TAOCP 7.2.2.1 Exercise 314

The core mistake in the proposed encoding is the failure to represent the two tilings as two independent exact-cover constraints.

taocpmathematicsalgorithmsvolume-4hard

Section 7.2.2.1: Dancing Links

Exercise 314. ▶ [28] [28] (G. Sicherman, 2008.) Can the four pentiamonds be used to make two 10-iamonds of the same shape? Formulate this question as an exact cover problem.

Verified: no
Solve time: 2m14s

The core mistake in the proposed encoding is the failure to represent the two tilings as two independent exact-cover constraints. In exact cover, every item must be covered exactly once, so any shared “triangle coverage” structure automatically forces unintended coupling between the two copies. The correct formulation must explicitly duplicate the geometric constraints for each copy.

We restart the construction from scratch and build a faithful exact cover model.

Correct formulation

We are given four fixed pentiamonds, each a polyform of area $5$. We want to know whether there exists a $10$-triangle region $S$ and two tilings of $S$ by the four pentiamonds such that both tilings use the same region $S$, possibly in different ways.

The key is to encode simultaneously:

  1. A choice of the region $S$ consisting of 10 unit triangles.
  2. A tiling of $S$ by the four pentiamonds for copy $A$.
  3. A tiling of the same $S$ by the four pentiamonds for copy $B$.

In an exact cover instance, this requires separate coverage constraints for each copy, plus a mechanism ensuring that both copies refer to the same chosen region.

Items

We introduce three types of primary items.

First, a single item selecting the shape:

$$ {S} $$

for each admissible $10$-triangle region $S$.

Second, triangle-coverage items for each copy. For each region $S$, each copy index $k \in {A,B}$, and each triangle $t \in S$, introduce an item:

$$ (S,k,t) $$

These enforce that each copy independently forms a full tiling of $S$.

Third, piece-usage items ensuring that each pentiamond is used exactly once in each copy. For each pentiamond $p \in {1,2,3,4}$ and each copy $k \in {A,B}$, introduce:

$$ (p,k) $$

Options

We now define the exact-cover options.

1. Shape selection

For each admissible region $S$, we include one option:

$$ {S} $$

This ensures exactly one region is chosen.

2. Pentiamond placements

Fix a region $S$, a copy $k \in {A,B}$, a pentiamond $p$, and a placement of $p$ inside $S$ that matches its geometry.

For each such placement, we create an option consisting of:

$$ {(p,k)} \cup {(S,k,t) : t \in \text{cells covered by the placement}} $$

This encodes that in copy $k$, each pentiamond covers exactly five triangles of $S$, and triangle coverage is enforced independently for each copy.

Correctness argument

We verify that exact covers correspond exactly to valid solutions.

From solution to exact cover

Suppose we have a region $S$ and two tilings of $S$ by the four pentiamonds.

We select the shape option for $S$, ensuring that exactly one region is active.

For copy $A$, each pentiamond placement contributes one option covering $(p,A)$ and the five corresponding items $(S,A,t)$. Since the tiling is valid, every triangle of $S$ is covered exactly once, and each pentiamond is used exactly once, so all items of the form $(S,A,t)$ and $(p,A)$ are covered exactly once.

The same holds independently for copy $B$. Thus we obtain a valid exact cover.

From exact cover to solution

Suppose we have an exact cover.

Exactly one shape item $S$ is chosen, since shape items are mutually exclusive.

Consider copy $A$. The selected placement options for copy $A$ cover each item $(S,A,t)$ exactly once. Since each placement corresponds to a pentiamond covering exactly five triangles, these placements form a tiling of $S$. The items $(p,A)$ ensure each pentiamond is used exactly once.

The same argument applies to copy $B$, producing a second tiling of the same region $S$.

Thus we obtain two valid tilings of the same $10$-triangle region.

Resolution of the original error

The previous construction failed because it used a single set of triangle items $(S,t)$, which incorrectly forced both tilings to contribute to the same exact-cover constraints, violating the “exactly once” requirement.

The correction is to duplicate triangle constraints by introducing separate item sets $(S,A,t)$ and $(S,B,t)$. This cleanly separates the two tilings while still coupling them through the shared choice of $S$.

This yields a correct exact cover formulation of the problem.