How the AI Evaluates Moves

On every AI turn, the engine generates every possible valid equation from the AI's current rack tiles, then evaluates each one based on where it could legally be placed on the board. Each potential placement is scored based on: the base equation answer, any bonus cells hit (green +5, blue +10, purple ×2), and complexity bonuses for equation length and operator variety. The engine then selects from this ranked list based on the difficulty mode.

The Three Modes Explained

• Easy: The AI generates valid moves but deliberately picks from the bottom 40% of options by score. It also introduces a 1–2 second artificial delay so it does not feel instant. Designed so beginners can win, build confidence, and learn the game without being overwhelmed.
• Medium: The AI uses weighted random selection from the top moves. It will almost always take a good move, but not always the optimal one. Games feel genuinely competitive — close outcomes are common, and a skilled player can reliably win.
• Hard: The AI always selects the single highest-scoring valid move, instantly. It specifically scans for ×2 cell coverage first. There is no artificial delay and no mercy. This is how the AI plays at its absolute best — expect to lose often until you have truly mastered the strategy.

Why the AI Sometimes Passes

If the AI has no valid equations from its current rack tiles and no tile swap is available, it passes its turn. This is rare but can happen in the late game when the board is dense and the remaining tile pool is nearly empty. Like a human player, the AI is constrained to the actual tiles it holds — it cannot invent equations from tiles it does not have.

Can You Outsmart the Hard AI?

Yes, but it requires thinking ahead. The Hard AI optimises locally — it always takes the best move available right now. A skilled human player can exploit this by setting traps: sacrifice a small score now to claim a ×2 cell that the AI has not yet prioritised, then score massively on the next turn. The AI will not anticipate this unless the ×2 cell is already its optimal current move.