The least wall for equal rooms

The most area a fence can hold answers the question for one enclosure: of all shapes with a given perimeter, the circle holds the most. Half a circle against a wall gives the enclosure a wall for free and finds the half-circle. That essay ends by asking what happens when every piece of wall is shared — when the plane is divided into many rooms of equal area, each wall separates two of them, and the aim is to use as little wall as possible in total.

The circle is useless here. Circles of equal area cannot fill the plane; the curved gaps between them are rooms of the wrong size. So the question is which shape, repeated, comes closest to the circle’s economy while leaving no gaps. Marcus Varro recorded the bees’ answer in the first century BC, Pappus of Alexandria argued for it in the fourth century AD, and the proof arrived in 1999, when Thomas Hales showed that the regular hexagonal honeycomb uses the least wall of any division of the plane into regions of equal area — curved walls included.

This essay draws the half of the argument that fits on a page: the case of straight walls, which László Fejes Tóth settled in 1943 with two facts, one about counting and one about convexity.

Three tilings, and the wall each needs #

Only three regular polygons tile the plane on their own — triangles, squares and hexagons — and each can be drawn with cells of the same area.

Charge every cell half of its own perimeter, since each wall is shared by two cells, and scale so that every cell has area one. A square cell then costs exactly $2$: its perimeter is $4$ and half of it is its own. A triangular cell costs $2.280$ and a hexagonal cell $1.861$. The disc of area one, with perimeter $2\sqrt\pi$, would cost $1.772$ if discs could tile, and they cannot; hexagons come within five per cent of that unreachable figure. Of the three tilings that exist, hexagons are cheapest by a clear margin: a honeycomb uses seven per cent less wall than a grid of squares holding the same area.

That settles the comparison among regular tilings and says nothing about the rest. The theorem is about every division of the plane into equal areas — cells of any shape, of different shapes, with any number of sides, straight or curved — and there are infinitely many of those.

Why only three regular polygons tile #

The three tilings are the only ones for a simple reason about angles. At a corner of a tiling by one regular polygon, some number $k$ of copies meet, and their angles must fill a full turn. A regular $n$-gon has interior angle $180(n-2)/n$ degrees, so $k \cdot 180(n-2)/n = 360$, which rearranges to $(n - 2)(k - 2) = 4$. The only ways to write four as a product of two positive whole numbers are $1 \times 4$, $2 \times 2$ and $4 \times 1$, giving triangles six at a corner, squares four at a corner and hexagons three at a corner. Pentagons fail because $108$ does not divide $360$; heptagons and beyond fail because three of them already exceed a full turn.

That is the same count, run on angles, that decides why only five solids are regular in three dimensions, where the angles at a corner must add to less than a full turn so that the corner can fold into a solid. In the plane the angles must add to exactly a turn, and there are three answers; on a sphere they add to less, and there are five. The hexagon is the last polygon that can tile the flat plane, three to a corner, and that — rather than any property of bees — is what makes it the candidate.

More sides need less wall #

The first fact is about single polygons.

Among all polygons with $n$ sides and a given area, the regular one has the least perimeter — the $n$-sided version of the isoperimetric inequality, proved by the same kind of improvement argument. Its half-perimeter at unit area is

$$f(n) = \sqrt{n \tan(\pi/n)},$$

and the figure shows it falling steadily towards $\sqrt\pi$ as $n$ grows. So if cells could have as many sides as they liked, more would always be better, and the cheapest cells would be nearly circular.

Two further properties of $f$ are what the argument uses. It is decreasing — every extra side helps. And it is convex — each extra side helps less than the one before, so the curve bends upward. That second property is what will turn a statement about averages into a statement about every division.

The numbers show both at once. Going from three sides to four saves $0.280$ of wall; from four to five, $0.094$; from five to six, $0.045$; from six to seven, $0.025$. The savings halve, roughly, with each side, which is the upward bend. It means that a room with one side too few loses more than a room with one side too many gains — so a mixture of fives and sevens, averaging six, always costs more than all sixes.

The rooms average six sides #

The second fact is a count, and it is the reason six is special.

Divide a square whose opposite edges are glued together — a torus, so that there are no edges to worry about — into $F$ rooms, with walls meeting three at a time at $V$ corners, along $E$ walls. Euler’s formula for the torus says $V - E + F = 0$. Every corner has three walls and every wall two ends, so $3V = 2E$. Every wall borders two rooms, so the rooms’ sides add up to $2E$. Put together: $E = 3F$, and the total number of sides is $6F$. The rooms average exactly six sides, whatever the division.

The figure’s forty random rooms have between four and eleven sides — a lot of fives and sevens, one eleven — and their sides add up to exactly 240. The count is not approximately six; it is six on the nose, in this division and in the two others checked the same way. For a large piece of the plane rather than a torus the same count gives an average that approaches six from below, the difference being a boundary effect. And if four or more walls meet at some corners, the average is smaller still. So: in any division of the plane, the rooms average at most six sides.

The same count on a sphere #

Run Euler’s count on a sphere instead of a torus and the answer changes by exactly twelve. On a sphere $V - E + F = 2$, and with three walls at every corner the same bookkeeping gives

$$\sum_{\text{rooms}} (6 - n) = 12:$$

the rooms fall short of six sides by twelve in total. A division of a sphere into pentagons and hexagons, three walls at every corner, must therefore have exactly twelve pentagons, however many hexagons it has — which is why a football has twelve black patches and a carbon cage has twelve five-sided rings whatever its size. On a torus the shortfall is nought, and every division averages exactly six. On a surface with more handles it is negative, and the rooms must average more than six sides — a two-holed surface divided three walls to a corner has rooms averaging above six, which is why tilings of the hyperbolic plane can use heptagons and octagons.

So the number six is a fact about flatness. It is what Euler’s formula says for a surface with no curvature, and the honeycomb’s optimality is, at bottom, the statement that the flat plane’s natural number of sides for a room is the number at which the regular polygons stop being able to tile it.

Averaging only helps: Fejes Tóth’s argument #

Now put the two facts together for rooms of equal area and straight walls.

A room with $n$ straight sides and area one needs at least $f(n)$ of its own wall, with equality only if it is a regular $n$-gon: every cell in the figure sits on or above the curve. So a division into $N$ rooms uses at least $f(n_1) + f(n_2) + \dots + f(n_N)$ units of wall. Because $f$ bends upward, the average of $f$ over the rooms is at least $f$ of the average — the curve lies above its chords — and the average number of sides is at most six. Because $f$ is decreasing, $f$ of something at most six is at least $f(6)$. So

$$\frac{1}{N}\sum_i f(n_i) \;\ge\; f\!\Bigl(\frac1N \sum_i n_i\Bigr) \;\ge\; f(6),$$

and the wall per room is at least the hexagon’s $1.861$. Equality needs every room to be a regular hexagon. That is the whole proof for straight walls, and every step of it can be seen in the figure: the cells above the curve, their average number of sides pinned at six, and the average cell sitting well above the hexagon’s point.

The figure’s cells are random, so their areas differ, and each has been scaled to unit area on its own; the argument needs equal areas to add the cells’ walls up in this form, and the picture shows the inequality it rests on rather than the equal-area case itself. It is still striking how far above the curve random cells sit: an average of $2.084$ against $1.861$, eleven per cent more wall than a honeycomb.

Curved walls, and why they took until 1999 #

Fejes Tóth’s argument assumes straight walls, and the restriction is not a technicality. A room with curved walls can have a shorter perimeter for its number of sides than any polygon: bulge each wall outward and a pentagon approaches a circle. The catch is that a wall bulging out of one room bulges into its neighbour, so what one room gains in roundness the other pays for. Whether a clever arrangement of bulges could beat the honeycomb on balance — some rooms with few sides, bulging outward, and others with many, bulging in — is exactly what the straight-wall argument cannot address.

Hales’s proof, circulated in 1999 and published in 2001, handles this with an inequality for a single room that charges it for its bulges. For a room of area one with $n$ sides, possibly curved, he proved that its perimeter is at least an expression built from $f(n)$ adjusted by the net area its walls bulge outward — with the adjustment chosen so that, summed over all the rooms, the bulges cancel, since every bulge out of one room is a bulge into another. The inequality is delicate because it must hold for every room, however irregular, and much of the proof is spent on rooms with few sides and large outward bulges. With it in hand the summing argument goes through as before, and the hexagons win again.

The bees’ version was also settled in principle, and settled against the bees. A real honeycomb cell is three-dimensional: a hexagonal prism closed at one end by three rhombuses that interlock with the cells on the other side of the comb. Fejes Tóth showed in 1964 that a different closing, made of two hexagons and two smaller rhombuses, uses slightly less wax — by well under one per cent of the cell’s surface — so the bees’ famous economy is excellent but not optimal.

The same question in three dimensions #

The three-dimensional version asks for the division of space into cells of equal volume with the least total wall, and it is open.

Lord Kelvin proposed an answer in 1887: truncated octahedra, the fourteen-faced solids that stack to fill space, with their faces very slightly curved so that walls meet at the angles soap films choose. It stood for more than a century. In 1993 Denis Weaire and Robert Phelan, simulating foams on a computer, found a structure made of two kinds of cell — irregular dodecahedra and fourteen-faced cells with two hexagonal faces — that uses about 0.3 per cent less surface than Kelvin’s. The Weaire–Phelan foam was the pattern of the outer skin of the swimming centre built for the 2008 Olympic Games in Beijing.

Nothing comparable to Fejes Tóth’s argument exists in three dimensions, because both of its facts fail. There is no single best polyhedron with a given number of faces to play the role of $f(n)$, and Euler’s formula in three dimensions does not fix the average number of faces of a cell the way it fixes six in the plane. Whether the Weaire–Phelan foam is optimal is not known, and neither is whether any optimum exists that is periodic at all.

What soap films already knew #

Plateau’s laws, found experimentally by the Belgian physicist Joseph Plateau in 1873 by dipping wire frames in soapy water, describe what soap films do when surface tension minimises their area: films meet three at a time along curves, at angles of 120 degrees, and those curves meet four at a time at points. The 120-degree rule is the planar honeycomb’s angle, and it is forced by the same local economy — at a corner where three walls meet, any other angle can be shortened by moving the corner, just as the point that minimises the total distance to three towns sees each pair at 120 degrees.

A two-dimensional foam of equal bubbles, squeezed between glass plates, settles into a honeycomb for the reason proved here: it is minimising total wall. A random scatter of seeds, divided by nearest seed as in the figure above, does not, because the rule that drew those walls was about distance and not about length. Moving each seed to the middle of its own cell and redrawing, again and again, also ends in a honeycomb, though it is minimising something different — the average squared distance of a cell’s points from its seed — and the two optimisation problems happen to share their answer.

What the pictures cannot show #

Curved walls. Every cell in every figure has straight walls, so the figures illustrate Fejes Tóth’s theorem and not Hales’s. The inequality that handles bulging walls is not drawn anywhere; it is the part of the proof that took until 1999.

Equal areas in the random divisions. The random rooms have different areas. The per-room inequality holds for each of them after scaling, and the average-of-six count holds exactly; the equal-area argument that combines them is stated in the text and illustrated rather than enacted.

Division by nearest point. The random rooms are drawn by giving each scattered point the region nearer to it than to any other — the division nearest neighbours make — which is a convenient way to produce many different straight-walled divisions and has nothing to do with wall length. They are witnesses to the count and the inequality, not candidates for the optimum.

The three-dimensional numbers. The foam figure’s values other than the cube’s are quoted from the calculations that found them, which required solving for the curved faces numerically; nothing here computes a foam.

Still open: the best foam #

Kelvin’s problem is open in the strongest sense: no structure has been proved optimal, no lower bound close to the Weaire–Phelan value has been proved, and there is not even a proof that the optimal division of space into equal volumes, if there is one, is periodic. The simplest lower bound comes from the single-cell isoperimetric inequality: no cell can have less surface than a ball of the same volume, which for volume one is $(36\pi)^{1/3} \approx 4.836$ — far below every structure in the figure, and the gap between that and $5.288$ is where the problem lives.

Computer searches since 1993 have found many foams within a few tenths of a per cent of Weaire and Phelan’s, several of them related to the atomic structures of crystals called clathrates, and none better. Whether that is because the Weaire–Phelan foam is optimal, or because the searches share assumptions that exclude the answer, is not known. In the plane the hexagon’s victory was decided by the fact that the rooms average six sides; in space there is no number that plays that role, and it is not clear what an argument would even count.