Thanks a lot! If I'm not mistaken, SODA elects A in this case. Which means that SODA is holding true to form and electing a beatpath winner*, even though in this case my proof that SODA does so does not hold. A assigns all of their delegated votes to approve C, and two of them to also approve B; then D assigns two to C; and then B is faced with a choice between A and C, so they choose A. I'll have to do some head-scratching to see if I can get my proof to cover this case. I think there may be a way to do it using double induction... The trouble is that my proof, as it stands, doesn't even prove monotonicity except in restricted cases. I'm pretty sure monotonicity holds but this problem is proving to be way more involved than you'd expect. I think that proving beatpath may actually be the easiest way to prove monotonicity (!!). *Note I say "a" beatpath winner, not "the" beatpath winner. With SODA, voters can express nondelegated, static approvals; and this can lead to candidates having mutual beatpaths against each other, for instance, in the chicken dilemma. That can lead to there being two or more potential beatpath winners, with the first candidate with enough delegated votes getting to choose which of those wins. 2015-10-05 16:54 GMT-04:00 Rob LeGrand <honky98@gmail.com>: > Jameson wrote: > > I can show (under reasonable strategic assumptions) that my delegated > > system SODA elects one of the candidates who beats all with a winning > vote > > beatpath of length 2 or less, if there is any such candidate. "A beats B > > with a winning vote beatpath of length 2" means: there is some C such > that > > the minimium of (votes for A against C, votes for C against B) is greater > > than votes for B against A. > > Obviously, if we allow any length of beatpath, and if there are no ties, > > there must be some beats-all winner. But is it possible that there is no > > winner with beatpaths of length 2 or less? Can anybody give an example of > > such an electorate? > > If I understand you correctly, this electorate does what you want: > > 29:A>C>B>D > 15:B>D>A>C > 16:D>C>B>A > > If beatpaths of all lengths are considered, as in the Schulze method, A > loses no beatpath comparisons. In particular, the beatpath A>C>B>D is > stronger than D>A. But A needs a beatpath that long to defeat D; the > beatpaths A>B>D and A>C>D are weaker than D>A. Using only beatpaths of > length 2 or less, every candidate loses a beatpath comparison. > > What would strategic SODA do in the above case? > > -- > Rob LeGrand > rob@approvalvoting.org >