Cooked studies

Incorrect studies? To be corrected.

5th Page. 81-100.

Here page-5 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 2005and also not in HHdbIV 2010.

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 administered in 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

100

Karstedt,M 1897 [+0000.33e3g3] Position after 4...b2

1. c3

(1. c4! {cook JU}) (1. Kd3! {cook JU}) (1. Kd2! {cook JU}) (1. Kd4! {cook JU})

1... bxc3 2. b4 axb4 (2... c2 3. Kd2) 3. a5 b3 4. Kd3 c2 5. Kd2 1-0

This study had already 4 cooks (duals) found by JU (Jan Ulrichsen?).

But white does not even win after:

4... b2! 5. Kc2 Kf4 6. a6 Ke3 7. a7 b1=Q+ 8. Kxb1 Kd2 9. a8=Q c2+ 10. Ka2 c1=Q =

Reaction of Ivica Mihoci from Croatia.

This study can be saved by playing 3.Kd3! (instead of 3.a5?).

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Wijnans,A 1938 [+0006.31e8e3] Position after: 2...Nxg2 End position.

Solution: 1... Nxa6! 2. g6! (2. gxh6? Nd3 3. h7 Ne5 4. h8=Q Nc7+!)

2... Nd3 3. g7 Nc7+ 4. Ke7! Nd5+ 5. Kf7 1-0

Black has 2...Nxg2 3. g7 Nc7+ 4. Ke7 Nd5+ 5. Kf7 Ngf4 (or Ndf4) 6. g8=Q h5= (EGTB)

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Markov,E 1991 [=3131.32a6a8] Position after 7...Bb6

Solution: 1. b7+ Kb8 2. Nc6+ Qxc6+ 3. dxc6 f2 4. c7+ Kxc7 5. b8=Q+ Kxb8 6. Rxh5 f1=Q+ 7.
Rb5+ Kc8 1/2-1/2

After: 7... Bb6 the stalemate is over and according to EGTB black wins in 38 moves.

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Troitzky,A 1898 [+0634.31h5f8] Position after 1...Rxf5+

Solution: 1. h8=Q Rh2+ 2. Kg5 Rxh8 3. Nd7+ Kg7 4. f6+ Kh7 5. dxe7 Ra8 6. Nf8+ Rxf8 7. exf8=N# 1-0

But black may escape with: 1... Rxf5+ 2. Kh6 Rxf6+ 3. Qxf6 Nf5+ 4. Qxf5 Re6+ 5. Kg5 Rxd6;

or 2.Kg4 Rxf6 3.Qxf6 Ng6 4.d7 Ne5+ 5.Qxe5 Rxe5 6.d8=Q+ Kg7

From Tantale, See geastbook: just move the black rook from f2 to e2 for example

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Amelung,F 1899 [=0046.21a1b3] Position after 3...Nc6

Solution: 1. Bxf7+ (1. e6? Ka3) 1... Nxf7 2. b6 (2. e6? Nd6 3. b6 Ka3) 2... Nd8 3. e6
Ka3 4. e7 Nxe7 5.b7 Nxb7 1/2-1/2

But black wins after: 3... Nc6! 4. b7 (or 4. e7 Nb4 5. e8=Q Nc2#) 4... Ka3 5. e7 Bf7 -+

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Krabbe,T 1976 [+4843.44h1g8] Position after 1...Nxb3

Solution: 1. e7 Rf1+ 2. Kh2 Rh1+ 3. Kxh1 Rf1+ 4. Kh2 Rh1+ 5. Kxh1 Qf1+ 6. Kh2 Ne2

7. Rb8+ Kh7 8. Rh8+ Kxh8 9. Qb8+ Kh7 10. Qh8+ Kxh8 11. Rb8+ Kh7 12. Rh8+ Kxh8

13. e8=Q+ Kh7 14. Bxc2+ 1-0

But black has: 1... Nxb3! 2. exf8=Q+ Rxf8 3. Qxb3 Bxb3 4. Rxb3 Rf1+ 5. Kh2 b1=Q -+

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McDonnell,A 1895 [=4286.64a1e7] Position after 4.Qb8!

Solution: 1. Bd8+ Nxd8 2. Qc7+ Kf6 3. Qxd8+ Ke5 4. Rh5+ f5 5. Rxf5+ Bxf5 6. Qd5+ Kf6

7. Qxf5+ Ke7 8. Qxc5+ Kd8 9. Qd6+ 1/2-1/2

But white wins after 4. Qb8+ Kf6 5. Rf1+ Bf4 6. Rxf4+ Bf5 7. Rxf5+ Ke7 8. Qd8+ Ke6 9. Re8#

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Mandler,A 1970 [=0400.11e6d3]

According to the Nalimov-tables / EGTB this is won for black.

After 1.Kd6 Kd4N wins fastest, but also the given 1...Rc2 wins.

Just enter the position in: http://www.k4it.de/index.php?topic=egtb&lang=de

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Goering,H 1895 [+4381.44h1b8]

Solution: 1. Bxf4! Qxf4 2. Qg1 Qxa4 (2... Bc5 3. Qxc5) 3. Qb6+ cxb6 4. c7# 1-0

But how after: 1... Bxg4 2. Bxe5 Bxd1 3. Bxc7+ Kc8 4. Bb6 Bd8 unclear

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Bondarenko,F Dolgov,V 1986 [=0440.01c2a3] Position after 1...Rh2!

Solution: 1. Rg6 Bd5 2. Kc3 Rh3+ 3. Kd4 Bf3 4. Be2 Bc6 5. Bb5 Bb7
6. Ba6 Ba8 7. Bd3 Rh6 8. Rg3 Rh3 9. Rg6 Rh6 10. Rg3 1/2-1/2

But black wins with: 1... Rh2! 2. Kc3 Bxb5 -+

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These two positions are both from Jan Timman 1970.

Left: [=0115.34a1d6] Right: [+0337.43e3h8] Starts with 1.d7.

As you may see they are almost the same, only with exchanged colours.

It seems that the second diagram is the newest, because it is a version.

The reader may judge which study is correct and which is not.

Here the two solutions in the database:

Left Diagram: 1. Nxd5 (1. Rxc4? dxc4 2. Ne4+ Ke5 3. Nxf2 e1=Q) (1. Ne4+? dxe4 2. Rxc4
e1=Q) 1... Kxd5 (1... f1=Q 2. Bxf1 exf1=Q 3. Rxc4 Qxc4 4. Ne3 Qc5 5. Nxa3!
Qxc3+! 6. Kb1) 2. Rxc4 Kxc4 (2... e1=Q 3. Ra4) 3. Bf1 exf1=N 4. Nd2+ Nxd2 1/2-1/2

Right Diagram: 1. d7 Nxe4 2. c8=Q! (2. Kxe4? Rxf5 3. Kxf5 Bc8! 4. dxc8=N Ne7+ 5. Nxe7=)
2... Bxc8 3. dxc8=Q Rxf5 4. Qxf5 Nd6 5. Qf4 Nxh6 6. Qxf6+! 1-0

Reaction:

Dear Mr. Boll,

As far as I remember, Jan Timman wrote in his book "Jan Timman analysiert
Studien und Partien" that the left diagram was intended to be a draw but
later he found black still can win, so he created the right diagram.

Best wishes,
Siegfried Hornecker
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89. Hervé Delboy from France reports that he found a way to correct the following study by which was cooked!

D. Gurgenidze - Tsereteli-150 JT 1991 1st Prize [=0700.21c4f6]

Look at his nice website and check it out http://hdelboy.club.fr/end_games_3.html Scroll down!

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Martin van Essen states that with the knowledge of today and the EGTB-tables to check, it is easy to find the real truth about these studies:

Fritz,J 1932.

De auteursoplossing is: 1. Pf4, Lg3 2. Pd3+, Kd2 3. Pc5, Ld6 4. Kb4, Le2 5. Ka5 = (5. …, Lxc5 pat)

Maar zwart staat gedurende het hele spelverloop, inclusief de slotstelling, gewonnen.

Met 2. …, Kd2 (beter Kc2) en zelfs 2. …, Ld6 (beter 2. …, Ke3) maakt hij het zich wel moeilijker dan nodig is, maar in de slotstelling zegt (Nalimov-EGTB)

http://www.k4it.de/index.php?topic=egtb&lang=de dat 5. ..,. Lc7+ in 70 zetten wint.

Ik kwam op het idee deze studie te bekijken doordat ik met het maken van andere studies er achter kwam dat twee lopers zelfs winnen als de paardpartij een pion, en soms zelfs twee pionnen heeft. Het blijkt nu inderdaad database-werk te zijn geworden inmiddels.

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Krabbe,T+Timman,J 1994 [+0810.66g2a4] Position after 2... Rxd2!?

Solution: 1. Bh6 (1. Bh8? dxe2 2. Rh1 Rh7) 1... dxe2 (1... e3 2. Rh3!) 2. Rh1 (2. Kf2
$2 e3+) 2... e3 3. Kf3 Rc1! (3... exd2 4. Kxe2 Re7+ 5. Re6) 4.
Rxc1 exd2 5. Rc4+! bxc4 6. Kxe2 c3 7. Rf1 Rb1! 8. Rxb1 c2 9. Rb4+! Kxb4
10. Kxd2 1-0

I was wondering how white will win after: 2... Rxd2!? I just give one line: 3. Re1 Rxa2 4. Rxd6 b4

5. a6 Rb5 6. Bxg5 b3 7. Bf6 b2 8. Bxb2 Rbxb2 9. Kf2 Kb5 = ?

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Prokes,L 1924 [+0447.54h2h5] Positon after 7...Nxf4+ End of analyses.

Solution: 1. g7 Nxg7 2. Kh3 Ng6! 3. Rf5+ Rxf5 4. Nf6+ Rxf6 5. g4+ Kh6 6. g5+ Kh5 7.
gxf6 Bc5 8. fxg7 Ne7 9. Ba3 Bxa3 10. a7 1-0

But black may try: 7... Nxf4+ 8. Kh2 Nge6 9. a7 Bxd2 10. a8=Q d5 11. Qg8 Bb4 12. Qxh7+
Kg4, how does white win this ending position?

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Amelung,F 1904 [=0342.14a8c8] Position after 4... Kc7!

Solution: 1. Bxa5 Bxa5 2. Ncb5 c6 (2... Rh7 3. Nxc4) (2... Bb6 3. Nxc4) (2... d5 3. Na7+
Kd8 4. Nc6+) 3. Nxc4 cxb5 4. axb5 Bd8 5. Nb6+ Bxb6 1/2-1/2

Black may be has: 2...Re7!? as a try.

But winning is: 4...Kc7! 5. Nxa5 Kb6 6. Nc4+ (or 6. Nc6 Rf7) 6... Kxb5 -+
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Horwitz,B 1880 [+0041.01e2h4] Position after 2...Bg2

Solution: 1. Be7+ Kg4 2. Ng5 Kh4 (2... h4 3. Bd8) 3. Kf3 Bf1 4. Kf4 Ba6 5. Bb4 1-0
But why not: 2... Bg2. May be the white king should on f2 in the startposition?

Because also 1...Kg3 is possible. Wrongly in the database?

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83.

Van_Reek,J 1967 [+0130.34f2h3] Position after 1...Bf7

Solution: 1. Kg1 f3 (1... fxg3 2. cxb4!) 2. Rb2 (2. Rxb4? f2+ 3. Kxf2 Kxh2 4. Rb1 Bb5 5. Rg1 Bc6)
2... b3 (2... bxc3 3. Rb6!) 3. Rd2 b2 (3... Bg6 4. Rd8) 4. Rxb2 Bd7 5. Rb8
Be8 6. Rb6 (6. Rxe8? f2+ 7. Kxf2 g1=Q+ 8. Kxg1) 6... Bc6 (6... Bg6 7. Rf6 Bh5
(7... f2+ 8. Rxf2!) 8. Rf8) 7. Rb2! Bb5 8. Ra2! Ba6 9. Ra5 Bb5 10. Ra8
Be8 11. Ra6 Bc6 12. Ra2 Bb7 (12... Ba4 13. c4) 13. Rb2 Bc6 14. c4 Bb5 15. c5
Bc6 16. Rd2 Bd5 17. c6 Bxc6 18. Rd6 1-0

But what after: 1... Bf7 2. Rxb4 fxg3 3. hxg3 Kxg3 4. c4 Kf4 5. c5+ Ke5 6. Rb6 Kd5

7. Rf6 Be8 8. Rf5+ Kd4 9. Kxg2 Bd7 10. Rg5 Bc6+ 11. Kg3 Bd5 12. Rxg4+ Kxc5

This is only one line. Does anyone see a winning line for white...

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82.

Ulrichsen,J and Hildebrand,A 1997 Position after 4...Ke4!?

[=0077.10b5f2] 1.hm tt22#2857

Solution: 1. Nd3+! Nxd3

(1... Kxg2 2. Nf4+ Kf3 3. Nxh5 Ba7 (3... Nd7 4. Kc6 Ne5+ 5. Kb7) 4. Kc6 Kg4 5. Nf6+)

2. d7 Bg4 3. dxc8=Q Bxc8 4. Be4 Nc5 5. Bb7 !! Nxb7 (5... Bxb7 6. Kxb6) 6. Kxb6 Nd8 (6... Nd6 7. Kc7) 7. Kc7 1/2-1/2

In the subline: 1... Kxg2 2. Nf4+ Kf3 3. Nxh5 Ba7 4. Kc6

Black could try 4... Ke4 After 5. Nf6+ (5. Ng7 Ke5) 5... Ke5 6. Ne8 Kd4 white gets in zugzwang:

7. Nc7 (7. d7 Ne7+ 8. Kd6 Nf5+ 9. Kc6 Ne6 10. Nc7 (10. Kb7 Bc5 11. Kc8 Be7 12. Nc7 Nd6+ 13.
Kb8 Nxc7 14. Kxc7 Nf7) 10... Nd8+ 11. Kb5 Ke5) 7... Bb8 8. Nb5+ Kc4 9. Na3+ Kb3
10. Nb5 (10. Kxc5 Bxd6+ 11. Kc6 Bxa3!) 10... Kb4 11. Nc7 Nxd6 12. Kxd6 Na6 13.
Kc6 Bxc7 14. Kb7 Kb5 and black wins.... This needs good checking...

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81.

Bondarenko,F 1990 [+0303.64b5c3] Position after 2...Nc4!

Solution: 1. bxc7 Na3+ 2. Kc5 Rc6+ (2... Ra5+ 3. Kb6 Nc4+ 4. Kxb7 Nd6+ 5. Kc6 Kd4 6. d8=Q
Ra6+ 7. Kd7) 3. Kd5 Rd6+ 4. Ke5 Nc4+ 5. Kf5 Ne3+ 6. Kg5 Rg6+ 7. Kh5 Nf1 8. c8=Q+ Kb3

9. Qxb7+ Kc2 10. Qc7+ Kb3 11. f5 1-0

But there is a win for black(!) with: 2... Nc4! (threathening Ra5#) 3. Kd5 e6+ 4. Kc5 Ra5#

To the first numbers 1 to 20.

To the second range 21 to 40.

To the 3rd range 41 to 60.

To the 4th range 61 to 80.

To the 6th range 101 to 120.

To the 7th range 121 to 140.

Peter Boll.

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