I fully agree with those who say that a weak Condorcet winner is in most
cases the best winner. The concern with the weak CW problem is, at least for
me, based on two different factors:
- 1. It may be that some of the voters who helped make that candidate a
CW would honestly regret their
Warren Smith wrote:
To reiterate and/or answer some questions:
1. there is no way to find the Kemeny winner that is much faster than
finding the full Kemeny ordering. More precisely, both are NP-complete tasks
so there is no poly-time algorithm for either unless P=NP.
2. Even if some
Kathy Dopp Sent: Friday, September 23, 2011 11:23 PM
My point is, that the two examples you gave IMO are very
*strong* Condorcet winners in the sense that the vast
majority of voters would prefer the Condorcet winner over one
or the other of the other two candidates which are far less
Toby Pereira wrote:
Most of the discusssion on this group is about single-winner methods and
while it's important to get things right for elections with single
winners, I don't think I can be alone in thinking that with
parliamentary elections, the gap (in quality) between any half-decent
PR
Andy Jennings wrote:
Very good example. Thanks.
Also note that even if IRV dominated Plurality on results, IRV fails
certain criteria that Plurality passes. IRV is not summable, but
Plurality is. Hence IRV could lead to worse results in places where
ballot tampering happen.
Further note. Burlington probably would not have happened under plurality.
Of course, the same ballots would have given the same results - but
Duverger's Law would have probably made voters act differently, so that the
Democrat would have stayed in the race.
General point: you can't just compare
Dear all,
I guess there could be some simple elimination of candidates before the
election, so that there will be a manageable set of candidates for the
Kemeny election, like 10 to 15.
I guess sometime the winner would be lost in the reduction, but I would
expect this to be extremely rare if
2011/9/24 Kristofer Munsterhjelm km_el...@lavabit.com
Warren Smith wrote:
To reiterate and/or answer some questions:
1. there is no way to find the Kemeny winner that is much faster than
finding the full Kemeny ordering. More precisely, both are NP-complete
tasks
so there is no poly-time
Dear all,
a small correction to my email below (a negation was forgotten):
I wrote: Yes, it might happen, that a good variable is eliminated, which
significantly improve the analysis, but this is very likely, since you
normally have so many other good variables.
The text should read: Yes, it
Ralph Suter Sent: Saturday, September 24, 2011 3:12 AM
1. Despite your own certainty about how the real world of partisan
politics functions, your opinion is entirely speculative with no basis
in historical events, since no Condorcet elections have ever been held
in any major public
On Sat, Sep 24, 2011 at 10:50 AM, Kristofer Munsterhjelm
km_el...@lavabit.com wrote:
Warren Smith wrote:
To reiterate and/or answer some questions:
1. there is no way to find the Kemeny winner that is much faster than
finding the full Kemeny ordering. More precisely, both are NP-complete
2011/9/24 Warren Smith warren@gmail.com
On Sat, Sep 24, 2011 at 10:50 AM, Kristofer Munsterhjelm
km_el...@lavabit.com wrote:
Warren Smith wrote:
To reiterate and/or answer some questions:
1. there is no way to find the Kemeny winner that is much faster than
finding the full
I like that title, but think it's a bit too long.
How about:
Declaration of Election-Method Reform Advocates -- A Shared Call to Improve
Democracy
JQ
2011/9/23 Richard Fobes electionmeth...@votefair.org
Based on the feedback about the declaration title -- including Andy's
comment below and
Peter Zbornik wrote:
Dear all,
I guess there could be some simple elimination of candidates before the
election, so that there will be a manageable set of candidates for the
Kemeny election, like 10 to 15.
I guess sometime the winner would be lost in the reduction, but I would
expect this
James Gilmour wrote:
Ralph Suter Sent: Saturday, September 24, 2011 3:12 AM
1. Despite your own certainty about how the real world of partisan
politics functions, your opinion is entirely speculative with no
basis in historical events, since no Condorcet elections have ever
been held in any
Warren Smith wrote:
--seems to me, KM is correct; if the Smith Set has less than about 20
members (and if this also is true recursively upon removing it), then
it always will be feasible to find the Kemeny order.
There are two related criteria in play here. First is plain old Smith.
If the
Kristofer MunsterhjelmSent: Saturday, September 24, 2011 10:22 PM
James Gilmour wrote:
You are right, so far as I am aware - there have never been any
Condorcet public elections anywhere in the world. That in itself
should tell us something as the Condorcet voting system has
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