Xref: utzoo sci.math:9037 comp.theory:126 Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!cs.utexas.edu!swrinde!zaphod.mps.ohio-state.edu!usc!ucsd!ucbvax!ernie.Berkeley.EDU!rice From: rice@ernie.Berkeley.EDU (Daniel S. Rice) Newsgroups: sci.math,comp.theory Subject: Re: Rubik's Cube Problem Message-ID: <33312@ucbvax.BERKELEY.EDU> Date: 19 Dec 89 21:13:10 GMT References: <5310@garfield.MUN.EDU> <4330@rtech.rtech.com> Sender: usenet@ucbvax.BERKELEY.EDU Reply-To: rice@ernie.Berkeley.EDU.UUCP (Daniel S. Rice) Organization: University of California, Berkeley Lines: 22 > == Mike Schilling >> == Chris Paulse >> If I had a solved Rubik's cube, and the colors on each face were >> just stickers on the black plastic surface, if I exchanged some >> of the stickers, would the cube still be solvable in the normal way? >> >No, there's three kinds of *parity* preserved by any legal move: [ Argues correctly that cube positions fall into 12 orbits] Mike's response correctly notes that *switching positions of cubies* can result in 11 families of unsolvable cubes. The original post asked about moving *stickers*. Clearly, there must be one side cubie that is blue and red, one that is blue and green, etc., etc., for the puzzle to even make sense. A corner cubie that is, say, all orange can never be part of a solved cube. So, strictly speaking, there are many more than 11 unsolvable cube families if replacement of stickers is allowed. I suppose this is a nitpick but it seemed as if others were answering the wrong (but mathematically much more interesting) question. Dan