[EM] Query for one and all

[Markus Schulze]

Dear John B. Hodges,

I wrote (2 Sep 2003):
> 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.

You wrote (3 Sep 2003):
> 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?

When e.g. the Green Party nominates 10 candidates then it
is a useful presumption that all voters prefer all these
candidates in an adjacent manner. But that doesn't mean
that the voters cannot differ these candidates.

******

"Reversal symmetry" says: When candidate X is the unique
winner then when the individual preferences of each voter
are inverted then candidate X must not be elected.

Example:

     19  A > C > B
     20  B > C > A
     1   C > A > B
     1   C > B > A
     1   B > A > C
     1   A > B > C

     Candidate C is the unique Bucklin winner. When all
     individual preferences are inverted then candidate C
     is still the unique Bucklin winner.

Markus Schulze