Selective Ranked Pairs
Here's a Smith method that seems to be monotonic, cloneproof and DMTC
burial resistant (at least in non-tie scenarios):
- List all the pairwise victories.
- For any XYZ cycle with X being the fpA-fpC winner, remove the Z>X
majority from the list. - 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.
(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.)
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):
- List all the pairwise victories.
- For any XYZ cycle with X being the fpA-fpC winner, remove the Z>X
majority from the list. - 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.
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
On Mon, Aug 14, 2023, 7:43 AM Kristofer Munsterhjelm km_elmet@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):
- List all the pairwise victories.
- For any XYZ cycle with X being the fpA-fpC winner, remove the Z>X
majority from the list. - 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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