Message #637
From: David Smith <djs314djs314@yahoo.com>
Subject: Re: [MC4D] Re: Parity on MC m^n
Date: Mon, 02 Feb 2009 17:49:17 -0800
Hi guys,
What a great discussion! I thought I would add my observations, based on my work
with the m^n formulas.
I think an important fact to realize, which I touched on earlier, is that on an m^n
cube, with n>=4, all types of pieces can only undergo even permutations, except for
the case when m is odd, in which the 2C and 3C pieces can be odd, but only
together (double odd). This is always true because of the nature of the faces
on a 4D or higher-dimensional cube. All 4-cycles of any pieces come in pairs,
hence resulting in even permutations for all piece types. Of course, as Levi
mentioned, when there are pieces that have identical colors and can occupy
the same position and orientation, even permutations and odd permutations
are basically the same, based on their definition. (Each can be converted
into the other by incorporating a swap of two identically-colored pieces.)
This leads to the immediate conclusion that both of the cases that Levi presented
in the log files are impossible, because each of them involves a swap of two
pieces without affecting the others. More generally, on an m^n cube, n>=4,
one can never swap two n-colored or two (n-1)-colored pieces without affecting
the others. (except for the double odd case when n=4. The reason I excluded
pieces with less colors is due to the fact I stated above regarding identically-colored
pieces.)
This is true regardless of the orientations of the pieces. In this interesting
discussion, Roice sometimes mentioned the possibility of the permutation
or orientation of a certain type of piece affecting the permutation or orientation
of another type of piece. (By "type of piece", I mean pieces of a particular
number of colors.) In fact, the only time any type of piece can affect the
permutation and orientation possibilities of another is the double odd case
concerning the permutations of 2C and 3C pieces on an odd cube! All other
permutations and orientations of a particular type of piece are completely
independent of the other types of pieces. Also, permutations never
affect orientations, either between different piece types or the same.
I hope this gives even more insight to what both of you were deducing together.
It was a great discussion to read, and I’m sorry I didn’t join in sooner! :)
By the way Roice, when you said,
P.S. David, I’m sure this was implicit in your statement, but when you
said "For example, on an n^5, a single corner can be in any possible
orientation while the rest of the cube is solved!" I thought I’d remark
explicitly that this excludes mirror orientations (enantiomorphs).
So e.g. you can twirl 3 of the stickers leaving 2 fixed, you can
cycle 4 of them leaving one fixed, etc., but you still can’t swap 2
stickers leaving 3 fixed.
I did realize that enantiomorphic orientations were impossible; I didn’t make that clear,
so thanks for mentioning it! Also, I thought I would mention that a 4-cycle of four
stickers on a 5D corner would itself be enantiomorphic (an odd permutation of
stickers), just like a swap of two stickers. You could have two swaps of two stickers,
though, so maybe this is what you meant. If not, I’m sure it was just an oversight! :)
All the best,
David
— On Mon, 2/2/09, Roice Nelson <roice3@gmail.com> wrote:
From: Roice Nelson <roice3@gmail.com>
Subject: Re: [MC4D] Re: Parity on MC m^n
To: 4D_Cubing@yahoogroups.com
Date: Monday, February 2, 2009, 12:25 AM
One last trimmed down reply for me as well :)
> It is interesting to note you can have an odd number of corner pair
> swaps with an even number of edge pair swaps if the orientations of
> the corners also come into play, the most simple example being a
> cube that is solved except for two corners. In this case I guess
> the double odd parity is shared with the orientations instead of
> the edges.
Are you refering to two corners that aren’t oriented correctly? If
not I’m not sure what you mean. On a 3^3, you cannot swap a single
pair of corners, regardless of orientation, unless an odd # of pairs
of edges are swapped as well.
I just learned something! And had to pull out my cube to sway myself :) After thousands of solves of the 3^3 over the years, I never realized the situation where there are two unsolved corners meant those corners had to be in their correct positions, but incorrect orientations. That’s what I love about the cube though, that it seems there are an endless stream of little facts like this to learn which I can then integrate into my cube "world view". And it is amazing what details one can gloss over. Sorry for the incorrect claim (often I learn by making assertions then seeing the counterexample, or having it pointed out to me).
> I think we can deduce the 4C case is impossible because corners are
> only permuted/oriented by outer twists (that is, by the same twist
> set as the 3^4), and this is not a possible configuration on the
> 3^4. The 3C case I’m unsure of, but I plan to experiment with it
> some.
I’m not so sure. The same thing is possible on the 4^3, yet not on
the 3^3….
Ah yes, there is a hole in the reasoning! I think it can be filled though with the extra observation you’ve made about the 4^4, which is that no twists create a single odd-parity condition. And a single set of swapped corners is odd, so it still looks impossible. What do you think?
In fact, this convinces me the 3C situation you uploaded is impossible as well. The manifestation of that would have to be the result of one of two cases:
(1) The two pieces are mirrored in place (impossible due to enantiomorphic constraints) .
(2) The pieces are exchanged and flipped, but a single swapped pair of edges is a single odd parity condition, again impossible.
The scenario in 4x4x4x4_roice2. log differs from both of these in that it is two pair of swaps, an even parity condition. So now it makes sense to me why that is possible whereas the above two situations are not.
> Since I had
> already mentally recategorized those sets of two as single merged
> pieces, the puzzle model I was trying to use to my advantage left
> me stuck.
Interesting. Essentially you went 4^4, to 4*3^3, to 3^4? This parity
case happened between 4*3^3 and 3^4? Is this close to accurate?
I’m not following what you mean by 4*3^3, but I was trying to apply the 4^3 pairing of 2C stickers approach/behavior to the 4^4 (the analogue was pairing up pairs of 2C stickers). And yep, this was an interim step in the effort to go from 4^4 -> 3^4).
Thanks for the lively discussion.
Absolutely. I really feel like I learned a lot the past few days reading the discussion and struggling over these scenarios. I’m really happy to have the concepts more crystallized in my mind, so thank you very much as well! And thanks for your patience with my mistakes.
Good night,Roice