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First solution in Clear category for Moore Neighbourhood by walkingpendulum
def count_neighbours(grid, row, col):
res = -1 * grid[row][col];
for i,j in [(i, j) for i in range(row - 1, row + 2) for j in range(col - 1, col + 2)]:
if 0 <= i < len(grid) and 0 <= j < len(grid[0]):
if grid[i][j]:
res += 1
else:
pass
return res
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"
March 25, 2015
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