[Help Thread] FMC Discussion

[Hssandwich]

Trying the FMC Asia scrambles:

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

Spoiler

U2 R2 L' U2 //2x2x2 4
(D' R2 D2 F' R) //2x2x3 9
(D R F' R') //F2L-1 13
(D F' R' D R) //ab4c 18

*U2 R2 L' U2 R' D' ^ R F D' R F R' D' R' F D2 R2 D
Insert at *: L' D' L U2 L' D L U2
Insert at ^: D R2 D' L' D R2 D' L

L' D' L U2 L' D R2 U2 R D' L' D R2 D' L R F D' R F R' D' R' F D2 R2 D //27

Couldn't find anything after the 222 and all other 222s led to nothing, eventually found this.

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

Spoiler

R2 F' D U2 L2 F D' //2x2x3 7
(L U L' U2 L' U') //F2L-1 13
(B L B' U2 B' U' B) //ab3c 20

R2 F' D U2 L2 * F D' B' U B U2 B L' B' U L U2 L U' L'
Insert at *: D' B D F D' B' D F'

R2 F' D U2 L2 D' B D F D' B2 U B U2 B L' B' U L U2 L U' L' //23

Really liked this scramble, had to play around a bit after the 223 though.

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

Spoiler

(B') F L2 R D2 // 2x2x2 5
U' (F2 L2 F' B' U B) F // F2L-1 13
(L2 B' U B U2 L2 U L') //ab4c 21

F L2 R * D2 U' F L U' L2 U2 B' U' ^ B L2 B' U' B F L2 F2 B
Insert at *: R D' R' U' R D R' U
Insert at ^: U B' D2 B U' B' D2 B

F2 L R2 D' R' U' R D R' D2 F L U' L2 U2 B2 D2 B U' B' D2 B2 L2 B' U' B F L2 F2 B //30

Ugh scramble

 

[coldsun0630]

Lee Choong-myoung's winning solutions from FMC Asia 2016.

/

<1st solve>
Scramble: R' U' F D L2 B2 U' F2 L2 F2 R2 U F2 D2 F' L' D2 B2 D' B L U' R' F' R' U' F
Solution: U2 R2 L' U2 F R' F D2 F2 D' L B L' F L B' L' F2 D2 R D' R F2 D R' (25)

Spoiler

U2 R2 L' U2 // 222 (4/4)
F R' F D2 // pseudo-223 (4/8)
(R D' F) // F2L-1 (3/11)
(F R' F') // pair (2/13)
(F D R' D' F') // Tripod (3/16)
(F D' F D F2) // AB3C (2/18)

Skeleton: U2 R2 L' U2 F R' F D2 F2 D' [1] F' D2 R D' R F2 D R' (18)
Insert at [1]: L B L' F L B' L' F' (7/25)

<2nd solve>
Scramble: R' U' F L2 F2 D2 U2 F2 R' D2 F2 L' F' R' D' R B' F' L' B' D U B R' U' F
Solution: R2 F' U D F L' F' L' D' L' B' L' B L2 D L D' L' U' L U L2 U D L2 F D' (27)

Spoiler

(D F' D') // pseudo-122 (3/3)
R2 F' // 222 (2/5)
(D L2 D' U') // pseudo-223 (2/7)
U // 223 (1/8)
(L2 U' L' U) // F2L-1 (4/12)
D F L' F' L' D' // EO+pair (6/18)
(L' B' L' B L2 D L D' L') // solve the rest (9/27)

rd solve>
Scramble: R' U' F L2 B2 U2 L R2 U2 L' D2 B2 D' B' F2 L B' D L2 D2 F' L' U' R' U' F
Solution: L' B2 R2 B R' F2 R B2 R' F2 L' U L2 F L2 F U D' L U L' D L U2 L' U2 L2 B (28)

Spoiler

(B') // pair (1/1)
L' B2 R2 B' R' L' // pseudo-223 (6/7)
(L2 U2) // 223 (2/9)
U L2 F2 // F2L-1 (3/12)
F' L2 F U L U' L' // AB4C (6/18)

Skeleton: L' B2 R2 B' [2] R' L' U L2 F L2 F U [1] L U' L' U2 L2 B (18)
Insert at [1]: D' L U L' D L U' L' (5/23)
Insert at [2]: B2 R' F2 R B2 R' F2 R (5/28)

25, 27, 28 = 26.67 mo3

 

[TheDubDubJr]

Here are the solutions to my 25.67 FMC Mean from Mega Minnesota Winter 2016.

Spoiler: Attempt #1 - 26

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

all on normal:
D' F L' D' B' // EO (5/5)
D2 R // 2x2x1 (2/7)
L F2 // 2x2x2 (2/9)
U' B2 R // 12 to L7C (3/12) [L3C & L4C separate cycles]

Skeleton: D' F L' D' B' D2 R L F2 U' B2 ([4] B2 R F2 R' B2 R F2 R') R // leaves 16 to Good L6C

D' F L' D' ([2]123 - D R2 D' L' D R2 D' L) B' ([2]456 - B L' B' R B L B' R') D2 R L F2 U' R F2 R' B2 R F2 // insertions together cancel 2 more moves

Solution: D' F L' R2 D' L' D R2 D' B' R B L B' R' D2 R L F2 U' R F2 R' B2 R F2 (26)

(Haven't checked IF because the site is down for me)

Spoiler: Attempt #2 - 26

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

start on normal
n U F' R' // EO (3/3)
switch
i U // 2x2x1 (1/4)
switch
n D' R2 B' // 2x2x2 (3/7)
D F D' L2 // 2x2x3 (4/11)
F D' F2 D2 F // 16 to L5C (5/16)

Skeleton: U F' R' D' R2 B' D ([4]512 - D' B U2 B' D B U2 B') F D' L2 F D' ([2]345 - D B U2 B' D' B U2 B') F2 D2 F U'

Solution: U F' R' D' R2 U2 B' D B U2 B' F D' L2 F B U2 B' D' B U2 B' F2 D2 F U' (26)

Spoiler: Attempt #3 - 25

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

start on inverse
i F B2 D' R' F' // 2x2x2 (5/5)
switch
n L2 D2 // 2x2x1 block (2/7)
switch
i L D' B // 2x2x3 + 2x2x1 block (3/10)
switch
n L2 D L2 B L' B' L2 D' L // 19 to L3C (9/19)

Skeleton: L2 D2 L2 D L2 B L' B' L2 D' L B' D L' F ([2] F' D' B D F D' B' D) R D B2 F'

Solution: L2 D2 L2 D L2 B L' B' L2 D' L B' D L' D' B D F D' B' D R D B2 F' (25)

Interesting fact, I have the most fully sub-27 means with 2. The only other 2 people that have 1 fully sub-27 means are
João Pedro Batista Ribeiro Costa at Worlds 2015 (25, 26, 26) and Marcel Peters WR Average at Schwandorf (24, 25, 24). (Obviously anyone can manipulate data to make themselves look better , but its interesting.)

 

[Kit Clement]

Want to try the other FMC Asia scrambles too, but this is what I got on the first one.

Spoiler: FMC Asia 2016 Scramble 1

All on inverse
R2 L' D R' L' //EO (5/5)
U F' U2 R2 //1x2x3 (4/9)
B2 D' F2 //+1x2x2 (3/12)
B U F //almost F2L-1 (3/15)
D' R2 D F D F' //L3C (6/21)

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

1: U' R D R' U R D R' (cancels 5, lol)

Final: F D' F' D' R2 D F' U' B' F2 D B2 R2 U2 F U' L U' R D R' U L R' (24)

 

[CxCxC]

Hello everyone,

I am new to FMC and I needed some help :

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

First, I created a 2x2x2 block by : [B] : F D R B' R'
and then a extended it to a 2x2x3 by : [2B] : R U L U' L' U R' F
So basically , [B+2B] : F D R B' U L U' L' U R' F creates a 2x2x3 block {2 moves got cancelled}

Then I didn't find anything that was convenient, so I switched to the inverse by : [B+2B]'(Scramble)'

Then I completed the first two layers by :
(U F B' U B F') (L' U2 L U' L' U L) and then OLL+PLL+AUF which gives 46 moves for total

But then, I found out :
(U F B' U B F') (L' U2 L U2 L' U L F U' F' L' U L) >OLL skip
and then I got a J-perm : 14 moves
this totalled up to 44 moves[/B]
{Actually, more moves cancel out while writing out the solution, but they are not too much}
[B]
Where can I reduce more moves? How to complete the cube after 2x2x3 efficiently? I was not able to get the last block for building F2L-1 in this scramble. On my first scramble, {it was a really easy one, I was able to achieve 31 moves : I was able to get F2L-1 in just 17 moves}

I need some advice : I am able to easily get a 2x2x3 in most of the cases but I generally get stuck up after that. How can I proceed?

Plus, can someone show how to use insertions in this solve?

Thanks in advance,
CxCxC

 

[Cale S]

Spoiler

R' U' F R2 U2 L2 U2 F D2 L2 F2 L2 F L' U' F2 R2 B' L U' R U2 L2 U' R' U' F

L (F D' R') // EO
(F U B U F2 D) // 3e4c

L D' F2 U' B' U' @ F' ^ R D F'
^ = L F' % U2 F U2 L'
@ = [U B' U', F]
% = [F, U' B' U]

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

25 Moves

for some reason Insertion Finder found my cool 6 move thing but not optimal L5C...

 

[xyzzy]

Where can I reduce more moves? How to complete the cube after 2x2x3 efficiently? I was not able to get the last block for building F2L-1 in this scramble.

You could… just use Petrus. The 2x2x3 block you started with is workable; doing EO cancels the last move. After that you get an easy F2L because two of the pairs happen to be formed.

F D R B' U L U' L' U R' // 223 (10/10)
L F2 U' F' // EO (4/14)
L U' L' U' L' // pseudo-F2L (5/19)

At this point you can either use a double Sune to end with an A perm (11-4 moves double Sune, 2 moves adjust, 9 moves A perm), or you can do an Antisune to end with L3C (7-3 moves Antisune, 2 moves adjust, 9 moves L3C). The latter has fewer moves, so if we were restricted to a linear solve, we'd choose to do that. But we're not, so let's do an insertion.

L U2 L' U' L' U' L U L2 // edges; 3 corners left (9-3/25)

Skeleton: F # D R B' U L U' L' U R' L F2 U' F' L U' L' U L' U' L U' L' U L2
Insert B' U2 B D B' U2 B D' at # to cycle the three corners and cancel two moves, leading to a 31-move solution. This insertion happens to be optimal.

With your original start, you can also cancel a lot of moves by doing a lefty J perm (it's actually a block commutator and you get to cancel away the setup moves).

F D R B' U L U' L' U R' F // 223 (11/11)
switch to inverse:
U F B' U B F' L' U2 L U2 L' U L F U' F' L' U L // F2L + OLL (19/30)
L' U' L F L' U' L U L F' L2 U L U // J perm (14-8/36)

 

[Cale S]

20 pb single for the German weekly competition

Spoiler

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

(F' L2 B' D') // EO
R' U2 R2 // 1x2x3
(F' B R' B2) // F2L - 1
(F' L F' L' F' L F L' F) // ayylmao

Solution: R' U2 R2 F' L F' L' F L F L' F B2 R B' F D B L2 F

20 moves

found in 20 minutes
made a 25.33 mean, 2nd solve should have been better...

 

[Parham Saeednia]

My solution for superflip scramble:
found in less than one hour

Spoiler

Scramble:U R2 F B R B2 R U2 L B2 R U' D' R2 F R' L B2 U2 F2
EO: F U R D B' L' F'
EO line: U B2
Square: R2 D2
2X2X3+square: R2 F2 R2 L+
AB5C: F2 R'
+: L' F R' F' L# F R F'
#: L D2 L' U2 L D2 L' U2
Final solution: F U R D B' L' F' U B2 R2 D2 R2 F2 R2 F R' F' L2 D2 L' U2 L D2 L' U2 F R F R'
(29 HTM)

 

[guysensei1]

Weekly FMC, 23 moves dahell

Spoiler

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

L D L2//lol 2x2x2
Inv
L' R2 D L B' D2 B'//2x2x3
D//F2L-1
Normal
R' F' R F//make pair
R' B' F' R F R' B R2//force OLL skip, get PLL skip too

Final L D L2 R' F' R F R' B' F' R F R' B R2 D' B D2 B L' D' R2 L (23) lolwat

 

[Parham Saeednia]

Weekly FMC

Spoiler

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

L2 U' D'+ B'//EO
L' U L U2 R2 U2//2X2X3
D' L' D R' D' L//F2L-1
D' R2 D R//AB3C

+: R2 B' L' B R2 B' L B

Final solution: L2 U' D' R2 B' L' B R2 B' U L U2 R2 U2 D' L' D R' D' L D' R2 D R
(24HTM)

 

[vm70]

This is my second (and worst) FMC solve. My other solve was 57 OBTM.

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

U B' U' B R F R // 2x2x2
L U2 L2 U L U2 L // 2x3x3
x' F R U R' U' F' y f R U R' U' f' F R U R' U' F' // Line & Pi 2LOLL
y R' U L' U2 R U' R' U2 R L U' // J[a] perm + AUF
(L' B L B' U B' U B U x') //F2L inverted
(U2 F2 U2 L F L') // F2L-1 (inverted

Spoiler

Solution with pre-moves:
https://alg.cubing.net/?setup=(R-_F_R_F-_U_F-_U_F_U_x)_//_F2L_pre&#45;moved (U2-_B2-_U2-_R_B_R-)_//_F2L&#45;1_pre&#45;moved R_F-_D_R_B2 F2_D2_F2_L_B R2_D-_U2_B2_L2 R_D_L2-_F_L2&title=FMC Test Normal&alg=(U_B-_U-_B)_(R_F_R)_//_2x2 (L_U2_L2_U_L_U2_L)_//_2x2x3 (x-_F_R_U_R-_U-_F-)_y_(f_R_U_R-_U-_f-_F_R_U_R-_U-_F-)_//_Line&#2b;Blinker_2Look_OLL (y_R-_U_L-_U2_R_U-_R-_U2_R_L)_U-_//J[a]_perm_&#2b;_AUF

Solution without pre-moves:
https://alg.cubing.net/?setup=R_F-_D_R_B2- F2-_D2-_F2-_L_B R2-_D-_U2-_B2-_L2- R_D_L2-_F_L2-&title=FMC Test Normal w/o Pre-Moves&alg=(U_B-_U-_B)_(R_F_R)_//_2x2 (L_U2_L2_U_L_U2_L)_//_2x3x3 (x-_F_R_U_R-_U-_F-)_y_(f_R_U_R-_U-_f-_F_R_U_R-_U-_F-)_//_Line_&_Blinker_2LOLL (y_R-_U_L-_U2_R_U-_R-_U2_R_L)_U-_//_J[a]_perm_&#2b;_AUF (L-_B_L_B-_U_B-_U_B_U_x-)_//F2L_(inverted) (U2_F2_U2_L_F_L-)_//_F2L&#45;1_(inverted)

I tried to use my understanding of FreeFOP for this solve. 58 HTM.
Go ahead and critique me. Please, I seriously had no idea what to do.

EDIT: Changed parentheses because I didn't know how to use them in FMC prior to @xyzzy telling me how to. Edit again for changing 2Look OLL names.

 

[xyzzy]

This is my second (and worst) FMC solve. My other solve was 57 OBTM.

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

(U B' U' B) (R F R) // 2x2
(L U2 L2 U L U2 L) // 2x3x3
(x' F R U R' U' F') y (f R U R' U' f' F R U R' U' F') // Line & Blinker 2LOLL
(y R' U L' U2 R U' R' U2 R L) U' // J[a] perm + AUF
(L' B L B' U B' U B U x') //F2L (inverted)
(U2 F2 U2 L F L') // F2L-1 (inverted)

Found this shorter 223:
B L B' // square (3/3)
D F2 D' // square (3/6)
R F R B // 223 (4/10)

Forming lots of squares (2x2x1 blocks) and hoping they join up nicely is a decent strategy in general. (There's actually a two-move square in the inverse, but continuing from there was kinda awkward.)

As for your solution, the most obvious thing to work on is to get rid of all the rotations. Also, don't write your moves within parentheses, because that's taken to mean that you're applying the moves on the inverse scramble (which you're not). If you don't know enough of OLL to one-look most cases, it usually pays off to do edge control or edge orientation before you reach the last layer. Even if you do know full OLL, having edges oriented allows you to mess with Sunes and such more effectively.

The shortest LL algs to flip two edges are 6 moves long and the shortest LL algs to flip four edges are 11 moves long, whereas in Petrus, you can usually flip two edges in 3 moves and four edges in 5.

Spoiler: A Petrus-style solve with the same 2x2x3

U B' U' B R F R // 222 (7/7)
L U2 L2 U L U2 L // 223 (7/14)
inv: R B' R' // EO (3/17)
inv: U2 B2 U2 B2 U' // F2L (5/22)
U' // adjust (1/23)
D' B' D B' D' B2 D // OCLL (7/30)
D B' U B2 D' B D B2 U' D' // J perm (10-3/37)

In this case, doing COLL is not a good idea because that takes 10 moves and you end up with a Z perm, which takes 12 moves (although you can always cancel at least two moves). ZBLL would be ideal, but I don't know this case.

Or actually, now that I'm going through your solution… why would you sledgehammer that last F2L pair in when all the edges are already oriented? (????)

Spoiler: what you did

U B' U' B R F R // 222 (7/7)
L U2 L2 U L U2 L // 223 (7/14)
inv: R B' R' U2 B2 U2 // F2L-1 (6/20)
inv: B' U' B' U // set up pair (4/24)
inv: B' U R' U' R // sledgehammer (5/29)

Spoiler: what you can try instead

U B' U' B R F R // 222 (7/7)
L U2 L2 U L U2 L // 223 (7/14)
inv: R B' R' U2 B2 U2 // F2L-1 (6/20)
inv: B' U' B' U // set up pair (4/24)
inv: B U' B2 U // insert (4/28)

This ends in a nice ZBLL case that's just a sexy move conjugated by a sledgehammer. (There's another optimal alg for this case that's something like wide sune + fruruf.)

inv: R' U R U' B2 R B' R' B' U R' U' R B // ZBLL (14/42)

You can also do a Niklas + R perm (7+14=21 moves) or Antisune + V perm (7+14=21 moves).

Spoiler: or a six-move F2L insert

U B' U' B R F R // 222 (7/7)
L U2 L2 U L U2 L // 223 (7/14)
inv: R B' R' U2 B2 U2 // F2L-1 (6/20)
inv: B2 R2 D B D' B' R2 // F2L (7/27)
inv: D' B2 U B' D B U' B D' B D // COLL (11/38)
// EPLL skip

(or Sune/Niklas + PLL, which would be 20-ish moves)

Or you can spot that you have a pair already formed and insert it to get a pseudo-F2L, like what I did in the first spoilerbox. Moving on to your last layer.

Spoiler: what you did

U R B R' B' U' // EO (6/6)
L B U B' U' L' R D B D' B' R' // CO (12/18)
J perm // (11/29)

Spoiler: alternatives

L' D L B L' D' L D B' D' // OLL (10/10)
U B' D B2 U' B U B2 U' D' B' // J perm (11/21)

inv: R' L2 U L' U L U2 L' U L' R // OLL (11/11)
inv: D' B U' B2 D B' U L R B2 L' R' // G perm (12/23)

U R B R' B' U' // EO (6/6)
D B' U' B D' B' U // CP (7/13)
U' B2 U B U' B U // CO (7-3/17)
U2 B L R' U2 L' R B U2 // U perm (9-1/25)

U R B R' B' U' // EO (6/6)
D B2 D2 B' D2 B' D2 B2 D // CO (9/15)
J perm // (11/26)

U R B R' B' U' // EO (6/6)
D' B2 D2 B D2 B D2 B2 D' // CO (9/15)
B' R2 D2 R U R' D2 R U' R // A perm (10/25)

(all of these kind of suck :v)

 

[xyzzy]

(loldoublepost)

I was practising earlier and I noticed that for corner 3-cycle insertions, stepping through every single move in the skeleton was kinda pointless; you could step through it two moves at a time and still be able to find all the insertions that cancel moves.

Is this something top FMCers already know, or is there some caveat to this shortcut?

 

[guysensei1]

(loldoublepost)

I was practising earlier and I noticed that for corner 3-cycle insertions, stepping through every single move in the skeleton was kinda pointless; you could step through it two moves at a time and still be able to find all the insertions that cancel moves.

Is this something top FMCers already know, or is there some caveat to this shortcut?

Perhaps in between the two moves the corners setup to a commutator with more than one 8 move solution, and you'd miss possible cancellations this way.

 

[EMI]

(loldoublepost)

I was practising earlier and I noticed that for corner 3-cycle insertions, stepping through every single move in the skeleton was kinda pointless; you could step through it two moves at a time and still be able to find all the insertions that cancel moves.

Is this something top FMCers already know, or is there some caveat to this shortcut?

I go through the skeleton from right to left and only check whether the first move of the commutator can be the current skeleton move to the left. This is equivalent to only checking each second spot as you suggest, and it is really fast because you always know with which move your commutator has to start with to cancel two moves (requires less thinking imo). A standard L3C insertion probably takes less than a minute this way.

 

[xyzzy]

Perhaps in between the two moves the corners setup to a commutator with more than one 8 move solution, and you'd miss possible cancellations this way.

Hmm. I was thinking that if you could insert an 8-move commutator that cancels at least two moves on one side (e.g. ... R [R', U L U'] F2 ...) then you can always "shift" the insertion location over the cancelled move (... [U L U', R] R F2 ...), but this doesn't seem to work if you cancel only one move (or one move on both sides).

I go through the skeleton from right to left and only check whether the first move of the commutator can be the current skeleton move to the left.

Wow, this sounds very efficient. I've been doing it in the opposite direction (left to right on skeleton, checking for cancellations to the right), but this might be considerably faster.

 

[Cale S]

(loldoublepost)

I was practising earlier and I noticed that for corner 3-cycle insertions, stepping through every single move in the skeleton was kinda pointless; you could step through it two moves at a time and still be able to find all the insertions that cancel moves.

Is this something top FMCers already know, or is there some caveat to this shortcut?

I've thought of this as well, but when insertions only cancel 1 move it's worth it to go move by move in that region

 

[vm70]

@xyzzy Thanks a lot, I really needed this! I'm still new to the idea of FMC; hopefully with more practice I'll get better solves.

I don't know COLL or even ZBLL; I guess I have a lot to work on from here.

 

[xyzzy]

I don't know COLL or even ZBLL; I guess I have a lot to work on from here.

You don't really need full OLL or COLL (or god forbid, full ZBLL); last layer algs that are longer than 10 moves are very rarely useful unless you one-look the last layer. The most useful LL alg is really just Sune/Antisune (R U R' U R U2 R'); you can use Sunes to solve the edges and then commutators to solve the last few corners. (Exception: if you get Z perm edges, you can't solve that with a single Sune.)

In fact, when you reach the last slot, what you should do isn't even so much to solve the whole pair (unless it's already formed), but to solve the last five edges. This leaves at most five corners unsolved, and you can proceed to writing out your skeleton (all the moves you've done so far in the correct order, including those on inverse) and stepping through it to insert commutators.

Something I forgot to mention in the previous post was that you really should use optimal algs for the easy cases. The pi OCLL case can be done with R U2 R2 U' R2 U' R2 U2 R or its mirror, which is shorter than the 12-move alg you used. Most PLLs are useless, but do learn optimal algs for at least U perm (9f), H perm (9f), and T perm (10f). (The standard A perm algs and the <R,U,L> J perm algs are already optimal.)