Given a boolean $0/1$ cube in $n$ dimensions with $2^{n-1}$ red and $2^{n-1}$ blue points can we complement the cube (blue becomes red and vice versa) in $\operatorname{poly}(n)$ transformations?


Here by transformation I mean the following.


1. Cutting the cube by $h=\operatorname{poly}(n)$ hyperplane inequalities each describable by $poly(n)$ bit coefficients into $r=\operatorname{poly}(n)$ pieces $P\_1,\dots,P\_r$.
2. Permuting coordinates of each piece (for example if $101$ was blue and $011$ was red then permuting first and second coordinate $011$ is blue and $101$ is blue).
3. Negating coordinates of each piece (for example if $101$ was blue and $111$ was red then negating first coordinate $111$ is blue and $101$ is blue).
4. Rotating, translating each piece.
5. Finally repasting all pieces so it becomes cube again.



Complementing the red and blue boolean cube?
There are at most $2^{\mathrm{poly}(n)}$ partitions $P\_1, \dots, P\_r$ satisfying condition 1. Each such partition has a part $P\_i$ of size at least $2^n/\mathrm{poly}(n)$. There are at most $2^{\mathrm{poly}(n)}$ transformations of $P\_i$ satisfying conditions 2-4.


Each such transformation $T$ of $P\_i$ can be represented by a one-to-one mapping from $P\_i$ to a subset of the hypercube, and a condition that the mapping preserves all colors or switches all colors. There are at most $2^{2^n}/2^{2|P\_i|/2}\le2^{2^n}/2^{2^n/\mathrm{poly}(n)}$ colorings of the hypercube satisfying these conditions for $T$. 


In total, the transformations satisfying conditions 1-4 can switch at most 
$$
2^{2^n-\frac{2^n}{\mathrm{poly}(n)}+\mathrm{poly}(n)}
$$
colorings of the hypercube, but the total number of colorings is $2^{2^n - \mathrm{poly}(n)}$, which is larger. Therefore, for sufficiently large $n$, there are colorings of the hypercube that cannot be switched (complemented) by the given transformations.


