From cube-lovers-errors@mc.lcs.mit.edu Wed Oct 1 13:18:18 1997 Return-Path: Received: from sun30.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.1/mc) with SMTP id NAA25712; Wed, 1 Oct 1997 13:18:15 -0400 (EDT) Precedence: bulk Errors-To: cube-lovers-errors@mc.lcs.mit.edu Mail-from: From ERCO@compuserve.com Wed Oct 1 01:56:54 1997 Date: Wed, 1 Oct 1997 01:52:24 -0400 From: Edwin Saesen Subject: Re: 4x4x4 solution -- [Digest v23 #159] Sender: Edwin Saesen To: CUBE Message-Id: <199710010152_MC2-225C-6120@compuserve.com> jbryan@pstcc.cc.tn.us wrote: >Your problem is one of parity. You have two edges cubies swapped >(this swap is visible) and two face center (centre) cubies of the >same color swapped (this swap is invisible). You have to have an >even number of swaps in the total cube. If you want an even number >in the edges (and you do), then you also have to have an even number >in the face centers, even if swaps in the face centers are invisible. I've had this problem as well. If I understand you correctly, this problem simply doesn't occur anymore as soon as you number (or mark in any other way) the center pieces which a) makes solving the cube a bit more difficult b) makes sure that you'll always get back to the original configuration of center pieces. I've had a similar problem on my 5x5x5 as well, and I assume that marking the nine center pieces might solve the problem as well. On my 4x4x4 I also had a problem of having two pairs of edges exchanged which simply can't happen on a 3x3x3. By experimenting with 3x3x3 moves I found a 24move solution to this, and I wonder if that's also sort of automatically solved by marking center pieces. Can anyone confirm this? Michael Ehrt --------------------------------------------- ERCO@compuserve.com