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White wins. Bishop intercepts queen wanna-be pawns and then helps check black king.
Fritz agrees with me.
To anon 11:06
If Bishop takes the f,g pawns fritz gives plus several points (+-) to white, though white can make no progress.
1. Bd5 f4
2. Be6 g3
3. Bd5 f3
4. Bxf3 g2
5. Bxg2 Ka8
6. Kd7 Kb8
7. Bf3 Ka8
8. Bxb7+ Kxb7
9. Kd6 Kb8
10. Kc6 Kc8
I don’t see how White can make progress. So my verdict is that this is a draw.
Fantastic win for white!!
Yes, white has a wondeful win in
her (I know it’s Susan’s blog!) hand.
It is amusing that my Shredder 11 with
tablebases up to 5 pieces didn’t find a
win for over 1 minute.
Even with his bishop, white cannot corner black sufficiently for a checkmate without stalemating him first; nor can he advance a pawn; he would lose his bishop and achieve the same result.
This is a draw!
This is not a draw.
White can win in a fabulous fashion!
Mate in 12!
The key strategy is to save one of black’s pawns to avoid stalemate, then take the a-pawn. If white takes the bishop, then move in with the king and mate, while black queens the pawn.
”…If white takes the bishop, then move in with the king and mate, while black queens the pawn.”
At that point the Black king should be on a8.
Maybe 1.Kd7 and now:
a) 1. … g3 2.Bd5 f4 3.Bf3 g2 4.Bxg2 f3 5.Bf1 f2 6.Kd8 Ka8 7.Bxa6 Kb8 8.Bf1 Ka8 9.Kd7 White takes f2 and wins
b) 1. … f4 2.Bxg4 f3 3.Bh3 f2 4.Bf1 Ka8 5.Bxa6 and like above.
Do I miss something or does this work?
To Arjan,
you got it right, i agree with you.
”1. … g3 2.Bd5 f4 3.Bf3 g2 4.Bxg2 f3 5.Bf1 f2 6.Kd8 Ka8 7.Bxa6 Kb8 8.Bf1 Ka8 9.Kd7 White takes f2 and wins.”
7.Bxa6 Kb8
8.Bf1 Ka8
9.Kd7 Kb8 [also wins 9.a6]
10.a6 wins.
[also wins 10.Bg2]
Pharaoh
There’s a lot going on in this gem of a study.
As several posters have pointed out, White can’t simply capture both the f- and g- pawns, because he can’t make any progress in the resulting ending. White has to capture the g-pawn before blockading the f-pawn on f1.
The bishop can’t leave the c8-h3 diagonal until one of Black pawns advances. For example, 1.Bd5? f4! 2.Be6 g3 3.Bd5 f3 (threatening 3…g2). Therefore:
1.Kd7! g3
1…f4 2.Bxg4 f3 3.Bh3! f2 (3…Ka8? 4.Kc8 f2 5.Bg2 f1=Q 6.Bxb7#) 4.Bf1 forces 4…Ka8, which loses (details below).
Of course not 1…Ka8? 2.Kc7! (or c8) followed by 3.Bd5 and 4.Bxb7#.
2.Bd5 f4 3.Bf3
Necessary, to stop 3…f3.
3…g2
4.Bxg2 f3
5.Bf1
5.Bh3 is also good enough. But not 5.Bxf3? Ka8 and White is stymied.
5…f2
6.Kd8! Ka8
7.Bxa6!
… and wins. If 7…Kb8 8.Bf1. Or 7…bxa6 8.Kc7 f1=Q 9.b7+ Ka7 10.b8=Q#.
Amazing, amazing, amazing!
This puzzle is stunning.