172
1. Incorrect studies? To be corrected!
Here a list of studies which are possibly incorrect and most were not yet discovered that way in the Harold van der Heijden database III 2005.
If in that database in the source of a study an "@" is found, it means that an incorrectness is already found.
The studies shown here are not yet that way. Also the term Busted is used.
You are invited to prove me wrong, or correct the studies in some way.
Many of them still have beautiful ideas in there!
Please react by using the guestbook Latest additions made on 6-Nov-2007.
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Contribution of Ivica Mihoci from Croatia.
He has made a chessbase-file with remarks (corrections and errors) in pawnendgamestudies.
He has checked 4413 pawn studies.
Studies are flagged with green (endgame) medal for found improvements or corrections, and with black (blunder) medal for found errors.
Please check them out yourself ....!
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172
Bondarenko,F Kakovin,A 1958 After 8...Qxe2
[=3530.87h5a7]
Solution: 1. Ra8+ Kxa8 2. b6 Rxd5 3. Rxd5 Bxe5 4. Rxe5 Qa4 5. b5! Qxb5 6. Rxb5 h1=Q
7.Re5! Qh2 8. g3! Qxg3 9. Re8+ Qb8 10. Rc8! a4 11.Rc5! 1/2-1/2
But black wins with: 8... Qxe2+ 9. g4 Qh2
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171
Amelung,F 1899 [=3012.01e7h8] After 1...Qd2
Solution: 1. Ne2 Qxe2+ (1... h4 2.
Nf4) 2. Kf8 Qe3 3. Nf7+ Kh7 4. Be4+ Qxe4 5. Ng5+ 1/2-1/2
May be black keeps winning chances with 1...Qd2!?
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170
Troitzky,A 1898 [+3031.30d5f4] After 3...Qxg5
Solution: 1. g3+ Kg5 2. Ne6+ Kg6 3. g5 1-0
But 3...Qxg5 only draws. May be white needs an extra pawn at a2...?
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169
Amelung,F 1899 [+0041.10e6h5] After 2...Kg5
Solution: 1. Bg7! Bxh7 2. Nf6+ Kg6 3. Bf8 Kg5 4. Nxh7+ 1-0
but black draws after 2...Kg5 3.Nxh7 Kg6
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168
Makhatadze,D 1988 [=0401.02a3d1] After 2...Rb5!
Solution: 1. Ne3+ Kc1 (1... Kd2 2.
Nxc2 b1=Q 3. Rxb4 Qc1+ 4. Ka2 Kc3 5. Rb3+
Kxc2 6. Rc3+ Kxc3) 2. Rxb4 b1=Q 3. Nxc2 Qxc2 4. Rc4 Qxc4 1/2-1/2
In the subline 1...Kd2 black has a win with 2.Nxc2 Rb5! -+
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167
Naef,W 1974 [=3000.63e4g5] After 1...Qg4+
Solution: 1. g7 Kf6 2. e8=N+ Kf7 3. d7 Qg4+ 4.
Ke3 Qe6+ 5.
Kf3 Ke7 6. c5 Qxa2 7. Nf6 Qb3+ 8. Ke2 Qc2+ 9. Ke1 Qc3+ 10. Ke2 Qe5+ 11. Kd1
Qd4+ 12. Ke2 Qc4+ 13. Ke1 Qh4+ 14. Ke2 Qh2+ 15. Kf3 Qh3+ 16. Kf2 Qf5+ 17. Ke1
Qe6+ 18. Kd2 1/2-1/2
Black wins after: 1... Qg4+ 2. Kd3 (2. Ke5? Qf4+ 3. Ke6 Qf5#) 2... Qe6 3.
g8=Q+ (3. c5
Kf6) 3... Qxg8 4. d7 Qh7+ 5. Kc3 Qxe7
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166
Kok,T 1933 [+0332.21b2d2] After 2...Be5+ !
Solution: 1. Nc4+ Ke2 2. Nd6 Bxd6 3. Nd4+ Rxd4 4. d8=Q Rd2+ 5. Kb3 Rd3+ 6. Ka4 1-0
But black draws with: 2... Be5+! 3. Ka2 b3+! 4. Kxb3 Rxd6 5. b8=Q Rd3+ 6. Kc4 Bxb8 7. Nd4+ Rxd4+ 8. Kxd4 Bc7 =
However there is probably a win for white after: 1. Nb3+ Ke2 2. Nc5 Bb8 3. Nxb4 Rd6 4. Kb3 +-
But this is not the idea what the composer ment.
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165
Voja,R 1958 [+0031.22d5g2] After 2...Bd3
Solution: 1. h7 Bxh7 (1... h3 2. h8=Q h2 3. Qg7 h1=Q 4. Qxg6+ Kf2 5. f7) 2. f7 f2 3. Nc4 f1=Q
(3... Bc2 4. Ne3+ Kf3 5. Kd4) 4. Ne3+ Kh3! 5.Nxf1 Bg8 6. fxg8=B! Kg2 7. Nd2!
(7. Ne3+ Kg1 8. Be6! {cook JN}) 7... h3 8. Kd4 h2 9. Bd5+ Kg1 10. Nf3+ Kg2 11. Ng5+ Kg1 12. Nh3+ 1-0
Although a dual found by JN (John Nunn?), there is refutation for black to draw, with:
2...Bd3 3. f8=Q f2 =
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164
Markovic,M 1976 [+0011.34d6c4] After 3...a1=N
Solution: 1. Na6 a2 2. Bd4 Kxd4 3. Nb4 a1=Q 4. Nc2+ Kc3 5. Nxa1 Kb2 6. Kxd7 Kxa1
7. Ke6
Kb2 8. d6 a3 9. d7 a2 10. d8=Q a1=Q 11. Qxf6+ 1-0
There is a try for black with: 3... a1=N 4. Kxd7 Kc4 5. d6 Nb3 6. Nc2 Nc5+ 7.
Kc6 Ne6
8. d7 Kc3 9. Na3 Kb4 unclear...
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163
Prigunov,V 1998 [=0116.23e7h3] After 6...Nhf7
Solution: 1. Rg8 (1. d8=Q? Nc6+ 2. Kf6 Nxd8 3. Bxf5+ Kh2 4. Ra1 (4. Ra2 Kg3 5.
Ra4 f2
6. Rg4+ Kf3 7. Be4+ Kxg4 8. Bxg2 Kxh4) 4... f2) 1... f2 2. d8=Q Nc6+ 3. Kf6!
Nxd8 4. Bxf5+ Kxh4 (4... Kh2 5. Rxg2+ Kxg2 6. Bd3) 5. Rxg2 (5. Rg4+? Kh3
6.
Rg8+ Kh2 7. Rxg2+ Kxg2 8. Bd3 Kf3 9. Ba6 Ke3 10. Bb5 Kd2 11. Bf1 Ke1 12. Bh3
Nc6 13. Kg7 Nd4 14. Kxh8 Ne2 15. Kg7 Nf4) 5... f1=Q 6. Rg4+ Kh5 7. Rg5+ Kh6 8.
Rg6+! Kh5 9. Rg5+ Kh4 10. Rg4+ Kh5 11. Rg5+ Kh6 12. Rg6+ Nxg6 1/2-1/2
In the subline 4... Kh2 5. Rxg2+ Kxg2 6. Bd3 black may continue with
6...Nhf7 7. h5 Nh6 8. Kg5 Ndf7+
9. Kf4 Ng4! 10. Ba6 Kh3 11. Be2 Kh4 12. Kf3 Ng5+ and wins....?
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161
Kassai,R 1958 [=0002.56a3a1] After 9...Kf4
Solution: 1. Nd2 h2 2. Nxb3+ Kb1 3. Nd2+ Kc2 4. Ndf3! h1=Q 5. Kb4 Kd3 6. Kc5 Ke4
7. Kd6
Kf4 8. Ke7 Kxg4 9. Kf6! Qxh5 10. Ne5+ Kf4 11. Nd3+ Ke4 12. Nc5+ Kf4 13. Nd3+ Kg4
14. Ne5+ 1/2-1/2
Black wins after: 9... Kf4 (or 9...d6) 10. Kg6 d6 11. Kxh6 Kg4 12. Kg6 Qxh5+
The version:
162
Kassai,R 1958 [=0002.56a3a1] After 9...c5
Solution: 1. Nd2 h2 2. Nxb3+ Kb1 3. Nd2+ Kc2 4. Ndf3! h1=Q 5. Kb4 Kd3 6. Kc5 Ke4
7. Kd6!
(7. Kxc6? Kf4) 7... Kf4 8. Ke6 Kxg4 9. Kf6! Qxh5
10. Ne5+ Kf4 11. Nd3+ Ke4 12. Nc5+ Kf4 13. Nd3+ Kg4 14. Ne5+ 1/2-1/2
It is at least unclear after: 9... c5 10. dxc5 Qxh5
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160
Bakcsi,G 1956 [=0834.43f6e8] After 7.Ra8! +- After 3...Bb3!
Solution: 1. d7+! Kf8 (1... Kxd7 2. Rd1+) (1... Rxd7 2. Re3+) (1... Bxd7 2. Nc7+
Kf8 3.
Rd1!) 2. Ra7! f1=Q (2... Be6 3. Rab7!) 3. Rb8! Qa6+! (3... Rxb8 4. d8=Q+ Rxd8 5.
Rf7+) 4. Nb6! (4. Rxa6? Rxb8 5. h6) 4... Qxb6+ 5. Rxb6 Be6 (5... f2 6. Rb8!)
6. Rxe6 (6. Rb8? Rxb8 7. d8=Q+ Rxd8 8. Rf7+ Bxf7) 6... f2 7. Re7
1/2-1/2
In the mainline white can even win with: 7. Ra8 Rxa8 8. Rc6 f1=Q 9. Rc8+
Rxc8 10. dxc8=Q#
But in the subline 2... Be6 3. Rab7! black has the surprising Bb3! 4.
R7xb3 f1=Q 5. Rb8 Qa6+ 6.
R1b6 Rxb8 7. Rxa6 Kg8 8. Rc6 Rf8+ 9. Ke7 Rf7+ 10. Ke6 Rxd7 11. Kxd7 f2 12. Rc8+
Kg7 13. h6+ Kxh6 14. Rxh8 f1=Q black wins!
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159
Grondijs,H 1996 [+0003.61c5f7] After 4...Nd7
Solution: 1. Kd4 (1. Kc6! {cook HH}
Ne8 (1... a4 2. Kd7 Nd5 3. cxd5 a3 4. g8=Q+) 2. Kb5
Nxf6 3. Kxa5 Ne4 4. f6! Nxf6 5. c5 Kxg7 6. c6 Nd5 7. c4 Ne7 8. Kb6 Kf6 9. Kb7
Kf5 10. c5! Ke6 11. c7 Kd7 12. c6+ Kd6 13. f5!) 1... a4 (1... Ne8 2. c5
Nxf6 3. c6 Ke7 4. Kc4 Kd6 5. Kb5 Kc7 6. Kxa5 Kxc6 7. Kb4 Kd5 8. c4+) 2. Kd3 Na6
3. Kc2 Nc5 4. Kb2 Kg8 5. Ka3 (5. Ka2? Kf7
6. Ka3 Kg8 7. Kb4 Kf7 8. Kxc5? a3 9. Kd6 a2 10. g8=Q+ Kxg8 11. Ke7 a1=Q 12.
f7+ Kh7 13. f8=Q Qa7+ (13... Qa3+? 14. Kf7!) (13... Qe1+? 14. Kd6!) 14.
Kf6 (14. Ke8 Qd7+ 15. Kxd7) 14... Qd4+ 15. Kg5 Qxf4+ 16. Kxf4) 5... Kf7 6. Kb4
Kg8 7. Kxc5 a3 8. Kd6 Kf7 9. g8=Q+ Kxg8 10. Ke7 1-0
Although a dual was already found, may be the original solution is not even sound.
If black plays 4... Nd7 5. Ka3 Nxf6 6. Kxa4 Kxg7 the situation is unclear. Who proves that white still wins?
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158
Troitzky,A 1896 [+0008.10h7c7] After 1...Kc8
Solution: 1. d6+ Kd7 2. Ndc5+ Ke8 3. d7+ Nxd7 4. Nd6+ Kd8
(4... Kf8 5. Ne6#) 5. Ne6# 1-0
But black draws by: 1... Kc8 2. dxe7 Kd7
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157
Timman,J 1997 [+0040.55f1b1] After 14...Bf8
Solution: 1. f3 g3 (1... gxf3 2. Bf2! (2. gxf3? Kc2 3. Bf2 Bg5) 2... Bg5 3. h4 Bh6
4. g4! fxg3 5. Bxg3 Be3 6. Bf2 Bh6 (6... Bf4 7. Bd4 Kc2 8. Kf2 Kd3 9. Kxf3)
7. Bg1! Kc2 8. Kf2 Kd3 (8... Bf4
9. Kxf3 Bxe5 10. Ke4 Bg7 11. Bd4 Bf8 12. Bf6!
Kb3 13. Ke5 Kc4 14. Kxe6 Kxb5 15. Kf7) 9. Kxf3 Kc4 10. Be3 Bg7 11. Bf4! Kxb5
12. Kg4 Kc4 13. Kg5 b5 14. Kg6 Bh8 15. Kh7 b4 16. Kxh8 Kc3 (16... b3 17. Bc1)
17. h5 Kc2 18. h6 b3 19. h7)
2. hxg3 Bxg3 (2... fxg3 3. Ke2 Kc2 4. f4 Kb3 5. Kf3 Kc4 6. Ba1! Kxb5 7. Kg4 Be7
8. Kxg3 Kc4 9. f5 exf5 10. Kf4 b5 11. Kxf5 b4 12. g4 Kb3 13. g5 Ka2 14. g6 Bf8
15. Bd4 c5 16. Bxc5) 3. Bf2 Bh2 4. Bg1 (4. Bh4? Kc2 5. g4 fxg3 6. Kg2 Kd3 7. Bd8 Kc4 8. Bxc7 Kxb5)
4... Bg3 5. Bh2!! Bxh2 6. g4 fxg3 7. Kg2 Kc2 8. f4 Kd3 9. f5 1-0
In the first (underlined) subline black has:
14... Bf8! 15. Kf7 Ba3 16. h5 b4 17.
h6 b3 18. h7 b2 19. h8=Q b1=Q =
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156
Liburkin,M 1946 [+0474.21f1a8] After 2...Ba6+
Solution: 1. g7 h2 2. Be3! (2. gxf8=Q? hxg1=Q+) 2... Rxg1+!3. Bxg1 Bh3+! (3... h1=Q 4. gxf8=Q)
4. Ke2 Bg4+ 5. Kd3 Bf5+ 6. Kc4 Be6+ 7. Kb5 Bd7+ 8. Ka6 Bc8+ 9. b7+! Bxb7+ 10. Kb5 Bc6+
11. Kc4 Bd5+ 12. Kd3 Be4+ 13. Ke2 Bf3+ 14. Kf1 Bg2+ 15. Kxg2 1-0
After 2...Ba6+ black is able to draw?: 2... Ba6+ 3. Kg2 hxg1=Q+ 4. Bxg1 Bb7+
5. Kf1 Ba6+ 6. Kxe1 Rxg1+ 7. Kd2 Rxg7 8. Rxg7 Bb5 9. Rg8 Kb7 10. Rxf8 Kxb6 =
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155
Farago,P 1958 [=0030.23b6d2] After 1...Be4
1. Kxb5 Bc6+ 2. Kxa5 Ke3 (2... Kc3 3. Kb6 Kd4 4. g4 Ke5 5. g5 Kf5 6. Kc5 Bf3
7. Kb6 Be4 8. Kc5 Kg4 9. Kd4 Bg6 10. Kc5 Kxh4 11. Kb6 Be4 12. g6)
3. g4 Kf4 4. g5 Kf5 5. Kb6 Kg6 6. Kc5 Kh5 7. Kd4 Be8 8. Kc5 Bg6 9. Kb5! (9. Kb6? Be4)
9... Be4 10. Kb6 Kxh4 11. g6 1/2-1/2
But Black starts with: 1... Be4 2. Kxa5 (or 2. g4 b6 3. h5 Bd3+ 4. Ka4 Ke3 5. g5
Kf4 6. g6
Kg5 7. g7 Bh7) 2... Ke3 3. Kb6 Kf4 4. h5 Kg5 5. g4 Bf3 white is in zugzwang -+
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154
Jespersen,J 1896 [+0051.54g2a7] After 1.Ne7
Solution: 1. Be7 c1=Q 2. Bc5+! Qxc5 3. Ne7 1-0
Black is able to continue with 3...Qd5+ and now 4. Nxd5 is stalemate
and not like 4. Kf2? Qd4+ 5. e3 Qd2+ 6. Kf3 Kb6 -+
But white is still able to win with (@1) 1. Ne7 Kb6 2. Nc8+ Kc5 3. Be7+ Kc4 4. Ba3 Kb3 5. Bc1 Kxa4 6. e4
But than it is not a study, but just another endgame.
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153
Seckar=M 1958 [+0400.22c6e8] After 1...Rh1
Solution: 1. Rc4! f2 2. Re4+ Re7 3. Rxe7+ Kf8 4. Re4! f1=Q 5. d7 Qf6+ 6. Kc7! Qf7
7.Kc8 Qf5 8. Rf4! Qxf4 9. d8=Q+ Kf7 10. Qc7+ 1-0
Black may try: 1... Rh1 2. Kc7 Rh7+ 3. Kc8 f2 4. Rf4 (4. Re4+ Kf7 5. Rf4+ Ke6)
4... Rd7 drawing?
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152
Korn,W 1944 [=3413.32a8c8] Ending position. There is a better version.
Solution: 1. Rf8+ (1. Bxf1? Nxf7 2.
exf7 Rxf7 3. Bd3 Rf1 4. b6 b1=Q 5. Bxb1 Rxb1 6. b7+
Kc7) (1. Rxf1? Rc1 2. b6 Kd8!) 1... Qxf8 2. e7+ Qf5 3. Bxf5+ Rd7 4. b6 b1=Q
5. e8=Q+! Nxe8 6. Bxb1 Nc7+ (6... Kd8 7. b7 Nc7+ 8. Ka7 Nd5 9. Be4 Nb4 10.
Ka8 Na6 11. b8=Q+ Nxb8 12. Kxb8 Rc7 13. Bd3) 7. bxc7 Rxc7 (7... Kxc7 8. Ka7 Rd4
9. Bc2) 8. Bf5+ Kd8 9. Bd3 1/2-1/2.
But the ending position can be continued with:
9... Kd7 10. Bb5+ Kd6 11. Kb8 Rg7 and black wins by grabbing pawn a4.
The 3rd diagram is a better version.
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151
Korn,W 1950 [+0340.22d5b8] After 1...Bxe7
Solution: 1. e7! (1. d7+? Ka7 2. Be3+
(2. d8=Q e1=Q) 2... Ka6 3. d8=Q e1=Q 4. Qa8+
Kb5 5. Qxb7+ Ka4 6. Qc6+ Kb3) 1... e1=Q (1... Rb5+ 2. Kc6 e1=Q 3. exf8=Q+ (3.
e8=Q+? Qxe8+ 4. d7+ Ka7 5. dxe8=Q a1=Q 6. Be3+ Bc5!) 3... Ka7 4. Qe7+)
2. e8=Q+!! (2. exf8=Q+? Ka7) 2... Qxe8 3. d7+ Ka7 4. dxe8=Q
a1=Q 5. Be3+ 1-0
But after 1... Bxe7 2. dxe7+ Ka7 3.
e8=Q Rb5+! 4. Qxb5 (4. Kc4 a1=Q) (4. Kc6 Rb6+ 5. Kc5
a1=Q) 4... e1=Q black draws.
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150
Rezvov,N 1958 [+0113.12c1b4] After 6...Nb4 7.c8=Q
Solution: 1. Kb1! Nd3! 2. Bd4 Kxa3 3.
c5 Kb4 (3... Nb4 4. Bb2#) (3... Nf4 4. c6 Ne6
5. c7 Nxc7 6. Bc5#) 4. c6 a3 5. Bc3+! (5. c7? a2+ 6. Ka1 Ne1! 7. Bc3+ Ka3!)
5... Kxc3 6. c7 a2+ 7. Ka1 Ne1 8. c8=Q+ Kd2 9. Qd8+!
(9. Qd7+? Kc1!) 9... Ke2 (9... Kc1
10. Qg5+) 10. Qe7+ Kd1 11. Qd6+ 1-0
But after 6... Nb4 7. c8=Q+ Kd4 white can't make progess (EGTB checked).
Adding two pawns like wh2-blh3 would repair it?
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149
Korn,W 1954 [=0141.27f1h1] After 14.Bc6!
Solution: 1. Ne2 (1. Nf3? e2+
(1... exf2 2. Kxf2) 2. Rxe2 dxe2+ 3. Kxe2 Kg2 4. Nxh2
Kxh2 5. Kf2! Bc4 6. g7 Bd5 7. Bd1 Bc4 8. Ke3! Kg1 9. Kxd2 Bd5 10. Ke3 h2
11. Bf3 Bxf3 12. g8=Q+ Bg2 13. Qa2 h1=Q 14. Qf2+ Kh2 15. Qh4+ Kg1 16. Qf2+)
1... dxe2+ 2. Rxe2 (2. Kxe2? Kg1 3. Rf1+ (3. Rxh2 Kxh2 4. g7 Kg1 5. Bc2
h2)
3... Kg2 4. Rd1 h1=Q 5. Rxh1 Kxh1 6. Kxe3 Kg1 7. Bd1 h2) 2... Bf5 3. g7 Bd3
4.
Bd1! (4. g8=B? Bxe2+ 5. Kxe2 Kg1 6. Bd5 h1=Q 7. Bxh1 Kxh1 8. Kxe3 Kg1 9.
Bd1 h2) (4. g8=Q? Bxe2+ 5. Kxe2 d1=Q+ 6. Kxd1 e2+ 7. Kxe2) (4. Bb3? Bc4 5.
g8=B Bxg8! (5... Bxb3? 6. Rxh2+! (6.Bxb3? d1=Q+ 7. Bxd1)
6... Kxh2 7. Bxb3 Kg3 8. Ke2 Kf4! 9. Bd5 Ke5! 10. Bxb7 Kd4! 11. Bf3 h2
12. Bh1 Kc3)) 4... Bc4 5. g8=B (5. g8=Q? Bxe2+ 6. Kxe2) 5... Bxe2+
(5... Bxg8 6. Rxe3 Bd5 7. Kf2 Be4 8. Ra3 Bd5 9. Ra1) 6. Bxe2 (6. Kxe2? Kg1
7. Bd5 h1=Q 8. Bxh1 Kxh1 9. Kxe3 Kg1 10. Bf3 h2 11. Kxd2 h1=Q 12. Bxh1 Kxh1
13. Ke3 Kg1 14. Ke4 Kf2 15. Kd4 Ke2 16. Ke4) 6... d1=Q+ 7. Bxd1 e2+ 8. Kxe2 Kg1
9. Bd5 h1=Q 10. Bxh1 Kxh1 11. Kf2
Kh2 12. Bg4 1/2-1/2
But why does this ends as equal? (wrong result in the database?)
White wins after: Kh1 13. Bf3+ Kh2 14.Bc6 bxc6 15. bxc6 b5 16. c7 b4 17. c8=Q b3 18. Qg4 b2 19. Qg1#
(The subline 4.Bb3 is still winning if white plays 4. ...Bc4 5. Bc2! Bd3 6. Bd1 Bc4 7. g8=B returning to the mainline)
The strange thing is that there are versions [+0140.26f1h1] from 1953! but there white wins ...!?
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148
Van_den_Ende,J 1962 [+0045.22d4g5] After 2...Nd7
Solution: 1. Nc3 Be6 2. Bb4 Nh7 3. hxg6 fxg6 4. Ne4+ Kh5 5. Nf3
Bxh3 6. Nfg5 Nxg5 7. Nf6+ Kh4 (7... Kh6 8. Bf8#) 8. Be1# 1-0.
Unclear is it after: 2... Nd7 3. hxg6 Kxg6
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147
Liburkin,M 1946 [+0460.10e1g7] After 5...Rh5+ !?
Solution: 1. b8=Q (1. bxc8=Q? Bh4+)
(1. Rxc8? Bh4+ 2. Ke2 Bg3) 1... Bh4+ 2. Kd2 Bg5+
3. Kc3 Bf6+ 4. Kb4 Be7+ 5. Ka5 Bd8+ 6. Kb5 Bd7+ 7. Kc4 (7. Ka6? Rh6+ 8. Ka7 Bb6+
9. Kb7 Bc6+ 10. Kxb6 Bxa8+ 11. Kc5 Bc6 12. Qc7+ Kf8 (12... Kg8? 13. Qe7) 13.
Qd8+ (13. Qf4+ Kg7) 13... Be8) 7... Be6+ 8. Kd3 Bf5+ 9. Ke2 Bg4+ 10. Kf1 Bh3+
11. Kg1 Bb6+ 12. Qxb6 Rxa8 13. Qb7+ 1-0
But black may try: 5... Rh5+ 6. Kb6 (6.Ka4? Bd7 7.Kb3 Rb5+=) Rh6+ 7. Kb5 Bd7+ 8. Kc4 Rc6+
9. Kd4 Rd6+ 10. Kc3 Rc6+ who investigates?
Reaction from Siegried Hornecker:
"I don't know if white can win (I think so) but he first should get out of the checks with 9.Kb3 Be6+ 10.Ka4 Rc4+ 11.Ka5 Rc5+ 12.Kb6.
This position is won, e.g. 12...Rd5 13.Ra7! Kf7 14.Qh2 leaving black with a totally lost position.
So there are only three more moves before (9,10,11) to test.
Still I think there's nothing better for black.
I even think Liburkin knew that this endgame (KQRKRBB) is won if white can get out of the checks."
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146
Amelung,F 1904 [=3432.52d4c7] After 2...Rd8
Solution: 1. d6+ Kxc6 2. d7 Bxd7 3. Nxc8! Qa8 4. Rc5+ Kb7 5. Nd6+ Kb8 6.
Ra5
Qc6 7. Rb5+ Kc7 8. Rc5 Kxd6 9. Rxc6+ Bxc6 10. b5 Bd5 11. a4 Kc7 12. a5 Kb7 13.
Ke3 Kc7 14. Kd4 Kd6 15. a4 Kc7 16. Ke3 Kd6 17. Kd4 Kd7 18. a6 Kc7 19. a5 Kd6
20. a7 Ba8 21. b6 Kc6 22. Kxe4 1/2-1/2
But black defends with: 2...Rd8! -+
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145
Gurvitch,A 1930 [+3011.23c3a5] After 4....dxc5
Solution: 1. b4+ Ka4 2. Nb2+ Ka3 3. Nc4+ Ka4 4. Bf5 b6 5. Bc2+ Kb5
6. Nxd6+ Kc6 7. b5+ Qxb5 8. Be4+ Kxc5 9. Nb7# 1-0
But black has: 4... dxc5 5. Bxd7+ b5 6. bxc5 Qf6+ -+
144
Same author, same year and look alike. After 1...Kb5
[+3011.24c3a5] Solution: 1. b4+ Ka4 2. Nb2+ Ka3 3. Nc4+ Ka4 4. Bc2+ Kb5
5. Nd6+ Kc6 6. b5+ Qxb5 7. Be4+ Kxc5 8. Nxb7# 1-0
But black may try: 1... Kb5 2. Bd3+ Kc6 3. Bxa6 bxa6 4. cxb6 Kxb6 5. Ne3
(5.Kc4 a5 6.b5 d5+=) Kb5 6. Nc4 d5 7. Nd6+ Ka4 8. Nb7 Kb5 =
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143
Rihay,E 1904 [+4300.20d8a8] After 1...Ra7! After 1.d7!
Solution: 1. Qa5+ Kb8 2. c7+ Kb7 3. Qb5+ Ka7 4. Qc5+ Ka6
(4... Kb7 5. c8=Q#) (4... Ka8 5.
c8=Q#) 5. c8=Q+ 1-0
1. Qa5? is not winning because black has 1...Ra7!, but 1.d7! will win.
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142
Korn,W 1964 [=0110.34c2e7] After 1...Kd6!
Solution: 1. Bc6 Ke6 2. Bxd7+ Kd6 3. Bc6 1/2-1/2 Black wins after Kxc6 and also
black wins after: 1... Kd6 2. Rxd7+ Kxc6 3. Rxh7 e1=Q 4. Rxh2 Qxg3.
I see no point in this study. Possibly the starting position is not correct entered?
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141
Salkind,L 1930 [+0300.42h1h3] After 1...Re2
Solution: 1. a8=R! (1. a8=Q? Re2! 2. Qa1 Re1+ 3. Qxe1 g2+ 4. Kg1) 1... Rb2 2. b8=R! 1-0
But after: 1...Re2 2. Ra1 Rh2+ 3. Kg1 Rg2+ 4. Kf1 Rb2 = (idea Kh2, g2, so 5.Kg1 Rg2 etc)
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Peter Boll.
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