Ranked Pairs and all of its variants that I am aware of abhor non-transitivity. Here is a variant of Ranked Pairs that embraces non-transitivity. Despite being non-transitive, it elects a unique winner, the Condorcet winner if there is one. In the cases I have looked at, the winner is also the Schulze winner. Is it always?
Candidates are classed in two categories: Winners and Losers. Initially, all candidates are Winners. Every candidate has an associated List of candidates that have defeated it. Every candidate initially has a List composed of itself and no other candidates. The method is so affirming of non-transitivity that it even treats each candidate as a non-transitivity loop unto itself. Winners are those candidates who have no Winners in their List aside from themselves. Rank the pairs in a strict order, in the same order one would use for your favorite strict order transitive variant of Ranked Pairs. Affirm each pair in order, from highest ranked to lowest. When A > B is affirmed, the List for candidate A is added to every List that includes candidate B (not just candidate B’s list). All Winners that now have other Winners in their List are reclassified as Losers. The count can be ended when only one Winner remains since affirming the remaining pairs cannot make the Winner a Loser. Provided that every pair is ranked into a strict ranking, and each pair expresses a definite ranking between the two candidates in the pair, there is guaranteed to be one Winner. Example election from: http://www.cs.wustl.edu/~legrand/rbvote/desc.html Brad > Erin 623, 298 Erin > Dave 610, 311 Dave > Brad 609, 312 Abby > Erin 511, 410 Abby > Dave 485, 436 Brad > Abby 463, 458 Abby > Cora 461, 460 Brad > Cora 461, 460 Dave > Cora 461, 460 Erin > Cora 461, 460 Each Candidate begins as a Winner with only itself on its List. Abby(W): Abby(W) Brad(W):Brad(W) Cora(W): Cora(W) Dave(W): Dave(W) Erin(W): Erin(W) The first affirmed pair is Brad> Erin. Brad’s List is added to Erin’s List, the only one that includes Erin. Abby(W): Abby(W) Brad(W):Brad(W) Cora(W): Cora(W) Dave(W): Dave(W) Erin(L): Erin(L), Brad(W) Erin is now a Looser. The next pair to be affirmed is Erin > Dave. Erin’s List is added to Dave’s List, the only one that includes Dave. Abby(W): Abby(W) Brad(W):Brad(W) Cora(W): Cora(W) Dave(L): Dave(L), Erin(L), Brad(W) Erin(L): Erin(L), Brad(W) Dave is now a Looser. The next pair to be affirmed is Dave > Brad. Dave’s List is added to Brad’s and Erin’s Lists, since both include Brad. Abby(W): Abby(W) Brad(W):Brad(W), Dave(L), Erin(L) Cora(W): Cora(W) Dave(L): Dave(L), Erin(L), Brad(W) Erin(L): Erin(L), Brad(W), Dave(L) Brad is still a Winner. The next pair to be affirmed is Abby > Erin. Abby’s List is added to Brad’s, Dave’s, and Erin’s Lists, since they all include Erin. Abby(W): Abby(W) Brad(L):Brad(L), Dave(L), Erin(L), Abby(W) Cora(W): Cora(W) Dave(L): Dave(L), Erin(L), Brad(L), Abby(W) Erin(L): Erin(L), Brad(L), Dave(L), Abby(W) Brad is now a Looser. The next pair to be affirmed is Abby > Dave. The Lists do not change. The next pair to be affirmed is Brad > Abby. Abby(W): Abby(W), Brad(L), Dave(L), Erin(L) Brad(L):Brad(L), Dave(L), Erin(L), Abby(W) Cora(W): Cora(W) Dave(L): Dave(L), Erin(L), Brad(L), Abby(W) Erin(L): Erin(L), Brad(L), Abby(W), Dave(L) Abby is still a Winner. The next affirmed pair is Abby > Cora. Abby(W): Abby(W), Brad(L), Dave(L), Erin(L) Brad(L):Brad(L), Dave(L), Erin(L), Abby(W) Cora(L): Cora(L), Abby(W), Brad(L), Dave(L), Erin(L) Dave(L): Dave(L), Erin(L), Brad(L), Abby(W) Erin(L): Erin(L), Brad(L), Abby(W), Dave(L) Cora is now a looser. Abby is the winner of the election.
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