Re: [EM] Are LIIA, majority, determinism, and neutrality incompatible?
The centrifugal margins solutions to this election seem like a
reasonable counterexample to me.
These are distributions over potential orderings. Alternatively, you can
think of these as if the method sometimes returns tied orderings it was
indifferent between and throws the decision to the tiebreaking mechanism
(in other words, the tiebreaking becomes external to the method instead
of internal).
{A, B, C}: 50% A > C > B, 50% C > B > A
{A, B}: 50% A > B, 50% B > A
{A, C}: 50% A > C, 50% C > A
{B, C}: 100% C > B
Centrifugal margins satisfies majority and Smith (the latter in the
"successive Smith set" way that ranked pairs does). I'm not sure whether
or not it always satisfies LIIA, though.
On 1/23/25 08:32, Kristofer Munsterhjelm wrote:
Someone check my reasoning here, because I might have found out why
LIIA methods like ranked pairs are so intertwined with their tiebreakers.
Suppose we have a method that passes majority and also LIIA. (It thus
has to pass Smith and ISDA.) Consider the following election:
4: A > B > C
5: A > C > B
1: B > A > C
4: B > C > A
1: C > A > B
5: C > B > A
A ties both B and C pairwise, and C beats B: A=B, A=C, C>B.
As a post-script, I realized at the last minute that you mentioned
determinism in the subject. The fact that my counterexample relies on a
distribution might actually be another piece of evidence in favor of
your argument.
On 1/30/25 8:51 PM, Joshua Boehme wrote:
The centrifugal margins solutions to this election seem like a
reasonable counterexample to me.
These are distributions over potential orderings. Alternatively, you can
think of these as if the method sometimes returns tied orderings it was
indifferent between and throws the decision to the tiebreaking mechanism
(in other words, the tiebreaking becomes external to the method instead
of internal).
{A, B, C}: 50% A > C > B, 50% C > B > A
{A, B}: 50% A > B, 50% B > A
{A, C}: 50% A > C, 50% C > A
{B, C}: 100% C > B
Centrifugal margins satisfies majority and Smith (the latter in the
"successive Smith set" way that ranked pairs does). I'm not sure whether
or not it always satisfies LIIA, though.
On 1/23/25 08:32, Kristofer Munsterhjelm wrote:
Someone check my reasoning here, because I might have found out why
LIIA methods like ranked pairs are so intertwined with their tiebreakers.
Suppose we have a method that passes majority and also LIIA. (It thus
has to pass Smith and ISDA.) Consider the following election:
4: A > B > C
5: A > C > B
1: B > A > C
4: B > C > A
1: C > A > B
5: C > B > A
A ties both B and C pairwise, and C beats B: A=B, A=C, C>B.
On 2025-01-31 02:56, Joshua Boehme wrote:
As a post-script, I realized at the last minute that you mentioned
determinism in the subject. The fact that my counterexample relies on a
distribution might actually be another piece of evidence in favor of
your argument.
Yes; my hunch is that the problem arises from ties at the top or bottom
adding too many constraints, and they can't all be satisfied.
E.g. having A=B>C>D means that, in addition to the bottom-end removal of
D, we must preserve the order both when A is removed and when B is
removed. But if the tie is broken beforehand, it only needs to stay
consistent with removing one of the candidates on the top end, not both
at once.
This could have more serious theoretical implications based on what's
possible to do with tiebreakers. For instance, I don't know if there
exists a summable cloneproof LIIA tiebreaker. The "random voter
hierarchy" tiebreaker is, as far as I know, not summable. Strictly
speaking, that means full ranked pairs isn't summable either (at least
when using RVH).
In practice, I don't think it's going to cause too much trouble for LIIA
methods, since large elections would probably have distinct counts for
every pairwise contest, so they'll be decisive. But it's a bit annoying.
It would be nice to have proof of whether a LIIA cloneproof tiebreaker
could be constructed, but I don't know where I would begin :-)
-km