frans@duvel.UUCP (Frans Meulenbroeks) (08/23/85)
Some weeks ago I aquired a 4x4x4 version of Rubik's cube.
It is somewhat more difficult than the 3x3x3 version.
Anyway, most of the time I can fix it by now.
I've only one problem left:
Is it possible to have a situation in which the two middle pieces of an
edge are twisted/turned upside down, with the remainder of the cube left
intact?
I sometimes encounter this situation, but have not yet found a way to
solve it.
Does anyone know if this is possible at all, or am I being fooled by one of
my friends?
Please, just answer yes or no. If it is possible I will try to find out
how by myself.
--
Frans Meulenbroeks, Philips Microprocessor Development Systems
...!{seismo|philabs|decvax}!mcvax!philmds!fransmer@prism.UUCP (08/27/85)
Not an answer to your question but your query made me think "wouldn't it be neat to have a 4x4x4x4 Rubik's cube", since that's the logical extension. I wonder how you'd solve it.
scs@wucs.UUCP (Steve Swope) (08/31/85)
In article <3600005@prism.UUCP>, mer@prism.UUCP writes: > > Not an answer to your question but your query made me think "wouldn't it > be neat to have a 4x4x4x4 Rubik's cube", since that's the logical > extension. I wonder how you'd solve it. Such a device exists. It's called "Rubik's Revenge", is manufactured by Ideal*, and (I've heard) is alse designed by Erno Rubik. I've also heard that he has developed 5^^3 and 6^^3 extensions, and is working on a 7^^3 extension (none of these have been marketed, to my knowlege *sigh*). I solved the 4^^3 version by gathering the centers (each 4 subcubes) and edges (each 2 subcubes) and applying my 3^^3 solution to that. Incidentally, Ideal* also makes "Pocket Rubik's Cube", a 2^^3 version. (It's equivalent to solving only the corners on a 3^^3.) *DISCLAIMER: Ideal is somebody's trademark, and I've never even considered working for them.