# Thread: "Parity on MC m^n"

From: "rev_16_4" <rev_16_4@yahoo.com>
Date: Wed, 28 Jan 2009 06:17:58 -0000
Subject: 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

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

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

From: David Smith <djs314djs314@yahoo.com>
Date: Tue, 27 Jan 2009 23:23:01 -0800 (PST)
Subject: Re: [MC4D] Parity on MC m^n

=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

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
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--0-1078339135-1233127381=:73875--

From: Roice Nelson <roice3@gmail.com>
Date: Wed, 28 Jan 2009 21:48:57 -0600
Subject: Re: [MC4D] Parity on MC m^n

--0015175cb896d05eb1046196f7b5
Content-Type: text/plain; charset=ISO-8859-1
Content-Transfer-Encoding: 7bit

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
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
>
> 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
>
>
> .
>
>
>

--0015175cb896d05eb1046196f7b5
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

Hi guys,

Thank you Levi.  This deepened my understanding, especially the d=
iscussion of what you termed "double odd", as well as the "w=
hy" of single corner swaps on the 4^3.  And thank you David for e=
numerating the parity conditions for d>=3D4.  I would like to go ov=
er what I took from this discussion, and hope the current state of my under=
standing 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 pa=
rities (even, odd, or double odd) that arise for the various piec=
e types.  In this sense odd parities indeed do not oc=
cur for n^d puzzles where n is even and d>=3D4 (sorry to differ on the l=
abels btw, I'm too set in my ways).

This discussion of parities of sticker permutations is enlightening, b=
ut I'm left with the feeling that saying "no parity conditions&nbs=
p;exist for the 4^4" misses some real effects the solver encounters wh=
en using the common reduction method on this puzzle.  It's been al=
most a decade, but I for one was pulling my hair out when working thro=
ugh 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 th=
e puzzle, we will never encounter a "parity problem&q=
uot;, by which I mean an unsolvable state.  Without trickery via disas=
sembly, the puzzle is always solvable (though the parity conditions do say =
something about how the solution must go).  What I'm coming to her=
e is that "odd parity condition" !=3D "par=
ity problem"mal; ">, and there is some confusion because the term "parity" is=
tunate 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 meanin=
g is clear.

So here is my take:  "Odd parities" will arise in vario=
us situations due to the way twists end up cycling stickers, and solving th=
ese requires an odd number of certain twists.  "Parity problems&q=
uot; 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 re=
strictions 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 unso=
lvable using only the limited 3^3 twist set - it is effectively a single sw=
ap of two edges, an odd parity condition.  In the context of unrestric=
ted 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 ter=
ms we could adopt for the distinction between these two uses of the word pa=
rity?

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 possi=
ble orientation while the rest of the cube is solved!" I thought I'=
;d remark explicitly that this excludes mirror orientations (p://en.wikipedia.org/wiki/Enantiomorph" target=3D"_blank">enantiomorphs=
).  So e.g. you can twirl 3 of the stickers leaving 2 fixed, you can c=
ycle 4 of them leaving one fixed, etc., but you still can't swap 2=
stickers leaving 3 fixed.

On 1/28/09, =
David Smith
<ank">djs314djs314@yahoo.com> wrote:
=20
0px 0.8ex;border-left:#ccc 1px solid">

<=
/div>
.
=3D"1" width=3D"1">
ote>

--0015175cb896d05eb1046196f7b5--

From: "rev_16_4" <rev_16_4@yahoo.com>
Date: Sun, 01 Feb 2009 07:20:16 -0000
Subject: Re: Parity on MC m^n

Roice, you bring up a very good point. I wasn't sure there were=20
positions on a 4^d that, using a reduction method, would generate=20
impossible positions on a 3^d (I'm going to switch to your notation,=20
it's been around longer). I thought it might be possible, seeing that=20
was the gereral consensus. But I hadn't experienced one myself. Can=20
someone email me a 4^4 log file with such a position?

I'm still retaining my nontraditional definition of parity errors=20
essentially as odd parity (and in my n^d solution double odd as=20
well). Ignoring this definition, check out the "single flipped"=20
parity error on a 4^3. It will take an odd number of inner slice=20
quarter twists to solve this.

On a 3^3, a single swapped pair has odd parity. This cannot happen=20
due to the even (aka double odd) parity of a single quarter twist on=20
a 3^3. However, using reduction, the "single swapped edge pair" (with=20
correct orientation) parity err on a 4^3 can seem to occur. This will=20
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=20
number of corner pair-swaps to perform, you will have an even number=20
of edge pair-swaps also. It took an even number of quarter twists to=20
generate this position, and it will take an even number of quarter=20
twists to solve. The same goes for odd. An odd # of Corner pair-swaps=20
will always be accompanied by an odd # of Edge pair-swaps and an odd=20
# of twists. This is my double odd parity.

Now with the case of n=3D4, d>3, this rule above is not the case. you=20
can generate any position with an even number of twists, or an odd=20
number of twist. I'd write out all the actual pair-swaps from a=20
single quarter twist, but I'm too lazy right now. In a nutshell there=20
are 6 corner pairs, 18 edge pairs, 18 face pairs, and 6 center pairs=20
swapped during an outer slice rotation. An inner slice rotation has 6=20
edge pairs, 18 face pairs, and 18 center pairs swapped. As you can=20
see, all are even, hence even parity (the faces and centers are=20
irrelevent). You can never generate an odd parity position, hence my=20
belief there are no parity errors for this puzzle.=20

With the reduction method, sometimes the pairs are swapped in such a=20
manner that a simple even parity position for the caging method I=20
use, if attempted to be solved using reduction, would result in an=20
unsolvable 3^4 position. (I'm trying to think of a position where=20
this could occur... SOMEONE PLEASE SEND ME AN EXAMPLE LOG FILE!) If=20
someone can show me a log with a "single 3C w/ two stickers flipped"=20
4^4 parity position, and how to generate it, I'd be grateful (and=20
completely shocked!). Other than that, I can't think of a 4^4 parity=20
that I think would be unsolvable with (non-caging) techniques similar=20
to a 3^4.

-Levi

--- In 4D_Cubing@yahoogroups.com, Roice Nelson wrote:
>
> Hi guys,
>=20
> Thank you Levi. This deepened my understanding, especially the=20
discussion
> of what you termed "double odd", as well as the "why" of single=20
corner swaps
> on the 4^3. And thank you David for enumerating the parity=20
conditions for
> d>=3D4. I would like to go over what I took from this discussion,=20
and hope
> the current state of my understanding can help clarify what we mean=20
by
> "parity".
>=20
> Levi preferred definitions of parity based on the permutations of=20
the pieces
> during twists, for which one can identify the possible cycle=20
parities (even,
> odd, or double odd) that arise for the various piece types. In=20
this sense
> odd parities indeed do *not* occur for n^d puzzles where n is even=20
and d>=3D4
> (sorry to differ on the labels btw, I'm too set in my ways).
>=20
> This discussion of parities of sticker permutations is=20
enlightening, but I'm
> left with the feeling that saying "no parity conditions exist for=20
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=20
for one
> was pulling my hair out when working through the 4^4 due to=20
encountering
> impossible 3^4 configurations.
>=20
> So what is going on here?
>=20
> We can figure out whether twists cycle pieces in an odd/even=20
fashion, but as
> long as we never limit the full set of twists that can be done on=20
the
> puzzle, we will never encounter a "parity *problem*", by which I=20
mean an
> unsolvable state. Without trickery via disassembly, the puzzle is=20
always
> solvable (though the parity conditions do say something about how=20
the
> solution must go). What I'm coming to here is that *"odd parity
> condition" !=3D "parity problem", and there is some confusion because=20
the term
is a bit
> unfortunate in my opinion (I've never nailed down the haziness I=20
> this term until now), but both uses of parity seem legitimate if=20
the meaning
> is clear.*
>=20
> So here is my take: "Odd parities" will arise in various=20
situations due to
> the way twists end up cycling stickers, and solving these requires=20
an odd
> number of certain twists. "Parity problems" or "parity errors"=20
will arise
> when we artificially limit the full set of twists used on a puzzle=20
midway
> through a solution, and solving these problems will necessarily=20
involve
> disavowing the artificial twist restrictions we tried to work with.
>=20
> To use a 4^3 example Levi mentioned, consider "two pair of joined=20
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=20
*is* a
> parity problem because it is unsolvable using only the limited 3^3=20
twist set
> - it is effectively a single swap of two edges, an odd parity=20
condition. In
> the context of unrestricted twisting, it is not a parity problem or=20
an odd
> parity condition (aside: assuming "even parity moves" means outer=20
twists,
> which are even for edges, the statement that you can solve this=20
state using
> even parity moves appears incorrect). In short, parity problems=20
(I'll never
> again just say parities!) occur when we use a limited 3^d model of=20
cycle
> parities on an n^d puzzle.
>=20
> I hope my thoughts on this are clear. Maybe there are better terms=20
we could
> adopt for the distinction between these two uses of the word parity?
>=20
> All the best,
> Roice
>=20
> P.S. David, I'm sure this was implicit in your statement, but when=20
you said
> "For example, on an n^5, a single corner can be in any possible=20
orientation
> while the rest of the cube is solved!" I thought I'd remark=20
explicitly that
> this excludes mirror orientations
> (enantiomorphs).
> So e.g. you can twirl 3 of the stickers leaving 2 fixed, you can=20
cycle 4 of
> them leaving one fixed, etc., but you still can't swap 2 stickers=20
leaving 3
> fixed.
>=20
>=20
> 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=20
parity, I
> > thought I would let you know what I have discovered about m^n=20
cubes.
> > You were absolutely right about parity conditions existing for=20
all m^3
> > cubes
> > where m>1. And you were also right about parity conditions not=20
existing on
> > m^n puzzles where n>=3D4 and m is even. When m is odd, the only=20
parity
> > condition
> > that exists is what you call Double Odd Parity, and only on the=20
central
> > groups
> > of 2-colored and 3-colored pieces.
> >
> > I would be more interested about what you have to say on=20
orientations,
> > whenever
> > you get the chance. Orientations can be very tricky when the=20
dimension is
> > greater than 3. For example, on an n^5, a single corner can be=20
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=20
pieces,
> > each of
> > which can have 4 different orientations, or a single piece could=20
have 4
> > different
> > orientations, the rest of the cube unaffected. And there is even=20
a group
> > of
> > 3-colored pieces that can have a pair of pieces in different=20
orientations
> > without
> > affecting the rest of the cube. I have pretty much figured out=20
all
> > possible permutation
> > and orientation possibilities for an m^n, so feel free to email=20
me if you
> > need
> > more information. The only reason I haven't figured out the m^n=20
formula
> > yet
> > is counting the number of groups (what you call identical type=20
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=20
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>=3D4 and using the reduction
> > method." A definition along these lines is perfectly suitable for=20
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=20
parity
> > condition can be corrected with a single quarter twist on the=20
parity
> > affected layer. (corrected, not solved...)
> >
> > I prefer a definition of parity more along the lines of "A=20
position
> > with an odd number of pairs of IDENTICAL TYPE pieces (without
> > duplicates) swapped. (Odd Parity)" This definition actually=20
allows a
> > parity condition on all m^3, including the standard 3^3. The=20
3^3's is
> > usually explained away because people say you're always swapping=20
an
> > even number of pairs of pieces on the final layer (Even Parity). I
> > disagree because sometimes its actually one pair of corners, and=20
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=20
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=20
degree
> > turn of a layer along an axis. All other turns and positions can=20
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=20
of
> > parity any number of these basic units have added together. We'll=20
use
> > this simple movement to start our analysis of parity. I'm also=20
going
> > to start with the pieces on a single face on a 2^3 puzzle, then=20
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=20
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=20
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=20
single
> > quarter turn can be written as (1,2)(1,3)(1, 4). A quarter twist=20
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'=20
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=20
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,=20
even is
> > an even parity position. Also note that the move (1,2)(1,3)=20
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=20
even
> > parity positions. It's just simple addition. However, since=20
there's
> > no physical way to swap a corner with an edge, I'd write the=20
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=20
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=20
odd
> > number of pairs of IDENTICAL TYPE pieces swapped." How do I=20
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=20
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=20
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!"=20
Let
> > me explain why I said even. The math is simpler. If I say "odd,"=20
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=20
it
> > defaults to an even parity position, even if it appears to be an=20
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=20
in
> >
> > Incidentally, this even edge parity, with odd corner parity, is=20
what
> > allows the 4^3 to exhibit the single corner swap positions=20
impossible
> > on the 3^3. (This is also why I don't consider the two pair of=20
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=20
inner
> > slice). This is the cause for the familiar "Single reverse=20
oriented
> > edge" parity condition. It is actually a single swapped pair of=20
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=20
see
> > why I say parity conditions do not exist for m^n puzzles for n>=3D4=20
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=20
using
> > the caging method I use does eliminate the typical manifestation=20
of
> > parity. It's simply fixed by a quarter twist of the appropriate
> > layer, then progress is continued on the pieces with fewer=20
stickers.
> >
> > The fact that I believe parity conditions cannot exist on pieces=20
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=20
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,=20
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=20
can
> > go into this later. This post is already quite long, so I'm not=20
going
> > to proceed here.
> >
> > Happy Hyper-cubing!
> >
> > -Levi
> >
> >
> > .
> >
> >=20
> >
>

From: Roice Nelson <roice3@gmail.com>
Date: Sun, 1 Feb 2009 12:16:59 -0600
Subject: Re: [MC4D] Re: Parity on MC m^n

--00163646d979a6b83f0461df710a
Content-Type: text/plain; charset=ISO-8859-1
Content-Transfer-Encoding: 7bit

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
>
> _._,___
>

--00163646d979a6b83f0461df710a
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

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 =3D"http://games.groups.yahoo.com/group/4D_Cubing/files/parity%20error%20lo=
gs/">a new folder in the files area of the group
showing the parity err=
or situations I encountered on my 4^4 solution.

quick aside on a program technical issue:  These weren&=
#39;t loading with the current java version (back then the 4^4 was only ava=
ilable 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 hav=
e.  Here is an example.

old:  MagicCube4D 1 0 857
new:  MagicCub=
e4D 2 2 844 4

Anyway, I'll do my best now to reconstr=
uct what looks like was going on - I can't remember last week, much les=
s the details of a decade ago :)

In the first file, I was trying to solve 2C pieces. &nb=
sp;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 reduct=
ion 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 flipp=
ed in relation to each other.  If placed on the same edge, the final t=
wo 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 la=
st 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 i=
n the context of a 3^4 reduction, again relative to 2C pieces.  The fi=
nal (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.  Ind=
eed, 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!

<=
div>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 the=
re 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, r>
it's been around longer). I thought it might be possible, seeing that <=
br>
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" <=
br>
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=3D4, d>3, this rule above is not the case. you r>
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 <=
br>
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&qu=
ot;

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 r>
that I think would be unsolvable with (non-caging) techniques similar

to a 3^4.

-Levi

=
le-span" style=3D"font-size: 13px; ">_._,___

=09
=09

=09

=09
=09
=09
=09
=09

--00163646d979a6b83f0461df710a--

From: Roice Nelson <roice3@gmail.com>
Date: Sun, 1 Feb 2009 14:43:32 -0600
Subject: Re: [MC4D] Re: Parity on MC m^n

--00163646d542c46c930461e17dd8
Content-Type: text/plain; charset=ISO-8859-1
Content-Transfer-Encoding: 7bit

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
>>
>> _._,___
>>
>
>

--00163646d542c46c930461e17dd8
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

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 man=
ually change that part).

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

Best,
Roice

quote">On Sun, Feb 1, 2009 at 12:16 PM, Roice Nelson wrote:<=
br>
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 =3D"http://games.groups.yahoo.com/group/4D_Cubing/files/parity%20error%20lo=
gs/" target=3D"_blank">a new folder in the files area of the group
show=
ing the parity error situations I encountered on my 4^4 solution.  v>

quick aside on a program technical issue:  These weren&=
#39;t loading with the current java version (back then the 4^4 was only ava=
ilable 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 hav=
e.  Here is an example.

old:  MagicCube4D 1 0 857
new:  MagicCub=
e4D 2 2 844 4

Anyway, I'll do my best now to reconstr=
uct what looks like was going on - I can't remember last week, much les=
s the details of a decade ago :)

In the first file, I was trying to solve 2C pieces. &nb=
sp;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 reduct=
ion 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 flipp=
ed in relation to each other.  If placed on the same edge, the final t=
wo 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 la=
st 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 i=
n the context of a 3^4 reduction, again relative to 2C pieces.  The fi=
nal (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.  Ind=
eed, 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!

<=
div>Roice

ss=3D"Wj3C7c">On Sun, Feb 1, 2009 at 1:20 AM, rev_16_4 &l=
t;rev_16_4@yahoo.co=
m
>
wrote:

Roice, you bring up a very good point. I wasn't sure the=
re 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, r>
it's been around longer). I thought it might be possible, seeing that <=
br>
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" <=
br>
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=3D4, d>3, this rule above is not the case. you r>
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 <=
br>
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&qu=
ot;

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 r>
that I think would be unsolvable with (non-caging) techniques similar

to a 3^4.

-Levi

style=3D"clear:both;color:#FFF;font-size:1px">">_._,___

=09
=09

=09

=09
=09
=09
=09
=09

--00163646d542c46c930461e17dd8--

From: "rev_16_4" <rev_16_4@yahoo.com>
Date: Sun, 01 Feb 2009 22:02:43 -0000
Subject: Re: Parity on MC m^n

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.
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
> 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
> >
> > _._,___
> >
>

From: "rev_16_4" <rev_16_4@yahoo.com>
Date: Sun, 01 Feb 2009 23:02:23 -0000
Subject: [MC4D] Re: Parity on MC m^n

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
> 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
> > 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
> >>
> >> _._,___
> >>
> >
> >
>

From: Roice Nelson <roice3@gmail.com>
Date: Sun, 1 Feb 2009 17:23:57 -0600
Subject: Re: [MC4D] Re: Parity on MC m^n

--0016364eed9c78a9940461e3bbdc
Content-Type: text/plain; charset=ISO-8859-1
Content-Transfer-Encoding: 7bit

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

--0016364eed9c78a9940461e3bbdc
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

'; font-size: 16px; ">

margin-bottom: 8px; margin-left: 8px; font: normal normal normal small/nor=
mal arial; ">
A little more roice spam, this time inline :)  (and a little of it fro=
m the earlier post).

=3D"margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.=
8ex; border-left-width: 1px; border-left-color: rgb(204, 204, 204); border-=
I'm still retaining my nontraditional definition of parity errors =

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
essentially as odd parity (and in my n^d solution double odd as
blockquote>
in-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width: 1=
px; border-left-color: rgb(204, 204, 204); border-left-style: solid; paddin=
g-left: 1ex; ">
well). Ignoring this definition, check out the "single flipped"&n=
bsp;
0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-lef=
t-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: sol=
parity error on a 4^3. It will take an odd number of inner slice
<=
/blockquote>
gin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width: =
1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; paddi=
ng-left: 1ex; ">
quarter twists to solve this.
ote" style=3D"margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margi=
n-left: 0.8ex; border-left-width: 1px; border-left-color: rgb(204, 204, 204=
); border-left-style: solid; padding-left: 1ex; ">

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
On a 3^3, a single swapped pair has odd parity. This cannot happen >
argin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width=
: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; pad=
ding-left: 1ex; ">
due to the even (aka double odd) parity of a single quarter twist on <=
br>
margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wid=
th: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; p=
a 3^3. However, using reduction, the "single swapped edge pair" (=
with
n-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; bord=
er-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-styl=
correct orientation) parity err on a 4^3 can seem to occur. This will =

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
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.

0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left=
-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: soli=
You can see a similar phenomenon with a 3^3. If you have an even
<=
/blockquote>
gin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width: =
1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; paddi=
ng-left: 1ex; ">
number of corner pair-swaps to perform, you will have an even number <=
br>
margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wid=
th: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; p=
of edge pair-swaps also. It took an even number of quarter twists to <=
br>
margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wid=
th: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; p=
generate this position, and it will take an even number of quarter >
argin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width=
: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; pad=
ding-left: 1ex; ">
twists to solve. The same goes for odd. An odd # of Corner pair-swaps =

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
will always be accompanied by an odd # of Edge pair-swaps and an odd <=
br>
margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wid=
th: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; p=
# of twists. This is my double odd parity.

=
It is interesting to note you can have an odd number of corner pair swaps w=
ith an even number of edge pair swaps if the orientations of the corners al=
so come into play, the most simple example being a cube that is solved exce=
pt for two corners.  In this case I guess the double odd parity is sha=
red with the orientations instead of the edges.

-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; borde=
r-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style=
I've also uploaded four log files to the same folder. Two of these =
;
x; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-w=
idth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid;=
are cases that I'd consider parity errors. I cannot see how these =

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
could be generated on a 3^4 or a 4^4. One involves a pair of 4C
blockquote>
in-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width: 1=
px; border-left-color: rgb(204, 204, 204); border-left-style: solid; paddin=
g-left: 1ex; ">
pieces swapped. The other, a pair of 3C swapped so they appear to
=
rgin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width:=
1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; padd=
ing-left: 1ex; ">
have 2 stickers flipped. I'd be shocked to see a solution for either&nb=
sp;
0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left=
-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: soli=
of these.

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

-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; borde=
r-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style=
The final log (Faked2CHalf) I believe is the other case you refered r>
margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-widt=
h: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; pa=
dding-left: 1ex; ">
to in 4_4Roice1.log. I would agree/argue that this isn't a parity =

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
error because you hadn't reached a 3^4 state yet. If this is a parity&n=
bsp;
0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-lef=
t-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: sol=
error, then any position before you reach a 3^4 is because they're =
;
x; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-w=
idth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid;=
all impossible on a 3^4. The spirit of parity errors are situations r>
margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-widt=
h: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; pa=
dding-left: 1ex; ">
you wouldn't know were a problem until you got to the end, and "Wa=
it
top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border=
-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style:=
a minute, I can't solve this position!" (like Roice2)
ote>

That is exactly what happened to me though ;)=
But using a mental parity model of a 4^3, not a 3^4.  Consideri=
ng 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 s=
olve it (because the solution necessitated breaking these already solved, a=
rtificially joined pieces back in two).  Since I had already mentally =
recategorized those sets of two as single merged pieces, the puzzle mo=
del I was trying to use to my advantage left me stuck.

Overall, I'd say the different usages of the word p=
arity is still clouding the discussion here.  I am conceding that my u=
se of the phrase "parity problem" is perhaps too general.  A=
fter all, a problem is simply a configuration you don't know how to sol=
ve 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 pari=
ties and deriving what that says about the solution.  I like the conce=
pt of "parity problem" defined in the context of mismatches betwe=
en the different parity characteristics of various puzzles, that they are m=
ental surprises (perceived impossibilities) when one tries to use solution =
methods across multiple puzzles.

All the best,
Roice

--0016364eed9c78a9940461e3bbdc--

From: "rev_16_4" <rev_16_4@yahoo.com>
Date: Mon, 02 Feb 2009 00:34:47 -0000
Subject: [MC4D] Re: Parity on MC m^n

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
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

From: Roice Nelson <roice3@gmail.com>
Date: Sun, 1 Feb 2009 23:25:18 -0600
Subject: Re: [MC4D] Re: Parity on MC m^n

--00163646d0c2be5f450461e8c7aa
Content-Type: text/plain; charset=ISO-8859-1
Content-Transfer-Encoding: 7bit

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.

>
> 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

--00163646d0c2be5f450461e8c7aa
Content-Type: text/html; charset=ISO-8859-1
Content-Transfer-Encoding: quoted-printable

One last trimmed down reply for me as well :)iv>

0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left=
-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: soli=
> It is interesting to note you can have an odd number of corner pair&nb=
sp;
0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left=
-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: soli=
> swaps with an even number of edge pair swaps if the orientations of&nb=
sp;
0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left=
-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: soli=
> the corners also come into play, the most simple example being a =

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
> cube that is solved except for two corners. In this case I guess =

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
> the double odd parity is shared with the orientations instead of =

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
> the edges.
margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex;=
border-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left=

p: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-l=
eft-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: s=
Are you refering to two corners that aren't oriented correctly? If =
;
x; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-w=
idth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid;=
not I'm not sure what you mean. On a 3^3, you cannot swap a single =
;
x; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-w=
idth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid;=
pair of corners, regardless of orientation, unless an odd # of pairs <=
br>
margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wid=
th: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; p=
of edges are swapped as well.

I just learne=
d something!  And had to pull out my cube to sway myself :)  Afte=
r thousands of solves of the 3^3 over the years, I never realized the situa=
tion 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 &qu=
ot;world view".  And it is amazing what details one can gloss ove=
r.  Sorry for the incorrect claim (often I learn by making assertions =
then seeing the counterexample, or having it pointed out to me).

style=3D"margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-le=
ft: 0.8ex; border-left-width: 1px; border-left-color: rgb(204, 204, 204); b=
> I think we can deduce the 4C case is impossible because corners are&nb=
sp;
0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left=
-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: soli=
> only permuted/oriented by outer twists (that is, by the same twist&nbs=
p;
px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-=
width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid=
> set as the 3^4), and this is not a possible configuration on the =

; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wi=
dth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; =
> 3^4. The 3C case I'm unsure of, but I plan to experiment with it&n=
bsp;
0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-lef=
t-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: sol=
> some.
n-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; bord=
er-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-styl=

p: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-l=
eft-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: s=
I'm not so sure. The same thing is possible on the 4^3, yet not on =
;
x; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-w=
idth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid;=
the 3^3....

Ah yes, there is a ho=
le in the reasoning!  I think it can be filled though with the extra o=
bservation 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 o=
dd, so it still looks impossible.  What do you think?

In fact, this convinces me the 3C situation you uploade=
d is impossible as well.  The manifestation of that would have to be t=
he 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.

div>
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 ar=
e not.

style=3D"margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-le=
ft: 0.8ex; border-left-width: 1px; border-left-color: rgb(204, 204, 204); b=
yle=3D"margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left:=
0.8ex; border-left-width: 1px; border-left-color: rgb(204, 204, 204); bord=
> already mentally recategorized those sets of two as single merged =
;
x; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-w=
idth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid;=
> pieces, the puzzle model I was trying to use to my advantage left =
;
x; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-w=
idth: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid;=
> me stuck.
argin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; =
border-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-=

p: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-l=
eft-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: s=
Interesting. Essentially you went 4^4, to 4*3^3, to 3^4? This parity <=
br>
margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-wid=
th: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; p=
case happened between 4*3^3 and 3^4? Is this close to accurate?
uote>

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/be=
havior 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).=

margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-widt=
h: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; pa=
dding-left: 1ex; ">
Thanks for the lively discussion.

Absol=
utely.  I really feel like I learned a lot the past few days reading t=
he discussion and struggling over these scenarios.  I'm really hap=
py to have the concepts more crystallized in my mind, so thank you very muc=
h as well!  And thanks for your patience with my mistakes.

Good night,
Roice

--00163646d0c2be5f450461e8c7aa--

From: David Smith <djs314djs314@yahoo.com>
Date: Mon, 2 Feb 2009 17:49:17 -0800 (PST)
Subject: Re: [MC4D] Re: Parity on MC m^n

=20=20=20=20
One last trimmed down reply for me as well :)=A0
> It is interesting to note you can have an odd number of corner pair=A0

> swaps with an even number of edge pair swaps if the orientations of=A0

> the corners also come into play, the most simple example being a=A0

> cube that is solved except for two corners. In this case I guess=A0

> the double odd parity is shared with the orientations instead of=A0

> the edges.

Are you refering to two corners that aren't oriented correctly? If=A0

not I'm not sure what you mean. On a 3^3, you cannot swap a single=A0

pair of corners, regardless of orientation, unless an odd # of pairs=A0

of edges are swapped as well.
I just learned something! =A0And had to pull out my cube to sway myself :) =
=A0After thousands of solves of the 3^3 over the years, I never realized th=
e situation where there are two unsolved corners meant those corners had to=
be in their correct positions, but incorrect orientations. =A0That'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 "wo=
rld view". =A0And it is amazing what details one can gloss over. =A0Sorry f=
or 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=A0

> only permuted/oriented by outer twists (that is, by the same twist=A0

> set as the 3^4), and this is not a possible configuration on the=A0

> 3^4. The 3C case I'm unsure of, but I plan to experiment with it=A0

> some.

I'm not so sure. The same thing is possible on the 4^3, yet not on=A0

the 3^3....

Ah yes, there is a hole in the reasoning! =A0I think it can be filled thoug=
h with the extra observation you've made about the 4^4, which is that no tw=
ists create a single odd-parity condition. =A0And a single set of swapped c=
orners is odd, so it still looks impossible. =A0What do you think?

In fact, this convinces me the 3C situation you uploaded is impossible as w=
ell. =A0The 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 edge=
s is a single odd parity condition, again impossible.
The scenario in=A04x4x4x4_roice2. log differs from both of these in that it=
is two pair of swaps, an even parity condition. =A0So now it makes sense t=
o me why that is possible whereas the above two situations are not.

> already mentally recategorized those sets of two as single merged=A0

> pieces, the puzzle model I was trying to use to my advantage left=A0

> me stuck.

Interesting. Essentially you went 4^4, to 4*3^3, to 3^4? This parity=A0

case happened between 4*3^3 and 3^4? Is this close to accurate?

I'm not following what you mean by=A04*3^3, but I was trying to apply the 4=
^3 pairing of 2C stickers approach/behavior to the 4^4 (the analogue was pa=
iring up pairs of 2C stickers). =A0And yep, this was an interim step in the=
effort to go from 4^4 -> 3^4).

Thanks for the lively discussion.

Absolutely. =A0I really feel like I learned a lot the past few days reading=
the discussion and struggling over these scenarios. =A0I'm really happy to=
have the concepts more crystallized in my mind, so thank you very much as =
well! =A0And thanks for your patience with my mistakes.

Good night,Roice

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