On Mar 3, 2008, at 1:45 , <[EMAIL PROTECTED]> <[EMAIL PROTECTED]>
wrote:
[EMAIL PROTECTED]:
>>Can you also clarify a bit how step 3 is counted when some
candidate X is beaten by two other candidates (Y and Z).
>>I find the proposed method interesting since it seems to aim at
electing good winners (using a function minimizes the problems
caused to the voters, from one point of view).
I'd be happy to try. Do you have an example election for me to play
with? I'm assuming you mean where I said
3. If there is no Condorcet winner, find the shortest distance (sum
of individual ranges) necessary to produce a Condorcet winner.
Sorry for some delay in replying. Here's one quick example.
1: A=10 B=2 C=1 D=0
1: A=10 C=7 B=6 D=0
1: B=10 C=6 A=5 D=0
3: C=10 D=5 A=1 B=0
3: D=10 B=4 A=3 C=0
C is now beaten by both A and B, and C has to win them both in order
to become a Condorcet winner. What is the "shortest distance (sum of
individual ranges)" for C in this example and how do you count it?
Juho
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