Here is a 2-player game played on a region of the
Gaussian integers, $\mathbb{Z}[i]$.
Initially four points are colored, opposite
corners of an $X$ by $Y$ rectangle:
$0 + 0i$ and $X + Yi$ are colored red,
and $0 + Yi$ and $X + 0i$ are colored blue.
A move by a player of color $C$ consists of selecting a red point and a blue
point, and coloring the previously uncolored "midpoint" color $C$,
where the midpoint of $z+w$ is
$\lfloor{ (z+w)/2 } \rfloor$.
The game ends when a player loses by coloring a Gaussian prime,
that is, a point $a + bi$, which, if either $a$ or $b$ is zero,
is a prime of the form $4n+3$, or otherwise if its norm $a^2+b^2$ is a prime.
Example. $X=51$, $Y=34$.
Red moves first and colors $25 + 0i$ red, the midpoint of
the points on the real axis.
Blue would not want to color the midpoint of $0 + 34i$ and $25 + 0i$,
because that is $12 + 17i$ which is prime (433).
So suppose Blue instead colors the midpoint of the points
on the imaginary axis, $0 + 0i$ and $0 + 34i$,
which is $0 +17i$.
Red could now select the midpoint of this blue point and $25 + 0i$,
which is $12 + 8i$.
And so on.
Will the game always end with a loser?
If both $X$ and $Y$ are of the form $2p$ or $2p+1$ with $p$
a prime $4n+3$, then Red is forced to lose on the first move.
Are there other values of $X$ and $Y$ for which the game
can be fully analyzed?
Is there any hope
analyzing who wins this game (under best play) for arbitrary
$X$ and $Y$?
This is original and quite possibly worthless, so caveat lector!
Edit1. See the suggested simplification by Michael Albert in the Comments: dispense with
colors, letting each move select any two points.
Edit2. Thanks for all the interesting comments.
It now seems to me this game is hopelessly complicated to analyze, perhaps PSPACE-complete in terms of complexity. The monochromatic version is much simpler but removes the
adversarial aspect that is the essence of a game. I don't think Milton-Bradley will be knocking on my door!
Best Answer
This is only a heuristics, but I would assume that the game either ends after a few steps, or the outcome is determined simply by the parity of $(X+1)(Y+1)$ minus the number of gaussian primes in the rectangle. This is certainly the case if at the end of the game all non-primes are taken. So the player for which this heuristics predicts a win should try to play in such a way that all points of the rectangle are midpoints of lines with endpoints of different colour. Since most gaussian integers are not prime one of the basic heuristics of additive combinatorics predicts that the only way to avoid this is to have some extremely large Fourier coefficient, that is, the board looks like having red and blue stripes. But by taking midpoints of red and blue points one of the player can prevent this from happening, provided he has a sufficient number of choices left. This means that either the game ends early, or it should continue until all non-primes are taken, in which case the winner is determined by the aforementioned parity.