Dear Nathan, I have attached a copy of my 8 June 1998 mail to Mike Ossipoff where I criticize Ossipoff's subcycle rule #2.
Markus Schulze > Mon Jun 08 16:40:30 1998 > To: Mike Ossipoff > From: Markus Schulze > Subject: Re: better letter > > Dear Mike, > > you wrote (8 Jun 1998): > > For the initial Condorcet count (by the previously-defined > > Condorcet(EM) ), ignore defeats within cycles. If the winner > > is in a cycle, then confine the election to the lowest order > > cycle that includes it (If cycle B is an element of cycle A, > > then cycle B is of higher order than cycle A). > > > > Do a count among the members of that cycle, disregarding > > defeats in its subcycles. Again, if the winner is in a higher > > order cycle contained in that cycle, then confine the election > > to the winner-containing cycle whose order is one greater than > > the order of the cycle to which the election was previously > > confined. And, as before, defeats in cycles of higher order > > than the one to which the election is confined are ignored. > > > > So then, each time, the election is confined to the > > winner-containing cycle of next higher order, ignoring > > defeats in its contained cycles of higher order. > > > > This continues till we get to the highest order cycle > > containing the winner so far, & then we solve that cycle > > to find the winner. I forgot to mention: Condorcet(EM) is > > used in each of these iterative elections. > > If I understand your suggestions correctly, then this > election method still fails to meet Pareto. > > ********************** > > Example (6 candidates; 100 voters): > > 24 voters vote D > A > F > C > B > E. > 20 voters vote B > E > C > D > A > F. > 20 voters vote B > E > F > C > D > A. > 20 voters vote A > C > B > E > F > D. > 11 voters vote D > F > C > B > A > E. > 5 voters vote A > F > C > B > E > D. > > The matrix of pairwise defeats looks as follows: > > A:B=49:51 > A:C=49:51 > A:D=25:75 > A:E=60:40 > A:F=69:31 > B:C=40:60 > B:D=65:35 > B:E=100:0 > B:F=60:40 > C:D=65:35 > C:E=60:40 > C:F=40:60 > D:E=35:65 > D:F=55:45 > E:F=60:40 > > As A > E > D > F > C > B > A, every candidate is in > the Smith Set. > > Step 1: > > As > (1) A > E > D > A, > (2) B > A, > B > D, > B > E, > (3) C > A, > C > D, > C > E, > (4) A > F, > D > F, and > E > F, > > A > E > D > A is a subcycle of the Smith Set. > > Thus, I ignore the defeats between the candidates of the > Smith Set and use the Condorcet(EM) tiebreaker. > > Thus, I get the following matrix of pairwise defeats: > > A:B=49:51 > A:C=49:51 > A:F=69:31 > B:C=40:60 > B:D=65:35 > B:E=100:0 > B:F=60:40 > C:D=65:35 > C:E=60:40 > C:F=40:60 > D:F=55:45 > E:F=60:40 > > As the worst defeat of candidate A is the smallest, > candidate A wins the tiebreaker. > > Step 2: > > As candidate A is in a subcycle, I have to use the > tiebreaker in the subcycle: > > A:D=25:75 > A:E=60:40 > D:E=35:65 > > As the worst defeat of candidate E is the smallest, > candidate E wins the elections. > > ********************** > > In the example above, candidate E is elected although > every voter prefers candidate B to candidate E. > > Of course, it can be questioned, whether Pareto is > important. But from the example above, you can also see, > that this election method fails to meet monotonicity. > > If those 11 voter, who have voted D > F > C > B > A > E, > change their opinion to D > F > C > B > E > A, then > we have: > > A:B=49:51 > A:C=49:51 > A:D=25:75 > A:E=49:51 > A:F=69:31 > B:C=40:60 > B:D=65:35 > B:E=100:0 > B:F=60:40 > C:D=65:35 > C:E=60:40 > C:F=40:60 > D:E=35:65 > D:F=55:45 > E:F=60:40 > > Now, there is no subcycle anymore. Candidate B or candidate > C wins the election, as their worst defeats are the smallest. > > ********************** > > I hope, that I haven't made any mistakes. > > Markus ---- Election-methods mailing list - see http://electorama.com/em for list info