Rubik's Dodecahedron

Introduction

Technically, this is neither a "Rubik" puzzle, or a cube. But it does share a lot in common with Rubik's Cube puzzle. Essentially it is a 12 sided polyhedron sliced up similarly to the Rubik's 5x5x5. You can take a look at this monstrosity here.

This page summarizes some of the techniques I use for solving this puzzle. It is not a step-by-step walkthough of a solution. It may help to first be familiar with the techniques I use for the 5x5x5 cube.

Notation

I haven't come up with a good notation for the faces yet. On a cube, there were intuitive names for the faces (Up, Down, Left, Right, etc), but we just don't have natural terminology for all 12 faces of a dodecahedron.

The following notation is awkward but precise:

Place the puzzle flat on a table and call the top face A. There will be a ring of 5 faces that touch this top one. Call the one facing you B and then proceed around clockwise to C, D, E, and F. All of these faces are on the top half of the puzzle. Now consider the bottom half of the puzzle. G will be the face opposite of F (left side facing slightly away from you), H will be opposite of E, I will be opposite D, etc until you end up with the bottom face L.

Turning a face clockwise 72 degrees is represented by the face's upper case letter (e.g. F). The inner slices are represented by lower case letters (f). A letter followed by a tick mark means to turn the face counter clockwise by 72 degrees. A letter followed by a number indicates that the move should be repeated multiple times (e.g. F2 would mean to turn face F twice).

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, ABF is the top-right-front corner.

Comparison to 5x5x5 Cube

This puzzle is remarkably similar to a 5x5x5 cube. Although the shapes are different, the same kinds of pieces are present. There are just a lot more of them.

20 Corners (example ABF)
30 Middle Edges (example AB)
60 Side Edges (example ABf)
60 Center Corners (example Abf)
60 Center Sides (example Ab)
12 Centers: these never move in relation to one another

Like with all cube puzzles, the spin of the corners always is equal to 0 mod 3, and flipping of middle edges must be done in pairs (total flips is equal to 0 mod 2).

Unlike the cube puzzles, the primitive move (turning a face or slice) is always an even permutation of each category of pieces. For example, turning a face will perform a 5-cycle on corners, which is an even permutation (even permutations are those that can be constructed from an even number of 2-element swaps). With the cube puzzles, the basic move is a 4-cycle (odd permutation), which means you can get into situations where you need to swap one pair of edges and one pair of corners. On a 5x5x5 cube you can have additional parity problems where it appears you need to swap a single pair of side edges. On the dodecahdron, none of these problem arise since the permutations of various types of pieces always must be even.

With 12 faces, the dodecahedron is a lot bigger (230 pieces not counting centers, compared to 20 for a 3x3x3 and 92 for a 5x5x5 cube). This means that you have a lot more rearranging to do in order to solve it, but you also have a lot more "scratch space" to work with. Overall I find that this makes the solution conceptually simpler but it takes longer to execute.

Faces A, B, F, and I work together similar to how U, F, R, and D work on a cube. Many of my basic cube operators translated to the dodecahedron using this mapping, perhaps because I focus on the URF corner and its surrounding pieces on a cube.

Solution Outline

Solve the 11 center pieces on each face.
Match Side Edge pieces with their corresponding Middle Edges. At this point the puzzle has been reduced to a 3x dodecahedron.
The puzzle has been reduced to a 3x dodecaderon (instead of 5x). Solve it without using any inner slice moves

Solving the Centers

This part of the solution is analagous to solving the centers on a 5x5x5 cube. It is easy to swap center edges and center corners between adjacent faces without disturbing center pieces on any other faces. So you can work on one face at a time and never disturb a previously completed face. You can completely ignore edge and corner pieces - just focus on the center 11-piece pentagons on each face.

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. The edge 3-cycle from the 4x4x4 and 5x5x5 cube works perfectly fine on the dodecahedron since it only relies on moves of two adjacent faces (using face F in place of R and A in place of U):

edge-3-cycle: f (A' F A) f' (A' F' A)

This will cycle ABf, BFa, and BIf. Actually, you can skip the last three moves because they 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.

Because all moves have even parity with respect to side edges, you will never get stuck in a situation where exactly two side edges need to be swapped (this was a rather annoying problem to solve on the 5x5x5 cube).

Solving the Dodecahedron

I'm still experimenting with different techniques here. Here's an approach that I feel works pretty well. It relies heavily on assembling pairs of a corner and adjacent edge, then placing that pair into position. In general this operation can be constrained to altering a single face, one edge that intersects that face, and the far corner from that edge. For example, if the desired corner and edge are in A, then it is pretty simple to juggle the corner and edge in ways that only affect the pieces in A plus edge BF, and corner BFI. Once you are comortable manipulating an edge an corner within this small "subspace" you should have no trouble with the rest of the solution.

By the way, from now on when I refer to an edge such as BF, I am really referring to the triple of BFa, BF, and BFi (the middle edge and its two side edges).

First, solve the bottom edges (LG, LH, LI, LJ, LK). This should be easy to do intuitively. You can alter anything else in the puzzle aside from the edges you have already places.

Second, word around the bottom layer placing pairs involving the bottom corner and its adjacent edge that doesn't lie in L. For example, place corner LHI along with edge HI. The top layer and all five upper faces can be trashed. Generally I assemble the pair in the upper face adjacent to the pair's final location, then move it into place. For example, when working on LHI/HI I would probably arrange the pair in face B then do something like IB'I' to move it down into place. At this point the bottom face and most of the lower faces (G-K) should be solved. All edges/corners in the upper faces and top face are still scrambled.

The third phase is to solve the 5 corners that touch 2 lower and 1 upper face (for example BHI) along with the two adjacent edges that have not yet been solved (BH, BI). Placing the first edge is easy since the related upper face (B) does not have to be preserved. Adding the remaining two pieces without disturbing the first edge is very similar to how pairs were placed in the second step. The top face is a good scratch space, as are the upper faces that you haven't worked on yet. But be careful not to destroy your work as you go. For example, once you place BHI/BH/BHI you should take care to preserve B. At this point 10 corners 20 edges have been solved.

Step four is to solve the remaining edges and corners that aren't in the top layer. Once again build a pair (such as BFI,BF) and then drop it into place. Use the top layer as scratch, but preserve everything else. After the completion of this phase everything except the 5 corners and 5 edges in the top layer will be solved. You're almost done.

There are actually four different things that now need to be addressed. Each can be done indepdenently (assuming you know the right operators). But moves are very constrained since almost all of the puzzle has been solved. Here is the order that I approach is, not because it makes the solution any simpler, but merely because it makes more visible impact right away and prevents me from getting lost during longer sequences.

Flip the edges. Some edges will already have the top color on their top side. But an even number (0 if you are lucky) will need to be flipped. The building block for this is a move that flips AB while leaving the rest of the top layer intact (but has some nasty side effects on the rest of the puzzle):
edge-flip: FC' B2I'H'B2 CF'
Since we want to preserve the rest of the cube we need to undo the side effects. We can do this by reversing the edge-flip, but in before we do that we'll twist A, thus flipping AB and AF:
double-edge-flip: (FC' B2I'H'B2 CF') A (FC' B'2HIB'2 CF') A'
It looks complicated but it isn't if you realize that the second moves in parentheses are just the inverse of the first group. The FC'/CF' pattern moves top layer corners out of the way. The rest (B2I'H'B2) just turns the AB edge around and moves it back.
Twist the corners. Some corners will need to be twisted clockwise or counter-clockwise in order for the top color to match the top face. If you count clockwise twists as +1 and counter-clockwise as -1 then the total amount of twisting will always be equal to 0 mod 3. You can always achieve this with pairs of clockwise/counter-clockwise twists on two corners. Just like with edge flipping we start with a move that twists a single corner while preserving A (but scrambling the rest of the puzzle):
single-corner-twist: F'IF I' F'IF, twists ABF clockwise.
We can then combine that operator with its inverse to twist one corner clockwise and another counter-clockwise without any side effects. I won't bother spelling out the entire sequence. Once you are finished with these first to steps the top face should all be one color but the rim around it will stil be scrambled. You might want to twist A a few times to find the position with the least amount out of place.
Fix edges. I don't have a large library of specialized edge movement operators. I just rely on a simple 3-cycle of edges. If exactly three edges are out of place, then you can manuever them into the magic locations and perform the 3-cycle. If you need to exchange two pairs, then you need to build it out of two three cycles. Here's the operator I use:
edge-2-cycle: (FC' B CF') A (FC' B' CF') A'. This will cycle AB, AF, CH. You can turn A multiple times in the middle as long as you reverse it with the corresponding number of reverse turns at the end. This will let you create cycles with other edges in A other than AF.
Fix corners. Again, I don't have any specialized moves for the various scenarios. I build everything from a simple 3-cycle of corners:
corner-3-cycle:F'I'F A F'IF A'. This will cycle ABF, AEF, BFI. As with the edge cycle, you can substitute multiple turns of A if you wish.

If your own edge/corner operations are prone to flipping/twisting the pieces then it would make sense to first place all of the top layer pieces in the right location, then cleanup any flips or twists. It think that is how most people work on a cube, but since my 3-cycle operators preserve flip and twist, I use the above order. It has the advantage of being able to easily spot the top layer pieces as they move about the puzzle and not get lost during a long operation.

Finally, in theory you could eliminate the first three phases and solve the entire 3x dodecahedron with the primitive flip/twist/cycle operations above. Personally I find it a lot easier to take advantage of the additional degrees of freedom in the earlier stages rather than conceive of everything as a series of 3-cycles.