Finding a vector in Euclidian space that minimizes a loss function subject to some constraints

I'm attempting to address the adhering to reduction trouble, and also I'm certain there have to be a typical technique that I can make use of, yet until now I could not locate any kind of excellent referrals. Please allow me recognize if you have anything in mind that can aid or any kind of referrals that you assume would certainly serve for tackling this trouble.

Intend you are offered $K$ factors, $p_i \in R^n$, for $i \in \{1,\ldots,K\}$. Think additionally that we are offered $K$ constants $\delta_i$, for $i \in \{1,\ldots,K\}$. We intend to locate the vector $x$ that decreases:

$\min_{x \in R^n} \sum_{i=1,\ldots,K} || x - p_i ||^2$

topic the adhering to $K$ restraints:

$\frac{ || x - p_i ||^2 } { \sum_{j=1,\ldots,K} ||x - p_j||^2} = \delta_i$

for all $i \in {1,\ldots,K}$.

Any kind of aid is exceptionally welcome!


modify: additionally, we understand that $\sum_{i=1,\ldots,K} \delta_i = 1$.

2022-06-07 14:30:26
Source Share
Answers: 2

You can attempt making use of Lagrange multiplier method - see wikipedia.

2022-06-07 14:57:41

Yes, this jobs. If $n \leq K-2,$ you have no warranty of any kind of lawful remedy, also when the $\delta_i$ amount to 1, as called for. It might be that the example factors, your $v_j$ and also $Y,$ remained in a Euclidean room of a lot lower measurement, nonetheless, that does not assure you can duplicate that item of good luck if the new $n$ in $\mathbf R^n$ is also tiny.

If $n = K -1,$ there need to be a solitary viable factor, "near" the simplex with the $K$ factors as vertices. No demand (or capacity) to decrease anything. In fact, unless the $\delta$ is are all equivalent, it shows up there is a 2nd viable factor away. If all angles in the simplex are intense, there is a viable factor in its inside.

So, my suggestions is, identify just how to locate a viable factor when $n=K-1.$ If condition pressures $n \geq K,$ revolve so the hyperplane having all the $p_i$ comes to be the hyperplane $x_1, x_2, \ldots, x_{K-1}, 0,0,\ldots,0,$ address the trouble there, after that revolve back.

At the same time, I see absolutely nothing incorrect with a mathematical method for locating the solitary viable factor near the simplex when $n=K-1.$ Easier than locating the junction of a lot of rounds and also aircrafts. Keep in mind that, when $n=K,$ the complete set of all viable factors is either a straight line (if all $\delta_i$ are equivalent) or, actually, a real circle. Go number. In either instance, fulfilling the hyperplane which contains the $p_i$ is orthogonally.

For that issue, your most convenient program is simply to address the trouble in the initial $v_i, Y$ area, that is, a mathematical method that locates the factor $Z$ near the $v_i$ simplex with the proper $\delta$'s. After that you can simply map $Z$ in addition to the $v_i.$

2022-06-07 14:46:10