El 25/05/15 15:21, ravas escribió:
I read an interesting comment:
"""
The coolest thing I've ever discovered about Pythagorean's Theorem is an
alternate way to calculate it. If you write a program that uses the distance
form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your
available precision because the square root operation is last. A more accurate
calculation is c = a * sqrt(1 + b^2 / a^2). If a is less than b, you should
swap them and of course handle the special case of a = 0.
"""
Is this valid? Does it apply to python?
Any other thoughts? :D
My imagining:
def distance(A, B):
"""
A & B are objects with x and y attributes
:return: the distance between A and B
"""
dx = B.x - A.x
dy = B.y - A.y
a = min(dx, dy)
b = max(dx, dy)
if a == 0:
return b
elif b == 0:
return a
else:
return a * sqrt(1 + (b / a)**2)
I don't know if precision lose fits here but the second way you gave to
calculate c is just Math. Nothing extraordinary here.
c = a * sqrt(1 + b^2 / a^2)
c = sqrt(a^2(1 + b^2 / a^2)) applying the inverse function to introduce
a inside the square root
c = sqrt(a^2 + a^2*b^2/a^2) then just simplify
c = sqrt(a^2 + b^2)
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