On Thu, Nov 19, 2009 at 4:51 AM, Brian Olson <[email protected]> wrote: > Oh, that is a problem. It gets the right answer if I use L1 norm instead of > L2. I think L2 norm is going to work better for single-seat IRNR but L1 norm > better for multi-seat. L2 inflates the amount of vote that winds up getting > applied to multiple choices.
The L1 norm mean that each voter always gets to cast exactly 1 vote (ratings add to 1). Thus the total number of votes cast is always constant. This means that a quota can be easily determined. You could use a different rule for eliminating than you use for electing (and I think that is a good idea anyway). For example, for electing, each ballot is scaled so that w(a)*r(a) + w(b)*r(b) + ..... = 1 All eliminated candidates have a w(x) = 0 and all non-elected candidates have a weighting of 1. Elected candidates have weighting so that they have exactly a quota of the votes. If any candidate meets the Droop quota, that candidate is declared elected and the next round is started. If no candidate is elected, a different rule is used, each ballot is scaled so that [w(a)*r(a)]^2 + [w(b)*r(b)]^2 + ... = 1 The running candidate the the lowest score is then eliminated. (The weights are based on the L1 calculation) This process has the nice feature that a group of voters equal to a Droop quota will decide their candidate using the L2 single seat (L2) version of the process. (This assumes that they rate all non-party candidates at zero and all voters outside the group rate their candidates at zero). Also, there is also a question if the weights assigned in step 1 will always yield a unique set of weights. Hopefully there is a Meek's method like proof. ---- Election-Methods mailing list - see http://electorama.com/em for list info
