[EM] Semiproportional Bucklin method

[Kristofer Munsterhjelm]

Raph Frank wrote:
> On Sat, Apr 25, 2009 at 4:09 PM, Kristofer Munsterhjelm
> <km-elmet at broadpark.no> wrote:
>> Clearly, it detects a Droop set when one exists, since the definition of the
>> Droop set is that at least a Droop quota vote these ahead of all others -
>> not necessarily in the same order, but Bucklin equalizes the order.
> 
> Right, so then it meets Droop, isn't that the definition?

Necessary but not sufficient. It fails if, say, there are two seats, and 
because of shadowing (or something), it elects something that isn't part 
of a Droop set, leaving only one candidate left, where the DPC would 
require it to elect two specific candidates.

Since the method finds Droop sets, this may appear as a contradiction. 
If so, you could say that the method either elects according to the DPC 
or fails entirely, but then that's a problem of the method.

A contradiction would be like this:

Apparently, 1 quota supports: A B
Apparently, 1 quota supports: C D
Apparently, 1 quota supports: E F

2 to elect.

> .. or by a Droop set, do you mean one that PR-STV will pick without
> the need to eliminate any candidate?

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.

>> For a more complex example
>> that may or may not be impossible, but which shows the point:
>>
>> 2 Droop quotas support A B C H
>> 1 Droop quota supports A C D H
>> 1 Droop quota supports E F G H
>>
>> If you elect A, B, and E, that satisfies all three support sets, because A
>> counts to satisfy both the first and second sets. Electing A, C, and E would
>> give 2 candidates to the second set, which is more than its fair share,
>> hence one can't elect both A and C. In effect, the candidates to be elected
>> that would satisfy all three sets would be one of:
>> 1. A B E
>> 2. A B F
>> 3. A B G
>> 4. B C E
>> 5. B C F
>> 6. B C G
>> 7. B H
> 
> Sounds reasonable.  In effect, you can elect candidates in any order
> you like as long as the group in question has fewer seats than it is
> entitled to and you cannot proceed to the next round until there are
> no more valid seats.
> 
> Would you consider A,H acceptable too?  Elect A from the first group,
> then group 2 is disabled, then elect H from the 3rd group?

No, because that would give the second group two candidates (A and H), 
which would lead to superproportional representation of that group.

Say H is a centrist, the first group is supported by people favoring 
decentralists, and the second group is supported by left-wingers. A is a 
decentralist left-wing candidate. Electing both A and H would represent 
the decentralist group in just the right amount (since they have two 
Droop quotas' support), but the left-wing group too much since the 
centrist already represents them.

One might have desirability rules that say which of the possible valid 
solutions one should pick. If the desirability rule is "pick the 
candidates that have wide support", B and H would be a very good choice, 
since H has wide support.
I haven't investigated which rules would be best, since the point of 
this method is that if it's truly proportional, it does something no 
other method (or method class) known to me does - it satisfies both the 
DPC and summability. To my knowledge, the "summable PR" rule that was 
the answer to the rangevoting puzzle didn't pass the DPC, but a weaker 
bloc criterion stating "if voters vote their party above all others, 
then each party should get their share proportional to the number of 
voters supporting that party" which doesn't need to deal with personal 
(non-party) votes.