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First solution in Clear category for Area of a Convex Polygon by mfrankowski
def checkio2(d1,d2,d3):
return checkio1((d2[0]-d1[0],d2[1]-d1[1]),(d3[0]-d1[0],d3[1]-d1[1]))
def checkio1(e1,e2):
return abs((e1[1]*e2[0]-e2[1]*e1[0])/2.0)
def checkio(data):
a=[data.pop(0)]
n=len(data)
b=sum(map(checkio2,a*(n-1),data[:n-1],data[1:n]))
return b
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"
Nov. 20, 2016
