From Sixteen Tiles to Fifty-One Kolams

Prelude

Kolam at the threshold

At daybreak, many thresholds in Tamil Nadu become temporary drawing surfaces. The ground is swept, sometimes dampened, and a kōlam is drawn with rice flour, rice paste, or mineral powder. Historically practised and transmitted largely by women, kōlam belongs to the meeting place between home and street: it may be a gesture of welcome, a sign of auspiciousness, a daily discipline, and a work of art intended to disappear beneath footsteps, wind, rain, and time before being made anew. Cultural sources ↓

Pulli means “dot”. In a pulli kōlam, a grid of dots provides the scaffold from which lines are joined or woven. In the sikku or chikku variety, one or more closed strands loop around the dots, producing patterns that invite questions about symmetry, continuity, topology, algorithms, and counting.

கோலம் · kōlam — A temporary drawing can carry a durable idea: local choices accumulating into global form.

Mathematics is only one way of seeing kōlam. The practice also carries aesthetic, social, ritual, ecological, and personal meanings. We will study one deliberately simplified square-tile model of a sikku kōlam, not a classification of the living tradition. The model keeps a small local grammar visible, turns every rule into an exact condition, and leads to a surprising count.

From six shapes to sixteen tiles

Take a square tile and mark its centre with a dot. Draw curve segments that surround the dot and meet the midpoint of any chosen subset of the four sides; if no side is chosen, draw a closed loop inside the tile. Up to rotation, this gives six shapes. Once orientation is fixed, they yield sixteen distinct tiles.

To turn the pictures into mathematical objects, record only one piece of information about each side: does the curve meet the midpoint of that side? Read the sides in the fixed order east, north, west, south, writing $1$ for “yes” and $0$ for “no”. Each tile therefore receives a binary string of length four. For example, $\mathtt{1011}$ has exits to the east, west, and south, but not to the north. The label $\mathtt{0000}$ is not an empty tile: its curve closes inside the square and meets no side.

Try the puzzle before reading on

Drag the sixteen tiles into the $4\times4$ frame in the Square Kolam Tile Challenge. The labels are fixed: you may move a tile, but you may not rotate it. Try to make every neighbouring pair agree and keep the fifteen nonzero tiles in one network.

A precise model of the board

We now give names to the board, the tile inventory, and the act of placing a tile. Let

$V=\{(x,y):x,y\in\{0,1,2,3\}\}$.

be the sixteen cells, with $x$ increasing eastwards and $y$ increasing southwards. Let $T=\{0,1\}^4$ be the set of sixteen $\mathrm{ENWS}$ words. For $t\in T$, write $t_E$, $t_N$,$t_W$, $t_S$ for its four bits. A placement is a function

$f:V\to T$,

where $f(x,y)$ is the globally oriented tile placed in cell $(x,y)$.

A successful board satisfies five precise conditions.

1. Exact inventory. The map $f$ is a bijection: every cell receives one tile and every word in $T$ occurs exactly once.
2. Horizontal matching. For $0\le x<3$ and $0\le y\le3$, $f(x,y)_E=f(x+1,y)_W$.
3. Vertical matching. For $0\le x\le3$ and $0\le y<3$, $f(x,y)_S=f(x,y+1)_N$.
4. Closed outer boundary. For every $x$ and $y$, $f(x,0)_N=f(x,3)_S=f(0,y)_W=f(3,y)_E=0$.
5. Connectivity. Form a graph $\Gamma_f$ whose vertices are the sixteen cells, joining neighbours when their common matched bits are both $1$. If $z_0=f^{-1}(\mathtt{0000})$, then the subgraph on $V\setminus\{z_0\}$ must be connected.

Conditions $1\text{--}4$ define a locally valid exact-inventory board. The fifth says that $\Gamma_f$ has exactly two components: the compulsory $\mathtt{0000}$ component and one component containing all fifteen nonzero tiles.

A small invariant with a large consequence

In each of the four bit positions, exactly eight of the sixteen words contain a $1$. The complete inventory therefore has $32$ exits. Because every exit is paired with exactly one neighbour, every locally valid board has precisely $16$ active connections.

After discarding the isolated $\mathtt{0000}$ vertex, every successful answer is a connected graph with $15$ vertices and $16$ edges. Its cyclomatic number is therefore $16-15+1=2$.

From a puzzle to satisfiability

A completely unrestricted placement has $16!=20{,}922{,}789{,}888{,}000$possibilities. Every condition can be encoded using Boolean variables, while the local rules give us an efficient way to prune the search. For each cell $v$ and tile $t$, introduce a variable

$X_{v,t}=1\iff\text{tile }t\text{ occupies cell }v$.

There are $16\times16=256$ such variables. The puzzle is now a Boolean satisfiability problem, usually abbreviated to SAT. Bijection gives the “one tile per cell” and “one cell per tile” clauses; matching and boundary conditions forbid incompatible placements. Connectivity is global, so it can be checked afterwards by graph search or encoded with auxiliary reachability variables.

1. One tile in every cell. Exactly one of the sixteen statements $X_{v,t}$ is true for each cell $v$.
2. Every tile used once. For each label $t$, exactly one cell variable is true.
3. Only compatible neighbours. Any pair of labels whose facing bits disagree is forbidden.
4. No open boundary. A tile with an outward-facing $1$ is forbidden in that position.
5. Reach all fifteen tiles. A breadth-first search from the unique $\mathtt{1111}$ tile must visit every nonzero tile.

A general SAT solver can enumerate the answers. For this $4\times4$ square, a row-by-row search makes the three reported counts particularly transparent.

$1{,}448$ horizontally legal rows

Call $R=(t_0,t_1,t_2,t_3)$ a horizontally legal row when its four labels are distinct, its left and right outer bits are $0$, and each pair of facing east–west bits agrees. Testing the $16\cdot15\cdot14\cdot13=43{,}680$ ordered four-tuples gives exactly $1{,}448$ such rows.

There is also a compact count. Record the five horizontal boundary and interface bits. The two outer bits are fixed at $0$, while the three inner bits are free, giving eight paths from $0$ back to $0$. For each transition, the north and south bits are free; correcting for repeated transitions so that labels remain distinct gives $24+3\cdot192+2\cdot256+144+192=1{,}448$.

$652$ locally valid boards

For each legal row, retain its four north bits, four south bits, and a sixteen-bit mask recording the labels it uses. Stack four rows only when adjacent north–south signatures match, the top and bottom boundary signatures are $\mathtt{0000}$, and the four masks are disjoint and together contain all sixteen labels. Exactly $652$ boards pass those local matching, boundary, and inventory tests. The search is exhaustive because every board has one unique ordered decomposition into four rows.

$408$ connected nonzero networks

Finally, construct $\Gamma_f$ for each of the $652$ boards and run a breadth-first search from $\mathtt{1111}$. Exactly $408$ searches reach all fifteen nonzero vertices. The other $244$ boards obey every local rule but split the nonzero tiles into two or more components.

When are two answers really the same?

Suppose you solve the puzzle and then turn the entire sheet through $90^\circ$. The picture has changed its orientation, but not its essential arrangement. The same is true if you reflect the whole square in a mirror.

A square has eight rigid symmetries. Together they form the dihedral group $D_4$: four rotations and four reflections.

The eight images of one board form its orbit. Counting “up to symmetry” means counting these orbits rather than individual labelled boards. But can we simply divide 408 by 8? Only if every orbit really has eight members. A highly symmetric board might have fewer, so we need a theorem that handles this possibility correctly.

Burnside’s lemma and the number 51

Let $X$ be the set of 408 accepted labelled boards. The eight symmetries of the square act on $X$ by moving a whole completed board. An orbit is the collection of boards obtainable from one another in this way, and our goal is to count these orbits.

Why not immediately divide 408 by 8? A board with its own symmetry could return to itself under more than the identity, making its orbit smaller than eight. Burnside’s lemma corrects for exactly this possibility. For a symmetry $g$, let $\operatorname{Fix}(g)$ be the accepted boards that $g$ leaves unchanged. Then

The average comes from a double count. Count pairs $(B,g)$ for which $g$ fixes$B$. Counting first by symmetry gives the sum above. If a board is fixed by$s$ symmetries, its orbit contains $8/s$ boards, and each is fixed by $s$symmetries. Thus every orbit contributes $(8/s)s=8$ pairs. Dividing the total by eight counts the orbits.

In our puzzle, the identity fixes all 408 boards. Both $\mathtt{0000}$ and $\mathtt{1111}$ are unchanged by every permutation of the compass directions. A symmetric board must therefore place both labels in cells fixed by the spatial symmetry. On an even $4\times4$ grid, a nontrivial rotation or a horizontal or vertical reflection has no fixed cell, so none can fix an accepted board.

A diagonal reflection does have four fixed cells, so it needs a final finite check. Each diagonal fixes two locally valid boards, but their component sizes are 9, 3, 3, and 1. None satisfies our connectivity rule.

Appendix

Catalogue of the 51 kolams

The catalogue below contains one canonical representative from every square-symmetry class in the enumeration.

build the 1,448 legal four-tile rows
index rows by their north and south signatures
join compatible rows with disjoint tile masks
keep the 652 exact-inventory boards
test reachability from the 1111 tile
keep the 408 accepted boards
replace each D₄ orbit by its least representative
verify: 51 orbits × 8 boards = 408