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.
翻译:我们考虑在斯拉特规则下决定选举获胜者的复杂性。 在这种背景下,我们得到了一场以美元=(V, A)$(V, A)的比赛,V的顶点代表候选人,而每个弧的方向表明两个端点中哪一个对多数选民来说是更可取的。 Slater 的顶点值是需要倒转的最低弧码数,以使T美元成为周期性的,而美元成为赢家。 在这种背景下,我们说,如果美元具有最低的Slater分,那么美元就是以美元计的Slater赢家。决定一个顶点在比赛中是斯拉特赢家,对于大多数选民来说早已为Slater赢家。然而,这一问题最已知的复杂程度是“$Theta_2 ⁇ p$”,这相当于与PNP或甲板平行接轨的超时速图灵机。 在这份论文中,我们缩小了这一差距,我们通过显示这一问题是$\Ta_2Q-plate suplifilations (Greal-plistations) subilstations) 。