Random Cubing Discussion

[Athefre]

OK, so can you use alg.garron.us to demonstrate it working? Ie, set the scramble to: R2 U' R2 U' D' R2 U' R2 F2 D2 ... and solve it, then post a link here.

The problem with that scramble is that it's not following my steps. Remember, after orientation, the two pairs you solve are supposed to be matching (if your cube is oriented based on Yellow and White on U/D and the first pair you solve is white, you have to solve the other white pair) and they are supposed to be placed diagonally - one at DL and the other at UR. Then you do a y rotation and use L2UR2U' to solve the two remaining pairs.

alg.garron.us solution

I'm getting errors when I try to link, maybe it's just my computer. I have to select the puzzle size and click "Test" for it to work.

 

[Weston]

I've been working on some F2L tricks such as
x U' R' U' F R' F' R U l
x r' U' F2 r U r' F L

Does anyone have more tricks like these that I could add to my list?

 

[Cride5]

The problem with that scramble is that it's not following my steps.

So if the cube ends up in that state after orientation, what is the solver supposed to do?

Your solution (like mine) breaks up existing pairs in order to build different pairs. Using it in this way requires more understanding than simply 'place complete pairs in DB and UF then apply alg to create new pairs'. The assumption is that solved pairs should remain solved. In order to use the technique like this your guide will need to explain a lot more, stating where it is desirable to break up existing pairs ... and I would argue that it's not very beginner friendly.

 

[Athefre]

This is extremely hard for me to explain but seems so very very simple in my mind. Did you go through all of my example solves? They all do the exact same thing:

Scramble with Yellow on top and Blue on front: F' R U2 R F2 U2 R' F' R'

-- Orient - x' F' U2RU'R'
-- Solve a pair and place it at DL. - UL2
-- What color is the pair you just placed at DL? It is Yellow. Now, solve the other Yellow pair and place it at UR. - U'R2U'R2UR2U'
You now have a solved pair at DL and a solved pair at UR. The U color at UR is Yellow and the D color at DL is Yellow.
- Do a y rotation. Hey, look, UB is already solved for you. L2UR2U' is meant to be used to solve pairs at UB. But, wait, the DF pair isn't solved so do x2 to put it at UB. Now you can continue - (L2UR2U')*4

 

[Athefre]

The pairs have to always be separated like this:

Notice that the Yellow bars are diagonal from each other on the cube and the White bars are diagonal from each other on the cube. This is so that the sequence will work. If someone just follows my steps and puts that second pair they solve at UR and does the y rotation, they will be fine.

Not like this:

And not like this:

 

[Athefre]

Maybe I could explain better through examples in chat.

 

[Cride5]

I totally understand the content of your examples ... that's what I was referring to as your 'original technique'. The scrambles I gave you can easily be solved using the 'original technique' ... but they can't be solved with:

A simpler way to do Step 3 is to not really have the pairs be completed pairs and use Step 4 to pair up all four uncompleted pairs. But, that's more moves and maybe not as fun. Maybe easier for a beginner though.

 

[Athefre]

I totally understand the content of your examples ... that's what I was referring to as your 'original technique'. The scrambles I gave you can easily be solved using the 'original technique' ... but they can't be solved with:

A simpler way to do Step 3 is to not really have the pairs be completed pairs and use Step 4 to pair up all four uncompleted pairs. But, that's more moves and maybe not as fun. Maybe easier for a beginner though.

But, it's the same situation. I've tried it many times and haven't run into problems but I guess it's possible that there are situations where it wouldn't work. In that quote of mine, the steps would be:

Step 2: Pair up two Red corners (for example) and place them at DL, the pair doesn't have to be a solved pair. Then, pair up the other two Red corners and place them at UR. As in the original, it's very important to have them separated at DL and UR.
Step 3: Same as the original Step 3. Except, when you have solved the UB and DF pairs, you do a y2 and and now the other unsolved pairs are at UB and DF.

So, you completely understand the original method? I didn't give enough detail in the original post, but maybe that little option I suggested isn't so good for a beginner. I mean, it doesn't seem so hard to explain and show to someone "A pair of Reds at DL and a pair of Reds at UR".

It doesn't matter I guess, the original is fine.

 

[qqwref]

If your method is getting to the point where people here have trouble understanding it, how do you intend to teach it to beginners?

 

[Athefre]

It looks like the only thing he didn't understand, or that I'm thinking about wrong, was the little optional step, not the method itself.

I think I would be able to teach someone this very quickly in person. I have trouble explaining things in writing. You know I'm doing something wrong when I can't even help people understand NMCMLL.

 

[Kirjava]

You know I'm doing something wrong when I can't even help people understand NMCMLL.

Lies. You explained it to me in one sentence.

 

[Athefre]

This is my favorite topic on speedsolving. I get to post things I've been working on without worrying if they are worthy of a new topic and I get to check out other people's ideas.

 

[Kirjava]

Yah, it's cool. I've put a bunch of stuff here that would otherwise be unreleased.

 

[Kirjava]

Example solve for Kyle;

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

1) U B' L F L F' L2 F2 L' F R U' x2
2) R' D2 L2 D F D R2 U2 R'
3) R2 B R' U B U2 R2 U R B' U y
4) R2 U2 F R F' U' R2 F' U' F U' R2 U2
5) R' U R U R' U2 R U' R' U' R U2

 

[irontwig]

I've been working on some F2L tricks such as
x U' R' U' F R' F' R U l
x r' U' F2 r U r' F L

Does anyone have more tricks like these that I could add to my list?

Found this with cube explorer l2' U l2 U R' D' r U2 M

 

[StachuK1992]

So. CPLL+ELL would be...how many algs? I want to say 84, 21 PLLs per 4 edge cases (all, adj, opp, none)

But Thom then said 6*48=288.

So then I looked again, and thought

21 (regular PLL)
21 (all edges flipped)
Those only get a set of PLL each, since the angle doesn't matter.

21*2 (opp edges)
One set where flipped on L/R, one on U/B.

21*4 (adj edges)
One set per AUF.

21 + 21 + 42 + 84 = 168.

Is this correct?

 

[Athefre]

Step 2: My orientation method (or another if there is something better).

I keep thinking about this, feeling terrible. So, I want to say I'm sorry if that was disrespectful. It's unlike me to post things like that.

 

[qqwref]

Example solve for Kyle;

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

1) U B' L F L F' L2 F2 L' F R U' x2
2) R' D2 L2 D F D R2 U2 R'
3) R2 B R' U B U2 R2 U R B' U y
4) R2 U2 F R F' U' R2 F' U' F U' R2 U2
5) R' U R U R' U2 R U' R' U' R U2

cool stuff...

how many algs would you estimate for each of the last four steps?

 

[Kirjava]

Roughly 50/12/20/10. I think those numbers can be cut down though - and this was a quick estimate with some guessing regarding AUFs and stuff.

 

[irontwig]

So. CPLL+ELL would be...how many algs? I want to say 84, 21 PLLs per 4 edge cases (all, adj, opp, none)

But Thom then said 6*48=288.

So then I looked again, and thought

21 (regular PLL)
21 (all edges flipped)
Those only get a set of PLL each, since the angle doesn't matter.

21*2 (opp edges)
One set where flipped on L/R, one on U/B.

21*4 (adj edges)
One set per AUF.

21 + 21 + 42 + 84 = 168.

Is this correct?

#Non-EPLL PLLs=22-5=17 (Don't forget pure flips)

ELL: 29
Regular PLL: 17
PLLEF: 17 (Like CLLEF)
Opp flip: 2*17-2=32 (N-perm overcount)
Adj flip: 4*17-2-6=60 (E&N-perm overcount)
Sum: 29+17+17+32+60=155

Please correct me if this is wrong.
Edit: There's actually only two adj flip Es.
Edit2: Of course all pure flips are ELLs, how could I forget that? And I wrote 19 instead of 29 in the sum, yikes.