bubbagrub
Member
Also, I may have missed it, but I don't think anyone has yet recommended the best tutorial:Nice, I'll try these!
http://fmcsolves.cubing.net/fmc_tutorial_ENG.pdf
Also, I may have missed it, but I don't think anyone has yet recommended the best tutorial:Nice, I'll try these!
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]
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. That gives 54 moves.
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)
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.
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]
** [L' D L, U2]
In the previous postIt's possible to get unlucky, but with a skeleton that long, it seems unlikely. Can you post the scramble...?
Doesn't seem to work...In the previous post
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.Doesn't seem to work...
OK -- I figured it out. Looks to me like you just got unlucky. I couldn't find any other commutators.Doesn't seem to work...
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.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
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
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)
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