In the paper "On the complexity of some coloring games", Bodlaender gives some open questions about the complexity of deciding if player 1 or 2 has a winning strategy in some graph coloring games. Does anyone know if they have been solved?

1) In one game, two players take turns selecting one vertex in a graph and coloring it properly with a color from a fixed finite set. The loser is the first player who is unable to color a vertex. In Schaefer's paper it is shown to be pspace-complete with 1 color and Bodlaender shows it to be pspace-complete with 2 colors but gives no answer with more color. Is it still open?

2) In another variation, the vertices have numbers 1..n. On a player's turn he must properly color the vertex with the lowest number that has not yet been colored. Again, they are using colors from a fixed set and the loser is the first player who is unable to color his vertex. Bodlaender shows it to be pspace-complete for general graphs. He asks who wins on trees, is it known?



2 Answers 2


The answer is yes, for the first game you list! This result was only established in 2019. Here is a link to the paper: Costa et al. 2019

Even more recently, some variants of the first game were proved to be PSPACE-complete. This result can be found here: Marcilon et al. 2019.


It sounds like this paper has some of what you're looking for: http://arxiv.org/abs/1202.5762

The general form of the first question is a really simple reduction: using colors {0, ..., n-1}, start with a Node Kayles instance and create a vertex for each of the colors from 1 to n-1 and connect them to each uncolored vertex. Now those colors can't be played and you're still playing the Node Kayles game.

  • $\begingroup$ Thanks for the link, I will take a look. In this question, we do not allow a 'pre-coloring' so we are not allowed to assume that some vertices already have a color. The game starts with all vertices uncolored. $\endgroup$
    – user32149
    Feb 20, 2015 at 14:28
  • $\begingroup$ That makes sense, but it changes the question of hardness. For many games, it's known which player has a winning strategy from an initial position, but it's not known which player has a winning strategy at a general position. Take Hex for example. Here the first player has a winning strategy. From a general position, determining whether the next player to move has a winning strategy is PSPACE-complete. $\endgroup$
    – Kyle
    Feb 20, 2015 at 17:24
  • $\begingroup$ Yes you are right, I should have clarified in the original question. I am talking about the computational complexity of determining who wins on a given graph before any vertices have been colored. $\endgroup$
    – user32149
    Feb 20, 2015 at 21:03
  • $\begingroup$ It's an interesting question, to be sure. Especially since you're talking about a general graph and not putting any requirements on its structure. I'll certainly be interested to know if you figure it out! $\endgroup$
    – Kyle
    Feb 21, 2015 at 16:44

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