[EM] Fwd: MMV and resolvability

[Michael Ossipoff]

 Better brief definition of MMV (which I formerly called DED-RP):

Keep every defeat that doesn't contradict (by being in a cycle with
them) a set of defeats that are all at least as strong as it.

[end of brief MMV definition]

It goes without saying that keeping a defeat means honoring it by not
electing its defeated candidate or alternative. Each defeat is a
public decision that someone shouldn't be elected.

With my previous definition, it would have been necessary to add
special wording to deal with equal defeats.

Steve suggested that my previous definition (posted before this
posting), as-is, implied MMV, which I was then calling RP with the
Drop-Equal-Defeats way of dealing with equal defeats. It seemed
plausible, and I assumed it to be so, but now I've stated, above, my
brief definition of MMV.

Michael Ossipoff




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