It is interesting to me that we can have a cycle of three candidates in which not every candidate is an approval winner for some allocation of approval cutoffs. However, I can prove that if there are no equal rankings (including truncations) then in a cycle of three candidates, each of them is an approval winner for some allocation of approval cutoffs.
In fact, it turns out that this is the case even if all voters are using above mean utility approval strategy based on the same winning probabilities. [Of course, to make C win you have to use different probabilities than those that make A win, etc.]
If anybody wants to see this proof, I would be happy to post it to the EM list.
One lesson to learn from this is that the process of symmetric completion can change an approval dominated candidate like C into a contender. In my view this is not desireable.
Remember in the spruce up process, we eliminate covered candidates because they are not serious contenders under sophisticated strategy. For the same reason we should also eliminate approval dominated candidates like C.
I think this would be easiest to do after having eliminated the pairwise covered candidates and having collapsed the beat clone sets, and certainly before any symmetric completion step that Chris might want to throw in as part of Weighted Median Approval, for example.
If after these three reductions we are still left with a cycle of three candidates, then why not use a non-deterministic method to resolve the cycle? I think that's our best hope for completely doing away with incentives for insincere order on the ballots.
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