Hans Walser, [20260707]

Sandpiles

Translated from German using AI.

1 What it's about

Modeling erosion

Counting problem

Figures

2 Description

Iteration: We imagine several stacks of squares placed side by side. We also choose an upper bound, for example, 3. If a stack is higher than the upper bound, we remove the top two squares and add one square each to the adjacent stacks on the left and right.

We start with a single stack of initial height n.

Interpreting the squares as grains of sand leads to a model of the erosion of a sandpile. Hence the title.

3 Example

We choose the initial height n = 11 and the upper bound 3. At the start, we have a stack of eleven squares (Fig. 1.0).

Fig. 1.0: Initial situation

Since 11 exceeds the upper bound 3, the stack is reduced by two squares. These are added to the (initially empty) stacks on the left and right (Fig. 1.1).

Fig. 1.1: Step 1

The height of the central stack (9) still exceeds the upper bound of 3. It is reduced again by two squares, which are added to the stacks on the left and right (Fig. 1.2).

Fig. 1.2: Step 2

The height of the central stack (7) still exceeds the upper bound of 3. It is reduced again by two squares, which are added to the stacks on the left and right (Fig. 1.3).

Fig. 1.3: Step 3

The height of the central stack (5) still exceeds the upper bound of 3. It is reduced again by two squares, which are added to the stacks on the left and right (Fig. 1.4).

Fig. 1.4: Step 4

Now the middle stack is within the limit, but the two stacks on the left and right are too high. They are each reduced by two squares. One square from the stack on the left is moved all the way to the left to the adjacent stack, and the other is moved to the stack in the middle. The two squares from the stack on the right are handled accordingly (Fig. 1.5).

Fig. 1.5: Step 5

Now the stack in the middle is too high again. It loses its two top squares. These go to the stacks to its left and right (Fig. 1.6).

Fig. 1.6: Step 6

Now everything is in order. The game is over. We need six steps to reach the final position.

The animation (Fig. 2) shows the steps every second.

Fig. 2: Steps

For n = 11 and upper bound 1, the steps are shown in Fig. 3. We need 21 steps.

Fig. 3: Upper Bound 1

For n = 11 and upper bound 2, we need 11 steps (Fig. 4).

Fig. 4: Upper Bound 2

4 Further Examples

4.1 Starting Height 12

Fig. 5: Starting Height 12

4.2 Starting Height 24

Fig. 6: Starting Height 24

5 Number of Steps

Table 1 shows the number of steps required depending on the starting height n and the upper bound s.

Table 1: Number of Steps

The column for s = 1 can be found here.

The table is only interesting for upper bounds that are less than one-third of the starting height. For larger upper bounds, only the central starting stack is reduced (Fig. 7 for n = 15 and s = 5).

Fig. 7: Reducing the Starting Stack

Weblinks

The On-Line Encyclopedia of Integer Sequences (OEIS)

https://oeis.org/A088804

Literatur

Labs, Oliver, und Skrodzki, Martin (2026): Abelian Sandpiles. Mitteilungen der deutschen Mathematiker-Vereinigung, dmvm-2026-0031, S. 132.DOI 10.1515