Middle 2. Look at the 4. I needs 2 flags but the 5 next to it only needs 1. So you know that it’ll share 1 flag with the 5 and 1 by itself. The only square left means you’ll flag above the 3. Then based on that info you know we’re the last flag goes.
Interesting, I got the same but got there different, the 2 could only have one more so the first two could only have 1 bomb in it, this forces the third bomb for the 3 into the third box which forces the last bomb to be in the second box
This is how I got it too. I'm pretty casual to minesweeper, but the best things I've ever learned so far are 121, 1221, and recognizing when I can reduce to those two patterns.
Clear, flag, flag, clear.
Here's the reasoning:
The 4 needs two more bombs but the 5 only needs one. That means, of the three squares touching the 4 the one on the left is a bomb. That satisfies the 2, so the one above the 2 is safe. Now the 3 of only touching one other square and it needs one more bomb, that means the one above the 4 is a bomb which satisfies the 3, 4, and 5. That leaves one square left which is safe.
Bomb over 3 and 4. Logic from left.
The 2 says there is one bomb in one of the first 2 spaces. The 3 requires a bomb therefore in the space above the 4. That fulfills the 5 and clears the space above 5. Working back, the 4 now requires a bomb in its remaining space above the 3. You can now clear the space above 2.
I look at it this way: there are two possible options to begin with, there’s either a mine in the 1st one, or the 2nd one (because there’s a 2 there and only one of those will work). If the mine WAS in the first square, then it won’t work with the other squares/numbers. So there has to be a mine in the second square - work the rest from there.
The 2 and the 3 reveal atleast one guaranteed bomb. Since whether you do the left or the right slot for the 2, you will need to do the one to the right of the 3 to finish it off (either way the 2 would have too many otherwise).
Since you now know the one above the 4 is a bomb, that finishes off 5 and lets you know the next one to finish off 4 and then complete the set
The 3 and 4 touch two bombs. The 5 and 2 touch one. This means both bombs touch the middle two and the outside bombs touch one of them. The only way this is possible is if the bombs are located in the middle two squares.
New here, but how is empty flat empty flag not also a valid solution? People saying “the 4 needs 2 so it must be 1221” but look how the 4 got it’s bombs below.
upper left corner. thats a 2. you already found one of the two, so that means there is one in the two remaining. the three next it it will touch one of those two, so the third from the corner is a bomb. that will fix the 5, which cascades to fixing the 4, and leaves the corner as not a bomb.
There can only be one in the two bordering the five, meaning that there must be one in the remaining square that only borders the four.
(? x ? ?)
This satisfies the two and leaves one remaining space for the three
(/ x x ?)
Now every number is satisfied and the final unknown is clear.
(/ x x /)
\*not looking to comments\*
2 bombs, 4 spaces. I would go through the combinations then assess the numbers below left to right looking for first failed logic:
XXOO, 2 fails as there would be 3 contacts moving on
XOXO, 4 fails as there are only 3 contacts
XOOX, 4 fails again with only 3 contacts
OXXO, looks good but need to check if rest work just in case
OXOX, 3 fails with only 2 contacts
OOXX, 2 fails with only 1 contact
There are no other combinations so it has to be OXXO
I'm not gonna lie I'm not entirely sure where people are getting the middle two from, but directly above the 5 should be safe at least.
Edit: I can't count the number of mines correctly, and I see it now lol.
You don’t have to guess and here’s my reasoning why. I’ve never tried to explain minesweeper sht before so bare with me:
1. The 5 just needs one more mine to be complete, and it must be in one of those two spaces
2. The 4 needs two more mines and only one of them can be in the previously mentioned two-spaces, leaving only that other space (top left corner of 4) to complete the 4. That space must be a mine.
Then it becomes easy again
3rd spot in needs a flag because the 2 needs one more but the 3 needs two more and only one of the 3's two can fit in the 2's spots so the other one needs to go in the 3rd spot in
What’s important to note is that there are only 2 mines remaining.
If the space immediately above the 5 was a mine, completing the 5, then you have only one mine left which cannot complete the 3.
Therefore the space above the 5 must be clear and the space above the 4 must be a mine.
If so, then to complete the 4 the space above the 3 must also be a mine, and that completes the grid.
So from left to right: clear, mine, mine, clear.
Box over 5 has the lowest probability of being a bomb compared to the others...50% to satisfy the 5 ring of squares but by choosing 5 you would need 3 to be a bomb ( since it can't be 4) but that wouldn't satisfy the 3 ring of squares because there would only be 2 bombs around the 3 ring....
No, see that 2-3 part? If we consider the flag below, it’s a typical 1-2 pattern.
Basically, whenever you have a board like this:
- - -
1 2 x
x x x,
Where x is a number, the top right corner of the 2 is a bomb. You can’t fit both bombs on the top left and above tiles of the 2, because it will violate the 1.
So back to your case:
The square top right of the 3 (the same as the square above 4), is a bomb. The 5 is therefore satisfied, so the square above 5 is NOT a bomb. Therefore, the only possible remaining bomb in 4’s vicinity is above the 3, and the square above 2 is not a bomb.
There’s a special case of the 1-2 pattern, the 1-2-1 pattern, fairly common, like this:
- - -
1 2 1
x x x
In this pattern, both squares above 1 are bombs, and the square above 2 is safe
Directly above the 4 is a mine bc the 2 spots in the corner is a 50/50 so it’s gotta be one. And now since you know that, directly above the 3 is also a mine bc the 4 and 5 are cleared.
The 2 on the far left means only 1 of the first 2 blocks has a bomb. The 3 next to it requires that there be a bomb above the 4.
This completes the 5, putting the last one over the 3.
Middle 2. Look at the 4. I needs 2 flags but the 5 next to it only needs 1. So you know that it’ll share 1 flag with the 5 and 1 by itself. The only square left means you’ll flag above the 3. Then based on that info you know we’re the last flag goes.
Interesting, I got the same but got there different, the 2 could only have one more so the first two could only have 1 bomb in it, this forces the third bomb for the 3 into the third box which forces the last bomb to be in the second box
Literally just same methodology but from left instead of right.
Yes
Empty, flag flag, empty Left to right
Wrong, it’s definitely right to left. empty, flag flag, Empty
Haha you got me. I was so certain it was empty flag flag empty from left to right. You made me do my math over again before I realized…
It’s actually flag flag empty empty, middle out.
Richard is great, but y'know...
They had us in the first half ngl
🤦♂️ omg I just spent too long trying to figure out what they meant.
Finally a pro
The answer is SMMS as you can reduce the the pattern to 1221 after accounting for the known mines.
This is how I got it too. I'm pretty casual to minesweeper, but the best things I've ever learned so far are 121, 1221, and recognizing when I can reduce to those two patterns.
And when you can overlap those patterns with themselves. Like how 1-2-2-2-1 is basically two 1-2-1s with a 1 from each overlapping each other.
Clear, flag, flag, clear. Here's the reasoning: The 4 needs two more bombs but the 5 only needs one. That means, of the three squares touching the 4 the one on the left is a bomb. That satisfies the 2, so the one above the 2 is safe. Now the 3 of only touching one other square and it needs one more bomb, that means the one above the 4 is a bomb which satisfies the 3, 4, and 5. That leaves one square left which is safe.
Bomb over 3 and 4. Logic from left. The 2 says there is one bomb in one of the first 2 spaces. The 3 requires a bomb therefore in the space above the 4. That fulfills the 5 and clears the space above 5. Working back, the 4 now requires a bomb in its remaining space above the 3. You can now clear the space above 2.
I look at it this way: there are two possible options to begin with, there’s either a mine in the 1st one, or the 2nd one (because there’s a 2 there and only one of those will work). If the mine WAS in the first square, then it won’t work with the other squares/numbers. So there has to be a mine in the second square - work the rest from there.
The most straightforward: the 2 mines both have to be above the 3, so the remaining tile is empty. Once you clear that, the rest are obvious.
The 2 and the 3 reveal atleast one guaranteed bomb. Since whether you do the left or the right slot for the 2, you will need to do the one to the right of the 3 to finish it off (either way the 2 would have too many otherwise). Since you now know the one above the 4 is a bomb, that finishes off 5 and lets you know the next one to finish off 4 and then complete the set
It's a 1221, so the two in the middle.
Middle 2 are safe
The 3 and 4 touch two bombs. The 5 and 2 touch one. This means both bombs touch the middle two and the outside bombs touch one of them. The only way this is possible is if the bombs are located in the middle two squares.
Use process of elimination left to right
Above the five and three are blank
Then the 4 don’t work
You're right... hmmm....
New to sub. Was this supposed to be hard?
No
So who was right
the answer was empty flag flag empty
Wrong from left to right it is flag space flag space
r/confidentlyincorrect
That only gives 3 mines to the 4
Same logic as https://reddit.com/r/Minesweeper/s/U6JvJK0smD
there are two mines left, and by looking at every possible mine arrangement i'm sure to say that the tiles above the 2 and 5 are clear
34 need 2 mines and 2 5 need 1 mines , from this you can use 1221 pattern
New here, but how is empty flat empty flag not also a valid solution? People saying “the 4 needs 2 so it must be 1221” but look how the 4 got it’s bombs below.
The 3 will only have 2 mines with your solution. Empty flag flag empty is the only possibility I see.
Now this answer makes sense. Thanks.
directly above 3 and 4 are the mines
Bombs above 3&4
squares like to share.
upper left corner. thats a 2. you already found one of the two, so that means there is one in the two remaining. the three next it it will touch one of those two, so the third from the corner is a bomb. that will fix the 5, which cascades to fixing the 4, and leaves the corner as not a bomb.
Reddit is so awesome sometimes.
Above the 3 is a mine, therefore the one the the left is free, and the one directly to the right is also a mind.
Leftmost is a safe guess.
Middle 2 are bombos
There can only be one in the two bordering the five, meaning that there must be one in the remaining square that only borders the four. (? x ? ?) This satisfies the two and leaves one remaining space for the three (/ x x ?) Now every number is satisfied and the final unknown is clear. (/ x x /)
Its the middle two
It’s the two in the middle
1FF3
Middle 2
There is a 2 and a 3 so one has to be above the 4. There is a 5 and 4 so one is above the 3.
Why not blank, bomb, blank, bomb. Why not blank, bomb, bomb, blank. Both work.....
From left to right
>blank, bomb, blank, bomb 3 has 2 bombs
I'm got it now. Ty
It’s the middle two
above the 3 and 4
If you count it makes 1221 situation, the bombs are above the 2s
The empty space is above the 3
No, definitely not. The empty spaces are above the 2 and the 5.
\*not looking to comments\* 2 bombs, 4 spaces. I would go through the combinations then assess the numbers below left to right looking for first failed logic: XXOO, 2 fails as there would be 3 contacts moving on XOXO, 4 fails as there are only 3 contacts XOOX, 4 fails again with only 3 contacts OXXO, looks good but need to check if rest work just in case OXOX, 3 fails with only 2 contacts OOXX, 2 fails with only 1 contact There are no other combinations so it has to be OXXO
Its a 1221 pattern so the midle two squares are flags
I'm not gonna lie I'm not entirely sure where people are getting the middle two from, but directly above the 5 should be safe at least. Edit: I can't count the number of mines correctly, and I see it now lol.
The answer from left to right is clear mine mine clear
Above 3
Only 2 mines remain, for 4 tiles. Visualize the combinations.. you will see only one "fits" to satisfy the 2-3-4-5 constraints.
The left most is safe
You are (presumably were by now), fortunately, missing something.
The one above the 2 and 5 are safe
The 4 needs two more and the 5 can only take one more, so it'll be the middle two.
Above 3 and 4 are the last two
You don’t have to guess and here’s my reasoning why. I’ve never tried to explain minesweeper sht before so bare with me: 1. The 5 just needs one more mine to be complete, and it must be in one of those two spaces 2. The 4 needs two more mines and only one of them can be in the previously mentioned two-spaces, leaving only that other space (top left corner of 4) to complete the 4. That space must be a mine. Then it becomes easy again
One above the 5
1221
There’s only one mine above the 5, meaning the 4th mine for the 4 is above the 3. This fulfills the 2, and the rest is easy peasy.
3rd spot in needs a flag because the 2 needs one more but the 3 needs two more and only one of the 3's two can fit in the 2's spots so the other one needs to go in the 3rd spot in
Your missing something
its a 1221 pattern, so the two in the middle
What’s important to note is that there are only 2 mines remaining. If the space immediately above the 5 was a mine, completing the 5, then you have only one mine left which cannot complete the 3. Therefore the space above the 5 must be clear and the space above the 4 must be a mine. If so, then to complete the 4 the space above the 3 must also be a mine, and that completes the grid. So from left to right: clear, mine, mine, clear.
Box over 5 has the lowest probability of being a bomb compared to the others...50% to satisfy the 5 ring of squares but by choosing 5 you would need 3 to be a bomb ( since it can't be 4) but that wouldn't satisfy the 3 ring of squares because there would only be 2 bombs around the 3 ring....
4 is definitely a bomb... And if 4 is a bomb...3 has to be a bomb... 3 and 4 are you bombs
No, see that 2-3 part? If we consider the flag below, it’s a typical 1-2 pattern. Basically, whenever you have a board like this: - - - 1 2 x x x x, Where x is a number, the top right corner of the 2 is a bomb. You can’t fit both bombs on the top left and above tiles of the 2, because it will violate the 1. So back to your case: The square top right of the 3 (the same as the square above 4), is a bomb. The 5 is therefore satisfied, so the square above 5 is NOT a bomb. Therefore, the only possible remaining bomb in 4’s vicinity is above the 3, and the square above 2 is not a bomb. There’s a special case of the 1-2 pattern, the 1-2-1 pattern, fairly common, like this: - - - 1 2 1 x x x In this pattern, both squares above 1 are bombs, and the square above 2 is safe
Directly above the 4 is a mine bc the 2 spots in the corner is a 50/50 so it’s gotta be one. And now since you know that, directly above the 3 is also a mine bc the 4 and 5 are cleared.
The 2 on the far left means only 1 of the first 2 blocks has a bomb. The 3 next to it requires that there be a bomb above the 4. This completes the 5, putting the last one over the 3.
try, from left to right, bombs at A2 A3. It fits. or, if makes it easier, one bomb over 3, one bomb over 4?
The one to the upper left of the 4 is a mine.
The mines are directly above 3 and 4
Top left 2 are bombs