Puzzler #2: Cube Dissection

The idea of interest today is, as the title states, cube dissection.

The question is, given a cube, can it be cut into a finite number of smaller cubes – all of different size?

Clearly, thinking in terms of a rubik’s cube for example, it is easy to cut a cube into a finite number of smaller cubes of the same size. One could imagine too, cutting a cube into an infinite number of infinitely tiny, though dissimilar, cubes. But can it be done in a finite number of different sized cubes? It’s like a Jigsaw Puzzle, where you can pick the shape of the pieces. You just don’t know if they actually form anything when put together.

Solution below the fold!

What we’re going to do is try to construct our dissection, working from the bottom up, and watch it fail. We’re going to assume the cube is of side length 1, just to simplify things later. Clearly it doesn’t matter as you can scale the cube and the subcube as you like, but it just makes things cleaner to let the side length be 1.

This is the bottom face of the cube. It looks about like what you would expect.

Now, of all the subcubes touching this face, one of them must be smallest. Looking at this one face as we are, it would look like the smallest square in the larger square above. Notice, this is not the smallest cube in the entire dissection, just the smallest cube touching the one face. The question then is where can it go.

For example, if we put the smallest cube in the corner, we get the situation below.

As the subcube in the corner is the -smallest- in this face, pressed up against one side of it there must be a subcube of a longer sidelength. But this creates the gap labeled in the picture, with same width as the corner subcube. Notice that the only way this gap can be filled (so as to be able to dissect the rest of the face), is with a cube of equal or lesser side length than the corner cube. Since we want all the cubes to be of different size, and we said explicitly that the one in the corner was the smallest subcube touching this face, this is clearly a contradiction, hence the fail. Therefore the smallest cube cannot be in a corner.

By a similar argument, looking at the picture below, the smallest subcube on this face cannot be sitting against an edge.

Indeed, using the gap argument, we can argue that the smallest cube is somewhere in the middle of the face, with other cubes arranged around it like so.

This establishes two important things. First and foremost, it establishes a bound on the size of the smallest cube. Looking at that picture, the smallest cube necessarily has two cubes on either side of it. If the total width of that face is 1, then the side length of the smallest cube is necessarily less than 1/3. If it equaled 1/3, or were greater, one of the cubes on either side would have to be of width less than 1/3, and the ’smallest cube’ wouldn’t actually be the smallest.

Second, what we’re looking at right now is the bottom of the smallest cube. Thinking about the top, the above arrangement shows that the smallest cube is surrounded on all sides by taller cubes. As such, the top of the smallest cube is bounded on all sides by the walls of the surrounding cubes, creating a sort of square shaped hole that needs to be filled in, to cover the top of the smallest cube.

This is where it starts to get interesting. This is also where I’m likely to lose readers, mosly for my lack of 3D pictures.

Because the top of the smallest cube needs to be covered, much the same way as we covered the bottom of the main cube. It must have, for example, a smallest cube sitting on it, somewhere. By the same arguments as before, we can establish it can’t be in a corner, it can’t be on a side. It must be in the middle, in the same arrangement. We can then bound its size as strictly less than 1/3 the width of the top of the cube – which we bounded above by 1/3. So this smallest cube is bounded above by 1/9.

We can keep going, creating this stack of smallest cubes, through the center of the main cube. And the width of each smallest cube is bounded by 1/3 the width of the previous. 1/3, 1/9, 1/27, etc.

So, let’s consider the height of this stack. Imagine a line from the top cube in this stack, straight down. We have a finite number of subcubes in total, so there must be a finite number of cubes in this stack. The width of the i’th cube is bound above by 1/3ⁱ, and we’ll say that there are k many cubes in the stack. We get

As this is a finite sum, though, we can bound it above by the infinite sum ^^

However, that sum is a common geometric series, and we can actually calculate its explicit value. This gives us the following bound,

What does this mean? Well, as we established, the bottom face of the cube must have a smallest cube touching it. And the top of that cube, bound by the taller walls of the cubes around it, must have a smallest cube touching it, and so on down the line. But we know this stack is finite, and we know this stack doesn’t even reach half way into the cube. So on top of this stack, there must be an i’th smallest cube, somewhere in the middle of the main cube, that, on its top face, bound by walls on all sides, does not have a smallest cube. And if it doesn’t have a smallest cube it can’t have any cube touching it, and we have a hole in the middle of the cube.

As this theoretical stack of smaller and smaller cubes is a necessity in the dissection, but leads to a contradiction with a hole in the middle of the larger cube, this shows that such a dissection cannot exist. You cannot cut a cube into a finite number of dissimilar cubes. Which I think is an interesting result.

Of course, if you allow an infinite number of cubes, not sure. Any thoughts?