Cooked studies

Incorrect studies? To be corrected.

7th Page.

Here the 7th page 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

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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?

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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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140

Prigunov,V 1999 [+0013.66g4c5] After 3...Nh3!

Solution: 1. d4+! (1. e6? Kd4 2. e7 c2 3. e8=Q c1=Q) 1... Kxd4 (1... Kc6 2. Kf3) 2.
exd6 (2. e6? Nf2+ 3. Kf3 Ne4) 2... Nf2+ 3. Kf3! (3. Kh4? Ne4) 3... Nd3!
(3... Ne4 4. Bxe4 dxe4+ 5. Ke2 e3 6. d7) 4. Bxd3 Kxd3 5. d7 c2 (5... Kc2 6. d8=Q Kb2 7. Qxd5
Kxa2 8. Ke3 Kb2 9. Qg2+ Kxb3 10. Kd3 a2 11. Qc2+ Ka3 12. Kc4) (5... d4 6. d8=R!

Kc2 7. Ke2 d3+ 8. Rxd3 Kb2 9. Rd2+ cxd2 10. Kxd2) 6. d8=R! (6. d8=Q? d4
7. Qc7 c1=Q 8. Qxc1) 6... d4 7. Rc8 Kd2 8. Ke4 c1=Q (8... d3 9. g4) 9. Rxc1
Kxc1 10. Kxd4 Kb2 11. Kd3! Kxa2 12. Kc2 Ka1 13. g4 Ka2 14. g5 hxg5 (14... Ka1
15. Kc1!) 15. h6 g4 16. h7 g3 17. h8=B! (17. h8=Q? g2 18. Qg7 g1=Q 19.
Qxg1) 17... g2 18. Bd4 g1=Q 19. Bxg1 1-0

Black has better 3... Nh3 4. d7 Ng5+ 5. Ke2 Ne6 6. Bf5 Nd8 7. g4 Ke5 8. Kd3 Nc6 =

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139

Cheron,A 1979 [+4618.38a3d5] After 10...c4

Solution: 1. Qg8+ Ke4 2. Qc4+ Kxe5 3. Qe6+ Kf4 4. Bh2+ Kf3 5. Qe3+ Kg2 6. Qg1+ Kf3 7.
Qf1+ Kg4 8. Qe2+ Rf3 9. Qe6+ Rf5 10. Nd7 1-0

The study ends, but how does white win after 10...c4..?

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138

Grondijs,H 1994 [=0406.42e7h8] After 1...Nxf4

Solution: 1. Kf6 (1. Kxd8? Nxf4 2. e7 Kxh7) (1. Rxd4? Nc6+ 2. Kf6 Rxf5+ 3. Kxf5 Nxd4+
4. Kxg4 Nxe6) 1... Rg7 (1... Nxf4 2. Kxg5 g3 (2... Nfxe6+ 3. fxe6 g3 (3...Nxe6+ 4. Kxg4 Kxh7 5. Kf5)

4. e7 Ne6+ 5. Kg4 Nc7 6. Kxg3) 3. e7 Nf7+ 4. Kxf4 Nd6 5. Kxg3) 2. Rxg4 Nxe6

(2... Rxg4 3. e7 Nb7 4. e8=Q+ Kxh7 5. Qe7+) 3. Rxd4

(3. fxe6? Rxg4 4. e7 Re4 5. Kf7 Ne5+ 6. Kf6 Nc6) 3... Nec5 (3... Nxd4)

(3... Rg6+? 4. fxg6 Nxd4 5. g7+ Kxh7 6. Kf7) 4. Rd8+ Kxh7 5. Rh8+ Kxh8 1/2-1/2

But after 1.Kf6 Nxf4 2.Kxg5 Nd5! black wins...

3.Kxg4 Kxh7 4.Kf3 Nc6 5.Ke4 Nf6+ 6.Kd3 Kg7 7.Kc4 Ng8! 8.Kd4 Ne7+

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137

Grondijs,H 1992 [=0320.44g4b2] After 5...Re2+!

Published in EBUR/1992/1/page 15.

Solution: 1. Bd3 f1=Q (1... Kc3 {main} 2. Bf1 Rxf1 3. Bxe5+ Kd3 (3... Kd2 4. Kxf3 Ke1 5.
Bg3) 4. Kxf3 Re1 5. Kxf2 Rxe3 6. Bf4 Re2+ 7. Kg3) 2. Bxf1 Rg1+ (2... Rxf1 3. b5 (3.
Bxe5+? Kc2 4. b5 Kd2 5. b6 Ke2 6. b7 Rb1) 3... Kc2 4. b6 Kd2 5. b7) 3. Bg2!
(3. Kxh3? Rxf1 4. b5 Kc2 5. b6 Kd2) 3... Rxg2+ 4. Kxf3 1/2-1/2

In the "main"line black has 5...Re2+ (instead of Rxe3) 6. Kf1 Rxe3 7. Kf2 Kxe4 8. Bd6 Rb3 9.
Be7 Rf3+ 10. Ke2 Rg3! and wins -+;

But how is the result after 6.Kg3 Rxe3+ 7.Kg4 Rxe4+ 8.Bf4 Rxb4!?

Who has an idea? Is it a win for the rook like Timman-Velimerovic?

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136

Korn,W 1932 [+0080.33g7e8] End analyses after 9.Bd4

1. a5 bxa5 (1... dxc4 2. Be4) 2. Ba4 Be6 (2... c6 3. bxc6 Bc7 4. Bd6 Kd8 5.
Bxc7+ Kxc7 6. Kxf7 dxc4 7. Ke6 c3 8. Kd5) 3. b6+ c6 (3... Bd7 4. b7) 4. Bxc6+
Bd7 5. b7 Bc7 6. Bxd5 (6. cxd5? Kd8 7. Bxd7 Kxd7) 6... Ba4 7. Bc5 Bb8 8. Kf6
Kd7 9. Bd4 1-0...

Here the analyses stop, but how does white win?

After 9... Bb3! 10. Be5 Bxe5+ 11. Kxe5 Kc7 12. Kd4 a4 13. Kc3
Ba2 14. Be4 (14.Kb4 a3! 15.Kxa3? Bxc4=) a3 15. c5 Be6 16. Kc2 Ba2 I see no win...
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135

Kok,T 1933 [+0331.44h8d1] After 3...Bd5!

Solution: 1. Nb4 Bh5 2. a7 Bf3 3. Nd3 Rg2 4. e7 Re2 5.Nf2+ Ke1 6. Ne4 1-0

But black has time for: 3... Bd5 4. e7 Rg8+ 5. Kxh7 Re8 avoiding interferences.

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134

Dobrescu,E Nestorescu,V 1966 After 4...Qe5!?

[=3011.22b3h6]

Solution: 1. g7 Qe3+ 2. Ka4!! (2. Kc4? Qf4+ 3. Kc5 Qxf6 4. Bc4 Qf2+) (2. Ka2? Qf2+
3. Kb1 Qxf6 4. Bb3 Qg6+ 5. Bc2 Qxg3) (2. Kb4? a5+ 3. Kxa5 Qe5+ 4. Ka6 Qxf6 5.
Ba2 c5+ 6. Kb5 Qb2+) 2... Qf4+ (2... Qd4+ 3. Kb3! Qxf6 4. Bd5!) (2... Qxg3
3. Bb3 Kh7 4. Bc2+ Kg8 5. Bb3+) 3. Ne4! Qxe4+ 4. Ka3! Qg6 (4... Qf3+ 5. Bb3)

(4... Qe5 5. Bb3 Qa1+ 6. Ba2 Qc1+ 7. Ka4 Qf4+ 8. Ka3 Qd6+ 9. Kb3 Kh7 10. Ka4
Qd4+ 11. Ka3 Qg4 12. Bf7 c5) 5. Bb3 Kh7 6. Kb4 Qg4+ 7. Kc3 1/2-1/2

But black may try: 4... Qe5 5. Bb3 Qa1+ 6. Ba2 Qc1+ 7. Ka4 Qf4+ 8. Ka3 Qd6+ 9. Kb3 Kh7 10. Ka4
Qd4+ 11. Ka3 Qg4 12. Bf7 c5 with winning chances....

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133

Bondarenko,F 1990 [=4634.56d1e7] After 7...Ke3+ !

Solution: 1. Ng8+ Kd6 2. Qc5+ Kd7 3. Qe7+ Rxe7 4. Nf6+ Kd6 5. Nxe4+ Kd5 6. Nc3+ Kxd4 7.
Ne2+ Kd3 8. Nc1+ Kd4 9. Ne2+ Kd5 10. Nc3+ Kd6 11. Ne4+ Kd7 12. Nc5+ 1/2-1/2

Black wins after 7... Ke3+. May be a white pawn at f2 is missing?

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132

Wijnans,A 1938 After 3...cxd5 After 1.dxc6

[+0031.65g4d1]

Solution: 1. Nf2+ Ke2 2. Nxd3 Kxd3 3. a5 Be4 4. d6 exd6 5. c5 dxc5 6. bxc5 Bd5
7. a6 Be6+ 8. Kf4 Bc8 9. a7 Bb7 10. Ke5 Kc4 11. Kd6 Kb5 12. Kc7 Ka6 13. Kb8 g6
14. h4 f6 15. g4 f5 16. h5 gxh5 17. g5!! h4 18. g6 h3 19. g7 h2 20. g8=Q h1=Q
21. Qa2+ Kb5 22. Qb2+ Kxc5 23. Kxb7 Qh7+ 24. Ka6 Qg8 25. Qb6+ 1-0

At move 3 black wins with:

3... cxd5 4. c5 f5+ 5. Kg5 Be8 6. a6 Kc4 7. Kxf5 d4 8. Ke6 d3 9. Kxe7 Bc6 10. Kd6 Kb5 -+

At move 1 white possibly wins with:

1. dxc6 d2 2. Kf3 e5 3. Nf2+ Ke1 4. c7 e4+ 5. Ke3 Bf5 6. b5 Bc8 7. b6 f5 +-

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130

Cheron,A 1970 [+0444.12h4c2] After 2...Rd2!

Solution: 1. d7 (1. Rxa3? Rh2+ 2. Kg5 Rxh8 3. Bf6 Rg8+ 4. Kf4 Bc8! 5. Ra8 Re8!)
1... Rd2 (1... Bc6? 2. Rc3+!) (1... Rh2+ 2. Kxg4 Bd5 (2... Rxh8 3. Bf6 Ra8
4. Rxa3) 3. Rc3+ Kb1 4. Rc1+ Ka2 5. Ra1+ Kb3 6. Rxa3+) 2. Rc3+ Kd1 (
2... Kxc3 3. Ba5+) (2... Kb2 3. Bf6) 3. Rc1+ Ke2 4. Re1+ Kf3 5. Re3+ Kg2 (5...
Kf4 6. Bg5+ Kf5 7. Re5+ Kxe5 8. Bxd2) (5... Kf2 6. Bb6) 6. Re2+! Rxe2 7. Ba5
1-0
Black is able to trick, in the subline: 1...Rh2+ 2.Kxg4 and now like the mainline

2....Rd2 3. Rc3+ (3. Bf6 Rxd7 4. Rxa3=) 3... Kb2 4. Bf6 Rxd7 5. Rd3+ Ka2 6. Rxd7 Bc8

all because the white king is on g4.

131. Therefore also the version: [+0444.13h4c2] which after 1.exd7 is exactly the same, is busted this way.

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Amelung,F 1899 [+0014.54g1c3] After 1...Ng4

1... Kxb4 2. c4! bxc4 3. bxc4 Kxc4 4. Nc2 1-0

Black wins with 1...Ng4 going for Nxb4, Nxc2

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De_La_Bourdonnais,C 1836 After 1...Rbd8!?

[+0700.66b1d6]

Solution: 1. e7 Kxe7 2. d8=Q+ Kxd8 3. Rxb8+ 1-0

What about 1...Rbd8! ...?

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De_Feijter,C 1932 [+0030.31g4c1] After 3...Be7

Solution: 1. Kf5 Bb6 2. Ke4 Bd8 3. Ke5 Bh4 (3... Bc7+ 4. Kd5 Bd8 5. h7 Bf6 6. Kc5)

4. h7 Be1 5. Kd4 Kb2 6. Kc5 Bc3 7. Kxb5 Kb3 8. h4 Bd4 9. h5 Bc3

10. h6 Bd4 11. Ka5 Bc3 12. h8=Q 1-0

After 3...Be7 4. h7 Bxb4 5. Kd4 Kc2! 6. h8=Q Bc3+ 7. Kc5 Bxh8 8. Kxb5 =

Possibly the composer has known this because there is a version: [+0030.21g4f1]

with the black king on f1, without the white pawn at h2, where this maneouvre is not possible..

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Almgren,S 1945 [+0031.11h8g2] Position after 4...Bg5

Solution: 1. Nc2 Bd6 2. Ne1+ Kg3 3. Nd3 Bf8 4. Kg8 Bh6 5. Kh7 Bf8 6. Ne5

(6. Nc5 {or}) 6... Kf4 7. Nd7 1-0

But in the subline 3...Be7 4. Kg7 black may continue with 4...Bg5 5. Kf7 (5. Ne5 c5 6. Kh7
Bf6 7. Nd7 Bc3 8. Nxc5 Kf4=) 5... Bh6 = and black has one tempo extra compared to the main line.

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Fischer,J 1936 [=0106.01d2b2] After 1. ... Ng3!

Solution: 1. Rh1 Nd1 (1... Ng3) 2. Rxd1 Nc3 3. Ra1 Nb1+ 4. Kd1 Kxa1 5. Kc2! 1/2-1/2

It seems black wins after 1...Ng3! Check it out yourself...

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Osintsev,S 2005 [=0147.02d2h1] After 4...h1=Q

2.hm ja13 te03

Solution: 1. Ne3 (1. Nxb6? Kg1) 1... Ne4+ 2. Ke1 Kg1 3. Rh5 Be2 4. Rh4! (4. Rxh2?
Kxh2 5. Nd1 Na4 6. Nb2 Nxb2 7. Bxd3 Bf3 8. Be2 Bh1) (4. Rh6? Ng3 5. Kd2 Bh5)
4... Ng3 5. Kd2 (5. Bxd3? Bxd3 6. Rg4 h1=Q 7. Rxg3+ Kh2+ 8. Kf2 Bf5! 9. Rg2+ Kh3

10. Rg3+ Kh4 11. Nxf5+ Kh5 12. Ng7+ Kh6 13. Nf5+ Kh7 14. Rg7+ Kh8) 5... Ne4+

6. Ke1 h1=Q 7. Rxh1+ Kxh1 8. Nd1 Na4 9. Nb2 Nxb2 10. Bxd3 Bf3 (10... Bxd3)

(10... Nxd3+ 11. Kxe2) 11. Be2 Bg2 12. Bf1 Bf3 13. Be2 1/2-1/2

Published recently in magazine EG-April 2005. But no line on:

4... h1=Q! 5. Rxh1+ Kxh1 6. Ba2 Kh2 7. Be6 Kg3 where black seems winning!

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Halberstadt,V 1929 [=0001.24a1a3] After 2...Kb4

Solution: 1. c6 dxc6 2. c5 e4 (2... b2+ 3. Kb1 e4 (3... Kb3 4. Nd3 a3 5. Nxb2 axb2) 4.
Nd3 exd3) 3. Nc2+ bxc2 1/2-1/2

But black has: 2... Kb4 3. Nd3+ Kb5! 4. Nxe5 a3 5. Nd7 Kc4 6. Kb1 Kc3 -+ or 6. Ne5+
Kxc5 and winning chances.

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Van_den_Ende,J 1941 [+0045.23e3b4] After 2...Ne5!?

Solution: 1. Be7+ (1. Nxf3? Bxf3 2. Be7+ Ka5) (1. Ne2? Nxg5) 1... Ka5 (1... Kc4 2.
Nxf3 Bxf3 3. Nd2+) (1... Ka4 2. Nc3+ Ka5 3. b4+ Kb6 4. Bc5+ Kb7 5. Nxf3) 2. Ne2
g1=Q+ (2... Nd4 3. b4+!) 3. Nxg1 Nxg1 4. Bc5! Ne2 ( 4... h5 5. Nc3 Nh3 6. b4#)

5. Nf6 h5 6. Nxg4 1-0

Unclear to see how white will win: 2... Ne5 3. Ng1 Bc8.

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