Enable Javascript in your browser and then refresh this page, for a much enhanced experience.
max(), min() solution in Clear category for Moore Neighbourhood by otonvm
def count_neighbours(grid, row, col):
n = 0
for y in range(max(0, row-1), min(row+2, len(grid))):
for x in range(max(0, col-1), min(col+2, len(grid[0]))):
if not (y == row and x == col):
n += grid[y][x]
return n
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"
Feb. 8, 2015
