Completion methods for Smith Sets

[Michael Rouse]

At 06:47 PM 6/18/2001 -0400, you wrote:

>D- Since Mr. Rouse is a new poster, I review some elementary points.
>
Actually, I'm not really a *new* poster, just a really, really, *really* 
infrequent one -- I posted something several months back (can't remember 
what it was offhand, though). I like to read the ideas expressed by the 
others in the group, though, and just wish more Americans would be 
interested in the "nuts and bolts" of how their concerns are reflected by 
the people they elect -- better a fervent IRVinger than a complacent 
Pluralidroid. I have to apologize if I annoy anyone with my sudden 
avalanche of posts or the opinions expressed therein; my only excuse is 
that my enthusiasm is sometimes matched only by my verbosity when I write 
about things that interest me (grin).

 >I repeat two of my earlier postings ---

>1. Vote YES or NO (default) on each choice and use Number Votes (1, 2, etc.)
>on each choice.
>2. Choices getting a YES majority go head to head using the Number Votes.
>3. If there is no Condorcet Winner (CW) using the Number Votes, then drop the
>choice with the least number of YES votes (i.e. the largest number of NO
>votes).
>4. Repeat steps 2 and 3, if necessary.
>
>Approval- Condorcet- Least Approval (ACLA)

I would prefer an inverted method -- run an Approval test on the Smith set 
(which simplifies to the Condorcet winner if available). Both methods are 
likely to choose the same winner, though, and far more often than 
Plurality. I'd vote for your method over Plurality or IRV, though, even if 
I prefer the calculations in a different order. Still, my mind is wide-open 
to a good argument, even if it is almost impervious to the more esoteric 
mathematical proofs. ;-)

Mike Rouse
mrouse at cdsnet.net