# Conway Game of Life in a single line of Python

posted on 2021-11-29T19:49:26Z · view page on GitHubOn Wikipedia, Conway's game of life is described as a 2D grid of cells for which:

- Any live cell with fewer than two live neighbours dies, as if by underpopulation.
- Any live cell with two or three live neighbours lives on to the next generation.
- Any live cell with more than three live neighbours dies, as if by overpopulation.
- Any dead cell with exactly three live neighbours becomes a live cell, as if by reproduction.

This is simple enough. Let's implement this in a single line of python. More specifically, we'll be able to represent the game as a single generator expression. This is possible since the introduction of the Walrus Operator :=.

Anyway... Not sure why I did this:

## Conway Game of life function¶

In [2]:

```
game_of_life = lambda initial_cells, num_steps: (cells := (set(initial_cells) if i == 0 else (cells.difference({(x, y) for (x, y) in cells if not 1 < len([(x + i, y + j) for i in (-1, 0, 1) for j in (-1, 0, 1) if (not i == j == 0 and (x + i, y + j) in cells)]) < 4 }).union({(x_, y_) for (x, y) in cells for (x_, y_) in ((x + i, y + j) for i in (-1, 0, 1) for j in (-1, 0, 1) if (not i == j == 0 and (x + i, y + j) not in cells)) if len([(x_ + i, y_ + j) for i in (-1, 0, 1) for j in (-1, 0, 1) if (not i == j == 0 and (x_ + i, y_ + j) in cells)]) == 3 }))) for i in range(num_steps))
```

## Visualize¶

In [3]:

```
X, Y = np.mgrid[0:50, 0:50]
grid = np.zeros_like(X, dtype=int)
initial_cells = {(2, 2), (2, 1), (2, 0), (1, 0), (0, 1)}
num_steps = 120
for cells in game_of_life(initial_cells, num_steps):
clear_output(wait=True)
grid[:, :] = 0
for x, y in cells:
grid[x, -y] = 1
plt.figure(figsize=(5, 5))
plt.axis('scaled')
p = plt.pcolormesh(X, Y, grid, cmap="Greys")
plt.xticks(X[:,0]+0.5, ["" for _ in X[:, 0]])
plt.yticks(Y[0,:]+0.5, ["" for _ in Y[0, :]])
plt.grid(True)
plt.show()
```

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