Since you are building this on the single-seat district tradition, three or four seats and 10 candidates is plenty. I'm used to numbers like 6 seats with 108 candidates, and 35 seats with 405 candidates, and at least eight parties in the parliament. (In that situation even ranking all of the candidates, or even all of the candidates of one's favourite party may be too tedious. One may however allow all votes (also short ones) to be counted for the party.)
What would be a good (non-limiting) number of candidates? Maybe something like (P * K1) * (S * K2), where P = current number of parties with representatives, K1 = 1.5 or 2, S = number of seats, K2 = 1. Juho On 6.11.2011, at 9.45, capologist wrote: > > On Nov 5, 2011, at 11:35 PM, [email protected] > wrote: > >> With two representatives per district this is a pretty good method, if we >> want a two-party system and if we accept the idea of having representatives >> with different weights. Spoiler and gerrymandering related problems are >> greatly reduced, and the method allows also third parties to grow. >> >> With more than two representatives (3, 4, 10,...) per district the systems >> becomes more multi-party oriented and more proportional. The proposed >> approach of using a Condorcet method to pick one of all possible candidate >> sets is however quite heavy computationally. For that reason the number of >> candidates (and representatives) in each district should be kept quite small. > > With three or four representatives per district, the computation is not > really all that expensive. For example, with four seats and 10 candidates, > there are 210 possible sets of winners. Running the Schulze Method on 210 > candidates is not as expensive as you might think. The average home PC can > compute the winner from the pairwise results in a minute. > > 10 representatives per district is getting a little ridiculous. It's not > necessary or desirable for every fringe movement and special interest group > to have its own representative in the legislature. Plus, if there are 10 > seats, there are likely to be 15, 20, or more candidates, and the voters > can't realistically know and rank them. > > Three or four per district would, I believe, suffice to broadly cover the > political landscape and give the vast majority of voters a reasonable choice > of representative. > ---- > Election-Methods mailing list - see http://electorama.com/em for list info ---- Election-Methods mailing list - see http://electorama.com/em for list info
