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

Colin Champion colin.champion at routemaster.app
Sun Dec 12 08:35:00 PST 2021

```Is "independence of cycles" not the same as Saari's "neutral Condorcet
requirement"? (mentioned in Risse's paper "Why the Count de Borda cannot
beat the Marquis de Condorcet"). The matter is discussed in Pacuit's
Stanford Encyclopedia article. He refers to ballots "cancelling
properly" and mentions a proof that "there is no Condorcet consistent
voting method that cancels properly".

CJC

On 12/12/2021 11:38, Kristofer Munsterhjelm wrote:
> 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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