坐标簇与沿单位矢量i的点之间的最小距离

Question

I have a set of 3D coordinates Q clustered into a crude sphere about an origin O, a unit vector i, and length d. Let p = c * i where c is a positive real number. Let M denote the set of distances between p and every q in Q. I need a robust way to compute the minimum value of c subject to the constraint that the minimum valve of M be bigger than d. Is there an elegant way to do this in python with scipy/numpy? I suspect that scipy.spatial.KDTree will be useful but it is unclear how I would efficiently find my constrained minimum.

Answer 1

You can solve the problem like this:

For a point q in Q, define:

C(q) = set of values of c >= 0 s.t. |q - c*i| >= d

and then define

C = intersection { q in Q } C(q)

and then the desired value c is:

c = inf C = minimum element of C

The sets C(q) are just the solution to a quadratic inequality and so will have one of the forms:

1) C(q) = [0, +infinity)
2) C(q) = [0, b] union [ e, +infinity )  (for some e > b)
3) C(q) = [e, +infinity )

You can safely discard sets of the first form.

If all of the remaining C(q) sets have form #2, then c = 0.

If instead they all have form #3, c is:

c = maximum { q in Q } e(q)

Otherwise things get more complicated.

For a given q, to compute C(q) note that:

|q - c*i| >= d
  <=> (q - c*i)^2 >= d*d
  <=> (q - c*i).(q - c*i) >= d*d
  <=> q.q + c*c - 2*c*(q.i) >= d*d
  <=> c*c - c*2*(q.i) + q.q - d*d >= 0

If the discriminant of this quadratic equation is < 0, then C(q) = [0, +infinity).

Otherwise let b and e be the roots, b <= e, and check these cases:

e < 0    -->  C(q) = [0, +infinity)
b < 0    -->  C(q) = [e, +infinity)
b >= 0   -->  C(q) = [0, b]  union [e, +infinity)

An example in 2-d which demonstrates the more complicated case:

Q = { q1, q2, q3, q4 }
  = { (1,0), (0,1), (4,4), (0,10) }
i = (1/sqrt 5, 2/sqrt 5)
d = 2

C(q1) = [ 2.236, +infinity )
C(q2) = [ 2.843, +infinity )
C(q3) = [ 0, 4.472 ] union [ 6.261, +infinity )
C(q4) = [ 0, +infinity )

The smallest element of the intersection of C(q1), C(q2) and C(q3) is 2.843.