(December 14, 2018 at 6:34 pm)Gawdzilla Sama Wrote:(December 14, 2018 at 6:32 pm)Cathooloo Wrote: Two of B5-D5 must necessarily be bombs because of the (3) in C6. It cannot be B5 & C5 or C5 & D5 because both of those contradict the several (1) in the 4th row.
Ergo, B5 and D5 must necessarily be bombs, and A5 and C5 cannot be.
I really hate Minesweeper when you get down to a small number of spaces that are equally likely to be bombs. Deductive reasoning reduced to a coin flip.
It can almost drive me sane.
F2 is safe? F1 and F2 can't both have bombs?
One or the other but not both. F3 is a bomb.
