• Welcome to SpeedSolving.com — the world’s largest puzzle community!

    You’re currently browsing as a guest, which means you have limited access to discussions, resources, member profiles, and community features.

    Join 50,000+ cubers and puzzle enthusiasts from around the world to ask questions, share solves, improve faster, and be part of the community.

    Registration is free, fast, and easy — create your account today.

    Already a member? Log in here to hide this message and start participating.

[Help Thread] FMC Discussion

Scramble: F' U B2 D L2 B2 D R F' D2 B U R2 F U' L2 B R U B

D2 L’ D2 R2 B L2 // 2x2x2
R F2 R D’ R2 F R2 F D2// AB3E2C

Any good way to finish this off? My best was 34, which felt kind of disappointing with that start.
 
Scramble: F' U B2 D L2 B2 D R F' D2 B U R2 F U' L2 B R U B

D2 L’ D2 R2 B L2 // 2x2x2
R F2 R D’ R2 F R2 F D2// AB3E2C

Any good way to finish this off? My best was 34, which felt kind of disappointing with that start.

D2 L’ D2 R2 B L2
R F2 R D’ R2 F R2 F * D2

* = F //Pseudo F2L
F L D L' D' F' //fruruf
F' D2 F D F' D F //antisune
D F' //Finish

28 HTM

or

D2 L’ * D2 R2 + B L2
R F2 R D’ R2 F R2 F D2
* = R B U B' U' R' U //fruruf
+ = R' U2 R D R' U2 R D'
 
D2 L’ D2 R2 B L2
R F2 R D’ R2 F R2 F * D2

* = F //Pseudo F2L
F L D L' D' F' //fruruf
F' D2 F D F' D F //antisune
D F' //Finish

28 HTM

or

D2 L’ * D2 R2 + B L2
R F2 R D’ R2 F R2 F D2
* = R B U B' U' R' U //fruruf
+ = R' U2 R D R' U2 R D'

Sorry to bother you again, but what are you doing there? Inserting sunes and frurufs? How does that work? :S
I feel like I don't know anything

not_kevin said:
I confess I still don't fully understand them, in the sense that I fully understand corner comms and the other common edge comms, but I was able to read through this description Ranzha linked me to on the Facebook group and make sense of how it roughly works:

https://www.speedsolving.com/forum/s...=1#post1054905

Thanks for the link, I think I've undertsood it
 
Last edited:
Sorry to bother you again, but what are you doing there? Inserting sunes and frurufs? How does that work? :S
I feel like I don't know anything

The first case is actually easy to see with premoves: finishing with ...R2 F2 instead of ...R2 F D2 gives you pseudo F2L, so you have a normal F2L by adding premoves F' D2. Not really an insertionn here.
The last one is a bit harder. After the first two moves (D2 L') if you've marked the 5 unsolved pieces, you can see that they all are on the same layer. You can "break" the solve at this point by adding as premoves all the rest (D2 R2 B L2 R F2 R D’ R2 F R2 F D2). I think I've called this "reverse NISS" in my guide.
 
I'm getting into FMC, and as expected, I'm a total nub. Tips?

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

switch to inverse:
2x2x2: U' R2 F' B D

switch to normal with premoves: D' B' F R2 U
2x2x3: F' U F2 R' F' R
F2L-1: D F' D' L' F' L2
Edges: L' F' L F D F D' F

Skeleton: * F' U F2 R' F' R D F' D' L' F' L2 L' F' L F D F D' F D' B' # F R2 U

Insert at #: R' D2 R U2 R' D2 R U2
Insert at *: R B' R' F' R B R' F

Final solution: R B' R' F' R B R' U F2 R' F' R D F' D' L' F' L F' L F D F D' F D' B' R' D2 R U2 R' D2 R U2 F R2 U (38 HTM)
 
How do you do L5C? What is the basis of block building to? F2L-1 and EO right?

technically, you can block build to anything you want and the last parts of the solve would use piece cycles to solve the rest of the cube.

Typically, however, an FMCer(?) will block build to a point where there are only 5 corners unsolved (hence L5C) and the work out the cycles to solve the pieces. The cuber will then attempt to find a point in which they could insert the alg they generated so that as many moves as possible are cancelled out by the skeleton (U by U', R by R' etc) alternatively, they will attempt to find a good point in which the corners could be cycled easily as well.

Thus the skeleton serves as the frame for which the final solution can be formed using.
 
The first case is actually easy to see with premoves: finishing with ...R2 F2 instead of ...R2 F D2 gives you pseudo F2L, so you have a normal F2L by adding premoves F' D2. Not really an insertionn here.
The last one is a bit harder. After the first two moves (D2 L') if you've marked the 5 unsolved pieces, you can see that they all are on the same layer. You can "break" the solve at this point by adding as premoves all the rest (D2 R2 B L2 R F2 R D’ R2 F R2 F D2). I think I've called this "reverse NISS" in my guide.

Grazie mille. Your help was very useful :)
 
Ok. Here goes.

Scramble (from qqtimer): U2 D' R' U L B U' F2 U2 L F2 B D2 F U2 L2 F2 B' R2 L2 D2

Solution:
D B U2 B2 R' B R U' R2 U R B' R2 L U' L' F' R F2

On inverse:
F2 R' F //pseudo 2x2x2
add premove D' to make it real 2x2x2
L U L' B // 2x2x3

Back to normal scramble with premoves B' L U' L' F' R F2
D // to correct the 2x2x3 block
B U2 B2 R' B R2 // AWSEOME MAGIC ENDING LEAVING ONLY A PAIR 3-CYCLE!!!
Lw' B' R2 B Lw B' R2 B // pair 3-cycle to solve the cube in 22 moves

BUT wait there is more:
If we write that 3-cycle without wide moves it becomes R' U' R2 U R B' R2 B
So now the first R' cancels with the R2
AND when we add our premoves to the end we also get B cancelling with B' resulting to a 19 move solution

D B U2 B2 R' B [R2 R'] U' R2 U R B' R2 [B B'] L U' L' F' R F2
 
Ok. Here goes.

Scramble (from qqtimer): U2 D' R' U L B U' F2 U2 L F2 B D2 F U2 L2 F2 B' R2 L2 D2

Solution:
D B U2 B2 R' B R U' R2 U R B' R2 L U' L' F' R F2

On inverse:
F2 R' F //pseudo 2x2x2
add premove D' to make it real 2x2x2
L U L' B // 2x2x3

Back to normal scramble with premoves B' L U' L' F' R F2
D // to correct the 2x2x3 block
B U2 B2 R' B R2 // AWSEOME MAGIC ENDING LEAVING ONLY A PAIR 3-CYCLE!!!
Lw' B' R2 B Lw B' R2 B // pair 3-cycle to solve the cube in 22 moves

BUT wait there is more:
If we write that 3-cycle without wide moves it becomes R' U' R2 U R B' R2 B
So now the first R' cancels with the R2
AND when we add our premoves to the end we also get B cancelling with B' resulting to a 19 move solution

D B U2 B2 R' B [R2 R'] U' R2 U R B' R2 [B B'] L U' L' F' R F2

Amazing!!

Know that the optimal solution to that was 18 moves xD (B' D2 L2 B R' F2 L B2 U' B2 R2 F2 B R2 F R2 U' L2 (18f*))

btw, post number 404, not found...
 
Hey, I did my first FMC attempt yesterday. 36 moves. Is this good for a beginner to FMC? What can I do to improve my skeleton building? Thanks :)

Scramble: B2 U' F2 D' R2 B2 F2 U' L2 R2 U B' U' F R F D2 L' U2 F2 U'

My solution: L R' U L' R' F D' B F R D' R' U' R D R' F R'F D B D B' D B D2 B2 D R' B R D B' D' B D2

L R' U L' R' F D' // 1x2x3 (7/7)
B F* [U'] F R' F // expand to 2x3x3 (6/13)
D B D B' D B D2 B2 D R' B R // F2L -1 (12/25)
D B' D' B D2 // Solved - 3c (5/30)


*Insertion: R D' R' U' R D R'
8 move commutator w/ 2 canceled moves.

L R' U L' R' F D' B F R D' R' U' R D R' (U U') F R'F D B D B' D B D2 B2 D R' B R D B' D' B D2
 
I want to share with everybody a 27 HTM corner-first FMC solve I got a couple of days ago. I've found it pretty quickly during the 1 hour time limit and than I've "submitted" a 25 HTM "standard" solution. Anyway, it's nice and quite simple to understand even if you are not familiar with corner-first solves.
The scramble is taken from the German forum weekly competition, even though from which week since I haven't fetched it myself.

Scramble: U' F2 U F2 D R2 B2 D2 F2 D2 U' B' L U' B2 F2 R2 D L' R
Solution: L' F' D R D2 B D' U R' D2 F' D' R2 D R' D' R D U' B' D B' U' L2 D2 B2 U2

L' F' on normal scramble orient corners. Solving corners by orienting first and then permuting not only gets you (usually) short solutions, but makes it somehow easy to solve edges in-between the two steps. I haven't exploited this advantage now, however. Note how with these 2 moves I also make one 2x2x1 square and 3 CE pairs (we can say that there are 5 edges solved in this case).
Then I switch to inverse scramble with premoves F L. Then U2 B2 D2 L2 U B2 D' B2 solves the permutation of corners and one more edge.
There are now 6 edges that need to be cycled. The permutation can be broken into 2 disjoint cycles:

UB->BR(->BU)
LF->FR->LD->FD(->FL)

The two cycles are both odd (made up by an even number of pieces; yeah I know, the terminology is confusing) and "flipped", which means I can't solve them separately; actually, one of the two conditions (either odd or flipped) is enough. However, as you will see, if we allow centres to be unsolved, we can solve an odd permutation (especially 4-cycles) on its own.
Disregarding centres (if at the end they are not solved, you can insert M E M' E' or similar moves somewhere and cancel lots of them, usually), we can cycle 4 edges by setup + single slice move + anti-setup. Before the last 3 moves (B2 D' B2) it's easy enough to solve the cycle UB->BR->LF->FR. This solves two more edges (FR and BR) and leaves the other 4 in a 4-cycle, which we can then solve with a similar technique.
How to solve them?
Setup: B D' B
Permute: E (or U D' y' if you don't know which way is E and which E')
Anti-setup: B' D B'
And one move even cancel with the B2 D' B2. Pay attention when re-writing the solution without rotations or slice moves (you can use something like Uw' U if you prefer). The solution up to now looks like:

On inverse scramble with premoves F L
U2 B2 D2 L2 U //Partial CP (3 moves left undone)
B D' B U D' R' D R D' R2 //Leaves 4 corners and 4 centres

Then we can solve the last 8 pieces:

Setup: D F D2 R
Solve: E'
Anti-setup: R' D2 F' D'

Which turns out to look like:

D F D2 R U' D B' D2 R' D'

And the cube is solved. Thanks to Attila for sharing many of his corner-first solution :)
 
My winning solutions from World Championship!

1. Scramble: L2 U' B2 L2 D' F2 L2 D U' L2 U2 B' R2 B' R B R' D' B' F2

U2 R' U2 R' // EO
B F2 U2 F // pseudo 2x2x3
(switch)
B L2 // 2x2x3
(switch)
B2 U' B2 U' B2 U' B2 U' B' U // L3C

Skeleton: U2 R' U2 * R' B F2 U2 F B2 U' B2 U' B2 U' B2 U' B' U L2 B'

*= U R' D2 R U' R' D2 R

Solution: U2 R' U' R' D2 R U' R' D2 B F2 U2 F B2 U' B2 U' B2 U' B2 U' B' U L2 B' (25 Moves)

2. Scramble: L2 U2 F' D2 B2 F' R2 B' U2 F' R D2 L D F' L U' B2 L' F2

(Inverse)
D' // Orient some edges
(switch)
L R D // EO
B2 U2 R' F // 2x2x2
(switch)
B2 F L' F' L' // F2L-1
R' B2 L B L' B2 R // L3C

Skeleton: L * R D B2 U2 R' F R' B2 L B' L' B2 R L F L F' B2 D

*= L' D2 L U2 L' D2 L U2

Solution: D2 L U2 L' D2 L U2 R D B2 U2 R' F R' B2 L B' L' B2 R L F L F' B2 D (26 moves)

3. Scramble: U2 B2 D2 L2 B2 U L2 D' L' F D' L F' D B' D2 F' R' U

B2 D2 B F R' L2 // Pseudo 2x2x3
(switch)
U' L' // 2x2x3
D2 L' D L // Pseudo F2L + EO
(switch)
B2 D2 // F2L
D B2 D' B' D B' D' // AB3E

B2 D2 B F R' L2 B2 D' B2 D' B' D * B' D' L' D' L D2 L U

*= D' B D' B' D' B' D' B D B D2

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

This solution could be 24 with optimal insertion, but i am terrible with edges insertion :p So i am ok with the 26.

Mean: 25 26 26 = 25.67 Sar! :D

I also did my first "official" avg12 = 27.10 :D
 
Thank you for sharing! Congratulations!! I hope I can learn a lot from these :)

Also, here's my 25 Oficial Solution from 3 weeks ago:

Scramble: D2 U' F2 L2 U2 F2 R2 D' R2 B' R2 D R B2 F L2 B' L' R B'

D F' D' R F' L2 - 2x2x2 (6/6)

B U F B' U2 - 3x2x2 (5/11)

F2 U F2 U' - F2L-1 (4/15)

Trying how to finish, I found this lucky ending:

R2 F R' F' R2 U' R' U - F2L (8/23)

U' R U R2 D' U' R U R' D - COLL with EPLL skip (10/33)

The funny thing is that the last 4 moves of the F2L pair cancel with the first 4 of the COLL, giving me a 25 move solution! (1 move away from NR):

D F' D' R F' L2 B U F B' U2 F2 U F2 U' R2 F R' F' D' U' R U R' D (25)

I posted an stopmotion video with a little surprise here: https://www.speedsolving.com/forum/showthread.php?54367-25-FMC-Solution-(Gernika-Open-2015)-surprise

:)
 
The permutation can be broken into 2 disjoint cycles:

UB->BR(->BU)
LF->FR->LD->FD(->FL)

The two cycles are both odd (made up by an even number of pieces; yeah I know, the terminology is confusing) ...

Perhaps I'm being a bit nitpicky, but when you talk about a cycle being even or odd, you should always be clear whether you're talking about the permutation effect of the cycle (which is what you intended) or the length of the cycle. What you are saying is that these two cycles, if considered as separate/individual permutations, are odd permutations.

In general, we talk about a cycle as being a component of a permutation. (And, of course, a permutation is called an odd permutation if, after being decomposed into cycles, the number of even-length cycles is an odd number.) When talking about a cycle by itself, it would be more proper to say the cycle represents an odd (or even) permutation, or that the permuation effect of the cycle is odd (or even).
 
Back
Top