TAOCP 7.2.2.1 Exercise 272

Section 7.2.2.1: Dancing Links

Exercise 272. [23] [23] Continuing exercise 271, explain how to find special kinds of packings: a) Those that decompose into $6 \times k$ and $6 \times (10-k)$. b) Those that have all twelve pentominoes touching the outer boundary. c) Those with all pentominoes touching that boundary except for V, which doesn't. d) Same as (c), with each of the other eleven pentominoes in place of V. e) Those with the minimum number of pentominoes touching the outer boundary. f) Those that are characterized by Arthur C. Clarke's description, as quoted below. That is, the X pentomino should touch only the F (aka R), the N (aka S), the U, and the V, no others.

Very gently, he replaced the titanite cross in its setting between the F, N, U, and V pentominoes, , ARTHUR C. CLARKE, Imperial Earth (1976)

Verified: no
Solve time: 2m40s

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Solution

In the exact cover formulation of pentomino packing, each option represents a placement of a specific pentomino, covering one item for the pentomino identity and five items for the occupied unit squares of the $6\times 10$ board. Algorithm X operates on this structure by selecting options and repeatedly covering incompatible items and options via (12)–(14).

Each modification below is achieved by restricting the option set and, when necessary, adding secondary constraints implemented as additional items or as controlled pruning during cover operations.

(a) Decompositions into $6\times k$ and $6\times (10-k)$

Fix an integer $k$ with $1\le k\le 9$. Introduce no new items; modify only the option set.

Each option corresponds to a placement of a pentomino on the $6\times 10$ board. Discard every option that occupies a cell in both columns ${1,\dots,k}$ and ${k+1,\dots,10}$. The remaining options define a standard exact cover problem on the same board, but with a hard geometric constraint forbidding straddling placements.

If a solution exists, every selected option lies entirely in one side or the other of the vertical cut. Since every board cell item must still be covered exactly once, no uncovered cell can lie on either side, hence the solution induces a decomposition into a tiling of $6\times k$ and a tiling of $6\times(10-k)$.

Conversely, any pair of tilings of these rectangles yields a valid solution of the restricted instance, since no placement crosses the boundary. Running Algorithm X separately for each $k$ produces all decompositions.

(b) All twelve pentominoes touching the outer boundary

In the standard formulation each option belongs to exactly one pentomino item. Let $P$ denote a pentomino, and let an option $O$ for $P$ be admissible only if at least one of its occupied cells lies on the boundary of the $6\times 10$ rectangle.

All options violating this condition are removed before Algorithm X begins. The remaining exact cover instance forces every chosen option to correspond to a boundary-touching placement, hence every pentomino in any solution touches the boundary.

No further modification of Algorithm X is required, since feasibility of a solution already implies that each pentomino item is covered exactly once, hence by a boundary-touching option.

(c) All pentominoes touch the boundary except $V$

Let $V$ denote the distinguished pentomino. Modify the option set as follows.

For every pentomino $P\ne V$, retain only those options in which at least one occupied cell lies on the boundary. For $V$, retain only those options in which no occupied cell lies on the boundary.

The resulting exact cover instance enforces the required condition directly, since Algorithm X selects exactly one option per pentomino item. Therefore every pentomino $P\ne V$ is forced to appear in a boundary-touching placement, while $V$ is forced to appear in a non-boundary-touching placement.

(d) Same condition with each pentomino playing the role of $V$

For each pentomino $P$ in turn, construct an instance identical to (c) but treating $P$ as the unique non-boundary-touching piece. This yields twelve independent exact cover instances.

Running Algorithm X once per instance enumerates all solutions satisfying the condition for the specified choice of exceptional pentomino. Taking the union over all choices of $P$ produces all configurations in which exactly one pentomino fails to touch the boundary.

(e) Minimum number of pentominoes touching the boundary

Let $t(O)$ be $1$ if option $O$ touches the boundary and $0$ otherwise. For any solution $S$, define

$T(S)=\sum_{O\in S} t(O).$

Algorithm X is modified into a branch-and-bound search on the same exact cover instance. During depth-first search, maintain the partial sum $T_{\mathrm{partial}}$ of already chosen options. For any partial solution, let $R$ be the set of remaining uncovered pentomino items. Since each pentomino must be covered exactly once, the smallest achievable contribution from $R$ is

$T_{\min}(R)=\sum_{P\in R} \min_{O\in \mathcal{O}(P)} t(O),$

where $\mathcal{O}(P)$ is the current set of admissible options for pentomino $P$.

Any completion of the partial solution satisfies

$T(S)\ge T_{\mathrm{partial}}+T_{\min}(R).$

During search, maintain the best value $T^*$ among complete solutions found. Any node satisfying

$T_{\mathrm{partial}}+T_{\min}(R)\ge T^*$

is pruned.

To establish minimality, let $S_0$ be a solution attaining the minimum over all valid tilings. The search explores a finite tree of all partial covers; since pruning only removes nodes whose lower bound is at least $T^$, and $T^$ is updated only upon completion of valid solutions, no branch containing a solution with fewer boundary-touching pentominoes can be discarded before its completion. Therefore the first attainment of $T^*$ equal to the global minimum identifies an optimal solution, and no solution with fewer touching pentominoes exists.

(f) $X$ touches only $F$, $N$, $U$, and $V$

Let $X$ be the distinguished pentomino. The condition restricts geometric adjacency between distinct pentominoes, hence it cannot be enforced solely by filtering single options.

Let $\mathcal{A}(O_1,O_2)$ denote the predicate that placements $O_1$ and $O_2$ are adjacent along at least one edge in the final tiling. The requirement is that if $O_X$ is the chosen option for $X$, then for every other chosen option $O_P$,

$\mathcal{A}(O_X,O_P)\Rightarrow P\in{F,N,U,V}.$

During Algorithm X, when an option $O_X$ for $X$ is chosen, an additional covering phase is executed before recursion. For every pentomino $P\notin{X,F,N,U,V}$ and for every option $O_P$ of $P$, if $\mathcal{A}(O_X,O_P)$ holds, then $O_P$ is hidden using the same mechanism as (13), namely by unlinking it from all vertical lists in which it participates. This is a direct application of hide(p) to all nodes belonging to forbidden options.

When backtracking, the corresponding unhide operations restore all removed options, preserving reversibility.

For correctness, any completed solution contains exactly one option for $X$ and one for each other pentomino. If any forbidden adjacency occurred, the corresponding option would have been removed prior to completion, contradicting feasibility of the completed cover. Conversely, any tiling satisfying the adjacency restriction is never pruned, since no forbidden $\mathcal{A}(O_X,O_P)$ arises.

This completes the construction of an exact cover formulation for all required restricted pentomino packings. ∎