[EM] Ideas for distance Schulze tie-breaking

Forest Simmons forest.simmons21 at gmail.com
Tue Mar 1 09:25:53 PST 2022

Define the total resistance of the losers in a beatpath as the sum of the
losing votes. Unfortunately, minimization of this total beatpath resistance
is a clone dependent method.

However, the method can be decloned.

Let L be the vector of losing votes along the beatpath. Let F be the vector
of corresponding favorite votes, i.e. F_k is the number of ballots on which
candidate k is designated as favorite.

The decloned resistance of the beatpath is the dot product of L and F.

Minimizing this dot product over beatpaths is a clone free method that
could be called de-cloned Min Resistance BeatPath (dcMinRBP).

Use this method to break ties.

Better yet, use dcMinRBP as the main method, and break ties with Schulze.

El dom., 27 de feb. de 2022 6:05 a. m., Kristofer Munsterhjelm <
km_elmet at t-online.de> escribió:

> Like minmax, Schulze sometimes has a lot of ties, particularly for
> elections with few voters. But perhaps this way could work to break ties
> in a consistent way:
>
> Let p[A, B] be the strength of the strongest beatpath going from A to B.
>
> Now do ext-Minmax on the p matrix. This is minmax (i.e. A's score is A's
> weakest pairwise contest vs B), but ties are broken by looking at second
> weakest pairwise contests, third weakest, etc.
>
> Since there always exists a candidate A who is the "Condorcet winner"
> according to p, and this is the same as the Schulze winner, and both
> Minmax and ext-Minmax pass Condorcet, this method should agree with
> Schulze when there are no ties.
>
> But do we lose any criteria from this? I don't know; it's just a thought
> that occurred to me as a way to make ext-Minmax more Schulze-like.
> Possibly there exist situations where there's a tie according to
> Schulze, and this breaks the tie in a way that (say) depends on clones;
> but I'm not sure how to construct such an example.
>
> Another possible drawback is that the social order may suffer, because
> minmax doesn't pass Condorcet loser. So it might be that the social
> ranking is A>B>C>D by Schulze, and this is strict (A has 3 strong
> beatpaths, B has 2, C has 1, D has none), but D's weakest beatpath is
> stronger than C's; and then the ext-Minmax modification returns a
> different social order even though there are no ties at any point.
>
> This might point to the idea being not entirely defensible. One could,
> of course, only break the ties that exist by ext-Minmax, but that feels
> rather like a hack.
>
> Perhaps it would be doable to augment Floyd-Warshall to maximize the
> leximin of the beatpath instead of just its minimum - among the paths
> with the strongest weakest link, find the one with the strongest
> second-weakest link, etc. Then comparing the full vectors of defeats
> along the beatpath thus recovered could resolve ties in a way more
> consistent with the Schulze method itself.
>
> I'm still only working by intuition, though; I haven't rigorously
> checked any of these ideas.
>
> -km
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