TAOCP 7.2.2.1 Exercise 268

Section 7.2.2.1: Dancing Links

Exercise 268. ▶ [21] [21] There are 1010 ways to pack the twelve pentominoes into a $5 \times 12$ box, not counting reflections. What's a good way to find them all, using Algorithm X?

Verified: yes
Solve time: 1m13s

Solution

The problem is an exact cover instance in the sense of (6)–(9): each legal placement of a pentomino on the $5\times 12$ board corresponds to one option, and a valid tiling corresponds to a set of options that cover each item exactly once.

Let the $60$ board squares be the primary items, indexed as ${s_{1},\dots,s_{60}}$. Introduce $12$ additional primary items ${p_{1},\dots,p_{12}}$, one for each pentomino, enforcing that each piece is used exactly once. Every option will contain exactly one $p_k$ and five square-items corresponding to the cells occupied by that placement of piece $k$. This produces a standard exact cover formulation in the sense of Section 7.2.2.1.

Each pentomino $k$ has a finite set of placements determined by translations and rotations (and, if allowed, reflections) inside the $5\times 12$ rectangle. For each such placement $\pi$, define an option $O(k,\pi)$ consisting of the six items

$$ O(k,\pi)={p_k}\cup{s_{i_1},s_{i_2},s_{i_3},s_{i_4},s_{i_5}}, $$

where $s_{i_1},\dots,s_{i_5}$ are exactly the covered squares. The full matrix $A$ of (5) is not stored explicitly; instead, the data structure of (10) is built with one node per incidence $s_i\in O(k,\pi)$ and one node per incidence $p_k\in O(k,\pi)$.

Algorithm X is then applied to this exact cover system using the dancing-links representation of (12)–(14). At each recursive step, the item $i$ selected in E1 is chosen according to the minimum column length heuristic, namely an item with minimal $\mathrm{LEN}(i)$ among all currently active items. This choice minimizes branching, since each option involving $i$ generates one recursive subproblem as in (9), and $\mathrm{LEN}(i)$ is exactly the number of such options.

The recursive search proceeds as follows in the language of (9). An item $i$ is selected; if $i$ is one of the piece-items $p_k$, the recursion immediately branches over all placements of pentomino $k$. If $i$ is a square-item $s_j$, the recursion branches over all placements covering cell $j$. Each branch applies $\mathrm{cover}(i)$ and then, for each option $O$ containing $i$, applies $\mathrm{cover}(j)$ for all other items $j\in O$, continuing recursively until all items are covered. The structure guarantees correctness because each recursive level reduces the active matrix to the residual exact cover problem described after (9).

To avoid counting mirror-image solutions, the search space is restricted by symmetry breaking on one designated chiral pentomino, for instance $F$. Among the two reflected forms of $F$, only one canonical orientation is admitted in the generation of placements $O(F,\pi)$. Since any global reflection of a complete tiling induces a reflection of every piece, forbidding one chirality class eliminates precisely one representative of each mirrored pair, ensuring that each solution is counted exactly once up to reflection.

The resulting algorithm enumerates all solutions by systematic backtracking on the DLX structure. Each complete solution corresponds bijectively to a terminal recursion state in which all $60+12$ items are covered, and every such state corresponds to a distinct tiling of the $5\times 12$ rectangle by the twelve pentominoes, with reflections excluded by the fixed canonical choice.

This completes the construction of a correct Algorithm X implementation for enumerating all $1010$ pentomino packings in a $5\times 12$ box up to reflection. ∎