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Solving the 1x1x1 Rubik's cube is probably a bit silly and whimsical, but

let's look at it anyway.

I was led in this direction by rereading some articles in the archives

from Dan Hoey and others concerning NxNxN Rubik's cubes. For example,

consider Dan's discussion "Cutism, Slabism, and Eccentric Slabism" from

1 June 83 19:39:00. Sometimes degenerate cases are slightly interesting.

I guess the 1x1x1 case is the most degenerate we have, unless you want to

consider the 0x0x0.

It seems to me that either cutism or slabism, as Dan calls them, reduce to

whole cube rotations for the 1x1x1 case. For example, a quarter turn face

turn or a quarter turn slice would be interpreted as a whole cube quarter

turn for the 1x1x1. Hence, the cube group for the 1x1x1 is simply C, the

group of 24 rotations of the cube.

By analogy with some of our previous work, I can think of essentially

three different ways to model the 1x1x1.

1) With the 2x2x2, we normally wish to consider the puzzle solved if

each face is all of one color. That is, whole cube rotations are

to be considered equivalent. With the Singmaster fixed face

center view of the 3x3x3, the issue of whole cube rotations does

not arise. But with the 2x2x2 we would normally consider

(for example) RL' equivalent to I. The most common way to

accomplish this type of equivalence is to fix one of the corners.

If we fix one of the corners of the 1x1x1, then we have a most

remarkable puzzle. There is only one state, nothing can ever

move, and the puzzle is always solved.2) A second way to model the 2x2x2 such that whole cube rotations are

considered to be equivalent is to consider the set of states to be

the set of cosets of C, that is, the set of all xC.

If we take this approach with the 1x1x1, then there is only one

coset, namely iC (or just C, if you prefer). The cube can rotate,

but all 24 states are considered to be equivalent and the puzzle

is always solved.3) Finally, if you model the 2x2x2 in such a way that whole cube

rotations are considered to be distinct, then you are really

modelling the corners of the 3x3x3. Indeed, a naive program

that simply modelled the permutations of the 2x2x2 facelets would

in fact unwittingly be modelling the corners of the 3x3x3.

If you take the same approach of modelling the permutations of the

1x1x1 facelets, then you in effect are considering whole

cube rotations to be distinct. You have a very easy problem,

but the problem is not totally trivial as it is with approach #1

or approach #2. The rest of this note will therefore consider the

problem of the 1x1x1 cube where whole cube rotations are considered

to be distinct.

Since we need to deal with whole cube rotations, I will use lower case

letters as our standard E-mail simulation of Frey and Singmaster's script

notation for whole cube quarter turns -- t for Top, r for Right, etc. We

need only three of the six letters because, for example, we have l=r',

d=t', b=f', etc. I will use t, r, and f.

We know before we start that there are 24 states. We also know before we

start that these 24 states form 5 M-conjugacy classes, where M is the set

of 48 rotations and reflections of the cube. (There are 10 M-conjugacy

classes of M, of which 5 are rotations and 5 are reflections.) Hence, any

discussion of God's algorithm will involve 5 conjugacy classes and 24

states.

The obvious searches to look at are for qturns only, and for qturns plus

hturns. We may generate the qturn case as C=<t,r,f>. We may generate

the qturn plus hturn case as C=<t,r,f,t2,r2,f2>.

Qturns OnlyDistance Conjugacy Positions

from Classes

Start0 1 1 {i} 1 1 6 {t,t',r,r',f,f'} 2 2 11 {tt,rr,ff},{tr,tr',tf,tf',t'r,t'r',t'f,t'f'} 3 1 6 {ttf,ttf',ffr,ffr',rrt,rrt'} --- ---- ---- Total 5 24

Qturns Plus HturnsDistance Conjugacy Positions

from Classes

Start0 1 1 {i} 1 2 9 {t,t',r,r',f,f'},{t2,r2,f2} 2 2 14 {tr,tr',tf,tf',t'r,t'r',t'f,t'f'}, {t2f,t2f',f2r,f2r',r2t,r2t'} --- ---- ---- Total 5 24

There are some additional problems we can look at. For an example, an

interesting problem on the 3x3x3 is variously called the stuck axle

problem or the five generator problem. In the case of the 1x1x1, we have

the "two generator problem" because we certainly can generate C as C=<t,f>

(Proof: r=tft'). But can we generate C with only one generator? The

answer is no. (Proof: Order(i)=1, Order(t)=4, Order(tt)=2, Order(tf)=3,

and Order(ttf)=2. All the orders are less than 24. Note that it suffices

to calculate the order for one representative of each conjugacy class.) I

will leave it as an exercise for the reader to determine the lengths of

each of the 24 positions if we generate C as <t,f>, and to determine the

appropriate conjugacy classes to take into account the symmetry of C

generated as <t,f>.

By the way, do we know the minimum number of generators required to

generate the 3x3x3? Here I do not mean the minimum number of quarter

turns. I am asking the question if we are permitted to use as generators

any elements of G.

Here is one final item about the 1x1x1. We do not know how many subgroups

of G there are for the 3x3x3. But we do know how many subgroups of C

there are. There has been much discussion of the 98 subgroups of M which

can be arranged in 33 conjugacy classes. The subgroups of C are simply

those subgroups of M which consist entirely of rotations. There are 30

such subgroups, and they may be arranged in 11 conjugacy classes.

= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us