[EM] Saari's Basic Argument
Forest Simmons
fsimmons at pcc.edu
Tue Jan 14 12:51:02 PST 2003
In EM archives message #8999 Alex recounts in his own words Saari's idea
of subtracting out the symmetrical part of a ballot pattern and then
deciding the winner on the basis of the residual ballots.
Let's do an example:
7 ABC, 5 ACB , 9 CAB, 3 CBA, 7 BCA, 8 BAC.
If we subtract three of each kind, we are left with
4 ABC, 2 ACB, 6 CAB, 0 CBA, 4 BCA, 5 BAC.
Now subtract four copies of the cycle {ABC, CAB, BCA} to get
0 ABC, 2 ACB, 2 CAB, 0 CBA, 0 BCA, 5 BAC.
Now subtract two copies of the symmetric pair {CAB, BCA}.
0 ABC, 2 ACB, 0 CAB, 0 CBA, 0 BCA, 3 BAC.
The residual ballot set is
2 ACB
3 BAC.
At this point we part company with Saari who would give the win to A
on the basis of the Borda Count.
>From the game theoretic point of view, if the majority winner B is not
awarded the win, there is too much incentive for insincere ranking and for
propping up candidates with dummy clones (like C in this residual
election).
The interesting thing that I just realized recently is that in three way
races removal of all the symmetry will always leave a residual in which at
least one of the candidates has no first place votes.
So the majority criterion is enough to decide which candidate wins the
residual election.
In fact, the residual will be restricted to one semicircle of the clock
face that I described in a recent posting, and which I include here:
Think of the preference orders ACB, CAB, CBA, BCA, BAC, and ABC positioned
on the face of a clock at the respective even hour marks 2, 4, 6, 8, 10,
and 12.
Notice that any three consecutive orders have only two candidates in first
place positions. In other words, when all of the (residual) factions fit
into one semicircle, the problem has been reduced to a choice between two
first placers.
[When all of the symmetry has been removed, there will be an empty
semicircle, so the remaining factions will be contained in a semicircle,
as required.]
This can be adapted to account for truncated rankings as well, if we agree
that A = .5 ABC + .5 ACB, for example.
In the case of full rankings of three candidates, this residual method
seems to always gives the same result as the Kemeny order, MinMax, Ranked
Pairs, SSD, etc. for the original problem.
So Saari's symmetry approach supports Condorcet more than Borda.
It would be interesting to see how far this approach could be taken in the
four candidate case.
In that case the adjacency graph of the 24 orderings fits in natural
symmetry on the surface of a sphere and as the vertices of a solid with 14
faces and 36 edges. [Six of the faces are squares, and the other eight are
hexagons.]
Forest
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