Re: Perl Translation

On Tue, 18 Jan 2005, Daniel Martin wrote:

> Now, the easiest group to impose on (0..N-2) is the group you get by
> using addition modulo N-1.  This then leads to:

> g_i(x) = (2*i - x) % (N-1)

If you can find a starter, a set of pairs whose differences are
all the differences (plus or minus) possible in the group, Z(N-1) once
and only once, to represent the first round, then you can
generate the rest of the rounds by adding one to each element,
except for the point at infinity.

Eg, 
Inf & 1, 0 & 2 (mod 3) has one difference of +-1
Inf & 2, 1 & 0
Inf & 0, 2 & 1

Z(5) has +-1, +-2
A starter is 
Inf & 4, 0 & 3, 1 & 2

The other rounds are:
Inf & 0, 1 & 4, 2 & 3
Inf & 1, 2 & 0, 3 & 4
Inf & 2, 3 & 1, 4 & 0
Inf & 3, 4 & 2, 0 & 1

The problem is finding the starter. I don't think any of the
solutions to the problem so far are such a cyclic series of
rounds.

-- 
Greg Matheson, Taiwan