Rubik's cube 4x4 parity moves
$begingroup$
Note: since asking this question I've purchased several other twisty puzzles and now regret taking the "look up algorithms online" approach to solving - in the end I get much more enjoyment from the puzzles where I worked out to solve them myself than the ones where I didn't. If you're stuck at this point I would recommend seeking resources to understand the concept of parity and then try to come up with a solution yourself. You won't end up with a super efficient algorithm, but you will understand it, which is ultimately more satisfying. Or it is for me at least. That said, answers to this question are still welcome.
I'm solving my 4x4 Rubik's cube (aka Rubik's revenge) for the first time, and I have the "OLL parity" case:
There are plenty of algorithms for this available online, but different people use different notation and nobody ever says which one they're using, so I risk messing up my cube if I guess wrongly.
So, please give me algorithms for solving the OLL parity, using the following notation:
R
: turn just the rightmost face clockwise
Rw
(R wide): turn the rightmost half of the cube clockwise, i.e. the rightmost face and the slice adjacent to it
r
: turn the rightmost slice clockwise, but not the rightmost face
R2
,Rw2
,r2
: as above but turning it twice
R'
,Rw'
,r'
: as above but turning it anticlockwise
L
,F
,B
,U
,D
: left, front, back, up, down. (All clockwise by default, can be modified as above)
x
: if you use this, please explain what it means, as I have no clue.
-
(hyphen): if you use this, please explain what it means, as I have no clue.
(
,)
(parentheses): if you use these, please explain what they mean, as I have no clue.
If you use anything else at all in your notation, please explain what it means. If you prefer a different notation from the above that's fine, but please explain it.
I'm holding my cube with the unsolved cubies facing me at the top. (i.e. the red face would be the front in the diagram above.) If I should start in some other orientation, please say so.
It would be helpful to have an algorithm that leaves all other pieces alone, as my cube is in exactly the almost-solved state above. But if algorithms that don't do this are significantly shorter and/or easier to remember then that would be fine too. (Please tell me which I can expect in your answer!)
rubiks-cube
$endgroup$
add a comment |
$begingroup$
Note: since asking this question I've purchased several other twisty puzzles and now regret taking the "look up algorithms online" approach to solving - in the end I get much more enjoyment from the puzzles where I worked out to solve them myself than the ones where I didn't. If you're stuck at this point I would recommend seeking resources to understand the concept of parity and then try to come up with a solution yourself. You won't end up with a super efficient algorithm, but you will understand it, which is ultimately more satisfying. Or it is for me at least. That said, answers to this question are still welcome.
I'm solving my 4x4 Rubik's cube (aka Rubik's revenge) for the first time, and I have the "OLL parity" case:
There are plenty of algorithms for this available online, but different people use different notation and nobody ever says which one they're using, so I risk messing up my cube if I guess wrongly.
So, please give me algorithms for solving the OLL parity, using the following notation:
R
: turn just the rightmost face clockwise
Rw
(R wide): turn the rightmost half of the cube clockwise, i.e. the rightmost face and the slice adjacent to it
r
: turn the rightmost slice clockwise, but not the rightmost face
R2
,Rw2
,r2
: as above but turning it twice
R'
,Rw'
,r'
: as above but turning it anticlockwise
L
,F
,B
,U
,D
: left, front, back, up, down. (All clockwise by default, can be modified as above)
x
: if you use this, please explain what it means, as I have no clue.
-
(hyphen): if you use this, please explain what it means, as I have no clue.
(
,)
(parentheses): if you use these, please explain what they mean, as I have no clue.
If you use anything else at all in your notation, please explain what it means. If you prefer a different notation from the above that's fine, but please explain it.
I'm holding my cube with the unsolved cubies facing me at the top. (i.e. the red face would be the front in the diagram above.) If I should start in some other orientation, please say so.
It would be helpful to have an algorithm that leaves all other pieces alone, as my cube is in exactly the almost-solved state above. But if algorithms that don't do this are significantly shorter and/or easier to remember then that would be fine too. (Please tell me which I can expect in your answer!)
rubiks-cube
$endgroup$
2
$begingroup$
Has a useful answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
$endgroup$
– Rubio♦
Apr 28 '18 at 13:32
add a comment |
$begingroup$
Note: since asking this question I've purchased several other twisty puzzles and now regret taking the "look up algorithms online" approach to solving - in the end I get much more enjoyment from the puzzles where I worked out to solve them myself than the ones where I didn't. If you're stuck at this point I would recommend seeking resources to understand the concept of parity and then try to come up with a solution yourself. You won't end up with a super efficient algorithm, but you will understand it, which is ultimately more satisfying. Or it is for me at least. That said, answers to this question are still welcome.
I'm solving my 4x4 Rubik's cube (aka Rubik's revenge) for the first time, and I have the "OLL parity" case:
There are plenty of algorithms for this available online, but different people use different notation and nobody ever says which one they're using, so I risk messing up my cube if I guess wrongly.
So, please give me algorithms for solving the OLL parity, using the following notation:
R
: turn just the rightmost face clockwise
Rw
(R wide): turn the rightmost half of the cube clockwise, i.e. the rightmost face and the slice adjacent to it
r
: turn the rightmost slice clockwise, but not the rightmost face
R2
,Rw2
,r2
: as above but turning it twice
R'
,Rw'
,r'
: as above but turning it anticlockwise
L
,F
,B
,U
,D
: left, front, back, up, down. (All clockwise by default, can be modified as above)
x
: if you use this, please explain what it means, as I have no clue.
-
(hyphen): if you use this, please explain what it means, as I have no clue.
(
,)
(parentheses): if you use these, please explain what they mean, as I have no clue.
If you use anything else at all in your notation, please explain what it means. If you prefer a different notation from the above that's fine, but please explain it.
I'm holding my cube with the unsolved cubies facing me at the top. (i.e. the red face would be the front in the diagram above.) If I should start in some other orientation, please say so.
It would be helpful to have an algorithm that leaves all other pieces alone, as my cube is in exactly the almost-solved state above. But if algorithms that don't do this are significantly shorter and/or easier to remember then that would be fine too. (Please tell me which I can expect in your answer!)
rubiks-cube
$endgroup$
Note: since asking this question I've purchased several other twisty puzzles and now regret taking the "look up algorithms online" approach to solving - in the end I get much more enjoyment from the puzzles where I worked out to solve them myself than the ones where I didn't. If you're stuck at this point I would recommend seeking resources to understand the concept of parity and then try to come up with a solution yourself. You won't end up with a super efficient algorithm, but you will understand it, which is ultimately more satisfying. Or it is for me at least. That said, answers to this question are still welcome.
I'm solving my 4x4 Rubik's cube (aka Rubik's revenge) for the first time, and I have the "OLL parity" case:
There are plenty of algorithms for this available online, but different people use different notation and nobody ever says which one they're using, so I risk messing up my cube if I guess wrongly.
So, please give me algorithms for solving the OLL parity, using the following notation:
R
: turn just the rightmost face clockwise
Rw
(R wide): turn the rightmost half of the cube clockwise, i.e. the rightmost face and the slice adjacent to it
r
: turn the rightmost slice clockwise, but not the rightmost face
R2
,Rw2
,r2
: as above but turning it twice
R'
,Rw'
,r'
: as above but turning it anticlockwise
L
,F
,B
,U
,D
: left, front, back, up, down. (All clockwise by default, can be modified as above)
x
: if you use this, please explain what it means, as I have no clue.
-
(hyphen): if you use this, please explain what it means, as I have no clue.
(
,)
(parentheses): if you use these, please explain what they mean, as I have no clue.
If you use anything else at all in your notation, please explain what it means. If you prefer a different notation from the above that's fine, but please explain it.
I'm holding my cube with the unsolved cubies facing me at the top. (i.e. the red face would be the front in the diagram above.) If I should start in some other orientation, please say so.
It would be helpful to have an algorithm that leaves all other pieces alone, as my cube is in exactly the almost-solved state above. But if algorithms that don't do this are significantly shorter and/or easier to remember then that would be fine too. (Please tell me which I can expect in your answer!)
rubiks-cube
rubiks-cube
edited Nov 12 '18 at 7:40
Nathaniel
asked Apr 21 '18 at 4:22
NathanielNathaniel
454619
454619
2
$begingroup$
Has a useful answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
$endgroup$
– Rubio♦
Apr 28 '18 at 13:32
add a comment |
2
$begingroup$
Has a useful answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
$endgroup$
– Rubio♦
Apr 28 '18 at 13:32
2
2
$begingroup$
Has a useful answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
$endgroup$
– Rubio♦
Apr 28 '18 at 13:32
$begingroup$
Has a useful answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
$endgroup$
– Rubio♦
Apr 28 '18 at 13:32
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Here is the algorithm I use (from http://www.rubiksplace.com/cubes/4x4/), but using your notation:
Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'
So it's basically a bunch of half cube turns and top face 180 degree turns.
The "x" means you rotate the cube clockwise along the x-axis. In other words, if the Red face was facing you, it will move to the top while the sides stay the same color.
$endgroup$
$begingroup$
It doesn't matter how carefully I perform this move, whether directly from this answer or elsewhere, it always completely messes up the cube. (Or to be precise, it does remove the parity, but it scrambles it as a 3x3 so I have to solve it again using 3x3 moves.) I must be making a mistake, but I've no idea what. Sorry, I thought asking for more precise notation would help.
$endgroup$
– Nathaniel
May 13 '18 at 9:12
$begingroup$
I just tried it again to make sure it works. Maybe watching someone do it will help: youtu.be/DX5aGS5k2o8
$endgroup$
– StevenWhite
May 14 '18 at 5:27
add a comment |
$begingroup$
I use the following that leaves all the other cubes alone:
r U2 r U2 r' U2 r U2 l' U2 r U2 r' U2 lr' r' U2 r'
If you look at it this way, the net effect is just an r' :
r U2 r
U2 r' U2 r U2 l'
U2 r U2 r' U2 l
r'
r' U2 r'
$endgroup$
add a comment |
$begingroup$
I can never remember any clean parity fix for this case, but there is a very easy to remember dirty one, which leaves only about 6 edge pieces out of place. Of course this is no good for speed solving, but it is so simple that I'll never forget it.
It is: R2 u R2 u R2 u R2 u R2 u R2
What this does is move the u
layer, while the R2
moves will swap out the centre pieces from that layer and put them back in a different place. The net effect is that the u
layer has been given one quarter turn, but the centre pieces have been restored. This is an odd permutation on the affected edge pieces.
If you start with the flipped pair at the FR
location, you then only have to perform a few 3-cycles to fix the last 6 edges.
$endgroup$
add a comment |
$begingroup$
Not technically an answer, but here is the meaning of the symbols:
x, y, z :
These are notation for rotation of the entire cube. they mean to rotate following the R, U, and F moves, respectively.
-:
I don't know what the hyphen means. Can you tell me where you saw the hyphen?
( )
These don't change what to do, they just tell you what moves are triggers, or sets of moves that repeat through many algorithms.You usually see them around sets of moves like (R U R' U') (R' F R F') and (R U2 R')
New contributor
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is the algorithm I use (from http://www.rubiksplace.com/cubes/4x4/), but using your notation:
Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'
So it's basically a bunch of half cube turns and top face 180 degree turns.
The "x" means you rotate the cube clockwise along the x-axis. In other words, if the Red face was facing you, it will move to the top while the sides stay the same color.
$endgroup$
$begingroup$
It doesn't matter how carefully I perform this move, whether directly from this answer or elsewhere, it always completely messes up the cube. (Or to be precise, it does remove the parity, but it scrambles it as a 3x3 so I have to solve it again using 3x3 moves.) I must be making a mistake, but I've no idea what. Sorry, I thought asking for more precise notation would help.
$endgroup$
– Nathaniel
May 13 '18 at 9:12
$begingroup$
I just tried it again to make sure it works. Maybe watching someone do it will help: youtu.be/DX5aGS5k2o8
$endgroup$
– StevenWhite
May 14 '18 at 5:27
add a comment |
$begingroup$
Here is the algorithm I use (from http://www.rubiksplace.com/cubes/4x4/), but using your notation:
Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'
So it's basically a bunch of half cube turns and top face 180 degree turns.
The "x" means you rotate the cube clockwise along the x-axis. In other words, if the Red face was facing you, it will move to the top while the sides stay the same color.
$endgroup$
$begingroup$
It doesn't matter how carefully I perform this move, whether directly from this answer or elsewhere, it always completely messes up the cube. (Or to be precise, it does remove the parity, but it scrambles it as a 3x3 so I have to solve it again using 3x3 moves.) I must be making a mistake, but I've no idea what. Sorry, I thought asking for more precise notation would help.
$endgroup$
– Nathaniel
May 13 '18 at 9:12
$begingroup$
I just tried it again to make sure it works. Maybe watching someone do it will help: youtu.be/DX5aGS5k2o8
$endgroup$
– StevenWhite
May 14 '18 at 5:27
add a comment |
$begingroup$
Here is the algorithm I use (from http://www.rubiksplace.com/cubes/4x4/), but using your notation:
Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'
So it's basically a bunch of half cube turns and top face 180 degree turns.
The "x" means you rotate the cube clockwise along the x-axis. In other words, if the Red face was facing you, it will move to the top while the sides stay the same color.
$endgroup$
Here is the algorithm I use (from http://www.rubiksplace.com/cubes/4x4/), but using your notation:
Rw U2 x Rw U2 Rw U2 Rw' U2 Lw U2 Rw' U2 Rw U2 Rw' U2 Rw'
So it's basically a bunch of half cube turns and top face 180 degree turns.
The "x" means you rotate the cube clockwise along the x-axis. In other words, if the Red face was facing you, it will move to the top while the sides stay the same color.
answered Apr 25 '18 at 22:58
StevenWhiteStevenWhite
347110
347110
$begingroup$
It doesn't matter how carefully I perform this move, whether directly from this answer or elsewhere, it always completely messes up the cube. (Or to be precise, it does remove the parity, but it scrambles it as a 3x3 so I have to solve it again using 3x3 moves.) I must be making a mistake, but I've no idea what. Sorry, I thought asking for more precise notation would help.
$endgroup$
– Nathaniel
May 13 '18 at 9:12
$begingroup$
I just tried it again to make sure it works. Maybe watching someone do it will help: youtu.be/DX5aGS5k2o8
$endgroup$
– StevenWhite
May 14 '18 at 5:27
add a comment |
$begingroup$
It doesn't matter how carefully I perform this move, whether directly from this answer or elsewhere, it always completely messes up the cube. (Or to be precise, it does remove the parity, but it scrambles it as a 3x3 so I have to solve it again using 3x3 moves.) I must be making a mistake, but I've no idea what. Sorry, I thought asking for more precise notation would help.
$endgroup$
– Nathaniel
May 13 '18 at 9:12
$begingroup$
I just tried it again to make sure it works. Maybe watching someone do it will help: youtu.be/DX5aGS5k2o8
$endgroup$
– StevenWhite
May 14 '18 at 5:27
$begingroup$
It doesn't matter how carefully I perform this move, whether directly from this answer or elsewhere, it always completely messes up the cube. (Or to be precise, it does remove the parity, but it scrambles it as a 3x3 so I have to solve it again using 3x3 moves.) I must be making a mistake, but I've no idea what. Sorry, I thought asking for more precise notation would help.
$endgroup$
– Nathaniel
May 13 '18 at 9:12
$begingroup$
It doesn't matter how carefully I perform this move, whether directly from this answer or elsewhere, it always completely messes up the cube. (Or to be precise, it does remove the parity, but it scrambles it as a 3x3 so I have to solve it again using 3x3 moves.) I must be making a mistake, but I've no idea what. Sorry, I thought asking for more precise notation would help.
$endgroup$
– Nathaniel
May 13 '18 at 9:12
$begingroup$
I just tried it again to make sure it works. Maybe watching someone do it will help: youtu.be/DX5aGS5k2o8
$endgroup$
– StevenWhite
May 14 '18 at 5:27
$begingroup$
I just tried it again to make sure it works. Maybe watching someone do it will help: youtu.be/DX5aGS5k2o8
$endgroup$
– StevenWhite
May 14 '18 at 5:27
add a comment |
$begingroup$
I use the following that leaves all the other cubes alone:
r U2 r U2 r' U2 r U2 l' U2 r U2 r' U2 lr' r' U2 r'
If you look at it this way, the net effect is just an r' :
r U2 r
U2 r' U2 r U2 l'
U2 r U2 r' U2 l
r'
r' U2 r'
$endgroup$
add a comment |
$begingroup$
I use the following that leaves all the other cubes alone:
r U2 r U2 r' U2 r U2 l' U2 r U2 r' U2 lr' r' U2 r'
If you look at it this way, the net effect is just an r' :
r U2 r
U2 r' U2 r U2 l'
U2 r U2 r' U2 l
r'
r' U2 r'
$endgroup$
add a comment |
$begingroup$
I use the following that leaves all the other cubes alone:
r U2 r U2 r' U2 r U2 l' U2 r U2 r' U2 lr' r' U2 r'
If you look at it this way, the net effect is just an r' :
r U2 r
U2 r' U2 r U2 l'
U2 r U2 r' U2 l
r'
r' U2 r'
$endgroup$
I use the following that leaves all the other cubes alone:
r U2 r U2 r' U2 r U2 l' U2 r U2 r' U2 lr' r' U2 r'
If you look at it this way, the net effect is just an r' :
r U2 r
U2 r' U2 r U2 l'
U2 r U2 r' U2 l
r'
r' U2 r'
edited Nov 12 '18 at 8:32
AHKieran
5,5241143
5,5241143
answered Nov 12 '18 at 5:33
Bradley ReedBradley Reed
211
211
add a comment |
add a comment |
$begingroup$
I can never remember any clean parity fix for this case, but there is a very easy to remember dirty one, which leaves only about 6 edge pieces out of place. Of course this is no good for speed solving, but it is so simple that I'll never forget it.
It is: R2 u R2 u R2 u R2 u R2 u R2
What this does is move the u
layer, while the R2
moves will swap out the centre pieces from that layer and put them back in a different place. The net effect is that the u
layer has been given one quarter turn, but the centre pieces have been restored. This is an odd permutation on the affected edge pieces.
If you start with the flipped pair at the FR
location, you then only have to perform a few 3-cycles to fix the last 6 edges.
$endgroup$
add a comment |
$begingroup$
I can never remember any clean parity fix for this case, but there is a very easy to remember dirty one, which leaves only about 6 edge pieces out of place. Of course this is no good for speed solving, but it is so simple that I'll never forget it.
It is: R2 u R2 u R2 u R2 u R2 u R2
What this does is move the u
layer, while the R2
moves will swap out the centre pieces from that layer and put them back in a different place. The net effect is that the u
layer has been given one quarter turn, but the centre pieces have been restored. This is an odd permutation on the affected edge pieces.
If you start with the flipped pair at the FR
location, you then only have to perform a few 3-cycles to fix the last 6 edges.
$endgroup$
add a comment |
$begingroup$
I can never remember any clean parity fix for this case, but there is a very easy to remember dirty one, which leaves only about 6 edge pieces out of place. Of course this is no good for speed solving, but it is so simple that I'll never forget it.
It is: R2 u R2 u R2 u R2 u R2 u R2
What this does is move the u
layer, while the R2
moves will swap out the centre pieces from that layer and put them back in a different place. The net effect is that the u
layer has been given one quarter turn, but the centre pieces have been restored. This is an odd permutation on the affected edge pieces.
If you start with the flipped pair at the FR
location, you then only have to perform a few 3-cycles to fix the last 6 edges.
$endgroup$
I can never remember any clean parity fix for this case, but there is a very easy to remember dirty one, which leaves only about 6 edge pieces out of place. Of course this is no good for speed solving, but it is so simple that I'll never forget it.
It is: R2 u R2 u R2 u R2 u R2 u R2
What this does is move the u
layer, while the R2
moves will swap out the centre pieces from that layer and put them back in a different place. The net effect is that the u
layer has been given one quarter turn, but the centre pieces have been restored. This is an odd permutation on the affected edge pieces.
If you start with the flipped pair at the FR
location, you then only have to perform a few 3-cycles to fix the last 6 edges.
answered yesterday
Jaap ScherphuisJaap Scherphuis
16.4k12772
16.4k12772
add a comment |
add a comment |
$begingroup$
Not technically an answer, but here is the meaning of the symbols:
x, y, z :
These are notation for rotation of the entire cube. they mean to rotate following the R, U, and F moves, respectively.
-:
I don't know what the hyphen means. Can you tell me where you saw the hyphen?
( )
These don't change what to do, they just tell you what moves are triggers, or sets of moves that repeat through many algorithms.You usually see them around sets of moves like (R U R' U') (R' F R F') and (R U2 R')
New contributor
$endgroup$
add a comment |
$begingroup$
Not technically an answer, but here is the meaning of the symbols:
x, y, z :
These are notation for rotation of the entire cube. they mean to rotate following the R, U, and F moves, respectively.
-:
I don't know what the hyphen means. Can you tell me where you saw the hyphen?
( )
These don't change what to do, they just tell you what moves are triggers, or sets of moves that repeat through many algorithms.You usually see them around sets of moves like (R U R' U') (R' F R F') and (R U2 R')
New contributor
$endgroup$
add a comment |
$begingroup$
Not technically an answer, but here is the meaning of the symbols:
x, y, z :
These are notation for rotation of the entire cube. they mean to rotate following the R, U, and F moves, respectively.
-:
I don't know what the hyphen means. Can you tell me where you saw the hyphen?
( )
These don't change what to do, they just tell you what moves are triggers, or sets of moves that repeat through many algorithms.You usually see them around sets of moves like (R U R' U') (R' F R F') and (R U2 R')
New contributor
$endgroup$
Not technically an answer, but here is the meaning of the symbols:
x, y, z :
These are notation for rotation of the entire cube. they mean to rotate following the R, U, and F moves, respectively.
-:
I don't know what the hyphen means. Can you tell me where you saw the hyphen?
( )
These don't change what to do, they just tell you what moves are triggers, or sets of moves that repeat through many algorithms.You usually see them around sets of moves like (R U R' U') (R' F R F') and (R U2 R')
New contributor
edited yesterday
Omega Krypton
4,9452544
4,9452544
New contributor
answered yesterday
FluxDigital1FluxDigital1
211
211
New contributor
New contributor
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$begingroup$
Has a useful answer been given? If so, please don't forget to $color{green}{checkmark smalltext{Accept}}$ it :)
$endgroup$
– Rubio♦
Apr 28 '18 at 13:32