Copeland's method: Difference between revisions
Content added Content deleted
mNo edit summary |
|||
Line 26:
Copeland's method also passes [[ISDA]]; since the Copeland winner is always in the Smith set, all candidates in the Smith set must have higher Copeland scores than all candidates not in the Smith set, and since by definition candidates in the Smith set have a pairwise victory against every candidate not in the Smith set, adding or removing any number of candidates not in the Smith set will only result in every candidate in the Smith set having that number of pairwise victories added or subtracted from their total; since the original Copeland winner must have had a higher Copeland score than all other Smith set candidates in order to win, they will still have a higher Copeland score and thus still win.
The Copeland ranking of candidates (the ordering of candidates based on Copeland score) is a [[Smith set ranking]]. This is because in general, a candidate in the n-th Smith set (if n is 1, this is the Smith set. If n is 2, this is the secondary Smith set, which is the set of candidates that would be in the Smith set if all the candidates in the Smith set were eliminated from the election, etc.) will have pairwise victories against at least all candidates in k-th Smith sets (for any value of k which is greater than n), and have pairwise defeats against at most all candidates in j-th Smith sets (for any non-negative value of j which is smaller than n) and all but two candidates in the n-th Smith set (since they can't be beaten by themselves, and must not be beaten by everyone else in their Smith set in order to be in it), while a candidate in any k-th Smith set will have pairwise victories against at most all candidates other than themselves in k-th Smith sets, and will have pairwise defeats against at least all candidates in n-th or j-th Smith sets. Mathematically, this can be represented as minimak/maximal Copeland scores of (K-J-N+2) and (K-J-N-1) respectively Therefore, the Copeland score of a candidate in the n-th Smith set is guaranteed to be at least 3 points higher than a candidate in any of the k-th (after or below n) Smith sets. Similar reasoning shows that a candidate in any of the j-th (before or above n) Smith sets is at least 3 points higher than any candidate in the n-th Smith set.
== See also ==
| |||