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First solution in Clear category for Area of a Convex Polygon by kamilinho20
def checkio(data):
result = 0
for i in range(len(data)):
try:
nextY = data[i+1][1]
except:
nextY = data[0][1]
try:
prevY = data[i-1][1]
except:
prevY = data[len(data)-1]
result += data[i][0]*(nextY - prevY)
result = abs(result/2)
print(result)
return result
if __name__ == '__main__':
#This part is using only for self-checking and not necessary for auto-testing
def almost_equal(checked, correct, significant_digits=1):
precision = 0.1 ** significant_digits
return correct - precision < checked < correct + precision
assert almost_equal(checkio([[1, 1], [9, 9], [9, 1]]), 32), "The half of the square"
assert almost_equal(checkio([[4, 10], [7, 1], [1, 4]]), 22.5), "Triangle"
assert almost_equal(checkio([[1, 2], [3, 8], [9, 8], [7, 1]]), 40), "Quadrilateral"
assert almost_equal(checkio([[3, 3], [2, 7], [5, 9], [8, 7], [7, 3]]), 26), "Pentagon"
assert almost_equal(checkio([[7, 2], [3, 2], [1, 5], [3, 9], [7, 9], [9, 6]]), 42), "Hexagon"
assert almost_equal(checkio([[4, 1], [3, 4], [3, 7], [4, 8], [7, 9], [9, 6], [7, 1]]), 35.5), "Heptagon"
Jan. 12, 2017
