Solve Rubik's Cube? You Liar!!

[Harry]

OK, that is my friend's line when he knew I solved it using algorithms and also with the one I took from internet (Pjk's one). I knew he just kidding, but......

Anyway, I don't know which topic this should be on, so I put it on off-topic discussion.

So, my question is, How do you solved it without using any algorithms or patterns, purely intuitive. Can you solved it?

 

[Dene]

That depends on your definition of what is intuitive, and what is an algorithm. For example, Lars Petrus, I understand, invented a few of what are now very common algorithms. But when he figured it out for himself it was "intuitive". Let's say, if you were to get a OLL+PLL skip, and did the F2L all intuitively it would be a completely lucky solve. If you could figure out last layer algorithms for yourself it would be intuitive.

 

[Karthik]

If you know the theory of commutators then, you can easily solve the cube "intuitively".

 

[cmhardw]

So, my question is, How do you solved it without using any algorithms or patterns, purely intuitive. Can you solved it?

I felt the same way about my solving at first, that I was just applying algorithms that I didn't understand. To be honest, for speedcubing I still feel that way, but yes I agree with Karthik, you can solve the cube intuitively if you understand commutators. Also you may be surprised, lots of COLL algs are just commutators. The A perm is just a simple commutator.

So to answer your question, when I speedsolve I am still just blindly applying memorized algs to solve, most of which I have no concept of why they work. But if you solve with commutators it can be 100% intuitive. Try learning BLD too, that really helps build your understanding of the cube on an intuitive level.

Chris

 

[Karthik]

There is a limit up to which one can look ahead and keep track of pieces while devising an algorithm.And since most of todays algorithms are computer generated, we shouldn't be feeling uneasy about not understanding "speed" algorithms.This is what I personally think, when I am learning algs for speedcubing.

 

[hait2]

Sure, just learn the theory of commutators, and the parity fix 'algorithm' (a single face turn) and you can solve the cube intuitively. it's pretty fun

 

[badmephisto]

indeed commutators will do this easily. Do F2L intuitively, then use a commutator to orient both corners and edges, and finally permutation of the pieces you dont even need commutator for. There is a beautiful way of doing it, using R' D2 R repeatedly for corners, and M U2 M' for edges.

for example for corners:
do:
x' R U' R D2 R' U R D2 R2 x (A perm)
to solve it intuitively you can just swap corners in and out, almost like in M2R2 BLD method. You will swap FUR and DBL always. So to solve this A perm just do
R' D2 R
U'
R' D2 R
U2
R' D2 R
U'
R' D2 R

done.
I also have an introduction to commutators video on YouTube if you are interested. Its in my channel.

 

[joey]

Badmephisto: That is basically a bastardized commutator!

 

[Harry]

Erm, guys and girls what is commutattors? A-perm?

 

[badmephisto]

Badmephisto: That is basically a bastardized commutator!

It seems like it should be and yet I cant figure out how. If commutator is ABA'B', then what are A and B in that approach :s I was confused by it the other day.
Its not REALLY a commutator. At least i dont think.


@Harry: Commutators are algorithms that only modify very small portions of the cube. And they are intuitive. For example an algorithm to twist two corners can be expressed as a commutator.

 

[Harry]

OKay, Mr Mephisto, where is your youtube cahnnel again?

 

[cmhardw]

for example for corners:
do:
x' R U' R D2 R' U R D2 R2 x (A perm)
to solve it intuitively you can just swap corners in and out, almost like in M2R2 BLD method. You will swap FUR and DBL always. So to solve this A perm just do

Also the A perm itself is just a simple commutator.

From the same angle you gave it after the x' turn:

A: R' U' R
B: D2

so the commutator is (R' U' R) (D2) (R' U R) (D2)
now add a R2 setup turn, and undo it at the end.

R2 (R' U' R) (D2) (R' U R) (D2) R2

Which becomes R U' R (D2) (R' U R) (D2) R2 since the R2 setup turn cancels with the first R'. Daniel and I call these 9 movers, and more specifically A9 movers since the cancellation from the setup turn cancels with the A part of the commutator.

So yes, A perm is fair game for intuitive solving ;-)

Chris

 

[badmephisto]

Chris thats awesome I knew that had to be commutator somehow because of its structure, but I never really tried to see it. I wish I knew what connection the R' D2 R thing has to commutators though

And Harry, here is the link to my Commutators video: (I hope you know F2L or this could get confusing at some points)
http://www.youtube.com/watch?v=7yZoDi_B1lI

or you could just search Commutators on Youtube, the first link is my video. (Which by the way is so cool i just found out about that lol)

 

[Harry]

Okay, Mr Badmephisto, I watch your videos already.... So you meant that I must invent my own commutators? How about if the cube is scrambled completely?

Could you tell me other perm? and the effect?

THanks everyone

 

[badmephisto]

Well... you invent your own commutators, you see what they do, and then you apply them on your cube depending on what you want to do. Instead of OLL for example, you could do the commutator that flips edges and then the commutator that twists corners... that will solve OLL. and then use the A perm and U perm to solve PLL, but of course, the A perm and U perm commutator version, which is a little longer but intuitive.

 

[hait2]

for example for corners:
do:
x' R U' R D2 R' U R D2 R2 x (A perm)
to solve it intuitively you can just swap corners in and out, almost like in M2R2 BLD method. You will swap FUR and DBL always. So to solve this A perm just do

Also the A perm itself is just a simple commutator.

From the same angle you gave it after the x' turn:

A: R' U' R
B: D2

so the commutator is (R' U' R) (D2) (R' U R) (D2)
now add a R2 setup turn, and undo it at the end.

R2 (R' U' R) (D2) (R' U R) (D2) R2

Which becomes R U' R (D2) (R' U R) (D2) R2 since the R2 setup turn cancels with the first R'. Daniel and I call these 9 movers, and more specifically A9 movers since the cancellation from the setup turn cancels with the A part of the commutator.

So yes, A perm is fair game for intuitive solving ;-)

Chris

i always thought of the Aperm as
setup: R U', setting up the 2nd and 3rd stickers to be interchangeable via U/U'
P = R D2 R'
Q = U
so final

(R U') (R D2 R') U (R D2 R') U' (U R')
where the last 4 cancel out to an R2

although your explanation makes sense as well i guess

 

[cmhardw]

i always thought of the Aperm as
setup: R U', setting up the 2nd and 3rd stickers to be interchangeable via U/U'
P = R D2 R'
Q = U
so final

(R U') (R D2 R') U (R D2 R') U' (U R')
where the last 4 cancel out to an R2

although your explanation makes sense as well i guess

I have to admit that I have never thought of the A perm that way, but now that I look at your example I can see it. That's really neat! I never would have thought to think of a 2 turn setup where one turn of the setup completely cancels out one turn of the commutator. I've always just tried to find 1 turn setups that combine with a turn of the commutator, but not cancel.

Neat!

Chris

 

[Harry]

OK2, so which way is faster your own commutator or the algorithms?

 

[hait2]

OK2, so which way is faster your own commutator or the algorithms?

I assume you're referring to algorithms in the sense of non-intuitive memorized sequences of moves (which may be commutators but that's beside the point)

Well, faster in what sense? Commutators usually aren't as finger trick friendly (although they can be optimized), but they're a lot more flexible. You can't learn an algorithm for every possible cycle because there's .. a lot of cycles.

Basically I'd say pre-learned algorithms are faster to solve a predictable and limited stage of the cube (such as PLL). But in just about every other situation in direct solving, you'd be better off with commutators (such as BLD for instance)

 

[Harry]

So, if I want to get faster, I should learn the algorithm first? Since I only undestand some of the Mr. Badmephisto's explanation......