Alex, --- Alex Small <[EMAIL PROTECTED]> a écrit : > Let's stick to 3 candidates and ignore equal rankings, to keep it simple. > There are 6 possible > types of voters, and so there are 6 variables to consider: The fraction of > the electorate with > each possible preference order. These variables live in a 6D space, but > because the fractions > add up to 1 we're dealing with a finite (and 5-dimensional) region of space. > Also, I assume > that the method treats all candidates equally, so that if every voter swapped > the winner (let's > call him candidate A) with the same other candidate (let's call him B) then B > would win.
Thanks to information from Forest, I was able to write a program that plots winners in a pyramid (a slice of which can be viewed at a time), for three candidates and 4 of the 6 ballot types at one time. > Conjecture #1: I conjecture that the methods which minimize the total area > of the boundaries > are those where the boundaries are "straight lines" (formally, 4D > hypersurfaces defined by a > linear equation, so there aren't any kinks or curves). The only method I've seen which produced curved boundaries in my program is Douglas Woodall's QLTD method, a Bucklin variant. The tie-breaker (when two candidates achieve a majority at the same rank) asks which candidate needs the lowest proportion of his last-rank votes to hit the 50% mark. > The notion of drawing "straight lines" in 5D might seem complicated, but the > methods that > correspond to "straight line" boundaries are easy to understand. They're > what Saari would call > "positional methods". Candidates get a certain number of points from each > voter depending on > how the voter ranked the candidates. > In one of my favorite positional methods (mostly > because of its importance in the strong FBC problem that I continue to pursue > on the side), a > candidate gets 1 point from each voter who ranks him 1st and 2nd, and 0 > points from each > candidate who ranks him 3rd. In my program, I have implemented this rule as "Approval." (There is no requirement to vote for a 2nd, correct?) > Anyway, I'm not saying these are the best methods to use, but simply posing a > question as a > warm-up for something that people on this list might find more practical: > > Question: Which Condorcet completion method minimizes the total area of the > boundaries? I think it's probably straight Approval (in the above sense), as in: If there is no CW, elect the Approval winner, even if he's not in the top tier. I get the impression because, if I take an arbitrary slice of a pyramid, with the four dimensions chosen arbitrarily and the static dimension set at 25% or 30%, I get the impression that "Approval" is much more likely to have huge win regions and shorter total boundary length than other methods. > My hunch is that it would be some sort of positional method, and that there > would be more than 1 > such positional method. However, it could just as easily turn out that > methods which compare > the magnitude of victory minimize the area of the boundaries and are > therefore simpler and less > prone to manipulation in some sense. No, I don't think so, since you're inviting "kinks" in the boundaries. You want a less sensitive method. Actually, "If no CW, Random Candidate" is probably the best. Kevin Venzke Découvrez le nouveau Yahoo! Mail : 250 Mo d'espace de stockage pour vos mails ! Créez votre Yahoo! Mail sur http://fr.mail.yahoo.com/ ---- Election-methods mailing list - see http://electorama.com/em for list info