[EM] A monotonic proportional multiwinner method

Raph Frank raphfrk at gmail.com
Mon Mar 15 07:48:44 PDT 2010

Ok, a slight restatement, which makes the operation less complex.
Also, it eliminates the need for a transition from the standard
process to the tie-breaking rule.

Sort conditions by

(total support)/(2*number of seats - 1)

i.e. Sainte-lague divisors


if (ABC)>(all others) has a support of 28, then

ABC-1 would have a score of 28/(2*1 - 1) = 28
ABC-2 would have a score of 28/(2*2 - 1) = 28/3 = 9.3
ABC-3 would have a score of 28/(2*3 - 1) = 28/5 = 5.6

When 2 conditions have the same score, they are processed together, as
a single step.

Go through the list and at each step, for each uneliminated result,
work out the number of conditions at that step that have been

After each step, eliminate all results except for the results that are
tied lowest for the number of contradictions so far.  At the start,
this will likely just eliminate all results, except those which meet
all the conditions.

Keep going until there is only 1 uneliminated result left.

Effectively, each result has a contradiction score

C(R, S) = number of contradictions for result R, in steps 1 to S

Moving a condition upwards (by voting for it), will decrease (or keep
equal) the score for results that are inconsistent with it and have no
effect on consistent results.  Thus, I think that ensures that it is
monotonic.  However, the elimination process might lead to

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