Created
November 25, 2018 00:02
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Game of Life in Python
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""" | |
Kyle's Game of Life Implementation | |
1) live cells die if they have 0, 1, or 4+ neighbors | |
2) empty cells have a birth if they have exactly three neighbors | |
""" | |
import numpy as np | |
# Create a blank board | |
board = np.zeros((5, 5)) | |
def iterate(board): | |
""" | |
This function takes the current board state and returns the next state. | |
""" | |
conv_board = np.zeros((7, 7)) | |
conv_board[1:6, 1:6] = board | |
conv = np.lib.stride_tricks.as_strided( | |
conv_board, | |
(5, 5, 3, 3), # view shape | |
(56, 8, 56, 8) # strides | |
) | |
# The new board | |
b = np.zeros((5, 5)) | |
for i in range(5): | |
for j in range(5): | |
# Count the number of neighbor live cells | |
if conv[i, j, 1, 1] == 1: | |
# Subtract itself from total count | |
b[i, j] = conv[i, j].sum() - 1 | |
else: | |
b[i, j] = conv[i, j].sum() | |
# Cells with 0, 1, or 4+ die | |
b[np.any([b <= 1, b >= 4], axis=0)] = 0 | |
# Living cells with 2 neighbors get to keep living | |
b[np.all([b == 2, board == 1], axis=0)] = 1 | |
# Dead cells with 2 neighbors stay dead | |
b[np.all([b == 2, board == 0], axis=0)] = 0 | |
# All cells with 3 neighbors live | |
b[b == 3] = 1 | |
# Return the new board state | |
return b | |
if __name__ == '__main__': | |
while input('Continue? [y/n] ') == 'y': | |
print(board) | |
board = iterate(board) |
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