# All values in a 6x6 Super Minesweeper grid

In a Super™ Minesweeper grid each cell is either a mine or a value. A value in row $$𝑟$$ and column $$𝑐$$ represents the total number of mines located in row $$𝑟$$ or column $$𝑐$$

Can you fill a 6x6 Super™ Minesweeper grid with mines such that every number from 0 to 8 appears at least once? Good luck!

Here is a similar question for 5x5: All values in a 5x5 Super Minesweeper grid

• I think there's a canonical way to do this by being greedy down the diagonal from the top left, putting $2n-4$ in the upper left and putting rows of $2n-2$ bombs in the top row and left column, leaving the corners empty. Continue down the diagonal, alternately shortening the column, then the row, bomb run, until you can't anymore (typically about halfway down. I've pushed this up to 12-14 without any problems. Nov 9, 2020 at 3:49
• Very nice Jeremy. I think that works. Nov 9, 2020 at 3:57
• I found solutions up to $n=28$ with $2n-4$ down to $n-1$ on the main diagonal, $n-2$ in the corners of the other diagonal, and no mines in the right column or bottom row. Nov 9, 2020 at 17:12

$$\begin{matrix} . & 5 & . & 3 & . & 6 \\ 5 & 3 & 4 & 1 & . & 4 \\ 4 & 2 & 3 & 0 & 4 & 3 \\ . & . & . & 4 & 8 & . \\ . & . & 7 & 4 & . & . \\ . & 6 & . & 4 & . & . \\ \end{matrix}$$
• Conjecture: $n \times n$ Super Minesweeper board for any $n \ge 5$ permits a configuration where all numbers from 0 to $2n-4$ inclusive appear at least once. Nov 9, 2020 at 2:04
• @Bubbler the conjecture is probably true. I was able to find solutions up to $n \leq 11$. Nov 9, 2020 at 2:30
• I found solutions up to $n=23$. Nov 9, 2020 at 4:34