# [EM] Ideas for distance Schulze tie-breaking

Forest Simmons forest.simmons21 at gmail.com
Tue Mar 1 13:38:18 PST 2022

```I wrote this in a hurry ... the basic method should be called
MinMaxLeastTotalResistanceBeatPath.

The total resistance TR of a beatpath BP is the sum of its losing votes.

For each pair of candidates (j,k) let TLR(j,k) be the total resistance of
the beatpath of least resistance from j to k. Elect the candidate j that
minimizes the maximum value of TLR(j,k) as k varies over the other
candidates.

To get the de-cloned version, weight each losing vote by the number of
ballots on which the pairwise loser is designated favorite.

Any electrical engineer should be able to design an "electo-meter" network
that has, for each pair of candidates (j,k), a diode from node j to node k
that allows current to move from j to k with resistance jointly
proportional to the number of ballots on which k is ranked ahead of j and
the number of ballots on which k is designated favorite.

Elect the candidate j that minimizes the maximum network resistance from
node j to node k as k varies over the other nodes. [Put the source probe of
a directional ohm meter at j while varying the sink probe over the other
nodes (k).]

One of ypu engineers correct me if my EE terminology isn't quite right.

-Forest

El mar., 1 de mar. de 2022 9:25 a. m., Forest Simmons <
forest.simmons21 at gmail.com> escribió:

> 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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>> info
>>
>
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