When beat path produces a tie, this method can produce a single winner unless
the tie is genuine. It is the same method I presented earlier except for the
addition of the Removing step, which resolves the ties.
Candidates are classed in two categories: Winners and
Losers. Initially, all can
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)
>
>
>
>
Markus is right.
One way of retaining monotonicity, I think, is to replace the Sets with objects
that record the number of times that a A has beaten B.
Then for the pair ordering
A>C, B>C
C>D
D>A, D>B, A>B
Affirming A>C and B> C
A(W):A(W)
B(W):B(W)
C(L):A(W),B(W),C(L)
D(W):D(W)
Affirming C>D
Dear Ross Hyman,
you wrote (28 Nov 2011):
> One way of retaining monotonicity, I think, is to replace
> the Sets with objects that record the number of times that
> a A has beaten B.
I guess that this tie-breaking strategy will violate
independence of clones.
Markus Schulze
Election-Meth
How so?
--
Dear Ross Hyman,
you wrote (28 Nov 2011):
> One way of retaining monotonicity, I think, is to replace
> the Sets with objects that record the number of times that
> a A has beaten B.
I guess that this tie-breaking strategy will violate
independence of clones.
Markus Schulze
Hallo,
in section 5 stage 3 of my paper, I explain how
Tideman's ranked pairs method can be used (without
having to sacrifice monotonicity, independence of
clones, reversal symmetry, or any other important
criterion) to resolve situations where the Schulze
winner is not unique:
http://m-schulze.
But is that the only monotonic clone independent method? The method I describe
elects D instead of A in accordance with D>A. But I don't see why it would
violate clone independence.
Consider the matrix in which the elements are the number of times the column
candidate has defeated the row c
Dear Ross Hyman,
you wrote (28 Nov 2011):
> One way of retaining monotonicity, I think, is to replace
> the Sets with objects that record the number of times that
> a A has beaten B.
Suppose A-C-B is the strongest path from candidate A to
candidate B. Suppose B-A is the strongest path from cand
