[EM] Completion methods for Smith Sets

Michael Rouse mrouse at cdsnet.net
Sun Jun 17 23:30:50 PDT 2001

Although this e-mail references "Completion Methods for Smith Sets," and 
they cover the same method, this is really about a stand-alone method. It 
has some good qualities: it should always choose the Condorcet winner, if 
there is one; if no Condorcet winner exists, it should always choose a 
winner from the Smith set. It can be used in single seat and multi-seat 
elections since it generates a ranking order. At least I *think* it should 
have those properties, though I'm a bit shaky on my mathematical proofs! ;-)

First, the ballot would be a standard ranked ballot. If a person has a tied 
preference -- say he or she likes two people equally for first place -- 
then each tied candidate gets a partial vote for each equivalent position, 
in this case a half point for first and a half point for second place. If 
someone has three people ranked at fourth place, then each has 1/3 of a 
point for fourth, 1/3 of a point for fourth, and 1/3 of a point for fifth. 
Truncated ballots would be handled the same way: if someone didn't rank 
candidates 8-11, then each would get 1/4 of a point for each place, in this 
case 8th, 9th, 10th, and 11th place. If one member of a tied pair is 
dropped, the remaining candidate(s) get the points. There would be no 
incentive to truncate a ballot or rank your preferred candidate first and 
everyone else last, while there *would* be an incentive (albeit small) to 
rank everyone you have a preference on. Unfortunately, there would still be 
a small incentive to rank you favorite candidate's chief rival at the 
bottom of the list, but that's hard to remove in any ranked ballot.

Now that we have the preference list, find the candidate with the highest 
number of first place votes and put it on another list. Then find the 
candidate with the highest number of first and second place votes combined, 
and if it isn't on the second list put it there. Then find the candidate 
with the highest combined total of first, second, and third place votes, 
and place it on the second list if it isn't already there. If there are tie 
votes and sufficient places remaining on the second list, simply put the 
tied names on the second list; if there aren't sufficient places remaining, 
use a pairwise comparison of the two candidates. If there is still room 
after the first pass, check the second-place "plurality" candidate against 
the list, then the #2 candidate on the combined first and second place 
score list, etc., until the second list has one fewer candidate that the 
first.Redo the preference orders using the remaining candidates, then start 
the process again. Continue until the number of candidates remaining equals 
the number of positions to be filled (in most cases this will be a single 
candidate, but I want to have a complete method no matter how many 
positions are open).

This method is somewhat like a "maximum approval" method -- you take the 
plurality winner, then the top candidate if people approved of their first 
and second choices, then the top candidate if people approved of their top 
three choices, and so on. To give an example, say there were eight 
candidates. You would find the winner with the following point scheme 
(plurality): 10000000. By this I mean all first place votes get one point, 
and all others get zero. Then the first two places get points: 11000000. 
Then 11100000, 11110000, 11111000, 11111100, and 11111110 --a kind of step 
function that moves over one place at a time. Ties can cut short the 
comparison -- it's possible that you might not get to check 11111110 -- but 
winner-repeats on the list extends it. Last place votes are generally the 
only votes that do not ever help.

Of course, this step function can go the other way as well. For a similar 
method with slightly different properties, invert the process: Find the 
candidate with the greatest number of last place votes and put it on a 
list; in other words, check antiplurality (00000001 in the above example). 
Then find the candidate with the greatest number of last and second-to-last 
combined votes (00000011). Continue until you have one less candidate than 
before, then start over. The final candidate is the loser candidate. Drop 
it from the list, redo the ranks of all of the other candidates, then start 
again. Stop when you have the appropriate number of candidates left.

One could even do both: during the calculations, calculate the "loser" 
using the standard and inverse method. If both agree, then drop that person 
and continue calculation with both methods. If they disagree, drop the 
pairwise loser between the two, recalculate the ranks, and continue the 
process until the appropriate number of candidates remain. My overall 
preferred method would be to find Condorcet winner, if present; otherwise, 
use this combined method on the Smith set to complete the rankings and 
generate the winner (unless it can be shown that this method alone is 
equivalent, in which case we can drop Condorcet altogether). It's 
complicated, but computer time is cheap, and if it does not have too many 
glaring flaws and generates a result generally recognized as fair, then 
that's all we can hope for under Arrow's theorem.

Because it tosses away losing candidates and recalculates, this method is 
not monotonic. My guess is that such ill-behavior will be rare, though, and 
it since it chooses the Condorcet winner (if any exists) or a member of the 
Smith set (if the Condorcet winner doesn't exist), it does have some 
attractive qualities plain Approval (let alone Borda, IRV and Plurality!) 
does not. I'm not sure what other deficiencies it might have, other than 
requiring a lot of calculation. If anyone has an example of ill behavior 
and/or voting strategies that this system allows, I'd love to see them -- 
the kinds of errors it generates let me know if the fault is in the method 
itself, or in non-rational circular ties.


Michael Rouse
mrouse at cdsnet.net

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