HD Tile

Visualization

@
$n=HD Tile
$dim=2
r=[0,1,1,0]
h1=-1*[-1,0]
h2=-1*[-2,-1]
h3=-1*[0,1]
h4=[-1,-1]
h5=[1,1]
h6=[-1,1]
h7=r*[0,-1]
h8=r*[1,0]
A=3^-1*(1|h1|h2|h3|h4|h5|h6|h7|h8)*A

Overview

The HD Tile is a self-similar plane tile generated by 9 affine maps, all with contraction ratio $\frac{1}{3}$ but with different linear parts — some are pure contractions, some include $180^\circ$ rotation, some an axis-swap reflection. Its Hausdorff dimension is $2$, so the tile has positive Lebesgue measure; as we verify below, it also tiles the plane under a crystallographic group.

The mix of orientations — central symmetry ($-I$) and axis-swap reflection ($r$) — makes the IFS look as if its images overlap. In fact, the Open Set Condition holds exactly: the neighbor graph has only 13 pairwise boundary components.

What makes the HD Tile exceptional is the combination of a tiny neighbor graph — only 13 pairwise boundary pieces — and a boundary dimension ≈ 1.993, the positive root of

$x^7 - 12x^6 + 27x^5 + 10x^4 - 61x^3 + 20x^2 + 9x - 18 = 0.$

For comparison, the Lévy C curve — widely regarded as one of the roughest self-similar boundaries — has dim ≈ 1.934 but requires 41 boundary pieces to encode its neighbor structure. The HD Tile exceeds that boundary dimension with fewer than a third of the pieces. The boundary is a near-plane-filling fractal curve of Hausdorff dimension $\approx 1.993$, and still has 2-dimensional Lebesgue measure zero.

Discovered by D. Mekhontsev using the IFStile search engine.

Structure of the Nine Maps

The attractor satisfies:

$A = \tfrac{1}{3}\bigl(1 \cup h_1 \cup h_2 \cup h_3 \cup h_4 \cup h_5 \cup h_6 \cup h_7 \cup h_8\bigr)(A)$

The nine maps fall into three groups by their linear part:

The translations are:

$h_1: [1,0], \quad h_2: [2,1], \quad h_3: [0,-1]$

$h_4: [-1,-1], \quad h_5: [1,1], \quad h_6: [-1,1]$

$h_7(x) = r\,x + [-1,0], \quad h_8(x) = r\,x + [0,1]$

Writing each map in the form $h_i(x) = \tfrac{1}{3}\,s_i(x + H_i)$ with $s_i \in \{I, -I, r\}$ and $H_i \in \mathbb Z^2$, the digit vectors $H_i$ are

$(0,0),\ (-1,0),\ (-2,-1),\ (0,1),\ (-1,-1),\ (1,1),\ (-1,1),\ (0,-1),\ (1,0)$

and their residues modulo $3\mathbb Z^2$ are all nine classes of $\mathbb Z^2 / 3\mathbb Z^2$, each appearing exactly once. By the Bandt–Gelbrich tiling criterion with symmetries, the attractor $A$ has non-empty interior, the OSC holds with $V = \operatorname{int}(A)$, $\dim_H A = 2$, and $A$ self-replicatingly tiles the plane under the crystallographic group $\mathbb Z^2 \rtimes \langle -I, r\rangle$.

A Hidden Polyomino

Behind the fractal boundary lies a remarkably clean combinatorial skeleton. If we drop the scale factor $1/3$, the nine maps $h_0 = \mathrm{id}, h_1, \ldots, h_8$ all send $\mathbb Z^2$ into $\mathbb Z^2$. Iterating them on a single unit square produces an exact polyomino at every stage:

$P_n = \bigcup_{i=0}^{8} h_i(P_{n-1}), \qquad P_0 = [0,1]^2.$

For the first three iterations the nine sub-pieces $h_i(P_{n-1})$ occupy pairwise disjoint sets of unit cells, so $|P_1|=9$, $|P_2|=81$, $|P_3|=729$ — no overlaps, no gaps. Already at $n=2$ the decomposition fully exposes all nine maps (identity, three central symmetries, three pure translations, two axis-swap reflections). What is surprising is that the IFS built from these very tame maps produces an attractor with a near-plane-filling fractal boundary.

Why the boundary dimension is so high. The recursion $P_n = \bigcup h_i(P_{n-1})$ is a universal identity — it says nothing about the shape of the attractor. What matters is how the silhouette of $P_n$ compares to $3 \cdot P_{n-1}$ (the would-be rep-tile). If they agreed exactly, the HD Tile would be a polyomino rep-tile with a polygonal boundary (dim = 1). They don’t — and the precise quantity that controls this disagreement is the spectral radius $\lambda$ of the boundary substitution: the linear recursion that tracks the number of boundary cells of $P_n$.

For any self-similar polyomino IFS with scaling 3 and nine digits, $\lambda$ lies in the half-open interval $[3, 9)$: $\lambda = 3$ corresponds to a true rep-tile with $\dim \partial A = 1$, and $\lambda \to 9$ is the degenerate limit where the tile loses its interior. Whenever $A$ is a genuine tile (non-empty interior, OSC), the boundary has Lebesgue measure zero, so $\dim \partial A < 2$ strictly. The dimension of the boundary is

$\dim_H(\partial A) = \log_3 \lambda.$

Holes emerge. Starting at $n=4$ the polyominoes $P_n$ are no longer simply connected: the first 18 holes appear inside $P_4$, and their number grows rapidly ($P_5$ has 216 holes, $P_6$ has 2 086). At the same stage sub-pieces $h_i(P_{n-1})$ begin to overlap (16 coincidences at $n=4$, 330 at $n=5$), so $|P_n|$ drops below $9^n$. The polyomino model is thus a faithful picture of the limit tile only for $n \le 3$.

Late collisions drive $\lambda$ to the ceiling. Each collision between two sub-pieces $h_i(P_{n-1})$ and $h_j(P_{n-1})$ — a shared cell or a trapped hole — removes a pair of boundary edges that would otherwise survive into $P_n$. While no collisions occur, the boundary grows by the full factor $9$ per step, and the boundary substitution acts indistinguishably from the full area substitution: nearly every pair of adjacent cells along the current boundary remains adjacent in the next generation, and new boundary pairs appear wherever they possibly can. The spectral radius $\lambda$ of the boundary substitution is therefore pushed toward its ceiling of $9$ by every generation in which collisions are absent, and the mismatch between $P_n$ and $3\cdot P_{n-1}$ gets spread along the whole boundary rather than concentrated in a thin collar. For the HD Tile, no collision happens at $n=1, 2, 3$ — three full generations of loss-free boundary growth — before the first 18 holes and 16 overlaps appear at $n=4$. That unusually long delay is the geometric reason $\lambda \approx 8.935$ sits so close to the theoretical maximum, and hence why $\dim_H(\partial A) = \log_3 \lambda \approx 1.993$ — a fractal curve that almost, but not quite, fills the plane.

The neighbor graph confirms the OSC and yields the 13 pairwise boundary pieces $i_1, \ldots, i_{13}$. Piece $i_{11}$ is a single-point contact (dim = 0); the remaining 12 pieces all have dimension governed by the same degree-7 polynomial as above, giving the final value $\dim_H(\partial A) = 2\log(8.935\ldots)/\log 9 \approx 1.993$.

Zoomed Interior Detail

With $\dim_H(\partial A) \approx 1.993$ the boundary is so rough that it is not obvious from the full-tile picture whether $A$ has interior at all. The following canvas is zoomed ≈ 716× into the tile (viewport radius $0.0013$ vs bounding radius $\approx 0.930$), centered at approximately $(0.134,\,-0.106)$. A large, roughly square solid yellow region — a piece of pure interior of $A$ — fills the middle of the view, surrounded by the near-planar fractal texture of $\partial A$ at the edges of the frame. This is a direct visual confirmation that $\operatorname{int}(A) \ne \emptyset$, the conclusion already guaranteed by the tiling criterion above but here made concrete.

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