On Tue, Oct 22, 2013 at 10:29 AM, Joerg Arndt <arndt@jjj.de> wrote:
* Andy Latto <andy.latto@pobox.com> [Oct 22. 2013 08:32]:
On Mon, Oct 21, 2013 at 3:52 AM, Joerg Arndt <arndt@jjj.de> wrote:
Pick's theorem (cf. http://en.wikipedia.org/wiki/Pick's_theorem):
Square grid, every second column shifted by a half unit: A = i/2 + b/2 - 1
where b counts only the boundary points whose local neighborhood is not concave (i.e., 90 degrees of outside, 270 degrees of inside).
What about a square of side 2? Doesn't this have i = 1 and b = 4?
I do not see which configuration you mean.
My numbers were wrong (I was confused as to what lattice you were using), but I think your formula still has a problem. Consider the square with corners at (0,0), (2,0), (0,2) and (2,2). b = 6 (because (0,1) and (2,1) are also on the boundary), and i = 2, because of (1, .5) and (1, 1.5), so your formula gives 3 instead of 4 for the area. I think that any formula that works will need to have the coefficient of i be twice the coefficient of b, so that if it holds for two regions that overlap only on a section of boundary, it will hold for their union. Andy