no, it's constant x + y + z + u + v = (x+y) + (-y) + (z+y) + u + v
On 5/25/06, greg heil <[EMAIL PROTECTED]> wrote:
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
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