Here's something close to what I had in mind originally: Let p be the probability distribution for a monotone, clone free, lottery like random favorite.
Define the distance between rating ballots b1 and b2 as d(b1, b2) = [sum (over the alternatives X) of the product p(X)*|b1(X)-b2(X)|^n]^(1/n), where we can choose n greater than or equal to one for various possibilities. Now for each alternative A, construct a representative ballot b as the average of all of the ratings ballots that rate A at the top. Then the (voter perceived) distance between two candidates is the distance between their representative ballots. The use of p to give weights to the "components" of the ballot vectors is what de-clones the metric. ----- Original Message ----- From: Kristofer Munsterhjelm Date: Monday, July 11, 2011 10:46 am Subject: Re: [EM] A distance based method To: [email protected] Cc: [email protected] > [email protected] wrote: > > First find a clone consistent way of defining distance between > candidates. > This could be an interesting algorithm problem in itself. It is > possible > to triangulate points in space (assuming Euclidean distances) if > you > have the exact distances; but what if you have only the rank > order of > distances? You get a (usually) underconstrained problem of the sort: > > Given v voters, n candidates, and a dimension integer k > 1, > find v + n coordinates in k-dimensional space so that the > Euclidean > distance from V's coordinate to C's coordinate for some voter V > and > candidate C is less than the Euclidean distance from V's > coordinate to > D's coordinate, for the same voter V and another candidate D, > iff V > ranks C ahead of D. Break ties by assigning coordinates so that > the sum > of the Euclidean distances to the candidates from the origin is > minimized.If it is not possible to make such an assignment, make > one that > contradicts as few candidate rankings as possible. > > I have no idea how you would actually do this, though, and it > would be > prone to overfitting. It might not be cloneproof, either, since > differences in clone rankings could eliminate some rotations > that would > otherwise be picked as the best choice by the tiebreaker. It > would > definitely not pass IIA, as the addition of "superbad" > candidates could > serve as anchors. > > > Then while two or more candidates remain > > of the two with the greatest distance from each other > > eliminate the one with the greatest pairwise defeat > > EndWhile. > > Or, if you have the voters' coordinates too, you could use > histograms, > kernel density estimation, or some other estimation to try to > reconstruct opinion space, and then pick a subset of the > candidates that > "reproduces" that opinion space as best as is possible. E.g. if > each > candidate (and voter) is a Gaussian in opinion space and you > want p > seats, find the p Gaussians where the difference between the > space given > by the sum of the p Gaussians (of the prospective council) and > of the v > Gaussians (of every voter), normalized, is minimized. > > Bandwidth selection would be a pain, as would finding the > "right" number > of dimensions. > > ---- Election-Methods mailing list - see http://electorama.com/em for list info
