[EM] MAM vs Schulze
juho.laatu at gmail.com
Sun Oct 9 14:14:22 PDT 2016
Ok, accepted. :-) Except that some methods like minmax look at pairwise defeats, but do not aim at breaking cycles (and forcing out a linear preference order). :-) Maybe at least all "path based" methods aim at breaking cycles. (All methods of course ignore at least one defeat if there is no Condorcet winner.)
> On 10 Oct 2016, at 00:05, Michael Ossipoff <email9648742 at gmail.com> wrote:
> If course breaking cycles isn't the only approach to finding the right winner. But it's the basis of MAM & CSSD, &, in general, the methods that look only at pairwise defeats, & disregard defeats to make someone unbeaten..
> So I brought it up in connection with the comparison between MAM & Beatpath.
> Cycle-breaking methods aren't among my favorite proposals.
> 3-Slot ICT
> Plain MMPO
> Michael Ossipoff
> On Oct 9, 2016 12:56 PM, "Juho Laatu" <juho.laatu at gmail.com <mailto:juho.laatu at gmail.com>> wrote:
>> On 09 Oct 2016, at 21:32, Michael Ossipoff <email9648742 at gmail.com <mailto:email9648742 at gmail.com>> wrote:
>> We want to disregard as few defeats as possible. Plainly, if it's necessary to disregard one of the defeats in a cycle, then it should be the weakest one.
> I almost agree with that but not quite. When picking the winner, the target is not to make the group opinion a linear opinion (where cycles have been broken, and some pairwise defeats disregarded). The idea is rather to accept the fact that group opinions may sometimes be cyclic, and identify a single best winner despite of that. One needs to identify one winner, but there is no need to break cycles of make the group opinion linear.
> It is a fact that if there is no Condorcet winner, there is some candidate that would beat the winner in a pairwise comparison. But that's about as far as we need to go in the direction of disregarding defeats.
> Some methods may well be based on a philosophy that is based on breaking cycles, but others need not do that. One simple example could be one that resorts to Approval if there is no Condorcet winner. That method quite clearly makes some sense, but doesn't break any cycles, or at least doesn't care about the preference strengths of the cycles, or the pairwise losses of the winner.
> Just trying to be very exact on what the single winner methods are supposed to do.
> BR, Juho
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