Actually, in a perfectly partisan world, you'd want to do this optimization within each party. Anything like n,n,n will probably be wasting most of its effort re-confirming that n consists of 2 from party X and 1 from party Y.
So one way to do it would be elect a full sequential slate, then partition that slate into subsets no bigger than 3 with high mutual correlation within each subset, then, for each subset, hold the rest of the seats constant, while finding the optimum 2-3 to take the place of that subset. But then, that's complicated enough that it's no longer anything like a "natural" method, but just a (probably less-than-perfect) optimization algorithm. And so that kinda points back to my original suggestion - anybody can submit a proposed optimization, and the best one wins. That lets anybody try any optimization algorithm they want. JQ 2011/6/20 <[email protected]> > ----- 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? >
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