side by side...?

[daniel0731ex]

hello there

i suppose that you all have tried solving the cube side by side when they first got it, right?

i wonder if solving it like that is possible?

 

[nitrocan]

Of course why not?

But I doubt that it'd be effective.

 

[d4m4s74]

if you solve two opposite sides first you could do the rest using OLLs PLLs and setup moves

though it's not really effective

 

[nitrocan]

if you solve two opposite sides first you could do the rest using OLLs PLLs and setup moves

though it's not really effective

Or you could do Roux.

 

[Fobo911]

Though that was totally how I thought a cube should have been solved all the way in the past (when I was a cube stooge), I've never seen anyone solve it like that before. Is there any video of it?

 

[blah]

I've never seen anyone solve it like that before.

As mentioned by nitrocan, classic Roux does exactly that. Just that the order in which the "first two sides" are done is kind of messed up. Does that count?

P/S: I said classic Roux because I believe there are more advanced Roux users who don't necessarily solve UL and UR first at the 6E4C stage.

 

[Ryanrex116]

I have done this a few times, but it is very slow.

Actually, you cant do 1,2,3,4,5,6. You have to do 1,2,3,6.

 

[fanwuq]

I've never seen anyone solve it like that before.

As mentioned by nitrocan, classic Roux does exactly that. Just that the order in which the "first two sides" are done is kind of messed up. Does that count?

P/S: I said classic Roux because I believe there are more advanced Roux users who don't necessarily solve UL and UR first at the 6E4C stage.

But in Roux, you actually solve 2 sides at a time, not side by side. Also, there's that EO step that would appear pointless to a noncubist.

 

[deco122392]

[/QUOTE]

But in Roux, you actually solve 2 sides at a time, not side by side. Also, there's that EO step that would appear pointless to a noncubist.[/QUOTE]

well not really, scence when i began cubing i did an intuitive cf method and the eo step seemed pretty nessessary when i had 2 or 4 fliped edges in the m layer

 

[Stefan]

Actually, you cant do 1,2,3,4,5,6. You have to do 1,2,3,6.

I don't think so.

 

[AvGalen]

Actually, you cant do 1,2,3,4,5,6. You have to do 1,2,3,6.

I don't think so.

That's not enough hint for Beginners Central Stefan. Maybe one more tiny hint (Stefan style) would help?

 

[ManuK]

How about try "R2 U2 R2 U2 _ _".

When I first started solving cube,I at first thought of solving it side by side.
So,when I learnt how to solve the cube, I wanted to try doing only 2 sides,3 sides,4 sides and then finally 6 sides. Doing 4 sides though took a bit of time to figure out.

 

[TMOY]

The very first time I solved a cube, it went that way. First one side, then two adjacent sides, then three, then four, then the whole cube. But I had no real method, it was completely trials and errors.

 

[Ryanrex116]

Actually, you cant do 1,2,3,4,5,6. You have to do 1,2,3,6.

I don't think so.

That's not enough hint for Beginners Central Stefan. Maybe one more tiny hint (Stefan style) would help?

Sorry. What I meant was: It is impossible to have 4 or 5 sides done. At the end, you will have to go from 3 sides done (correctly permutated) To solved.

In fact, it is impossible to have 3 sides done IF they share a corner. (ex. Blue, Orange and Yellow.)

 

[not_kevin]

Actually, you cant do 1,2,3,4,5,6. You have to do 1,2,3,6.

I don't think so.

That's not enough hint for Beginners Central Stefan. Maybe one more tiny hint (Stefan style) would help?

Sorry. What I meant was: It is impossible to have 4 or 5 sides done. At the end, you will have to go from 3 sides done (correctly permutated) To solved.

In fact, it is impossible to have 3 sides done IF they share a corner. (ex. Blue, Orange and Yellow.)

Um... (U2R2)*3 is an exception? And so is (M2U2)*2?

 

[pcharles93]

Actually, you cant do 1,2,3,4,5,6. You have to do 1,2,3,6.

I don't think so.

That's not enough hint for Beginners Central Stefan. Maybe one more tiny hint (Stefan style) would help?

Sorry. What I meant was: It is impossible to have 4 or 5 sides done. At the end, you will have to go from 3 sides done (correctly permutated) To solved.

In fact, it is impossible to have 3 sides done IF they share a corner. (ex. Blue, Orange and Yellow.)

Um... (U2R2)*3 is an exception? And so is (M2U2)*2?

Peel off a center sticker...

 

[Ryanrex116]

Actually, you cant do 1,2,3,4,5,6. You have to do 1,2,3,6.

I don't think so.

That's not enough hint for Beginners Central Stefan. Maybe one more tiny hint (Stefan style) would help?

Sorry. What I meant was: It is impossible to have 4 or 5 sides done. At the end, you will have to go from 3 sides done (correctly permutated) To solved.

In fact, it is impossible to have 3 sides done IF they share a corner. (ex. Blue, Orange and Yellow.)

Um... (U2R2)*3 is an exception? And so is (M2U2)*2?

Those DO get four sides done, but they are not correctly permuted.

 

[not_kevin]

Actually, you cant do 1,2,3,4,5,6. You have to do 1,2,3,6.

I don't think so.

That's not enough hint for Beginners Central Stefan. Maybe one more tiny hint (Stefan style) would help?

Sorry. What I meant was: It is impossible to have 4 or 5 sides done. At the end, you will have to go from 3 sides done (correctly permutated) To solved.

In fact, it is impossible to have 3 sides done IF they share a corner. (ex. Blue, Orange and Yellow.)

Um... (U2R2)*3 is an exception? And so is (M2U2)*2?

Those DO get four sides done, but they are not correctly permuted.

Ah. Thank you for pointing that out.

 

[TMOY]

It depends on what you"re calling a side.
For me, solving a side is putting all stickers of the same color on it, disregarding permutation. for example, doing a Fridrich OLL is solving the U side.
If you count permutation as well, you're solving the layer, not only the side.

 

[Stefan]

If you count permutation as well, you're solving the layer, not only the side.

Exactly.

In fact, it is impossible to have 3 sides done IF they share a corner. (ex. Blue, Orange and Yellow.)

And wow, this is wrong even if you replace "side" with "layer".