SNTV is not a very good multiwinner method, but it is proportional under
strategy. It is also monotone, which is a property few multiwinner
methods have. Thus, it may be useful to try to improve upon SNTV so as
to fix its flaws, while retaining the useful monotonicity aspect. If we
can manage to construct a truly proportional monotone method, that would
be new, indeed.
What are the flaws of SNTV? There are two, mainly:
- First, it suffers from its own sort of vote-splitting problem. If a
party with less than 1/x support fields more than seats/x candidates, it
is at risk of losing seats because the votes are spread out too thin. In
effect, this means candidate selection must be coordinated, which leads
to centralized parties, which is not something we want.
- Second, all voters have to vote for the various candidates in equal
order. If voters predominantly vote for a single of the party's
candidates, the excess votes are lost and the party loses seats.
While I have not been able to find a solution to the first problem, I
think I have found one for the second. I'll mention other solutions to
the first, and how I don't think they would work (at least not without
further research).
-
My solution to the second problem requires a change of ballots, a
cumulative vote version of SNTV if you will. Unlike plain cardinal
ratings, the cumulative vote doesn't have the rich party effect that has
been mentioned earlier.
First count the totals for all the candidates, then, using a divisor
method, either directly (as Webster's) or indirectly (as by
Sainte-Laguë), apportion as many seats as you want. If every candidate
that has a nonzero count of seats has exactly one seat, you're done.
This is what would happen if everybody were careful about spreading
their votes.
However, if some candidate has more than one seat, there's a problem.
What is the problem? Well, no candidate can occupy more than one seat
(the "simple" solution of weighted votes notwithstanding). But now
consider: if, say, candidate A has two seats out of three, what does
that mean? It means that voters who voted for A should have the power to
determine who gets the other seat.
Hence, we get to what I call diminuation factors. Each candidate has a
diminuation factor which starts at 1 and is always greater than zero.
How do the diminuation factors work? It's quite simple: before counting
a cumulative ballot, first multiply each candidate's rating by its
diminuation factor, then rescale the entire ballot so it has the same
power as any other cumulative ballot within the system. Deweighting a
candidate like this means the "power" granted to that candidate is
redistributed to all the other candidates (by the renormalization process).
In the case of a bullet vote, the renormalization will cancel out any
deweighting, but that just means the excess power will flow to other
voters for that candidate, through the form of a lesser diminuation factor.
Do the count - it's probably easier to do it indirectly (Sainte-Laguë)
than directly - and then slightly decrease the diminuation factor of the
candidates that got more than one seat, proportionally to how much more
they got. That is, if there are 5 seats and one got 2, the other 3,
deweight the former 1.5 times the latter. I'm not sure which function
would modify the diminuation factors in the best way, so let's go with
division - divide them by a number very close to 1.
Doing so might alter the divisor method's optimal divisor, which is why
I say that doing it indirectly is probably easier. This is also the
reason we only adjust the factors slightly: deweighting some candidate
might lead another to get more than one seat.
There are also two ways to se the goal criterion for adjusting the
diminuation factors. Say that candidate A has 2.4 seats according to the
least divisor that gives the right rounded total number of seats. Then
one may adjust the diminuation factor until A has some value < 1.5, or
until A has exactly one seat. The latter is more "greedy" than the
former, and I'm not sure about the implications - the former seems more
right, but it may make voters regret their vote to a greater degree.
Let's show that again, as an example. Say the sums are:
A: 730, B: 240, C: 200
Two seats. Then, when using Webster, the least integer divisor that
leads to two seats is slightly above 480 (say 480.0001). The unrounded
results are:
A: 1.52, B: 0.499... C: 0.42
The greedy approach would be to adjust A's diminuation factor until his
unrounded result is 0.5001 (possibly changing the least divisor in the
process). The less greedy approach would stop somewhere below 1.5. The
less greedy approach seems better, but I'm showing that there is an
ambiguity here.
In terms of Sainte-Laguë, the less-greedy approach stops when no
candidate gets more than one seat. I'm not sure how the greedy approach
would work, so again, indirect seems better.
-
What's the point of all that? Well, now the voters can rate the
candidates however they want, because if one gets an excess, it's not
wasted, but redistributed to other candidates. I haven't actually
checked, but it appears to me that this modification also retains
monotonicity, because increasing the power given to a candidate can't
make that candidate lose.
It can, however, make other candidates you preferred lose, because the
other voters who like that candidate may give power to other candidates
than those you like. In plain terms: if you're a right-wing voter and
give more of your ballot to a specific left-wing candidate, then that
will cause his diminuation factor to decrease, meaning that other
left-wing voters get more power, possibly thwarting your right-wing
candidates. But that shouldn't be of great concern because you already
"thwarted" your right-wing candidates by moving power away from them
over to the left-wing candidate.
-
In any event, I think that solved the second problem. The first one
remains, as one can construct examples where no deweighting ever occurs.
For instance:
67: A1 (10) A2 ( 8) A3 ( 8) B1 ( 1) B2 ( 1) B3 ( 0) power: 28
33: A1 (1) A2 ( 1) A3 ( 0) B1 (10) B2 ( 8) B3 ( 8) power: 28
sum: A1 (703) A2 (569) A3 (536) B1 (397) B2 (331) B3 (264)
For three seats, we'd reason that the majority would get two and the
minority one. But that's not what happens. The divisor is 800, and:
A1: 0.88, A2: 0.71, A3: 0.67, B1: 0.497, B2: 0.41, B3: 0.33
the three As get a seat each. If B had been just one candidate, his
count would have been 992 and he would have got a seat (divisor 1080):
A1: 0.65, A2: 0.53, A3: 0.496, B: 0.92.
I can see two solutions, but they have their own problems. The first is
to use coalitions (like DAC and DSC), but it's not obvious how to do
that for a cumulative vote, and even when ranked ballots are used, such
methods (like Setwise Highest Average) can be nonmonotone.
One way to investigate these would be to set up constraints. For
instance, if the ballot is
10: A>B>C
20: B>C>A
24: C>B>A
and the coalitions are:
54: A B C
44: B C
10: A B
24: C
20: B
10: A
then the constraints would be
pick at least round(54/x) from {A, B, C}
pick at least round(44/x) from {B, C}
pick at least round(10/x) from {A, B}
pick at least round(24/x) from {C}
pick at least round(20/x) from {B}
pick at least round(10/x) from {A}
with x the minimum that doesn't produce a contradiction. Those
constraints would then give a bunch of output sets - candidate councils
that pass the constraints. While a multiwinner method can't elect a set
of councils, if a way could be found to pick one of the set so that
raising A can never cause the method to go from a set containing A to
one not containing A, then the method might be monotone. On the other
hand, if the constraints themselves can remove all sets containing A
when the voter raises A (possibly due to a change of minimum x), then
there's no way this kind of method can be monotone.
The second solution arises from observing that the problem with the
cumulative vote SNTV is vote-splitting. Therefore, if only one candidate
was "exposed" at a time, there would be no vote-splitting: there could
never be votes spread too thin, only concentrated too heavily - in which
case one would "expose" more candidates.
That would lead to a solution similar to Meek's. As one starts, consider
just the highest rated candidate. However, to do so could easily lead
into good old nonmonotonicity city; it does so with Meek. Apparently
there are ways of making IRV monotone, but at the cost of almost never
electing the CW (http://www.mcdougall.org.uk/VM/ISSUE15/P2.HTM).