Random Cubing Discussion

[blade740]

It'll be cut in half due to mirrors. Sub1000 algs is learnable.

 

[rachmaninovian]

i has 4by4 :3

F*** that's small!

ballcore?

hahahaha no. turning kinda stinks..but its turnable.

the mech is liek 3x3 bandaged into 2x2 extended to a 4x4. 3D printing hurts my pocket...cost me 40 bucks including shipping..

 

[Athefre]

Simpler?

Step 1: Three corners of a face (Matching or opposite).
Step 2: My orientation method (or another if there is something better).
Step 3: Solve a pair of corners and place them at DL. Then, solve the other pair of corners that belong with the pair at DL. Don't place this newly created pair at DR. Place it at UR. You'll have, for example, a correct pair of red corners at DL and the other correct pair of red corners at UR.
Step 4: Do a y rotation so that your completed pairs are at DB and UF. Now, do L2UR2U' until the remaining two uncompleted pairs are solved. If you notice one uncomplete pair left at DF, place it at UB (x2 rotation or F2U2F2) and perform the sequence until it is solved. Align the completed pairs and you are finished.

[size=-2]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.[/size]

 

[qqwref]

Maybe I should have been more clear.

This:

MUM' U2 MUM'
l F'LF l' F'L'F

is a solution to this:

LR'l'URU'lrUR'U2L'Ur'l'U'LUL'URU'lUR'U'R

 

[Cride5]

Simpler?

...[orientation method]...

Step 3: Solve a pair of corners and place them at DL. Then, solve the other pair of corners that belong with the pair at DL. Don't place this newly created pair at DR. Place it at UR. You'll have, for example, a correct pair of red corners at DL and the other correct pair of red corners at UR.

Step 4: Do a y rotation so that your completed pairs are at DB and UF. Now, do L2UR2U' until the remaining two uncompleted pairs are solved. If you notice one uncomplete pair left at DF, place it at UB (x2 rotation or F2U2F2) and perform the sequence until it is solved. Align the completed pairs and you are finished.

Niiice! This is like sexy move for 2x2

Just some quick analysis. The move counts (excluding setup/alignments) are:
Solved: 1/6 - 0
Opp: 1/6 - 24
Adj-1: 2/6 - 16
Adj-2: 1/6 - 8
Adj-3: 1/6 - 32
Avg: (24+2*16+8+32)/6 = 16

If also using mirror: R2 U' L2 U
Avg: (24+2*16+8+8)/6 = 12

So not too bad for a beginner method, thanks for sharing!

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.

It's not possible to solve this way if you have an adjacent-swap on 1 D-layer pair and 1 U-layer pair. No 3-cycle of these pieces will ever create another pair.

 

[Athefre]

Niiice! This is like sexy move for 2x2

I've actually been thinking of it more as a half-turn version of L'URU'LUR' (or RU'L'UR'U'L) because of the way it moves corners around on a 3x3 without changing edges.

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.

It's not possible to solve this way if you have an adjacent-swap on 1 D-layer pair and 1 U-layer pair. No 3-cycle of these pieces will ever create another pair.

I guess I just assumed it would work since I had never run into problems.

 

[Athefre]

Hey, can I have an example of a case where it wouldn't work? I don't think I understand.

 

[Cride5]

R2 U' R2 U' D' R2 U' R2 D2

 

[Athefre]

The point of the sequence is to have the pairs separated diagonally on the cube. That's what I meant for the main Step 3 and the option I had in little writing. It's hard for me to explain, as usual. Maybe these pictures will show it:

Step 3 Option (with all pairs not completely solved):

And of course with the regular Step 3, the UF and DB pairs would be complete.

 

[hawkmp4]

I find that method to be more complicated than teaching someone how to make a first layer (which most people, I've found, grasp pretty intuitively) and then using R'D'RD commutator for orientation of the LL and RDR' for permutation. I've had good luck with teaching people the concept of commutators, even those who haven't solved a twisty puzzle before.

 

[Cride5]

The point of the sequence is to have the pairs separated diagonally on the cube.

OK, so how about:
R2 U' R2 U' D' R2 U' R2 F2 D2

 

[Athefre]

The point of the sequence is to have the pairs separated diagonally on the cube.

OK, so how about:
R2 U' R2 U' D' R2 U' R2 F2 D2

Doing that forwards, the color pairs aren't diagonal. Doing that backwards, you have a completed pair at UB and an unsolved pair at DF and UF. Remember, you are supposed to put unsolved pairs at UB.

So, the solution to that would be x2(L2UR2U')*4 z2 (L2UR2U')*4

 

[Cride5]

no workey

 

[Athefre]

Heh That was just to complete the pairs. Of course, add an F2 at the end to align the pairs and solve the cube.

EDIT: Wait, I'm not sure what's happening in that link.

EDIT 2: Oh, you put the forwards version in.

I really have to get off, I'll explain more tomorrow.

 

[Cride5]

OK, so applying the alg an odd number of times will break up your UF/DB pairs, so we can't do that. An even number of applications of the alg performs a 3-cycle. The case I presented cannot be solved by a 3-cycle unless some of the solved pairs are broken up. Trust me, it won't work - it's theoretically impossible.

By breaking up all the pairs it can be solved with:
U2 (L2 U R2 U')*2 x2 (L2 U R2 U')*2 z2 (L2 U R2 U')*4 F2

... but breaking up pairs isn't in your specification. Moreover, if you're breaking up pairs without an understanding of what you're doing it will involve a lot of trial and error!

 

[KboyForeverB]

qqTimer can be more accurate than a stackmat.
because you can't get a 0.00 with a stackmat!

this photo is not made by photoshop or paint.
sometimes I just keep hitting the space button so fast that the timer doesn't even stop.

That's very true, that's why few moths ago I stopped using cubetimer and used qqtimer. Now I use CCT (FTW!!)

 

[Athefre]

OK, so applying the alg an odd number of times will break up your UF/DB pairs, so we can't do that. An even number of applications of the alg performs a 3-cycle. The case I presented cannot be solved by a 3-cycle unless some of the solved pairs are broken up. Trust me, it won't work - it's theoretically impossible.

Doing your case forwards, of course it doesn't work because it's not following my steps. Doing it backwards, it does work.

... but breaking up pairs isn't in your specification. Moreover, if you're breaking up pairs without an understanding of what you're doing it will involve a lot of trial and error!

I'm just not seeing how this is trial and error. You do L2UR2U' until you see that the UB pair is solved, and of course you could let someone know that it takes an even number of times for it to correct itself to have your UF and DB pairs back to where they were. That's not exactly trial and error in my mind. So, obviously, instead of being taught to do L2UR2U' they could be taught to do that twice then check the UB pair. If it isn't solved, they just have to do that sequence twice again. Like you said, similar to "sexy move" where you repeat it until what you need done is done.

For now, let's forget about the Step 3 option I had in my original post and just go with the first idea of having two completed pairs. Now, here are some examples (Scramble with Yellow on top and Blue on front):

Scramble: F U' R U' R2 U' R2 U2 F2

Orient - z' F2 U'F' UR F'UF
Pair + Place at DL - L2
Other Pair (notice your pair at DL is red, so you now have to make the other red pair and place it at UR) - U2R2U'
Complete remaining two pairs - y (L2UR2U')*2
All pairs are solved, now finish with F2U'.

Scramble: R2 U F U2 R' F U2 R'

Orient - x'y U'F' F'U'F
Pair + Place at DL - R2D'
Other Pair (Notice your pair at DL is yellow, so solve the other yellow pair and place it at UR) - UR2U'
Complete remaining two pairs - y (L2UR2U')*4 x2 (L2UR2U')*4
Finish with F2U'
[size=-2]In this example, the pairs could have been completed with (R2U'L2U)*2. But that's not what I've been teaching. Similar to not telling a beginner that they could have done the inverse of "sexy move".[/size]

Scramble: F R U R2 F' R F U' R2 U2

Orient - x UR F' RU'R'
Pair + Place at DL - Already there
Other Pair (Your DL pair is green, solve the other green pair) - Already there
Complete remaining two pairs - y (L2UR2U')*4
Finish with F2U

Scramble: F' R U2 R F2 U2 R' F' R'

Orient - x' F' U2RU'R'
Pair + Place at DL - UL2
Other Pair (Your DL pair is yellow, solve the other yellow pair) - U'R2U'R2UR2U'
Complete remaining two pairs - y x2 (L2UR2U')*4
Finish with F2U

Scramble: F' U R F' R2 F R' F R' U'

Orient - z2 y U'F' U2RU'R'
Pair + Place at DL - L2
Other Pair (Your DL pair is orange) - U'R2U
Complete remaining two pairs - y, UB and DF are already complete
Finish with F2U2

 

[Zyrb]

A tip to everyone, dont throw your cube against a wall, you may be be unfortunate like me and have one of your center pieces unscrew itself and make the cap stuck onto the cubie *sigh* thats why I have multiple cubes

 

[hawkmp4]

A tip to everyone, dont throw your cube against a wall, you may be be unfortunate like me and have one of your center pieces unscrew itself and make the cap stuck onto the cubie *sigh* thats why I have multiple cubes

A very thin blade works for caps that are remarkably tight. You might want to try that before you give up.

 

[Cride5]

OK, so applying the alg an odd number of times will break up your UF/DB pairs, so we can't do that. An even number of applications of the alg performs a 3-cycle. The case I presented cannot be solved by a 3-cycle unless some of the solved pairs are broken up. Trust me, it won't work - it's theoretically impossible.

Doing your case forwards, of course it doesn't work because it's not following my steps. Doing it backwards, it does work.

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.

... but breaking up pairs isn't in your specification. Moreover, if you're breaking up pairs without an understanding of what you're doing it will involve a lot of trial and error!

I'm just not seeing how this is trial and error. You do L2UR2U' until you see that the UB pair is solved, and of course you could let someone know that it takes an even number of times for it to correct itself to have your UF and DB pairs back to where they were. That's not exactly trial and error in my mind. So, obviously, instead of being taught to do L2UR2U' they could be taught to do that twice then check the UB pair. If it isn't solved, they just have to do that sequence twice again. Like you said, similar to "sexy move" where you repeat it until what you need done is done.

yup ... that works if you solve to D-layer pairs first, but not for solving all pairs ... unless of course you can demonstrate a solve of the case I outlined above

For now, let's forget about the Step 3 option I had in my original post and just go with the first idea of having two completed pairs.

I have no problem with the original technique, it works quite nicely actually.