[next] [prev] [up] Date: Fri, 31 May 96 15:41:34 -0400
~~~ [prev] [up] From: Buddha Buck <phaedrus@dreamscape.com >
~~~ [prev] [up] Subject: Re: An Ultimate Cube

On the subject of an ultimate cube...

There can not possibly be an ultimate cube just like (because) there is
no "ultimate", ie., largest, integer. But if you can solve the
(N+1)x(N+1)x(N+1) cube then you can surely solve the NxNxN cube. You
would simply reduce the (N+1)x(N+1)x(N+1) permutation to one that is in
the NxNxN group and continue from there like MBW did when looking for
God's algorithm in the 3x3x3 group.

I would not necessarily agree with the assertion that if one can solve
an N^3 cube, one can solve an (N-1)^3 cube.

Your construction, of solving the N^3 cube into a subgroup that is
homeomorphic to the (N-1)^3 cube, assumes that one already knows how to
solve a (N-1)^3 cube.

The group of an NxNxN cube is a proper subgroup of an (N+1)x(N+1)x(N+1)
cube. For example, the 2x2x2 cube group is the 3x3x3 group minus the
edge moves and the center cubie orientation moves - that is, as
Singmaster pointed out, it is just the corners of the 3x3x3 cube. Adding
the 3rd cut added 2 additional types of cubies to the 2x2x2 cube, the
edges and the centers, and along with them came the edge moves (to form
the group of the 3x3x3 cube) and the center orientations (to form the
3x3x3 super-group). The edge moves alone are a proper subgroup of the
cube group and the cube group is a proper subgroup of the super-group.

True, and I will conceed that if you know how to solve a 3x3x3, you can
solve a 2x2x2.
> A similar situation occurs when you go from the 3x3x3 cube to the 4x4x4
> cube. If you constrain the cube so that the central 2 slices can not be
> moved independently of one another then the 2 central edge pieces act
> exactly like the edges of a 3x3x3 cube and the 4 face center pieces act
> exactly like the face centers of the 3x3x3 cube. When the central slices
> are allowed to move independently of one another permutations are added
> to the 3x3x3 group and super-group to make up the 4x4x4 group and
> super-group. Thus the 3x3x3 groups are proper subgroups of the 4x4x4
> groups.

Yes, the 3x3x3 groups are proper subgroups (or, probably more
accurately, homeomorphic to proper subgroups) of the 4x4x4 groups, but
that doesn't mean that knowing how to solve the 4x4x4 allows one to
solve the 3x3x3.

For instance, I can solve a 4x4x4. However, my solution to the 4x4x4
involves slice moves that don't exist on a 3x3x3 cube, through all
stages of my solution, including the final stage. I cannot directly
apply my 4x4x4 solution to a 3x3x3 cube. (I can to the 2x2x2 cube,
since the techniques for solving the corners are applicable to cubes of
all order N). If my solution for solving the 4x4x4 involved reducing
it to the subgroup of the 4x4x4 generated by face turns only, then yes,
I could directly solve a 3x3x3 by the methods I use for a 4x4x4, but I

The pattern continues as the value of N increases with the N+1 group
being larger than the N group and properly containing the N group. So
the answer is no, there is no ultimate cube.

The question originally asked (by Aaron Wong) was "Is there an ULTIMATE
Rubik's cube that, if an algorithm for it was known, it would contain
an algorthm for ANY Rubik's cube?"

There might be no answer to the general question of if -any- algorithm
was known for the U^3 cube, than an algorithm could be derived for any
N^3 cube. For instance, few here would argue the assertion that if you
can solve a 3x3x3, you can solve a 2x2x2, but from the discriptions
I've heard of it, I wonder how well Thistlewaite's algorithm would work
on a 2x2x2 cube.

A Thistlewaite type algorithm for a (2N)^3 cube might very well reduce
the (2N)^3 cube to the subgroup that is equivilant to a 2^3 cube in its
final stages. Such an algorithm would be totally unsuited for solving
a (2M+1)^3 cube, because there would be no way to reduce that to a 2^3
cube. (In general, I would guess that any algorithm for an n^3 cube
that involved reducing it to an m^3 cube, where n = km, would be
unsuited for solving a l^3 cube, where l does not have n or m as a

However, I think the question can be divided into two parts, if we look
at it differently (requiring the existance of an algorithm with the
stated property for order U^3 cubes, rather than requiring that all
algorithms for order U^3) cubes have the stated property): First, is
there a general algorithm that can be used to solve cubes of all
orders? I think the answer is "yes". Second, what is the smallest
order U^3 cube requiring a complete description of the algorithm? I
think the answer is U=5.

My current solution for the 4^3 cube is very closely related to my current solution for the 3^3. There are only minor changes in one stage, major changes in another, (both to deal with the split edge pieces) and the addition of a completely new stage to handle the centers, which aren't in the 3^3 at all. Transforming this algorithm to the 5^5 and higher is relatively easy, once I have the 3^3 and 4^4 down. All the important components of the two lower order solutions are needed for the 5^5, and nothing really new is added. The same goes for the higher orders. The tedium of solving increases, but not the real difficulty.

I have been thinking (but haven't done much yet) of writing a collection of web pages describing my general solution (at least, for the 2^3, 3^3, and 4^4 cubes).

Buddha Buck bmbuck@acsu.buffalo.edu
"She was infatuated with their male prostitutes, whose members were
like those of donkeys and whose seed came in floods like that of
stallions." -- Ezekiel 23:20

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