For an even number of cells, codex responds with the cell obtained by rotating the previous red move by degrees. There is no fixed cell, so this is always legal. Rotation then pairs every all-red placement with an all-blue placement, giving a tie. If one dimension is , pair consecutive cells , leaving the last cell unpaired for odd length. codex takes the mate of the previous move if it is uncolored, and any uncolored cell otherwise. Every pair ends with at least one blue cell. Every consecutive triple contains a complete pair, so no red tromino exists. codex wins. Otherwise both dimensions are odd and at least . coxld first takes the center, then responds by -degree rotation. Placements avoiding the center cancel in red-blue pairs. Among the four neighbors of the center, opposite cells have opposite colors. Exactly two neighbors are red, and they lie in perpendicular directions from the center. Together with the red center they form a red L tromino. Thus coxld wins strictly. Subtask 1 uses ternary board states and minimax. With placements, it takes time and memory. The full solution takes time and auxiliary space per case. There are at most placements; compute area products in signed 64-bit arithmetic. Solution written by GPT6.1