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DGCubes

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@GenTheThief Oh okay, got it now. That seems long, manual seems much easier. Thanks!

Just for reference, that algorithm was referring to the OLL case you'd get after twisting a corner. It's impossible to twist a single corner with an algorithm; it happens because a corner was accidentally twisted at some point earlier in the solve, so it needs to be twisted back manually (mathematically, it doesn't matter which corner you twist, as long as you twist it in the opposite direction of the initial twist). This does not mean you should twist corners in other situations, and you'll only ever need to twist a single one (competitors have gotten in trouble for twisting more than one corner per solve in competitions, and for twisting corners when their cube was solvable without it).
 

RedJack22

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Just for reference, that algorithm was referring to the OLL case you'd get after twisting a corner. It's impossible to twist a single corner with an algorithm; it happens because a corner was accidentally twisted at some point earlier in the solve, so it needs to be twisted back manually (mathematically, it doesn't matter which corner you twist, as long as you twist it in the opposite direction of the initial twist). This does not mean you should twist corners in other situations, and you'll only ever need to twist a single one (competitors have gotten in trouble for twisting more than one corner per solve in competitions, and for twisting corners when their cube was solvable without it).
Man, you seem so smart with the math-side of cubes :)
 

Conor Chin

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Index finger+Middle finger=Fast U2
Pinky finger+Ring finger=Fast D2

Everything else is slow. Make sure you can do these with your left hand too.

This is mostly used for OH. Example with left hand:
Your index finger flicks from the Fru sticker to the Bru position. U2
There's also Blu to Fru position [U2'], but that requires more extension.

And also Frd to Bru [R2'] pinky flick.
Or Pinky finger+Ring finger.
Or you can do a slight regrip and do the double finger double flicks above.
Thank you!
 

Duncan Bannon

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This was very recently answered by @Tabe

For brackets, those are "triggers". Essentially, these are really common sets of moves that show up in lots of algs. They are grouped together with brackets to aid with memorization. The idea is you see that set of moves and know "oh, that's a sexy move" rather than trying to memorize 4 more moves. So a simple alg becomes like "F [sexy move] F'" or something. Much easier to remember that way.
 

iheartgeo

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I'm confused about the diagrams in the CxLL page. What I see in the link below looks like it should swap a pair of corners diagonally, but the given algorithm F - RUR'U' - F' does not do this (it swaps two pairs of adjacent corners). Am I reading the diagram incorrectly or this is an error on the page itself?

https://www.speedsolving.com/wiki/index.php/CxLL_U_D
 

CLL Smooth

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I'm confused about the diagrams in the CxLL page. What I see in the link below looks like it should swap a pair of corners diagonally, but the given algorithm F - RUR'U' - F' does not do this (it swaps two pairs of adjacent corners). Am I reading the diagram incorrectly or this is an error on the page itself?

https://www.speedsolving.com/wiki/index.php/CxLL_U_D
I’m pretty sure you’re mistaken
 

DGCubes

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I'm confused about the diagrams in the CxLL page. What I see in the link below looks like it should swap a pair of corners diagonally, but the given algorithm F - RUR'U' - F' does not do this (it swaps two pairs of adjacent corners). Am I reading the diagram incorrectly or this is an error on the page itself?

https://www.speedsolving.com/wiki/index.php/CxLL_U_D

On 2x2, swapping 2 pairs of adjacent corners is equivalent to a diag-swap (just do a U or U' to see what I mean).
F R U R' U' F' with a U or U' appended to the beginning or end will do a diag-swap, but those generally aren't included in algs because you may or may not have to do them depending on the solve.
 

Tom Joad

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Hello everybody,

So, I’m doing my f2l and after solving two pairs in the front right slot and back left slot, I have a situation which I can only solve really inefficiently.

The other two corner pieces are both solved but the front left edge piece is in the back right slot and vice versa.

All I need to do is swap the front left edge piece and back right edge piece. Is there a beautifully efficient way to do this?

edit: I’ve just realised that there are still many situations that may arise depending on the orientation of the edge pieces...

Thanks
 

Tom Joad

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When I am drilling plls, one thing has always confused me: the number of times I have to repeat the pll to get back to a solved cube.

Now take an H perm for example. To me this makes total sense. I have a solved cube, I do the algorithm once to get an H perm then I do it again to solve the cube. So far, so good. I even give this type of perm a name: mod 2.

Now I start to drill my Gd perm. But I am not very bright and based on the logic of the previous sentence, I have a premonition. I expect it to be a mod 3 perm. I can’t really explain why, perhaps this is not an extrapolation of my mod 2 experience, perhaps it is purely intuitive: probably something to do with those three interchanging pieces on all the G perm pictures.

But whatever the reason for my intuition, like so many that have gone before it, it is WRONG. The Gd perm is mod 4!

No doubt the explanation for this is really straightforward. Anyone care to enlighten me?
 

Thom S.

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No doubt the explanation for this is really straightforward. Anyone care to enlighten me?

It is. If you do a Ua Perm(Cycles 3 pieces, mod 3) your case is a Ub Perm. Now, if you do a U move, do you still have a Ub Perm - yes. But can you solve it with the same Ub Perm - no.
An Algorithm cycles pieces in their location regardless of where they are.
thus Gd Perm cycles pieces in the positions UBL-UBR-UFL and UF-UB-UL but the algorithm leaves the Last Layer in a state where different pieces will be cycled each time. If you do a U' after a Gd Perm, all the pieces will be in their correct location.
Technically that U' is part of the Algorithm.

Long answer, could have said this in half the text but whatever.
 

Tom Joad

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It is. If you do a Ua Perm(Cycles 3 pieces, mod 3) your case is a Ub Perm. Now, if you do a U move, do you still have a Ub Perm - yes. But can you solve it with the same Ub Perm - no.
An Algorithm cycles pieces in their location regardless of where they are.
thus Gd Perm cycles pieces in the positions UBL-UBR-UFL and UF-UB-UL but the algorithm leaves the Last Layer in a state where different pieces will be cycled each time. If you do a U' after a Gd Perm, all the pieces will be in their correct location.
Technically that U' is part of the Algorithm.

Long answer, could have said this in half the text but whatever.

Aha! Now I get it!

Thanks
 

GenTheSnail

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Hello everybody,

So, I’m doing my f2l and after solving two pairs in the front right slot and back left slot, I have a situation which I can only solve really inefficiently.

The other two corner pieces are both solved but the front left edge piece is in the back right slot and vice versa.

All I need to do is swap the front left edge piece and back right edge piece. Is there a beautifully efficient way to do this?

edit: I’ve just realised that there are still many situations that may arise depending on the orientation of the edge pieces...

Thanks
4 cases: none flipped, right flipped, left flipped, both flipped
2 cases per case: with or without any other slots solved

If neither edges are flipped:
Front left and back right slots are unsolved: D' R L U2 R' L' D
Front left or back right slot solved: (R U R') (L U' L') (R U' R') U (L U L')

E: worked out other cases

Back left is flipped
LF/RB slots solved: R L U' R' (F R' F R) L' U' (L U2 L') U' (L U L')

Front right is flipped
LF/RB slots solved: (F R' F R) U' L U (R U2 R') U' L' U (R U' R' )

Both are flipped
LF/RB slots solved: (L R U' L' R') (F Lw' U' Lw2) U L U' M'

I didn't feel like figuring out cases using empty slots.
Also note that I use ZZ, never really deal with flipped edges, and don't know if there are any fancy CFOP tricks that others might know.

@Smiles ?

E2: crud, I just realized that I did this for the wrong pair of slots...
Cases should be able to be mirrored though

E3: or you can do a y/y' and use these
(I did say that I use ZZ. I can't think like a CFOPer)
 
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