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[Help Thread] FMC Discussion

I don't think it's reasonable to expect a sub-30 with Fridrich.
I think it is and I base this on experience. I use Fridrich and I think more than half of the solutions I've ever done in my entire life (about 6/10) were sub 30.
Wow. That's better than what most can do using FMC techniques. I take that back, then.

I actually went back and checked my FMC submissions and I was overestimating a bit. It's probably actually a bit less than half (4/11 or something). I edited my other post.

Please allow me do doubt your claims about fridrich for FMC. Pure fridrich with 2-look LL would take for the last layer alone about 20 turns. Clearly you do a non-standard fridrich f2l in order to achieve such low move counts:cool: In fact i strongly doubt that ANY standard speed cubing method could reach so low move counts on average ...

Per
 
I don't think it's reasonable to expect a sub-30 with Fridrich.
I think it is and I base this on experience. I use Fridrich and I think more than half of the solutions I've ever done in my entire life (about 6/10) were sub 30.
Wow. That's better than what most can do using FMC techniques. I take that back, then.

I actually went back and checked my FMC submissions and I was overestimating a bit. It's probably actually a bit less than half (4/11 or something). I edited my other post.

Please allow me do doubt your claims about fridrich for FMC. Pure fridrich with 2-look LL would take for the last layer alone about 20 turns. Clearly you do a non-standard fridrich f2l in order to achieve such low move counts:cool: In fact i strongly doubt that ANY standard speed cubing method could reach so low move counts on average ...

Per

I normally make a extended or (or double extended) cross, then mess about with the F2L pairs until I can either engineer an OLL skip or something that will lead to an easy LL. If possible, I also use insertions if I have a 3 cycle or something. Sometimes I don't put the last edge/corner in and include it in my LL insertion or something like that too, but I don't think I've ever done a 2LLL so maybe it's not pure Fridrich. Here's one of my best ever solutions as an example.

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

Cross on R

2x2x1 (3): D2 B D2
Double X-Cross (6): U' R' F2 D' F R'
3rd Pair (7): U' L' U L' B' L B
Last Pair + LL (12): B L B' L' U' L U L' U' L2 U L2
B + B cancels to B2 to give
D2 B D2 U' R' F2 D' F R' U' L' U L' B' L B2 L B' L' U' L U L' U' L2 U L2 (27)
 
Ok, here's a scramble for you FMC guys. It's the one from the UK Masters:
L B2 F2 R2 D2 R' F2 L U2 R' F2 L' F L2 B2 F' R' B2 F' U' L2

I thought this scramble was horrible, and I'm pretty annoyed my only competetion FMC solution was sup 30.. my solution was pretty poor. Can any of you find a better one?

My solution was:
X-Cross: x2 U L' U2 F L' F' D R D2 F2 (10/10)
2nd Pair: U' L' U2 L U2 L' U2 L (8/18)
3rd Pair: U2 y R U' R' (4/22)
4th Pair: U' R' U' R (4/26)
LL: R U R' U R U2 R' U' (8-1/33)



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

U2 B2 U B2 U' R D F2 U2 block 2x3 (9)
D2 B F' D'* F Pse-f2l minus 1 slot (14)
B2 L B L2 D L D2 leave 3 corner (21)
*insert: [DR2D',L'] (27)

I love this 'horrible' scramble:D, 27 move is my PB. I have done some weeks competition, only get two 29 move solution with lucky two 3-cycles insetion.
I think this is better in block building,which is my weaknesses.
 
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fmc:U2 B2 U B2 U' R D F2 U2 D2 B F' R2 D' L' D R2 D' L F B2 L B L2 D L D2 (27)

U2 B2 U B2 U' R D F2 U2 block 2x3 (9)
D2 B F' D'* F Pse-f2l minus 1 slot (14)
B2 L B L2 D L D2 leave 3 corner (21)
*insert: [DR2D',L'] (27)

May I know how you came up with this? I've been trying it over and over again but all I see is just random moves and suddenly there are just 3 corners left :eek:
 
This week's scramble 5 for 3x3x3 speedsolve: U2 R2 U' R2 D B2 D2 U R2 U F2 R' B F R2 D B R' F' L B2

1x2x3: B U R B2 R B2 L (7/7)
Opposite 1x2x2: D' F' D' (3/10)
3 corners and 5 edges left: F D' . F D (3/13)
Corner 3-cycle insertion at .: D L D' R' D L' D' R (7/20)

I need help with inserting the remaining 5 edges. Anyone got a good solution?

(One of my shortest skeletons so far :))
 
fmc:U2 B2 U B2 U' R D F2 U2 D2 B F' R2 D' L' D R2 D' L F B2 L B L2 D L D2 (27)

U2 B2 U B2 U' R D F2 U2 block 2x3 (9)
D2 B F' D'* F Pse-f2l minus 1 slot (14)
B2 L B L2 D L D2 leave 3 corner (21)
*insert: [DR2D',L'] (27)

May I know how you came up with this? I've been trying it over and over again but all I see is just random moves and suddenly there are just 3 corners left :eek:

it's Pse-f2l, premove D' will help you, it's so easy ending like RUR'...
I am lazy for scrambleing again, so just do it without premove to see whether
anything good exists. Seems I am lucky this time:)
 
fmc:U2 B2 U B2 U' R D F2 U2 D2 B F' R2 D' L' D R2 D' L F B2 L B L2 D L D2 (27)

U2 B2 U B2 U' R D F2 U2 block 2x3 (9)
D2 B F' D'* F Pse-f2l minus 1 slot (14)
B2 L B L2 D L D2 leave 3 corner (21)
*insert: [DR2D',L'] (27)

May I know how you came up with this? I've been trying it over and over again but all I see is just random moves and suddenly there are just 3 corners left :eek:

Think of it as just trying to solve the edges, and the corners miraculously come out close. If you think of it as trying to solve the edges, the moves in bold all make pretty much sense. At least, that's how I might have come up with this. I could see myself getting this, but the corners being just a 3-cycle would be lucky. I often try this on solves, though, just to see how bad the corners come out.
 
fmc:U2 B2 U B2 U' R D F2 U2 D2 B F' R2 D' L' D R2 D' L F B2 L B L2 D L D2 (27)

U2 B2 U B2 U' R D F2 U2 block 2x3 (9)
D2 B F' D'* F Pse-f2l minus 1 slot (14)
B2 L B L2 D L D2 leave 3 corner (21)
*insert: [DR2D',L'] (27)

May I know how you came up with this? I've been trying it over and over again but all I see is just random moves and suddenly there are just 3 corners left :eek:

Think of it as just trying to solve the edges, and the corners miraculously come out close. If you think of it as trying to solve the edges, the moves in bold all make pretty much sense. At least, that's how I might have come up with this. I could see myself getting this, but the corners being just a 3-cycle would be lucky. I often try this on solves, though, just to see how bad the corners come out.

yes, I don't kown many LL algs to manage all the pieces especially corners, only 4 OLL and 4 Pll...:(

So I just try to solve egdes in accepted moves, sometimes insert 1-2 moves to get lucky. Then all I do is pray:o
 
Ok, here's a scramble for you FMC guys. It's the one from the UK Masters:
L B2 F2 R2 D2 R' F2 L U2 R' F2 L' F L2 B2 F' R' B2 F' U' L2
WOOT! 33-move solution in 27 minutes! (Incidentally, that's a 33-move solution with 33 minutes left to go :p) This is easily my best solution by far in terms of its (solution length) : (time taken) ratio :D

Inverse scramble: L2 U F B2 R F B2 L2 F' L F2 R U2 L' F2 R D2 R2 F2 B2 L'

1x2x3: F2 L2 U' L D2 B2 (6/6)
Opposite 1x2x3: B R2 B' L2 R U' R (6/12)
CxLL: R U R' U R U2 R' (6/18)
All but 6 edges: U2 . L' (2/20)
Corner 3-cycle insertion at .: U R L' B2 .. R' L U (6/26)
Corner double-2-cycle insertion at ..: B R2 L2 D2 R2 L2 U2 B' (7/33)

Inverse solution: F2 L2 U' L D2 B' R2 B' L2 R U' R2 U R' U R U2 R' U' R L' B' R2 L2 D2 R2 L2 U2 B' R' L U L'
Solution: L U' L' R B U2 L2 R2 D2 L2 R2 B L R' U R U2 R' U' R U' R2 U R' L2 B R2 B D2 L' U L2 F2 (33, it's 29 STM by the way :))

Isn't that an edge cycle? You might want to correct that. Great solution though!
 
Ok, here's a scramble for you FMC guys. It's the one from the UK Masters:
L B2 F2 R2 D2 R' F2 L U2 R' F2 L' F L2 B2 F' R' B2 F' U' L2
WOOT! 33-move solution in 27 minutes! (Incidentally, that's a 33-move solution with 33 minutes left to go :p) This is easily my best solution by far in terms of its (solution length) : (time taken) ratio :D

Inverse scramble: L2 U F B2 R F B2 L2 F' L F2 R U2 L' F2 R D2 R2 F2 B2 L'

1x2x3: F2 L2 U' L D2 B2 (6/6)
Opposite 1x2x3: B R2 B' L2 R U' R (6/12)
CxLL: R U R' U R U2 R' (6/18)
All but 6 edges: U2 . L' (2/20)
Corner 3-cycle insertion at .: U R L' B2 .. R' L U (6/26)
Corner double-2-cycle insertion at ..: B R2 L2 D2 R2 L2 U2 B' (7/33)

Inverse solution: F2 L2 U' L D2 B' R2 B' L2 R U' R2 U R' U R U2 R' U' R L' B' R2 L2 D2 R2 L2 U2 B' R' L U L'
Solution: L U' L' R B U2 L2 R2 D2 L2 R2 B L R' U R U2 R' U' R U' R2 U R' L2 B R2 B D2 L' U L2 F2 (33, it's 29 STM by the way :))

Isn't that an edge cycle? You might want to correct that. Great solution though!
Oh yeah, thanks for spotting that :)
 
All 6-move edge 5-cycles:

U R U' D B' D' (6f*)
U R L' B' R' L (6f*)
U R' U' D B D' (6f*)
U R' L F' R L' (6f*)
U F U' D R' D' (6f*)
U F B' R' F' B (6f*)
U F' U' D R D' (6f*)
U F' B L' F B' (6f*)
U D' R U' D B' (6f*)
U D' R' U' D B (6f*)
U D' F U' D R' (6f*)
U D' F' U' D R (6f*)
U D' L U' D F' (6f*)
U D' L' U' D F (6f*)
U D' B U' D L' (6f*)
U D' B' U' D L (6f*)
U L U' D F' D' (6f*)
U L' U' D F D' (6f*)
U B U' D L' D' (6f*)
U B' U' D L D' (6f*)
U' R U D' F' D (6f*)
U' R L' B R' L (6f*)
U' R' U D' F D (6f*)
U' R' L F R L' (6f*)
U' F U D' L' D (6f*)
U' F B' R F' B (6f*)
U' F' U D' L D (6f*)
U' F' B L F B' (6f*)
U' D R U D' F' (6f*)
U' D R' U D' F (6f*)
U' D F U D' L' (6f*)
U' D F' U D' L (6f*)
U' D L U D' B' (6f*)
U' D L' U D' B (6f*)
U' D B U D' R' (6f*)
U' D B' U D' R (6f*)
U' L U D' B' D (6f*)
U' L' U D' B D (6f*)
U' B U D' R' D (6f*)
U' B' U D' R D (6f*)

I looked deeper into them, and found that there are 3 types of "algorithms":

1. S F S' F'
2. F S F' S'
3. Conjugated S F S' F'
(S = quarter slice turn; F = quarter face turn, which obviously cannot be parallel to S)

The first two are pretty straightforward. I found the third one pretty interesting, I'll describe it with an example:

U D' R U' D B' is a Type 1 "algorithm". Using D as a setup move, we get D (U D' R U' D B') D' = U R U' D B' D', which is a Type 3 "algorithm". That's it, that's what all the Type 3 algs are.

I doubt knowing 5-cycles are gonna be very useful in FMC, but at least knowing all the 6-movers can turn a nasty situation into a very nice one if you're lucky enough :)
 
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All 6-move edge double-2-cycles:

U2 R2 U2 R2 U2 R2 (6f*)
U2 R2 U2 D2 L2 D2 (6f*)
U2 R2 U2 L2 D2 L2 (6f*)
U2 R2 F2 B2 R2 D2 (6f*)
U2 R2 D2 R2 U2 L2 (6f*)
U2 R2 D2 L2 D2 R2 (6f*)
U2 R2 L2 U2 F2 B2 (6f*)
U2 R2 L2 F2 B2 D2 (6f*)
U2 R2 L2 D2 R2 L2 (6f*)
U2 F2 U2 F2 U2 F2 (6f*)
U2 F2 U2 D2 B2 D2 (6f*)
U2 F2 U2 B2 D2 B2 (6f*)
U2 F2 R2 L2 F2 D2 (6f*)
U2 F2 D2 F2 U2 B2 (6f*)
U2 F2 D2 B2 D2 F2 (6f*)
U2 F2 B2 U2 R2 L2 (6f*)
U2 F2 B2 R2 L2 D2 (6f*)
U2 F2 B2 D2 F2 B2 (6f*)
U2 D2 R2 U2 D2 L2 (6f*)
U2 D2 R2 F2 B2 R2 (6f*)
U2 D2 R2 L2 F2 B2 (6f*)
U2 D2 F2 U2 D2 B2 (6f*)
U2 D2 F2 R2 L2 F2 (6f*)
U2 D2 F2 B2 R2 L2 (6f*)
U2 D2 L2 U2 D2 R2 (6f*)
U2 D2 L2 F2 B2 L2 (6f*)
U2 D2 B2 U2 D2 F2 (6f*)
U2 D2 B2 R2 L2 B2 (6f*)
U2 L2 U2 R2 D2 R2 (6f*)
U2 L2 U2 D2 R2 D2 (6f*)
U2 L2 U2 L2 U2 L2 (6f*)
U2 L2 F2 B2 L2 D2 (6f*)
U2 L2 D2 R2 D2 L2 (6f*)
U2 L2 D2 L2 U2 R2 (6f*)
U2 B2 U2 F2 D2 F2 (6f*)
U2 B2 U2 D2 F2 D2 (6f*)
U2 B2 U2 B2 U2 B2 (6f*)
U2 B2 R2 L2 B2 D2 (6f*)
U2 B2 D2 F2 D2 B2 (6f*)
U2 B2 D2 B2 U2 F2 (6f*)

I have no idea how these work. Can anyone help me with this? I've found 7 types :eek: I'm pretty sure it can be reduced to only 3 or 4 types but I don't know how :(

(A and B are adjacent faces. a is the face opposite A. b is the face opposite B.)

1. A2 B2 A2 B2 A2 B2
2. A2 a2 B2 A2 a2 b2 = S2 F2 S2 F2
3. A2 B2 A2 a2 b2 a2 = a2 + type 2 + a2
4. A2 B2 A2 b2 a2 b2 = type 3 + type 1
5. A2 B2 a2 B2 A2 b2 = b2 + type 4 + b2
6. A2 B2 a2 b2 a2 B2 = A2 + type 4 + A2
7. A2 B2 b2 a2 B2 b2 = F2 S2 F2 S2
 
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All 6-move edge 3-cycles:

1. S F2 S' F2
2. F2 S F2 S'
3. Conjugated S F2 S' F2 (read two posts above for explanation)
(S = quarter slice turn; F2 = half face turn)

----------

I don't know how handy this'll be, but... all 6-move edge double-3-cycles:

1. S2 F S2 F'
2. F S2 F' S2
3. Conjugated S2 F S2 F' (read two posts above for explanation)
(S2 = half slice turn; F = quarter face turn)
 
All 6-move edge 5-cycles:

U R U' D B' D' (6f*)
U R L' B' R' L (6f*)
U R' U' D B D' (6f*)
U R' L F' R L' (6f*)
U F U' D R' D' (6f*)
U F B' R' F' B (6f*)
U F' U' D R D' (6f*)
U F' B L' F B' (6f*)
U D' R U' D B' (6f*)
U D' R' U' D B (6f*)
U D' F U' D R' (6f*)
U D' F' U' D R (6f*)
U D' L U' D F' (6f*)
U D' L' U' D F (6f*)
U D' B U' D L' (6f*)
U D' B' U' D L (6f*)
U L U' D F' D' (6f*)
U L' U' D F D' (6f*)
U B U' D L' D' (6f*)
U B' U' D L D' (6f*)
U' R U D' F' D (6f*)
U' R L' B R' L (6f*)
U' R' U D' F D (6f*)
U' R' L F R L' (6f*)
U' F U D' L' D (6f*)
U' F B' R F' B (6f*)
U' F' U D' L D (6f*)
U' F' B L F B' (6f*)
U' D R U D' F' (6f*)
U' D R' U D' F (6f*)
U' D F U D' L' (6f*)
U' D F' U D' L (6f*)
U' D L U D' B' (6f*)
U' D L' U D' B (6f*)
U' D B U D' R' (6f*)
U' D B' U D' R (6f*)
U' L U D' B' D (6f*)
U' L' U D' B D (6f*)
U' B U D' R' D (6f*)
U' B' U D' R D (6f*)

I looked deeper into them, and found that there are 3 types of "algorithms":

1. S F S' F'
2. F S F' S'
3. Conjugated S F S' F'
(S = quarter slice turn; F = quarter face turn, which obviously cannot be parallel to S)

The first two are pretty straightforward. I found the third one pretty interesting, I'll describe it with an example:

U D' R U' D B' is a Type 1 "algorithm". Using D as a setup move, we get D (U D' R U' D B') D' = U R U' D B' D', which is a Type 3 "algorithm". That's it, that's what all the Type 3 algs are.

I doubt knowing 5-cycles are gonna be very useful in FMC, but at least knowing all the 6-movers can turn a nasty situation into a very nice one if you're lucky enough :)

So nice to share these. I don't realize it's edges 5-cycles even if I have done this moves before:fp

I'm intersting about how you get all this algs? :)
 
So nice to share these. I don't realize it's edges 5-cycles even if I have done this moves before:fp

I'm intersting about how you get all this algs? :)

Cube Explorer :p I only know such primitive means to get things done :o
 
Cube Explorer :p I only know such primitive means to get things done :o
?

How it works? I have Acube insead.

Just choose 5 edges to solve? how do you know it will be 6 moves.
I'm noob in this program.:(
 
Cube Explorer :p I only know such primitive means to get things done :o
?

How it works? I have Acube insead.

Just choose 5 edges to solve? how do you know it will be 6 moves.
I'm noob in this program.:(
Gray out all the edges. Solve. Sort all the 6-movers manually.

//order 3: 3-cycle or double-3-cycle.
//order 2: Double-2-cycle or some random useless large n-cycle.
(nothing): 5-cycle.

Took me 2 or 3 minutes to sort them out, and another 5 or so to classify them into types. It's no big task, really :)
 
Took me 2 or 3 minutes to sort them out, and another 5 or so to classify them into types. It's no big task, really :)

Good job, big task for me:D,I'm slow almost in everything:(,so did't learn cfop for speedcubing, do fmc just for fun. I really think you'll be the fmc master someday~
 
Took me 2 or 3 minutes to sort them out, and another 5 or so to classify them into types. It's no big task, really :)

Good job, big task for me:D,I'm slow almost in everything:(,so did't learn cfop for speedcubing, do fmc just for fun. I really think you'll be the fmc master someday~
I think it's already clear that I had a hell lot of beginner's luck when I started :o To be honest, I average ~40 now despite doing 2 or 3 solves a day :( So I don't think I'll ever get close to being an FMC master, but still, it's a fun event that I enjoy doing and one that I'll probably never give up because it's different from speed events in that it doesn't take much practice to maintain one's standard in FMC :)

On the other hand, I think it's great to be able to "do FMC just for fun" and still be so good in it like you are!
 
Hey guys. Is it humanly possible to utilize a variant of the two-phase algorithm (used from Cube Explorer) during FMC. I'm thinking of this, and I'm currently looking for papers about the two-phase, but I can only find descriptions, not any papers.
 
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