Need help solving my Shenshou 8x8x8

[sainclair]

Hello,

I recently got an 8x8x8 Shenshou cube and I've managed to solve all the centers but the last two, and I'm having a hard time understanding how to solve these last two centers. Unless someone has solved an 8x8x8 and can tell me that it's markedly different from solving the 7x7x7's edges I don't think I'll need it.

I've searched far and wide and cannot find a guide to solve an 8x8x8, in fact most places don't even have discussions for it. So can anyone give me some pointers on how to solve the last two centers on an 8x8x8, and maybe the edges too?

 

[drewsopchak]

Form lines. Solve it like the last two 7x7 centers. Use commutators if you have to in order to finish it off.

 

[moralsh]

Hello,

I recently got an 8x8x8 Shenshou cube and I've managed to solve all the centers but the last two, and I'm having a hard time understanding how to solve these last two centers. Unless someone has solved an 8x8x8 and can tell me that it's markedly different from solving the 7x7x7's edges I don't think I'll need it.

I've searched far and wide and cannot find a guide to solve an 8x8x8, in fact most places don't even have discussions for it. So can anyone give me some pointers on how to solve the last two centers on an 8x8x8, and maybe the edges too?

Edges or centers?

I don't own a 8x8x8 but I don't think there's nothing markedly different as in other major cube on solving centers. Could you post a pic or a sketch of your problem?

 

[Goosly]

Once you can solve a 6x6, you should be able to solve anything higher than that

 

[ben1996123]

I do them exactly the same as 7x7 last 2 centres.

Label each layer of the centres from 1-6 left to right, I solve in this order on the F centre: 3, 2, 1, 4, 5, 6

If you do this, you might find that you can't solve one of the blue, red or green pieces (you might have 1 red and 1 blue left for example. By red and blue, I mean the type of piece in the image, not the actual colour of the sticker)

Spoiler

 

[Uberzj]

I call my method, which is probably the common commutator method, "Cross-out".
We will use red and blue for simplicity.
Place the red that needs to be moved onto the blue face relative to a blue piece on the red side.
Meaning if the red piece is 3 over and 3 up, the blue piece must be 3 over and 3 up on the other side.

Move the slice with the red piece to be moved onto the blue.
Move the blue side clockwise or counter, doesn't matter.
Move the slice containing that piece up again.
This creates a cross pattern of red later in the progression.
Turn the blue side reverse of earlier.
Turn down the first slice turn performed.
Turn the blue side the same way as first.
Turn down the second slice turn performed.

This is a very logical way of solving the last two centers.

 

[ben1996123]

I call my method, which is probably the common commutator method, "Cross-out".
We will use red and blue for simplicity.
Place the red that needs to be moved onto the blue face relative to a blue piece on the red side.
Meaning if the red piece is 3 over and 3 up, the blue piece must be 3 over and 3 up on the other side.

Move the slice with the red piece to be moved onto the blue.
Move the blue side clockwise or counter, doesn't matter.
Move the slice containing that piece up again.
This creates a cross pattern of red later in the progression.
Turn the blue side reverse of earlier.
Turn down the first slice turn performed.
Turn the blue side the same way as first.
Turn down the second slice turn performed.

This is a very logical way of solving the last two centers.

it's also really slow.

 

[sainclair]

I call my method, which is probably the common commutator method, "Cross-out".
We will use red and blue for simplicity.
Place the red that needs to be moved onto the blue face relative to a blue piece on the red side.
Meaning if the red piece is 3 over and 3 up, the blue piece must be 3 over and 3 up on the other side.

Move the slice with the red piece to be moved onto the blue.
Move the blue side clockwise or counter, doesn't matter.
Move the slice containing that piece up again.
This creates a cross pattern of red later in the progression.
Turn the blue side reverse of earlier.
Turn down the first slice turn performed.
Turn the blue side the same way as first.
Turn down the second slice turn performed.

This is a very logical way of solving the last two centers.

Admittedly it is slow, but it worked! Thank you to everyone who replied. Now I'm going to go find out how to solve the edges on this thing.
Thanks!

 

[sainclair]

Strange 8x8 parity

I imagine that it's rude to do this, but I felt that this was unrelated enough to post in a new thread.

I've solved all the centers and all the edges, except for the last two, and I don't quite understand how to solve this.

I can't use any reversing parity algorithms to fix the swapped red/yellow pieces, and I don't know how to replace the two pieces that are on the wrong row without disturbing another edge I've already built.

And the explanation for why I don't know any of these parity algorithms is because I only have two cubes: the 3x3 and this 8x8, and only because a friend who can't solve cubes gave me this one do I have it.

So, thanks!

 

[Sa967St]

Thread merge + title change. Might as well make this a general 8x8x8 help thread.

 

[Cool Frog]

(post got lost but here it is again)

you have "parity" on the outer two edges, that can be solved by doing the OLL parity algorithm gripping the outer 3 layers, it then should be solvable by doing 3 cycles (Slice, flip edge, slice back)

 

[moralsh]

You can also move away one of the missoriented edges and then insert it back in the place of the other missoriented edge, you will have to unsolve another edge but with a little bit of planning you shouldn't have any problems solving afterwards.

then when solving the 3x3x3 you may find parities that you can solve with the 4x4x4 algs just gripping half of the cube instead of 2 layers, at least that's how I do it on my 6x6x6, as I don't own a 8x8x8

 

[Mikel]

Once you can solve a 6x6, you should be able to solve anything higher than that

I disagree. Once you can solve a 4x4 you should be able to solve anything higher than that. I didn't learn any new algs to solve my 5,6 or 7x7 cubes.

 

[moralsh]

I'd say "Once you can solve a 4x4 and a 3x3, you should be able to solve anything higher than that" which I agree although I can understand people who have problems with solving centers on bigger cubes

 

[ben1996123]

Hold your cube like this

Spoiler

R2 u2 R' U2 R B U B' u2 R2 z' 4R U2 4R U2 4R' U2 4R U2 4L' U2 4L x U2 4R' U2 4R' x' U2 4R'

u2 = top 2 layers
4R = 4th slice from the right only

 

[sainclair]

Okay, so it's been a long time, but since this was designated as the help thread for this cube size, I'll revive it.

I let the cube go, and finally I decided to solve it again, and this time I managed, without parity, to reduce it to a 3x3x3, but the problem now is that I seem to have solved the centers' locations incorrectly with respect to each other. That is, there are centers with shared edges that I have managed to put on the opposite side of the puzzle. I see no clear way to solve this, and it's troubling me.

 

[XTowncuber]

4r U2 4r2 F2 4r. That should switch the U and F centers without messing up the edges. For opposite centers do 4r2 U2 D2 4r2 (switches U and D). With those algs you should be able to get the centers into the proper color scheme.