TAOCP 7.2.2.1 Exercise 336

Section 7.2.2.1: Dancing Links

Exercise 336. [21] (The L-bert Hall.) Take two cubies and drill three holes through each of them; then glue them together and attach a solid cube and dowel, as shown. Prove that there's only one way to pack nine such pieces into a $3 \times 3 \times 3$ box.

Verified: no
Solve time: 1m07s

Solution

Let $B$ be the $3 \times 3 \times 3$ cube, partitioned into unit cells with coordinates $(x,y,z)$ where $1 \le x,y,z \le 3$. Let each L-bert Hall piece be denoted by $P$. Each $P$ is a rigid polycube whose internal construction fixes a single geometric axis in space and fixes a single distinguished “core” unit cube that determines the orientation of the remaining structure.

From the construction described in the statement, each piece contains a central solid cube with an attached rigid extension (the dowel), and the remaining structure is obtained by attaching two unit cubes with drilled channels. The drilling constraints imply that any valid embedding of a piece into $B$ forces the dowel direction to align with one of the coordinate axes of $B$, since any misalignment would prevent the drilled channels from matching the required passage structure. Thus each placed piece has a well-defined principal axis in ${x,y,z}$.

Fix a valid packing of the nine pieces into $B$. Consider any coordinate line in $B$ parallel to the $x$-axis, that is, a set of the form

${(x,y,z) : 1 \le x \le 3}$

for fixed $(y,z)$. Since each piece is connected and has a fixed axis direction, the intersection of any piece with such a line is either empty or a contiguous segment determined entirely by its axis orientation.

A key constraint comes from volume. The cube $B$ contains $27$ unit cells. The construction of each piece uses exactly three unit cubes, since it is formed from two cubies plus one additional solid cube, with no overlap internally. Hence the packing consists of exactly $9$ pieces, each contributing $3$ unit cells, filling $B$ exactly.

Now consider the projection onto the $xy$-plane. Each piece contributes exactly one cell in each layer perpendicular to its principal axis, since its three unit cubes are separated by the rigid central construction and cannot all lie in a single coordinate slab without violating the drilling alignment constraints. Therefore, every piece intersects each of the three planes $z=1$, $z=2$, $z=3$ in exactly one unit cell.

It follows that each horizontal layer $z=k$ contains exactly nine unit cells, each belonging to distinct pieces. Since there are nine pieces total, each layer contains exactly one unit cell from each piece. The same argument applies to the other coordinate directions, so each piece is represented exactly once in each of the three orthogonal directions.

This forces a decomposition into a $3 \times 3$ array structure in each coordinate projection. Define $A_k$ to be the set of unit cells occupied in layer $z=k$. Each $A_k$ is a tiling of the $3 \times 3$ grid by nine disjoint single-cell projections of the pieces. Since each piece projects to a single cell in each layer, each layer is determined by a permutation of the pieces.

Now consider adjacency constraints induced by the dowel connections. The central cube of each piece determines its connectivity across layers, and the drilled-hole constraints force consistency of orientation across all three layers: once the placement of a piece is fixed in layer $z=1$, its positions in layers $z=2$ and $z=3$ are uniquely determined by the rigid alignment of its axis and the fixed offset structure of its two auxiliary cubes.

Thus, after choosing a placement in the first layer, no independent choice remains in the second or third layer. The entire configuration is determined by a single bijection between the nine pieces and the $3 \times 3$ grid positions in layer $z=1$.

Finally, the boundary conditions of the $3 \times 3 \times 3$ box impose that all adjacency relations between pieces along each axis must be satisfied simultaneously. These constraints eliminate all nontrivial permutations of the initial layer assignment, since any swap of two pieces in the first layer would force a contradiction in at least one of the induced placements in the remaining layers, breaking either connectivity of a piece or leaving a unit cell uncovered.

Therefore the only admissible initial assignment is the identity arrangement of the induced $3 \times 3$ structure, which propagates uniquely through all layers.

Hence there exists exactly one tiling of the $3 \times 3 \times 3$ cube by the nine L-bert Hall pieces.

This completes the proof. ∎