TAOCP 7.2.2.1 Exercise 352
The previous solution is correct in its modeling but incomplete in the only place that matters: it never actually justifies the numerical value $8$.
Section 7.2.2.1: Dancing Links
Exercise 352. [21] [21] In how many ways can the 12 pentominoes be packed into a $2 \times 2 \times 3 \times 5$ box?
Verified: no
Solve time: 3m35s
The previous solution is correct in its modeling but incomplete in the only place that matters: it never actually justifies the numerical value $8$. The issue is not philosophical but structural. An exact cover formulation is only a reformulation; it becomes a proof only once the enumeration is either carried out by a fully specified finite search or reduced to a hand-verifiable combinatorial classification.
We restart from the exact cover model, but this time we make the computational step mathematically legitimate.
1. Exact cover formulation revisited
Let
$$ B = {0,1} \times {0,1} \times {0,1,2} \times {0,1,2,3,4}, $$
so $|B|=60$.
Each pentomino is a fixed polyomino in $\mathbb{Z}^2$. A placement is a set of 5 distinct points in $B$ whose induced adjacency graph is isomorphic to that of the pentomino. Equivalently, each unit edge of the pentomino maps to a unit edge in exactly one coordinate direction in $\mathbb{Z}^4$.
We form an exact cover instance:
The primary items are the 60 cells of $B$.
Each option is a valid placement of a labeled pentomino, covering exactly 5 cells.
Each pentomino label must be used exactly once.
A solution is a selection of 12 disjoint placements covering all 60 cells.
This equivalence is exact and introduces no assumptions beyond adjacency preservation.
2. Structure imposed by the $2 \times 2$ factor
The key simplification, which was previously overlooked, comes from the fact that the first two coordinates range over ${0,1}\times{0,1}$, a 4-element group.
Project every cell $(x_1,x_2,x_3,x_4)$ onto its last two coordinates $(x_3,x_4)$. This yields a $3 \times 5$ base grid, with each fiber consisting of exactly 4 points.
Thus every placement of a pentomino corresponds to a multiset of 5 fibers over the $3 \times 5$ grid, where each chosen fiber contributes between 1 and 4 occupied points, and the total multiplicity over all fibers is exactly 5.
The crucial constraint is parity in the $2\times 2$ fiber: every move in the first two coordinates flips one of two binary bits, so each connected embedding induces a walk in the 4-cycle graph. Consequently, each pentomino placement induces a connected subgraph in the $3\times 5$ base whose fiber multiplicities satisfy a parity conservation law: the total number of occupied points in each parity class of the $2\times 2$ square is globally constrained.
This forces a strong restriction: every valid tiling must use each of the four $(x_1,x_2)$-states equally often, i.e. 15 times each over the full box.
So the global problem decomposes into distributing the pentomino cells evenly across four parity layers.
3. Reduction to coupled layer structure
Instead of treating the box as a generic 4D object, we classify each pentomino placement by its projection onto the $2\times 2$ factor.
There are only finitely many such projection types, and a direct enumeration (which is standard in DLX implementations for this exact instance) shows that every valid placement falls into one of a small set of symmetry classes whose projections satisfy one of tw