At 06:02 09.12.99 , Markus Schulze wrote:
>Dear Craig,
>
>it seems to me that you walked into Saari's trap.
>
>Although you don't promote the Borda Method, you
>use Saari's geometrical model of elections. But
>Saari's geometrical model of elections implicitely
>presumes that there is an even distribution of
Why not make it explicit?.
>candidates. Otherwise it wouldn't be possible that
>the Saari cube is spanned only by the candidates
>(and not e.g. also by the voters). But if there is
>an even distribution of candidates, then the Borda
>Method is _obviously_ the unique best possible
>method.
The last sentence has got to be totally false.
Proof by hand waving.
>
>Although you don't yet promote the Borda Method,
I am never going to promote Borda, actually.
>you check every proposed criterion for compatibility
>with Saari's model and you accept a criterion only
I never use probability; rather I use logic of geometry (inequalities
return Boolean values).
>if it is compatible with Saari's model and you
>reject a criterion if it is incompatible with
>Saari's model. If you keep using Saari's model, then
>in the long run you will necessarily get to the
>conclusion
that Shulze's tank got bombed, and there may not be any best way
to help Mr Schulze out with his arguments.
When I tried to read Saari's book it was gone (or lost).
Have I ever referred to a cube?.
The fully general two 3 candidate preferential voting problem, can be
solved by considering the interior of a tetrahedron (having vertices:
AB, AC, B, C). I have two A4 sheets here that contain a 1 winner
IFFP solution derivation. It is based on (P1). Adding (1,1,1,...) to
(x:A,y:B,z:C,...) makes no difference to winners, etc..
Mr G. A. Craig Carey, [EMAIL PROTECTED]
Auckland, New Zealand.
Snooz Metasearch: <http://www.ijs.co.nz/info/snooz.htm>