For Christmas, my sister-in-law received a wooden block puzzle consisting of fifty-four solid “T” tetracubes, the puzzle being to assemble them into a 6×6×6 cube.

This turns out to be surprisingly difficult, despite the simplicity of the parts. It’s a puzzle that Berlekamp, Conway and Guy might approve of:

In our view the good puzzles are those with simple pieces but difficult solutions. Anyone can make a hard puzzle with lots of complicated pieces but how can you possibly make a hard puzzle out of a few easy pieces? (Winning Ways, volume 2, page 736)

The difficulty with this puzzle is that of finding a way to break it down into manageable subsections without accidentally making it impossible. My first idea is to divide the 6×6×6 block into eight 3×3×3 sub-blocks, but of course 3×3×3 = 27 and can’t be made out of tetracubes. My second idea is to divide the puzzle into six 6×6×1 layers: 6×6×1 = 36 so it might be packable with nine tetracubes. But a simple colouring argument shows that this is impossible.

When the 6×6 grid is given a checkerboard colouring as shown, no matter how the “T” tetracube is placed in the grid, it covers one square of one colour, and three squares of the other. In particular, it covers an odd number of squares of each colour. So nine tetracubes, however placed, will cover an odd number of squares of both colours (because the sum of nine odd numbers must be itself odd). However, in the grid there are 18 squares of each colour. So a complete packing is impossible.

But the 3×3×3 idea might succeed with a small modification. I can’t build a 3×3×3 cube out of tetracubes. But if I build some 3×3×3 cubes with tabs (sticky-out bits) and some others with slots, I might be able to fit the tabs in the slots and so assemble the whole puzzle.

With seven tetracubes I can make a 3×3×3 cube with one tab. Call this assembly A:

With six tetracubes I can make a 3×3×3 cube with three slots. Call this assembly B:

Now, if I make three copies of assembly A, I can fit the three tabs into the three slots of assembly B, like this:

And if I do this twice, it’s easy to fit the two parts together and solve the puzzle.

The colouring argument shows that I can’t fit 9 tetracubes into a 6×6×1 layer. But there’s no obvious reason why I can’t fit 18 tetracubes into a 6×6×2 slab. This turns out be just small enough for me to solve by solve by hand.

There are only a small number of ways to fill each of the four corners. But most of these arrangements quickly prove impossible to complete. If you fill two adjacent corners in opposite (mirror image) ways, that leaves two adjacent gaps on the edge (see diagram below, top left), and there’s only one way to fill those gaps (below, top right) that doesn’t immediately lead to disaster. This leaves a gap of length four which must contain a piece occupying three cubes (the grey piece, bottom left; this could be one cube to the right but that’s just a mirror image). That forces the brown piece (bottom right), but then there are three cubes in the corner that cannot all be filled (indicated in red with a dotted outline).

I think that the only arrangement of the corners that has any prospect of success is to fill them all in the same way. With eight pieces in place, it doesn’t take long to fit in the remaining ten. See the diagram below for assembly instructions.

In the final figure (bottom right), the red line shows the axis of symmetry, which passes diagonally along the plane between the two 6×6×1 layers. Give the assembly a half turn about this axis and it remains unchanged (colours excepted).

Stack three of these 6×6×2 slabs and you’ve got my second solution to the puzzle.

Keiichiro Ishino gives 11 solutions, and claims that there are more than 20,000 solutions. (I presume he stopped his computer search at that point.)

It’s natural to ask, for which other tetracubes can you assemble 54 of them into a 6×6×6 cube?

Four of them are easy. Each of these can be paired to make a 2×2×2 cube, 27 of which can be assembled into a 6×6×6 cube.

Six “L” tetracubes pack into a 6×2×2 cuboid, nine of which can be assembled into a 6×6×6 cube.

The “I” tetracube can be shown to be impossible with a colouring argument.

With the above arrangement of colours, the “I” tetracube always covers one square of each colour, however it is placed. So in order for all 54 “I” tetracubes to be packed into the cube, there must be 54 squares of each colour. But that’s not the case: there are 53 each of pink and green, and 55 each of blue and orange. So no arrangement of “I” tetracubes can fill a 6×6×6 cube.

Which leaves …

… which is even harder to pack into 6×6×6 cube than the “T” tetracube. Best left for another post … once I’ve found a way to do it.