This puzzle consists of 36 tiles. Each tile has a digit and may or may not have a mine.
The goal is to assemble a 6×6 Minesweeper grid — in other words, each digit must equal the number of mines adjacent to it (including diagonals). Unlike in traditional Minesweeper, the mines themselves have digits on them, but they only count the surrounding mines, not themselves. It’s therefore possible to have a mine with a 0 on it.
Two example puzzles and a possible solution for each are shown below. The solutions may unfortunately may not be unique (even beyond just rotation and reflection).
Here’s what we found so far:
- There are 𝑐 mines in the corners, each of which is seen by 3 tiles
- There are 𝑒 mines on the edges, each seen by 5 tiles
- There are 𝑖 mines on the inside, each seen by 8 tiles
- Therefore, 3𝑐 + 5𝑒 + 8𝑖 = 𝑠, where 𝑠 is the sum of all digits on all tiles
- Also, 𝑐 + 𝑒 + 𝑖 = 𝑛, where 𝑛 is the number of mines
- This gives rise to a set of Diophantine equations, which usually have only 1 or 2 solutions because 0 ≤ 𝑐 ≤ 4.
Beyond this, we haven’t found much else to go by. Is there a way to approach this puzzle that doesn’t amount to literal trial and error?
Any additional deductions would be helpful, even if they don’t solve the whole puzzle.
