Solving Corners Last On Large Cubes

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I have just recently been experimenting with solving the 5x5x5 via reduction: centers, tredges, corners last. I notice some of these corners are much longer to solve than I anticipated. The shortest solution for any such arrangement seems to be 8 moves.

visualcube.php


R2 B L2 B' R2 B L2 B'

Question 1: Are there any cases where fewer than 8 moves can solve a large cube if only the corners are incorrect?
Question 2: Is there a "known longest" corners-only-unsolved position (and what is its solution)?
Question 3: Given any sufficiently scrambled cube, isn't the longest solution to solve all of the corners first only 11 moves?
Question 4: Would it be wiser to solve the cube in the order: corners, tredges, centers... if the objective was to build a method that has the fewest moves to solve on average?
 
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1. The shortest "Corners Only" alg I know of is Niklas (and all of its mirrored/inversed iterations): R U' L' U R' U' L (7 moves).

2. There are lots of different types of "corners only" algs. But if you're asking for an area where they're commonly used today, then what you're asking for are Corner Commutators. They're ubiquitous in advanced FMC and BLD methods. The longest commutators used in BH/3-Style for BLD (which cycles 3 corners at a time) are 12 moves. I may defer to an FMC or BLD expert to talk about if there are known cases which cycle more than 3 corners. I don't think anyone uses any method to solve all 8 corners in one shot... I just can't fathom how that could be broken down into easy to recognize cases that preserve the rest of the cube. https://www.speedsolving.com/wiki/index.php/Commutators http://www.speedcubing.com/chris/bhcorners.html

3. If you think of "corners only" solutions to be the equivalent of solving a 2x2, in which case God's Number is 11, then yes... 11. But once again, on big cubes, preserving the corners to the end would be an enormous pain without a proper method. And I don't think one exists.

4. I don't think a method like that has been explored because 'direct solving' on big cubes, efficient though it could be, is not very conducive to speedsolving. Recognition is atrocious and algs are not friendly to muscle memory, so speedsolving is out. However, using center commutators is very easy. Then maybe something like Freeslice for edges could also be efficient, and then corner commutators. But my gut is telling me that would not be very efficient.

I'm not sure what kind of answer you're looking for here, because what's "wiser" could be subjective. The method you describe sounds interesting from an FMC perspective, but from a speedsolving perspective, just painful. As a speedsolver, I would say it's wiser to use a method more akin to Reduction: Solve the centers, Pair the tredges (placement is arbitrary), and then just solve the whole thing like a normal 3x3. Reduction winds up being fairly efficient, and I'd actually really like to see if anyone knows of an already existing method that is even more so.

Have you spent any time checking out any more common methods for solving big cubes? K4 and Cage has some elements of what you're describing, though maybe not in the same order you listed them here. They also might not be as "direct" as what you're hoping for.
 
Thanks for the elaborations; very informative.

1. The shortest "Corners Only" alg I know of is Niklas (and all of its mirrored/inversed iterations): R U' L' U R' U' L (7 moves).

That's very close to what I was referring to, but the "tredges" are not fully solved. They're in the right orientation but not the correct position. See below:

R U' L' U R' U'


2. There are lots of different types of "corners only" algs. But if you're asking for an area where they're commonly used today, then what you're asking for are Corner Commutators. They're ubiquitous in advanced FMC and BLD methods. The longest commutators used in BH/3-Style for BLD (which cycles 3 corners at a time) are 12 moves. I may defer to an FMC or BLD expert to talk about if there are known cases which cycle more than 3 corners. I don't think anyone uses any method to solve all 8 corners in one shot... I just can't fathom how that could be broken down into easy to recognize cases that preserve the rest of the cube. https://www.speedsolving.com/wiki/index.php/Commutators http://www.speedcubing.com/chris/bhcorners.html

I was thinking of something along these lines:

U2 R U' F' D' R' U' R' B D' F2 L F' U' B' L' F' L' D B'

I thought somewhere I saw a 22-move with just two opposite corners twisted on an otherwise completely solved cube. I will continue hunting.
But is there a shorter solution to the corner position shown above? That one consists of 20 moves.

3. If you think of "corners only" solutions to be the equivalent of solving a 2x2, in which case God's Number is 11, then yes... 11. But once again, on big cubes, preserving the corners to the end would be an enormous pain without a proper method. And I don't think one exists.

Yes, I'm thinking about having my own solving program switch to solve corners first, so it would kind of be like a 2x2 solve, only I need a center reference on the 5x5x5. It turns out that solving with respect to the center does not change the distance; it's still at a max of 11. Thanks for verifying.

4. I don't think a method like that has been explored because 'direct solving' on big cubes, efficient though it could be, is not very conducive to speedsolving. Recognition is atrocious and algs are not friendly to muscle memory, so speedsolving is out. However, using center commutators is very easy. Then maybe something like Freeslice for edges could also be efficient, and then corner commutators. But my gut is telling me that would not be very efficient.

Ahhhh, ok, that makes sense now. It's easier to execute some well known algs fast rather than use time to derive a shorter solution with such a narrow scope of applicability.

I'm not sure what kind of answer you're looking for here, because what's "wiser" could be subjective. The method you describe sounds interesting from an FMC perspective, but from a speedsolving perspective, just painful. As a speedsolver, I would say it's wiser to use a method more akin to Reduction: Solve the centers, Pair the tredges (placement is arbitrary), and then just solve the whole thing like a normal 3x3. Reduction winds up being fairly efficient, and I'd actually really like to see if anyone knows of an already existing method that is even more so.

The results that I might achieve with the program would probably not be worth it for the human speedsolvers. I'm hunting for a fast FMC-type solution for the program. Currently the more time I invest, the lower the movecount drops, but the corners-last approach is not bearing fruit. I'd need to pre-compute all 88,179,840 corner positions and perform a lookup on the result. With the tredges + centers already solved, this would take a great deal of time.

Have you spent any time checking out any more common methods for solving big cubes? K4 and Cage has some elements of what you're describing, though maybe not in the same order you listed them here. They also might not be as "direct" as what you're hoping for.

This will pretty much evolve into a Cage-solving method. I have a very quick way to deal with the cage once it is set up.

Thanks again, I'll re-read the links you posted.
 
Since outer-layer turns on the 5x5x5 reduces the puzzle to a 3x3x3, then God's number for the 3x3x3 of 20 moves can be applied. So the 22-move algorithm you described must be sub-optimal.

Also, your 20-move alg is easily proven to be sub-optimal:
 
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Since outer-layer turns on the 5x5x5 reduces the puzzle to a 3x3x3, then God's number for the 3x3x3 of 20 moves can be applied. So the 22-move algorithm you described must be sub-optimal.

Also, your 20-move alg is easily proven to be sub-optimal:
https://www.youtube.com/watch?v=CBBr8q-T-x0

The larger cubes can create corner and edge scenarios that can't be created on the 3x3x3 though. Correct? Case in point: the single "dedge" flip on the 4x4x4:

2R2 D2 2R' F2 2R F2 D2 2R D2 2L' F2 2L F2 D2 2R2


I was trying to look for a way to create this corner position in fewer moves for the 4x4x4:

U2 F2 U' L' U2 L F U R' U2 R Uw F' U2 F Uw' F Fw' R U2 Fw' U2 Fw' U2 Fw' U2 Fw2 R' Fw z' 2R2 D2 2R' F2 2R F2 D2 2R D2 2L' F2 2L F2 D2 2R2 y'x

Edit: Here it is, and you can see why this is probably not possible on the 3x3x3:

r2 2F2 U2 b2 D' r2 U2 b2 U f2 R2 D2 B2 l2 U


My question is: Can this corner orientation be created on the 3x3x3? My supposition at this point is that it can't. If it can, I'd like to see the 3x3x3 solution. If it can't, I'd like to see the fewest moves to the 4x4x4 solution.
 
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The larger cubes can create corner and edge scenarios that can't be created on the 3x3x3 though. Correct? Case in point: the single "dedge" flip on the 4x4x4:

2R2 D2 2R' F2 2R F2 D2 2R D2 2L' F2 2L F2 D2 2R2

If we're talking about 5x5, all 3x3 states that come up after reduction (once OLL parity which does not represent a 3x3 stage and can easily be recognised as such) can be solved exactly like a 3x3 as the 5x5 can be viewed as a 3x3 when it comes to the middle slice edge and the corners. This is why you can't get PLL parity on a 5x5 and OLL parity is an error in the wings rather than a single tredge flip.

In addition, all CO states possible on big cubes can be directly created on a 3x3 or 2x2.

However, even layered cubes are slightly different as they don't have edges which directly correspond to 3x3 edges and can therefore have 2-swaps.

I was trying to look for a way to create this corner position in fewer moves for the 4x4x4:

U2 F2 U' L' U2 L F U R' U2 R Uw F' U2 F Uw' F Fw' R U2 Fw' U2 Fw' U2 Fw' U2 Fw2 R' Fw z' 2R2 D2 2R' F2 2R F2 D2 2R D2 2L' F2 2L F2 D2 2R2 y'x

Edit: Here it is, and you can see why this is probably not possible on the 3x3x3:

r2 2F2 U2 b2 D' r2 U2 b2 U f2 R2 D2 B2 l2 U


My question is: Can this corner orientation be created on the 3x3x3? My supposition at this point is that it can't. If it can, I'd like to see the 3x3x3 solution. If it can't, I'd like to see the fewest moves to the 4x4x4 solution.

I think when it comes to big cubes, you can essentially have 3 sets of pieces: (corners and middle edges), (wings) and (centres).

You are correct that the above position cannot be created on a 3x3 though the same pattern cannot be replicated (ie opposite corner swap with no other pieces affected) on any odd layered cube (such as 5x5, 7x7 etc) as you cannot create PLL parity on these either.

To summarise, you can get things like the above on even layered cubes as the are no middle edges but you can't get them on odd layered cubes.
 
In the block turn metric, there are 6-move maneuvers, but I don't think the OP meant to allow that metric.

For example (SiGN notation):
U2 2-4r2 2-4f2 D2 2-4r2 2-4f2 (same algorithm to scramble or solve)

Using only single-layer turns, I can't think of any positions having only corners unsolved that take less than 8 turns.
 
Edit: Here it is, and you can see why this is probably not possible on the 3x3x3:

r2 2F2 U2 b2 D' r2 U2 b2 U f2 R2 D2 B2 l2 U


My question is: Can this corner orientation be created on the 3x3x3? My supposition at this point is that it can't. If it can, I'd like to see the 3x3x3 solution. If it can't, I'd like to see the fewest moves to the 4x4x4 solution.
I won't go into trying to explain (again) what the laws of the cube are, but Clément Gallet recently found some pretty brief algorithms for this case and the adjacent case in your preferred move metric. I added these to the 4x4x4 parity algorithms wiki about a month ago.

Adjacent (17 moves):
F2 R2 B' D' B R2 F' U 2F2 F' L2 2F2 2L2 L2 2F2 2L2 U'

Diagonal (18 moves)
2R2 F2 L' U2 L2 2L2 F2 R' D2 R2 U2 B2 R D2 B2 R2 2R2 U2


May I also say that:


In addition, there are no odd permutation "corners only" positions on the odd nxnxn cube (which includes a 2 corner swap only). So you only have to consider 88,179,840 corner only positions for the even nxnxn cube, but 44,089,920 for the odd nxnxn cube.

EDIT:
And the two corner swap position you asked about is not referred to as a "corner orientation", but a "corner permutation". Corner orientation is a term used to describe how corners are twisted, not how they are moved around.
 
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You are correct that the above position cannot be created on a 3x3 though the same pattern cannot be replicated (ie opposite corner swap with no other pieces affected) on any odd layered cube (such as 5x5, 7x7 etc) as you cannot create PLL parity on these either.

In addition, there are no odd permutation "corners only" positions on the odd nxnxn cube (which includes a 2 corner swap only). So you only have to consider 88,179,840 corner only positions for the even nxnxn cube, but 44,089,920 for the odd nxnxn cube.

Thank you both. I'm working on determining what the longest "corners only" position is on the 5x5x5. If they're as long as some of these 4x4x4 positions,
my program will have to discover another means of dealing with them.

In the block turn metric, there are 6-move maneuvers, but I don't think the OP meant to allow that metric.

For example (SiGN notation):
U2 2-4r2 2-4f2 D2 2-4r2 2-4f2 (same algorithm to scramble or solve)

Using only single-layer turns, I can't think of any positions having only corners unsolved that take less than 8 turns.

I'm doing a full-width search out to depth 8 on the 5x5x5. If all of the tredges and centers are solved, I'm capturing the sequence.
Depth 7 already completed (93,510,461,296 nodes cumulative, including the 1 node @ depth 0) and there are 0 corner positions, for certain.

Depth 8 has found only 400 so far (2.1 trillion and counting). That's a pretty sparse distribution.
My supposition at this point is that there will be some very deep corners-only positions for the 5x5x5.

R U R' D' R U' R' D
R U R' D R U' R' D'
R U R' D2 R U' R' D2
R U L' U' R' U L U'
R U L U' R' U L' U'
R U L2 U' R' U L2 U'
R U' R' D' R U R' D
R U' R' D R U R' D'
R U' R' D2 R U R' D2
R U' L' U R' U' L U
R U' L U R' U' L' U
R U' L2 U R' U' L2 U
R U2 R' D' R U2 R' D
R U2 R' D R U2 R' D'
R U2 R' D2 R U2 R' D2
R D' R' U R D R' U'
R D' R' U' R D R' U
R D' R' U2 R D R' U2
R D' L' D R' D' L D
R D' L D R' D' L' D
R D' L2 D R' D' L2 D
R D R' U R D' R' U'
R D R' U' R D' R' U
R D R' U2 R D' R' U2
R D L' D' R' D L D'
R D L D' R' D L' D'
R D L2 D' R' D L2 D'
R D2 R' U R D2 R' U'
R D2 R' U' R D2 R' U
R D2 R' U2 R D2 R' U2
R F R' B' R F' R' B
R F R' B R F' R' B'
R F R' B2 R F' R' B2
R F L' F' R' F L F'
R F L F' R' F L' F'
R F L2 F' R' F L2 F'
R F' R' B' R F R' B
R F' R' B R F R' B'
R F' R' B2 R F R' B2
R F' L' F R' F' L F
R F' L F R' F' L' F
R F' L2 F R' F' L2 F
R F2 R' B' R F2 R' B
R F2 R' B R F2 R' B'
R F2 R' B2 R F2 R' B2
R B' R' F R B R' F'
R B' R' F' R B R' F
R B' R' F2 R B R' F2
R B' L' B R' B' L B
R B' L B R' B' L' B
R B' L2 B R' B' L2 B
R B R' F R B' R' F'
R B R' F' R B' R' F
R B R' F2 R B' R' F2
R B L' B' R' B L B'
R B L B' R' B L' B'
R B L2 B' R' B L2 B'
R B2 R' F R B2 R' F'
R B2 R' F' R B2 R' F
R B2 R' F2 R B2 R' F2
R' U R D' R' U' R D
R' U R D R' U' R D'
R' U R D2 R' U' R D2
R' U L' U' R U L U'
R' U L U' R U L' U'
R' U L2 U' R U L2 U'
R' U' R D' R' U R D
R' U' R D R' U R D'
R' U' R D2 R' U R D2
R' U' L' U R U' L U
R' U' L U R U' L' U
R' U' L2 U R U' L2 U
R' U2 R D' R' U2 R D
R' U2 R D R' U2 R D'
R' U2 R D2 R' U2 R D2
R' D' R U R' D R U'
R' D' R U' R' D R U
R' D' R U2 R' D R U2
R' D' L' D R D' L D
R' D' L D R D' L' D
R' D' L2 D R D' L2 D
R' D R U R' D' R U'
R' D R U' R' D' R U
R' D R U2 R' D' R U2
R' D L' D' R D L D'
R' D L D' R D L' D'
R' D L2 D' R D L2 D'
R' D2 R U R' D2 R U'
R' D2 R U' R' D2 R U
R' D2 R U2 R' D2 R U2
R' F R B' R' F' R B
R' F R B R' F' R B'
R' F R B2 R' F' R B2
R' F L' F' R F L F'
R' F L F' R F L' F'
R' F L2 F' R F L2 F'
R' F' R B' R' F R B
R' F' R B R' F R B'
R' F' R B2 R' F R B2
R' F' L' F R F' L F
R' F' L F R F' L' F
R' F' L2 F R F' L2 F
R' F2 R B' R' F2 R B
R' F2 R B R' F2 R B'
R' F2 R B2 R' F2 R B2
R' B' R F R' B R F'
R' B' R F' R' B R F
R' B' R F2 R' B R F2
R' B' L' B R B' L B
R' B' L B R B' L' B
R' B' L2 B R B' L2 B
R' B R F R' B' R F'
R' B R F' R' B' R F
R' B R F2 R' B' R F2
R' B L' B' R B L B'
R' B L B' R B L' B'
R' B L2 B' R B L2 B'
R' B2 R F R' B2 R F'
R' B2 R F' R' B2 R F
R' B2 R F2 R' B2 R F2
R2 U L' U' R2 U L U'
R2 U L U' R2 U L' U'
R2 U L2 U' R2 U L2 U'
R2 U' L' U R2 U' L U
R2 U' L U R2 U' L' U
R2 U' L2 U R2 U' L2 U
R2 D' L' D R2 D' L D
R2 D' L D R2 D' L' D
R2 D' L2 D R2 D' L2 D
R2 D L' D' R2 D L D'
R2 D L D' R2 D L' D'
R2 D L2 D' R2 D L2 D'
R2 F L' F' R2 F L F'
R2 F L F' R2 F L' F'
R2 F L2 F' R2 F L2 F'
R2 F' L' F R2 F' L F
R2 F' L F R2 F' L' F
R2 F' L2 F R2 F' L2 F
R2 B' L' B R2 B' L B
R2 B' L B R2 B' L' B
R2 B' L2 B R2 B' L2 B
R2 B L' B' R2 B L B'
R2 B L B' R2 B L' B'
R2 B L2 B' R2 B L2 B'
L' U R U' L U R' U'
L' U R' U' L U R U'
L' U R2 U' L U R2 U'
L' U L D' L' U' L D
L' U L D L' U' L D'
L' U L D2 L' U' L D2
L' U' R U L U' R' U
L' U' R' U L U' R U
L' U' R2 U L U' R2 U
L' U' L D' L' U L D
L' U' L D L' U L D'
L' U' L D2 L' U L D2
L' U2 L D' L' U2 L D
L' U2 L D L' U2 L D'
L' U2 L D2 L' U2 L D2
L' D' R D L D' R' D
L' D' R' D L D' R D
L' D' R2 D L D' R2 D
L' D' L U L' D L U'
L' D' L U' L' D L U
L' D' L U2 L' D L U2
L' D R D' L D R' D'
L' D R' D' L D R D'
L' D R2 D' L D R2 D'
L' D L U L' D' L U'
L' D L U' L' D' L U
L' D L U2 L' D' L U2
L' D2 L U L' D2 L U'
L' D2 L U' L' D2 L U
L' D2 L U2 L' D2 L U2
L' F R F' L F R' F'
L' F R' F' L F R F'
L' F R2 F' L F R2 F'
L' F L B' L' F' L B
L' F L B L' F' L B'
L' F L B2 L' F' L B2
L' F' R F L F' R' F
L' F' R' F L F' R F
L' F' R2 F L F' R2 F
L' F' L B' L' F L B
L' F' L B L' F L B'
L' F' L B2 L' F L B2
L' F2 L B' L' F2 L B
L' F2 L B L' F2 L B'
L' F2 L B2 L' F2 L B2
L' B' R B L B' R' B
L' B' R' B L B' R B
L' B' R2 B L B' R2 B
L' B' L F L' B L F'
L' B' L F' L' B L F
L' B' L F2 L' B L F2
L' B R B' L B R' B'
L' B R' B' L B R B'
L' B R2 B' L B R2 B'
L' B L F L' B' L F'
L' B L F' L' B' L F
L' B L F2 L' B' L F2
L' B2 L F L' B2 L F'
L' B2 L F' L' B2 L F
L' B2 L F2 L' B2 L F2
L U R U' L' U R' U'
L U R' U' L' U R U'
L U R2 U' L' U R2 U'
L U L' D' L U' L' D
L U L' D L U' L' D'
L U L' D2 L U' L' D2
L U' R U L' U' R' U
L U' R' U L' U' R U
L U' R2 U L' U' R2 U
L U' L' D' L U L' D
L U' L' D L U L' D'
L U' L' D2 L U L' D2
L U2 L' D' L U2 L' D
L U2 L' D L U2 L' D'
L U2 L' D2 L U2 L' D2
L D' R D L' D' R' D
L D' R' D L' D' R D
L D' R2 D L' D' R2 D
L D' L' U L D L' U'
L D' L' U' L D L' U
L D' L' U2 L D L' U2
L D R D' L' D R' D'
L D R' D' L' D R D'
L D R2 D' L' D R2 D'
L D L' U L D' L' U'
L D L' U' L D' L' U
L D L' U2 L D' L' U2
L D2 L' U L D2 L' U'
L D2 L' U' L D2 L' U
L D2 L' U2 L D2 L' U2
L F R F' L' F R' F'
L F R' F' L' F R F'
L F R2 F' L' F R2 F'
L F L' B' L F' L' B
L F L' B L F' L' B'
L F L' B2 L F' L' B2
L F' R F L' F' R' F
L F' R' F L' F' R F
L F' R2 F L' F' R2 F
L F' L' B' L F L' B
L F' L' B L F L' B'
L F' L' B2 L F L' B2
L F2 L' B' L F2 L' B
L F2 L' B L F2 L' B'
L F2 L' B2 L F2 L' B2
L B' R B L' B' R' B
L B' R' B L' B' R B
L B' R2 B L' B' R2 B
L B' L' F L B L' F'
L B' L' F' L B L' F
L B' L' F2 L B L' F2
L B R B' L' B R' B'
L B R' B' L' B R B'
L B R2 B' L' B R2 B'
L B L' F L B' L' F'
L B L' F' L B' L' F
L B L' F2 L B' L' F2
L B2 L' F L B2 L' F'
L B2 L' F' L B2 L' F
L B2 L' F2 L B2 L' F2
L2 U R U' L2 U R' U'
L2 U R' U' L2 U R U'
L2 U R2 U' L2 U R2 U'
L2 U' R U L2 U' R' U
L2 U' R' U L2 U' R U
L2 U' R2 U L2 U' R2 U
L2 D' R D L2 D' R' D
L2 D' R' D L2 D' R D
L2 D' R2 D L2 D' R2 D
L2 D R D' L2 D R' D'
L2 D R' D' L2 D R D'
L2 D R2 D' L2 D R2 D'
L2 F R F' L2 F R' F'
L2 F R' F' L2 F R F'
L2 F R2 F' L2 F R2 F'
L2 F' R F L2 F' R' F
L2 F' R' F L2 F' R F
L2 F' R2 F L2 F' R2 F
L2 B' R B L2 B' R' B
L2 B' R' B L2 B' R B
L2 B' R2 B L2 B' R2 B
L2 B R B' L2 B R' B'
L2 B R' B' L2 B R B'
L2 B R2 B' L2 B R2 B'
U R U' L' U R' U' L
U R D2 R' U' R D2 R'
U R' U' L2 U R U' L2
U R' D R U' R' D' R
U R' D2 R U' R' D2 R
U R2 U' L' U R2 U' L
U L' U' R2 U L U' R2
U L' D L U' L' D' L
U L' D2 L U' L' D2 L
U L U' R' U L' U' R
U L D2 L' U' L D2 L'
U L2 U' R' U L2 U' R
U F U' B U F' U' B'
U F U' B2 U F' U' B2
U F D' F' U' F D F'
U F D F' U' F D' F'
U F' U' B' U F U' B
U F' U' B U F U' B'
U F' U' B2 U F U' B2
U F' D' F U' F' D F
U F' D2 F U' F' D2 F
U F2 U' B' U F2 U' B
U F2 U' B U F2 U' B'
U F2 U' B2 U F2 U' B2
U B' U' F U B U' F'
U B' U' F' U B U' F
U B' U' F2 U B U' F2
U B' D' B U' B' D B
U B' D2 B U' B' D2 B
U B U' F U B' U' F'
U B U' F2 U B' U' F2
U B D' B' U' B D B'
U B D B' U' B D' B'
U B2 U' F U B2 U' F'
U B2 U' F' U B2 U' F
U B2 U' F2 U B2 U' F2
U' R U L2 U' R' U L2
U' R D' R' U R D R'
U' R D2 R' U R D2 R'
U' R' U L U' R U L'
U' R' D2 R U R' D2 R
U' R2 U L U' R2 U L'
U' L' U R U' L U R'
U' L' D2 L U L' D2 L
U' L U R2 U' L' U R2
U' L D' L' U L D L'
U' L D2 L' U L D2 L'
U' L2 U R U' L2 U R'
U' F U B' U' F' U B
U' F U B U' F' U B'
U' F U B2 U' F' U B2
U' F D F' U F D' F'
U' F D2 F' U F D2 F'
U' F' U B' U' F U B
U' F' U B2 U' F U B2
U' F' D' F U F' D F
U' F' D F U F' D' F
U' F2 U B' U' F2 U B
U' F2 U B U' F2 U B'
U' F2 U B2 U' F2 U B2
U' B' U F' U' B U F
U' B' U F2 U' B U F2
U' B' D' B U B' D B
U' B' D B U B' D' B
U' B U F U' B' U F'
U' B U F' U' B' U F
U' B U F2 U' B' U F2
U' B D B' U B D' B'
U' B D2 B' U B D2 B'
U' B2 U F U' B2 U F'
U' B2 U F' U' B2 U F
U' B2 U F2 U' B2 U F2
U2 R D' R' U2 R D R'
U2 R D R' U2 R D' R'
U2 R D2 R' U2 R D2 R'
U2 R' D' R U2 R' D R
U2 R' D R U2 R' D' R
U2 R' D2 R U2 R' D2 R
U2 L' D' L U2 L' D L
U2 L' D L U2 L' D' L
U2 L' D2 L U2 L' D2 L
U2 L D' L' U2 L D L'
U2 L D L' U2 L D' L'
U2 L D2 L' U2 L D2 L'
U2 F D' F' U2 F D F'
U2 F' D F U2 F' D' F
U2 B' D B U2 B' D' B
U2 B D' B' U2 B D B'
D' R U' R' D R U R'
D' R U2 R' D R U2 R'
D' R D L2 D' R' D L2
D' R' U2 R D R' U2 R
D' R' D L D' R D L'
D' R2 D L D' R2 D L'
D' L' U2 L D L' U2 L
D' L' D R D' L D R'
D' L U' L' D L U L'
D' L U2 L' D L U2 L'
D' L D R2 D' L' D R2
D' L2 D R D' L2 D R'
D' F U F' D F U' F'
D' F U2 F' D F U2 F'
D' F D B' D' F' D B
D' F D B D' F' D B'
D' F D B2 D' F' D B2
D' F' U F D F' U' F
D' F' U' F D F' U F
D' F' D B' D' F D B
D' F' D B2 D' F D B2
D' F2 D B' D' F2 D B
D' F2 D B D' F2 D B'
D' F2 D B2 D' F2 D B2
 
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I'm doing a full-width search out to depth 8 on the 5x5x5. If all of the tredges and centers are solved, I'm capturing the sequence.
Depth 7 already completed (93,510,461,296 nodes cumulative, including the 1 node @ depth 0) and there are 0 corner positions, for certain.

Depth 8 has found only 400 so far (2.1 trillion and counting). That's a pretty sparse distribution.
My supposition at this point is that there will be some very deep corners-only positions for the 5x5x5.
Why are you giving your computer unnecessary stress? The corner only positions on the 5x5x5 are identical to the corner only positions on the 3x3x3. If you want to do such an analysis, do it on the 3x3x3.

Have you ever solved a big cube with the 3x3x3 reduction method to know that this is true, or have you just solved the 4x4x4 and up by using a form of cage?
 
Depth 8 has found only 400 so far

I count essentially 11 types of corner 3-cycles that have 8-move commutators. There are 48 symmetry variations of each type for 11*48 = 528 positions. Of the 11 types, 7 have two separate 8-move commutators. That makes a total of (11+7)*48 = 864 8-move maneuvers, if you're counting maneuvers rather than positions.
 
I count essentially 11 types of corner 3-cycles that have 8-move commutators. There are 48 symmetry variations of each type for 11*48 = 528 positions. Of the 11 types, 7 have two separate 8-move commutators. That makes a total of (11+7)*48 = 864 8-move maneuvers, if you're counting maneuvers rather than positions.

The program has gone through 35 of the 45 moves through depth 8. The most recent count shows 496. That's roughly ~640 if they're evenly distributed. I'm only counting unique cube positions. If a different set of moves produces the same cube, it does not get counted thereafter. We'll have to see what the final tally shows.

EDIT: Here are all of the various counts so far:


[table="class: grid"]
[tr]
[td]Depth[/td]
[td]Cumulative Node Count[/td]
[td]Only Centers Unsolved[/td]
[td]Only Tredges Unsolved[/td]
[td]Only Corners Unsolved[/td]
[/tr]
[tr]
[td]0[/td]
[td]1[/td]
[td]0[/td]
[td]0[/td]
[td]0[/td]
[/tr]
[tr]
[td]1[/td]
[td]46[/td]
[td]0[/td]
[td]0[/td]
[td]0[/td]
[/tr]
[tr]
[td]2[/td]
[td]1,666[/td]
[td]0[/td]
[td]0[/td]
[td]0[/td]
[/tr]
[tr]
[td]3[/td]
[td]59,086[/td]
[td]0[/td]
[td]0[/td]
[td]0[/td]
[/tr]
[tr]
[td]4[/td]
[td]2,095,696[/td]
[td]366[/td]
[td]30[/td]
[td]0[/td]
[/tr]
[tr]
[td]5[/td]
[td]74,331,496[/td]
[td]1,044[/td]
[td]84[/td]
[td]0[/td]
[/tr]
[tr]
[td]6[/td]
[td]2,636,431,696[/td]
[td]20,481[/td]
[td]628[/td]
[td]0[/td]
[/tr]
[tr]
[td]7[/td]
[td]93,510,461,296[/td]
[td]140,917[/td]
[td]2,452[/td]
[td]0[/td]
[/tr]
[tr]
[td]8[/td]
[td]3,316,682,338,696[/td]
[td]1,707,431[/td]
[td]32,715[/td]
[td]528[/td]
[/tr]
[tr]
[td]9[/td]
[td]4,172,450,898,004,695[/td]
[td](no data yet)[/td]
[td]165,602[/td]
[td](no data yet)[/td]
[/tr]
[tr]
[td]10[/td]
[td]147,990,866,406,820,695[/td]
[td](no data yet)[/td]
[td]1,722,997[/td]
[td](no data yet)[/td]
[/tr]
[/table]

By "only tredges unsolved" I mean if any middle edge or wing edge is not solved, while the rest of the cube is solved, I include it in the count.

Why are you giving your computer unnecessary stress?

I've got 8 CPU cores. The Intel i7-5960X overclocked to 4.6 GHz outperforms every computer on the planet. It feels no pain.

I'm enumerating corners-only, tredge-only, and centers-only disturbed positions. The one with the fewest instances per depth is the worst to solve last. As it turns out, this will be the corners.

Theoretically, solving corners first, tredges next, and centers last should produce, on average, the lowest move counts per solve.
 
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In addition, there are no odd permutation "corners only" positions on the odd nxnxn cube (which includes a 2 corner swap only). So you only have to consider 88,179,840 corner only positions for the even nxnxn cube, but 44,089,920 for the odd nxnxn cube.

But there are sequences with an odd number of moves that can create positions with only corners being unsolved, right? Or is this not what you meant.

Example:

R U' R D2 R' U R D2 R2
 
But there are sequences with an odd number of moves that can create positions with only corners being unsolved, right? Or is this not what you meant.

Example:

R U' R D2 R' U R D2 R2

He means on odd layered cubes you can't get a single 2-swap of corners in the way you demonstrated with a 4x4 above. These can only exist on even layered cubes due to PLL parity. (essentially, there must always be an even number of 2 swaps to have even "parity").
 
I'm enumerating corners-only, tredge-only, and centers-only disturbed positions. The one with the fewest instances per depth is the worst to solve last. As it turns out, this will be the corners.

Theoretically, solving corners first, tredges next, and centers last should produce, on average, the lowest move counts per solve.
I'm not sure if this way of looking at things is right or not, but I do know it doesn't take all factors into account. When talking about what order to solve groups of pieces in, you also want to consider what kinds of moves and move sequences are allowed, and what are not, based on what you have already solved and must keep in place. The question to consider for each piece type is, how many moves do I lose due to having to respect the earlier steps?

Of course, usually this number can't be known for sure, but it is something that can be reasoned about. For instance, although there are 4-move centers only cases, they only affect six centers out of 48 (and in a very specific pattern, too), whereas you could otherwise do something like r U b and affect many centers at once in a few moves - giving a pattern that would require many more moves to solve if you know you can't disturb the edges or corners. So one could argue that the restriction of having all the edges and corners in place could potentially cost a lot of moves. With corners, on the other hand, you are looking at 11 moves to solve them first and maybe 18-20 to solve them last. Are those 7-9 moves worth that restriction in movement for the whole rest of the solve?

However - corners last should really never be considered. That's because they should always be solved while solving the middle edges. I'll explain.

There are really two fundamentally different types of edges, and lumping them together as a "tredge" can obscure some important points. I'll call them middles (the one in the center of each edge, which only shows up on odd-layered cubes) and wings (the rest). The two big distinctions are (a) middles have two possible orientations (and thus can be flipped) whereas wings only have one; and (b) like corners, middles can only be moved using turns of the outer layer (relative to the fixed centers). The net effect is that middles can be treated like 3x3x3 edges.

This matters because, as any 3x3x3 solver knows, solving corners first and then edges, or edges first and then corners, is slower and takes more moves than just solving the whole cube directly. So if you think of the tredges as two separate steps - solve the middle, and, pair the tredges (i.e. get each middle next to its two corresponding wings) - then it immediately makes sense to combine the "solve the middle" step with solving the corners, in a 3x3x3 stage, like we do in reduction methods. Then really the three steps are solving the centers, pairing the tredges, and the 3x3x3 stage, and now it makes more sense to argue about how to order them. And of course, the 3x3x3 stage can be done in the same number of moves at any point in the solve, so you might as well do it last.
 
So, 132 moves + parity + center 3-cycle + corners? Not very efficient for a computer solver; humans have even done shorter FMC solves. But it's definitely a good start.

In case you didn't read my post in detail and just angrily skimmed it, I'll point out here that solving pieces in any order is entirely possible, and with computing power and cleverness most of them can be more efficient than a human. My point was that some piece orders will be much more efficient than others, and if you want the best order, you can rule out corners last.
 
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