[EM] Semiproportional Bucklin method
Kristofer Munsterhjelm
kmelmet at broadpark.no
Sat Apr 25 05:26:54 PDT 2009
First, I have been trying to construct proofs that various methods pass
mutual majority. So far, I've managed to make proofs for elimination
methods where the base method passes Majority (I've mentioned that proof
before), for methods that are Smith (really just showing that Smith is a
subset of the mutual majority set), and for Bucklin.
This suffices to cover almost all the methods condorcet.org mentions as
passing mutual majority: Bucklin, Coombs (elimination with majority),
IRV (ditto), pairwise elimination (ditto), KemenyYoung (elects from the
Smith set), Ranked Pairs, Schulze, and Smith//Minmax (all elect from the
Smith set).
That leaves Bordaelimination (Baldwin) and Nanson, both of which are
elimination methods based on Borda. Borda doesn't meet majority, and so
one can't use that proof. However, if there is a Condorcet winner, he'll
always have aboveaverage Borda score. This secures the Smith set from
being completely eliminated, therefore the Borda elimination methods
meet Smith (and hence mutual majority).
This would suggest that, as I've mentioned before, for loser elimination
methods where the base method that elect candidates with property X, the
loser eliminatino method passes the criterion for all sets that reduce
to the candidate with property X when there's only one left. If X is
majority, the set property is mutual majority. If X is Condorcet, then
the set property is Smith.

From investigating the various methods, I've found a semiproportional
method based on Bucklin. It seems to often elect according to the
constraints of the Droop Proportionality criterion. My question is, does
it always do that?
The definition of the method is:
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's that simple.
Let's try it, using the PSCCLE example:
33 A>D1>D2>D3>B>C
33 B>D1>D2>D3>A>C
32 C>D1>D2>D3>A>B
2 D1>D2>D3>A>B>C
100 voters, 2 to be elected, so the Droop quota is 100/3 = 33.333^. For
simplicity, I'll use the "method of first difference" to determine who
gets picked within a set. The method of first difference is simply this:
if two or more candidates are tied as far as support goes, pick the one
who first scored above the others.
Round 1: support is: A: 33, B: 33, C: 32, D1: 2. No set of card 1 has
reached the quota.
Round 2: support is: A: 33, B: 33, C: 32, D1: 100. No set of card 2 has
reached the quota. Although D1 has reached the quota, it can't be paired
with any other candidate to make a set that reached the quota, and
therefore it can't be used yet.
Round 3: support is: A: 33, B: 33, C: 32, D1: 100, D2: 100. D1 and D2
are above quota, but can't make a set of cardinality 3.
Round 4: support is: A: 35, B: 33, C: 32, D1: 100, D2: 100, D3: 98
Now we have the set {D1 D2 D3 A} which is supported by 35,
since the weakest candidate in the set, A, is supported by 35. 35 >
33.333, so we can elect a single candidate from the set. The method of
first difference says to elect A, because the first round where A had a
different score than any others was the first round, where A had the
most support.
Round 5: support is: A: 100, B: 68 C: 32, D1: 100, D2: 100, D3: 100.
We have the set {A B D1 D2 D3} which is supported by 68.
That's two Droop quotas (68 > 2/3), but we've already elected one from
that set, so we can only elect one more. That candidate is B, since it
had a greater score than the others in the first round.
All done, and the winners are A and B. The method of first difference
may not be very good, but it shows the point.
Does this method (or group of methods) pass the DPC? If not, when does
it not? If it does, that's quite interesting, because the method is
summable.
If it does not pass the DPC, I think the reason would be something like
what I'd call "shadowing". Consider the following situation for
singlewinner Bucklin, where the set in question is the mutual majority set:
2: A > B > ...
2: B > A > ...
2: A > ... > B
1: B > ... > A
1: C > B > ...
1: D > C > ...
9 voters
At round 2, we have
7: A, 6: B, 2: C
This suggests that {A, B} is a mutual majority set, but it is in fact
not so. Shadowing is not a problem in singlewinner Bucklin, since it
implies that there is no mutual majority set, in which case Bucklin is
free to pick whatever it wants. However, in the multiwinner version, one
might have a shadowing candidate (that is, one that's picked due to
shadowing) occupy the place that the DPC demands should be given to
another candidate  that is, unless shadowing requires enough voters
that you couldn't make this work *and* set up other Droop sets at the
same time.
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