[EM] The resistant set

[Filip Ejlak]

If we have a cycle scenario

7: A>B>C
7: B>C>A
6: C>A>B

then the resistant set is {A}, so A is the winner. But if we change one BCA
voter to an ABC voter (mono-raising A), then the resistant set is {A, C} -
and now every majoritarian method will choose C. So Resistant//? or
Smith//Resistant//? will be non-monotonic, no matter what the "?" method
is, right?




On 8/16/23 13:38, Kristofer Munsterhjelm wrote:
>
> > So point three, the "no-expansion property": a candidate B could be
> > admitted if some C who was previously barring B stops either having
> >  >1/|S| support in every restricted election involving both of them, or
> > stops beating B pairwise. The latter is obviously impossible. But the
> > former... that's what has to be shown, and why I'm not certain.
> >
> > You would think that set growing would be possible by a variant of
> > Kevin's counterexample to method X. Let A bar C from the set, then let
> > the base method outcome be C>B>A so that B wins. Then raising B on an
> > ABC ballot could reduce A below the threshold, after which C wins. But
> > somehow I can't get that to work...
> >
> > It gives an idea of where you could start if you have a monotonicity
> > checker, though.
>
> The reason I can't prove that is because it isn't true. Here's a 6
> candidate Yee diagram of Resistant,max A>B. (Sorry about the
> near-identical shades of green on the lower left.)
>
> If you can draw a line from a region of one color to the origin of that
> color, and that line passes over a region of a different color, then the
> method is nonmonotone (if I understand correctly). And that's clearly
> the case here.
>
> I kind of feel like I'm advancing up Forest's homotopy method sequence,
> with method X being f_0, and the resistant set compositions being f_1.
> Can we ever reach high enough to get monotonicity? :-)
>
> (Note that it's still possible that *some* composition is monotone. So
> I'd still be interested if you were to check.)
>
> -km----
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