Question 40
A and B play a game on a Cartesian coordinate plane. First, A chooses a position $(x_0, y_0)$ and places the first stone there. Then, starting with B, A and B take turns placing stones:
i) If the stone A placed in the previous turn is at $(x, y)$, then B may only place a stone at one of $(x + 2, y + 1)$, $(x + 2, y - 1)$, $(x - 2, y + 1)$, $(x - 2, y - 1)$.
ii) If the stone B placed in the previous turn is at $(x, y)$, then A may only place a stone at one of $(x + 1, y + 2)$, $(x + 1, y - 2)$, $(x - 1, y + 2)$, $(x - 1, y - 2)$.
Moreover, if $a \equiv c \pmod{n}$, $b \equiv d \pmod{n}$ and there is already a stone at $(a, b)$, then neither player may place a stone at $(c, d)$. The first player who cannot place a stone anywhere loses.
(1) When $n = 2018$, who has a winning strategy?
(2) When $n = 2019$, who has a winning strategy?
i) If the stone A placed in the previous turn is at $(x, y)$, then B may only place a stone at one of $(x + 2, y + 1)$, $(x + 2, y - 1)$, $(x - 2, y + 1)$, $(x - 2, y - 1)$.
ii) If the stone B placed in the previous turn is at $(x, y)$, then A may only place a stone at one of $(x + 1, y + 2)$, $(x + 1, y - 2)$, $(x - 1, y + 2)$, $(x - 1, y - 2)$.
Moreover, if $a \equiv c \pmod{n}$, $b \equiv d \pmod{n}$ and there is already a stone at $(a, b)$, then neither player may place a stone at $(c, d)$. The first player who cannot place a stone anywhere loses.
(1) When $n = 2018$, who has a winning strategy?
(2) When $n = 2019$, who has a winning strategy?