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Euclidean distance

The Euclidean distance of two points x = (x1,...,xn) and y = (y1,...,yn) in Euclidean n-space is computed as
It is the "ordinary" distance between the two points that one would measure with a ruler, which can be proven by repeated application of the Pythagorean theorem. By using this formula as distance, Euclidean space becomes a metric space (even a Hilbert space).

Table of contents
1 Two-dimensional distance
2 Three-dimensional distance

Two-dimensional distance

For two 2D points P=[px,py] and Q=[qx,qy], the distance is computed as


A fast approximation of 2D distance based on an octagonal boundary can be computed as follows. Let dx = |px-qx| (absolute value) and dy = |py-qy|. If dydx, aproximated distance is 0.41dx+0.941246dy. (If dy<dx, swap these values.) The difference from the exact distance is between -6% and +3%; more than 85% of all possible differences are between -3% to +3%.

The following Waterloo Maple code implements this approximation and produces the plot on the right, with a true circle in black and the octagonal approximate boundary in red:
fasthypot :=
         dx, dy):
hypot := unapply(sqrt(x^2+y^2), x, y): plots[display](
 plots[implicitplot](fasthypot(x,y) > 1, 
 plottools[circle]([0,0], 1),

Other approximations exist as well. They generally try to avoid the square root, which is an expensive operation in terms of processing time, and provide various error:speed ratio. Using the above notation, dx + dy - 2*min(dx,dy) yields error in interval 0% to 12%. (Attributed to Alan Paeth.)

Three-dimensional distance

For two 3D points P=[px,py,pz] and Q=[qx,qy,qz], the distance is computed as