[EM] Three rounds

[Juho Laatu]

Is Schwartz set specifically "Condorcetian"? Also methods like minmax could be said to be strongly "Condorcetian" (although they do not necessarily elect from the Schwartz/Smith sets).

Also party lists could be used. One approach to breaking cycles and identifying clone sets would be to not to count the votes of fellow party members against the candidates (not before all candidates of the competing parties/branches have been eliminated). How did you use D'Hondt without parties?

Sorry about making only "opposite proposals" :-). Schwartz and partyless approaches may be ok too. And use of some sequential approach to break a Condorcet cycle as well.

Juho


--- On Tue, 11/11/08, Kristofer Munsterhjelm <km-elmet at broadpark.no> wrote:

> From: Kristofer Munsterhjelm <km-elmet at broadpark.no>
> Subject: Re: [EM] Three rounds
> To: "Raph Frank" <raphfrk at gmail.com>
> Cc: juho4880 at yahoo.co.uk, election-methods at lists.electorama.com
> Date: Tuesday, 11 November, 2008, 8:50 PM
> Raph Frank wrote:
> > On Mon, Nov 10, 2008 at 4:05 PM, Juho Laatu
> <juho4880 at yahoo.co.uk> wrote:
> >> One could e.g. force supporters of the
> "eliminated" candidates to approve more than one
> candidate (at least one of the "remaining"
> candidates) (instead of just bullet voting their second
> preference). On possible way to terminate the algorithm
> would be to stop when someone has reached >50% approval
> level.
> >> 
> >> Also in "non-instant" runoffs one could
> e.g. force the voters to approve at least one on the
> "remaining" candidates. (One could eliminate more
> than one candidate at different rounds.)
> > 
> > That is kinda like Bucklin, though without the
> approval threshold
> > changing in each round for all voters.
> > 
> If you're going to have an advanced runoff method, why
> not do something explicitly more Condorcetian? Perhaps
> something like:
> 
> Determine the Schwartz set. If it is singular, the
> candidate wins,
> otherwise: the two highest ranked members of the Schwartz
> set, according to some Condorcet rule, advance to the
> runoff.
> 
> Another option would be to use D'Hondt without lists,
> based on a good Condorcet method, to elect the two
> candidates for the runoff. But that's too complex, I
> think.