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Second solution in Clear category for Moore Neighbourhood by yarmak_vladislav
from itertools import product
neighbours = [ (dy,dx) for dx,dy in product([-1,0,1],repeat=2) if dx or dy ]
def count_neighbours(grid, row, col):
return sum( grid[row+dy][col+dx] for dy, dx in neighbours
if 0 <= row + dy < len(grid) and 0 <= col + dx < len(grid[0]) )
if __name__ == '__main__':
#These "asserts" using only for self-checking and not necessary for auto-testing
assert count_neighbours(
((1, 0, 0, 1, 0),
(0, 1, 0, 0, 0),
(0, 0, 1, 0, 1),
(1, 0, 0, 0, 0),
(0, 0, 1, 0, 0),), 1, 2) == 3, "1st example"
assert count_neighbours(((1, 0, 0, 1, 0),
(0, 1, 0, 0, 0),
(0, 0, 1, 0, 1),
(1, 0, 0, 0, 0),
(0, 0, 1, 0, 0),), 0, 0) == 1, "2nd example"
assert count_neighbours(((1, 1, 1),
(1, 1, 1),
(1, 1, 1),), 0, 2) == 3, "Dense corner"
assert count_neighbours(((0, 0, 0),
(0, 1, 0),
(0, 0, 0),), 1, 1) == 0, "Single"
Oct. 12, 2014
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