Efficient 1-Space Bounded Hypercube Packing Algorithm

Abstract

A space bounded $O(d/\log d)$-competitive hypercube packing algorithm with one active bin only is presented. As a starting point we give a simple 1-space bounded hypercube packing algorithm with competitive ratio $(3/2)^{d}+O((21/16)^d)$, for $d\ge 3.$

1 Introduction

In the bin packing problem, we receive a sequence S of items of different sizes that must be packed into a finite number of bins in a way that minimizes the number of bins used. When all the items of S are accessible, the packing method is called offline. The packing method is called online, when items arrive one by one and each item has to be packed irrevocably into a bin before the next item is presented.

One can consider an online method with t bins available for packing at each point in time. It is called t-space bounded. There are three types of bins: active, open and closed. At each point in time, exactly t bins are declared active. At the beginning, t bins are declared active and the remaining bins are open (there are no closed bins). Each incoming item is packed into one of the active bins; the remaining open bins are not available at this moment. We can decide to close an active bin. The most frequent reason for doing this is not enough space to pack an incoming item, however there may be other reasons based on the packing algorithm. When an active bin is closed, a new bin from among open bins is declared active. None of the closed bins is used again. It is natural to expect a packing method to be less efficient with fewer number of active bins. An unbounded space model does not impose any limits on the number of active bins.

Let S be a sequence of items, let A(S) be the number of bins used by algorithm A and let OPT(S) be the minimum possible number of bins used to pack items from S. The asymptotic competitive ratio for algorithm A is defined as:

$$\begin{aligned} R^{\infty }_A = \limsup _{n \rightarrow \infty } \ \sup _{S} \ \left\{ \frac{A(S)}{OPT(S)} \ | \ OPT(S) =n \right\} . \end{aligned}$$

Online bin packing is a classical problem studied for more than forty years. One-dimensional bin packing was first investigated in [27] (see also [20]), where the performance ratio of the First Fit algorithm was proved to be 17/10. The Next Fit algorithm with performance ratio not greater than 2 was discussed in [19]. Revised First Fit presented in [30] has performance ratio 5/3. The article also gives the lower bound 3/2 on the competitive ratio. The result was then improved in [3, 22] (the lower bound not smaller than 1.53635) and in [28], where the reader can find the lower bound 1.54014. First Fit and Best Fit algorithms can be found in [7]. The authors of the article [23] improve the upper bound to 1.61217 and give the lower bound 1.58333 for the class of Modified Harmonic algorithms. Seiden in [24] further improved the upper bound to 1.58889. Moreover, the upper bound 1.5813 was proved by Heydrich and van Stee (see [14]). Recently, the lower bound on the asymptotic competitive ratio of any online algorithm for bin packing was improved to 1.54278 (see [2]). Furthermore, an algorithm AH (Advanced Harmonic) whose asymptotic competitive ratio does not exceed 1.57829 was presented in [1].

Coppersmith and Raghavan in [5] presented the 2-dimensional online bin packing algorithm with competitive ratio 3.25. The result was later improved in [6] to 3.0625 and in [13] to 2.7834. Further improvements can be found in [25], where the authors show the upper bound of 2.66013 of the asymptotic competitive ratio. Currently the upper bound stands at 2.5545 (see [12]).

The classical 1-dimensional result in bin packing comes from Lee and Lee [21]. The authors presented an online bounded-space algorithm called Harmonic with the lower bound with the competitive ratio $\varPi _{\infty} \approx 1.69103$. The authors also showed that there is no bounded space algorithm with performance ratio below $\varPi _{\infty }$.

In the Harmonic algorithm and its improvements when the asymptotic competitive ratio approaches the optimal value, the number of active bins diverges to infinity. A question arises: What asymptotic competitive ratio can be achieved when the number of active bins is bounded above by a small natural number? This question was addressed by Woeginger in [29] whose Simplified Harmonic 6-space bounded online algorithm has competitive ratio beneath 17/10.

Let $d\ge 3$. We focus on the problem of packing d-dimensional hypercubes of the edge lengths not greater than 1 into a bin (a hypercube of the edge length 1).

The problem of multidimensional bin packing is discussed in [4]. Although previous studies used space bounded models, the number of active bins was usually large (for example, greater than 9 in [8]). The paper [8] by Epstein and van Stee gives a space bounded multidimensional hypercube packing algorithm with competitive ratio $O(d/\log d)$, however the number of active bins is about $d/\log d$. In this paper we describe a hypercube packing algorithm with competitive ratio $O(d/\log d)$ and with one active bin only.

Articles [15, 17] provide optimal estimates for online packing of hypercubes to a single bin for $d\ge 4$: any sequence of d-dimensional hypercubes of total volume not greater than $2^{1-d}$ can be packed online into a unit hypercube. The first paper concerns the case $d\ge 5$, the second $d=4$. Online packing of hyperboxes into a single bin is studied in [18], where the following upper bound is presented: any sequence of d-dimensional hyperboxes of the edge length smaller than or equal to 1 with total volume not greater than $(3 - 2\sqrt{2})\cdot 3^{-d}$ can be packed online into the d-dimensional unit hypercube.

An algorithm with competitive $2^{d+1}$ that uses only one active bin is presented in the article [31]. In the same paper the authors provide 1-space bounded algorithm for hyperbox packing with competitive ratio $4^d$. Algorithms with smaller competitive ratios can be found in [11], where the authors give hyperbox packing methods with ratios $(3.5)^d$ and $12\cdot 3^d$. 1-space bounded 2-dimensional online packing algorithms were studied in [32] (a 4.3-competitive algorithm) and [10] (a 3.888-competitive algorithm). The 2-space bounded 3.8165-competitive algorithm can be found in [16]. The 3-space bounded 3.577-competitive algorithm is presented in [9].

In the first part of the paper we provide the online algorithm tt(d) (with two types of small items) that uses one active bin only with competitive ratio not greater than $(3/2)^{d}+O((21/16)^d)$ which is a significant improvement of the previous result ($2^{d+1}$, see [31]). The algorithm distinguishes three types of hypercubes: big, 2-small and 3-small. Big hypercubes are packed alone, while the last two types are packed starting from the opposite corners of the bin. Counting the number of 2- and 3-small hypercubes that could be packed together led us to Gould’s sequence and a sequence of its partial sums linked with odd entries in Pascal’s triangle, which we consider to be a fact worth noting.

The second part of the paper contains the algorithm har(d) in which about $d/\log d$ types of items are distinguished. The algorithm is a generalization of the tt(d) algorithm and is based on the well-known Harmonic algorithm. Instead of using weights we give a direct proof that also in 1-space bounded model the competitive ratio $O(d/\log d)$ is achievable.

2 The tt(d) Algorithm for $d\ge 3$

2.1 Types of Items

Given an item (a d-dimensional hypercube) $C_i$, denote by $a_i$ its edge length. Items are divided into types:

• 1. $C_i$ is big, provided $a_i > 1/2$;

• 2. $C_i$ is small, provided $\ a_i \le 1/2$;

  • 2.1 a small $C_i$ is 2-small if $a_i$ belongs to $\bigcup _{n=0}^\infty (1/3\cdot 2^{-n},1/2\cdot 2^{-n}]$;

  • 2.2 a small $C_i$ is 3-small if $a_i$ belongs to $\bigcup _{n=0}^\infty (1/4\cdot 2^{-n},1/3\cdot 2^{-n}]$.

Each small item is either 2-small or 3-small. Items from these classes will be packed starting from the opposite vertices of the hypercube.

Any bin ${\mathscr {B}}$ can be partitioned into $2^d$ congruent hypercubes called 2-cubes as well as into $3^d$ congruent hypercubes called 3-cubes (see Fig. 1). Moreover, in the course of packing every k-cube can be partitioned into $2^{d\,n}$ congruent hypercubes with the edge lengths $1/ (k\cdot 2^n)$, called k-subcubes, for $\ k \in \{2,3 \}$ and $n\in \{0,1,2,\ldots \}$. 2-cubes [3-cubes] are also called 2-subcubes [3-subcubes, respectively].

Fig. 1

Partition into 2-cubes and 3-cubes

We lose no generality in assuming that $\ {\mathscr {B}}=[0,1]^d$. We will define an ordering of 2-cubes in ${\mathscr {B}}$. For $d=1$ the bin ${\mathscr {B}}$ is the interval [0, 1] and the 2-cubes are numbered from left to right: 1 and 2. When the order is defined for all dimensions up to $d-1$, we define the order in ${\mathscr {B}}=[0,1]^d$ in such way that all d-dimensional 2-cubes of $[0,1]^{d-1}\times [0,1/2]$ are numbered as in dimension $d-1$: from 1 to $2^{d-1}$. Then the d-dimensional 2-cubes of $[0,1]^{d-1}\times [1/2,1]$ are numbered from $2^{d-1}+1$ to $2^{d}$ in the order borrowed from the dimension $d-1$ (see Fig. 2, where $d=3$).

Fig. 2

Numbering of some 2-cubes and 2-subcubes (left) and 3-cubes (right)

The order of 3-cubes in ${\mathscr {B}}$ is defined in an analogous way.

Furthermore, each subcube has a number assigned to it.

Any bin ${\mathscr {B}}$ is partitioned into congruent 2-subcubes with the edge length 1/2, numbered from 1 to $2^d$. If S is the subcube with the edge length 1/2 and with number $\lambda _1$ ($\lambda _1 \in \{1,\ldots , 2^d\}$), then all 2-subcubes with the edge length $1/2^2$ (the partition of S) are numbered in an arbitrary order from $2^d(\lambda _1-1)+1$ to $2^d\lambda _1$ (see Fig. 2). Moreover, if S is the subcube with the edge length $1/2^2$ and with number $\lambda _2$ ($\lambda _2 \in \{1,\ldots , 4^d\}$), then all 2-subcubes with the edge length $1/2^3$ (the partition of S) are numbered in an arbitrary order from $2^d(\lambda _2-1)+1$ to $2^d\lambda _2$. Generally, $2^d$ congruent 2-subcubes forming the partition of S are numbered in an arbitrary order from $\ 2^d(\lambda -1)+1$ to $\ 2^d\lambda$.

Similarly, any bin ${\mathscr {B}}$ is partitioned into congruent 3-subcubes with the edge length 1/3, numbered from 1 to $3^d$. If S is the subcube with the edge length 1/3 and with number $\mu _1$ ($\mu _1 \in \{1,\ldots , 3^d\}$), then all 3-subcubes with the edge length $1/(2\cdot 3)$ (the partition of S) are numbered in an arbitrary order from $2^d(\mu _1-1)+1$ to $2^d\mu _1$ (see Fig. 2). Moreover, if S is the subcube with the edge length $1/(2\cdot 3)$ and with number $\mu _2$ ($\mu _2 \in \{1,\ldots , 6^d\}$), then all 3-subcubes with the edge length $1/(2^2\cdot 3)$ (the partition of S) are numbered in an arbitrary order from $2^d(\mu _2-1)+1$ to $2^d\mu _2$. Generally, $2^d$ congruent 3-subcubes into which the 3-subcube with number $\mu$ is divided are numbered in an arbitrary order from $\ 2^d(\mu -1)+1$ to $\ 2^d\mu$.

2.2 Packing Algorithm

If $C_i$ is a k-small item, then denote by $K_i$ the smallest hypercube of the edge lengths from the set $\ \{\frac{1}{k}, \frac{1}{2k},\frac{1}{4k},\ldots \}$ into which $C_i$ can be packed, $k\in \{2,3\}$. For example, if $a_i=10/81$ ($C_i$ is 2-small), then $K_i$ is a hypercube of the edge length 1/8; if $a_i=10/31$, then the edge length of the smallest hypercube $K_i$ containing the 3-small item $C_i$ equals 1/3.

A k-subcube of ${\mathscr {B}}$ is empty if its interior has an empty intersection with any item packed so far.

Algorithm tt(d) for packing of $C_i$ into ${{\mathscr {B}}}$.

• 1. If $C_i$ is big, then we close the active bin, open a new bin, pack $C_i$ and close the bin. Then we open a new active bin.

• 2. If $C_i$ is a 2-small item, then we pack $C_i$ into the empty 2-subcube congruent to $K_i$ with the smallest number. If there is no empty subcube, we close the active bin and open a new bin to pack $C_i$.

• 3. If $C_i$ is a 3-small item, then we pack $C_i$ into the empty 3-subcube congruent to $K_i$ with the greatest number. If there is no empty subcube, we close the active bin and open a new active bin to pack $C_i$.

Example

We need to pack a sequence of four items $C_1,C_2,C_3,C_4$ according to the tt(3) algorithm, see Fig. 3. The edge lengths of arriving items are 1/9, 1/2, 10/81 and 10/31.

The first hypercube is a 3-small item. Moreover $C_1=K_1$ and thus it is packed into the empty 3-subcube with the greatest number (inside the 3-cube numbered 27, see Fig. 2). The hypercube $C_2=K_2$ is packed into the first 2-cube. For $C_3$ the smallest 2-subcube into which it can be packed is congruent to a cube $K_3$ of the edge length 1/8. $C_3$ is packed into $K_3$ and then $K_3$ is packed into a 2-subcube of the second 2-cube. Finally, $C_4$ is packed into a cube $K_4$ of the edge length 1/3, which is then packed into the empty 3-cube with the greatest number, which would be the 3-cube with number 26.

Fig. 3

Four items packed according to the tt(3) algorithm

Denote by |A| the d-dimensional volume of an item A.

Lemma 1

Suppose a 2-small item $C_i$ was packed into a 2-subcube congruent to $K_i$. At least $\bigl ( 2/3\bigr )^d$ units of volume of the 2-subcube is occupied, i.e., $\ |C_i| \ge \bigl ( 2/3\bigr )^d |K_i|$.

Proof

Recall that $K_i$ is congruent to the smallest 2-subcube into which $C_i$ can be packed. $C_i$ is 2-small thus its edge length is greater than $\ 3^{-1}\cdot 2^{-n}$ and not greater than $2^{-n-1}$ for some non-negative integer n. We use that lower bound in calculations.

$$\begin{aligned} \frac{|C_i|}{|K_i|} > \frac{ ( 3^{-1} \cdot 2^{-n})^d}{(2^{-n-1})^d} = \Bigl ( \frac{2}{3}\Bigr )^{d} . \end{aligned}$$

$\square$

We now give the analogous lemma for 3-small items.

Lemma 2

Suppose a 3-small item $C_i$ was packed into a 3-subcube congruent to $K_i$. At least $\bigl ( 3/4\bigr )^d$ of the 3-subcube is occupied, i.e., $\ |C_i| \ge \bigl ( 3/4\bigr )^d |K_i|$.

Proof

$$\begin{aligned} \frac{|C_i|}{|K_i|} > \frac{ (2^{-n-2})^d}{(3^{-1} \cdot 2^{-n})^d} = \Bigl ( \frac{3}{4}\Bigr )^{d} . \end{aligned}$$

$\square$

Let $\ \rho _2=\bigl ( 2/3\bigr )^d$, $\rho _3= \bigl ( 3/4\bigr )^d$ and let either $k= 2$ or $k=3$. We say that a k-subcube is used for packing, provided a k-small item was packed into it.

Lemma 3

Assume that $C_i$ is the first k-small item that cannot be packed into a k-cube of a bin ${\mathscr {B}}$ by our algorithm. Denote by $m_k$ the number of k-cubes (of the edge lengths $k^{-1}$) used for packing. Then the total volume of all k-small items preceding $C_i$ packed into ${\mathscr {B}}$ is greater than $\ \rho _k m_k \cdot k^{-d} -|C_i|$.

Proof

Let $C_i$ be a k-small item. By the assumption, there is no empty k-subcube greater than or congruent to $K_i$, i.e., all empty k-subcubes in $\mathscr {B}$ are smaller than $K_i$. Furthermore, for each $\ i=1,2,\dots$ there are at most $2^d-1$ empty k-subcubes congruent to $2^{-i}K_i$. Notice that if there were $2^d$ empty k-subcubes congruent to $2^{-i}K_i$, they would form a whole empty k-subcube congruent to $2^{-i+1}K_i$ (recall that the algorithm imposes packing to the subcube with the smallest number).

The total volume of empty k-subcubes in $\mathscr {B}$ is smaller than

$$\begin{aligned} (2^d-1)\cdot (2^{-d} + 4^{-d} + \cdots ) \cdot |K_i| =|K_i|. \end{aligned}$$

By Lemmas 1 and 2 we know that the total volume of k-small items packed into ${\mathscr {B}}$ is greater than

$$\begin{aligned} \rho _k (m_k \cdot k^{-d}-|K_i|) = \rho _k m_k \cdot k^{-d}- \rho _k|K_i| > \rho _k m_k \cdot k^{-d}-|C_i|. \end{aligned}$$

$\square$

2.3 2-Cubes Versus 3-Cubes

Denote by $n_2$ the number of 2-cubes used for packing and by $U_{n_2}$ the union of 2-cubes used for packing in a bin ${\mathscr {B}}$. Moreover, denote by $n_3^-$ the number of 3-cubes contained in $\ {{\mathscr {B}}} {\setminus } U_{n_2}$.

For $d=1$ we have $n_3^-=3$ for $n_2=0$, $n_3^-=1$ for $n_2=1$ and $n_3^-=0$ for $n_2=2$ (see Fig. 4).

Fig. 4

2-cubes and 3-cubes for $d=1$

We want to have a relation between $n_3^-$ and $n_2$. In Lemma 4 we will present a recursive formula. Clearly, $n_3^-$ also depends on d, thus we will use the notation $\ n_3^-=y_d(n_2)$ (see Fig. 4).

The number of empty 3-cubes can be calculated directly for $d=2$ (see Fig. 5):

• $y_2(0)=3^2$ (all 3-cubes are empty),

• $y_2(1)=5$ (exactly one 2-cube is packed),

• $y_2(2)=3^1$ (only one row of 3-cubes is empty),

• $y_2(3)=1$,

• $y_2(4)=0.$

If $\ {{\mathscr {B}}} = [0,1]^d$, then all 2-cubes of ${{\mathscr {B}}}$ are of the form

$$\begin{aligned} C_2(\alpha _1, \dots , \alpha _d) = [\alpha _1, \alpha _1+1/2]\times \cdots \times [\alpha _d, \alpha _d+1/2], \end{aligned}$$

where $\ \alpha _i \in \{ 0, 1/2\}$ for $\ i=1,2,\ldots , d$. Moreover, all 3-cubes of ${{\mathscr {B}}}$ are of the form

$$\begin{aligned} C_3(\beta _1, \dots , \beta _d) = [\beta _1, \beta _1+1/3]\times \cdots \times [\beta _d, \beta _d+1/3], \end{aligned}$$

where $\ \beta _i \in \{ 0, 1/3, 2/3\}$ for $\ i=1,2,\ldots , d$. We will use the notation $\ C_2[\alpha _d] = C_2(\alpha _1, \dots , \alpha _d)$ and $\ C_3[\beta _d]=C_3(\beta _1, \dots , \beta _d)$.

Fig. 5

2-cubes, 3-cubes for $d=2$ visualised as layers

For $d\ge 2$ the values of the sequence $y_d$ can be calculated inductively (see the bottom row of Fig. 5). There is a 1-1 correspondence between 3-cubes in a d-dimensional unit hypercube and 3-cubes in three $(d-1)$-dimensional unit hypercubes. Assign all $3^{d-1}$ hypercubes $C_3[0]$ to 3-cubes in the first $(d-1)$-dimensional unit hypercube, all $3^{d-1}$ hypercubes $C_3[1/3]$ to 3-cubes in the second $(d-1)$-dimensional unit hypercube and finally all hypercubes $C_3[2/3]$ to 3-cubes in the third $(d-1)$-dimensional unit hypercube (see Fig. 6, where $d=3$).

Fig. 6

3-cubes $C_3[0]$, $C_3[1/3]$ and $C_3[2/3]$ in a three-dimensional cube visualised as 3 layers of two-dimensional 3-cubes in a square

If $\ n_2 \le 2^{d-1}$, i.e., if no hypercube $C_2[1/2]$ was used for packing, then all $3^{d-1}$ hypercubes $C_3[2/3]$ remain empty. The number of empty hypercubes $C_3[1/3]$ as well as the number of empty hypercubes $C_3[0]$ equals $y_{d-1}(n_2)$. The number of all empty 3-cubes is equal to $\ 3^{d-1} + 2y_{d-1}(n_2)$.

If $\ n_2 > 2^{d-1}$, then all $2^{d-1}$ hypercubes $C_2[0]$ were used for packing. There is no empty hypercube $C_3[0]$ as well as there is no empty hypercube $C_3[1/3]$. Let m be an integer such that $\ n_2 = 2^{d-1}+m$. Obviously, exactly m hypercubes $C_2[1/2]$ were used for packing. The number of empty 3-cubes is exactly the same as in the dimension $d-1$ and equals $\ y_{d-1}(m)$.

Consequently, we get the following inductive formula

$$\begin{aligned} y_d(0)& = 3^d, \\ y_d(n_2)& = 3^{d-1}+2\cdot y_{d-1}(n_2),\quad {\text{for}}\,\, n_2<2^{d-1} \end{aligned}$$

(1)

$$\begin{aligned} y_d(n_2)& = y_{d-1}(m),\quad {\text{for}}\,\, n_2=2^{d-1}+m, \quad m\le 2^{d-1}. \end{aligned}$$

(2)

Note that the sequence $y_d(n_2)$ is directly related to Gould’s sequence, what is proved in “Appendix 1”.

Now we give the relation between the number $n_2$ of 2-cubes used for packing and the number $\ n_3^{-}=y_d(n_2)$.

Lemma 4

Let $n_2$ be the number of 2-cubes already used for packing, let $U_{n_2}$ be the union of all 2-cubes used for packing and let $n_3^-$ be the number of 3-cubes contained in $\ {{\mathscr {B}}} {\setminus } U_{n_2}$. For $\ d\ge 1$ and $\ n_2\le 2^d$ we get

$$\begin{aligned} n_24^d+3^d n_3^- \ge 8^d - \frac{5}{8}\cdot 7^d. \end{aligned}$$

Proof

Let

$$\begin{aligned} R=\frac{1}{3^d} \left( 8^d-\frac{5}{8} \cdot 7^d -n_24^d\right) \end{aligned}$$

and let $n_3^-=y_d(n_2)$. We will show that

$$\begin{aligned} y_d(n_2)\ge R. \end{aligned}$$

The inequality is obvious for $\ n_2=0,\, 2^{d-1},\, 2^d$ and arbitrary d.

Notice that for $d=1$ the inequality holds true.

• For $\ n_2=0,\ y_1(0)=3$ we have $\ R=\frac{1}{3}(8-\frac{5}{8}\cdot 7-0\cdot 4)=\frac{29}{24}<3=y_1(0)$.

• For $\ n_2=1,\ y_1(1)=1$ we have $\ R=\frac{1}{3}(8-\frac{5}{8}\cdot 7-1\cdot 4)<0<1=y_1(1)$.

• For $\ n_2=2,\ y_1(2)=0$ we have $\ R=\frac{1}{3}(8-\frac{5}{8}\cdot 7-2\cdot 4)<0=y_1(2)$.

The rest of the proof goes by induction. Let $d\ge 2$.

• Case 1 $\ n_2<2^{d-2}$. By the inductive assumption,

  $$\begin{aligned} y_d(n_2)& = 3^{d-1}+2y_{d-1}(n_2) \\ &\ge 3^{d-1}+2\cdot \frac{1}{3^{d-1} }\Bigl ( 8^{d-1} - \frac{5}{8} \cdot 7^{d-1}-n_24^{d-1}\Bigr )\\& = R+\frac{1}{3^d}\Bigl (3^{2d-1}-2\cdot 8^{d-1} - 2n_24^{d-1}+\frac{5}{8} \cdot 7^{d-1}\Bigr ) \\ & > R+\frac{1}{3^d}\Bigl (3^{2d-1}-2\cdot 8^{d-1} - 2\cdot 2^{d-2}\cdot 4^{d-1}\Bigr ) \\& = R+\frac{1}{3^d}\Bigl (3\cdot 9^{d-1}-3\cdot 8^{d-1}\Bigr ) >R. \end{aligned}$$

• Case 2 $\ n_2=2^{d-2} .$

  $$\begin{aligned} y_d(n_2)& = y_d(2^{d-2})=3^{d-1}+2y_{d-1}(2^{d-2})=3^{d-1}+2\cdot 3^{d-2} = \frac{5}{9} \cdot 3^d \\ &> \Bigl ( \frac{8}{3} \Bigr )^d - \frac{5}{8} \cdot \Bigl ( \frac{7}{3} \Bigr )^d = \frac{1}{3^d}\Bigl (8^{d}- \frac{5}{8} \cdot 7^{d}\Bigr )>R. \end{aligned}$$

• Case 3 $\ 2^{d-2}< n_2 <2^{d-1}.$ We can assume that $d\ge 3$ (if $d=2$, then there is no $n_2$ such that $\ 2^0<n_2<2^1$).

Clearly, $\ n_2=2^{d-2}+l$ for some $\ 0< l<2^{d-2}$. Consequently,

$$\begin{aligned} y_d(n_2)& = 3^{d-1}+2y_{d-1}(n_2)\\& = 3^{d-1}+2y_{d-1}(2^{d-2}+l)\\& = 3^{d-1}+2y_{d-2}(l)\\ &\ge 3^{d-1}+2\cdot \frac{1}{3^{d-2} }\bigl ( 8^{d-2} - \frac{5}{8} \cdot 7^{d-2}-l4^{d-2}\bigr ) \\& = 3^{d-1}+2\cdot \frac{1}{3^{d-2} }\Bigl ( 8^{d-2} - \frac{5}{8} \cdot 7^{d-2}-(n_2-2^{d-2})4^{d-2}\Bigr )\\& = R+\frac{1}{3^{d}}\left( 3\cdot 9^{d-1}-\frac{7}{2}\cdot 8^{d-1} +\frac{155}{8}\cdot 7^{d-2}-2n_24^{d-2}\right) \\ &> R+\frac{1}{3^{d}}\left( 3\cdot 9^{d-1}-\frac{9}{2}\cdot 8^{d-1} +2\cdot 7^{d-1}\right) >R. \end{aligned}$$

• Case 4 $\ 2^d-2 \le n_2 \le 2^{d}$.

If $n_2=2^d$, then $\ y_d(n_2)=0>R$.

If $\ n_2 \in \{ 2^d-2, 2^d-1 \}$, then $\ y_d(n_2)\ge 1 > \frac{1}{3^d} \Bigl ( 8^d - \frac{5}{8} \cdot 7^d - (2^d-2)\cdot 4^d \Bigr ) \ge R$.

• Case 5 $\ 2^{d-1} \le n_2 \le 2^d-3$.

Observe that there are integers j and k such that $\ 1\le j\le d-2$, $\ 0\le k< 2^{d-j-1}$ and that

$$\begin{aligned} n_2=2^{d-1}+2^{d-2}+\cdots +2^{d-j}+k = 2^d(1-2^{-j})+k. \end{aligned}$$

Using the inductive definition of the sequence (2), i.e., $\ y_d(2^{d-1}+m)=y_{d-1}(m)$, for $\ 0<m\le 2^{d-1}$, we get

$$\begin{aligned} y_d(n_2)& = y_d(2^{d-1}+2^{d-2}+\cdots +2^{d-j}+k) \\& = y_{d-1}(2^{d-2}+\cdots +2^{d-j}+k) \\& = y_{d-2}(2^{d-3}+\cdots +2^{d-j}+k) \\& = \cdots = y_{d-j}(k) \\ &\ge y_{d-j}(2^{d-j-1}-1)\\& = 3^{d-j-1} +2y_{d-j-1}(2^{d-j-1}-1)\\& = 3^{d-j-1} +2. \end{aligned}$$

Since $\ n_2 \ge 2^d(1-2^{-j})$, to prove $\ y_d(n_2)\ge R$, is suffices to show the following inequality

$$\begin{aligned} 3^{d-j-1}+2 \ge \frac{1}{3^d} \left( 8^d-\frac{5}{8}7^d -2^d(1-2^{-j})4^d\right) , \end{aligned}$$

i.e.,

$$\begin{aligned} \frac{9^d}{3^{j+1}} - \frac{8^d}{2^j} + \frac{5}{8}\cdot 7^d +2\cdot 3^d\ge 0. \end{aligned}$$

This is equivalent to check that

$$\begin{aligned} f_d(j)\ge 0, \end{aligned}$$

where

$$\begin{aligned} f_d(x) = \frac{9^d}{3^{x+1}} - \frac{8^d}{2^x} +\frac{5}{8}\cdot 7^d+2\cdot 3^d. \end{aligned}$$

The function $f_d$ has a local minimum at

$$\begin{aligned} x_{\min }= \log _{2/3}\left( \Bigl (\frac{8}{9}\Bigr )^d3\log _32 \right) =d\log _{2/3}\frac{8}{9}+\log _{2/3}(3\log _32), \end{aligned}$$

and

$$\begin{aligned} 0.29d-1.58<x_{\min }<0.3d-1.57. \end{aligned}$$

Moreover, $f_d$ is continuous, decreasing on the interval $(-\infty , x_{\min }]$ and increasing on $[x_{\min },+\infty )$.

First assume that $\ d \in \{ 2,3, \dots , 11\}$. It is easy to verify that $\ f_d(1) = 9^{d-1} - \frac{8^d}{2} +\frac{5}{8}\cdot 7^d+2\cdot 3^d>0$ as well as $\ f_d(2) = \frac{9^{d-1}}{3} - \frac{8^d}{4} +\frac{5}{8}\cdot 7^d+2\cdot 3^d>0$. Since $\ x_{\min }<0.3\cdot 11-1.57<2$, it follows that

$$\begin{aligned} f_d(j)\ge \min \bigl ( f_d(1), f_d(2)\bigr ) >0 \end{aligned}$$

for $\ j \in \{ 1,2,\ldots , d-2 \}$.

Now assume that $d\ge 12$. For $\ j \in \{ 1,2 \ldots , d-2\}$ we get

$$\begin{aligned} f_d(j) &\ge f_d(x_{min}) = \frac{9^d}{3\cdot 3^{x_{\min }}}-\frac{8^d}{2^{x_{\min }}} +\frac{5}{8}\cdot 7^d +2\cdot 3^d \\ &> \frac{9^d}{3\cdot 3^{0.3d}\cdot 3^{-1.57}}- \frac{8^d}{2^{0.29d}\cdot 2^{-1.58}} +0.625\cdot 7^d \\ &> 1.87\cdot (6.473)^d - 2.99 \cdot (6.544)^d + 0.625\cdot 7^d \\ &> (6.473)^d \cdot \Bigl ( 1.87 - 2.99 \cdot (1.011)^d + 0.625\cdot (1.081)^d \Bigr ) \\ &\ge (6.473)^d \cdot \Bigl ( 1.87 - 2.99 \cdot (1.011)^{12} + 0.625\cdot (1.081)^{12}\Bigr )>0 \end{aligned}$$

$\square$

2.4 Competitive Ratio

Lemma 5

Assume that only small items are in the sequence S. The total volume of small items packed into any closed bin ${\mathscr {B}}$ is greater than

$$\begin{aligned} \rho = \left( 1-5/8 \cdot \left( 7/8\right) ^d\right) \cdot \left( 2/3\right) ^d -2^{-d} -3^{-d} \end{aligned}$$

for $d\ge 3$.

Proof

By Lemma 3 we know that the total volume of 2-small items packed into ${\mathscr {B}}$ is greater than $\ \rho _2 n_2 \cdot 2^{-d} -2^{-d}$. Moreover, the total volume of 3-small items packed into ${\mathscr {B}}$ is greater than $\ \rho _3 n_3 \cdot 3^{-d} -3^{-d}$. Consequently, the sum of the volumes of packed small items is greater than

$$\begin{aligned}&(2/3)^d n_2 (1/2)^d + (3/4)^d n_3 (1/3)^d -1/2^d -1/3^d \\& = 12^{-d} \cdot ( n_2\cdot 4^d +n_3\cdot 3^d) -1/2^d -1/3^d. \end{aligned}$$

This value, by Lemma 4, is greater than

$$\begin{aligned} 12^{-d} \cdot \Bigl ( 8^d -5/8 \cdot 7^d\Bigr ) -1/2^d -1/3^d = \rho . \end{aligned}$$

$\square$

Theorem 1

The asymptotic competitive ratio of the tt(d) algorithm is not greater than

$$\begin{aligned} \ (3/2)^{d}+O((21/16)^d). \end{aligned}$$

Proof

Let S be a sequence of items of the total volume v, let $\lambda _1$ denote the number of big items in S and let $\beta$ be the number of bins used to pack items from S according to the tt(d) algorithm.

Obviously, $\ OPT(S) \ge v$ as well as $\ OPT(S) \ge \lambda _1$. By Lemma 5 and Rule (1) from the description of the tt(d) algorithm we have

$$\begin{aligned} v > \frac{1}{2^d}\cdot \lambda _1 +\rho \cdot (\beta -2\lambda _1-1), \end{aligned}$$

i.e.,

$$\begin{aligned} \beta < \frac{1}{\rho } v +\left( 2-\frac{1}{2^d \rho } \right) \lambda _1+1. \end{aligned}$$

Let $\ \mu = \max (v, \lambda _1)$. Since $\ OPT(S) \ge \mu$, we get

$$\begin{aligned} \frac{\beta }{OPT(S)} \le \frac{\beta }{\mu } < \frac{\frac{1}{\rho } \mu +\bigl (2-\frac{1}{2^d \rho } \bigr )\mu +1}{\mu } = \frac{1}{\rho } \bigl (1-\frac{1}{2^d}\bigr )+ 2+\frac{1}{\mu }. \end{aligned}$$

Consequently, the asymptotic competitive ratio for tt(d) algorithm is not greater than

$$\begin{aligned} \frac{1}{\rho } \left( 1-\frac{1}{2^d}\right) +2 = \left( \frac{3}{2} \right) ^d+ \left( \frac{21}{16} \right) ^d\cdot \left( \frac{\frac{5}{8}+(\frac{6}{7})^d}{1-\frac{5}{8} \cdot (\frac{7}{8})^d-(\frac{3}{4})^d-(\frac{1}{2})^d}+2\left( \frac{16}{21} \right) ^d\right) . \end{aligned}$$

$\square$

3 The har(d) Algorithm for $d\ge 5$

In this section we describe a one-space bounded algorithm for packing d-dimensional hypercubes with competitive ratio $\ O(d/\log d)$. The har(d) algorithm is based on the well-known Harmonic algorithm as well as on the tt(d) algorithm and thus some definitions, notations and even lemmas are similar to those related to the tt(d) algorithm. For instance, if all the hypercubes have the edge lengths from the set $\ \{ 1/2,1/4,1/8,$ $1/16, \dots \}$, then the algorithm works exactly the same as the tt(d) algorithm. Some of the presented bounds are not sharp and can be improved; we preferred non-optimal constants rather than complicated calculations.

By $\ a_1 \times a_2 \times \cdots \times a_d$ we mean a hyperbox such that its edges parallel to the k-th axis of the coordinate system are of the length $a_k$, for $\ k=1,2,\ldots , d$. Furthermore, we will write $\ a^{d-q}\times b^q$ instead of $\ \underbrace{a\times \cdots \times a}_{(d-q)\ \small {\text{ times }}} \times \underbrace{b \times \cdots \times b}_{q\ \small {\text{ times }}}$.

3.1 Intuition About How the Algorithm Works

Different types of items are packed into different active bins by the Harmonic algorithm. Since any unit d-dimensional hypercube can be divided into $2^d$ hypercubes of the edge length 1/2 (we called such hypercubes 2-cubes, see Sect. 2.2), our first idea for packing with one active bin only was to place different items into different 2-cubes instead of into many active bins. Unfortunately, the losses were too large.

For example, any unit hypercube can be partitioned into hypercubes of the edge length 1/5, but a 2-cube cannot. Any 2-cube contains only $2^d$ hypercubes of the edge length equal to 1/5. Consequently, only $(4/5)^d$ of any 2-cube is occupied by such hypercubes and this ratio is close to zero for large dimensions d. If hypercubes were packed into $\ B_1= 1^{d-1}\times (1/2)$, then 4/5 of $B_1$ would be occupied. Unfortunately, each unit hypercube contains only two hyperboxes congruent to $B_1$. Therefore items should be packed into hyperboxes smaller than $B_1$ but larger than 2-cubes.

In the online version of packing we do not know anything about the size of incoming items, so we should prepare an empty space to pack items of different sizes. For the items of the edge length 1/t, where $\ t = 5,7,9,11,\ldots$ we will reserve an empty space in special hyperboxes $\ 1^{d-q}\times (1/2)^{q-1} \times (1/t)$, called layers of the height 1/t. Let us mention that $\ q=q(t)$ depends on t and that $\ q(t_1)\ge q(t_2)$, provided $\ t_1 \ge t_2$.

For example, the items of the edge length 1/5 will be packed into layers $\ 1^{d-4}\times (1/2)^{3} \times (1/5)$. Such a layer contains $5^{d-4}\cdot 2^3$ hypercubes of the edge length 1/5. This means that 64/125 of the layer is occupied by the hypercubes and this ratio does not depend on d. The items of the edge length 1/t, for $\ t \in \{ 7,9 \}$, will be packed into layers $\ 1^{d-4}\times (1/2)^{3} \times (1/t)$. Since $\ \frac{1}{5} + \frac{1}{7} + \frac{1}{9}<\frac{1}{2}$, it is possible to create three layers (of pairwise disjoint interiors) of the height 1/5, 1/7 and 1/9 in any hyperbox $\ 1^{d-4}\times (1/2)^4$. At least 64/125 of any layer of the height 1/t is occupied by hypercubes of the edge length 1/t. Similarly, it is possible to create eight layers $\ 1^{d-5}\times (1/2)^{4} \times (1/t)$, for $\ t=11,13,\ldots , 25$, in any hyperbox $\ 1^{d-5}\times (1/2)^5$ (we have $\ \frac{1}{11}+\frac{1}{13}+\cdots + \frac{1}{23}+ \frac{1}{25} < \frac{1}{2}$) . In a similar way, layers $\ 1^{d-6}\times (1/2)^{5} \times (1/t)$, for $\ t=27,29,\ldots , 69$, will be created in $\ 1^{d-6}\times (1/2)^6$. Generally, layers of small height will be created in hyperboxes with comparatively large number of edges of the length 1/2. The number of such hyperboxes in the active bin is relatively large. In any case, despite many edges of the layer are of the length 1/2, at least 64/125 of any layer of the height 1/t is occupied by hypercubes of the edge length 1/t.

Items (d-dimensional hypercubes) will first be packed into hypercubes, called m-cubes, of the edge lengths $\ 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, \dots$ in our main method. For each m-cube we describe a place in the bin into which this hypercube will be packed. First, we will assign an open B-box (the union of some 2-cubes). Then we will determine the right place in the open B-box to pack the hypercube. Some of m-cubes will be contained in layers created in open B-boxes. The hypercubes of the edge length 1/2 will be packed into open 2-cubes. The hypercubes of the edge length 1/3 will be placed into open B-boxes $\ 1^{d-3}\times (1/2)^3$. Moreover, the items of the edge lengths 1/(2t) will be packed next to the items of the edge lengths 1/t, for $\ t= 2,3,\ldots$.

All open hyperboxes have pairwise disjoint interiors. Moreover, all created layers have pairwise disjoint interiors.

Since hypercubes will be packed into layers $\ 1^{d-q}\times (1/2)^{q-1} \times (1/t)$, where $\ q\ge 4$, we will assume that

$$\begin{aligned} d\ge 5 \end{aligned}$$

in this section. Let us add that it is possible to modify the algorithm and create layers also in smaller dimensions. However, such a modified algorithm would not be effective. Even for $d=5$ the upper bound presented in Theorem 2 is distinctively worse than the one given in the proof Theorem 1. The aim of this paper is to present a one-space bounded algorithm efficient for large d.

Fig. 7

Items in the bin

Example

Incoming items will be packed into open hyperboxes of a proper size. Let $d=5$. At the beginning of the packing process three hyperboxes

$$\begin{aligned} H_1=[0,1/2]^5, \ H_2=[0,1]\times [1/2,1]\times [0,1/2]^{3}\ \ \mathrm{and}\ \ H_3=[0,1]^2\times [1/2,1]\times [0,1/2]^{2} \end{aligned}$$

are open in the active bin; the place of the bin into which these hyperboxes are contained is determined by the box(d) algorithm described in Sect. 3.2. $H_1$ will be called $BC_2$-box in Sect. 3.6; hypercubes of the edge length $\ 1/2, 1/4, 1/8, \ldots$ will be packed into it. $H_3$ will be called $BC_3$-box; hypercubes of the edge length $\ 1/3, 1/6$, $1/12, \ldots$ will be packed into it. $H_2$ will be called $BL_4$-box; three layers of the height 1/5, 1/7 and 1/9 are created in it.

The items of the edge lengths

$$\begin{aligned} 1/4, 1/3, 1/8, 1/5, 1/7, 1/2, 10/51 \end{aligned}$$

are packed as on Fig. 7. The first item is placed into $H_1$. We will use the $aux^+(5)$ algorithm described in Sect. 3.4: the first item is packed in the first 2-subcube of the edge length 1/4 contained in $H_1$. The second item is placed into $H_3$. We will use the $aux^+(5)$ algorithm described in Sect. 3.4: the item is packed in the first subcube of the edge length 1/3 contained in $H_3$. The third item is packed into $H_1$ by the $aux^+(5)$ algorithm; it is packed into the first empty 2-subcube of $H_1$ of the edge length 1/4 (see Fig. 7, left). The fourth item is packed in $H_2$ into the layer $\ [0,1] \times [1/2,1] \times [0,1/2]^2 \times [0,1/5]$ of the height 1/5 (see Fig. 7, right). The place in the layer into which this item is packed is determined by the $aux^+(5)$ algorithm described in Sect. 3.4, just the item is packed into the first hypercube of the edge length 1/5 contained in this layer. The next item is placed into the layer $\ [0,1]\times [1/2,1]\times [0,1/2]^2 \times [1/5, 1/5+1/7]$ contained in $H_2$. The place in the layer into which this item is packed is determined by the $aux^+(5)$ algorithm: just the item is packed into the first hypercube of the edge length 1/7 contained in this layer (see Fig. 7, right). There is no empty space in $H_1$ to pack the penultimate item of the edge length 1/2. We open a new $BC_2$-box $\ H_4= [1/2,1]\times [0,1/2]^{4}$ (this place is determined by the box(d) algorithm) and pack the item into $H_4$. The last item is packed into the layer $\ [0,1]^{d-4}\times [1/2,1] \times [0,1/2]^2 \times [0,1/5]$ contained in $H_2$. The place in the layer into which this item is packed is determined by the har(5) algorithm; just the item is packed into the second hypercube of the edge length 1/5 contained in this layer.

Let us add that if all subsequent items in the sequence (eighth, ninth, ...) have the edge lengths 1/7, then $7\cdot 3^3-1$ of them are packed in the layer $\ [0,1]\times [1/2,1]\times [0,1/2]^2 \times [1/5, 1/5+1/7]$ contained in $H_2$. Then a new $BL_4$-box is opened and three layers of the height 1/7 are created in it. Details of this method are described in the next subsections.

3.2 B-Boxes

As it was said in Sect. 3.1, items will be packed into open hyperboxes with pairwise disjoint interiors. Any hypercube of the edge length 1/2 was packed in the empty 2-cube with the smallest number by the tt(d) algorithm. We will present a similar method for hyperboxes that are the union of 2-cubes.

By a $B_q$-box (a B-box, for short), where $\ q \in \{ 0,\dots , d \}$, we mean $\ 1^{d-q}\times (1/2)^q$ (see Fig. 8). Any $B_q$ box is the union of $2^{d-q}$ many 2-cubes.

Fig. 8

$B_q$-boxes in the bin

For each $\ q \in \{ 0,\dots , d \}$ the bin ${\mathscr {B}}$ is partitioned into $2^q$ congruent $B_q$-boxes that are numbered as follows. The entire bin $\ {\mathscr {B}}=[0,1]^d$ is a $B_0$-box and is numbered with 1. Two $B_1$-boxes contained in ${\mathscr {B}}$, i.e., hyperboxes $\ [0,1]^{d-1}\times [0,1/2]$ and $\ [0,1]^{d-1}\times [1/2,1]$, are numbered 1 and 2, respectively. Finally, let $k< d$ and let

$$\begin{aligned} X=[0,1]^k \times [\gamma _{k+1}, \gamma _{k+1}+1/2] \times \cdots \times [\gamma _{d}, \gamma _{d}+1/2], \end{aligned}$$

where $\ \gamma _{k+1}, \ldots \gamma _d \in \{0,1/2\}$, be a $B_k$-box numbered $\lambda$. X is partitioned into two congruent $B_{k+1}$-boxes:

$$\begin{aligned}&[0,1]^{k-1} \times [0,1/2] \times [\gamma _{k+1}, \gamma _{k+1}+1/2] \times \cdots \times [\gamma _{d}, \gamma _{d}+1/2],\\&[0,1]^{k-1} \times [1/2,1] \times [\gamma _{k+1}, \gamma _{k+1}+1/2] \times \cdots \times [\gamma _{d}, \gamma _{d}+1/2] \end{aligned}$$

with numbers $\ 2\lambda -1$ and $\ 2\lambda$, respectively.

For example, $\ H_1=[0,1]^{d-1}\times [0,1/2]$ on Fig. 8 is the $B_1$-box with number 1 and number 2 is assigned to the other $B_1$-box ${\mathscr {B}} {\setminus } H_1$. Moreover, $H_1$ contains: two $B_2$-boxes with numbers 1 and 2 as well as four $B_3$-boxes with numbers 1, 2, 3 and 4, eight $B_4$-boxes with numbers $1,2,\dots , 8$ and so on. $H_2=[0,1]^{d-2}\times [0,1/2] \times [1/2,1]$ is the $B_2$-box with number 3 and number 4 is assigned to the $B_2$-box ${\mathscr {B}} {\setminus } (H_1\cup H_2)$. $H_3=[0,1]^{d-3}\times [0,1/2]\times [1/2,1]^2$ is the $B_3$-box with number 7 and number 8 is assigned to the $B_3$-box ${\mathscr {B}} {\setminus } (H_1\cup H_2\cup H_3)$.

Incoming items will be packed into open B-boxes by the har(d) algorithm described in Sect. 3.8. If there is no empty space, then we will open a new $B_q$-box in the place described by the following algorithm.

Let $\ H_1, H_2, \dots$ be a sequence of B-boxes. A B-box of ${\mathscr {B}}$ is called empty if its interior is disjoint with any open B-box.

Algorithm box(d)

• the first open B-box is the first B-box of ${\mathscr {B}}$ that is congruent to $H_1$;

• if B-boxes of ${\mathscr {B}}$ congruent to $\ H_1,\ldots , H_{j-1}$ are open, then the the next open B-box is the empty B-box of ${\mathscr {B}}$ with the smallest number that is congruent to $H_j$.

For example, $B_d$-box $H_1$, $B_{d-1}$-box $H_2$, $B_{d-2}$-box $H_3$ and $B_d$ box $H_4$ are open in the active bin by box(d) in places shown on Fig. 7 ($d=5$). $H_1$ is the first $B_d$-box, $H_2$ is the second $B_{d-1}$-box, $H_3$ is the second $B_{d-2}$-box and $H_4$ is the second $B_d$-box in the active bin.

Another example is illustrated on Fig. 8, where hyperboxes: $B_1$-box $H_1$, $B_2$-box $H_2$ and $B_3$-box $H_3$ are open in places described by box(d). $H_1$ is the first $B_1$-box, $H_2$ is the third $B_2$-box and $H_3$ is the seventh $B_3$-box in the active bin.

Lemma 6

Let $\ H_1, H_2, \dots$ be a sequence of B -boxes. If B-boxes congruent to $\ H_1$, $\ldots , H_{z-1}$ are open in ${\mathscr {B}}$ and if there is no empty space in ${\mathscr {B}}$ to open a new B-box congruent to $H_z$, then $\ \sum _{i=1}^{z-1} |H_i| >1-|H_z|$.

Proof

$H_z$ is a $B_q$-box for some $\ q \in \{ 0,\dots , d \}$. Obviously, $|H_z|=2^{-q}$. Since it is not possible to open a B-box congruent to $H_z$, there are no empty $B_k$-boxes in ${\mathscr {B}}$ for $k\le q$. There can be at most one empty $B_{q+1}$-box. Otherwise suppose there are two empty $B_{q+1}$-boxes $X_1$ and $X_2$. These $B_{q+1}$-boxes of ${\mathscr {B}}$ are not contained in one $B_q$-box, because that would make a whole empty $B_q$-box in ${\mathscr {B}}$. Therefore $X_1$ and $X_2$ lay in different $B_q$-boxes, say $X_2$ is in a $B_q$-box with a higher number than $X_1$ is. Let $X_2'$ be the complementary half of $X_2$ in a $B_q$-box they lay in. Since there is no empty $B_q$-box, $X_2'$ contains an open B-box. It means that a $B_{q+1}$-box $X_2'$ contains an open B-box and has higher number than $X_1$ that is empty. This contradicts the box(d) algorithm. Therefore there is at most one empty $B_{q+1}$-box in ${\mathscr {B}}$.

Using the above argument we obtain at most one empty $B_k$-box of ${\mathscr {B}}$ for each $k\in \{q+1,\dots , d \}$ with the interior disjoint with other empty B-boxes of ${\mathscr {B}}$.

Finally, the empty space in ${\mathscr {B}}$ does not exceed

$$\begin{aligned} \sum _{k=q+1}^d2^{-k}=2^{-q}-2^{-d}<2^{-q}=|H_z|. \end{aligned}$$

$\square$

3.3 Layers and m-Cubes

To prepare an empty space for incoming items of different sizes, we will create layers of the height 1/t, for $\ t=5,7,9,\ldots$ in open $B_q$-boxes. Items of the edge lengths 1/t, 1/(2t), $1/(4t), \ldots$ will be placed in layers of the height 1/t.

Let $\ 10\le m \le 2^{d-1}$ be an even number and let $\ q \in \{ 1,\dots , d \}$.

The integer m determines the number of layers of different heights. For fixed m only layers of the height $1/5, 1/7, 1/9, \dots , 1/(m-1)$ can be created. For example, $d=5$ and $m=10$ on Fig. 7 and only layers of the height 1/5, 1/7 and 1/9 were created. Ultimately, we will take m close to $4d/\log d$ in the proof of Theorem 2.

By an m-cube $K^+$ we mean a hypercube of the edge length $\ 1/(t\cdot 2^p)$, where $\ p \ge 0$ is an integer and $\ t \in \{ 1,2, \dots , m \}$. We say then that $K^+$ is a cube of type (t, p). When the value of p can be arbitrary we say that $K^+$ is a cube of class t.

For instance, 6-cubes have the edge lengths from the set

$$\begin{aligned} \Big \{ 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{8}, \frac{1}{10}, \frac{1}{12}, \frac{1}{16}, \frac{1}{20}, \frac{1}{24}, \dots \Big \} \end{aligned}$$

while cubes of class 6 have the edge lengths from the set $\Big \{ \frac{1}{6}, \frac{1}{12}, \frac{1}{24}, \dots \Big \}$.

Let us note that any cube of type $(2^n\cdot t,p)$ is also of type $(t,p+n)$. This fact will be used in the description (Rule 5) of the $har^+(d)$ algorithm. For instance, the 6-cube of the edge length $\ \frac{1}{32}=\frac{1}{2^1\cdot 2 \cdot 2^3}=\frac{1}{2\cdot 2^{3+1}}$ is of type $(2^1\cdot 2,3)$ as well as $(2, 3+1)$.

Fig. 9

Layers filled with m-cubes

By a (t, q)-layer we mean $\ 1^{d-q}\times (1/2)^{q-1}\times (1/t)$; the value 1/t is called the height of the layer (see Fig. 9).

Clearly, each (2, q)-layer is a $B_q$-box. We emphasize the fact that in the process of packing we will only use layers for t being odd. In the main packing method q depends on t (for detailed description see Sect. 3.5): if $t\in \{5,7,9\}$, then $q=4$; if $t=11$, then $q=5$; if $t=111$, then $q=6, \ldots$.

For example, (5, 4)-layer, (7, 4)-layer and (9, 4)-layer were created on Fig. 7. The first layer contains $5\cdot 2^3$ hypercubes of the edge length 1/5. The first and the second hypercube are used for packing. The second layer contains $7\cdot 3^3$ hypercubes of the edge length 1/7. The first hypercube is used for packing.

It is easy to see that for an odd number t any (t, q)-layer contains $\ \bigl ( \frac{1}{2}t - \frac{1}{2} \bigr )^{q-1} \cdot t^{d-q}$ cubes of type (t, 0) (see Fig. 9, where (5, 4)-layer contains $2^3\cdot 5^{d-4}$ hypercubes of the edge length 1/5, while (7, 4)-layer contains $3^3\cdot 7^{d-4}$ hypercubes of the edge length 1/7 ).

Lemma 7

If $\ q \ge 4$ and if $\ t\ge 2^{q-2}+1$, then at least 64/125 of any (t, q)-layer is occupied by cubes of type (t, 0).

Proof

$\bigl ( \frac{1}{2}t-\frac{1}{2}\bigr )^{q-1}\cdot t^{d-q}$ cubes of type (t, 0) are contained in each (t, q)-layer. The volume of any (t, q)-layer equals $\ 2^{1-q} \cdot t^{-1}$. Observe that

$$\begin{aligned} \frac{\bigl ( \frac{1}{2}t-\frac{1}{2}\bigr )^{q-1}\cdot t^{d-q}\cdot t^{-d}}{2^{1-q}\cdot t^{-1}} =\Bigl (1-\frac{1}{t}\Bigr )^{q-1}. \end{aligned}$$

Since $\ t \ge 2^{q-2}+1$, we get

$$\begin{aligned} \Bigl ( 1-\frac{1}{t}\Bigr )^{q-1} \ge \Bigl (1-\frac{1}{2^{q-2}+1}\Bigr )^{q-1}. \end{aligned}$$

If $\ q=4$, then $\ (1-\frac{1}{2^2+1})^3 = \frac{64}{125}$. If $\ q \ge 5$, then

$$\begin{aligned} \bigl (1-\frac{1}{2^{q-2}+1}\bigr )^{q-1} = \frac{1}{\bigl ( 1+ \frac{1}{2^{q-2}}\bigr )^{q-1}}> e^{-\frac{q-1}{2^{q-2}}} \ge e^{-\frac{4}{8}} > \frac{64}{125}. \end{aligned}$$

This implies that at least 64/125 of any (t, q)-layer is occupied by cubes of type (t, 0). $\square$

3.4 Auxiliary Algorithm $aux^+(d)$

If all items have the edge lengths from the set $\ \{ 1/2, 1/4, 1/8, \ldots \}$, then we will use the tt(d) algorithm described in Sect. 2. Each item will be packed into the proper subcube with the smallest number contained in the union of hypercubes of the edge length 1/2. Similarly, each item of the edge length slightly smaller than either 1/2 or 1/4 or 1/8, ...will be packed into the proper subcube with the smallest number contained in the union of hypercubes of the edge length 1/2. In the main packing method the items of the edge length slightly smaller than either 1/t or 1/(2t) or 1/(4t), ...will be placed into layers of the height 1/t (for odd $t\ge 5$). More precisely, the items will be packed into the union of hypercubes of the edge length 1/t contained in layers of the height 1/t. Therefore we need an algorithm that packs items into hypercubes. Our algorithm is similar to the tt(d) algorithm.

Let $\ 10\le m \le 2^{d-1}$ be an even number and let either $\ 3 \le t \le m-1$ be an odd number or $\ t=2$. Moreover let p be a non-negative integer.

Consider the union $U_T$ of cubes of type (t, 0) with pairwise disjoint interiors. All cubes are numbered with successive positive integers (compare Fig. 2, left, where $t=2$). Furthermore, $2^d$ cubes of type (t, p) contained in the cube of type $(t,p-1)$ with number $\eta$ are numbered in an arbitrary order from $\ 2^d(\eta -1)+1$ to $\ 2^d\eta$.

Consider a sequence $\ K_1^+, K_2^+, \dots$ of cubes of class t. We say that a cube of type (t, p) of $U_T$ is empty if its interior has an empty intersection with any cube packed so far.

Algorithm $aux^{+}(d)$ for packing of $K^{+}_{i}$ into ${U_T}$.

• pack $K_i^+$ into the empty hypercube of $U_T$ with the smallest number that is congruent to $K_i^+$.

If $U_T$ is the unit hypercube partitioned into the union of $2^d$ many 2-cubes and if $t=2$, then $aux^+(d)$ works exactly the same as the tt(d) algorithm. For instance, the first item from the Example of Sect. 3.1 is packed by the $aux^+(d)$ algorithm into the first hypercube of the edge length 1/4; the third item is packed into the first empty hypercube of the edge length 1/8. The second item is packed by the $aux^+(d)$ algorithm into the first hypercube of the edge length 1/3 contained in $H_3$. The third item is packed by the $aux^+(d)$ algorithm into the first hypercube of the edge length 1/5 contained in the proper layer (see Fig. 7).

Lemma 8

Let $S_t$ be a sequence of cubes of class t. If $K^+_z$ is the first hypercube from $S_t$ that cannot be packed in $U_T$ by $aux^{+}(d)$ algorithm, then the total volume of hypercubes packed in $U_T$ plus the volume of $K^+_z$ is greater than $|U_T|.$

Proof

We proceed as in the proof of Lemma 3. Since $K^+_z$ is a cube of class t, there is $\ p \ge 0$ such that $K_z^+$ is a cube of type (t, p). Clearly, there is no empty cube of type (t, q) in $U_T$ for any $\ q \le p$. Furthermore, for each $\ q> p$ there are at most $2^d-1$ empty cubes of type (t, q). The total volume of empty cubes in $U_T$ is smaller than

$$\begin{aligned} \ (2^d-1)\cdot (2^{-d} + 4^{-d} + \cdots ) \cdot \bigl (t \cdot 2^p\bigr )^{-d} =\bigl (t \cdot 2^p\bigr )^{-d}. \end{aligned}$$

This implies that the sum of the volumes of all cubes packed into $U_T$ is greater than $\ |U_T|-\bigl (t \cdot 2^p\bigr )^{-d}.$ By $\ |K^+_z| = (t\cdot 2^p)^{-d}$, it follows that the total volume of items packed in $U_T$ is greater than $\ |U_T|- |K^+_z|.$ $\square$

3.5 Layers in B-Boxes

Since the series $\ \sum _{j=2}^{\infty } \frac{1}{2j-1}$ is divergent, there is no space in the bin to create too many layers of different heights.

Clearly, the sum of the heights of layers created in any $B_q$-box (of the height 1/2) is not greater than 1/2.

Since $\ \frac{1}{5}+\frac{1}{7}+\frac{1}{9}\le \frac{1}{2}$, it follows that three layers: (5, q), (7, q) and (9, q) can be created in one $B_q$-box. Similarly, $\ \frac{1}{11}+\frac{1}{13}+ \cdots + \frac{1}{25} \le \frac{1}{2}$ implies that eight layers: (11, q), (13, q), $\dots , (25,q)$ can be created in one $B_q$-box.

Now we will estimate the number of layers of different heights that can be created in $B_q$-boxes.

Let $\ n_3= 3$ and, for $\ k=4,5 \dots$, let $\ n_k$ be the greatest odd number such that

$$\begin{aligned} \frac{1}{n_{k-1}+2}+\frac{1}{n_{k-1}+4} + \cdots +\frac{1}{n_k}\le \frac{1}{2}. \end{aligned}$$

This means that the layers of the height $\ n_{k-1}+2, n_{k-1}+4, \ldots , n_k$ can be created in one $B_q$-box for any $\ k=3,4,\ldots$. It is easy to check that $\ n_4=9,\ n_5=25$, $\ n_6=69$, $\ n_7=189$, $\ n_8=515$, $\ n_9=1401$ and $\ n_{10}=3809$. By the definition of $n_k$ we know that the layers: $\ (5,q),\ (7,q), \ \dots , (n_k,q)$ can be accommodated into the union of $k-3$ many $B_q$-boxes. It is enough now to give a lower bound for $n_k$ (see Lemma 9).

During the packing process (t, q(t))-layers (for odd $t\ge 5$) will be created in $B_{q(t)}$-boxes (see Sect. 3.6). Let us add that for large t the integer $\ q=q(t)$ will be comparatively large, i.e., hypercubes of the edge length 1/t will be packed into layers with comparatively large number of edges of the length 1/2. The connection between t and q(t) is as follows. Given an odd number $t\ge 5$, let q(t) be an integer such that

$$\begin{aligned} n_{q(t)-1} < t \le n_{q(t)}. \end{aligned}$$

For example,

• $q(5)=q(7)=q(9)=4$ ($n_{4-1}=3<5<7 < 9=n_4$);

• $q(11)=q(13)=\cdots = q(25)=5$ ($n_{5-1}=9<11<\cdots <25=n_5$),

• $q(27)=\cdots = q(69)=6$ ($n_6=69$).

Lemma 9

For any integer $l\ge 4$ we get $\ n_l \ge 2^{l-1}-1$.

Proof

$$\begin{aligned} \sum _{j=3}^{\infty }{\frac{1}{2j-1}}& = \left( \frac{1}{5} + \frac{1}{7} \right) + \left( \frac{1}{9} + \frac{1}{11} + \frac{1}{13} + \frac{1}{15} \right) + \left( \frac{1}{17} + \cdots + \frac{1}{31} \right) + \cdots \\&\quad+ \left( \frac{1}{2^{l-1}+1} + \cdots + \frac{1}{2^{l}-1} \right) + \cdots \\ &< \Big (\underbrace{\frac{1}{4} + \frac{1}{4}}_{2^1 \ \text{ times }} \Big ) + \Big (\underbrace{\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}}_{2^2 \ \text{ times }} \Big ) +\Big (\underbrace{\frac{1}{16} + \cdots + \frac{1}{16}}_{2^3 \ \text{ times }} \Big )+ \cdots \\&\quad +\, \Big (\underbrace{\frac{1}{2^{l-1}} + \cdots + \frac{1}{2^{l-1}}}_{{2^{l-2}} \ \text{ times }} \Big ) + \cdots = \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots \end{aligned}$$

This implies that: $\ n_4\ge 7=2^3-1, \ n_5 \ge 15=2^4-1, \dots ,\ n_{l+1}\ge 2^{l}-1$. $\square$

3.6 Description of the $har^+(d)$ Algorithm

The main algorithm is presented in two steps. First, we give the algorithm $har^+(d)$ that packs only m-cubes into the unit cube ${\mathscr {B}}$. Then we provide the final algorithm har(d): a method of packing items of arbitrary sizes into ${\mathscr {B}}$. Simply we will pack any item into the smallest possible m-cube and then this m-cube will be packed into ${\mathscr {B}}$.

Let $\ d \ge 5$. Moreover, let $\ 10\le m \le 2^{d-1}$ be an even number.

Since some m-cubes will be packed into $B_{q(m-1)}$-boxes, i.e., hypercubes $\ 1^{d-q(m-1)}\times (1/2)^{q(m-1)}$, first we show that

$$\begin{aligned} q(m-1)\le d. \end{aligned}$$

By the choice of q(t) (see Sect. 3.5) we know that $\ n_{q(m-1)-1} < m-1 \le n_{q(m-1)}$. By Lemma 9 we get

$$\begin{aligned} m-1> n_{q(m-1)-1} >2^{q(m-1)-2}-1. \end{aligned}$$

Since $\ 2^{d-1} \ge m$, we have $\ 2^{d-1} > 2^{q(m-1)-2}$, i.e., $\ q(m-1)<d+1$. Consequently, $\ q(m-1)\le d.$ This implies that it is possible to create a $B_{q(m-1)}$-box in ${\mathscr {B}}$.

We distinguish two classes of B-boxes based on the size of items packed into them:

• BC-boxes (see Figs. 10 and 11);

• BL-boxes in which layers will be created (see Fig. 12).

Each time a B-box is opened it is assigned to one of those classes and we mention whether it is a BC- or a BL-box.

At the beginning of the process of packing the following B-boxes are open in ${\mathscr {B}}$:

• two $\ BC$-boxes: one $B_d$-box called $BC_2$-box into which hypercubes of the edge lengths $\ 1/2, 1/4,$ $1/8 , \dots$ will be packed and one $B_3$-box called $BC_3$-box into which hypercubes of the edge lengths $\ 1/3, 1/6,$ $1/12 , \dots$ will be packed;

• $q(m-1)- 3$ many BL-boxes: one $B_k$-box called $BL_k$-box, for each $\ k = 4,5,$ $\dots , q(m-1)$.

The place in the bin into which these $q(m-1)-1$ many B-boxes are contained is described by the box(d) algorithm. Just we open: one $B_d$-box, one $B_3$-box, one $B_4$-box, ...and one $B_{q(m-1)}$-box in ${\mathscr {B}}$ in places determined by the box(d) algorithm (see Fig. 7, where $m=10$ and $\ q(m-1)=q(9)=4$).

Fig. 10

$B_d$-box ($BC_2$-box)

Fig. 11

$B_3$-box ($BC_3$-box)

Fig. 12

Layers in $BL_4$-boxes for $d=5$

Fig. 13

Layers in three $BL_4$-boxes

Into $BL_4$-boxes the hypercubes of the edge lengths $\ 1/5, 1/7, 1/9 , 1/10, 1/14,$ $1/18, 1/20, 1/28, 1/36,\dots$ will be packed. The hypercubes of the edge lengths $\ 1/11,$ $1/13,\dots , 1/25, 1/22, 1/26, \dots$ will be placed into $BL_5$-boxes. Into $BL_{s}$-boxes hypercubes of the edge lengths

$$\begin{aligned} 1/(n_{s-1}+2), 1/(n_{s-1}+4),\dots , 1/n_s, 1/(2n_{s-1}+4), 1/(2n_{s-1}+8),\dots , 1/(2n_s), \dots \end{aligned}$$

will be packed, for $\ s=4,5, \dots , q(m-1)$. The number of open B-boxes increases during the packing process.

Note that it is possible to pack m-cubes of types (2, 0) and (3, 0) into one B-box, as in the tt(d) algorithm, however we will pack these cubes separately to simplify the calculations.

Three layers are created in the first open $BL_4$-box: one (5, 4)-layer, one (7, 4)-layer and one (9, 4)-layer as on Fig. 12. Let us note that $\ 5=n_3+2$ and $\ 9=n_4$. Then, depending on the size of incoming items it is possible that in the second open $BL_4$-box only two (5, 4)-layers were created as on Fig. 12, right. It is also possible that in the second open $BL_4$-box only one (5, 4)-layer and one (9, 4)-layers were created (see Fig. 13).

Similarly, in the first $BL_k$-box ($k\in \{4,5,\dots , q(m-1)\}$) the following layers are created: one $\ (n_{k-1}+2,k)$-layer, one $\ (n_{k-1}+4,k)$-layer, $\dots$, one $\ (n_k,k)$-layer.

We will use the algorithm $aux^+(d)$, therefore we enumerate m-cubes contained in layers. All cubes of type (t,0) contained in the created (t, q)-layer with number $\xi$ are numbered in an arbitrary order from $\ \bigl ( \frac{1}{2}t - \frac{1}{2} \bigr )^{q-1} \cdot t^{d-q}\cdot (\xi -1)+1$ to $\ \bigl ( \frac{1}{2}t - \frac{1}{2} \bigr )^{q-1} \cdot t^{d-q}\cdot \xi$. Furthermore, $2^d$ congruent hypercubes into which the cube of type (t, p) with number $\lambda$ can be partitioned are numbered in an arbitrary order from $\ 2^d(\lambda -1)+1$ to $\ 2^d\lambda$.

Algorithm $har^{+}(d)$ for packing of $K^{+}_{i}$ into ${{\mathscr {B}}}$.

Let $K_i^+$ be a cube of type (t, p). All $B_q$-boxes of ${{\mathscr {B}}}$ are numbered as in Sect. 3.2.

• 1. If $(t,p)=(1,0)$, i.e., $K_i^+$ is a hypercube of the edge length 1, then we close the active bin, open a new bin, pack $K^+_i$ and close the bin. Then we open a new active bin.

• 2. If $(t,p)=(2^j,n)$, for some $j\ge 1$ and $n\ge 0$, then $K_i^+$ is packed into the union of cubes of type (2, 0) contained in the union of open $BC_2$-boxes according to the $aux^+(d)$ algorithm.

  If it is impossible, we open in ${\mathscr {B}}$ a new $BC_2$-box: an empty $B_d$-box with the smallest number (compare the box(d) algorithm). If there were v open $BC_2$-boxes so far, then the number assigned to the cube of type (2, 0) that is the open $BC_2$-box is $v+1$. We pack $K_i^+$ into this cube by the $aux^+(d)$ algorithm.

• 3. If $(t,p)=(2^j\cdot 3,n)=(3,j+n)$, for some $j,\,n\ge 0$, then $K_i^+$ is packed into the union of cubes of type (3, 0) contained in open $BC_3$-boxes according to the $aux^+(d)$ algorithm.

  If it is impossible, we open in ${\mathscr {B}}$ a new $BC_3$-box: an empty $B_3$-box with the smallest number (compare the box(d) algorithm). If there were w open $BC_3$-boxes so far, then the cubes of type (3, 0) contained in the open $BC_3$-box are numbered from $\ w\cdot 3^{d-3}+1$ to $\ (w+1)\cdot 3^{d-3}$. We pack $K_i^+$ by the $aux^+(d)$ algorithm.

• 4. If $\ t$ is an odd number greater then 3 and $p\ge 0$ is arbitrary, then $K_i^+$ is packed into the union of cubes of type (t, 0) contained in the union of (t, q(t))-layers created in open $BL_{q(t)}$-boxes by $aux^+(d)$ algorithm.

  If it is impossible, we create a new (t, q(t))-layer in an open $BL_{q(t)}$-box (as low with respect to $x_d$-axis as possible). If there were u created layers of the height 1/t so far, then the cubes of type (t, 0) contained in the created layer are numbered from $\ u\cdot \bigl ( \frac{1}{2}t - \frac{1}{2} \bigr )^{q(t)-1} \cdot t^{d-q(t)}+1$ to $\ (u+1)\cdot \bigl ( \frac{1}{2}t - \frac{1}{2} \bigr )^{q(t)-1} \cdot t^{d-q(t)}$. We pack $K_i^+$ by the $aux^+(d)$ algorithm. If it is impossible, we open a new $BL_{q(t)}$-box: an empty $B_{q(t)}$-box with the smallest number (compare the box(d) algorithm) and create at the bottom of this box (with respect to $x_d$-axis) a new (t, q(t))-layer to pack $K_i^+$.

• 5. If $\ t$ is an even number but not a power of 2 and $p\ge 0$ is arbitrary, then there is an odd number $t_1$ and an integer s such that $K_i^+$ is also of type $(t_1,s+p)$. If $\ t_1=3$, then $K_i^+$ is packed as described in Rule (3) into $BC_3$-boxes. If $\ t_1\ge 5$, then $K_i^+$ is packed as described in Rule (4) into $BL_{q(t_1)}$-boxes.

• 6. If there is no empty m-cube in open B-boxes to pack $K^+_i$ and if there is no empty space in ${\mathscr {B}}$ to open a new B-box, we close the active bin and open a new bin to pack $K^+_i$.

3.7 Packing Density

In this subsection we will show that the total volume of m-cubes packed into each closed bin is greater than 1/9.

Assume that m-cubes were packed into ${\mathscr {B}}$ by the algorithm $aux^+(d)$ and that ${\mathscr {B}}$ is closed.

Lemma 10

Denote by $v_{bc}$ the total volume of BC-boxes (contained in a closed bin ${\mathscr {B}}$) into which at least one m-cube was packed. The total volume of m -cubes packed in BC-boxes is greater than $\ \frac{8}{27}\bigl ( v_{bc}-2^{-3}-2^{-d}\bigr )$.

Proof

Each $BC_2$-box is a cube of type (2, 0) (see Fig. 10). Each $BC_3$-box (of the volume $2^{-3}$) contains $3^{d-3}$ cubes of type (3, 0) of the total volume $\ 3^{d-3}\cdot 3^{-d} = 3^{-3}=\bigl ( \frac{2}{3}\bigr )^3 \cdot 2^{-3}$ (see Fig. 11).

Denote by $b_2$ the number of open $BC_2$-boxes and by $b_3$ the number of open $BC_3$-boxes. From among open $BC_2$-boxes the first $\ b_2-1$ boxes are called basic, provided $\ b_2 \ge 2$. Moreover, from among open $BC_3$-boxes the first $\ b_3-1$ boxes are called basic, provided $\ b_3 \ge 2$. Clearly, $\ v_{bc}=b_2\cdot 2^{-d} + b_3\cdot 2^{-3}$. The total volume of basic boxes is greater than or equal to $\ (b_2-1)\cdot 2^{-d} + (b_3-1)\cdot 2^{-3}.$ The total volume of cubes of type (2, 0) and (3, 0) contained in basic boxes is greater than

$$\begin{aligned} (b_2-1)\cdot 2^{-d} + \Bigl ( \frac{2}{3} \Bigr )^3 \cdot (b_3-1)\cdot 2^{-3} &\ge \Bigl ( \frac{2}{3} \Bigr )^3 \cdot \bigl ( (b_2-1)\cdot 2^{-d} + (b_3-1)\cdot 2^{-3} \bigr ) \\& = \Bigl ( \frac{2}{3} \Bigr )^3 \cdot \bigl ( v_{bc}-2^{-d}-2^{-3}\bigr ). \end{aligned}$$

By Lemma 8, the sum of the volumes of m-cubes packed in basic boxes plus the volumes of the first m-cubes packed in the last $BC_2$- and $BC_3$-boxes is greater than the total volume of basic boxes. Consequently, the total volume of all m-cubes packed in BC-boxes is greater than $\ \bigl ( \frac{2}{3} \bigr )^3 \cdot \bigl ( v_{bc}-2^{-d}-2^{-3}\bigr )$. $\square$

Lemma 11

Denote by $v_{bl}$ the total volume of BL -boxes (contained in a closed bin ${\mathscr {B}}$) into which at least one m-cube was packed. Then the total volume of m-cubes packed in BL-boxes is greater than $\ \frac{192}{625}v_{bl}-\frac{64}{625}$.

Proof

Each BL-box is a $B_k$-box for some $\ k \in \{ 4,5 \dots , q(m-1) \}$.

Since $\ n_{q(t)-1} < t \le n_{q(t)}$ and both t and $n_{q(t)-1}$ are odd numbers, it follows that $\ t \ge n_{q(t)-1}+2$. By Lemma 9 we get

$$\begin{aligned} t \ge n_{q(t)-1}+2> 2^{q(t)-2}-1+2 = 2^{q(t)-2}+1. \end{aligned}$$

This means, by Lemma 7, that at least 64/125 of any (t, q(t))-layer is occupied by cubes of type (t, 0).

For $\ k \in \{ 4,5,\dots , q(m-1)\}$ denote by $b_k$ the number of open $BL_k$-boxes. If $b_k\ge 2$, then the first $b_k-1$ boxes are called basic. Observe that at least 3/5 of any basic $BL_k$-box is occupied by layers. The reason is that the height of the highest layer that can be created in any $B_q$-box is not greater than $\ 1/5$. Since in any basic $BL_k$-box there is no space to create a new layer, the sum of the heights of all layers contained in any basic $BL_k$-box is greater than $\ \frac{1}{2} -\frac{1}{5} = \frac{3}{10}$ (see Fig. 13). This implies that the total volume of all layers created in a $BL_k$-box is greater than $\ \frac{3}{10}\cdot (\frac{1}{2})^{k-1}$. This volume divided by the volume $2^{-k}$ of a $BL_k$-box gives the desired ratio.

Finally we estimate the total volume of packed m-cubes.

Let $\ t \in \{ 4,5,\dots , q(m-1)\}$. Among all (t, q(t))-layers created in BL-boxes, the last (t, q(t))-layer is called not-full and the remaining layers are full. Since there is at most one not-full layer of each size, by the choice of $n_k$ we deduce that the sum of volumes of not-full layers is smaller than

$$\begin{aligned}&\Bigl ( \frac{1}{5} + \frac{1}{7} + \frac{1}{n_4} \Bigr )\cdot 2^{-3} + \Bigl ( \frac{1}{11}+ \cdots + \frac{1}{n_5} \Bigr )\cdot 2^{-4} + \cdots \\&\qquad +\Bigl ( \frac{1}{n_{q(m-1)-1}+1} + \cdots + \frac{1}{n_{q(m-1)}} \Bigr )\cdot 2^{-q(m-1)} \\&\quad<\frac{1}{2}\cdot 2^{-3}+\frac{1}{2}\cdot 2^{-4}+ \cdots + \frac{1}{2}\cdot 2^{-q(m-1)} < \frac{1}{8}. \end{aligned}$$

The total volume of basic BL-boxes is greater than

$$\begin{aligned} v_{bl} - (2^{-4}+2^{-5}+\cdots +2^{-q(m-1)}) > v_{bl}-1/8. \end{aligned}$$

The total volume of created layers is greater than $\ \frac{3}{5}(v_{bl}-1/8)$. The total volume of full layers is greater than $\ \frac{3}{5}(v_{bl}-1/8) - \frac{1}{8}$. Since at least 64/125 of any (t, q(t))-layer is occupied by cubes of type (t, 0), by Lemma 8 we deduce that the total volume of packed m-cubes is greater than

$$\begin{aligned} \Bigl ( \frac{3}{5}(v_{bl}-1/8) - \frac{1}{8}\Bigr ) \cdot \frac{64}{125} = \frac{192}{625}v_{bl}-\frac{64}{625}. \end{aligned}$$

$\square$

Lemma 12

If $\ d \ge 5$, then the total volume of m -cubes packed into each closed bin ${\mathscr {B}}$ is greater than 1/9.

Proof

By Lemma 6 we get that

$$\begin{aligned} v_{bc}+v_{bl} >1-2^{-3}=7/8, \end{aligned}$$

where $v_{bl}$ denotes the sum of the volumes of all open BL-boxes and $v_{bc}$ denotes the sum of the volumes of all open BC-boxes (there is no empty space in ${\mathscr {B}}$ to open a $B_k$-box for some $\ k \ge 3$).

If $\ v_{bc} > \frac{13}{24}$, then by Lemma 10 we obtain that the total volume of packed m-cubes is greater than

$$\begin{aligned} \frac{8}{27} \Bigl ( \frac{13}{24} -\frac{1}{8}-\frac{1}{32}\Bigr ) >\frac{1}{9}. \end{aligned}$$

If $\ v_{bc} \le \frac{13}{24}$, by Lemmas 10 and 11 we deduce that the total volume of packed items is greater than

$$\begin{aligned}&\frac{8}{27} \bigl ( v_{bc} -\frac{1}{8}-\frac{1}{32}\bigr ) + \frac{192}{625}v_{bl}-\frac{64}{625}> \frac{8}{27} v_{bc} -\frac{5}{108}+ \frac{192}{625}\Bigl ( \frac{7}{8} - v_{bc}\Bigr )-\frac{64}{625}\\&\quad = -\frac{184}{16875}v_{bc} - \frac{5}{108} + \frac{104}{625} \ge -\frac{184}{16875} \cdot \frac{13}{24} - \frac{5}{108} + \frac{104}{625}> \frac{1}{9}. \end{aligned}$$

$\square$

3.8 Algorithm har(d)

Let $\ 10\le m \le 2^{d-1}$ be an even number. Moreover let $\ C_1, C_2, \dots$ be a sequence of items (hypercubes). For each $C_i$ from the sequence let $K^+_i$ be the smallest m-cube containing $C_i$. For example, if $C_i$ is a hypercube of the edge length greater than 1/2, then $K_i^+$ is the unit hypercube (a cube of type (1, 0)).

An m-cube $K_i^+$ is big, provided its edge length is greater than 1/m; in that case we say also that the item $C_i$ is big. Clearly, any big cube $K_i^+$ is a cube of type (t, 0) for $\ t \in \{ 1,2, \dots , m-1 \}$; we say also that in that case $C_i$ is a big t-item.

An m-cube $K_j^+$ is small, provided its edge length is not greater than 1/m; in that case we say also that the item $C_j$ is small.

For example: big 4-cubes have the edge lengths from the set $\ \{ 1, 1/2, 1/3\}$ while small 4-cubes have the edge lengths from the set $\ \{ 1/4, 1/6, 1/8, 1/12, 1/16, \dots \}.$ Moreover, for $m=4$ each item $C_i$ of the edge length greater than 1/4 is big (the smallest 4-cube containing $C_i$ has the edge length from the set $\ \{1,1/2,1/3\}$) while each item of the edge length not greater than 1/4 is small. Another example: big 6-cubes have the edge lengths from the set $\{ 1, 1/2, 1/3, 1/4, 1/5\}$ while small 6-cubes have the edge lengths from the set

$$\begin{aligned} \{ 1/6, 1/8, 1/10, 1/12, 1/16, 1/20, 1/24, 1/32, 1/40, \dots \}. \end{aligned}$$

For $m=6$ each item $C_i$ of the edge length greater than 1/6 is big.

Lemma 13

If $C_i$ is a small item, then $\ |C_i|/|K_i^+|> \bigl ( m /(m+2)\bigr )^d$.

Proof

Big m-cubes have the edge lengths from the set

$$\begin{aligned} \left\{ 1, \frac{1}{2}, \dots , \frac{1}{m/2}, \frac{1}{m/2+1}, \frac{1}{m/2+2},\dots , \frac{1}{m-1} \right\} . \end{aligned}$$

Small m-cubes have the edge lengths of the form

$$\begin{aligned}&\frac{1}{m} = \frac{1}{2\cdot \frac{m}{2}},\ \frac{1}{m+2}=\frac{1}{2\bigl ( \frac{m}{2} +1\bigr )}, \ \frac{1}{m+4}=\frac{1}{2\bigl ( \frac{m}{2} +2\bigr )}, \ \dots ,\\&\frac{1}{2m},\ \frac{1}{2m+4}= \frac{1}{4 \bigl ( \frac{m}{2}+1 \bigr )},\ \frac{1}{2m+6},\ \dots \end{aligned}$$

Since $K^+_i$ is the smallest small m-cube containing $C_i$, it follows that

$$\begin{aligned}\frac{|C_i|}{|K_i^+|}> \frac{( m+2)^{-d}}{m^{-d}} = \Bigl ( \frac{m }{m+2}\Bigr )^d. \end{aligned}$$

$\square$

Algorithm har(d) for packing of $C_i$ into ${{\mathscr {B}}}$.

• First find the smallest m-cube $K_i^+$ containing $C_i$. Then pack $K_i^+$ together with $\ C_i \subset K_i^+$ by the algorithm $har^+(d)$.

3.9 Competitive Ratio $\ O(d/\log D)$

Let $\ 10\le m \le 2^{d-1}$ be an even number.

Lemma 14

If $\ d\ge 5$, then the asymptotic competitive ratio of the har(d) algorithm is not greater than $\ 9 \Bigl ( \bigl ( \frac{m+2}{m}\bigr )^d +m -1 \Bigr )$.

Proof

Let S be a sequence of items of the volume v, let $\lambda _j$ denote the number of big j-items in S and let $\beta$ be the number of bins used to pack items from S according to the har(d) algorithm. In particular, $\lambda _1$ is equal to the number of hypercubes of the edge lengths greater than 1/2 (big 1-items) and $\lambda _2$ denotes the number of hypercubes of the edge lengths greater than 1/3 but not greater than 1/2.

Two items of the edge lengths greater than 1/2 cannot be packed into one bin. This implies that $\ OPT(S) \ge \lambda _1$. Moreover, $2^d+1$ items of the edge lengths greater than 1/3 cannot be packed into one bin also. For example, if $\lambda _2=2^d+1$, then $\ OPT(S) > 1$; if $\lambda _2=2\cdot 2^d+1$, then $\ OPT(S) > 2$; Thus $\ OPT(S) \ge \lambda _2 / 2^d$. Finally, $\ j^d+1$ items of the edge lengths greater than $1/(j+1)$ cannot be packed into one bin. Consequently, $\ OPT(S) \ge \lambda _j\cdot j^{-d}$ for $\ j\ge 1$.

Let

$$\begin{aligned} \mu =\max (v, \lambda _1, \lambda _2 \cdot 2^{-d}, \lambda _3 \cdot 3^{-d}, \dots , \lambda _{m-1} \cdot (m-1)^{-d}). \end{aligned}$$

Since $\ OPT(S) \ge v$ as well as $\ OPT(S) \ge \lambda _j\cdot j^{-d}$ for $\ j\ge 1$, it follows that

$$\begin{aligned} OPT(S) \ge \mu . \end{aligned}$$

If $\ \beta -2\lambda _1-1\le 0$, then

$$\begin{aligned} \frac{\beta }{OPT(S)} \le \frac{2 \lambda _1+1}{\mu } \le \frac{2 \mu +1}{\mu } = 2 +\frac{1}{\mu }. \end{aligned}$$

Consider the case when $\ \beta -2\lambda _1-1> 0$.

Each big t-item was packed into a big cube of type (t, 0). The total volume of big cubes of type (t, 0) containing a big t-item equals $\ \lambda _t \cdot t^{-d}$ while the total volume of big t-items is greater than $\ \lambda _t \cdot (t+1)^{-d}$.

The last bin from among $\beta$ bins can be almost empty. Only one bin (from among $\beta$ bins) can be almost empty. By the description of the har(d) algorithm (Rule 1) we deduce that in two consecutive bins only one small item and only one 1-big item can be packed. Consequently, in $2\lambda _1$ bins the average occupation can be close to $2^{-d-1}$. We will omit the volume of items packed into these $2\lambda _1+1$ bins in our computations.

By Lemma 12 we know that the sum of the volumes of packed m-cubes is greater than

$$\begin{aligned} \frac{1}{9}(\beta -2\lambda _1-1). \end{aligned}$$

From among these m-cubes there are big cubes as well as small cubes. The total volume of big m-cubes of edges not greater than 1/2 is equal to

$$\begin{aligned} v_C = \lambda _2 \cdot 2^{-d} + \lambda _3 \cdot 3^{-d} + \cdots + \lambda _{m-1} \cdot (m-1)^{-d}. \end{aligned}$$

If $\ \beta \le 9\lambda _1 +9v_C+1$, then

$$\begin{aligned} \frac{\beta }{OPT(S)} &\le \frac{9\lambda _1+9v_C+1}{\mu } \\& = \frac{9 \mu +9\bigl (\lambda _2 \cdot 2^{-d} + \lambda _3 \cdot 3^{-d} + \cdots + \lambda _{m-1} \cdot (m-1)^{-d}\bigr )+1}{\mu }\\ &\le \frac{9\mu +9(\mu +\cdots +\mu )+1}{\mu } = 9(m-1)+\frac{1}{\mu }. \end{aligned}$$

Consider the case when $\ \beta > 9\lambda _1 +9v_C+1$.

The total volume of big items is greater than

$$\begin{aligned} v_B= \ \lambda _1\cdot 2^{-d} +\lambda _2 \cdot 3^{-d} + \cdots + \lambda _{m-1} \cdot m^{-d}. \end{aligned}$$

This value is relatively small compared to $\ 2\lambda _1+v_C$ and we will omit it in our computations.

The total volume of small m-cubes is greater than

$$\begin{aligned} \ \frac{1}{9}(\beta -2\lambda _1-1) - v_C. \end{aligned}$$

By Lemma 13 we deduce that the total volume of small items is greater than

$$\begin{aligned} \Bigl ( \frac{m}{m+2} \Bigr )^d \cdot \Bigl ( \frac{1}{9}(\beta -2\lambda _1-1) - v_C \Bigr ). \end{aligned}$$

Consequently, the total volume v of packed items is greater than

$$\begin{aligned} v> & v_B+\Bigl ( \frac{m}{m+2} \Bigr )^d \cdot \Bigl ( \frac{1}{9}(\beta -2\lambda _1-1) - v_C \Bigr ) \\\ge & \Bigl ( \frac{m}{m+2} \Bigr )^d \cdot \Bigl ( \frac{1}{9}(\beta -2\lambda _1-1) - \lambda _2 \cdot 2^{-d} - \lambda _3 \cdot 3^{-d} - \cdots - \lambda _{m-1} \cdot (m-1)^{-d} \Bigr )\\\ge & \Bigl ( \frac{m}{m+2} \Bigr )^d \cdot \bigl (\frac{1}{9} \cdot (\beta -1) - \lambda _1 -\lambda _2 \cdot 2^{-d} - \lambda _3 \cdot 3^{-d} - \cdots - \lambda _{m-1} \cdot (m-1)^{-d} \bigr ), \end{aligned}$$

i.e.,

$$\begin{aligned} \beta< & 9 \cdot \Bigl ( \Bigl ( \frac{m+2}{m} \Bigr )^d v + \lambda _1 + \lambda _2 \cdot 2^{-d} + \lambda _3 \cdot {3}^{-d} + \cdots + \lambda _{m-1} \cdot (m-1)^{-d} \Bigr ) +1. \end{aligned}$$

As a consequence,

$$\begin{aligned} \frac{\beta }{OPT(S)} \le \frac{\beta }{\mu } \le \frac{9 \cdot \Bigl ( \bigl ( \frac{m+2}{m} \bigr )^d \mu +\mu + \cdots + \mu \Bigr ) +1}{\mu } = 9 \cdot \Bigl ( \bigl ( \frac{m+2}{m}\bigr )^d +m -1\Bigr )+\frac{1}{\mu }. \end{aligned}$$

This means that the asymptotic competitive ratio of the har(d) algorithm is not greater than $\ 9 \Bigl ( \bigl ( \frac{m+2}{m}\bigr )^d +m -1\Bigr )$. $\square$

Theorem 2

If $\ d\ge 5$, then the asymptotic competitive ratio of the har(d) algorithm is not greater than $\ 9 \bigl (\sqrt{d} +\frac{4d}{\log d} +1\bigr )$.

Proof

Similarly as in the proof of Theorem 2.2 in [8] take as m the even number such that

$$\begin{aligned} 4d/\log d <m \le 2+ 4d/\log d. \end{aligned}$$

By Lemma 14 we get that the asymptotic competitive ratio of the har(d) algorithm is not greater than

$$\begin{aligned} 9 \Bigl ( \bigl ( 1+ \frac{2}{m}\bigr )^d +m -1\Bigr ) &< 9 \Bigl ( \bigl ( 1+ \frac{2\log d}{4d}\bigr )^d +2+\frac{4d}{\log d} -1 \Bigr ) \\ &< 9 \Bigl (e^{(\log d) /2} +\frac{4d}{\log d}+1 \Bigr ) \\& = 9 \Bigl ({(e^{\log d}})^{1/2} +\frac{4d}{\log d} +1 \Bigr ) \\& = 9\bigl (\sqrt{d} +\frac{4d}{\log d} +1 \bigr ). \end{aligned}$$

$\square$

Corollary 1

The asymptotic competitive ratio of the har(d) algorithm is $\ O(d/ \log d)$.

Appendix 1: The Reverse 2-Cubes Versus 3-Cubes Sequence and Its Increments

Let $n\ge 0$, we define y(n) to be an increasing sequence made from $\ y_d(2^d-n)$, i.e.,

$$\begin{aligned} y(0)=0,\quad y(1)=1, \end{aligned}$$

and for arbitrary $n\ge 2$ let d be the smallest integer greater than or equal to $\log _2n$, then

$$\begin{aligned} y(n)=y_d(2^d-n). \end{aligned}$$

Let x(n) be the sequence of increments of y(n)

$$\begin{aligned} x(n)=y(n)-y(n-1) \quad \text{ for } \,n\ge 1. \end{aligned}$$

We will prove that the sequence x(n) is a Gould’s sequence, see [26].

Fact 1

The sequence x(n) starts with 1 and can be defined inductively:

When the first $2^d$ elements of the sequence x(n) are given, the next $2^d$ elements are defined

$$\begin{aligned} x(2^d+l)=2\cdot x(l),\quad {\text{for}}\,\, 0< l\le 2^d. \end{aligned}$$

Proof

Let $d\ge 0$ and $0< l\le 2^d$.

$$\begin{aligned} x(2^d+l)& = y(2^d+l)-y(2^d+l-1)\\& = y_{d+1}(2^{d+1}-(2^d+l))-y_{d+1}(2^{d+1}-(2^d+l-1))\\& = y_{d+1}(2^d-l)-y_{d+1}(2^{d}-(l-1))\\& = 3^d+2\cdot y_d(2^d-l)-3^d-2\cdot y_d(2^d-(l-1))\quad \text{ by } \text{(1) }\\& = 2\cdot (y_d(2^d-l)-y_d(2^d-(l-1)))=2\cdot (y(l)-y(l-1))\\& = 2\cdot x(l). \end{aligned}$$

$\square$

The immediate consequence of the fact is that the sequence x(n) can be created in the following way

$1\quad$ the first element is 1, multiply by 2 and combine

$1\; |\;2\quad$ multiply it by 2 and combine

$1\; 2\; |\;2\; 4\quad$ multiply by 2 and combine

$1\; 2\; 2\; 4\; |\;2\; 4\; 4\; 8\quad$ multiply by 2 and combine

$1\; 2\; 2\; 4\; 2\; 4\; 4\; 8\; |\;2\; 4\; 4\; 8\; 4\; 8\; 8\; 16 \quad \ldots$

The above pattern can also be recognised in Pascal’s triangle as in Fig. 14, see also [26].

Fig. 14

The parity of elements of Pascal’s triangle: a circle indicates an odd number, a filled circle—an even number