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The Namer-Claimer game

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 Added by Ben Barber
 Publication date 2018
  fields
and research's language is English
 Authors Ben Barber




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In each round of the Namer-Claimer game, Namer names a distance d, then Claimer claims a subset of [n] that does not contain two points that differ by d. Claimer wins once they have claimed sets covering [n]. I show that the length of this game is of order log log n with optimal play from each side.

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We analyze a coin-based game with two players where, before starting the game, each player selects a string of length $n$ comprised of coin tosses. They alternate turns, choosing the outcome of a coin toss according to specific rules. As a result, the game is deterministic. The player whose string appears first wins. If neither players string occurs, then the game must be infinite. We study several aspects of this game. We show that if, after $4n-4$ turns, the game fails to cease, it must be infinite. Furthermore, we examine how a player may select their string to force a desired outcome. Finally, we describe the result of the game for particular cases.
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Consider equipping an alphabet $mathcal{A}$ with a group action that partitions the set of words into equivalence classes which we call patterns. We answer standard questions for the Penneys game on patterns and show non-transitivity for the game on patterns as the length of the pattern tends to infinity. We also analyze bounds on the pattern-based Conway leading number and expected wait time, and further explore the game under the cyclic and symmetric group actions.
In a Maker-Breaker game there are two players, Maker and Breaker, where Maker wins if they create a specified structure while Breaker wins if they prevent Maker from winning indefinitely. A $3$-AP is a sequence of three distinct integers $a, b, c$ such that $b-a = c-b$. The $3$-AP game is a Maker-Breaker game played on $[n]$ where every round Breaker selects $q$ unclaimed integers for every Makers one integer. Maker is trying to select points such that they have a $3$-AP and Breaker is trying to prevent this. The main question of interest is determining the threshold bias $q^*(n)$, that is the minimum value of $q=q(n)$ for which Breaker has a winning strategy. Kusch, Rue, Spiegel and Szabo initially asked this question and proved $sqrt{n/12-1/6}leq q^*(n)leq sqrt{3n}$. We find new strategies for both Maker and Breaker which improve the existing bounds to [ (1+o(1))sqrt{frac{n}{5.6}} leq q^*(n) leq sqrt{2n} +O(1). ]
While the game chromatic number of a forest is known to be at most 4, no simple criteria are known for determining the game chromatic number of a forest. We first state necessary and sufficient conditions for forests with game chromatic number 2 and then investigate the differences between forests with game chromatic number 3 and 4. In doing so, we present a minimal example of a forest with game chromatic number 4, criteria for determining the game chromatic number of a forest without vertices of degree 3, and an example of a forest with maximum degree 3 and game chromatic number 4.
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