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First solution in Clear category for Moore Neighbourhood by jorgedgb
def count_neighbours(grid, row, col):
count = 0
for par in ((-1,-1), (-1,0), (-1,1), (0,-1), (0,1), (1,-1), (1,0), (1,1)):
i = col + par[0]
j = row + par[1]
if 0 <= i < len(grid[0]) and 0 <= j < len(grid):
count += grid[j][i]
return count
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"
April 8, 2015
