8.33 Two players, A and B, take turns using a chisel to pierce through a wooden board with $n \times n$ unit squares. The chisel marks must start from the edge of the board or from a node that has already been chiseled, and each player chisels through one side of a unit square along the grid lines. If a player causes the board to break apart after their move, that player loses. If A starts first and both play with the correct strategy, who will win?
[Solution] Clearly, on the edges of the board and inside it, there are a total of $(n+1)^{2}$ nodes. Nodes on the edges of the board or connected to a notch are called "occupied," while the others are called unoccupied. When two occupied nodes are first connected by a notch, the board will break. Therefore, when it is someone's turn and they cannot find any unoccupied node, they will inevitably lose. Since there are $(n-1)^{2}$ unoccupied nodes at the start of the game, if both players play correctly, it is certain that someone will lose on the $(n-1)^{2}+1$-th step. Thus, the outcome of the game is uniquely determined by the parity of $n$. When $n$ is odd, player B wins; when $n$ is even, player A wins.