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<div>Right ... we need strong mono-raise ... when I said Plurality for setting the agenda, I was actually thinking of Equal Top Approval, where the approval cutoff (virtual candidate) is positioned immediately below equal top.</div>
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<div>There's a superficial similarity, but a subtle (and crucial) difference, especially in the context of monotonicity.</div>
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<div>Mono-raise of X from below to above the approval cutoff is not supposed to knock some other candidate Y from above to below the approval cutoff, for example.</div>
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<div>......</div>
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<div>You wrote..</div>
<div>> in general, it seems that in a three-cycle, ABB elects the candidate who</div>
beats the winner of the base method pairwise.
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<div>True, as long as "winner" means least promising ... as in least approval, for example.<br>
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<div style="font-size:85%;color:#575757" dir="auto">Sent from my MetroPCS 4G LTE Android Device</div>
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<div>-------- Mensaje original --------</div>
<div>De: Kristofer Munsterhjelm <km_elmet@t-online.de> </div>
<div>Fecha: 4/8/21 9:16 a. m. (GMT-08:00) </div>
<div>A: Susan Simmons <suzerainsimmons@outlook.com>, election-methods@lists.electorama.com
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<div>Asunto: Re: [EM] Agenda Based Banks </div>
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<div class="PlainText">On 04.08.2021 00:29, Susan Simmons wrote:<br>
> Actually, in the burial example below, ABB performs strictly better than<br>
> DMC because under DMC, the burial of C by A either rewards A with a win,<br>
> or leaves intact A's sincere 2nd choice as winner ... a zero risk gambit.<br>
> <br>
> By way of contrast, if A's gambit fails under ABB, then A's least<br>
> preferred, B, wins ... a big risk, because A's raising of B enough to<br>
> introduce the necessary cycle might also be enough to move B to the<br>
> necessary agenda level (if it was not already there).<br>
> <br>
> In summary, ABB has both the best burial resistance and the best chicken<br>
> attack resistance of any agenda based Condorcet method that we know of<br>
> (and we know of oodles!).<br>
> <br>
> It seems doubtful that any simpler, monotone, clone free, ISDA, burial<br>
> and chicken resistant method exists ... except perhaps Asset Voting,<br>
> which is too far ahead of its time (and our time ...first promoted only<br>
> 150 years ago by Charles L Dodgson) even for Steven J Brams the great<br>
> academic promoter of Approval, another simple method too far ahead of<br>
> its time.<br>
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As the perennial cold shower man, I must ask if this is actually<br>
monotone for all monotone base methods :-)<br>
<br>
Suppose that the base method is Plurality. In this election:<br>
<br>
5: A>B>C<br>
2: A>C>B<br>
6: B>C>A<br>
1: C>A>B<br>
1: C>B>A<br>
<br>
there's an ABCA cycle and the Plurality order is A>B>C. If I understand<br>
correctly, ABB would first deposit A and then find the candidate that<br>
beats A pairwise, i.e. C, and then terminate, giving C as the winner.<br>
<br>
Now let 3 of the A>B>C voters raise C to C>A>B:<br>
<br>
2: A>B>C<br>
2: A>C>B<br>
6: B>C>A<br>
4: C>A>B<br>
1: C>B>A<br>
<br>
Now the Plurality order is B>C>A. B is first deposited, and then the<br>
candidate who beats B (i.e.) A is deposited and wins.<br>
<br>
So raising C made C lose. Looks like we need strong mono-raise.<br>
<br>
With Plurality as the base method, I think ABB reduces to Eivind<br>
Stensholt's BPW - or in my ABCA jargon, the method where candidate A's<br>
score is "-fpC". If we just had a way of making it fpA-fpC, then all<br>
would be well, at least in the case of Smith set equals three!<br>
<br>
(Here's an ugly hack: The inner procedure takes a "tentative first<br>
candidate" A. It inserts A, then it inserts the candidate that beats<br>
everybody in the list, whose [first preference count - the last inserted<br>
candidate's first preference count] is maximized. The inner procedure is<br>
over once nobody beats everybody inside pairwise; the candidate last<br>
inserted is the winner, and his score is the (first pref - last guy's<br>
first pref) count.[1]<br>
The outer procedure runs the inner procedure with every candidate as<br>
tentative first, and elects the winning outcome with the greatest score.<br>
But it's extremely ugly and I have no idea if it's strategy resistant or<br>
monotone beyond a Smith cycle of three.)<br>
<br>
In general, it seems that in a three-cycle, ABB elects the candidate who<br>
beats the winner of the base method pairwise.<br>
<br>
-km<br>
<br>
[1] One could imagine other score functions as well, as long as they<br>
specialize properly. E.g. Q's score is min P: fpQ - fpP, where P is a<br>
candidate from the current chain. But I suspect that properly<br>
generalizing fpA-fpC (so that it preserves DMTBR) will require some kind<br>
of solid coalition logic -- or eliminations.<br>
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