From mindcrf!peabody.mindcraft.com!ronnie@decwrl.dec.com Fri May 10 19:16:47 1991 Return-Path: Received: from uucp-gw-1.pa.dec.com by life.ai.mit.edu (4.1/AI-4.10) id AA24712; Fri, 10 May 91 19:16:47 EDT Received: by uucp-gw-1.pa.dec.com; id AA29291; Fri, 10 May 91 16:16:33 -0700 Received: by mindcrf.mindcraft.com (AIX 2.1 2/4.03) id AA10398; Mon, 6 May 91 11:02:49 PDT Received: by peabody.mindcraft.com (AIX 1.3/4.03) id AA22955; Mon, 6 May 91 11:14:02 -0700 Date: Mon, 6 May 91 11:14:02 -0700 From: mindcrf!ronnie@peabody.mindcraft.com (Ronnie Kon) Message-Id: <9105061814.AA22955@peabody.mindcraft.com> To: @mindcrf.pa.dec.com:decwrl!ai.ai.mit.edu!Cube-Lovers Subject: 5by cubes As far as I can tell, if you can solve the order 3 and order 4 cube, you should be able to solve the order 5 with no additional fiddling, even if you only know cookbook solutions. Spoiler follows: I solve the off-center edges first (just like in the order 4 cube-- the transformations are identical), then the corners (exactly like all other orders, from 2 through 4), then the center edges (exactly like the order 3 cube, just treat the two edge faces as attached and you have an order 3 cube). All that's left are the eight centers. Four of these can be solved exactly as in the order 4, and if you can't generalize your cookbook solution to solve the remaining 4 you have no business cubing. I suspect this is why there are (and will probably never be) cubes of orders greater than 5. I believe (though have not proved) that the 5 cube contains all the complexity that is possible. Adding more cubies would only increase the amount of time needed to solve. On the other hand, I would be willing to pay a fair amount of money for an order 21 cube. :-) Ronnie