# Message #160

From: "marc.guegueniat@libertysurf.fr" <marc.guegueniat@libertysurf.fr>

Subject: Re:[MC4D] Re: new member

Date: Thu, 30 Jun 2005 17:44:29 +0200

>This just gave me an idea of a method:

>1. Solve everything but one face of the cube

>2. Orient all pieces of that face (let it be face A) so that their

>stickers that belong to face A are in the face A

>3. Solve that face like a 3x3x3 cube.

>I’m going to try this right now.

>Sebastian

I once thought of this method too.

But I found two points that are interesting.

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.

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. Otherwise, the last face will be ok as expected but not the rest.

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.

Here is an other point:

i doubt the last 3*3*3 is a classic 3*3*3.

Think of the 3D cube. Take two 2-colors piece that are on the same 3*3*1 but opposite. In 3d you can’t just exchange them (and leave the rest of the cube done).

But in 4D you can! Even if the stickers on the last face are with the good colors.

What are your results so far?

Please let me know if you think I’m mistaken or if you have precisions.

As I come to think about it, maybe my 0 mod 4 condition from above can be down to 0 mod 2…hum… I will check.

Marc.

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