[Help Thread] FMC Discussion

[Kit Clement]

I love it - hopefully we can gain some empirical evidence for expected numbers soon :3

A couple quick things about the form:
- Given that there's distinction for the various kinds of L4C, as well as distinction between the 2-comm L6C and the 3-comm L6C, why is there no similar separation for L5C? For example, I got a pair of 2-cycles plus a twist in a solve I did earlier this week, and I feel like it probably shouldn't be lumped with 2-comm L5Cs.
- I feel like having a comments section would be nice to note anything different you did. (As an easy example, in the L5C I did earlier, I used triple sexy to solve 3 corners and cancel 5 moves, which allowed me to beat IF's estimate.)
- For EO, does this refer to having an explicit EO step at any point in the solve, or only if you do it at the beginning? I often do EO Petrus-style, after 2x2x3; does that count?

Added the option for L5C with three insertions, as well as a comment section.

I also added more options for the EO question as well to differentiate between going for EO at the very start and going for EO in the middle.

 

[not_kevin]

Added the option for L5C with three insertions, as well as a comment section.

I also added more options for the EO question as well to differentiate between going for EO at the very start and going for EO in the middle.

Great - thanks!

A very very minor note - it looks like 3e4c (pure corner 2-cycles) got placed funny, as did L5C (3 comms) - the 3e4c is after L5C (5-cycle), while L5C is after 3e5c.

Also, maybe have an "other" category, for other kinds of insertions? (Thinking of 4 twisted corners, 4e4c (pure 4-cycles or pure 2-cycles), etc.)

 

[obelisk477]

If you go from a 2x2x3 to L6C, what are the odds that those corners are solvable in two insertions (so two three-cycles)?
'

 

[not_kevin]

If you go from a 2x2x3 to L6C, what are the odds that those corners are solvable in two insertions (so two three-cycles)?
'

If you're just considering the probability that being left with 6 pre-defined corners is going to be solvable in 2 insertions (that is, your extension from 2x2x3 to L6C doesn't influence the distribution of possibilities for L6C other than specifying which two corners are solved), then it's a "basic" counting problem.

The number of two-comm L6C cases is equal to choosing which two sets of corners you have (6 choose 3 = 20), choosing the direction of each cycle (2 * 2 = 4), and choosing the twisted-ness of each cycle's corners (3*3 for each cycle, because they must both be completely solvable on their own, for a total multiplier of 81), so there are 20 * 4 * 81 = 6480 two-comm L6C.

Calculating the number of total L6C cases is a bit more tricky, but there a couple ways of doing it. The one I find easiest is using the inclusion-exclusion principle, where I can isolate what I want by overcounting and undercounting carefully:

• There are 6! * 3^5 / 2 = 87480 ways to throw those 6 corners anywhere in the puzzle. Obviously, this number is much larger than the one we want (for example, it includes the case where all the corners are solved). So we subtract the total number of cases where at least one piece is solved:
• There are 6 * 5! * 3^4 / 2 = 29160 ways to make sure that when you throw those 6 corners in, one is guaranteed to be solved. The first 6 represents which corner we want to be solved; the 5! * 3^4 / 2 is the total number of ways to throw the remaining 5 corners. But we've definitely overcounted here - eg, the case where all corners are solved is counted 6 times (once for each of its solved corners). So we have to add in the cases where at least 2 pieces are solved:
• There are (6 choose 2) * 4! * 3^3 / 2 = 4860 cases for guaranteeing at least two solved pieces (6 choose 2 for choosing which two are solved, and 4! * 3^3 / 2 for (you guessed it ) throwing in the rest. But again we've overcounted, so onto guaranteeing 3 pieces:
• There are (6 choose 3) * 3! * 3^2 / 2 = 540 cases for guaranteeing 3 solved pieces,
• And (6 choose 4) * 2! * 3^1 / 2 = 45 for guaranteeing 4 solved pieces,
• And (6 choose 5) * 1! * 1 = 6 for guaranteeing 5 solved pieces,
• And finally (6 choose 6) = 1 way to guarantee all 6 pieces solved.

So, we combine: 87480 - 29160 + 4860 - 540 + 45 - 6 + 1 = 62680 total L6C cases.

Therefore, the probability is 6480 / 62680 ~ 10.34%.

 

[not_kevin]

My winning solution from Utah Megacomp 2015:

Scramble: L2 D2 U2 F2 L2 B' D2 B F2 U' L2 R' D' F' R' F2 D' L2
Solution: D L U' L' D' L U L2 B2 L' F U R' D2 R' U R D2 R U2 R2 U2 R2 D (24 moves)

Spoiler

L' B2 L' // EO + 1x1x3 [3/3]
F // wait why are all the edges solving [1/4]
U R2 // build 1x2x3 pseudo 2x-cross [2/6]
U R2 U2 R2 U2 R2 D // solve edges while preserving UB(l) pair, L5C [7/13]

Skeleton: (2) L' B2 L' F U R2 (1) U R2 U2 R2 U2 R2 D
Insert at (1): R D2 R' U' R D2 R' U' // cancels 4
Insert at (2): D L U' L' D' L U L' // cancels 1

I found that this scramble was pretty bad for block-building, so I decided to try some EO stuff. F2/B2 EO can be done with L2 F' (which also makes a pair), but I didn't find anything much out of that; around 15 or 20 minutes, I found L' B2 L to give L2/R2 EO with a block. This led into some cute skeletons, but nothing great (a whole bunch of 24 -> L4C, for example). Around 30 minutes in, I accidentally did L' B2 L (note to self: EO is really flexible sometimes ), and was liek "uh k", and just kinda went with it. The more intuitive finish (doing L' B2 L' -> F -> U' R2 U2 R2 D U' for 10 -> L6C [4 + 2]) was a bit too ambitious for me to play with, since I didn't immediately find an amazing cancel anywhere. IF says that it's optimally a 23, 'tho

But based on this skeleton, I'm wondering if there's some way to finish it with two-phase or something...

 

[okayama]

My solution for FMC round 89, missing in the site.

Scramble: U' B2 R' F2 U' B' R U' F2 D' B L2 F2 U2 B D' U2 B D2 L'
Solution: B2 U2 F R' U' B2 F2 L U' L U L2 U' L2 U2 L U2 L U' L F2 U2 L2 (23 moves)

Spoiler

(Normal)

Orient edges: B2 U2 F
2x2x2 block: R' U' B2

(Inverse)
Pre-scramble: U F2 B2 U R F' U2 B2

2x2x3 block: L2 U2 F2
Finish F2L: L' U L' U2 L' U2
LL: L2 U L2 U' L' U L' U'
Correction: U F2 B2 U R F' U2 B2

 

[Cale S]

My winning solution from Utah Megacomp 2015:

Scramble: L2 D2 U2 F2 L2 B' D2 B F2 U' L2 R' D' F' R' F2 D' L2
Solution: D L U' L' D' L U L2 B2 L' F U R' D2 R' U R D2 R U2 R2 U2 R2 D (24 moves)

Spoiler

L' B2 L' // EO + 1x1x3 [3/3]
F // wait why are all the edges solving [1/4]
U R2 // build 1x2x3 pseudo 2x-cross [2/6]
U R2 U2 R2 U2 R2 D // solve edges while preserving UB(l) pair, L5C [7/13]

Skeleton: (2) L' B2 L' F U R2 (1) U R2 U2 R2 U2 R2 D
Insert at (1): R D2 R' U' R D2 R' U' // cancels 4
Insert at (2): D L U' L' D' L U L' // cancels 1

I found that this scramble was pretty bad for block-building, so I decided to try some EO stuff. F2/B2 EO can be done with L2 F' (which also makes a pair), but I didn't find anything much out of that; around 15 or 20 minutes, I found L' B2 L to give L2/R2 EO with a block. This led into some cute skeletons, but nothing great (a whole bunch of 24 -> L4C, for example). Around 30 minutes in, I accidentally did L' B2 L (note to self: EO is really flexible sometimes ), and was liek "uh k", and just kinda went with it. The more intuitive finish (doing L' B2 L' -> F -> U' R2 U2 R2 D U' for 10 -> L6C [4 + 2]) was a bit too ambitious for me to play with, since I didn't immediately find an amazing cancel anywhere. IF says that it's optimally a 23, 'tho

But based on this skeleton, I'm wondering if there's some way to finish it with two-phase or something...

You can get to L6C in a slightly different way for a 22 move solution:

Spoiler

L' B2 L' // EO
F D U2 B2 U2 B2 // to L6C

with IF insertions:
L' B L' F L2 F2 L' B L F2 L2 D' U2 B' U' B D2 B' U B' U2 B2 (22)

 

[Wilhelm]

My 25 from German Nationals

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

Pre: D' (1/1)
EO: R2 D R (3/4)
2x2x2: U' F U (3/7)
2x2x3: D2 R2 D' R2 (4/11)
To L5C: D' F' D F' D2 F' (6/17)

Skeleton: D F D2 F D' F D R2 D* R2 D2 U' F' U# R' D' R2

*= D' L D R2 D' L' D R2 (8-5)
#= U' F' D2 F U F' D2 F (8-3)

Solution: D F D2 F D' F D R2 L D R2 D' L' D' U' F2 D2 F U F' D2 F R' D' R2
Optimal Insertions

I'm really happy with the solution. Overall I got pretty lucky ^^. I think the 2x2x3 is quite obvious and I was surprised that nobody else used it. The continuation initially was a meant to be a backup solution but ended up being the best option I could find. At first I didn't count how many moves I did to get to L5C. I was going to build a Pseudo F2l-1 and wrote down a D' instead of a D. The Pseudo F2L's that I tried were terrible because they left me with L5C with a twisted corner in either 17 or 18 moves if I remember correctly. Switching didn't help.
When I started inserting after roughly 30 minutes I wasn't expecting too much because I needed to cancel 4 moves to tie my old PB. When I found the 5 move cancel I had 15 minutes left and wanted to play it safe and started doing my second insertion. At the time I had a 28 move solution without any further cancellations. I was really happy when I found a 1 move cancel right at the beginning. I was pretty sure that I wouldn't find much more as I only had 5 moves left to check. I was hoping to podium with that. But then I tried to cancel two moves at the U which didn't work at first but then I managed to do it. I was really excited and when writing down my solution I found that I could cancel even 1 more move ^^.
I didn't expect to win at all because I think the first part of my solution was easy to find and I didn't think I would cancel 8 moves! I just recently started doing FMC again and beat my overall PB by one move with this. I probably won't beat the 25 in a long time . Now I got to work on my official mean

 

[not_kevin]

You can get to L6C in a slightly different way for a 22 move solution:

Spoiler

L' B2 L' // EO
F D U2 B2 U2 B2 // to L6C

with IF insertions:
L' B L' F L2 F2 L' B L F2 L2 D' U2 B' U' B D2 B' U B' U2 B2 (22)

Yeah, Kit showed the scramble to Mark, who found the same skeleton. I feel dumb for not seeing that

 

[Joey VOV]

Utah Megacomp
Scramble: L2 D2 U2 F2 L2 B' D2 B F2 U' L2 R' D' F' R' F2 D' L2
Inverse: L2 D F2 R F D R L2 U F2 B' D2 B L2 F2 U2 D2 L2

on normal:
premoves: F D U B2 U'
B D B' D L2 / 223 + 221 BLOCK (10)
R D2 R' B2 / TO L5C (14)

skeleton: B D ^B' D L2 R D2 R' B2 F U D B2 U'
insert at ^: U' F2 U B' U' F2 U B
updted skeleton: B D U' F2 U B' U' F2 U D L2 R D2 R' *B2 F U D B2 U'
Insert at *: R' B' L' B R B' L B

final solution: B D U' F2 U B' U' F2 U D L2 R D2 R2 B' L' B R B' L B' F U D B2 U' (26)

Very happy with the skeleton, VERY unhappy with the insertions. 32 different ways to cancel 4 moves, but no more than 4, so my 26 is optimal for a 14 move skeleton :/

 

[AlphaSheep]

I got this 29 move solution on the African Cubing facebook group's comp a couple weeks ago, which I was quite proud of at the time. I thought I'd post it here for some critique and advice.

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

F2 D' F U F' // 2x2x2
U R' B' U (R') // 2x2x3
(R) B' R' (B') // F2L-1
(B) L U2 L' B' // EO
R' U R // Noticed the 3 edge + 3 corner cycles to finish. Wasn't intentional, but went with it

Skeleton: * F2 D' F U F' U R' B' U B' R' L U2 L' B' R' U R (18)

Insert at *: D2 ^ R' L B2 R L' (only one I tried)
Insert at ^: D R' U2 R D' R' U2 R (Cancels 3 moves... Last one I tried since I start from the end and work backward through the skeleton)

Solution: D' R' U2 R D' R' U2 L B2 R L' F2 D' F U F' U R' B' U B' R' L U2 L' B' R' U R (29)

 

[CyanSandwich]

Weekly FMC spoilers:

Spoiler

Got a 24 mover. PB by 5 moves.

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

inv
D U F' L2 R2 F2 R2 U' //2x2x3 (8)

switch
F' R2 F R' U' R' D R2 D' U F2 //L3C (11/19)

Skeleton: F' R2 F R' U' R' D R2 D' U F2 (1) U R2 F2 R2 L2 F U' D'

(1) = F' D' F U F' D F U'

Solution: F' R2 F R' U' R' D R2 D' U F D' F U F' D F R2 F2 R2 L2 F U' D'

 

[porkynator]

Weekly FMC spoilers:

Spoiler

Got a 24 mover. PB by 5 moves.

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

inv
D U F' L2 R2 F2 R2 U' //2x2x3 (8)

switch
F' R2 F R' U' R' D R2 D' U F2 //L3C (11/19)

Skeleton: F' R2 F R' U' R' D R2 D' U F2 (1) U R2 F2 R2 L2 F U' D'

(1) = F' D' F U F' D F U'

Solution: F' R2 F R' U' R' D R2 D' U F D' F U F' D F R2 F2 R2 L2 F U' D'

Spoiler

You could use one fewer move for L3C; after switching, very similar to your solution:
F' R2 F R' U'
R U D' F2 D
But IF says you would still get a 24

 

[CyanSandwich]

Spoiler

You could use one fewer move for L3C; after switching, very similar to your solution:
F' R2 F R' U'
R U D' F2 D
But IF says you would still get a 24

Oh cool. I feel silly for missing that, but it might have been worse if I didn't.

 

[okayama]

My winning solution from Tohoku Open 2015.

Scramble: R' D2 F2 R U2 B2 L2 B2 R2 B U F U2 F' D2 R2 D' U' B' U
Solution: L' R B' L B R B' L' F2 L D2 U F' D' U2 L U' F' L F2 L' F' L' F' L' F2 L' (27 moves)

Spoiler

30 min backup solution.

2 c/e pairs: L'
1st square: R2 * F2
2nd square: B' L
2x2x3 block: D U2
Orient edges: D U' F' D' U2 L U'
All but 3 corners: F' L F2 L' F' L' F' L' F2 L'

Insert at *: R' B' L B R B' L' B

On inverse, F2 or L2 also makes 2 c/e pairs, but I couldn't find any better solution.

 

[Cale S]

23 move solution for the ongoing weekly FMC, one move from pb

Spoiler

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

(L' D2) // square
(F U' F2) // 2x2x2
D' L' B2 // 2x2x3
(F' D' F) // EO
D' L' D2 L2 F' D' F // F2L
F' L' D' L D F // to L3C

skeleton: D' * L' B2 D' L' D2 L2 F' D' L' D' L D2 F' U F' D2 L
insert at *: [F, L' B L] // L3C, cancels 3

Final solution: D' F L' B L F' L' B D' L' D2 L2 F' D' L' D' L D2 F' U F' D2 L (23)


Interestingly, both my 22 pb and this both have no R moves in the solution...

also makes a 25.33 mo3: 27, 26, 23
and a 27.00 avg5: 28, 28, 27, 26, 23 (hopefully I can roll this next week)

 

[Mark Boyanowski]

Now that I've gotten into FMC a lot more I've been looking at scrambles from comps I was at before I knew all the techniques I know now. I remembered the third scramble from Another FMC 2015, which had a 2x2x1 already done. I got a 30 on it officially, here's what I got just now (moves in parenthesis done on inverse):

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

L // 2x2x2
F' R F' R2 // 2X2X3
F // Pseudo F2L -1
(F) // Fix pseudo
(L' U2 L2 F' L' F) // EO
(F U F') // 3C

Skeleton (1 move cancels with NISS): L F' R F' R2 F2 U' F2 L % F L2 U2 L F'

% = U B' U' F U B U' F'

Final solution, 20 moves: L F' R F' R2 F2 U' F2 L U B' U' F U B U' L2 U2 L F'

Not sure if I'm happy or really upset

 

[Cale S]

Haha I got 23 for both of my weekly competitions this week

Spoiler

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

(U' L') // EO
(F R2) // square
R2 D // 2x2x2
(L2 B2 U' F U2 F') // 2x2x3
(U B2) // F2L - 1
(U' B' U2 B U' B' U2 B) // to L3C

skeleton: R2 D B' U2 B U B' U2 B U B2 U' F U2 % F' U B2 L2 R2 F' L U
%=[U B2 U', F'] // L3C, cancels 7

final solution: R2 D B' U2 B U B' U2 B U B2 U' F U' B2 U' F' U L2 R2 F' L U (23)


Initially I did a block comm on normal to get to L3C which resulted in a 27, but then I tried this (inserting the pair and cancelling into a sune) which was a different 22 move skeleton giving a much better insertion.

My past 3 attempts have been 26, 23, 23 = 24.00 mean (also would make a 25.33 avg5 and 27.2 avg12) but I'm only counting Weekly FMC solves for my average pb's

 

[EMI]

My 25 from German Nationals

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

Pre: D' (1/1)
EO: R2 D R (3/4)
2x2x2: U' F U (3/7)
2x2x3: D2 R2 D' R2 (4/11)
To L5C: D' F' D F' D2 F' (6/17)

Skeleton: D F D2 F D' F D R2 D* R2 D2 U' F' U# R' D' R2

*= D' L D R2 D' L' D R2 (8-5)
#= U' F' D2 F U F' D2 F (8-3)

Solution: D F D2 F D' F D R2 L D R2 D' L' D' U' F2 D2 F U F' D2 F R' D' R2
Optimal Insertions

I'm really happy with the solution. Overall I got pretty lucky ^^. I think the 2x2x3 is quite obvious and I was surprised that nobody else used it.

I just checked my solution sheet, and I did actually find your start in the last five or so minutes.

On inverse:
Your 2x2x3 + EO: R2 D R U' F U D2 R2 D' R2
F2l-ish: D' F' D' F2 D'

I didn't have any time left to look further into this, so I went with my backup solution:

Start on normal,
Pseudo-2x2x2: D R' @ F U' F
2x2x3: D' R2 D R2 U
Switch, Pseudo-F2L + EO: F' R D' F D F
Switch, L3C: F2 D' F' R' F R D

Lucky insertion: @ = R B R' F R B' R' F'

Total: D B R' F R B' R' U' F D' R2 D R2 U F2 D' F' R' F R D F' D' F' D R' F (27)

 

[okayama]

My solution from Japan Open 2015 (got 2nd place).

Scramble: R2 U2 B2 L B2 D2 U2 R2 D2 U' F2 U F R2 D' B' D2 R D2
Solution: F' L U L2 F' U2 D' B2 D F2 D' B2 D F' U F' U F' U F U' F U' F' U' L F2 L2 U2 D2 F2 D' R' (33 moves)

Spoiler

20 min backup solution.

(Normal)
Pre-scramble: L' D2 F2 D' R'

2x2x2 block: (ready)
2x2x3 block: F' L U L2

(Inverse)
Pre-scramble: L2 U' L' F

2x2x2 block: R D F2 D2
2x2x3 block: L
Orient edges: L' U2 L2 F2 L'
Finish F2L: U F U F' U F' U' F2
All but 3 corners: F' U' F U' F' * U2 F
Correction: L2 U' L' F

Insert at *: F2 D' B2 D F2 D' B2 D

In the remaining time, I investigated the following start:

(Inverse)
1st square: D' L R' F'
Orient edges: D
c/e pair: L'

[1] D2 R' B L2 B' R2 D2 R L U2
[2] D2 B R2 F R L2 B2 L
[3] D2 B R2 F R U2 R' L' U2

but I couldn't finish any of them.
After [3], R B' R B' leaves 5 corners, which should have resulted in 29 moves.

EDIT:
After the competition, Fumiya Matsui found the following skeleton almost linearly.

Spoiler

(Inverse)

2x2x2 block: R D F2 D2 (he didn't find this in the competition, but Yuki Tanaka told him)
2 c/e pairs: F U2 F2
More square: U' F L' F2
Expand the block: B
Finish F2L: L B' L2 F2 R' F R L'
All but 3 corners: L U * F U' F' L'

Insert at *: U B U' F U B' U' F'

which results in 28 moves.