[EM] clone independent modification of Baldwin

[Ross Hyman]

Oops.  Need to show that if S_A is negative, one of the S_Aj will be more negative.  Haven't done so.  Without that, clone independence isn't proved.  Back to the drawing board.   

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

> From: Ross Hyman <rahyman at sbcglobal.net>
> Subject: Re: clone independent modification of Baldwin
> To: election-methods at electorama.com
> Date: Sunday, February 3, 2013, 9:22 PM
> 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.  
> > 
> > 
> >
>