[EM] Enhanced DMC

[C.Benham]

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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