[EM] Hello again -- and a new method for you!

[Markus Schulze]

Hallo,

here is an example where the river method violates mono-add-top:

   3 ABDC
   4 ADBC
   6 BADC
   4 BCAD
   6 CABD
   2 CBAD
   5 DBCA
   6 DCAB

   A:B=19:17
   A:C=13:23
   A:D=25:11
   B:C=22:14
   B:D=21:15
   C:D=12:24

   The winner of the river method is candidate A.

   When I add 3 ABCD voters then the pairwise defeats are:

   A:B=22:17
   A:C=16:23
   A:D=28:11
   B:C=25:14
   B:D=24:15
   C:D=15:24

   Now, the winner of the river method is candidate B.

Here is an example where the river method violates mono-remove-bottom:

   10 ABDC
    4 ADBC
   12 ADCB
    6 BCAD
   20 BDCA
    2 CABD
    8 CADB
    6 CBAD

   A:B=36:32
   A:C=26:42
   A:D=48:20
   B:C=40:28
   B:D=44:24
   C:D=22:46

   The winner of the river method is candidate A.

   When I add 7 DCAB voters then the pairwise defeats are:

   A:B=43:32
   A:C=26:49
   A:D=48:27
   B:C=40:35
   B:D=44:31
   C:D=22:53

   Now, the winner of the river method is candidate B.

For the sake of completeness: Also Tideman's ranked pairs
method violates mono-add-top and mono-remove-bottom already
in the 4-candidate case.

Markus Schulze