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| (1) Posted by Hauke Reddmann [Saturday, Jun 20, 2009 15:24] |
Waiting for an intro
Obviously I wasn't the first to notice that in Kc6 Bd5 Pa6/Kb8 Pa7 a second
bishop on White would be very beneficial. It wasn't hard either to set up
Kh1 Be8 Pa6 Pg7 - Kh6 Pa7 Ph7 (1.g8L!). A bit of recherche found exactly (?)
this position which might have benefitted from an intro.
Do you know other, even better, examples where White has to promote to
another bishop of the same color? Maybe even three?
Hauke
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| (2) Posted by Vlaicu Crisan [Saturday, Jun 20, 2009 18:40] |
I had the same idea in an endgame published some time ago in Diagrammes. However it has a very short introduction, but perhaps worth exploring further.
The position is: Kb3 Rb8 Bd8 pc6 h6 / Ka6 Ra1 pa2 a7 h7.
Solution: 1.c7! Rb1+ 2.Kxa2 Rxb8 3.cxb8=B!! (i) etc.
(i) 3.cxb8=S? Kb7 4.Sd7 Kc8 5.Sf6/Sf8 Kxd8 6.Sxh7 Ke7 7.Sg5 Kf6 8.h7 Kg7 draw
There are also two tries - first trivial, second more interesting:
1.Bc7? Rb1+ 2.Kxa2 Rb2+ 3.Ka3 Rb3+ perpetual check or stalemate
1.Rc8!? Rb1+ 2.Kxa2 Rb8 3.Rc7 Rxd8 4.Rxh7 Rd6 5.c7 Kb7 6.Rh8 Ra6+ 7.Kb3 Rb6+ 8.Kc4 Rc6+ 9.Kd5 Rxc7 10.h7 Ka6 draw (at that time there was no Thompson database to check this line)
However, as I learnt later, this position was first shown by Pal Benko. I don't remember the source (Chess Life in sixties?), but I hope some better documented lectors will help you.
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| (3) Posted by Arpad Rusz [Saturday, Jun 20, 2009 19:45] |
Here are some studies (there are many more):
A.Troitzky (1925)
 (= 5+4 )
Win
1.b7 Rg4+ 2.Kf2 Rg8 3.d6 Kc4 4.d7 Kb5! 5.d8Q Rxd8 6.Bxd8 Ka6 7.b8B!! [7.b8N+? Kb7 8.Nd7 Kc8 9.Nf6 Kxd8 10.Nxh7 Ke7! 11.Ng5 (11.Kf3 Kf7=) 11...Kf6 12.h7 Kg7=; 7.b8Q? stalemate] 7...Kb7 8.Be5 Kc8 9.Be7 Kd7 10.B7d6 Ke6 11.Kf3 a5 12.Kf4 a4 13.Ba3 Kf7 14.Kf5 Kg8 15.Kf6 Kh8 16.Kf7#
V.Korolkov (1929)
 (= 6+9 )
Draw (Cooked!)
1.f7! a1Q 2.h8B+!! [2.h8Q+? Kd3 3.Qxa1 Bd7+ 4.Kb4 a5+ 5.Kb3 Be6+ 6.Kb2 Bxf7 7.Qxb1 cxb1Q+ 8.Kxb1] 2...Kd3 3.Bxa1 Bd7+ [but 3...Be6! 4.f8Q Bc4+ 5.Kb4 a5#] 4.Kb4 a5+ 5.Kb3 Be6+ 6.Kb2 Bxf7 stalemate
H.Ginninger (1933)
 (= 8+6 )
Win
1.d8B! [1.d8Q? Nf1 2.Qxc7 Ne2 3.b8B Nf4+ 4.Qxf4 Ng3+ 5.Qxg3 stalemate] 1...Nf1 2.Bxc7 Ne2 3.b8B! [3.b8Q? Nfg3+ (3...Neg3+) 4.Bxg3 Nxg3+ 5.Qxg3 stalemate] 3...fxg6+ 4.Kxg6 1–0
V.Smyslov (1936)
 (= 9+7 )
Win
1.Bb1! a1Q+ 2.Kb5 Bg3 [2...Qa3 3.g7] 3.g7 Bb8! 4.g8B!! [4.g8Q? Qa4+ 5.Kxa4 stalemate] 4...Bf4 5.Bga2! Bxd2 6.f6! Bf4 7.f7 Bd6 8.Kc6 Bf8 9.Kc7 1–0
V.Troitzky, V.Korolkov (1938)
 (= 11+6 )
Win
1.Qc3 Nxc3 2.d8B! [2.d8Q? f1N 3.Qxc7 Nde4 (3...Ne2) 4.Bf4 Ne2 5.b8B Nd6! 6.Qxd6 Neg3+ 7.Bxg3 Nxg3+ 8.Qxg3 stalemate] 2...Ne2 3.Bxc7 f1N! 4.b8B! [4.b8Q? Ne4=] 4...Ne4 5.B1f4 Nfg3+ 6.Bxg3 N2xg3+ 7.Bxg3 Nxg3+ 8.Bxg3 1–0
V.Korolkov (1941)
 (= 10+7 )
Win
1.a8B!! [1.a8Q? Nc4 2.dxc4 d3 3.e5 Nc6 (3...Nd5) 4.Qxc6 Kh2 5.Qxf3 stalemate] 1...Nc4! 2.dxc4 d3 3.e5 Bxa8 4.c8B!! [4.c8Q? Bf3 5.Qb7 Nc6 (5...Nd5) 6.Qxc6 Kh2 7.Qxf3 stalemate] 4...Bf3! 5.Bb7 Nc6! 6.Bxc6 Bxc6 7.g8B!! [7.g8Q? Bf3 8.Qd5 Kh2! 9.Qxf3 stalemate] 7...Bf3 8.Bd5 Kh2 9.Bxf3 1–0
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| (4) Posted by Hauke Reddmann [Sunday, Jun 21, 2009 16:54] |
<hits himself repeatedly with chessboard>
How could I forget the last position when asking? I'm getting old :-)
Hauke
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| (5) Posted by Joost de Heer [Sunday, Jun 21, 2009 18:59] |
Some more (I found these in Nieuwe Schaakcuriosa by Tim Krabbe):
C. Bent, W. Veitch
 (= 5+4 )
Win
1. Sh6 Kf8 2. c6 Rb4 3. c7 Rc4 4. Bc4 Bc2 5. Kh8 Bf5 6. Sf5 Se8 7. c8=B! and white wins
A. Troitski
 (= 6+7 )
Win
1. Qg3 Qg3 2. ab8=B Kxd5 3. Bg3 (2. ab8=Q? 3... e1=Q!) Ke6 4. Ka5 (4. Ka4? Kd7 5. Ka5 Kc8 6. Kb6 e1=Q 7. Be1 Kb8=) Kd7 5. Kb6 Kc8 6. Ka7 Kd8 7. Kb8 Kd7 8. Kb7 Kd8 9. Kc6 Kc8 10. Bcd6 Kd8 11. Bh4 Kc8 12. Bh2 e1=Q 13. Be1 Kd8 14. Kb7 Ke7 15. Bh4 Kd7 16. Bg5 Ke6 17. Kc6 Kf5 18. Kb5 Ke6 19. Ka4 Kd7 20. Kb5 Kc8 21. Kb6+
V. Korolkov, A. Doluchanov
 (= 7+6 )
Win
1. g8=Q ef2 2. Kf1 Rg8 3. hg8=B (3. hg8=Q? Bc4 4. Qc4=). In this position, black will draw if he can reach c7 and keep the bishop on the board. There are only two serious defences, in other cases the bishop is caught easily (e.g. 3... Bc8? 4. Bge6 Bb7 5. Bcd5).
A) 3... Bd7 4. Kf2 Kb4 5. Kf3 Kc5 6. Bge6 Be8 7. Kf4 Kd6 8. Kg5 Ke5 9. b3 Kd6 10. Bh3 Ke5 11. Bg4 Kd6 12. Kf6 Kc7 13. Bce2 Kd6 14. Bd3 Kc7 15. Ke7+
B) 3... Bf5 4. Kf2 Kb4 5. Kf3 Kc5 6. Kf4 Bh3 7. Ke5 Bd7 8. b3 Bg4 9. Bge6 Bd1 10. Bd7 Bf3 11. Ke6 Bg4 12. Ke7 Bf3 13. Bc6 Bg4 14. Be6 Bd1 15. Kd7 Be2 16. Kc7 Bd1 17. Kb7 (17. b4? Kb4 18. Kb6 Ba4!=) Bc2 18. Ka6 Bd1 19. Bc4 Bc2 20. B6d5 Bd1 21. Bd3!! zz Kd5 22. Kb6 Bb3 23. Kc7 Kc5 24. b6 Bd5 25. Ba6 Bf3 26. Bb7 Be2 27. Bg2 Ba6 28. Bf1 Bc8 29. Be2
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| (6) Posted by Sarah Hornecker [Sunday, Jun 21, 2009 19:12] |
http://rankzero.de/?p=4048
Not to forget this study by Teschke 2008
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| (7) Posted by Joost de Heer [Sunday, Jun 21, 2009 21:00] |
And a Troitski joke. Not entirely what OP wanted (the extra bishops are already present in the diagram), but white needs all 5 bishops to win!
A. Troitski
 (= 6+2 )
Win
1. Bce5 a5 2. Ba1 a4 3. Bbe5 a3 4. Kd2 Ka2 5. Kc3 Ka1 6. Kb3 Kb1 7. Ba1 a2 8. Kc3 Ka1 9. Kc2
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