Message #159

From: Sebastian Dumitrescu <>
Subject: Re: new member
Date: Tue, 28 Jun 2005 20:03:17 -0000

> So, I would like to know, which algorithm for the 3d-cube did you
expand for
> solving the 4d-cube?

First of all, the method. I used the Petrus method which you can
find here: The idea of this method is the
construction of blocks of pieces that do not obstruct moving other
faces, so that more and more of the cube gets solved. When only one
face is left to be solved I apply algs which I know beforehand and
which I know how they will affect the cube. The algs I used for the
3x3x3x3 are the ones called Sune and Niklas in the Petrus method.

I used the Sune to permute centres and edges, and the Niklas to
permute the corners. There are two ways of executing an algorithm.
The first way, that which I used for permuting the centres, is
simply pretending that two adjacent faces are two-dimensional, and
the effect of the algorithm is that groups of 3 pieces in 4
dimensions correspond to one piece in three dimensions, that is one
centre piece and two edge pieces that touch that centre piece and
are opposite to each other correspon to an edge piece in three
dimensions, and a column of one edge piece and two corners
corresponds to a corner in three dimensions.

When I used this kind of algorithm, I only cared about permuting the
centres. The second way of doing it, the effect of which looks
exactly like the algorithm done in three dimensions, was used for
the edges. The idea is to pretend we only have to solve the last
face of a three-dimensional cube. We will simulate turning two 3x3
faces of the same 3x3x3 face of the 4D cube. We will turn one of
them by turning a 3x3x3 face (let this face be B) adjacent to the
3x3x3 face in question (let this face be A). Next, we move face A so
that the 3x3 face that we want to move next is oriented towards face
B. Then we move face B. And so on. Only faces A and B will be moved,
and so we can simulate any sequence on moves on the 3x3x3 cube. If
such an alg is done, the centres are not moved at all, and the
colour of all pieces that are in face A that is at the beginning in
face A stays in face A.

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.