, 2021

Answer: Yes, there exists a colouring that yields a winning strategy for the hunter.

A central idea is that several colourings $C_{1}, C_{2}, \ldots, C_{k}$ can be merged together into a single product colouring $C_{1} \times C_{2} \times \cdots \times C_{k}$ as follows: the colours in the product colouring are ordered tuples $(c_{1}, \ldots, c_{n})$ of colours, where $c_{i}$ is a colour used in $C_{i}$, so that each cell gets a tuple consisting of its colours in the individual colourings $C_{i}$. This way, any information which can be determined from one of the individual colourings can also be determined from the product colouring.

Now let the hunter merge the following colourings:
- The first two colourings $C_{1}$ and $C_{2}$ allow the tracking of the horizontal and vertical movements of the rabbit.
The colouring $C_{1}$ colours the cells according to the residue of their $x$-coordinates modulo $3$, which allows to determine whether the rabbit moves left, moves right, or moves vertically. Similarly, the colouring $C_{2}$ uses the residues of the $y$-coordinates modulo $3$, which allows to determine whether the rabbit moves up, moves down, or moves horizontally.
- Under the condition that the rabbit's $x$-coordinate is unbounded, colouring $C_{3}$ allows to determine the exact value of the $x$-coordinate:
In $C_{3}$, the columns are coloured white and black so that the gaps between neighboring black columns are pairwise distinct. As the rabbit's $x$-coordinate is unbounded, it will eventually visit two black cells in distinct columns. With the help of colouring $C_{1}$ the hunter can catch that moment, and determine the difference of $x$-coordinates of those two black cells, hence deducing the precise column.
Symmetrically, under the condition that the rabbit's $y$-coordinate is unbounded, there is a colouring $C_{4}$ that allows the hunter to determine the exact value of the $y$-coordinate.
- Finally, under the condition that the sum $x+y$ of the rabbit's coordinates is unbounded, colouring $C_{5}$ allows to determine the exact value of this sum: The diagonal lines $x+y=\text{const}$ are coloured black and white, so that the gaps between neighboring black diagonals are pairwise distinct.
Unless the rabbit gets stuck, at least two of the three values $x, y$ and $x+y$ must be unbounded as the rabbit keeps moving. Hence the hunter can eventually determine two of these three values; thus he does know all three. Finally the hunter works backwards with help of the colourings $C_{1}$ and $C_{2}$ and computes the starting cell of the rabbit.