# [EM] Semiproportional Bucklin method

Sat Apr 25 08:09:40 PDT 2009

```Raph Frank wrote:
> On Sat, Apr 25, 2009 at 1:26 PM, Kristofer Munsterhjelm
>> Start as Bucklin. Call each step a "round". A set of q candidates has a
>> support equal to the support of that candidate in the set that has the least
>> support. So, for instance, the set {A, B, C} has support 20 if the
>> individual support of each candidate is A: 20, B: 30, C: 98.
>> At round p, a candidate has support equal to the number of voters who placed
>> that candidate in rank 1..p inclusive, where rank 1 is first preference,
>> rank 2 second preference, etc.
>>
>> At round p, if there's some set of cardinality p that's supported by a Droop
>> quota more than the number of members we've already elected from it, pick a
>> candidate from that set (use any method you want to decide which).
>> Otherwise, go on to round p+1.
>
> It seems like it would meet Droop to me.

It seems to me that it does in the vast majority of the cases, but I'm
not good enough to determine whether it may have false positives.

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.

The question that remains is whether there is some case when a set would
appear to be supported by a Droop quota, yet not really be so. Hence my
example of "shadowing" for single-winner Bucklin.

> The only problem I can think of is if only 1 candidate can be elected
> per round (I am not clear if  that is a requirement).  You could end
> up with 2 parties with a Droop quota each "colliding" in round N, if
> they both ran N candidates.
>
> This is not a problem if more than 1 candidate can be elected if 2
> distinct groups manage to met the election condition in the same
> round.

Yeah, I may have described that situation vaguely. Electing more than
one candidate in a single round is permitted - both between sets and
inside a single set. For instance, if you have:

2 Droop quotas support A B C
1 Droop quota supports D E F
1 Droop quota supports G H I

Then you should elect two from the first set and one from each of the
sets below, unless they already got what they wanted. 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

That assumes none of the candidates have been elected in an earlier
round. If A had already been elected, you would be limited to options 1
to 3. If H had already been elected, you would be forced to elect B.

```