Deterministic Impartial Selection with Weights
Javier Cembrano, Svenja M. Griesbach, Maximilian J. Stahlberg
TL;DR
This work advances the theory of impartial selection by providing the first deterministic, weighted-impartial mechanism for inexact $(n,k)$-selection with a nontrivial approximation ratio, achieved via robust partition systems and hypergraph duality. The Select$_k$ mechanism attains an explicit α-optimal bound $\alpha = \frac{k - k\bmod 2}{k \lceil \frac{2n}{k - k\bmod 2} \rceil}$ for the regime $k - k\bmod 2 \ge 2\sqrt{n}$, and an extension lemma broadens applicability to general $(n,k)$. The approach generalizes to impartial assignment across $m$ jobs, incurring a factor-$\tfrac12$ loss in the approximation. These results improve deterministic guarantees beyond the classic $1/k$ bound in the unweighted setting for large $k$, and hinge on a novel robust partition framework rooted in 2-regular, b-uniform, linear hypergraphs and their graph duals.
Abstract
In the impartial selection problem, a subset of agents up to a fixed size $k$ among a group of $n$ is to be chosen based on votes cast by the agents themselves. A selection mechanism is impartial if no agent can influence its own chance of being selected by changing its vote. It is $α$-optimal if, for every instance, the ratio between the votes received by the selected subset is at least a fraction of $α$ of the votes received by the subset of size $k$ with the highest number of votes. We study deterministic impartial mechanisms in a more general setting with arbitrarily weighted votes and provide the first approximation guarantee, roughly $1/\lceil 2n/k\rceil$. When the number of agents to select is large enough compared to the total number of agents, this yields an improvement on the previously best known approximation ratio of $1/k$ for the unweighted setting. We further show that our mechanism can be adapted to the impartial assignment problem, in which multiple sets of up to $k$ agents are to be selected, with a loss in the approximation ratio of $1/2$.
