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The FMC thread

DGCubes

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Speedsolving weekly competition (week 52):

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

(B' L' U' L' B' L2 B) // pseudo-2x2x3
(F' R2 B R F) // AB3E5C
(D2 F' B R2 B' F) // AB3C (technically an insertion, but I basically used this as part of my skeleton)

I could've canceled a move with the 3 edges, but I had already done corner insertions and it would've messed with one of those, so I started with the D2 to cancel with one of the corner insertions instead.

Skeleton (18 to AB5C): F' B R2 B' F D2 F' * R' B' R2 F B' L2 B L U L B $

* = F D' F' U2 F D F' U2 (cancels 3)
$ = B R2 B' L B R2 B' L' (cancels 1, kind of a last resort)

Unfortunately, IF found a 26 with this skeleton. I knew there was a lot more potential than I was able to find, but I was working until the last second and didn't get to check every corner insertion. :(

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

Anyone have any general tips for how to get faster at insertions? For this attempt I had 30 minutes for 2 insertions and still didn't have time to check everything.
 

obelisk477

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Speedsolving weekly competition (week 52):

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

(B' L' U' L' B' L2 B) // pseudo-2x2x3
(F' R2 B R F) // AB3E5C
(D2 F' B R2 B' F) // AB3C (technically an insertion, but I basically used this as part of my skeleton)

I could've canceled a move with the 3 edges, but I had already done corner insertions and it would've messed with one of those, so I started with the D2 to cancel with one of the corner insertions instead.

Skeleton (18 to AB5C): F' B R2 B' F D2 F' * R' B' R2 F B' L2 B L U L B $

* = F D' F' U2 F D F' U2 (cancels 3)
$ = B R2 B' L B R2 B' L' (cancels 1, kind of a last resort)

Unfortunately, IF found a 26 with this skeleton. I knew there was a lot more potential than I was able to find, but I was working until the last second and didn't get to check every corner insertion. :(

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

Anyone have any general tips for how to get faster at insertions? For this attempt I had 30 minutes for 2 insertions and still didn't have time to check everything.
How do you do your stickering for 5c? What is your process for checking each insertion?

Also, a pretty good tip I've found is to write down every single possible insertion for a cancellation that you find up until the point that you've found your first insertion, and also write down the cycle that that insertion solves. Then, after you've found your first insertion, look and see which corners are still stickered after you remove the stickers for the newly solved corners. It will be one of the possible 5 three cycles. At this point, you can fast forward to the first insertion you did when checking for the second, knowing that (say you have the 2-3-4 cycle remaining) you aren't missing out on anything because you would have already written down possible cancellations for all 2-3-4 cycle insertions prior to the asterisk. Then check the rest of the skeleton (and within the first commutator) to see if there are any better cancellations than the ones you have written down.

EDIT1: Also, this of course only applies to pure 5 cycles of corners, as in your example above. Other AB5C would use a different process.

EDIT2: I checked insertion finder and it only found 28 move solutions, not 26 move solutions. So you were only 2 moves off of optimal
 
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DGCubes

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How do you do your stickering for 5c? What is your process for checking each insertion?

Also, a pretty good tip I've found is to write down every single possible insertion for a cancellation that you find up until the point that you've found your first insertion, and also write down the cycle that that insertion solves. Then, after you've found your first insertion, look and see which corners are still stickered after you remove the stickers for the newly solved corners. It will be one of the possible 5 three cycles. At this point, you can fast forward to the first insertion you did when checking for the second, knowing that (say you have the 2-3-4 cycle remaining) you aren't missing out on anything because you would have already written down possible cancellations for all 2-3-4 cycle insertions prior to the asterisk. Then check the rest of the skeleton (and within the first commutator) to see if there are any better cancellations than the ones you have written down.

EDIT1: Also, this of course only applies to pure 5 cycles of corners, as in your example above. Other AB5C would use a different process.

EDIT2: I checked insertion finder and it only found 28 move solutions, not 26 move solutions. So you were only 2 moves off of optimal
It just clicked... your point about only checking the rest of the skeleton makes so much sense! I've been wasting a lot of time going through each move for my second insertions in the past. I'll definitely do it the way you recommended from now on. Thanks so much! :)

To answer your questions, though:
I put five stickers on the cube, one on each corner in the cycle, and number them 1-5. I go through my skeleton move by move and check all 5 possible 3-cycles. On my sheet of paper, I put a caret between each move and write down the optimal number of moves for any insertion there (if there is one 8 moves or less), and then move to the next move.

Also, I should've added that I checked insertion finder for the 12-move skeleton (the one without the three edges solved at the end). So had I tried to do three insertions, I could've had a 26 (although seeing as I could barely fit in two, it was pretty unlikely for that to happen).

Thanks again for your advice; it seems so obvious now but I never considered it before. :)
 

obelisk477

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It just clicked... your point about only checking the rest of the skeleton makes so much sense! I've been wasting a lot of time going through each move for my second insertions in the past. I'll definitely do it the way you recommended from now on. Thanks so much! :)

To answer your questions, though:
I put five stickers on the cube, one on each corner in the cycle, and number them 1-5. I go through my skeleton move by move and check all 5 possible 3-cycles. On my sheet of paper, I put a caret between each move and write down the optimal number of moves for any insertion there (if there is one 8 moves or less), and then move to the next move.

Also, I should've added that I checked insertion finder for the 12-move skeleton (the one without the three edges solved at the end). So had I tried to do three insertions, I could've had a 26 (although seeing as I could barely fit in two, it was pretty unlikely for that to happen).

Thanks again for your advice; it seems so obvious now but I never considered it before. :)
Another trick to speed up insertion checking is that instead of checking all 5 possible 3 cycles for each move of the skeleton, do this instead: check all 6 faces for interchange moves involving adjacent numbers, and then check only those cycles which involve that valid interchange, as those are the only ones which will be solveable with an 8 move commutator
 

DGCubes

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Another trick to speed up insertion checking is that instead of checking all 5 possible 3 cycles for each move of the skeleton, do this instead: check all 6 faces for interchange moves involving adjacent numbers, and then check only those cycles which involve that valid interchange, as those are the only ones which will be solveable with an 8 move commutator
Oooh, that's really helpful. Thanks for all the advice! :)
 

NykoCuber1

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// F U' F2 D' B U R' F' L D' R' U' L U B' D2 R' F U2 D2

- 29 - :p

(L F L') // Paires (3)
L' F' U R // Blocks (4)
(D' F' D2) // 223 (3)
(L' U L2) // Blocks (3)
(B2 L U' L2) // PF2L (4)
B // F2L (1)
R B L' U' L U' L' U2 L U2 B' R' // LL (12) :)
 

mazh

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Here is my personal best solution.

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

--18--

R2 B' U' //EO(3/3)
R2 F' D2 L # B' //222+122(5/8)
R' F' R U2 R2 //L4C(5/13)
# = R' * F R B' R' F' R B(8-5/21-5)
* = L' F R F' L F R' F'(8-6/24-6)

My friend used CE to calculate this scramle,and CE says 18 is already the best.
 
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mazh

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Speedsolving weekly competition (week 52):

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

(B' L' U' L' B' L2 B) // pseudo-2x2x3
(F' R2 B R F) // AB3E5C
(D2 F' B R2 B' F) // AB3C (technically an insertion, but I basically used this as part of my skeleton)

I could've canceled a move with the 3 edges, but I had already done corner insertions and it would've messed with one of those, so I started with the D2 to cancel with one of the corner insertions instead.

Skeleton (18 to AB5C): F' B R2 B' F D2 F' * R' B' R2 F B' L2 B L U L B $

* = F D' F' U2 F D F' U2 (cancels 3)
$ = B R2 B' L B R2 B' L' (cancels 1, kind of a last resort)

Unfortunately, IF found a 26 with this skeleton. I knew there was a lot more potential than I was able to find, but I was working until the last second and didn't get to check every corner insertion. :(

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

Anyone have any general tips for how to get faster at insertions? For this attempt I had 30 minutes for 2 insertions and still didn't have time to check everything.
When you find the first insertion of 5C or 4C, do not pay too much attention on how to solve each possible of three-cycle insertion.
First you should notice what is the next move in your skeleton, and then consider how to cancelled that move by your insertion. Finally, you just need to check whether you have this sort of insertion to cancelled next move. This method will save your much time. I usually use 10~15 minutes to do 2 insertions by this method.

Example
Scramble:
R' U' F L2 D2 U2 F D2 R2 F2 L2 U2 B2 U' B' L R' D2 F D2 L2 U B R' U' F
Skeleton:
R2 B' U' R2 F' D2 L B' R' F' R U2 R2 //L4C​
When I found insertion between "L" and "B'", I noticed my last move of insertion must be "B" to cancelled "B'", then I found the insertion "R' F R B' R' F' R B" quickly.​

Another suggestion: do not write down every possible insertion when you find the first insertion of 5C or 4C.
Because it will waste too much time to write down one by one. I just write down the insertion with 3 or more than 3 moves cancelled, and only record the position of the insertion with 2 moves cancelled but not write down the moves.

I'm not good at English,so I don't know if I express what I mean clearly, but I hope this tips are helpful for you.:)
 
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mazh

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My solution of weekly competition:

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

start on inverse

B' L' # U' //EO(3)
L' * B' L2 //222+122(3)
& D2 B D2 B R' //L5E3C(5)

# = F2 U' D R2 U D'(6-2)
* = L' B2 D2 F2 R' F2 D2 B2(8-2)
& = R U R' D R U' R' D'(8-1)

IF gave a 25:https://fewestmov.es/cube/if/2c367013b0f45645d94488477b03a064.cube
I tried to find insertion like "R U R' d' R U' R' d" or "M' U' M U" to make 3E or 3C, but all failed.
 
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Bubtore

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Just found a 17 moves solution on a scramble from a yt video


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

R B' U' * // 222 + EO (3)
L U2 L D2 L' // Domino reduction (+5 = 8)
U F2 D F2 U // L4E (+5 = 13)

* U2 R2 D2 R2 U2 L2

Solution // R B' U R2 D2 R2 U2 L' U2 L D2 L' U F2 D F2 U - 17 Moves

 

xyzzy

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dadams weekly #212

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

Found two 31-move solutions.

First one:
U D B' // square (3/3)
(L' B' R' B R2 U L2) // 223 (7/10)
L D' L2 F L // EO (5/15)
(D F D F2) // pseudoF2L-c (4/19)
L' F U2 B' R B U2 F2 L D' // edges; ab3c (10-3/26)
Skeleton: U D B' L D' L2 F2 @ U2 B' R B U2 F2 L D' F2 D' F' D' L2 U' R2 B' R B L
@ = B D2 B' U2 B D2 B' U2 // 3 cancel (8-3/31)

Second one:
B' U' D' F // EO (4/4)
D' R2 L // 222 (3/7)
U' R' U B2 // 223 (4/11)
(R2 U2 R2 U2 R U' R) // F2L (7/18)
(L' U2 L U L' U L) // edges; ab3c (7/25)
Skeleton: B' U' D' F D' R2 L U' @ R' U B2 L' U' L U' L' U2 L R' U R' U2 R2 U2 R2
@ = U' L U R2 U' L' U R2 // 2 cancel (8-2/31)

(IF says there's an insert on the second one that'd cancel 3 moves. Might've missed it because I was rushing through the insertion.)

Is there something nicer that can be done with the second start? 11 moves EO+223 is nice, but I couldn't get a nice continuation from there.

Edit: messing with Cube Explorer, I find this (not very good) 3c skeleton, but it almost looks legit!
B' U' D' F // EO (4/4)
D' R2 L // 222 (3/7)
U' R' U B2 // 223 (4/11)
R' U L U2 R' U2 R U R U R2 L' U2 // magic?? (13/24)
 
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obelisk477

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dadams weekly #212

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

Found two 31-move solutions.

First one:
U D B' // square (3/3)
(L' B' R' B R2 U L2) // 223 (7/10)
L D' L2 F L // EO (5/15)
(D F D F2) // pseudoF2L-c (4/19)
L' F U2 B' R B U2 F2 L D' // edges; ab3c (10-3/26)
Skeleton: U D B' L D' L2 F2 @ U2 B' R B U2 F2 L D' F2 D' F' D' L2 U' R2 B' R B L
@ = B D2 B' U2 B D2 B' U2 // 3 cancel (8-3/31)

Second one:
B' U' D' F // EO (4/4)
D' R2 L // 222 (3/7)
U' R' U B2 // 223 (4/11)
(R2 U2 R2 U2 R U' R) // F2L (7/18)
(L' U2 L U L' U L) // edges; ab3c (7/25)
Skeleton: B' U' D' F D' R2 L U' @ R' U B2 L' U' L U' L' U2 L R' U R' U2 R2 U2 R2
@ = U' L U R2 U' L' U R2 // 2 cancel (8-2/31)

(IF says there's an insert on the second one that'd cancel 3 moves. Might've missed it because I was rushing through the insertion.)

Is there something nicer that can be done with the second start? 11 moves EO+223 is nice, but I couldn't get a nice continuation from there.

Edit: messing with Cube Explorer, I find this (not very good) 3c skeleton, but it almost looks legit!
B' U' D' F // EO (4/4)
D' R2 L // 222 (3/7)
U' R' U B2 // 223 (4/11)
R' U L U2 R' U2 R U R U R2 L' U2 // magic?? (13/24)
After the nice 11 move start, just do U'. That leaves 5c3e, which is just 3 insertions, and IF found a pretty easy to recognize 27
 

Robert Marik

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I miss the most obvious things sometimes! Thanks.
I submitted the same 2x2x3. Then I did orientation with respect to the second axis, as suggested in December in this forum https://www.speedsolving.com/forum/threads/introducing-a-variation-for-fewest-moves.67299/(4 moves). The rest was easy. This produced 29 moves solution with 2x2 edge cycle and 3-corners cycle at the end (could be 28 if the corner cycle is inserted properly). I look forward to see other solution when the competition finishes.
 
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mazh

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Weekly Competition 2018-03
R' U' F B2 D2 B' U2 L2 B2 L2 F' R2 B' D2 R' D2 B R2 U' B' L R U' F2 R' U' F
NISS
B' U B' U B //222
L //EO
B D2 B' D //block
F' D2 F D' //block
F' U' F' U B2 F' D' F' D B2 //ALL
 

G2013

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Just cancelled 7 moves in an insertion... o_O
Scramble: R' U' F B2 R' F2 R2 B2 R F2 R' D2 R F' R U2 R' D' U' F U R' D R' U' F
Inverse: F' U R D' R U' F' U D R U2 R' F R' D2 R F2 R' B2 R2 F2 R B2 F' U R

R B' R' B D F' //Pseudo 2x2x3
[D2 L2 //f2l-2 pseudo
B D B' D' B D B' D' //pair
F L' F' L D' L D L' F' //some sort of VLS?
R F R B2 R' F R B2 R2 //to Backup in 34 moves]

Skeleton: R B' R' B D F' D2 L2 B D B' D' B (F2 D B' D' F2 D B D') D B' D' F L' F' L D' L D L' F' (25)

Final solution: R B' R' B D F' D2 L2 B D B' D' B F2 D B' D' F' L' F' L D' L D L' F' (26!!!!!! 7 moves cancelled!!!!)
Solution was 26 moves, PB is 25... these two attempts were done in a row (with some weeks in between), so I probably got PB mo3 without noticing it :p
Of course the insertion was optimal af, according to IF

EDIT: the solve previous to the 25 was a 35, giving a (35,25,26) = 28.66 PB mo3!
 

guysensei1

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My solutions for Medan Open 2018, 28.67 NR mean :)

1)R' U' F R2 D' F2 L2 U B2 U' B2 R B F U' B2 L B U F2 R B' F R' U' F, 27 moves

(U B' R U2)//2 squares (4/4)
B2 U' F U//2x2x3 (4/8)
(B D B2 L' B)//EO (5/13)
(D2 L)//F2L-1 (2/15)
(D' L D' L')//leaves 3 pairs (4/19)
(D B L B' D' B L' B')//pairs comm (8/27)

Final: B2 U' F U B L B' D B L' B' D' L D L' D L' D2 B' L B2 D' B' U2 R' B U' (27)

Alternate pairs comm that could be used: (B' D L D' L' D' B D), gives same solution length though.

2)R' U' F U2 L2 B' R2 B F2 R2 D' F2 R B F R D' B L' U2 F2 U' R' U' F, 30 moves

L2 F2 D L2//2x2x2 (4/4)
(L B' R' B L' D B R2 D R)//f2l (10/14)
(B' D' B' L B L' D B D')//random LL alg, leaves 3C (9/23)

Skeleton: L2 F2 D L2 D B' D' L B' L' B D * B R' D' R2 B' D' L B' R B L'

*= D2 L U L' D2 L U' L'

Final: L2 F2 D L2 D B' D' L B' L' B D' L U L' D2 L U' L' B R' D' R2 B' D' L B' R B L' (30)

A bit disappointed that only 1 cancels, and it was optimal too. Also disappointed that I got such a poor solution considering the 14 move F2L. Can anyone find a better solution with my F2L?

3)R' U' F L2 D2 U' R2 F2 R2 U' F2 D2 F U' L2 B' U2 B' D B D R' B' F2 R' U' F, 29 moves

(R' D2 B)//square (3/3)
(B' F R F B)//another square+preserve pair (5-2/8-2)
(U B' R B)//pseudo 2x2x2+2 squares (4/10)
U2//pseudo correction (1/11)
(U R2 D R D')//make a 1x2x3 (5/16)
(R U' R2 D R')//F2L (5/21)
(R' B2 R2 D' R' D R' B2 R)//LL (9-1/30-1)

Final: U2 R' B2 R D' R D R2 B2 R2 D' R2 U R' D R' D' R2 U' B' R' B U' B' F' R' F' D2 R (29)

Found this in the first 15 minutes but didnt find anything better in the remaining time...
 

mazh

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My solutions for Medan Open 2018, 28.67 NR mean :)

1)R' U' F R2 D' F2 L2 U B2 U' B2 R B F U' B2 L B U F2 R B' F R' U' F, 27 moves

(U B' R U2)//2 squares (4/4)
B2 U' F U//2x2x3 (4/8)
(B D B2 L' B)//EO (5/13)
(D2 L)//F2L-1 (2/15)
(D' L D' L')//leaves 3 pairs (4/19)
(D B L B' D' B L' B')//pairs comm (8/27)

Final: B2 U' F U B L B' D B L' B' D' L D L' D L' D2 B' L B2 D' B' U2 R' B U' (27)

Alternate pairs comm that could be used: (B' D L D' L' D' B D), gives same solution length though.

2)R' U' F U2 L2 B' R2 B F2 R2 D' F2 R B F R D' B L' U2 F2 U' R' U' F, 30 moves

L2 F2 D L2//2x2x2 (4/4)
(L B' R' B L' D B R2 D R)//f2l (10/14)
(B' D' B' L B L' D B D')//random LL alg, leaves 3C (9/23)

Skeleton: L2 F2 D L2 D B' D' L B' L' B D * B R' D' R2 B' D' L B' R B L'

*= D2 L U L' D2 L U' L'

Final: L2 F2 D L2 D B' D' L B' L' B D' L U L' D2 L U' L' B R' D' R2 B' D' L B' R B L' (30)

A bit disappointed that only 1 cancels, and it was optimal too. Also disappointed that I got such a poor solution considering the 14 move F2L. Can anyone find a better solution with my F2L?

3)R' U' F L2 D2 U' R2 F2 R2 U' F2 D2 F U' L2 B' U2 B' D B D R' B' F2 R' U' F, 29 moves

(R' D2 B)//square (3/3)
(B' F R F B)//another square+preserve pair (5-2/8-2)
(U B' R B)//pseudo 2x2x2+2 squares (4/10)
U2//pseudo correction (1/11)
(U R2 D R D')//make a 1x2x3 (5/16)
(R U' R2 D R')//F2L (5/21)
(R' B2 R2 D' R' D R' B2 R)//LL (9-1/30-1)

Final: U2 R' B2 R D' R D R2 B2 R2 D' R2 U R' D R' D' R2 U' B' R' B U' B' F' R' F' D2 R (29)

Found this in the first 15 minutes but didnt find anything better in the remaining time...
My solutions for this competition. 26 26 29=27.00

1)R' U' F R2 D' F2 L2 U B2 U' B2 R B F U' B2 L B U F2 R B' F R' U' F , 26 moves
NISS
U B2 R U2 // pseudo-222+2x122(4/4)
D' L' D' L2 B2 // DEC(5/9)
NISS
F2 L F L' // TEC(4/13)
D' F' D // L3E4C(3/16)

Skeleton: * F2 L F L' D' F' D B2 L2 D L # D U2 R' B2 U'
# = B' F' U' F U B (6/22)
* = L B' L' F2 L B L' F2 (8-4/30-4)

2)R' U' F U2 L2 B' R2 B F2 R2 D' F2 R B F R D' B L' U2 F2 U' R' U' F , 26 moves
L2 F2 D L2 // 222(4/4)
R B' D' B' R // DEC(5/9)
B2 L' D # L D' // TEC(5/14)
U' * B2 U B' // L5C(4/18)
# = D R' D' L D R D' L' (8-4/26-4)
* = U' F' U B2 U' F U B2 (8-4/30-4)

And anthor better skeleton(but IF says 20+8-1):
L2 F2 D L2 // 222(4/4)
NISS
L B' R' L' // block(4/8)
B D B R B R2 // block(6/14)
B' D B' D' B D2 // L3C(6/20)

3)R' U' F L2 D2 U' R2 F2 R2 U' F2 D2 F U' L2 B' U2 B' D B D R' B' F2 R' U' F , 29 moves
NISS
R' # L' F' L' F' B2 // 2x122(6/6)
D2 R U F' * U' F D R' // DEC+122(8/14)
U' B U' B' U' // L5C(5/19)
# = B2 L' F2 L B2 L'F2 L (8-3/27-3)
* = F2 U' B U F2 U' B' U (8-3/32-3)
 
D

Daniel Lin

Guest
I wanna get good at FMC so I can be cool like @Cale S :D. I currently suck, but I got a pretty lucky solve with no NISS. Tied PB

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

U2 L' U B' R' D R // 2x2x2 (7/7)
F2 U2 F L' U L' // F2L - 1 (6/13)
F2 L' F' L F' L' F L // 3c (8/21)

Skeleton: * U2 L' U B' R' D R F2 U2 F L' U L' F2 L' F' L F' L' F L
* = [U B U', F2] (8)

Solution: U B U' F2 U B' U' F2 U2 L' U B' R' D R F2 U2 F L' U L' F2 L' F' L F' L' F L (29)
 
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