On Fri, Sep 16, 2011 at 5:27 PM, Warren Smith <[email protected]> wrote:
> On Fri, Sep 16, 2011 at 8:21 PM, <[email protected]> wrote: > > You're right, I forgot that Kemeny only needed the pairwise matrix. And > according to Warren > > Dodgson is summable. I don't see how. > > --if "Dodgson" minimizes the total travel distance for all candidates > on all ballots to "travel" from their current position to the > output-permutation's position, > and "position" means "rank" then all you need to know is the total > number of times candidate X is in rank Y on any input ballot, for all > (X,Y). > > That count-info is publishable by each precinct. For N candidates this > is N^2 different counts published by each precinct. > > Right? I think we just have to find the minimal travel distance to make someone a Condorcet winner. We don't have to make all of the ballots identical.
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