[EM] Selective Ranked Pairs

[Forest Simmons]

On Mon, Aug 14, 2023, 7:43 AM Kristofer Munsterhjelm <km_elmet at t-online.de>
wrote:

> On 8/14/23 13:12, Filip Ejlak wrote:
> > Here's a Smith method that seems to be monotonic, cloneproof and DMTC
> > burial resistant (at least in non-tie scenarios):
> >
> > 1. List all the pairwise victories.
> > 2. For any XYZ cycle with X being the fpA-fpC winner, remove the Z>X
> > majority from the list.
> > 3. Proceed with the Ranked Pairs method.
> >
> > I'm not sure how to handle ties in order to get total criteria
> > compliance. Also, I will be thankful for some independent checks.
>
> Ties are known to be a point of trouble even for ordinary Ranked Pairs.


Here's where strong [dis]approval gives the desired high degree of
resolution ...

Ranked Pairs with the pairwise defeat of Y by X gauged by the following sum;

Total Strong Approval of X
Plus
Total Strong Disapproval of Y.

Default strong approval of X is the number of ballots on which no candidate
outranks X.


Default strong disapproval of Y is the number of ballots on which Y
outranks nobody.



This gauge is so decisive that pairs tied for strength are statistically
negligible ... and deferring to the first term could be resorted to in that
once in a lifetime event.

>
> It's NP-complete to determine if some candidate X can win with a
> particular tiebreaking order:
> https://webspace.maths.qmul.ac.uk/felix.fischer/publications/bf_ranked.pdf
>
> The standard fix is to use a random tiebreaker that is itself
> cloneproof, usually the random voter hierarchy: choose a random voter
> and use his preference ranking. Keep drawing new random ballots and fill
> out unspecified preferences with them until every preference is set.
> (E.g. if the first voter voted A=B>C, then you draw ballots until you
> find one that rankes A and B differently, then use that to break its tie.)
>
> > (I've also looked into IRV with donations, but my implementation was
> > painfully clone-dependent - when the winner is cloned, then another
> > candidate may get the opportunity to donate their votes so that they
> > become the winner, as the previous winner's clones have too few votes on
> > their own to influence the elimination process.)
>
> Right. I think there's a way to deal with this that uses IRV's own clone
> independence, but I'm not entirely sure how to do it. Something along
> the lines of: if it's a three candidate election (for the sake of
> simplicity) and A and B get into the final, and usually B would win but
> you can donate to C to make the final A and C instead, then if any
> candidate is cloned, you defer donations until the final round, where
> all the clones will have been eliminated anyway. In effect, using the
> O(phi^n) algorithm for devising strategy against IRV, but only for
> donations instead of arbitrary strategy.
>
> But there could be snags that I'm not seeing right now, of course.
>
> -km
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>
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