Duals

2. Duals. To be corrected!

A dual does not always mean that the study is incorrect, but that there are more ways to get to the stipulation.

If a dual is found at move one, the value of a study is lost, if it is a dual at the end, the study keeps a lot of value.

But "only one way" to the stipulation is what composers are fond of'.

You are invited to correct the study in some way or prove me wrong.

Many of them still have beautiful ideas in there!

Please react by using the guestbook.

Latest changes made on 14-aug-2007.

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Korn,W 1941 [+0401.13b6b8] After 4...Re7 After 1.Rf4

Solution: 1. Rd4 (1. Re4? Rh8 2. Ne5 h2 3. Nc6+ Kc8 4. Ne7+ Kd7)

(1. bxa6? f2 2. a7+ Ka8 3. Rf4 Rh8 4. Rxf2 (4. Nxf2 Rh6+) 4... h2 5. Rxh2 (5. Nxh2
Rh6+) 5... Rxh2 6. Nxh2) 1... Rh8 (1... Kc8 2. bxa6 Rxg4 3. Rxg4 f2 4. Rc4+ Kd7 5. Rc1) 2.
Ne5 f2 (2... Kc8 3. Rc4+ Kb8 (3... Kd8 4. Nf7+) 4. Nd7+ Ka8 5. Kxa6 f2 6. Rc7
Rh6+ 7. Nb6+ Rxb6+ 8. Kxb6 Kb8 9. Rf7) (2... axb5 3. Nc6+) 3. Nc6+ Kc8 (3...
Ka8 4. Rd7) 4. Na7+ Kb8 5. Rd7 Rh6+ 6. Nc6+ Rxc6+ 7. Kxc6 axb5 (7... f1=Q 8.
Rd8+ Ka7 9. b6#) (7... Kc8 8. b6) 8. Rd1 1-0

The 1st subline with 1.Re4 also wins after 4.Re7!;

And also white wins with: 1. Rf4 Rh8 2. Ne5 Kc8 3. Rf7 Kd8 4. bxa6 Rh6+ 5. Nc6+
Ke8 6. Rb7

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Dikusarov,A 1989 [+0010.22f8d1] After 1...Ke7

Solution: 1. Be7! Kc2 2. Ba3
Kb3 3. Ke7 Kxa4 4. Kd6 b5 5. Kc5 b4 6. Kc4 (6. Bxb4? axb4 7. Kc4 Ka5 8. b3
Ka6 9. Kxb4 Kb6) 6... bxa3 7. b3# 1-0

White probably also wins with: 1. Ke7 Kc2 2. Kd6 Kxb2 (or 2... Kb3 3. Kd5 Kxa4 4. Kc4 b5+ 5. Kc3 b4+
6. Kc4 b3 7. Kc3 Kb5 8. Kxb3 a4+ 9. Ka3) 3. Kc6 Kb3 4. Kb5 Kc3 5. Be3 Kd3 6.
Bxb6 Ke4 7. Bxa5 Kd5 8. Bc7 Ke6 9. a5 Kd7 10. Kb6 Kc8 11. a6

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Marwitz,J 1958 [=3050.52a4d8] After 6.c3!?

Solution: 1. f8=Q+ Qxf8 2. Ba5+ Ke8 3. Bg6+ Kd7 4. Bf5+ Kc6 5. Be4+ Kc5 6. Bc3! (6.
Kxa3? Kb5+) 6... Kb6 (6... Bc1? 7.Bb4+) (6... Qf7 7. Kxa3) 1/2-1/2

[May be 6...Qg8 7.Kxa3 Qxg4 is still better for black!?]

May be white also can draw after: 6. c3!? Qf4 7. Kxa3 Qxg4 8. Bb4+ Kb6 ...

Therefore mr. JMR.Dalip from France suggests to move the white pawn from c2 to e2!

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54 Asaba,E 1973 [+0100.33d8b8]

Solution: 1. Rf6! b1=Q 2. Rf8 Qh7 3. c6 Qg7 4. Ke8! Qh7 5. f3! Qg7

6. f4 Qh7 7. f5 Qg7 8. f6 Qh7 9. f7 1-0
But White has faster: 1.Rxc7 and 2.Rc8 mate.

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Jaroslavtsev,A 1950 [+4040.10e6b5] After 1.Bd3

Solution: 1. Qe5+ Kb6 2.Qb8+ Kc5 3. Qa7+ Kc4 4. Qa4+ Kc5 5. Qa5+ Kc4 6. Bd3+ 1-0

At move 3 white has a mate in 4: 3. Qd6+ Kb5 4. Bd3+ Ka5 5. Qa6+ Kb4 6. Qa3#

At move 1 white has a mate in 9: 1. Bd3+ Kc6 2. Qd6+ Kb7 3. Be4+ Ka7 4. Qc7+ Ka6 5. Bd3+ etc.

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Wijnans,A 1937 [+0340.20g3a5] After 4.Bd2

Solution: 1. c7 Rg4+ (1... Rc4 2. g7) 2. Kf2! (2. Kh3? Rc4 3. g7 Bg4+) (2. Kh2? Rh4+

3. Kg3 Rh8) 2... Rc4 3. g7 Bb3 4. g8=Q Rc2+ 5. Bd2+ 1-0
White has a dual win after 4.Bd2+ Kb6 5.g8=Q Kxc7 6.Qg3+ and Qxb3.

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Birnov,Z 1929 [=0032.02h7f8] After 6.Nb1

Solution:
1. Ng6+ Bxg6+ 2. Kxg6 Ke7 3. Kf5 Kd6 4. Ke4 Kc5 5. Kd3 Kb4 6. Kc2 Kxa3 7. Kb1 Kb3 1/2-1/2

In the end white has a dual with 6.Nb1 a3 7.Kc2 a2 8. Kxb2

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Wijnans,A 1936 [+3001.65f7h5] After 9.Kd7

Solution: 1. Ng7+ Kg5 2. Ne6+ Kf5 3. Nd4+ Ke5 (3... Kg5 4. gxh4+ Kxh4 5. Nxb3 Kxh3 6. Kg7)

4. Nxb3 hxg3 5. Nc1 Kf4 6. Ne2+ Kf3 7. Nxg3 Kxg3 8. Ke6 Kxh3 9. b5 h5 10. b6 cxb6

11. Kxd6 h4 12. Ke5 Kg3 13. d6 h3 14. d7 h2 15. d8=Q h1=Q 16. Qg5+ Kf3 17. Qf4+ Ke2
18. Qe4+ Qxe4+ 19. dxe4 (19. Kxe4? b5) 19... Kxd2 20. Kd4 h5 21. e5 h4 22. Ke4 1-0.

White also wins with: 9. Kd7 h5 10. Kxc7 h4 11. b5 Kg2 12. b6 h3 13. b7 h2 14. b8=Q h1=Q

15. Qg8+ Kf2 16. Qf8+ Ke2 17. Qxd6 Kxd3 18. Qg3+ Kxd2 19. Qf4+ Kd3 20. d6 +-

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Kok,T 1938 [+0014.12h4f1] After 1.Ba6+

Solution: 1. Kg3 Ke1 2. Ba6 Kd2 3. Kf2 (3. Bd3! {cook JU} Ke3 (3... g5 4. Bf5) 4. Nc6 Nb3 5. Bxg6)

3... Nxc2 4. Nb3+ Kd1 5. Be2# 1-0

Although already a dual was found, some more comments are to be made.

First black can draw with: 2... Nxc2 3. Nxc2+ Kd2

But white wins more easily with: 1. Ba6+ Ke1 2. Bd3 Kd2 3. Bxg6; or 1...Kf2 2.Bd3 Ke3 3.Nc6

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Bondarenko,F Neidze,V 1969 [+3041.62a3c5] After 2.Kxa2

Solution: 1. d7+ Kxc4 2. b3+ Kxb5 3. Nxa2 Qa5+ 4. Kb2 Bb6 5. Bb4 Qxa6 6. c4# 1-0
But also white wins with: 2. Kxa2 Kxb5 3. d8=Q Qxa6+ (3... Qxd8 4. Bxd8) 4. Kb1 Qa4
5. b4 +-

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Enderlein,G 1912 [+0040.45f3d6] After 5.g5

Solution: 1. fxg3!! bxc4 2. Ke4 Kc6 (2... c3 3. Kd3 c2 4. Kxc2 Kd5 5. Kd3 g5

6. g4 g6 7. Kc3 f5 8. Kd3) (2... g5 3. g4!) 3. Kd4 Kb5 4. g4 f6 5. Ke4 g5 6. Kd4 f5

7. Kc3 f4 8. Kd4 1-0.
White has also: 5. g5 fxg5 6. g4; And also winning is: 7. gxf5 g4 8. g3 c3 9. Kxc3

An extra subline is: 4... g5 5. Ke4 g6 6. Kd4 f5 7. Kc3 fxg4 8. g3 +-

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Zachodjakin,G 1930 [=0014.22a2h8] After 6.Nf7+ =

Solution: 1. g7+ Nxg7 (1... Kg8 2. Ng4) 2. Nf7+ Kg8 3. Bc5! f1=Q 4. Nh6+ Kh8 5. Bd6!
Qe1 (5... Qa6+ 6. Kb2 Qxd6 7. Nf7+) (5... Qe2+ also does not win) 6. Be5 1/2-1/2
White has also a draw with 6.Nf7+ and 7.Nh6+.

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Loyd,S 1859 [+0401.10b6b8] After 1.Rd4

Solution: 1. Rc4 Rc3 2.Rc8+! Rxc8 3. Na7! Rc1 4. Nc6+ Rxc6+ 5. Kxc6 Ka7 6. Kc7 1-0

White has some other first rook moves like 1.Rd4 Rd3 2. Rd8+ Rxd8 3. Na7 Rd6+ 4. Nc6+ Rxc6

Ending up in the same solution. Although duals at move 1, still minor duals.

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Bordeniuk,M 1969 [+0401.11c4g4] After 1.Ne5+

Solution: 1. Kd3 Rg3+ 2. Ke4 Kh5 3. Ne5 Kxh4 4. Rh6+ Kg5 5. Rg6+ Kh4 6. Nf3+ Kh3

7. Rh6+ Kg2 8. Rh2+ Kf1 9. Ke3 Rg2 10. Rh1+ Rg1 11. Rxg1# 1-0

But white also has a win with: 1. Ne5+ Kh5 2. Nf3 Rg3 3. Rf6

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Salkind,L 1929 [+0315.12e4c1] After 2.Nd4 After 1.Ne8

Solution: 1. Bxf4+ Kc2 2. Ne8 Rxe8 3. dxe8=Q Nf6+ 4. Ke5 Nxe8 5. Kxe6 Kd3

6. Ng3 (6. Bb8! {cook JU} Ke4 7. Ng3+ Kf3 8. Kf7) 6... Ng7+ 7. Kf6 Ne8+ 8. Kf7 1-0

Although already a dual known at move 6 there are two more earlier duals:

2. Nd4+ Kc3 3. Ne8 Rxe8 4. dxe8=N Ne7 5. Nxe6 (EGTB) and

1. Ne8 Rxe8 (1... exf5+ 2. Kxf4) 2. Bxf4+ Kb1 3. dxe8=Q Nf6+ 4. Ke5 Nxe8 5. Kxe6

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Troitzky,A 1898 [+4007.32b5d4] After 1.Qf6+

Solution: 1. g7 Qxg7 2. Qe1 Qg4 (2... Qa7 3. Qg1+ Kc3 4. Na4+)
3. Ne6+ Kc3 4. Qa1+ Kc2 5. Nd4+ 1-0

White seems to win also with: 1. Qf6+ Ke3 2. g7

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Isenegger,S Signer,R 1961 [+0110.04h7h4] After 2.Rd7

Solution: 1. Rd4+! Kg5! 2. Bb4!! c5! 3. Bxc5 c2 4. Rxd2 c1=Q 5. Be3+ Kh4 6.
Rd4+ 1-0

White also wins with: 2. Rd7 e5 (c2 3.Bh6) 3. Bb4 e4 4. Bxc3 e3 5. Bxd2 exd2 6.
Rxd2 Kf5 7. Rc2 Ke5 8. Rxc6

Possible correction suggested by JMR.Dalip: Move the white king from h7 to g7.

1. Rd4+ Kg5 2. Bb4 (2. Kh7 $143 Kf5 3. Bh6 c5 4. Rd8 Ke4 5. Rd6 Kf3 6. Bxd2
cxd2 7. Rxd2 $11) 2... c5 3. Bxc5 c2 4. Rxd2 c1=Q 5. Be3+ Kg4 6. Rd4+ 1-0

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Bondarenko,F 1986 [+0012.34f5c6] After 1.Bxe7

Solution: 1. Nd8+ (1. Bxe7 a1=Q (1... Kxd7 2. Bf6) 2. e6) 1... Kxd7 2. e6+ Kxd6 3. Nb7+
Kc6 4. Na5+ Kc5 5. Nb3+ Kxc4 6. Na1 Kc3 7. Ke4 Kb2 8. Kd3 Kxa1 9. Kc2 1-0

But after: 1. Bxe7 a1=Q (1... Kxd7? 2. Bf6 and 3.e6) 2. e6 white wins with 3 pieces against the queen.

40. Therefore also the version: Bondarenko,F 1988 [+0012.56f5c6] is cooked the same way:

Solution: 1. Nd8+ Kxd7 2. e6+ Kxd6 3. Nb7+ Kc6 4. Na5+ Kc5 5. Nb3+ Kxc4 6. Na1 Kc3

7. Ke4 c5 8. h5 Kb2 9. Kd3 c4+ 10. Kd2 c3+ 11. Kd1 Kxa1 12. Kc2! a5 13. h6 a4

14. h7 a3 15. Kc1 c2 16. h8=Q# 1-0

Again 1. Bxe7 a1=Q 2. e6

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Kitsigin,V Bondarenko,F [+0501.36a8b6] Position after 2.Nxf5

1988

Solution: 1. Rb7+ Kc5 2. b4+ 2... Kd5 3. Rd7+ Rd6 4. Rxd6+ exd6 5. Nxf5
a1=Q 6. Ne7# 1-0
But white wins also with: 2. Nxf5 exf5 (2... a1=Q? 3. Nh6+ e5 4. Rxe5+ Kd6 5. Nf7#) 3.
Rxf5+ Kd6 4. Ra5 exd3 5. Rb4 e5 6. Rb3 e4 7. Rb4 Kc7 8. Rxa2 Rd6 9. Rb7+ Kc6
10. Rxa6+ Kc5 11. Rxd6 Kxd6 12. Rb4 +-

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Wortel,N 1942 [+4001.03g5h8] After 3.Nf7+

Solution: 1. Kf6! a2 (1... Kh7 2. Qc7+ Kxh6 3. Qg7+ Kh5 4. Qg5#) 2. Qd4! Qd7!

3. Qxd7 a1=Q+ 4. Kg6 Qa6+ (4... Qb1+ 5. Nf5) 5. Kg5 Qa5+

(5... Qa2 6. Qe8+ Kg7 7. Nf5+) 6. Nf5 Qa1 (6... Qe5 7. Kg6) 7. Qe8+

(7. Qc8+! {cook HH} Kh7 8. Qb7+ Kg8 9. Kg6) 7... Kh7 8. Qe7+ Kh8 (8... Kg8 9.
Nh6+ Kh8 10. Qf8+ Kh7 11. Qg8#) 9. Kh6! (9. Qf8+! {cook HH} Kh7 10. Qf7+
Kh8 11. Qh5+ Kg8 12. Nh6+ Kh8 13. Qe8+ Kg7 14. Qg8#) 9... Qc1+! {cook}

(9...Qa6+ 10. Nd6) (9... Qh1+ 10. Nh4 Qc6+ 11. Ng6+) 1-0

Already cooked by HH (Harold van der Heijden?!) at move 7, but there is an earlier cook for white with:

3. Nf7+ Kg8 4. Qxd7 a1=Q+ 5. Ne5 Qa6+ 6. Kf5 Qa2 7. Kg6 Qa6+ 8. Nc6! Qxc6+ 9. Qxc6 Kf8 10. Qd7 g3 11. Qf7#

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Kok,T 1992 [+0000.33f8d7] After 1.exd5

Solution: 1. e6+ Kxe6 2. b7 e2 3. b8=Q e1=Q 4. Qe8+ Kd6 (4... Kf6 5. Qe7+ Kg6 6. exf5+)

5. Qe7+ Kc6 6. exd5+ 1-0

White also wins after: 1. exd5 e2 2. e6+ Kd6 3. b7 e1=Q 4. b8=Q+ Kxd5 5. e7

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Grondijs,H 1980 [=0430.22g2d7] After 4.Re4+

Solution: 1. Rd5+ Ke6 2. Rxd4 Rh2+ 3. Kf1 Kxe5 4. Rd2 4... Bxf3 5. Rd3 (5. Rc2? Rh1+) 5...
Bd5 (5... Bg4 6. Re3+ Kd6 7. Rd3+ Kc5 8. Rc3+ Kd6 9. Rd3+ Ke7 10. Re3+ Be6 11.
Re2) 6. Re3+ Kd4 7. Re2 Bf3 8. Rd2+ Ke4 9. Rd4+ Ke5 10. Rd3 1/2-1/2

After 4. Re4+ Kd5 5. Rg4 Bxf3 6. Rf4= or 4... Kf5 5. Rf4+ Ke6 (Kxf4 =) 6. Rg4=

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Mann,C 1991 [+4004.00d3b6] After 4. Nc6+

Solution: 1. Nc4+ Kb7 (1... Kc7 2. Qa7+ Kc6 3. Qb6+ Kd7 4. Qb7+ Ke6 5. Qe4+ Kd7 6. Ne5+
Kc7 7. Qc6+ Kb8 8. Nd7+) 2. Qd7+ Kb8 3. Na5 Qb2 4. Qd8+ (4. Nc6+ Ka8 5. Qa7#)
4... Ka7 5. Nc6+ Ka6 6. Qa8+ Kb6 7. Qb8+ 1-0

Faster win is: 4.Nc6+ Ka8 5. Qa7#

Also the version:

Mann,C 1991 [+4004.00d3a5] After 4.Nb6+

Solution: 1. Qa4+ Kb6 2. Nc4+ Kb7 3. Qd7+ Ka8 4. Na5 Qb2 5. Qd8+ Ka7 6. Nc6+ Ka6

7. Qa8+ Kb6 8. Qb8+ 1-0
The black queen is earlier won by: 4. Nb6+ Kb8 5. Nd5 Qe5 6. Qb5+ K... 7.N..+ and 8.Qxe5.

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Kok,T 1992 [+0000.45f2h1] Position after 5.g7!

Solution: 1. d4 cxd4 (1... a5 2. dxc5 a4 3. c6 a3 4. c7 a2 5. c8=Q a1=Q 6. Qxh3#) 2. g4
a5 3. g5 a4 4. g6 a3 (4... fxg6 5. f7 a3 6. f8=Q a2 7. Qf3+) 5. gxf7 a2 6. f8=Q a1=Q 7. Qa8+

(7. Qg8? Qd1! 8. Qd5+ Kh2 9. Qe5+ Kh1 10. Qe4+ Kh2 11. Qf4+ Kh1)

7... Kh2 8. Qb8+ Kh1 9. Qb7+ Kh2 10. Qc7+ Kh1 11. Qc6+ Kh2 12. Qd6+ Kh1

13. Qd5+ Kh2 14. Qe5+ Kh1 15. Qe4+ Kh2 16. Qf4+ Kh1 17. Qf3+ Kh2 18. Qg3+ Kh1 19. Qxh3# 1-0

White wins faster with: 5. g7 a2 6. g8=Q a1=Q 7. Qg3 (idea Qxh3#).

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Mann,C 1923 [+4001.03f1h2] Position after 1.Qg1

Solution: 1. Ne5 Qc8 (1... Qg7 {main} 2. Nf3+ Kh3 3. Nd4+ Qg3 4. Qh6+ Qh4

5. Qe3+ Qg3 6. Qe4 b3 7. Ne2 Qg5 8. Ng1+ Kg3 9. Qf3+) 2. Nf3+

(2. Qf4+! {cook EP} Kh3 3. Qh6+ Kg3 4. Qg5+ Kh3 5. Qg2+ Kh4 6. Nf3+)

2... Kg3 3. Ng1+ Kh4 4. Qh6+ Kg3 5. Ne2+ Kg4 6. Qg6+ Kf3 7. Qd3+ Kg4 8. Qg3+ Kh5

9. Nf4+ Kh6 10. Qg6# 1-0

White has mate in 4 with: 1. Qg1+ Kh3 2. Ng5+ Kh4 3. Nf3+ Kh3 4. Qg2#

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Kok,T 1933 [+0000.21b7a2] Position after 1.h6

Solution: 1. Ka8! e3 2. b7 e2 3. b8=Q e1=Q 4.Qa7+ Kb1 5. h6 Kc1 6. Qc7+

(6. h7? Qe8+) (6. Qa1+! {cook JP}) 6... Kd1 (6... Kb2 7. h7 Qa1+ 8. Qa7)

7. Qd7+ Kc1 (7... Ke2? 8. Qe7+) 8. h7 Qa5+ 9. Kb8 Qb4+ 10. Qb7 Qh4

11. Qc8+ Kd1 12. h8=Q Qb4+ 13. Qb7 Qf4+ 14. Qc7 Qb4+ 15. Kc8 1-0

After: 1. h6 e3 2. h7 e2 3. h8=Q e1=Q EGTB says +-

See http://www.k4it.de/index.php?topic=egtb&lang=de

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Hanstein,W Harrwitz,D 1848 [+4444.23g1b8] Position after 3.Bc7+

Solution: 1. Qd6+ Ka8 2. Ra4 Rc6 (2... Rc1+ 3. Be1+) 3. Bb6+ Ba6 4. Rxa6+ Kb7 5. Ra7+ Kxb6 6. Qb8+ Kc5
7. Ra5+ Kc4 8. Qxf4+ 1-0

White wins faster at move 8 with: 8. Qb5+ Kd4 9. Qd3#

But even faster at move 3. Bc7+ Ba6 4. Rxa6+ Kb7 5. Rb6+ Rxb6 (Ka7 6.Qa3) 6. Qxb6+ Kc8 7. Qb8#

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Kling,J 1849 [+0130.10c4c8] Position after 4.Rg7

1. Kc5 Bd2 2. c7 Kb7 3. Kd6 Ba5 4. c8=Q+ Kxc8 5. Kc6 1-0

The end position in this study let's black play 5...Kd8 =

But white can play better a waiting move like: 4. Rg7! (to get the black bishop to b6) Bb6 5. c8=Q+ Kxc8 6. Kc6 +-

(but it leaves white with many rook alternatives on 7th rank).

Wrongly entered in the database!?

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Kekely,L 1985 [+0301.42f1f3] After 5.e8=Q End Position

Solution: 1. Ke1 Ke3 2. Kd1 Kd3 3. Kc1 Kc3 4. Nb5+ Kb4 5. Nd6 Kc3 6. Kd1 Kd3 7. Ke1 Ke3
8. Nc4+ Kd4 9. Nxb6 Ke3 10. Nc4+ Kd4 11. Nd6 (11. Na5! {cook} Ke3 12. Kd1 Kd3
13. Kc1 Kc3 14. Kb1 Kb4 15. b8=Q+ Rxb8 16. Nc6+) 11... Ke3 12. Kd1 Kd3 13. Kc1
Kc3 14. Kb1 (14. Nb5+! {cook} Kb4 15. e8=Q Rxe8 16. Nc7) 14... Kb3 15. b8=Q+
Rxb8 16. e8=Q Rxe8 17. Nxe8 1-0

Although already cooked twice, a possible more early cook with:

5. e8=Q Rxe8 6. Nc7 Rg8 7.b8=Q Rxb8 8. Na6+ Kc4 9. Nxb8 Kd4 10. Nd7 Kxe4 +-

Author wrote: Because of duals was known in 1986, it was eliminated from that tourney.

Besides duals, which are publicated in mentioned section, there is next dual founded in 1986:

5.Nd4 Kc3 6.Ne2+ Kd3 7.Nf4+ Kxe4 8.Nxe6 and 9.Nd8 +-. I did not repair this study yet.

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26. Kessling,G 1897 [+0001.14f6g8] In the end white wins faster with 9.Qe6 Kc5 10.Qd6#

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Salvioli,C 1880 [+1600.44e1g8] Position after 1.Kf1

Solution: 1. e6 Re3+ 2. Kf1 Rxe6 3. Qg2+! Rxg2 4. b8=Q+ Kh7 5. Kxg2 Rxf6 6. Qe5 Re6

7. Qxc5 Re2+ 8. Kf3 1-0

White may try (dual at move 1): 1. Kf1 Rg4 2. Qh3 Rf4+ 3. Kg1 Rb1+ 4. Kg2 Rb2+ 5. Kg3 Rd4 6. Qf5 +-

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Amelung,F 1899 [+3414.12c6b8] Position after 1.Na6+

Solution: 1. Kb6 Qf7 (1... Nc8+ 2. Bxc8 Kxc8 (2... Rf6+ 3. Be6 Rxe6+ 4. fxe6)
3. Rc7+ Kd8 4. Nc6+ Ke8 5. Re7#) (1... Nc4+ 2. Bxc4 Qf3 3. Na6+ Kc8 4. Be6+) (
1... Rf7 2. Na6+ Ka8 3. Bd5+ Nb7 4. Bxb7#) (1... Ka8 2. Ra7+ Kb8 3. Nc6#) 2.
Na6+ Ka8 3. Bd5+ Qxd5 4. Ra7# 1-0

But sadly at move 1 a dual with: 1. Na6+ Ka8 2. Bd5! Nb7 3. Rxb7 Rf6+ 4. Kc5 Rxa6

5. Rxb5+ Ka7 6. Rb7+ Ka8 7. Rh7+

Correction suggested by Tantale: putting a white pawn at h4 and a black pawn at d4, see guestbook.

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Rinck,H 1899 [+1300.02e1g1] Position after 3.Qe3

Solution: 1. Qg6+ Rg2 2. Qb6+ Kh1 3. Qc6 Kg1 4. Qc5+ Kh1 5. Qd5 Kg1 6.
Qd4+ Kh1 7. Qe4 Kg1 8. Qe3+ Kh1 9. Qf3 e6 10. Kf1 e5 11. Qxg2# 1-0

But black wins faster with: 3.Qe3 Rg1+ 4. Kf2 Rg2+ 5. Kf3 Rg4 6. Qe1+ Rg1
7. Qe4 Rg2 8. Kf4 e5+ 9. Ke3 Kg1 10. Qb1#. This is less beautiful.

May be the composer has known this, because he made in the same year also the better [+1300.01e1h1]

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Lobron,E Timman,J 1994 [+0031.20b8h8] Position after 3.g7

Solution: 1. Nh4 (1. h6? Bf8 2. h7 Bg7 3. Ne3 Be5+ 4. Kc8 Kg7)

1... Kg7 2. Nf5+ Kf6 3. Nh6! Bf8 4. Nf7 Bg7 5. Kc8! (5. Kc7? Ke7 6. Kc6 Ke6 7. Kc5
Bb2! 8. Kc4 (8. h6 Kf6 9. Nh8 Ba3+ 10. Kd5 Bf8) 8... Bg7 9. Kd3 Kf5) 5... Ke7
6. Kc7 Ke6 (6... Bf8 7. Nd6 Ke6 8. Ne4 Ke7 9. Ng3! Ke6 10. Kd8 Bg7 11. Ke8
Kf6 12. Nf5) 7. Kd8 Bf8 8. Ke8 Bg7 9. Ne5! Kf6 10. Nf3 Bh6 (10... Ke6 11.
Nd4+! Kf6 12. Nf5!) 11. Nd4 Kg5 12. Kf7 Kxh5 13. Nf5 Bf8 14. Ng7+! Kh6
15. Ne6 1-0

But after 3.g7 white also wins. Check with EGTB: http://www.k4it.de/index.php?topic=egtb&lang=de

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Olimpiev,B 2001 [+0031.12c5g4] Position after 3.Nc2

Solution: 1. Kd5 Bd6 2. Kxd6 a3 3. a6 a2 4. Nc2 f3 5. a7
a1=Q 6. Nxa1 f2 7. Nc2 Kf4 8. a8=R! (8. a8=Q? f1=Q 9. Qf8+ Ke4 10. Qxf1)
8... f1=Q 9. Rf8+ Ke4 10. Rxf1 1-0

Marcel van Herck found the dual 3. Nc2 f3 4. a6 f2 5. a7 Kf4 (f1Q 6.Ne3+) 6.a8=Q f1=Q 7.Qf8+

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