[EM] A more elegant three-candidate fpA-fpC phrasing, inspired by Heaviside formulation

[Daniel Carrera]

On Thu, Jan 20, 2022 at 5:21 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> f(A, e) = sum over B != A: H(A>B) * (2 fpA + fpB).
>
> Unlike IRV, here the H term is a pairwise one. This makes it hard to
> construct a recursive version. Now we could of course just generalize
> the "sum over all B that A beats" concept to more than three candidates,
> but I don't think the resulting method would pass DMTBR, and I also
> think it would fail clone dependence (vote splitting with a cycle among
> the clones).
>

How does H being pairwise make it hard to write a recursive version? Let's
try the easiest form of recursion:is:

f(A,e) = sum over B!=A: H(A>B) * f(A,e/B)

where "e/B" is the election you get if you remove candidate B.

I tried to analyze this, but I didn't get very far. I think I figured that
it's monotone. I struggled to prove whether it's clone independent. I don't
know how to begin investigating DMTBR.

-- 
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
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