>>I call a subset of candidates a quasi-clone set, if: >> >>1. they don't make up the whole set of candidates >>2. for every candidate out of the set they are in >>the same winning relation with (all beat / all tie / >>all lose)
>This is similar to Forest Simmons' "beat clone sets" he uses in his >Dec. 2004 "sprucing up" process idea. >your proposal is very similar to Mike Ossipoff's subcycle rule. I apologise for not reading the whole archive, but it's volume is bigger now than that of the Holy Bible. I realized that the mentioned proposals are really close to mine, but not exactly the same. But I have to admit that my proposal still fails the Pareto criterion. Well, I think it's just natural that if you are new in the field, you dream of working out something new. So, I kept thinking in the last weeks, hoped for a new method, then for at least a new criterion, and finally got a paradox - only hope that it's new and interesting for some extent. The beginning is: if in a Condorcet voting three candidates are in circular ambiguity, let's say A > B > C > A, it only can be caused by circular sets of votes: ABC BCA CAB We can call a set of ballots a circular set, if there is a permutation wich if applied on each ballot, we get the same set of ballots. Here an appropriate permutation is: move the first element to second place; move the second element to third place; move the third element to first place. So it makes ABC into CAB, makes BCA into ABC, and makes CAB into BCA, which are the same ballots, only in different order. It seems obvious that these ballots give wholly symmetrical positions to the candidates, so we can eliminate them. Unfortunately, this usually works only on three candidates. These three ballots ABCDE CDEAB EABCD also makes a circular ambiguity, but there is no circular set to eliminate - it would need 5 ballots. So as a method it fails, but still, I hoped for a criterion: eliminating a circular set must not change the result. Disappointment again. Let's see the following 9 ballots: "the nice group" BCDA CDAB DABC "the ugly group" DCAB BADC CDBA "the completter" ABCD "the two others" DCBA CDBA The "nice group" and the "completter" together makes a circular set: BCDA CDAB DABC ABCD so we can eliminate them, and remains: DCAB BADC CDBA DCBA CDBA in which D is the Condorcet and also the Instant Runoff winner. But the "ugly group" and the "completter" also makes a circular set: DCAB BADC CDBA ABCD which is not so easy to see, but use the following permutation: move the first element to third place, the second to fourth, the third to second, and the fourth to first. It transforms each ballot into the next one in this set. Removing them, remains: BCDA CDAB DABC DCBA CDBA in which C is the Condorcet and also the Instant Runoff winner. So this criterion seems logical at the first sight but useless in the second one (Borda complies with it, but otherwise works poorly). Peter Barath ____________________________________________________________________ Tavaszig, most minden féláron! ADSL Internet már 1 745 Ft/hó -tól. Keresse ajánlatunkat a http://www.freestart.hu oldalon! ---- election-methods mailing list - see http://electorama.com/em for list info
