Dear Ross Hyman, you wrote (27 Nov 2011):
> A and B are both winners. A is in B's set and B is in A's set. > So A is deleted from B's set and B is deleted from A's set. > > A(W): A(W), C(L), D(L) > > B(W): B(W), C(L), D(L) > > C(L): A(W), B(W), C(L), D(L) > > D(L): A(W), B(W), C(L), D(L) > > > > affirm A > B > > A(W):A(W), C(L), D(L) > > B(L): A(W), B(L), C(L), D(L) > > C(L): A(W), B(L), C(L), D(L) > > D(L): A(W), B(L), C(L), D(L) > > B was reclassified as a Loser since A(W) is in its set. When I understand your proposal correctly, then you are basically saying that, when contradicting beatpaths have the same strength, then they are cancelling each other out and the next strongest beatpath decides. I believe that your proposal can lead to a violation of monotonicity. Let's say that there is one beatpath from candidate X to candidate Y of strength z and two beatpaths from candidate Y to candidate X of strength z. Then these beatpaths cancel each other out. However, if one of the two beatpaths from candidate Y to candidate X is weakened, then this beatpath decides that candidate Y is ranked ahead of candidate X in the collective ranking. (This is problematic especially when the weakened beatpath was the direct comparison Y vs. X.) By the way: In my paper, I also recommend that the ranked pairs method should be used to resolve situations where the Schulze winner is not unique. However, the precise formulation is important. See section 5 stage 3 of my paper: http://m-schulze.webhop.net/schulze1.pdf Markus Schulze ---- Election-Methods mailing list - see http://electorama.com/em for list info
