[EM] Agenda Based Banks

[Kristofer Munsterhjelm]

On 04.08.2021 00:29, Susan Simmons wrote:
> Actually, in the burial example below, ABB performs strictly better than
> DMC because under DMC, the burial of C by A either rewards A with a win,
> or leaves intact A's sincere 2nd choice as winner ... a zero risk gambit.
> 
> By way of contrast, if A's gambit fails under ABB, then A's least
> preferred, B, wins ... a big risk, because A's raising of B enough to
> introduce the necessary cycle might also be enough to move B to the
> necessary agenda level (if it was not already there).
> 
> In summary, ABB has both the best burial resistance and the best chicken
> attack resistance of any agenda based Condorcet method that we know of
> (and we know of oodles!).
> 
> It seems doubtful that any simpler, monotone, clone free, ISDA, burial
> and chicken resistant method exists ... except perhaps Asset Voting,
> which is too far ahead of its time (and our time ...first promoted only
> 150 years ago by Charles L Dodgson) even for Steven J Brams the great
> academic promoter of Approval, another simple method too far ahead of
> its time.

As the perennial cold shower man, I must ask if this is actually
monotone for all monotone base methods :-)

Suppose that the base method is Plurality. In this election:

5: A>B>C
2: A>C>B
6: B>C>A
1: C>A>B
1: C>B>A

there's an ABCA cycle and the Plurality order is A>B>C. If I understand
correctly, ABB would first deposit A and then find the candidate that
beats A pairwise, i.e. C, and then terminate, giving C as the winner.

Now let 3 of the A>B>C voters raise C to C>A>B:

2: A>B>C
2: A>C>B
6: B>C>A
4: C>A>B
1: C>B>A

Now the Plurality order is B>C>A. B is first deposited, and then the
candidate who beats B (i.e.) A is deposited and wins.

So raising C made C lose. Looks like we need strong mono-raise.

With Plurality as the base method, I think ABB reduces to Eivind
Stensholt's BPW - or in my ABCA jargon, the method where candidate A's
score is "-fpC". If we just had a way of making it fpA-fpC, then all
would be well, at least in the case of Smith set equals three!

(Here's an ugly hack: The inner procedure takes a "tentative first
candidate" A. It inserts A, then it inserts the candidate that beats
everybody in the list, whose [first preference count - the last inserted
candidate's first preference count] is maximized. The inner procedure is
over once nobody beats everybody inside pairwise; the candidate last
inserted is the winner, and his score is the (first pref - last guy's
first pref) count.[1]
The outer procedure runs the inner procedure with every candidate as
tentative first, and elects the winning outcome with the greatest score.
But it's extremely ugly and I have no idea if it's strategy resistant or
monotone beyond a Smith cycle of three.)

In general, it seems that in a three-cycle, ABB elects the candidate who
beats the winner of the base method pairwise.

-km

[1] One could imagine other score functions as well, as long as they
specialize properly. E.g. Q's score is min P: fpQ - fpP, where P is a
candidate from the current chain. But I suspect that properly
generalizing fpA-fpC (so that it preserves DMTBR) will require some kind
of solid coalition logic -- or eliminations.