Talk:Resolvability criterion

2005 & 2006 comments

This definition of "resolvability" seems to be due to Markus Schulze. Steve Eppley however, calls this instead "reasonable determinism" and uses "resolvability" to indicate a different criterion [1]. It would be nice if the people could agree on the terminology. -- Dissident (Talk) 13:16, 26 December 2005 (UTC)

It is nothing special that definitions differ slightly from paper to paper. As long as the authors define their terms properly, I don't see any problem with it. Markus Schulze 22:03, 26 December 2005 (UTC)

It's a problem in the sense that it makes it harder to settle on the terminology used by Wikipedia (I assume you got dibs here based on your famous paper). So what should we call Steve Eppley's resolvability criterion then? The "tiebreaker criterion"? -- Dissident (Talk) 23:19, 26 December 2005 (UTC)

Interesting. From what I've gathered it seems Nicolaus Tideman himself already used resolvability in 1987 and he defined it as Steve Eppley did. I might change the article now to reflect this. -- Dissident (Talk) 02:27, 26 February 2006 (UTC)

Where is the proof Schulze satisfies Tideman's resolvability? It doesn't seem to be trivial. -- Dissident (Talk) 02:18, 27 February 2006 (UTC)

Well, you can choose the strongest paths from candidate A to every other candidate in such a manner that these paths form an arborescence. (See section 2.5.2 of my paper!) You can choose a ballot in such a manner that each link of this arborescence is increased. When you add this ballot, then the strongest path from candidate A to each other candidate is increased. Markus Schulze 08:00, 27 February 2006 (UTC)

If it's true that the weakest links in the strongest paths towards the other co-winners do not form a cycle (and thus a ballot increasing all of them is constructible), that's not enough. By adding that ballot the strongest paths towards A must not increase as well. This might very well be the case, but you didn't say so. You might want to include a formal proof of Tideman's resolvability in your latest paper (just calling it resolvability and renaming Woodall's version to discrimination). -- Dissident (Talk) 19:25, 27 February 2006 (UTC)

Dear Dissident, thank you for your suggestions. You can now find a detailed proof in section 4.1.2 of my paper. Markus Schulze 00:18, 5 March 2006 (UTC)

A minor point perhaps, but in the introduction of 4.1, you talk about resolvability in the case of more than one candidate, but Tideman's version is stronger than that insofar that it also implies mono-add-plump. This doesn't pertain to 4.1.2, where you drop the more than one candidate requirement in the proof. -- Dissident (Talk) 01:18, 5 March 2006 (UTC)