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Simple iteration over neighbours solution in Clear category for Moore Neighbourhood by Tedium
def count_neighbours(grid, row, col):
count = 0
for i in range( max(row-1, 0), min(row+2, len(grid)) ):
for j in range( max(col-1, 0), min(col+2, len(grid[0])) ):
count += grid[i][j]
count -= grid[row][col]
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"
Dec. 11, 2015
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