IRL_ABooklet_2023

We claim that the answer for $n^3$ small cubes, for any $n>1$, is $2^{3n^2}$. In particular, for $1000$ cubes, the answer is $2^{300}$.

First note that, up to rotation, the polarisations of all small cubes are identical: three faces with polarisation N meet at one corner of the cube. Therefore, instead of rotating an already polarised small cube, we may simply repolarise its faces, making sure that opposite faces have opposite polarisations. Hence, there are $2^3=8$ possible ways to polarise a non-moving small cube, i.e. there are $8$ different rotational positions of a polarised small cube.

Below, we will build the megacube by placing the small cubes one after the other. If a small cube is placed adjacent to one or two small cubes that were placed before, its rotational position is constrained along one or two axes. There is only one way to place a small cube adjacent to three small cubes that were placed before.

If a small cube is constrained along one axis, i.e. if the polarisation of one pair of opposite faces is fixed, it has $2^2=4$ possible configurations. For example, if the Left-Right polarities are set, we can freely choose the Top-Bottom polarities in two ways and, independently, the Front-Back polarities in two ways.

If a small cube is constrained along two axes, it has $2$ possible configurations. To see this, suppose for instance that the Left-Right and Top-Bottom polarities are set. The Front-Back polarities can then be set in two ways.

Let us now give coordinates to the cubes in the megacube with $n>1$ cubes on each side: for $1\le i,j,k\le n$, $(i,j,k)$ indicates the cube that is in the $i$-th layer from left to right, the $j$-th layer from top to bottom, and the $k$-th layer from front to back. For instance, $(1,1,1)$ indicates the cube $C$ in the left-top-front corner of the megacube.

We now build the megacube in stages. We first place $C$. This is unconstrained so we get $2^3$ possible configurations. Next, we add all other cubes on the three edges of the megacube that are incident on $C$, i.e. cubes with coordinates $(x,1,1)$, $(1,y,1)$, or $(1,1,z)$ for $2\le x,y,z\le n$. We do so in increasing order of $x$, $y$, and $z$. This constrains them along one axis, so each of these gives a factor $2^2$ to the number of configurations.

Next, we fill in the three primary faces, meaning the faces that are each incident on two of the three edges that we filled in above. We fill in these faces in a similar fashion, row by row, in bibliographic order of the coordinates. For instance, for the front face, i.e. the one with cubes $(x,y,1)$, we have already added these cubes for $x=1$, so we add $(2,y,1)$ in increasing order of $y\ge 2$, then $(3,y,1)$ in increasing order of $y\ge 2$, etc. Doing so, we note that the polarities are fixed along two coordinate directions, so each cube contributes a factor $2$ to the number of configurations.

We finally fill in all remaining cubes $(x,y,z)$ in bibliographic order, noting that each cube is constrained along all three axes, so these contribute nothing to the number of configurations. So choosing the polarity for the three faces of the megacube that include a face of $C$ fixes the configuration. In particular, the rule stating that we cannot sense the bottom face is irrelevant.

It remains to add up the number of factors of $2$ above. This is made easier once we realise that the number of factors for each cube equals the number of faces of that cube that are contained in a primary face of the megacube. Each such face has $n^2$ cubes, so our claim follows.