[EM] Pairwise Margins
Forest Simmons
fsimmons at pcc.edu
Mon Jan 7 12:56:13 PST 2002
So far Blake Cretney (or his clone, Blake Cretney) holds the record of 20
factions or fewer for generating an arbitrary 5X5 pairwise margin matrix.
If someone as skeptical as Blake about the value of this exercise can do
that well, surely someone else can beat it.
In Blake's method the number of factions for an n by n matrix has a second
degree polynomial growth law. Is there a constant k such that the number
of factions needed is less than k*n*log(n) ?
Forest
On Sun, 6 Jan 2002, Blake Cretney wrote:
> Blake Cretney wrote:
>
> > I don't know how many are needed by the optimal procedure, but I do
> > know how many are needed by my procedure, 2*(n^2-n)
> >
> > So, for 5X5, that is 40
>
> Blake makes some good points, but I have to quibble with that statement.
> You only need two distinct factions for every pair of candidates (not
> every cell). So, the formula is actually n^2-n.
> So, for 5X5, that gives 20.
>
> ---
> Blake Cretney
>
>
>
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