Filling in an 8x8 minesweeper grid with mines (Day 3: Zen Garden Intro)

Question

_{This day is inspired by Day 20 of the Minesweeper Advent Calendar on heptavegeesimal.com}

_{rot13(ab jnl! vzntrf jbex ba gnoyrf! :Q)}

_{okay hopefully this is my easiest Minesweeper puzzle so far}

---

Today we have 2 new conditions for today, other than the mines that are present. Here are what they are:

Image	New condition name	How it works(1)
	Blue marble	This tile has a mine exactly one mine knights path away from it.(2)
	Cactus	For any tile adjacent to the cactus, that tile has a mine if and only if the corresponding tile on the opposite side of the cactus has a mine.(3)

⁽¹⁾As mentioned (and slightly rewritten) on Day 20 - Zen Garden

---

⁽²⁾If you're confused by how "Blue Marble" works, here's a quick rundown of how it works:

On Day 6 - Knight's Path, it was noted that tiles would signify the amount of mines that could be "seen" from how a knight moves in chess. For example, if you're unfamiliar with how a knight in chess works, here's how it can legally move (the "kn" represents where the knight is and the "x"s represent where it can legally go):

	x		x
x				x
		kn
x				x
	x		x

So a blue marble on R4C4 could have a mine at:

1. R3C2
2. R2C3
3. R5C2
4. R6C3
5. R6C5
6. R2C5
7. R3C6
8. R5C6

If one of the aforementioned tiles has a mine, all of the other tiles that could have a mine are safe automatically. However, it is allowed for there to be a mine on say, R5C4 (assuming a blue marble on R4C4) since a knight cannot legally move there in one move.

---

⁽³⁾As mentioned by Gareth McCaughan ♦.

---

_{Original rewrite of mine: Out of the 4 diagonal tiles, only the opposite tiles can have mines. (I wasn't aware of orthogonal cactus mines)}

---

An example: Say we have a cactus on R4C4. Then we can have mines at either

1. R3C4 and R5C4
2. R4C3 and R4C5
3. R3C3 and R5C5
4. R3C5 and R5C3

---

Here's the puzzle:

	-2				1		2
			-4	-3	-1		1
0	-1		-3		-1		0
	1		-3	-2			-2
	2		-3		-3		-1
	1		-3	-3			0
	3			-2	0	1
2				-1	1		2

Plain text version (b = blue marble, c = cactus):

b	-2	b			1		2
	c		-4	-3	-1		1
0	-1	c	-3		-1		0
	1		-3	-2	b	c	-2
c	2	c	-3		-3		-1
	1		-3	-3		b	0
	3	b		-2	0	1
2			b	-1	1		2

Mines used: Anti-mine (-1), single mine (1). There are 12 anti-mines and 10 single mines.

Note: Blue Marble and Cactus only detect a mine's position, but not what the value of a mine is.

Answer 1

Solution:

---

Step by step:

1:

Starting in the top left, the -2 must have two anti-mines. The -4 at the top can also be solved simply, as can all the -3s in the middle column.

2:

The 3 bottom left must have all remaining cells as 1s. The 2 bottom right, and top right as well. The 0 at the top right means we can place an anti-mine, which confirms the cactus below it. Finally the cactus on the left gives the remaining mine

However note:

Some of the clues in column 2 don't work, which seems to be a mistake. I've highlighted the clues below in red: