Random Cubing Discussion

[Berkmann18]

I can only use <r, U> though.

What are you trying to do exactly ?

 

[PenguinsDontFly]

I can only use <r, U> though.

RCFOP.

I just invented it. For every move that isnt <r,U>, you ROTATE. Its an OH method for people with tiny hands.

 

[TDM]

What are you trying to do exactly ?

Reddit's weekly competition has <r, U> as one of the events.

 

[Berkmann18]

Have you tried to do like ZZ but with only the two gen you specified or something like that ?

 

[qaz]

This is the method I use for <r, U>:
1. solve bottom center and DR edge (intuitive)
2. solve the back F2L pair (intuitive)
3. solve DB edge
- misorient edge and put at UL, then do (r' U r) U (r' U r) U' (r' U' r)
4. orient edges (setup moves may be needed)
- r U' r2 U' r2 U r2 U r' U r2 U r2 U' r2 - orients UL and UF
5. solve U-layer edges one by one
- r2 U' r2 U' r U r2 U' r2 U2 r2 U r2 U' r U r2 - cycles DF-UB-FR (moves edge at DF to UB)

sub-1 definitely possible with this

 

[TDM]

This is the method I use for <r, U>:
1. solve bottom center and DR edge (intuitive)
2. solve the back F2L pair (intuitive)
3. solve DB edge
- misorient edge and put at UL, then do (r' U r) U (r' U r) U' (r' U' r)
4. orient edges (setup moves may be needed)
- r U' r2 U' r2 U r2 U r' U r2 U r2 U' r2 - orients UL and UF
5. solve U-layer edges one by one
- r2 U' r2 U' r U r2 U' r2 U2 r2 U r2 U' r U r2 - cycles DF-UB-FR (moves edge at DF to UB)

sub-1 definitely possible with this

Thank you, this is (almost) perfect!

One question: How do you solve the last five corners? You don't seem to do that at all.

 

[qaz]

Oh whoops, I forgot that step!

1. solve bottom center and DR edge (intuitive)
2. solve the back F2L pair (intuitive)
3. solve DB edge
- misorient edge and put at UL, then do (r' U r) U (r' U r) U' (r' U' r)
4. solve corners <r, U> like a 2x2 without messing up the back F2L (solve DFR then rest with sune/antisune)
5. orient edges (setup moves may be needed)
- r U' r2 U' r2 U r2 U r' U r2 U r2 U' r2 - orients UL and UF
6. solve U-layer edges one by one
- r2 U' r2 U' r U r2 U' r2 U2 r2 U r2 U' r U r2 - cycles DF-UB-FR (moves edge at DF to UB)

 

[TDM]

Oh whoops, I forgot that step!

1. solve bottom center and DR edge (intuitive)
2. solve the back F2L pair (intuitive)
3. solve DB edge
- misorient edge and put at UL, then do (r' U r) U (r' U r) U' (r' U' r)
4. solve corners <r, U> like a 2x2 without messing up the back F2L (solve DFR then rest with sune/antisune)
5. orient edges (setup moves may be needed)
- r U' r2 U' r2 U r2 U r' U r2 U r2 U' r2 - orients UL and UF
6. solve U-layer edges one by one
- r2 U' r2 U' r U r2 U' r2 U2 r2 U r2 U' r U r2 - cycles DF-UB-FR (moves edge at DF to UB)

Thank you!

Any advice for step 5? I find myself having to use the step 6 alg to get pieces to the U layer, which slows me down a lot.

Edit: Just realised I could do r setup moves, so I think that should help.

 

[qaz]

Thank you!

Any advice for step 5? I find myself having to use the step 6 alg to get pieces to the U layer, which slows me down a lot.

Edit: Just realised I could do r setup moves, so I think that should help.

Yeah, that'd what I do as well. I only came up with this method a few days ago, so if you see something that you think could be improved chances are you're right.

 

[TDM]

Yeah, that'd what I do as well. I only came up with this method a few days ago, so if you see something that you think could be improved chances are you're right.

I was talking to ben about this and he uses an F2L -> 3-look LL method. It makes EP much faster, and there are no misoriented edges not on U. I think that could be a bit faster. Most of the LL was repeating an L-shape OLL, except for the U perm alg, which I can't remember at the moment.

 

[qaz]

I was talking to ben about this and he uses an F2L -> 3-look LL method. It makes EP much faster, and there are no misoriented edges not on U. I think that could be a bit faster. Most of the LL was repeating an L-shape OLL, except for the U perm alg, which I can't remember at the moment.

How fast is Ben with his method? I've gotten a 36.xx ao12 with mine, but if his is significantly faster I'll probably just use it instead...

 

[TDM]

How fast is Ben with his method? I've gotten a 36.xx ao12 with mine, but if his is significantly faster I'll probably just use it instead...

From what he's posted on FB (though he might have beaten these since then):
15.849, 16.677, 18.556, 14.551, 21.246, 14.503, 23.028, (23.388), 14.029, 15.607, (13.943), 14.498= 16.854
(23.388), 14.029, 15.607, (13.943), 14.498 = 14.711
9.549 single

 

[United Thought]

Hang on, with this <r, U> method, can't you just use Roux where you do CMLL earlier on or something? (Using r R' = M', etc, obviously)

---

Also,

What is the minimum number of solves in an average you need for it to be called a global average?

 

[PenguinsDontFly]

Hang on, with this <r, U> method, can't you just use Roux where you do CMLL earlier on or something? (Using r R' = M', etc, obviously)

---

Also,

What is the minimum number of solves in an average you need for it to be called a global average?

I would say 1000 solves is considered a global average.

 

[Username]

What is the minimum number of solves in an average you need for it to be called a global average?

I consider global average to be the average you get when your amount of solves is infinite. As this is impossible to accomplish, just use some smaller number that still isn't all too small (like 500+ atleast)

 

[PenguinsDontFly]

I consider global average to be the average you get when your amount of solves is infinite. As this is impossible to accomplish, just use some smaller number that still isn't all too small (like 500+ atleast)

well its dependent on the event as well. I think for 3x3, 1000 is the right number. something like 2x2 or skewb I would say like 5000, but like 250 for 4x4.

 

[TDM]

I simply don't count my PBs as my global average. What I say is my global average is what I think I average. I guess you could use PB Aox if you don't do Aoxs often though (and x is large).

 

[qaz]

From what he's posted on FB (though he might have beaten these since then):
15.849, 16.677, 18.556, 14.551, 21.246, 14.503, 23.028, (23.388), 14.029, 15.607, (13.943), 14.498= 16.854
(23.388), 14.029, 15.607, (13.943), 14.498 = 14.711
9.549 single

Ok then, I guess his method is a lot better haha

Hang on, with this <r, U> method, can't you just use Roux where you do CMLL earlier on or something? (Using r R' = M', etc, obviously)

---

Also,

What is the minimum number of solves in an average you need for it to be called a global average?

You can't do R moves in this solving style, only r, so no.

 

[United Thought]

You can't do R moves in this solving style, only r, so no.

Oh, ok. Should have guessed that from <r, U>

Thanks to everyone who replied to my question about global average(too many posts to quote)

 

[Berkmann18]

Hang on, with this <r, U> method, can't you just use Roux where you do CMLL earlier on or something? (Using r R' = M', etc, obviously)

---

Also,

What is the minimum number of solves in an average you need for it to be called a global average?

At least 100 solves, But 500+ is preferable.