[Help Thread] FMC Discussion

MustCube round 450:

25 moves all but 6 corners..

I know how to cycle 3 corners.. How do I go about with 6 corners? One is on it's place but flipped. Do I start with cycling this one with two random other misplaced corners?

No, not two random misplaced corners.
On your first cycle you want to solve one misplaced corner, break out the twisted one and the 3rd piece is indeed a random misplaced corner.

But typically you only use this for 4 corners of which one is flipped. --> need 2 cycles
When you have 5 including a twist --> need 3 cycles
6 including a twist --> don't even go there :)
 
So I sticker one corner that I will solve with number 1, the flipped with no 2 and a random misplaced corner with no 3?

nope :)
This 3-sticker approach is limiting your options.
sticker all 6 corners
1 goes to 2 goes to 3 goes to 4 goes to 5 (5 will be another face on the cubie that already has the "1" sticker)
sticker the twisted cubie with "t" or "x"or whatever you like.

now the insertion hunting goes as mentioned before :)
solve one, break out "t" and include another one.
examples:
1 - 2 - t (here 1 is solved and 2 is the other one by default)
4 - t - 3 (here 3 is solved )
etc

The nice thing is that the orientation of t does not matter; just include any face of the "cubie" that contains the t sticker.

But really; typically you use this for 4 corners left and in very rare cases for 5.
For 6 including a twist I suggest building another skeleton :)
 
No! I got very good results already with 6 or 7 corners left needing 3 insertions (including the currently leading solution of this weeks forum competition). Just don't be so afraid! ;)

just out of curiosity, how do you go about doing more than one insertion? do you have a general strategy to try and find the most cancelations or do you just try a bunch of different options?
 
just out of curiosity, how do you go about doing more than one insertion? do you have a general strategy to try and find the most cancelations or do you just try a bunch of different options?

I wrote about this not too long ago: http://www.speedsolving.com/forum/showthread.php?13599-The-FMC-thread&p=842689&viewfull=1#post842689

I know you have used three insertions before but I don't think you did this on a 25 HTM skeleton :)

well, true. ;)
 
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For some reason I can't see my solution for FMC #450 on Fredlund's site, so I'll just write it here:

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

F2 R' F' U' B L' D' L D B' L' F' D2 B F L R D2 L' D' R' D F2 (23)

F2 R' F' U' [Pseudo 2x2x2]
B L' D' L D B' L' B [F2L-1]
.R D' R' D F2 [Leaving four edges]

At dot insert H-perm: B' F' D2 B F L R D2 L' R' (Four moves cancel)

iirc found in about 30-40 minutes.
 
For some reason I can't see my solution for FMC #450 on Fredlund's site, so I'll just write it here:

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

F2 R' F' U' B L' D' L D B' L' F' D2 B F L R D2 L' D' R' D F2 (23)

F2 R' F' U' [Pseudo 2x2x2]
B L' D' L D B' L' B [F2L-1]
.R D' R' D F2 [Leaving four edges]

At dot insert H-perm: B' F' D2 B F L R D2 L' R' (Four moves cancel)

iirc found in about 30-40 minutes.

Something is very wrong with posting the results. Nice solution you found! I had a 36 moves solution. 25 move all but 6 corners and 3 inserts with 13 moves cancelled. Round 451 is a nice scramble I think. I have found a great start in two minutes.
 
Can anyone help me with the following:

I want to cycle two edges in the same face. Let's say solved cube white on U and green on F:

R U R' U R U2 R' U' R U' R' U2 R U' R'

Now white/orange and white/red have been swapped. Is there a shorter way to do this?

not quite sure what you are getting at but: F2 U2 F2 U2 F2 U2 swaps the same edges
side effect is another edge swap though.

ah...
you have finished F2L-1, oriented edges and are now facing an edge swap :)

There is a 10 mover: y R' F R' u2 R F' R' u2 R2 F' that swaps the edges and two U face corners

there is also the optimal T perm

You can also try to insert these in the skeleton somewhere and the advantage is that you can also use the optimal J-perms
 
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Can anyone help me with the following:

I want to cycle two edges in the same face. Let's say solved cube white on U and green on F:

R U R' U R U2 R' U' R U' R' U2 R U' R'

Now white/orange and white/red have been swapped. Is there a shorter way to do this?

Tperm would be better than that

I'm not sure what is optimal for it. I try avoid it completely I hate this case for solving edges

edit: ninja'd
 
Thanks CubeNovice and Cuberoots. Let me show you..

Warming, MustCube 451 OnGoing
MustCube 451: L' F2 U R2 F2 U D' B' U' L D2 U' F2 R2 B R' D2 L R D2 U' B U' F2 U2 R' U' F R' L

pre move R'

L2 U L' R U' Pseudo 2 X 2 X 2
D2 B L' U B' U' Huh.. looks good
R D' R' D B Ahh only these two freaking egdes..

CubeNovice , can you explain a little further how J-perm can be of use? Oh wait... I think I get it. When I sticker the four egdes, I could find them earlier in the solve and then ... no .. I don't get it. How is the J-perm (no two edges swap) going to help?
 
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Thanks CubeNovice and Cuberoots. Let me show you..

CubeNovice , can you explain a little further how J-perm can be of use? Oh wait... I think I get it. When I sticker the four egdes, I could find them earlier in the solve and then ... no .. I don't get it. How is the J-perm (no two edges swap) going to help?

oooh Marcel!
My eyes!

Ongoing competitions must be under a spoiler tag and carry a warning :)

T-perm swaps opposite edges (and two corners)
J-perms swaps adjecent edges (and two corners)

For inserting the edge swap you would sticker the two edges and (in this particular case) the one corner that is solved.
You can now go trough your solution and find a place where you have the edges opposite (insert optimal T perm or the alg I posted) or adjecent (insert optimal J-perm). In you particular case you want to make sure that the stickered corner is not involved in the swapping alg: you do not want to "unsolve" it.

BTW your skeleton also has two corners twisted in place so you still have a lot to solve!
but you have two swapped corners in there so with a little luck the right T-perm or J-perm you can solve four pieces at one time.

Cheers
 
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