Maximum Distance Between Slater Orders and Copeland Orders of Tournaments

Abstract

Given a tournament T = (X, A), we consider two tournament solutions applied to T: Slater's solution and Copeland's solution. Slater's solution consists in determining the linear orders obtained by reversing a minimum number of directed edges of T in order to make T transitive. Copeland's solution applied to T ranks the vertices of T according to their decreasing out-degrees. The aim of this paper is to compare the results provided by these two methods: to which extent can they lead to different orders? We consider three cases: T is any tournament, T is strongly connected, T has only one Slater order. For each one of these three cases, we specify the maximum of the symmetric difference distance between Slater orders and Copeland orders. More precisely, thanks to a result dealing with arc-disjoint circuits in circular tournaments, we show that this maximum is equal to n(n − 1)/2 if T is any tournament on an odd number n of vertices, to (n 2 − 3n + 2)/2 if T is any tournament on an even number n of vertices, to n(n − 1)/2 if T is strongly connected with an odd number n of vertices, to (n 2 − 3n − 2)/2 if T is strongly connected with an even number n of vertices greater than or equal to 8, to (n 2 − 5n + 6)/2 if T has an odd number n of vertices and only one Slater order, to (n 2 − 5n + 8)/2 if T has an even number n of vertices and only one Slater order.

An updated survey on the linear ordering problem for weighted or unweighted tournaments

“…6.1); then this ratio does not exceed 5. But the distance between a Copeland order of T and a Slater order of T can be very large, as shown in Charon and Hudry (2009c):…”