Kristofer Munsterhjelm wrote (29 June 2013):
"The combined method would go like this:
1. Run the ballots through RP (or Schulze, etc). Reverse the outcome ordering
(or the ballots; these systems are reversal symmetric so it doesn't matter).
Call the result the elimination order.
2. Distribute seats using Sainte-Laguë.
3. Call parties that receive no seats "unrepresented". If there are
unrepresented parties, remove the unrepresented party that is listed first in
the elimination order.
4. Go to 2 until no party is unrepresented.
This should help preserve parties that are popular as second preferences but
not as first preferences, because the elimination order will remove parties
that hide the second preferences before it removes the party
that is being hidden, thus letting the second-preference party grow in support
before it is at risk of being eliminated.
Note that this doesn't solve the small-council problem. If we have:
46: L > C > R
44: R > C > L
10: C > R > L
1 seat,
then the first seat goes to L just like in Plurality. The elimination order
never enters the picture."
Kristopher,
I don't see this. Your elimination order is obviously L, R,C. R and C are
"unrepresented" so we eliminate R.
Then we have
46: L
54: C
Then we redistribute the seat to C and then eliminate L and confirm the final
redistribution.
But I'm not on board with the spirit of this method, because it seems to give a
say to voters who are efficiently represented a say in which party/candidate
will represent other voters.
Chris Benham
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