It has been pointed out from time to time that when the voters and candidates all lie along the same one-dimensional spectrum of opinion, then the Condorcet winner is the top preference of the medium voter. [This assumes that there are an odd number of voters, or that the two voters next to the empty middle position agree. Otherwise, there is a tie to be broken.] So (in the odd number case) one way to find the Condorcet winner is to remove outer pairs of ballots until there remains only one. The first choice of that one ballot is to be elected. This suggests that when the issue space is not just one dimensional, we could define the median ballot by removing maximal diametric pairs until fewer than three remain. The main difficulty with this approach is coming up with an appropriate metric for determining the distance between two ballots. Most naive attempts at defining a metric turn out to be clone dependent. When a candidate is replaced with a clone set with identical ratings, that clone set rating has more influence on the distance between ballots than the original candidate did, so the distances become distorted with a "teaming" or "clone loser" effect. Hereafter assume that we are using range (i.e. cardinal rating) style ballots. We can think of each range ballot B as a function from the from the set K of candidates into the allowable interval [a, b] of range values. So the metrics that first come to mind are the common metrics used on the function spaces L_1, L_2, and L_infinity . None of these work, unless we declone them. For example, here's how to declone the L_1 metric, sometimes called the Hamming metric: Let f and g be two members of a set S of range ballots, defined on a set K of candidates. The Hamming distance between f and g is just the sum Sum over k in K of |f(k)-g(k)| . To declone this we throw in an appropriate factor p(k) : d(f, g) = Sum over k in K of p(k)*|f(k)-g(k)| . For each candidate k, the number p(k) is the probability that a random ballot process would single out k. Which random ballot process? It seems to me that the outer candidates should be given more weight. If that is so, then we should use random ballot for the lowest rated candidate. If each ballot gives only one candidate the minimum rating, then p(k) is just the percentage of ballots on which candidate k has that minimum rating. Why should we give the outer candidates more weight? Because when you are trying to triangulate from several reference points, it is best to have the reference points widely spaced. Suppose that you were trying to locate your position inside a circle by comparing distances to three fixed points. Which would you prefer? All three of those points near the center of the circle or all near the circumference? Apart from these considerations, I believe that this definition of p(k) has a better chance of preserving monotonicity, though I do not know yet if it does. I gotta go. Bye for now. Forest
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