What is the maximum number of the colors what can be used to color the vertices of a eight-dimensional hypercube, such that for every vertex of the cube, every color is used as the color of a neighbour vertex?
(Two vertices are neighbours if they are the endpoints of an edge).
I transformed it into a coding problem. I noticed that the vertices are vectors in $\mathbb{F}_2^{8}$.
On
Take a 7-dimensional face --- say $x_8 = 0$, which has $2^7$ vertices, of the form $b_1 b_2 \ldots b_7 0$.
Each is joined to the vertex $$ b_1 b_2 \ldots b_7 1 $$ by an edge. So you can color each such pair a different color, and they'll satisfy your criterion. So the answer is "at least $2^7$ colors." I'll let you work out why it's at most $2^7$, and is therefore exactly $2^7$.
On
So you want to color the vertices such that every vertex is neighbor to at least one vertex of the same color, and you want to maximize the number of colors used?
Let the first 7 coordinates of each point determine its color, with $2^7=128$ different colors used. The last coordinate is not used, so for any vertex, flipping that coordinate between 0 and 1 will produce a neighbor of the same color.
On the other hand, your restriction clearly implies that every used color is used by at least two vertices, so with only 256 vertices in total, it is impossible to have more than 128 colors.
Here is an article about this exercise.