Painting game on graphs Xuding Zhu Zhejiang Normal University th Shanghai Conference on Combinatorics

There are six teams, each needs to compete with all the others. Each team can play one game per day How many days are needed to schedule all the games? Answer: 5 days A scheduling problem:

1st day

2nd day

3rd day

4th day

5th day

This is an edge colouring problem. Each day is a colour.

There are six teams, each needs to compete with all the others. Each team can play one game per day How many days are needed to schedule all the games? Answer: 5 days Each team can choose one day off 7 days are enough A scheduling problem: 7 days are needed

Each edge misses at most 2 colours There are 7 colours Each edge has 5 permissible colours I do not know any easy proof

List colouring conjecture: For any graph G, However, the conjecture remains open for Haggkvist-Janssen (1997)

There are six teams, each needs to compete with all the others. Each team can play one game per day How many days are needed to schedule all the games? Answer: 5 days Each team can choose one day off 7 days are enough The choices are made before the scheduling A scheduling problem

There are six teams, each needs to compete with all the others. Each team can play one game per day How many days are needed to schedule all the games? 7 days are enough Each team can choose one day off A scheduling problem is allowed not to show up for one day On each day, we know which teams haven’t shown up today but we do not know which teams will not show up tomorrow We need to schedule the games for today

On-line list colouring of graphs We start colouring the graph before having the full information of the list

f-painting game (on-line list colouring game) on G is the number of permissible colours for x Two Players: ListerPainter Each vertex v is given f(v) tokens. Each token represents a permissible colour. But we do not know yet what is the colour. Reveal the list Colour vertices

At round i is the set of vertices which has colour i as a permissible colour. Painter chooses an independent subset of vertices in are coloured by colour i. Lister choose a set of uncoloured vertices, removes one token from each vertex of

If at the end of some round, there is an uncolored vertex with no tokens left, then Lister wins. If all vertices are coloured then Painter wins the game.

G is f-paintable if Painter has a winning strategy for the f-painting game. G is k-paintable if G is f-paintable for f(x)=k for every x. The paint number of G is the minimum k for which G is k-paintable.

choice number Painter start colouring the graph after having the full information of the list List colouring: On-line list colouring: before

Theorem [Erdos-Rubin-Taylor (1979)] is 2-choosable. is not 2-paintable

Lister wins the game is not 2-paintable

Theorem [Erdos-Rubin-Taylor,1979] A connected graph G is 2-choosable if and only if its core is or However, if p>1, then is not 2-paintable. Theorem [Zhu,2009] A connected graph G is 2-paintable if and only if its core is or

A recursive definition of f-paintable Assume. Then G is f-paintable, if (1) or (2)

For any question about list colouring, we can ask the same question for on-line list colouring Planar graphs and locally planar graphs Chromatic-paintable graphs Partial painting game Complete bipartite graphs b-tuple painting game and fractional paint number For any result about list colouring, we can ask whether it holds for on-line list colouring

Some upper bounds for are automatically upper bounds for

Theorem [Galvin,1995] If G is bipartite, then Upper bounds for ch(G) proved by kernel method are also upper bounds for

Upper bounds for ch(G) proved by Combinatorial Nullstellensatz are also upper bounds for

Brooks’ Theorem paintable [Hladky-Kral-Schauz,2010]

Upper bounds for ch(G) proved by induction Theorem [Thomassen, 1995] Every planar graph is 5-choosable Planar graphs [ Schauz,2009 ] paintable

Thomassen proves a stronger result: paintable Basically, Thomassen’s proof works for f-paintable.

Locally planar embedded in a surface contractible non-contractible edge-width of G length of shortest non-contractible cycle edge-width is large Theorem [Thomassen, 1993] For any surface, there is a constant, any G embedded in with edge-width > is 5-colourable.

Locally planar embedded in a surface contractible non-contractible edge-width of G length of shortest non-contractible cycle edge-width is large Theorem [Thomassen, 1993] For any surface, there is a constant, any G embedded in with edge-width > is 5-colourable. DeVos-Kawarabayashi-Mohor 2008 choosable Han-Zhu paintable

Find a subgraph H G-H is planar Each piece in H is planar Apply strategy for planar graphs on the pieces of H and on G-H, one by one + some other nice properties

Chromatic-paintable graphs

A graph G is chromatic choosable if Conjecture: Line graphs are chromatic choosable. Conjecture: Claw-free graphs are chromatic choosable. Conjecture: Total graphs are chromatic choosable. Ohba Conjecture: Graphs G with are chromatic choosable. Theorem [Noel-Reed-Wu,2013] paintable Conjecture: Graph squares are chromatic choosable. [Kim-Park,2013] paintable

A graph G is chromatic choosable if Conjecture: Line graphs are chromatic choosable. Conjecture: Claw-free graphs are chromatic choosable. Conjecture: Total graphs are chromatic choosable. Ohba Conjecture: Graphs G with are chromatic choosable. paintable Conjecture: Graph squares are chromatic choosable. [Kim-Park,2013] paintable NO! Question

is not 3-paintable.

Lister

Lister Painter 23 33

Lister Painter Lister

Lister Painter Lister

Lister Painter Lister Painter

Lister Painter Lister Painter Lister

Lister Painter Lister Painter Lister Painter Lose Painter Lose Painter Lose is not 3-paintable {123} {1}{2}{3}

Theorem [Kim-Kwon-Liu-Zhu,2012] For k>1, is not (k+1)-paintable.

Ohba Conjecture: Graphs G with are chromatic choosable. paintable On-line versionHuang-Wong-Zhu 2011 To prove this conjecture, we only need to consider complete multipartite graphs.

Theorem [Huang-Wong-Zhu,2011] is n-paintable The first proof is by using Combinatorial Nullstellensatz A second proof gives a simple winning strategy for Painter The proof uses induction.

Theorem [Kozik-Micek-Zhu,2014] On-line Ohba conjecture is true for graphs with independence number at most 3. The key in proving this theorem is to find a “good” technical statement that can be proved by induction.

Partition of the parts into four classes ordered

G is f-paintable

Theorem [Kozik-Micek-Zhu,2014] On-line Ohba conjecture is true for graphs with independence number at most 3. Theorem [Chang-Chen-Guo-Huang,2014+]

Theorem [Erdos,1964] Theorem[Zhu,2009] If G is bipartite and has n vertices, then Complete bipartite graphs probabilistic proof

Painter colours, double the weight of each vertex in A B Initially, each vertex x has weight w(x)=1 Assume Lister has given set If

If x has permissible colours, Painter will be able to colour it. A B The total weight of uncoloured vertices is not increased. If a vertex is given a permissible colour but is not coloured by that colour, then its weight doubles. If x has been given k permissible colours, but remains uncoloured, then

Theorem [Carraher-Loeb-Mahoney-Puleo-Tsai-West]

3-choosable complete bipartite graphs Theorem Mahadev-Roberts -Santhanakrishnan, 1991 Furedi-Shende-Tesman, 1995 O-Donnel,1997

3-paintable complete bipartite graphs Theorem [Chang-Zhu,2013]

Theorem [Erdos, 1964] and Theorem [Zhu, 2010] Erdos-Lovasz Conjecture Maybe

Theorem [RadhaKrishnan-Srinivasan,2000] Probabilistic mothed. and

Theorem [Duray-Gutowksi-Kozik,2014+] Theorem [Gerbner-Vizer, 2014+]

?

b-tuple list colouring b-tuple on-line list colouring G is (a,b)-choosable if |L(v)|=a for each vertex v, then there is a b-tuple L-colouring. If each vertex has a tokens, then Painter has a strategy to colour each vertex a set of b colours.

(a,b)-choosable Conjecture [Erdos-Rubin-Taylor] (am,bm)-choosable On-line version (a,b)-paintable (am,bm)-paintable Theorem [Tuza-Voigt, 1996] 2-choosable (2m,m)-choosable Theorem [Mahoney-Meng-Zhu, 2014] 2-paintable (2m,m)-paintable

Theorem[Alon-Tuza-Voigt, 1997][Gutowski, 2011] Infimum attainedInfimum not attained

Theorem[Alon-Tuza-Voigt, 1997][Gutowski, 2011] Infimum attainedInfimum not attained Every bipartite graph is (2m,m)-choosable for some m Theorem [Mahoney-Meng-Zhu,2014] For any m, a connected graph G is (2m,m)-paintable if and only if its core is or

Partial painting game Partial f-painting game on G same as the f-painting game, except that Painter’s goal is not to colour all the vertices, but to colour as many vertices as possible.

Fact: Conjecture [Albertson]: Conjecture [Zhu, 2009]:

Theorem [Wong-Zhu,2013]

Question: Can the difference be arbitrarily large ?

Nine Dragon Tree Thank you

An easy but useful lemma: Corollary: If G is k-degenerate, then G is (k+1)-paintable If f(v) >d(v), then G is f-paintable iff G-v is f-paintable. Painter uses his winning strategy on G-v. Colour v only if v is marked (i.e., the current colour is permissible for v), and none of its neighbours used the current colour. v will be coloured when all its tokens are used gone, if not earlier.

Case 1 u w

Painter apply his winning strategy on with, obtain independent set Case 1 Then Painter apply his winning strategy on with obtain independent set u w

Case 2 defined as follows: Painter apply his winning strategy on with input obtains add if possible.