, 2021

We colour the board in three colours in such a way that no neighbouring tiles are of the same colour. We can give each hexagon a coordinate using $\vec{e}_1 = (1, 0)$ and $\vec{e}_2 = (\cos(120^\circ), \sin(120^\circ)) = (\frac{-1}{2}, \frac{\sqrt{3}}{2})$ as basis. Let the center square be the origin. Then each hexagon has center at $a \cdot \vec{e}_1 + b \cdot \vec{e}_2$, $(a, b) \in \mathbb{Z}^2$. The tuple $(a, b)$ is the coordinate for a given hexagon; its neighbours are $(a + 1, b)$, $(a + 1, b + 1)$, $(a, b + 1)$, $(a - 1, b)$, $(a - 1, b - 1)$ and $(a, b - 1)$.

We colour the hexagon with coordinates $(a, b)$ with colour number $(a+b) \pmod{3}$. It is clear that neighbouring hexagons do not share a colour. (In fact this is the only three colouring of a hexagonal tiling). See figure 8.

We see that if $n \equiv 1 \pmod{3}$, the token begins in a space in the same colour as the center hexagon, let that colour be grey. By regarding a few cases, we see that whatever Anne does, Ellie and Milo can end their turns by getting the token to a prescribed grey hexagon of the closest grey hexagons. Therefore they can get the token to the center.

If $n \not\equiv 1 \pmod{3}$, the token does not begin on the same grey colour as the center. Say the stone begins on a white tile, and say the third colour is black. Anne can always move the stone to a grey hexagon that is not on the same horizontal/diagonal line as the center hexagon. Then Anne moves the stone to a white or black hexagon. After Milo moves the token is still again on a white/black hexagon. Anne can continue this indefinitely, with the token never reaching the center.