Re: Combinatoric Problem

On Sun, 04 May 2008 20:32:00 +0100, I wrote:

(0 1 2 3) (0 1 4 7) (0 1 10 11)
(0 2 5 8) (0 2 10 11) (0 3 6 9)
(0 3 10 11) (0 4 5 9) (0 4 6 8)
(0 5 6 7) (0 7 8 9) (1 2 4 7)
(1 2 5 8) (1 3 4 7) (1 3 6 9)
(1 5 9 10) (1 6 8 11) (2 3 5 8)
(2 3 6 9) (2 4 9 11) (2 6 7 10)
(3 4 8 10) (3 5 7 11) (4 5 6 10)
(4 5 6 11) (7 8 9 10) (7 8 9 11)

(Some blocks are repeated in the design, but we don't need that.)

So we know that the answer is <= 27.

In fact, it's less than that, possibly quite a lot less! Some of
the games in the above list are redundant, from the point of view
of the present problem (ignoring for the moment the uncertainty -
or my uncertainty! - as to whether pairing is involved). Starting
from a "design" that solves a different problem might turn out to
be a very bad idea, and one might get better results just making
up solutions in an /ad hoc/ way. It's probably best to ignore all
this ...

--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril
.

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