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[Help Thread] FMC Discussion

So, how does this inverse scramble stuff work?
Like, do you invert the scramble, apply moves, and then apply the inverse of those moves to the original scramble?

take a solved cube.

scramble it.

solve it.

it is now solved.

the sequence "scramble" + "solution" moved the cube from SOLVED to SOLVED.

that sequence is essentially a really long alg that does nothing to the cube.

just like the Reverse of the T-perm is also a T-perm, the reverse of the sequence above will also bring the cube from SOLVED to SOLVED.

you can also keep switching back and forth to the inverse and normal (NISS)


take a look at my solution on the weekly fmc:

scramble: B R' L' D2 F B2 U2 D2 B' R2 F B2 R2 U L' D2 F2 L U' R2
F2L-1: B’ D L’ U2 F R’ B R2 B’ R’ B

so the move sequence "scramble"+"F2L-1" leaves F2L-1 intact and just screws up the rest of the pieces. basically, it's a really long alg that does random CRAP to the pieces that are not in the F2L-1.

so if I do that WHOLE SEQUENCE backwards, i'll still have F2L-1. but i'll have DIFFERENT random crap everywhere else. that gives me 2x the chance to get lucky (that's what im up all night for anyway).

so then, imagine the scramble is (backwards of F2L-1) + (backwards of scramble) and just solve the cube from there. in the old weekly FMC, I do

(F' D' B L' B' D F2 U F' U') on the inverse to solve all but 3 corners.

so basically, that's just
scramble: (B' R B R2' B' R F' U2' L D' B) (R2' U L' F2' D2' L U' R2' B2' F' R2' B D2' U2' B2' F' D2' L R B')
ab3c: F' D' B L' B' D F2 U F' U'


so the total sequence is actually:

(B' R B R2' B' R F' U2' L D' B) (R2' U L' F2' D2' L U' R2' B2' F' R2' B D2' U2' B2' F' D2' L R B') (F' D' B L' B' D F2 U F' U')

so that whole long thing...is actually just the SAME thing as a 3 corner cycle.

if we reverse the whole thing, and shift the scramble to the left, we have:

B R' L' D2 F B2 U2 D2 B' R2 F B2 R2 U L' D2 F2 L U' R2 B’ D L’ U2 F R’ B R2 B’ R’ B U F U' F2 D' B L B' D F

that whole thing is still basically just a 3-corner cycle...just different corners.

so now insert an insertion and solve. i got lucky and found @ = [B L’ B’, R2] // insertion (4)

and put it right between the underlined B and R2 in the giant thing above.


and thats how to do it
 
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take a solved cube.

scramble it.

solve it.

it is now solved.

the sequence "scramble" + "solution" moved the cube from SOLVED to SOLVED.

that sequence is essentially a really long alg that does nothing to the cube.

just like the Reverse of the T-perm is also a T-perm, the reverse of the sequence above will also bring the cube from SOLVED to SOLVED.

you can also keep switching back and forth to the inverse and normal (NISS)


take a look at my solution on the weekly fmc:

scramble: B R' L' D2 F B2 U2 D2 B' R2 F B2 R2 U L' D2 F2 L U' R2
F2L-1: B’ D L’ U2 F R’ B R2 B’ R’ B

so the move sequence "scramble"+"F2L-1" leaves F2L-1 intact and just screws up the rest of the pieces. basically, it's a really long alg that does random CRAP to the pieces that are not in the F2L-1.

so if I do that WHOLE SEQUENCE backwards, i'll still have F2L-1. but i'll have DIFFERENT random crap everywhere else. that gives me 2x the chance to get lucky (that's what im up all night for anyway).

so then, imagine the scramble is (backwards of F2L-1) + (backwards of scramble) and just solve the cube from there. in the old weekly FMC, I do

(F' D' B L' B' D F2 U F' U') on the inverse to solve all but 3 corners.

so basically, that's just
scramble: (B' R B R2' B' R F' U2' L D' B) (R2' U L' F2' D2' L U' R2' B2' F' R2' B D2' U2' B2' F' D2' L R B')
ab3c: F' D' B L' B' D F2 U F' U'


so the total sequence is actually:

(B' R B R2' B' R F' U2' L D' B) (R2' U L' F2' D2' L U' R2' B2' F' R2' B D2' U2' B2' F' D2' L R B') (F' D' B L' B' D F2 U F' U')

so that whole long thing...is actually just the SAME thing as a 3 corner cycle.

if we reverse the whole thing, and shift the scramble to the left, we have:

B R' L' D2 F B2 U2 D2 B' R2 F B2 R2 U L' D2 F2 L U' R2 B’ D L’ U2 F R’ B R2 B’ R’ B U F U' F2 D' B L B' D F

that whole thing is still basically just a 3-corner cycle...just different corners.

so now insert an insertion and solve. i got lucky and found @ = [B L’ B’, R2] // insertion (4)

and put it right between the underlined B and R2 in the giant thing above.


and thats how to do it

This is the most concise and useful explanation of NISS I've seen! Thank you for taking the time to write it out :)
 
Thanks for the explanation. However, when I apply this knowledge to a solve, it doesn't accomplish the same thing. Say I have a scramble:R U' R2 F2 R2 F2 D2 L2 D' R2 B2 U2 L' U' B U2 L' B2 R U2 F'
and I apply the moves :U' R' L' D' To solve a 2x2x1 block.
However, on the inverse scramble, if I apply the inverse of that, or D L R U, It doesnt accomplish the same thing. Now, the same thing happens on the inverse scramble. If on the inverse scramble I apply the moves: L' D' F' U L' in order to solve a 2x2x1, and apply the inverse of that, or L D F U' L, to the normal scramble, it doesn't accomplish the same thing. Why is this?
Perhaps I'm misunderstanding?
 
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Thanks for the explanation. However, when I apply this knowledge to a solve, it doesn't accomplish the same thing. Say I have a scramble:R U' R2 F2 R2 F2 D2 L2 D' R2 B2 U2 L' U' B U2 L' B2 R U2 F'
and I apply the moves :U' R' L' D' To solve a 2x2x1 block.
However, on the inverse scramble, if I apply the inverse of that, or D L R U, It doesnt accomplish the same thing. Now, the same thing happens on the inverse scramble. If on the inverse scramble I apply the moves: L' D' F' U L' in order to solve a 2x2x1, and apply the inverse of that, or L D F U' L, to the normal scramble, it doesn't accomplish the same thing. Why is this?
Perhaps I'm misunderstanding?

The moves DLRU (the inverse of the block) you should do as premoves of the inverse scramble. This means: DLRU + inverse of the scramble. After scrambling you should get your first block. And if you create a block on the inverse of the scramble, you inverse these moves (LDFU'L) and do them before making original scramble. After finding the solution you add these moves to the solution in the end of the sequences. Correct me if I am wrong. I am still learing NISS.
 
I noticed that about 30 minutes after his last post, MiDipalma was banned...

Anyway, an attempt to do an FMC solve

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

F' U D2 F//2x2 (4/4)

L' B' D L' D' U2 R2 U' L2 R' U2 R L' U B' U2 B L U' R B' R' B U'// F2L-1 (24/28)

B L B' U F2 D U2 L F' L' D' U2 F2//THE CUBE (12/40)

Final Solution:
F' U D2 F L' B' D L' D' U2 R2 U' L2 R' U2 R L' U B' U2 B L U' R B' R' B U' B L B' U F2 D U2 L F' L' D' U2 F2
 
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Learning FMC, it's awesome.
Got my first sub-30 trying last week's weekly comp scramble:

Scramble: U' D2 L2 B2 R' F2 U2 D2 R' L2 D B L' F R' U' L' D' B' F
F B @ D R2 B' R // 2x2x2
L' R F' R' * // 2x2x3
{F L' U L U' L2} // F2L-1 on inverse
{D F D'} // All but 5 corners on inverse
@: B' R F' R' B R F R'
*: R U' R' D R U R' D'

// Final solution
F R F' R' B R F R' D R2 B' R2 L' F' U' R' D R U R' F' D' L2 U L' U' L F' (28)

It sort of doesn't count because I used Insertion Finder for the corner 5-cycle (hadn't learned 5-cycles properly yet so I used it to figure out what was going on, that thing finds crazy good cancellations). Participated in this week's FMC competition for the first time (found a 6 move 3-cycle insertion on my own for that one, no cheating :p)
 
i finished my first FMC solve using NISS today!

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

D R' L' U2 L' // 2x2x2

switch to inverse scramble (with premoves L U2 L R D')
F' R2 F2 R' // pseudo 2x2x3

switch back to regular scramble (with premoves R F2 R2 F)
D R' L' U2 L' R // 2x2x3

realised i could edit this (added F' for EO, R and L' are interchangeable) to get
D R' L' U2 R F' L' // 2x2x3 + some EO

U2 L' U L F' U' F leaves 2+2 edges and three corners

ended up just doing U' F2 U2 F2 U2 F2 U' to solve the edges because i couldn't find an obvious place to insert the 4-cycle

leaves me with the skeleton (for scramble with premoves R F2 R2 F)
D R' L' U2 R F' L' U2 L' U L F' U' F U' F2 U2 F2 U2 F2 U'

inserted [F R2 F', L'] at the beginning (because i'm bad at finding insertions) to get 29 moves

add premoves R F2 R2 F at the end to get the final solution

F R2 F' L' F R2 F' L D R' L' U2 R F' L' U2 L' U L F' U' F U' F2 U2 F2 U2 F2 U' R F2 R2 F
33 moves.

Looked for corner insertion with IF, could only cancel 1 move apparently... It found a 26 when i asked it to insert the 2+2 edge cycle also, but I don't think I would have ever found that :P
 
Scramble: D' R2 L2 F2 L2 U2 L2 D L2 U2 B2 L' F' R2 U' L F D U2 B2 R2

D2 F' R F //2x2 (4/4)

R' U' B U R L D' L B L2 B2//2x2x3 (9/13)

L2 D L2 F L F' L //F2L-1 (7/20)

z2 F' U F R U' R' U F' U2 F' U F// I dont know the name of this step in heise :p (12/32)

U2 L' D2 L U2 L' D2 L// corner cycle (8/40)

Final Solution: D2 F' R F R' U' B U R L D' L B L2 B2 L2 D L2 F L F' L z2 F' U F R U' R' U F' U2 F U F' U F U2 L' D2 L U2 L' D2 L
40 Moves HTM. Almost sub-40!
 
That's what a move-cancelling insertion looks like... My 22 Move Solution on the 3rd scramble at Velbert Open 2015

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

On Inverse:
L' D' R' F U D' $ B' D2 (8/8) - 2x2x3 except 1 Corner
U F' U (3/11) - F2L-1 except 1 Corner
R' F % R F' (4/15) - all but 3 Corners and 3 Edges

$ = U' D L' U L U D' B' U' B (10-8/17) - Edges, 8-move cancellation!
% = F R B2 R' F' R B2 R' (8-3/22) - Corners

Solution: F B2 R' F R B2 R' F2 R U' F D2 B D U' L' U' L F' R D L (22 Moves)
 
My winning 30.67 mean from Indiana 2015:

N: D F2 L \\Two 1x2x2s (3/3)
I: L F2 L' B2 \\2x2x3 (4/7)
N: R2 B' R B R2 U2 \\F2L-1 (6/13)
B' R' B R U R2 U' R2 \\L4C (8/21)

Skel: D F2 L R2 B' R B R2 U2 B' R' B (2) R U R2 U' R2 B2 (1) L F2 L' (21)

(1): B2 L F' L' B2 L F L' (cancels 5)
(2): D R' U' R D' R' U R (cancels 1)

Final: D F2 L R2 B' R B R2 U2 B' R' B D R' U' R D' R' U R2 U R2 U' R2 L F' L' B2 L F' L' (31)
Comments: I found this skeleton in about 10 minutes, and decided to save it for about 25 minutes left to go. Unfortunately, there wasn't much else that I could find, so I did start looking for insertions around that mark. I put a lot of time into the first insertion, which paid off with the 5 move cancellation. Looking through this far into the skeleton took a lot of time though, and I was a bit rushed on the resulting L3C. As a result, I was worried that I might have missed a second insertion that cancelled 2 moves, but insertion finder told me otherwise. :)

I: R D \\1x2x2 (2/2)
N: R' D2 R \\2x2x2 (3/5)
F2 L F' \\2x2x3 (3/8)
I: U2 L F' L2 F L2 U' \\F2L-1 (7/15)
B' U B U' L' U' L' U L U' L U L2 \\L3C (13/28)

Skel: R' D2 R F2 L F' L2 U' L' U L' U' L U L U B' U' B U L2 F' L2 F (1) L' U2 D' R'

(1): D R2 D' L' D R2 D' L (cancels 3)

Final: R' D2 R F2 L F' L2 U' L' U L' U' L U L U B' U' B U L2 F' L2 F D R2 D' L' D R2 U2 D2 R' (33)
Comments: 9 one-move pairs on both the scrambles, and I thought they were all crap. This scramble turned out to be the hardest for everyone, and I think it had a lot to do with the fact that there were a plethora of mediocre starts that people were prone to spending too much time looking at. I was "lucky" enough to find a safety L3C skeleton, which bought me more time to look for something better rather than start inserting. Ended up a little lucky with the optimal insertions on this one, cancelling 3 on an L3C.

N: R' \\pair (1/1)
I: U' L' B' D' B2 \\2x2x2 (5/6)
F2 L2 F' L' F' \\2x2x3 (5/11)
L2 U L' U2 \\F2L-1 (4/15)
(U') F U F2 L' F L' \\L4C (6/21)

Skel: R' L (1) F' L F2 U' F' U' L U' L2 (2) F L F L2 F2 B2 D B L U

(1): L B L' F' L B' L' F (cancels 5)
(2): L' B L F L' B' L F' (cancels 4)

Final: R' L2 B L' F' L B' F2 U' F' U' L U' L B L F L' B' L2 F L2 F2 B2 D B L U (28)
Comments: If the last skeleton had lucky insertions, then this skeleton won the lottery. It wasn't going well for a majority of the time, though. I spent the first 8 minutes inspecting the 6 one-move pairs on normal, and at that time, nature called, and I had to make a restroom run. 30-45 seconds later, I came back to find there were no pairs at all on inverse, which was very strange in contrast to the normal scramble. None of the pairs on normal looked any good, so I started seeing if those pairs were any better after switching to inverse. By 30 minutes, I had several meh starts, but no skeletons. I happened to stumble across this one when I decided to go back to this 2x2x2, saw the pair made with F2, and decided to roll with it. That led to this skeleton with about 20 minutes to go, which is a bit of a time crunch for me doing two insertions. Luckily enough, I cancelled 5 only two moves into the skeleton, and that made up a ton of time for me, giving me enough time to find the next insertion that cancelled 4.
 
i did my first one hour NISS attempt yesterday!

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

On inverse:
F' D B' D R2 U R U' F D' // 2x2x3 (10/10)
U2 L' U2 L2 U2 L2 // F2L - 1 corner (6/16)
R U B U' B' R' // edges (6/22)

leaves 5 corners in one cycle, so 2 insertions needed

Skeleton (on inverse):
F' D B' D R2 U R U' F D' U2 (1) L' U2 L2 U2 L2 R U B U' B' R'

(1) - insert (2) B R2 B' L' B R2 B' L to cancel 2 moves (8-2/28)
(2) - insert L' F' L B2 L' F L B2 to cancel 1 move (8-1/35)

Solution on inverse: F' D B' D R2 U R U' F D' U2 L' F' L B2 L' F L B' R2 B' L' B R2 B' U2 L2 U2 L2 R U B U' B' R' (35)
Invert to get: R B U B' U' R' L2 U2 L2 U2 B R2 B' L B R2 B L' F' L B2 L' F L U2 D F' U R' U' R2 D' B D' F (35)

I thought the insertions weren't very good but it turns out I was only 2 moves from optimal, so I'm pretty pleased with that :)
2nd best 1 hour attempt ever, and my best was a 34 move CFOP hack :P
question: when finding insertions, do you guys tend to start from the beginning, or the end of your solution? i've been starting from the beginning, but i've realised after a couple of attempts that i can be prone to making mistakes when searching through the skeleton... i feel like if i started from the back, it'd be a lot more obvious if i made a wrong move (and it's a lot easier to check that) and wondered if anybody else did this :)
 
^ I start from the back, it seems more logical. You have to put the insertion stickers on anyway, and you can easily check if you did a mistake by just performing the moves in the correct order again at the end.
 
I make more errors working from the back myself, just because having to invert moves throws me off more.

If the skeleton is L4C, I usually look at the opportunities to cancel a ton of moves first, like in between L F L' F' or something, and if there's nothing good, I will re scan all the other points I did not check.
 
FMC advice please! Thanks!
Scramble: R D2 B2 F2 R F2 D2 R' U2 R' B' R D' B2 L R2 D F' R2 D2

On the normal:
F B' L' F' D' F2 D' R D // Pseudo 2x2x3

On the inverse:
D2 U B2 U B' // F2L-1

On the normal: (premove B U' B2 U' D2)
L' B L B' // F2L
B' U F U' B U F' U2 // All but 3 edges
F R B' R' F' B U B U' B' // last 3 edges (in case you were wondering I checked insertion finder afterwards and it still gave me 32 moves.)

Final solution:
F B' L' F' D' F2 D' R D L' B L B2 U F U' B U F' U2 F R B' R' F' B U B U2 B2 U' D2 (32 moves)
 
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FMC advice please! Thanks!
Scramble: R D2 B2 F2 R F2 D2 R' U2 R' B' R D' B2 L R2 D F' R2 D2

On the normal:
F B' L' F' D' F2 D' R D // Pseudo 2x2x3

On the inverse:
D2 U B2 U B' // F2L-1

On the normal: (premove B U' B2 U' D2)
L' B L B' // F2L
B' U F U' B U F' U2 // All but 3 edges
F R B' R' F' B U B U' B' // last 3 edges (in case you were wondering I checked insertion finder afterwards and it still gave me 32 moves.)

Final solution:
F B' L' F' D' F2 D' R D L' B L B2 U F U' B U F' U2 F R B' R' F' B U B U2 B2 U' D2 (32 moves)

Notice that after you insert the F2L pair, instead of jumping to solving the corners right then, you can just do a U', which leaves 3 corners (along with the 3 edges you had, which unfortunately are still only solvable in 7 optimal). In this case, your skeleton is:
F B' L' F' D' F2 D' R D L' B L B' U' (*) B U' B2 U' D2 (19 moves)

Throwing in (*): U' F' U B U' F U B' cancels 5 moves, making the overall a 29 :D
 
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Notice that after you insert the F2L pair, instead of jumping to solving the corners right then, you can just do a U', which leaves 3 corners (along with the 3 edges you had, which unfortunately are still only solvable in 7 optimal). In this case, your skeleton is:
F B' L' F' D' F2 D' R D L' B L B' U' (*) B U' B2 U' D2 (19 moves)

Throwing in (*): U' F' U B U' F U B' cancels 3 moves, making the overall a 29 :D

Is it conventional to simply do two insertions once the next step is OLL (one corner, one edge)?
 
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