[EM] Condorcet completion with two tiebreakers

[mrouse1 at mrouse.com]

I was wondering if anyone -- ha, okay, scratch that, I was wondering *WHO*
has looked into using two tie breakers for Condorcet cycles, rather than
just one? By that, I mean

1. Find the Condorcet winner
2. If there is a Condorcet cycle, take the Smith (or Schwarz) set
3. Out of this set, find a single winner with method A, and find a single
winner with a completely different method B.
4. If both methods pick the same candidate, this  person wins, otherwise
compare candidates A and B to find the overall winner.

Perhaps I should backtrack a bit. I was looking at Yee diagrams today for
the umpteenth time (they are a lot of fun). When I was looking at the
diagrams for Borda and Plurality, it struck me that if Borda tends to pick
centrists, and Plurality tends to pick extremists, it might be interesting
to find the winners with each method, and then compare them against each
other to choose the overall winner. (As a side note, it would be fun to
look at the Yee diagrams of such pairs to see which combined pairs came
closest to Voronoi diagrams.)

Using the framework above, you'd have:


1. Find the Condorcet winner.
2. If there is a Condorcet cycle, take the Smith set.
3. Find the Plurality winner out of this set.
4. Find the Borda winner out of this set.
5. If both methods pick the same candidate, this  person wins, otherwise
compare the Borda winner with the Condorcet winner. (In other words, in a
three-person tie, you'd drop the candidate that was neither the Borda nor
the Plurality winner, and compare the remaining two.)

Of course, you could use other methods. I picked Plurality and Borda
because they were very simple and acted in opposite ways in certain
situations. Plus, it's likely someone has looked into them some time in
the past. You might be able to come up with better pairs of methods that
have opposing strategies, making strategic voting less useful (or at least
more difficult).

Michael Rouse