[EM] Descending Solid Coalitions Variant for de-cloning Copeland

Forest Simmons forest.simmons21 at gmail.com
Mon May 9 22:02:07 PDT 2022


Let's say that a subset S of candidates is "uninterrupted" on ballot B  iff
no candidate that is not a member of S is ranked between two members of S.

An uninterrupted set that contains some top ranked candidate is "top
tethered." Similarly, a "bottom tethered"
uninterrupted set has at least one candidate that does not outrank any
candidate.

Every "solid coalition" of Woodall is an example of a top tethered
uninterrupted set. Each completely ranked ballot of n candidates has n of
these (non-empty) solid coalitions, as also n bottom tethered uninterrupted
sets, and many more untethered uninterrupted sets. In fact, the total
number of uninterrupted sets on a fully ranked ballot of n candidates would
have to be C(n+1,2) or n(n+1)/2, since it takes two cutoffs to delineate an
uninterrupted set, and there are n+1 slots for those boundary marks.

Let beta be a set of ballots. Then for each subset S of candidates, let
UI(S) be the number of ballots in beta on which S is uninterrupted.

For each ballot B we determine a representative candidate K(B) by
considering the uninterrupted sets S in order of decreasing UI(S).

When a set S is considered, every candidate not in the set becomes
ineligible to represent ballot B, unless this would cause all candidates to
be ineligible, in which case that set is ignored.

When only one candidate is still eligible to represent ballot B, that
candidate is selected as K(B).

For each candidate k, let N(k) be the number of ballots for which k=K(B).

De-cloned Copeland:

Elect the candidate X  with the greatest sum (over those k that do not
defeat X) of N(k).
Does that work?

Really, there are at least three versions ... top tethered, bottom
tethered, and untethered ... not to mention acquiescing variants.

If the descending uninterrupted sets end in a tied set of candidates T(B)
to represent ballot B, then B contributes to each of their N(k) values
1/#T(B).

Is this the right way to adapt Woodall's idea for this context?

-Forest
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