[EM] Oops, 2 fpA + fpB busted

[Kristofer Munsterhjelm]

Some time ago, I said that fpA-fpC could be rephrased in a more elegant
manner as simply:

A's score is the sum over other candidates B that he pairwise defeats,
of the function f(A,B) = 2 fpA + fpB.

But now I see that this formulation is not Condorcet. Consider the
limiting case of the LCR scenario that traditionally gives IRV such trouble:

2e: A>B>C
1/2-e: B>A>C
1/2-e: C>A>B.

Without loss of generality, I've set the total number of voters to 1,
and the weights are all real valued. Here for any arbitrarily small
epsilon > 0, A remains the Condorcet winner. However, the scores are:

A: 2(4e + 1/2-e) = 6e + 1
B: 3/2 - 3e
C: 0

Since 3/2 > 1, B wins even though A is the Condorcet winner. The CW
stops being the winner when A has less than 1/6 of the total first
preferences.

Note that this problem remains if we change the sum to min or max. If we
change the function to be summed, it must have the property so that in
an ABCA cycle, it's just fpA-fpC. Changing it to some other linear
function of the voting vector won't work.

More complicated modifications might still save the method, but then we
lose its elegance. E.g. recursive:

g(A) = sum over B s.th. A>B: <v(A),x> + g(B)

where v(B) is the voting vector rotated so that B is now A. Perhaps
there exists a vector x so that in an ABCA cycle, g(A) is an affine
scaling of fpA-fpC. For instance, if my calculations are right,

g(A) = sum over B s.th. A>B: 12/7 fpA - 3/7 fpB + 2g(B)

should work for three-candidate elections. But the constants seemingly
come from nowhere, which makes it extremely hard to generalize. In
addition, the g values would have to be solved by a set of linear
equations when cycles exist, and that would be much more complex than
the original "CW otherwise fpA-fpC".

-km