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[Help Thread] FMC Discussion

Why is this not working? Is there some special rule I can't find? Could someone please tell me what I am doing wrong so I can stop bothering this thread.

I can't figure out what you're trying to accomplish with those insertions. It looks like you're moving the corners around, but not to places that actually would make any sense. How are you working out your insertions?

You need to start with the finished state, after you've done your whole solve (minus the insertions, with the corners still bad), and then pick three corner stickers. Then you need to find a place in the solve where those three stickers set up well for an insertion, and add the algorithm there.

For the insertions you have above, it looks almost like you're picking random corners to move around. I'm sure that's not what you're actually doing, but I can't figure out what you're trying to accomplish. Perhaps you could do something like, at each insertion, say "this solves the corners LUF-FDL-RUB for the finished cube", so we'll know what you think you're trying to accomplish. (I picked that example because, to me, that looks like what you solved with your first insertion. Which is strange, because it's a slow, 9 move insertion and it only solves one corner - if you're going to be only solving one corner because you have 6 to solve, I'd think you'd want a really good insertion with both 8 moves and a bunch of cancellations - the single-piece-fix is always going to be your best insertion.
 
Why is this not working? Is there some special rule I can't find? Could someone please tell me what I am doing wrong so I can stop bothering this thread.

You can't just write down algs, which solve your corners. And then insert them somewhere in your solution. Your algs only work at the end of your skeleton.

An insertion works this way. If you want to cycle 3 corners, you have to cycle the actual pieces and not only the positions.

An example:
Scramble: R' F R' B2 R F' R' B2 R2 L2 U R2 (WG-orientation)
Solution:
All but 3 corners: R2 U' L2 (3 moves)
You have to cycle UBL->UFR->URB. (=White-Green-Red -> White-Red-Blue -> White-Blue-Orange)
You can just do the cycle at the end with an A-perm: R2 B2 R F R' B2 R F' R, which would give you a 3+9=12 move solution (R2 U' L2 R2 B2 R F R' B2 R F' R)
Your idea of insertion (which ist wrong) would be to insert the alg right after R2, so your solution would be (R2 R2 B2 R F R' B2 R F' R U' L2 = B2 R F R' B2 R F' R U' L2 = 10 moves.
But this doesn't work. Why? Just scramble the cube with R' F R' B2 R F' R' B2 R2 L2 U R2 and apply R2. If you would insert the A-Perm right here, you would cycle UBL->UFR->URB which are now the Yellow-Orange-Blue, White-Blue-Orange, Yellow-Green-Orange. But you don't want to cycle these. You want to cycle the White-Green-Red, White-Red-Blue and White-Blue-Orange. These corners are now on the positions DFL, ULF and UFR. So now you have to cycle DFL, ULF and UFR. Which isn't an 8-mover. Normally you only have to look for 8-mover. There just (much) shorter.
Next try to insert them after U':
Your 3 corners (White-Green-Red, White-Red-Blue and White-Blue-Orange) are now on the positions DFL, UFR and URB. These 3 corners you can cycle with F2 R B R' F2 R B' R' (8 moves). So your solution would be: R2 U' F2 R B R' F2 R B' R' L2 (11 moves), which is the optimum for this scramble.
With a bit of searching you can find in your skeleton more 8-move insertion which can cancel more moves. This 3-move skeleton is a bad example for cancellation.

The important thing is: Trace the corners you want to cycle.
Note: Most people use stickers for tracing. This is also allowed on official attempts.

By the way, the optimum insertion for your skeleton would be:
L2 B F2 [@1] L' U' L D' L' R2 B R' B' D R D R
Insert at @1: D2 L' U' L D2 L' U L
After the 1st insertion: L2 B F2 [@2] D2 L' U' L D L' R2 B R' B' D R D R
Insert at @2: B D F' D' B' D F D'
After the 2nd insertion: L2 B2 F2 D [@3] F' D' B' D F D L' U' L D L' R2 B R' B' D R D R
Insert at @3: D2 F' U' F D2 F' U F
Fewest moves: 28. 12 moves cancelled
The final solution: L2 B2 F2 D' F' U' F D2 F' U D' B' D F D L' U' L D L' R2 B R' B' D R D R
 
my official 27 moves FMC at Bandung Open 2012 :D really happy with the result, coz usually I averages around 37 moves

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

Solution: U' F U F R' U' B2 D z U L F R' F' L' R U R U' R' z' F' D F2 D' L2 U' F' U'
//explanation:
//2x2x3 block: U' F U F R' U' B2 D
//(switch to inverse scramble)
//Finish F2L: U F U L2 D F2 D' F
//OLL: z R U R' U' R' L R' F R F' L'
//AUF: U'
//inverse all to get solution
 
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Ugg, I am just not getting it.
Is there some special rule I can't find? Could someone please tell me what I am doing wrong so I can stop bothering this thread.

Here's a small exercise for you :)

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

B' U' B' U2 B2 - cross
D2 F' D2 F - 1st pair
B D B' F D' F' - 2nd pair
D2 R D' R' - 3nd pair
L' D L D' - 4th pair leaves 3C at 23 HTM

Makes skeleton B' U' B' U2 B2 D2 F' D2 F B D B' F D' F' D2 R D' R' L' D L D'

Now physically put stickers labelled 1 - 2 - 3 or A - B - C onto the three remaining corners.
Starting with an arbitrary face of a corner, you are now going to sticker the cycle.

For ease of discussion/help lateron please sticker according the example below:

sticker 1 - place on the orange face of the of orange-green-yellow corner
Where should this orange sticker be? What sticker is occupying its spot?
This is where sticker 2 goes...

sticker 2 - place on the orange face of the of orange-yellow-blue corner
repeat
Where should this sticker be? What sticker is occupying its spot?
This is where sticker 3 goes...

sticker 3 - place on the red face of the red-yellow-green corner

check your cycle: where does this sticker belong? sticker 1 should be there: cycle is closed

Now for insertion-hunting you can either work backwards through the solution or re-scramble and work forwards. For beginners I recommend to go forward.

Now go hunting!!!

Don't click the spoiler below until you have found at leas 4 possible insertions!

. B' U' B' U2 B2 : D2 F' D2 + F B D B' # F @ D' $ F' D2 R D' R' L' D L D'

. R B2 R' F' R B2 R' F - 8 mover cancel 0
: R U2 R' D2 R U2 R' D2 - 8 mover cancel 2
+ R' B2 R F R' B2 R F' - 8 mover cancel 2
# L' F' R' F L F' R F - 8 mover cancel 1
@ F U F' D F U' F' D' - 8 mover cancel 2
$ D F U F' D F U' F' D2 - 9 mover cancel 3

$ is just a longer version of @ giving the exact same end result.
It is very unlikely* that there are no 8 move commutators in a skeleton

*but not impossible, I think i have only seen 2 occasions
 
Got a decent start where about half of the cube is solved in 13 moves

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

U' L D' R U2 L2 U B U' B'
F' U F2

Having a hard time in finding finishes.

How possibly could I proceed from here?
 
Got a decent start where about half of the cube is solved in 13 moves

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

U' L D' R U2 L2 U B U' B'
F' U F2

Having a hard time in finding finishes.

How possibly could I proceed from here?
After U' L D' R U2 L2 U B U' B', U2 makes the last F2L pair, but if you first separate the pair & corner with L' U L you get:
L' U L U2 F' U2 F2
last F2L pair: U R' [F R F'
COLL: F R' F'] r U R U' r' = 6 moves canceled

24 moves total
 
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D' L2 R' F D' L2 B U' L' B2 L' B2 D' R U B' L2 B2 U B' R L' F2 B2 U'
(stupid qqtimer scramble)
Double x-cross: x2 y D R U R' F2 U2 R2
f2l1: U’ R U’ R’ U R
f2l2: y’ R U’ R’ U R U R’
COLL → EPLL skip: y2 R’ U R2 D r’ U2 r D’ R2’ U’ R

can someone find a better solution?
 
found a solution with a LL-Skip. But you can't do this in a speedsolve:

x2 y D R U R' L F2 L' - pseudocross
F R' F R' F' U' F' D2 B U' B' D2 - TXCross
U L' U R' U L U' R U2 L' U2 L U - F2L-Pair4 + forced LL-Skip
=41 Moves :D
 
Crazy Scramble...

Crazy start: D R F' R' F U D R2

5 corners: F D F D' F2

optimal insertion bring 7 move cancelation --> 22

Or crazy finish: D F2 D' U R F2 R' F2 U' F R F' R'

--> 21 moves
 
Here're the scrambles of PKU FM 2012.

1. R2 B D2 B' D2 F2 L' U R2 F' D' R' D' R2 B' R2 F2 U B2
2. L B2 R2 U2 F2 R2 D' F' R U' B2 U2 L' B' U L D2 F2
3. B2 F' D2 L2 R2 F2 D2 L2 B D2 F' D' B' U2 R' B2 D2 B D'

I luckily got a solution of 25 moves and broke the AsR in the first try. The following is my solution:

222: F2 L B' R' F
Inverse it and go on building block:
B L' D' L2 B2 D' B' D' L D L D' B D
Remained 3 corners:
D' L U' L ' D L U L'
Final solution:
F2 L B' R' F L U' L' D' L U L' B' D L' D' L' D B D B2 L2 D L B'
 
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doesnt work for me with the scramblie you provided.

Edit: Taking the inverse of your solution as scramble your insertion is optimal. Did you search within the whole skeleton?
 
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doesnt work for me with the scramblie you provided.

Edit: Taking the inverse of your solution as scramble your insertion is optimal. Did you search within the whole skeleton?

Sorry I mixed the scrambles. The first two should be exchange. I've corrected them now.

When I found the skeleton, there's only about 10mins to find the insertions and obvious the insertion canceled 2 moves so I wrote the solution down immediately. After that I tried to find a better insertion but failed. In fact, I used the Insertion Finder to search the optimal insertion and found out it's the same as mine.
 
B' U R U B2 R B U' L' D L2 D R2 B2 D R2 L2 B2 U'

. F L , U' B U2 B' R' U2 F2 U' // 2x2x3 (10/10)
R D2 B' D R' D R B D' // F2L (19/19)
F' U F D' F' U' F D R // L3E (28/28)

Insert B' R2 B F' U2 F @ "." to cancel 1 move.

At this point I ran out of time, but soon after I found a 3-corner insertion to replace the underlined moves: insert L' D R D' L D R' D' @ "," to cancel 2 moves.


Final solution: B' R2 B F' U2 F2 L U' B U2 B' R' U2 F2 U' R D2 B' D R' D R B D' F' U F D' F' U' F D R (33 HTM)

With extra insertion: B' R2 B F' U2 F2 D R D' L D R' D' U' B U2 B' R' U2 F2 U' R D2 B' D R' D R B D' R (31 HTM)
 
First attempt on THIS scramble got me a 30 FMC.
D' L2 B2 D' L2 U R2 D2 L2 F2 R' B U F' L2 U' B2 U R' U2

y F2 R' F' R2 B L' // Rouxblock [6]
U D F U2 x2 // 2x2x3 [4/10]
y' R' F R F' R U2 R' U' R // EOF2L [9/19]
y R U2 R' U2 L' U R U' R' L // COLL [10/29]
U // EPLL [1/30]

Took about five minutes. Almost linear.
 
Next: U B2 F' R2 D' B' R' D2 R2 B2 L' D' F R L2 F2 R B F2 D' F L2 F B D'

x2 // insp
F2 D' L R2 D2 // 2x2x2
y' R' U R U2 R // 2x2x3
y' U' R U R' U2 R2 U' // pairs
F R' F' R' U' // setup
y R U2 r' F M' U // fin
alg.garron
28 HTM

This was one of my solves off the example solve thread.
 
i didn't want to ask this in the q&a thread because i thought id get a better answer here:

what does "NISS" stand for?
 
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