so the ordering is B>C>A no matter what x is within the allowed range.
I've been thinking about particular types of three-candidate Condorcet
methods. What I've been doing suggests that for a continuous method to
pass UM and Condorcet, it must return an A=B=C tie for every pairwise
tie election (i.e. where A=B, A=C, B=C). Furthermore, adding ballots to
turn A=B and A=C into A>B and B>C must make A win with certainty.
The reason is that if we're somewhere on the "A is the CW" side of the
(A>B, B>C, A=C) boundary, there exists an election where changing an
epsilon of a B>A>C ballot into B>C>A will get us *onto* that boundary.
Such a change leads to an UM violation if A doesn't win with certainty
(It also looks like the number of elections near the boundary with no
B>A>C ballots is so small compared to the number of elections with, that
any election that makes A win with certainty on the area of the boundary
you can reach by turning B>A>C into B>C>A must make A win with certainty
along the whole boundary. At least if it's continuous, although I've
only properly investigated a form of linear method.)
"Center-squeezy" methods like fpA-fpC fail by not returning an A=B=C tie
on every pairwise tied election. Methods like minmax pass the boundary
condition, but fail inside the ABCA cycle region itself.
What I'd have to show to have an actual impossibility proof is that it's
impossible to construct a method that both respects UM when going from
the "A always wins" region to the ABCA cycle region, and also respects
UM inside the region. I don't have that (yet, at least). But if I were
to guess, I think Condorcet and UM are incompatible. This is unfortunate
because it would otherwise be a good criterion to use to find
manipulation-resistant Condorcet methods.
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