[EM] Enhanced DMC

[fsimmons at pcc.edu]

Chris,

It could happen that the lowest approval candidate X that covers all higher
approved candidates is covered by an even lower candidate Y that beats all
higher approved candidates but doesn't cover them all.

In that case X, even though X is the "Covering DMC" winner, some candidate with
less approval will be the Enhanced DMC winner.

I like Enhanced DMC better than Covering DMC because the latter is more
vulnerable to compromise: a weak favorite which is not covered by compromise can
block compromise's chances by merely getting higher approval.  In the former
version favorite has to definitively defeat compromise to ruin her chances.

----- Original Message -----
From: Chris Benham

> I  like this.  Regarding how approval is inferred, I'm also
> happy with Forest's idea of using Range
> (aka Score) type ballots (on which voters give their most
> preferred candidates the highest numerical
> scores) and interpreting any score above zero as approval and
> breaking approval ties as any score
> above 1 etc.  Or any other sort of  multi-slot ratings ballot
> where all except the bottom-most slot is
> interpreted as approval.
>  
> Another idea is to enter above-bottom equal-ranking between any
> 2 candidates in the pairwise matrix
> as a whole vote for both candidates, and then take each
> candidate X's highest single pairwise score
> as X's approval score.
>  
> Here are a couple of examples to demonstrate how this method
> varies from some other Condorcet
> methods.
>
> 48: A
> 01: A>D
> 24: B>D
> 27: C>B>D
>  
> D is the most approved candidate and in the Smith set, and so
> Smith//Approval elects D.
> Forest's "Enhanced DMC" or  "Covering DMC"  (and your suggested
> "SARR" implementation)
> elects B.
>
> B covers D and to me looks like a better winner. This method has
> a weaker truncation incentive
> than Smith//Approval.
>
> 25: A>B
> 27: B>C
> 26: C>A
> 22: C
>
> Approvals: C75,  B52, A51.    A>B 51-49,   B>C 52-48,   C>A 75-25
>  
> Plain DMC and using MinMax or one of the algorithms that is
> equivalent to it when there are three
> candidates (such as Schulze and Ranked Pairs and River) and
> weighing defeats either by Winning
> Votes or Margins all elect B.
>  
> If  5 of  the 22 C voters change to A those methods all elect C
> (a failure of  Woodall's "mono-sub-delete"
> criterion).
>  25: A>B
> 27: B>C
> 26: C>A
> 17: C
>
> 05: A  (was C)
>  
> Approvals: C70,  A56,  B52.    A>B 56-49,   B>C 52-48,   C>A 70-30.
>  
> In both cases our favoured method (like Smith//Approval) elects
> C, the  positionally dominant candidate. It
> seems those other methods are more vulnerable to Push-over strategy.
>  
> (To be fair, Woodall has demonstrated that no Condorcet method
> can meet mono-raise-delete.)
>  
>  
> Chris Benham
>   
>
>
> ________________________________
> From: Ted Stern
> To: election-methods at lists.electorama.com
> Cc: Forest Simmons ; Chris Benham
> Sent: Wednesday, 5 October 2011 8:35 AM
> Subject: Re: [EM] Enhanced DMC
>
> After some private email exchanges with Forest and Chris, I'm
> proposing a simple way of implementing Enhanced DMC, plus a new name,
> Strong Approval Round Robin Voting (SARR Voting).
>
> Ballot:
>
> Ranked Voting, all explicitly ranked candidates considered approved.
> Equal ranking allowed.  I'm basing this on recommendation from Chris
> Benham.  I'm open to alternatives, but it seems to be the
> easiest way
> to do it for now, and the most resistant to burying strategies.
>
> Tallying:
>
> Form the pairwise matrix, using the standard Condorcet
> procedure.  In
> the diagonal entries, save total Approval votes.
>
> For N candidates, the list of candidates in order from highest to
> lowest approval is
>
>   X_0, X_1, ..., X_k, X_{k+1}, ..., X_{N-1}
>
> Initialize the Strong set to the empty set
>
> Initialize the Weak set to the empty set.
>
> For k = 0 to N-1,
>
>   If X_k is already in the Weak set, continue iterating.  (X_k is
>   defeated by a higher approved candidate.  This is called being
>   "strongly defeated".)
>
>   If X_k loses to a member of the Weak set, continue iterating.  (X_k
>   may defeat all higher approved candidates, but is "weakly defeated"
>   by at least one of them.)
>  
>   If we're still here in the loop, X_k defeats all candidates in the
>   Strong Set and all candidates in the Weak set.  (X_k "covers" all
>   previously added members of the Strong set.)
>
>   Add X_k to the Strong set and add all of X_k's defeats to the Weak
>   set.
>
>   Set the provisional winner to X_k.
>
> The last provisional winner (the last candidate added to the Strong
> set) is the winner of the election.
>
> Note:
>
> The first member of the Strong Set will be X_0.
>
> It is easiest to do this by hand if you first permute the pairwise
> array so that it follows the same X_0, ..., X_{N-1} ordering.
>
> As an example election, consider the one on this page:
>
>   http://wiki.electorama.com/wiki/Marginal_Ranked_Approval_Voting
>
> Iterating through E, A, C, B, D, we find
>
>   E:  Strong and Weak Sets are empty, so E has no losses to either.
>
>       Strong set = {E};          Weak set = {C, D}
>  
>       Provisional winner set to E.
>
>   A:  A defeats Strong set {E} and Weak set {C, D}.
>
>     => Strong set = {E, A};      Weak set = {C, D}
>
>       Provisional winner set to A.
>
>   C:  in Weak set, not added to Strong set.
>
>   B:  Defeats A, but is defeated by D from Weak set (and is therefore
>       "weakly defeated" by A).
>
>   D:  in Weak set, not added to Strong set.
>
> A is the last candidate added to the Strong set, so A wins.
>
> Ted
> --
> araucaria dot araucana at gmail dot com
>
> On 26 Sep 2011 11:44:13 -0700, Chris Benham wrote:
> >
> > 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: "fsimmons at pcc.edu"
> > To: C.Benham
> > Cc: election-methods-electorama.com at 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.com at 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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