Kevin Venzke wrote: > The defeats are A>B, B>C, A=C. What reasoning do you use to find that > B and C are in the Landau set? I gather I don't have a complete > understanding of what Landau refers to, but I'm very surprised if the > definition is such that a Landau winner can fail to be a Schwartz > winner. This makes Landau seem less worthwhile to me, since Schwartz > is more intuitive.
I'm using what I believe is Markus Schulze's definition of Landau winners: "Candidate A is a Landau winner iff for every other candidate B at least one of the following two statements is correct: (1) A >= B. (2) There is a candidate C such that A >= C >= B." where >= means "beats or ties pairwise". It's the same thing as Smith except that the beatpaths can be of length at most two. You could easily define a "Schwartz-Landau" set that may give you want you were expecting by changing "beats or ties pairwise" in the above definition to "beats pairwise". Such a set would always be a subset of the Landau set and of the Schwartz set. -- Rob LeGrand, psephologist [EMAIL PROTECTED] Citizens for Approval Voting http://www.approvalvoting.org/ ____________________________________________________________________________________ Check out the hottest 2008 models today at Yahoo! Autos. http://autos.yahoo.com/new_cars.html ---- Election-Methods mailing list - see http://electorama.com/em for list info
