Copeland's method: Difference between revisions

The reasoning for why Copeland's method is Smith-efficient is as follows: every candidate in the [[Smith set]] has a pairwise victory over every candidate not in the Smith set by definition, and at most has a pairwise defeat against all but one candidate other than themselves in the Smith set, which is reduced to no pairwise defeats when the Smith set has fewer than 3 candidates (since, they can't be pairwise defeated by themselves, and if they had a pairwise defeat against all candidates other than themselves in the Smith set, then they themselves would not be in the Smith set by definition, and when there are fewer than 3 Smith candidates, either there is only one Smith candidate, who thus can't be beaten by anyone, or 2 Smith candidates, who must pairwise tie each other), so all candidates in the Smith set have a Copeland score of at least (NS - S + 1), fully written out as ((number of candidates not in the Smith set) - ((number of candidates in the Smith set) - 1). Every candidate not in the Smith set has a pairwise defeat against every candidate in the Smith set by definition, and can at most have pairwise victories against every candidate other than themselves not in the Smith set, thus their Copeland score can at most be (NS - S - 1), which is ((number of candidates not in the Smith set) - 1) - (number of candidates in Smith set). Thus, the members of the Smith set will always have a Copeland score at least 2 points higher than the candidates not in the Smith set.

Copeland's method also passes [[ISDA]]; since the Copelandprevious winnerparagraph isproves always in the Smith set,that 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 (and thus no pairwise defeat) 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 (with no change to their number of pairwise defeats); 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. If n is 3, this is the tertiary Smith set, the set of candidates that would be in the Smith set ignoring all candidates in the Smith set and secondary Smith set, 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, which reduces to no defeats against any candidates in the n-th Smith set when the n-th Smith set is smaller than 3 candidates (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, and there are guaranteed to be no pairwise defeats for Smith candidates when the Smith set is one or two candidates large), 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 minimakminimal/maximal Copeland scores of (K-J-N+1) and (K-J-N-1) respectively, with K being the number of candidates in all Smith sets "after" the n-th Smith set, J for all Smith sets "before" the n-th Smith set, and N for the candidates in the n-th Smith set. Therefore, the Copeland score of a candidate in the n-th Smith set is guaranteed to be at least 2 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 2 points higher than any candidate in the n-th Smith set.