I've been trying to come up with a combinatorial game even simpler than Hex with non-trivial gameplay and been failing dismally. Currently, my idea is that players sequentially lay pieces on a toroidal grid, meaning that squares at the bottom are adjacent to the corresponding square at the top, and squares on the left are adjacent to the corresponding square on the right. Both of the side lengths of the grid are odd and the side lengths are at least $3$ (see why in the background). The only restriction is that players cannot place a piece next to a piece their opponent has already placed (diagonally adjacent is fine). When one player can no longer make any moves, the other player may fill in any remaining legal spaces, and then the game ends. The player with the most pieces wins. Any idea as to whether or not this game has an easy-to-compute solution?
Background
I started off with the grid not being toroidal. This is easy to solve. On an $n\times m$ grid with $n$ odd, player $1$ places a piece at $\left(\left\lceil\frac{n}{2}\right\rceil,\frac{m}{2}\right)$ if $m$ is even, or $\left(\left\lceil\frac{n}{2}\right\rceil,\left\lceil\frac{m}{2}\right\rceil\right)$ if $m$ is odd. Thereafter, if player $2$ plays $(a,b)$, player $1$ plays the $180$-degree rotation $(n-a+1,m-b+1)$ on the next turn. This is always legal. Under the involution $f(a,b)=(n-a+1,m-b+1)$, the only squares which $(a,b)$ which are adjacent to or coincident with $f(a,b)$ are $\left(\left\lceil\frac{n}{2}\right\rceil,\left\lceil\frac{m}{2}\right\rceil\right)$ if $m$ is odd, and $\left(\left\lceil\frac{n}{2}\right\rceil,\frac{m}{2}\right)$ and $\left(\left\lceil\frac{n}{2}\right\rceil,\frac{m}{2}+1\right)$ if $m$ is even, both of which player $1$ blocked on the first turn. So if player $2$ plays $(a,b)$ and $(n-a+1,m-b+1)$ is illegal on the next turn, it's because it is blocked by a piece player $2$ already played $(c,d)$, other than $(a,b)$. But the state of the board when $2$ played $(a,b)$ was player $1$'s first piece, plus the values under the adjacency-preserving involution of all player $2$'s pieces. So player $2$ couldn't have played $(a,b)$, because it was already blocked by $f(c,d)$. So player $2$ runs out of moves first, so player $1$ automatically wins, having played one more piece than $2$ at that time.
For an $n\times m$ grid with both $n$ and $m$ even, there are no squares $(a,b)$ adjacent to or coincident $f(a,b)$. Therefore player $2$ can always play $f(a,b)$ on the turn after player $1$ plays $(a,b)$. Then player $1$ runs out of moves first. But the arrangement of pieces on the board is symmetric under $f$, so clearly player $2$ cannot place any extra pieces after that, so player $2$ has only forced a draw, as they have played an equal number of pieces. However, player $1$ can essentially play the same strategy but picking an arbitrary new square on the first move and whenever player $2$ copies them, which at best results in a draw.
If we think of this game as being played on a general finite graph $G$, with pieces placed at the nodes, and a rule against placing on a node that your opponent has already placed a piece adjacent to, then this method generalises somewhat. If there is a graph automorphism $f$ of $G$ with $f^2=f$ such that there are no nodes $v\in G$ such that $f(v)=v$ or $f(v)$ is adjacent to $v$ (notice that this forces the number of nodes in $G$ to be even), then player $2$ has no choice but to force a draw by playing $f(v)$ on the turn after player $1$ plays, resulting in a draw. If, however, there is a graph automorphism $g$ with $g^2=g$, and the subgraph of nodes that are adjacent to or coincident with themselves under $g$ is non-empty and can be covered by a single node in the subgraph (not necessarily unique), then player $1$ starts with such a node, and then player $g(v)$ after player $2$ plays $v$ and wins. Incidentally, this is a very strange proof that no $G$ may simultaneously have an $f$ and a $g$, or as we'll say, a draw-involution and a win-involution. For non-trivial gameplay we need to look at graphs with neither kind of involution.
So I made the grid toroidal. That is, that $(n,b)$ is adjacent to $(1,b)$ for all $1\leq b\leq m$ and $(a,m)$ is adjacent to $(a,1)$ for all $1\leq a\leq m$. The case that $n>2$ is even ends in a draw because of $f(a,b)=f{\left(a+\frac{a}{2},b\right)}$. If $n=2$, $180$-degree rotation is a win for $1$ if $m$ is odd, and a draw if $m$ is even. For a square grid, if $n=m$ is odd. If $n=1$ and $m=1$ or $3$, then obviously the identity is a win-involution. But now there are no more win or draw-involutions. The general $n=1$ case also has a fairly simple winning strategy for player $1$, and non-win/draw involutions play a role in it (whenever you have any involutions, you have to ensure that you cover some of the special points before your opponent gets them all), but I doubt it would generalise, and, surprisingly, it's apparently more complex than the cases just covered. So, can we solve the case that $n$ and $m$ are odd and $n,m\geq3$?