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?
the sum monotonically decreases ... eventually you will go under 0. i presume that is end game? ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
