# Message #2134

From: David Vanderschel <DvdS@Austin.RR.com>

Subject: Faulty Logic Counting States for 4D Center-Edge Cubies

Date: Thu, 10 May 2012 21:51:55 -0500

With the renewed interest in counting states which

Charlie Mckiz’s attempt stirred up, I have taken a closer look

at one aspect of the problem

I am talking here about cubies of the 4D puzzle which have 3

stickers and lie in a center slice. To my knowledge, there

are three persons who have publicly addressed the issue of

counting the number of distinct ways in which these cubies

can be arranged:

Eric Balandraud here:

http://www.superliminal.com/cube/permutations.html

Charlie Mckiz here:

http://games.groups.yahoo.com/group/4D_Cubing/message/2095

Brandon Enright here:

http://games.groups.yahoo.com/group/4D_Cubing/message/2094

They all reach the same valid conclusion; but I do not think

any of them have defended it in a rigorous manner. The good

news is that there is a simple argument which is rigorous,

and I will get to that later after I indicate what I find

lacking in the presentations I am aware of.

Eric wrote:

```
Every 3-colored can have 3! positions on one place,<br>
except for the last one, which can only have 3<br>
positions, ...
```

Brandon wrote:

```
The orientation of the 32nd 2-color edge can only be in<br>
half (3) of the orientations.
```

The quotes from Eric and Brandon above are correct

statements, but they are inadequately defended in my

opinion. Brandon did go on to add:

```
The only "tricky" portion of the calculation is counting<br>
the orientations for the last 3-color and the last<br>
4-color piece. I can create a macro to demonstrate how<br>
to put a single 3-color piece or a single 4-color piece<br>
in these reduced orientations if you'd like.
```

But that does not prove much as relates to a fully scrambled

puzzle. What requires proof here is not what you can do,

but what you cannot do.

Charlie wrote:

```
The net permutation of rotation of all 3-color pieces is<br>
even.
```

What Charlie wrote does not make much sense. He has a

semantic problem using the word "rotation", as that does not

mean much when addressing orientations of cubies in a

general sense in 4D. (E.g., there exist reorientations

which have no fixed axis at all and reorientations which

could be viewed as simultaneous rotation in 2 2D planes.)

It would be more meaningful had he used the word

"orientation" instead of "rotation". However, even then

there are difficulties. When one of these cubies remains in

its home position, then it is meaningful to talk about the

permutation of its stickers because we have a reference for

the unpermuted state. However, when the cubie is in some

arbitrary other position in the pile, we lose our reference

and it is not possible to say whether its stickers are

permuted or not. (Or, if it is possible, it won’t be easy

and Charlie obviously did not do it.) If you cannot talk

about permutation of ‘rotation’ of an individual cubie, it

makes little sense to talk about the net permutation of

‘rotation’.

What I suspect: The statements are all based on folks’

experience with attempting to solve the puzzle. In

particular, it is possible get the puzzle into a state in

which it is solved except for one of these cubies, that

cubie is in its home position, and the permutation of its

stickers is even. However, the fact that no one has ever

seen that last cubie with an odd permutation of its stickers

does not constitute _proof_ that the sticker permutation on

the one disoriented cubie cannot be odd. I claim that such

proof is needed. Furthermore, I do not think that there is

any such proof which follows in an obvious manner from what

the above 3 have written.

The above reasoning is based just on the situation when all

the cubies are in their initial positions. When the puzzle

is completely scrambled, the constraint, whatever it is,

becomes even less clear. (I pointed to some of the

difficulties in my discussion of Charlie’s statement.) In

fact, it might be that the 32nd cubie does happen to be in

its home position (where we do have a reference for talking

about the permutation of the stickers on that cubie) and

have an _odd_ permutation of its stickers relative to their

positions in initial state. More is needed to make the

argument rigorous.

If a more rigorous argument has been presented somewhere,

I would appreciate learning about it.

A valid argument:

There are 96 sticker positions in the pile occupied by

stickers from these 32 cubies. When we scramble the puzzle,

we are permuting not only the cubies but the individual

stickers themselves. If you look at the permutation of the

cubies in terms of its disjoint cycles, the even length

cycles correspond to odd permutations of the cubies. So the

parity of the cubie permutation is the parity of the number

of even length cycles. Each of those even length cubie

permutation cycles corresponds to 3 even length permutation

cycles of the stickers. Thus the parity of the sticker

permutation is equal to the parity of the cubie permutation.

Now we can come back to the placement of 31 of the cubies,

each with any of 6 possible orientations. The permutation

parity of the cubies themselves is now determined and so is

that of the stickers. Of the six possible orientations for

the 32nd cubie only half of them will lead to the one

possible overall sticker permutation parity value that is

possible for whatever the cubie permutation parity is.

Without working out the actual parity of the cubie

permutation, we don’t know what that parity is. Indeed, it

can go either way; but, whichever way that is, only half of

the 6 possible orientations will lead to the correct

corresponding overall sticker permutation parity.

Considerations for any transposable set of 2-sticker cubies:

The analogous issue arises for 2-sticker cubies as well.

Leaving the order unspecified, my concept of "transposable"

refers to sets of cubies which can all occupy the same set

of positions in the pile. (For orders larger than 4, there

are multiple such distinct sets.)

Even for the transposable sets which admit an odd

permutation of the cubies, the permutation of the stickers

on each of them is always even. Similar logic to the above

for center-edge cubies applies - resulting in only one

possible way of aligning the stickers for the placement of

the last cubie in any such set. (For the 2-sticker cubies

which lie in 2 center slices, the cubie has more than one

orientation relating to axes for which it is in a center

slice and for which you cannot see that orientation. I.e.,

multiple invisible orientations which do not bear on the

sticker alignments.)

Regards,

David V.