., D t−1 .Note that when Lister marks the set M t , some vertex will have been marked 2b + b/k + 1 times.In particular t 2b + b/k .If a vertex v i is marked in the set M j , then Painter strategy implies that v i−1 was also marked in M j .Since v i−1 is colored at most b times, there must be at least b/k + 1 rounds where both v i and v i−1 are marked and not colored.This only happens once every 2k + 1 rounds when both vertices are marked.Thus the number of rounds where all vertices are marked is at least b/k (2k + 1) + 1, which is greater than 2b + b/k .So there are at elast 2b + b/k + 1 rounds in which all vertices are marked.If any of these rounds were preceded by a round in which not all vertices were marked, then one vertex would be marked 2b + b/k + 1 times earlier than round t.Thus the first 2b + b/k + 1 marked sets must all be V (C 2k+1 ).After marking all vertices just 2b + b/k times, Painter's strategy ensures that every vertex is colored b times.Thus there is no vertex for Lister to mark on round 2b + b/k + 1, and thus there is no way for Lister to win the game against this Painter strategy.

Planar Graphs
The following lemma is a generalization of Lemma 2.2 in [5].
Proof.Let S be a winning strategy for Painter in the (f, g)-paintability game on G.In the (f , g )-paintability game on G ∪ uv, whenever Lister marks u, we sacrifice a token on v by having Painter respond to the marked set M − v.At most f (u) tokens are sacrificed on v.In rounds when u is not marked, Painter may respond according to S because any response in G is an independent set in G ∪ uv.At least f (v) tokens are available for moves of this type, so g(v) colors will be assigned to v by playing according to S.
The following lemma is a generalization of Lemma 2.5 in [5].
Proof.We use induction on g(v).For the basis step, if g(v) = 0, then G is trivially (f, g)-paintable.Now consider g(v) > 0. Let M be the set marked by Lister.For i ∈ {1, 2}, let S i be a winning strategy for Painter in G i under token assignment f i , and let M i = M ∩V (G i ).Let D 1 be the response to M 1 in G 1 according to S 1 .In G 2 , Painter responds to the marked set M 2 − (T − D 1 ) according to S 2 .We interpret vertices of (M − D 1 ) ∩ T as having lost a token in To make use of the induction hypothesis, we define the following functions: Because D 1 and D 2 were chosen according to a winning strategies in G 1 and in G 2 , we have that G i is (f i , g i )-paintable for i ∈ {1, 2} and f 2 (v) = g 2 (v) = g 1 (v) for all v ∈ T .Since M = ∅, we may assume that D 1 ∪ D 2 = ∅.Thus g(v) decreases, and by induction this yields G * is (f * , g * )-paintable where f g 2 (v), otherwise .This proves that Painter's response to M is a winning move, and thus G is (f, g)-paintable.
We now prove the main theorem of this section.
Proof.We proceed using an argument mirroring that of Thomassen [6] and of Schauz [5].First, we restrict our attention to weak triangulations of planar graphs since adding edges only makes coloring the graph more difficult for Painter.Let G be a planar graph of order n with vertices v 1 , . . ., v p in clockwise order on the unbounded face.By induction on n, we prove a stronger result: . the electronic journal of combinatorics 24(4) (2017), #P4.26

Theorem 1 . 3 .
For k 1, the following are equivalent: (a) C 2k+1 is (a, b)-paintable.the electronic journal of combinatorics 24(4) (2017), #P4.26(b) C 2k+1 is (a, b)-choosable.(c) a 2b + b/k .Proof.(a) ⇒ (b) Always (a, b)-paintability implies (a, b)-choosability.(b) ⇒ (c) Originally proved in [8], we provide a short proof for completeness.Suppose C 2k+1 is (a, b)-choosable.Consider the list assignment where L(v) = {1, . . ., a} for each vertex v.Each color can be used on at most k vertices.Since each vertex must receive b colors, we have that the lists must have size at least (2k + 1)b/k.(c) ⇒ (a) Give the cycle a consistent orientation, and label the vertices v 0 , . . ., v 2k .Consider all indices modulo 2k + 1.Let M be the set Lister marks.If |M | < 2k + 1, then the graph induced by the marked set is a linear forest.Painter colors vertices greedily along each path starting at the tail.