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

Toby Pereira tdp201b at yahoo.co.uk
Sun Dec 12 02:04:49 PST 2021

```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>C2 voters: B>A>C
Plus
2 voters: A>B>C2 voters: B>C>A2 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. In the above example it would be:
B beats A by 2 votes to 1B beats C by 2 votes to 0A beats C by 2 votes to 0
If a candidate wins every head-to-head in every triplet, they are elected (the equivalent of a Condorcet winner for this method).
I will presume it's possible for there to be no such winning candidate, and if so, there would need to be some sort of tie-break. But we could simply base it on existing Condorcet methods - e.g. the Ranked Pairs method. Find the single biggest head-to-head win across all the triplets, and lock that in. Then find the next biggest that won't contradict a locked in position and lock that in and so on.
I think the idea of removing tie cycles like this has some merit, although it would be interesting to see what you all think.
Toby
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