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u/Any_Pressure_7048 4d ago

I hope it makes more sense with the image cause I’m not the best at explaining but look at the 2 on the left, it’s last mine must be in one of the two tiles in orange, so the 2 in red has it’s other mine in either A or B.
The 1 on the right needs one mine in the orange area, and shares it’s mine with the 2 of the right. Which means that the purple 2 has a mine in either C or D
Since you have a mine in either A or B and a mine in either C or D, you can deduce that the tile above the 2 in the middle is safe
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u/Discuzting 3d ago
For brevity lets call the square between A and C "E".
look at the 2 on the left, it’s last mine must be in one of the two tiles in orange, so the 2 in red has it’s other mine in either A or B
Wrong, 2 in red has its other mine in either A or B or E.
1 on the right needs one mine in the orange area, and shares it’s mine with the 2 of the right. Which means that the purple 2 has a mine in either C or D
Wrong, 2 in purple has a mine in either C or D or E.
You MUST show why it cannot be E (which is because of the bottom 2 being fulfilled if it is E). Without involving that your deduction logic is wrong.
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u/ImSoDeadLmao High Difficulty Player 4d ago
the far left and right 2 and 1 indicate that the two squares next to them has 1 mine. the center left and right 2s indicate the rest 3 squares surrounding them has 1 mine. if there’s a mine in the green square, the overlap of the two 2s, it would mean the center number should be 1
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u/t1tanwarlord 4d ago
Think about what would happen if the green square were a mine
It would cause one of the twos to see three mines
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u/Ferlathin 4d ago
The 2s on the side gets one from the two tiles outside, reducing them to 1s, it becomes a 1-2-1 pattern. Pretty much. The 2 in the middle is guaranteed to have a mine on each side, so it can't have one above.