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.