Kathy Dopp wrote:
On Sun, Feb 1, 2009 at 6:01 AM, Raph Frank <[email protected]> wrote:
Party list systems are (mostly) monotonic.
Do not know what "Party list systems" are, but all plurality elections
are monotonic.
A party list system works like this. You have one vote. Vote for a
party. The number of votes are counted, and then each party gets a share
of the assembly proportional to the number of votes it got. Each party
has a list, and if a party gets (say) four seats in the assembly, the
first four candidates on the list are elected to the assembly.
Here's an example: there are three parties, the assembly is of size ten.
We'll use Webster's method, since it's most fair (possibly with the
exception of Warren's dynamic method).
The first party fields: A1, A2, A3, A4, A5, A6, A7, A8, A9, A10. The
same for the second and third parties, except their candidates start
with B and C respectively.
Run the election. Say the vote counts are:
Party A: 847 votes = v_1
Party B: 300 votes = v_2
Party C: 640 votes = v_3
Now we need to pick x so that round(v_1 / x) + round(v_2 / x) +
round(v_3 / x) = 10 (size of the assembly).
It's rather easy to find x by just trying[1]. In this case, x = 160
works, and you get:
Party A: round(847 / 160) = 5 seats
Party B: round(300 / 160) = 1 seat
Party C: round(640 / 160) = 4 seats
which sums up as desired. To find out which candidates the parties got,
just read off the list. There are five from the first party: A1 A2 A3 A4
A5; then there is one from the second party: B1; then there are four
from the third party: C1 C2 C3 C4.
The assembly is A1 A2 A3 A4 A5 B1 C1 C2 C3 C4.
Open list PR works like this, but also lets the voters influence the
order of the lists.
When comparing STV to party list, I don't like party list PR that much
(since it makes parties formally a part of the process), but party list
*is* both simple and proportional, and the results are much better than
those given by a two-party system.
-
[1] the function (size of assembly) - (round(v_1 / x) + round(v_2 / x) +
round(v_3 / x)) crosses zero at the desired value of x, and it's also
rather linear and certainly monotone, so we can use root finding with
little trouble, or just try it manually. As an aside, it might be
interesting to determine the asymptotic runtime of the fastest algorithm
that finds x.
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