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The FMC thread

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Finally a sub-40, but I know it could've been much better, considering I was at F2L-1C after 14 moves. Any suggestions?

Scramble: R2 D2 B' D2 F2 D2 U2 L2 F L2 F' U B' F2 R2 U R B F' D U2


F' U B U' B' U [2x2x2]
L D L' D2 [2x2x3]
F' D' F2 D [F2L -1C]

Now the part I have most issues with: edges on LL:

B R U R' U' B' [orientation of edges]
B R B' R B R2 B' R2 [permutation of edges, for L5C]

Skeleton to L5C:
F' U B U' B' U L D L' D2 F' D' F2 D B R U R' U' * R B' R B R2 B' R2

Insert at *:
U R U' L2 U # R' U' L2 (6 cancellations)

Insert at #:
U' B' U F U' B U F' (2 cancellations)


Final solution:
F' U B U' B' U L D L' D2 F' D' F2 D B R L2 B' U F U' B U F' R' U' L2 R B' R B R2 B' R2 [34 moves]
 
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Finally a sub-40, but I know it could've been much better, considering I was at F2L-1C after 14 moves. Any suggestions?

Scramble: R2 D2 B' D2 F2 D2 U2 L2 F L2 F' U B' F2 R2 U R B F' D U2


F' U B U' B' U [2x2x2]
L D L' D2 [2x2x3]
F' D' F2 D [F2L -1C]
F' U B U' B' U // 2x2x2
L D L' D2 // 2x2x3
//F' D' F2 D
U F' U' D' F2 D // F2L
B' R' D' R D B // edges; ab3c

I tried inserting a U and U' around the F' in your solution to flip two edges, and while there are still two misoriented edges in the LL, now you can fix it with a single fruruf. This leaves 3 unsolved corners.

Edit: Here are some other alternatives.

You can use the eight-move pure OLL alg R U R2 F R F2 U F to flip the two edges. Fun fact: this alg is actually three sledgehammers cancelled into each other: (U' R U R') (R' F R F') (F' U F U').

Or you can switch to the inverse, use a sledgehammer to orient the edges, then reinsert the F2L edge. There are two ways to sledgehammer, leading to two different skeletons.

… F' D' F2 D // F2L-1 (4/14)
(F D' F' D R' D R D' R2) // edges; ab5c (9/23)

… F' D' F2 D // F2L-1 (4/14)
(D' F D F' R F' R' F R2) // edges; ab4c (9/23)

But in both of these cases, there are some other things you can do after the sledgehammer. This first one is pretty obvious: rather than inserting only the edge, you pair it up with the corner and then insert.

… F' D' F2 D // F2L-1 (4/14)
(F D' F') // EO (3/17)
(D2 R' D' R D R' D' R') // F2L; ab3c (8/25)

The second one is weirder. After the sledgehammer, you'll see that there's actually a bunch of pairs lying around, and, you know, wouldn't it be a waste to destroy them?

… F' D' F2 D // F2L-1 (4/14)
(D' F D F' R2) // EO + align (5/19)

At this point, you need to cycle the three pairs. You can't use an 8-move commutator here, but you can set it up to an 8-move case with a single move.

… (D' F D F' R2) // EO + align (5/19)
(F Dw2 L' U2 L Dw2 L' U2 L F') // block commutator (10/29)

… (D' F D F' R2) // EO + align (5/19)
(F2 R B' Lw2 B R' B' Lw2 B F2) // block commutator (10/29)

(stealth edit: fixed the move counts of the last few examples)
 
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Ok, I tried my first fmc ever. I watched Speedcubereview's youtube tutorials to get started. I ended up with 55 moves and was thinking if someone could help me especially after f2l-1

Scramble: B2 U2 R2 D L2 F2 U2 R2 U L2 U F L' U' B' F D F U2 R' D' U'
alg.cubing.net

Solution:
(L D B F L2 B L' B') 2x2x2 block
(U D R D' U B' R2 B) 2x2x3 block
(F L F L2 U L) F2L-1

Ok and then things got pretty inefficient. I tried some inverse solving and eo before last f2l pair but ended up with normal cfop ending. I had unlucky e-perm which gave 16 moves (optimal is 14 i guess). Still my movecount was lower than with some EO tricks which I tried and definitely didn't quite get.

(F' U' F' U F U' F' U) Last F2L pair
(U F U' F' U' R U R') OLL
(U L' U' R U L U' R' U L U' R U L' U' R') E-perm
F2

So if someone could help me with EO and LL. I also know only 25 OLL so without EO I'm gonna end up with 2-look. I think my f2l-1 was pretty good for beginner but the long algorithms in the end irritate me.



EDIT: Oh I didn't even notice that after last f2l pair I have U2. :D That gives 54 moves.
 
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Ok, I tried my first fmc ever. I watched Speedcubereview's youtube tutorials to get started. I ended up with 55 moves and was thinking if someone could help me especially after f2l-1

Scramble: B2 U2 R2 D L2 F2 U2 R2 U L2 U F L' U' B' F D F U2 R' D' U'
alg.cubing.net

Solution:
(L D B F L2 B L' B') 2x2x2 block
(U D R D' U B' R2 B) 2x2x3 block
(F L F L2 U L) F2L-1

Ok and then things got pretty inefficient. I tried some inverse solving and eo before last f2l pair but ended up with normal cfop ending. I had unlucky e-perm which gave 16 moves (optimal is 14 i guess). Still my movecount was lower than with some EO tricks which I tried and definitely didn't quite get.

(F' U' F' U F U' F' U) Last F2L pair
(U F U' F' U' R U R') OLL
(U L' U' R U L U' R' U L U' R U L' U' R') E-perm
F2

So if someone could help me with EO and LL. I also know only 25 OLL so without EO I'm gonna end up with 2-look. I think my f2l-1 was pretty good for beginner but the long algorithms in the end irritate me.



EDIT: Oh I didn't even notice that after last f2l pair I have U2. :D That gives 54 moves.
I'm no expert, but here are some thoughts:

First, best to avoid using parentheses, as that implies the moves are done on the inverse scramble.

Next, after your 2x2x3 block, notice that F U' is two moves to a pseudo-F2L. In other words, it would be F2L if on that white 1x2x3 block, all the non-white colours were inverted (i.e., orange <--> red, green <--> blue). You can fix this by doing U2 as a premove -- i.e., do U2 before you do the scramble. Then once you've finished your solve, you add U2 to the end of your scramble. It sounds like magic, but it works.

So this gives you a fairly quick F2L (19 moves, including the pre-move), but it gives you a bad last layer. Still, U F L U' F' L' cancels some moves and orients some edges. If you then do L D F D' F' L' you cancel a couple more moves and end up with 3 corners and 2 edges. That's not a particularly nice case (particularly with the twisted corner) but in theory you could then look for an insertion to solve it with cancellations. It might be that there's some other way to approach that last layer that leads to a nicer insertion.

I'm not saying this is the right way to approach your solve, but I wanted to give an idea of what a typical continuation from 2x2x3 might look like.
 
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Thanks a lot about the pseudo tip! I looked that I had a 3x2x1 block ready but didn't know how could I use it because it was upside down/didn't fit my f2l.

I have really no idea of these cancellations and insertions and commutators so I ended up doing cfop ending again. I got to 52 moves even with 2-look oll (dot case) and r-perm. I guess I need to watch these last layer/EO tutorials and try to start using some tricks.
 
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F' U B U' B' U // 2x2x2
L D L' D2 // 2x2x3
//F' D' F2 D
U F' U' D' F2 D // F2L
B' R' D' R D B // edges; ab3c

I tried inserting a U and U' around the F' in your solution to flip two edges, and while there are still two misoriented edges in the LL, now you can fix it with a single fruruf. This leaves 3 unsolved corners.

Edit: Here are some other alternatives.

You can use the eight-move pure OLL alg R U R2 F R F2 U F to flip the two edges. Fun fact: this alg is actually three sledgehammers cancelled into each other: (U' R U R') (R' F R F') (F' U F U').

Or you can switch to the inverse, use a sledgehammer to orient the edges, then reinsert the F2L edge. There are two ways to sledgehammer, leading to two different skeletons.

… F' D' F2 D // F2L-1 (4/14)
(F D' F' D R' D R D' R2) // edges; ab5c (9/23)

… F' D' F2 D // F2L-1 (4/14)
(D' F D F' R F' R' F R2) // edges; ab4c (9/23)

But in both of these cases, there are some other things you can do after the sledgehammer. This first one is pretty obvious: rather than inserting only the edge, you pair it up with the corner and then insert.

… F' D' F2 D // F2L-1 (4/14)
(F D' F') // EO (3/17)
(D2 R' D' R D R' D' R') // F2L; ab3c (8/25)

The second one is weirder. After the sledgehammer, you'll see that there's actually a bunch of pairs lying around, and, you know, wouldn't it be a waste to destroy them?

… F' D' F2 D // F2L-1 (4/14)
(D' F D F' R2) // EO + align (4/18)

At this point, you need to cycle the three pairs. You can't use an 8-move commutator here, but you can set it up to an 8-move case with a single move.

… (D' F D F' R2) // EO + align (4/18)
(F Dw2 L' U2 L Dw2 L' U2 L F') // block commutator (10/28)

… (D' F D F' R2) // EO + align (4/18)
(F2 R B' Lw2 B R' B' Lw2 B F2) // block commutator (10/28)
Cheers! There are some really good pointers in there, thank you for that :)

I'll defo have a look at re-inserting edges to influence orientation of LL instead of trying to do it purely with OLL and/or sunes.
Block commutators may be a step to far for me at the moment, as this was only my second solve where I actually used insertions, but I'll defo start playing around with it a bit to get a feel for it, and who knows I'll come across a solve to actually use it :)
 
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Sorry if spamming but I really want to understand this. So, I've started to figure out how commutators and insertions work and I would like to know how could I use them. So, here is a 46 move solution I found. It left me with an A-perm (So a corner 3-cycle, shouldn't that be the easiest of them all?). I've tried to find a place to insert it but my question is: how do I recognise if the corner is flipped right or wrong? Anyone could give me an example insertion for this solve?

Scramble: U B2 L2 D U2 R2 U R2 L2 F2 U’ R B2 F’ L’ D U2 F’ D F’ L U2
Solution: F D F’ D2 R’ L2 D’ L2 (2x2x2) R’ U2 B R B’ (2x2x3) U R U R’ U’ R U R U2 (F2L-1) R2 B’ R2 B U R2 U’ (F2L) B’ R’ B R’ B’ R2 B (OLL) D’ F D’ B2 D F’ D’ B2 D2 R (PLL)


And additionally, if there are twisted corners or multi-cycles, how could I use multiple commutators?

This was a pretty lucky solve, I didn't use any pseudo-blocks or NISS, just tried both f2l orientations after 2x2x3. I'd just like to understand these insertions and then try to add all of these methods to solutions to get lower movecounts.
 
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Sorry if spamming but I really want to understand this. So, I've started to figure out how commutators and insertions work and I would like to know how could I use them. So, here is a 46 move solution I found. It left me with an A-perm (So a corner 3-cycle, shouldn't that be the easiest of them all?). I've tried to find a place to insert it but my question is: how do I recognise if the corner is flipped right or wrong? Anyone could give me an example insertion for this solve?

Scramble: U B2 L2 D U2 R2 U R2 L2 F2 U’ R B2 F’ L’ D U2 F’ D F’ L U2
Solution: F D F’ D2 R’ L2 D’ L2 (2x2x2) R’ U2 B R B’ (2x2x3) U R U R’ U’ R U R U2 (F2L-1) R2 B’ R2 B U R2 U’ (F2L) B’ R’ B R’ B’ R2 B (OLL) D’ F D’ B2 D F’ D’ B2 D2 R (PLL)

And additionally, if there are twisted corners or multi-cycles, how could I use multiple commutators?

This was a pretty lucky solve, I didn't use any pseudo-blocks or NISS, just tried both f2l orientations after 2x2x3. I'd just like to understand these insertions and then try to add all of these methods to solutions to get lower movecounts.
OK -- so if you do R after your OLL, you just have three wrong corners. You're right that the best case is when you have solved everything about three corners, but it doesn't need to be an A-Perm -- it can be any 3 corners in a cycle, anywhere on the cube. Once you've got to that point, you need some little removable stickers. Put one sticker on each side of the three wrong corners. Pick one, at random, and label it 1A. Now work around that corner clockwise, and label the next one 1B and the third one 1C.

Now, take a look at what piece 1 is, and see where it needs to go. So let's say it's the yellow-red-blue corner, and you labeled red as 1A. Now you want to find where 1A goes, which is where the red sticker of red-yellow-green. So you mark red as 2A, and work around again, 2B, 2C. Now you look again and see where 2A wants to go, which is where red-green-white now is. So you put 3A on the red sticker of that corner, and so on.

Now, invert the solution on your cube, to get back to the scrambled state.

Now, work through the solution, one move at a time, and look for places where you can do a commutator. To recognise it, you want two As, two Bs or two Cs both on the same face, and then you want the third one with the same letter to be on a corner on the opposite side, but not on the opposite face (e.g., if 1A and 2A are both on D, then you want 3A to be on F, R, L or B, on a corner on the top of the cube, but not on the U face). To get a cancellation, ideally, the next move in the solution after your insertion would be D, D' or D2, as your commutator will probably end with D, if 1A and 2A are on D.

For multiple cycles, there are two approaches. The way I recommend starting with, is to do your insertion and then you write up your new skeleton -- your original skeleton plus your commutator. Now remove the stickers that are now solved, and start again, working just with whatever is left. E.g., if you have 5 corners, you might do a commutator that permutes 2, 3 and 4. That will mean 2 is now solved and 3 is now solved, and you'll be left with 1, 4, 5 which you solve in that order -- 1-->4-->5-->1.

That's it. Hope that all makes sense -- let me know if any of it is not clear. Again, I really recommend porkynator's tutorial which covers all this very clearly.
 
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Now, work through the solution, one move at a time, and look for places where you can do a commutator. To recognise it, you want two As, two Bs or two Cs both on the same face, and then you want the third one with the same letter to be on a corner on the opposite side, but not on the opposite face (e.g., if 1A and 2A are both on D, then you want 3A to be on F, R, L or B, on a corner on the top of the cube, but not on the U face). To get a cancellation, ideally, the next move in the solution after your insertion would be D, D' or D2, as your commutator will probably end with D, if 1A and 2A are on D.
I found only two possible cases and not near cancellations. Is it normal or is there usually more of them?

F D F’ D2 R’ L2 D’ L2 R’ U2 B R B’ U R U R’ U’ R U R U2 R2 B’ R2 B * U R2 U’ B’ R’ B R’ ** B’ R2 B R

* D' F D B2 D' F' B2
** L' D L U2 L' D' L U2
 
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I found only two possible cases and not near cancellations. Is it normal or is there usually more of them?

F D F’ D2 R’ L2 D’ L2 R’ U2 B R B’ U R U R’ U’ R U R U2 R2 B’ R2 B * U R2 U’ B’ R’ B R’ ** B’ R2 B R

* [D' F D, B2]
** [L' D L, U2]
It's possible to get unlucky, but with a skeleton that long, it seems unlikely. Can you post the scramble...?
 
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Doesn't seem to work...
Should work. Anyway, thanks for helping me out, I'm starting to get the idea. I learned also edges insertion and got a 43 moves solve. Too bad I didn't know how to do the edges today when doing the weekly comp. Instantly after that I watched how to do them and cancelled my weeklies solve (46) to 43. I could start practising NISS too.
 
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I found only two possible cases and not near cancellations. Is it normal or is there usually more of them?

F D F’ D2 R’ L2 D’ L2 R’ U2 B R B’ U R U R’ U’ R U R U2 R2 B’ R2 B * U R2 U’ B’ R’ B R’ ** B’ R2 B R

* D' F D B2 D' F' B2
** L' D L U2 L' D' L U2
Insertion finder (http://fewestmov.es) is an awesome tool for checking how you did on the insertions. Often after I've finished a solve I'll use insertion finder to see if there were any obvious insertions that I missed.

In your case there is only one better insertion that can cancel a move.

F D F' D2 R' L2 D' L2 R' U2 B R B' U R U R' U' R U R U2 R2 B' R2 [@1] B U R2 U' B' R' B R' B' R2 B R
Insert at @1: L' F2 L B2 L' F2 L B2
Fewest moves: 44. 1 moves cancelled
The final solution: F D F' D2 R' L2 D' L2 R' U2 B R B' U R U R' U' R U R U2 R2 B' R2 L' F2 L B2 L' F2 L B' U R2 U' B' R' B R' B' R2 B R
http://fewestmov.es/cube/if/6554f47212252f4dd124d9c4bfa147a0.cube
 
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Block commutators may be a step to far for me at the moment, as this was only my second solve where I actually used insertions, but I'll defo start playing around with it a bit to get a feel for it, and who knows I'll come across a solve to actually use it :)
In my experience, block comms aren't that useful. If you have a 3c3e skeleton and want to use a block comm insertion, you need to set up the corners to line up with the edges, which uses so many moves that it's better to do separate edge and corner insertions.

On the other hand, there are a few algs you might use that are actually block commutators. For instance, R U R' U' R' F R F' can be written as Lw F R' F' Lw' F R F'. Some Winter Variation algs are also block comms, like L' Dw2 L U' L' Dw2 L.

Essentially, if you know how to construct corner 3-cycles on the fly, then when you encounter a pair 3-cycle case, you should be able to adapt your knowledge to work with pairs instead of single corners. The tricky part is to correctly convert the wide moves into single-layer moves.

Scramble: U B2 L2 D U2 R2 U R2 L2 F2 U’ R B2 F’ L’ D U2 F’ D F’ L U2
Solution: F D F’ D2 R’ L2 D’ L2 (2x2x2) R’ U2 B R B’ (2x2x3) U R U R’ U’ R U R U2 (F2L-1) R2 B’ R2 B U R2 U’ (F2L) B’ R’ B R’ B’ R2 B (OLL) D’ F D’ B2 D F’ D’ B2 D2 R (PLL)
If you insert your final F2L pair differently, you can skip OLL:

F D F' D2 R' L2 D' L2 // (2x2x2)
R' U2 B R B' // (2x2x3)
U R U R' U' R U R U2 // (F2L-1)
R2 B' R2 B R' U R' U' // (F2L)
U' R D' R2 U R' U' R2 U D R // J perm

40 moves if you use the optimal J perm alg (from an angle that cancels one move), 41 moves if you use an optimal alg that doesn't cancel moves, 44 moves if you use the alg that starts with L' U' L F.
 
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so here are my solves from my 29,67 mean of 3 from Hamburg open 2017

solve 1: 34

scramble R' U' F D2 B2 F2 L B2 D2 F2 R' F2 D' B R' B' F U' B' R2 U L' B2 R' U' F

2x2x2 R2 F2 B' D F
Switch
pair R2
Switch
P2x2x3 R D2 R' B2 pre B'
F2L-1 B R2 B' pre R2 B'
Switch
F2l B R' B' R D B R' B' R
ZBLL L D2 L' B' D L D L' D' B D'

final solution
R2 F2 B' D F R D2 R' B' R2 B' D B' D L D' L' D' B L D2 L' R' B R B' D' R' B R B' R2 B' R2

i found a way to get it to 29 later.
2x2x2 R2 F2 B' D F
Switch
pair R2
Switch
P2x2x3 R D2 R' B2 pre B'
F2L-1 B R2 B' pre R2 B'
Switch
to3c B R' B' R2 D2 R' D' R2 F' R' F R'

R2 F2 B' D F R D2 R' B' R2 @ B' R F' R F R2 D R D2 R2 B R B' R2 B' R2
Insert at @: R2 B L2 B' R2 B L2 B'
8-5/29


solve 2: 25

scramble R' U' F R2 B2 R2 F D2 F D2 F' D2 F D' U2 F D' L R' D2 F' D2 L D' R' U' F

Switch
2x2x2 R D' L B' D'
2x2x3 U2 R2 U2 B U2 B'
F2L-1 F2 U' L' U2 L
to3c U' R U' R' U2

U2 R U R' U L' U2 L U F2 B U2 B' U2 R2 U2 D B L' @ D R'
insert at @ R' U R D R' U' R D'

final soultion U2 R U R' U L' U2 L U F2 B U2 B' U2 R2 U2 D B L' R' U R D R' U'

solve 3: 30

scramble R' U' F L2 F2 D' L2 D R2 D F2 U' L' B R' D' F D' F L' D2 B2 F2 R' U' F

2x2x2 L' R2 U' R B'
2x2x3 R F U2 R2 U' @
block R' U F2 U'
to3c D' F2 D F D' F' D R F
insert at @ U' F D' F' U F D F'

final solution L' R2 U' R B' R F U2 R2 U2 F D' F' U F D F' R' U F2 U' D' F2 D F D' F' D R F

34, 25, 30=29.67 mean of 3
 
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cool solution for weekly German FMC
9 move skeleton :p
R' U' F L2 D2 B2 D F2 U F2 L D F' D2 L' D B F R' B' U F' R' U' F

L B' (D' B') // EO
(D2 B2 R U' D') // 3c6e

L B' D U R' B2 ^ @ D2 B D
^ = B2 U' B2 L2 F2 D' F2 L2
@ = L' D' U B2 % U' D L'
% = [B', D F' D']

Solution: L B' D U R' U' B2 L2 F2 D' F2 L D' U B D F' D' B D F U' L' D2 B D (26)
 
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