Let we have $n$ points $C_i$, and also have $n$ vectors $d_i$ corresponding to each points.
I want to find a point $X$ that satisfies $f(X,C_i)=d_i$, where $f$ is a simple Euclidean distance function.
Problem is, the vectors $d_i$ include sort of errors.

I think this problem is equivalent with approximating the intersection point of multiple circles.

We have multiple circles that intersect at a single point.
We already know the centers perfectly, but radii includes some error. (Around 10%. Not so much.)

My question is, how can I approximate the intersection points in this case?
I tried to formulate a kind of optimization problem, but it includes sum of absolute values...

And I also want to ask second problem...
Remember our original problem (finding point).
In fact, I want to find multiple points with multiple distances.
So, the actual goal is finding $X_i$s satisfying $f(X_i,C_j)=d_{ij}$.
Luckily, we have additional constraints here: $f(X_i,X_j)=l_{ij}$
(Of course, these $l_{ij}$ also include errors)

Any kind of advise will be really appreciated.

  • $\begingroup$ But $d_i$ are vectors or scalars? If $f$ is a distance I'd expect them to be scalars. $\endgroup$ Jul 8, 2020 at 16:19

1 Answer 1


You could obtain an estimate of $X$ by searching for the minimum of the weighted sum of the quadratic difference between $f(X,C_i)$ and $d_i$: $$ F(X)=\sum_i\left({f(X,C_i)-d_i\over \sigma_i}\right)^2, $$ where $\sigma_i$ is the uncertainty on $d_i$.


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