Kevin Venzke wrote: > Do you mean that after you found the center of mass for the complete > triangle, you would elect the candidate corresponding to the corner > that the center of mass is nearest to? >
That, and it gives the entire rank order depending on which area it lands in -- for example, in the triangle marked B>C>A, with B winning, C coming in second, and A losing. With 4 candidates, you could use a tetrahedron to place them, and place the votes in the appropriate volume (or volumes, in the case of truncated or equal votes) out of the 24 defined. Of course, this becomes unwieldy for higher-order dimensions (10 candidates would have 10! boxes to worry about), but there probably is an easier way to calculate the "center of mass" as you keep adding votes. Anyway, there should be a existing voting system that is equivalent to this method. What would really be interesting is if it fails IIA (or in this case, IIM -- Independence of Irrelevent Masses :) ) since there would be an apparent conflict between how things balance and Arrow's impossibility theorem. Michael Rouse [EMAIL PROTECTED] ---- election-methods mailing list - see http://electorama.com/em for list info
