# No subject

Tue May 6 19:13:21 PDT 2014

```difference if this matrix is processed according to the rules of MAM,
CSSD, or MinMax; the outcome will be the same: if there is a cycle, then
it will be broken at the weakest link.  Otherwise the finalist that beats
both of the others wins.

Who has the best idea on how to narrow down to three finalists?

Here are some that have already been proposed:

1.  Take the top three approval scorers.

2.  Take the IRV winner along with the last two that IRV would have
eliminated.

3.  Take the Buckley winner, the Nanson winner, and the Coombs winner.

4.  Take the MinMax winner along with the two MinMax runnerups.

5.  List the candidates in order of approval. Then (from strongest
discrepancy to weakest discrepancy) transpose adjacent candidates in the
list until there is no order discrepancy among adjacent pairs. Take the
top three candidates from this list.

This last suggestion has the advantage of always choosing from the Smith
Set, without ever having to explicitly enumerate that set.  Indeed, after
the adjacent discrepancies have been removed, the Smith Set will
automatically head the list, as surely as the cream rises to the top in
Brown Cow yogurt.

The discrepancy removal process is just the familiar process of
sorting by height a motley line of recruits before undertaking military
drill:

WHILE any recruit is taller than the one to his/her immediate right,

DO switch the adjacent recruits with the greatest such height discrepancy.

END

This version of Top Three Condorcet is summable, since it only requires
the pairwise matrix and the approval scores (which for convenience can be
incorporated as diagonal entries in the pairwise matrix).

[For ranking purists (who don't like approval cutoffs) substitute Borda
Scores for Approval Scores in Method 5.]

Any other ideas for Top Three Condorcet?

Forest

```