Let $D$ be the digraph on $2^d$ vertices with $d2^d$ edges that we obtain by directing each edge of the $d$-dimensional hypercube in both directions.



> 
> Can we partition the edges of $D$ into $d$ directed Hamiltonian cycles?
> 
> 
> 


An alternative formulation would be to ask whether the hypercube has a so-called directed double cover by Hamiltonian cycles.


For $d\le 2$ the statement holds but quite embarrassingly I'm stuck already at $d=3$.



Can we find 3 disjoint directed Hamiltonian cycles in the cube?
It looks like the titled question (d=3) is not directly answered: I will hint at how to
show the answer is no.


At each vertex, there are two ways that the decomposition can go. I like
to call them black or white, or clockwise and counterclockwise. If we look
at a cube face and pick all of one type or the other, we induce a path
around that face, so not a Hamiltonian path. Similarly, choosing three
vertices on a face with the same color/orientation leads to a non
Hamiltonian decomposition. So each face has two vertices of
each orientation.


However, if we have a path that goes through a white then a black
then a white vertex, it will have gone tbrough exactly three vertices
of a face, and again will not yield a Hamiltonian path. So each face
has to have two adjacent vertices of each color.


But if we have two adjacent faces sharing two vertices of the
same color, say black, then a third face adjacent to both will
have a white black white coloring which obstructs the
decomposition. So not all faces can have the required coloring,
so there is no such decomposition into three Hamiltonian paths.


