Finally I got my 3x3x6....very nice and smooth puzzle...are there any directions to solve it?...it seems quite difficult...many parity problems..any help should be very appreciated...thanks all!!!

Hi,

I would do it inside-out:
start with the two middle layers, edges only as if they were domino-corners
then layers 2 and 5
then the outer layers
then the (floating) edges

you can practice on a 3x3x4 I would think. I don't have ay real alg's, I usually manage by fiddlig around...
Good luck!

Maarten

Parity Alg: (Uu)2 F R u2 R F (Uu)2 =(centre fix) the rest is domino algs as Maarten said.

Cheers,
Burgo.

Thanks Guys!!! I'll try...I just solved the 3x3x4...it seems that the 3x3x6 is more difficult...I'll see....

I would personally try it by solving the top and bottom layers like a domino, then the middle 4 layers similar to a 6x6. middles, then edges. the same parity algorithms from the larger x by x by x cubes would work. I am sure there is a more efficient way of doing it, but that should get the job done.

Are you aware that this puzzle has four faces that can be turned by 180 degrees only? In Burgo’s post:the R and F moves are a shorthand for F2 and R2, because nothing else can be done.
I do not understand how the analogy with 6x6x6 centre pieces helps?

If you can solve a 3x3x4, you can solve this 3x3x6 as well. As pointed out already, you have just 3 Dominos in one puzzle.
The 3x3x4 can be viewed as two Dominos. A parity problem occurs on the 3x3x4 as well.
So, solving a 3x3x6 should be straight forward, if you can solve a 3x3x4. Maybe it is a little bit more confusing.
BTW, if you want to use Burgo's parity algorithm you have to adapt it for a parity on the outer (U/D white/yellow) layers. The current WCA notation does not have a notation for third layer tutns, therefore I introduce 3u for the layer under u (and u is under U. WCA has 3U for turning three layers together.
It becomes: 3U2, F2, R2, (u3u)2, R2, F2, 3U2

Centre fix= centre `layers` fix.
It was a hint, you will know parity when you get it.
(It is the centre pieces that you are fixing though.. but that sounds confusing unless you can see it, it looks like corners -if you are doing 3 dominos `some 3x3x6 edges` act as `domino corners` at times-). No matter which way you go, outside in or inside out the parity is essentially the same.
Cheers,
Burgo .

Yes. Its the way I solve my 3x3x4, and I'm sure it isn't the best way, but it works. Surprisingly you you can use the normal parity algorithms on the cube, as all the U, B and F turns are U2, B2 or F2, at least for the algorithms that I learned when I learned the 4x4 and up.

If you do Burgo's parity fix on a solved 3x3x6 (orange = F; white = U), you'll get this:
This can be transformed by a few obvious turns to: where two centres are wrong. Which kind of parity you will see, depends on the order of solving steps.
My terminolgy would be:
- centre pieces with one sticker; correspond to Domino edges
- middle layer edges with two stickers; correspond to Domino corners

What causes the "parity" is the fact that centre pieces can be swapped without recognizing this. Each centre piece has an undistinguishable double.

EDIT: @Noreg89: I understand that a 6x6x6 parity algorithm can come in quite handy. (Actually, Burgo's parity algorithm is one of that kind.) As a general method, it seems not so easy to grasp. I believe that viewing this puzzle as a set of three Dominos is much easier.

DONE!!!! It was more difficult then 3x3x4.....I had to fix so many parity problems.....but now is done!!!!
BTW...I hate parity problems....
Thanks all!!!

Hi Superciuk,
I am happy that you have solved your cube.

You should only have to deal with 1 parity problem per solve. If you are careful with your placement with the domino alg application (when you perform domino algs in a different layer, make them only flip the same pieces in the solved layers over and over). And leave the parity till last: If you get 2 cases you can do them together: One application= It should be painless .

Cheers,
Burgo.

It is not really more difficult. If you start from the central layers. you just work it out like the 3x3x4, then you consider the intermediate layers as they were the outer layers of the 3x3x4, then you do the outer layers as in the 3x3x4.

If you start from outside, it is also very similar. It has in fact the same difficulty level as the 3x3x4, it just takes longer...

Roberto