Filling in an 8x8 minesweeper grid with mines (Day 4: Looking Up)

Question

We have a sort of weird condition for the tiles today. What is that, you ask? Well, we actually have 2:

1. The visible neighbors are different. Here is what they will be for today (the "x" represents a tile that can be seen, the "[]" represents the tile):

x	x	x
x	x	x
	[]

2. This is a Mobius Strip Y. What this means is that a tile at, for example, R1C4 on an 8x8 Minesweeper grid will have its value affected by R8C3, R8C4, R8C5, R7C3, R7C4, and R7C5.

Here is the puzzle:

-2	-2	0	1	0	-3
			-2		-2		1
	-1			1	3
-3	-1			2
			3			3	4
$\color{white}{.}$
4	4				-3		-2
			6				-3

There are 23 single mines (1 mine) and 17 anti mines (-1 mines) if I remember correctly - I have lost the key but luckily I had already created the puzzle here so I didn't have to recreate it. (I started creating the puzzle at around 7pm UTC and wrote this paragraph as of around 1:20pm UTC)

Answer 1

Solution:

(Note the 3 in the middle is an error and should be a 4)

---

Step by step:

1:

To start, we can shadow the bottom two rows at the top to help with our deductions.

The 6 is an easy start point, it must have all mines above it. The -3 to the right must now have all anti-mines remaining. The -2 and -3 bottom right can also have all anti-mines filled in above them:

2:

The 4 on the bottom left must have all mines above it. The -3 a few cells above must also have all anti-mines above it. The -2 top left means there must be anti-mines bottom left. The 1 at the top middle also gives anti-mines at the bottom.

3:

The -3 at the top now also gives anti-mines at the bottom. The 1 top right must have the remaining cells as mines, and the 4 on the side at the right similarly.

4:

Finally, the 2 in the middle gives 2 more mines, the incomplete 4 bottom left must have an anti-mine in the remaining slot, the -1 to the right of the -3 on the left gives two mines. The 3 top right has a remaining mine, the 3 next to the 4 on the right has a remaining anti-mine.

The remaining two cells can't be solved, as the 3 is a mistake. However, from the fact there should be 23 mines, and that there is 21 currently placed, we can assume they should be mines and that the 3 should actually be a 4, and we get the answer: