[EM] Manual Construction of Smith Set

[Kristofer Munsterhjelm]

On 26.01.2022 20:30, Richard, the VoteFair guy wrote:
> My conclusion is that ALWAYS eliminating EVERY non-Smith-set candidate
> is too difficult to do as a visible process on an auditorium stage.  As
> Kristofer says:
> 
> On 1/16/2022 10:31 AM, Kristofer Munsterhjelm wrote:
>> But you still need to decide when to stop, which can be pretty
>> difficult. At first it'd seem to be obvious: if the bottom-most
>> candidate beats anyone else pairwise, he'd be in the Smith set, no?
>> But that doesn't quite work. Consider something like this: ...
> 
> It's much easier, and almost as "good," to eliminate pairwise losing
> candidates as they occur during elimination rounds.
> 
> As a reminder, a "pairwise losing candidate" is the candidate who would
> lose every one-on-one match against each and every remaining
> (not-yet-eliminated) candidate.
> 
> That would eliminate MOST, but not all, non-Smith-set candidates in MOST
> cases.
> 
> Thanks to those who helped get us close to a process that could be
> understood by an audience of typical voters.

The problem is, of course, that there's not always a Condorcet loser.
And so you may eliminate a Smith set member because the method's
fallback metric fails to exclude them from elimination.

Meanwhile, both the "move to the right of the line" and the sorting
proposals will succeed in identifying the Smith set. And they're both
simple, IMHO, so I would disagree with your conclusion.

If you need only a single winner, there are plenty of alternatives that
pass Smith. Copeland elimination seamlessly handles the pairwise loser
elimination step, because a Condorcet loser is also a Copeland loser;
and since Smith set members have above average Copeland score whenever
non-Smith members exist, you'll eliminate every non-Smith member before
you eliminate every Smith member. It shouldn't be too difficult to
phrase, either: instead of "eliminate the candidate who loses every
pairwise contests", it's "eliminate the candidate who loses the most
pairwise contests".

Or Forest's sequential pairwise elimination, which in a sense mimics
legislative procedure and so should also be pretty easy to explain.
First set up an ordering, then go up it and say "does the next proposal
beat my current proposal? If so, switch it out, otherwise keep my
current proposal".

Or even BTR. If your elimination procedure, instead of saying "eliminate
the worst by some measure X" says "eliminate the candidate of the second
worst and worst who beats the other one pairwise". This also passes Smith.

So I disagree. I don't think simplicity requires you to throw away Smith
compliance.

-km