Hallo, Situation 1:
04 ABCDEF 02 ABFDEC 04 AEBFCD 02 AEFBCD 02 BFACDE 02 CDBEFA 04 CDBFEA 12 DECABF 08 ECDBFA 10 FABCDE 06 FABDEC 04 FEDBCA A:B=40:20 A:C=30:30 A:D=30:30 A:E=30:30 A:F=24:36 B:C=34:26 B:D=30:30 B:E=30:30 B:F=38:22 C:D=36:24 C:E=22:38 C:F=30:30 D:E=42:18 D:F=30:30 E:F=32:28 The winner is candidate A. Situation 2: 3 AEFCBD voters are added. A:B=43:20 A:C=33:30 A:D=33:30 A:E=33:30 A:F=27:36 B:C=34:29 B:D=33:30 B:E=30:33 B:F=38:25 C:D=39:24 C:E=22:41 C:F=30:33 D:E=42:21 D:F=30:33 E:F=35:28 Now, the winner is candidate D. Thus the 3 AEFCBD voters change the winner from candidate A to candidate D. ****** John wrote (23 May 2003): > instead, the per-option quorum will throw out the IDW in > favour of a less-favoured option due to quorum requirements. > > R=15 > 10 ABD > 5 BDA I suggest that one should at first calculate the ranking of the candidates according to the beat path method and then, of those candidates whose beat path to the default option meets the quorum, that candidate should be elected who is ranked highest in the ranking of the beat path method. That's the maximum that you can get without undermining the intention of super-majority requirements. In Situation 1, for example, the beat paths have the following strengths: A:B=40:36 A:C=34:32 A:D=34:32 A:E=34:32 A:F=38:36 B:C=34:32 B:D=34:32 B:E=34:32 B:F=38:36 C:D=36:38 C:E=36:38 C:F=32:34 D:E=42:36 D:F=32:34 E:F=32:34 Therefore, the ranking according to the beat path method is ABFDEC. Suppose that, for example, the default option is C and the quorum is 38. Then the winner is candidate D. Markus Schulze (not Martin Schulze)
