Packing 2D disks into a 3D container

In this article, we consider the problem of finding in three dimensions a minimum-volume axis-parallel box into which a given set of unit-radius disks can be packed under translations. The problem is neither known to be NP-hard nor to be in NP. We give a constant-factor approximation algorithm based on reduction to finding a shortest Hamiltonian path in a weighted graph. As a byproduct, we can show that there is no finite size container into which all unit disks can be packed simultaneously.

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1 Introduction

Packing a set of geometric objects in a non-overlapping way into a minimum-size container is an intriguing problem. Because of its practical significance, many variants of the problem have been investigated, see the surveys [1, 8] and the references therein. Even a simple variant of packing a set of rectangles into a rectangular container turns out to be NP-complete [5]. In many cases, not much is known about the true complexity of the problem.

Constant-factor approximation algorithms of polynomial running time have been found for many variants of packing problems, in particular for finding minimum-size rectangular or convex containers for a set of convex polygons under translations [2], that is, the objects may be translated but rotations are not allowed. Also, approximation algorithms for rigid motions (translations and rotations) are known in this case, see for instance [10].

In three dimensions, most previous results are concerned with “regular” packing problems where objects to be packed are axis-parallel boxes, see for instance [4, 7]. In particular, approximation algorithms for packing rectangular cuboids or convex polyhedra into minimum-volume rectangular cuboids or convex containers under rigid motions are known [3]. Whether this is possible under only translations remains open.

In this paper, we give a positive answer for a restricted set of possible objects, namely disks of unit radius and axis-parallel box containers (see Figure 1) by describing a constant-factor approximation algorithm. Our approximation factor is high, but it is of theoretical interest that the problem, whose decision version is neither known to be NP-hard nor to be in NP, can be approximated within a constant factor.

The input to the problem is a set of unit-radius disks. Since we are allowed to freely translate them, we can assume each disk to be centered at the origin, so that the disk is fully defined by its normal vector. Two disks are

identical if their normal vectors are multiples of each other (in particular, and  define the same disk). A set of distinct disks is a set where no two disks are identical.

Two disks are in a position where they touch if their intersection contains exactly one point. Two disks intersect if their intersection consists of more than one point. By non-overlapping, we mean a placement of disks such that no two disks intersect, but where disks are allowed to touch.

Our main problem is as follows (see Figure 1):

Definition 1 (DiskPacking).

Given a set of distinct unit-radius disks in  by their normal vectors, find

  • an axis-parallel box of minimum volume such that all disks can be translated into the disk without overlap;

  • and the actual packing of the disks inside the box.

Figure 1: Input and solution for the DiskPacking-problem.

We will approximately solve DiskPacking by arranging subsets of disks such that their centers lie on a line and they are packed as densely as possible. Let be a given ordering of a set of unit-radius disks, and let  be a vector. We define the length of  with respect to direction  as follows: let  be a line with direction vector . Place the disks such that their centers lie on  and appear in the ordering  when traversing the line in direction , and such that two consecutive disks touch but do not overlap. Then the the length of the ordering with respect to is the distance from the center of the first disk to the center of the last disk. Figure 2 illustrates the definition in two dimensions, where the disks are line segments of length two.

Figure 2: The length of an ordering of unit-radius disks illustrated in two-dimensions.

We can now define the DiskStabbing-problem as follows (see Figure 3):

Definition 2 (DiskStabbing).

Given a set of distinct unit-radius disks by their normal vectors, and an additional vector defining the direction of a line, find an ordering of the disks that minimizes the length with respect to .

Figure 3: Input and solution for the DiskStabbing-problem

In Section 2, we study the geometry of the DiskStabbing-problem and define a metric on the family of unit-radius disks. This will allow us to prove that an optimal solution to DiskStabbing implies an approximate solution to DiskPacking. We prove this first for disks with similar normal vectors in Section 3, then for the general case in Section 4.

In Section 5 we then show how to approximately solve DiskStabbing by (approximately) computing a shortest Hamiltonian path in a complete weighted graph, implying a constant-factor approximation for DiskPacking.

In two dimensions, all (infinitely many) unit-length line segments can be packed into a rectangle of area  (Figure 4), and even smaller containers are possible. In Section 6 we show that a similar result does not hold in three dimensions: there is no bounded-size container into which all unit-radius disks can be packed.

Figure 4: All unit-length segments can be packed into a container of area .

In Section 7 we briefly consider packing other fixed shapes like squares and mention some open problems.

2 A metric on unit-radius disks

Given a vector , we define the -distance  between two unit disks  and  with centers  and  as the length of the ordering  with respect to . In other words, it’s the distance  when the disks touch and the ray  has direction . When and  are identical, we define .

Theorem 3.

For any , the distance  is a metric on the set of unit-radius disks.

Proof.

If and are not identical, it is clear that . Otherwise, by definition.

Symmetry holds since a point reflection keeps both disks identical (their normal vectors are negated), but reverses the order in which a line with direction  meets them.

It remains to prove the triangle inequality . It clearly holds when any two of the three disks are identical, so we will assume for a contradiction that there are three distinct disks  with centers  such that . We place in this order on a line  with direction  such that (so the disks touch but do not overlap) and that  (so these disks touch in a point  and do not overlap). Since , the disks  and  do not intersect at all.

We consider the situation in the plane  containing the line  and the point . Since , the intersection  is a segment of length two centered at , for . We choose a coordinate system for  where  is the -axis and , see Figure 5(top). By assumption, the segment  has length two, its midpoint is , and it does not intersect  or . In other words, it is contained in the wedge bounded by the rays  and  (light blue in Figure 5).

Figure 5: Proof of Theorem 3.

We first observe that we can move  to the left and  to the right while keeping  and  unchanged, until . This can only grow the wedge, so it will still contain , see Figure 5(bottom).

As a result, we now have  and , for . Setting , we have , and so .

We pick a point  on the segment  and point  on the segment  such that is parallel to  and is parallel to . We have and . It follows that the parallelogram  lies strictly in the interior of the disk of radius one around , and so  can only intersect this parallelogram in , see Figure 5(bottom).

It follows that the line  supporting  intersects the ray  in a point  and the ray  in a point . Breaking symmetry, we can assume that lies on the segment . Let , and let , see see Figure 6. We will let  decrease from  down to , and observe how  and  change. The distance  decreases monotonically as  decreases. If , then the distance  decreases until . During this decrease, we obviously have , so  cannot support . Once , the distance  increases monotonically while  decreases.

Figure 6: Proof of Theorem 3 (continued).

The minimum  is therefore maximized when the two distances are equal. Let  be the angle where this happens, and let  be the orthogonal projection of  on the -axis, see Figure 7. Since , the triangles  and  are congruent, which implies that also  and  are congruent. This implies that , and so  (since ).

Figure 7: Proof of Theorem 3 (final step).

It follows that for any , we have , and so the line  cannot support the segment , a contradiction. ∎

Consider now a configuration of two disks  and  that realize the -distance  for a given direction vector . We let  denote the normal of disk ,  its center, and  the plane supporting . Let  be the line  (its direction vector is ). The two disks must touch in a point . Let denote the half-plane of  delimited by  that contains .

We can distinguish three cases:

  • is tangent to  in , lying entirely in , and ;

  • is tangent to  in , lying entirely in , and ;

  • and  intersects both disks in a segment of positive length. These segments have disjoint interior.

We have the following lemma.

Lemma 4.

Let  and  be two disks whose normal vectors form an angle . Then .

Proof.

We first observe that the angle formed by  and  is either  or . Since , we can assume the former. The claim is then obvious in case (i), as  has distance  from the plane , and case (ii) is symmetric. In case (iii) we consider  and set . Since , we have ∎∎

We next discuss how to compute , given the normal vectors  and  for the disks and the direction vector . We pick the point  to be the origin, so that the planes  and  and the line  are all determined. The direction vector  tells us which half-planes of  and  are  and . We then check the three cases above:

In case (1), the location of  is fully determined. We intersect the line with direction  through  with  to obtain . If , this is the solution.

Otherwise, we symmetrically test case (ii).

If neither case holds, then we are in case (iii). We now know that lies on a circle of radius one around  in . Since  has direction , we construct the elliptic cylinder  whose axis is the line of direction  through the origin, and that intersects the plane  in the unit-radius circle around the origin. The center  lies both on  and on the unit-radius circle around  in . It suffices to test the at most two intersection points.

Seiferth’s Ph.D. thesis [9] explains how to perform these computations algebraically, and how to compute  exactly using rational arithmetic when  are given with rational coordinates.

3 Stabbing helps to pack similar disks

We fix the angle . This is the angle between the diagonal and an edge of the cube. In this section we consider a set  of unit-radius disks whose normal vector makes an angle of at most  with the positive -axis, that is, with the vector . We will show that the DiskStabbing and DiskPacking problems are related by proving that an optimal solution to DiskStabbing for  with respect to direction  provides a -approximation to DiskPacking for . In the next section we will then extend this to arbitrary sets of disks.

Let us first define  to be the maximum extent of all the disks. Formally, let  be the smallest axis-parallel box that contains all the disks if we place them with their center at the origin, and let , , and  be the dimensions of . We have and symmetrically .

Let now  denote a minimum-volume axis-parallel box into which  can be packed—that is, an optimal solution to DiskPacking—and denote its dimensions by . Its volume is .

When we project  onto the -plane, we obtain a rectangle of size . We place a grid of  points inside this area at interval , starting at the point at distance  from the left or upper boundary, and stopping when we have passed the same distance from the right or lower boundary, see Figure 8. We have

and symmetrically .

Figure 8: Placing a grid on a face of the optimal container .

We observe that since the disks lie entirely in , the projection of a disk center on the rectangle must have distance at least  from the rectangle boundary (that is, it must lie in the shaded region of Figure 8), so it has distance at most  to the nearest grid point.

For each of the  grid points, we consider the segment of length  obtained by intersecting  with a line parallel to the -axis through the grid point. By our construction, every disk center has distance at most  to one of the segments. This means that the segment stabs the disk in a point  at distance at most  from the disk center. The smaller disk of radius  around  lies entirely inside the original disk.

It follows that we can place all segments behind each other and obtain a disk stabbing of length  for , where the disks have been scaled down to radius . Simply enlarging this stabbing by factor  results in a solution to DiskStabbing for  of length .

Theorem 5.

Let  be a family of unit-radius disks whose normals make an angle of at most  with the -axis. An optimal solution to DiskStabbing for  with respect to the -direction has length at most , where opt is the volume of an optimal solution of DiskPacking for . The DiskStabbing-solution thus provides a -approximation for DiskPacking.

Proof.

Since an optimal solution to DiskPacking implies a stabbing of length at most , an optimal stabbing has at most this length, implying the first claim.

The smallest axis-parallel box enclosing the stabbing is a valid packing of the disks (here we make use of the triangle inequality from Theorem 3: if consecutive disks do not intersect, then no pair of disks can intersect). If  is the length of the stabbing, its volume is at most  (the -term is needed because we measure the length of a stabbing only from the first to the last center). Since and , this is at most

(making use of the fact that ). ∎

4 Stabbing helps to pack general disks

We now address the general DiskPacking problem. Given a set of disks , we partition it into three subsets , , and , depending on which coordinate axis the normal vector makes the smallest angle with (this corresponds simply to the highest absolute coordinate value in the normal vector). We then consider an optimal stabbing for  with respect to the -axis, an optimal stabbing for  with respect to the -axis, and an optimal stabbing for  with respect to the -axis. Note that the normals for each subfamily make an angle of at most  with its corresponding axis.

Theorem 6.

Optimal stabbings for , , and  with respect to the appropriate axis provide a -approximation for DiskPacking for .

Proof.

Again we define  (of dimension ) to be the extent of all the disks (in the entire set ). Let , , and  be the length of an optimal stabbing for the sets , and , respectively. Without loss of generality we assume that  is the largest among these.

If , then we are done by Theorem 5, so we assume this is not the case. This implies that for all three extent dimensions we have

We now consider the segment of length  that stabs the disks in , and partition it into six equal pieces of length . For each piece, we take the smallest axis-parallel box containing the disks whose centers lie on the piece. Each box has dimensions at most . We stack them in -direction to obtain a single box of dimensions at most  that packs all disks of . These are the blue pieces in Figure 9.

Figure 9: (left) Arranging the pieces for . (right) Combining into a single container.

Let now

Consider the segment of length  that stabs the disks in , and partition it into  equal pieces of length . For each piece, we take the smallest axis-parallel box containing the disks whose centers lie on the piece. Each box has dimensions at most . We arrange them in three layers of  boxes to obtain a single box of dimensions at most  that packs all disks of . These are the green pieces in Figure 9.

Similarly, we partition the segment of length  that stabs the disks of  into  equal pieces of length . For each piece, we obtain a box of dimensions at most . We arrange them in three columns of  boxes to obtain a single box of dimensions at most  that packs all disks of . These are the orange pieces in Figure 9.

We finally arrange the three boxes into a single container  of dimensions , see Figure 9(right). For the -direction, we make use of the fact that , so that , and obtain

It remains to bound the volume  in terms of opt. We use as well as and . By Theorem 5 we have . We thus have

5 Approximating the optimal disk stabbing and solving everything

It remains to show how to actually solve the DiskStabbing problem.

Theorem 7.

A -approximation for DiskStabbing can be computed in polynomial time.

Proof.

Given a set of  unit disks by their normal vectors and a direction vector , we construct the complete weighted graph  on  vertices , where the weight of the edge  is . The Hamiltonian paths in this graph correspond one-to-one to the orderings of the disks, and the length of the path in the graph is equal to the length of the ordering with respect to . It follows that the optimal ordering is given by the shortest Hamiltonian path in .

Christofides’ approximation algorithm applies to our problem, since by Theorem 3 the triangle inequality holds in . We can therefore compute a Hamiltonian path for  whose length is at most  times the optimal [6]. ∎

We then have the following corollary.

Corollary 8.

A -approximation to DiskPacking can be computed in polynomial time.

Proof.

In the proof of Theorem 6, we now have the weaker bound , and so

6 There is no bounded-size container for all unit-radius disks

While in two dimensions all (infinitely many) unit-length line segments can be packed into a single fixed rectangle (Figure 4), we now show that this is not true in three dimensions.

Theorem 9.

Packing a set of  unit disks requires a container of volume in the worst case.

Proof.

We construct a set  of  unit-radius disks as follows: we pick a a -square in the -plane that is centered at the origin, for some constant . We partition the square into grid cells with side length , and project each grid point vertically up onto the unit-radius sphere, see Figure 10. These  points define the normal vectors for our set  of disks. When  is small enough, all normals make an angle of less than  with the -axis.

Figure 10: Projecting a grid onto the unit sphere.

We observe next that the angle made by any two of the normal vectors is at least . Indeed, since any two grid points  have distance at least , so do their projections  and . The angle  is the length of the great circle arc between  and  on , so it is longer than , see Figure 11.

Figure 11: Two grid points and their projection onto .

This implies that for any two disks  we have by Lemma 4. This implies that any stabbing of  has length at least . By Theorem 5, this implies that any packing of  has volume . ∎

From Theorem 9, we obtain immediately the following corollary.

Corollary 10.

There is no container of bounded size into which all unit-radius disks in  can be packed.

7 Other objects and open problems

Our approximation algorithm can be extended to any arbitrary fixed planar shape , provided that  can be enclosed by some disk  (that is, it is bounded) and contains another disk  (that is, it has nonempty interior). More precisely, if we are given a finite set of congruent copies of  in three dimensions, we can approximate the smallest axis-parallel box into which the set can be packed by translations.

This can be done by just applying our algorithm to the corresponding set of copies of . Since it gives a constant-factor approximation of the optimal packing of the ’s it also gives an approximation of the optimal packing of the ’s. Observe however, that the approximation factor is multiplied by  where is the ratio between the radii of  and . Since the optimal packing of the ’s provides some packing of the ’s its container must be at least as large from which we obtain an approximation for the ’s.

The approximation factor obtained this way depends on the shape of . For standard shapes such as squares (), equilateral triangles () etc., we can directly compute it from the approximation factor of Section 4.

We have given a constant-factor approximation for a special case of objects, unit-radius disks. It remains an open problem whether an optimal packing of disks of different radii can be efficiently approximated.

Approximating the packing of arbitrarily oriented boxes or convex polyhedra seems to be much more difficult.

We showed further that there is no bounded-size container for all unit-radius disks in three dimensions. Does this generalize to higher dimensions, for instance to three-dimensional balls in four dimensions?

References

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