Rich writes: << Gosper's packing puzzle is to fit as many 4x4x1 bricks as possible into a 7x7x7 cube. I'm assuming an "orthogonal, registered" fit, with each brick oriented parallel to a cube face, and placed at integral subcube boundaries. The answer is 18 bricks. [Here is a packing of 18.] My proof is the same as Michael Reid's: Consider the (orthogonal) lines going through the center of the 7x7x7 cube. Take the union, remove the center cell. Resulting collection of subcubes has 18 cells. Any 4x4x1 brick must occupy at least one of these cells. Note that the simple volume constraint gives an upper bound of 7^3/4^2 = 343/16 = 21.4375.
Cute proof. As usual I'm interested in the torus version of this cube puzzle. QUESTION: Suppose we identify the opposite faces of this 7x7x7 cube, to get a 7x7x7 three-torus T. How many registered* 4x4x1 bricks will fit ? (Of course from the cube solution & the volume constraint, the answer lies in {18, 19, 20, 21}. And the proof of Michael & Rich for the cube no longer works for the torus.) --Dan ____________________________________________ * To register your bricks, just send me a check for $50.