[EM] STAR Challenge
Forest Simmons
forest.simmons21 at gmail.com
Mon Jun 27 16:21:51 PDT 2022
Let's start with a list L of candidates listed from top to bottom in order
of their score totals ... high scores listed above low scores
Initialize a second list L' with the the top member of L at its head.
Then incorporate the next member of L into L', above or below its initial
member depending on whether or not the new member of L' is scored above its
initial member on more ballots than not.
So far the list L' is the Score Then Automatic Runoff "STAR" finish order.
In general, while some member X of L can be incorporated into L' without
defeating any member of L' listed above it and without being defeated by
any member of L' listed below it, so incorporate the highest such X from L
into the list L'.
Then elect the candidate that ends up at the top of L'.
Example:
Suppose the score order list L is
S1
S2
S3
and that the pairwise defeats are
S3 defeats S2 defeats S1defeats S3.
Then L' is initialized as S1, and S2 is incorporated above S1. So far the
list L' is
S2
S1
Can S3 be incorporated?
Not at the head, because that would entail defeat by a lower listed
candidate S1.
Not at the middle or bottom because then it would defeat a candidate (S2)
listed above it.
So the completed list L' is
S2
S1
The head S2 is elected from the top of the list L'.
Since the method is independent from Smith dominated alternatives, and
since Smith is highly unlikely to have more than three members, we can
easily boil it down to three cases (barring unlikely ties):
I. Smith has one member: Then that member will be the head of L'.
II. The cyclic order counters the score order as in the above example. As
we have just seen, the Smith member with the 2nd highest score is elected
in this case.
III. The cyclic defeat order does not counter the score order. In this
case it is easy to see that the first two members of L' are S1 and S2, and
that S3 cannot be incorporated. So the head of L' is the score winner S1.
So we see that in public elections this (Banks!) method reduces to the
simple rule "Elect the lowest score candidate that defeats every higher
score candidate," which is another formulation of Score,Benham as well as
Ranked Pairs, River, and BeatPath,etc when defeat strength is measured by
winning score.
Which of these equivalent formulations will be easiest to sell?
I mention Banks only to add prestige to the method, and to show how to
generalize it to specialized, non-public elections that are likely to have
more interesting Smith structure (twisted prism, etc).
-Forest
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