[EM] Monotonic Burial Resistant Method

[Kristofer Munsterhjelm]

On 7/6/23 22:30, Forest Simmons wrote:

> Then  the MMPO method would have detected & restored C as the sincere CW 
> in the A vs C sincere final ... while our new method would fave still 
> elected B ... punishing the burier A by electing its sincere antifavorite.
> 
> The equality in the B>A=C cycle confounds things a little with two 
> equally weak links B>C and C>A.

Unless it's due to a bug in my linear program, there seems to be 
something about the structure that makes it impossible to get B elected 
after burial both by this new method and by (pre-tiebreaker) minmax. I 
suspect that it's a matter of too few degrees of freedom and that thus 
it'd be possible to produce such an election with four candidates.

It's possible to make the ordering strict, though, e.g.

4: A>B>C
1: B>C>A (honest: B>A>C)
3: C>A>B
1: C>B>A

which should give a result of B>C>A.

> How about the monotonicity and Landau Efficiency of the new method?

I'm working on a post showing how to use differentiation to determine 
monotonicity, with this method as an example. But the short answer is: 
no, it's not monotone. (I don't think so, at least.)

Suppose C and A have an equal number of last preferences, so that A and 
B are tied. Then raising A on a B>C>A ballot can increase C's last 
preferences, helping B and making him the argmin candidate:

1: A>B>C
1: B>A>C
2: B>C>A
3: C>A>B

Penalties:
	A = lpC = 2
	B = lpA = 2
	C = lpB = 3

so A=B>C.

Now raise A on a B>C>A ballot:

1: A>B>C
2: B>A>C
1: B>C>A
3: C>A>B

Penalties:
	A = lpC = 3
	B = lpA = 1
	C = lpB = 3

B > A = C.

You can probably add strategic epsilons to the ballot counts to break 
the outcome ties; I haven't done so here.

It's kind of the reverse of the IRV problem where raising someone 
decreases someone else's first preference count, so presumably there's 
an analog of fpA-fpC that restores monotonicity. But then you could just 
use fpA-fpC itself!

-km