When some weeks ago I started to read electo pages, I was both fascinated an overwhelmed by the information. One day at home thinking I realized I didn't know whether in "clone-proof" and other phrases "clone" means a candidate which relates (wins/ties/loses) to every other candidate as its "clone-companion", or (which would be a narrower definition) a candidate which is placed next to its companion on every vote.
The next day at my workplace by continuing the surf I realized that the latter case is true, but it was too late. I created a (maybe) new variant for single-winner elections, which I name quasi-cloning. (Let me note that for single winner, I very quickly became pro-Condorcet.) 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) (You can ask why to make the subsets at all, but I think this Rubicon is already crossed with the Smith-set, which is a special kind of quasi-clone sets.) When evaluating a vote, we make up the most coarse quasi-clone distribution of the whole candidate set. Then compute the winner quasi-clone set. Then consider the winner set as a whole candidate set and if it contains more than one candidate, do the procedure again for it. And so on, while it is necessary. To run the contest between the quasi-clone sets, I propose the Schwarz method. The number of members in a quasi-clone set, of course, can vary from set to set, so to measure winning stengths, we may use some average of margins, winning votes, or proportions or whatever we want. It's a question whether the most coarse (the one with the least number of members) distribution is unique. I found that in most cases it is. Let's take a case where candidate x is a member of both A and B (different) quasi-clone sets. If one of the two is a strict part of the other, only the bigger one "has the right" to take part in the most coarse distribution. If they both have at least one member outside the other, their union have to be related uniformly to outsiders, so, it "has the right". But what if the union of A and B constitutes the whole candidate set? In that case: let's call Ap the part of A outside B; let's call Bp the part of B outside A and let's call M the common part. Let a be a candidate in Ap. If x (which is in M) beats a, then (beacuse M is part of B) anyone in B must beat a, therefore anyone in Bp must beat anyone in A. In this case the most coarse distribution is not unique, but in any case the winner subset will be Bp sooner or later. The case is similar with a beating x. If a ties x, it makes unsolvable ties anyway. (If all this is hard to follow, draw it on paper.) Peter Barath _______________________________________________________________________________ Kiadós akció! 40% engedménnyel kínáljuk az Európa Könyvkiadó több mint hétszáz kötetét webáruházunkban kedden és szerdán. http://ad.adverticum.net/b/cl,1,6022,158433,222557/click.prm ---- election-methods mailing list - see http://electorama.com/em for list info
