## 5 Three-Dimensional Tiling with H and T  
One can extend this solution of the puzzle to a tiling of 3D Euclidean space with H and T. First, we can arrange six copies of T so that they form a hexagonal pyramid hole to which H fits in as Figure 11(a,b) show. This configuration can be extended as Figure 11(c) shows, and we can form a trihexagonal tiling pattern if we continue it infinitely. By adding horizontally-flipped copies of T, we have a tiling of a region between two parallel planes. Thus, by repeating it infinitely, we can form a tiling of the three-dimensional space. As this construction shows, there are infinitely many ways of tiling the space with H and T by sliding each region between parallel planes arbitrary. Among them, we are only interested in the case where triangular tessellation patterns on the planes coincide. Note that this three-dimensional tiling has a 6-fold symmetry around a line connecting the two v-vertices of a copy of H. The symmetry group of this tiling is the product of the infinite dihedral group and the symmetry group of the trihexagonal tiling. As Figure 12 shows, a block of pieces with three H and six T taken from this tiling forms the solution of our puzzle. Figure 11: A tiling of the 3D Euclidean space with H and T.  
We view this tiling more directly as an extension of our solution of the puzzle. For integers \(x,y,z\), we denote by \(C(x,y,z)\) the unit cube with a pair of diagonal vertices \((x,y,z)\) and \((x+1,y+1,z+1)\). We consider a cubic lattice each of whose unit cubes contains one copy of \(H\), \(T_{1}\), and \(T_{2}\). Copies of \(H\) are put in \(C(x,y,z)\) such that \(x+y+z=3k\) for an integer \(k\). Copies of \(T_{1}\) and \(T_{2}\) are assigned in \(C(x,y,z)\) such that \(x+y+z=3k+1\) and \(x+y+z=3k+2\) for an integer \(k\), respectively. With this assignment, at each vertex of the lattice, eight v-vertices in the surrounding cubes meet or no v-vertices exist. V-vertices meet at those lattice vertices on planes \(x+y+z=3k\) for integers \(k\), which are indicated in Figure 13 with red marks. Since an e-vertex of a piece exists on a cube-edge both of whose endpoints are not v-vertices, for each cube-edge, all the four cubes surrounding it share an e-vertex on the middle point or there is no e-vertex  
Figure 12: **(a)** Three H and five T pieces are in this picture. We consider a block obtained by adding one more copy of T at the bottom with the same direction as the copy on the top; **(b)** We can put this block into the puzzle box. Figure 13: V-vertices of the three dimensional tiling with H and T with \(T_{3}\), \(T_{4}\), and \(H\) pieces. on it. Therefore, at those lattice vertices without v-vertices, we have an octahedral open space, which has the form of T. There are two directions of them. We have holes of the direction \(T_{3}\) at \((x,y,z)\) with \(x+y+z=3k\), and of the direction \(T_{4}\) which is the opposite of \(T_{3}\) at \((x,y,z)\) with \(x+y+z=3k+1\). Therefore, by filling the holes with copies of T, we have the tiling. A solution of the puzzle is obtained by selecting the pieces contained in the \(2\times 2\times 2\)-box with a pair of diagonal vertices \((1,0,0)\) and \((3,2,2)\). Thus, we have two explanations for the same three-dimensional tiling. One important thing is that the tiling has a 6-fold symmetry whereas the cubic lattice has a 3-fold symmetry. The line \(x=y=z\) is an axis of the 6-fold symmetry, and by the 60-degree rotation around this axis, a \(T_{2}\)-piece is mapped to a \(T_{3}\) piece, a \(T_{1}\) piece is mapped to a \(T_{4}\) piece, and a \(H\) piece is mapped to a \(H\) piece. We denote by \(L_{1}\) the cubic lattice of the centers of the unit cubes. We call the center of the surrounding cube of a piece the center of the piece. In this tiling, centers of \(H\), \(T_{1}\), and \(T_{2}\) pieces form the lattice \(L_{1}\). If we rotate this lattice by 60-degrees around the axis \(x=y=z\), then we have a cubic lattice of centers of \(H\), \(T_{3}\), and \(T_{4}\) pieces which we denote by \(L_{2}\).