Re: [EM] Query for one and all

[John B. Hodges]

From: Markus Schulze <[EMAIL PROTECTED]>
Subject: Re: [EM] Query for one and all
Dear Kevin,

you wrote (2 Sep 2003):
 I think MCA meets Clone Independence and Participation,
 but I'd like to hear reasoning to the contrary.
Situation 1:

   2   A > B > C
   3   B > C > A
   4   C > A > B
   The winner is candidate C.

Situation 2:

   Replacing C by C1, C2, and C3 gives:

   2   A  > B  > C2 > C1 > C3
   3   B  > C3 > C2 > C1 > A
   4   C1 > C2 > C3 > A  > B
   The winner is candidate B.

Markus Schulze
(JBH) Good, simple demonstration. But, remember that in MCA ties are 
allowed; so, shouldn't the voters rank C1, C2, C3 equally?

        2  A > B > (C2, C1, C3)
        3  B > (C3, C2, C1) > A
        4  (C1, C2, C3) > A > B
The winner is a tie between C1, C2, C3; use tie-breaking method to pick one.
If the voters do not rank them equally when the option is allowed, 
doesn't that show they are not true clones?
--
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John B. Hodges, jbhodges@  @usit.net
Do Justice, Love Mercy, and Be Irreverent.
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