I buy a 4x4x4 octahedron and now I don´t have any idea of how to solve it!! Any hellp?

http://cube4you.com/images/IMG_4717.jpg

It may take some guess-and-check to solve the centers, but with logic you can finish the rest of the puzzle fairly easily like a normal 4x4x4.

There are two ways I've come up with for this puzzle.

The first is to solve it like a 4x4, using an "edges-first" approach. This way, you end up doing the "centers" (actually the corners) last. This isn't too difficult because of all the duplicated pieces, but you'll need to learn some new algorithms.

The second method, that I haven't actually tried yet, is "one-color-at-a-time". Once you've got one color solved, you can treat the puzzle a little like a 2x2 and rotate individual faces, because you can use the complete face to "hide parity". Then, you can (ideally) solve each color in a similar manner to the centers on a big cube - slide in, rotate, and slide out.

I solved all the corners except for two that need to swap places. How can I do it?
For example in the picture below I need to change the blue and purple corners.

http://cube4you.com/images/IMG_4717.jpg

Swap any two identical edges, and the corners should be solvable.

This picture does not show your problem.
I assume that you mean by "corner" the pieces that look like a corner (three rhombi per face) but are the centres of the underlying 4x4x4. (The real corners of the 4x4x4 are the little triangles in the middle of a face.)
I solve it face by face. A face is a "subcube" of the 4x4x4: One corner (little triangle), three edges (trapezoid), three centres (rhombus that looks like a "corner". If you view an ordinary 4x4x4 as a 2x2x2 such a face is equivalent to the corner of the 2x2x2).
If you know an algorithm that exchanges the centres of a 4x4x4 and leaves the rest of the cube unchanged, it should be easy.
It is not too hard to find a pure 3-cycle of three centres for two adjacent faces on a normal 4x4x4.
It's late now and I'm going to bed, but if these hints are not sufficient, you may ask again.

I try to solve as a normal 4x4x4. Centers first (in this case corners) edge and for last the corners (little triangles). But look what happens. How can i fix it? All the others pieces are in the correct position!

Isn't that OLL parity?

Yeah. That's when an edge on a 4x4x4 is flipped.

I think that is parity too. But my alg for this case, that I use for a normal 4x4x4, scramble others pieces. I dont know how to solve it.

When you solve it with your strategy, you handle it like a 4x4x4 Supercube. (The centres - which are the corners here - have a strong order!)
This situation cannot happen on a normal 4x4x4 Supercube: When you have solved the centres, this kind of parity is impossible. What is different here is the fact that three edges (the three trapezoids on one face) are identical.
You should use a pure 3-cycle that exchanges 3 edges: 1. white -> 2. white -> purple -> 1.white
This exchanges virtually one pair of white and purple.
Here is a 3-cycle that leaves everything else on a 4x4x4 Supercube untouched:

in WCA notation, please highlight:
{ l' U R U' l U R' U' This commutator does a pure 3-cycle: lUF -> lUB -> RuF -> lUF}
You need to do a setup of your 3 edges first.
This shouldn't be very confusing, as you can turn around "cube halfs" (4 faces around a corner of the octahedron) freely without doing anything to the solved faces.
Whatever your preferred color scheme for the octahedron may be, you can easily restore it after having virtually swapped the two edges.
I'm using a similar commutator when solving the "corners". { l d' l' U l d l' U' }
Good luck!

WOW!!! Fantastic!! I finally solve it. Thankyou so much for your help.

Glad to hear that! Congratulation!

I just got this puzzle and I've tried solving it this way. It actually works, until you get down to the last 3 faces, then there isn't enought room to swap edges around.

Why did you bump the topic?

The answer is right there

Anyway, that's exactly what happened to me. I solved 5 faces, but I didn't know how to solve the last 3 faces. After that I tried the usual reduction method, but got confused when it came to the last layer and the orientation of the pieces
But as has been discussed here, both methods are perfectly feasible... I guess in the beginning it's just a bit difficult to picture it as a 4x4.

It's like a pictured 4x4x4.



I first solved centers (corners on this puzzle) , then edges and finaly corners (small triangular pieces).
I use Harwick's method for 4x4x4 , and 2 algorithms for pictured 3x3x3.
It's just uneasy to see pieces as they should be on a cubic 4x4x4.

I think it's, with the crazy 4x4x4, one of the most interesting and amazing mass-produced puzzles of the moment.

I think, the more natural way solving it, is face by face. There is so much redundancy (three identical edges, three identical corners (centres on the 4x4x4), that even the last faces are no real problem.
I prefer this method, because there is no natural color scheme for the 8 faces relative to each other.
Regarding Crazy 4x4x4: I think the type II is a bigger challenge than type I and really a good puzzle.
If you are looking for a great challenge that is mass-produced, try one of the harder Crazy 3x3x3, like Neptune, Earth or Mars.

I agree 92% cause I still haven't solved any of my 8 crazy 3x3x3 (Smaz Mod, 8 in 1).
the crazy 4x4x4 type I is quite boring but type II is very funny.

Haha, 1 % for each of the 8 variants

Ah, there's the trick.

Remember that you can rotate each face individually as long as you have at least one completed face! So, you can "push in", rotate the face, and then "pull back".

There is really a lot of flexibility in solving this puzzle. It's turned into one of my all-time favorites.

I don't have an FTO, I have a CTO.

And the advice still holds.

Actually, I find this puzzle pretty easy, but I find your advice not easy at all
Is it just me?
What do you mean by "face" here?
A solved face on this cube consists out of 7 pieces: one little triangle = corner on the 4x4x4, three trapezoids in the same color = edges on the 4x4x4; 3 rhomboids in the same color = centres on the 4x4x4
A face represents 1/8 of the corresponding 4x4x4.
If you turn a face without changing it, you turn half of the cube.

What do you mean by "Remember that you can rotate each face individually as long as you have at least one completed face!"
If you mean an algorithm on the corresponding 2x2x2, I can follow, but this would be a complex paradigm and not very helpful, in my view.
Anyway, an advice that is a puzzle in itself, seems not very helpful to me

*brain explosion*

Do you know of some site that might show a picture or a video of the last three faces? I swear to God I'm trying really hard, but I can't seem to keep track of all the faces.

I do not know a better place than this thread
Have you had a look at the hints from June 30th I had given to fermf above that helped to solve it?
(You need to highlight my hints. The notation corresponds to a 4x4x4. You hold the Octahedron that one corner (of the complete Octahedron) becoms U (up), one R, one L; l and r are the adjacent inner layers)
It's not so hard really, if you look at it the right way.
The two pure 3-cycles leave all other pieces untouched besides three edges or three corners of the ocathedron.
EDIT:
Here is the Edges 3 Cycle:
l' U R U' l U R' U'
and the picture shows the change on a solved Octahedron


here is a net for this:


and here a "centre" 3 cycle

l d' l' U l d l' U'

the solved cube after this algorithm:



and the net for this

That's what I meant - use 2x2 methods to rotate faces (octants). Hold the puzzle with the uncompleted faces on top and the face you want to rotate facing you. If the tips to its lower left and right are called L and R, just do (L R' L' R) x2 or (R L' R' L) x2, treating the puzzle as a 2x2. This rotates the upper face. The lower faces stay solved but change positions.

...Well, it works for me, anyway.

Thanks konsassen! I finally solved it. I am happppppppppppyyyyyyyyyyyyyyyyyyyyyy!

May I ask, if the pictures were helpful?
If yes, the photos or the nets?

The photos were very helpful. I can now solve my CTO without having to look at any algorithms at all.

Because of feature of the form and colours of sides, any parity of edges on Octahedron 4х4х4 can be solved with the help of simple eight-running 3-cycles, using only intermediate turns:

How to change 2 centres on Octahedron 4х4х4.

Variant 1 (14 turns):



Variant 2 (10 turns):