I'm guessing the answer is no because if x is 0 and integer y < 0, and you're allowed to pick whatever negative number you want, it looks like you could keep flipping between a couple of states. In fact, I think this demonstrates this:
1 1 0,0 _1 0,:0 1 1 NB. Matrix to flip to next state.... 1 1 0 0 _1 0 0 1 1 flipXYZ=: 3 : 'y. +/ . * ~ 1 1 0,0 _1 0,:0 1 1' flipXYZ 0 _1 0 _1 1 _1 flipXYZ^:2 ] 0 _1 0 NB. Back where you started... 0 _1 0 On 5/24/06, Miodrag Milenkovic <[EMAIL PROTECTED]> wrote:
There's an integer on each vertex of the pentagon. If there is a negative integer, y on one of the vertices, flanked by x and z, replace x, y, z by x+y, -y, z+y. Sum of all the numbers on the pentagon is > 0. If there is more than one negative number, pick whichever one you want. Does this procedure have to end? ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
-- Devon McCormick ^me^ at acm. org is my preferred e-mail ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
