[Help Thread] FMC Discussion

[Brest]

2011-10 weekly competition.
Scramble: B R B' D F' R2 B' L' B2 L B L' U2 L B U2
Inverse: U2 B' L' U2 L B' L' B2 L B R2 F D' B R' B'

I've broken this up with spoilers as I have written a wall of text.

The solution I submitted:
U L' U' R B R L F B2 R' B' R D2 F D' F' R' F' B R2 F B' D L2 F' D' F L2 B' U B' U' B (33)

Spoiler

Pre move: B'
Scramble: B R B' D F' R2 B' L' B2 L B L' U2 L B U2
Pseudo 2x2x3: U L' U' R B R L F B (find pre move B')
EO & 2 pairs: B R' B' R D2 F D' F' R'
Insert Edge: F' B R2 F B'
J Perm: D L2 F' D' F L2 B' U B' U'
Cancel pre move: B

I'm sure that there is a better solution from that pseudo 2x2x3, but this is what I found within the time limit. I wasn't really happy with the ending, and I needed the practice so I continued. Just after the time limit expired, I realized I could cancel three moves just be using a different J perm. :fp

D L2 F' D' F L2 B' U B' U' B becomes B' D F' D2 B D' B' D2 F B

Drop the D which is just an ADL (adjust D layer), first move cancels with the move before, last move cancels with the premove.

U L' U' R B R L F B2 R' B' R D2 F D' F' R' F' B R2 F B2 D F' D2 B D' B' D2 F (30)

So silly to miss that...

More practice, this time with a different start.

Spoiler

I was frustrated with missing those easy cancels, and obviously I need more practice, so I looked for a different start.
Here is a 19 move F2L, which was fun to build, but I didn't find a great ending for it:
R2 F R' D L' B D2 R F D' F D B R' B' L2 B' L' B

So instead I made this 2x2x3 pseudo block:
(Pre-move L) R2 F R' D' L' B R B R2 B'

which also creates another two CE pairs, so I was fairly happy with it. I decided to try using the inverse scramble with the 2x2x3 as premoves. This is NISS, right?

(B R2 B' R' B' L D R F' R2) U2 B' L' U2 L B' L' B2 L B R2 F D' B R' B'

Which revealed some really sweet orientations. The original pre move (L') aligns the 1x2x2; however a D' before will create another 1x2x2 block and F will complete F2L minus one slot. To try and preserve the last pair as an easy insertion, I settled on these moves (D' L' D2 F) which replaced the original pre-move when returning to the scramble.

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

Insert the 4th pair
L D L'

Upon insertion I recognized that the OLL would create a PLL skip!
B' D' B' L B L' D B

Sequence together:
R2 F R' D' L' B R B R2 B'
L D L'
B' D' B' L B L' D B
F' D2 L D
(25)

This was all mad skillz, no luck whatsoever. :rolleyes: Still, I'm really happy finding this.


Upon a closer inspection of the inverse, there is a really sweet finish to the F2L:

(B R2 B' R' B' L D R F' R2) U2 B' L' U2 L B' L' B2 L B R2 F D' B R' B'

D' L' D R' D2 R F

This leaves 3 corners after ADL!

D2 L' F' R F L F' R' F

Invert and sequence:
R2 F R' D' L' B R B R2 B' F' R F L' F' R' F L D2 F' R' D2 R D' L D (26)

After a bunch of help from you guys, especially from Cubenovice and irontwig, I found my first commutator insertion! Yeah!

Leave 3 corners:
R2 F R' D' L' B R B R2 B' D2 F' R' D2 R D' L D

Spoiler

R2 F (1) R' D' L' (2) B (3) R B R2 B'
D2 (4) F' (5) R' D2 (6) R D' (7) L D

Insertions:
1 F U' F' D' F U F' D
2 L' F' R' F L F' R F
3 F L F' R F L' F' R'
4 R F L F' R' F L' F'
5 F' U' F D' F' U F D
6 F' U' F D' F' U F D
7 B' L' F L B L' F' L

I did not find an insertion which canceled more than a single move, so the best solution I found in all this is still 25 HTM; however this solution seems a lot less lucky!

Can anyone find an insertion with more than a single cancellation?
After the 2x2x3 block, are there any other improvements to this solution?

If you read all that, thank you for your time. Any help or constructive comments are appreciated.

 

[Cubenovice]

Hi Brest, you're doing a great job!

Don't worry about missing those cancellations and the long J perm.
You missed a few moves in this solve but due to this learning experience you wil cancel lots and lots in the future.

Do not have time to look for insertions but what I see:
- you use both R U R' and R U' R' style which is good
- you do not have any double turns in your commutators, they can be in either the A or B part, perhaps there where none, perhaps you did not check?

 

[irontwig]

Your (3) insertion or R F' L F R' F' L' F after your first move both cancel two moves resulting in 24 moves.

 

[Kenneth]

Fellow communists, there are two terms I miss in the wiki, both essential for advanced FMC; [wiki]Skeleton[/wiki] and [wiki]Insertion[/wiki].

Any FMC expert who cares to explain these?

 

[irontwig]

Fellow communists, there are two terms I miss in the wiki, both essential for advanced FMC; [wiki]Skeleton[/wiki] and [wiki]Insertion[/wiki].

Any FMC expert who cares to explain these?

http://www.speedsolving.com/wiki/index.php/Fewest_Moves_Techniques

 

[Kenneth]

Ok, redirecting is an option of course...

 

[Kenneth]

Yay, I had 28 in FMC classic 338, no luck, just skillz and a nice LL case

I won't post my solution now, the comp is still ongoing...

 

[Attila]

I believe, maybe useful, if i write a longer explanation to my last solution (mustcube 337. round, classic).
Scramble: U' F2 D' F' L' B U' B2 U B' U' L' F' U' F2 L2 B' R2 F R2 F U' B' R U' R' D2 U L R2
Solution: B2DR’B’D2U2FLD’U’R’U2R’BF2D2L2U’DRU’R’LB2
First time i need a short corners-algo, usually i search something premove, if the scramble it seems too hard. In this case i found this premoves for normal scramble: DB
B2D’B’D2B (Guimond)
B2D’R2F2
Corners solve, without premoves:
B2D’B’D2B’D’R2F2DB
Corners solve for inverse scramble: (inverse of previous solve)
B’D’F2R2DBD2BDB2
difference from the previous solve , to be more edges solved:
B’ D’ F2 R2 D B D2 B D B2 only the corners solve
B2Mb d’ M’ F2 R2 d2D’ b d2 B corners -2 moves and 5 edges solve,
Then i write this, without centers move:
B2L’RUR’D’UL2D2F2B’RU2R.
The second move (M) does not change the first five edges position, but must to insert, to be lucky end.
The next step is obvious, solve more 3 edges:
UD setup moves , L’F’U2D2BR 3 edges algo, then a lucky E slice skip, therefore had previous M move.

 

[Cool Frog]

I don't know where to ask this but. What would you do for the Commutator that does this...
FUR>BUL>FDL
My solution is a 4 move setup, 8 move commutator then undo setup.

 

[Mike Hughey]

I don't know where to ask this but. What would you do for the Commutator that does this...
FUR>BUL>FDL
My solution is a 4 move setup, 8 move commutator then undo setup.

That's what Chris Hardwick calls a "Per Special". Optimal in HTM is 12 moves. The solution he does for that particular one would be something like F U2 F' U2 F' L2 F U2 F U2 F' L2.

Edit: If you want to understand how it works, he explains it here.

 

[irontwig]

You won't ever need to be able to solve that case in a FMC solve though.

 

[Cubenovice]

A very sexy (But as anti-optimal as can be) solution:

F - L' B' L B - L' B' L B - L' B' L B - F2 - L' B' L B - L' B' L B - L' B' L B - F2 - F'

But offcourse I would perform a cube rotation so the sexy move parts execute as L' U' L U or R U R' U'

Ehm yes, I actually do use this as it is mindless execution, you don't even need to inverse the sexy move part ;-)


EDIT: silly me, by starting with B and going all D' B' D B I can cancel an additional move, this is the FMC thread afterall!

 

[marco.garsed]

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

Solve:
B2 F R' F L B [U'] (' F U' B U F' [U']) (x-cross + one f2l couple + 3-cycle insertion)
D F' U2 F (third f2l)
U' B L B' (fourth f2l + one cross-edge)
R U R' F2 D' L U' L' U L' D F2 (G perm :S)
U2 (auf)

32 HTM

6 moves canceled!

 

[Erzz]

Wide turns are counted as one move right?

First try at FMC, using TF and trying to optimize as much as possible

Scramble: B2 U' B2 R2 B2 D' L2 U F2 L2 U' F' D L U' L2 D' U2 B R U'
Solution: y2 F' L2 U' L' U' F' U' F U' L U2 L' U' B u2 b U b' y2 M' U2 M y U F2 M' U' M U F2 y' R U' R2 U2 R U R' U R2 U R' D2 y' x' L2 D2 L U L' D2 L U' L (53)

Spoiler

y2 F' L - 2x2x1 block (2)
L U' L' U' F' U' F - Add a corner plus solve a middle layer edge (8)
U' L U2 L' U' B - Finish the triangle and solve a middle layer edge (14)
u2 b U b' - Finish the middle layer (18)
This seemed good so far, but then I didn't know how to proceed, so proceeded normally.
y2 M' U2 M - Insert edge (23, 21)
y U F2 M' U' M U F2 - Insert other edge and orientate edges (32, 28)
y' R U' R2 U2 R U R' U R2 U R' - Insert corner and orientate corners (43, 39)
D2 - Fix D layer (44, 40)
y' x' L2 D2 L U L' D2 L U L - A-perm (53, 49)

 

[Mike Hughey]

You won't ever need to be able to solve that case in a FMC solve though.

True - no matter how awful it is, surely there's a way to get rid of some of those moves. I'm bad enough at insertions sometimes that I've had a case where the best I could do with 3 corners was 9 moves, though. (Embarrassing, I know.)

 

[Cubenovice]

Wide turns are counted as one move right?

Right!

I do not know how the average movecount of TF compares to other speedsolving methods but for FMC you typically use a low movecount method. In practice this (most of the time) means blockbuilding methods like Petrus and Heise.

Two other comments on your solve:
- Try to write down your solution without cube rotations ans double layer turns. You may have some hidden cancellations isn there
- Instead of performing the A-perm at the end you can look for an insertion of the corner cycle, possibly cancelling moves.

 

[Kenneth]

Solution: y2 . y2 . y . y' . y' x' . (53)

- Try to write down your solution without cube rotations ans double layer turns.

The trick is to look at the centre of the face that is rotating. This way you can orient the cube any how but always be able to know wich side you are turning.

Examples, assuming white U and green F (official colour sheme and orientation), cube rotate: x' y z2 y' z' (pretend you did some steps and oriented like this).

Then do R' but look at the centre at the R-side, it is blue, that is the B-face (B as in blue, R is red =) So you notate it as a B'
Next move you do is a f (wide), then it is the face at the back that is turning, in this case it is white = U, so write U
Your third move is again a R', but after the F-wide that centre is not blue anymore but red = R-side, notate R'

So, you did R' f R' but write B' U R'

Test : 1x1x1 scramble #94: z y2 z' y2 z x' y' z' x' z'

1; After that, what do you got for U and F? (the rest are fixed, we only need two adjacent sides to know the orientation)
2; If you now do d' b (wide moves), how would you notate that?

Make sure you know both answers before looking at the spolier =)

Spoiler

1; green and red = F and R
2; green is on top, (the opposite side of what is currently on D) and then white moves in on front, (opposite of B), notate F' U

 

[MaeLSTRoM]

That's what Chris Hardwick calls a "Per Special". Optimal in HTM is 12 moves. The solution he does for that particular one would be something like F U2 F' U2 F' L2 F U2 F U2 F' L2.

Edit: If you want to understand how it works, he explains it here.

Can't you just use [D - Aperm w/o first or last R turn - D'] - 10HTM
or for the other way of cycling, [D' - CCW Aperm w/ first or last R turn - D]

 

[Tim Reynolds]

All right, can someone explain to me how to do an insertion of more than 3 corners? Specifically, can you find something for the below? (Warning: this is a skeleton to the in-progress weekly competition.)

Spoiler

Scramble: U2 B D' B R U2 R U' R2 B D R2 U L B R D' R
Skeleton: L D L' D / B2 R2 B' R2 U2 / R' U R2 U' / B R B R' (18)

I'll post my full solution in the weekly comp thread (2011-12) when I finish some BLD stuff.

So I know the idea is to do 2 insertions. I tried this and basically flailed for 15 minutes not succeeding. For a cycle of the form (A B C A+) (D-), where the first cycle means it should go to A but one spot clockwise, one insertion has to be like (D A B)--the point is that it has to involve D and solve some other piece. So I guess that means there are 9 different options for the first cycle. Hmm, I guess writing all this out while the clock was not running wasn't too hard. Any thoughts from people more experienced in insertions on more than 3 pieces?

 

[Mike Hughey]

I'd love to see what the experts say about this. Since, in my experience, the best insertions come with cycles that are already good to begin with, I'd probably first check all of the 9 possible cycles and see if any of them are 8 moves, and focus on just those (to cut down on the possibilities). And I'd probably check them one at a time. And if there were several, I might simply look at the 8 move result for each one and see how good the remaining cycle is. If it's an 8 mover followed by another 8 mover, my past experience (which admittedly isn't much) seems to indicate that that particular sequence would be a really good one to look for insertions with.

For yours, I see that U' L' D' L U L' D L gives a nice 8 move second one: R' U' L U R U' L' U. I saw that one right away. So I'd probably first work on that set of insertions and see what I could find.