Alice and Bob are playing the following game: They have a 4x4 empty grid and take turns coloring one square each, starting with Alice, both using the same color. Whoever completes any 2x2 area on the grid (after having made his move) is the loser. Is there any winning strategy for any of the two players?
I have played the game several times and can't see a clear strategy for any of the two. It seems to me that Bob will loose.
Any ideas?
If Alice selects a square $(i,j)$, Bob should choose $(i+2,j)$. In this way he will never complete a square before Alice does, and hence Bob will always win. Of course, modulo 4 is required with this strategy.
Likewise a strategy $(i,j+2)$ would have worked. Note that these strategies are always possible, because they fill the full grid with particular pairs.
In the case of a $n \times n$ grid for odd values of $n=2 m + 1$, there is a winning strategy for Alice. If she starts with $(m,m)$ as the first square and responds to any choice $(i,j)$ of Bob by playing $(2 m -i,2 m-j)$, Bob will make the first $2\times2$ square. This approach can also be generalised to rectangular grids, and will result in a winning strategy for Alice when both dimensions are odd.
Thanks to @Philip for pointing out that the pairing $(i,j)$ and $(i+2,j+2)$ would not work in the $4\times4$ grid, because the choosing of the 4 corners by Alice would create a central square made by Bob.
Thanks to @Carmeister for pointing out that a generalisation for even values $n=2m$ by pairing the square $(i,j)$ of Alice by Bob taking $(i+m,j)$, would also not be successful for the same reason. By choosing suitable squares (this time not the corners) Alice could force Bob to make a central square as well if he would stick to this particular strategy.
So the question of the existence/absence of a winning strategy for even sided square grids larger than $4\times4$ is still open.