[EM] Advanced Placement Chain Climbing
Forest Simmons
forest.simmons21 at gmail.com
Tue Jul 5 21:52:39 PDT 2022
I'm setting aside Ranked Pairs Chain Building for awhile to keep my head
from exploding, and to explore an AP Chain Climbing idea that is less apt
to overload my little brain.
Customary chain Climbing starts at the least promising end of the agenda
(like traditional Sequential Pairwise Elimination does) and climbs as far
as possible towards the promising end ... stopping only when no candidate
outside the chain defeats every chain member.
Working from less promising to more promising is the simplest way we know
of to achieve monotonicity. But it seems like if we started part way along
the path to the promising end, we might make it higher up the ladder, and
perhaps get more Voter Satisfaction Efficiency.
How far along?
As far as possible as long as the resulting chain cannot be dominated by
any agenda member outside of the chain.
Suppose the agenda is from least to most approval C<B<A.
If the cyclic defeat order is ABCA and the chain climbing starts at C, then
the resulting chain is C<B. The remaining agenda item A cannot be added
because it is defeated by C.
If on the other hand, we start at B, then the resulting chain is B<A, which
still has the maximal property since the candidate C (the only agenda
member outside of the chain) cannot be added since it is defeated by B,
which is a member of the chain.
What if the cyclic defeat order is CBAC?
Then starting the climb at C results in the chain C<A, which is maximal
because B is defeated by C.
If we started an AP chain climb at B, then B would be the only member of
the chain, which would not be maximal, because C defeats B.
So the AP method yields maximal chains A>B and A>C in the two respective
cyclic orders. When the top cycle has just three members, the AP Score
Chain Climbing method will yield the top score member of the cycle.
The method seems to be monotonic.
If so, we have an excellent contender for the STAR challenge ... AP Score
Chain Climbing.
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