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Technical discussion of election methods

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A sorta IFPP-like method

KM
Kristofer Munsterhjelm
Tue, Jul 11, 2023 11:24 PM

Here's an idea of a possible method inspired by my MCAB and IFPP:

Say that a set of candidates is eliminable if their combined first
preference count is less than or equal to a third of the total number of
voters, but adding any candidate would make the total exceed 1/3.

For one candidate: A's score is the number of voters.
For two candidates: A's score is A>B.
For three or more: A's score is the maximum score obtained by choosing
some eliminable set, eliminating it, and recursing. The set is chosen so
as to maximize A's score.

The idea here is that suppose A is a DMT candidate, but burial reversed
A>C into C>A. Then it's likely that A>C was weaker than A>B, so that A's
score was A>B. So A's score isn't affected by the burial. Suppose X is
not a DMT candidate. Then A must be retained because just adding A would
push the set over capacity. So X's score must be X>A which (given the
assumption above that C>A is weak) must be weak; either A beats X or X
is C and X just barely beats A pairwise.

That the set is maximizing A's score should (hopefully?) retain
monotonicity because shuffling candidates who are part of an eliminable
set around won't affect the outcome. Maybe there's a possibility for
nonmonotonicity involving some other candidate going out of or into the
eliminable set? But I can't quite see it.

So yeah, this is all kinda sloppy, but I thought the idea would be
interesting. The recursion would make it a pain to program, though;
maybe dynamic programming tricks can be used to make it polytime.

-km

Here's an idea of a possible method inspired by my MCAB and IFPP: Say that a set of candidates is eliminable if their combined first preference count is less than or equal to a third of the total number of voters, but adding any candidate would make the total exceed 1/3. For one candidate: A's score is the number of voters. For two candidates: A's score is A>B. For three or more: A's score is the maximum score obtained by choosing some eliminable set, eliminating it, and recursing. The set is chosen so as to maximize A's score. The idea here is that suppose A is a DMT candidate, but burial reversed A>C into C>A. Then it's likely that A>C was weaker than A>B, so that A's score was A>B. So A's score isn't affected by the burial. Suppose X is not a DMT candidate. Then A must be retained because just adding A would push the set over capacity. So X's score must be X>A which (given the assumption above that C>A is weak) must be weak; either A beats X or X is C and X just barely beats A pairwise. That the set is maximizing A's score should (hopefully?) retain monotonicity because shuffling candidates who are part of an eliminable set around won't affect the outcome. Maybe there's a possibility for nonmonotonicity involving some other candidate going out of or into the eliminable set? But I can't quite see it. So yeah, this is all kinda sloppy, but I thought the idea would be interesting. The recursion would make it a pain to program, though; maybe dynamic programming tricks can be used to make it polytime. -km
KM
Kristofer Munsterhjelm
Wed, Jul 12, 2023 10:00 AM

On 7/12/23 01:24, Kristofer Munsterhjelm wrote:

Here's an idea of a possible method inspired by my MCAB and IFPP:

Say that a set of candidates is eliminable if their combined first
preference count is less than or equal to a third of the total number of
voters, but adding any candidate would make the total exceed 1/3.

For one candidate: A's score is the number of voters.
For two candidates: A's score is A>B.
For three or more: A's score is the maximum score obtained by choosing
some eliminable set, eliminating it, and recursing. The set is chosen so
as to maximize A's score.

Oh, I forgot to say: if there doesn't exist any eliminable sets without
A in it, then A's score is 0. For instance, if A is ranked last by
everybody, no matter who's left in the running, every eliminable set
will contain A.

-km

On 7/12/23 01:24, Kristofer Munsterhjelm wrote: > Here's an idea of a possible method inspired by my MCAB and IFPP: > > Say that a set of candidates is eliminable if their combined first > preference count is less than or equal to a third of the total number of > voters, but adding any candidate would make the total exceed 1/3. > > For one candidate: A's score is the number of voters. > For two candidates: A's score is A>B. > For three or more: A's score is the maximum score obtained by choosing > some eliminable set, eliminating it, and recursing. The set is chosen so > as to maximize A's score. Oh, I forgot to say: if there doesn't exist any eliminable sets without A in it, then A's score is 0. For instance, if A is ranked last by everybody, no matter who's left in the running, every eliminable set will contain A. -km