Hyperbolic Tiling, Explained

After reading this you will know why five squares fit around a point in the hyperbolic plane, how the Poincaré disk squeezes an infinite surface into a finite circle, and how to read a {p, q} tiling without mistaking the shrinking rim for tiles of different size.

What it is and why the corners are crowded

A regular tiling covers a surface with identical regular polygons, meeting edge to edge and corner to corner, the same number at every vertex. Notation \{p, q\} means p-sided polygons with q of them meeting at each vertex. The flat plane has exactly three: \{4, 4\} (squares, four per corner), \{3, 6\} (triangles, six per corner) and \{6, 3\} (hexagons, three per corner). That is the complete list. Try to put five squares around a point on a flat table and you fail: four squares already use all 360 degrees.

Hyperbolic geometry has room for the rest. There the angles of a polygon are smaller than in flat space, so more polygons can crowd around a corner. Five squares around a vertex is the tiling \{4, 5\}. Seven triangles is \{3, 7\}. These cannot exist on any flat sheet, but they tile the hyperbolic plane perfectly, and the Poincaré disk lets you see them.

M. C. Escher met this geometry through the mathematician H. S. M. Coxeter and used it in his four Circle Limit woodcuts. The fish or angels shrink toward the boundary circle. In the hyperbolic metric they do not shrink at all: every fish is the same size. The shrinking is what happens when you flatten an infinite surface into a disk.

When the tiling is hyperbolic, flat, or spherical

The angle sum decides everything. Around one vertex of a regular tiling you have q corners of p-gons meeting. In flat space each interior angle of a regular p-gon is (p-2)\cdot 180^\circ / p, and q of them must sum to exactly 360^\circ. Set that equal and simplify and you get the clean condition below.

(p-2)(q-2) = 4

Here p is the number of sides per tile and q is the number of tiles per vertex. When the product equals 4 the tiling is flat. The only whole-number solutions are (4,4), (3,6) and (6,3), the three flat tilings. When the product is less than 4 the tiling is spherical, and the five solutions are the Platonic solids seen as tilings of the sphere. When the product is greater than 4 the tiling is hyperbolic.

(p-2)(q-2) \gt 4

Check a few. For \{4, 5\}: (4-2)(5-2) = 2 \cdot 3 = 6 \gt 4, hyperbolic. For \{3, 7\}: 1 \cdot 5 = 5 \gt 4, hyperbolic. For \{5, 4\}: 3 \cdot 2 = 6 \gt 4, hyperbolic. For \{6, 3\}: 4 \cdot 1 = 4, flat. Use this explorer whenever the product exceeds 4. For the flat cases you want ordinary graph paper, and for the spherical cases the Platonic solids explorer is the right tool.

The same inequality appears in the footnote on the tool page. It is not a rule of thumb: it is the exact break-even between too little angle to close a vertex flatly (spherical) and too much (hyperbolic).

The Poincaré disk model in one page

The hyperbolic plane is infinite and cannot sit inside a Euclidean circle without distortion. The Poincaré disk accepts the distortion in exchange for two nice properties: angles are preserved (the model is conformal), and straight lines of the hyperbolic plane become simple curves.

The whole infinite plane maps to the open unit disk. The boundary circle is infinitely far away, so a tile drawn near the rim covers an enormous hyperbolic distance while occupying a thin Euclidean sliver. A hyperbolic straight line, called a geodesic, appears as a circular arc that meets the boundary circle at a right angle. Diameters through the center count too, as arcs of infinite radius.

Distance stretches as you approach the rim. The hyperbolic distance element in the disk is

ds = \frac{2\,\lvert dz \rvert}{1 - \lvert z \rvert^2}

where z is a point in the disk and \lvert z \rvert is its Euclidean distance from the center. At the center the factor 2/(1-\lvert z\rvert^2) equals 2. At \lvert z \rvert = 0.9 it equals 2/(1-0.81) \approx 10.5. At \lvert z \rvert = 0.99 it is about 100. So a step of the same hyperbolic length looks about 50 times shorter near the rim than at the center. That single factor is the entire reason the tiles appear to shrink.

How the explorer builds the tiles by reflection

You do not compute every tile from scratch. You build the central polygon once, then reflect it in its own edges to make neighbors, then reflect those, and so on. In the Poincaré disk a reflection across a geodesic is a circle inversion in the arc that represents that geodesic. Reflecting is exact and repeatable, so the tiling grows outward layer by layer without accumulating drift.

The central polygon is regular and centered at the disk center, where the model looks almost Euclidean. Its p vertices sit at equal angles. The key number is the radius r at which to place the vertices so that the interior angle comes out to exactly 2\pi/q, which is what a clean vertex figure requires. That radius is

r = \sqrt{\frac{\cos(\pi/p + \pi/q)}{\cos(\pi/p - \pi/q)}}

Here p is sides per tile and q is tiles per vertex, and r is the Euclidean distance from the disk center to each vertex of the central tile. The formula only gives a real answer when \pi/p + \pi/q \lt \pi/2, which rearranges to (p-2)(q-2) \gt 4. The geometry refuses to build a central tile unless the tiling is genuinely hyperbolic.

Reflections are their own inverse. Reflect a tile across an edge and you get its neighbor; reflect the neighbor across the same edge and you return to the original. That is how the explorer avoids drawing a tile twice: it tracks which edges have already been crossed.

Worked example: the {4, 5} tiling from the defaults

The demo uses the defaults, which build five squares around each vertex: p = 4, q = 5. Follow the numbers.

Confirm it is hyperbolic: (4-2)(5-2) = 6 \gt 4. Yes.
Interior angle needed at each vertex: 360^\circ / 5 = 72^\circ. Five squares of 72 degrees each close the corner exactly.
Compare with flat: a flat square has 90 degree corners, so each hyperbolic square gives up 18 degrees per corner. The four corners lose 4 \times 18^\circ = 72^\circ of angle. That missing angle is the angle defect, and it is proportional to the tile's hyperbolic area.
Vertex radius: \pi/p = \pi/4 = 0.7854, \pi/q = \pi/5 = 0.6283. Then \cos(0.7854 + 0.6283) = \cos(1.4137) = 0.1564 and \cos(0.7854 - 0.6283) = \cos(0.1571) = 0.9877.
So r = \sqrt{0.1564 / 0.9877} = \sqrt{0.1583} = 0.3979. The four vertices of the central square sit at Euclidean radius about 0.398 from the center, at angles 45, 135, 225 and 315 degrees.
Reflect the central square across each of its four edges to get four neighbors, then continue outward. Each new layer packs more tiles into less Euclidean room, so tile counts roughly multiply by a fixed factor per layer.

The hyperbolic area of one tile follows the Gauss-Bonnet theorem. For a tile with angle sum S the area is (p-2)\pi - S in curvature units. Here S = 4 \times 72^\circ = 288^\circ = 1.6\pi, so the area is 2\pi - 1.6\pi = 0.4\pi \approx 1.257. Every square in the picture has this same area, no matter how small it looks.

Reading the picture without being fooled

The single most common misreading is treating the rim tiles as smaller. They are not. In the hyperbolic metric they are congruent to the central tile. What changes toward the rim is only the map's scale factor, the 2/(1-\lvert z \rvert^2) term from earlier. The table below shows how many tiles land in each successive layer of the \{4, 5\} tiling and how the shrink factor climbs.

Growth of the {4, 5} tiling by layer, with the disk scale factor at each layer's outer radius
Layer	Tiles in layer	Cumulative tiles	Outer radius	Scale factor
0	1	1	0.398	2.37
1	4	5	0.67	3.64
2	16	21	0.83	6.42
3	52	73	0.91	12.3
4	172	245	0.955	22.9

Two things stand out. The tile count grows faster than geometrically, which is why hyperbolic tilings feel bottomless. And the scale factor is climbing toward infinity, so no finite number of layers ever reaches the boundary. A rendered picture always stops at some cutoff radius near 0.99, beyond which the tiles are smaller than a pixel.

The scale factor is near 2 across most of the disk, then shoots up near the rim. That vertical wall is why the boundary is infinitely far away in hyperbolic distance.

Common mistakes

Swapping p and q. The tiling \{4, 5\} (five squares per vertex) and \{5, 4\} (four pentagons per vertex) are different tilings. Both are hyperbolic, and they are duals of each other, but they look nothing alike. Read the notation as sides first, count second.

Believing the rim tiles are tiny. They are congruent. The map projection distortion tool shows the same idea for the sphere: any flat map of a curved surface must stretch something, and Tissot's circles make the stretching visible.

Expecting a whole-number layer count to fill the disk. It never does. There are infinitely many layers, and the tool draws until tiles fall below a pixel. If your picture looks unfinished at the edge, that is correct, not a bug.

Trying flat values. Feed in \{6, 3\} or \{4, 4\} and the vertex radius formula returns zero or a non-real value, because (p-2)(q-2) \le 4. Those tilings belong on flat paper.

Do not read the Poincaré disk as a photograph taken from above a dome. It is not perspective. The boundary is not a horizon at finite distance: it is genuinely at infinite hyperbolic distance, and nothing in the tiling ever reaches it.

Related tools on this site

If aperiodic tilings interest you, the Penrose tiling generator covers the flat plane with two rhombi that never repeat, a different way of escaping the three regular flat tilings. For symmetry play in the flat plane, the kaleidoscope mirrors your strokes into n-fold mandalas, and Truchet tiles build weaves and mazes from one rotated square.

The circle-inversion machinery behind hyperbolic reflections also drives fractal boundaries. The Mandelbrot explorer and the Julia set explorer live in the same complex plane, and Ford circles pack the number line with mutually tangent circles in a way that quietly uses the modular group, a close cousin of the symmetry group of the \{3, \infty\} tiling.

Frequently asked questions

Why can't five squares meet at a point on flat paper?

A flat square corner is 90 degrees. Four of them use 4 \times 90 = 360 degrees, filling the plane exactly. Five would need 450 degrees, which overflows. In hyperbolic space each square corner shrinks to 72 degrees, and 5 \times 72 = 360 works.

Are the tiles near the edge really the same size as the middle one?

Yes, in hyperbolic distance. Every tile of \{4, 5\} has hyperbolic area about 1.257 curvature units. The apparent shrinking comes only from the disk model's scale factor 2/(1-\lvert z\rvert^2), which climbs from 2 at the center toward infinity at the rim.

What do the curved edges mean?

They are straight lines in hyperbolic geometry. Every polygon edge is a geodesic, and geodesics in the Poincaré disk appear as circular arcs that meet the boundary at a right angle. Edges through the center look straight because a diameter is a degenerate arc of infinite radius.

How many tiles does the explorer draw?

As many as fit above the pixel cutoff. In \{4, 5\} the first five layers already hold 245 tiles, and each further layer roughly multiplies the count. A dense render can carry several thousand tiles before they vanish into the rim.

Is this how Escher made the Circle Limit prints?

Escher worked by hand from a grid Coxeter sent him, not by computing circle inversions, but the underlying geometry is exactly this. Circle Limit III is close to a \{8, 3\} structure with the fish following lines that are not quite geodesics, which is why mathematicians still enjoy analyzing it.

Try it yourself in the tool