[EM] "non-cyclic" pairwise loss?

[Kevin Venzke]

Hi Robert,
Yes. Only one of B>D or C>D (you don't mention whether perhaps A>D is also the case) can be locked against D. If B>D is stronger than C>D, then when we encounter C>D, we'll throw it out, even if it wouldn't cause a cycle. (D has already lost the election.)
Kevin
      De : robert bristow-johnson <rbj at audioimagination.com>
 À : election-methods at lists.electorama.com 
 Envoyé le : Lundi 13 avril 2015 18h49
 Objet : Re: [EM] "non-cyclic" pairwise loss?
   
On 4/13/15 6:56 PM, Kevin Venzke wrote:


>
> "Non-cyclic pairwise loss" just means a loss that wouldn't create a 
> cycle of locked wins (at the time in the process that you consider the 
> loss). RP has this concept just like River does. What makes River seem 
> similar to a Minmax method is that only one loss (the strongest 
> non-cyclic one) will get counted for any particular candidate. Once 
> you lock a win against somebody, you won't lock any more against them.
>

quick question: if A beats B and A beats C and both B and C beat D, does 
River open that loop?  it ain't a cycle.  A has two beat-paths against 
D.  does River prevent one of them?

just trying to grok the basic.

-- 

r b-j                  rbj at audioimagination.com

"Imagination is more important than knowledge."



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