Abstract
We consider biased $(1:b)$ Avoider-Enforcer games in the monotone and strict versions. In particular, we show that Avoider can keep his graph being a forest for every but maybe the last round of the game if $b \geq 200 n \ln n$. By this we obtain essentially optimal upper bounds on the threshold biases for the non-planarity game, the non-$k$-colorability game, and the $K_t$-minor game thus addressing a question and improving the results of Hefetz, Krivelevich, Stojaković, and Szabó. Moreover, we give a slight improvement for the lower bound in the non-planarity game.
Highlights
Avoider-Enforcer games can be seen as the misere version of the well-known MakerBreaker games
We may assume that Avoider is always the first player since the choice of the player who is making the first move does not have an impact on our results
It was proved by Hefetz, Krivelevich, Stojakovic, and Szabo [7] that for every the electronic journal of combinatorics 22(1) (2015), #P1.60 k 3, Avoider can win the strict (1 : b) “non-k-colorability” game N Cnk against any bias larger than
Summary
Avoider-Enforcer games can be seen as the misere version of the well-known MakerBreaker games (studied first by Lehman [10], Chvatal and Erdos [6] and Beck [1, 3]). Define N Cnk to be the set consisting of the edge sets of all non-k-colorable graphs on n vertices It was proved by Hefetz, Krivelevich, Stojakovic, and Szabo [7] that for every the electronic journal of combinatorics 22(1) (2015), #P1.60 k 3, Avoider can win the strict (1 : b) “non-k-colorability” game N Cnk against any bias larger than. For n sufficiently large and b 200n ln n, Avoider can ensure that in the monotone/strict (1 : b) Avoider-Enforcer game by the end of the game his graph is planar, k-colorable for k 3, and does not contain a Kt-minor for t 4. Krivelevich, Stojakovic, and Szabo conjectured in [7] that the Avoider-Enforcer non-planarity, non-k-colorability and the Kt-minor games should be asymptotically monotone as n tends to infinity That is their upper and lower threshold should be of the same order, i.e. fF−n = Θ(fF+n).
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