Improve my 3D puzzle to have a single solution

Question

3D View of one solutions http://autode.sk/2Dg4FWx

I designed a 4x4x4 soma cube in AutoCad and then built it with wood cubes. There are twelve pentacubes and one tetracube. Each piece is unique and if one is in the set and you can get a different piece by mirroring it, that one is also in the set.

I don't know how many solutions there are, but there are many (Maybe you can tell me how many?). It is still really hard to put it together because you have so many options to place a piece.

What I want is to change it in a way so that there is only one solution left, but it is still hard to put together. Coloring faces or whole cubes could be an option but maybe you would have better ideas!

Answer 1

Your puzzle has 7790 solutions (and their mirror images). I used my own polyform solver (here), which took about 11 minutes. It is slow because it is rather flexible, but you can use faster solvers that are more specific for polycubes, such as the Polycube Puzzle Solver already mentioned in the comments above.

I doubt that there is any 4x4x4 puzzle with a unique solution when you use pieces as small as pentacubes. You are likely going to need one or more larger pieces, or some other way to restrict the placement of the pieces. You could use a colouring scheme (e.g. checkerboard colouring).

I'll do a bit of playing around with my solver to see if I can find something interesting.

Edit: I decided to use 10 hexacubes instead of 12 pentacubes, to make it easier to find sets of pieces with few solutions. My best effort is the following set.

The 10 flat hexacubes consisting of a 1x1x4 spine with two cubes attached to the spine:

A     B     C     D     EE
AAAA  BBBB  CCCC  DDDD  EEEE
A      B      C      D

F F   G  G   HH    I     J
FFFF  GGGG  HHHH  IIII  JJJJ
                   I      J

Plus the L tetracube:

KKK
K

This has only 3 solutions (and their mirror images).

In the comments below, theonetruepath writes:

Pack a 2x3x4 with any 12 pentacubes and any one tetracube, in all ways. Include all symmetric packings. Count how many times each piece occurs. Make a list of the 12 pentacubes that occur least often plus the tetracube that occurs least often. This set of pieces does not pack the 4x4x4.

Unfortunately I was unable to reproduce this. The set of pentacubes I get still has many solutions when combined with any tetracube. I therefore still stand by my view that pentacube pieces are too small to produce a puzzle with a unique solution.

Answer 2

This is the best I could do. As Jaap Scherphuis points out, the 'tilability' of the pieces is a bit too high to yield unique packings easily. It's still possible (or even likely) that there are sets that will work, but the time to find them might be too high. A complete search would take the rest of my life, assuming I double the speed of my PC every year. A more directed search might turn up a solution in days or months... or it might not.

However if we slant the deck in our favour as follows we can come up with a suitable set:

1. Pack any four pentacubes and any one tetracube into a 2x3x4 in all possible ways.
2. Order the pentacubes and the tetracubes into tilability order based on how often each appears in packings of this shape.
3. Make a list of the 12 least-tilable pentacubes plus the least-tilable tetracube. This set of pieces has lots of packings.
4. Remove the most tilable pentacube from the list and replace it with a copy of the least tilable. So now you have two X-pentacubes. Still too many solutions.
5. Remove the second most tilable pentacube and add a copy of the second-least tilable. Now you have two X- and two Z-pentacubes. Solution count drops to three.
6. Try switching out the tetracube. Turns out the third-most-tilable tetracube gives exactly one solution.

The two dark purples are the X-pentacube, the two dark blues are the two Z-pentacubes.