[EM] Centrifugal Margins

[Joshua Boehme]

Here are some additional tidbits about the method...


When we use examples with minimal sets of ballots, it's not unusual for the centrifugal margin solution to come out the same as random ballot. However, here's a quick example to show that 100% of the weight can end up on orderings that don't appear on any ballots:

1 B>A>C>D
1 A>C>B>D
1 A>B>D>C

The solution is 100% A>B>C>D. This puts all the non-losing edges at 100%, so no other solution can possibly beat it (every excess is at its absolute maximum).



The approach I'm using for the higher-order steps follows two rules.

1. A k-tuple is non-losing if it has at least 1/k! strength on the actual ballots. For an edge, that's 1/2, exactly as we'd expect

2. Resolve the edges first, then the 3-tuples, then the 4-tuples...

The reason for those specific rules is that every variant I've tested that modified either/both of them failed to be cloneproof, and typically it was relatively easy to find a counterexample. With those two rules, I have yet to find any failures, though that doesn't mean it's cloneproof for sure



Here's another set of examples that blew my mind when I saw the results. Consider an election where every candidate ties head to head. Centrifugal margins handles some straightforward scenarios exactly as we'd intuitively expect.

If all four candidates are clones (A=B=C=D) then the solution is an equal 25% toss-up between the four.

If there are two equal factions, A > B=C=D and vice versa, then it splits 50% for A and 1/6 for each of B, C, and D

If the electorate is equally split between A, B, and C=D (i.e., 1/6 for each of the six distinct orderings), then it's 1/3 A, 1/3 B, 1/6 C, 1/6 D.

Random ballot can do that too, but it's not Condorcet or Smith. Maximal lottery would allow each of those respective solutions, but it also allows literally every other possible distribution in each case.



On 11/26/23 17:31, Joshua Boehme wrote:
> 
> Hello everyone. I've lurked in the archives off and on for a while. I finally decided to join since I've been playing around with a method and I'm at a point where outside feedback/criticism could help. Also, someone might have already described this in some paper I don't know about.
> 
> I'm calling it centrifugal margins for now, and it stochastically chooses an ordering based on ordinal ballots. (Since single-outcome methods can be contentious, assume multiple outcomes wouldn't make sense in the particular context or we're satisfying some external constraint.) Centrifugal margins generally requires ballot-level detail, but sometimes the head to head margins suffice.
> 
> The reason for the name is that it tries to maximize winning comparisons' margins and minimize them for losing ones. Majority voting with 2 candidates is the prototype: a candidate with >50% of the votes wins 100% of the time. Although cycles can prevent us from reaching 100%, the method tries to push non-ties away from the tie point. Ties remain perfectly balanced, like a pencil on its tip.
> 
> Let E be the set of non-losing edges over the ballots B. Centrifugal margins looks for new ballots B' that leximax (the margins of E over B') - (the margins of E over B). Note that E is determined solely by the actual ballots. For simple elections this suffices. Otherwise, we do the same for 3-candidate subgraphs after leximaxing the edges, then 4-candidate subgraphs, etc. The final ballots B' give the distribution.
> 
> The actual calculation resembles the nucleolus in game theory (a big inspiration) and has similar pitfalls. It involves iterated linear programming problems and using the duals to lock constraints.
> 
> Centrifugal margins satisfies Smith, and orderings with nonzero weight should have successive Smith sets in order, similar to ranked pairs. I think it's possibly cloneproof for ballot-level clones. Every ordering can have nonzero weight when all candidates tie, so worst-case complexity for n candidates is at least n!
> 
> Notably, centrifugal margins fails the blank ballot criterion. I think that's defensible, though it's probably a discussion for another time.
> 
> I'll leave it there for now to keep this brief, but questions are welcome! No promises that I have answers yet, though.