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" <[EMAIL PROTECTED]> Subject: [EM] is the Raynaud method Smith-efficient? To: election-methods-electorama.com@electorama.com Message-ID: <[EMAIL PROTECTED]> 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
---- Election-methods mailing list - see http://electorama.com/em for list info
