This puzzle has 10 congruent pieces, each an affine transformation of a cube. The length of the edges are all the same.

Consider the set of vectors: V = {(−1, −1, −1), (−1, 1, 1), (1, −1, 1), (1, 1, −1)}, the vertices of a regular tetrahedron whose center is (0, 0, 0). Consider a subset {p, q, r} of V. Consider {ap + bq + cr | 0 < a, b, c < 1}. That shape is one of our pieces with edge length √3. Depending on which subset you take you get one of four pieces whose union is the rhombic dodecahedron which is the convex hull of the 8 points (±1, ±1, ±1) and the 6 permutations of (0, 0, ±2). The four pieces are pairwise disjoint. See a turnable version.

If we translate the rhombic dodecahedron by (1, 1, 1) we get another and the intersection of the two is one of our pieces. The assembled puzzle is the union of the following translations of the rhombic dodecahedron: (0, 0, 0), (1, 1, 1), (2, 2, 2), (3, 3, 3).

The pieces have magnets on some of their faces. Each magnet is naturally oriented and some pairs of piece faces repel and some attract, as is the wont of magnets. Four of the pieces have magnets on three faces next adjoining an obtuse vertex. The other six pieces have magnets on four faces—all but two adjoining faces. Here is the table of polarities:

An older awkward discovery of shape: We adopt an oblique corner of the piece as an origin and the 3 edges therefrom as basis vectors for a coordinate system.

Branch with bug

An awkward derivation of shape.