[EM] Voting systems theory and proportional representation vssimple representation. (Abd ul-Rahman Lomax)

[Kristofer Munsterhjelm]

Raph Frank wrote:
> On Mon, Mar 15, 2010 at 12:04 AM, Terry Bouricius
> <terryb at burlingtontelecom.net> wrote:
>> Ralph is describing the open list system used in such places as the
>> Netherlands (nation-wide) and Finland (regionally).
> 
> I was thinking of a system with 1 list per candidate, rather than 1
> list per party.  This gives the voters more choice.
> 
> Open list works the same, except that there is only 1 list per party.
> 
> However, combining candidate lists in a monotonic way would be difficult.

Transform a party list vote of the type "Party A > Party B > Party C" 
into a ranked ballot of the type: A1 > A2 > ... > An > B1 > ... > Bn > 
C1 > ... > Cn.

Then just dump that into my method, since that is monotonic. Also, when 
all voters simply vote for a single list, the method reduces to 
Webster's: each bloc (necessarily disjoint, so no constraints) gets as 
many candidates as Webster's method would have given it.

The problem with my method is that (currently) it uses exponential 
space. As a consequence, it uses exponential time as well, but if the 
exponential space could be fixed, it would only be "exponential time in 
terms of candidates, not in terms of voters", and thus similar to STV in 
that respect.

Perhaps your Sainte-Lague version (if it works out to always elect the 
same council) will let us get it down to polyspace.

> The tree method would have seats distributed between the parties and
> then within the various "wings" of the parties based on how many votes
> each party/wing/candidate obtained.

That could also be handled behind the scenes. If the tree is defined as:
   B is root
    BA has A as ancestor
    BB has A as ancestor
     BBA has BA as ancestor

and someone votes for BBA, then the proper ballot might be BBA1 > ... > 
BBAn > BB1 > ... > BBn > B > ... > Bn.
I'm not sure where BA would fit. Perhaps
  BBA1 > ... > BBAn > BB1 = BA1 > BB2 = BA2 > ... > BBn = BAn > B > ... 
 > Bn.
Even that leaves out where to put a descendant of BA.