[EM] ranked pair method that resolves beath path ties.

[Markus Schulze]

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