[EM] Dodgson and Kemeny "done right"?

[Andy Jennings]

On Fri, Sep 16, 2011 at 5:27 PM, Warren Smith <warren.wds at gmail.com> wrote:

> On Fri, Sep 16, 2011 at 8:21 PM,  <fsimmons at pcc.edu> 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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