A lower bound for online rectangle packing

Abstract

We slightly improve the known lower bound on the asymptotic competitive ratio for online bin packing of rectangles. We present a complete proof for the new lower bound, whose value is above 1.91.

Appendix A: Proofs of technical lemmas

1.1 Appendix A.1: Proof of Lemma 4.1

The first part holds since $\varepsilon < \frac{1}{7224}$, and the second part holds by definition.

Consider the third part. For $1 \le t \le k-2$, we have

$$\begin{aligned} \sum _{i=1}^{t} w_{1i}= & {} (1+\delta )\sum _{i=1}^t \frac{1}{5^{k-i-1}}= (1+\delta )\frac{1}{5^{k-t-1}}\sum _{i=1}^t \frac{1}{5^{t-i}}= (1+\delta )\frac{1}{5^{k-t-1}}\sum _{j=0}^{t-1} \frac{1}{5^{j}}\\= & {} (1+\delta )\frac{1}{5^{k-t-1}}\frac{1-\frac{1}{5^t}}{1-\frac{1}{5}}=(1+\delta )\frac{1}{4 \cdot 5^{k-t-2}}\left( 1-\frac{1}{5^t}\right) \\= & {} (1+\delta )\left( \frac{1}{4 \cdot 5^{k-t-2}}-\frac{1}{4 \cdot 5^{k-2}}\right) <\frac{1-2^{42}\delta }{4 \cdot 5^{k-t-2}} , \end{aligned}$$

by $2^{42}+1<2^{43}$ and since

$$\begin{aligned} \frac{2^{43}\delta }{4 \cdot 5^{k-t-2}} < \frac{1+\delta }{4 \cdot 5^{k-2}} \end{aligned}$$

holds by $t \le k-2$ and

$$\begin{aligned} \frac{4 \cdot 5^{k-2}}{4 \cdot 5^{k-t-2}} = 5^{t} \le 5^{k-2} \text{ while } \frac{1+\delta }{2^{43}\delta }> \frac{1}{2^{43}\delta }> 2^{3k+7} > 8^k . \end{aligned}$$

In particular for $t=k-2$, the total width is below $\frac{1}{4}$. Thus, we also have

$$\begin{aligned} \sum _{i=1}^{k-1} w_{1i}< \frac{1-2^{42}\delta }{4} + \frac{1+2^{40}\delta }{4}< \frac{1}{2} \text{ and } \sum _{i=1}^{k} w_{1i}< \frac{1-2^{42}\delta }{4} + 3 \cdot \frac{1+2^{40}\delta }{4} < 1 . \end{aligned}$$

The fourth and fifth parts hold by definition. The sixth part holds by definition and since $3 \cdot w_{20} + w_{1(k-1)} = 3\cdot (\frac{1}{4}-2^{32}\delta )+\frac{1+2^{40}\delta }{4} > 1 - 2^{34}\delta +2^{38}\delta$. The seventh part holds by definition and since $3 w_{(j+1)0}+ w_{j1} = 3(\frac{1}{4}-2^{52-10j}\delta )+(\frac{1}{4}+2^{60-10j}\delta ) >1$. $\square$

1.2 Appendix A.2: Proof of Lemma 4.4

• 1. For every $j \in \{2,3,4\}$, the type $\ell _{j0}$ dominates $\ell _{j1}$ since the width of the former type is smaller, and their heights and weights are equal. Type $\ell _{j1}$ dominates type $\ell _{j2}$ since the width of the former is twice as small, their heights are equal, and the weight ratio satisfies $v_{j2}/v_{j1}=2$.

• 2. For $1 \le i \le k-3$, type $\ell _{1i}$ dominates type $\ell _{1(i+1)}$ as their heights are equal, and $\frac{w_{1(i+1)}}{w_{1i}}=\frac{v_{1(i+1)}}{v_{1i}}=5$. Type $\ell _{1(k-2)}$ dominated type $\ell _{1(k-1)}$ since their heights are equal, $w_{1(k-2)}<w_{1(k-1)}$ and $v_{1(k-2)}=v_{1(k-1)}$. Type $\ell _{1(k-1)}$ dominated type $\ell _{1k}$ since their heights are equal, $2 \cdot w_{1(k-1)} = w_{1(k-1)}$ and $2 \cdot v_{1(k-1)}=v_{1k}$.

• 3. The height of item type $\ell _{1(k-2)}$ is $\frac{1}{43}+\varepsilon$, and the height of item type $\ell _{20}$ is $\frac{1}{7}+\varepsilon$. We have $6(\frac{1}{43}+\varepsilon )<\frac{1}{7}+\varepsilon$ as $\varepsilon <0.0001$. The width of item type $\ell _{1(k-2)}$ is $\frac{1+\delta }{5}$, and the width of item type $\ell _{20}$ is $\frac{1}{4}-2^{32}\delta$. We have $\frac{1+\delta }{5}<\frac{1}{4}-2^{32}\delta$ as $2^{64}\delta <1$. As the weight of six items of type $\ell _{1(k-2)}$ is 6, while the weight of one item of type $\ell _{20}$ is 4, the domination holds.

• 4. Item type $\ell _{20}$ has height $\frac{1}{7}+\varepsilon$ while item type $\ell _{30}$ has height $\frac{1}{3}+\varepsilon$, and we have $2(\frac{1}{7}+\varepsilon )<\frac{1}{3}+\varepsilon$, as $\varepsilon < 0.0001$. Item type $\ell _{20}$ has smaller width than item type $\ell _{30}$. The weight of two items of type $\ell _{20}$ is 8 while the weight of one type $\ell _{30}$ item is 6. Thus, the domination holds.

• 5. Item type $\ell _{30}$ has both smaller height and smaller width than an item of type $\ell _{40}$ and they have the same weights. Thus, the domination holds.

$\square$

1.3 Appendix A.3: Proof of Lemma 5.3

We start with the numerator. We use

$$\begin{aligned} \sum _{i=1}^{k-2} v_{1i}= \sum _{i=1}^{k-2}\frac{1}{5^{k-i-2}} =\sum _{j=0}^{k-3} \frac{1}{5^{j}}=\frac{1-\frac{1}{5^{k-2}}}{4/5}=1.25-\frac{1}{4\cdot 5^{k-1}} , \end{aligned}$$

and $\sum _{i=k-1}^k v_{1i}+\sum _{j=2}^4 \sum _{i=0}^2 v_{ji}=67$, to get

$$\begin{aligned} \sum _{i=1}^k v_{1i}+\sum _{j=2}^4 \sum _{i=0}^2 v_{ji}=68.25-\frac{1}{4\cdot 5^{k-1}} . \end{aligned}$$

Next, we consider the denominator. We have

$$\begin{aligned} Q\le & {} \frac{1}{168} \cdot \left( 42\cdot \left( 5-\frac{1}{5^{k-3}}\right) /5^{k-3}+\sum _{i=2}^{k-2}42\cdot \left( 5-\frac{1}{5^{k-i-2}}\right) \right. \cdot \left( \frac{1}{5^{k-i-2}}-\frac{1}{5^{k-(i-1)-2}}\right) \\&+\,1\cdot 126+2\cdot 112+6\cdot 96\\&\left. \,+6\cdot 72+12\cdot 68+14\cdot 48+14\cdot 42+28\cdot 36+21\cdot 24+21\cdot 18+42\cdot 12\right) . \end{aligned}$$

Since

$$\begin{aligned}&\sum _{i=2}^{k-2}\left( 5-\frac{1}{5^{k-i-2}}\right) \cdot \left( \frac{1}{5^{k-i-2}}-\frac{1}{5^{k-(i-1)-2}}\right) =\sum _{i=2}^{k-2}\left( 5-\frac{1}{5^{k-i-2}}\right) \cdot \frac{4}{5^{k-i-1}}\\&\quad =4\cdot \sum _{i=2}^{k-2}\frac{1}{5^{k-i-2}}-0.8 \cdot \sum _{i=2}^{k-2}\frac{1}{25^{k-i-2}}=4\cdot \sum _{u=0}^{k-4}\frac{1}{5^{u}}-0.8\cdot \sum _{u=0}^{k-4}\frac{1}{25^{u}}\\&\quad =4\cdot \left( \frac{1-(1/5)^{k-3}}{0.8}\right) -0.8\cdot \left( \frac{1-(1/25)^{k-3}}{0.96}\right) =\frac{25}{6}-\frac{1}{5^{k-4}}+ \frac{1}{6 \cdot 5^{2k-7}} \ \end{aligned}$$

we have,

$$\begin{aligned} Q\le & {} \frac{1}{168} \cdot \left( 42/5^{k-4}-42/5^{2k-6}+42\left( \frac{25}{6}-\frac{1}{5^{k-4}}+ \frac{1}{6 \cdot 5^{2k-7}}\right) +5828\right) \\= & {} \frac{6003-\frac{7}{5^{2k-6}}}{168} . \end{aligned}$$

$\square$