[EM] Fixing IRV
Richard Moore
rmoore4 at home.com
Thu Aug 16 18:39:03 PDT 2001
Anthony Simmons wrote:
> About defining an elimination method, doesn't that sound like
> kind of a hopeless task? It's pretty simple to define one in
> terms of the definition, such as it is, that I mentioned
> earlier. One simply constructs a sequence of election
> methods, En, En-1, ... E1 (please excuse the clumsy
> subscripts). The first has n candidates, the second has n-1,
> etc., etc. They all have the same number of ballots (same
> meaning that you remove one candidate from ballots and close
> up the holes to get the next in the sequence). I don't have
> a clue how that could be expressed in a form that would suit
> your purposes. And it suffers from the same difficulty in
> actually working with it that you note below. (Actually,
> only the first and last elements in the sequence are needed
> in order to define "elimination method".)
I now realize what a can of worms that was opening.
I was trying to think of a recursive definition (along the
lines of your approach) but it was turning out to be an
awkward approach.
If we try defining what elimination by what it is not, we
will probably find some methods that aren't excluded from
the definition but still are monotonic, so to fit the
theorem (if theorem it is; without a formal statement or
proof I'm led to wonder if Riker was just stating a
generalization) we would have to modify our definition to
exclude those methods too. Eventually we will arrive at a
definition that satisfies the statement, but at that point
we may end up with a tautology, as Markus already suggested.
So right now I can only say that I can't define elimination
but I (sometimes) know it when I see it.
One final thought on the subject: Maybe Riker only meant to
include elimination that isn't based on pairwise
comparisons. IRV fits this restriction. Note that in IRV,
since you are going by place votes, you have to retabulate
the ballots after each round, and that seems to be at the
root of IRV's non-monotonicity. This would also explain why
pairwise single-elimination doesn't seem to be covered.
Richard
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