find out whether a given point is inside or outside a given polyhedron in matlab

main.m
function flag = is_inside_polyhedron(point, vertices, faces)
    % Determine whether a point is inside a polyhedron.
    % point: 3x1 vector of the point coordinates.
    % vertices: Nx3 matrix of the polyhedron vertices coordinates.
    % faces: Mx3 matrix of the polyhedron faces, where each row
    %        contains the indices of the vertices that form the face.

    % Initialize flag to outside.
    flag = false;

    % Check each face of the polyhedron.
    for i = 1:size(faces, 1)
        % Get the vertices of the face.
        face_vertices = vertices(faces(i,:), :);

        % Compute the normal vector to the face.
        normal = cross(face_vertices(2,:) - face_vertices(1,:), ...
                       face_vertices(3,:) - face_vertices(1,:));

        % Compute the distance from the face to the point.
        d = dot(normal, point - face_vertices(1,:));

        % If the distance is positive, the point is outside the polyhedron.
        if d > 0
            continue
        end

        % Compute the intersection of the ray with the face.
        t = -dot(normal, point - face_vertices(1,:)) / dot(normal, normal);

        % If the intersection is behind the point, the point is outside
        % the polyhedron.
        if t < 0
            continue
        end

        % Compute the coordinates of the intersection point.
        intersection = point + t*(point - face_vertices(1,:));

        % Find the edges of the face.
        edges = [face_vertices; face_vertices(1,:)];
        d = zeros(size(edges, 1) - 1, 1);

        % Compute the distances from the intersection point to each edge.
        for j = 1:size(edges, 1)-1
            d(j) = norm(cross(edges(j+1,:) - edges(j,:), ...
                              intersection - edges(j,:)));
        end

        % If the point is on the edge, consider it inside the polyhedron.
        if any(d < eps)
            flag = true;
            return
        end        

        % If the number of intersections is odd, the point is inside
        % the polyhedron.
        if mod(nnz(t > 0 & t < 1), 2) == 1
            flag = true;
            return
        end
    end
end
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