"If QLTD isn't cloneproof (and it isn't), then the result won't be
either, hence we could just as well go with first preference Copeland
(unless that has a flaw I'm not seeing)."
What is supposed to be the attraction of "first preference Copeland"?
And how do you define it exactly?
The attraction of FPC (which is what I call Simmons' supposed Cloneproof
extension of Copeland, but that wasn't cloneproof after all) is that
it's extremely hard to do burial with it.
The definition of first preference Copeland is:
The candidate for which those who beat him pairwise gather fewest
first-place votes, in sum, is the winner.
Simmons invented the method, I just use that name instead of "Simmons
cloneproof method", as it isn't cloneproof. It's an extension of
Copeland since the only information it takes from the pairwise matrix is
binary; in this case, whether some candidate Y beats X, and in
Copeland's case whether some candidate Y is beaten by X; thus "first
preference Copeland".
I could also just call it "Simmons", I suppose, since the other Simmons
methods I know of have defined names of their own and so wouldn't be
confused with it.
First preference Copeland would be vulnerable to the situation where
multiple candidates have equal first place rival scores. One way to
solve this would be to use Schwartz, or Schwartz//. Another would be to
use a positional system that counts second, third, etc, place votes also
but only very weakly, like Nauru-Borda (or something going 1/10^p, p =
0..n); yet another would be to use Bucklin (if there are any ties, count
first and second place votes of rivals, etc), and even another would be
to have an approval cutoff and use Approval instead of first preferences
(unless everybody bullet votes).
I haven't tested the positional or Bucklin variants here so I don't know
if those solutions would be any good. I'm not sure if it's possible to
make a situation where two candidates are in the Schwartz set yet none
of their rivals rank first or the rivals' ranked-first sum is equal for
all. Perhaps that's possible if you make "dummy candidates" that
collectively hog all the first place votes, but who each are ranked
below the various other candidates enough times that they don't beat any
of them? Something like
Q1 > A > B > C > Q2 > Q3
Q2 > B > C > A > Q3 > Q1
Q3 > B > A > C > Q1 > Q2
..
In the general case, such scaffolding will work (block winners) with
"Schwartz,". It won't work with Schwartz// unless you can somehow get
all the Qs inside the Schwartz set yet still have them "cover" each
candidate, first-preference wise, equally. But, (reading the "cloneproof
Copeland" thread even as I'm writing this,) Schwartz//FPC would not be
summable.
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