[EM] Semiproportional Bucklin method

[Kristofer Munsterhjelm]

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.