[EM] MMV and resolvability

[Michael Ossipoff]

Anders & EM:

I'd been referring to MMV as the Drop-Equal-Defeats (DED) version or
Ranked-Pairs, for when a tied result is acceptable, and the use of
randomization in the count is unacceptable (as could be the case in
some Internet votes). But the name MMV is as good as any.

Ranked-Pairs seems to be taken, at EM, when there are no pairwise ties
or equal defeats, to mean:

Keep every defeat that doesn't contradict a set of stronger kept
defeats (by being in a cycle with them).

[end of Ranked-Pairs definition]

Methods more recent thanTtidemans seem to agree on allowing
equal-ranking, and truncated rankings, and in measuring
defeat-strength by "winning-votes" (the number of ballots ranking X
over Y is the measure of the strength of the X>Y defeat).

Those more reecent methods seem to disagree only in regards to how
pairwise-ties and equal defeats are dealt with.

Eppley is probably right, that MAM is the best. But, when a count
without mid-count randomization is desired, and it's ok to have a tied
outcome, then I prefer the DED tiebreaker,  the Maximum-Majorities
Voting (MMV) method.

And, actually, the RP definition that I stated above, in this posting,
when applied literally, implies MMV.

But, in the example that you gave, the fact that D>A contradicts a set
of stronger defeats disqualifies D>A from being kept.

So MMV, at least as I interpret it, doesn't have the problem that you described.

Michael Osspoff


On Sat, Dec 7, 2013 at 6:34 AM, Anders Kaseorg <andersk at mit.edu> wrote:
> Maximum Majority Voting exhibits a silly failure of Tideman’s
> resolvability criterion in the following four-candidate election:
>   3: A > B > C > D
>   1: B > D > A > C
>   2: C > D > B > A
>   1: D > A > B > C
>
> According to my reading of
> http://radicalcentrism.org/resources/maximum-majority-voting/, and also
> according to Eric Gorr’s calculator at http://condorcet.ericgorr.net/, MMV
> proceeds as follows.  The defeats are sorted:
>   5/2: A > C, B > C, C > D
>   4/3: A > B, B > D, D > A
> In the first step, MMV affirms the three 5/2 defeats.  In the second step,
> it ignores all three 4/3 defeats because they form a cycle.  The final
> result is A = B > C > D.

>
> However, there’s no way to add another ballot to make B the unique winner,
> in violation of resolvability.
>
> This failure is silly because, if the three 4/3 defeats had been sorted in
> _any_ strict order, D > A would have been ignored for already forming a
> cycle with the 5/2 defeats, and the final result would unambiguously be A
>> B > C > D.
>
> This feels like an oversight in the MMV definition, and although it’s
> highly unlikely to matter in practice, fixing it is simple enough.  At
> each step, when considering a set of equally strong defeats, we should
> immediately discard any defeats that would complete a cycle with strictly
> stronger affirmed defeats, so that they are not considered for use in
> cycles with equally strong defeats.
>
> Anders