[EM] Enhanced DMC (correction)

Thu Oct 6 16:26:50 PDT 2011

I got the Woodall monotonicity criterion garbled.  Instead of  "mono-sub-delete" and then
"mono-raise-delete", in both instances I meant  *mono-sub-plump*.  (I've made the corrections
in the text below.)

It means that candidate x shouldn't be harmed (i.e. have his probability of election reduced)
by swapping some ballots that don't have x top-ranked for some that rank x alone in top place
and ignore (or vote equal-bottom) all the other candidates.

In the context of a universe of  ballots that allow only strict ranking from the top, with truncation
(but no other equal-ranking) allowed, Woodall defined it:

<snip>
A candidate x should not be harmed if:
(
are replaced by ballots that have 
choice;

<snip>

http://f1.grp.yahoofs.com/v1/4CSOTg18Jla8JeZD7lGf-L15LhewtDexi1BvIN9JQ79d6fDKQfZlI5ygNNqdMM_8b3XPbatc01XxYUjo1LgxMq9WirJdubA/wood1996.pdf

mono-sub-plump) some ballots that do not have x topx top with no second
Also I referred to "Push-over" strategy.  That refers to the strategy of raising some weak candidate to enable 
some other candidate to win (or to have a better chance of winning).

It was coined with methods like Top-Two Runoff and IRV in mind. Perhaps "Push-over  *like* strategy" was 
more apt for the methods I referred to.

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-plump"
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-sub-plump)

Chris Benham

________________________________
From: Ted Stern <araucaria.araucana at gmail.com>
To: election-methods at lists.electorama.com
Cc: Forest Simmons <fsimmons at pcc.edu>; Chris Benham <cbenhamau at yahoo.com.au>
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" <fsimmons at pcc.edu>
> To: C.Benham <cbenhamau at yahoo.com.au>
> 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?
>> 
>> 
>> 
>> 
> -------------- next part --------------
> An HTML attachment was scrubbed...
> URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20110926/a10be831/attachment.htm>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20111006/5b340707/attachment-0004.htm>