Compatibility of ULC, IIB, and IIA

Previously, @parker_friedland told me that it is unknown whether ULC and IIB are compatible.

I have determined that they are, by adding constraints to optimal Monroe in a way that forces it to pass. However, these constraints introduce other flaws, such as failing, at the very least, an optimal variant of monotonicity, where adding an approval for a candidate can lower the scores of subsets containing that candidate. This means it probably fails standard monotonicity as well. Also, it’s kind of cheating.

Anyway, the rules are the same as optimal Monroe, except first each ballot giving all candidates in a set the same rating are divided equally between each candidate. Then if any candidates are universally approved on the unspent ballots, these ballots contribute equal amounts of weight to the ULCs. This spending is locked in, and the remaining ballots are distributed to the remaining candidates in a way that maximizes the total Monroe score. Ties between winner sets are broken by total unweighted sum score. This makes sure that ULCs are actually elected (since Monroe needs tiebreaker specifications to pass Pareto), without having the tiebreaker violate IIA or IIB.

I think it’s possible to break ties in optimal Monroe in a way that passes ULC factions, in which case only the first (IIB) step would be necessary for passing IIB and ULC factions.
To pass monotonicity, constraints would need to be added that prevent the addition of a ballot from increasing the Monroe score of a winner set by more than that ballot’s highest score for a candidate in the winner set. This may make a perfect score achievable only when all ballots approve all winners in a set.

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Passing Monotonicity

These changes make the method significantly less “Monroe”, despite the original method being a constricted version of Monroe. The modified method also no longer requires “cheating steps” to ensure that it actually passes IIB and ULC. It is still optimization based.

To score a winner set, first, each ballot is split between each candidate in the winner set. The Monroe score for this division is the “base score”. Notice that it is just the

average of the unweighted scores for each candidate in the winner set.

Next: ballots (or portions of ballots) are switched between candidates. For a switch to be allowed, the following conditions must be met:

  • Each candidate gains as many ballots as it gives.
  • Each ballot (or portion of a ballot) that is switched gets reassigned to a candidate that it strictly prefers to its old designation.
    • Notice that because of this rule, irrelevant ballots cannot be switched.
    • Also notice that ULCs cannot trade any of their ballots.

The score for each switch is half the improvement of the ballot portion with the smallest gain from reassignment times the total weight of the switched ballots. The ballots must be switched in a way that maximizes the sum of the switch scores, while following the rules for switching. The total score of the winner set is the base score plus the switch scores.
The tiebreaker preferrably passes ULC factions. The base score inclusion is sufficient to pass Pareto.

It’s not at all obvious that this method is proportional. I have proven that in 2-party scenarios where the smaller party has exactly 1 Droop quota of support, it ties with the larger party for the final seat, for any number of total seats. While this is not sufficient to show that it generally passes PR, it suggests that it is possible.

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While I haven’t had time to try to fully prove the proportionality of this method, I will note a property that again suggests that it is. When all voters vote in a partisan manner, no candidates are from multiple parties, and each party gets an integer number of Hare quotas, changing a proportional set by one candidate to a nonproportional set must lower the set score. This is because replacing one candidate with another can swing the base score by no more than a Hare quota of points (and that only happens when a unanimously disapproved candidate is replaced by a unanimously approved one. So replacing a candidate from a party with enough support to have a proportionality guarantee cannot increase the base score by as much as a Hare quota.)

In the very specific scenario I described, it is possible to swap every ballot onto a candidate that they approved when the winner set is proportional. However, if a party then loses a candidate, then a Hare quota of that party’s voters will become newly “stranded” by the change, as party voters can only swap onto a member of their party. Thus the swap score will fall by a Hare quota of points, more than the base score could possibly rise.

While I could probably justify a stronger claim with this sort of reasoning, I don’t have the time right now.

Very interesting! (Now you can surpass the 3 consecutive comments limit)