# [EM] A monotonic proportional multiwinner method

Raph Frank raphfrk at gmail.com
Sun Mar 14 05:12:10 PDT 2010

```On Sun, Mar 14, 2010 at 10:26 AM, Kristofer Munsterhjelm
<km-elmet at broadpark.no> wrote:
> But that doesn't narrow it down to {AC}, since both AC and BC share "at
> least one of AC", namely C. So am I doing it wrong here? You would have to
> go to ABC (at least 3 of these) then "one from A" before it reduces to {AC}.

Ahh true, I misapplied the method.  As you said, I needed to continue
to the "A" rule.

>> Tie between: AC-1, BC-1, B (5)
>> {None}
>>
>> Tie break, continue with allocation:
>> ... C (4)
>> ... {AC-1 and BC-1} remain
>> ... {C} wins
>
> If the tiebreak was broken in favor of B, B would win. However, in my
> method, C wins (period). You'd have to break the tie both ways and then
> somehow gather the margins data in order to exclude B.

The point was that the tie break is accomplished by continuing down
the list.  I think this is what your margins does.  It scans for new
conditions that weren't quite activated.

Actually, what it does is stop before the contradiction and then go
into tie- break mode.

With your second example:

ABC              12
AB                2
AC                5
BC                5
A                 3
B                 5
C                 4

ABC (12 -> 4)
{All}

Tie between: AC-1, BC-1, B (5)

Tie break mode
All solutions which share a common winner with any of the current
coalition are
{A,B,C}

Tie break, continue with next coalition:
C (4)
{C} wins

I think also if you hit a step which would eliminate all remaining
coalitions, then you skip that step.

Another option would be to just have the rule that you run through all
the conditions.  If the current conditions (i.e. a set of conditions
with the same score) will eliminate all the remaining coalitions, then
instead only eliminate coalitions that fail to meet any of the current
conditions.  If that still eliminates all remaining coalitions, then
skip the step.

```