[EM] Completion methods for Smith Sets

LAYTON Craig Craig.LAYTON at add.nsw.gov.au
Sun Jun 17 22:27:44 PDT 2001

Hi Michael,

>Once you have the candidates in the Smith set, rank them, starting with the

>candidate that has the highest number of first place votes (the plurality 
>winner), then the winner of the combined first and second place votes, then

>the winner of the first, second, and third place votes together, continuing

>until you have one fewer candidate on the list than you started with. Don't

>count last place rankings; if there are remaining places on the "winner" 
>list, start with the second candidate on the "plurality" list and continue 
>down the line. Ties are ok unless there is one spot remaining; in that 
>case, choose the pairwise winner between the two candidates. Throw out the 
>loser, adjust the rankings, and begin the process again. Continue until you

>have one candidate.

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
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.

I'm not sure about my step 2).  The alternatives would be to adjust the
rankings, taking the #1 ranked candidate out of the race, or to find the
candidate with the most first and second place votes OTHER THAN the #1
ranked candidate.  Could you clarify this?

>For a multi-candidate Condorcet completion method, find the largest inner 
>unbeaten set and the smallest unbeaten outer set that encloses the number 
>of positions available -- in other words, C<=N for the inner set (where C 
>is the number of candidates, and N is the number of positions in the winner

>set) and C>=N for the outer set.
>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.

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