Cooked studies

Incorrect studies?

4Th Page. 61-80.

Here page-4 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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80.

Ashworth,G 1927 Positions after 2...Nf3+! and after 2.Bxc8!

[+4884.66g1a8]

Solution: 1. d4 Rc8 (1... Nf3+ 2. Rxf3 Bxf3 3. Qxa6+ bxa6 4. Bc6+ Kb8 5. Bd8+)

2. Ra1 Kb8 3. Qxa6 bxa6 4. Bd8+ Bb4 5. Rxb4+ 1-0

But with 2...Nf3+ black wins: 2... Nf3+ 3. Rxf3 Bxf3 4. Rxa6+ Kb8 5. Ba7+ Kc7

black threatens Qg2# and Qxh2 followed by Rh8-h1.

White should play and win: 2.Bxc8! Nf3+ 3. Rxf3 Bxf3 4. Bxb7+! Kxb7 (Kb8 5.Nc6+)

5. Ba5+ Ka8 6. Qxa6# but then that's a complete other solution.

This is more a game-like combination.

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79.

Afanasjev,G 1927 Positions after 5...Ke5 and after 7...Kxg6.

[=0144.23h3e4]

Solution: 1. Re6+ Kxd5! 2. Rxe2 d1=Q 3. Rd2+ Qxd2 4. Bf7+ Ke4 5. Bg6+ Kf3 6. Bxh5+ Ke4 7. Bg6+ 1/2-1/2

But is it really a draw? after: 5... Ke5 6.Nc4+ Kf6 7. Nxd2 Kxg6 Black wins !?

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78.

Bania,R 1926 [+0040.34b1b7] Position after 3... Kc8!?

Solution: 1. e7 Bf5+ 2. Bc2 Bd7 3. c6+ Bxc6 (3... Kxc6 {main} 4. Ba4+) 4. Be4! 1-0

Black could try: 3... Kc8 4. cxd7+ Kxd7 5. Bf5+ Kxe7 6. Kc2
d5 7. Kd3 c5 8. Bh3 Kd6 and white can't make progress...

or 4. Be4 Be8 5. Bf5+ Kb8 6. Bd7 Bg6+ 7. Kb2 Ka7 seems equal.

Better 7.Kc1! Ka7 8.Kd2 Kb6 9.Ke2! h3 10.Kf2+-

But black can also try 6...h3! 7.Bxh3 Bxc6 or 7.Bxe8 h2 8.Bd7 h1Q+

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77.

Birnov,Z 1929 [=0046.20a1a3] After 6... Ka3!

Solution: 1. Be1! Nb3+ 2. Kb1 Bxe2 3. Bb4+ Kxb4 4. h7 Bd3+ 5. Ka2! Nc1+ (5... Bxh7=)

(5... Nc3+ 6. Kb2 Nd1+ (6... Bxh7=) 7. Ka2 Nc1+ (7... Bxh7=) 8. Ka1 Bxh7=) 6. Ka1
Bxh7 Stalemate 1/2-1/2

But black wins simple with: 6... Ka3 7. h8=Q Nb3# mate

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76.

Berben,A 1927 [+0100.56d2b1] After 5... b5!

Solution: 1. Re1+ Kb2 2. Ra1 Kxa1 (2... Ka3 3. Kc1 c6 4. c4 bxc4 5. bxc4 b6 6. c3 Kb3 7.
Kd2 c5 8. Kc1) 3. Kc1 c5 4. c4 b4 5. c3 bxc3 6. Kc2 b6 7. Kc1 h6 8. h5 1-0

But Black has a beautiful pawnbreak with 5... b5! 6. cxb5 c4 7.bxc4 b3 or 6. cxb4 bxc4 7.bxc4 cxb4 and now black wins!!

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75.

Markovics,B 1976 [+0002.35b3c6] Position after 2...f2

Solution: 1. Nd7! (1. h6? f2) 1... f2 (1... c2 2. Ne5+! Kd5 3. Kxc2; 1... f6 2. h6)

2. Ne5+ Kb5 (2... Kd5 3. Ng4) 3. Nc7+ Ka5 4. Nc4# 1-0

The subline stops after 1... f6 2. h6 But continuing with 2 ... f2 3. h7 f1=Q 4. h8=Q Qd1+ 5. Kxc3 Qe1+ this seems equal

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74.

Liburkin,M 1941 [+0044.22b2d4] Position after 3...Bb1

Solution: 1. Nc5 Nc2 2. Kxc2 e3+ 3. Kd1 Bc2+ (3... Bb1 4. Na4 Bxa2 5. Bb2+ c3 6. dxc3+
Kc4 7. Nb6+ Kd3 8. Nd7 Bb3+ 9. Ke1 Kc2 10. Ba3 Kxc3) 4. Ke2 exd2 5. Kxd2 Bb1 6.
Na4 Bxa2 7. Nc3 Bb3 8. Bd6 1-0

What happens after: 3... Bb1 4. Na4 Bxa2 5. Bb2+ c3! 6. dxc3+ Kc4 7. Nb6+ Kd3 8. Nd7 Bb3+

9. Ke1 Kc2 10. Ba3 Kxc3 and no win...

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73.

Liburkin,M 1939 [=0816.43h5b8] Position after 3...Nf3!

Solution: 1. c7+ Kxc7 (1... Kc8 2. Rxg8+! Nxg8 3. Kh4! Nf4 (3... Ng5 4. Re8+ Kxc7 5.
Rxg8 Nf3+ 6. Kh3 Rxg8) (3... Rh1 4. Re8+ Kxc7 5. Rxg8 Nf2+ 6. Kg3 Rg1+ 7. Kh2
Rxg8) 4. Re8+ Kxc7 5. Rxg8 Rxg8) 2. Rxg1! Nxg1 3. Re1! Nf5 (3... Ng4 4.Rxg1 Nf6+

5. Kh6! Rxg1) (3... Rh8 4. Rxg1 Nf7+ 5. Kg6 Rg8+ 6. Kh7 Rxg1) 4.Rxg1 Rxg1 1/2-1/2

In the database there is a new version dated 1953: [=0816.43h5d7] which is sound.

So it looks as if the composer already found himself a cook? What would it have been?

But in the database there is no @.

I found that after 3...Nf3! 4. Re7+ Kd6 5. Re3 Nf5 6. Rxf3 Ke5 7. Rf2 Kf6 8. Rg2 Rh8+ 9. Kg4 Ne3+;

or 4. Re3 Nd4 5. Kxh6 Nf5+ 6. Kh7 Nxe3 7. Kxg8 Nd1 8. Kf7 Nb2 9. Ke6 Nxa4-+

black wins.

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72.

Kekely,L 1985 [=0031.21b5c3] Position after 3... Qc5+

3.hm tn34#1655

Solution: 1. Nd4 Kxd4! (1... c1=Q 2. Ne2+) 2. b7 c1=Q (2... c1=R 3. Kb6 (3. Ka6? Bc8)) 3. b8=Q (3. Ka6? Bc8!)

3... Qb2+ 4. Kc6! Qxb8 (4...Bg2+ 5. Kc7) 1/2-1/2

Sorry, but Black wins after: 3... Qc5+ 4. Ka4 Bd7+ 5. Kb3 Qc3+ 6.Ka2 Be6+ 7. Kb1 Qxd3+ 8. Kc1 Qc3+

9. Kb1 Qe1+ 10. Kc2 (10. Kb2 Qd2+ 11. Kb1 Bf5+ 12. Ka1 Qc1+ 13. Ka2 Be6+) 10... Bf5+ 11. Kb3 Qb1+.

Author reported that this is an old version:

New repaired version (with turned of colours) was publicated in Slovak magazine Mat-Pat in nr. 30/1991.

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71.

Liburkin,M 1939 [+0311.11d3h6] Position after 2...Rh3

Solution: 1. Kd2 Rg2 (1... Rh5 2. Bh2 (2. Bd4? Rd5! (2... Kxh7? 3. e7 e1=Q+ 4. Kxe1
Rh6 5. Bf6!) 3. e7 Rxd4+ 4. Ke1 Re4) 2... Kg7 3. e7 Kf7 (3... Rd5+ 4. Kxe2 Kf7 5.
e8=Q+ Kxe8 6. Nf6+) 4. e8=Q+ Kxe8 5. Nf6+) 2. e7! Rxg1 3. Kxe2 Rg8 4. Nf8 Rg5
5. Ng6! (5. e8=Q? Re5+ 6. Qxe5) 1-0

In the subline 1...Rh5 2.Bh2 Black could try Rh3 3. Bf4+ Kg6 4. Ng5
e1=Q+ 5. Kxe1 Rh1+ 6. Ke2 Kf6 7. Be3 Rb1=
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70.

Lasker,Ed 1905 [=4007.25d3a8] Position after 1...e5

Solution: 1. Qb2 Qe8 2. Nd5 exd5 3. Qh8 Qxh8 1/2-1/2
But what after: 1... e5 2. Qxa1 (2. Nd5 e4+) 2... Qb5+ Black seems winning...

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Prokes,L 1906 [+3101.46b6c8] Position After 7... c5

Solution: 1. Rf7 Qxh2 (1... Qc1 2. Nxd6+ Kd8 3. Nxb7+ Ke8 4. Nd6+ Kd8 5. Kxc6 Qh1+ 6. f3)
(1... Qh4 2. Nf6 Kd8 3. Kxb7! Qh3 4. Rd7+ Qxd7+ 5. Nxd7 Kxd7 6. f4) (1... Qe5
2. f4! Qe6 3. Rf8+ Kd7 4. Nf6+ Ke7 5. Re8+ Kxf6 6. Rxe6+ Kxe6 7. Kxb7) (1... Qg4 2. Rf8+ Kd7 3. Nf6+) 2.
Nf6 Kd8 3. Rd7+ Kc8 4. Re7! 1-0
But it seems to me that in the bold/underlined subline, at the end continuing with

7....c5 8.Kc6 h5 9. h4 Kf5 10. Kxd6 Kxf4 11. Kxc5 Ke3 this is a draw.
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68.

Vandiest,J 1954 [+0344.43d2a4] Position after 3... Kxa2

Solution: 1. b3+ Ka3 2. Nxc4+ Nxc4+ 3. bxc4 Bb4 4.Bxb4+ axb4 5. g6 hxg6 6. e5 Kxa2 7. e6 f2

8. Ke2 b3 9. e7 b2 10. e8=Q f1=Q+ (10... b1=Q 11. Qa4+ Kb2 12. Qb4+) 11. Kxf1 b1=Q+ 12. Qe1 Qb3 13. Qa5+ Kb2 14.
Qb5 Qxb5 15. cxb5 1-0

The following variation with a beautiful ending position, needs to be checked further. Is this really a draw?

3... Kxa2!? 4. Bxa5 Bc5 5. e5 Kb3 6. Bd8 Kxc4 7. e6 Kd5 8.e7 Bxe7 9.Bxe7 Ke4 10.Bc5 Kf4 (or Kf5) 11.Be3+ Ke4!

and white can only play the bishop after which black plays Kf4; Positional draw.

(Or 10.Ke1 Kf5 11.Kf2 Kg6 12.Kxf3 h6=; and not 10.Bf8 Kf4 11.Bh6? Kg3-+).

Another try for white is: 4.Bf6 Kb3 5. e5 a4 6. e6 a3 7. e7 Bxe7 8. Bxe7 a2 9. Bf6 Kxc4 10. Ke3 Kd5 11. Kxf3 Ke6 = (12. Kg4, h6! =) just in time!!

By the way not winning is 7. ... Kd4 because of 8.Bb6! (as Martin van Essen pointed out. After 8. e7? black has the beautiful Bb4+! 9. Kd1 Bxe7 10. Bxe7 Ke3 11. Ke1 f2+ 12. Kf1 Kf3=)

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Grondijs,H 1991 [=0311.53b5e8] Position after 2.Bd3!

Solution: 1. Nf5 Rg5 2. Ka5 (2. Kc6? h3! (2... Rxf5? 3. b5 h3 4. b6 axb6 5. Kd6!
h2 6. a7 Ra5 7. Bb5+ Kf8 8. e7+ Kg7 9. e8=Q) 3. Bd3 h2 (3... Rxf5?? 4. Bxf5 h2
5. Kd6 h1=Q 6. Bg6+ Kf8 7. e7+ Kg7 8. e8=Q) 4. Kd6 Rg6!) 2... Rxf5+ 3. Bb5+
Kd8 4. e7+ Kxe7 Stalemate 1/2-1/2

But white wins with: 2. Bd3 h3 3. Kc5! Rxf5+ (or 3. ... h2 4.Bb5+ Kd8 5.e7+ Kc7 6.e8=Q Rxf5+

7.Kd4 Rxb5 8.Qf7+ Kd8 9.Qg8+ Ke7 10.Qh7+ Kd6 11.axb5 +-)

4. Bxf5 h2 5. Kd6 h1=Q 6. Bg6+ Kf8 7. e7+ Kg7 8. e8=Q +-

Funny that I forgot that I already did mention this study in EBUR/1994/2/page 21, but I then did not find this cook.

I only did show the 2.Kc6? line.

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Meijer,H 1904 [+1363.22h3g5] Position after 1...Bxg2+!

Solution: only: 1.Qc3 but what after 1...Bxg2 2.Kxg2 Bf6 and later Nc7 may be black is better!?

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Tarrasch,S 1904 [=0131.13c8a8] Position after 3.Rh3!

1. Nxf4 d2 2. Rh1 Be1 3. Rh6 Ka7 4. Rh7+ Ka6 5. Rh6+ Ka5 6. Rh5 d1=Q 7. Rxd5+
1/2-1/2

White can win with 3.Rh3 Ka7 4.Nxd5 and the knight goes to c3.

Possibly wrongly entered in the database? Should the white pawn be on a3?

Reaction of Harold van der Heijden about this:

Ik heb er de originele bron even op nageslagen (Deutsche Schachzeitung 1904). Bij de oplossing staat daar de volgende uitleg: "Damit dieser zweite Remiskombination [HH er wordt verwezen naar een studie van Rinck (#63647-9)] allein zur Geltung gelange, haben wir den weissen Bauern von a3 nach a2 versetzt; der Bauer sollte aber auf f3 stehen, damit der Zug 3.Th1-h3 verhindert wird". Kortom 3.Th3 is al meteen ontdekt in 1904, maar door mij blijkbaar overzien. Of niet begrepen, want dat van pion f3 snap ik ook nu nog niet. De bedoeling van Tarrasch was dus om alleen diens nevenoplossing in Rinck's HHdbIII#63647 te laten werken.

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Liburkin,M 1930 [+0305.10a7h4] Position after 6... Rh1

Solution: 1. Nf5+ Kh3 (1... Kh5 {main} 2. Nf6+ Kg6 (2... Kg5 3. Nd6 Nb5+ 4. Kb8! Rxd6 (
4... Nd4 5. Nfe4+! Kg6 6. Nc3 Nc6+ 7. Kc7 Rxd6 8. Kxd6 Nd8 9. Kc7
Nf7 10. Ne4 Kg7 11. Nd6 Ng5 12. Kc8 Ne6 13. Nf5+ Kf6 14. Nd4) 5. Ne4+) 3. Nd5
Nb5+ 4. Kb6! Rxd5 5. Ne7+) 2. Ng5+ Kh2 3. Nd4 (3. Nf3+? Kh1!! 4. N3d4 Nb5+
5. Kb6 Nxd4 6. d8=Q Rb1+ 7. Ka6 Ra1+ 8. Kb7 Rb1+ 9. Kc8 Rb8+! 10. Kxb8 Nc6+)
3... Nb5+ 4. Kb6! Rxd4 (4... Nxd4 5. d8=Q Rb1+ 6. Kc5 Rc1+ 7. Kd5) 5. Nf3+1-0

In the subline: 1. Nf5+ Kh5 2. Nf6+ Kg5 3. Nd6 Nb5+ 4. Kb8! Nd4 5. Nfe4+! Kg6 6. Nc3

Now black can play 6. ... Rh1! because after 7.d8=Q then Nc6+ is still available and Ne6 is also possible later.

This should be a draw.
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Afek,Y 1994 [=3111.12f6f8] Position after 2. ... e1=Q

1.p tm54 {c}

Solution: 1. Ne5! (1. Kxg6+? Kg8 2. Be4 Qe6+) 1... Qa6+ (1... Qa1 2. Rxe2) (1... Qxa8
2. Nxg6+ Kg8 3. Ne7+ Kf8 4. Ng6+) 2. Bc6! Qxc6+ 3. Nxc6 e1=Q 4. Rf3!! (4. Ra2? Qf1+ 5. Ke5 Qf5+) (
4. Rb2? Qc3+) (4. Rc2? Qf1+ 5. Ke5 Qf5+) (4. Rh2? Qf1+ 5. Kxg6 Qf7+ 6.
Kh6 Qg7+ 7. Kh5 Qh7+) (4. Rf4? Qg3! 5. Rc4 Qd3!) 4... Qd1 5. Nd4!! Qxd4+
6. Ke6+! Kg8 7. Rf8+ Kg7 8. Rf7+ Kg8 9. Rf8+ Kxf8 1/2-1/2

But what after: 2... e1=Q 3. Nxg6+ Kg8 4.Ne7+ Qxe7+ 5. Kxe7 Qa7+ winning the rook.

Reaction of the composer: Afek,Y.

A correction of the study was published in 1994 (and is also in the database and Fide-Album):

1.p Martin {c}

Solution: 1. Ne7+ Kf8 (1... Kh8 2. Nxg6+ Kg8 3. Nf4) 2. Bc6 Qxc6+ (2... Qxf1 3. Nxg6+ Kg8
4. Bd5+ Kh7 5. Nf8+) 3. Nxc6 e2 4. Rxf2 e1=Q 5. Rf3 (5. Ra2? Qf1+ 6. Kxg6
Qf7+) (5. Rc2? Qf1+ 6. Kxg6 Qd3+) (5. Rh2? Qf1+ 6. Kxg6 Qf7+ 7. Kh6 Qg7+) (
5. Rf4? Qe3 6. Rf1 Qd3) 5... Qd1 6. Nd4 Qxd4+ 7. Ke6+ Kg7 8. Rf7+ Kg8 9. Rf8+
Kxf8 1/2-1/2 (Brilliant).

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Calvi,I 1847 [+0040.22g1f3] Position after 4. ... f2+

Solution: 1. e4! Bc8 (1... Kxe4 2. Bg6!) 2. e5 Kg3 3. Bc4 f3 4. e6 Bb7 5. Kf1 f2 6. Ke2 Kg2 7. Ke3 f1=Q 8. Bxf1+ Kxf1
9. e7 g3 10. e8=Q g2 11. Qf8+ Kg1 12. c8=Q 1-0

However black makes a draw with: 4... f2+ 5.Kf1 Kh2 6. Kxf2 g3+ 7. Ke2 g2=
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Wortel,N 1939 [+4301.01b4g2] Position after 1.Nf4++ !

Solution: 1. Ne3+ Kf2 (1... Kg3 2. Qg2+ Kf4 3. Qg4+ Kxe3 4. Qg5+) 2. Qg2+
Ke1 3. Qf1+ Kd2 4. Nc4+ Kc2 5. Qe2+! (5. Na3+? Kb2 6. Qe2+ Ka1) (5. Qf5+?
e4! 6. Qf2+ Kb1 7. Nd2+ Ka2!) 5... Kb1 6. Kb3! Ra6 (6... Rb6+ 7. Nxb6) 7.
Nd2+ Ka1 {eg} 8. Qxa6+ 1-0

White wins immediately with 1.Nf4++ The forks Ne2 and Nd3 are killing. After 1...Kf1 2.Qg2 wins.

Reaction Martin van Essen:

The first move Nf4 is possible to repair by changing the position to (Qg7,Ng4):

After 1. Ne3†† black should play 1. …, Kf2 because after 1. …, Kf3 2. Dg2†, Kf4 3. Dg4† the queen is lost.

After 1. …, Kf2 The solution is known, but there is a second problem, there is a dual with:

3. Qh1† (instead of 3. Df1†) Kd2 4. Nc4†, Kc2 5. Qe4†, Kd1 6. Qf3†, Ke1 (6. ..,Kc2 6. Qb3‡) 7. Qh1† winning the queen.

Reaction Harold van der Heijden:

If we transport all pieces one line to the right, the dual at move 3 is gone and the "New" study is born!!

Wortel,N+Van_Essen,M+Van_der_Heijden,H 2005

To the first numbers 1 to 20.

To the second range 21 to 40.

To the 3rd range 41 to 60.

To the 5th range 81 to 100.

To the 6th range 101 to 120.

Peter Boll.

Homepage ARVES.