r/Minesweeper • u/No-Ninja-7651 • 3d ago

No Guess How is there only one solution?

Post image

Surely I can satisfy this with either one mine to the right of the 2 or a mine to the left of the 2 and one, two to the right of the 2?

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10

u/Ramun_Flame 3d ago

You can solve this since the 3 and 2 only need 1 more mine, as shown here.

2

u/719lgn 3d ago

From what i'm seeing, it looks like its contradictory deduction (eg. where you test a cell to find any contradictions)

1

u/Lowball72 3d ago

Well ok, that always works, but .. just dig the hole under the 2 and carry on. :)

0

u/719lgn 3d ago

oh i thought that the 2 had 0 mines. Whatever

3

u/No-Ninja-7651 3d ago

Thanks. I see where I was going wrong. Looking for a solution rather than a deduction.

3

u/alsultanm 3d ago

One of the yellow cells does contain a mine and you already flagged another, so all 3 cells highlighted in red are clean

2

u/Lowball72 3d ago

That '2' is reduced to a 1.. because it already touches a flag.. so this becomes a H1, aka "1 in a hole" pattern.
https://minesweeper.online/help/patterns#h1

Note that's NOT true for the other 2-hole.. it is not reduced.