[EM] Methods of elimination in quota preferential STV

[Craig Carey]

The maths I have done tends to suggest that the use of quotas to
reject losers is exactly what would turn up in an ideal method that
lacks clear arbitrariness.

The slopes of the quotas is done by P1, at least in IFPP. P1 is defined
on http://www.ijs.co.nz/ifpp.htm P1 only specifies slopes, and the
1/3 in the 3 candidate election comes from the line touching the
centre of the triangle with P1 saying it is parallel to an edge.

There is a need to consider STV's algorithm or else there won't be any
quotas.
I'd say it is a good idea to patch up STV.

There was an attempt and a first attempt 4 candidate guess I tried,
failed P1. So guessing is not useful (it might be later).

Another problem was that to get STV as a whole to have good properties,
it is possible that its stages may fail to have that good property.
That was the case with the duality property (invariance on negating
votes and swapping winners with losers). So with this quota problem,
it might occur that it is unhelpful to regard STV as a repeated
applying of a single method. This could be quite a problem. But it
could sort of vanish if IFPP is solved.

I attempted to make progress on the problem similar to putting
quotas into some Alternative Vote using REGLOG/REDUCE code that was
using real numbers. It turned out that the algorithm had trouble finding
the defect regions and much faster code was needed. Some of that program
is online at http:/www.ijs.co.nz/polytopes.htm  (It seemed to find
nothing of value).

The sort of quotas expected in 4 candidate STV would be (I guess):
  * a 1/4 quota limiting effects of the 3rd preferences; and
  * 4 x 1/3 quotas would limit the effects of the 2nd preferences.

While STV could be tested numerically, I doubt that people have got
a good chance of guessing how to alter the STV algorithm. I am quite
confident that the values of the quotas for losers would be 1/3, 1/4,
1/5, etc.. It quite possible that the redesign of STV is not at all
obvious. In preferential voting, guessing does not lead to much.
Few would have guessed the B wins 4W1BCD formula I just sent in.

I suggest computerises symbolic guessing as a way to a solution.
I will have the answer out for 4 candidates out before Oct 2001.



At 12:11 04.10.00 +1000 Wednesday, LAYTON Craig wrote:
 >I'm new to this mailing list (so hi).
 >
 >I was wondering if anyone had encountered any serious analysis of the best
 >method of eliminating candidates in a multi-member election counted using
 >quota preferential STV (Hare-Clark here in Australia).  The system is
 >usually advocated as eliminating candidates in much the same way as IRV, so
 >at first glance (without the skills to model electoral systems) it seems
 >that all the same problems could apply, albeit with much less significant
 >consequences.  Is there any scope for using the principles of some of the
 >more advanced methods in single candidate elections for eliminating

Just use my IFPP axioms as best as can be done. Especially P1.
It is approximately P1 that creates the correct quotas, at least in IFPP.
In STV it should still be P1 (and P2) but there would be some unavoidable
appropriateness (STV has curved boundaries, etc.).

 >candidates in multi-member elections?
 >
 >Oh, and in regards to the question of strategic voting; strategic voting (to
...
There is an initial constraint that the method be quite like STV, and
truncation resistance simplifies the problems and leads to a problem
where the problem is too simple to get rid of undesirable defects.
Then with it not being removed, the defect is present in all solutions,
before the "bad and undesirable" complainers have begun to speak. I.e.
it is a little trivial.


 >
 >Craig Layton

This is a problem that probably should be solved.

Two ways:
(a) solve IFPP and then alter STV to make it like IFPP. That can be
     done by using numerical examples.
(b) guess at adjustments for STV and try to improve its compliance
     with P1. This could be very much harder. P1 says nothing about
     which candidate ought win. I prefer path (a). Path (b) is not
     so good for thesis students either, since the final finding could
     be seemingly arbitrary.