发表机构
TU Wien; KTH Royal Institute of Technology; Freie Universität Berlin（维也纳工业大学; 瑞典皇家理工学院; 柏林自由大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文提出预算版Maker Breaker三角形博弈，证明其阈值偏差精确为⌈√(2n-2)-3/2⌉，并首次给出K_4博弈的显式上下界。
AI 中文摘要
Maker Breaker三角形博弈涉及两个玩家，Maker和Breaker，他们分别轮流在K_n中认领1条边和q条边。Maker的目标是认领任意三角形的所有三条边，而Breaker的目标是阻止这一行为。阈值偏差，即Breaker获胜的最小q，已知位于⌈√(2n-2)-3/2⌉和(√(8/3)+o(1))√n之间。确定其精确值是一个长期未解决的问题。在本文中，我们引入了该博弈的一个新版本，其中Breaker每轮可以认领少于q条边，以积累预算，供其在后续轮次中使用。凭借Breaker的这一额外能力，我们确定了对于所有n，阈值偏差精确为⌈√(2n-2)-3/2⌉，与已知下界匹配。这是Maker Breaker三角形博弈中第一个已知精确阈值偏差的版本。即使不允许Breaker将预算用于威胁，我们也证明了阈值偏差仍为(√2+o(1))√n。此外，我们研究了其他Maker Breaker博弈的预算版本。具体而言，对于K_4博弈，我们在原始版本和预算版本中均证明了阈值偏差的首个显式下界和上界。
英文摘要
The Maker Breaker Triangle Game involves two players, Maker and Breaker, who alternately claim 1 and $q$ edges of $K_n$, respectively. Maker's goal is to claim all three edges of any triangle, whereas Breaker's goal is to prevent this. The threshold bias, i.e. the minimum $q$ such that Breaker wins, is known to lie between $\big\lceil\sqrt{2n-2}-\frac{3}{2}\big\rceil$ and $\big(\sqrt{8/3}+o(1)\big)\sqrt{n}$. Determining its exact value is a longstanding open problem. In this paper, we introduce a novel version of the game in which Breaker may claim fewer than $q$ edges per round to build up a budget that he can spend in later rounds. With this additional power for Breaker, we determine the threshold bias to be precisely $\big\lceil\sqrt{2n-2}-\frac{3}{2}\big\rceil$ for all $n$, matching the known lower bound. This is the first version of the Maker Breaker Triangle Game for which the exact threshold bias is known. Even if Breaker is not allowed to use the budget for threats, we prove that the threshold bias is still $\big(\sqrt{2}+o(1)\big)\sqrt{n}$. Furthermore, we study the budget version of other Maker Breaker Games. Specifically, for the $K_4$-Game, we prove the first explicit lower and upper bounds on the threshold bias in both the original and the budget version of the game.
