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style="font-family: -moz-fixed; font-size: 13px;" lang="x-western">James
G-A (and interested others), <br>
<br>
As I mentioned in my last message (in the "Condorcet completed by IRV" thread),
in September this year I described and <br>
recommended Schwartz // SC-WMA. (The letters stand for "Symmetrically Completed-Weighted
Median Approval"). <br>
<br>
<blockquote type="cite">Voters rank the candidates, truncation ok. Non-last
equal preferences also ok (but a version that doesn't allow them is also
good <br>
and might be a more practical proposition). <br>
Eliminate the non-members of the Schwartz set (and henceforth continue
as though they had never stood). <br>
Symmetrically complete the ballots. <br>
Now apply the "Weighted Median Approval" method to pick the winner, thus:
<br>
Each (remaining) candidate is assigned a "weight" which is equal to the
number of first-preferences they get. The sum of the "weights" <br>
is equal to the total number of non-empty ballots. <br>
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. <br>
And so on until each ballot has approved at least half the candidates "by
weight". <br>
The candidate with the highest total (thus derived) approval score wins.
<br>
</blockquote>
<br>
<br>
If there are three candidates, all in the Schwartz set, then each ballot
approves the two highest-ranked candidates and those <br>
ballots that only rank one candidate approve that candidate and half-approve
the other two. <br>
<br>
At the time I wrongly believed that this method is Clone Independent, but
since then Douglas Woodall has showed that it fails <br>
Clone-Winner. This can only arise when the Schwartz set contains more than
three members (maybe at least two more). <br>
Here is his demonstration. <br>
<br>
<blockquote type="cite">Consider this example: <br>
<br>
42: A B|C <br>
28: B C|A <br>
30: C A|B <br>
<br>
Here the weights are A 42, B 28, C 30, total 100. The sum of any two <br>
candidates' weights is greater than 50, and so the cutoffs are as shown, <br>
giving A 72, B 70, C 58, and A wins. Now suppose we clone A: <br>
<br>
14: A1 A2 A3 B|C <br>
14: A2 A3 A1 B|C <br>
14: A3 A1 A2 B|C <br>
28: B C|A1 A2 A3 <br>
10: C A1 A2|A3 B <br>
10: C A2 A3|A1 B <br>
10: C A3 A1|A2 B <br>
<br>
Here the weights are A1 14, A2 14, A3 14, B 28, C 30, total 100. The <br>
cutoffs are in the analogous positions to those in the previous election,
<br>
except that since 30+14+14 = 58 > 50, the last 30 voters approve only <br>
3 candidates each instead of the 4 that one might expect. Now the approval
<br>
votes are A1 62, A2 62, A3 62, B 70, C 58, and B wins. <br>
<br>
So WMA fails clone-winner. <br>
</blockquote>
<br>
<br>
CB: Clone Independence has been one of my "essential" criterion compliances,
but in general, as a practical matter, <br>
I don't consider problems that can only arise when there are more than three
(or four?) candidates in the Schwartz set to be <br>
very serious. <br>
<br>
But I do have an idea for a bizarre "patch" to fix the clone problem! <br>
<br>
It seems to me that lots of methods have clone (and/or sometimes other) problems
that only <br>
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. <br>
<br>
Use some rule to determine the loser in each triowise comparison, and a way
of scoring each <br>
loser (so that it can be compared with other losers). Repeatedly eliminate
the worst loser <br>
in any of the triowise comparisons among remaining candidates, until three
(or two) candidates <br>
remain. <br>
Three possible simple rules spring to mind. Based on the symmetrically completed
ballots <br>
(I prefer, but alternatively not): (1) the number of top preferences <br>
(2) the number of top plus middle preferences (quasi-Bucklin) <br>
(3) the number of top minus bottom preferences (Borda) <br>
<br>
A couple of pure and simple examples of complete methods that employ one
of these "triowise trimmers". <br>
(1) "SC-TT(tp)//IRV" Based on the symmetrically completed ballots, repeatedly
eliminate the candidate with the fewest <br>
top preferences in any of the triowise comparisons among the remaining candidates
until two <br>
remain. Elect the winner of the pairwise comparison between these two. <br>
<br>
This reduces to IRV when there are three candidates, but it fixes (at least
the worst of) the <br>
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 <br>
comparison, and so able to win. <br>
<br>
(2) "SC-TT(Borda)//Borda-Elimination" <br>
Based on the symmetrically completed ballots,repeatedly eliminate the candidate
with the lowest <br>
top-minus-bottom preferences score in any of the triowise comparisons among
remaining candidates <br>
until two remain. Elect the winner of the pairwise comparison between these
two. <br>
<br>
This fixes the clone problem with Borda-Elimination. <br>
<br>
My patch for Schwartz//SC-WMA ? Eliminate non-members of the Schwartz set.
Symetrically complete the ballots, and then repeatedly <br>
eliminate the candidate with the lowest top-plus-middle preferences score
in any of the triowise <br>
comparisons among remaining candidates, until three remain. Ignoring eliminated
candidates, elect <br>
the WMA/Bucklin winner among these three. <br>
<br>
(WMA and Bucklin are equivalent when there are three candidates). This does
look a bit makeshift <br>
and inelegant. The method meets (at least 3-candidate) Minimal Defense. If
the field is trimmed <br>
to two candidates like in the other two methods, then it doesn't. <br>
<br>
<br>
Chris Benham <br>
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