It's still 18. Note first that each 7x7x1 plane can contain at most one entire brick, so in order to fit more than 18 bricks, one of the directions must have one entire brick in each plane; call these the base set. Looking at each of the other dimensions, there are 28 1x4 cross-sections of the base set distributed amongst the 7 planes. Since no plane can have more than 7 of these (one from each brick in the base set), there must be at least two planes with at least 4 such cross-sections (5*3+6+7=28). A little fiddling will show that it is impossible to fit a brick and 4 parallel cross-sections into a single plane (they have to be parallel because they come from bricks aligned in the same dimension). So each of the other dimensions can have at most 5 bricks, for 17 total, contrary to our assumption. Franklin T. Adams-Watters -----Original Message----- From: dasimov@earthlink.net To: math-fun <math-fun@mailman.xmission.com> Sent: Tue, 14 Feb 2006 18:09:52 -0500 (EST) Subject: Re: [math-fun] AB=B.AA; Gosper's packing problem 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. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ___________________________________________________ Try the New Netscape Mail Today! Virtually Spam-Free | More Storage | Import Your Contact List http://mail.netscape.com