I mentioned in a previous e-mail that I like Nanson's method, where you take a Borda count of all the candidates, drop the candidate with the lowest score, and repeat. It's simple, takes into account people's higher preferences, and picks the winner from the Smith set (whether that set consists only of the Condorcet winner or of three or more candidates). It isn't perfect, but it has nice features. That got me to thinking: I wonder if anyone has applied different Borda weights to a method like Nanson's? I know Borda has a bad reputation on this list (grin), but it has some bearing on the "Crosscut" method I'm working on. ***Important Note*** Most of what follows is me thinking out loud, so feel free to ignore it unless you are interested in Borda weights and their use in Nanson-type functions. It is pretty boring even then (grin), but it has some bearing on my proposed method. Anyway, I tried to think of what limits we would put on weights. The first limit I could think of is that higher ranks must have equal or higher scores than low ranks (this is reversed in the inverse method but the result is the same). For example, if we have eight candidates ordered A-H, where A is in first place, B is in second place, and so on, until we get to H is in last place, then: A>=B>=C>=D>=E>=F>=G>=H (order of the scores in a Borda-count method) In addition, the highest rank *must* have a higher score than the lowest rank -- A>H -- otherwise all Borda scores would be equal and there would be no way to calculate the outcome of an election. I couldn't think of any other absolute requirements (if I've missed any, please let me know). With these requirements, I've come up with the following "well-behaved" graphs for Borda-like methods: Step function: Single value until it hits a certain point, then it drops to another value and stays there. With Plurality and eight candidates, this is equivalent to 10000000. Anti-Plurality uses 00000001. Approval uses a variable cutoff. A center cutoff method would use 11110000. Standard Borda: rank and points are related by a linear function of the form N-1, N-2, N-3...2, 1, 0. Most common method, equivalent to 7,6,5,4,3,2,1,0 for eight voters. Slope does not change. Well-behaved and easy to figure out. Power of Two: First place is one point, second place is one-half point, third place is one-quarter point, etc., or 2^(1-N). For eight voters, you have (1)(1/2)(1/4)(1/8)(1/16)(1/32)(1/64)(1/128). Steep function with a peak near the top ranking, almost horizontal at the bottom ranking. Tends to emphasize top ranking. "Reflected" Power of Two: of the form 1-2^(N-M), where N is the rank of the candidate and M is the number of candidates. For eight candidates this is (1-(2^-7))(1-(2^-6))(1-(2^-5))(1-(2^-4))(1-(2^-3))(1-(2^-2))(1-(2^-1))(1-(2^0)), or 127/128, 63/64, 31/32, 15/16, 7/8, 3/4, 1/2, 0. Rounded "peak," dropping more and more rapidly as you go down the ranking. Emphasizes effect of bottom ranking. Power of Two + "Reflected" Power of Two: (1/2)+(2^-N)-(2^(N-M-1)) = (255/256), (95/128), (39/64), (17/32), (15/32), (25/64), (33/128), (1/256). Sharp peak near the top ranking, sharp dip near the bottom ranking, gentle slope in the middle. Emphasizes extremes, de-emphasizes middle. Zipf law: 1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8... A bit more gentle than the slope from powers of two. I'm not sure if there are any special properties, but I saw it mentioned somewhere. As I said, pretty boring even if you are interested in Borda-like counts, but it may be useful when I explain different graph types (other than step functions) to use in a "Crosscut" method. If you've seen other functions used, let me know. Mike Rouse [EMAIL PROTECTED]
