[EM] Dodgson and Kemeny "done right"?

[Warren Smith]

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?