Voting Profiles Admitting All Candidates as Knockout Winners

A set of $2^n$ candidates is presented to a commission. At every round, each member of this commission votes by pairwise comparison, and one-half of the candidates is deleted from the tournament, the remaining ones proceeding to the next round until the $n$-th round (the final one) in which the final winner is declared. The candidates are arranged on a board in a given order, which is maintained among the remaining candidates at all rounds. A study of the size of the commission is carried out in order to obtain the desired result of any candidate being a possible winner. For $2^n$ candidates with $n \geq 3$, we identify a voting profile with $4n -3$ voters such that any candidate could win simply by choosing a proper initial order of the candidates. Moreover, in the setting of a random number of voters, we obtain the same results, with high probability, when the expected number of voters is large.

💡 Research Summary

The paper investigates a fundamental question in knockout (single‑elimination) tournaments: how large a voting committee must be in order to guarantee that any of the (2^{n}) candidates can be made the eventual winner simply by choosing an appropriate initial ordering (bracket). The authors answer this question constructively for all (n\ge 3).

1. Majority graphs as the underlying structure
Each voter submits a strict ranking of all candidates. For any pair ((a,b)) the direction of the edge ((a,b)) in the majority graph (G_R) is determined by whether a strict majority of voters rank (a) ahead of (b). This directed graph encodes the outcome of every possible pairwise match, irrespective of the exact vote counts. The winner of a knockout tournament is therefore a deterministic function of the majority graph together with the initial permutation (\pi) of the candidates on the board; we denote this function by (w_n(G,\pi)).

2. A recursive family of “universal” majority graphs
The authors first exhibit an explicit graph (G_3) on eight candidates. By listing eight carefully chosen permutations (\pi(1),\dots,\pi(8)) they show that each candidate can be forced to win under (G_3). This establishes the base case: (w_3(G_3,\Pi(3))=