Packing rectangles into a large square

Abstract

If $c>5$ and if $x$ is sufficiently large, then any collection of rectangles of sides of length not greater than $1$ with total area smaller than $\ x^2-cx^{5/6}\$ can be packed into a square of side length $x$.

1 Introduction

First publications related to packing of rectangles or squares appeared over fifty years ago. In 1957 Kosiński [6] proved, among others, that any sequence of rectangles of total area $V$ and with sides of length not greater than $D$ can be packed into a rectangle of side lengths $3D$ and $\ (V+D^2)/D$. This result was improved in [4, 7, 8]. Other problems related to this subject were outlined in the sixties of the last century by L. Moser [9]. He asked, for example, “Can every set of rectangles of total area 1 and maximal side 1 be accommodated in a square of area 2?” (the answer is positive [5]) or “What is the smallest number A such that any set of squares of total area 1 can be packed into some rectangle of area A?” (some bounds are given in [3, 10, 11]). The question of packing of equal squares into a square as small as possible was posed in [2].

Let $I_x$ be a square of side length $x$. We say that a collection $\ R_1, R_2, \dots \$ of rectangles can be packed into $I_x$, if it is possible to apply translations and rotations to the sets $R_i$ so that the resulting translated and rotated rectangles are contained in $I_x$ and have mutually disjoint interiors. Denote by $s(x)$ the greatest number such that any collection of rectangles of sides of length not greater than $1$ with total area smaller than $s(x)$ can be packed into $I_x$.

Groemer [4] proved that $s(x) \ge (x-1)^2\$ provided $x\ge 3$. By Remark 3 of [5] we know that $\ s(x) \ge x^2-2x+2 \$ for $\ x\ge 2$. The aim of this note is to show that $\ s(x) \ge x^2 - O \bigl ( x^{5/6} \bigr )$. It is an open question whether the exponent $5/6$ may be lessened in the above-presented estimation.

If all rectangles are unit squares, then $\ s_{unit}(x) \ge x^2 -O \bigl ( x^{(3+\sqrt{2} )/7} \log x \bigr )$ (see [1]). Also in this case we do not know whether the exponent $(3+\sqrt{2} )/7$ may be lessened. On the other hand, by [12] we know that $s_{unit}(x)$ is smaller than $\ x^2 - 10^{-100} \sqrt{x|x- \lfloor x+1/2 \rfloor | } \$ provided $\ x(x-\lfloor x \rfloor ) > 1/6$.

2 Preliminaries

Let $\mathcal R$ be a finite collection of rectangles $\ R_1, R_2, \dots , R_z \$ of sides of length not greater than $1$. Denote by $w_i$ the width and by $h_i$ the height of $R_i$. Furthermore, assume that $\ w_i \le h_i\$ for any $\ i=1,\dots , z\$ and that $\ h_1\ge h_2 \ge \dots \ge h_z$.

Let $S$ be a rectangle of width $\ a \ge 1 \$ and height $\ d\ge 1$. Denote by $p$ a vertex of $S$. Moreover, let $\ S_1, \dots , S_r\$ be a collection of rectangles $S_i$ of width $\ v_i <a\$ and height smaller than $d$ such that $\ p\in S_i\$ and $\ S_i \subset S\$ for $\ i =1,\dots , r$. Then $\ S \setminus \bigcup _{i=1}^r S_i\$ is called a $\sigma -polygon$ of base $a$, top $\ a-\max (v_1, \dots , v_r)\$ and height $d$ (see Fig. 1). The rectangle $S$ is also called a $\sigma$-polygon.

Fig. 1

$\sigma$-polygon

Lemma 2.1

Let $\ a\ge 1$, $\ d\ge 1\$ and assume that the total area of rectangles in $\mathcal R$ is not smaller than $\ (a+1)(d+1)$. There exist integers $\ j_1< \dots < j_k\$ such that the following conditions are fulfilled:

• $a\le b_{i} < a +1$, where $\ b_{i} = w_{j_{i-1}} + \dots + w_{j_i-1}\$   for $\ i=1, \dots , k\$ $(j_0=1)$;

• $d \le h_1+h_{j_1}+\dots +h_{j_{k-1}} <d+1$;

• the rectangles $\ R_1, \dots , R_{j_k-1}\$ can be packed into the union $\ \bigcup _{i=1}^k L_i\$ of rectangles $($with mutually disjoint interiors$)$ $L_i$ of sides of length $b_i$ and $h_{j_{i-1}}$;

• the area of the uncovered part of $\ \bigcup _{i=1}^k L_i\$ is smaller than $\ a+1$.

Proof

Denote by $j_1$ the smallest integer such that $\ w_1+w_2+\dots +w_{j_1-1} \ge a$. Moreover, denote by $j_2$ the smallest integer satisfying $\ w_{j_{1}}+\dots +w_{j_2-1} \ge a\$ and so on. Let $k$ be the smallest integer such that $\ h_1+h_{j_1}+\dots +h_{j_{k-1}} \ge d$. Clearly, the rectangles $\ R_1, \dots , R_{j_k-1}\$ can be packed into the union $\ \bigcup _{i=1}^k L_i\$ of rectangles $($with mutually disjoint interiors$)$ $L_i$ of sides of length $b_i$ and $h_{j_{i-1}}$ (see Fig. 2). The area of the uncovered part in each $L_i$ does not exceed $\ b_i(h_{j_{i-1}}-h_{j_i-1} )$. Consequently, the area of the uncovered part of $\ \bigcup _{i=1}^k L_i\$ (the waste in this packing) does not exceed

$$\begin{aligned} \omega _0&= (h_1-h_{j_1-1})b_1 +(h_{j_1}-h_{j_2-1})b_2+\dots +(h_{j_{k-1}}-h_{j_k-1})b_k \\&\le (h_1-h_{j_1} +h_{j_1}-h_{j_2}+ \dots +h_{j_{k-1}}-h_{j_k}) \cdot \max (b_{1}, \dots , b_{k}) \\&= (h_1-h_{j_k})\cdot \max (b_{1}, \dots , b_{k}) \\&< \max (b_{1}, \dots , b_{k}) \\&< a+1. \end{aligned}$$

$\square$

By the proof of Lemma 2.1 we deduce the following two results (see Figs. 2, 3, 4).

Fig. 2

$L_1, \ L_2, \dots$

Fig. 3

$A_i$

Fig. 4

$L_m$

Lemma 2.2

Let $\ a\ge 1$, $\ d\ge 1\$ and let $n$ be a positive integer. Assume that the total area of rectangles in $\mathcal R$ is not smaller than $\ n(a+1)(d+1)$. There is an integer $k$ and there are $n$ mutually disjoint $\sigma$-polygons $A_i$ (for $\ i = 1, \dots , n$) of base $a_i$, top $\ a_i - \uplambda _i\$ and height $d_i$, where $a \le a_i < a+1$, $\ d \le d_i < d+1\$ and $\ \sum _{i=1}^n \uplambda _i <1$ such that the following conditions are fulfilled:

• the rectangles $\ R_1, \dots , R_k\$ can be packed into $\ \bigcup _{i=1}^n A_i;$

• the area of the uncovered part of $\ \bigcup _{i=1}^n A_i\$ is smaller than $a+1$.

Lemma 2.3

Let $\ a\ge 1$. There is an integer $m$ and there are $m$ rectangles $L_i$ $($with mutually disjoint interiors$)$ of height not greater than $1$ and width $b_i$, where $\ a\le b_{i} < a +1$, such that:

• the rectangles from $\mathcal R$ can be packed into $\ \bigcup _{i=1}^{m} L_i$;

• the area of the uncovered part of $\ \bigcup _{i=1}^{m} L_i\$ is smaller than $\ 2a+1$.

In the following lemma we will describe how to efficiently pack rectangles $L_i$.

Lemma 2.4

Let $B$ be a $\sigma$-polygon of base $b$, top $\ b-\uplambda \$ and height $h$, where $\ h \ge b\ge 27\$ and $\ 0\le \uplambda <1$. Furthermore, let $L_i\,(\hbox {for } i=1, \dots , m )$ be a rectangle of width $b_i$ and height $t_i$, where

$$\begin{aligned} b\le b_1 \le \dots \le b_{m} <b+1 \end{aligned}$$

and where $\ t_i \le 1\$ for $\ i=1, \dots , m$. Put $\ \mu = b_m - b \$ and

$$\begin{aligned} v(b,h,\uplambda , \mu )= (b^{2}+h+1)b^{-1/2} \bigl [ (2\uplambda +2\mu )^{1/2}+3b^{-1/4}\bigr ].\end{aligned}$$

If $b$ is sufficiently large and if

$$\begin{aligned}\sum _{i=1}^{m} area(L_i) \le area(B) - v(b,h,\uplambda , \mu ),\end{aligned}$$

then $\ L_1, \dots , L_{m}\$ can be packed into $B$.

Proof

Assume that $\ b\ge 27\$ and that the sum of the areas of rectangles $\ L_1, \dots , L_{m}\$ is not greater than $\ area(B) - v(b,h,\uplambda , \mu )$.

Put

$$\begin{aligned} \vartheta _i = \arctan \frac{t_i}{b_i} + \arccos \frac{b}{\sqrt{b_i^2 + t_i^2 } } \end{aligned}$$

(see Fig. 5). Without loss of generality we can assume that $\ \vartheta _1 \le \dots \le \vartheta _{m}$.

Fig. 5

$\vartheta _i$

We pack the rectangles $\ L_1, L_2, \dots \$ into $B$ as in Fig. 6. Contrary to the statement suppose that the rectangles cannot be packed. We show that this leads to a contradiction. Let $L_{\kappa }$ be the first rectangle which cannot be packed into $B$.

Fig. 6

$\ L_i \subset B$

By $\ b_{\kappa } \le b+\mu < b+1\$ and $\ u>b-\uplambda -1 >b-2\$ in Fig. 6 we have

$$\begin{aligned} \theta \le \tan \theta = \frac{\sqrt{b_{\kappa }^2 -u^2}}{u} < \frac{\sqrt{(b+1)^2 -(b-2)^2}}{b-2} < \frac{\sqrt{6b}}{b-2}< \sqrt{\frac{7}{b}}. \end{aligned}$$

We need a more precise estimation. Since $\ t_{\kappa } \le 1\$ and $\ \sin \theta \le \theta <\sqrt{7/b}$, it follows that $\ u = b - \uplambda - t_{\kappa } \sin \theta > b - \uplambda - \sqrt{7/b}$. By $\ b_{\kappa } \le b +\mu \$ we obtain

$$\begin{aligned} \tan \theta&= \frac{\sqrt{b_{\kappa }^2 -u^2}}{u} \\&< \frac{\sqrt{ (b+\mu )^2-(b-\uplambda - \sqrt{7/b} )^2}}{b-\uplambda -\sqrt{7/b}} \\&= \sqrt{\frac{ 2b(\uplambda +\mu )+ \mu ^2-\uplambda ^2-7/b+2(b-\uplambda )\sqrt{7/b}}{b^2-2\uplambda b+\uplambda ^2+7/b-2(b-\uplambda )\sqrt{7/b}}} \\&< \sqrt{\frac{2b( \uplambda + \mu ) +1 +6\sqrt{b}}{b^2-2\uplambda b -6\sqrt{b}}}. \end{aligned}$$

It is easy to check that

$$\begin{aligned} \frac{2b( \uplambda + \mu ) +1 +6\sqrt{b}}{b^2-2\uplambda b -6\sqrt{b}} < 2(\uplambda + \mu )b^{-1} + 7b^{-3/2},\end{aligned}$$

for sufficiently large $b$.

Since $\ \sqrt{\alpha _1+\alpha _2} \le \sqrt{\alpha _1}+\sqrt{\alpha _2}\$ for non-negative values $\alpha _1$ and $\alpha _2$, it follows that

$$\begin{aligned}\theta \le \tan \theta < f(b,\uplambda , \mu ),\end{aligned}$$

where

$$\begin{aligned} f(b, \uplambda , \mu )=(2\uplambda + 2\mu )^{1/2}b^{-1/2} + 7^{1/2}b^{-3/4}.\end{aligned}$$

The uncovered dark shaded part on the left side of $B$ in Fig. 7 consists of a number of triangles. The total length of the left sides of the triangles is smaller than $h+1$. The height of each such triangle (the height parallel to the bottom of $B$) is not greater than $\sin \theta$. Consequently, the uncovered dark shaded part on the left side of $B$ in Fig. 7 is of the area

$$\begin{aligned} \omega _l < \frac{1}{2} (h+1) \sin \theta < \frac{1}{2}(h+1) \theta < \frac{1}{2}(h+1) f(b,\uplambda , \mu ). \end{aligned}$$

Similarly we estimate the area $\omega _r$ of the uncovered dark shaded part on the right side of $B$:

$$\begin{aligned} \omega _r \le \omega _l < \frac{1}{2}(h+1) f(b,\uplambda , \mu ).\end{aligned}$$

Since the distance between $p$ and $q$ is equal to $\uplambda$ and the height of each $L_i$ is not greater than $1$, it follows that the non-shaded uncovered part on the right side of $B$ in Fig. 7 is of the area

$$\begin{aligned}\omega _u \le \uplambda < 1.\end{aligned}$$

Denote by $\omega ^+_s$ the area of a right triangle of legs of length $b$ and $b \tan \theta$. Moreover, denote by $\omega _s$ the area of the light shaded uncovered part of $B$ in Fig. 7. By $\ \eta _1 + \eta _2 + \dots + \eta _{\kappa } = \theta \$ (see Fig. 6) we deduce that

$$\begin{aligned} \omega _s \le \omega ^+_s = \frac{1}{2} b^2 \tan \theta < \frac{1}{2} b^2 f(b,\uplambda , \mu ). \end{aligned}$$

The uncovered non-shaded part on the top of $B$ in Fig. 7 is of the area

$$\begin{aligned} \omega _t < \frac{1}{2}b^2 \tan \theta + area(L_{\kappa }) < \frac{1}{2}b^2 f(b,\uplambda , \mu ) +b+1. \end{aligned}$$

Consequently, the area of the uncovered part of $B$ does not exceed

$$\begin{aligned} \omega&= \omega _l+\omega _r+\omega _u+\omega _s+\omega _t \\&< ( b^2+h+1)f(b,\uplambda , \mu ) +b+2\\&= ( b^2+h+1) \bigl [(2\uplambda + 2\mu )^{1/2}b^{-1/2} + 7^{1/2}b^{-3/4}\bigr ] +b+2 \\&< ( b^2+h+1) \bigl [(2\uplambda + 2\mu )^{1/2}b^{-1/2} + 3b^{-3/4}\bigr ], \end{aligned}$$

for sufficiently large $b$. This implies that

$$\begin{aligned} \sum _{i=1}^{\kappa -1} area(L_i) \ge area(B) - \omega > area(B) - v(b,h,\uplambda , \mu ), \end{aligned}$$

which is a contradiction. $\square$

Fig. 7

Wasted area

3 Packing into a large square

In the main packing method $I_x$ will be partitioned into a number of $\sigma$-polygons. Next, rectangles from $\mathcal R$ will be packed into adequate $\sigma$-polygons.

Theorem 3.1

Let $\ \epsilon >0$. Any collection of rectangles of sides of length not greater than $1$ with total area smaller than $\ x^2 -(5+\epsilon ) x^{5/6}$ can be packed into $I_x$, for sufficiently large $x$.

Proof

Assume that $\ \epsilon > 0 \$ and that $\ x > (5+\epsilon )^{6/7}$. Consider a collection $\mathcal C$ of rectangles $P_i$ of sides of length not greater than $1$ with total area smaller than $\ x^2 - (5+\epsilon ) x^{5/6}$. If $\mathcal C$ is finite, then put $\ {\mathcal R}={\mathcal C}\$ and denote by $z$ the number of rectangles in $\mathcal R$. Otherwise, we can assume that $\ area(P_1)\ge area(P_2) \ge \dots$. There is an integer $z$ such that $\ \sum _{i=z}^{\infty } area(P_i) < \frac{1}{2}$. By [5] we know that rectangles $\ R_z, R_{z+1}, \dots \$ can be packed into $I_1$. Let $\mathcal R$ be a collection of rectangles $R_i$, where $\ R_1 = I_1\$ and $\ R_i = P_{i-1}\$ for $\ i=2, \dots , z$.

We show that rectangles from $\mathcal R$ can be packed into $I_x$ provided $x$ is sufficiently large. Clearly,

$$\begin{aligned} \sum _{i=1}^z area(R_i) < x^2 - (5+\epsilon )x^{5/6}+1.\end{aligned}$$

We can assume that the width $w_i$ of $R_i$ is not greater than its height $h_i$ for $\ i=1,\dots , z\$ and that $\ h_1\ge \dots \ge h_z$. Put

$$\begin{aligned} n&= \lfloor x^{1/6} \rfloor ,\\ a&= x/n - x^{1/2}\end{aligned}$$

and

$$\begin{aligned}d=x-x^{1/2}.\end{aligned}$$

It is easy to verify that $\ n(a+1)(d+1) < (x-1)^2.$ If $\ \rho < n(a+1)(d+1)$, then, by [4], all rectangles from $\mathcal R$ can be packed into $I_x$. Otherwise, by Lemma 2.2 we deduce that there is an integer $k_1$ and there are $n$ mutually disjoint $\sigma$-polygons $A_i$ of base $a_i$, top $\ a_i - \uplambda _i\$ and height $d_i$, where

$$\begin{aligned} a\le a_i < a+1, \ \ \ d \le d_i < d+1 \end{aligned}$$

(for $\ i=1, \dots , n$) and where $\ \sum _{i=1}^n \uplambda _i <1\$ such that $\ R_1, \dots ,R_{k_1}\$ can be packed into $\ \bigcup _{i=1}^n A_i\$ and that the waste in this packing (i.e., the area of the uncovered part of $\ \bigcup _{i=1}^n A_i$) is at most

$$\begin{aligned} \omega _1= a+1 = x/n-x^{1/2}+1.\end{aligned}$$

Clearly,

$$\begin{aligned}\sum _{i=1}^{k_1} area(R_i) \ge \sum _{i=1}^n area (A_i) - \omega _1.\end{aligned}$$

We lose no generality in assuming that $\ d_1 \ge d_2 \ge \dots \ge d_n$.

$I_x$ will be divided into: $n$ polygons $A_i$ and $n+1$ other $\sigma$-polygons. Then $\ R_1, \dots , R_{k_1}\$ will be packed into $\ \bigcup _{i=1}^n A_i.$ The remaining rectangles from $\mathcal R$ will be first packed into larger rectangles $L_i$ or $L_i'$. Next, $L_i$ and $L_i'$ will be packed into $\ I_x \setminus \bigcup _{i=1}^n A_i$.

We apply Lemma 2.3 for packing $\ R_{k_1+1}, \dots , R_z$. There is an integer $m$ and there are rectangles $L_i$ (for $\ i=1, \dots , m\$) of width $b_i$, where

$$\begin{aligned} x^{1/2} \le b_i < x^{1/2}+1 \end{aligned}$$

and height not greater than $1$ such that $\ R_{k_1+1}, \dots , R_{z} \$ can be packed into $\ \bigcup _{i=1}^m L_i\$ and that the waste in this packing is no more than

$$\begin{aligned} \omega _2 = 2x^{1/2}+1. \end{aligned}$$

There is no loss of generality in assuming that $\ b_1 \le b_2 \le \dots \le b_m.$

We divide $I_x$ into: $n$ rectangles $D_i$ (for $\ i=1,\dots , n$) of width $e_i$ and height $d_i$ and one $\sigma$-polygon $B_{n+1}$ of base $\ x-d_n\$ and height $x$ (as in Fig. 8). Now we will describe how to choose proper values $\ e_1, \dots , e_n$. This action depends on the width of some rectangles $L_i$.

Fig. 8

Partition of $I_x$

Put $\ e_1 = a_1-\uplambda _1+b_{1}$. Clearly, $D_1$ can be divided into the $\sigma$-polygon $A_1$ and a $\sigma$-polygon $B_1$ of base $b_{1}$, top $\ b_1-\uplambda _1\$ and height $d_1$ (see Fig. 9). Denote by $m_1$ the greatest integer such that

$$\begin{aligned} \sum _{i=1}^{m_1} area(L_i) \le area(B_1) -v(b_1,d_1,{\uplambda }_{i}, b_{m_1}-b_1). \end{aligned}$$

By Lemma 2.4 we know that $\ L_1, \dots , L_{m_1} \$ can be packed into $B_1$, for sufficiently large $x$. Obviously,

$$\begin{aligned}\sum _{i=1}^{m_1+1} area(L_i) > area(B_1) - v(b_1,d_1,{\uplambda }_{i}, b_{m_1+1}-b_1).\end{aligned}$$

Consequently,

$$\begin{aligned}\sum _{i=1}^{m_1} area(L_i) > area(B_1) - (x^{1/2}+1)- v(b_1,d_1,{\uplambda }_{i}, b_{m_1+1}-b_1).\end{aligned}$$

We proceed in a similar way for $\ i=2, \dots , n-1$. Put

$$\begin{aligned} e_i = a_i - {\uplambda }_{i} +b_{m_{i-1}+1}\end{aligned}$$

for $\ i=2, \dots , n-1$. Each $D_i$ is divided into the $\sigma$-polygon $A_i$ and a $\sigma$-polygon $B_i$ of base $b_{m_{i-1}+1}$, top $\ b_{m_{i-1}+1}-{\uplambda }_i \$ and height $d_i$. Denote by $m_i$ the greatest integer such that

$$\begin{aligned}\sum _{i=m_{i-1}+1}^{m_i} area(L_i)\le area(B_i) - v(b_{m_{i-1}+1},d_i,{\uplambda }_{i},b_{m_i}-b_{m_{i-1}+1}),\end{aligned}$$

By Lemma 2.4 we know that $\ L_{m_{i-1}+1}, \dots , L_{m_i} \$ can be packed into $B_i$ provided $x$ is sufficiently large. Moreover,

$$\begin{aligned} \sum _{i=m_{i-1}+1}^{m_i} area(L_i) > area(B_i) - (x^{1/2}+1)- v(b_{m_{i-1}+1},d_i,{\uplambda }_{i}, b_{m_i+1}-b_{m_{i-1}+1}).\end{aligned}$$

Clearly, if $\ m_i=m\$ for some integer $i$, then all rectangles from $\mathcal R$ were packed into $I_x$.

Fig. 9

Partition of $D_i$

Denote by $m_n$ the greatest integer such that

$$\begin{aligned}\sum _{i=m_{n-1}+1}^{m_n} area(L_i)\le area(B_{n+1}) -v(x-d_n,x,1,1).\end{aligned}$$

By Lemma 2.4 we know that $\ L_{m_{n-1}+1}, \dots , L_{m_n} \$ can be packed into $B_{n+1}$. Moreover,

$$\begin{aligned}\sum _{i=m_{n-1}+1}^{m_n} area(L_i) > area(B_{n+1}) - (x^{1/2}+1)- v(x-d_n,x,1,1).\end{aligned}$$

Finally, put

$$\begin{aligned} e_n = x -\sum _{i=1}^{n-1} e_i .\end{aligned}$$

The rectangle $D_n$ is divided into the $\sigma$-polygon $A_n$ and a $\sigma$-polygon $B'_n$ of height $d_n$ and base

$$\begin{aligned} b' = e_n - a_n + {\uplambda }_{n} = x - \sum _{i=1}^n a_n + \sum _{i=1}^n {\uplambda }_{i} - \sum _{i=1}^{n-1} b_{m_{i-1}+1} \end{aligned}$$

$(m_0 = 0)$. Since

$$\begin{aligned}&\sum _{i=1}^n {\uplambda }_{i} < 1,\\&x-nx^{1/2} = na \le \sum _{i=1}^n a_i < n(a+1)= x-nx^{1/2} +n\end{aligned}$$

and

$$\begin{aligned} (n-1)x^{1/2} \le \sum _{i=1}^{n-1} b_{m_{i-1}+1} < (n-1)(x^{1/2}+1), \end{aligned}$$

it follows that

$$\begin{aligned}x^{1/2} -2n + 1 \le b' < x^{1/2}+1.\end{aligned}$$

Denote by $R_{k_2}$ the last rectangle packed in $L_{m_{n}}$. The rectangles $\ R_1, \dots , R_{k_1}$ were packed into $\ A_1 \cup \dots \cup A_n$. The rectangles $\ R_{k_1+1}, \dots , R_{k_2}\$ were packed into $\ B_1 \cup \dots \cup B_{n-1} \cup B_{n+1}$. The remaining rectangles $\ R_{{k_2}+1}, \dots ,$ $R_z\$ will be packed into $B_n'$. Unfortunately, it is possible that $\ L_{m_n+1}, \dots , L_m\$ are too large to apply Lemma 2.4 and therefore we need to repack $\ R_{{k_2}+1}, \dots , R_z\$ into other larger rectangles. We apply Lemma 2.3. There is an integer $l$ and there are rectangles $L'_i$ (for $\ i=1, \dots , l\$) of width $b'_i$, where

$$\begin{aligned}b' \le b'_i < b'+1\end{aligned}$$

and height not greater than $1$ such that $\ R_{k_2+1}, \dots , R_{z} \$ can be packed into $\ \bigcup _{i=1}^l L'_i\$ and that the waste in this packing is no more than

$$\begin{aligned} \omega _3 = 2b'+1 < 2x^{1/2}+3.\end{aligned}$$

It remains to check that

$$\begin{aligned} \sum _{i=1}^l area(L'_i) \le area(B_n') -v(b', d_n, {\uplambda }_{n},1), \end{aligned}$$

(*)

for sufficiently large $x$ (then, by Lemma 2.4, $\ L'_1, \dots , L'_l\$ and, consequently, the rectangles $\ R_{k_2+1}, \dots , R_{z} \$ can be packed into $B_n'$).

Put $\ \mu _1 = b_{m_1+1}-b_1 \$ and $\ \mu _i = b_{m_i+1}-b_{m_{i-1}+1}\$ for $\ i=2, \dots , n-1$. Obviously, $\ \sum _{i=1}^{n-1} \mu _i <1$. By

$$\begin{aligned} \sum _{i=1}^{k_1} area(R_i) \ge \sum _{i=1}^n area(A_i) - \omega _1,\\ \sum _{i=k_1+1}^{k_2} area(R_i) \ge \sum _{i=1}^{m_n} area(L_i) -\omega _2 \end{aligned}$$

and

$$\begin{aligned} \sum _{i=k_2+1}^{z} area(R_i) \ge \sum _{i=1}^l area(L'_i) - \omega _3, \end{aligned}$$

we have

$$\begin{aligned} \sum _{i=1}^{z} area(R_i)&\ge \sum _{i=1}^n area(A_i) - \omega _1 + \sum _{i=1}^{n-1} area(B_i) \\&- (n-1)(x^{1/2}+1)- \sum _{i=1}^{n-1} v(b_{m_{i-1}+1}, d_i, {\uplambda }_{i}, \mu _i) \\&+ area(B_{n+1}) -(x^{1/2}+1)- v(x-d_n,x,1,1) -\omega _2 \\&+ \sum _{i=1}^{l} area(L'_i) - \omega _3. \end{aligned}$$

Consequently,

$$\begin{aligned} x^2-(5+\epsilon ) x^{5/6} +1&> x^2 - area(B'_n) - \omega _1 -\omega _2 -\omega _3 -n(x^{1/2}+1)\\&- \sum _{i=1}^{n-1} v(b_{m_{i-1}+1}, d_i, {\uplambda }_{i}, \mu _i) - v(x-d_n,x,1,1) \\&+ \sum _{i=1}^{l} area(L'_i). \end{aligned}$$

To prove the inequality $(*)$ we show that

It is easy to check that

$$\begin{aligned} \omega _1 + \omega _2 + \omega _3 +1 +n(x^{1/2}+1) \le x/n + 3x^{1/2} +6 +n(x^{1/2}+1) < (1+ \epsilon /4 )x^{5/6}, \\ v(x-d_n, x, 1,1) \le v(x^{1/2}, x,1,1) = (2x+1)(2x^{-1/4} +3x^{-3/8})< \epsilon x^{5/6}/4\end{aligned}$$

and

$$\begin{aligned} v(b', d_n, {\uplambda }_{n}, 1)< \epsilon x^{5/6}/4, \end{aligned}$$

for sufficiently large $x$. This implies that

$$\begin{aligned} \zeta&< (1+3\epsilon /4) x^{5/6}+ \sum _{i=1}^{n-1} \frac{ b_{m_{i-1}+1}^{2}+x +1 }{ b_{m_{i-1}+1}^{1/2} } \cdot \bigl [ (2{\uplambda }_{i} + 2\mu _i)^{1/2}+3b^{-1/4}_{m_{i-1}+1} \bigr ] \\&< (1+3\epsilon /4) x^{5/6}+ \frac{ (x^{1/2}+1)^{2}+x +1}{ (x^{1/2})^{1/2} } \cdot \sum _{i=1}^{n-1} \bigl [ (2{\uplambda }_{i} + 2\mu _i)^{1/2}+3x^{-1/8} \bigr ]. \end{aligned}$$

The arithmetic mean of a list of non-negative real numbers is smaller than or equal to the quadratic mean of the same list. Therefore

$$\begin{aligned} \alpha _1^{1/2} + \dots +\alpha _{n-1}^{1/2} \le \bigl [(n-1)(\alpha _1 + \dots + \alpha _{n-1}) \bigr ]^{1/2} \end{aligned}$$

for non-negative numbers $\ \alpha _1, \dots , \alpha _{n-1}$. Since $\ \sum _{i=1}^{n-1} ({\uplambda }_{i}+\mu _i) < 2$, it follows that

$$\begin{aligned} \sum _{i=1}^{n-1} ( 2{\uplambda }_{i} + 2\mu _i)^{1/2} < 2(n-1)^{1/2}. \end{aligned}$$

Hence

$$\begin{aligned} \zeta&< (1+3\epsilon /4)x^{5/6} + \frac{2x+2x^{1/2} +2}{x^{1/4} } \cdot \bigl [ 2 (n-1)^{1/2} + 3(n-1)x^{-1/8} \bigr ] \\&< (1+3\epsilon /4)x^{5/6} + (2x^{3/4}+2x^{1/4} +2x^{-1/4} ) \cdot \bigl [ 2(x^{1/6})^{1/2} +3x^{1/6}x^{-1/8} \bigr ] \\&= (5+3\epsilon /4)x^{5/6}+6x^{19/24}+4x^{1/3} + 6x^{7/24}+4x^{-1/6}+6x^{-5/24}. \end{aligned}$$

Consequently,

$$\begin{aligned} \zeta < (5+\epsilon )x^{5/6}, \end{aligned}$$

for sufficiently large $x$. $\square$