[EM] Manual Construction of Smith Set

[Forest Simmons]

Single Pairwise Elimination has been done this way for centuries:

Line up the candidates on the stage in order of Score, lowest score on the
left.

At each stage the head-head loser between the two left most candidates is
dismissed and the remaining candidates close ranks.

In small groups the head-head decisions can be resolved by acclamation...
confirming with a hand count when close.

In large groups the previously tabulated head-head results are posted
alphabetically by name of pairwise victor before the elimination show
begins.

For a deluxe show, do Score Sorted Margins:

While there are any adjacent candidates out of order pairwise, among these
swap the members of the pair that are closest in score (i.e.larger minus
smaller score difference closest to zero).



El jue., 27 de ene. de 2022 12:13 p. m., Richard, the VoteFair guy <
electionmethods at votefair.org> escribió:

> On 1/27/2022 2:30 AM, Kristofer Munsterhjelm wrote:
>  > So I disagree. I don't think simplicity requires you to
>  > throw away Smith compliance.
>
> I agree.  Yet remember that my goal is to make it easy for voters to
> watch vote counting on a stage and understand what's going on.
>
> Kristofer suggests River and some other mathematically good methods, but
> all of them require using the pairwise counts in ways that are not as
> easy to understand as asking "which number is bigger?, that's the winner
> in this pair."
>
> Yes Copeland is simple, but ...
>
> The popularity of IRV has shown that vote counting is easier to
> understand when candidates are eliminated one at a time.  Most watchers
> won't trust a process that ends so quickly, and fails to also
> (sequentially) reveal who was least popular, who was second-least
> popular, etc.
>
> In a different message ... Kristofer Munsterhjelm wrote:
>  > So "on stage", Smith//IRV may not be a good method. I don't
>  > think that IRV itself would be a good method for on-stage
>  > voting, but if you had to do it then I would imagine either
>  > Benham (if you need strategy resistance) or BTR (otherwise)
>  > would be better than Smith//IRV.
>
> Actually one of the big selling points of IRV is it can be done on stage
> simply by piling ballots into stacks.  The stacks are named by
> candidate, and the ballot goes to the stack of that ballot's currently
> highest-ranked candidate.  The height of the stacks make some
> comparisons very clear.  When the heights are similar, simple counting
> can confirm which stack has fewer ballots.
>
> Near the end of the process the tallest stack can be counted to see if
> it contains more than half the ballots.  For added clarity, the ballots
> in all the other stacks can be counted to show there are fewer ballots
> in all those other stacks.
>
> With IRV it's even easier -- but not fairer -- (than other methods)
> because after each elimination the only ballot stack that needs to be
> re-allocated is the stack that supports the candidate just eliminated.
>
> I'll repeat that I strongly dislike IRV!!!  So I'm not defending it!
>
> I'm just trying to point to something better that also can be counted in
> a way where watchers can easily verify the (relative) fairness of each
> step.
>
> When a method requires subtraction -- such as for calculating margins --
> or sorting pairwise counts, the ease of understanding disappears.  (Yes
> subtracting can be done by hand on a white board, but following the
> subtraction process many times would make some watcher's head hurt. Yes
> a calculator is usually trustworthy, but a watcher doesn't want to watch
> lots of numbers being entered on the calculator and then transcribed
> onto signs, and then named, etc.  Yes sorting can be followed, but it's
> not easy to trust if the process also involves something else going on
> too.)
>
> Yes, BTR-IRV does fit the requirement of being better, but IMO it's not
> good enough.  It elects the Condorcet winner, but in a way that would
> prompt lots of voters to ask "why are we protecting the Condorcet winner
> when it has the shortest stack of ballots?"
>
> In contrast, it's usually easy to follow eliminations that are based on
> something that's easy to understand, such as eliminating a candidate who
> loses every one-on-one match against the remaining candidates.
>
> I believe such methods exist.  They may not have established names.
> They must be easy for non-math-savvy watchers to follow.  Yes they won't
> have superb mathematical characteristics.  But IRV too easily yields
> flawed results, so getting better results shouldn't be difficult to
> achieve, even with the can-be-followed-on-stage constraint.
>
> Richard Fobes
> The VoteFair guy
>
>
> On 1/27/2022 2:30 AM, Kristofer Munsterhjelm wrote:
> > 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
> >
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