[EM] Completion methods for Smith Sets
Michael Rouse
mrouse at cdsnet.net
Sun Jun 17 23:56:16 PDT 2001
At 03:27 PM 6/18/2001 +1000, you wrote:
>Hi Michael,
>
>Is this what you mean?;
>
>1) Find the candidate with the most first place votes. That candidate is
>ranked #1
>2) Find the candidate with the most first and second place votes. If that
>candidate is different to the candidate ranked #1, then that candidate is
>ranked #2.
>3) Continue process (with 1st, 2nd & 3rd place votes, then adding 4th place
>votes) until you have ranked all of the candidates but one. Eliminate that
>candidate.
>4) Adjust the rankings so that 2nd preferences of the eliminated candidate
>becomes 1st preferences etc. Return to step 1) until there is only one
>candidate remaining.
You've got it! -- and clearer than my explanation (grin). I really need to
start using bullets or numbers or Roman numerals. I've e-mailed a second
description, including adding an inverse method (find the worst candidate
and drop it) and a combined method (find the worst candidate two different
ways, tie-break discrepancies, drop the worst one, then repeat). Feel free
to rewrite my description there, especially to clarify it :-)
>If the inner and outer set are the same, those candidates are elected, and
> >we are done. If they are different, the ones in the inner set are
> >automatically elected, and the ones between the inner and outer set are run
>
> >through the single-member method above, dropping one candidate at a time
> >until the number of candidates is reduced to its final number.
>
>So this method can't always elect a fixed number of candidates? This might
>be a problem.
Actually, the method by itself can elect a fixed number of candidates,
which is why it would make a good completion rule for Condorcet winning
groups. Say you might have 5 positions to fill. You find the Smith set with
three candidates and the next "winning" Condorcet set (where no candidate
from outside the set can win against a candidate inside the set) of six
candidates. The Smith set candidates should fill the first three spots, and
you can use the method above on those remaining in the winning set to find
out who should fill the final two spots. As a side note, I think the
combined standard/inverse method for "completing" Condorcet groups (you
might check my other note for a description of both) would yield a method
that is generally fair and well-behaved most of the time, without too many
illogical or strange results.
Michael Rouse
mrouse at cdsnet.net
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