Aug 1, 2007 at 4:51 PM
Edited Oct 31, 2007 at 8:14 PM

Regarding the maximum number of blocks on a board:
John Tromp <john.tromp@gmail.com> wrote:
> which confirm yours. we also found a general formula n^2  floor((n^2+4n16)/5)
Gunnar Farneback <gunnar@lysator.liu.se>wrote:
> The formula can also be written floor(4n(n1)/5+4) for a slightly more compact expression.
This may be useful when optimizing the tracking of adjacement blocks.



I don't remember creating any generic formula for the maximum number of blocks on the board. That formula may be the work of someone else.



Thanks for the correction; it was actually John Tromp <john.tromp@gmail.com> that posted the formula on the computergo mailing list.

