0
$\begingroup$

I have a problem where I have two vectors a and b representing a list of angles.

I need to find a transformation T where T(a,b) = T(b,a), where T has a distance metric to compare two transformations, and that is inversible : I should be able to retrieve a and b (swapped is ok) from T(a,b)

For example: a+b, cos(a+b), sin(a)+sin(b), cos(a)*cos(b) or any combinations of these qualify

a.b doesn't qualify because I don't have a metric to compare square angles.

a-b doesn't qualify because it is not commutative

Do you think this is possible?

What I have tried that didn't work:

T(a,b) = {cos(a)*cos(b), cos(a)+cos(b), sin(a)*sin(b), sin(a)+sin(b)}

And solve the system for cos(a) cos(b) sin(a) sin(b). However because it is a quadratic equation, I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending when the determinant of the equations reaches zero.

This was just one idea of transformation that was symmetric so it would satisfy my conditions, but I couldn't get back my original vectors.

Thank you!

$\endgroup$
0
$\begingroup$

Notation convention: the $k$th element of an array, such as $a$, will be denoted with a subscript, as in $a_k$.

... I get two solutions and my two arrays a and b are not consistent anymore, I got two new vectors that have either values from a or from b depending ...

If you're working with functions of the form $c_k=f(a_k,b_k)$ for some symmetric $f$, this is unavoidable. We can swap $a_1$ and $b_1$, leave all of the other $a_k$ and $b_k$ alone, and the $c_k$ will still be the same. Repeat with $d_k=g(a_k,b_k)$ and the same thing happens; adding more functions will never solve the problem.

So then, we need functions that cross over, and use more than one $k$. Here's an idea:

\begin{align*}c_1 = a_1+b_1\quad d_1=\cos(a_1-b_1) &\\ c_2 = a_2+b_2\quad d_2=\cos(a_2-b_2) &\quad e_2=\cos(a_1+a_2-b_1-b_2)\\ c_3 = a_3+b_3\quad d_3=\cos(a_3-b_3) &\quad e_3=\cos(a_1+a_2+a_3-b_1-b_2-b_3)\end{align*} and so on. By adding in the third vector there, we know how the angle differences combine, and can tell the difference between the likes of $((x,y),(u,v))$ and $((x,v),(y,u))$ since $|x+y-u-v|\neq |x+v-u-y|$ in general.

$\endgroup$
  • $\begingroup$ This is what was missing!! Thanks a lot! $\endgroup$ – user5441518 Mar 14 at 6:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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