The Professor Cube (5x5x5 Rubik's Cube)

Introduction

The Professor Cube is a 5x5x5 version of a Rubik's cube.  If you already have a solid understanding of the 3x3x3 and 4x4x4 versions, then the 5x5x5 should be easy enough to solve (this page gives some hints for the 4x4x4).

This page summarizes some of the techniques I use for solving the 5x5x5 cube.  It is not a step-by-step walkthough of a solution.

Notation

The cube has six faces, which will be called Up, Down,Front, Back, Right, and Left.  Turning a face 90 degrees clockwise is represented by the first letter of the face's name, in upper case.  For example, "R" means to turn the right face 90 degrees clockwise (such that the front right pieces become the top right pieces).  The inner slices are represented by lowercase letters for the face that they are closest to.  For example "r" means to turn the layer just inside of the right face.  There is no notation for the center slices.  A letter followed by a tick mark means to turn the face counter clockwise by 90 degrees (e.g. R').  A letter followed by the number 2 means to turn it twice (180 degrees).  Parentheses are sometimes uses to group moves together for easier readability.

Pieces are described by the faces they are closest to.  An upper case letter means the piece is actually on the face, while a lower case letter indicates that it is one space in from the face.  When a piece is in the middle, no letter is used for that particular dimension.  So, UFR is the top-right corner on the front face.  Moving to the left, we have UFr, UF, UFl, and UFL.

Types of Pieces

The cube contains the following types of pieces:

Solution Outline

  1. Solve the 9 center pieces on each face (see below)
  2. Match Side Edge pieces with their corresponding Middle Edges (see below)
  3. Solve the cube as a 3x3x3 cube using only face moves (no inner slice moves)

Solving the Centers

If you are already know some good operators for moving 4x4x4 centers, then you can solve the 5x5x5 centers by first solving the Center and Center sides for each face, then using the 4x4x4 operators to move the Center Corners into position. 

Solving the Center Sides without preserving any other pieces is actually quite simple.  The following two sequences should be enough:

r' F r : moves uF to Ur
r2 D r2 : moves Df to Ur

Remember that faces can be turned without disturbing any center pieces and you can also turn the entire cube to face any way you want, so it is quite easy to arrange things such that one of the above two sequences solves a Center Side piece.  Do this at most 24 times and the Center Sides will be complete.

The Center Corners tend to take a few more moves.  Initially, I used the following operators such as:
These are really handy on the 4x4x4 cube if you solve the centers last because they don't change edges or corners at all.  However, for the 5x5x5 cube we are solving the centers first, so those operators are overkill. 

Here's a faster operator for fixing a single center corner.
 
center-corner-move: d R d' R d R2 d' - move Rdf to Fdr

Note that it has some side effects.  It cycles some side edges on R and cycles side corners on R, but since we don't care about permuting the side corners within a single face and we haven't fixed the edges yet, these side effects are harmless.

Solving the Centers without Operators

The above solution, although effective, can be a bit tedious.  Just like on a 3x3x3 cube most people solve the first layer or two intuitively before resorting to general purpose operators for the last layer, it is possible to solve the centers without brute forcing each individual center side and center corner.

It is quite east to work with 2 faces at a time without distrubing the centers on the other 4 faces.  The key is to do mostly face moves and always match a slice move with its inverse.  Between the slice and its inverse you have to be careful to only turn a face that consists entirely of pieces that you are willing to move.  For example, when working on the U and F centers you could do something like this:

U2 F r' F2 r U

Actually, the stuff before r' and after r is irrelevant - it can be any combination of U and F moves that you want.  Since we're only concerned with the centers, the order of the moves doesn't even matter since upper centers are only affected by U and front centers are only affected by F.  The important part is that between r and r', you restrict yourself to turning F.

Matching Up Edges

The goal here is to form edge groups by matching each middle edge with the two corresponding side edges.  One very important thing to keep in mind is that face moves always move middle and side edges as a group.  Thus, any number of face moves can be made without disturbing previously matched up edges.  However, you still need a way to move side edges independent of middle edges, and the 4x4x4 cube's edge-3-cycle is your best friend here:

edge-3-cycle: r (U' R U) r' (U' R' U)

This will cycle DFr, UFr, FRu.  Actually, you can skip the last three moves and just do r U' R U r', because the last three moves are face moves which have no effect on matching of edges.  Their only purpose is to restore edge groups and corners to their original position.  Since we only care about matching up the edges, and not where the edge group (or corners) wind up, we can skip those last moves.

I usually just pick a piece at DFr, and use face moves to move the corresponding middle edge to UF.  Make sure you get UF flipped the right way (the front face of DFr should match the up face of UF).  Then examine UFr and put the corresponding middle edge at FR (the F face of UFr should match the R face of FR).  Then execute the edge-3-cycle and you are 2 pieces closer to matching all of the edge groups.

It is possible to wind up with only two unmatched side edges that need to be swapped with one another.  No amount of manipulation with edge-3-cycles will be able to solve this problem.  Another operatoin is required.  Basically, you need to create an odd permutation of the edges.  Once you do this, you should then be able to perform one or more 3 cycles to clean things up.  My preferred way of solving this problem is to put the two problematic edges at uFR and uBR, then do this:

edge-4-cycle:  R2 u  R2 u R2 u R2 u R2 u R2. 

This cycles uFR, uBR, uBL, and uFL.  Since uFR and uBR needed to be swapped, after the 4-cycle, exacly three side edges will be out of place and an edge-3-cycle should be able to fix them.