Jameson, I'm surprised that you consider a Condorcet method to be too extremist
or apt to suffer center
squeeze.
Think Yee diagrams; all Condorcet methods yield identical diagrams, while
center squeeze shows up
clearly in methods that allow it.
Of course if we have a multiwinner method, we
On 7/13/2011 11:14 AM, fsimm...@pcc.edu wrote:
Jameson, I'm surprised that you consider a Condorcet method to be too extremist
or apt to suffer center
squeeze.
Think Yee diagrams; all Condorcet methods yield identical diagrams, while
center squeeze shows up
clearly in methods that allow it.
2011/7/13 fsimm...@pcc.edu
Jameson, I'm surprised that you consider a Condorcet method to be too
extremist or apt to suffer center
squeeze.
Hmm... you're right, I hadn't recognized that your remove one of closest
pair method was Condorcet-compliant, as any pairwise method.
Think Yee
Bob Richard wrote:
On 7/13/2011 11:14 AM, fsimm...@pcc.edu wrote:
Jameson, I'm surprised that you consider a Condorcet method to be too
extremist or apt to suffer center
squeeze.
Think Yee diagrams; all Condorcet methods yield identical diagrams,
while center squeeze shows up
clearly in
2011/7/13 Kristofer Munsterhjelm km_el...@lavabit.com
fsimm...@pcc.edu wrote:
Of course if we have a multiwinner method, we don't want all of the
winners concentrated in the center of the population. That's why we
have Proportional Repsentation.
Also the purpose of stochastic single
fsimm...@pcc.edu wrote:
Trying to build a metric from a set of ranked ballots is fraught with
difficulties, and your outline of a procedure for doing it is
interesting to me.
The simplest, least sophisticated idea I have so far that seems to
have some use is to define the distance between two
After looking up some old email threads, it now seems to me that I made
a significant mistake in the post below. It is true that the model
underlying Yee diagrams guarantees that there will always be a Condorcet
winner. But apparently that has nothing to do with the two dimensions
being
- Original Message -
From: Kristofer Munsterhjelm
fsimm...@pcc.edu wrote:
...
There may also be another scenario where Majority Judgement (or
median
ratings, for that matter) would do better than ranked methods.
If it's
possible for the voters to agree on what, say, Good means
Actually, any centrally symmetric distribution will do, no matter how many
dimensions.
The property that we need about central symmetry is this: any plane (or
hyper-plane in higher
dimensions) that contains the center of symmetry C will have equal numbers of
voters on each side of
the
2011/7/13 fsimm...@pcc.edu
- Original Message -
From: Kristofer Munsterhjelm
fsimm...@pcc.edu wrote:
...
There may also be another scenario where Majority Judgement (or
median
ratings, for that matter) would do better than ranked methods.
If it's
possible for the voters
That proof assumes a euclidean distance metric. With a non-Euclidean one,
the planes could have kinks in them. I believe I have heard that the
result still holds with, for instance, a city-block metric, but I cannot
intuitively demonstrate it to myself by imagining volumes and planes as in
this
If we abandon the Euclidean metric, then we also abandon Voronoi Polygons; the
corresponding idea for
more general metrics is that of a Dirichlet region.
It would be amusing to see Yee diagrams based on L_1 and L_infinity metrics
Of course, Yee uses the L_2 metric to make his pictures
I doubt it's monotonic, though it's probably not a practical problem. That
is, it would probably be totally impractical to try to use the
nonmonotonicity for anything strategic, and it wouldn't even lead to Yee
diagram ugliness.
2011/7/13 fsimm...@pcc.edu
Here's a simpler version that is
fsimm...@pcc.edu wrote:
Here's a simpler version that is basically the same:
Make use of cardinal ratings so that the rating of candidate X on
ballot b is given by b(X).
Define the closeness of candidate X to candidate Y as the dot product
Sum b(X)*b(Y)
where the sum is taken over all b in
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