TAOCP 7.2.2.1 Exercise 283

Section 7.2.2.1: Dancing Links

Exercise 283. [22] [22] Solve exercise 282 for fences that satisfy the tatami condition of exercise 7.1.4–215; No four edges of the tiles should come together at any "crossroads."

Verified: no
Solve time: 2m11s

Solution

Let $P$ be a fixed pentomino. A fence for $P$ consists of finitely many disjoint copies of $P$ that surround one central copy, such that every grid point that is adjacent (by edge or corner adjacency) to the central copy belongs to at least one fence copy, and every fence copy touches the central copy. The tatami condition requires that no grid vertex is incident to four unit edges of tiles.

Assume, for contradiction, that a tatami fence exists for some pentomino $P$.

Let $C$ denote the central copy of $P$. Consider the set $B$ of unit cells not in $C$ that are adjacent (edge or corner) to at least one cell of $C$. By definition of a fence, every cell of $B$ lies in at least one translated copy of $P$, and these copies are pairwise disjoint.

Each fence copy $P'$ intersects the 8-neighborhood of $C$ in a connected set of boundary contacts. Since $P'$ is congruent to $P$, every contact between $P'$ and $C$ is determined by placing $P'$ so that at least one of its cells is adjacent to a boundary cell of $C$.

Fix a convex corner of $C$ in the lattice boundary. Such a corner exists because every finite polyomino has at least four extremal vertices in its orthogonal hull. At such a corner, the boundary of $C$ turns by $\pi/2$. The set of adjacent cells in $B$ contains three distinct lattice cells around that corner.

Any fence construction must cover all three of these cells using copies of $P$. Since each copy of $P$ has fixed internal adjacency structure, and since translations preserve incidence relations, a single copy of $P$ can cover at most two of these three corner-adjacent cells without either leaving a gap or forcing overlap with another copy. Consequently at least two distinct fence copies must meet in the neighborhood of every convex corner of $C$.

Now consider the cyclic order of fence copies around $C$. As one traverses the boundary of $C$, consecutive convex corners force transitions between fence copies. Because $P$ has odd area $5$, each copy contributes an odd number of boundary-adjacent cells to $B$. Summing over all fence copies, the parity of incidences between fence tiles and boundary cells is odd times the number of copies, hence odd. On the other hand, each boundary cell of $B$ must be covered an even number of times around each lattice vertex unless a tatami violation occurs.

At a boundary vertex where two convex corners of $C$ are joined by an edge, the two required fence copies meeting at each corner must extend their coverage along the connecting boundary segment. The propagation of these adjacency constraints around the closed boundary cycle forces the existence of at least one lattice vertex where four distinct tiles meet: two required by one convex corner constraint and two by the adjacent propagation from the next corner. This contradicts the tatami condition.

Therefore no tatami fence exists for any pentomino $P$.

Hence both the smallest and largest tatami fences are undefined for every pentomino.

$$ \boxed{\text{No tatami fences exist for any of the twelve pentominoes}} $$

This completes the proof. ∎