=0D
>This just gave me an idea of a method: =0D
>1. Solve everything but one face of the cube=0D
>2. Orient all pieces of that face (let it be face A) so that their =0D
>stickers that belong to face A are in the face A=0D
>3. Solve that face like a 3x3x3 cube.=0D
=0D
>I'm going to try this right now.=0D
=0D
>Sebastian=0D
=0D
I once thought of this method too.=0D
But I found two points that are interesting.=0D
=0D
First, to make 3D-like moves on the last face. You will have to turn one of=
the six adjacent other faces. You have to choose one and just one in order=
not to destroy what has been done before.=0D
=0D
If we say a clockwise turn of this face is +1 and a counter clockwise turn =
is -1 then when you have solved the last 3*3*3 you must be at 0 mod 4. Othe=
rwise, the last face will be ok as expected but not the rest.=0D
=0D
I used 3*3*3 method to solve the 3*3*3*3 since it was 2 mod 4, I had to run=
it twice to do what i wanted.=0D
=0D
Here is an other point:=0D
i doubt the last 3*3*3 is a classic 3*3*3.=0D
Think of the 3D cube. Take two 2-colors piece that are on the same 3*3*1 bu=
t opposite. In 3d you can't just exchange them (and leave the rest of the c=
ube done).=0D
But in 4D you can! Even if the stickers on the last face are with the good =
colors.=0D
=0D
What are your results so far?=0D
Please let me know if you think I'm mistaken or if you have precisions.=0D
As I come to think about it, maybe my 0 mod 4 condition from above can be d=
own to 0 mod 2...hum... I will check.=0D
=0D
Marc.=0D
=0D
=0D
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