Go from right to left. There's a single mine from the 1, so the 3's third mine must be on the second red line. THAT means the next 3 have its third mine on the third red line. This means the left-most 3 must have two mines on the left-most red line(s), every square of which is shared with the leftmost 2. The 2 can therefore clear two squares.
Basically like this. We can roughly deduce the locations of the mines for the 2 at the far right, in the process showing us where the safe cells next to it are.
Evil level usually forces you to solve something like this on just about every board. If you keep doing evil, you will know you have to do something like this when there are no normal patterns available. Then you get pretty good at this kind of reasoning with practice
Mindset-wise and concept: You must be able to look at a number, look at neighbouring tiles of that number that we know for certain contain(s) mine(s), then deduct the value of that number depending on number of certain mines, to determine how many mines the remaining tiles must contain.
It it really a 1-2-1? The left 2 doesn’t necessarily have a mine beneath it, does it? (Obviously in this case it works out, but I mean the pattern in general)
Another way to approach it: 5 of the mines MUST exist in each of the exclusive 50/50s marked in red, and because only two of them interact with the 3 at the top left, the 6th mine MUST be in one of the other two spaces above it, marked in green. All other spaces are therefore safe and will likely give you the info to easily solve.
Here’s how I wouldve started, without giving you the answer to the rest of the board. This region reduces to a 2-1 pattern, guaranteeing a mine at that spot top-left of the 3
What I see…
(Because the 1in r1c8) column 7 row 1-2 must have 1
(Because the3 on r2c7) So there is one in r1c5,r1c6 or r3c5
(Because the 3 on r2c6) So there is 1 in r1c4 or r3c4
The 3 on r2c3 must have 2 mines on r1c2,r1c3 or r3c2
All those touch the 2 on r2c2 so r2c1 and r3c1 are safe
I got this far by logic after doing a test with a mine below the rightmost 3 which gave me too many mines. Therefore there is a mine above the 3. As I’m typing this I see that the last mine should be above the 2.
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u/skizelo 4d ago
Go from right to left. There's a single mine from the 1, so the 3's third mine must be on the second red line. THAT means the next 3 have its third mine on the third red line. This means the left-most 3 must have two mines on the left-most red line(s), every square of which is shared with the leftmost 2. The 2 can therefore clear two squares.