Thread: "Parity on MC m^n"

I was hoping mentioning parity would get a discussion started on=20
this, as parity was my biggest hang-up on announcing my m^n solution.=20
A traditional definition of parity, with regards to a m^n puzzle,=20
usually is along the lines of "A position that cannot occur on a=20
standard 3^n, when solving a m^n with m>=3D4 and using the reduction=20
method." A definition along these lines is perfectly suitable for the=20
m^3 puzzles. I don't particularly like this definition because it=20
never really explains parity. I wish I could remember the site I=20
first read and UNDERSTOOD parity. What's crazy is once you=20
reach a complete understanding of parity, you'll realize any parity=20
condition can be corrected with a single quarter twist on the parity=20
affected layer. (corrected, not solved...)

I prefer a definition of parity more along the lines of "A position=20
with an odd number of pairs of IDENTICAL TYPE pieces (without=20
duplicates) swapped. (Odd Parity)" This definition actually allows a=20
parity condition on all m^3, including the standard 3^3. The 3^3's is=20
usually explained away because people say you're always swapping an=20
even number of pairs of pieces on the final layer (Even Parity). I=20
disagree because sometimes its actually one pair of corners, and one=20
pair of edges. (Double Odd Parity as I call it.) I also ignore=20
orietation as far as parity is concerned which I hope will become=20
clear why later. Another definition for parity I like is "Any=20
position that cannot be solved using only even parity moves for each=20
unique type of piece."

Let me explain this all a little better. What is the basic unit of=20
movement on a m^3 (and the m^n in general)? I consider it a 90 degree=20
turn of a layer along an axis. All other turns and positions can be=20
created from a sequence of these basic units. So if we can figure=20
what type of parity each basic unit has, we can figure what type of=20
parity any number of these basic units have added together. We'll use=20
this simple movement to start our analysis of parity. I'm also going=20
to start with the pieces on a single face on a 2^3 puzzle, then show=20
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=20
position 1 with the piece in position 2."

Let's say I swapped another single pair of corners, 2 and 3.=20

---------
| 3 | 1 |
---------
| 4 | 2 |
---------

Since 2 is in position one, I would write this as (1,3). "Swap piece=20
in pos 1 with piece in pos 3."

Finally, I perform one last swap, 3 and 4.=20

---------
| 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=20
quarter turn can be written as (1,2)(1,3)(1,4). A quarter twist on a=20
2^3 is an odd parity position.

Now let me show you a different sequence: (I removed the 4 for=20
clarification) (Here's the sequence if you have a cube handy: R' F R'=20
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.=20
As shown above, it's generated from two pair-swaps, aka an even=20
parity position. (1,2)(1,3)

Now I'm sure some of you are asking "What happens when I perform a=20
second quarter twist?" I'll let you verify for yourselves, but the=20
end results is an even parity position. Parity works like basic=20
addition. Add the number of pair-swaps performed, odd is odd, even is=20
an even parity position. Also note that the move (1,2)(1,3) required=20
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=20
the 4^3, I need to perform two different quarter twists due to the=20
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=20
issues come in with solving the parity problems using the 4^3=20
reduction method. You CANNOT generate odd parity positions using even=20
parity positions. It's just simple addition. However, since there's=20
no physical way to swap a corner with an edge, I'd write the position=20
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=20
double odd positions to generate a single odd position - the math is=20
still there (2 odds equals an even). However I CAN'T use a=20
combination of even parity moves (like 2 corner swaps and 2 edge=20
swaps) to generate a double odd position.=20

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=20
identical, and from my definition of parity: "A position with an odd=20
number of pairs of IDENTICAL TYPE pieces swapped." How do I determine=20
if two pieces are identical? Well here's where my math knowledge=20
fails me, but if you can physically rotate a puzzle so that one piece=20
occupies the same space as another, then they are identical type=20
pieces.=20

I'd actually call an outer 4^3 quarter twist odd corner, even edge,=20
even face parity. I know a couple of you are saying "Hey, dummy!=20
Based on what you've told us so far, the faces have odd parity!" Let=20
me explain why I said even. The math is simpler. If I say "odd," that=20
means I MUST use an odd number of pair swaps to fix it. If you can=20
generate a position with an even number of pair swaps, then I say it=20
defaults to an even parity position, even if it appears to be an odd=20
number of pair swaps. You can swap an odd number of faces using an=20
even number of swaps because some of the faces are colored=20
identically. ((RED,RED)(RED,BLUE) - the first swap could just as=20
easily not happened.) Anthony, this is exactly what you mentioned in=20
your post.

Incidentally, this even edge parity, with odd corner parity, is what=20
allows the 4^3 to exhibit the single corner swap positions impossible=20
on the 3^3. (This is also why I don't consider the two pair of joined=20
edges swapped that look like a single pair swapped on a 3^3 during=20
the 4^3 reduction method a parity condition. This is actually even=20
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=20
slice). This is the cause for the familiar "Single reverse oriented=20
edge" parity condition. It is actually a single swapped pair of edges=20
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=20
why I say parity conditions do not exist for m^n puzzles for n>=3D4 and=20
all m that are even. I do think they exist on the m^3 for ALL m>1.=20
They are just not the major problem people think they are. And using=20
the caging method I use does eliminate the typical manifestation of=20
parity. It's simply fixed by a quarter twist of the appropriate=20
layer, then progress is continued on the pieces with fewer stickers.=20

The fact that I believe parity conditions cannot exist on pieces with=20
duplicates is exactly why I used the caging method for my m^n=20
solution. After you reach the pieces with n-2 stickers, the solution=20
is trivial for puzzles with an even m. With an odd m, it takes a=20
little more work. A quarter twist affecting only the pieces w/o=20
duplicates (the center pieces) and w/ odd parity is all that's=20
required (plus several 100's of twists, an even number of course, to=20
fix any additional pieces repositioned).

I reread this post and realized I never explained why I don't=20
consider orientation when considering parity. If anyone wants I can=20
go into this later. This post is already quite long, so I'm not going=20
to proceed here.

Happy Hyper-cubing!

-Levi

=20=20=20=20
I was hoping mentioning parity would get a discussion started o=
n=20

this, as parity was my biggest hang-up on announcing my m^n solution.=20

A traditional definition of parity, with regards to a m^n puzzle,=20

usually is along the lines of "A position that cannot occur on a=20

standard 3^n, when solving a m^n with m>=3D4 and using the reduction=20

method." A definition along these lines is perfectly suitable for the=20

m^3 puzzles. I don't particularly like this definition because it=20

never really explains parity. I wish I could remember the site I=20

first read and UNDERSTOOD parity. What's crazy is once you=20

reach a complete understanding of parity, you'll realize any parity=20

condition can be corrected with a single quarter twist on the parity=20

affected layer. (corrected, not solved...)



I prefer a definition of parity more along the lines of "A position=20

with an odd number of pairs of IDENTICAL TYPE pieces (without=20

duplicates) swapped. (Odd Parity)" This definition actually allows a=20

parity condition on all m^3, including the standard 3^3. The 3^3's is=20

usually explained away because people say you're always swapping an=20

even number of pairs of pieces on the final layer (Even Parity). I=20

disagree because sometimes its actually one pair of corners, and one=20

pair of edges. (Double Odd Parity as I call it.) I also ignore=20

orietation as far as parity is concerned which I hope will become=20

clear why later. Another definition for parity I like is "Any=20

position that cannot be solved using only even parity moves for each=20

unique type of piece."



Let me explain this all a little better. What is the basic unit of=20

movement on a m^3 (and the m^n in general)? I consider it a 90 degree=20

turn of a layer along an axis. All other turns and positions can be=20

created from a sequence of these basic units. So if we can figure=20

what type of parity each basic unit has, we can figure what type of=20

parity any number of these basic units have added together. We'll use=20

this simple movement to start our analysis of parity. I'm also going=20

to start with the pieces on a single face on a 2^3 puzzle, then show=20

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=20

position 1 with the piece in position 2."



Let's say I swapped another single pair of corners, 2 and 3.=20



---------

| 3 | 1 |

---------

| 4 | 2 |

---------



Since 2 is in position one, I would write this as (1,3). "Swap piece=20

in pos 1 with piece in pos 3."



Finally, I perform one last swap, 3 and 4.=20



---------

| 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=20

quarter turn can be written as (1,2)(1,3)(1, 4). A quarter twist on a=20

2^3 is an odd parity position.



Now let me show you a different sequence: (I removed the 4 for=20

clarification) (Here's the sequence if you have a cube handy: R' F R'=20

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.=20

As shown above, it's generated from two pair-swaps, aka an even=20

parity position. (1,2)(1,3)



Now I'm sure some of you are asking "What happens when I perform a=20

second quarter twist?" I'll let you verify for yourselves, but the=20

end results is an even parity position. Parity works like basic=20

addition. Add the number of pair-swaps performed, odd is odd, even is=20

an even parity position. Also note that the move (1,2)(1,3) required=20

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=20

the 4^3, I need to perform two different quarter twists due to the=20

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=20

issues come in with solving the parity problems using the 4^3=20

reduction method. You CANNOT generate odd parity positions using even=20

parity positions. It's just simple addition. However, since there's=20

no physical way to swap a corner with an edge, I'd write the position=20

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=20

double odd positions to generate a single odd position - the math is=20

still there (2 odds equals an even). However I CAN'T use a=20

combination of even parity moves (like 2 corner swaps and 2 edge=20

swaps) to generate a double odd position.=20



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=20

identical, and from my definition of parity: "A position with an odd=20

number of pairs of IDENTICAL TYPE pieces swapped." How do I determine=20

if two pieces are identical? Well here's where my math knowledge=20

fails me, but if you can physically rotate a puzzle so that one piece=20

occupies the same space as another, then they are identical type=20

pieces.=20



I'd actually call an outer 4^3 quarter twist odd corner, even edge,=20

even face parity. I know a couple of you are saying "Hey, dummy!=20

Based on what you've told us so far, the faces have odd parity!" Let=20

me explain why I said even. The math is simpler. If I say "odd," that=20

means I MUST use an odd number of pair swaps to fix it. If you can=20

generate a position with an even number of pair swaps, then I say it=20

defaults to an even parity position, even if it appears to be an odd=20

number of pair swaps. You can swap an odd number of faces using an=20

even number of swaps because some of the faces are colored=20

identically. ((RED,RED)(RED, BLUE) - the first swap could just as=20

easily not happened.) Anthony, this is exactly what you mentioned in=20

your post.



Incidentally, this even edge parity, with odd corner parity, is what=20

allows the 4^3 to exhibit the single corner swap positions impossible=20

on the 3^3. (This is also why I don't consider the two pair of joined=20

edges swapped that look like a single pair swapped on a 3^3 during=20

the 4^3 reduction method a parity condition. This is actually even=20

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=20

slice). This is the cause for the familiar "Single reverse oriented=20

edge" parity condition. It is actually a single swapped pair of edges=20

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=20

why I say parity conditions do not exist for m^n puzzles for n>=3D4 and=20

all m that are even. I do think they exist on the m^3 for ALL m>1.=20

They are just not the major problem people think they are. And using=20

the caging method I use does eliminate the typical manifestation of=20

parity. It's simply fixed by a quarter twist of the appropriate=20

layer, then progress is continued on the pieces with fewer stickers.=20



The fact that I believe parity conditions cannot exist on pieces with=20

duplicates is exactly why I used the caging method for my m^n=20

solution. After you reach the pieces with n-2 stickers, the solution=20

is trivial for puzzles with an even m. With an odd m, it takes a=20

little more work. A quarter twist affecting only the pieces w/o=20

duplicates (the center pieces) and w/ odd parity is all that's=20

required (plus several 100's of twists, an even number of course, to=20

fix any additional pieces repositioned) .



I reread this post and realized I never explained why I don't=20

consider orientation when considering parity. If anyone wants I can=20

go into this later. This post is already quite long, so I'm not going=20

to proceed here.



Happy Hyper-cubing!



-Levi




=20=20=20=20=20=20

=20=20=20=20
=20=20=20=20
=09
=09=20
=09
=09








=09


=09
=09


=20=20=20=20=20=20
--0-1078339135-1233127381=:73875
Content-Type: text/html; charset=us-ascii

Thanks for your definition of parity and detailed explanation!  Everything you said was right on.  Based on your permutation-based definition of parity, I thought I would let you know what I have discovered about m^n cubes. You were absolutely right about parity conditions existing for all m^3 cubes where m>1.  And you were also right about parity conditions not existing on m^n puzzles where n>=4 and m is even.  When m is odd, the only parity condition that exists is what you call Double Odd Parity, and only on the central groups of 2-colored and 3-colored pieces. I would be more interested about what you have to say on orientations, whenever you get the chance.  Orientations can be very tricky when the dimension is greater than 3.  For example, on an n^5, a single corner can be in any possible orientation while the rest of the cube is solved!  The non-central 4-colored pieces behave like the corners on an n^4; there can be a single pair of pieces, each of which can have 4 different orientations, or a single piece could have 4 different orientations, the rest of the cube unaffected.  And there is even a group of 3-colored pieces that can have a pair of pieces in different orientations without affecting the rest of the cube.  I have pretty much figured out all possible permutation and orientation possibilities for an m^n, so feel free to email me if you need more information.  The only reason I haven't figured out the m^n formula yet is counting the number of groups (what you call identical type pieces) for all possibilities.  They get horrendously complicated for large n. I hope I was able to help, and good luck with your project! David --- On Wed, 1/28/09, rev_16_4 <rev_16_4@yahoo.com> wrote: From: rev_16_4 <rev_16_4@yahoo.com> Subject: [MC4D] Parity on MC m^n To: 4D_Cubing@yahoogroups.com Date: Wednesday, January 28, 2009, 1:17 AM 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

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Hi guys,

Thank you Levi. This deepened my understanding, especially the discussion
of what you termed "double odd", as well as the "why" of single corner swaps
on the 4^3. And thank you David for enumerating the parity conditions for
d>=4. I would like to go over what I took from this discussion, and hope
the current state of my understanding can help clarify what we mean by
"parity".

Levi preferred definitions of parity based on the permutations of the pieces
during twists, for which one can identify the possible cycle parities (even,
odd, or double odd) that arise for the various piece types. In this sense
odd parities indeed do *not* occur for n^d puzzles where n is even and d>=4
(sorry to differ on the labels btw, I'm too set in my ways).

This discussion of parities of sticker permutations is enlightening, but I'm
left with the feeling that saying "no parity conditions exist for the 4^4"
misses some real effects the solver encounters when using the common
reduction method on this puzzle. It's been almost a decade, but I for one
was pulling my hair out when working through the 4^4 due to encountering
impossible 3^4 configurations.

So what is going on here?

We can figure out whether twists cycle pieces in an odd/even fashion, but as
long as we never limit the full set of twists that can be done on the
puzzle, we will never encounter a "parity *problem*", by which I mean an
unsolvable state. Without trickery via disassembly, the puzzle is always
solvable (though the parity conditions do say something about how the
solution must go). What I'm coming to here is that *"odd parity
condition" != "parity problem", and there is some confusion because the term
"parity" is being overloaded with both meanings. This overloading is a bit
unfortunate in my opinion (I've never nailed down the haziness I felt about
this term until now), but both uses of parity seem legitimate if the meaning
is clear.*

So here is my take: "Odd parities" will arise in various situations due to
the way twists end up cycling stickers, and solving these requires an odd
number of certain twists. "Parity problems" or "parity errors" will arise
when we artificially limit the full set of twists used on a puzzle midway
through a solution, and solving these problems will necessarily involve
disavowing the artificial twist restrictions we tried to work with.

To use a 4^3 example Levi mentioned, consider "two pair of joined edges
swapped that look like a single pair swapped on a 3^3 during the 4^3
reduction method". In the context of restricted twisting, this *is* a
parity problem because it is unsolvable using only the limited 3^3 twist set
- it is effectively a single swap of two edges, an odd parity condition. In
the context of unrestricted twisting, it is not a parity problem or an odd
parity condition (aside: assuming "even parity moves" means outer twists,
which are even for edges, the statement that you can solve this state using
even parity moves appears incorrect). In short, parity problems (I'll never
again just say parities!) occur when we use a limited 3^d model of cycle
parities on an n^d puzzle.

I hope my thoughts on this are clear. Maybe there are better terms we could
adopt for the distinction between these two uses of the word parity?

All the best,
Roice

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.


On 1/28/09, David Smith wrote:
>
> Thanks for your definition of parity and detailed explanation!
> Everything you
> said was right on. Based on your permutation-based definition of parity, I
> thought I would let you know what I have discovered about m^n cubes.
> You were absolutely right about parity conditions existing for all m^3
> cubes
> where m>1. And you were also right about parity conditions not existing on
> m^n puzzles where n>=4 and m is even. When m is odd, the only parity
> condition
> that exists is what you call Double Odd Parity, and only on the central
> groups
> of 2-colored and 3-colored pieces.
>
> I would be more interested about what you have to say on orientations,
> whenever
> you get the chance. Orientations can be very tricky when the dimension is
> greater than 3. For example, on an n^5, a single corner can be in any
> possible
> orientation while the rest of the cube is solved! The non-central
> 4-colored pieces
> behave like the corners on an n^4; there can be a single pair of pieces,
> each of
> which can have 4 different orientations, or a single piece could have 4
> different
> orientations, the rest of the cube unaffected. And there is even a group
> of
> 3-colored pieces that can have a pair of pieces in different orientations
> without
> affecting the rest of the cube. I have pretty much figured out all
> possible permutation
> and orientation possibilities for an m^n, so feel free to email me if you
> need
> more information. The only reason I haven't figured out the m^n formula
> yet
> is counting the number of groups (what you call identical type pieces) for
> all
> possibilities. They get horrendously complicated for large n.
>
> I hope I was able to help, and good luck with your project!
>
> David
>
> --- On *Wed, 1/28/09, rev_16_4* wrote:
>
> From: rev_16_4
> Subject: [MC4D] Parity on MC m^n
> Date: Wednesday, January 28, 2009, 1:17 AM
>
> 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
>
>
> .
>
>
>

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Sorry, just reread this after going to lunch with a friend and saw a couple
things I wanted to clarify/correct.
- The reason for the different move counts in the old/new headers was that
re-saving the log file with the new program corrected the move numbers (I
didn't manually change that part).

- As I'm sure you noticed Levi, the second problem I encountered was not
the example you were looking for since it applied to 2C pieces instead of 3C
ones. Nonetheless, these effects seem legitimate examples of what many
think of as parity problems...

Best,
Roice

On Sun, Feb 1, 2009 at 12:16 PM, Roice Nelson wrote:

> I dug up old cd backups I had and found my log files from April 2000! I
> save solutions along the way out of paranoia, and luckily I had files at the
> problem points I saw. I just uploaded 2 log files to the a new folder in
> the files area of the groupshowing the parity error situations I encountered on my 4^4 solution.
> quick aside on a program technical issue: These weren't loading with the
> current java version (back then the 4^4 was only available in the linux
> version). I altered the header line to look more current, and they seems to
> load fine now. However, since I don't know the format, I'm unsure of what
> ramifications the editing might have. Here is an example.
> old: MagicCube4D 1 0 857
> new: MagicCube4D 2 2 844 4
>
> Anyway, I'll do my best now to reconstruct what looks like was going on - I
> can't remember last week, much less the details of a decade ago :)
>
> In the first file, I was trying to solve 2C pieces. When matching up sets
> of four, I found the very final set (orange/yellow) had two in one
> orientation and two in another, as in the puzzle state of the log file.
> This is not a parity problem in the context of a reduction to a 3^4 since
> the reduction hasn't even happened yet. Rather it is a parity problem in
> the context of a 4^3! Because when pairing up the 2Cs on a 4^3, you will
> never encounter the situation where all are matched up to form single
> 3^3-like edges except that the last two are flipped in relation to each
> other. If placed on the same edge, the final two will always be in the same
> orientation. I'm glad I pulled this out again, because this feels more
> subtle that what I had written in my last email. In essence, the problem is
> the same however. I saw an "impossible" configuration when using a simpler
> mental model of parities on a more complicated puzzle. I hope this made
> sense because it is a really interesting effect to me.
>
> In the second file, the situation is a parity problem in the context of a
> 3^4 reduction, again relative to 2C pieces. The final (pink/red) 2C was
> flipped as a whole, so I believe this is the case you wanted to see an
> example of. Unfortunately, the log file doesn't easily show how to generate
> this position from a pristine state. Indeed, the possibility of it may rely
> on the permutations/orientations of 3C pieces, so it may not be possible
> without scrambled 3C pieces?
>
> Let me know what you think!
>
> Roice
>
> On Sun, Feb 1, 2009 at 1:20 AM, rev_16_4 wrote:
>
>> Roice, you bring up a very good point. I wasn't sure there were
>> positions on a 4^d that, using a reduction method, would generate
>> impossible positions on a 3^d (I'm going to switch to your notation,
>> it's been around longer). I thought it might be possible, seeing that
>> was the gereral consensus. But I hadn't experienced one myself. Can
>> someone email me a 4^4 log file with such a position?
>>
>> I'm still retaining my nontraditional definition of parity errors
>> essentially as odd parity (and in my n^d solution double odd as
>> well). Ignoring this definition, check out the "single flipped"
>> parity error on a 4^3. It will take an odd number of inner slice
>> quarter twists to solve this.
>>
>> On a 3^3, a single swapped pair has odd parity. This cannot happen
>> due to the even (aka double odd) parity of a single quarter twist on
>> a 3^3. However, using reduction, the "single swapped edge pair" (with
>> correct orientation) parity err on a 4^3 can seem to occur. This will
>> alway take an even number of quarter twists to solve.
>>
>> You can see a similar phenomenon with a 3^3. If you have an even
>> number of corner pair-swaps to perform, you will have an even number
>> of edge pair-swaps also. It took an even number of quarter twists to
>> generate this position, and it will take an even number of quarter
>> twists to solve. The same goes for odd. An odd # of Corner pair-swaps
>> will always be accompanied by an odd # of Edge pair-swaps and an odd
>> # of twists. This is my double odd parity.
>>
>> Now with the case of n=4, d>3, this rule above is not the case. you
>> can generate any position with an even number of twists, or an odd
>> number of twist. I'd write out all the actual pair-swaps from a
>> single quarter twist, but I'm too lazy right now. In a nutshell there
>> are 6 corner pairs, 18 edge pairs, 18 face pairs, and 6 center pairs
>> swapped during an outer slice rotation. An inner slice rotation has 6
>> edge pairs, 18 face pairs, and 18 center pairs swapped. As you can
>> see, all are even, hence even parity (the faces and centers are
>> irrelevent). You can never generate an odd parity position, hence my
>> belief there are no parity errors for this puzzle.
>>
>> With the reduction method, sometimes the pairs are swapped in such a
>> manner that a simple even parity position for the caging method I
>> use, if attempted to be solved using reduction, would result in an
>> unsolvable 3^4 position. (I'm trying to think of a position where
>> this could occur... SOMEONE PLEASE SEND ME AN EXAMPLE LOG FILE!) If
>> someone can show me a log with a "single 3C w/ two stickers flipped"
>> 4^4 parity position, and how to generate it, I'd be grateful (and
>> completely shocked!). Other than that, I can't think of a 4^4 parity
>> that I think would be unsolvable with (non-caging) techniques similar
>> to a 3^4.
>>
>> -Levi
>>
>> _._,___
>>
>
>

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Sorry, just reread this after going to lunch with a friend and saw a couple=
things I wanted to clarify/correct.

Thanks, Roice. Impressive you found CD's from 2000! I'll take a look=20
at these, and see what I think. (Actually, I'm loading them right=20
now...)

I've also uploaded four log files to the same folder. Two of these=20
are cases that I'd consider parity errors. I cannot see how these=20
could be generated on a 3^4 or a 4^4. One involves a pair of 4C=20
pieces swapped. The other, a pair of 3C swapped so they appear to=20
have 2 stickers flipped. I'd be shocked to see a solution for either=20
of these.=20

The third is the 2C case I believe you're refering to in=20
4_4Roice2.log. I've included my solution to this case in the Faked2C=20
file. I solved this using the same technique I'd use to solve a pair=20
of fliped 2C's on the 3^4. The only difference is I had a 4th layer=20
on the U(?) axis to work with instead of 3, which allowed for a quick=20
conjugate. I use this same sort of technique on the 4^3 for the=20
single edge pair-swap parity.

The final log (Faked2CHalf) I believe is the other case you refered=20
to in 4_4Roice1.log. I would agree/argue that this isn't a parity=20
error because you hadn't reached a 3^4 state yet. If this is a parity=20
error, then any position before you reach a 3^4 is because they're=20
all impossible on a 3^4. The spirit of parity errors are situations=20
you wouldn't know were a problem until you got to the end, and "Wait=20
a minute, I can't solve this position!" (like Roice2)

However, I consider this one slightly more difficult (at least=20
initially) than the last. It's also a key to understanding why any=20
piece with at least 1 identical piece has no parity to me, so I'll=20
explain. This is still solvable with two pair-swaps (before or after=20
you've reduced it to a 3^4). I used an algorithm that swaps a set of=20
three pieces. This is the same algorith I use for all (d-2)C on all=20
n>3. I needed a conjugate to set it up (first and last six twists),=20
but the rest was the algorith (I was in a hurry and used at least 8=20
more twists than my algorith needed for normal solving). Swap an=20
identical pair, swap the pair with the problem.

I've got a major inspection at work all week, so it might be a few=20
days before any additional responses. Who am I kidding, I'm gonna=20
need the down time, so I'm sure I'll be on here!

-Levi

--- In 4D_Cubing@yahoogroups.com, Roice Nelson wrote:
>
> I dug up old cd backups I had and found my log files from April=20
2000! I
> save solutions along the way out of paranoia, and luckily I had=20
files at the
> problem points I saw. I just uploaded 2 log files to the a new=20
folder in
> the files area of the
> group20error%20logs/>showing
> the parity error situations I encountered on my 4^4 solution.
> quick aside on a program technical issue: These weren't loading=20
with the
> current java version (back then the 4^4 was only available in the=20
linux
> version). I altered the header line to look more current, and they=20
seems to
> load fine now. However, since I don't know the format, I'm unsure=20
of what
> ramifications the editing might have. Here is an example.
> old: MagicCube4D 1 0 857
> new: MagicCube4D 2 2 844 4
>=20
> Anyway, I'll do my best now to reconstruct what looks like was=20
going on - I
> can't remember last week, much less the details of a decade ago :)
>=20
> In the first file, I was trying to solve 2C pieces. When matching=20
up sets
> of four, I found the very final set (orange/yellow) had two in one
> orientation and two in another, as in the puzzle state of the log=20
file.
> This is not a parity problem in the context of a reduction to a=20
3^4 since
> the reduction hasn't even happened yet. Rather it is a parity=20
problem in
> the context of a 4^3! Because when pairing up the 2Cs on a 4^3,=20
you will
> never encounter the situation where all are matched up to form=20
single
> 3^3-like edges except that the last two are flipped in relation to=20
each
> other. If placed on the same edge, the final two will always be in=20
the same
> orientation. I'm glad I pulled this out again, because this feels=20
more
> subtle that what I had written in my last email. In essence, the=20
problem is
> the same however. I saw an "impossible" configuration when using a=20
simpler
> mental model of parities on a more complicated puzzle. I hope this=20
made
> sense because it is a really interesting effect to me.
>=20
> In the second file, the situation is a parity problem in the=20
context of a
> 3^4 reduction, again relative to 2C pieces. The final (pink/red)=20
2C was
> flipped as a whole, so I believe this is the case you wanted to see=20
an
> example of. Unfortunately, the log file doesn't easily show how to=20
generate
> this position from a pristine state. Indeed, the possibility of it=20
may rely
> on the permutations/orientations of 3C pieces, so it may not be=20
possible
> without scrambled 3C pieces?
>=20
> Let me know what you think!
>=20
> Roice
>=20
> On Sun, Feb 1, 2009 at 1:20 AM, rev_16_4 wrote:
>=20
> > Roice, you bring up a very good point. I wasn't sure there were
> > positions on a 4^d that, using a reduction method, would generate
> > impossible positions on a 3^d (I'm going to switch to your=20
notation,
> > it's been around longer). I thought it might be possible, seeing=20
that
> > was the gereral consensus. But I hadn't experienced one myself.=20
Can
> > someone email me a 4^4 log file with such a position?
> >
> > I'm still retaining my nontraditional definition of parity errors
> > essentially as odd parity (and in my n^d solution double odd as
> > well). Ignoring this definition, check out the "single flipped"
> > parity error on a 4^3. It will take an odd number of inner slice
> > quarter twists to solve this.
> >
> > On a 3^3, a single swapped pair has odd parity. This cannot happen
> > due to the even (aka double odd) parity of a single quarter twist=20
on
> > a 3^3. However, using reduction, the "single swapped edge pair"=20
(with
> > correct orientation) parity err on a 4^3 can seem to occur. This=20
will
> > alway take an even number of quarter twists to solve.
> >
> > You can see a similar phenomenon with a 3^3. If you have an even
> > number of corner pair-swaps to perform, you will have an even=20
number
> > of edge pair-swaps also. It took an even number of quarter twists=20
to
> > generate this position, and it will take an even number of quarter
> > twists to solve. The same goes for odd. An odd # of Corner pair-
swaps
> > will always be accompanied by an odd # of Edge pair-swaps and an=20
odd
> > # of twists. This is my double odd parity.
> >
> > Now with the case of n=3D4, d>3, this rule above is not the case.=20
you
> > can generate any position with an even number of twists, or an odd
> > number of twist. I'd write out all the actual pair-swaps from a
> > single quarter twist, but I'm too lazy right now. In a nutshell=20
there
> > are 6 corner pairs, 18 edge pairs, 18 face pairs, and 6 center=20
pairs
> > swapped during an outer slice rotation. An inner slice rotation=20
has 6
> > edge pairs, 18 face pairs, and 18 center pairs swapped. As you can
> > see, all are even, hence even parity (the faces and centers are
> > irrelevent). You can never generate an odd parity position, hence=20
my
> > belief there are no parity errors for this puzzle.
> >
> > With the reduction method, sometimes the pairs are swapped in=20
such a
> > manner that a simple even parity position for the caging method I
> > use, if attempted to be solved using reduction, would result in an
> > unsolvable 3^4 position. (I'm trying to think of a position where
> > this could occur... SOMEONE PLEASE SEND ME AN EXAMPLE LOG FILE!)=20
If
> > someone can show me a log with a "single 3C w/ two stickers=20
flipped"
> > 4^4 parity position, and how to generate it, I'd be grateful (and
> > completely shocked!). Other than that, I can't think of a 4^4=20
parity
> > that I think would be unsolvable with (non-caging) techniques=20
similar
> > to a 3^4.
> >
> > -Levi
> >
> > _._,___
> >
>

I was looking at another site. This gives a similar insight to my=20
position on parity, and the whole even/odd issue.=20

http://www.ryanheise.com/cube/parity.html

My n^d solution is entirely composed of conjugates and commutators,=20
with the odd single quarter twist thrown in for parity. What I meant=20
to say before was "There are no positions that cannot be solved with=20
just conjugates and commutators on a n^d, with d>3 and even n. This=20
is not the case with all other n^d." These other cases are parity=20
errors, in my opinion.

The first statement implies that on a n^d, with d>3 and even n, you=20
can solve any position with an even number of quarter twists. This=20
implies that a single quarter twist can be solved with an even number=20
of twists. This implies that any position can be solved with an even=20
or odd number of twists. This logic sounds awefully circular...

On a side note, I think most people when they develop their first=20
solution to the 3^3, and get to the bottom layer realize the need for=20
a quarter twist. This puts the problem back in a corners then edges=20
situation. However I think people have more of a disconnect with the=20
4^3's extra layer's required quarter twist. I think at the point you=20
realize you need it you have so much of the rest of the puzzle=20
solved, you don't want to perform this twist and have to redo it all.=20
This is where the need for solving parity comes in.=20

-Levi

--- In 4D_Cubing@yahoogroups.com, Roice Nelson wrote:
>
> Sorry, just reread this after going to lunch with a friend and saw=20
a couple
> things I wanted to clarify/correct.
> - The reason for the different move counts in the old/new headers=20
was that
> re-saving the log file with the new program corrected the move=20
numbers (I
> didn't manually change that part).
>=20
> - As I'm sure you noticed Levi, the second problem I encountered=20
was not
> the example you were looking for since it applied to 2C pieces=20
instead of 3C
> ones. Nonetheless, these effects seem legitimate examples of what=20
many
> think of as parity problems...
>=20
> Best,
> Roice
>=20
> On Sun, Feb 1, 2009 at 12:16 PM, Roice Nelson wrote:
>=20
> > I dug up old cd backups I had and found my log files from April=20
2000! I
> > save solutions along the way out of paranoia, and luckily I had=20
files at the
> > problem points I saw. I just uploaded 2 log files to the a new=20
folder in
> > the files area of the=20
group20error%20logs/>showing the parity error situations I encountered on=20
my 4^4 solution.
> > quick aside on a program technical issue: These weren't loading=20
with the
> > current java version (back then the 4^4 was only available in the=20
linux
> > version). I altered the header line to look more current, and=20
they seems to
> > load fine now. However, since I don't know the format, I'm=20
unsure of what
> > ramifications the editing might have. Here is an example.
> > old: MagicCube4D 1 0 857
> > new: MagicCube4D 2 2 844 4
> >
> > Anyway, I'll do my best now to reconstruct what looks like was=20
going on - I
> > can't remember last week, much less the details of a decade ago :)
> >
> > In the first file, I was trying to solve 2C pieces. When=20
matching up sets
> > of four, I found the very final set (orange/yellow) had two in one
> > orientation and two in another, as in the puzzle state of the log=20
file.
> > This is not a parity problem in the context of a reduction to a=20
3^4 since
> > the reduction hasn't even happened yet. Rather it is a parity=20
problem in
> > the context of a 4^3! Because when pairing up the 2Cs on a 4^3,=20
you will
> > never encounter the situation where all are matched up to form=20
single
> > 3^3-like edges except that the last two are flipped in relation=20
to each
> > other. If placed on the same edge, the final two will always be=20
in the same
> > orientation. I'm glad I pulled this out again, because this=20
feels more
> > subtle that what I had written in my last email. In essence, the=20
problem is
> > the same however. I saw an "impossible" configuration when using=20
a simpler
> > mental model of parities on a more complicated puzzle. I hope=20
this made
> > sense because it is a really interesting effect to me.
> >
> > In the second file, the situation is a parity problem in the=20
context of a
> > 3^4 reduction, again relative to 2C pieces. The final (pink/red)=20
2C was
> > flipped as a whole, so I believe this is the case you wanted to=20
see an
> > example of. Unfortunately, the log file doesn't easily show how=20
to generate
> > this position from a pristine state. Indeed, the possibility of=20
it may rely
> > on the permutations/orientations of 3C pieces, so it may not be=20
possible
> > without scrambled 3C pieces?
> >
> > Let me know what you think!
> >
> > Roice
> >
> > On Sun, Feb 1, 2009 at 1:20 AM, rev_16_4 wrote:
> >
> >> Roice, you bring up a very good point. I wasn't sure there were
> >> positions on a 4^d that, using a reduction method, would generate
> >> impossible positions on a 3^d (I'm going to switch to your=20
notation,
> >> it's been around longer). I thought it might be possible, seeing=20
that
> >> was the gereral consensus. But I hadn't experienced one myself.=20
Can
> >> someone email me a 4^4 log file with such a position?
> >>
> >> I'm still retaining my nontraditional definition of parity errors
> >> essentially as odd parity (and in my n^d solution double odd as
> >> well). Ignoring this definition, check out the "single flipped"
> >> parity error on a 4^3. It will take an odd number of inner slice
> >> quarter twists to solve this.
> >>
> >> On a 3^3, a single swapped pair has odd parity. This cannot=20
happen
> >> due to the even (aka double odd) parity of a single quarter=20
twist on
> >> a 3^3. However, using reduction, the "single swapped edge pair"=20
(with
> >> correct orientation) parity err on a 4^3 can seem to occur. This=20
will
> >> alway take an even number of quarter twists to solve.
> >>
> >> You can see a similar phenomenon with a 3^3. If you have an even
> >> number of corner pair-swaps to perform, you will have an even=20
number
> >> of edge pair-swaps also. It took an even number of quarter=20
twists to
> >> generate this position, and it will take an even number of=20
quarter
> >> twists to solve. The same goes for odd. An odd # of Corner pair-
swaps
> >> will always be accompanied by an odd # of Edge pair-swaps and an=20
odd
> >> # of twists. This is my double odd parity.
> >>
> >> Now with the case of n=3D4, d>3, this rule above is not the case.=20
you
> >> can generate any position with an even number of twists, or an=20
odd
> >> number of twist. I'd write out all the actual pair-swaps from a
> >> single quarter twist, but I'm too lazy right now. In a nutshell=20
there
> >> are 6 corner pairs, 18 edge pairs, 18 face pairs, and 6 center=20
pairs
> >> swapped during an outer slice rotation. An inner slice rotation=20
has 6
> >> edge pairs, 18 face pairs, and 18 center pairs swapped. As you=20
can
> >> see, all are even, hence even parity (the faces and centers are
> >> irrelevent). You can never generate an odd parity position,=20
hence my
> >> belief there are no parity errors for this puzzle.
> >>
> >> With the reduction method, sometimes the pairs are swapped in=20
such a
> >> manner that a simple even parity position for the caging method I
> >> use, if attempted to be solved using reduction, would result in=20
an
> >> unsolvable 3^4 position. (I'm trying to think of a position where
> >> this could occur... SOMEONE PLEASE SEND ME AN EXAMPLE LOG FILE!)=20
If
> >> someone can show me a log with a "single 3C w/ two stickers=20
flipped"
> >> 4^4 parity position, and how to generate it, I'd be grateful (and
> >> completely shocked!). Other than that, I can't think of a 4^4=20
parity
> >> that I think would be unsolvable with (non-caging) techniques=20
similar
> >> to a 3^4.
> >>
> >> -Levi
> >>
> >> _._,___
> >>
> >
> >
>

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A little more roice spam, this time inline :) (and a little of it from the
earlier post).

> I'm still retaining my nontraditional definition of parity errors
>
essentially as odd parity (and in my n^d solution double odd as
>
well). Ignoring this definition, check out the "single flipped"
>
parity error on a 4^3. It will take an odd number of inner slice
>
quarter twists to solve this.
>

> On a 3^3, a single swapped pair has odd parity. This cannot happen
>
due to the even (aka double odd) parity of a single quarter twist on
>
a 3^3. However, using reduction, the "single swapped edge pair" (with
>
correct orientation) parity err on a 4^3 can seem to occur. This will
>
alway take an even number of quarter twists to solve.
>

Furthermore, we can say that some of those twists will have to be of the
inner slice. In fact, both inner and outer twists will be required, an even
number of both. I can provide my reasoning for this conclusion if desired.

You can see a similar phenomenon with a 3^3. If you have an even
>
number of corner pair-swaps to perform, you will have an even number
>
of edge pair-swaps also. It took an even number of quarter twists to
>
generate this position, and it will take an even number of quarter
>
twists to solve. The same goes for odd. An odd # of Corner pair-swaps
>
will always be accompanied by an odd # of Edge pair-swaps and an odd
>
# of twists. This is my double odd parity.


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.

I've also uploaded four log files to the same folder. Two of these
>
are cases that I'd consider parity errors. I cannot see how these
>
could be generated on a 3^4 or a 4^4. One involves a pair of 4C
>
pieces swapped. The other, a pair of 3C swapped so they appear to
>
have 2 stickers flipped. I'd be shocked to see a solution for either
>
of these.


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.

The final log (Faked2CHalf) I believe is the other case you refered
>
to in 4_4Roice1.log. I would agree/argue that this isn't a parity
>
error because you hadn't reached a 3^4 state yet. If this is a parity
>
error, then any position before you reach a 3^4 is because they're
>
all impossible on a 3^4. The spirit of parity errors are situations
>
you wouldn't know were a problem until you got to the end, and "Wait
>
a minute, I can't solve this position!" (like Roice2)
>

That is exactly what happened to me though ;) But using a mental parity
model of a 4^3, not a 3^4. Considering two adjacent, identically colored
and oriented pieces as behaving like a single edge piece on a 4^3, I reached
this state and did not know how to solve it (because the solution
necessitated breaking these already solved, artificially joined pieces back
in two). 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.

Overall, I'd say the different usages of the word parity is still clouding
the discussion here. I am conceding that my use of the phrase "parity
problem" is perhaps too general. After all, a problem is simply a
configuration you don't know how to solve using a tool set of sequences.
And parity just means even or odd, which can be applied in a number of
ways. I promise it's clear in my mind though :) I like the reductionist
approach of analyzing parities and deriving what that says about the
solution. I like the concept of "parity problem" defined in the context of
mismatches between the different parity characteristics of various puzzles,
that they are mental surprises (perceived impossibilities) when one tries to
use solution methods across multiple puzzles.

All the best,
Roice

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'; font-size: 16px; ">

Ok, one last reply today. I'm going to trim this down to the points=20
I'm commenting on (apologies if this is in bad form):

>> On a 3^3, a single swapped pair has odd parity. This cannot happen
>> due to the even (aka double odd) parity of a single quarter twist=20
>> on a 3^3. However, using reduction, the "single swapped edge pair"=20
>> (with correct orientation) parity err on a 4^3 can seem to occur.=20
>> This will alway take an even number of quarter twists to solve.

> Furthermore, we can say that some of those twists will have to be=20
> of the inner slice. In fact, both inner and outer twists will be=20
> required, an even number of both. I can provide my reasoning for
> this conclusion if desired.

I agree completely.

>> You can see a similar phenomenon with a 3^3....

> It is interesting to note you can have an odd number of corner pair=20
> swaps with an even number of edge pair swaps if the orientations of=20
> the corners also come into play, the most simple example being a=20
> cube that is solved except for two corners. In this case I guess=20
> the double odd parity is shared with the orientations instead of=20
> the edges.

Are you refering to two corners that aren't oriented correctly? If=20
not I'm not sure what you mean. On a 3^3, you cannot swap a single=20
pair of corners, regardless of orientation, unless an odd # of pairs=20
of edges are swapped as well.

> I think we can deduce the 4C case is impossible because corners are=20
> only permuted/oriented by outer twists (that is, by the same twist=20
> set as the 3^4), and this is not a possible configuration on the=20
> 3^4. The 3C case I'm unsure of, but I plan to experiment with it=20
> some.

I'm not so sure. The same thing is possible on the 4^3, yet not on=20
the 3^3....

>>[..."Wait a minute! This position is unsolvable!"]

> That is exactly what happened to me though ;) But using a mental=20
> parity model of a 4^3, not a 3^4. Considering two adjacent,=20
> identically colored and oriented pieces as behaving like a single=20
> edge piece on a 4^3, I reached this state and did not know how to=20
> solve it (because the solution necessitated breaking these already=20
> solved, artificially joined pieces back in two). Since I had=20
> already mentally recategorized those sets of two as single merged=20
> pieces, the puzzle model I was trying to use to my advantage left=20
> me stuck.

Interesting. Essentially you went 4^4, to 4*3^3, to 3^4? This parity=20
case happened between 4*3^3 and 3^4? Is this close to accurate?

> Overall, I'd say the different usages of the word parity is still=20
> clouding the discussion here. I am conceding that my use of the=20
> phrase "parity problem" is perhaps too general. After all, a=20
> problem is simply a configuration you don't know how to solve using=20
> a tool set of sequences.
> And parity just means even or odd, which can be applied in a number=20
> of ways. I promise it's clear in my mind though :) I like the=20
> reductionist approach of analyzing parities and deriving what that=20
> says about the solution. I like the concept of "parity problem"=20
> defined in the context of mismatches between the different parity=20
> characteristics of various puzzles, that they are mental surprises=20
> (perceived impossibilities) when one tries to use solution methods=20
> across multiple puzzles.

I agree, the terminology is annoying. I was typing some of these=20
posts thinking "Do I really need to say 'single edge pair-swap parity=20
error' every time?" I figured it was too ambiguous otherwise. I like=20
your definition for traditional "parity problems" above. It's exactly=20
what's happening, regardless of how you solve the puzzle.=20

This was the question I kept having while trying to develop a=20
solution across all n^d. How do you address a possible parity=20
condition that might only happen on d>10, but has no relevence on=20
realistic puzzles? 4_4Roice2.log is a good example. This can't happen=20
on a 4^3 (or 3^4). Are there similar issues on the 4^5? I don't know.=20
Not with the caging method I settled on. And not with even n, d>3 if=20
the only limitation you self impose is "only conjugators and=20
commutators are legal." I suppose this is the spirit of the no parity=20
problem comment I made. I'm also confident that given a 4^100, and=20
time was no issue, I could solve any given solvable parity case with=20
the solution I have. The only reason time is an issue is because each=20
algorith would be somewhere around 14 * 2^97 quarter-twists!

Thanks for the lively discussion.

-Levi

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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

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