# Bounded positioning of hypercubes

I have the following question:

Let be (C_n) a sequence of m-dimensional hypercubes such that the series over the volumes of this cubes converges. I assume that it's possible to place those cubes in the m-dimensional space in such a way that those are pairwise disjoint and contained in a bounded set. However, I do not see how to prove that. Does somebody know a referecne or a proof?

• Can you solve the 2-dimensional case where there is a square of side $1/n$ for each $n$? The total area is $\pi^2/6$. Commented Oct 30, 2022 at 0:05
• The case $m=2$ is discussed in Stefan Hougardy, On packing squares into a rectangle, Computational Geometry, Volume 44, Issue 8, October 2011, Pages 456-463, sciencedirect.com/science/article/pii/S0925772111000319 Commented Oct 30, 2022 at 0:34
• Abstract We prove that every set of squares with total area $1$ can be packed into a rectangle of area at most $2867/2048=1.399\dots$ . This improves on the previous best bound of $1.53$. Also, our proof yields a linear time algorithm for finding such a packing. Commented Oct 30, 2022 at 0:36
• Thank you very much, the paper is amazing. Commented Oct 30, 2022 at 11:28

## 1 Answer

This can be generalized:

Let $$(C_n)$$ be a sequence of sets of $$\mathbb{R}^m$$ such that $$\sum_n \text{diam}(C_n)^m=S<\infty$$. Then we can find disjoint sets $$(D_n)_n$$ such that $$D_n$$ is a translate of $$C_n$$ and $$\bigcup_{n=1}^\infty D_n$$ is bounded.

To prove that, reorder the sets $$C_n$$ so that the sequence $$d_n:=\text{diam}(C_n)$$ is decreasing. Now consider a ball $$B\subseteq\mathbb{R}^m$$ of volume $$4^m S$$. Then we can find by recursion a sequence of points $$p_n\in B$$ such that for any $$i we have $$d(p_i,p_j)>2d_i$$. To do this we just need to choose $$p_n\in B$$ outside the closed balls $$B(p_1,2d_1),B(p_2,2d_2),\dots,B(p_n,2d_n)$$: this is possible because, if $$m$$ is Lebesgue measure, $$m(B)=4^mS=\sum_{i=1}^\infty(2\text{diam}(C_i))^m\geq\sum_{i=1}^{n-1}(4\text{diam}(C_i)^m)>\sum_{i=1}^{n-1}m(B(p_i,2d_i))$$.

Then choose $$D_n$$ to be some translate of $$C_n$$ containing $$p_n$$. Clearly $$\bigcup_n D_n$$ is bounded, and the $$D_i$$ are disjoint because for any $$i, $$d(p_i,p_j)>2\text{diam}(C_i)\geq \text{diam}(C_i)+\text{diam}(C_j)$$.

• Thank you very much, now I understand. Commented Oct 30, 2022 at 14:33