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Maximum Distance Between Slater Ordersand Copeland Orders of Tournaments

机译:比赛的Slater订单和Copeland订单之间的最大距离

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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- l)/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- l)/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.
机译:给定一个锦标赛T =(X,A),我们考虑将两个锦标赛解决方案应用于T:Slater解决方案和Copeland解决方案。 Slater的解决方案包括确定通过反转T的最小有向边以使T可传递而获得的线性阶数。 Copeland应用于T的解决方案根据T的顶点的递减度对其进行排名。本文的目的是比较这两种方法提供的结果:它们可以在多大程度上导致不同的顺序?我们考虑三种情况:T是任何锦标赛,T是紧密相连的,T只有一个Slater订单。对于这三种情况中的每一种,我们指定Slater阶和Copeland阶之间的对称差异距离的最大值。更确切地说,由于有一个关于处理圆形比赛中弧形不连续电路的结果,我们证明,如果T是在奇数个顶点n上的任何比赛,则此最大值等于n(n-1)/ 2,即(n〜如果T是偶数个顶点n上的任何锦标赛,则为2-3n + 2)/ 2,如果T与奇数个顶点上的n个强连通,则为(n〜2-3n),则为n(n-1)/ 2 -2)/ 2(如果T与偶数个大于或等于8的顶点强连接),则为(n〜2-5n + 6)/ 2,如果T具有奇数个顶点且只有一个Slater阶,如果T的顶点数为偶数n并且只有一个Slater阶,则为(n〜2-5n + 8)/ 2。

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