The conjecture that the 2x3x1s make the puzzle strictly harder is still alive. (I.e., there are significantly fewer solutions, harder to find.) But the stronger conjecture, that you might as well glue the 2x3x1s back into a 2x2x3, is false. http://gosper.org/7x7x7.jpg shows a solution where they're not even touching. (Whitish, open-faced.) --rwg On Thu, Nov 1, 2012 at 9:37 PM, Bill Gosper <billgosper@gmail.com> wrote:
I sent this to the kids last week. Neil 3D printed two 2x3x1s and searched manually, with Burrtools, and with a too-slow Mathematica program, and was unable to find a solution in which the new pieces relaxed the constraints on the other key pieces. (Not firm) conclusion: Replacing the three 2x2x1s by two 2x3x1s makes the puzzle harder (for anyone who doesn't already know how to solve it). --rwg But replacing the the 2x2x2 and 2x2x1 by two 2x3x1s makes Conway's 5x5x5 *much* easier.http://gosper.org/7%5E3leg.pdf
---------- Forwarded message ---------- From: Bill Gosper <billgosper@gmail.com> Date: Wed, Oct 24, 2012 at 12:00 PM To: Neil Bickford <techie314@gmail.com> Cc: Julian Ziegler Hunts <julianj.zh@gmail.com>, Jon Ziegler < jonathan.zh@gmail.com>, Corey Ziegler Hunts <corwin.zh@gmail.com>, Michael Beeler <mikebeeler@verizon.net>, rcs@xmission.com, stan@isaacs.com
Working Stan [Isaac]'s copy at the Gardner thing, I was reminded that it always seems easy to permute
a solution so that the three 2x2x1s form a 2x2x3. The puzzle might be harder if we force this configuration by replacing the 2x2x1s with two 2x3x1s (e.g. by truncating two spare 2x4x1s), assuming that the temptation to use a 2x3x1 singly in the solution proves both frequent and fatal. If nonfatal, it would probably make the puzzle way too easy. --Bill