# 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