On Mon, May 25, 2015 at 1:21 PM, ravas <ra...@outlook.com> wrote: > 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
This branch is incorrect because a could be negative. You don't need this anyway; the a == 0 branch is only there because of the division by a in the else branch. > else: > return a * sqrt(1 + (b / a)**2) Same issue; if a is negative then the result will have the wrong sign. -- https://mail.python.org/mailman/listinfo/python-list
