From cube-lovers-errors@mc.lcs.mit.edu Fri Dec 4 14:26:23 1998 Return-Path: Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38]) by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id OAA05692 for ; Fri, 4 Dec 1998 14:26:23 -0500 (EST) Precedence: bulk Errors-To: cube-lovers-errors@mc.lcs.mit.edu To: cube-lovers@ai.mit.edu From: whuang@ugcs.caltech.edu (Wei-Hwa Huang) Subject: Method for Solving the 4x4x4 Date: 4 Dec 1998 17:26:00 GMT Organization: California Institute of Technology, Pasadena Message-Id: <7495v8$3nv@gap.cco.caltech.edu> References: I have been told that my solution for the 5x5x5 includes knowing how to solve the 4x4x4, which is of course not trivial. With the post asking about the parity problem, I thought I might as well post my solution to the 4x4x4. Yes, the biggest barrier is the parity problem where two adjacent edge cubies are flipped. My earliest attempt at a 4x4x4 solution was the following: 1. Match all the centers. 2. Match all the edges. 3. You now have a 3x3x3. Solve. Unfortunately, with the parity problem you can often end up with an unsolvable 3x3x3 by the time you get to step 3. Any simple moves that fix the parity problem tend to mess up the rest of the cube quite badly -- I wrestled with this problem a long time until I realized one thing: Most solutions of the 3x3x3 treat the centers as static, using them as "anchors" for the entire cube. But this is entirely unnecessary! If you solve the 3x3x3 while IGNORING the centers, you will eventually get a solved cube where the centers are either in the "6 dots" or "4 dots" situation well known to cubists -- and these have rather simple solutions, essentially consisting of a slice turn conjugated with another slice turn. So, my most favorite 4x4x4 solution is now: 1. Match all the edges. 2. Solve the parity problem, if necessary (postpone until after step 3 if desired). 3. Ignore the centers and treat the cube as a 3x3x3. Solve. 4. Solve the centers. Okay. Now to qualify the solution. Part 1 is simple and can be done anyway you wish (the move rF2r'F2 will be rather useful in the later stages). Part 3 is simple, with the caveat that you may be treating the "centers" in the wrong manner! Part 2 stems from the fact that the cube apparently has an "even" permutation (a 2-cycle involving two edge pieces), an apparent paradox since 2-cycles should not exist (e.g., on the 3x3x3 it is impossible to swap exactly two edges). The reason this is only an "apparent" paradox, however, is because of the misassumption that the centers of the 4x4x4 are static, which they certainly are not! In fact, just rotate one slice incident on your 2-cycle, and you have magically turned the 2-cycle into a 5-cycle, which is perfectly solvable! Personally, I solve the 5-cycle by two or more 3-cycles, which generally take on the form: FR'F' r FRF' r' This move performs a cycle on the three edges fUR, FUr, and FDr, without disturbing the corners, but doing rather annoying things to the centers. (This move is an extension of the perhaps-not-so-well-known sequence for the 3x3x3: FR'F'LR'DRD'L'R that rotates 3 edges.) For FR'F' you may substitute any sequence of moves that brings your desired edge piece (in this case fUR) to the FUr position, as long as it does not disturb any other edges on the r slice (specifically, the edges FDr, BUr, and BDr). You will also have to substitute the inverse of your sequence for FRF'. As an example, the move F'L2F r F'L2F r' cycles BdL, FUr, and FDr. You may also use r2 instead of r and r', which means that BDr is affected instead of FDr. And finally, step 4: the centers. This is solved by a generalization of the "6-dots" rule. This move creates "6 dots": u'r'ur This permutes bUr, Fur, and buR, as well as their "opposites" fDl, Bdl, and fdL, while affecting no other cubies. This is two 3-cycles on 6 faces, which is rather unwieldy, so I conjugate (is that the right word?) it with a simple face turn to get u'r'ur F r'u'ru F' which permutes Fur, Fdr, and buR in a simple 3-cycle. Both Fur and Fdr are on the same face, which makes this move rather easy to deal with. Especially, if one of those is the right color already, it can be involved in the 3-cycle without "increasing error." I think I may have extended my personal jargon a bit more into this post -- if you wish to understand anything in this post or, conversely, would like to teach me more "Standard" jargon, please e-mail me. -- Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/ --------------------------------------------------------------------------- O*e T*o: "Thre* *our fi*e s*x; se*en *ight *ine, *en!"