Sounds like a classic cellular automaton with a twist—like a Frost‑Pythagoras game. You’ll need to decide what constitutes a “frosty” neighbor. Maybe count adjacent cells that are already ice; if that count is between one and three you freeze the cell, otherwise it stays dry. The recursion can then be a simple depth‑first expansion from the seed, but you have to guard against infinite loops by marking visited cells or limiting recursion depth. Don’t forget the edge cells: you could wrap the board to make it toroidal, or simply treat out‑of‑bounds as dry—makes the crystal shape more jagged. Also, if you want realism, consider a stochastic component: a small probability that a cell fails to freeze even if the rule is satisfied, to mimic imperfect deposition. Once you have the rule, just iterate until no new cells change state; that’s a good termination condition. And if you’re curious about performance, try memoizing neighbor counts or using a queue instead of pure recursion—Python’s recursion depth is limited, but a simple loop will do the trick without the stack overflow drama.

Got it—I'll ping you as soon as the grid starts showing that unmistakable branching frost. In the meantime, keep an eye on the seed cells; if they’re too isolated, the whole thing might collapse into a single block. Happy coding!