[EM] Completion methods for Smith Sets

[Michael Rouse]

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