Duals.                   To be corrected!

1st page 1-20.

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.

20               

Afek,Y 1982 [+4010.11f3h1]        Position after 1.Qxc2

Solution: 1. Bg2+ Kh2 2. Qxc2 (2. Qe3? e1=N+ 3. Qxe1 Qf2+ 4. Qxf2)

2... e1=N+ 3. Kf2 Nxc2 4. b5 Na3 5. b6 Nc4 6. b7 Nd6 7.b8=R (7. b8=Q?) 1-0

Does white win also after: 1. Qxc2 exf1=Q+ 2. Kg4 Qg1+ 3. Kh5

NO, as Marcel Van Herck checked this on http://www.k4it.de/index.php?topic=egtb&lang=de

Black makes a draw.

On this website you can enter any endgames with a maximum of 6 pieces (including kings) and it will give you the outcome.

This database is called the Nalimov-Tables (EGTB).

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19.               

Wijnans,A 1937 [+0134.00e2f4] Position after 1.Kf1     Position after 9...Kf4

Solution: 1. Kd3 Nc3 (1... Nb4+ 2. Kc4 Kg3

(2... Nc6 3. Rh1 Ne5+ 4. Kb3 Bd4 5. Ne6+ Ke3 6. Rh3+)

3. Rh1 Nc2 4. Kb3 Nd4+ 5. Ka4 Bc3! 6. Ne4+ Kg2 7. Rc1 Bb2 8. Re1 Kf3 9. Nd2+)

2. Nb3 Kg3 3. Rd2 Nb1 4. Rd1 1-0

At move 1 white can try  1. Kf1 Nc3 (1... Nb4 2. Rh4+ Kf5 3. Rxb4) 2. Nb3 winning a piece.

But also not clear is in the subline 1.Kd3 Nb4+ 2. Kc4 Kg3 (2... Nc6 3. Rh1 Ne5+ 4. Kb3 Bd4 5. Ne6+ Ke3 6. Rh3+)

3. Rh1 Nc2 4.Kb3 Nd4+ 5. Ka4 Bc3! 6. Ne4+ Kg2 7. Rc1 Bb2 8. Re1 Kf3 9. Nd2+ As marcel van Herck comments in the guestbook.

Because what after 9...Kf4 how will white win?    It seems that only 1.Kf1 is winning...!?

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18.                

Becker,K 1927 [+4040.22g4g7]     Position after 3.Be4+

Solution: 1. Bd5 Qf8 2. Qa1+ Kg6

(2... Bf6 3. Qxf6+ Qxf6 4. gxf6+ Kxf6 5. Kh5 Kg7 (5...Ke5 6. Bg8 Kf4 7. g4 h6 8. Be6 b4 9. Kxh6 b3 10. g5 b2 11. Ba2)

6. Kg5 b4 7.Bb3 Kh8 8. Kf6 h6 9. Bc2 b3 10. Bb1 b2 11. Kg6)

3. Qb1+ Kg7 4. Qb2+ Kg6 5. Qc2+ Kg7 6. Qc3+ Kg6 7. Qd3+ Kg7 8.Qd4+ Kg6 9. Qe4+ Kg7 10. Qe5+ Kg6 11. Be4+ Kf7

12. Qf5+ Ke8 13. Bc6+ Kd8 14.Qd7# 1-0
 

White has faster: 3. Be4+ Kf7 4. g6+! hxg6 (4... Ke6 5. g7)

5. Bd5+ Ke8 6. Qa8+ Kd7 (6... Bd8 7. Qc6+ Ke7 8. Qe6#) 7.Qc6+ Kd8 8. Be6 +-

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17.             

Bania,R 1926 [+0170.01d6f8] Position after 1.Re2!?  Position after 3.Ra8+

Solution: 1. Bb3 Be4 (1... Bb7 2. Rf2+ Ke8 3. Bf7+ Kd8 4. Be6)

(1... Bh1 2.Rh2 Be4 3. Rh4 Bd3 4. Rg4)

2. Ra4 Bh7 3. Rf4+ Kg7 4. Rf7+ Kg6 5. Re7 Bg7 6. Bc2+ Kf6 7. Re6+ 1-0

White has 2 duals in this study: 1.Re2 and also 3. Ra8+ Kg7 4. Ra7+ Kg6 5. Bc2+

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16.                 

Lazard,F 1912 [=0140.26c4a8]        Position after 6.Ra7+

3.p tl20#398

Solution: 1. Kb5 g2 2. Bc6 bxc6+ 3. Ka6 g1=Q 4. a5 Bd6! 5. cxd6 Qg8

6. d7 Qb8 7. Rf8 Qxf8 8. d8=Q+ Qxd8 1/2-1/2

White can also give perpetual check with: 6. Ra7+ Kb8 7. d7 h2 8. Rb7+ Ka8 9. Ra7+ Kb8 10. Rb7+ =

Comment: Tantale in the guestbook. (5.1.2007)

Maybe this can be corrected by removing the pawn e6 and moving the point h6 to g7.

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15.                 

Havasi,A 1922 [+0410.23f2a4]              After 2.Ra1

Solution: 1. Rh1 Kb5 2. d6 Kc6 (2... a1=Q 3. Rxa1Rxa1 4. d7 Ra8 5. Be8 Kb6 6. h6!

(6. Kf3? Kc7 7. Kxf4 Ra3 8. Kxf5 Rh3 9. Kg6 Rg3+ 10. Kh7 Kd8 11. h6 Ke7)

6... Kc7 7. h7 Ra2+ 8. Kg1 Ra1+ 9. Kh2 Ra2+ 10.Kh3 Ra1 11. d8=Q+ Kxd8 12. Bc6)

3. h6 a1=Q 4. Rxa1 Rxa1 5. h7 Rh1 6. Bd5+ 1-0
 

This study was already critisized in EBUR/1994/1/page 25 and later saved by Marcel van Herck see EBUR/1994/3/page 25.

But it looks as if white has a dual with 2. Ra1 Ra7 3. Be6 Kc5 4. Kf3 Kd6 5.Kxf4 and winning?

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14.                 

Vandiest,J Missiaen,R 2000           Position after 2.Rd1

[=0406.10a5c6]

Solution: 1. Rb5!

(1. Kxa6? Rb2! 2. b8=Q 2... Rxb8 3. Ka7 Rb7+ 4. Ka6 Rb6+)

(1. b8=Q? Nxb8 2. Rb5! (2. Rd8 Kc7)
2... Na6! 3. Kxa4 (3. Kxa6 Nc5+) 3... Nc5+) (1. Rd8? Ra2! 2. b8=N+! Kc7!

3. Nxa6+ (3. Rd7+ Kxb8 4. Kxa6 Nc5+) 3... Kxd8 4. Nb4 Ra1 5. Nc2 Rc1) 1...
Ra2! (1... N4c5 2. b8=N+!) (1... Nc3 2. b8=N+ Nxb8 3. Rxb8 Ra2+ 4. Kb4 Rb2+
5. Ka3! Rxb8) 2. b8=N+! Nxb8 3. Rxb8 Nb6+ (3... Nc5+ 4. Kb4 Rb2+ (4... Na6+
5. Kb3) 5. Ka5! Rxb8) (3... Nc3+ 4. Kb4 Rb2+ 5. Ka3! Rxb8) 4. Kb4 Rb2+ 5.
Ka3! (5. Ka5? Rb5+ 6. Ka6 Rb4!) 5... Nc4+ 6. Ka4 Rxb8 1/2-1/2

In the subline 1.Kxa6 Rb2 what about: 2. Rd1!? Nc3 3. Ra1 Rb6+ 4. Ka7 Rxb7+ 5.Ka8=

NO, as Tantale in the guestbook arguments: 2...Nc5+ 3.Ka7 Rxb7 or Kc7

can be checked at Nalimov: on http://www.k4it.de/index.php?topic=egtb&lang=de

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13.                 

Liburkin,M 1933 [+0161.13a3a8]      After 1.Re1.

Solution: 1. Kb2 c1=Q+ 2.Kxc1 Bg5+ 3. Kc2! f4+ 4. Kd2! Bxh6 5. Re6 f3+ 6. Ke1 f2+!

(6... Bxf7 7.Rxh6) 7. Kxf2 Be3+! 8. Kxe3 d4+! 9. Kxd4 Bxf7 10. Rh6 Bb3 11. Rb6 1-0
 

What after: 1. Re1 Be7+ 2. Kb2 c1=Q+ 3. Rxc1 Kb7 4. Rg1 Bh5 5. Rg8 (or also 5.Nxf5) +-!?

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12.                 

VanDiest,J 1950 [+0410.35g5h8]     Position after 4.f5

Solution: 1. Bd5+ Rg8 2. Rxg8+ Kh7 3. Rc8 f2 4. Bg2 a2 5. Rc7 f1=Q (5... a1=Q 6. Rxd7+ Kg8 7.
Kxg6) 6. Bxf1 a1=Q 7. Rxd7+ Kg8 8. Kxg6 Qa5 9. Rb7! Qh5+! 10. Kxh5 c1=Q 11. d7 Qd1+

12. Kh6 Qxd4 13. Bc4+ Kf8 14. Rb8+ Ke7 15. d8=Q+ Qxd8 16. Rxd8 Kxd8 17. f5 1-0

White has also the faster win: 4. f5 gxf5 5. Kf6 f1=Q 6. Bg8+ Kh6 7. Bf7 Kh7 8. Bg6+ Kh6 9. Rh8#

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11.                  

Grondijs,H [+0131.36a6a4] 1995     Position after 1.Nxf4

Solution: 1. Rd7 e5 2. Nxf4 (2. Ng7? h5!) 2... Bxf4 3. c3 Bc1 4. Ra7 Be3

5. Rb7 Bc5 6. Rb3 g2 7. Ra3+ Bxa3 8. b3# 1-0
 

White also wins with: 1. Nxf4 Bxf4 2. c3 Bd2 3. Rxh7

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10.                   

Gurgenidze,D+Kralin,N                     Position after 3.b8Q+

[+0444.11c5c7] 1990

3/4.p Tsjernobyl Ty Vestnik Chernobila

Solution: 1. Rf7+ Kb8 2. Nd7+ Ka7 3. bxa8=Q+  Rxa8 (3... Kxa8 4. Nb6+ Kb8 5. Rb7#) 4. Kb5! Be2+

(4... Bh5 5. Ne5+ Bxf7 6. Nc6#) 5. Ka5 Bxa6 6. Ne5+! Kb8 (6... Bb7 7.Nc6#) 7. Kb6 c1=Q 8. Rf8+ Bc8 9. Nd7# 1-0

Also winning is 3. b8=Q+ Rxb8 4. Nxb8+ Kxb8 5. Rb7+ Kc8 6. Kd6 Be2 7. Rb5+! Kd8 8. Rb8# 

or 6...Nc7 7.Rxc7++ Kb8 8.Bd3+-

Tantale (see guestbook) suggests to put an black pawn at e3, which seems to work.

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10b. In Bent,C [+0070.54b6b8] 1959 there is a dual in the mainline at move 6 white can play better

6.Ne7! which mates faster Bb3 7.Ka6 a4 8.Nc8 and Nb6 mate follows.

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9                      

Platov,V [+3110.43g3c6]  1903          Position after 7.Re5+!

There is also a version to [+3110.33g3c6] where pawn a3 is missing, with the same solution.

1. Bg2 Qc8 (1... Qa6 2. Rd3+ Kxc5 3. Rd5+ Kxc4 4. Bf1+ Kxd5 5. Bxa6 Ke5 6. Bc4
Kxf6 7. Kf4) 2. Rf5+ Kd7 3. Rd5+ Ke8 4. Rh5 Qa6 5. c6 Qxa3+ 6. Kg4 Qf8 7. Rb5 Qd6 8. Rb8+ 1-0

There is better move and so a dual in: 7. Re5+ Kd8 8. Rb5 Kc8 9. Be4 Qc5 10. Bf5+ Qxf5+ 11. Kxf5 a6 12. Ra5 Kd8 13.
Rxa6 Ke8 14. Ra8# because it forces a mate. I don't think it changes much about the beauty of this study.

(Correction suggestion see guestbook)

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8.                    

Jakimtsjik,V [=0430.11h5g8] 1975.  Position after 3. Rg5+

Solution: 1. Rc5! Rf5+! 2. Rxf5 c2 3. e7! Bf7+ 4. Kh6! c1=Q+ 5. Rg5+ Kh8 6. e8=Q+ Bxe8 1/2-1/2
White has a dual in 3. Rg5+ After Kf8 4. Rg1 Black possibly planned Be2+? (better Bxe6!) but then 5. Kg5 Bd1? 6. Kf6 (+-)

c1=Q 7. e7+ Ke8 8. Rg8+ Kd7 9. e8=Q+ Kd6 10. Qe7+ Kd5 11. Rd8+ and white wins. 

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7.                    

Gaggiottini,M [+0400.43c3a4] 1983 Position after 1.b3+

Solution: 1. Rc4+ Ka5 2. b4+ Kb5 3. a4+ Kxa4 4. b5+ Ka5 5. Ra4+ Kxa4 6. bxa6 1-0
Also winning is:

1. b3+ Kb5 (Ka3 2.Rc4 Kxa2 3.Kc2) 2. a4+ Ka5 3. Rd7 Rc6+ 4. Kb2 Rb6 5. Ka3 Rb4 6. Rf7 d5 7. Rd7 +-

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6.                 

Bent,C 2000 [+3047.24a1e8]         Position after 3.Ng7+!

Solution: 1. a8=Q+ Kd7 2. Qxd5+ Ke8 3. Qa8+  Kd7 4.Be6+ Kxe6 5. Nh6! Qc4 6. Qg8+ 1-0

Dual: a Forced mate is possible after: 3. Ng7+!! Qxg7 4. Qc6+ Kd8 5. Be6 and 6.Qd7#
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5.                 

Van_Tets,A 1968 [+1311.03e3h1] Position after 1.Ke2!?

Solution: 1. Ne4  1... Rb3 2. Qxb3 cxb3 3. Bb7 Kg1 4. Nd2 b2 5. Be4 b1=Q 6.
Bxb1 Kh1 7. Be4 Kg1 8. Nf3+ Kh1 9. Kf2 g1=Q+ 10. Nxg1# 1-0

Dual at move 1. Ke2 Rxc8 (1... g1=Q 2. Ne4) 2. Ne4 g1=Q 3. Nf2+ Qxf2+ 4. Kxf2 Rf8+
5. Kg3 Rg8+ 6. Kh3 +-

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4.                  

Grondijs,H 1980 [+0074.22d7g3]    Position after 2. Kd8!?

Solution: 1. bxa7 Be6+ 2. Kxe6  Bd4 3. Ne4+ Kf4 4. a8=N Ke3 5. Bf1 Nb4 6.
Nc7 Nxd3 7. Nd5+ Kxe4 8. Bg2# 1-0

After 2.Kd8 white wins also and fast.

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3.                  

Grondijs,H 1994 [+3115.44h1a2]    Position after 2. e6

1. Nec3+ Ka1 2. Rd7 (2. Bc1 Nd3 3. Bb2+ Nxb2 4. Nc5 Nc4)  2... Qxd7
3. Bc1 Nd3 4. Bb2+ Nxb2 5. Nc5 Qf7 6. e6 1-0

There is a strong dual in 2. e6.

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Small note:

2. R. Missiaen [+0041.01b6a8] has already duals, an extra dual 6.Nc5 Bc4 7.Nd7+ Kc8 8.Ne5+ Kd8 9.Nxc4+-

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1.

Pallier,A + Van_der_Heijden,H 1998. [+0000.76c6g8]

White plays and wins.

Authors-solution: 1. c3! (1. cxb3 ? fxg5 2. f5 exf5 3. gxf5 g4 ! 4. Kd6 g3 5. Ke7 g2 6. f6
gxf6 7. h7+ Kg7) 3. f5 exf5 4. cxd5 fxg4 5. d6 g3 6. d7 g2 7. d8=Q#) 1... fxg5 (1... a5 2. h7+
Kh8 3. gxf6 gxf6 4. Kd6 a4 5. Ke7 a3 6. Kf7) 2. fxg5 ! (2. f5 ? exf5 3. Kxd5
fxg4 4. Ke4 a5 ! 5. c4 a4 6. c5 a3 7. c6 axb2 8. c7 b1=Q+) 2... a5 (2... e5 3.
Kxd5 e4 4. Kd4 !) 3. Kb5 e5 4. Ka4 ! e4 5. Kxb3 a4+ 6. Kc2 Kh8 7. Kd2 Kg8 8.
Ke3 Kh8 9. Kd4 Kg8 10. c4 ! dxc4 11. Kxc4 ! 1-0

Dual at move 1: 1. h7+ Kh8 2. c4! fxg5 (or 2... d4 3. gxf6 gxf6 4. Kd6 d3 5. Ke7
d2 6. Kf7 d1=Q 7. g7+ Kxh7 8. g8=Q+ Kh6 9. Qh8# (9. Qg7#; 9. Qg6#)) (or 2... dxc4
3. gxf6 gxf6 4. Kd6 c3 5. Ke7 cxb2 6. Kf7 b1=Q 7. g7+ Kxh7 8. g8=Q+ Kh6 9. Qg7#)

3.f5! exf5 4.cxd5 and ... d8Q#

so the study is correct but this variation also wins and faster.

With a dual at move one, very much of the beauty is gone.

After 2. c4!

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To newest duals.

Peter Boll.

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