A lemmy community for the logic puzzle game minesweeper.
This is guess-free, and 14 mines left, all under those uncovered squares.
Any thoughts on how to proceed ?
EDIT - never mind. Tried brute force with the hope that the path I choose would be the right one - and lost.
But there's always a next ~~time~~ mine, lol.
And it was indeed heartening to see the Minesweeper community in Lemmy is not dead.
The squares around the top left most 4 has 1 chance out of 3 to have a mine. All the others have one chance out of 2 or worse. If you take a chance, I'd take it there.
Edit: never mind. How did the flag above that 4 get there?
Because the two covered squares to the left of the "4-with-flag-on-top-of-it" can have only 1 mine (because of the "2" there), which means the "3" needs one more mine, which can only be on top of that "4".
At least that's my reasoning.
I see what you mean. That means that the covered square that is top left of that 4 has 1/2 chance to be a mine. While the other two squares on its right each have 1/4 chance to have a mine in them since they require a 1/2 chance condition for it to be possible for a mine to be in either of those two remaining squares. Edit: legiblity
Sadly, it appears that only trying one of them successfully will still leave you stuck with 1/2 chance guesses so it doesn't help much still.
Bingo. I guess you got my reason behind flagging the square above the "4".
But I am still stuck.... May be brute force is the only way out...
The 3 4 3 section in the middle is where you want to start.
I think so too, but so have been unable to reason anything. Heading in brute-force is obviously the last resort for me, but was looking for something logical before I do that.
I believe the two squares immediately above and below the right 3 must be mines. Let me see if my reasoning makes sense: The 4 has five remaining squares around it, but three of those overlap with the 3 to its left. That 3 already has one mine marked, so those three squares that neighbor both it and the 4 can have at most two mines in them. That means the two squares that don't overlap must both have mines.
I was more going for the right 3 having the two rightmost squares be free of mines. Because the 4 has five squares surrounding it and four of those five squares are also in contact with the 3 on the right, no matter how the mines are arranged all three mines would be to the left or center of the 3. This leaves the two right squares unable to be mined.
Your reasoning makes perfect sense, and now I wish I didn't brute-force it head-first. Thanks for pointing that out !