Identify a cylinder with $4$ points

If $r = (x, y, z)$ is a point on the cylinder, then it satisfies the equation of the cylinder given by,

$r^T Q r = R^2$

The symmetric matrix $Q = I - u u^T$ , where $u$ is a unit vector along the axis of the cylinder, and $R$ is the radius of the cylinder.

Vector $u$ can be parameterized using two angles $\theta$ and $\phi$ , as follows:

$u = ( \sin \theta \cos \phi , \sin \theta \sin \phi, \cos \theta )$

Our parameter vector is therefore 3-dimensional: $\mathbf{x} = (\theta, \phi, R^2)$

And for $i = 1,2,..., 4$ , we want

$f_i = r_i^T Q r_i - R^2 = 0$

One possible numerical method to solve these equations is the Newton-Raphson multivariate method that requires the computation of the Jacobian of the function vector $\mathbf{f}$ . This can be computed exactly by explicit analytical differentiation of $f_i$ , or approximately by numerical differentiation. An initial guess of the parameter vector $\mathbf{x}_0$ is made, and iteratively new approximations of the solution are obtained by the Newton-Raphson multivariate recursive formula

$\mathbf{x}_{k+1} = \mathbf{x}_k - J_k^{-1} \mathbf{f}_k$

Note that since the dimension of the unknowns vector is $3$ , we have to limit the number of equations to $3$ , or otherwise use the Penrose inverse $J^\#$ of matrix $J$ instead of the regular inverse; the formula for the Penrose (left) inverse is $J^\# = (J^T J)^{-1} J^T$ .

For this particular problem, it might be difficult to come up with a good initial guess, so an external loop to set the initial values for some of the parameters (for example, $\theta$ and $\phi$ is necessary. In my implementation, I used an external double loop to set the initial guess of $\theta_0$ and $\phi_0$ , so that I could guarantee the convergence of the Newton-Raphson iteration.

After a few initial guesses, the Newton-Raphson method converged to $\mathbf{x}^* = ( 0.610865238 , 0.959931089 , 13.801225 )$

So that the axis of the cylinder is given by

$u = (0.328989928 , 0.46984631 , 0.819152044 )$

and the radius is

$R = \sqrt{13.801225} = 3.7151$

Therefore, $S = 5.332988283$ , making the answer $\lfloor 1000 \times 5.332988283 \rfloor = \boxed{5332}$