[EM] No luck with simple monotone Heaviside methods, but a surprising idea

[Forest Simmons]

"To boldly go where no man has gone before!"

It takes a very tenacious mind... with bulldog tenacity... to pursue an
idea this far into the wild unknown!

On Wed, Dec 21, 2022, 10:18 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> So I was trying to find some methods that would have impartial culture
> manipulability less than 100% of the time for four candidates as the
> number of voters approaches infinity. (I should strictly speaking say
> that I don't know if any methods stay below 100% here, but I had the
> impression that IRV does.)
>
> I did some brute forcing of methods of the type
>
> A's score = f(A) = sum over other candidates B != A:
>         A>B *
>         product over other candidates C, D not A nor B:
>                 H(x_1 * [fpA fpB fpC fpD]) *
>                 H(x_2 * [fpA fpB fpC fpD]) *
>                 H(x_3 * [fpA fpB fpC fpD])
>
> highest score wins, with x_1, x_2, x_3 being vectors, and fpX being the
> first preference count of candidate X. This is a generalization of the
> Contingent vote,
>
> f(A) = sum over B
>         A>B *
>         product over C:
>                 H(fpA - fpB) * H(fpB - fpC)
>
> where H(x) as usual is the function
>         H(x) = 0   if x < 0
>                 1/2 if x = 0
>                 1   otherwise
>
> As a simple monotonicity check I used the possibly too generous
> assumption that if we write the first term
>         H(x_1 * [fpA fpB fpC fpD])
> as
>         H(x_11 fpA + x_12 fpB + x_13 fpC + x_14 fpD)
> then if x_11 >= max(x_12, x_13, x_14), then a voter who ranks say C
> first, can't harm A by raising A to top, because the benefit to the term
> inside the H can never decrease by doing so. So the method must be
> monotone since the A>B term is also monotone, hence the check x_11 >=
> max(x) is sufficient (but probably not necessary) for monotonicity.
>
> The results show a big nope: every such method (with the values of x
> integer between 2 and -2 inclusive) seem to be 100% manipulable in the
> limit. However, it did come up with this, which seemed to converge less
> quickly, and might be an interesting idea on its own:
>
> f(a) = sum (or max) over B:
>         A>B * H(fpA + fpB - fpC - fpD)
>
> I.e. for four candidates, there'll always exist at least one pair of
> candidates who have more than majority combined first preference support
> (unless there's a four-way tie). Give each candidate in this pair his
> pairwise defeat strength against the other.
>
> (In the case where there are multiple, e.g. a single candidate has
> majority, using sum would give a different method than using max.)
>
> -km
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>
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