Message #622

From: rev_16_4 <rev_16_4@yahoo.com>
Subject: Parity on MC m^n
Date: Wed, 28 Jan 2009 06:17:58 -0000

I was hoping mentioning parity would get a discussion started on
this, as parity was my biggest hang-up on announcing my m^n solution.
A traditional definition of parity, with regards to a m^n puzzle,
usually is along the lines of "A position that cannot occur on a
standard 3^n, when solving a m^n with m>=4 and using the reduction
method." A definition along these lines is perfectly suitable for the
m^3 puzzles. I don’t particularly like this definition because it
never really explains parity. I wish I could remember the site I
first read and UNDERSTOOD parity. What’s crazy is once you
reach a complete understanding of parity, you’ll realize any parity
condition can be corrected with a single quarter twist on the parity
affected layer. (corrected, not solved…)

I prefer a definition of parity more along the lines of "A position
with an odd number of pairs of IDENTICAL TYPE pieces (without
duplicates) swapped. (Odd Parity)" This definition actually allows a
parity condition on all m^3, including the standard 3^3. The 3^3’s is
usually explained away because people say you’re always swapping an
even number of pairs of pieces on the final layer (Even Parity). I
disagree because sometimes its actually one pair of corners, and one
pair of edges. (Double Odd Parity as I call it.) I also ignore
orietation as far as parity is concerned which I hope will become
clear why later. Another definition for parity I like is "Any
position that cannot be solved using only even parity moves for each
unique type of piece."

Let me explain this all a little better. What is the basic unit of
movement on a m^3 (and the m^n in general)? I consider it a 90 degree
turn of a layer along an axis. All other turns and positions can be
created from a sequence of these basic units. So if we can figure
what type of parity each basic unit has, we can figure what type of
parity any number of these basic units have added together. We’ll use
this simple movement to start our analysis of parity. I’m also going
to start with the pieces on a single face on a 2^3 puzzle, then show
a 3^3, and finally a 4^3. Here’s my Up face on my solved 2^3:

———
| 1 | 2 |
———
| 4 | 3 |
———

Let’s say I could somehow swap just a single pair of corners, 1 and 2.

———
| 2 | 1 |
———
| 4 | 3 |
———

I would write this as (1,2). This would signify "Swap the piece in
position 1 with the piece in position 2."

Let’s say I swapped another single pair of corners, 2 and 3.

———
| 3 | 1 |
———
| 4 | 2 |
———

Since 2 is in position one, I would write this as (1,3). "Swap piece
in pos 1 with piece in pos 3."

Finally, I perform one last swap, 3 and 4.

———
| 4 | 1 |
———
| 3 | 2 |
———

3 is again in position one, so I would write this as (1,4).

after a single clockwise quarter turn on my Up face, I have:

———
| 4 | 1 |
———
| 3 | 2 |
———

This looks just like my position after three pair-swaps. So a single
quarter turn can be written as (1,2)(1,3)(1,4). A quarter twist on a
2^3 is an odd parity position.

Now let me show you a different sequence: (I removed the 4 for
clarification) (Here’s the sequence if you have a cube handy: R’ F R’
B2 R F’ R’ B2 R2)

——— ———
| 1 | 2 | | 3 | 1 |
——— -> ———
| | 3 | | | 2 |
——— ———

This is a common move, known by almost anyone who can solve the cube.
As shown above, it’s generated from two pair-swaps, aka an even
parity position. (1,2)(1,3)

Now I’m sure some of you are asking "What happens when I perform a
second quarter twist?" I’ll let you verify for yourselves, but the
end results is an even parity position. Parity works like basic
addition. Add the number of pair-swaps performed, odd is odd, even is
an even parity position. Also note that the move (1,2)(1,3) required
12 odd parity quarter twists to generate an even parity position.

I’m going to skip to the meat and potatoes of the 3^3 and 4^3. For
the 4^3, I need to perform two different quarter twists due to the
inner Up layer.

3^3
————- ————-
| 1 | 2 | 3 | | 7 | 8 | 1 |
————- ————-
| 8 | X | 4 | -> | 6 | X | 2 |
————- ————-
| 7 | 6 | 5 | | 5 | 4 | 3 |
————- ————-
(1,3)(1,5)(1,7)(2,4)(2,6)(2,8)

This is traditionally called even parity. This is also where the
issues come in with solving the parity problems using the 4^3
reduction method. You CANNOT generate odd parity positions using even
parity positions. It’s just simple addition. However, since there’s
no physical way to swap a corner with an edge, I’d write the position
above as:
(1,3)(1,5)(1,7) and (2,4)(2,6)(2,8)

Hence my terminology "Double Odd." I cannot use a combination of
double odd positions to generate a single odd position - the math is
still there (2 odds equals an even). However I CAN’T use a
combination of even parity moves (like 2 corner swaps and 2 edge
swaps) to generate a double odd position.

4^3out
—————– —————–
| 1 | 2 | 3 | 4 | | a | b | c | 1 |
—————– —————–
| c | d | e | 5 | | 9 | g | d | 2 |
—————– -> —————–
| b | g | f | 6 | | 8 | f | e | 3 |
—————– —————–
| a | 9 | 8 | 7 | | 7 | 6 | 5 | 4 |
—————– —————–
(1,4)(1,7)(1,a)(2,5)(2,8)(2,b)(3,6)(3,9)(3,c)(d,e)(d,f)(d,g)

I’d write this as:
(1,4)(1,7)(1,a) and (2,5)(2,8)(2,b)(3,6)(3,9)(3,c) and (d,e)(d,f)(d,g)

Why did I only make three groups? Simply, all the edges are
identical, and from my definition of parity: "A position with an odd
number of pairs of IDENTICAL TYPE pieces swapped." How do I determine
if two pieces are identical? Well here’s where my math knowledge
fails me, but if you can physically rotate a puzzle so that one piece
occupies the same space as another, then they are identical type
pieces.

I’d actually call an outer 4^3 quarter twist odd corner, even edge,
even face parity. I know a couple of you are saying "Hey, dummy!
Based on what you’ve told us so far, the faces have odd parity!" Let
me explain why I said even. The math is simpler. If I say "odd," that
means I MUST use an odd number of pair swaps to fix it. If you can
generate a position with an even number of pair swaps, then I say it
defaults to an even parity position, even if it appears to be an odd
number of pair swaps. You can swap an odd number of faces using an
even number of swaps because some of the faces are colored
identically. ((RED,RED)(RED,BLUE) - the first swap could just as
easily not happened.) Anthony, this is exactly what you mentioned in
your post.

Incidentally, this even edge parity, with odd corner parity, is what
allows the 4^3 to exhibit the single corner swap positions impossible
on the 3^3. (This is also why I don’t consider the two pair of joined
edges swapped that look like a single pair swapped on a 3^3 during
the 4^3 reduction method a parity condition. This is actually even
parity, and is solvable using even parity moves.)

4^3in
—————– —————–
| 1 | 2 | 3 | 4 | | a | b | c | 1 |
—————– —————–
| c | | | 5 | | 9 | | | 2 |
—————– -> —————–
| b | | | 6 | | 8 | | | 3 |
—————– —————–
| a | 9 | 8 | 7 | | 7 | 6 | 5 | 4 |
—————– —————–

(1,4)(1,7)(1,a)(2,5)(2,8)(2,b)(3,6)(3,9)(3,c)

I’d write this as:
(1,4)(1,7)(1,a) and (2,5)(2,8)(2,b)(3,6)(3,9)(3,c)

This is odd for edges and even for faces (remember this is the inner
slice). This is the cause for the familiar "Single reverse oriented
edge" parity condition. It is actually a single swapped pair of edges
that can’t help but be disoriented.

In summary, on each puzzle:
A quarter twist here: Causes odd parity here:
2^3 Corner
3^3 Corner and Edge
4^3 Out Corner
4^3 In Edge

Easy, huh? I’ll let you guys work on the higher order puzzles to see
why I say parity conditions do not exist for m^n puzzles for n>=4 and
all m that are even. I do think they exist on the m^3 for ALL m>1.
They are just not the major problem people think they are. And using
the caging method I use does eliminate the typical manifestation of
parity. It’s simply fixed by a quarter twist of the appropriate
layer, then progress is continued on the pieces with fewer stickers.

The fact that I believe parity conditions cannot exist on pieces with
duplicates is exactly why I used the caging method for my m^n
solution. After you reach the pieces with n-2 stickers, the solution
is trivial for puzzles with an even m. With an odd m, it takes a
little more work. A quarter twist affecting only the pieces w/o
duplicates (the center pieces) and w/ odd parity is all that’s
required (plus several 100’s of twists, an even number of course, to
fix any additional pieces repositioned).

I reread this post and realized I never explained why I don’t
consider orientation when considering parity. If anyone wants I can
go into this later. This post is already quite long, so I’m not going
to proceed here.

Happy Hyper-cubing!

-Levi