On 11/13/06, Chrilly <[EMAIL PROTECTED]> wrote:
There is of course the question how sure this number is. Is it some sort of proove or just an example the author has found?
A simple upper bound can be calculated by number_of_intersections*4/5, which gives 288 for the 19x19 board. Each empty intersection can be adjacent to at most 4 strings and all strings need at least one liberty. To maximize the number of strings they should all have the minimal size of 1 (larger strings can always be shrunk to one without affecting the string count). So for every 5 intersections we will have at best one liberty and 4 strings. If you take the edge into account this number will go down a bit, but apparently not enough to fit in 8 bits :-( Erik _______________________________________________ computer-go mailing list [email protected] http://www.computer-go.org/mailman/listinfo/computer-go/
