Originally posted by wittywonkaEDIT: on second thought, I was wrong.
-White to move, mating in two.
-There are four possibilities...can you find them all?
[fen]4K3/8/2B3B1/3pBp2/3BkB2/3pBp2/2B3B1/8[/fen]
EDIT2 : on third thought, there is no mate in two. Whatever white does one of the bishops is going to be taken by one of the black pawns.
Originally posted by hypotheticalTwo of these are not solutions. Can you tell why?
1. Bb3 d2
2. Bc2#
or
1. Bd1 d2
2. Bc2#
or
1. Bf1 f2
2. Bg2#
or
1. Bh3 f2
2. Bg2#
People who come up with stupid problems like this must be either extremely bored or don't spend much time coming up with them...
(But then on the other hand I'm stupid for posting on this topic. 🙄 😛 🙄)
Originally posted by wittywonka
-White to move, mating in two.
-There are four possibilities...can you find them all?
[fen]4K3/8/2B3B1/3pBp2/3BkB2/3pBp2/2B3B1/8[/fen]
White mates in 2
This achieves the same idea as your problem without all the useless Bishops, and with tons less pieces on the board, and it's actually sound (there really is only one solution). In general, a composer ought to strive for economy.
Originally posted by BigDoggProblemBh2 b6
[fen]2K5/kB6/1p6/8/8/8/8/6B1[/fen]
White mates in 2
This achieves the same idea as your problem without all the useless Bishops, and with tons less pieces on the board, and it's actually sound (there really is only one solution). In general, a composer ought to strive for economy.
Bg1#
Obvious solution. I haven't checked it for a while, but I think ilywrin's profile puzzle has the same sort of theory...