Fast Verification of Convexity of Piecewise-linear Surfaces

09/23/2003 ∙ by Konstantin Rybnikov, et al. ∙ UMass Lowell 0

We show that a realization of a closed connected PL-manifold of dimension n-1 in n-dimensional Euclidean space (n>2) is the boundary of a convex polyhedron (finite or infinite) if and only if the interior of each (n-3)-face has a point, which has a neighborhood lying on the boundary of an n-dimensional convex body. No initial assumptions about the topology or orientability of the input surface are made. The theorem is derived from a refinement and generalization of Van Heijenoort's theorem on locally convex manifolds to spherical spaces. Our convexity criterion for PL-manifolds implies an easy polynomial-time algorithm for checking convexity of a given PL-surface in n-dimensional Euclidean or spherical space, n>2. The algorithm is worst case optimal with respect to both the number of operations and the algebraic degree. The algorithm works under significantly weaker assumptions and is easier to implement than convexity verification algorithms suggested by Mehlhorn et al (1996-1999), and Devillers et al.(1998). A paradigm of approximate convexity is suggested and a simplified algorithm of smaller degree and complexity is suggested for approximate floating point convexity verification.



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1 van Heijenoort-Alexandrov’s Theorem
for Spaces of Constant Curvature

Throughout the paper denotes , , or . Following the original proof of van Hejeenoort’s, we will now show that his theorem holds in a somewhat stronger form for for . van Hejeenoort’s theorem does not hold for unbounded surfaces in . We will give three different kinds of counterexamples and pose a conjecture about simply connected locally compact embeddings of manifolds in .

Imagine a ”convex strip” in 3D which is bent in the form of handwritten so that it self-intersects itself, but not locally. Consider the intersection of this strip with a ball of appropriate radius so that the self-intersection of the strip happens to be inside the ball, and the boundary of the strip outside. Regarding the interior of the ball as Klein’s model of we conclude that the constructed surface is strictly locally convex at all points and has a complete metric induced by the immersion into the hyperbolic space. This gives an example of an immersion of a simply connected manifold into which does not bound a convex surface. Notice that in this counterexample the surface self-intersects itself.

Consider the (affine) product of a non-convex quadrilateral, lying inside a unit sphere centered at the origin, and a line in . The result is a non-convex polyhedral cylindrical surface . Pick a point inside the sphere, but outside the cylinder, whose vertical projection on the cylinder is the affine center of one of its facets. Replace this facet of with the cone over The part of the resulting polyhedral surface, that lies inside the sphere of unit radius, is indeed a PL-surface embedded in (in Klein’s model). The surface if locally convex at every point and strictly convex at . However, it is not the boundary of a convex body. Notice that this surface is not simply connected.

Consider a locally convex spiral, embedded in the -plane with two limiting sets: circle and the origin. That is this spiral coils around the origin and also around (from inside) the circle. Let be the double cone over this spiral with apexes at and , intersected with the unit ball . This non-convex surface is obviously simply connected, embedded, locally convex at every point, and strictly convex at all point of the spiral.

A locally compact realization of is a realization such that for any compact subset of is compact. The question remains:

Problem 3

Is it true that any locally compact embedding of a simply connected surface in is convex?

It remains an open question whether van Heijenoort’s (1952) criterion works for embedded unbounded surfaces in . We conjecture that it is, indeed, the case. The proof of the main theorem makes use of quite a number of technical propositions and lemmas. The proofs of these statements for by most part can be directly repeated for , but in some situations extra care is needed. If the reader is referred to van Heijenoort’s paper for the proof, it means that the original proof works without any changes.

Notation: The calligraphic font is used for sets in the abstract topological manifold. The regular mathematics font is used for the images of these sets in , a space of constant curvature. The interior of a set is denoted by , while the closure by . The boundary of S is denoted by . Since this paper is best read together with van Heijenoort’s (1952) paper, we would like to explain the differences between his and our notations. van Heijenoort denotes a subset in the abstract manifold by , while denoting its image in by ; an interior of a set in is denoted in his paper by .

The immersion induces a metric on by

where stands for the greatest lower bound, and for the length of an arc joining and on , which is the -image of an arc joining these points on . We will call this metric -metric.

Lemma 4

(van Heijenoort) Any two points of can be connected by an arc of a finite length. Thus is not only connected, but also arcwise connected.

Lemma 5

(van Heijenoort) The metric topology defined by the -metric is equivalent to the original topology on .

Lemma 6

(van Heijenoort) is closed in for any closed subset of .

Lemma 7

(van Heijenoort) If on a bounded (in -metric) closed subset mapping is one-to-one, then is a homeomorphism between and .

The proofs of the last two lemmas have been omitted in van Heijenoort (1952), but they are well known in topology.

Theorem 8

Let () be a Euclidean, spherical, or hyperbolic space. Let be an immersion of an -manifold in , such that is bounded in . Suppose that satisfies the following conditions:

1) is complete with respect to the metric induced on by the immersion ,

2) is connected,

2) is locally convex at each point,

4) is strictly convex in at least one point,

Then is a homeomorphism from onto the boundary of a compact convex body.

Proof. Notice that our theorem for directly follows from van Heijenoort’s proof of the Euclidean case. Any immersion of into can be regarded as an immersion into the interior of a unit ball with a hyperbolic metric, according to Klein’s model. If conditions 1)-4) are satisfied for the hyperbolic metric, they are satisfied for the Euclidean metric on this ball. Geodesics in Klein’s model are straight line segments and, therefore, for a bounded closed surfaces in , that satisfies the conditions of the theorem, the convexity follows from the Euclidean version of this theorem.

The original Van Heijenoort’s proof is based on the notion of convex part. A convex part of , centered at a point of strict convexity , , is an open connected subset of that contains and such that: (1) , where is a hyperplane in , not passing through , (2) lies on the boundary of a closed convex body bounded by and . We call the lid of the convex part. Let be a supporting hyperplane at . We call the open half-space defined by , where the convex part lies, the positive half-space and denote it by . We call the -preimage of a convex part in an abstract convex part, and denote it by . In van Heijenoort’s paper is required to be parallel to the supporting hyperplane of at , but this is not essential. In fact, we just need a family of hyperplanes such that: (1) they do not intersect in the positive half-space, (2) the intersections of these hyperplanes with the positive half-space form a partition of the positive half-space, (3) all these hyperplanes are orthogonal to a line , passing through . Let us call such a family a fiber bundle of the positive half-space defined by and . (In the case of

it is a vector bundle.) In fact, it is not necessary to assume that

passes through , but this assumption simplifies our proofs. Here denotes the distance, along the line , between the hyperplane in this family and . We will call the height of .

Proposition 9

A convex part exists.

Proof. van Heijenoort’s proof works for , , without changes.   

Denote by the least upper bound of the set of heights of the lids of convex parts centered at and defined by some fixed fiber bundle . Since is bounded, then .

Consider the union of all convex parts, centered at . We want to prove that this union is also a convex part. Let us depart for a short while (this paragraph) from the assumption that is bounded. may only be unbounded in the hyperbolic and Euclidean cases. As shown by van Heijennort (1952), if and , must be bounded even when is allowed to be unbounded. If and , can be unbounded, and this is precisely the reason why van Heijenoort’s theorem does not hold for unbounded surfaces in hyperbolic spaces.

Since in this theorem is assumed to be bounded, is bounded. Let us presume from now on that (the case of is considered in the beginning of the proof). belongs to the hyperplane and is equal to . bounds a closed bounded convex set in . Two mutually excluding cases are possible.

Case 1: . Then, following the argument of van Heljenoort (Part 2: pages 239-230, Part 5: page 241, Part 3: II on page 231), we conclude that is the homeomorphic pre-image of an -sphere . Since is connected, , and is a convex surface.

Case 2: . The following lemma is a key part of the proof of the main theorem. Roughly speaking, it asserts that if the lid of a convex part is of co-dimension 1, then either this convex part is a subset of a bigger convex part, or this convex part, together with the lid, is homeomorphic to via mapping .

Lemma 10

Suppose . Let be a convex part centered at a point and defined by a hyperplane from a fiber bundle . Suppose is the boundary of an -dimensional closed convex set in . Either is the -image of an -disk in and , where , or is a proper subset of a larger convex part, defined by the same fiber bundle .

Proof. Using a perturbation argument, we will prove this lemma by reducing the spherical case to the Euclidean one. Since is -dimensional and belongs to one of the hyperplanes in the fiber bundle , is either empty or -dimensional. If it is non-empty, must have a point other than and its opposite. The closure of a convex set in is convex. Since is convex, if it contains a point of other than and its opposite, it contains some geodesic segment lying in . Since is a point of strict convexity, there is a neighborhood of on all whose points, except for , are not points of , which contradicts to the choice of .

So, is definitely empty. Since, by Lemma 4, is arcwise connected, all of , except for the point , lies in the positive subspace. Therefore, there is a hyperplane in such that lies in an open halfspace defined by . We can regard as a standard sphere in . defines a hyperplane in . Consider an -dimensional plane in parallel to this hyperplane and not passing through the origin. Central projection of on obviously induces an immersion of a submanifold of into .

This submanifold is defined as the maximal arcwise connected open subset of such that (1) all points of this subset are mapped by to , and (2) it contains . It is obviously a manifold. Let us prove that it exists. Consider the union of all open arcwise connected subsets that contain . It is open and is acrwise connected, since it contains . Let .

The immersion obviously satisfies Conditions 2-4 of the main theorem 8. defines a metric on . Any Cauchy sequence on under this metric is also a Cauchy sequence on under the metric induced by . Therefore is complete and satisfies the conditions of the main Theorem 8. The central projection on maps a spherical convex part of on a Euclidean convex part of ; it also maps the fiber bundle to a fiber bundle in the Euclidean -plane . van Heijenoort (1952) proved Lemma 10 for the Euclidean case. Therefore, either , or is a proper subset of a larger convex part centered at , and defined by the same fiber bundle .   

The second alternative ( is a subset of a larger convex part) is obviously excluded, since is the convex part corresponding to the height which is the least upper bound of all possible heights of convex parts. Therefore in Case 2 is the boundary of a convex body which consists of a maximal convex part and a convex -disk, lying in the hyperplane .


2 Locally convex PL-surfaces

Theorem 11

Let be a realization map from a compact connected manifold of dimension into () such that is complete with respect to the -metric. Suppose that is locally convex at all points. Then is either strictly locally convex in at least one point, or is a spherical hyper-surface of the form , where is a convex cone in , whose face of the smallest dimension contains the origin (in particular, may be a hyperplane in ).

Proof. The proof of this rather long and technical theorem will be included in the full length paper of Rybnikov (200X).   

Theorem 12

Let be a realization map from a closed connected -dimensional manifold , with a regular CW-decomposition, in or () such that on the closure of each cell of map is one-to-one and lies on a subspace of dimension equal to . Suppose that is strictly locally convex in at least one point. The surface is the boundary of a convex polyhedron if and only if each -face has a point with an -neighborhood which lies on the boundary of a convex -dimensional set.

Proof. is locally convex at all points of its -cells. Suppose we have shown that is locally convex at each -face, . Consider a -face . Consider the intersection of with a sufficiently small -sphere centered at some point of and lying in a subspace complimentary to . is locally convex at if and only if the hypersurface on the sphere is convex. Since is locally convex at each -face, this hypersurface is locally convex at each vertex. By Theorem 11 is either the intersection of the boundary of a convex cone with or has a point of strict convexity on the sphere . In the latter case the spherical generalization of van Heijenoort’s theorem implies that is convex. Thus is convex at and therefore at all points of .

This induction argument shows that must be locally convex at all vertices. If is locally convex at all vertices, it is locally convex at all points. We assumed that had a point of strict convexity. The metric induced by is indeed complete. By van Heijenoort’s theorem and Theorem 8, is the boundary of a convex polyhedron.   

3 New Algorithm for Checking Global
Convexity of PL-surfaces

Idea: check convexity for the star of each -cell of .

We present an algorithm checking convexity of PL-realizations (in the sense outlined above) of a closed compact manifold .

The main algorithm uses an auxiliary algorithm C-check. The input of this algorithm is a pair , where is a one vertex tree with a cyclic orientation of edges and is its rectilinear realization in 3-space. This pair can be thought of as a PL-realization of a plane fan (partition of the plane into cones with common origin) in 3-space. The output is 1, if this realization is convex, and 0 otherwise. Obviously, this question is equivalent to verifying convexity of a plane polygon. For the plane of reference we choose a plane perpendicular to the sum of all unit vectors directed along the edges of the fan. The latter question can be resolved in time, linear in the number of edges of the tree (e.g. see Devillers et al (1998), Mehlhorn et al (1999)).

Input and Preprocessing: The poset of faces of dimensions , and of and the equations of the facets, OR the poset of faces of dimensions , and of and the positions of the vertices. We assume that we know the correspondence between the rank of a face in the poset and its dimension. There are mutual links between the facets (or vertices) of in the poset and the records containing their realization information. All -faces of are put into a stack . There are mutual links between elements of this stack and corresponding elements of the face lattice of .

Output: YES, if is the boundary of a convex polyhedron, NO otherwise.

1. while is not empty, pick an -face from ;
2. compute the projection of , and of all -faces incident to ,
   onto an affine 3-plane complimentary to ; denote this projection by ;
3. compute the cyclicly ordered one-vertex tree , whose edges
   are the -faces of and whose vertex is ;
4. Apply to the algorithm C-check
     if C-check then remove from the stack
else Output:=NO; terminate
5. Output:=YES
Remark 13

The algorithm processes the stars of all -faces independently. On a parallelized computer the stars of all -faces can be processed in parallel.

Proof of Correctness. The algorithm checks the local convexity of at the stars of all -cells. is compact and closed by Krein-Milman theorem (or Lemma 11) has at least one strictly convex vertex. By Theorem 12 local convexity at all vertices, together with the existence of at least one strictly convex vertex, is necessary and sufficient for to be the boundary of a convex body.

Complexity estimates

Denote by the number of -faces of , and by – the number of incidences between -faces and -faces in . Step 1 is repeated at most times. Steps 2-4 take at most arithmetic operations for each , where does not depend on . Thus, steps 2-4, repeated for all -faces of , require operations. Therefore, the total number of operations for this algorithm is .

Remark 14

The algorithm does not use all of the face lattice of .

Remark 15

The algorithm requires computing polynomial predicates only. The highest degree of algebraic predicates that the algorithm uses is , which is optimal (see Devillers et al, 1998).

From a practical point of view, it makes sense to say that a surface is almost convex, if it lies within a small Hausdorf distance from a convex surface that bounds an -dimensional convex set . In this case, the measure of lines, that pass through interior points of and intersect in more than 2 points, will be small, as compared to the measure of all lines passing through interior points of

. These statements can be given a rigorous meaning in the language of integral geometry, also called “geometric probability” (see Klain and Rota 1997).

Remark 16

If there is a 3-dimensional coordinate subspace of such that all the subspaces spanned by -faces are complementary to , the polyhedron can be projected on and all computations can be done in 3-space. This reduces the degree of predicates from to 3. In such case the boolean complexity of the algorithm does not depend on the dimension at all. therefore, for sufficiently generic realizations the algorithm has degree 3 and complexity not depending on .

Remark 17

This algorithm can also be applied without changes to compact PL-surfaces in or .


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