Forest, "I think in general that if the approval scores are at all valid I would go for the enhanced DMC winner over any of the chain building methods we have considered. I think other considerations over-ride the importance of being uncovered." I agree. I think the chain building method in comparison seems a bit arbitrary and less philosophically justified. Also the method has a fairly straight-forward description that doesn't need to mention "Smith set" or "the Condorcet winner". So of these similar methods (that include Smith//Approval and all elect the same winner if the Smith set contains 3 members or 1 member), I think this is my favourite. Maybe it could use a new name? :)
Chris From: "fsimm...@pcc.edu" <fsimm...@pcc.edu> To: C.Benham <cbenha...@yahoo.com.au> Cc: election-methods-electorama....@electorama.com Sent: Monday, 12 September 2011 8:50 AM Subject: Re: Enhanced DMC Very good Chris. I tried to build a believable profile of ballots that would yield the approval order and defeats of this example without success, but I am sure that it is not impossible. I think in general that if the approval scores are at all valid I would go for the enhanced DMC winner over any of the chain building methods we have considered. I think other considerations over-ride the importance of being uncovered. ----- Original Message ----- From: "C.Benham" Date: Sunday, September 11, 2011 10:08 am Subject: Enhanced DMC To: election-methods-electorama....@electorama.com Cc: Forest W Simmons > Forest Simmons wrote (15 Aug 2011): > > >Here's a possible scenario: > > > >Suppose that approval order is alphabetical from most approval > to least A, B, C, D. > > > >Suppose further that pairwise defeats are as follows: > > > >C>A>D>B>A together with B>C>D . > > > >Then the set P = {A, B} is the set of candidates neither of > which is pairwise > >beaten by anybody with greater approval. > > > >Since the approval winner A is not covered by B, it is not > covered by any > >member of P, so the enhanced version of DMC elects A. > > > >But A is covered by C so it cannot be elected by any of the > chain building > >methods that elect only from the uncovered set. > > > > Forest, > > Is the "Approval Chain-Building" method the same as simply > electing the > most approved uncovered candidate? > > I surmise that the set of candidates not pairwise beaten by a > more > approved candidate (your set "P", what I've > been referring to as the "Definite Majority set") and the > Uncovered set > don't necessarily overlap. > > If forced to choose between electing from the Uncovered set and > electing > from the "DM" set, I tend towards > the latter. > > Since Smith//Approval always elects from the DM set, and your > suggested > "enhanced DMC" (elect the most > approved member of the DM set that isn't covered by another > member) > doesn't necessarily elect from the Uncovered set; > there doesn't seem to be any obvious philosophical case that > enhanced > DMC is better than Smith//Approval. > > (Also I would say that an election where those two methods > produce > different winners would be fantastically unlikely.) > > A lot of Condorcet methods are promoted as being able to give > the > winner just from the information contained in the > gross pairwise matrix. I think that the same is true of these > methods > if we take a candidate X's highest gross pairwise > score as X's approval score. Can you see any problem with that? > > > Chris Benham > > > > > ----- Original Message ----- > From: > Date: Friday, August 12, 2011 3:12 pm > Subject: Enhanced DMC > To: election-methods at lists.electorama.com, > > > > 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. > > > > Here's a possible scenario: > > Suppose that approval order is alphabetical from most approval > to least > A, B, C, D. > > Suppose further that pairwise defeats are as follows: > > C>A>D>B>A together with B>C>D . > > Then the set P = {A, B} is the set of candidates neither of > which is > pairwise > beaten by anybody with greater approval. > > Since the approval winner A is not covered by B, it is not > covered by any > member of P, so the enhanced version of DMC elects A. > > But A is covered by C so it cannot be elected by any of the > chain building > methods that elect only from the uncovered set. > > > Forest Simmons wrote (12 June 2011): > > > I think the following complete description is simpler than anything > > possible for ranked pairs: > > > > 1. Next to each candidate name are the bubbles (4) (2) (1). The > > voter rates a candidate on a scale from > > zero to seven by darkening the bubbles of the digits that add > up to > > the desired rating. > > > > 2. We say that candidate Y beats candidate Z pairwise iff Y > is rated > > above Z on more ballots than not. > > > > 3. We say that candidate Y covers candidate X iff Y pairwise beats > > every candidate that X pairwise > > beats or ties. > > > > [Note that this definition implies that if Y covers X, then Y > beats X > > pairwise, since X ties X pairwise.] > > > > Motivational comment: If a method winner X is covered, then the > > supporters of the candidate Y that > > covers X have a strong argument that Y should have won instead. > > > > Now that we have the basic concepts that we need, and > assuming that > > the ballots have been marked > > and collected, here's the method of picking the winner: > > > > 4. Initialize the variable X with (the name of) the > candidate that > > has a positive rating on the greatest > > number of ballots. Consider X to be the current champion. > > > > 5. While X is covered, of all the candidates that cover X, > choose the > > one that has the greatest number of > > positive ratings to become the new champion X. > > > > 6. Elect the final champion X. > > > > 7. If in step 4 or 5 two candidates are tied for the number of > > positive ratings, give preference (among the > > tied) to the one that has the greatest number of ratings > above level > > one. If still tied, give preference > > (among the tied) to the one with the greatest number of > ratings above > > the level two. Etc. > > > > Can anybody do a simpler description of any other Clone Independent > > Condorcet method? > > > >
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