[EM] Mr.SODA

[fsimmons at pcc.edu]

----- Original Message -----
From: Jameson Quinn
Date: Sunday, June 19, 2011 7:27 am
Subject: Re: [EM] Mr.SODA
To: Kristofer Munsterhjelm
Cc: fsimmons at pcc.edu, election-methods at lists.electorama.com

> 2011/6/19 Kristofer Munsterhjelm
>
> > fsimmons at pcc.edu 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?