[EM] ranked pair method that resolves beath path ties.
Markus Schulze
markus.schulze at alumni.tu-berlin.de
Sun Nov 27 23:01:58 PST 2011
Dear Ross Hyman,
you wrote (27 Nov 2011):
> A and B are both winners. A is in B's set and B is in A's set.
> So A is deleted from B's set and B is deleted from A's set.
>
> A(W): A(W), C(L), D(L)
>
> B(W): B(W), C(L), D(L)
>
> C(L): A(W), B(W), C(L), D(L)
>
> D(L): A(W), B(W), C(L), D(L)
>
>
>
> affirm A > B
>
> A(W):A(W), C(L), D(L)
>
> B(L): A(W), B(L), C(L), D(L)
>
> C(L): A(W), B(L), C(L), D(L)
>
> D(L): A(W), B(L), C(L), D(L)
>
> B was reclassified as a Loser since A(W) is in its set.
When I understand your proposal correctly, then you are
basically saying that, when contradicting beatpaths have the
same strength, then they are cancelling each other out and
the next strongest beatpath decides.
I believe that your proposal can lead to a violation of
monotonicity. Let's say that there is one beatpath from
candidate X to candidate Y of strength z and two beatpaths
from candidate Y to candidate X of strength z. Then these
beatpaths cancel each other out. However, if one of the two
beatpaths from candidate Y to candidate X is weakened, then this
beatpath decides that candidate Y is ranked ahead of candidate X
in the collective ranking. (This is problematic especially when
the weakened beatpath was the direct comparison Y vs. X.)
By the way: In my paper, I also recommend that the ranked
pairs method should be used to resolve situations where the
Schulze winner is not unique. However, the precise formulation
is important. See section 5 stage 3 of my paper:
http://m-schulze.webhop.net/schulze1.pdf
Markus Schulze
More information about the Election-Methods
mailing list