[EM] Re: is the Raynaud method Smith-efficient

[Forest Simmons]

James, your proof is sound, but here's a shorter one based on Kevin's 
comment:

Raynaud eliminates candidates one by one until there is only one candidate 
left.  At some stage the Smith set must have only one candidate A left. 
This candidate A is not beaten pairwise by any of the remaining 
candidates, so A will never be eliminated, and is therefore the Raynaud 
winner.

Forest


> From: "James Green-Armytage" <jarmyta at antioch-college.edu>
> Subject: [EM] is the Raynaud method Smith-efficient?
> To: election-methods-electorama.com at electorama.com
> Message-ID: <fc.005b8ffc0129ebf7005b8ffc0129ebf7.129ecc8 at mcgregor.edu>
> Content-Type: text/plain; charset=ISO-8859-1
>
>
> 	A brief definition of Raynaud:
> Eliminate the candidate with the strongest defeat against him or her,
> until no defeated candidates remain. (Note that when a given candidate is
> eliminated, his victories over other candidates (if he
> has any) are removed from consideration.)
> http://userfs.cec.wustl.edu/~rhl1/rbvote/desc.html
>
> 	A brief definition of Smith-efficiency:
> Dominant set: A set of candidates such that every candidate inside the set
> pairwise-beats every candidate outside the set.
> Minimal dominant set (Smith set): A dominant set that doesn't contain
> smaller
> dominant sets.
> A Smith-efficient method always selects a member of the minimal dominant
> set.
> http://lists.electorama.com/pipermail/election-methods-electorama.com/2005-February/014742.html
>
>   ***
>
> 	It seems to me that Raynaud is Smith-efficient. I'm not good at
> constructing proofs, but I think a proof of this would loosely revolve
> around the following line of reasoning:
>
>
> Assume that some candidates are in the Smith set, and other candidates are
> not in the Smith set. (If all candidates are in the Smith set, then no
> method can fail Smith-efficiency in that case.)
> 1. If Raynaud eliminates a candidate that is not in the Smith set, then
> the Smith set will not change. Why not? All of the candidates in the Smith
> set still defeat all of the remaining non-Smith set candidates.
> 2. If Raynaud eliminates a candidate that is in the Smith set, no
> candidate who was not in the initial Smith set will enter the new Smith
> set. Why not? All of the remaining candidates within the initial Smith set
> still defeat all of the other candidates.
> 3. Because initially non-Smith set candidates cannot enter subsequently
> revised Smith sets as a result of candidate eliminations, the winner must
> come from the initial Smith set.
>
> 	Does this make sense? Does anyone with a more academic bent know how one
> would convert this into a more rigorous proof? (I could use some practice
> in this, since I plan to enter an economics PhD program in the fall.)
> 	I recognize that Raynaud fails monotonicity, but personally I don't
> consider that to be a big deal. Raynaud is arguably the most intuitively
> obvious pairwise tally method. I'm not willing to argue that Raynaud is
> superior to defeat-dropping pairwise methods, but I do feel that it should
> continue to be part of the conversation.
>
> my best,
> James Green-Armytage
> http://fc.antioch.edu/~james_green-armytage/voting.htm
>