[EM] Semiproportional methods 1: "Droop Judgment" fails proportionality

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[EM] Semiproportional methods 1: "Droop Judgment" fails proportionality

Kristofer Munsterhjelm-3
I've been on-and-off considering the following generalization of MJ:

Elect the council that makes the voter who's worst off, best off, after
eliminating a Droop quota's worth of voters so as to maximize this measure.

A voter's satisfaction with a council is equal to that of his highest
ranked/graded candidate. So eliminating a Droop quota's worth of voters
is as simple as eliminating a Droop quota of ballots that score the
council the worst.

(The easiest way to handle ties is to consider the next-to-worst-off
voter, etc.)

It's pretty obvious to see that it's monotone: if someone decides to
rank a candidate A higher and the winning council contains A, either
that has no effect on the worst grade for A, or the voter was the
worst-off voter and the council's grade becomes better.

I initially thought it would be proportional because "if you eliminate a
Droop quota, then voter who is worst off must have belonged to at least
a Droop quota of voters, and so maximizing his preference satisfies a
Droop constraint. And since that holds for every Droop constraint, then
proportionality ensues". But I couldn't manage to rigorously prove it.

After trying to prove it proportional, I ended up coding a brute force
test for Droop proportionality violations. It showed that the method is
proportional for three candidates, but fails with more than three. (So
no wonder I couldn't prove it proportional - because it wasn't.)

Here's one example:

58: A>C>B>D
18: B>C>A>D
21: C>A>B>D
32: D>B>A>C

(two to elect, Droop quota is 43)

The solid coalition {A,B,C} is supported by 97 voters, which is more
than two Droop quotas. Hence, by the Droop proportionality criterion,
the council should have two of these candidates on it. But the Droop
Judgment winner is AD.

Comparing AD to the Droop-proportional council that scores highest by
Droop Judgment, AB:

AB:
The 58 A>C>B>D voters give this council rank 1
The 18 B>C>A>D voters give this council rank 1
The 21 C>A>B>D voters give this council rank 2
The 32 D>B>A>C voters give this council rank 2

So the number of voters at each rank is 76 53 0 0. After removing 43
votes, the result is 76 voters rank the council first, and 10 rank it
second.

AD:
The 58 A>C>B>D voters give this council rank 1
The 18 B>C>A>D voters give this council rank 3
The 21 C>A>B>D voters give this council rank 2
The 32 D>B>A>C voters give this council rank 1.

So the number of voters at each rank is 90 21 18 0. After eliminating 43
voters, every remaining voter ranks the council first.

I think the problem lies in that while Droop Judgment gives each Droop
quota someone from the smallest coalition it supports, it may give two
Droop quotas the same someone, i.e. the same candidate. In the example
above, the {A,B,C} voters get represented by A and the remaining group
gets represented by D.

To put another way, it doesn't respect surplus transfers. This failure
creates an SNTV-ish incentive to spread the votes around, though not as
severe as in SNTV itself. It also explains why it works in the
three-candidate case, because then a coalition can't have more than two
Droop quotas' worth of support.

But maybe there is a way to salvage it without making it too hard to
prove monotonicity. On the other hand, ranked (Borda) Monroe is
nonmonotonic while Borda Chamberlin-Courant is monotone, so perhaps it's
precisely that surplus constraint that makes monotonicity difficult.

I tried to add a hack by considering groups of two candidates and
eliminating two Droop quotas' worth, and then breaking ties with the one
Droop quota elimination and single candidates idea of above, but that
didn't work. At least the obvious way of handling groups of two
candidates by considering the rank given to the second highest ranked
candidate on the council didn't work. Maybe a more sophisticated scoring
function could work?

In any case, my main reason for posting this, besides just showing what
I've been doing, is to give others ideas for monotone proportional
methods by showing something that "almost" worked :-) I had another idea
that didn't quite work, too, but I'll have to write down all its details
first.
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