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

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.

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.

• okay I really am sorry about the random edit that is about to take place but I just noticed a blue marble that is illegally placed unless I make a modification to the board, sorry about not seeing that. Oct 26 at 19:05
• For the clues I highlighted in my answer, I think they can all be corrected easily without anything else breaking, seem like just a miscount. Also the number of anti-mines and mines needs to be updated after the edit Oct 26 at 19:43

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:

• Ah, good job! Thanks for telling me which solutions don't work, I will definitely keep this in mind when creating future Minesweeper puzzles! However, your solution is in fact, correct. Oct 26 at 19:45
• @CrSb0001 no problem, I'd also say this was probably on the easier side, I think it is actually full solvable without using the cactus or blue marbles at all, so this could definitely be made harder! Maybe in future having it so there can be blank cells like in normal minesweeper could make it harder, as then not every cell is necessary full Oct 26 at 19:49