120
Incorrect studies? To be corrected.
6th Page. 101-120.
Here page-6 with a list of studies which are possibly incorrect and most were not yet discovered that way according to 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.
You are invited to prove me wrong, or correct the study in some way.
Many of them still have beautiful ideas in there!
Please react by using the guestbook
120
Troitzky,A 1896 [=3104.54g2h8] After 3... Qe2!
Solution: 1. Rxh7+ Kxh7 2. d7 Qxa2+ 3. Kh3 Qa5 4. d8=Q Qxd8 5. Ng5+ Qxg5 1/2-1/2
Black wins after: 3... Qe2 4. d8=Q (or 4.Ng5+ Kh6 5.d8=Q Kh5!)
Qxf3+ 5. Ng3 (5. Kh4? Qh5#) 5... fxg3 -+
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119
Fischer,J 1938 [=0340.20e8e6] Position after 2...Kf5
1. Be7 Ra8+ 2. Bd8 Rc8 (2... Bf8
{main} 3. Kxf8 Rxd8+ 4. Kg7 Ke7 5. Kh7 Kf6 6.
f8=Q+! Rxf8 7. g7 Rf7 8. Kh8 Rxg7) (2... Bf6 {main} 3. f8=N+ Kf5 4. Kf7 Rxd8
(4... Bxd8 5. g7 Ra7+ 6. Ke8) 5. g7 Ra8 6. g8=Q (6. g8=N {or}) 6... Ra7+ 7. Nd7
$1 (7. Ke8? Re7+ 8. Kd8 Rg7+)) (2... Kf5) 3. f8=Q Bxf8 4. g7 Bxg7 1/2-1/2
After 2...Kf5 black wins.
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118
Nestorescu,V 1953 [+0014.12b7b4] Position after 7.Nd5+
Solution: 1. Nc3 Nd2 2. Be7+ Ka5 3. Bd8+ Kb4 4. a5 Nb3 5. a6 Nc5+ 6. Kb6 Nxa6 7. Ne4! Nc5
(7... c3! {cook} 8. Kxa6 c2 9. Bg5 h5 10. Bc1 h4 11. Nf2 Kb3 =) 8. Nxc5 1-0
This studie was found incorrect after 7.Ne4 c3!
But the study seems still correct! with: 7. Nd5+ Kb3 8. Kxa6 c3 9. Nf4! c2 10. Ne2 Kb2 11. Bg5 h5
12. Kb5 h4 13. Bc1+ Kb1 14. Ba3 h3 15. Nc3+ Ka1 16. Kc4 h2 17. Kb3 c1=Q 18. Bxc1 h1=Q 19. Bb2#
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117
Nestorescu,V 1953 [+0440.32h1h8] Position after 3...Bf8! After 2.Rd7
1. c6 Rc2 2. Bc5 2... Bxc5 3.
Rd8+ Kg7! (3... Bf8 4. Rxf8+
Kg7 5. Re8 Rxc6 6. Rxe4 Rc1+ 7. Kg2 Rc2+ 8. Kf3 Rxa2) 4. c7 Rc1+ 5. Kg2 Rc2+ 6.
Kh3! Rc3+ 7. Kg4 Bf2 8. c8=Q Rg3+ 9. Kh4 Rc3+ 10. Kg5 Rg3+ 11. Qg4 1-0
Black has a defence with: 3... Bf8! 4. Rxf8+ Kg7 5. Re8 Rxc6 6. Rxe4 Rc1+ 7. Kg2 Rc2+ 8. Kf3 Rxa2.
Although white still can win with: 2.Rd7N Bf8 3. c7 the idea of the study (Bc5) is gone.
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116
Kok,T 1992 [+0000.33c3e5] Position after 2...Kf4; After 7...Kg6
Solution: 1. Kb4 c6 2. g4 (2. h4? Kf5) (2. h3? Kf5) 2... Kf6 3. h4 Kg6 4. g5 Kh5 5. g3!
(5. g4+? Kg6 6. Kc3 c5 7. Kb3 b4 8. Kc4 b5+ 9. Kb3 Kg7 10. h5 Kh7 11. g6+ Kh6
12. g5+
Kg7 13. Kb2 c4 14. Kc2 b3+ 15. Kc3 b4+ 16. Kb2 Kh8 17. h6 Kg8 18. g7 Kh7 19.
g6+ Kg8 20. Kb1 c3 21. Kc1 b2+ 22. Kc2 b3+ 23. Kb1 c2+ 24. Kxb2 c1=Q+ 25. Kxc1
b2+ 26. Kc2 b1=Q+ 27. Kxb1) 5... Kg6 6. g4 Kg7 7. h5 Kh7 8. g6+ Kh6 9. g5+ Kg7
10. Kb3 c5 11. Kc3 b4+ 12. Kc4 b5+ 13. Kb3 Kg8 14. h6 Kh8 15. g7+ Kh7 16. g6+
Kg8 17. Kb2 c4 18. Kc2 b3+ 19. Kc3 b4+ 20. Kb2 c3+ 21. Kxb3 c2 22. Kxc2 b3+ 23.
Kb1 b2 24. h7+ Kxg7 25. Kxb2 1-0
But black should try: 2... Kf4 3. h3 Kg5 4. g3 Kg6 5. h4 Kf6 6. g5+ Kf5 7. g4+ Kg6.
And now the white pawns are stopped at 'lower' ranks and this brings a draw. Check yourself.
This study could be corrected by putting the black king on e6. May be wrongly in
the database?
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115
Bondarenko,F 1940 [=0131.23e8a6] Positions after 7...Kd5 and 1.Ne4!
Solution: 1. Nf3! a2 2. Kxf7!
Kb7 3. Re8! a1=Q 4. Re7+ Kb6 5. Re6+ Kb5 6. Re5+ Kc4 7. Re4+ Kd3 (
8. Re3+ Kc2 9. Re2+ 1/2-1/2
It seems though that after: 7... Kd5 8. Re5+ Qxe5 9. Nxe5 Kxe5 black wins.
But white has at move 1 the better: Ne4 Bxe4 2.Rg3 or 1... a2 2. Nc5+ Kb5 3. Nb3 Ka4 4. Rg3 +-
There is a better one with GBR-code [=0131.12f7a6], so possibly these issues were known by the composer.
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114
Mann,C 1913 [+0004.13c5f4] Position after 1...Kg4
Solution: 1. Nd5+ Kf5 2. b8=Q h1=Q 3.
Qf4+ Kg6
4. Qf6+ Kh7 5. Qg5 Kh8 6. Nf6 1-0
Beside a strong dual at move 4 with 4.Ne7+, I see no win after 1...Kg4.
Who checks this line: 1... Kg4! 2.
b8=Q h1=Q 3. Qf4+ Kh3 4. Qh6+ Kg2 5. Nf4+ Kg1 6. Ne2+
Kg2 7. Qc6+ Kh2 8. Qd6+ Kg2 9. Qd5+ Kh2 10. Qh5+ Kg2 11. Nf4+ Kg1 12. Qd1+ Kh2
13. Qxc2+ Kg3 14. Ne2+ Kg4! 15. Qa4+ Kg5 16. Qxa5 Qe4 =
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113
Bogatirev,L 1929 [=0642.12f8h5] Position after 5...Rg8!?
1. Bf7+ (1. Nxg2? Rc8+ 2. Kf7 (2. Ke7
Rxg8) 2... Rxb8) 1... Rg6 (1... Kg5 2.
Nxg2 Rc8+ 3. Ke7 Rxb8 4. Bd5) 2. Nc2! Rxc2 (2... Bc5+ 3. Ke8) 3. a7 Rc8+ 4.
Be8 Bc5+ 5. Kf7 Rc7+ 6. Nd7 Rxa7 1/2-1/2
What will happen after 5...Rg8!?
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112
Kok,T 1933 [+0341.20h4g2] Position after 3...Rd4!
Solution: 1. Ne3+ Kf2 (1... Kg1 2.
Nd5 Rh2+ (2... Rxd5 3. a8=Q Rd4+ 4. Be4) 3. Kg3 Rh8 4.
Kxf3) 2. Be4 Bxe4 3. Nd5 Rxd5 (3... Bxd5 4. d8=Q) 4. a8=Q 1-0
But Black has a surprising defence in: 3... Rd4!! 4. d8=Q Bg2+ 5. Kg5 Rxd5+ 6. Qxd5 Bxd5 =.
On the other hand white has (a dual?) better: 2.Ng4+ Kg1 3.Ne5 Ba8 4.Bd3 +- or 2...Ke2 3.Ne5 Rd4+ 4.Kg5 Ba8 5.Bf5 +-
but in these lines the whole idea (interference) of the study is lost.
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111
Paoli,E 1986 [+4400.00a2d8] Position after 5... Qe6
Solution: 1. Rd4+ Kc8 2. Rc4+ Kd8 3. Qc7+ Ke8 4. Ka3 Rf3+ 5. Ka4 Qd5
6. Qb8+ Kf7 7. Rc7+ Kg6 8. Qb6+ Rf6 9. Qg1+ Kh6 10. Qh2+ Kg6 11. Qh7+ Kg5 12.
Rg7+ Kf4 13. Qh2+ Ke4 14. Qe2+ (14. Qh1+! {cook MC} Kd4 (14... Rf3 15. Rg4+
Ke3 16. Qe1+ Kd3 17. Qd1+) 15. Qa1+) 14... Kf4 15. Qg4+ (15. Qh2+! {cook MC}
Kf5 16. Qh3+ Ke5 17. Rg5+) 15... Ke3 16. Re7+ (16. Qg1+! {cook MC} Rf2 17.
Qe1+ Kd3 18. Qd1+ Rd2 19. Rg3+ Ke4 20. Qf3+ Ke5 21. Rg5+) 16... Kd2 17. Qb4+
Kd3 18. Qb1+ 1-0
Although this study has already some cooks (duals) it seems incorrect after all.
Marcel van Herck found: 5... Qe6! 6. Qb8+ Kf7 7. Qb7+ Kg8 8. Rc8+ Rf8 9. Qg2+ Kh7
10. Rxf8 Qc4+ 11. Ka3 Qc5+ 12. Kb2 Qxf8 =
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110
Bania,R 1927 [+0161.01f1e8] Position after 2...Bf4
Solution: 1. Ne5 Bh5! (1... Ba5 2. Rb8+ Ke7 3. Rb7+ Kf6 4. Nxf7)
2. Rh6 Bd1 3. Rd6 Bf4 4. Rxd1 Bxe5 5. Re1 1-0
Black is save after: 2...Bf4
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109
Karstedt,M 1905 [+0344.43a3f3] After 4...Kf2
Solution: 1. a7 Bb7 2. Bd5 Bxd5 (2... Nf6 3. Bxb7 Rh8 4. a8=Q) 3. Ne5+ Kxe3 4. Nc6 1-0
But why can't black play on like: 4...Kf2 5. a8=Q Rxh3+ 6. Kb4 e3 7. Qf8+ Nf6
(or 7... Bf3)
Can white win this position?
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108
Minerva,E Campioli,M 2003 [+0004.31g5d2] After 1...Nxe1 and 3...Nb4!
Solution: 1... Nb4 2. Kxf4 Kc3 3. Ke5 Kxc4 4. Kd6 Kb5 5. Nc2 Na6 6. c6 Kxa5 7. Ne3 Kb6
8. Nd5+ Ka7 9. Nc7 Nb8 10. Nb5+ Kb6 11. c7 Kb7 12. Na7 Kxa7 13. c8=Q 1-0
Checking the straightforward: 1... Nxe1 2. Kxf4 Nd3+ 3. Ke4 Nb4!
(not 3... Nxc5+ 4.Kd5 Na6 5.Kc6 Kc3 6.Kb6 Nb4 7.Kb5 wins)
4. Ke5 Kc3 5. Kd6 Kxc4 (or 5... Na6) 6. c6 Na6 7. c7 Nxc7 8. Kxc7 Kb5 =
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107
Bondarenko,F Sidorov,B 1991 [=3151.04g1e1] Position after 5...Qa1!?
Solution: 1. Bb4+ Ke2+ 2. Bd1+ Ke3 3. Bc5+ Kd2 4. Bb4+ Kc1 5. Ba3 Qa2 6. Rc8+ Kd2
7. Rd8 Qb1 8. Rb8 Qa2 9. Ra8 Kc1 10. Rc8+ Kd2 11.Ra8 1/2-1/2.
The reader is asked to analyse 5... Qa1!?
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106
Caro,H 1899 [+0440.42e4d6] Position after 3.Be6!
Solution: 1. c7 (1. Rxf6? gxf6 2.
e7 Rg1!) 1... Kxc7 2. Rxf6 gxf6
3. e7 Kd7 (3... Rg1 4. Be6 Re1+ 5. Kf5) 4. Be6+ 1-0
Although the subline 1.Rxf6 is given as if white does NOT win, it has a
beautiful win after all.
So: 1. Rxf6! gxf6 2. e7 Rg1! 3. Be6! Kxe7 4. c7 Rc1 5.c8=Q and win.
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105
Liburkin,M 1960 [=0056.01d5a8] Position after 4...c6+
Solution: 1. Bc6+ Kb8 2. Bb5! Kb7 3. Bc6+ Kb6 4. Bd7! Kb7 5. Bc6+ 1/2-1/2
But: 4... c6+ 5. Bxc6 (or 5. Ke5 Bb8+) 5... Nc7+ and black wins?
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104
Jespersen,J 1905 [=4570.28a5b2] Position after 2...Bd2
Solution: 1. Qd1 bxa3 2. Be5 Bxf6 3. Rxc3 Bd8+ 4. Rc7+ Rxe5 5. Qxb3+ Kxb3 1/2-1/2
But what after 2...Bd2! -+
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103
Amelung,F 1899 [+3101.23f6h8] Position after 2...Qb8!?
Solution: 1. Ne5 Qb2
(1... Kh7 2. Rd7+ Kg8 3. Rg7+ Kf8 4. Ng6+ Ke8 5. Rg8+ Kd7 6. Ne5+ Kd6 7. Nc4+)
(1... Kg8 2. Rd8+ Kh7 3. Rd7+) (1... Qf8+ 2. Nf7+)
2. g5 Qb6+ 3. Kf7 Qb8 4. Ng6+ Kh7 5. Nf8+ Kh8 6. g6 1-0
What will happen after 2...Qb8!?
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102
Markov,K Pogosjants,E 1978 Position after 2... Bd7
[+0080.11g6e7]
1. f6+! Kf8! (1... Kd8 2. f7 Bd6 3. Bf6+ Kd7 4. Bb5+ Kc7 5. Be5) 2. Bf5!
(2. f7? e5 3. Bc3 Bd6 4. Bd2 Ke7 5. Bg5+ Kd7 6. Bf5+ Kc7 7. Bxc8 Kxc8 8. Kf5
Kd7) 2... exf5 (2... Bd6 3. f7 e5 4. Bxc8) (2... Bd7 3. f7 Be8 4. Bxe6 Bxf7+
5.Bxf7)
3. f7! 1-0
But black can try: 2... Bd7 3. f7 Be8 4. Bxe6 Bxf7+ 5.Bxf7 =
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101
Traxler,K+ Halprin 1898 [+1103.02b6g1] Position after 2...g1=B !
Solution: 1. Qg3 Kh1 2. Qxf2 g1=Q
3. Rxf1 1-0
But black has the beautiful refutation 2... g1=B! and draw!
But also white has better with a dual at move 1:
1.Qe3! Kh1 2.Qh6+ Kg1 3.Re2 Ng3 4.Rxf2 Kxf2 5.Qf4+ Kg1 6.Qxg3 etc.
Peter Boll.
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