Question: Later Edit: I uploaded here a sample of my original data. It's actually a segmentation image in the DICOM format. The volume of this structure as it is it's ~ 16 mL, so I assume the inner ellipsoid volume should be smaller than that. to extract the points from the DICOM image I used the following code: import os import numpy as np import SimpleITK as sitk def get_volume_ml(image): x_spacing, y_spacing, z_spacing = image.GetSpacing() image_nda = sitk.GetArrayFromImage(image) imageSegm_nda_NonZero = image_nda.nonzero() num_voxels = len(list(zip(imageSegm_nda_NonZero[0], imageSegm_nda_NonZero[1], imageSegm_nda_NonZero[2]))) if 0 >= num_voxels: print('The mask image does not seem to contain an object.') return None volume_object_ml = (num_voxels * x_spacing * y_spacing * z_spacing) / 1000 return volume_object_ml def get_surface_points(folder_path): """ :param folder_path: path to folder where DICOM images are stored :return: surface points of the DICOM object """ # DICOM Series reader = sitk.ImageSeriesReader() dicom_names = reader.GetGDCMSeriesFileNames(os.path.normpath(folder_path)) reader.SetFileNames(dicom_names) reader.MetaDataDictionaryArrayUpdateOn() reader.LoadPrivateTagsOn() try: dcm_img = reader.Execute() except Exception: print('Non-readable DICOM Data: ', folder_path) return None volume_obj = get_volume_ml(dcm_img) print('The volume of the object in mL:', volume_obj) contour = sitk.LabelContour(dcm_img, fullyConnected=False) contours = sitk.GetArrayFromImage(contour) vertices_locations = contours.nonzero() vertices_unravel = list(zip(vertices_locations[0], vertices_locations[1], vertices_locations[2])) vertices_list = [list(vertices_unravel[i]) for i in range(0, len(vertices_unravel))] surface_points = np.array(vertices_list) return surface_points folder_path = r"C:\Users\etc\TTT [13]\20160415 114441\Series 052 [CT - Abdomen WT 1 0 I31f 3]" points = get_surface_points(folder_path)  I have a set of points (n > 1000) in 3D space that describe a hollow ovoid like shape. What I would like is to fit an ellipsoid (3D) that is inside all of the points. I am looking for the maximum volume ellipsoid fitting inside the points. I tried to adapt the code from Minimum Enclosing Ellipsoid (aka outer bounding ellipsoid) by modifying the threshold err > tol, with my logic begin that all points should be smaller than < 1 given the ellipsoid equation. But no success. I also tried the Loewner-John adaptation on mosek, but I didn't figure how to describe the intersection of a hyperplane with 3D polytope (the Ax <= b representation) so I can use it for the 3D case. So no success again.  The code from the outer ellipsoid: import numpy as np import numpy.linalg as la import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D pi = np.pi sin = np.sin cos = np.cos def plot_ellipsoid(A, centroid, color, ax): """ :param A: matrix :param centroid: center :param color: color :param ax: axis :return: """ centroid = np.asarray(centroid) A = np.asarray(A) U, D, V = la.svd(A) rx, ry, rz = 1. / np.sqrt(D) u, v = np.mgrid[0:2 * np.pi:20j, -np.pi / 2:np.pi / 2:10j] x = rx * np.cos(u) * np.cos(v) y = ry * np.sin(u) * np.cos(v) z = rz * np.sin(v) E = np.dstack((x, y, z)) E = np.dot(E, V) + centroid x, y, z = np.rollaxis(E, axis=-1) ax.plot_wireframe(x, y, z, cstride=1, rstride=1, color=color, alpha=0.2) ax.set_zlabel('Z-Axis') ax.set_ylabel('Y-Axis') ax.set_xlabel('X-Axis') def mvee(points, tol = 0.001): """ Finds the ellipse equation in "center form" (x-c).T * A * (x-c) = 1 """ N, d = points.shape Q = np.column_stack((points, np.ones(N))).T err = tol+1.0 u = np.ones(N)/N while err > tol: # assert u.sum() == 1 # invariant X = np.dot(np.dot(Q, np.diag(u)), Q.T) M = np.diag(np.dot(np.dot(Q.T, la.inv(X)), Q)) jdx = np.argmax(M) step_size = (M[jdx]-d-1.0)/((d+1)*(M[jdx]-1.0)) new_u = (1-step_size)*u new_u[jdx] += step_size err = la.norm(new_u-u) u = new_u c = np.dot(u,points) A = la.inv(np.dot(np.dot(points.T, np.diag(u)), points) - np.multiply.outer(c,c))/d return A, c folder_path = r"" # path to a DICOM img folder points = get_surface_points(folder_path) # or some random pts A, centroid = mvee(points) U, D, V = la.svd(A) rx_outer, ry_outer, rz_outer = 1./np.sqrt(D) # PLOT fig = plt.figure() ax1 = fig.add_subplot(111, projection='3d') ax1.scatter(points[:, 0], points[:, 1], points[:, 2], c='blue') plot_ellipsoid(A, centroid, 'green', ax1)  Which gives me this result for the outer ellipsoid on my sample points:  The main question: How do I fit an ellipsoid (3D) inside a cloud of 3D points using Python?  Is it possible to modify the algorithm for the outer ellipsoid to get the maximum inscribed (inner) ellipsoid? I am looking for code in Python ideally.

Answer: Problem statement Given a number of points v₁, v₂, ..., vₙ, find a large ellipsoid satisfying two constraints:  The ellipsoid is in the convex hull ℋ = ConvexHull(v₁, v₂, ..., vₙ). None of the points v₁, v₂, ..., vₙ is within the ellipsoid.  I propose an iterative procedure to find a large ellipsoid satisfying these two constraints. In each iteration we need to solve a semidefinite programming problem. This iterative procedure is guaranteed to converge, however it not guaranteed to converge to the globally largest ellipsoid. Approach Find a single ellipsoid The core of our iterative procedure is that in each iteration, we find an ellipsoid satisfying 3 conditions:  The ellipsoid is contained within ConvexHull(v₁, v₂, ..., vₙ) = {x | Ax<=b}. For a set of points u₁, ... uₘ (where {v₁, v₂, ..., vₙ} ⊂ {u₁, ... uₘ}, namely the given point in the point clouds belongs to this set of points u₁, ... uₘ), the ellipsoid doesn't contain any point in u₁, ... uₘ. We call this set u₁, ... uₘ as "outside points". For a set of points w₁,..., wₖ (where {w₁,..., wₖ} ∩ {v₁, v₂, ..., vₙ} = ∅, namely none of the point in v₁, v₂, ..., vₙ belongs to {w₁,..., wₖ}), the ellipsoid contains all of the points w₁,..., wₖ. We call this set w₁,..., wₖ as "inside points".  The intuitive idea is that the "inside points" w₁,..., wₖ indicate the volume of the ellipsoid. We will append new point to "inside points" so as to increase the ellipsoid volume. To find such an ellipsoid through convex optimization, we parameterize the ellipsoid as {x | xᵀPx + 2qᵀx ≤ r}  and we will search for P, q, r. The condition that the "outside points" u₁, ... uₘ are all outside of the ellipsoid is formulated as uᵢᵀPuᵢ + 2qᵀuᵢ >= r ∀ i=1, ..., m  this is a linear constraint on P, q, r. The condition that the "inside points" w₁,..., wₖ are all inside the ellipsoid is formulated as wᵢᵀPwᵢ + 2qᵀwᵢ <= r ∀ i=1, ..., k  This is also a linear constraint on P, q, r. We also impose the constraint P is positive definite  P being positive definite, together with the constraint that there exists points wᵢ satisfying wᵢᵀPwᵢ + 2qᵀwᵢ <= r guarantees that the set {x | xᵀPx + 2qᵀx ≤ r} is an ellipsoid. We also have the constraint that the ellipsoid is inside the convex hull ℋ={x | aᵢᵀx≤ bᵢ, i=1,...,l} (namely there are l halfspaces as the H-representation of ℋ). Using s-lemma, we know that a necessary and sufficient condition for the halfspace {x|aᵢᵀx≤ bᵢ} containing the ellipsoid is that ∃ λᵢ >= 0, s.t [P q -λᵢaᵢ/2] is positive semidefinite. [(q-λᵢaᵢ/2)ᵀ λᵢbᵢ-r]  Hence we can solve the following semidefinite programming problem to find the ellipsoid that contains all the "inside points", doesn't contain any "outside points", and is within the convex hull ℋ find P, q, r, λ s.t uᵢᵀPuᵢ + 2qᵀuᵢ >= r ∀ i=1, ..., m wᵢᵀPwᵢ + 2qᵀwᵢ <= r ∀ i=1, ..., k P is positive definite. λ >= 0, [P q -λᵢaᵢ/2] is positive semidefinite. [(q-λᵢaᵢ/2)ᵀ λᵢbᵢ-r]  We call this P, q, r = find_ellipsoid(outside_points, inside_points, A, b). The volume of this ellipsoid is proportional to (r + qᵀP⁻¹q)/power(det(P), 1/3). Iterative procedure. We initialize "outside points" as all the points v₁, v₂, ..., vₙ in the point cloud, and "inside points" as a single point w₁ in the convex hull ℋ. In each iteration, we use find_ellipsoid function in the previous sub-section to find the ellipsoid within ℋ that contains all "inside points" but doesn't contain any "outside points". Depending on the result of the SDP in find_ellipsoid, we do the following  If the SDP is feasible. We then compare the newly found ellipsoid with the largest ellipsoid found so far. If this new ellipsoid is larger, then accept it as the largest ellipsoid found so far. If the SDP is infeasible, then we remove the last point in "inside points", add this point to "outside point".  In both cases, we then take a new sample point in the convex hull ℋ, add that sample point to "inside points", and then solve the SDP again. The complete algorithm is as follows  Initialize "outside points" to v₁, v₂, ..., vₙ, initialize "inside points" to a single random point in the convex hull ℋ. while iter < max_iterations: Solve the SDP P, q, r = find_ellipsoid(outside_points, inside_points, A, b). If SDP is feasible and volume(Ellipsoid(P, q, r)) > largest_volume, set P_best = P, q_best=q, r_best = r. If SDP is infeasible, pt = inside_points.pop_last(), outside_points.push_back(pt). Randomly sample a new point in ℋ, append the point to "inside points", iter += 1. Go to step 3.  Code from scipy.spatial import ConvexHull, Delaunay import scipy import cvxpy as cp import matplotlib.pyplot as plt import numpy as np from scipy.stats import dirichlet from mpl_toolkits.mplot3d import Axes3D # noqa def get_hull(pts): dim = pts.shape[1] hull = ConvexHull(pts) A = hull.equations[:, 0:dim] b = hull.equations[:, dim] return A, -b, hull def compute_ellipsoid_volume(P, q, r): """ The volume of the ellipsoid xᵀPx + 2qᵀx ≤ r is proportional to power(r + qᵀP⁻¹q, dim/2)/sqrt(det(P)) We return this number. """ dim = P.shape[0] return np.power(r + q @ np.linalg.solve(P, q)), dim/2) / \ np.sqrt(np.linalg.det(P)) def uniform_sample_from_convex_hull(deln, dim, n): """ Uniformly sample n points in the convex hull Ax<=b This is copied from https://stackoverflow.com/questions/59073952/how-to-get-uniformly-distributed-points-in-convex-hull @param deln Delaunay of the convex hull. """ vols = np.abs(np.linalg.det(deln[:, :dim, :] - deln[:, dim:, :]))\ / np.math.factorial(dim) sample = np.random.choice(len(vols), size=n, p=vols / vols.sum()) return np.einsum('ijk, ij -> ik', deln[sample], dirichlet.rvs([1]*(dim + 1), size=n)) def centered_sample_from_convex_hull(pts): """ Sample a random point z that is in the convex hull of the points v₁, ..., vₙ. z = (w₁v₁ + ... + wₙvₙ) / (w₁ + ... + wₙ) where wᵢ are all uniformly sampled from [0, 1]. Notice that by central limit theorem, the distribution of this sample is centered around the convex hull center, and also with small variance when the number of points are large. """ num_pts = pts.shape[0] pts_weights = np.random.uniform(0, 1, num_pts) z = (pts_weights @ pts) / np.sum(pts_weights) return z def find_ellipsoid(outside_pts, inside_pts, A, b): """ For a given sets of points v₁, ..., vₙ, find the ellipsoid satisfying three constraints: 1. The ellipsoid is within the convex hull of these points. 2. The ellipsoid doesn't contain any of the points. 3. The ellipsoid contains all the points in @p inside_pts This ellipsoid is parameterized as {x | xᵀPx + 2qᵀx ≤ r }. We find this ellipsoid by solving a semidefinite programming problem. @param outside_pts outside_pts[i, :] is the i'th point vᵢ. The point vᵢ must be outside of the ellipsoid. @param inside_pts inside_pts[i, :] is the i'th point that must be inside the ellipsoid. @param A, b The convex hull of v₁, ..., vₙ is Ax<=b @return (P, q, r, λ) P, q, r are the parameterization of this ellipsoid. λ is the slack variable used in constraining the ellipsoid inside the convex hull Ax <= b. If the problem is infeasible, then returns None, None, None, None """ assert(isinstance(outside_pts, np.ndarray)) (num_outside_pts, dim) = outside_pts.shape assert(isinstance(inside_pts, np.ndarray)) assert(inside_pts.shape[1] == dim) num_inside_pts = inside_pts.shape[0] constraints = [] P = cp.Variable((dim, dim), symmetric=True) q = cp.Variable(dim) r = cp.Variable() # Impose the constraint that v₁, ..., vₙ are all outside of the ellipsoid. for i in range(num_outside_pts): constraints.append( outside_pts[i, :] @ (P @ outside_pts[i, :]) + 2 * q @ outside_pts[i, :] >= r) # P is strictly positive definite. epsilon = 1e-6 constraints.append(P - epsilon * np.eye(dim) >> 0) # Add the constraint that the ellipsoid contains @p inside_pts. for i in range(num_inside_pts): constraints.append( inside_pts[i, :] @ (P @ inside_pts[i, :]) + 2 * q @ inside_pts[i, :] <= r) # Now add the constraint that the ellipsoid is in the convex hull Ax<=b. # Using s-lemma, we know that the constraint is # ∃ λᵢ > 0, # s.t [P q -λᵢaᵢ/2] is positive semidefinite. # [(q-λᵢaᵢ/2)ᵀ λᵢbᵢ-r] num_faces = A.shape[0] lambda_var = cp.Variable(num_faces) constraints.append(lambda_var >= 0) Q = [None] * num_faces for i in range(num_faces): Q[i] = cp.Variable((dim+1, dim+1), PSD=True) constraints.append(Q[i][:dim, :dim] == P) constraints.append(Q[i][:dim, dim] == q - lambda_var[i] * A[i, :]/2) constraints.append(Q[i][-1, -1] == lambda_var[i] * b[i] - r) prob = cp.Problem(cp.Minimize(0), constraints) try: prob.solve(verbose=False) except cp.error.SolverError: return None, None, None, None if prob.status == 'optimal': P_val = P.value q_val = q.value r_val = r.value lambda_val = lambda_var.value return P_val, q_val, r_val, lambda_val else: return None, None, None, None def draw_ellipsoid(P, q, r, outside_pts, inside_pts): """ Draw an ellipsoid defined as {x | xᵀPx + 2qᵀx ≤ r } This ellipsoid is equivalent to |Lx + L⁻¹q| ≤ √(r + qᵀP⁻¹q) where L is the symmetric matrix satisfying L * L = P """ fig = plt.figure() dim = P.shape[0] L = scipy.linalg.sqrtm(P) radius = np.sqrt(r + q@(np.linalg.solve(P, q))) if dim == 2: # first compute the points on the unit sphere theta = np.linspace(0, 2 * np.pi, 200) sphere_pts = np.vstack((np.cos(theta), np.sin(theta))) ellipsoid_pts = np.linalg.solve( L, radius * sphere_pts - (np.linalg.solve(L, q)).reshape((2, -1))) ax = fig.add_subplot(111) ax.plot(ellipsoid_pts[0, :], ellipsoid_pts[1, :], c='blue') ax.scatter(outside_pts[:, 0], outside_pts[:, 1], c='red') ax.scatter(inside_pts[:, 0], inside_pts[:, 1], s=20, c='green') ax.axis('equal') plt.show() if dim == 3: u = np.linspace(0, np.pi, 30) v = np.linspace(0, 2*np.pi, 30) sphere_pts_x = np.outer(np.sin(u), np.sin(v)) sphere_pts_y = np.outer(np.sin(u), np.cos(v)) sphere_pts_z = np.outer(np.cos(u), np.ones_like(v)) sphere_pts = np.vstack(( sphere_pts_x.reshape((1, -1)), sphere_pts_y.reshape((1, -1)), sphere_pts_z.reshape((1, -1)))) ellipsoid_pts = np.linalg.solve( L, radius * sphere_pts - (np.linalg.solve(L, q)).reshape((3, -1))) ax = plt.axes(projection='3d') ellipsoid_pts_x = ellipsoid_pts[0, :].reshape(sphere_pts_x.shape) ellipsoid_pts_y = ellipsoid_pts[1, :].reshape(sphere_pts_y.shape) ellipsoid_pts_z = ellipsoid_pts[2, :].reshape(sphere_pts_z.shape) ax.plot_wireframe(ellipsoid_pts_x, ellipsoid_pts_y, ellipsoid_pts_z) ax.scatter(outside_pts[:, 0], outside_pts[:, 1], outside_pts[:, 2], c='red') ax.scatter(inside_pts[:, 0], inside_pts[:, 1], inside_pts[:, 2], s=20, c='green') ax.axis('equal') plt.show() def find_large_ellipsoid(pts, max_iterations): """ We find a large ellipsoid within the convex hull of @p pts but not containing any point in @p pts. The algorithm proceeds iteratively 1. Start with outside_pts = pts, inside_pts = z where z is a random point in the convex hull of @p outside_pts. 2. while num_iter < max_iterations 3. Solve an SDP to find an ellipsoid that is within the convex hull of @p pts, not containing any outside_pts, but contains all inside_pts. 4. If the SDP in the previous step is infeasible, then remove z from inside_pts, and append it to the outside_pts. 5. Randomly sample a point in the convex hull of @p pts, if this point is outside of the current ellipsoid, then append it to inside_pts. 6. num_iter += 1 When the iterations limit is reached, we report the ellipsoid with the maximal volume. @param pts pts[i, :] is the i'th points that has to be outside of the ellipsoid. @param max_iterations The iterations limit. @return (P, q, r) The largest ellipsoid is parameterized as {x | xᵀPx + 2qᵀx ≤ r } """ dim = pts.shape[1] A, b, hull = get_hull(pts) hull_vertices = pts[hull.vertices] deln = hull_vertices[Delaunay(hull_vertices).simplices] outside_pts = pts z = centered_sample_from_convex_hull(pts) inside_pts = z.reshape((1, -1)) num_iter = 0 max_ellipsoid_volume = -np.inf while num_iter < max_iterations: (P, q, r, lambda_val) = find_ellipsoid(outside_pts, inside_pts, A, b) if P is not None: volume = compute_ellipsoid_volume(P, q, r) if volume > max_ellipsoid_volume: max_ellipsoid_volume = volume P_best = P q_best = q r_best = r else: # Adding the last inside_pts doesn't increase the ellipsoid # volume, so remove it. inside_pts = inside_pts[:-1, :] else: outside_pts = np.vstack((outside_pts, inside_pts[-1, :])) inside_pts = inside_pts[:-1, :] # Now take a new sample that is outside of the ellipsoid. sample_pts = uniform_sample_from_convex_hull(deln, dim, 20) is_in_ellipsoid = np.sum(sample_pts.T*(P_best @ sample_pts.T), axis=0)\ + 2 * sample_pts @ q_best <= r_best if np.all(is_in_ellipsoid): # All the sampled points are in the ellipsoid, the ellipsoid is # already large enough. return P_best, q_best, r_best else: inside_pts = np.vstack(( inside_pts, sample_pts[np.where(~is_in_ellipsoid)[0][0], :])) num_iter += 1 return P_best, q_best, r_best if __name__ == "__main__": pts = np.array([[0., 0.], [0., 1.], [1., 1.], [1., 0.], [0.2, 0.4]]) max_iterations = 10 P, q, r = find_large_ellipsoid(pts, max_iterations)  I also put the code in the github repo Results Here is the result on a simple 2D example, say we want to find a large ellipsoid that doesn't contain the five red points in the figure below. Here is the result after the first iteration . The red points are the "outside points" (the initial outside points are v₁, v₂, ..., vₙ), the green point is the initial "inside points". In the second iteration, the ellipsoid becomes . By adding one more "inside point" (green dot), the ellipsoid gets larger. This gif shows the animation of the 10 iteations.