[EM] Re: New (? ) Condorcet method: Schwartz PC

Ernie Prabhakar drernie at mac.com
Sat Jan 10 11:19:09 PST 2004

On Jan 9, 2004, at 6:53 AM, MIKE OSSIPOFF wrote:
> The method that you described is Schwartz PC.

Ah, okay.  After experimenting with all these advanced proposals, I end  
up "rediscovering" Plain Condorcet.  :-)

> Of course Plain Condorcet could be reworded to say:
> "The winner is the candidate whose greatest defeat is the least."

So, I guess I need to re-learn how these methods differ from each other.

On Jan 9, 2004, at 6:53 AM, MIKE OSSIPOFF wrote:
> Here's how it differs from SSD:
> You sequentially drop weakest defeats from the _initial_ Schwartz set.  
> SSD sequentially drops weakst defeats from the _current_ Schwartz set.  
> Dropping defeats removes candidates from the Schwartz set, because a  
> candidate can't be in the Schwartz set merely by tying one of its  
> members.
> In public elections the initial Schwartz & Smith sets are identical,  
> because there are no pairwise ties.

Okay, let me see if I understand.  If there are no pairwise ties:
- The Schwartz & Smith sets are identical
- PC, Smith PC, and Schwartz PC all give the same result
- SD, SSD, CSSD (Beatpath) are identical to each other, but may differ  
from * PC
- Tideman still differs from *SD and *PC [is that true?]

If there are pairwise ties:
- The Smith set (beat all outside) is larger than the Schwartz  
(smallest unbeaten set)
- PC, Smith PC, and Schwartz PC can give different results
That is, the Plain Condorcet winner - with the least greatest defeat -  
may NOT be in the Schwartz or Smith set

So, back to the algorithm I proposed earlier:
- count the number of pairwise contests each candidate wins  
- identify the candidate(s) with the greatest number of wins
- pick the candidate that beats the other candidates with the same  
number of wins (LB)
    if cyclic tie, pick one at random

Is that in fact sure to pick a member of the Schwartz set?  And of the  
Smith set?

Similarly, if this algorithm gives the Schwartz set:
- choose all the candidates which beat LB, directly or indirectly

Then, does this algorithm give the Smith set:
- choose all the candidates which beat OR TIE LB, directly or indirectly

Thanks for your help.

-- Ernie P.

> The Smith set is more briefly defined, and so I prefer Smith PC to  
> Schwartz PC as a public proiposal.
> I haven't dropped Smith PC as a public proposal. It has criterion  
> advantages over PC, mostly cosmetic criterion differences, but they  
> could matter in a campaign. It all depends on what people are found to  
> like. I'd start by offering SSD & MAM, then SD, Smith PC, then PC.
> Mike Ossipoff
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