----- Original Message ----- From: Jameson Quinn Date: Sunday, June 19, 2011 7:27 am Subject: Re: [EM] Mr.SODA To: Kristofer Munsterhjelm Cc: [email protected], [email protected]
> 2011/6/19 Kristofer Munsterhjelm > > > [email protected] wrote: > > > >> In the discussion of a proportional representation version of > SODA it > >> was lamented that the non- sequential version of PAV was > >> computationally hard, and was suggested to make use of the > PAV measure of > >> goodness to pick the winning slate from all of the slates > >> that anybody cared to nominate. > >> > >> While that would certainly be feasible and very likely near > optimal,>> another possibility is to use non- sequential PAV to > choose the first > >> three members of the slate, and then choose the remaining members > >> sequentially, conditioned on the membership of the first > three as > >> well as those chosen subsequently. > >> > >> The number of slates of size three is only n*(n-1)*(n-2)/6 , which > >> is less than five million when there are (n=)three hundred > >> candidates. > >> > >> If the members of the senate were chosen this way, the first three > >> could be a kind of triumvirate presidency of the senate. > >> > > > > You could also do this to get a hybrid of sequential and non- > sequential> PAV. Decide on a chunk size k, then you calculate > the first k winners > > non-sequentially (n choose k for n candidates). Lock these > (i.e. decide > > they'll win), then try every council of size min(k*2, desired > size) with > > these locked. Then min(k*3, desired size), min(k*4, desired > size), etc, > > until you have elected as many as you want. > > > > Sequential PAV is just this kind of PAV with k=1. k=2 or k=3 > should be > > feasible even with very large values of n, and would > approximate true > > PAV better than would ordinary sequential PAV. > > > > So, for instance, if you want to elect eight winners out of A- > Z with > > k=3, you'd first do > > > > elect ??? -> say the optimum here is ABC, then > > elect ABC??? -> say the optimum is ABCDFG, then > > elect ABCDFG?? -> say the optimum is ABCDFGKQ, and then that's your > > outcome. > > > > > > > I like this idea, but instead of electing 3,3,2 in this example > as you > suggest, you should elect 2,3,3 (that is, AB, CDF, GKQ, assuming > it comes > out the same, which it usually would). That concentrates the broadest > optimization nearer to the threshold, where it matters more. (In > fact, you'd > probably get the same results with less computing effort by electing > 1,1,1,1,1,3). > > JQ > Great idea, Kristofer. How about 3, 2, 3? Wouldn't this give better balance? ---- Election-Methods mailing list - see http://electorama.com/em for list info
