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The FMC thread

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#41
This is from this week's forum competition scramble.

Premoves + Scramble: B2 F D' + F R2 F2 R2 B' D2 R2 B' L' B2 F' D F U2 F U' R' F U2 L'
1x2x3: L' U' B' R D2 L2 (6/6)
Triple x-cross: F' D' F' D' B2 (5/11)
11 + 3 = 14-move triple x-cross. How would you solve the rest of this?
F2L + OLL: F' U' F R U' R2 U' R2 U' R2 U2 R U2 (13)
Left with a 3-cycle of corners though.
 
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#42
I've used post-its a lot. And yeah, definitely I have the same problem. Sometimes I add tape to the back of the post-its, and that helps for a little while. But I think the best thing would be to get some extra Cubesmith stickers or something, and use those. I keep meaning to do that, but I never get around to it.
I've had an idea, but I don't know if it's practical, I'd like to know what you guys think of it since you've definitely had more experience with other weird methods to track insertions.

Since I'm allowed three cubes, and I've been doing fine with only one so far, I thought I could leave the third cube blank, i.e. unstickered, and always have a set of Cubesmith stickers ready. The simplest example I can think of to describe this is an A-perm: sticker the three U positions yellow, the three clockwise positions blue, and the three counterclockwise positions red, now write 1, 2 and 3 on each of those stickers (which denotes their respective orders in the cycle). In case it wasn't clear enough, each sticker on the same piece must share the same number.

Since every interchangeable sticker is of same color and labeled with numbers, it's very easy to see which two pieces are interchangeable instantly (especially on a blank cube) because you'd have two same colors on the same face. My problem with insertions is that I take a while to realize that 2 pieces are interchangeable if the specific stickers I'm tracing are not on the same face. E.g. FRU and LFU are interchangeable, but I take a while to notice this, it'd be much easier to notice this if I see that UFR and ULF share the same color.

So with this approach, at any point in the solution when 2 same-colored stickers are not on the same face, you know instantly that an 8-mover is not possible. Obviously, when 3 are on the same face, an 8-mover is also not possible.

Besides, with three "angles" to look at the commutator, you'd see it more quickly than with only one "angle".

So, what do you think?

As for preparation, obviously you could have a blank cube with 5 corners (and edges) labeled and numbered prepared before competition.

P/S: In case you were wondering, I'd use the second cube for the inverse scramble ;)
I seem to recall blank cubes being discussed somewhere, and that it seems to be a clear violation of WCA regulations as currently stated. Personally I think such an exception for fewest moves solving is perfectly reasonable, but I think the next round of regulations should be more explicit about what's legal for fewest moves.

Personally, I like to have the cube stickered because it helps me to keep track of what moves I've made on the cube.

I find that mailing labels or stickers that come with video tapes or other media stick better and last longer than Post-It Notes.
 
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Thread starter #43
Then I start trying OLLs as fast as I can, to see if one can solve edges and leave just 3 corners to make a good insertion. This one was really lucky because a 6-move OLL actually gave me one. You can't beat that - any time I get one of those, I'm elated. If I were any good, I wouldn't go beyond 7-move OLLs with this, but I stink, so I often go as far as the 9 or 10 movers, figuring if I get a good insertion, it still at least won't be terrible. But I'm sure I waste too much time doing that - the 6- and 7-movers are probably all that are really worth trying.
I did a quick study on this.

2 adjacent edges oriented: 2 6-movers, 4 7-movers, 24 8-movers.
2 opposite edges oriented: 4 6-movers, 0 7-movers, 12 8-movers.

These include mirrors, but not AUFs, i.e. I've discounted the algs with leading or trailing U/U'/U2 turns. I've also discounted stupid algs like D-conjugated ones.

Pros and cons of each case are pretty straightforward. With 2 adjacent edges oriented, you get more chances at a short solution. With 2 opposite edges oriented, you get more chances at an optimal solution.

I don't know about the 8-movers yet, but it's pretty obvious that for the adjacent case, the 2 6-movers are F U R U' R' F' and its mirror, while the 4 7-movers are the M-conjugated Sunes (or double-layered Sunes as some might call it); for the opposite case, the 4 6-movers are F R U R' U' F' and its mirrors.

So Mike, I think the 8-movers are pretty worthwhile. Just by adding one move, you get 36 more opportunities at an easy insertion (which will definitely have to cancel more than 1 move anyway) :)

Edit: Here it is in case anyone's interested:

2 adjacent edges oriented:
L' U' B' U B L (6f*)
B U L U' L' B' (6f*)
R B L' B L B2 R' (7f)
F' L' B L' B' L2 F (7f)
L' B2 R B R' B L (7f)
B L2 F' L' F L' B' (7f)
R U R' F' L' U' L F (8f)
R U B' R B R' U' R' (8f)
R U2 R' F2 L F L' F (8f)
R B L U' L' U B' R' (8f)
R' U2 R U2 R B' R' B (8f)
F U2 F' U2 F' L F L' (8f)
F' U2 F R2 B' R' B R' (8f)
F' U' F R B U B' R' (8f)
F' U' L F' L' F U F (8f)
F' L' B' U B U' L F (8f)
L F U' F' L' B' U B (8f)
L' U' L2 F' L' F2 U' F' (8f)
L' B L F' L2 B' L2 F (8f)
L' B L B' U2 B' U2 B (8f)
L' B L B' U' B' U B (8f)
L' B L' B' L2 F U2 F' (8f)
L' B2 R B2 L B' R' B (8f)
B U B2 R B R2 U R (8f)
B L2 F' L2 B' L F L' (8f)
B L' B L B2 R' U2 R (8f)
B L' B' R B2 L B2 R' (8f)
B L' B' L U L U' L' (8f)
B L' B' L U2 L U2 L' (8f)
B' R' U R B L U' L' (8f)

2 opposite edges oriented:
F R U R' U' F' (6f*)
F' L' U' L U F (6f*)
B L U L' U' B' (6f*)
B' R' U' R U B (6f*)
R U R' U' R' F R F' (8f)
R' U' R U R B' R' B (8f)
F U F R' F' R U' F' (8f)
F R U' B U B' R' F' (8f)
F' U' F' L F L' U F (8f)
F' L' U B' U' B L F (8f)
L U L' U' L' B L B' (8f)
L' U' L U L F' L' F (8f)
B U B L' B' L U' B' (8f)
B L U' F U F' L' B' (8f)
B' U' B' R B R' U B (8f)
B' R' U F' U' F R B (8f)
 
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#44
Thanks blah for that tip on EJ pairs, and explaining speed blockbuilding and FMC blockbuilding.
Say you build a 2x2x2: A B C D E (5), but can't find a good expansion into a 2x2x3, and even Johannes solver tells you the optimal solution to expand into a 2x2x3 is say, 7 moves. You get depressed :eek:

Here's a nice trick I use very often: Look for a point (which may or may not exist) within the first 5 moves where all 4 2x2x2 pieces are not on any one (or more) face(s), do an X/X'/X2 turn on that face and proceed with your original solution. Now you have 9 more opportunities to expand into a 2x2x3 efficiently.

This is, in my opinion, the main difference between speed-blockbuilding and FMC-blockbuilding. Tell me if my explanation isn't clear enough.
Well, no it's not. The main difference lies in that in FMC you can explore many options. No need/time for that when speeding. The more options you can explore the more likely to get "lucky". In it's nature speed solcing is linear, not taking back/undoing any turns. In FMC you can take bck turns and explore many paths. Normally you cannot know (early in the solve anyway) which path is better ...

Per
 

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#45
So Mike, I think the 8-movers are pretty worthwhile. Just by adding one move, you get 36 more opportunities at an easy insertion (which will definitely have to cancel more than 1 move anyway) :)
Huh. I guess I didn't think about how many 8-movers there were. For some reason I was expecting there to be fewer. I'm probably missing some of them when I'm trying them - I should go over them so I know exactly what they are.

And yeah, I have had a couple of pretty nice solves with 8-movers before, so I guess you're right. So I take back what I said.
 
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Thread starter #46
MAYDAY. I really suck at insertions.

This week's forum competiton's scramble 3 for 3x3x3 speedsolve.

Premoves + inverse scramble: B F' + F' L' D L D2 B' R F R2 U2 F2 D F2 D2 R2 L2 F2 D B2 D' L2

2x2x2: L2 U' *B2 R' F2 (5/5) (*B2 inserted to simplify 2x2x3 expansion)
2x2x3: D R' B2 (3/8)
Triple x-cross: B R B' D' R F' R' F (7/15)
Last slot + AUF: D B' D B R' B R B' + D2 (9/24)
Undo premoves: B F' (2/26)
Corner 3-cycle remaining.

Pseudo-solution for inverse scramble: L2 U' B2 R' F2 D R' B' R B' D' R F' R' F D B' D B R' B R B' D2 B F'

I went through the whole solution twice, and I couldn't find a single insertion point with at least one cancellation! :( :( :( Can anyone help me?

Question: Is there any difference in looking for insertion points in any of the following: Inverse solution without premoves, inverse solution with premoves, forward solution? What I'm trying to say is, for example, if I look for insertion points in the inverse solution with premoves and don't find one at all, will I find one in the inverse solution without premoves, or the forward solution? Please tell me if my question isn't clear enough.
 
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Thanks blah for that tip on EJ pairs, and explaining speed blockbuilding and FMC blockbuilding.
Say you build a 2x2x2: A B C D E (5), but can't find a good expansion into a 2x2x3, and even Johannes solver tells you the optimal solution to expand into a 2x2x3 is say, 7 moves. You get depressed :eek:

Here's a nice trick I use very often: Look for a point (which may or may not exist) within the first 5 moves where all 4 2x2x2 pieces are not on any one (or more) face(s), do an X/X'/X2 turn on that face and proceed with your original solution. Now you have 9 more opportunities to expand into a 2x2x3 efficiently.

This is, in my opinion, the main difference between speed-blockbuilding and FMC-blockbuilding. Tell me if my explanation isn't clear enough.
Well, no it's not. The main difference lies in that in FMC you can explore many options. No need/time for that when speeding. The more options you can explore the more likely to get "lucky". In it's nature speed solcing is linear, not taking back/undoing any turns. In FMC you can take bck turns and explore many paths. Normally you cannot know (early in the solve anyway) which path is better ...

Per
You're right.

I guess I was thinking of the difference between "normal" FMC and linear FMC :eek:
 
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Thread starter #48
So Mike, I think the 8-movers are pretty worthwhile. Just by adding one move, you get 36 more opportunities at an easy insertion (which will definitely have to cancel more than 1 move anyway) :)
Huh. I guess I didn't think about how many 8-movers there were. For some reason I was expecting there to be fewer. I'm probably missing some of them when I'm trying them - I should go over them so I know exactly what they are.

And yeah, I have had a couple of pretty nice solves with 8-movers before, so I guess you're right. So I take back what I said.
I'm guessing it's possible to memorize how all 46 of those OLLs affect edge permutation (you already know how it affects orientation) with relatively little effort as compared to say, learning full OLL. This'll probably save a lot of time wasted trial-and-error-ing. What do you think?
 
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#50
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)
 
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#51
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..
Horrible? There are two ce-pairs that can be used easily. Annoyed? Your solution is really lucky, and I don't think it's reasonable to expect a sub-30 with Fridrich.

You also wrote it down wrong. The first U2 in 2nd pair should be U'.
 

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#52
I'm guessing it's possible to memorize how all 46 of those OLLs affect edge permutation (you already know how it affects orientation) with relatively little effort as compared to say, learning full OLL. This'll probably save a lot of time wasted trial-and-error-ing. What do you think?
One of the FMC masters (I think it was Guus) was acting surprised in another thread when I told him I hadn't already done that. I think it's obvious that if you want to be really good at FMC, you should. It would REALLY help for this trick. I've just been too lazy to work that hard yet at it.
 
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Thread starter #54
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)
Edge 3-cycle insertion at .: U R L' B2 .. R' L U (6/26)
Edge 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 :))

It's nowhere near my shortest solution, but it's one of my better ones, and I'm especially satisfied with this because I don't think I could've bettered any of the steps even with 33 minutes to go :) So I just spent the rest of the hour smiling to myself and kept doing and redoing the solution to make sure it worked :p

I spent 2 or 3 minutes with the original scramble, gave up, then tried the inverse. I got to the "all but 6 edges" stage in 10 minutes. Rest of the time was spent hunting for insertion points. Told you I suck at insertions :p

The only reason I was able to finish up to step 3 of Roux so quickly was because my solution was almost linear. Try it yourself, it's very straightforward! So I guess this shouldn't really count as a legitimate solution because it was really, really lucky :eek: I'm sure any competent Roux user will be able to get a much better solution with the inverse scramble :)

That said, this solution marks many firsts for me:
1. First solution under 30 minutes.
2. First (non-DNF) Roux solution.
3. First solution with edge cycle insertions.
4. First solution with a double-2-cycle insertion.
5. First solution with 2 insertions completed in time.
6. First solution in which I didn't have a backup and just went with my original starting moves :p Risky!

Would've been an AsR :rolleyes: I hope I get to a competition soon. I really love FMC. I don't want it to end up like BLD: I had my first (and only) competition after I quit BLD, so I got a DNF and a lousy time :(

Edit: Now that I've seen Jude's solution, I feel kinda lousy that my solution was only as good as a Fridrich solution :( Guess I've still got a lot of work to put into my FMC :)
 
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#57
For other adjacent corners or centrally opposite corners, use the same alg with a setup move such that you get a cancellation between the setup move (or undo setup move) with the alg itself.
Cyclical shifts of that same algorithm is also useful:eek:

Per
 
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#58
Horrible? There are two ce-pairs that can be used easily.
And I couldn't continue either of them into anything nice, hence I found it horrible.

Annoyed? Your solution is really lucky.
Yes, annoyed. The point is that it's lucky and still bad. My solution is sup 30 even WITH a skip (Although, it is 'engineered' - I found an F2L in 4 moves less which left a U perm. In fact, I spent almost 20 minutes TRYING to find an F2L which also solved the last 3 U edges too, so I'm not sure I'd call it "really lucky").

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 this is my 2nd worst solve ever. In fact, I think I've done just over 10 solves in my whole life (all with Fridrich) and almost half of them were sub 30.

You also wrote it down wrong. The first U2 in 2nd pair should be U'.
Thanks, I fixed that now.
 
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#60
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.
 
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