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Re: Maybe Schulze is decisive.

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  • Markus Schulze
    Dear Norman, the aim of this e-mail is to prove, that Schulze wins cannot be cyclic. It is sufficient to prove that if Candidate A defeats candidate B via
    Message 1 of 2 , Aug 31, 1998
      Dear Norman,

      the aim of this e-mail is to prove, that Schulze wins
      cannot be cyclic. It is sufficient to prove that if
      Candidate A defeats candidate B via beat-paths and
      candidate B defeats candidate C via beat-paths, then
      candidate A defeats candidate C via beat-paths.

      **********

      First: Suppose, that candidate A defeats candidate B
      via beat-paths ab:ba. Suppose, that candidate B defeats
      candidate C via beat-paths bc:cb.

      ( 1) ab > ba
      ( 2) bc > cb

      Furthermore, I suppose, that the strength of the
      strongest beat-path from candidate A to candidate C
      is ac and that the strength of the strongest beat-path
      from candidate C to candidate A is ca.

      *****

      Second: I get the following equations:

      ( 6) min{ab;bc} <= ac
      ( 7) min{ac;cb} <= ab
      ( 8) min{ba;ac} <= bc
      ( 9) min{bc;ca} <= ba
      (10) min{ca;ab} <= cb
      (11) min{cb;ba} <= ca

      Example: If min{ab;bc} > ac, then I could find a
      stronger beat-path from candidate A to candidate C
      by going from candidate A to candidate B and then
      from candidate B to candidate C. This would be a
      contradiction to the assumption, that ac is already
      the strength of the strongest beat-path from
      candidate A to candidate C.

      *****

      Third: Case X: (12a) ab >= bc.

      With (12a) ab >= bc and (2) bc > cb I get:

      (13a) ab > cb.

      When I insert (13a) into (10), I get:

      (14a) ca <= cb.

      When I insert (12a) into (6), I get:

      (15a) bc <= ac.

      With (14a), (2), and (15a) I get:

      (16a) ca <= cb < bc <= ac.

      Thus: ca < ac. Candidate A defeats candidate C
      via beat-paths.

      *****

      Fourth: Case Y: (12b) bc > ab.

      With (12b) bc > ab and (1) ab > ba I get:

      (13b) bc > ba.

      When I insert (13b) into (9), I get:

      (14b) ca <= ba.

      When I insert (12b) into (6), I get:

      (15b) ab <= ac.

      With (14b), (1), and (15b) I get:

      (16b) ca <= ba < ab <= ac.

      Thus: ca < ac. Candidate A defeats candidate C
      via beat-paths.

      **********

      I have demonstrated, that if candidate A defeats candidate B
      via beat-paths and candidate B defeats candidate C via
      beat-paths, then candidate A defeats candidate C via
      beat-paths. Thus: Schulze wins cannot be cyclic.

      Markus Schulze
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