For our purposes a generalized median voting method is any method that elects the candidate that minimizes the total distance from the ballots or voters to the candidate to be elected. If the "candidates" are proposed locations for a community center , the distances to the voters are easy to visualize. Here's a rather general way of specifying the distance from a voter ballot B to a candidate X: d(B, X)=Sum {P(Y)| Y both defeats X pairwise AND outranks X on B}, where P(Y) is the percentage of ballots on which Y is the lowest ranked candidate that covers every candidate ranked above it. If a candidate covers every candidate ranked above it, then it will either be the top ranked candidate or it will cover the top ranked candidate as well as any other candidates ranked between them. For each candidate X let T(X) be the total Sum over B of d(B,X). Then elect argminT(X), the candidate X that minimizes T(X), the total distance from all of the ballots to X. Compare this to Kemeny-Young. K-Y minimizes the sum of Kendall-tau distances from the ballots to all possible finish orders of the candidates, instead of to the nominated individual candidates themselves ... a lot of un needed computation if all you need is one winner. -Forest

The only problem with this method is that it requires two passes through the ballots to get the P percentages. We can simplify it greatly by just using the percentages of first place preferences, and all of the candidates ranked above X, not just the ones defeating X. Then let M(j,k) be the number of ballots on which j outranks k, and let F be the row matrix whose j-th column element is the first choice count for candidate j. Let V be a variable initialized as the row vector F*M. Update V by sorting its columns from smallest to largest, and then replacing each entry by the name of the candidate it represents ... i.e. candidate k if the value came from column k of the pre-sorted V. At this point, V is the finish order of de-cloned Borda. Now we uncover the de-cloned Borda winner: Until the left most element of V is uncovered, update V by rotating into first place the left most element that covers it. When the dust settles, elect the candidate whose name ended up in the left most column of V. How's that for clean? IMHO this is the one-pass, monotone, clone free RCV method with the most VSE potential. But, of course that conjecture needs checking. Also, because it is the uncovered Generalized Median winner it should be fairly resistant to strategic order reversal temptations. ... which also needs checking by simulation, as well as the scrutiny of astute observers like Kevin, Kristofer, Chris Benham, etc. (The academic hotshots are no good at this ... they don't seem to have a clue.) -Forest On Wed, Nov 2, 2022, 9:00 PM Forest Simmons <forest.simmons21@gmail.com> wrote: > For our purposes a generalized median voting method is any method that > elects the candidate that minimizes the total distance from the ballots or > voters to the candidate to be elected. > > If the "candidates" are proposed locations for a community center , the > distances to the voters are easy to visualize. > > Here's a rather general way of specifying the distance from a voter ballot > B to a candidate X: > > d(B, X)=Sum {P(Y)| Y both defeats X pairwise AND outranks X on B}, where > P(Y) is the percentage of ballots on which Y is the lowest ranked candidate > that covers every candidate ranked above it. > > If a candidate covers every candidate ranked above it, then it will either > be the top ranked candidate or it will cover the top ranked candidate as > well as any other candidates ranked between them. > > For each candidate X let T(X) be the total > > Sum over B of d(B,X). > > Then elect argminT(X), the candidate X that minimizes T(X), the total > distance from all of the ballots to X. > > Compare this to Kemeny-Young. K-Y minimizes the sum of Kendall-tau > distances from the ballots to all possible finish orders of the candidates, > instead of to the nominated individual candidates themselves ... a lot of > un needed computation if all you need is one winner. > > -Forest > > >