3d point closest to multiple lines in 3D space - algorithm

Question

I search for non iterative, closed form, algorithm to find Least squares solution for point closest to the set of 3d lines. It is similar to 3d point triangulation (to minimize re-projections) but seems to be be simpler and faster?
Lines can be described in any form, 2 points, point and unit direction or similar.

Answer 1

Let the i th line be given by point ai and unit direction vector di. We need to find the single point that minimizes the sum of squared point to line distances. This is where the gradient is the zero vector:
Expanding the gradient,
Algebra yields a canonical 3x3 linear system,
where the k'th row (a 3-element row vector) of matrix M is
with vector ek the respective unit basis vector, and
It's not hard to turn this into code. I borrowed (and fixed a small bug in) a Gaussian elimination function from Rosettacode to solve the system. Thanks to the author!
#include <stdio.h>
#include <math.h>
typedef double VEC[3];
typedef VEC MAT[3];
void solve(double *a, double *b, double *x, int n); // linear solver
double dot(VEC a, VEC b) { return a[0]*b[0] + a[1]*b[1] + a[2]*b[2]; }
void find_nearest_point(VEC p, VEC a[], VEC d[], int n) {
MAT m = {{0, 0, 0}, {0, 0, 0}, {0, 0, 0}};
VEC b = {0, 0, 0};
for (int i = 0; i < n; ++i) {
double d2 = dot(d[i], d[i]), da = dot(d[i], a[i]);
for (int ii = 0; ii < 3; ++ii) {
for (int jj = 0; jj < 3; ++jj) m[ii][jj] += d[i][ii] * d[i][jj];
m[ii][ii] -= d2;
b[ii] += d[i][ii] * da - a[i][ii] * d2;
}
}
solve(&m[0][0], b, p, 3);
}
// Debug printing.
void pp(VEC v, char *l, char *r) {
printf("%s%.3lf, %.3lf, %.3lf%s", l, v[0], v[1], v[2], r);
}
void pv(VEC v) { pp(v, "(", ")"); }
void pm(MAT m) { for (int i = 0; i < 3; ++i) pp(m[i], "\n[", "]"); }
// A simple verifier.
double dist2(VEC p, VEC a, VEC d) {
VEC pa = { a[0]-p[0], a[1]-p[1], a[2]-p[2] };
double dpa = dot(d, pa);
return dot(d, d) * dot(pa, pa) - dpa * dpa;
}
double sum_dist2(VEC p, VEC a[], VEC d[], int n) {
double sum = 0;
for (int i = 0; i < n; ++i) sum += dist2(p, a[i], d[i]);
return sum;
}
// Check 26 nearby points and verify the provided one is nearest.
int is_nearest(VEC p, VEC a[], VEC d[], int n) {
double min_d2 = 1e100;
int ii = 2, jj = 2, kk = 2;
#define D 0.01
for (int i = -1; i <= 1; ++i)
for (int j = -1; j <= 1; ++j)
for (int k = -1; k <= 1; ++k) {
VEC pp = { p[0] + D * i, p[1] + D * j, p[2] + D * k };
double d2 = sum_dist2(pp, a, d, n);
// Prefer provided point among equals.
if (d2 < min_d2 || i == 0 && j == 0 && k == 0 && d2 == min_d2) {
min_d2 = d2;
ii = i; jj = j; kk = k;
}
}
return ii == 0 && jj == 0 && kk == 0;
}
void normalize(VEC v) {
double len = sqrt(dot(v, v));
v[0] /= len;
v[1] /= len;
v[2] /= len;
}
int main(void) {
VEC a[] = {{-14.2, 17, -1}, {1, 1, 1}, {2.3, 4.1, 9.8}, {1,2,3}};
VEC d[] = {{1.3, 1.3, -10}, {12.1, -17.2, 1.1}, {19.2, 31.8, 3.5}, {4,5,6}};
int n = 4;
for (int i = 0; i < n; ++i) normalize(d[i]);
VEC p;
find_nearest_point(p, a, d, n);
pv(p);
printf("\n");
if (!is_nearest(p, a, d, n)) printf("Woops. Not nearest.\n");
return 0;
}
// A linear solver from rosettacode (with bug fix: added a missing fabs())
#define mat_elem(a, y, x, n) (a + ((y) * (n) + (x)))
void swap_row(double *a, double *b, int r1, int r2, int n)
{
double tmp, *p1, *p2;
int i;
if (r1 == r2) return;
for (i = 0; i < n; i++) {
p1 = mat_elem(a, r1, i, n);
p2 = mat_elem(a, r2, i, n);
tmp = *p1, *p1 = *p2, *p2 = tmp;
}
tmp = b[r1], b[r1] = b[r2], b[r2] = tmp;
}
void solve(double *a, double *b, double *x, int n)
{
#define A(y, x) (*mat_elem(a, y, x, n))
int i, j, col, row, max_row, dia;
double max, tmp;
for (dia = 0; dia < n; dia++) {
max_row = dia, max = fabs(A(dia, dia));
for (row = dia + 1; row < n; row++)
if ((tmp = fabs(A(row, dia))) > max) max_row = row, max = tmp;
swap_row(a, b, dia, max_row, n);
for (row = dia + 1; row < n; row++) {
tmp = A(row, dia) / A(dia, dia);
for (col = dia+1; col < n; col++)
A(row, col) -= tmp * A(dia, col);
A(row, dia) = 0;
b[row] -= tmp * b[dia];
}
}
for (row = n - 1; row >= 0; row--) {
tmp = b[row];
for (j = n - 1; j > row; j--) tmp -= x[j] * A(row, j);
x[row] = tmp / A(row, row);
}
#undef A
}
This isn't extensively tested, but seems to be working fine.

Answer 2

Let base point of line is p and unit direction vector is d.
Then distance from point v to this line might be calculated using cross product
SquaredDist = ((v - p) x d)^2
Using Maple packet symbolic calculation, we can get
d := <dx, dy, dz>;
v := <vx, vy, vz>;
p := <px, py, pz>;
w := v - p;
cp := CrossProduct(d, w);
nrm := BilinearForm(cp, cp, conjugate=false); //squared dist
nr := expand(nrm);
//now partial derivatives
nrx := diff(nr, vx);
//results:
nrx := -2*dz^2*px-2*dy^2*px+2*dz^2*vx+2*dy^2*vx
+2*dx*py*dy-2*dx*vy*dy+2*dz*dx*pz-2*dz*dx*vz
nry := -2*dx^2*py-2*dz^2*py-2*dy*vz*dz+2*dx^2*vy
+2*dz^2*vy+2*dy*pz*dz+2*dx*dy*px-2*dx*dy*vx
nrz := -2*dy^2*pz+2*dy^2*vz-2*dy*dz*vy+2*dx^2*vz
-2*dx^2*pz-2*dz*vx*dx+2*dy*dz*py+2*dz*px*dx
To minimize sum of squared distances, we have to make system of linear equations for zero partial derivatives like this:
vx*2*(Sum(dz^2)+Sum(dy^2)) + vy * (-2*Sum(dx*dy)) + vz *(-2*Sum(dz*dx)) =
2*Sum(dz^2*px)-2*Sum(dy^2*px) -2*Sum(dx*py*dy)-2*Sum(dz*dx*pz)
where
Sum(dz^2) = Sum{over all i in line indexes} {dz[i] * dz[i]}
and solve it for unknowns vx, vy, vz
Edit: Old erroneous answer for planes instead of lines, left for reference
If we use general equation of line
A * x + B * y + C * z + D = 0
then distance from point (x, y, z) to this line is
Dist = Abs(A * x + B * y + C * z + D) / Sqrt(A^2 + B^2 + C^2)
To simplify - just normalize all line equations dividing by Norm's
Norm = Sqrt(A^2 + B^2 + C^2)
a = A / Norm
b = B / Norm
c = C / Norm
d = D / Norm
now equation is
a * x + b * y + c * z + d = 0
and distance
Dist = Abs(a * x + b * y + c * z + d)
and we can use squared distances like LS method (ai, bi, ci, di are coefficients for i-th line)
F = Sum(ai*x + bi*y + ci * z + d)^2 =
Sum(ai^2*x^2 + bi^2*y^2 + ci^2*z^2 + d^2 +
2 * (ai*bi*x*y + ai*ci*x*z + bi*y*ci*z + ai*x*di + bi*y*di + ci*z*di))
partial derivatives
dF/dx = 2*Sum(ai^2*x + ai*bi*y + ai*ci*z + ai*di) = 0
dF/dy = 2*Sum(bi^2*y + ai*bi*x + bi*ci*z + bi*di) = 0
dF/dz = 2*Sum(ci^2*z + ai*ci*x + bi*ci*y + ci*di) = 0
so we have system of linear equation
x * Sum(ai^2) + y * Sum(ai*bi) + z * Sum(ai*ci)= - Sum(ai*di)
y * Sum(bi^2) + x * Sum(ai*bi) + z * Sum(bi*ci)= - Sum(bi*di)
z * Sum(ci^2) + x * Sum(ai*ci) + y * Sum(bi*ci)= - Sum(ci*di)
x * Saa + y * Sab + z * Sac = - Sad
x * Sab + y * Sbb + z * Sbc = - Sbd
x * Sac + y * Sbc + z * Scc = - Scd
where S** are corresponding sums
and can solve it for unknowns x, y, z

Answer 3

I needed this for a sketch in Processing, so I ported Gene's answer. Works great and thought it might save someone else a little time. Unfortunately PVector/PMatrix don't have array accessors for vectors or matrices so I had to add these as local functions.
float getv(PVector v, int i) {
if(i == 0) return v.x;
if(i == 1) return v.y;
return v.z;
}
void setv(PVector v, int i, float value) {
if (i == 0) v.x = value;
else if (i == 1) v.y = value;
else v.z = value;
}
void incv(PVector v, int i, float value) {
setv(v,i,getv(v,i) + value);
}
float getm(float[] mm, int r, int c) { return mm[c + r*4]; }
void setm(float[] mm, int r, int c, float value) { mm[c + r*4] = value; }
void incm(float[] mm, int r, int c, float value) { mm[c + r*4] += value; }
PVector findNearestPoint(PVector a[], PVector d[]) {
var mm = new float[16];
var b = new PVector();
var n = a.length;
for (int i = 0; i < n; ++i) {
var d2 = d[i].dot(d[i]);
var da = d[i].dot(a[i]);
for (int ii = 0; ii < 3; ++ii) {
for (int jj = 0; jj < 3; ++jj) {
incm(mm,ii,jj, getv(d[i],ii) * getv(d[i],jj));
}
incm(mm, ii,ii, -d2);
incv(b, ii, getv(d[i], ii) * da - getv(a[i], ii) * d2);
}
}
var p = solve(mm, new float[] {b.x, b.y, b.z});
return new PVector(p[0],p[1],p[2]);
}
// Verifier
float dist2(PVector p, PVector a, PVector d) {
PVector pa = new PVector( a.x-p.x, a.y-p.y, a.z-p.z );
float dpa = d.dot(pa);
return d.dot(d) * pa.dot(pa) - dpa * dpa;
}
//double sum_dist2(VEC p, VEC a[], VEC d[], int n) {
float sum_dist2(PVector p, PVector a[], PVector d[]) {
int n = a.length;
float sum = 0;
for (int i = 0; i < n; ++i) {
sum += dist2(p, a[i], d[i]);
}
return sum;
}
// Check 26 nearby points and verify the provided one is nearest.
boolean isNearest(PVector p, PVector a[], PVector d[]) {
float min_d2 = 3.4028235E38;
int ii = 2, jj = 2, kk = 2;
final float D = 0.1f;
for (int i = -1; i <= 1; ++i)
for (int j = -1; j <= 1; ++j)
for (int k = -1; k <= 1; ++k) {
PVector pp = new PVector( p.x + D * i, p.y + D * j, p.z + D * k );
float d2 = sum_dist2(pp, a, d);
// Prefer provided point among equals.
if (d2 < min_d2 || i == 0 && j == 0 && k == 0 && d2 == min_d2) {
min_d2 = d2;
ii = i; jj = j; kk = k;
}
}
return ii == 0 && jj == 0 && kk == 0;
}
void setup() {
PVector a[] = {
new PVector(-14.2, 17, -1),
new PVector(1, 1, 1),
new PVector(2.3, 4.1, 9.8),
new PVector(1,2,3)
};
PVector d[] = {
new PVector(1.3, 1.3, -10),
new PVector(12.1, -17.2, 1.1),
new PVector(19.2, 31.8, 3.5),
new PVector(4,5,6)
};
int n = 4;
for (int i = 0; i < n; ++i)
d[i].normalize();
PVector p = findNearestPoint(a, d);
println(p);
if (!isNearest(p, a, d))
println("Woops. Not nearest.\n");
}
// From rosettacode (with bug fix: added a missing fabs())
int mat_elem(int y, int x) { return y*4+x; }
void swap_row(float[] a, float[] b, int r1, int r2, int n)
{
float tmp;
int p1, p2;
int i;
if (r1 == r2) return;
for (i = 0; i < n; i++) {
p1 = mat_elem(r1, i);
p2 = mat_elem(r2, i);
tmp = a[p1];
a[p1] = a[p2];
a[p2] = tmp;
}
tmp = b[r1];
b[r1] = b[r2];
b[r2] = tmp;
}
float[] solve(float[] a, float[] b)
{
float[] x = new float[] {0,0,0};
int n = x.length;
int i, j, col, row, max_row, dia;
float max, tmp;
for (dia = 0; dia < n; dia++) {
max_row = dia;
max = abs(getm(a, dia, dia));
for (row = dia + 1; row < n; row++) {
if ((tmp = abs(getm(a, row, dia))) > max) {
max_row = row;
max = tmp;
}
}
swap_row(a, b, dia, max_row, n);
for (row = dia + 1; row < n; row++) {
tmp = getm(a, row, dia) / getm(a, dia, dia);
for (col = dia+1; col < n; col++) {
incm(a, row, col, -tmp * getm(a, dia, col));
}
setm(a,row,dia, 0);
b[row] -= tmp * b[dia];
}
}
for (row = n - 1; row >= 0; row--) {
tmp = b[row];
for (j = n - 1; j > row; j--) {
tmp -= x[j] * getm(a, row, j);
}
x[row] = tmp / getm(a, row, row);
}
return x;
}