[EM] clone independent modification of Baldwin

[Ross Hyman]

Just realized silly mistake. S_A1+SA2+SA3 etc = S_A does not mean that some S_Aj have to be above S_A and some below S_A.  Does not effect clone independence.   

--- On Sun, 2/3/13, Ross Hyman <rahyman at sbcglobal.net> wrote:

> From: Ross Hyman <rahyman at sbcglobal.net>
> Subject: clone independent modification of Baldwin
> To: election-methods at electorama.com
> Date: Sunday, February 3, 2013, 8:42 PM
> Here is a clone independent
> modification of Baldwin.
> Has this been discussed before?  
> 
> V_A>B is the number of ballots that rank A above B.
> V_A is the number of ballots that rank A at the top.
> 
> S_A = sum_B (V_A>B - V_B>A)V_A V_B is the score for
> candidate A.  The V_AV_B factor makes it a modification
> of Baldwin.
> 
> Eliminate the candidate with lowest score.  Recalculate
> V_A's and S_A's.  Repeat until one candidate remains.
> 
> Like Baldwin, if there is a Condorcet winner it will have a
> positive score.  Also like Baldwin sum_A S_A =0 so that
> if there is a Condorcet winner it is guaranteed that there
> will be at least one other candidate with negative score so
> the Condorcet winner will not be eliminated.
> 
> It is clone independent because S_A does not change if one
> of the other candidates is cloned.  If A is cloned to
> A1,A2 etc. then S_A1+SA2+SA3 etc = S_A so some of the clones
> will have a higher score than the original A and some
> less.  This might mean that one of the clones of A
> would be eliminated before A would have been, but since
> other clones of A remain, and we are eliminating just one at
> a time, everything is ok.  
> 
> I do not think that the Nanson version of this would always
> be clone independent, but I haven't checked. I think that
> for Nanson it might be possible that S_A is negative so
> would be eliminated but when cloned, one of the clones could
> have positive score and remain after the elimination step
> and possibly win the election.  
> 
> 
>