[EM] Re: Schwartz//SC-WMA

[Forest Simmons]

It's interesting to me that before I opened my email today I was already 
thinking about proposing Banks//WMA.

Now that Chris has brought up this clone problem, maybe I'll hold off.

Some thoughts on other possible ways of de-cloning WMA.

1. Proceed as in my de-cloning of Copeland.  (Use an "irreducible beat 
clone set hierarchy structure" to modify the weights used in WMA.)

2. Does the problem persist in the version of WMA that uses fractional 
approval at the level where half of the weight is used up?

3. Break down and use Cardinal Ratings style ballots along with weighted 
mean instead of weighted median.

With regard to (1):  A four candidate election always gives rise to what I 
recently described under the moniker of "a proper beat clone set."

If the election has a CW then the other three candidates form a proper 
beat clone set, because they are all beaten by evey (i.e. the only) 
candidate outside the set.

A similar analysis takes care of the case of a Condorcet Loser.

If there is neither a Condorcet Loser nor a CW (and assuming no pairwise 
ties) then there will be two candidates C1 and C2 such that both of these 
beat a third candidate B, and both are beaten by the other candidate A. 
[To see this, try putting arrows on the edges of a tetrahedron in such a 
way that no vertex has an in-degree or out-degree of three. All such 
arrangements of arrows are isomorphic.]

The set {C1,C2} is a proper beat clone set, and this should be taken into 
account when assigning the weights for WMA, or better yet, just eliminate 
the member C that is beaten by the other C, i.e. keep the one that is 
in the Banks set.

[De-cloned Copeland would give a tie to A, B, and the remaining C.]

To summarize in slightly different terms, we can say that a four candidate 
set can never have more than three uncovered candidates.

So "top four" is just as easy to handle as "top three" and might have even 
less incentive for "burying" than "top three."  [If I am not mistaken, 
the main advantage of "top three" over "top two" is that it has less 
incentive for burying.]

I wonder how common it is that a Ranked Pairs winner (for example) will 
not be the winner in all of its four candidate supersets.  If this is 
extremely uncommon, then giving the win to the candidate that wins in all 
of its four candidate supersets (if there is such a candidate), else 
random ballot among the Banks set candidates, would be a nice method.

I'm not sure where these ideas will lead; I haven't sorted them all out, 
but here's some more related ideas.

The idea of WMA is to automate near optimal approval strategy based on a 
set of ordinal ballots in order to compensate for Approval's Achilles' 
heel (bad information can lead one to make unfortunate approval cutoffs).

>From what I understand, the only positive probability winners under 
perfect information with sophisticated player strategies would be the 
members of the Banks set.  So keeping in mind the purpose of WMA, why not 
restrict to the Banks set in the first place.

Here's a five candidate example that is pleasant to contemplate:

Start with the transitive closure of A beats B beats C beats D, before 
adding in E such that E beats A and C but is beaten by B and D.

There are no proper beat clone sets in this example, so there is a beat 
path from every candidate to every other candidate.  But C and D are both 
covered by B, so they cannot be sophisticated strategy winners, and hence 
should be eliminated.

WMA could then be applied to the cycle A beats B beats C beats A, to give 
near optimal approval cutoffs, at least much better than the voters could 
determine from typical quality horse race information.

And remember, if the voters have a strong approval or disapproval feeling 
that they want to have respected by WMA, they can always use the equal 
ranking options at the top and bottom of their ordinal ballots.

[To respect this feature, we modify Chris' rule of completely throwing out 
the unworthy candidates; we'll keep them on the ballots that rank them at 
the top, by ranking them equal to the top worthy candidate.  Also 
symmetric completion might has to be modified to completely respect these 
strong preferences; yes, use the weights from the symmetric completion, 
but apply those weights in the ballots in their form immediately before 
the symmetric completion.]

The other line of thought that brought me to this topic was this:  random 
ballot to choose among Banks set members might be a reasonable improvement 
on plain old random ballot.  The weights in plain WMA are proportional to 
the probabilities of winning under random ballot.  So why not make the 
weights proportional to winning under the improved random ballot method, 
i.e. under random ballot Banks?

I believe that random ballot Banks is clone free.  If so, it would make 
sense for the fractional weight version of Banks//WMA to be clone free as 
well.

I hope that these musings stimulate some more thought along the lines that 
Chris and Jobst have been advocating.

Forest

Chris Benham wrote ...

> From: Chris Benham <chrisbenham at bigpond.com>
>
> James  G-A  (and interested others),
>
> As I mentioned in my last message  (in the "Condorcet completed by IRV"
> thread), in September this year I described and
> recommended  Schwartz // SC-WMA. (The letters stand for "Symmetrically
> Completed-Weighted Median Approval").
>
>> Voters rank the candidates, truncation ok. Non-last equal preferences
>> also ok (but a version that doesn't allow them is also good
>> and might be a more practical proposition).
>> Eliminate the non-members of  the Schwartz set (and henceforth
>> continue as though they had never stood).
>> Symmetrically complete the ballots.
>> Now apply the "Weighted Median Approval" method to pick the winner, thus:
>> Each (remaining) candidate is assigned a  "weight" which is equal to
>> the number of  first-preferences they get. The sum of the "weights"
>> is equal to the total number of  non-empty ballots.
>> Each ballot approves the candidate they rank in first-place. If  the
>> weight of candidates so far approved by a ballot sums to less than
>> half the total  weight of all the candidates, then that ballot also
>> approves the candidate they rank second.
>> And so on until each ballot has approved at least half the candidates
>>  "by weight".
>> The candidate with the highest total  (thus derived)  approval score
>> wins.
>
>
>
> If  there are three candidates, all in the Schwartz set, then each
> ballot approves the two highest-ranked candidates  and those
> ballots that only rank one candidate approve that candidate and
> half-approve the other two.
>
> At the time I wrongly believed that this method is Clone Independent,
> but since then Douglas Woodall has showed that it fails
> Clone-Winner. This can only arise when the Schwartz set contains more
> than three members (maybe at least two more).
> Here is his demonstration.
>
>> Consider this example:
>>
>> 42: A B|C
>> 28: B C|A
>> 30: C A|B
>>
>> Here the weights are A 42, B 28, C 30, total 100.  The sum of any two
>> candidates' weights is greater than 50, and so the cutoffs are as shown,
>> giving A 72, B 70, C 58, and A wins.  Now suppose we clone A:
>>
>> 14: A1 A2 A3 B|C
>> 14: A2 A3 A1 B|C
>> 14: A3 A1 A2 B|C
>> 28: B C|A1 A2 A3
>> 10: C A1 A2|A3 B
>> 10: C A2 A3|A1 B
>> 10: C A3 A1|A2 B
>>
>> Here the weights are A1 14, A2 14, A3 14, B 28, C 30, total 100.  The
>> cutoffs are in the analogous positions to those in the previous election,
>> except that since 30+14+14 = 58 > 50, the last 30 voters approve only
>> 3 candidates each instead of the 4 that one might expect.  Now the
>> approval
>> votes are A1 62, A2 62, A3 62, B 70, C 58, and B wins.
>>
>> So WMA fails clone-winner.
>
>
>
> CB: Clone Independence has been one of my "essential" criterion
> compliances, but in general, as a practical matter,
> I don't consider problems that can only arise when there are more than
> three (or four?) candidates in the Schwartz set to be
> very serious.
>
> But I do have an idea for a bizarre "patch" to fix the clone problem!
>
> It seems to me that lots of methods have clone (and/or sometimes other)
> problems that only
> arise when there are many or more than three candidates. It occurred to
> me that, as an interesting alternative to pairwise comparisons, triowise
> comparisons could be used as a clone-proof way to "trim" the field to
> three (or two) candidates.
>
> Use some rule to determine the loser in each triowise comparison, and a
> way of scoring each
> loser (so that it can be compared with other losers). Repeatedly
> eliminate the worst loser
> in any of the triowise comparisons among remaining candidates, until
> three (or two) candidates
> remain.
> Three possible simple rules spring to mind. Based on the symmetrically
> completed ballots
> (I prefer, but alternatively not): (1) the number of top preferences
> (2) the number of top plus middle preferences (quasi-Bucklin)
> (3) the number of top minus bottom preferences (Borda)
>
> A couple of pure and simple examples of complete methods that employ one
> of these "triowise trimmers".
> (1) "SC-TT(tp)//IRV" Based on the symmetrically completed ballots,
> repeatedly eliminate the candidate with the fewest
> top preferences in any of the triowise comparisons among the remaining
> candidates until two
> remain. Elect the winner of the pairwise comparison between these two.
>
> This reduces to IRV when there are three candidates, but it fixes (at
> least the worst of) the
> horror IRV examples with many candidates (like Adam Tarr's "Lucky Right"
> example). A candidate could have no first preferences at all, and yet
> not be last in any triowise
> comparison, and so able to win.
>
> (2) "SC-TT(Borda)//Borda-Elimination"
> Based on the symmetrically completed ballots,repeatedly eliminate the
> candidate with the lowest
> top-minus-bottom preferences score in any of the triowise comparisons
> among remaining candidates
> until two remain. Elect the winner of the pairwise comparison between
> these two.
>
> This fixes the clone problem with Borda-Elimination.
>
> My patch for Schwartz//SC-WMA ? Eliminate non-members of the Schwartz
> set. Symetrically complete the ballots, and then repeatedly
> eliminate the candidate with the lowest top-plus-middle preferences
> score in any of the triowise
> comparisons among remaining candidates, until three remain. Ignoring
> eliminated candidates, elect
> the WMA/Bucklin winner among these three.
>
> (WMA and Bucklin are equivalent when there are three candidates). This
> does look a bit makeshift
> and inelegant. The method meets (at least 3-candidate) Minimal Defense.
> If the field is trimmed
> to two candidates like in the other two methods, then it doesn't.
>
>
> Chris Benham
>
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