[EM] Some unusual Condorcet STV methods

Kristofer Munsterhjelm km_elmet at t-online.de
Sun Sep 13 13:34:19 PDT 2015


Here are some Condorcet/STV hybrids that I thought of while thinking
about proportional representation:

Method one:

1. Count first preferences of the (remaining) candidates.
2. While any single candidate has more than a Droop quota of first
preference support, elect him and redistribute the surplus.
3. If all seats are elected, we're done.
4. Otherwise, perform a Ranked Pairs election with the whole electorate
and eliminate the loser.
5. Go to 1.

This is a variant of STV with Condorcet loser elimination, and so the
winners will be biased towards the center of the whole electorate, which
isn't ideal. However, it should be more well-behaved than ordinary STV
because Ranked Pairs passes local IIA. Suppose that there are no
candidates with a Droop quota's worth of first preferences; then the RP
loser will be eliminated, but that won't alter the Ranked Pairs ranking
itself. So the rule above is equivalent to:

1. Count first preferences of the (remaining) candidates.
2. While any single candidate has more than a Droop quota of first
preference support, elect him and redistribute the surplus.
3. If all seats are filled, we're done.
4. Otherwise, perform a Ranked Pairs election with the whole electorate
and keep eliminating the losers until some candidate has more than a
Droop quota of first preference support.
5. Go to 1.

It should also be more clone resistant because the Ranked Pairs order
list the clones in sequential order; they're thus also eliminated one
after another. The Ranked Pairs order can still (and probably will)
change after a candidate is elected, however.

The loser elimination method should be one that passes LIIA, so that the
elimination order doesn't change during the elimination itself.

-

The second one is based on a Hare quota and is a pretty straightforward
port of the idea of clustering. It's sort of a greedy clustering
approach. For simplicity's sake, let's disallow equal rank since it's
going to use coalitions and I don't want to get bogged down in whether
to use acquiescing or solid ones.

1. Count the strengths of every solid coalition.
2. Among the coalitions with more than a Hare quota's worth of
support, let the one with least number of candidates be the "smallest
strong Hare coalition". Break ties in favor of the coalition with most
votes.
3. Perform a Condorcet (say Ranked Pairs) election limited to the
ballots that make up the smallest strong Hare coalition.
4. Elect the winner and eliminate him, and redistribute the surplus.
5. If all seats are filled, we're done.
6. Otherwise go to 1.

There are no loser eliminations here, so it should be more well-behaved
with respect to monotonicity. But it is Hare-based and so may fail to
find enough winners to begin with.

Going further along this line of reasoning brought me to what I'd call
"expansion methods". But I'll mention them later - no need to clutter up
one post with too many methods :-)


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