Query Complexity of Tournament Solutions

TL;DR

This work studies the problem of computing standard tournament solutions from a tournament accessed only via edge-orientation queries, with the goal of minimizing the number of queries in the worst case. It establishes tight lower bounds for several solutions: Condorcet non-losers require $2n-\lfloor \log n \rfloor -2$ queries (tight with the Condorcet-winner bound), while deterministic (and corresponding randomized in expectation) approaches for Copeland, Slater, and Markov need $\binom{n}{2}$ queries, and the bipartisan, uncovered, Banks, and top-cycle solutions require $\Omega(n^2)$ queries; moreover, randomized extensions preserve these bounds. A positive result shows that if the top cycle size is at most $k$, one can compute all listed solutions with $O(nk + \frac{n\log n}{\log(1-\frac{1}{k})})$ queries, leveraging a partitioning strategy and structural reductions. Together, these findings delineate the inherent cost of eliciting tournament solutions and reveal that small top cycles can dramatically reduce query complexity, informing practical decision-making in settings with costly edge evaluations.

Abstract

A directed graph where there is exactly one edge between every pair of vertices is called a {\em tournament}. Finding the "best" set of vertices of a tournament is a well studied problem in social choice theory. A {\em tournament solution} takes a tournament as input and outputs a subset of vertices of the input tournament. However, in many applications, for example, choosing the best set of drugs from a given set of drugs, the edges of the tournament are given only implicitly and knowing the orientation of an edge is costly. In such scenarios, we would like to know the best set of vertices (according to some tournament solution) by "querying" as few edges as possible. We, in this paper, precisely study this problem for commonly used tournament solutions: given an oracle access to the edges of a tournament T, find $f(T)$ by querying as few edges as possible, for a tournament solution f. We first show that the set of Condorcet non-losers in a tournament can be found by querying $2n-\lfloor \log n \rfloor -2$ edges only and this is tight in the sense that every algorithm for finding the set of Condorcet non-losers needs to query at least $2n-\lfloor \log n \rfloor -2$ edges in the worst case, where $n$ is the number of vertices in the input tournament. We then move on to study other popular tournament solutions and show that any algorithm for finding the Copeland set, the Slater set, the Markov set, the bipartisan set, the uncovered set, the Banks set, and the top cycle must query $Ω(n^2)$ edges in the worst case. On the positive side, we are able to circumvent our strong query complexity lower bound results by proving that, if the size of the top cycle of the input tournament is at most $k$, then we can find all the tournament solutions mentioned above by querying $O(nk + \frac{n\log n}{\log(1-\frac{1}{k})})$ edges only.

Figures (4)

• Figure 1: Schematic diagram of the proof of \ref{['thm:bps_lb']}. The oracle pretends to have the tournament on the left if the output of the algorithm contains any vertex from $\mathcal{B}$; otherwise the oracle pretends to have the tournament on the right.
• Figure 2: Schematic diagram of the proof of \ref{['thm:ucs_lb']}. All the edges between $\mathcal{A}$ and $\mathcal{B}$ need to be queried to find whether $x$ is uncovered or not.
• Figure 3: Schematic diagram of the proof of \ref{['thm:banks_lb']}. $b_1, b_2, \ldots, b_n$ is a transitive order of the tournament induced on $\mathcal{B}$. All the edges between $\mathcal{A}$ and $\mathcal{B}$ need to be queried to find whether the Banks set of the input tournament contains $x$ or not.
• Figure 4: Schematic diagram of the proof of \ref{['thm:topcycle_lb']}. There are two cycles namely (i) $a_1 \rightarrow a_2 \rightarrow \cdots \rightarrow a_i \rightarrow a_{i+1} \rightarrow \cdots \rightarrow a_n \rightarrow a_1$ and (ii) $b_1 \rightarrow b_2 \rightarrow \cdots \rightarrow b_i \rightarrow b_{i+1} \rightarrow \cdots \rightarrow b_n \rightarrow b_1$. All the edges between $\mathcal{A}$ and $\mathcal{B}$ need to be queried to determine the top cycle of the input tournament.

Theorems & Definitions (17)

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