[EM] Median bifurcation using pairwise matrix

Rob LeGrand honky1998 at yahoo.com
Mon Aug 13 16:35:04 PDT 2001

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


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

Do You Yahoo!?
Make international calls for as low as $.04/minute with Yahoo! Messenger

More information about the Election-Methods mailing list