AI 中文总结
该研究驳斥了三个关于锦标赛可诱导性的猜想，涉及Kemeny中位数问题的复杂性、最小权反馈弧集的性质及特定锦标赛的可预测性，其中5名选民无法诱导顶点数43的Paley锦标赛。
AI 中文摘要
Kemeny中位数问题要求找到一个线性序，使其与给定的n个选项的m个排名的总成对不一致性最小；对于所有偶数m≥4和所有奇数m≥7，该问题是NP难的，而m=3和m=5的情况仍未解决。对多数锦标赛的每条弧按其优势幅度加权，可将该问题简化为最小权反馈弧集（FAS）问题。诱导出一个锦标赛所需的最少选民数称为其McGarvey数，其可预测性α*(T)是T可被诱导的最大超级多数阈值。我们驳斥了三个关于可诱导性的猜想：(i) 对于任意锦标赛，每个最小FAS都是有向3-环的极小击中集，这强化了Milosz、Hamel和Pierrot的定理；他们的两个猜想均不成立：对于所有奇数m≥5的3-环扩展猜想，以及n=11时FAS=HS₃的等式猜想。(ii) Shepardson和Tovey提出的阈值猜想在m=3时在边界上失效（可预测性=2/3）。(iii) 对于m=5时该猜想严格失效：顶点数为43的Paley锦标赛，其可预测性为181/301>3/5，无法由5名选民的多数票产生，这是首个规模适中的超出5名选民可达范围的明确锦标赛。
英文摘要
The Kemeny median problem asks for a linear order minimizing the total pairwise disagreement with $m$ given rankings of $n$ options; it is NP-hard for every even $m \ge 4$ and every odd $m \ge 7$, while $m = 3$ and $m = 5$ remain open. Weighting each arc of the majority tournament by its margin reduces the problem to minimum-weight feedback arc set (FAS). The fewest voters inducing a tournament is its McGarvey number, and its predictability $α^{*}(T)$ is the largest supermajority threshold at which $T$ is inducible. We refute three conjectures on inducibility. (i) In any tournament, every minimum FAS is a minimal hitting set of the directed 3-cycles, strengthening a theorem of Milosz, Hamel and Pierrot; both of their conjectures fail: the 3-cycle extension for all odd $m \ge 5$, and the equality $\mathrm{FAS} = \mathrm{HS}_3$ at $n = 11$. (ii) The threshold conjecture proposed by Shepardson and Tovey fails for $m = 3$, exactly on the boundary (predictability $= 2/3$). (iii) For $m = 5$ it fails strictly: the Paley tournament on 43 vertices, with predictability $181/301 > 3/5$, is not the majority of any 5 voters, making it the first explicit tournament of modest size beyond the reach of five voters.
Comments29 pages, 7 figures
