2
$\begingroup$

I'm minimizing distances between two 6 dimensional vectors. I have been using manhattan distance so far and it works ok but my problem would benefit from discriminating between the following two cases

case 1. 
v1 = [1, 1, 1, 1, 1, 1]
v2 = [1, 1, 1, 1, 1, 4]
manhattan(v1, v2) = 3

case 2.
v1 = [1, 1, 1, 1, 1, 1]
v2 = [1.5, 1.5, 1.5, 1.5, 1.5, 1.5]
manhattan(v1, v2) = 3

I want to have a distance measure that prefers (smaller distance) case 2 because it ensures that v1 - v2 is small in all dimensions

$\endgroup$

1 Answer 1

4
$\begingroup$

There is an entire family of norms, the $p$-norms:

$$ ||x||_p := \big(\sum_{k=1}^n|x_k|^p\big)^{1/p}, $$

which give rise to a distance through $d_p(x,y):= ||x-y||_p$.

The Manhattan distance is the case $p=1$. The Euclidean distance is the case $p=2$.

You may be looking for the maximum norm and its associated distance metric, which you get by letting $p\to\infty$:

$$||x||_{\max} := \max \{|x_k|, k=1, \dots n\}, \quad d_{\max}(x,y)=\max\{|x_k-y_k|, k=1, \dots, n\}.$$

This will be small if all entries of your two vectors are close together.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.