31: A>B
32: B>C
37: C>A
Approvals: B63, A68, C69. A>B>C>A.
TACC elects A, but C is positionally the dominant candidate and
pairwise beats A.
For a Condorcet method with pretension to mathematical elegance, I don't
see how that
can be justified.
Chris Benham
PS: Could someone please refresh our memories: What is the "Banks Set"?
From Jobst Heitzig (March 2005):
ROACC (Random Order Acrobatic Chain Climbing):
--------------------------------------------------------------
1. Sort the candidates into a random order.
2. Starting with an empty "chain of candidates", consider each
candidate in the above order. When the candidate defeats all
candidates already in the chain, add her at the top of the chain.
The last added candidate wins.
The good thing about ROACC is that it is both
- monotonic and
- the winner is in the Banks Set,
in particular, the winner is uncovered and thus the method is Smith-,
Pareto-, and Condorcet-efficient.
Until yesterday ROACC was the only way I knew of to choose an
uncovered candidate in a monotonic way. But Forest's idea of needles
tells us that it can be done also in another way.
The only difference is that in step 1 we use approval scores instead
of a random process:
TACC (Total Approval Chain Climbing):
------------------------------------------------
1. Sort the candidates by increasing total approval.
2. Exactly as above.
The main differences in properties are: TACC is deterministic where
ROACC was randomized, and TACC respects approval information where
ROACC only uses the defeat information.
And, most important: TACC is clone-proof where ROACC was not! That was
something Forest and I tried to fix without violating monotonicity but
failed. More precisely, ROACC was
only weakly clone-proof in the sense that cloning cannot change the
set of possible winners but can change the actual probabilites of
winning. With TACC, this makes no difference since it
is deterministic and so the set of possible winners consists of only
one candidate anyway.
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