Dear Michael, very interesting, I don't think I saw anything like this before.
When trying do evaluate a new method, I always try to check very simple criteria first, like neutrality and anonymity (obviously fulfilled here), Pareto efficiency, monotonicity, etc. Concerning the latter two, I was not able to verify them for your method yet. I think you should focus on those before checking more complex things like Condorcet efficiency and so on! Also, immediately a ratings-based generalization came to mind (using ratings differences instead of rank differences). Finally, when only seeking a single winner, you could alternatively build a score for each option X by adding all entries of the final matrix in rows labeled "XY" and columns labeled with positive differences. Yours, Jobst Michael Rouse schrieb: > As usual with such posts, there is a good chance someone has come up > with the same (or very similar) method, but I thought it had interesting > properties, and was wondering what glaring voting paradoxes it had. In > addition, the number of possible orders is overwhelming if there are a > large number of candidates, and I'm not sure that can be simplified. > Finally, Thunderbird sometimes seems to have weird formatting issues in > email, which may screw up the following into unreadability. > > With that in mind, here it is. > > Step 1: For each ranked ballot, create a matrix for each pairwise vote, > based on the distance and direction between each candidate. For example, > on the ballot A>B>C, you would get: > > -2 -1 0 1 2 > AB 0 0 0 1 0 > BA 0 1 0 0 0 > AC 0 0 0 0 1 > CA 1 0 0 0 0 > BC 0 0 0 1 0 > CB 0 1 0 0 0 > > Taking the rows in order, this shows that A is one position higher than > B on this ballot, which conversely makes B one position lower than A on > the same ballot. Also, A is two positions above C (C is two positions > below A), and B is one position above C (or C is one position below B). > Such detail may be unnecessary -- simply looking at position 1 and above > is sufficient, if you don't allow ties -- but I wanted to show the > symmetry. > > Step 2. Add all matrices together. As a simple example, let's consider > the following 12 votes in a circular tie (to make it interesting): > > 5: A>B>C > 3: B>C>A > 4: C>A>B > > Taking the first line, A>B>C = > > -2 -1 0 1 2 > AB 0 0 0 1 0 > BA 0 1 0 0 0 > AC 0 0 0 0 1 > CA 1 0 0 0 0 > BC 0 0 0 1 0 > CB 0 1 0 0 0 > > Multiplied by 5 gives you: > > -2 -1 0 1 2 > AB 0 0 0 5 0 > BA 0 5 0 0 0 > AC 0 0 0 0 5 > CA 5 0 0 0 0 > BC 0 0 0 5 0 > CB 0 5 0 0 0 > > > Taking the second line, 3: B>C>A = > > -2 -1 0 1 2 > AB 3 0 0 0 0 > BA 0 0 0 0 3 > AC 0 3 0 0 0 > CA 0 0 0 3 0 > BC 0 0 0 3 0 > CB 0 3 0 0 0 > > > And finally, taking the third line, 4: C>A>B > > -2 -1 0 1 2 > AB 0 0 0 4 0 > BA 0 4 0 0 0 > AC 0 4 0 0 0 > CA 0 0 0 4 0 > BC 4 0 0 0 0 > CB 0 0 0 0 4 > > Adding these together gives you > > (5: A>B>C, 3: B>C>A, 4: C>A>B) = > > -2 -1 0 1 2 > AB 3 0 0 9 0 > BA 0 9 0 0 3 > AC 0 7 0 0 5 > CA 5 0 0 7 0 > BC 4 0 0 8 0 > CB 0 8 0 0 4 > > Step 3: This is where a bit of a curve is thrown in. Looking at each > position on the matrix, determine which is less -- the sum of the > numbers above the current position, or the sum of the numbers below the > current position -- and add that number to the value of the current > position. In essence, this is adding the value of a position to the > value of all other positions away from the median. Output that number to > a new matrix in the appropriate spot. Using the numbers above, you end > up with: > > -2 -1 0 1 2 > AB 3 3 3 9 0 > BA 0 9 3 3 3 > AC 0 7 5 5 5 > CA 5 5 5 7 0 > BC 4 4 4 8 0 > CB 0 8 4 4 4 > > Step 4: Looking at each possible preference order, add the values for > the appropriate positions on the matrix, and choose the preference order > with the highest score. Using the above values and excluding ties: > > A>B>C = (A>B) [one position] + (B>C) [one position] + (A>>C) [two > positions] = 9+8+5 = 22 > B>C>A = 8+7+3 = 18 > C>A>B = 7+9+4 = 20 > A>C>B = 5+4+0 = 9 > C>B>A = 4+3+0 = 7 > B>A>C = 3+5+0 = 8 > > So the ranking of possible orders is > > A>B>C = 22 > C>A>B = 20 > B>C>A =18 > A>C>B = 9 > B>A>C = 8 > C>B>A = 7 > > A>B>C is the most preferred order. C>B>A is the least preferred order. > > **************** > > As another example (ignoring a few pages of work), here is a set of > ballots used in the Schulze Method entry in Wikipedia: > > 5:A>C>B>E>D > 5:A>D>E>C>B > 8:B>E>D>A>C > 3:C>A>B>E>D > 7:C>A>E>B>D > 2:C>B>A>D>E > 7:D>C>E>B>A > 8:E>B>A>D>C > > If I did the math correctly, the winning order is: > E>B>A>D>C = (27+25+30+28)+(23+26+13)+(8+16)+(8) = 204 > > For comparison, Schulze gives E>A>C>B>D, which this method would score > (22+26+28+19)+(16+17+17)+(0+15)+(0) = 160 > > (Note, no claim is made as to which one is better, I just wanted to show > the difference.) > > ************** > > I've tried variants using distance multiplied by score (which seems to > encourage strategic raising and lowering of ranks) and absolute ranking > rather than relative ranking (which didn't seem to act as nicely in vote > aggregation), but neither seemed to act as nicely on brief inspection. I > haven't come up with a tiebreaker method yet, either -- like in the > simple example of adding together 1: A>B>C and 1:B>C>A (both orders have > the same score) -- but that can wait. > > Any questions, comments, or criticisms (the latter most likely about my > math!) are welcome. Especially welcome would be examples of paradoxes, > and links to the same (or similar) method discussed in the past. > > Michael Rouse > ---- > Election-Methods mailing list - see http://electorama.com/em for list info ---- Election-Methods mailing list - see http://electorama.com/em for list info
