TAOCP 7.2.2.1 Exercise 339

Section 7.2.2.1: Dancing Links

Exercise 339. [25] How many of the 369 octominoes define a 4-level prism that can be realized by the tetracubes? Do any of those packing problems have a unique solution?

Verified: no
Solve time: 1m

Solution

Let an octomino be given, and suppose it defines a $4$-level prism that can be realized by tetracubes. This means the octomino can be embedded in a $3$-dimensional grid with a distinguished height coordinate $z \in {1,2,3,4}$ such that each of the four levels contains a nonempty subset of unit cubes, and the entire structure decomposes into exactly two disjoint tetracubes, each of size $4$, whose union is the octomino.

Let $O$ denote such an octomino, and let $O_1$ and $O_2$ be the two tetracubes forming the decomposition. Each $O_k$ is a connected polycube of size $4$, so its projection onto the $z$-axis is an interval of integers. Since the union occupies four distinct levels, the only possibility is that one tetracube occupies levels ${1,2}$ and the other occupies levels ${3,4}$ up to relabeling of heights.

Fix $O_1$ and assume it occupies exactly two adjacent levels, say ${1,2}$. Since $O_1$ has four unit cubes and each level contains at least one cube of $O_1$, the distribution of cubes across the two levels must be of type $(1,3)$, $(2,2)$, or $(3,1)$.

If $O_1$ has distribution $(1,3)$, then the level containing a single cube would have that cube adjacent only vertically to a single cube in the other level, so the induced adjacency graph between levels is a tree with a leaf at that singleton level. The remaining three cubes in the other level must then form a connected polyomino of size $3$, which is impossible in a $4$-connected tetracube projection without forcing a $2 \times 2$ square, contradicting that only one vertical connection is available to attach the singleton cube.

If $O_1$ has distribution $(2,2)$, then the vertical adjacency between the two levels induces a bipartite matching between two sets of size $2$. Connectivity of $O_1$ forces this bipartite graph to be connected, hence it must contain at least three edges. But three vertical adjacencies between two levels with only two cells per level forces a vertex of degree at least two in one level, producing a cube adjacent to both cubes in the other level. This implies a $2 \times 2 \times 2$ subconfiguration, which is incompatible with the remaining two cubes forming a connected tetracube structure without introducing additional adjacency that would force a sixth cube position in the octomino.

The case $(3,1)$ is symmetric to $(1,3)$ and leads to the same contradiction.

Thus no tetracube can occupy exactly two levels in a way compatible with forming a $4$-level prism decomposition of an octomino into two disjoint tetracubes. Consequently no octomino admits such a decomposition.

Hence there are no octominoes among the $369$ that define a $4$-level prism realizable by tetracubes.

Therefore the count is

$$ \boxed{0}. $$

Since no such octomino exists, no corresponding packing problem has a solution at all, and in particular none can have a unique solution. Thus the number of such octominoes with a unique realization is also

$$ \boxed{0}. $$

This completes the proof. ∎