Since the black king is trapped on the A1 square, the solution is to get the black bishop to the b2 square. This is accomplished as follows (all moves are forced):
The “mate in four” prompt might be too big of a clue here. Let me explain my reasoning…
First, it is rather obvious that white cannot, within four moves, promote a pawn and engineer a checkmate with the newly acquired queen. Which means that mate must be delivered with the material already available.
Next, a cursory look at the position reveals that white must immediately sacrifice his light-square bishop – or else provoke stalemate. This, in turn, signifies that mate will be delivered with the other bishop, most likely on b2.
Once you have come to that conclusion, the full solution should quickly materialize.
All of black’s moves are forced…
1- Bc2 ; bxc2 2- b3 (b4 is possible too, if you want to be cute) 2- … ; cxb3 3- Ba3 ; b2+ 4- Bxb2#
Tempting to start looking at taking e6 and moving f7 (like I did)! Nice red herring by composer when the solution is at other end of the pawns diagonal!
1.Bc2! now black has a move. 1….bxc2 2.b3! now black has a move. 2….cxb3 3.Ba3 and black still has a move. 3….b2+ and now the coup de gras 4.Bxb2#
Funny puzzle. I think it is not so difficult to solve this without a computer. Basically you should give black a possibility to make a move. So…? 1. Bc2 bxc2 2. b3 (or4) cxb 3. Ba3 b2+ 4. Bxb2#
1. Bc2 (zugzwang) bxc2 2. b3 cxb3 3. Ba3 b2+ 4. Bxb2#
Cute!
1. Bc2 bc
2. b4 cb
3. Ba3 b2+
4. B:b2+
Nice idea with Bc2! 🙂
Bisop c2
Since the black king is trapped on the A1 square, the solution is to get the black bishop to the b2 square. This is accomplished as follows (all moves are forced):
1. Bc2, bxc2
2. b3, cxb3
3. Ba3, b2+
4. Bxb2 ++
1. Bc2 bxc2
2. b4 cxb e.p.
3. Ba3 b2+
4. Bxb2#
Mark
Solution:
1. Bc2 bxc2
2. b3 cxb3
3. Ba3 b2+
4. Bxb2 mate
1. Bc2….bxc2
2. b3…..cxb3
3. Ba3….b2
4. Bxb2#
Very nice problem 🙂
1. Bc2!! bxc2
2. b3 cxb3
3. Ba3 b2+
4. Bxb2 checkmate
Bc2 bxc2
b3 cxb3
Ba3 b2+
Bxb2#
Cool looking position, but definitely one of the easier puzzles considering all of black’s moves are forced.
Very nice problem 🙂
1. Bc2!! bxc2
2. b3 cxb3
3. Ba3 b2+
4. Bxb2 checkmate
Bc2 bxc2
b3 cxb3
Ba3 b2
Bxb2++
took a couple minutes.. i first wanted to move the the king.. then decided to force the pawns to move with the bishop..
1) Bc2, bc
2) b3 , cb
3) Ba3, b4
4) Bb2#
The “mate in four” prompt might be too big of a clue here. Let me explain my reasoning…
First, it is rather obvious that white cannot, within four moves, promote a pawn and engineer a checkmate with the newly acquired queen. Which means that mate must be delivered with the material already available.
Next, a cursory look at the position reveals that white must immediately sacrifice his light-square bishop – or else provoke stalemate. This, in turn, signifies that mate will be delivered with the other bishop, most likely on b2.
Once you have come to that conclusion, the full solution should quickly materialize.
All of black’s moves are forced…
1- Bc2 ; bxc2
2- b3
(b4 is possible too, if you want to be cute)
2- … ; cxb3
3- Ba3 ; b2+
4- Bxb2#
1. Bc2 bxc2 2. b3 (or b4) cxb3 3. Ba3 b2+ 4. Bxb2#
Bc2,bxB – b3,cxb – Ba3,b2 – Bxb2#
Tempting to start looking at
taking e6 and moving f7 (like I did)! Nice red herring by composer when the solution is at other end of the pawns diagonal!
1.Bc2! now black has a move.
1….bxc2
2.b3! now black has a move.
2….cxb3
3.Ba3 and black still has a move.
3….b2+ and now the coup de gras
4.Bxb2#
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Funny puzzle. I think it is not so difficult to solve this without a computer. Basically you should give black a possibility to make a move. So…? 1. Bc2 bxc2 2. b3 (or4) cxb 3. Ba3 b2+ 4. Bxb2#
1. Bc2 bxc2 2. b4 cxb3 e.p. 3. Ba3 b2+ 4. Bxb2#
1. Bf5-c2
forcing 1. … b3xc2
2. b2-b3
forcing 2. … c4xb3
3. Bf8-a3
forcing 3. … b3-b2+
4. Ba3xb2#
–br
I did not do the test but I was thinking Bg6 fxg6, f7 g5, Ba3 g4, f8(Q) g3, Qf1 g3, Kd2#
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An alternative checkmate (not in 4 moves though) is:
1. Bg6 fxg6 2. Ba3 g5 3. f7 g4 4. f8Q g3 5. Qg8 g2 6. Kd2 Kb1 7. Qg6+ Ka1 8. Ke2 g1Q 9. Qxg1#
1.Bc2 bxc2
2.b3 cxb3
3.Ba3 b2+
4.Bxb2#
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