A more elegant three-candidate fpA-fpC phrasing, inspired by Heaviside formulation
For three candidates: Let A be the current candidate, then over all
other candidates B who A beats, add 2 fpA + fpB to A's score. Highest
score wins.
This works because in an A>B>C>A three-cycle, fpA - fpC = |V| + fpA -
fpC where |V| is the number of voters, because we're just adding a
constant. And |V| = fpA + fpB + fpC, so fpA - fpC = fpA + fpB + fpC +
fpA - fpC. The fpCs cancel and we get 2 fpA + fpB.
Now suppose that A is the Condorcet winner. Then A's score is 4 fpA +
fpB + fpC, i.e. 3 fpA + |V|. This exceeds the score any of the other
candidates can obtain, and so this version of fpA-fpC still passes
Condorcet.
In the case of a two-way tie, if we use the symmetric step function,
H(0) = 0.5, then we should have A's score be fpA + fpB/2 in such a case.
This would in effect break a two-way tie in favor of the Plurality
winner of the two.
f(A, e) = sum over B != A: H(A>B) * (2 fpA + fpB).
Unlike IRV, here the H term is a pairwise one. This makes it hard to
construct a recursive version. Now we could of course just generalize
the "sum over all B that A beats" concept to more than three candidates,
but I don't think the resulting method would pass DMTBR, and I also
think it would fail clone dependence (vote splitting with a cycle among
the clones).
-km
On Thu, Jan 20, 2022 at 5:21 AM Kristofer Munsterhjelm km_elmet@t-online.de
wrote:
f(A, e) = sum over B != A: H(A>B) * (2 fpA + fpB).
Unlike IRV, here the H term is a pairwise one. This makes it hard to
construct a recursive version. Now we could of course just generalize
the "sum over all B that A beats" concept to more than three candidates,
but I don't think the resulting method would pass DMTBR, and I also
think it would fail clone dependence (vote splitting with a cycle among
the clones).
How does H being pairwise make it hard to write a recursive version? Let's
try the easiest form of recursion:is:
f(A,e) = sum over B!=A: H(A>B) * f(A,e/B)
where "e/B" is the election you get if you remove candidate B.
I tried to analyze this, but I didn't get very far. I think I figured that
it's monotone. I struggled to prove whether it's clone independent. I don't
know how to begin investigating DMTBR.
--
Dr. Daniel Carrera
Postdoctoral Research Associate
Iowa State University
On 20.01.2022 14:12, Daniel Carrera wrote:
On Thu, Jan 20, 2022 at 5:21 AM Kristofer Munsterhjelm
<km_elmet@t-online.de mailto:km_elmet@t-online.de> wrote:
f(A, e) = sum over B != A: H(A>B) * (2 fpA + fpB).
Unlike IRV, here the H term is a pairwise one. This makes it hard to
construct a recursive version. Now we could of course just generalize
the "sum over all B that A beats" concept to more than three candidates,
but I don't think the resulting method would pass DMTBR, and I also
think it would fail clone dependence (vote splitting with a cycle among
the clones).
How does H being pairwise make it hard to write a recursive version?
Let's try the easiest form of recursion:is:
f(A,e) = sum over B!=A: H(A>B) * f(A,e/B)
where "e/B" is the election you get if you remove candidate B.
I tried to analyze this, but I didn't get very far. I think I figured
that it's monotone. I struggled to prove whether it's clone independent.
I don't know how to begin investigating DMTBR.
I think you're right about it being monotone.
The problem with pairwise preferences is that at the outset, they're
manipulable by burial. Suppose A wins and voters who prefer some other B
bury A under C. Then they can manipulate A>C to their advantage for any
other C. However, they can't manipulate any of the first preference
counts by burial, because they're already voting B>A, so their first
preference isn't A.
(They could use compromising or a more general strategy to alter fpC for
some C that's neither A nor B.)
At the same time, if we're going to Smith-complete the method later on
and want this to pass DMTBR, the method itself should pass DMT: since
Smith implies DMT, then otherwise you could have a situation where
there's a CW who automatically wins and is part of the innermost DMT
set; then creating a cycle exposes an instance of the underlying method
where it fails DMT, which thus means that the combined method fails DMTBR.
The easiest recursive way to do DMT and DMTBR is to ensure that if, at
any point, there's only a remaining candidate that is a CW and has >1/3
first preferences, that candidate doesn't get eliminated - or more
generally, always has a higher score than other candidates. Using
pairwise counts makes that quite difficult. (The alternative would be to
use solid coalitions somehow, I think.)
But granted, your suggestion only uses A>B once (and then B is
eliminated), so it might work since B>A buriers can't affect the
magnitude of A>B anyway. On the other hand, the contribution of H(A>C) *
f(A, e/C) to A's sum might make burial possible. It's hard to tell... I
really need an easy to program simulator, or to proceed in a more
rigorous fashion. Perhaps both.
(Making it harder still is the fact that no Condorcet method is
absolutely unburiable. So the "greater than 1/3 first preferences" must
somehow come into play, and just saying "can't look at A vs C when
considering whether A or B should win" idea can't work, because if that
were possible, the Condorcet method would be unburiable ouright, which
is impossible. Or put another way: when a Condorcet method checks if A
is the CW, it must clearly look at both A>B and A>C: nothing can escape
that.)
So I guess I should be more clear: I didn't mean that making some
recursive method would be hard, but that getting a recursive formulation
close to DMTBR would be hard. But at least I'm feeling a sense of
progress! The recursive function phrasing is useful.
-km