[EM] Median bifurcation using pairwise matrix

[Rob LeGrand]

Roy wrote:
> First, notice that if you sum all the rows, then sum those totals,
> you'll get the same number as if you sum all the columns and sum
> *those* numbers. (Let's all say it together: Duh.) The rows indicate
> wins for each candidate, the columns indicate losses. If a candidate
> has a higher total in his row than in his column, he's in the upper
> half. And because the rows and columns total out the same, there's
> going to be a lower half -- candidates whose row totals are less than
> or equal to their column totals. (Note: "half" refers to median rank
> here. With N candidates, as many as N-1 can be in one "half". In a
> very special case, all of them could have row_total = column_total,
> calling for a special rule.)
>
> Make a table for each new group. Compute row and column totals, and
> find the new winners and losers in each table by the same method,
> until you've got all "groups" of one.

FYI, I believe this procedure will give the same results as the procedure
called Nanson in Vandercruyssen's paper "Analysis of voting procedures in
one-seat elections: Condorcet efficiency and Borda efficiency".  The paper also
describes the procedure that Blake's site calls Nanson, but calls it Baldwin. 
See the paper at

http://www.econ.kuleuven.ac.be/ew/admin/Publications/DPS99/DPS9911.pdf

I've only received one ballot for the political party poll.  Please participate
when you have a few minutes whether you live in the U.S. or not; I promise
there's a point to it.  Thank you!

--
Rob LeGrand
honky98 at aggies.org
http://www.aggies.org/honky98/

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