Xref: utzoo sci.math:9045 comp.theory:128
Path: utzoo!utgpu!jarvis.csri.toronto.edu!cs.utexas.edu!sun-barr!rutgers!cbmvax!amix!ag
From: ag@amix.commodore.com (Keith Gabryelski)
Newsgroups: sci.math,comp.theory
Subject: Re: Rubik's Cube Problem
Message-ID: <216@amix.commodore.com>
Date: 19 Dec 89 22:04:53 GMT
References: <5310@garfield.MUN.EDU>
Reply-To: ag@amix.commodore.com (Keith Gabryelski)
Followup-To: sci.math
Organization: Commodore Amix Development
Lines: 66

In article <5310@garfield.MUN.EDU> chris2@garfield.MUN.EDU (Chris Paulse)
writes:
>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?

Negative.  There are certain arangment of squares that are not
``reachable'' from the solved Rubik's form.

For instance, there will never be a Rubik's cube, that started from
the solved form and was made by normal turns of the cube, in the a
form were all cubes are in their correct place and only one edge piece
is flipped:


	OOO
	OOO
	OOO

	BBB RRR GGG YYY
	BBB RRR GGG YYY
	BWB RRR GGG YYY

	WBW
	WWW
	WWW


*------------------------------*

As I remember, this is also ``illegal''.

	OOO
	OOO
	OOO

	BBB RRR GGG YYY
	BBB WRR GGG YYY
	BBB RRR GGG YYY

	WRW
	WWW
	WWW

One corner pieces flips are ``illegal'', also:

	OOO
	OOO
	OOB

	BBR ORR GGG YYY
	BBB RRR GGG YYY
	BBB RRR GGG YYY

	WWW
	WWW
	WWW

This is all from ~10 year old memory, but I am pretty sure of my
assertions.

Pax, Keith

Ps, follow-ups to sci math in hopes someone has a proof.
-- 
ag@amix.commodore.com        Keith Gabryelski          ...!cbmvax!amix!ag