[EM] Schwartz // SC-WMA (from "reccomendations" thread)

[Chris Benham]

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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