Kolams on Octahedron

Prelude

From a square board to a closed surface

In From Sixteen Tiles to Fifty-One Kolams, we placed sixteen square tiles inside a square boundary. Every tile carried four bits, neighbouring bits had to agree, and a computer-assisted enumeration reduced the valid boards to 51 symmetry classes.

What changes if the tiles are equilateral triangles?

A triangle has three sides, so its local information is shorter. There are only eight possible binary labels. Eight equilateral triangles are also the faces of a regular octahedron. The board can therefore close around itself: there is no outer boundary, and every side meets another tile.

We will ask for arrangements that use each of the eight tiles exactly once, match across every edge, and draw exactly two curve components: the small loop on $\mathtt{000}$, and one connected kolam through the other seven tiles.

The surprise is not merely that the answer is three. It is that no exhaustive search is needed. Once the octahedron is replaced by its dual cube, the tile inventory forces a forest, the forest forces a subdivided Y, and the Y has only three possible sets of arm lengths.

As in the companion article, this is a deliberately simplified mathematical model inspired by sikku or chikku kōlam. It is not a classification of the living tradition, which carries artistic, social, ritual, ecological, and personal meanings beyond the model. Cultural sources ↓

Eight triangular tiles

Fix three side names on an equilateral triangle. In the upward-facing drawings below, read them in the order

$X$

base

$Y$

right side

$Z$

left side

Write $1$ when the kolam curve meets the midpoint of that side, and $0$ when it does not. Each tile receives a word $XYZ$ of length three. Thus $\mathtt{101}$ meets the $X$- and $Z$-sides but not the $Y$-side. The word $\mathtt{000}$ is not blank: its curve is a small closed loop that meets no side.

Each bit has two possible values, so the total number of globally oriented tiles is

$2^3=8$.

The orientation is global. We may move an entire completed octahedron by a symmetry, but we do not rotate one labelled tile independently after placing it.

Why an octahedron?

Suppose a closed Platonic surface uses one triangular face for each tile. Then the number of faces is

$F=8$.

Every face has three edges, and every edge belongs to two faces. Counting face–edge incidences in two ways gives

$3F=2E$,

so $E=12$. Euler’s formula for a polyhedral sphere now gives

$V-E+F=2$,

and hence

$V=2+12-8=6$.

There are $3F=24$ face–vertex incidences. Because a Platonic solid looks the same at every vertex, four triangles meet at each of the six vertices. The resulting type is $\{3,4\}$: triangular faces, four around each vertex. This is the regular octahedron.

The solid is therefore not an arbitrary container chosen after the tiles were made. Among the Platonic solids, the eight-face inventory points directly to the octahedron.

The cube hidden inside the octahedron

The curves are drawn on octahedral faces, but the proof becomes simpler when we record only which faces touch.

Place one vertex at each face centre and join two centres when their faces share an edge. This is the dual graph of the octahedron. It is a cube: the octahedron’s eight faces become the cube’s eight vertices, and its twelve edges become the cube’s twelve edges.

There is also an intrinsic way to name the three side directions. Group the six octahedron vertices into three opposite pairs

$X_+,X_-;\quad Y_+,Y_-;\quad Z_+,Z_-$.

Every triangular face contains one vertex from each pair. We label it by a sign triple, or equivalently by a word in $\{0,1\}^3$. On a face, the $X$-side is the side opposite its $X$-vertex, and similarly for $Y$and $Z$. Crossing an $X$-side changes only the $X$-coordinate of the face label. The same is true in the other two directions.

Thus the dual graph is precisely the three-dimensional cube

$Q_3=\{0,1\}^3$.

Let $T=\{0,1\}^3$ be the set of tile words. A placement is a bijection

$f:Q_3\to T$.

The bijection is the exact-inventory rule. Every cube vertex receives one tile, and every tile is used exactly once.

If $x\oplus e_i$ is the cube neighbour obtained by changing coordinate $i$, then matching across the shared octahedron edge means

$f(x)_i=f(x\oplus e_i)_i$.

Now retain only the active shared edges: include the cube edge $\{x,x\oplus e_i\}$ when the common $i$th bit is $1$. Call the resulting graph $\Gamma_f$. The label of a vertex tells us exactly which active directions meet there, so

$\deg_{\Gamma_f}(x)=\text{number of 1s in }f(x)$.

Each tile motif is connected within its triangular face. Consequently, two face motifs belong to the same drawn curve component exactly when their vertices belong to the same component of $\Gamma_f$. The graph keeps all the connectivity information we need.

Twelve exits become six edges

The eight binary words have a rigid weight pattern.

Consequently, the degree multiset of $\Gamma_f$ is $0,1,1,1,2,2,2,3$. There are twelve $1$s in the full inventory, since $0+3\cdot1+3\cdot2+3=12$.

Each active octahedron edge is seen from both of its incident faces, so those twelve incidences pair up to give

This small count already tells us almost everything about the global kolam.

Two components, automatically

We intended to require two components. In fact, matching and exact inventory force them.

First we show that $\Gamma_f$ has no cycle. A vertex on a cycle has degree at least two. Our inventory contains only four such vertices: the three weight-two tiles and $\mathtt{111}$. Any cycle can therefore use at most four vertices.

The cube is bipartite, so it has no triangles. A cycle would have to be a four-cycle using all four vertices of degree at least two. Every four-cycle in a cube alternates between two directions, say $i$ and $j$. At each of its three degree-two vertices, the two active directions would then be exactly $\{i,j\}$.

But that would give all three vertices the same tile label. Exact inventory requires the degree-two labels to be the three different words $\mathtt{110}$, $\mathtt{101}$, and $\mathtt{011}$. This contradiction shows that no cycle exists.

Therefore $\Gamma_f$ is a forest. A forest with eight vertices and six edges has $8-6=2$ components. The tile $\mathtt{000}$ is the unique degree-zero vertex, so it is one isolated component. Every other tile must lie in the second component.

This is the second appearance of Euler characteristic. First, $V-E+F=2$identified the surface. Now, for a forest, $V-E$ counts its components.

A subdivided Y

The seven-vertex component is a tree with degree sequence $3,2,2,2,1,1,1$.

Its unique degree-three vertex is the face carrying $\mathtt{111}$. Remove that central vertex. What remains is three paths: the three arms of a Y.

Let their lengths be positive integers $\ell_1$, $\ell_2$,$\ell_3$. The tree has six edges in total, so

Up to reordering, there are only three positive partitions of 6 into three parts.

We now have three possible abstract tree shapes. One final issue remains: could the same tree shape sit inside the cube in several essentially different ways?

Three tree shapes are not yet three kolams

The cube remembers not only which faces are joined, but also the directions of those joins.

Name the cube directions $1,2,3$. Starting at the $\mathtt{111}$ face, read the edge directions along each arm. Three facts are forced by the inventory:

1. The first directions of the arms are $1,2,3$, because the central tile is $\mathtt{111}$.
2. The last directions are $1,2,3$, because the leaves are the three one-bit tiles.
3. At the three internal degree-two vertices, the direction pairs are $12,23,31$, because the tiles are $\mathtt{110}$, $\mathtt{011}$, and $\mathtt{101}$.

The three degree-two vertices are exactly the three transitions between consecutive symbols. Their distribution among the arms is forced by the arm lengths.

The bars separate the arms. Relabelling the directions applies one global permutation of $1,2,3$. Reversing the cyclic order gives a mirror image, which will be identified when reflections of the octahedron are allowed.

Thus every arm-length type has one realisation up to the full octahedral symmetry group. The directional argument is the step that turns three candidate tree shapes into three actual symmetry classes.

Symmetry and the final count

The octahedron and cube are dual, so they have the same symmetries. A full symmetry may move a chosen face to any one of eight faces. Once that face is fixed, it may permute its three side directions in any of $3!=6$ways. Therefore the full symmetry group has

$8\cdot6=48$

elements. On the dual cube, the group can be written

$C_2^3\rtimes S_3$.

The factor $C_2^3$ complements any chosen subset of the three cube coordinates, while $S_3$ permutes their directions.

Every symmetry preserves the multiset of arm lengths. The three types$(2,2,2)$, $(3,2,1)$, and $(4,1,1)$ therefore cannot become equivalent to one another. Conversely, the direction argument shows that every placement of a fixed type is related to its canonical representative by a face move and a direction permutation.

There are $112$ globally oriented matching placements before quotienting by symmetry. The three full-symmetry orbits have sizes $16,48,48$, and $16+48+48=112$.

The balanced type has a threefold stabilising rotation; the other two have trivial stabilisers in the full group.

A proof without exhaustive computation

The companion square-tile problem and the octahedral problem begin in the same way: binary edge data, exact inventory, local matching, global connectivity, and symmetry. They end by very different mathematical routes.

We may enumerate the $8!=40{,}320$ raw placements as an independent audit, but the proof never needs to inspect them. The number 112 is a consequence and a consistency check, not the source of the theorem.

This is a useful mathematical contrast. Sometimes computation certifies a finite landscape that remains too large to see all at once. Sometimes an invariant is strong enough to make the landscape collapse into a handful of inevitable forms.

Fold the three kolams

The three representatives below are drawn on connected octahedron nets. Select a net, fold it continuously into the solid, rotate the completed octahedron, and optionally reveal the active graph used in the proof.