Tournaments determined by three and five voters

2026-07-29

AI Summary

This study investigates the inducibility of majority tournaments under odd numbers of voters—specifically three and five—and their connection to the Kemeny median problem. By reformulating the Kemeny median as a minimum-weight feedback arc set (FAS) problem, the authors employ techniques from combinatorial optimization, graph theory, and explicit constructions to systematically characterize how majority edge weights and supermajority thresholds govern the realizability of voting configurations. Key contributions include disproving three longstanding conjectures: demonstrating that a minimum FAS need not coincide with a minimal hitting set of directed 3-cycles; showing that the Shepardson–Tovey threshold conjecture fails precisely at its boundary when m = 3; and constructing, for the first time, a 43-vertex Paley tournament with predictability exceeding 3/5 that cannot be induced by any profile of five voters, thereby establishing strict limits on tournament inducibility for m = 3 and m = 5.

Abstract

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.

Problem

• Kemeny median
• majority tournament
• McGarvey number
• feedback arc set
• inducibility

Innovation

• Kemeny median
• feedback arc set
• majority tournament
• McGarvey number
• Paley tournament