Raph Frank wrote:
On Sat, Apr 25, 2009 at 11:15 PM, Kristofer Munsterhjelm
<km-el...@broadpark.no> wrote:
A Droop set is a set from which, according to the DPC, at least one
candidate must be elected. It's analogous to a solid coalition, only with a
Droop quota.

Ok, I think I have a example of it failing.

I see the problem. It sees {B, C, D} as a distinct set from {A, B, C, D}, where this is not really the case. By adding more candidates, you can make the teaming effect more serious.

Could this be fixed by a look-back? At each step, don't just construct sets of cardinality p, but of p or larger. Then elect the candidate from cardinality p sets unless this would contradict the >p sets. E.g.

(your example, made a bit more formal)

14: B C D E F A
 5: A B C D E F
 5: A C D E F B
 5: A D E F B C
 5: A E F B C D
 5: A F B C D E
 7: X1 X2 X4 X3 F E     (party X)
 7: X2 X3 X4 X1 E D
 7: X3 X4 X1 X2 D F

Quota is 20.

Round 1:
 A: 25 B: 14 X1: 7 Q1: 3 (split)

Sets (only those > one DQ will be mentioned):
 {A}: 25        to be elected: 1, already elected: 0
 No greater cardinality sets can be constructed.

Round 2:
 A: 25, B: 19, C: 19, D: 5, X1: 7, X2: 14, X3: 14
 No sets eligible

Round 3:
 A: 25, B: 24, C: 24, D: 24, X1: 14, X2: 14, X3: 14, X4: 14

Now, {B C D} has quota, but
        {A B C D}       to be elected: 1, already elected: 1

so we can't elect from {B C D}, because if we do, {A B C D} would have two representatives, which is more than it has earned.

There is a problem with this, however. It's not internally consistent, because you could say that at the first round (for instance),

 Round 1:
  {A}: 25       to be elected: 1, already elected: 0
but
  {A B}: 14     to be elected: 0, already elected: 0

which is a contradiction. Now, we could easily get around that by only considering sets of at least a Droop quota's worth, but that suggests that we could hide smaller sets with larger ones, something like:

DQ: A > B > C > D > X1
DQ/2: X1 > B > C > D
DQ/2: X2 > B > C > D
DQ/2: X3 > B > C > D
DQ/2: X4 > B > C > D

so that the "to be elected: 0, already elected: 1" of {A B C D} hides that {B C D} has two Droop quotas. The example above doesn't quite work, since the X* voters don't make any Droop sets, but perhaps a variant on it would?

I'm not sure.
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