[EM] "Monotonified" IRV: DMT?

[Kristofer Munsterhjelm]

The "monotone fixed" IRV-like is:

f(A, e) = A>B for a two candidate election e, and

f(A, e) = g over all candidates B not equal to A:
		H(fpA-fpB) * f(A, e\B)
for an election e with more than two candidates.

Does this pass DMT in the three candidate case? For ease of analysis,
explicitly write the three candidate case as

f(A, e) = g(H(fpA-fpB) * (A>C), H(fpA - fpC) * (A>B))

Suppose that A has more than 1/3 fpp and beats everybody else pairwise,
but that B is the plurality winner. Since the Plurality loser can't have
more than 1/3 first preferences, the Plurality order must be B>A>C. Then
for A we get:

f(A, e) = g(H(<0) * A>C, H(>0) * A>B),
        = g(0, A>B)

and for B:

f(B, e) = g(H(fpB-fpA) * B>C, H(fpB-fpC) * B>A),
        = g(B>C, B>A)

If we're using pairwise opposition, then it might well be that (A>B) <
(B>A + B>C). In this case, with g being the sum operator, the method
would fail DMT.

I then had the thought that letting g be the max operator would salvage
things. Then we have A's score being max(0, A>B), and B's score being
max(B>C, B>A). Since A is the CW, A>X is greater than X>A for any X, so
(A>B) > (B>A). This narrows the scope of a DMT failure, but we still
have one if B>C > A>B. So max doesn't work either.

Using wv instead of parwise opposition imparts a max-like logic to the
sum operator. Then wv(B>A) = 0, so that we get DMT failure if (B>C) >
(A>B) with either the sum or max operator.

Just to be sure, I checked if that can actually happen. The constraints are:
	(B>C) > (A>B)
	fpA > 1/3 |V|
	(A>B) > (B>A)
	(A>C) > (C>A)
and	fpB > fpA > fpC.

A solution is:

2: A>B>C
2: A>C>B
2: B>A>C
3: B>C>A
2: C>A>B.

So the simple monotone modification does not appear to pass DMT with g
being either max or plus. It's not going to be that easy.

-km