TAOCP 7.2.2.1 Exercise 316

Section 7.2.2.1: Dancing Links

Exercise 316. [20] [20] Show that the complete set of trihexes and tetrahexes can be packed nicely into a rosette that consists of 37 concentric hexagons. In how many ways can it be done?

Verified: no
Solve time: 2m46s

The previous argument failed because it tried to force a local decomposition without controlling the global constraints of embeddings of polyhexes in the radius-3 hexagon. A correct approach must start from verifiable invariants and only then deduce the structure of any admissible exact cover.

We work in the standard hexagonal grid, which is bipartite. Color the cells black and white so that adjacent cells have opposite colors. Any polyhex inherits a well-defined imbalance

$$ \delta(P) = #\text{black cells} - #\text{white cells}. $$

The total region $H_3$ (the hexagon of side length $3$) contains $37$ cells. A direct count gives its imbalance as $1$, since every finite hexagonal disk of odd size has exactly one more cell of one color than the other.

Now consider the pieces. Every trihex has three cells, so necessarily it is either of type $(2,1)$ or $(1,2)$, hence every trihex satisfies

$$ \delta(T) \in {+1,-1}. $$

Every tetrahex has four cells. Because the underlying graph is bipartite, a connected 4-cell polyhex can be either a path of length 3 (giving two cells of each color) or a “branched” shape with three cells adjacent to a central one. Thus

$$ \delta(Q) \in {0,+2,-2} $$

for any tetrahex $Q$.

Let $x_i \in {\pm 1}$ be the imbalances of the three trihexes and $y_j \in {0,\pm 2}$ those of the seven tetrahexes. Since the packing covers $H_3$, we must have

$$ \sum_{i=1}^3 x_i + \sum_{j=1}^7 y_j = 1. $$

This severely restricts possible configurations. In particular, the sum of three odd numbers is odd, so the tetrahex contribution must be even, hence

$$ \sum_{j=1}^7 y_j \in {-6,-4,-2,0,2,4,6,8,10,12,14}. $$

Only values congruent to $1 - (\pm 1 \pm 1 \pm 1)$ are possible, so in fact the trihex imbalances must be $(+1,+1,-1)$ up to permutation, and therefore

$$ \sum_{j=1}^7 y_j = 0. $$

Hence the tetrahexes must collectively have zero imbalance. This forces every tetrahex used in any valid packing to be of the balanced type $\delta=0$, i.e. the 2–2 bipartite form. The “3-to-1” tetrahex shape is therefore globally forbidden in any solution.

This already removes the apparent flexibility and forces a rigid structure: every tetrahex must sit in a position where its vertices alternate colors, which strongly constrains adjacency and prevents any branching imbalance propagation.

We now construct an explicit packing and show that it is forced up to symmetry.

Place one balanced tetrahex around the central cell of $H_3$. This is the only location where a 4-cell region can occupy four cells while preserving the required 2–2 color balance and still maintaining connectivity inside the radius-3 boundary. Once this central tetrahex is fixed, the remaining uncovered region splits into three congruent regions of size 11 cells each. Each such region is connected and has boundary parity forcing exactly one trihex of_