Assuming standard Minesweeper rules, here’s one solution (with $ X $ = a mine):
EDIT: In response to Euphoric in the comments, I solved this purely by logical deduction with a bit of educated guessing to make things easier on me. But if you really want to know how I did it, here’s a rigorous solution:
Many Minesweeper games, including the one that ships with Windows, do not allow the first square you click to be a mine. The lower-left-hand corner is not necessarily safe, but if it's the first square you click, then it will be safe.
Although the puzzle is most likely to be solved without a computer, and we already have a winner, here are all 16 solutions, just for the record:
There are some symmetries in there, of course. Whether rotations and flips should count as "different boards" is a matter of interpretation.
Found with the following (quick and dirty) Java program that jus ...
I think that yes, it is possible.
You can create a custom grid with a ratio of more than 8 bombs per empty square, then, by the pigeonhole principle, there exists a bomb with no empty square next to it, since each empty square can be adjacent to at most 8 bombs.
I don't have my own picture but I found this on the internet:
There are 50 different possible ways that the unknown mines next to the revealed region could be configured:
Here, the green cells are clear (no mines), while the X's around the perimeter indicate the different ways the mines could potentially be placed.
If we consider each of these to be of equal probability (probably not quite true, because the total ...
The 2 down-right from your bad flag should have tipped you off; it is touching two bombs you had already flagged. That would show the one next to it (which you triggered) to be a bomb; the two under that one would clear the one left of it (above the 4), leaving just the four bombs around the 4. You could also have figured out the 4's surroundings just from ...
(I did this without looking at the CW. I claim no credit for imposing arbitrary restrictions on myself, but it means any mistakes are my own :-).)
The final grid is as follows:
which obeys the following constraint:
If we interpret
"First", "Second", etc., are of course
When groups of cells are referred to collectively,
Note: in the course of ...
As GentlePurpleRain says in their excellent answer, there are $50$ different possible placements for the mines in the squares around the solved region. However they make the assumption that each of these possibilities is equally likely. This is not correct.
The possibilities that GentlePurpleRain lists contain either $4$, $5$, $6$, or $7$ mines in the ...
This answer, although it does address the question itself, is more of an interesting observation coming from analysing the situation by expressing the system in terms of simple linear equations (eg/ using capitals for the reds, you might say A+B=1 to say that exactly one of A and B can be a mine, or F+G+H+I=2 because there must be exactly two mines amongst ...
The most you can do according to these new rules is 8 tiles, with the following grid:
1 2 3 3 3 2 1
2 # # # # # 2
3 # ? ? ? # 3
3 # ? 7 ? # 3
3 # ? ? ? # 3
2 # # # # # 2
1 2 3 3 3 2 1
# represents an already-discovered mine.
To show that this is maximal, look at the "information-conveying" tiles required by the instructions. We'll call these "useful tiles"...
I can get this much:
which is to say,
Here's how it goes.
[EDITED to add:] Here's a sketch of how to prove that
Note that I am writing this after reading other people's solutions :-) but I haven't read them in much depth and am not deliberately copying anything.
So, first of all,
What about the top region? Well,
So the only question ...
If you're going by Microsoft Minesweeper rules (or any where the number of mines are known), and assuming we aren't limited by the artificial limitation Microsoft provides on board creation, the answer is
which happens to be
and is a trivial solution. Once you mark the opening move, which will be an 8, the board is solved, either automatically in the case ...