[EM] Electing Cabinets/Executive Committees

[Filip Ejlak]

Perhaps this improvement would make the algorithm monotone:

1. Calculate the proportional ranking, but start assigning ministries from *the
last place*, not the first one.
2. When a party/candidate chooses a ministry, they can choose either from
among the *not selected ones* or the *ones selected by someone in a lower
place*.
3. If a new ministry was chosen, it's the next higher place's turn.
4. If a previously chosen ministry was chosen again, the lower place loses
the ministry and it is now their turn to choose.

Of course the backtracking can happen automatically if the participating
parties just state their preferred ministries instead of making
turn-by-turn choices.


śr., 6 mar 2024 o 12:01 Kristofer Munsterhjelm <km_elmet at t-online.de>
napisał(a):

> On 2024-03-06 04:26, Closed Limelike Curves wrote:
> > I assume there's methods for this, but I don't know the search terms:
> > say I want to proportionally elect an inhomogeneous committee, like a
> > Cabinet or a set of executive officers. What methods handle this?
>
> Steven Brams considers this exact problem in the book "Mathematics and
> Democracy" (Chapter 9, "Allocating Cabinet Ministries in a Parliament").
> He first gives the following algorithm:
>
> For i = 1...n:
>         - Let the "current party" be the party next in line according to
> Sainte-Laguë based on votes in the most recent parliamentary election.[1]
>         - Ask the current party which of the remaining positions it would
> like
> to claim (e.g. PM, minister of defense, etc.)
>         - Assign the chosen position to the party.
>
> He then shows that this method, although proportional, is nonmonotone in
> the sense that sometimes a party might want to be asked later. So he
> introduces a "trading step" where at each i, the current party may ask
> some other party if it wants to go first instead. He finally shows that
> this does not completely eliminate tactical voting problems, but that
> "some of the problems of cabinet selection can be ameliorated if not
> solved" (section 9.8).
>
> If you don't have parties, you might want to look into the matrix vote:
> https://en.wikipedia.org/wiki/Matrix_vote. I'm not aware of any
> generalization to methods other than Borda; the obvious multi-way Range
> method[2] is not proportional.
>
> -km
>
> [1] Section 9.3. discusses the different divisor methods, and Brams
> argues that whether you'd prefer D'Hondt or Sainte-Laguë is a matter of
> preference - whether you think large-party bias is worth it for
> stability or not. IMHO, the way to solve kingmaker problems with party
> list parliamentarism is to use a slightly consensus-biased ranked
> method. If that method is house monotone, then it could be used as a
> replacement here. (Note that this is not the same as using D'Hondt, due
> to parallels to center squeeze if only first preferences are used.)
>
> [2] Each voter rates each candidate for each position, then the method
> selects candidates so that the sum of chosen candidate-position pairs is
> maximized. This can be done by linear programming.
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