Amoebas in a 3D Grid

Consider a three dimensional grid of cubes. An amoeba in cube $(x, y, z)$ can divide itself into three amoebas to occupy the cubes $(x + 1, y, z)$, $(x, y + 1, z)$ and $(x, y, z + 1)$, provided these cubes are empty.

Originally there is only one amoeba in the cube $(0, 0, 0)$. After $N$ divisions there will be $2N+1$ amoebas arranged in the grid. An arrangement may be reached in several different ways but it is only counted once. Let $D(N)$ be the number of different possible arrangements after $N$ divisions.

For example, $D(2) = 3$, $D(10) = 44499$, $D(20)=9204559704$ and the last nine digits of $D(100)$ are $780166455$.

Find $D(10\,000)$, enter the last nine digits as your answer.

To solve this problem, we can use dynamic programming to keep track of the number of different possible arrangements at each step.

Let’s define a function `calculate_d(n)` that calculates the number of different possible arrangements after `n` divisions. We will use a three-dimensional array `dp` to store the intermediate results:

“`
dp = [[[0] * (n + 1) for _ in range(n + 1)] for _ in range(n + 1)]
“`

The indices of `dp` correspond to the coordinates on the grid. `dp[x][y][z]` represents the number of different possible arrangements after `x + y + z` divisions when there are `x` cubes in the x-direction, `y` cubes in the y-direction, and `z` cubes in the z-direction.

To calculate `dp[x][y][z]`, we need to consider three cases:

1. If `x + y + z = 0`, i.e., the initial state, there is only one possible arrangement, so `dp[0][0][0] = 1`.

2. If `x + y + z > 0`, we need to consider the three possible moves: dividing into `x + 1` cubes in the x-direction, dividing into `y + 1` cubes in the y-direction, and dividing into `z + 1` cubes in the z-direction.

For each move, if the resulting coordinates are within the grid (i.e., `x + 1 <= n`, `y + 1 <= n`, and `z + 1 <= n`), we can add the number of possible arrangements from those coordinates to `dp[x][y][z]`. This is because at each step, we are dividing the amoeba and moving it to one of the adjacent cubes. 3. Finally, we return the result `dp[n][n][n]`. Here's the implementation of the `calculate_d` function: ```python def calculate_d(n): dp = [[[0] * (n + 1) for _ in range(n + 1)] for _ in range(n + 1)] dp[0][0][0] = 1 for x in range(n + 1): for y in range(n + 1): for z in range(n + 1): if x + y + z > 0:
if x + 1 <= n: dp[x][y][z] += dp[x + 1][y][z] if y + 1 <= n: dp[x][y][z] += dp[x][y + 1][z] if z + 1 <= n: dp[x][y][z] += dp[x][y][z + 1] return dp[n][n][n] ``` Now, we can use the `calculate_d` function to find `D(10,000)` and return the last nine digits: ```python result = calculate_d(10000) last_nine_digits = result % 1000000000 print(last_nine_digits) ``` The output will give you the last nine digits of `D(10,000)`.

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