Revisions to How many unit squares can you pack into a rectangle with nearly integer side lengths?

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edit approved Oct 9, 2017 at 3:38

jeq

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Earlier today, somebody asked what looks like a homework problem, but admits the following reading which I think is interesting:

Suppose $a_1,\dots, a_n$ are positive integers, and $\varepsilon$ is a positive real number which you can take to be as small as you like. How many non-overlapping unit hypercubes can you fit into an $n$-dimensional rectangular solid with side lengths $a_i-\varepsilon$?

It's clear that you can't fit $\prod_i a_i$ and that you can fit at least $\prod_i (a_i-1)$. A little playing around shows that you can sometimes fit strictly more than $\prod_i (a_i-1)$. For example, here's a packing of three unit squares into a $(2-\varepsilon)\times(3-\varepsilon)$ rectangle:

squares squares http://i39.tinypic.com/2vwvei0.jpg(source)
If

If I haven't made a mistake, you can take $\varepsilon$ to be as large as $1-\frac 23 \sqrt 2$.

Does anybody know how to find good lower bounds on this number? Using the trick in the above picture, you can effectively get a layer of hypercubes whose length in a given direction is $\sqrt 2 -\frac 12$ instead of 1$1$. Is there a higher-dimensional version of this trick which does better?
Does anybody know how to get good upper bounds on this number? In particular, is there an easy way to see that it's never possible to get $\prod_i a_i -1$?

Earlier today, somebody asked what looks like a homework problem, but admits the following reading which I think is interesting:

Suppose $a_1,\dots, a_n$ are positive integers, and $\varepsilon$ is a positive real number which you can take to be as small as you like. How many non-overlapping unit hypercubes can you fit into an $n$-dimensional rectangular solid with side lengths $a_i-\varepsilon$?

It's clear that you can't fit $\prod_i a_i$ and that you can fit at least $\prod_i (a_i-1)$. A little playing around shows that you can sometimes fit strictly more than $\prod_i (a_i-1)$. For example, here's a packing of three unit squares into a $(2-\varepsilon)\times(3-\varepsilon)$ rectangle:
squares http://i39.tinypic.com/2vwvei0.jpg
If I haven't made a mistake, you can take $\varepsilon$ to be as large as $1-\frac 23 \sqrt 2$.

Does anybody know how to find good lower bounds on this number? Using the trick in the above picture, you can effectively get a layer of hypercubes whose length in a given direction is $\sqrt 2 -\frac 12$ instead of 1. Is there a higher-dimensional version of this trick which does better?
Does anybody know how to get good upper bounds on this number? In particular, is there an easy way to see that it's never possible to get $\prod_i a_i -1$?

Earlier today, somebody asked what looks like a homework problem, but admits the following reading which I think is interesting:

Suppose $a_1,\dots, a_n$ are positive integers, and $\varepsilon$ is a positive real number which you can take to be as small as you like. How many non-overlapping unit hypercubes can you fit into an $n$-dimensional rectangular solid with side lengths $a_i-\varepsilon$?

It's clear that you can't fit $\prod_i a_i$ and that you can fit at least $\prod_i (a_i-1)$. A little playing around shows that you can sometimes fit strictly more than $\prod_i (a_i-1)$. For example, here's a packing of three unit squares into a $(2-\varepsilon)\times(3-\varepsilon)$ rectangle:

squares (source)

If I haven't made a mistake, you can take $\varepsilon$ to be as large as $1-\frac 23 \sqrt 2$.

Does anybody know how to find good lower bounds on this number? Using the trick in the above picture, you can effectively get a layer of hypercubes whose length in a given direction is $\sqrt 2 -\frac 12$ instead of $1$. Is there a higher-dimensional version of this trick which does better?
Does anybody know how to get good upper bounds on this number? In particular, is there an easy way to see that it's never possible to get $\prod_i a_i -1$?

replaced http://mathoverflow.net/ with https://mathoverflow.net/

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edited Apr 13, 2017 at 12:58

Community Bot

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Earlier today, somebody asked what looks like a homework problem a homework problem, but admits the following reading which I think is interesting:

Suppose $a_1,\dots, a_n$ are positive integers, and $\varepsilon$ is a positive real number which you can take to be as small as you like. How many non-overlapping unit hypercubes can you fit into an $n$-dimensional rectangular solid with side lengths $a_i-\varepsilon$?

It's clear that you can't fit $\prod_i a_i$ and that you can fit at least $\prod_i (a_i-1)$. A little playing around shows that you can sometimes fit strictly more than $\prod_i (a_i-1)$. For example, here's a packing of three unit squares into a $(2-\varepsilon)\times(3-\varepsilon)$ rectangle:
squares http://i39.tinypic.com/2vwvei0.jpg
If I haven't made a mistake, you can take $\varepsilon$ to be as large as $1-\frac 23 \sqrt 2$.

Does anybody know how to find good lower bounds on this number? Using the trick in the above picture, you can effectively get a layer of hypercubes whose length in a given direction is $\sqrt 2 -\frac 12$ instead of 1. Is there a higher-dimensional version of this trick which does better?
Does anybody know how to get good upper bounds on this number? In particular, is there an easy way to see that it's never possible to get $\prod_i a_i -1$?

Earlier today, somebody asked what looks like a homework problem, but admits the following reading which I think is interesting:

Suppose $a_1,\dots, a_n$ are positive integers, and $\varepsilon$ is a positive real number which you can take to be as small as you like. How many non-overlapping unit hypercubes can you fit into an $n$-dimensional rectangular solid with side lengths $a_i-\varepsilon$?

It's clear that you can't fit $\prod_i a_i$ and that you can fit at least $\prod_i (a_i-1)$. A little playing around shows that you can sometimes fit strictly more than $\prod_i (a_i-1)$. For example, here's a packing of three unit squares into a $(2-\varepsilon)\times(3-\varepsilon)$ rectangle:
squares http://i39.tinypic.com/2vwvei0.jpg
If I haven't made a mistake, you can take $\varepsilon$ to be as large as $1-\frac 23 \sqrt 2$.

Does anybody know how to find good lower bounds on this number? Using the trick in the above picture, you can effectively get a layer of hypercubes whose length in a given direction is $\sqrt 2 -\frac 12$ instead of 1. Is there a higher-dimensional version of this trick which does better?
Does anybody know how to get good upper bounds on this number? In particular, is there an easy way to see that it's never possible to get $\prod_i a_i -1$?

Earlier today, somebody asked what looks like a homework problem, but admits the following reading which I think is interesting:

Suppose $a_1,\dots, a_n$ are positive integers, and $\varepsilon$ is a positive real number which you can take to be as small as you like. How many non-overlapping unit hypercubes can you fit into an $n$-dimensional rectangular solid with side lengths $a_i-\varepsilon$?

It's clear that you can't fit $\prod_i a_i$ and that you can fit at least $\prod_i (a_i-1)$. A little playing around shows that you can sometimes fit strictly more than $\prod_i (a_i-1)$. For example, here's a packing of three unit squares into a $(2-\varepsilon)\times(3-\varepsilon)$ rectangle:
squares http://i39.tinypic.com/2vwvei0.jpg
If I haven't made a mistake, you can take $\varepsilon$ to be as large as $1-\frac 23 \sqrt 2$.

Does anybody know how to find good lower bounds on this number? Using the trick in the above picture, you can effectively get a layer of hypercubes whose length in a given direction is $\sqrt 2 -\frac 12$ instead of 1. Is there a higher-dimensional version of this trick which does better?
Does anybody know how to get good upper bounds on this number? In particular, is there an easy way to see that it's never possible to get $\prod_i a_i -1$?