[EM] Query for one and all

[John B. Hodges]

>From: Markus Schulze <markus.schulze at alumni.tu-berlin.de>
>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?
-- 
----------------------------------
John B. Hodges, jbhodges@  @usit.net
Do Justice, Love Mercy, and Be Irreverent.