Which shapes tile by themselves

The eleven uniform tilings use several regular polygons at once, and their duals use one irregular one each. Underneath both sits a question that is easier to state and much harder to answer.

Given a single convex polygon and nothing else, can copies of it cover the plane with no gap and no overlap?

The answer is a classification with an unusual shape. Three sides: always. Four sides: always, and convexity is not even required. Six sides: three families. Seven sides: never, for a reason that takes three lines. Five sides: fifteen families, the last found in 2015, and the list closed in 2017 by an exhaustive computer search that this collection does not reproduce.

Two of those five lines are about existence and three are about non-existence, which is the usual shape of a classification on this site. What is unusual is the fifth: a case where the non-existence claim — there is no sixteenth pentagon — arrived eighty years after the existence claims and by a completely different kind of work.

Every triangle, and the trick that does it #

That single observation is the whole of the triangle and quadrilateral cases. Half-turn the tile about the midpoint of each of its edges; half-turn each copy about the midpoints of its edges; keep going. The copies pile up into a patch, and for a triangle the patch is a tiling.

The reason it works is visible in the count. The three angles of a triangle add to 180°, so twice the three angles adds to 360°: at every vertex of the tiling, six triangles meet, two showing each angle. Nothing has to be arranged and no special triangle is needed.

The quadrilateral works for the same reason with the arithmetic one step shorter: its four angles already add to 360°, so four tiles meeting at a point close up exactly. And the argument never used convexity.

Every quadrilateral tiles the plane, and the non-convex ones are not a special case to be handled separately — they are the same case. That is worth pausing over, because “which shapes tile” is usually asked about convex ones and the restriction is doing less work here than it looks.

Run the same construction on a quadrilateral with a reflex angle and nothing changes. The four angles still add to 360°, four tiles still meet at every vertex, and the sampled disc still shows no gap and no overlap: the reflex corner of one tile is exactly the notch another fits into, because the two are related by the half turn that put them there. Convexity was never used — the argument needs the angle sum and the edge midpoints, and a reflex angle affects neither. That is unusual enough to state plainly, since every other line of the classification below is about convex polygons and stops being true without the word.

What the construction is, as a group #

It is worth naming what the half-turn construction produces, because the answer is one of the seventeen.

Half-turns about the midpoints of a tile’s edges generate a group. It contains no reflection, since a half-turn preserves handedness, and it contains no rotation of order more than two, since composing two half-turns gives a translation. A plane group with two-fold rotations and nothing else is p2 — the second of the seventeen — and the tile is its fundamental domain.

That reframes the result. “Every quadrilateral tiles” is the statement that every quadrilateral is a fundamental domain of p2, and the reason is that p2’s fundamental domain is a region bounded by four half-turn-related pieces of boundary, which a quadrilateral supplies with an edge each. The triangle is the same statement with two of its edges made from the two halves of one.

So the shapes that tile most easily are the ones that fit a group with nothing in it but half-turns, and the shapes that do not tile are the ones no group can accommodate. The classification below is a question about tiles that keeps turning into a question about groups.

Seven sides is impossible, and not by trying #

The bound at the top of the list needs no search over heptagons. In any tiling of the plane, every edge borders two tiles and at least three edges meet at every vertex, and Euler’s relation — with the plane’s characteristic of zero rather than a sphere’s two — then gives

$$\sum_n (6-n)\, p_n = 0.$$

That is the identical accounting that puts twelve pentagons on any closed cage, with the two on the right-hand side replaced by nothing. In a tiling by one shape there is only one term in that sum, so either 6 − n = 0 or there are no tiles at all. A tile has at most six sides.

The accounting is the proof and no picture is offered as one. What a picture can do is be refused, and the census at the end of this essay does exactly that: a regular heptagon is fed to the same construction that tiles the triangle and the quadrilateral, and it overlaps itself immediately — points inside the sampled disc that lie in two tiles at once. That failure carries no weight as an argument, since one heptagon failing under one construction says nothing about all heptagons under all constructions. What it does is confirm that the machinery which reports success on six shapes is capable of reporting failure at all, which is the only thing a positive result from it is worth.

And the same accounting settles what a vertex may look like. A tiling by one convex shape has an average vertex degree of at least three and an average tile size of at most six, and those two averages are the same statement twice: the tiles and the vertices are counted by the same edges. A tiling of triangles has six at a vertex and a tiling of hexagons has three, and everything between is a trade one against the other. The tiling of quadrilaterals has four at a vertex, and that is why the quadrilateral case is as easy as it is.

Six sides, in three ways #

Six is the boundary case — the only side count where 6 − n vanishes without vanishing trivially — and Reinhardt settled it in 1918: three families and no more.

The first family is the centrally symmetric hexagon, and there is nothing to prove about it: opposite sides equal and parallel means the three alternate angles sum to 360°, and translation alone tiles. This collection has met that condition twice already, under two other names — as the Wigner–Seitz cell of a plane lattice, which is either a hexagon of this kind or a rectangle, and as the two-dimensional case of Fedorov’s five parallelohedra.

The third family is the one whose construction is not the general one, and that is why it is here. Its three 120° corners are three-fold centres of the tiling: three copies of the hexagon meet at each of them and close up exactly, because three lots of 120° is a full turn. The half-turn construction does not tile with it at all; the rotation construction does, with no gaps and no overlaps in the sample.

The remaining family needs one pair of opposite sides equal and three consecutive angles summing to 360°, and is not built here. Naming what is not built is the point of naming it.

Reinhardt’s argument for three and no more is not reproduced here either, and it is worth saying what kind of argument it is: a case analysis on which vertices of the hexagon are vertices of the tiling and which lie in the middle of a neighbour’s edge. It is the same shape of reasoning as the classification proof for the seventeen — finite, elementary and long — and it was published seven years after Fedorov’s death and forty years before anybody found the tenth pentagon.

Five sides, where the story is #

A pentagon’s angles add to 540°, which divides no whole number of times into 360°, so pentagons cannot meet corner-to-corner in the tidy way triangles and quadrilaterals do. What they do instead is pair up.

Half-turn a pentagon about the midpoint of an edge whose two end angles add to 180°. The two corners at that edge become straight angles and stop being corners, so the union of two pentagons is a hexagon rather than an eight-sided figure — and because the union is invariant under the half-turn that made it, that hexagon is centrally symmetric. It is a parallelogon, and parallelogons tile by translation.

That is the first of the fifteen types, and it is the one with the shortest story. The other fourteen have blocks of three, four, six or twelve pentagons, some with no translational relation between the members of a block at all, and several were found by people looking rather than by anybody deducing. Type 10 was found by Richard James in 1975; types 9, 11, 12 and 13 by Marjorie Rice, an amateur working from a magazine article, between 1976 and 1977; the fifteenth by a computer search in 2015; and the search that closed the list ran in 2017.

This collection does not reproduce that search and says so. What it does instead is what it always does with a count it has not derived: exhibit an instance, state where the number comes from, and be explicit about the gap. The fifteen appears in this essay as a fact from the literature; the tiling above is a construction.

It is worth being exact about what the 2017 search actually closed, because “there is no sixteenth pentagon” is a stronger sentence than the theorem. A convex pentagon that tiles does so with its copies falling into finitely many orbits under the tiling’s own symmetry group, and the number of orbits — the isohedral number — is bounded: Rao’s argument enumerates every way a small number of pentagons can meet around a vertex, which is a finite list of angle equations, and solves each. So the theorem is that no convex pentagon tiles with three or fewer orbits beyond the known families, and the reduction of the general case to that bound is the part done by hand. Neither half is short and neither is checkable by drawing anything.

And the shape of the classification is worth reading against the rest of the site. Three sides and four sides are settled by an angle sum, seven and above by Euler’s relation, and six by a case analysis a person can follow. Five sits between two arguments that both run out: the angle sum gives 540°, which is no help, and the Euler bound permits five without saying anything about it. That is exactly the position the wallpaper groups would be in if the crystallographic restriction had left a case the trace argument could not decide — and it is why the pentagon list took eighty years while the seventeen took one paper.

The regular pentagon is the other refusal in that census, and it is the sharper of the two. It is convex, five-sided, and a member of no one of the fifteen types; fed to every construction here it leaves gaps. Being convex, five-sided and beautiful is not enough, and the fifteen types are not a list of shapes but a list of conditions — relations among the angles and edges that let a block of pentagons close up. A pentagon satisfying none of them tiles nothing, and the regular one satisfies none, which is the failure this collection has a whole essay about.

The history is the point rather than a decoration. Nearly every count in this collection was closed by an argument: seventeen by a case analysis, thirty-two by a search over subgroups, two hundred and thirty by an enumeration of extensions. The pentagons were closed by a computer doing an exhaustive search over a parameter space nobody could exhaust by hand, after eight decades in which the list grew by discovery — and one of the discoverers was not a mathematician. That is a different relationship between a classification and its proof from any other on this site, and it is the reason this essay names the search instead of running it.

Two failures, on purpose #

A grower that tiles with everything proves nothing by tiling. So the two shapes that must fail are put through the same machinery as the six that must not, and the sampling is asked the same question of all eight: does every point of a disc lie in exactly one tile?

The heptagon fails by overlapping: copies placed at its edges run into one another, because seven angles of 128.6° cannot be assembled around a point — two make 257°, three make 386°, and there is no whole number of them that comes to a turn. The regular pentagon fails by leaving gaps: three of its 108° angles come to 324° and a fourth will not fit, which is the refusal this collection opened with seen from the other side. Two shapes, two different modes of failure, one test.

The asymmetry between the two failures is worth a sentence. An overlap is a local contradiction: two tiles occupy the same point and the construction has already gone wrong. A gap is not local at all — every individual placement was legal and the region simply cannot be closed, which is why a gap is the harder failure to see by eye and the one a sampling test earns its place on. Every gate this collection has asks whether what is drawn is right; this one asks whether what is not drawn should have been.

What the classification does not say #

It is about one tile and not about one shape. A tiling by a single shape may still use that shape in many attitudes — the pentagon above appears both ways up, the triangle appears in six orientations — and nothing here counts them. Whether a tile can be made to tile only in ways with few orientations is a different question, and the pinwheel is the extreme answer to it.

It is a classification of convex tiles. The non-convex quadrilateral above tiles, but the general non-convex question has no classification at all and cannot have a short one: a shape with enough notches and tabs can be made to tile or not to tile almost at will, and deciding whether a given one does is a problem with no known general procedure. Convexity is what makes the list short, and dropping it does not extend the list — it destroys it.

It says nothing about which tilings, only whether. A quadrilateral tiles, and it tiles in more than one way; the classification asks for existence. The finer question — how many essentially different tilings a shape admits — is a classification of tilings rather than of tiles, and the seventeen groups are what it is asked in terms of.

Nothing here is about the plane groups the tilings have. A quadrilateral’s tiling is p2; the parallelogon’s is p1 plus whatever the shape happens to allow; the three-fold hexagon’s is p3. Those are facts about particular constructions rather than about the shapes, and a shape that tiles in one group usually tiles in others as well. The tile fixes a fundamental domain and the fundamental domain fixes a group, but neither determines the other on its own.

And the sampling is a check, not a proof. Every patch here is finite and every verdict is about a disc. That a construction which works out to a disc of some radius works out for ever is an argument about the group generating the copies, not about the sample — and it is available for the triangle, the quadrilateral, the parallelogon and the three-fold hexagon, because each of those is the orbit of a plane group and a plane group’s orbit is infinite by construction. Where the essay says “tiles”, it means the construction is a group orbit and the sample confirms it; where it says “does not”, it means one finite check found gaps or overlaps, and that is enough to refuse.

One dimension up, where the same question is barely started #

The classification here is complete for convex tiles in the plane. The three-dimensional analogue is not, and the shape of what is known says a great deal about why the plane’s answer was reachable.

The parallelohedra are classified. A convex polyhedron that tiles space by translations alone — no rotations, no reflections — is one of five: the cube, the hexagonal prism, the rhombic dodecahedron, the elongated dodecahedron and the truncated octahedron. That is Fedorov’s result of 1885, and its plane analogue is the two parallelogons, the parallelogram and the centrally symmetric hexagon.

Everything beyond that is open. Which convex polyhedra tile space when rotations and reflections are allowed is not classified, and there is no counterpart to Reinhardt’s hexagon families or to the fifteen pentagon types. Even the count of faces is not bounded in the way seven sides is bounded in the plane, because the Euler relation that forbids the heptagon has no such consequence for a polyhedron: a tiling of space imposes no analogous constraint on the average number of faces.

So the plane’s classification rests on a two-dimensional accident — the vanishing of 6 − n at six — and the third dimension does not supply one. That is the honest reason the pentagon problem took a century and the space-filling problem has not been attempted, and it is worth setting beside the pentagon story: the plane was hard, and it was hard inside a framework that at least bounded the question.

What convexity is doing #

The restriction to convex tiles is stated as a scope and it is closer to a precondition, because dropping it changes the kind of question being asked.

A convex polygon is described by a few numbers — its side lengths and angles — and a tiling condition is a set of equations among them, which is why the classification comes out as families with parameters. A non-convex tile has a boundary that can be as complicated as one likes, and that boundary can encode arbitrary structure. There is no small parameter list, so there is no system of equations, and there is nothing for a classification to be a classification of.

The consequences are exactly the ones the neighbouring essays report. Whether a set of tiles covers the plane is undecidable, and the tiles in that theorem are non-convex. Whether a single shape tiles is not known to be decidable, and the shapes at issue are polyominoes rather than convex polygons. And a single aperiodic tile is a thirteen-sided non-convex shape, which is the strongest possible statement that convexity was doing the work: no convex tile is aperiodic, since every convex tile that tiles at all tiles periodically by the constructions on this page.

That is the boundary this classification sits against. Convexity is not a simplifying assumption; it is what makes the question finite.