Determining a Slater Winner is Complete for Parallel Access to NP

TL;DR

This paper proves that determining a Slater winner in an election modeled by a tournament is complete for the complexity class ^p, establishing its computational difficulty even with only 7 voters, and closing a long-standing gap.

Contribution

It establishes the ^p-completeness of the Slater winner problem, showing its inherent computational hardness in a parallel NP access setting.

Findings

The problem is ^p-complete.

Hardness holds even with preferences aggregated from 7 voters.

Closes the gap between NP-hardness and ^p upper bound.

Abstract

We consider the complexity of deciding the winner of an election under the Slater rule. In this setting we are given a tournament $T = (V, A)$, where the vertices of V represent candidates and the direction of each arc indicates which of the two endpoints is preferable for the majority of voters. The Slater score of a vertex $v\in V$ is defined as the minimum number of arcs that need to be reversed so that $T$ becomes acyclic and $v$ becomes the winner. We say that $v$ is a Slater winner in $T$ if $v$ has minimum Slater score in $T$. Deciding if a vertex is a Slater winner in a tournament has long been known to be NP-hard. However, the best known complexity upper bound for this problem is the class $\Theta_2^p$, which corresponds to polynomial-time Turing machines with parallel access to an NP oracle. In this paper we close this gap by showing that the problem is $\Theta_2^p$-complete, and that this hardness applies to instances constructible by aggregating the preferences of 7 voters.