[EM] MMV and resolvability

[Michael Ossipoff]

 The only thing wrong with my ordered-procedure MMV definition was
that I left something out:

MMV (ordered-procedure definition):

A defeat contradicts a set of other defeats if it is in a cycle
consisting of only it and them.

In order of stronger first, consider each defeat as follows:

Keep it unless it contradicts a set of stonger defeats, or a set
consisting of defeats equal to it and defeats stronger than it.

[end of ordered-procedure MMV definition]

Of course an alternative wins if it doesn't have a kept defeat. If
there are more than one winner, and if it is desired to choose just
one alternative, then random-ballot should be used for choosing among
the winners.

By the way, my brief definition doesn't do wrong in Marcus' examples.
It merely doesn;t give any result. "No result" can be construed as a
tie.

Michael Ossipoff


On Sun, Dec 8, 2013 at 12:05 AM, Markus Schulze
<markus.schulze at alumni.tu-berlin.de> wrote:
> Hallo,
>
> here is another example to illustrate MMV's violation
> of monotonicity.
>
> Situation 1:
>
>   A>B, B>C, C>D, D>A, D>E, E>A each have the same
>   strength and are stronger than every other pairwise
>   defeat.
>
>   The other pairwise defeats are (sorted by their strength
>   in a decreasing order):
>
>   A>C
>
>   C>E
>
>   E>B
>
>   B>D
>
>   MMV skips A>B, B>C, C>D, D>A, D>E, and E>A, since they
>   form a directed cycle.
>
>   Then it locks A>C,C>E,E>B, and B>D, so that A is the
>   unique winner.
>
> Situation 2:
>
>   Some voters rank candidate A higher (relatively to
>   candidate D), so that the pairwise defeat D>A becomes
>   weaker.
>
>   A>B, B>C, C>D, D>E, E>A each have the same
>   strength and are stronger than every other pairwise
>   defeat.
>
>   The other pairwise defeats are (sorted by their strength
>   in a decreasing order):
>
>   A>C
>
>   C>E
>
>   E>B
>
>   D>A
>
>   B>D
>
>   MMV skips A>B, B>C, C>D, D>E, and E>A, since they
>   form a directed cycle.
>
>   Then it locks A>C,C>E,E>B, and D>A, so that D is the
>   unique winner.
>
> Thus, by ranking candidate A higher candidate A is changed
> from a winner to a loser.
>
>
> Markus Schulze
>
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