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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

don't.

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

factor).

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