[EM] Enhanced DMC

[fsimmons at pcc.edu]

> From: "C.Benham" 
> To: election-methods-electorama.com at electorama.com
> Subject: [EM] Enhanced DMC

> Forest,
> The "D" in DMC used to stand for *Definite*.

Yeah, that's what we finally settled on.

> 
> I like (and I think I'm happy to endorse) this Condorcet method 
> idea, 
> and consider it to be clearly better than regular DMC
> 
> Could this method give a different winner from the ("Approval 
> Chain 
> Building" ?) method you mentioned in the "C//A" thread (on 11 
> June 2011)?

Yes, I'll give an example when I get more time.  But for all practical purposes they both pick the highest approval Smith candidate.

> 
> >Initialize a variable X to be the candidate with the most approval.
> >
> >While X is covered, let the new value of X be the highest 
> approval candidate that covers the old X.
> >
> >Elect the final value of X.
> >
> >For all practical purposes this is just a seamless version of 
> C//A, i.e. it avoids the apparent 
> >abandonment of Condorcet in favor of Approval after testing for 
> a CW.
> >
> >
> >Assuming cardinal ballots, candidate A covers candidate B, iff 
> whenever B is rated above C on more 
> >ballots than not, the same is true for A, and (additionally) A 
> beats (in this same pairwise sense) some 
> >candidate that B does not.
> > 
> >
> 
> Your newer suggestion ("enhanced DMC") seems to have an 
> easier-to-explain and justify motivation.
> 
> Chris Benham
> 
> 
> Forest Simmons wrote (12 July 2011):
> 
> >One of the main approaches to Democratic Majority Choice was 
> through the idea that if X beats Y and 
> >also has greater approval than Y, then Y should not win.
> >
> >If we disqualify all that are beaten pairwise by someone with 
> greater approval, then the remaining set P 
> >is totally ordered by approval in one direction, and by 
> pairwise defeats in the other direction. DMC 
> >solves this quandry by giving pairwise defeat precedence over 
> approval score; the member of P that 
> >beats all of the others pairwise is the DMC winner. 
> >
> >The trouble with this solution is that the DMC winner is always 
> the member off P with the least approval 
> >score. Is there some reasonable way of choosing from P that 
> could potentially elect any of its members?
> >
> >My idea is based on the following observation: 
> >
> >There is always at least one member of P, namely the DMC 
> winner, i.e. the lowest approval member of 
> >P, that is not covered by any member of P.
> >
> >So why not elect the highest approval member of P that is not 
> covered by any member of P?
> >
> >By this rule, if the approval winner is uncovered it will win. 
> If there are five members of P and the upper 
> >two are covered by members of the lower three, but the third 
> one is covered only by candidates outside 
> >of P (if any), then this middle member of P is elected.
> >
> >What if this middle member X is covered by some candidate Y 
> outside of P? How would X respond to 
> >the complaint of Y, when Y says, "I beat you pairwise, as well 
> as everybody that you beat pairwise, so 
> >how come you win instead of me?"
> >
> >Candidate X can answer, "That's all well and good, but I had 
> greater approval than you, and one of my 
> >buddies Z from P beat you both pairwise and in approval. If Z 
> beat me in approval, then I beat Z pairwise, 
> >and somebody in P covers Z. If you were elected, both Z and 
> the member of P that covers Z would have 
> >a much greater case against you than you have against me."
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