The Loud Zero
An evening with the Abelian sandpile. Its identity element is a pile of sand that changes nothing when you add it to another pile, and it has a fractal drawn all over it.
- 1.44Mtopplings set off by adding the identity to a 64 × 64 pile. Afterwards every cell held exactly what it held before.
- 5/12of the side is taken by the flat plateau of 2s at the centre of a square grid's identity. I measured it at every size up to 320, and at 512, 768 and 1024.
- 11/8is the aspect ratio at which a rectangle's plateau disappears, then comes back standing on end.
- 0 grains
- 1
- 2
- 3
Four grains and it falls over
Put grains of sand on a grid. A cell holding four or more grains topples: it sends one grain to each of its four neighbours. Grains that go over the edge are lost. Keep toppling until every cell holds between 0 and 3 grains.
One toppling can set off others, so a single grain can start an avalanche. The final pile is the same whatever order you topple in. That property is why the model is called abelian, and everything else on this page depends on it.
- On the grid
- 0
- Lost over the edge
- 0
- Topplings
- 0
- 0
- 1
- 2
- 3
- 4 or more
A zero that makes noise
Call a stable pile recurrent if you can reach it from any other pile by adding sand. There are finitely many of them, and under the rule "add the two piles, then topple" they form a group. Every group has an identity: an element you can add to anything without changing it.
The sandpile identity takes one line to build. Put 6 grains on every cell and topple; call the result s. Then put 6 − s on each cell and topple again. The pile you end up with is the identity.
On the right is a random recurrent pile on a 100 × 100 grid. Add the identity and watch the count of cells that differ from where they started.
Checking the size of the group
The number of recurrent piles on a grid should equal its number of spanning trees, which is the determinant of the grid's reduced Laplacian. I tested every possible pile on the small grids below and counted the recurrent ones.
| Grid | Piles tested | Recurrent | det(L) |
|---|---|---|---|
| 1 × 1 | 4 | 4 | 4 |
| 1 × 2 | 16 | 15 | 15 |
| 2 × 2 | 256 | 192 | 192 |
| 2 × 3 | 4,096 | 2,415 | 2,415 |
| 3 × 3 | 262,144 | 100,352 | 100,352 |
- Topplings
- 0
- Cells different from the start
- 0
The plateau is five twelfths of the side
The identity of a square grid has a flat square of 2s in the middle. I measured its width for every grid side from 8 to 320, then at 512, 768 and 1024. The width stays within about two cells of 5n/12. At 768 it is exactly 320. At 1024 it is 428, where 5n/12 is 426.7.
A straight-line fit over sides 60 to 320 gives a slope of 0.41669 for even sides and 0.41668 for odd sides. Five twelfths is 0.41667.
Show the numbers
Odd grids carry a seam
On a grid with an odd side, a cross one cell wide cuts the plateau into quarters. It is a line of 1s through the 2s, with a single 0 at the centre, and it carries on as a line of 2s through the collar of 3s out to the edges.
On an even grid the mirror lines of the square run between cells. On an odd grid they run through a row and a column of cells, and the identity marks them.
The grid picks the plateau's shape
Change the region and the identity changes completely. A disc is almost all 2s, divided by thin ribbons. A diamond, which is the square turned 45°, has no plateau at all. A hexagon's plateau is a rectangle, because the grid's own axes decide its shape.
Pick a region to build its identity. The first stage topples the sixes, the second subtracts from six and topples again. Your browser computes it with the same sweeps I used.
- 0
- 1
- 2
- 3
- 4 or more
Past 11/8, the plateau flips
Stretch the square into a rectangle r rows tall and c columns wide. At first the plateau gets wider and flatter. Its height reaches zero at c = 11r/8, and there the plateau is gone: the middle is solid 3s. Stretch further and it comes back as a band running the full height.
I tried ten heights from 64 to 240. For nine of them the plateau is empty at exactly c = 11r/8, and after that the band's width is exactly c − 11r/8. At r = 240 everything happens two cells early.
Show the numbers
What I don't know
- Whether any of this is new. The sandpile identity has been studied for decades. Papers I'd read next, cited from memory and not checked: Le Borgne and Rossin on the identity of the sandpile group; Caracciolo, Paoletti and Sportiello on an explicit description of it; Alevy and Mkrtchyan on the identity of an ellipse, which I believe proves the mostly-2s picture for discs. The 5/12 and the 11/8 may well be in one of them.
- Why 5/12 and 11/8. I have measurements, not a derivation.
- What is special about r = 240. It is the one height where the flip lands two cells early.
How I computed it
- I started with the slow, obvious method: topple every unstable cell, repeat. Every faster version was checked against it and matched cell for cell.
- The inner loop is in C. It sweeps the grid in place and alternates direction, so an avalanche can cross the whole grid in one pass.
- Most of the toppling became one linear solve. Final heights are at most 3, so the number of times each cell topples, u, satisfies Lu ≥ h − 3, where L is the grid Laplacian. The solution of Lw = h − 3 is therefore a lower bound on u. Toppling each cell ⌊w⌋ times at once is safe, and ordinary toppling finishes the job. The 1024 × 1024 identity took about three minutes.
- The live pictures on this page are computed in your browser by the same sweeps, in JavaScript.