Cooked studies

Incorrect studies? To be corrected.

Third Page.

Here the 3rd 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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Van der Holst,V 1936 [+0000.45b6g5] Position after 5...Kxh3!

De Schaakwereld.

Solution: 1. h6 (1. Kxc5? f4 2. Kd4 Kh4 3. Kd3 f3 4. Ke3 Kg3) 1... gxh6 (1...
Kxh6 2. Kxc5 f4 3. Kd4 Kg5 4. Kd3 Kh4 5. Ke2 Kg3 6. Kf1) 2. a5 f4
3. a6 f3 4. a7 f2 5. a8=Q f1=Q 6. Qg8+ Kf6 (6... Kh4 7. Qg4#) 7. Qf8+ 1-0.

In the subline with 1...Kxh6 black has 5... Kxh3 and has winning chances e.g.

6.Kf2 (6.Kf1 Kh2!) g5 7.a5 g4 8.a6 g3+ 9.Kg1 f3 10.a7 f2+ 11.Kf1 g2+ 12.Kxf2 Kh2 13.a8=Q g1=Q+ and black has an advantage.

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Mees,W 1998 [+0040.47a3e8] Position after 8. ... b2+

Solution: 1. Bc1 Kd7 2. Kb2 Ke8 3. Ka1 Kd7 4. Ba3 Ke8 5. Kb2 Kd7 6. Kc1 Ke8 7. Bb2 Kd7 8.
Bxd4 Ke8 9. Bb2 Kd7 10. Ba3 Ke8 11. Kb2 d4 12. Kc1 Kd7 13. Bb2 Ke8 14.
Bxd4 Kd7 15. Bb2 Ke8 16. Ba3 Kd7 17. Kb2 c1=Q+ 18. Kxc1 b2+ 19. Kxb2 Bd1 20.
Kc3 Be2 (20... Bc2 21. Kd4 Ke8 22. Kc5 Ba4 23. Kc6) 21. Kd4 Bf1 22. Kc5 1-0

But when black plays: 8... b2+ 9. Bxb2 Bb3 10. Ba3 Bc4 11. Kb2 d4 12. Kc1 Bd5 13. Bb2 Bc6
14. Bxd4 Ke6 the white king can't become dangerous, so =

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Kralin,N [+3113.42d4f8] 1985 Positons after 5...Qa8! Position after 8.Bxe8 Stalemate.

1.p Moscow Champ

Solution: 1. e7+ Ke8 2. Rc8 Qxc8 3. Bb5+ Nc6+ 4. Kd5 (4. Kc5? Kd7 5. Kd5 Qg8+) 4... Kd7
5. Kc5 Ke6 6. Bxc6 Qb8 7. Kc4 Kxd6 8. e8=Q Qc7 9. Qd7+ Qxd7 10. Bxd7 Kxd7 11. Kd5 Ke7 12. Kc6 Ke6 13. f3 Ke7
14. Kc7 Ke6 15. Kd8 1-0

Black can surprisingly escape with 5... Qa8! 6. e8=Q+ Qxe8 7. Bxc6+ Ke6 8. Bxe8 Stalemate!

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57.

Pervakov,O Kralin,N Position after 6. ... e4!?

[=1037.25d5b5] 1992

2.hm Tidskrift för Schack

Solution: 1. c4+ Ka5 2. Qa2 Nxc4! 3. Nxc5 (3. Qa1? Be6+) (3. Nxe5? Be6+ 4. Kxc5
Nb3+ 5. Qxb3 d6+) 3... Be6+! 4. Nxe6 dxe6+ 5. Kc5! (5. Ke4? Nb3 6. Kf3
Nbd2+ 7. Ke2 b3 8. Qa1 a3) 5... Nb3+ (5... Nd2 6. Kd6 N4b3 7. Kxe5)
6. Kxc4 e4 7. Qxa4+ Kxa4 1/2-1/2

In the subline 5. ...Nd2 6.Kd6 black has 6... e4!? i.e. 7.Ke5 b3 8.Qa1 Nc2 9.Qd1 e3 and black has good winning chances.

Also after 7.Qa1 Nc4+ 8.Kc7 e3 9.b3 Nxb3 10.Qh1 Nd4 11.Qa8+ Kb5 12.Qb7+ Kc5 13.Qa7+ Kd5 14.Qa8 Ke5 15.Qxa4 e2 -+

But this needs analyses!

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56.

Bent,C 1993 [=0318.00h1d5] Position after 2...Re8!

1. Be3 Re7 (1... Rh4+ 2. Kg2 Rh7 3. Nd8 Rd7 4. Bg5) (1... Rb4 2. Nd8 Rh4+ 3.
Kg2 Rh8 4. Nf7) (1... Kc6 2. Nd8+ Kd7 3. Nf7) 2. Nd8 Rd7 (2... Re8 3. Nf7 Rf8
4. Ng5 Rf1+ 5. Kg2 Rxd1) 3. Bg5 Rh7+ 4. Kg2 Rg7 5. Kf3 Rxg5 6. Ne3+ 1/2-1/2
Black has: 2... Re8! 3. Nf7 Rf8 4. Ng5 Rf1+ 5. Kg2 Rxd1 and wins.

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55.

Bent,C 1993 [=0048.27g8f6] Position after 3.Kf7!

Solution: 1. Neg6+ Ne7+ 2. Bxe7+ Kf5 3. Nh4+ Ke5 4. Nhg6+ Kf5 5. Nh4+ Kxf4 6. Bd6+ Kg5 7. Be7+ Kh5

8.Ng2 Kg6 9. Nh4+ Kh5 10. Ng2 Ne2 11. Bd8 (11. a4? a5) 1/2-1/2
It seems that even white can win with: 3. Kf7 e2 (or 3... Ne2 4. Ne6 Nf4 5. Ng7#) 4.Nh5 e1=Q 5. Ng7#

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54.

Behting,C [+0014.23a1b5] 1920. Position after 2....Kxa5

1.p Kagan's Neueste Schachnachrichten#2

Solution: 1. Kb2 Nxd3+

(1... Ne2 2. Bh2 h3 3. Be5 a3+ 4. Kxa3 Nc1 5. Ne1 Kxa5 6. Kb2 Ne2 7. Nf3 Kb5 8. Kc2 Kc5 9. Kd2)

(1... a3+ 2. Kb1 Ne2 3. Bh2 Kb4 4. Be5)

2. Kc3 Nc1 (2... Nf4 3. Nd4+ Kxa5 4. Kc4 a3 5. Nb3+ Ka4 6. Nc5+ Ka5 7. Bh2!)

(2... Nb4 3. Nd4+ Kxa5 4. Kc4) 3. Nd4+ Kxa5 4. Nc6+ Kb5 5. Na7+ Ka5 6. Be3 h3
7. Nc6+ Kb5 8. Nd4+ Ka5 9. Bxc1 h2 10. Bf4! h1=Q 11. Bc7# 1-0
First it has to be analysed why 1.d4 is not winning.

Second: in the subline 1. ... Ne2 2.Bh2 Kxa5 seems better. After 3.Nxh4 Nd4 what will the outcoming of this be?

Is this endgame-theory?

Third: The subline 1. ... a3+ should not be refuted by 2.Kb1 because of Nxd3. Better simple 2.Kxc1.

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53.

Rinck,H [+4004.12b1d1] 1906. Position after 3...Kxc3

1.p Bohemia

Solution: 1. Nc3+ Ke1 2. Qg3+ Kd2 (2... Kf1 3. Qf3+ Kg1 4. Ne2+) 3. Qh2+ Kd3
4. Nd5 Qf7 (4... Qd8 5. Qc2+ Kd4 6. Qc3+ Ke4 7. Qe3+ Kxd5 8. Qd3+) 5. Qc2+ Kd4
6. Qc3+ Ke4 7. Qe3+ Kxd5 8. Qb3+ 1-0

It is hard to believe that a first price winner is so easy to cook this with 3...Kxc3.

I think that the beginning position is wrongly in the database.

Possibly the black knight is on a different place so not defending the black queen so that white has Qb2.

Also 1...Kd2 is a try for black.

Who can check this out with one of the famous books of the great Henri Rinck?

Reaction Harold van der Heijden:

Yes, incorrectly in the database (thanks). I have consulted all the 10 secundairy sources (ie. 1414#0098) and in all the black Knight should be at f8.

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52.

Jakimtsjik,V Korolkov,V Position after 3.Rh7!

[=0740.33a7h8] 1977

Solution: 1. Bf6+ Kg8 2. Rg7+ Kf8 3. Rf7+ 3.Ke8 4. Re7+ Kd8 5. Rf7+ Kc8 6.
Rf8+ Kc7 7. Rf7+ Kd6 8. Be7+ Ke5 (8... Kc7 9. Bc5+ Kc8 10. Rf8+ Bd8 11. Bb6) 9.
Bf6+ Kf4 10. Bd4+ Kg5 11. Bf6+ Kh6 12. Bg7+ Kh7 13. Bf6+ 1/2-1/2

Sadly white wins with 3.Rh7 and the mating threat Rh8 can't be stopped.

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51.

Brogi,G [+3111.01c2a6] 1929. Position after 5... Qa1+!

Solution: 1. Be2+ Ka5 (1... Ka7 2. Rb7+ Ka8 3. Bf3 Qg3 (3... f4+ 4. Kb2 Qf6+ 5. Ka2 Qe6+
6. Rb3+ Ka7 7. Nb5+ Ka6 8. Nc7+) 4. Rb3+ Ka7 5. Nc8+ Ka6 6. Bb7+) 2. Nc4+ Ka4
3. Bd1 Qg2+ {eg} (3... f4+ 4. Kc1+) 4. Kb1+ 1-0

In the subvariation 1...Ka7 and 3... f4+ black has a stalemate-trap.

So after: 1... Ka7 2. Rb7+ Ka8 3. Bf3 f4+ 4. Kb2 Qf6+ 5. Ka2 Qa1+ 6.Kb3 (6.Kxa1=) Qf1 7.Be4 Qf3+ =

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50.

Brogi,G [+3015.20d1d3] 1929. Position after 6. ... Kxf3

Solution: 1. Ne5+ Kd4 2. Nf3+ Kd3 3. Ne1+ Kd4 4. Be5+ Kxe5 5. Nf3+ Ke6 (5... Kd5 6. Nc7+)
(5... Ke4 6. Nd6+) (5... Kf5 6. Nd4+) 6. Nd4+ 1-0

Sadly after 5. ... Ke4 6.Nd6+ Black makes the draw simply by taking Kxf3 7.Nxb5 Ne6 or Kxg3=

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49.

Cuppini,A [=0342.45h2c5] 1976. Position after 1.Ne4+ Kxb5 2.Nd4+ Ka5 3.Nxc2 Bxg2!

In this study there were already some cookes found, but sadly it is incorrect in two more ways:

Solution in database: (in bold and italics my comments)

1. Ne4+ (1. d4+? Kxb5 2. Nd6+ Ka4 3. h7 c1=Q 4. h8=Q Rxg2+ 5. Kh1 Rxg3) 1...
Kd5

(1... Kxb5 2. Nd4+ (2. Nfd6+? Ka6 [Ka4!]) 2... Ka5 (2... Ka6 3. Nxc2 Rxc2 4. h7
Rxg2+ 5. Kh1 Bxd3 6. h8=Q Bxe4 7. Qc8+ Kb5 8. Qxe6) 3. Nxc2 (3. h7 ? c1=Q 4.
h8=Q Qf4+) 3... Rxc2 (3... Bxg2! 4. h7 cxd3 5. Nd2 Bd5 6. Ne3 Rxd2+
7. Kg3 Re2 8. Nxd5 d2 9. h8=Q d1=Q 10. Qa8+ Kb5 11. Qe8+ Kc4 12. Qc6+ Kb3 13.
Qxb6+ Kc2 -+)

4. h7 Rxg2+ (4... Bxg2 5. Kg3 Bxe4 6. dxe4 Rxc3+ 7. Kg2)
5. Kh1 Bxd3 6. h8=Q Bxe4 7. Qa8+ Bxa8) 2. Ne7+ (2. Ne3+ Ke5 3. Nxc2 (3. h7 ! {
cook MC} Ra8 4. d4+ Kf4 5. Nxf6) 3... cxd3 4. Ne3 Bxg2 5. Nxg2 Kxe4) 2... Ke5
3. Kg3 (3. Ng6+ ? Kf5 4. Nh4+ Kf4 5. Ng6+ Ke3) 3... Ra7 (3... cxd3 4. Nc6+ Kf5
5. Ne7+ Ke5 6. Nc6+ Kd5 7. Nb4+ Kc4 8. Nxa2 and White wins!) 4. Ng6+ (4. d4#! {cook MC}) 4...
Kf5 5. Nh4+ Ke5 6. Ng6+ (6. d4+! {cook MC} Kd5 7. Nxf6+ Kd6 8. Ne8+ Ke7 9. h7
c1=Q 10. h8=Q Qg5+ 11. Kh2 Bxg2 12. Nxg2 Ra2 13. Qg7+ Qxg7 14. Nxg7 Rc2 15. d5
exd5 16. Nf5+ Kf7 17. Nfe3 Rxc3 18. Nxd5 Rb3 19. Nxb6 c3 20. Ne3 Rxb5 21. Na4
Re5 22. Nxc3 Rxe3 23. Nd1 Re2+ 24. Kg3) 6... Kd5 7. Nf4+ 1/2-1/2

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Bent,C 2001 [=0074.11d6a4] Position after 4...Be3!?

Solution: 1. Nc5+ Kb5 2. Nxe6 (2. Nxd7? Bxd7 3. Kxd7 Kxb6 4. e6 (4. Ke8 Kc6 5. Kf7 Bh4 6. Kg7 Ng5)

4... Nf8+ 5. Kd6 Ng6) 2... dxe6 3. Bc7! (3. Kxe6 ? Kxb6 4. Kf7 Bh4 5. e6 Ng5+ 6. Kf6 Nf3+)

3... Nf8 4. Bd8! Nh7 (4... Bxd8=) (4... Bh6 5. Be7) 5. Bc7 (5. Bxg5? Nxg5) 5... Nf8 6. Bd8 Bxd8 1/2-1/2

After: 4... Be3!? 5. Ke7 Bc5+ 6. Kf7 Kc4 7. Be7 Bxe7 8. Kxe7 Kd5 black wins....

Martin van Essen suggests to put an extra white pawn at e4.

After which 4...Be3 only draws.

He suggests the following position with a better start:

1. b6, axb6 2. Bxb6, Kb5 3. Bc7! etc.

On 2. …, Nf8 3. Kc6!! draws. This keeps out Kb4 and Kb5 and threatens Bc5 followed by Kd7

with attack on Knight and Bishop when black tries to defend its pawn at e6 with the Knight at f4 or f8.

On 3. ... Be7 then 4.Be3! with plan Bh6

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Pogosjants,E [=0345.00h5h8] 1985. Position after 2... Rg5!

Solution: 1. Nc6 Be2+ 2. Kh6 Bxd1 3. Ne7 Re8 4. Ng6+ Kg8 5. Be6+ Rxe6 1/2-1/2

But better: 2... Rg5! 3. Nf2 Rh5+ 4. Kg6 Rd5 5. Ne7 Rd6+ 6. Kf5
Rf6+ 7. Ke5 Rxf2 -+

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Richter,E [=0400.21h3d1] 1984. Position after 5. ... Rb7.

Solution: 1. Kg2 Rxf7 2. b6 Ke2 3. Re5+ Kd3 4. Rd5+ Kc3 5. Rd8 Rf8 6. Rd7 Rf7 7. Rd8 Rb7 8. Rc8+ Kd3 9. Rd8+
Ke2 10. Re8+ Kd1 11. Re6 Kc2 12. Rc6+ 1/2-1/2

But Black may play 5... Rb7! 6. Rc8+ Kd3
7. Rd8+ Ke2 8. Re8+ Kd1 9. Re6 Rd7 (this position also can be reached at move 11 in the solution) and wins.
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Bent, C [=0347.20d8c6] 1985 Position after 5... Ne7!

Solution: 1. g8=Q Rb8+ 2. Bc8 Rxc8+ 3. Kxc8 Ne7+ 4. Kb8 Nxg8 5. Nd3 Ba3 (5... Bd2 6. Ne5+
Kd6 (6... Kb6 7. Nc4+) 7. Nc4+) (5... Be3 6. Ne5+ Kd6 (6... Kb6 7. Nc4+) 7.
Nc4+) 6. Ne5+ Kb6 (
6... Kd6 7. Nc4+) 7. Nc4+ 1/2-1/2

But possibly black may win with: 5... Ne7 (or Ba3!?) 6. Nxc1? Kb6 7. Nd3 Na6+ 8. Ka8 Nc6 9. Ne5 Nc7#

But also after other 6th moves of white black is winning, eg. 6.Ne5+ Kd6 7.Nc4 Kd7 8.Ne5+ Kd8 9.Nf7 Ke8-+

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44.

Van_Tets,A [=3001.20g5e4] 1970 Position after 2... Qe5!

Solution: 1. Ne2 (1. h8=Q ? Qf4+ 2. Kh5 Qh2+) 1... Qxe2

(1... Qf5+ 2. Kh6 Qf8+ 3. Kg5 Qf5+ (3... Ke5 4. Ng3 Qf4+ 5. Kh5 Qf8 (5... Qxg3 6. h8=Q+) 6. Kg5 Qf6+
7. Kh6) (3... Kf3 4. Nd4+ Ke4 5. Ne2 Kf3 6. Nd4+ Ke3 7. Nf5+ Kf3 8. g7) 4. Kh6
Qf6 5. Ng3+ Ke5 6. Nh5) (1... Qe3+ 2. Kf6 Qh6 (2... Qxe2 3. g7) 3. Kf7 Ke5 4.
Nd4 Kxd4 5. Kg8 Qxg6+ 6. Kh8) (1... Qh2 2. Kf6 Qe5+ 3. Kf7 Kf5 4. g7 Qf6+ 5.
Kg8 Kg6 6. h8=N+ Kf5 7. Ng3+ Ke6 8. Nh5 Qg5 9. Ng6) 2. g7 (2. h8=Q ? Qd2+ 3.
Kg4 (3. Kf6 Qc3+) (3. Kh5 Qh2+) 3... Qg2+ 4. Kh4 Qh2+) 2... Qb5+ (2... Qg2+ 3.
Kf6 Qf2+ 4. Ke7 Qh4+ 5. Kf8 Qf6+ 6. Kg8 Qe6+ 7. Kf8) 3. Kh4 (3. Kg6 ? Qf5+ 4.
Kh6 Qf6+) 1/2-1/2

But after 1... Qf5+ 2. Kh6 Qe5! Black will win! 3.Ng3+ Kf4 4.Nh5 Kg4

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Erlin,K [=3011.32d7b6] 1895. Position after 2... Ka6!

1.p Rigaer Tageblatt

Solution: 1. c4 Qxd4 2. c5+ Qxc5 (2... Kxc5 3. Bf2 Qxf2 4. Ne4+) 3. Bxa5+ Kxa5 (3... Qxa5 4. Nc4+)

(3... Ka6 4. b4) 4. Nb7+ 1/2-1/2

But better: 2... Ka6 3. c6 Qg7+ 4.Ke6 Kb6 and black wins?

A repair with an extra black pawn at a6 does work but then a complete new study is born because White then starts with

1. Bh4 (threat Bd8#) Qa4+ 2.bxa4 Stalemate.

The Erlin-line does not work anymore because at the end of the mainline black continues with

4... Kb6 5.Nxc5 Kxc5 6.Kc7 a4! 7.bxa4 a5 and wins.

If this is a new study, then it is mine... Peter Boll.

Martin van Essen suggests to change the starting position by adding two pawns at f5 and f6:

But he comments:

"Not a beautiful improvement.

It keeps 2...Ka6 with Qg7+ out of the position. The solution is almost the same, although the mainline now should be 3...Ka6 4.b4 because 3...Kxa5 4.Nb7+ wins.

But after 1.Bd2 it is unclear if that is also a winning line? White mostly can win the pawn at f6 and it will hard to proove that black can win in these lines."

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Simkovitch,F [=0461.36a1e8] 1996. Position after 2. ... Bxb8

Solution: 1. Re7+ Kxe7 2. g7 Bxc5 3. gxh8=Q Kf7 4. Nd7 Bd4+ 5.Ka2

(5. Kb1 f4 6. Ka2 f3 7. Kxa3 f2 8. b4 Bb2+ 9. Kxb2 f1=Q) 5... f4

(5...b4 6. Nb8 f4 7. Nc6 bxc6 8. Qxh7+ Bxh7) 6. Kxa3 b4+ (6... f3 7. b4 f2 8. Ne5+
Bxe5 (8... fxe5 9. Qg7+ Kxg7) 9. Qg7+ Kxg7) 7. Ka2 f3 8. Ne5+ (8. Nf8 f2 (
8... Kxf8 9. Qxh7 Bxh7) 9. Nxh7 f1=Q) 8... Bxe5 9. Qxh7+ Bxh7 1/2-1/2

But after: 2... Bxb8 3. gxh8=Q Kf7 Black wins.

Martin van Essen suggest a correction by changing the starting-position to:

The solution is about the same: 2...Bxe3 3. gxh8=Q Kf7 4. Nd7 Bd4+ etc.

He also points out that at move 8.Nxf6 is a dual Bxf6 9.Qxh7 Bxh7= (but that's all in line with the ideas).

Bravo!

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Pogosjants,E 2001 [=0163.00b1a3] Position after 1... Nb4!

Solution: 1. Rg5 Bf4 2. Rf5 (2. Rg4 Nb4 3. Rxf4 Bg6+ 4. Ka1 (4. Kc1 Nd3+) 4... Nc2+ 5. Kb1 Nd4+ 6. Ka1 (6. Kc1 Ne2+)

6... Nb3#) 2... Bg6 3. Rf6 Bh7 4. Rf7 Bg6 5. Rf6 Bd3 6. Rxf4 Nd4+ 7. Kc1 Ne2+ 8. Kd2 Nxf4 9. Ke3 1/2-1/2

But with 1... Nb4! 2. Rxe5 Ba2+ 3. Kc1 (3.Ka1? Nc2#) Nd3+ black wins.

Martin van Essen has found a way to correct this study by changing the startingposition (See guestbook):

Solution: 1. Re5 Bf4 2.Rf5 and the same position after the 2nd move as above has been reached.

So this is now a new Pogosjants,E Van Essen, M 2005 study!

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To the first numbers 1 to 20.

To the second range 21 to 40.

To the 4th range 61 to 80.

To the 5th range 81 to 100.

To the 6th range 101 to 120.

To the 7th range 121 to 140.

Peter Boll.

Homepage ARVES.