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Improved Pseudo-Polynomial-Time Approximation for Strip Packing

We study the strip packing problem, a classical packing problem which generalizes both bin packing and makespan minimization. Here we are given a set of axis-parallel rectangles in the two-dimensional plane and the goal is to pack them in a vertical strip of a fixed width such that the height of the obtained packing is minimized. The packing must be non-overlapping and the rectangles cannot be rotated. A reduction from the partition problem shows that no approximation better than 3/2 is possible for strip packing in polynomial time (assuming P≠NP). Nadiradze and Wiese [SODA16] overcame this barrier by presenting a (7/5+ϵ)-approximation algorithm in pseudo-polynomial-time (PPT). As the problem is strongly NP-hard, it does not admit an exact PPT algorithm. In this paper, we make further progress on the PPT approximability of strip packing, by presenting a (4/3+ϵ)-approximation algorithm. Our result is based on a non-trivial repacking of some rectangles in the empty space left by the construction by Nadiradze and Wiese, and in some sense pushes their approach to its limit. Our PPT algorithm can be adapted to the case where we are allowed to rotate the rectangles by 90^∘, achieving the same approximation factor and breaking the polynomial-time approximation barrier of 3/2 for the case with rotations as well.

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02/12/2022

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1 Introduction

In this paper, we consider the strip packing problem, a well-studied classical two-dimensional packing problem [6, 15, 32]. Here we are given a collection of rectangles, and an infinite vertical strip of width in the two dimensional (2-D) plane. We need to find an axis-parallel embedding of the rectangles without rotations inside the strip so that no two rectangles overlap (feasible packing). Our goal is to minimize the total height of this packing.

More formally, we are given a parameter and a set of rectangles, each one characterized by a width , , and a height . A packing of is a pair for each , with , meaning that the left-bottom corner of is placed in position and its right-top corner in position . This packing is feasible if the interiors of the rectangles are pairwise disjoint in this embedding (or equivalently rectangles are allowed to overlap on their boundary only). Our goal is to find a feasible packing of minimum height .

Strip packing is a natural generalization of one-dimensional bin packing [14] (when all the rectangles have the same height) and makespan minimization [13] (when all the rectangles have the same width). The problem has lots of applications in industrial engineering and computer science, specially in cutting stock, logistics and scheduling [32, 21]. Recently, there have been a lot of applications of strip packing in electricity allocation and peak demand reduction in smart-grids [40, 31, 36].

A simple reduction from the partition problem shows that the problem cannot be approximated within a factor for any in polynomial-time unless P=NP. This reduction relies on exponentially large (in ) rectangle widths.

Let denote the optimal height for the considered strip packing instance , and (resp. ) be the largest height (resp. width) of any rectangle in . Observe that trivially . W.l.o.g. we can assume that . The first non-trivial approximation algorithm for strip packing, with approximation ratio 3, was given by Baker, Coffman and Rivest [6]. The First-Fit-Decreasing-Height algorithm (FFDH) by Coffman et al. [15] gives a 2.7 approximation. Sleator [38] gave an algorithm that generates packing of height , hence achieving a 2.5 approximation. Afterwards, Steinberg [39] and Schiermeyer [37] independently improved the approximation ratio to 2. Harren and van Stee [22] first broke the barrier of 2 with their 1.9396 approximation. The present best -approximation is due to Harren et al. [21].

Recently algorithms running in pseudo-polynomial time (PPT) for this problem have been developed. More specifically, the running time of a PPT algorithm for Strip Packing is , where 222For the case without rotations, the polynomial dependence on can indeed be removed with standard techniques.. First, Jansen and Thöle [28] showed a PPT -approximation algorithm, and later Nadiradze and Wiese [35] overcame the -inapproximability barrier by presenting a PPT -approximation algorithm. As strip packing is strongly NP-hard [19], it does not admit an exact PPT algorithm.

1.1 Our contribution and techniques

In this paper, we make progress on the PPT approximability of strip packing, by presenting an improved approximation. Our approach refines the technique of Nadiradze and Wiese [35], that modulo several technical details works as follows: let be a proper constant parameter, and define a rectangle to be tall if . They prove that the optimal packing can be structured into a constant number of axis-aligned rectangular regions (boxes), that occupy a total height of inside the vertical strip. Some rectangles are not fully contained into one box (they are cut by some box). Among them, tall rectangles remain in their original position. All the other cut rectangles are repacked on top of the boxes: part of them in a horizontal box of size , and the remaining ones in a vertical box of size (that we next imagine as placed on the top-left of the packing under construction).

Some of these boxes contain only relatively high rectangles (including tall ones) of relatively small width. The next step is a rearrangement of the rectangles inside one such vertical box (see Figure 2(a)), say of size : they first slice non-tall rectangles into unit width rectangles (this slicing can be finally avoided with standard techniques). Then they shift tall rectangles to the top/bottom of , shifting sliced rectangles consequently (see Figure 2(b)). Now they discard all the (sliced) rectangles completely contained in a central horizontal region of size , and they nicely rearrange the remaining rectangles into a constant number of sub-boxes (excluding possibly a few more non-tall rectangles, that can be placed in the additional vertical box).

These discarded rectangles can be packed into extra boxes of size (see Figure 2(d)). In turn, the latter boxes can be packed into two discarded boxes of size , that we can imagine as placed, one on top of the other, on the top-right of the packing. See Figure 0(a) for an illustration of the final packing. This leads to a total height of , which is minimized by choosing .

(a) Final packing obtained by
Nadiradze & Wiese [35].

(b) Final packing obtained in this work.
Here is a small constant depending on .
Figure 1: Comparison of final solutions.

Our main technical contribution is a repacking lemma that allows one to repack a small fraction of the discarded rectangles of a given box inside the free space left by the corresponding sub-boxes (while still having many sub-boxes in total). This is illustrated in Figure 2(e). This way we can pack all the discarded rectangles into a single discarded box of size , where is a small constant depending on , that we can place on the top-right of the packing. The vertical box where the remaining rectangles are packed still fits to the top-left of the packing, next to the discarded box. See Figure 0(b) for an illustration. Choosing gives the claimed approximation factor.

We remark that the basic approach by Nadiradze and Wiese strictly requires that at most tall rectangles can be packed one on top of the other in the optimal packing, hence imposing . Thus in some sense this work pushes their approach to its limit.

The algorithm by Nadiradze and Wiese [35] is not directly applicable to the case when

rotations are allowed. In particular, they use a linear program to pack some rectangles. When rotations are allowed, it is unclear how to decide which rectangles are packed by the linear program. We use a combinatorial

container-based approach to circumvent this limitation, which allows us to pack all the rectangles using dynamic programming. This way we achieve a PPT -approximation for strip packing with rotations, breaking the polynomial-time approximation barrier of 3/2 for that variant as well.

1.2 Related work

For packing problems, many pathological lower bound instances occur when is small. Thus it is often insightful to consider the asymptotic approximation ratio. Coffman et al. [15] described two level-oriented algorithms, Next-Fit-Decreasing-Height (NFDH) and First-Fit-Decreasing-Height (FFDH), that achieve asymptotic approximations of 2 and 1.7, respectively. After a sequence of improvements [20, 5], the seminal work of Kenyon and Rémila [32] provided an asymptotic polynomial-time approximation scheme (APTAS) with an additive term . The latter additive term was subsequently improved to by Jansen and Solis-Oba [27].

In the variant of strip packing with rotations, we are allowed to rotate the input rectangles by (in other terms, we are free to swap the width and height of an input rectangle). The case with rotations is much less studied in the literature. It seems that most of the techniques that work for the case without rotations can be extended to the case with rotations, however this is not always a trivial task. In particular, it is not hard to achieve a approximation, and the hardness of approximation extends to this case as well [27]. In terms of asymptotic approximation, Miyazawa and Wakabayashi [34] gave an algorithm with asymptotic performance ratio of 1.613. Later, Epstein and van Stee [16] gave a asymptotic approximation. Finally, Jansen and van Stee [29] achieved an APTAS for the case with rotations.

Strip packing has also been well studied for higher dimensions. The present best asymptotic approximation for 3-D strip packing is due to Jansen and Prädel [24] who presented a 1.5-approximation extending techniques from 2-D bin packing.

There are many other related geometric packing problems. For example, in the independent set of rectangles problem we are given a collection of axis-parallel rectangles embedded in the plane, and we need to find a maximum cardinality/weight subset of non-overlapping rectangles [2, 10, 11]. Interesting connections between this problem and the unsplittable flow on a path problem were recently discovered [4, 7, 9]. In the geometric knapsack problem we wish to pack a maximum cardinality/profit subset of the rectangles in a given square knapsack [3, 17, 30]. One can also consider a natural geometric version of bin packing, where the goal is to pack a given set of rectangles in the smallest possible number of square bins [8]. We refer the readers to [12, 33] for surveys on geometric packing problems.

Subsequent Progress: Since the publication of our extended abstract [18], new results have appeared. Adamaszek et al. [1] proved that there is no PPT -approximation algorithm for Strip Packing unless . On the other hand, Jansen and Rau [26] independently showed a PPT -approximation algorithm with running time for the case without rotations. Very recently, new results have been announced [23, 25] claiming to give a tight -approximation algorithm.

1.3 Organization of the paper

First, we discuss some preliminaries and notations in Section 2. Section 3 contains our main technical contribution, the repacking lemma. Then, in Section 4, we discuss a refined structural result leading to a packing into many containers. In Section 5, we describe our algorithm to pack the rectangles and in Section 6 we extend our algorithm to the case with rotations. Finally, in Section 7, we conclude with some observations.

2 Preliminaries and notations

Throughout the present work, we will follow the notation from [35], which will be explained as it is needed.

Recall that denotes the height of the optimal packing for instance . By trying all the pseudo-polynomially many possibilities, we can assume that is known to the algorithm. Given a set of rectangles, will denote the total area of rectangles in , i.e., , and (resp. ) denotes the maximum height (resp. width) of rectangles in . Throughout this work, a box of size means an axis-aligned rectangular region of width and height .

In order to lighten the notation, we sometimes interpret a rectangle/box as the corresponding region inside the strip according to some given embedding. The latter embedding will not be specified when clear from the context. Similarly, we sometimes describe an embedding of some rectangles inside a box, and then embed the box inside the strip: the embedding of the considered rectangles is shifted consequently in that case.

A vertical (resp. horizontal) container is an axis-aligned rectangular region where we implicitly assume that rectangles are packed one next to the other from left to right (resp., bottom to top), i.e., any vertical (resp. horizontal) line intersects only one packed rectangle (see Figure 1(b)). Container-like packings will turn out to be particularly useful since they naturally induce a (one-dimensional) knapsack instance.

2.1 Classification of rectangles

Let , and assume for simplicity that

. We first classify the input rectangles into six groups according to parameters

satisfying and , whose values will be chosen later (see also Figure 1(a)). A rectangle is

Small

Medium

Vertical

Tall

Medium

Medium

Medium

Tall

Horizontal

Medium

Large

Large
(a) Each rectangle is represented
as a point on the plane with
(resp., ) coordinate indicating
its width (resp., height).
(b) Example of vertical container. Every vertical line intersects at most one rectangle.

Box
(c) Gray rectangles are nicely cut by , dashed ones are cut but not nicely cut by , and light gray one is not cut by .
Figure 2: Illustration of some of the definitions used in this work.
  • Large if and .

  • Tall if and .

  • Vertical if and ,

  • Horizontal if and ,

  • Small if and ;

  • Medium in all the remaining cases, i.e., if , or and .

We use , , , , , and to denote large, tall, vertical, horizontal, small, and medium rectangles, respectively. We remark that, differently from [35], we need to allow and due to some additional constraints in our construction (see Section 5).

Notice that according to this classification, every vertical line across the optimal packing intersects at most two tall rectangles. The following lemma allows us to choose and in such a way that and ( and , respectively) differ by a large factor, and medium rectangles have small total area.

Lemma 1.

Given a polynomial-time computable function , with , any constant , and any positive integer , we can compute in polynomial time a set of many positive real numbers upper bounded by , such that there is at least one number so that by choosing , , and .

Proof.

Let . Let , and, for each , define . Let . For each , let and similarly . Observe that is disjoint from (resp. is disjoint from ) for every , and the total area of rectangles in ( respectively) is at most . Thus, there exists a value such that the total area of the elements in is at most . Choosing , , , verifies all the conditions of the lemma. ∎

Function and constant will be chosen later. From now on, assume that and are chosen according to Lemma 1.

2.2 Next-Fit-Decreasing-Height (NFDH)

One of the most common algorithms to pack rectangles into a box of size is Next-Fit-Decreasing-Height (NFDH). In this algorithm, the first step is to sort rectangles non-increasingly by height, say . Then, the first rectangle is packed in the bottom-left corner, and a shelf is defined of height and width . The next rectangles are put in this shelf, next to each other and touching each other and the bottom of the shelf, until one does not fit, say the -th one. At this point we define a new shelf above the first one, with height . This process continues until all the rectangles are packed or the height of the next shelf does not fit inside the box.

This algorithm was studied by Coffman et al. [15] in the context of strip packing, in order to bound the obtained height when all the rectangles are packed into a strip. The result obtained can be summarized in the following lemma.

Lemma 2 (Coffman et al. [15]).

Given a strip packing instance , algorithm NFDH gives a packing of height at most .

One important observation is that each horizontal shelf can be thought of as a vertical container. Another important property of the algorithm is that, if a given set of rectangles needs to be packed into a given bin, and all of them are relatively small compared to the dimensions of the bin, then NFDH is very efficient even in terms of area. This result is summarized in the following lemma.

Lemma 3 (Coffman et al. [15]).

Given a set of rectangles with width at most and height at most , if NFDH is used to pack these rectangles in a bin of width and height , then the total used area in that bin is at least (provided that there are enough rectangles so that NFDH never runs out of them).

2.3 Overview of the algorithm

We next overview some of the basic results in [35] that are required for our result. We define the constant , and w.l.o.g. assume .

Let us forget for a moment small rectangles

. We will pack all the remaining rectangles into a sufficiently small number of boxes embedded into the strip. By standard techniques, as in [35], it is then possible to pack (essentially using NFDH in a proper grid defined by the above boxes) while increasing the total height at most by . See Section 5.1 for more details on how to pack small rectangles.

The following lemma from [35] allows one to round the heights and positions of rectangles of large enough height, without increasing much the height of the packing.

Lemma 4.

[35] There exists a feasible packing of height where: (1) the height of each rectangle in is rounded up to the closest integer multiple of and (2) their -coordinates are as in the optimal solution and their -coordinates are integer multiples of .

We next focus on rounded rectangle heights (i.e., implicitly replace by their rounded version) and on this slightly suboptimal solution of height .

The following lemma helps us to pack rectangles in .

Lemma 5.

If in Lemma 1 is chosen sufficiently large, all the rectangles in can be packed in polynomial time into a box of size and a box of size . Furthermore, there is one such packing using vertical containers in and horizontal containers in .

Proof.

We first pack rectangles in using NFDH into a strip of width . From Lemma 2 we know that the height of the packing is at most . Since and , because of Lemma 1, the resulting packing fits into a box of size . As , the number of shelves used by NFDH is at most , and this also bounds the number of vertical containers needed.

We next pack into a box of size . Recall that . Note that, for each , we have and . By ideally rotating the box and the rectangles by , we can apply the NFDH algorithm. Lemma 2 implies that we can pack all the rectangles if the width of the box is at least . Now observe that

and also, since ,

where the last inequality is true for any . Similarly to the previous case, the number of shelves is at most . Thus all the rectangles can be packed into at most horizontal containers. ∎

We say that a rectangle is cut by a box if both and are non-empty (considering both and as open regions with an implicit embedding on the plane). We say that a rectangle (resp. ) is nicely cut by a box if is cut by and their intersection is a rectangular region of width (resp. height ). Intuitively, this means that an edge of cuts along its longest side (see Figure 1(c)).

Now it remains to pack : The following lemma, taken from [35] modulo minor technical adaptations, describes an almost optimal packing of those rectangles.

Lemma 6.

There is an integer such that, assuming , there is a partition of the region into a set of at most boxes and a packing of the rectangles in such that:

  • each box has size equal to the size of some (large box), or has height at most (horizontal box), or has width at most (vertical box);

  • each is contained into a large box of the same size;

  • each is contained into a horizontal box or is cut by some box. Furthermore, the total area of horizontal cut rectangles is at most ;

  • each is contained into a vertical box or is nicely cut by some vertical box.

Proof.

We apply Lemma 3.2 in [35], where we set the parameter to . Recall that ; by requiring that , and since rectangles with height in are in , we have that .

Let be the set of horizontal rectangles that are nicely cut by a box. Since rectangles in satisfy , at most of them are nicely cut by a box, and there are at most boxes. Hence, their total area is at most , which is at most , provided that . Since Lemma 3.2 in [35] implies that the area of the cut horizontal rectangles that are not nicely cut is at most , the total area of horizontal cut rectangles is at most . ∎

We denote the sets of vertical, horizontal, and large boxes by and , respectively. Observe that can be guessed in PPT. We next use and to denote tall and vertical cut rectangles in the above lemma, respectively. Let us also define and .

Using standard techniques (see e.g. [35]), we can pack all the rectangles excluding the ones contained in vertical boxes in a convenient manner. This is summarized in the following lemma.

Lemma 7.

Given as in Lemma 6 and assuming , there exists a packing of such that:

  1. all the rectangles in are packed in ;

  2. all the rectangles in are packed in plus an additional box of size ;

  3. all the rectangles in are packed as in Lemma 6;

  4. all the rectangles in are packed in an additional vertical box of size .

Proof.

Note that there are at most rectangles in and at most rectangles in , since at most tall rectangles can be nicely cut by the left (resp. right) side of each box; this is enough to prove points (1) and (3).

Thanks to Lemma 6, the total area of horizontal cut rectangles is at most . By Lemma 2, we can remove them from the packing and pack them in the additional box using NFDH algorithm, proving point (2).

At most rectangles in can be nicely cut by a box; thus, in total there are at most nicely cut vertical rectangles. Since the width of each vertical rectangle is at most , they can be removed from the packing and placed in , piled side by side, as long as , which is equivalent to . This proves point (4). ∎

We will pack all the rectangles (essentially) as in [35], with the exception of where we exploit a refined approach. This is the technical heart of this paper, and it is discussed in the next section.

3 A repacking lemma

We next describe how to pack rectangles in . In order to highlight our contribution, we first describe how the approach by Nadiradze and Wiese [35] works.

It is convenient to assume that all the rectangles in are sliced vertically into sub-rectangles of width each333For technical reasons, slices have width in [35]. For our algorithm, slices of width suffice.. Let be such sliced rectangles. We will show how to pack all the rectangles in into a constant number of sub-boxes. Using standard techniques it is then possible to pack into the space occupied by plus an additional box of size . See Lemma 11 for more details.

We next focus on a specific vertical box , say of size (see Figure 2(a)). Let be the tall rectangles cut by . Observe that there are at most such rectangles ( on the left/right side of ). The rectangles in are packed as in Lemma 7. Let also and be the tall rectangles and sliced vertical rectangles, respectively, originally packed completely inside .

They show that it is possible to pack into a constant size set of sub-boxes contained inside , plus an additional box of size . Here denotes the region inside not contained in . In more detail, they start by considering each rectangle . Since by assumption, one of the regions above or below cannot contain another tall rectangle in , say the first case applies (the other one being symmetric). Then is moved up so that its top side overlaps with the top boundary of . The sliced rectangles in that are covered this way are shifted right below (note that there is enough free space by construction). At the end of the process all the rectangles in touch at least one of the top and bottom side of (see Figure 2(b)). Note that no rectangle is discarded up to this point.

Next, we partition the space inside into maximal height unit-width vertical stripes. We call each such stripe a free rectangle if both its top and bottom side overlap with the top or bottom side of some rectangle in , and otherwise a pseudo rectangle (see Figure 2(c)). We define the -th free rectangle to be the free rectangle contained in stripe .

Note that all the free rectangles are contained in a rectangular region of width and height at most contained in the central part of . Let be the set of (sliced vertical) rectangles contained in the free rectangles. Rectangles in can be obviously packed inside . For each corner of the box , we consider the maximal rectangular region that has as a corner and only contains pseudo rectangles whose top/bottom side overlaps with the bottom/top side of a rectangle in ; there are at most such non-empty regions, and for each of them we define a corner sub-box, and we call the set of such sub-boxes (see Figure 2(c)). The final step of the algorithm is to rearrange horizontally the pseudo/tall rectangles so that pseudo/tall rectangles of the same height are grouped together as much as possible (modulo some technical details). The rectangles in are not moved. The sub-boxes are induced by maximal consecutive subsets of pseudo/tall rectangles of the same height touching the top (resp., bottom) side of (see Figure 2(d)). We crucially remark that, by construction, the height of each sub-box (and of ) is a multiple of .

By splitting each discarded box into two halves and , and replicating the packing of boxes inside , it is possible to pack all the discarded boxes into two boxes and , both of size .

A feasible packing of boxes (and hence of the associated rectangles) of height is then obtained as follows. We first pack at the base of the strip, and then on top of it we pack , two additional boxes and (which will be used to repack the horizontal items; see Section 7 for details), and a box (which will be used to pack some of the small items). The latter boxes all have width and height . On the top right of this packing we place and , one on top of the other. Finally, we pack , and on the top left, one next to the other. See Figure 0(a) for an illustration. The height is minimized for , leading to a approximation.

The main technical contribution of this paper is to show how it is possible to repack a subset of into the free space inside not occupied by sub-boxes, so that the residual sliced rectangles can be packed into a single discarded box of size (repacking lemma). See Figure 2(e). This apparently minor saving is indeed crucial: with the same approach as above all the discarded sub-boxes can be packed into a single discarded box of size . Therefore, we can pack all the previous boxes as before, and on the top right. Indeed, the total width of , and is at most for a proper choice of the parameters. See Figure 0(b) for an illustration. Altogether the resulting packing has height . This is minimized for , leading to the claimed approximation.

(a) Original packing in a vertical box after removing . Gray rectangles correspond to , dark gray ones to and light gray ones to .
(b) Rectangles in are shifted vertically so that
they touch either the top
or the bottom of box , shifting also slices in accordingly.

(c) Classification in . Crosshatched stripes correspond to pseudo rectangles, empty stripes to free rectangles, and dashed regions to corner sub-boxes.

(d) Rearrangement of pseudo and tall rectangles to get sub-boxes, and additional packing of as in [35].

good indexes

(e) Our refined repacking of according to Lemma 8: some vertical slices are repacked in the free space.
Figure 3: Creation of pseudo rectangles, how to get constant number of sub-boxes and repacking of vertical slices in a vertical box .

It remains to prove our repacking lemma.

Lemma 8 (Repacking Lemma).

Consider a partition of into unit-width vertical stripes. There is a subset of at least such stripes so that the corresponding sliced vertical rectangles can be repacked inside in the space not occupied by sub-boxes.

Proof.

Let denote the height of the -th free rectangle, where for notational convenience we introduce a degenerate free rectangle of height whenever the stripe inside does not contain any free rectangle. This way we have precisely free rectangles. We remark that free rectangles are defined before the horizontal rearrangement of tall/pseudo rectangles, and the consequent definition of sub-boxes.

Recall that sub-boxes contain tall and pseudo rectangles. Now consider the area in not occupied by sub-boxes. Note that this area is contained in the central region of height . Partition this area into maximal-height unit-width vertical stripes as before (newly free rectangles). Let be the height of the -th newly free rectangle, where again we let if the stripe does not contain any (positive area) free region. Note that, since tall and pseudo rectangles are only shifted horizontally in the rearrangement, it must be the case that:

Let be the (good) indexes where , and be the bad indexes with . Observe that for each , it is possible to pack the -th free rectangle inside the -th newly free rectangle, therefore freeing a unit-width vertical strip inside . Thus it is sufficient to show that .

Observe that, for , : indeed, both and must be multiples of since they correspond to the height of minus the height of one or two tall/pseudo rectangles. On the other hand, for any index , , by the definition of . Altogether

We conclude that . The claim follows since by assumption . ∎

4 A refined structural lemma

The original algorithm in [35] uses standard LP-based techniques, as in [32], to pack the horizontal rectangles. We can avoid that via a refined structural lemma: here boxes and sub-boxes are further partitioned into vertical (resp., horizontal) containers. Rectangles are then packed into such containers as mentioned earlier: one next to the other from left to right (resp., bottom to top). Containers define a multiple knapsack instance, that can be solved optimally in PPT via dynamic programming. This approach has two main advantages:

  • It leads to a simpler algorithm.

  • It can be easily adapted to the case with rotations, as discussed in Section 6.

The goal of this section is to prove the following lemma that summarizes the aforementioned properties.

Lemma 9.

By choosing , there is an integer such that, assuming and , there is a packing of