# [EM] Witnessed Pairs (simple fpA-fpC)

Filip Ejlak tersander at gmail.com
Wed May 24 13:55:37 PDT 2023

```Well, Ranked Pairs-like completion has this important quality that X1/X2/X3
can win even if (X1>X2)-score, (X2>X3)-score, and (X3>X1)-score are all
very low - it doesn't matter as long as the clones' defeats against the
other candidates are strong enough. And I think that cloning X can't
actually decrease X1/X2/X3's strength against any candidate Y, because if
X1 and X2 are in the same pairwise relation with Y (as they should be
because of how we define clones), then they cannot form a cycle, so there
is no new fpA-fpC score to potentially replace the current strength value.

niedz., 21 maj 2023 o 18:34 Kristofer Munsterhjelm <km_elmet at t-online.de>
napisał(a):

> On 5/21/23 10:42, Filip Ejlak wrote:
> > And here I have a simpler fpA-fpC generalization (haven't tested it yet):
> >
> > When making an X vs Y comparison with Z as a witness, X's strength
> > against Y equals:
> > - X's fpA-fpC score if there's a {X,Y,Z} cycle
> > - the number of X>Y votes if there's no cycle
> >
> > Final strength of X against Y equals the minimum possible strength, i.e.
> > the strength with such a witness that the strength value is minimized.
> >
> > Create a comparison matrix with final strength values. Proceed with a
> > defeat-dropping Condorcet method.
>
> I've been thinking about methods that try to generalize fpA-fpC by using
> max, but I think similar problems may exist with methods that use min.
>
> Suppose we clone a candidate (call him A) into a three-cycle,
> A1>A2>A3>A1. Then if we just say, we're going to make a method that's
>
> score(A) = max over B, C: fpA-fpC(restricted to A, B, C)
>
> then it's possible to arrange the clones' first preferences so that the
> clones grant A a very high maximum score. (The same trick works for the
> method where A's score is just the margin of the greatest landslide A is
> involved in.)
>
> So suppose that we define a matrix D so that (X>Y)_D = min over Z:
> fpA-fpC restricted to X,Y,Z.
>
> Then there needs to be some additional structure so that cloning X
> doesn't lead fpA-fpC restricted to X1, X2, X3 to be very unfavorable to
> X, so that X loses. (I think an analogous observation is why Minmax
> fails clone dependence.)
>
> Perhaps this additional structure would come from an observation of the
> type: suppose X1, X2, and X3 are clones; then it may be impossible for
> (X1>X2)_D, (X2>X3)_D, and (X3>X1)_D to *all* be small. Thus if X used to
> win, at least one of the clones will succeed through Ranked Pairs (or
> Schulze, etc) applied to D, and thus X will keep winning.
>
> Clearly it's impossible for all the clones' inter-clone victories to be
> small if they have the same witness. But would it be possible to
> carefully craft a ballot where say
>         (X1>X2)_D has witness A, with low score for X1
>         (X2>X3)_D has witness B, with low score for X2
>         (X3>X1)_D has witness C, with low score for X3?
>
> Perhaps at least Ranked Pairs deals with this by disregarding such low
> scores until the higher scores have all been locked in...
>
> -km
>
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