Minimize Euclidean distance from sets of points in 3-d

问题

Let's look at four (m) points in 3-d space- I want to generalize to n-d, but 3 should suffice for a solution ( Part 1).

a= (x1, y1, z1)

b= (x2, y2, z2)

c= (x3, y3, z3)
.
.

p= (x , y , z)

Find point q = c1* a + c2* b + c3* c + ..

where c1 + c2 + c3 +.. = 1
and  c1, c2, c3, .. >= 0
s.t.
euclidean distance pq is minimized.

What algorithms can be used ? Idea or pseudocode is enough.

Part 2: solve for m points in n-dimensions :

I thought it would be trivial to generalize to m points in n dimensions, but turns out it is not straightforward. I created another problem for the general problem here: minimize euclidean distance from sets of points in n-dimensions

回答1:

I think your question in 3D can be reduced to a simple affine 2D geometry problem by projecting the point P on the plane defined by the three points A, B, C, or the two vectors AB and AC (or another combinations of AB, AC, and BC).

At first sight, it seems likely that the 3+1 points problem generalizes to N dimensions (3 points always defining a triangle and a plane).
However, it is not immediately clear if this approach would work for more points that would not be coplanar.

1- reduction to 2D by projecting P to a point P'on the plane defined by vectors AB, and AC.

2- understand that the position of P' is determined by only one coefficient t in the Reals s.t. P' is an affine combination of AB and AC :
P' = t * AB + (1-t) * AC

3- from there, P' can be in 3 distinct locations:

• (a) inside the triangle ABC: in that case, Q = P'

• (b) in the areas delimited by an orthogonal outwards projection of one of the segments; in that case Q is the orthogonal projection of P' on the closest segment.

• (c) not in (a) or (b); in that last trivial case, Q is the closest of A, B, or C

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来源：https://stackoverflow.com/questions/52129598/minimize-euclidean-distance-from-sets-of-points-in-3-d