当你来到国王时，你最好不要错过
When you come at the kings you best not miss
摘要：
竞赛图是完全图的定向图。如果存在从$x$到$y$的长度至多为2的有向路径，则我们说锦标赛中的一个顶点$x$ $\vec T$控制另一个顶点$y$。一个顶点被称为国王，如果它控制比赛的每个顶点。众所周知，每场比赛都有一个国王。我们遵循沈，盛，吴（SIAM J.计算，2003）在研究寻找国王的查询复杂度时，即，为了确定地将至少一个顶点标识为国王，必须知道$\vec T$中的弧的数量。前面提到的作者表明，总是必须至少查询$Ω(n^{4/3})$弧，并提供了一个最多查询$O(n^{3/2})$的策略。虽然这个上界还没有针对原始问题进行改进，但Biswas等人（Frontiers in Mongolia，2017）证明了使用$O(n^{4/3})$查询可以识别半国王，这意味着控制至少一半顶点的顶点。我们的贡献是一种新的策略，提高了控制顶点的数量：使用$O(n^{4/3} \operatorname{polylog} n)$查询，我们可以确定一个$(\frac{1}{2}+\frac{2}{17})$ -国王。为了实现这一目标，我们使用一个新的结构结果的比赛。
英文摘要：
A tournament is an orientation of a complete graph. We say that a vertex $x$ in a tournament $\vec T$ controls another vertex $y$ if there exists a directed path of length at most two from $x$ to $y$. A vertex is called a king if it controls every vertex of the tournament. It is well known that every tournament has a king. We follow Shen, Sheng, and Wu (SIAM J. Comput., 2003) in investigating the query complexity of finding a king, that is, the number of arcs in $\vec T$ one has to know in order to surely identify at least one vertex as a king. The aforementioned authors showed that one always has to query at least $Ω(n^{4/3})$ arcs and provided a strategy that queries at most $O(n^{3/2})$. While this upper bound has not yet been improved for the original problem, Biswas et al. (Frontiers in Algorithmics, 2017) proved that with $O(n^{4/3})$ queries one can identify a semi-king, meaning a vertex which controls at least half of all vertices. Our contribution is a novel strategy which improves upon the number of controlled vertices: using $O(n^{4/3} \operatorname{polylog} n)$ queries, we can identify a $(\frac{1}{2}+\frac{2}{17})$-king. To achieve this goal we use a novel structural result for tournaments.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
