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Forest Simmons wrote (15 Aug 2011):<br>
<br>
<blockquote type="cite">
<pre>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.</pre>
</blockquote>
<br>
Forest,<br>
<br>
Is the "Approval Chain-Building" method the same as simply electing
the most approved uncovered candidate?<br>
<br>
I surmise that the set of candidates not pairwise beaten by a more
approved candidate (your set "P", what I've<br>
been referring to as the "Definite Majority set") and the Uncovered set
don't necessarily overlap.<br>
<br>
If forced to choose between electing from the Uncovered set and
electing from the "DM" set, I tend towards<br>
the latter.<br>
<br>
Since Smith//Approval always elects from the DM set, and your suggested
"enhanced DMC" (elect the most <br>
approved member of the DM set that isn't covered by another member)
doesn't necessarily elect from the Uncovered set; <br>
there doesn't seem to be any obvious philosophical case that enhanced
DMC is better than Smith//Approval. <br>
<br>
(Also I would say that an election where those two methods produce
different winners would be fantastically unlikely.)<br>
<br>
A lot of Condorcet methods are promoted as being able to give the
winner just from the information contained in the<br>
gross pairwise matrix. I think that the same is true of these methods
if we take a candidate X's highest gross pairwise<br>
score as X's approval score. Can you see any problem with that?<br>
<br>
<br>
Chris Benham<br>
<br>
<br>
<br>
<br>
----- Original Message -----<br>
From:<br>
Date: Friday, August 12, 2011 3:12 pm<br>
Subject: Enhanced DMC<br>
To: election-methods at lists.electorama.com,<br>
<br>
<blockquote type="cite">> From: "C.Benham" <br>
> To: election-methods-electorama.com at electorama.com<br>
> Subject: [EM] Enhanced DMC<br>
<br>
> Forest,<br>
> The "D" in DMC used to stand for *Definite*.<br>
<br>
Yeah, that's what we finally settled on.<br>
<br>
> <br>
> I like (and I think I'm happy to endorse) this Condorcet method <br>
> idea, and consider it to be clearly better than regular DMC<br>
> <br>
> Could this method give a different winner from the ("Approval <br>
> Chain Building" ?) method you mentioned in the "C//A" thread (on
11 <br>
> June 2011)?<br>
<br>
Yes, I'll give an example when I get more time. But for all practical
purposes they both pick the highest approval Smith candidate.<br>
</blockquote>
<br>
<br>
Here's a possible scenario:<br>
<br>
Suppose that approval order is alphabetical from most approval to least
A, B, C, D.<br>
<br>
Suppose further that pairwise defeats are as follows:<br>
<br>
C>A>D>B>A together with B>C>D .<br>
<br>
Then the set P = {A, B} is the set of candidates neither of which is
pairwise<br>
beaten by anybody with greater approval.<br>
<br>
Since the approval winner A is not covered by B, it is not covered by
any<br>
member of P, so the enhanced version of DMC elects A.<br>
<br>
But A is covered by C so it cannot be elected by any of the chain
building<br>
methods that elect only from the uncovered set.<br>
<br>
<br>
Forest Simmons wrote (12 June 2011):<br>
<br>
> I think the following complete description is simpler than
anything <br>
> possible for ranked pairs:<br>
><br>
> 1. Next to each candidate name are the bubbles (4) (2) (1). The <br>
> voter rates a candidate on a scale from<br>
> zero to seven by darkening the bubbles of the digits that add up
to <br>
> the desired rating.<br>
><br>
> 2. We say that candidate Y beats candidate Z pairwise iff Y is
rated <br>
> above Z on more ballots than not.<br>
><br>
> 3. We say that candidate Y covers candidate X iff Y pairwise
beats <br>
> every candidate that X pairwise<br>
> beats or ties.<br>
><br>
> [Note that this definition implies that if Y covers X, then Y
beats X <br>
> pairwise, since X ties X pairwise.]<br>
><br>
> Motivational comment: If a method winner X is covered, then the <br>
> supporters of the candidate Y that<br>
> covers X have a strong argument that Y should have won instead.<br>
><br>
> Now that we have the basic concepts that we need, and assuming
that <br>
> the ballots have been marked<br>
> and collected, here's the method of picking the winner:<br>
><br>
> 4. Initialize the variable X with (the name of) the candidate
that <br>
> has a positive rating on the greatest<br>
> number of ballots. Consider X to be the current champion.<br>
><br>
> 5. While X is covered, of all the candidates that cover X, choose
the <br>
> one that has the greatest number of<br>
> positive ratings to become the new champion X.<br>
><br>
> 6. Elect the final champion X.<br>
><br>
> 7. If in step 4 or 5 two candidates are tied for the number of <br>
> positive ratings, give preference (among the<br>
> tied) to the one that has the greatest number of ratings above
level <br>
> one. If still tied, give preference<br>
> (among the tied) to the one with the greatest number of ratings
above <br>
> the level two. Etc.<br>
><br>
> Can anybody do a simpler description of any other Clone
Independent <br>
> Condorcet method?<br>
<br>
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