this post was submitted on 13 May 2024
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minesweeper

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This was a rare moment where I actually had to think to find it. Can you?

Screenshot is from a solver I found online because I didn't think to take a screenshot until I already opened the safe square.

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[–] Euphorazine@lemmy.world 1 points 5 months ago (1 children)

The corner of 1,2,4 on the left must be a bomb.

In the bottom gap, either side of the 1 being a bomb causes the 1,2,4 corner to be a bomb.

If the bomb is right of the 1, then the square above and below the 4 must be bombs because of the 2.

If the bomb is to the left of 1, the 2,2,3,3 forces one of the squares below the second 2 to be a bomb, which means the square below the 4 must be safe, so the square above 4 must be a bomb.

[–] brian@lemmy.ca 1 points 5 months ago (1 children)

If the bomb is to the left of the one, there could still be another bomb directly next to it, no?

[–] Euphorazine@lemmy.world 2 points 5 months ago* (last edited 5 months ago) (1 children)

No, the 3 below the two 1s has a shared partner with the 2 to it's left. So because it has a bomb directly above it, it must have one more below it. Therefore the space under the 4 would be empty and the space above it would have to be a bomb for the upper 2.

[–] brian@lemmy.ca 1 points 5 months ago

Got it now, you're very right. I didn't evaluate that 3 correctly at all.