On Tue, Mar 06, 2007 at 07:57:51PM +0100, Markus Schulze wrote:
>The MinMax method satisfies this criterion.

Concrete counterexample:

3       B > C > D > A
4       D > B > C > A
6       D > A > B > C
8       C > D > A > B

Gives the following defeats:

11      C > D   -> greatest defeat of D
13      B > C   -> greatest defeat of C
14      A > B
15      C > A
18      D > B   -> greatest defeat of B
21      D > A   -> greatest defeat of A

Candidate D is elected by MinMax(wv).

If D is removed, then C is elected; but C pairwise defeats D

This violates my criterion.

-- 
Matthijs van Duin  --  May the Forth be with you!
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