Filling in an 8x8 minesweeper grid with mines (Day 10: Double Neighbors)

Question

_{Previous Minesweeper puzzles}

_{Based off of Day 10 of the Minesweeper Advent Calendar on heptaveegesimal.com}

_{Difficulty (approximate): ★★★★☆}

_{Contains: Single mines (1 mine), Anti-mines (-1 mines), Double mines (2 mines)}

---

To get the $\color{green}✓$:

1. Solve the 8x8 Minesweeper grid (the bonus puzzle is just for more practice)
2. Show the steps used to fill in the grid.

So today, we have a pretty cool gimmick. What is the gimmick, you might ask? Today, the direct neighbors are counted twice, however the diagonal neighbors are counted once as always.

There are going to be no missing numbers from the actual puzzle since the actual puzzle is a really difficult puzzle on its own already.

An example:

Take this grid (contains only single mines) (Missing: 8x2, 9x1):

4
			12
	10
				$\color{white}.$
5

Now, how is everything placed in this example? Here's how:

1. Note that mines that are orthogonal to a numbered cell are counted twice. Since there is only one way to add up to 4 and 5 using this info, we can place some single mines right away (m represents a single mine, x represents where a mine cannot go):

4	m
m	x		12
	10
m	m			$\color{white}.$
5	m

2. There is also only one way to get 12:

4	m	m	m	m
m	x	m	12	m
	10	m	m	m
m	m			$\color{white}.$
5	m

3. We can then deduce that there are mines at R3C1 and R4C3 because the total number of mines add up to 10 mines which is the only way to satisfy the 10 since we can't have a mine at R2C2 which would make the 4 unsatisfiable, which means that 9 goes at R2C2:

4	m	m	m	m
m	9	m	12	m
m	10	m	m	m
m	m	m		$\color{white}.$
5	m

4. Due to the gimmick, there is only one way to place mines to get 8 (3 orthogonal mines having a total value of 6 mines added to 2 diagonal mines to get a total of 8 mines) so we can fill in the rest of the grid to get the final unique solution:

4	m	m	m	m
m	9	m	12	m
m	10	m	m	m
m	m	m	m	8
5	m	8	m	m

And that is the final and unique solution to our example grid.

Bonus example for more practice (not required for the green checkmark):

Contains: Single mine (1 mine), Double mines (2 mines)

Mine count: Single mine x13, Double mine x11

Missing numbers: N/A

	8			12
		13
7				13
		18		12
10					9
6		11	7

The (actual) puzzle:

Contains:

+---------------+---+
|Single mine    |x10|
+---------------+---+
|Double mine    |x20|
+---------------+---+
|Anti mine      |x14|
+---------------+---+
|Missing numbers|x0 |
+---------------+---+

8				5		14
	14	8
				7		24
9	16		6
					7		7
13		7		-11
						-6	0
	8		-2		-7

Hint:

There are anti-mines at R2C4 and R2C5

Answer 1

The completed grid:

The solve path:

First, note that we have no missing numbers. So every square contains a mine or anti-mine.

We need double-mines in all the free spaces around the 24 and the 8 in R1C1.

Then we need anti-mines in all the free spaces around the -7 and the -11.

The -2 in R8C4 has -5 in known mines. We need 3 in its remaining unknown neighbours, which must both be single mines.

The -6 in R7C7 has -5 in known mines, so we need -1 in its remaining unknown neighbours. The adjacent 0 in R7C8 has -1 in known mines, so needs +1 in its remaining unknown neighbours. These are the same three squares, so there must be an anti-mine orthogonally adjacent to the -6 (in R6C7). (And the remaining unknowns in R6C8 and R8C8 must be a double-mine and an anti-mine, but we don't know which way round yet.)

The 7 in R5C6 has 1 in known mines. We need 6 in its remaining unknown neighbours, which must both be double-mines.

And that gives us 9 in known mines around the the 7 in R5C8. So the remaining unknown must be an anti-mine. (And R8C8 must be the double mine to balance the 0.)

The 7 in R5C3 has 12 in known mines. We need -5 in its remaining unknown neighbours, which must all be anti-mines.

That is all 14 anti-mines accounted for, so we know all the remaining unknowns are single- or double- mines.

The 8 in R8C2 has 3 in known mines. We need 5 in its remaining unknown neighbours, which must all be single-mines.

The 13 in R6C1 has 3 in known mines. We need 10 in its remaining unknown neighbours, which must all be double-mines.

The 7 in R6C3 has only a single unknown neighbour, which must be a single-mine.

The 9 in R4C1 has 6 in known mines. We need 3 in its remaining unknown neighbours, which must all be single-mines.

And similar forced choices around the 16, the 14 and along the top edge complete the grid.