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How many $n$-dimensional unit cubes are needed to cover a cube with side lengths $1+\epsilon$ for some $\epsilon>0$?

For n=1, the answer is obviously two. For n=2, the drawing below shows that three unit cubes suffice, but it is impossible using only two cubes. In general, a total of $n+1$ cubes is enough, as is shown in Agol's answer, but is this the smallest number possible?

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Why does it seem that n+1 is sufficient? This is not obvious to me even for n = 3. –  user332 Dec 16 '10 at 18:47
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Yes, $n+1$ is OK. With one unit cube, you can cover an $(n-1)$-dimensional face, just as shown in the figure. Then select one vertex $S$ of the larger cube. Use $n$ unit cubes to cover the $(n-1)$-faces attached to $S$. There remains to cover a compact set which contains only the opposite vertex, which can be covered with one unit cube. –  Denis Serre Dec 16 '10 at 20:49
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@Denis Serre: How do you cover one n-1-dimensional face with one cube? What is the largest square that will fit inside a cubical box? If you look at cross-sections of the unit n-cube that are perturbation cross-sections parallel to a face, the shape is determined by the slope, a linear function: the unit n-1-cube can be stretched by any linear function. You can't get a larger n-1 cube that way. For n=3, I see how to cover one of the n-1 faces except for a neighborhood of one of its edges. With 3 cubes, you can cover 2- faces attached to S, so 4 works. –  Bill Thurston Dec 16 '10 at 21:11
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For the n=3 case, see mathworld.wolfram.com/PrinceRupertsCube.html for how to cover a (1+epsilon) square by a single cube. After three faces of the large cube adjacent to a single vertex are covered in this way, the remaining part of the large cube can be covered by a fourth unit cube, exactly as in the n=2 case. –  David Eppstein Dec 17 '10 at 0:03
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For the n=3 case, I found the following picture useful : mathworld.wolfram.com/CubeSquareInscribing.html –  François Brunault Dec 17 '10 at 23:28

1 Answer 1

I'll expand my comment into an answer. Following Serre's suggestion in the comments, to show that $n+1$ unit cubes cover an $n$-cube of side lengths $1+\epsilon$ for some $\epsilon > 0$, it suffices to show that one may fit a unit $n-1$-cube in the interior of a unit $n$-cube. If you can do this, then you can fit a $1+\epsilon$ $n-1$-cube in the interior of a unit $n$-cube for some $\epsilon$. As Serre suggests, you may then take a vertex of the $1+\epsilon$ $n$-cube, and cover each of the $n$ $n-1$-cube faces containing it by a unit $n$-cube. Then one may use another $n$-cube to cover the antipodal vertex and the rest of the cube, possibly for a smaller $\epsilon$. This case $n=2$ is in the picture given in the question. As Bill points out in the comments, one needs an explanation as to why an $n-1$-cube fits in the interior of the $n$-cube.

To see that a unit $n-1$-cube fits in the interior of an $n$-cube, one may proceed by induction. This is true for $n=1$. Let $I=[0,1]$. By induction, suppose we have an embedding $f: I^{n-1}\to int(I^n)$. Then we have an embedding $f\times Id: I^{n-1}\times I \to int(I^n)\times I$. This gives a map which touches the boundary of $I^{n+1}$ only along $f(I^{n-1})\times \{0,1\}$. Think of this as a map $f×e:I^{n−1}\times I\hookrightarrow f(I^{n−1})\times D^2\subset f(I^{n−1})\times I\times \mathbb{R}$ ($\mathbb{R}$ is just the normal bundle to $f(I^{n-1})\times I \subset I^n\times I = I^{n+1}$). Then rotate the interval $I$ in $D^2$ by a small angle $\theta$ to get a map $e_{\theta}:I\to D^2\subset I\times \mathbb{R}$. The endpoints no longer lie on $\{0,1\}\times \mathbb{R}$. Then the map $f \times e_{\theta}: I^{n-1}\times I \to f(I^{n-1})\times D^2$ will have image in $int(I^{n+1})$ for small enough $\theta$.

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Thanks Agol, this is a nice argument! This shows that $n+1$ cubes is always enough. The question whether $n+1$ is the smallest possible number is still open. –  J.C. Ottem Dec 18 '10 at 1:48
    
@Agol. It seems that the construction proposed by Denis and you implies the following : for sufficiently small $\epsilon>0$, it is possible to arrange $n$ unit hypercubes in the $(1+\epsilon)$-hypercube in such a way that the region not covered by them is arbitrarily small. Is it true? –  François Brunault Dec 18 '10 at 13:09
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@Francois: I don't think so, at least this doesn't follow from the argument. The point is that if you can cover the $n$ faces adjacent to a vertex of a $1+\epsilon$ cube with unit cubes, then they cover some $\delta$ neighborhood of those faces. So you know that there will be a $1+\epsilon-\delta$ cube left over, and by adjusting $\epsilon$, you can make sure that a unit cube can cover what is left over. –  Ian Agol Dec 18 '10 at 18:52
    
@Agol : Oh, I see now, thanks... For $n=3$ at least, I got more or less convinced that 3 cubes are sufficient to cover everything but a small neighborhood of $S'$, the vertex opposite to $S$. But I wouldn't swear to it, because my 3D perception is limited... –  François Brunault Dec 18 '10 at 22:18
    
Francois (forgive my limited typeset), I had a visualization difficulty too, until I cut out the rectilinear cube and saw that I just had a union of thin prisms adjacent to one vertex left to cover. It makes me wonder if I can use a "volume-greedy" cover which involves 2 rectilinear cubes and (n-2) other cubes to cover a large enough neighborhood which surrounds the uncovered ring of dimension (n-2) "edges". Gerhard "Ask Me About System Design" Paseman, 2010.12.18 –  Gerhard Paseman Dec 18 '10 at 23:27

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