[EM] Semiproportional Bucklin method
Kristofer Munsterhjelm
km-elmet at broadpark.no
Sun Apr 26 02:09:05 PDT 2009
Raph Frank wrote:
> On Sat, Apr 25, 2009 at 11:15 PM, Kristofer Munsterhjelm
> <km-elmet at 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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