[EM] Easy fix to Alaska's ranked-choice voting

[Forest Simmons]

It may surprise you that given the same agenda, SPE  (Sequential Pairwise
Elimination) and Highest Uncovered are extremely hard to differentiate.

There has probably never been an actual ballot set where they would elect a
different candidate from the same agenda order.

Certainly none of Kevin's seven taxonomical examples could do it because
they all involve Smith sets of only three members ... in which case it is
easy to show that the two methods will always agree.

Here 's the demonstration:

First, both methods are ISDA.

SPE is IPDA because nonSmith candidates have no influence on the pairwise
sorting of Smith, which always comes out solid ahead of the other
candidates no matter what the sorting priority rule ... top, bottom,
margins, etc.

Highest Uncovered is ISDA because only Smith members can be uncovered ...
so electing the highest uncovered agenda item is the same as electing the
highest uncovered Smith item on the agenda.

This brings us the next point: when there are only three Smith members, all
of them are uncovered. This is because all Smith members have beatpaths to
all other Smith members, and when there are few Smith members those
beatpaths must be short (or unnecessarily repetitious)... in this case only
one or two steps are possible without unnecessary repetition.

The only remaining questionis which member of a three member Smith does SPE
elect.

Suppose the three members are  X , Y and Z,  and that their cyclic order is
XYZX. If the agenda order is in agreement with cyclic order, pairwise
sorting will not change it, so the highest agenda Smith candidate is the
SPE winner.

What if the agenda order is not the pairwise cyclic order?

Then, since SPE goes from bottom to top, it will switch the order of the
bottom two Smith members. That switch puts Smith into cyclic order so no
further change in the order of Smith. The highest agenda Smith candidate
gets elected, as in Highest Uncovered.

In sum, when Smith has three members, SPE and Highest Uncovered both elect
the top Smith candidate in the sgenda.

Since SPE and Highest Uncovered are both equally easy to compute, the main
question is what is the best agenda?

Colin proposed basing the agenda on first place votes, but that is subject
to note splitting (our only good reason for abandoning simple, traditional
FPP Plurality) Here's an example that shows the vote splitting problem:

21 A1>B>C
19 A2>B>C
35 B>C>A1
33 C>A2>A1

The agenda order is B>C>A1>A2.

B is both the SPE winner and the Highest Uncovered agenda winner, as well
as the winner of Ranked Pairs, River, Schulze, and many other methods
including IRV.

If A had not been split into A1 and A2, the agenda would have been A>B>C,
and the winner of all the aforementioned methods would have been A.

So that's why I suggested Implici Approval: If they are not going to allow
us to rank more than three candidates, then the simplest agenda method that
avoids vote splitting is the implicit approval agenda ... the candidates
most favored by the agenda are the ones unranked on the fewest ballots.
Since you cannot rank all of the candidates leave the candidates that least
approve of unranked.

Although that's the simplest acceptable agenda in this context (of forced
truncations) there may be another that is worth the extra complexity.

-Forest


On Sat, Nov 12, 2022, 12:00 AM Forest Simmons <forest.simmons21 at gmail.com>
wrote:

> Evidently the IRV proposers hacve had to settle for ránking only there or
> tour candidateson each ballot ... better than nothing.
>
> One pass through the ballots to get the pairwise information and the
> number of truncations for each candidate ... the exact same work as the SPE
> method you propose ... but an agenda of first place votes breeds lots of
> vote splitting unless you expect the voters to have lots of equal first
> rankings ... not a good idea.
>
> Electing the uncovered candidate unranked on the fewest ballots is a
> simpler Condorcet method than SPE ... and it is guaranteed to elect an
> uncovered member of the Smith Set.
>
> But SPE is also good ... if the agenda is clone independent, like implicit
> approval.
>
> -Forest
>
>
> On Fri, Nov 11, 2022, 5:26 PM Colin Champion <
> colin.champion at routemaster.app> wrote:
>
>> On 11/11/2022 20:49, Forest Simmons wrote:
>> >
>> > Since almost all RCV implementations limit the number of candidates
>> > that can be ranked on a ballot, the simplest decent  RCV method is ...
>> > Elect the uncovered candidate that is unranked on the fewest ballot
>> >
>> Does anyone know why this truncation is imposed? If it’s to limit the
>> amount of work needed to count the ballots, wouldn’t it make sense for
>> Condorcet supporters to advocate a method which was countable in linear
>> time? In practice this would presumably be Sequential Pairwise
>> Elimination with an FPTP pre-ranking. If you insist on a quadratic time
>> method and accept the corollary of ballot truncation, I don’t imagine it
>> will work very well. Or am I missing something?
>>
>> CJC
>> ----
>> Election-Methods mailing list - see https://electorama.com/em for list
>> info
>>
>
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