[EM] MMV and resolvability

Michael Ossipoff email9648742 at gmail.com
Sat Dec 7 20:01:22 PST 2013

Markus wrote:

suppose that there are 4 candidates and that all pairwise
defeats A > B, B > C, C > D, D > A, B > D, and C > A have the
same strength. Then all 4 candidates are potential winners.

MMV returns a tie in which all of the defeats are disregarded, and all
of the alternatives win.

However, suppose that some voters rank candidate A higher so
that the pairwise defeats A > B, B > C, C > D, D > A, and B > D
still have the same strength and C > A is weaker than the other
pairwise defeats.

Ok, say they previously ranked C over A, but now they switch the rank
positions of C and A.

Then you would reject A > B, B > C, C > D,
D > A, and B > D and keep C > A

There are no defeats that aren't in a cycle in which all the other
defeats are at least as strong.
When I wrote my definition, I didn't consider such a situation. I
assumed that there would be some defeats kept because they weren't in
a cycle with defeats that were all at least as strong.

In your example, my definiton doesn't give a result. That brief
definition doesn't work in that example. When it isn't possible to say
which defeats are or aren't kept, that sounds like a tie among all the
alternatives. So I'd say that the method, as i define it, returns a
tie, before and after those voters switch A and C in their rankings. A
tie because of indeterminability.

I don't have a copy of Prabhakar's definition right now, but if he
said to consider the defeats one at a time, stronger ones first, and
keep them if they aren't in a cycle consisting only of them and
stronger defeats, and to otherwise throw them out, then the method
would first keep all of the defeats except for A>C. Next A>C would be
considered, and thrown out. But A wouldn't win, because D>A is kept..
It's still a tie.

By the way, do you have a set of ranings that gives the pairwise
defeats in your example? Shouldn't an example state the rankings, in
order to avoid examples consisting of sets of pairwise vote-totals not
consistent with any set of rankings?

Your example seems to show that my definition can result in not having
a determinable result at all. A tie in the broadest sense, but not the
kind that one would prefer.

Maybe it would be better to define the method in the usual
ordered-considering way.

But, for the MMV/DED way of handling equal defeats:

Say that if several equal defeats, when considered (at the same time,
of course) are all in a cycle consisting only o them and stronger
defeats, than all of those equallly weaker ones are rejected together.
...and if they're all in a cycle only with eachother, then too are
they're all rejected together.

That will be my ordered-procedure definition of MMV/DED.

Applying that method to your example, all of the defeats except for
C>A are kept. Next, C>A is rejected. But A still has its D>A defeat,
and so (as I described above), there is no winner because every
alternative has a kept defeat. The switching of A and C didn't change
the outcome, and so there isn't a nonmonotonic result.

Michael Ossipoff

On Sat, Dec 7, 2013 at 1:07 PM, Markus Schulze
<markus.schulze at alumni.tu-berlin.de> wrote:
> Hallo,
> in my opinion, an even worse problem of MMV
> is its violation of monotonicity. See:
> http://lists.electorama.com/pipermail/election-methods-electorama.com/2003-December/011519.html
> Markus Schulze
> ----
> Election-Methods mailing list - see http://electorama.com/em for list info

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