I was working in my project when I was struck by the question of whether it would be necessary, or at least cautious, prevent overflow and underflow in the calculation of these two distances.

I remembered that there is an implementation of the calculation of the hypotenuse to prevent this. Most languages ​​implementers, and is known for Hypot

The calculation of the Euclidean distance remains the same "pattern" and I thought that if Hypot() controls the overflow and underflow should also beware of the Euclidean distance. I've disappointed to note that the language we use, and others, do not control the overflow and underflow for calculating distance. Will not worth spend this "additional effort"?

I did a searchs and came to a question in Math.StackExchange

There is no definitive answer to this issue and is somewhat old. The first thing I wondered is: Will okay? I think that yes, seeing that is a generalization of the same procedure that performs Hypot().

I decided to extrapolate this concept to the Mahalanobis distance. The original is as follows:

$$D_M(X,Y,L) = \sqrt{\sum_{i=1}^{n}\left(\frac{X_i-Y_i}{L_i}\right)^2}$$

Since $L$ is the vector of eigenvalues.

And my proposal is this:

$$D_M(X,Y,L) = C\sqrt{\sum_{i=1}^{n}\left(\frac{X_i-Y_i}{L_i }\frac{1}{C}\right)^2}$$

That is the same to:

$$D_M(X,Y,L) = C\sqrt{\sum_{i=1}^{n}\left(\frac{X_i-Y_i}{L_i C}\right)^2}$$

And $C$ is the max value from the $|(X_i-Y_i)/L_i|$:

$$C = \max_{i}\left(\mid\frac{X_i-Y_i}{L_i}\mid\right)$$

Is it okay?

  • $\begingroup$ Did anyone could indicate whether this is correct? $\endgroup$ – Delphius Mar 14 '15 at 19:09

Assuming the $L_i$ are nonzero, such an algorithm will probably suffice in mitigating overflow issues, by analogy to the analysis done for norms; the Mahalanobis distance is like a weighted norm.

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  • $\begingroup$ I've come to think that actually being a weighted norm. It would not be necessary to do. What remains is whether or not I doubt the theoretical development that looks at link I posted is fine. I notice that I got confused when calculating c. Must be the maximum modulus. $\endgroup$ – Delphius Mar 16 '15 at 17:39

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