On Sep 1, 2008, at 0:49 , Kristofer Munsterhjelm wrote:
Juho wrote:
This particular example resembles Hylland free riding (it is an
optimized version of it) but there could be also many other
examples, some of which resemble e.g. Woodall free riding. Only in
some special cases this
On Sep 1, 2008, at 1:10 , Raph Frank wrote:
On Sun, Aug 31, 2008 at 10:18 PM, Juho [EMAIL PROTECTED] wrote:
It may well be the best strategy to rank A below D in the example
above if A
will be elected almost certainly since the voter has an interest to
guarantee that D will be elected rather
On 9/3/08, Juho [EMAIL PROTECTED] wrote:
I think you should rank A as high as possible but behind at least 1
candidate who you are reasonably sure will be elected.
I didn't quite get this part of the mail. Usually candidates that are sure
to be elected would go down in the rankings.
1) Every odd year, an 'election' is held but voters vote for parties
2) based 1), seats are distributed using d'Hondt between the parties
3) Also based on 1), districts are modified so that they are
gerrymandered to give each party the correct number of seats
4) The following year, those
On Sep 3, 2008, at 12:28 AM, Juho wrote:
I hope this speculation provided something useful. And I hope I got
the Meek's method dynamics right.
Meek completely fixes Woodall free riding. That strategy takes
advantage of the fact that most STV methods (to the extent we're in a
STV/Meek/etc
Kristofer Munsterhjelm wrote:
On the other hand, approximating may make strategy more difficult. I
think Rob LeGrand wrote something about how approximations to minimax
Approval obscured the strategy that would otherwise work.
Yes, you're thinking of
On Wed, Sep 3, 2008 at 1:57 PM, Brian Olson [EMAIL PROTECTED] wrote:
I checked my code and I'm not doing the expensive square root. It's not
quite the second though, it's actually:
((dx*dx + dy*dy) * weight)
The weight gets nudged by multiplying by 1.01 or 0.99. Squaring the weight
or not
Raph Frank wrote:
On Tue, Sep 2, 2008 at 11:00 PM, Kristofer Munsterhjelm
[EMAIL PROTECTED] wrote:
The reasonable thing to use would be Euclidean distance, since that makes
sense, given the geometric nature of the districting problem. If you want to
be even more accurate, you can use great
I haven't been following this thread in great detail, but I do have a
question: what is the distance function actually trying to measure and
minimize? What exactly are we trying to optimize when we minimize
distance, by whatever measure? I might be close in a crow's-flight
sense to a
On Wed, Sep 3, 2008 at 8:35 PM, Kristofer Munsterhjelm
[EMAIL PROTECTED] wrote:
If you use a Mercator projected map, you're just hiding the quantization.
All maps have some distortion, and since the map projection uses
trigonometric functions, you can just use the Haversine distance directly.
On Wed, Sep 3, 2008 at 8:46 PM, Jonathan Lundell [EMAIL PROTECTED] wrote:
OK, we could solve that in principle (though not too quickly) by using
Google Maps driving time, or the like. But what does driving time have to do
with grouping voters (unless we're drawing a precinct and measuring
Raph Frank wrote:
1) Every odd year, an 'election' is held but voters vote for parties
2) based 1), seats are distributed using d'Hondt between the parties
If you're going to have D'Hondt, or PR in general, why bother with the
districting? Just use open list or a party-neutral proportional
Jonathan Lundell wrote:
On Sep 3, 2008, at 12:28 AM, Juho wrote:
I hope this speculation provided something useful. And I hope I got
the Meek's method dynamics right.
Meek completely fixes Woodall free riding. That strategy takes advantage
of the fact that most STV methods (to the extent
Brian Olson wrote:
I guess my time in Computer Science land has left me pretty comfortable
with the idea that there are lots of problems that are too hard to ever
reliably get the best solution. I don't know if there's a short-short
popularizing explanation of how finding a good solution is
Hello electorama fans,
regarding that last comment, I invite those interested in non-geographical
district
to consider astrological district. The idea is to obtain equivalent samples
of the electorate
in term of any distribution (age, geography, profession, language,
religion,...) like
poll
In message [EMAIL PROTECTED] (on 3 September
2008 22:01:24 +), [EMAIL PROTECTED]
(=?iso-8859-1?B?U3TpcGhhbmUgUm91aWxsb24=?=) wrote:
Hello electorama fans,
regarding that last comment, I invite those interested in non-geographical
district to consider astrological district.
How about
On Wed, Sep 3, 2008 at 10:27 PM, Kristofer Munsterhjelm
[EMAIL PROTECTED] wrote:
Raph Frank wrote:
1) Every odd year, an 'election' is held but voters vote for parties
2) based 1), seats are distributed using d'Hondt between the parties
If you're going to have D'Hondt, or PR in general, why
On Wed, Sep 3, 2008 at 10:51 PM, Kristofer Munsterhjelm
[EMAIL PROTECTED] wrote:
In general then, any method that acts like Z had never run (when Z is
eliminated) would be resistant to Woodall free-riding.
Right, you can get that benefit from alot of methods. For example,
you could do hand
Why not self-chosen districts ?
Because then the last half of voters would be able to pick
between district already composed of majoritarians ideologies.
Again the least organized and the smallest group would finish splitted
between several districts where they would be in minority.
Do you
On Sep 3, 2008, at 10:02 PM, Stéphane Rouillon wrote:
STV-PR suffers from three principal problems that are exacerbated
when trying to push the proportionality limit. They are all caused
by the large number of candidates:
1) A pre-selection occurs within each party, in order for the star
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