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

Toby Pereira tdp201b at yahoo.co.uk
Tue Dec 14 15:10:30 PST 2021

``` I'm not sure I understand the question.
On Tuesday, 14 December 2021, 21:11:39 GMT, Forest Simmons <forest.simmons21 at gmail.com> wrote:

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> escribió:

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>DB>C>D>AC>D>A>BD>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>Cx: A>C>Bx: 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> 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

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
----
Election-Methods mailing list - see https://electorama.com/em for list info

-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://lists.electorama.com/pipermail/election-methods-electorama.com/attachments/20211214/37a987bc/attachment-0003.html>
```