[EM] "Independence of cycles" and a possible new method.

Forest Simmons forest.simmons21 at gmail.com
Tue Dec 14 13:11:26 PST 2021

How does this idea relate to "Break Cycle"?

El mar., 14 de dic. de 2021 4:55 a. m., Toby Pereira <tdp201b at yahoo.co.uk>

> A perfectly balanced cycle isn't neutral in its influence, but it is
> neutral in terms of deciding which is the "best" candidate when viewed
> alone. So arguably it should be neutral in its influence. But then we can
> add another set of ballots where one candidate would clearly be the winner
> (A in my example), but the winner of all ballots together would be a
> different candidate (B in my example).
> If I was just looking at the ballots afresh without any method in mind, I
> might decide that candidate A should be the winner. But Condorcet dictates
> B should be the winner. I don't subscribe to the view (that others might)
> that Condorcet methods are the best by default and not to be questioned,
> although if you devise a method in order to get the result you want in one
> particular election, it may have glaring faults elsewhere.
> The method I devised was in any case arguably over-simplistic as it only
> deals with three-candiate cycles. I looked up "Neutral Condorcet Criterion"
> and found it discussed by Mathias Risse - e.g. here
> http://www.math.buffalo.edu/~dhemmer/UE141F08/Projectpossibilities/RisseBordaCount.pdf
> The criterion requires neutrality for larger cycles as well - e.g. with
> the following four ballots:
> A>B>C>D
> B>C>D>A
> C>D>A>B
> D>A>B>C
> It might be that in order to pass exactly it requires the Borda Count, but
> another method that might go a decent way to passing it would be to
> eliminate any cycles involving all remaining candidates, then eliminate the
> "worst" candidate (e.g. ranked pairs loser) and continue until you have a
> winner.
> So if there are four candidates, take out any four-way cycles as above
> (and any going the other way) and then eliminate the last placed candidate.
> Then with the three remaining candidates, eliminate all three way cycles,
> and eliminate the losing candidate. Then pick the winner.
> Toby
> On Monday, 13 December 2021, 07:35:19 GMT, Forest Simmons <
> forest.simmons21 at gmail.com> wrote:
> For what it's worth, three way tie of the opposite cyclic order would
> cancel the above tied cycle and reverse it for x>2.
> ...
> x:B>A>C
> x: A>C>B
> x: C>B> A
> A perfectly balanced cycle is not neutral in its influence ... it exerts
> torsion.
> The torsion from the original "tied" cycle reinforces the smaller faction
> order, but goes against the order of the larger faction ... weakening its
> influence.
> If we were electing cyclic social orders instead of individual candidates
> this would make more sense ... add two of these neutral cycles together ...
> if they were absolutely neutral they wouldn't be able to affect each other.
> Let's have an election to decide democratically if rock/paper/scissors is
> the true cyclic order or its reverse. Some people may believe that rocks
> are stronger than paper and that paper can cover scissors and that scissors
> can chip away at rocks, for example.
> El dom., 12 de dic. de 2021 3:38 a. m., Kristofer Munsterhjelm <
> km_elmet at t-online.de> escribió:
> On 12/12/21 11:04 AM, Toby Pereira wrote:
> > If you look at the example here:
> > https://www.rangevoting.org/TobyCondParadox.html
> > <https://www.rangevoting.org/TobyCondParadox.html> having a tie cycle
> in
> > a Condorcet election can change the result. The ballots are essentially:
> >
> >
> > 1 voter: A>B>C
> > 2 voters: B>A>C
> >
> > Plus
> >
> > 2 voters: A>B>C
> > 2 voters: B>C>A
> > 2 voters: C>A>B
> >
> > A is the Condorcet winner in this election, beating both B and C by 5
> > votes to 4. However, in the top section of ballots, B is the Condorcet
> > winner, and the bottom section is a three-way tie as it is a perfect
> > cycle. Adding in tie cycles to change the result could be seen as a
> > failure of "independence of cycles" (or perhaps there is a better name).
> > This is also a specific case of failure of consistency. Or maybe a weak
> > failure because B doesn't win both the sub-elections. B would have an
> > outright win from the top ballots and tie from the bottom ballots, but
> > loses to A when all are combined.
> >
> > This example, by the way, is a simplified version of something that
> > Donald Saari used to promote the Borda Count.
> >
> > The problem is that in an election with more than three candidates, you
> > couldn't simply remove the cycles and calculate the result. Ballots and
> > candidates would potentially be involved in many intertwined cycles, so
> > there would be no straightforward way of doing it.
> >
> > But what you can do is compare every possible triplet of candidates
> > (like Condorcet methods compare pairs). For each triplet, all tie cycles
> > are removed and you look at the head-to-heads.
> You could do this, but as I understand the example, the method would no
> longer be a Condorcet method. You could also define the irrelevance of
> cycles criterion, perhaps something like:
> Removing a constant number of voters who together form an exact tied
> Condorcet cycle should not modify the output.
> Though I'm not sure what the implications would be - or if it's possible
> to pass by any method that fails IIA.
> As for using triplets instead of pairs, I think Stensholt suggested that
> doing so might be a way to generalize his BPW method, which is only
> defined for three candiates. Similarly, it might be a way of
> generalizing my fpA-fpC, though I'm again unsure how to do so and
> preserve the desired properties of DMTBR and monotonicity.
> -km
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