Given the distance between more than 3 points and no prerequisite for coordinate values, can the distance formula be reversed? Very similar to this, but the accepted solution given only works for 3 points, not n. 
I have a filled-out logic grid of distances between 16 points and I would like to plot a plausible configuration of them in 3D space. I don't care where in space this configuration lives, just that all distances are visually represented. Can this be done? If so, how?
 A: Supposing that you have the complete matrix of mutual distances between any couple of $n$ points in 3D.
Supposing also that the distances have been measured on $n$  physical points, i.e. we are sure that
they represent an actual euclidean geometric layout.
If not, we shall check at least that any three of them respect the triangle inequality.  More precisely 
we can check that for any four of them the Cayley-Menger determinant
is positive (the volume of the tetrahedron is real). 
That premised, the problem falls in the field of trilateration-multilateration so common
in surveying, air navigation, radio astronomy, satellite positioning, etc.
Most probably an algorithm has been developed in one of those disciplines to deal with the problem you propose. 
However the underlying concept is simple:
 - choose three points, verifying that they are not aligned, anf fix the position of the relevant triangle (e.g. on the $x-y$ plane);
 - choose a fourth point, verifying it is not coplanar with the previous,  and fix its position (e.g. sign of $z$) as the vertex of a base tetrahedron;
 - the tetrahedron  will allow to univocally fix the position of a fifth point;
 - proceed with the other points using the base terahedron or any of the newly fixed.
