CorrectBut

3. Additional analyses to studies.

These studies are still correct but some extra analyses or additions were done to show the study is correct.

Sometimes it looses beauty by this, but not always.

Please sign the guestbook and let us know what you search for or what should be improved.

Please react by using the guestbook.

Latest addition made on 23-may-2006.

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Hermann,R 1898 [+1330.02c1a2] After 7...Rh4 After 6.Qe6+!

Solution: 1. Qc3 Rb3 (1... Rb8 2. Qxc5 Kb3 3. Qe3+) 2. Qxc5 Rh3 3. Qa7+ Kb3 (3... Ra3 4.
Qf7+ Rb3 5. Qd5) 4. Qb7+ Ka3 5. Qe7+ Ka2 6. Qf7+ Ka3 (6... Ka1 7. Qg7+) 7. Qd5 1-0

But the given mainline does not win after 7...Rh4;

Instead white wins with 6. Qe6+ Rb3 7. Qd5 Ka1 8.Qxb3 Ba2 9. Qb2#

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Pervakov,O 1994 [+0143.12e1h1] After 1.Rh3...2...Bd4!

Solution: 1. Re7! (1. Rg3? Ba5 2. Bg7 Ne4+) (1. Rd3? Ba5 2. Bg7 Nb3+ 3. Kxe2 Nc1+)
(1. Rh3+? Kg2 2. Rh6 Nf3+ 3. Kxe2 Bd4!) (1. Re8? Ba5 2. Bh8 a1=Q+ 3. Bxa1
Nb3+ 4. Kxe2 Nxa1 5. Ra8 Bc3) (1. Re6? Nf3+ 2. Kxe2 Nd4+) 1... Ba5 2. Bh8!
a1=Q+ (2... Ne4+ 3. Kxe2 Nc3+ (3... Bc3 4. Bxc3 Nxc3+ 5. Kf2 Nd1+ 6. Kg3!) 4.
Kf3! a1=Q 5. Rh7+ Kg1 6. Bd4+) 3. Bxa1 Nb3+ 4. Kxe2 Nxa1 5. Ra7! Bc3 6. Kf1!

(6. Kd3! {cook} Bf6 (6... Be5 7. Ra5 Bf6 8. Rf5) 7. Rf7 Be5 8. Rf5 Bb2 9.
Rb5 Bf6 10. Rb1+ Kg2 11. c3) 6... Kh2 (6... Be5 7. Ra5 Nxc2 8. Rxe5 Kh2 9. Re2+
Kh1 10. Re4) 7. Ra2 $3 Be5! (7... Kh1 8. Ra3) (7... Nxc2 8. Rxc2+) 8. c3+!
Kg3 9. Rxa1 Bxc3 10. Ra3 1-0

Although this study has a dual at move 6, mr. Odd Øivind Bergstad found that in the subline:

1.Rh3 the given line was not OK. 1. Rh3+? Kg2 2. Rh6 Nf3+ 3. Kxe2 Bd4!

The line stops here but after 4. Rg6+ Kh2 5. Bxd4 Nxd4+6. Kf2 Kh3 7. Rh6+!! Kg4 8. Ra6 Nxc2 9. Rxa2 +-

white wins the Knight; He also found the right refutation for black: 2....Bd4! (instead of Nf3)

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Kasparjan,G 1958 [+0183.21e6a6] After 2...Bxd5

Solution: 1. Bh1 (1. Bf1+? Kb7) 1... Kb7 2. Bg3! (2. Be5? Nxf3 3. Bxf3 Kc6)

(2. Bd6?) 2... Nxf3! (2... Kc6 3. f4 Ng6 4. Kf6) 3. Bxf3 Kc6 4. Ke5 Bb2+ 5. Rd4+ Kc5 6. Bf2
h4! (6... Bf7 7. h4! Be8 8. Be4 Bc3 9. Bf5 Bc6 10. Bg6 Bf3 11. Be4! Bd1
12. Bd5! Bc2 13. Bf7 Bd1 14. Ke4 Bc2+ 15. Rd3+) 7. Ke4 Bc2+ 8. Rd3+ Kc4 9.
Be2 h3! (9... Bf6 10. h3 Be7 11. Be3 Bb1 12. Bf4 Bc5 13. Bg5 Bf2 14. Kf3) 10.
Ke3 Bc1+ 11. Rd2+ Kc3 12. Be1 Bf5 13. Bf3 Bd3 14. Bg4 Bf1 15. Be2 Bg2 16. Ba6
Bc6 17. Ke2 1-0

There is no subline about: 2... Bxd5+ 3. Kxd5 Nf5 and 4. f4 Nxg3
(4... Kc7 5. Be1 Kd7) 5. hxg3 Kc7 6.Bf3 h4 7.gxh4 Ke7 is equal...

Marcel van Herck states that after 3.Be5! (white should keep his bishop-pair) white has enough advantages

to win: extra pawn, active king, weak black pawn at h5 and bishop-pair.

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Dobrescu,E 1966 [+1330.01b3a6] Position after 3...Ka5 and 4...Ba3

Solution: 1. Qf3 (1. Qg6+? Rd6) (1. Kc2? Rd6 2. Qa3+ Kb7 3. Qb3+ Kc8) 1... Rb1+ 2.
Kc2 (2. Ka4? Rb4+) (2. Ka2? Rb8) (2. Kc3? Rb8) 2... Rb8 3. Qc6+ Ka7! (
3... Ka5 4. Kc3! e4 (4... Ba3 5. Qc7+ Rb6 6. Qa7+ Ra6 7. Qb7 Bc5 8. Kc4 Bb6 9. Qc6) 5. Qc7+ Rb6 6. Kc4 Bb4
7. Qa7+ Ra6 8. Qb7 Rb6 9. Qd5+ Ka4 10. Qd1+ Ka5 11. Qa1+) 4. Qc7+ Ka8 (4... Rb7
5. Qa5+ Kb8 6. Qd8+) 5. Qd7! e4 (5... Rb7 6. Qc8+ Rb8 7. Qa6#) 6. Qc6+ Ka7 7.
Qc7+ Ka8 8. Qd7! e3 9. Kd3 Rb3+ 10. Ke4! (10. Ke2? Rb2+ 11. Kf3 e2) 10...
Rb4+ (10... e2 11. Qa4+) (10... Rb8 11. Kf3) 11. Kf3! (11. Kxe3? Bc5+ 12.
Kf3 Ba7) 11... Rb8 12. Qc6+ Ka7 13. Qc7+ Ka8 14. Qd7! e2 {eg} (14... Rb2 15.
Qd5+ Ka7 16. Qd4+ Rb6 17. Qa4+ Kb7 18. Qd7+ Ka8 19. Qd5+ Rb7 20. Qa5+) 15. Kxe2
Rb2+ 16. Kd3 1-0

In the subline: 3...Ka5 4.Kc3 Ba3 white should not play 5.Qc7? because of: Rb6 6. Qa7+ Ra6 7. Qb7 and now 7... Rh6! N

Better is however: 5. Qd5+! Rb5 6. Qa8+ Kb6 7. Qxa3 +- keeping the study sound.
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Von_Holzhausen,W 1898 After 1.Nb4 After 3...h3 -+

[=0101.03h7a8]

Solution: 1. Nb8! Ka7! (1... a1=Q?? 2. Nc6!) 2. Rb4 a1=Q 3. Nc6+ Ka6

4.Nb8+ Ka5 5. Nc6+ 1/2-1/2

Also not working for white is: 1. Nb4 Ka7!

(1... a1=Q? 2. Nc6 Qa7+ 3. Nxa7 Kxa7 4. Rg6+-) 2. Ra6+ Kb7 3. Rxa2 h3 -+

And the two pawns are stronger than rook+knight. White is not able to save the knight!

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17.

Missiaen,R 1954 [+0342.21e4h4] After 1... d5+ End-position

Solution: 1. g6! Kxh5! 2. g7! d5+ 3. Kf5 Rg6! 4. g8=Q Rxg8 5. Nxg8 Bxh2 6. Bf2
Bb8 7. Nf6+ Kh6 8. Bd4 Bc7 9. Be3+ Kg7 10. Ne8+ 1-0

Black has also: 1... d5+ 2. Kf5 (2.Kxd5 Kxh5) Be5 3. Kxe5 Rxb6 4. g7 Rb8 5. g8=Q Rxg8 6.
Nxg8 Kxh5 7. Kd4 (see end position).

The black pawn is still more than one square before the 'Troitzky-Line'.

So a Knight at d3 can take over from the king and that means white still can win.

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16.

Marwitz,J [=0740.30h7c5] 1948. Positions after 4...Rg5!N and 4.Kg6!N

Solution: 1. Be6! (1. Rg7? Rh6+ 2. Kg8 Re8+) 1... Rh5+ (1... Rb7 2. Rc3+ Kb6 3. Rc6+
Kxb5 4. Bf7! (4. Bd7? Re7+! 5. Kg6 Bxc7! 6. Rxc7+ Rxd7!) 4... Rh5+ 5. Bxh5 Kxc6 6. Bf3+) 2. Kg6 Rh6+! (2... Bxc7
3. Rc3+ Kd4 4. Rc4+) 3. Kg7 Bxc7! 4. Rg5+ (4. Rc3+? Kd4 5. Rc4+ Kd3 6. Kxh6
Rxe6+) 4... Kd4! 5. Rg4+ (5. Kxh6? Bf4) 5... Bf4! 6. Rxf4+ Ke5 7. Rf5+
Kxe6 8. g4!! 1/2-1/2

In the subline 1...Rb7 black can win after 4.Bf7? with 4... Rg5! 5. c8=Q Rxf7+ 6. Kh6 Rxg2+ 7. Kh5 Rh7+ 8. Rh6 Rxh6#

But white has better 4. Kg6! Rg5+ 5. Kf6 Bxc7 6. Rxc7 Rxc7 7. Kxg5 and makes a draw.

So the study is correct, but needs a better subline.

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The following study has a subline (bold) which is not clear:

15.

R. Missiaen [+0071.22f5h4] 1953 Position after 5. ... Kh6

Solution: 1. Be1+ Kxh5 2. Bf2 Bg6+ (2... Bb8 3. d7 Bxd7+ 4. Nxd7 Bc7 (4... Bh2 5. Nf6+
Kh6 6. Ng4+) 5. Nf6+ Kh6 6. Be3+ Kg7 7. Ne8+) 3. Kf6 Be8 4. d7 Bxd7 5. Nxd7 Kg4
(5... Kh6 6. Kf5 Kh5 7. Nf6+ Kh6 8. Nd7) (5... h6 6. Kf5) 6. Ne5+ Kf4 (6... Kh3
7. Nc6 h5 8. Kg5) 7. Nc6 Kf3 (7... b5 8. Bxa7! (8. Nxa7? b4 9. Nc6 b3 10.
Bd4 h5 11. Na5 h4 12. Nxb3 Kf3)) 8. Be1 Kg4 9. Nxa7 h5 10. Nc6 h4 11. Ne5+ 1-0

The subline with 5. ... Kh6 only results in repeating positions, so that brings a draw?

Martin van Essen has pointed out that the study is still correct!! See the ARVES-guestbook.

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Some small notes:

14. R. Missiaen [+3120.01f7g3] Extra line 2. ... Qf1 3.Kf6 h5 4.Ke7 h4 5.Rxh4 Ke3 6.Re4+ Kf3 7.Kf6 +- (Zugzwang)

13. L. Kubbel [+0030.43h3a7] The solution move 5.Nb6? is wrong because of Be2+ and Kxa7.

But 5. a8=Q+ Kxa8 6.Nb6+ and Nxc4 wins. Possibly wrongly entered in the database?

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The following study ends a bit faster after the direct 5. Qd8 Mate.

12.

R. Missiaen [+0303.20h6f8] 1982 And now better 5. Qd8#

DatabaseSolution: 1. h8=Q+ Ke7 2. Kg6 (2. Qg7+? Nf7+ 3. Kh7 Rh5+ 4. Kg8 Rg5 (4... Rh8+? 5. Qxh8) 5. Qxg5+ Nxg5 6. a6 Ne6 7. a7 Nc7) (2. Qc3? Nf7+ 3. Kh7 Rh5+ 4. Kg6 Rg5+ 5. Kh7 Rh5+) 2... Rxa5 3. Qc3 (3. Qg7+? Kd6) (3. Qf6+? Kd7 4. Qd4+ (4. Qc3 Rd5)

4... Kc6) 3... Rb5 (3... Rd5 4. Qc7+ Ke8 (4... Ke6 5. Qc6+ Ke5 6. Kxg5 ) (4... Rd7 5. Qc5+) (4... Kf8 5. Qg7+ Ke8 6. Qg8+) 5. Qc6+) 4. Qc7+ Kf8 5. Qd6+? Ke8 (5... Kg8 6. Qd8#) 6. Qc6+ 1-0
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11.

Havasi, A. 1925 [+0302.11h2e1]

Author solution only: 1. Ne5 Rxe5 2. Nf6 Re2+ 3. Kg3 Kf1 4. h7
Rg2+ 5. Kh3 Kg1 6. Nh5 Rh2+ 7. Kg4 Rg2+ 8. Kf3 Rf2+ 9. Ke3! 1-0

"Marcel van Herck reports that during the Belgian solving contest Guy Baete found the following try:

After 1.Nf6 Rb8 2.h7? Rh8 will be a draw because Rxh7 and the f-pawn can't be blocked on f5.

Position after 1. Nf6.

But Guy Baete suggests 2.Ne5!.

It looks as if white wins after 2. ... Rh8 (white was threatening Nxf7 ) 3.Neg4 Ke2 4.Kg3 Kd3 5.Kf4
the plan is : Kf5, h7, and then Ng4 to h5 before the black king reaches g7and that's just possible because the black king can't go the 4th row because of Ne3! followed by Nf5, the ideally square for the knight and black has nothing against h7, Kg5 to h6.
But one day later Marcel reports: better 3. ... Kd2! 4.Kg3 Kc3! 5.Kf4 Kb4! black is making a sort of Reti manoeuvre to f8 if white plays Kf5, h7 and N .. to h5. First I thought 6.Kg5 wins but black has: 6. ...Kc5 7.Ne3 Kd6 8.Nf5 Ke5! ( Ke6 loses) 9.h7 Ke6 10.Ng7 Ke5 11.Ngh5 Rxh7! 12.Nxh7 f6! ( f5? Nf4 +- ) and white can't block the f-pawn from advancing to f5 and to f4.
If all this is right the study is correct but a very beautiful try is found, not published before"
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10.

Queckenstadt, W. 1917. [+0400.55f2h8]

White wins!?!?

Solution author: 1. f5 (1. Rh6 ? Rg8 2. f5 Ra8 3. fxe6 fxe6 4. Rxe6 Rxa6) 1... exf5 (1... Rg8
2. e5 Ra8 (2... Kg7 3. f6+ Kg6 4. exd6) (2... dxe5 3. f6 Ra8 4. Rxb4 h5 5. g6
fxg6 6. Rb7) (2... exf5 3. exd6) 3. f6 Rxa6 (3... dxe5 4. Rxb4 h6 5. g6 fxg6 6.
Rb7) (3... h5 4. g6) 4. Rc4 Ra8 5. g6) 2. a7 (2. exf5 ? Rxg5 3.
Rf4 Rh5 4. Kg2 Rg5+ 5. Kf3 Rg8 6. Rxb4 Ra8 7. Rb6 d5 8. Kf4 Kg7 9. Ke5 Kf8)
2... Rg8 3. g6 fxg6 (3... Ra8 4. Rxh7+ Kg8 5. Rxf7 b3 6. exf5 b2 7. Rb7) (3...
Kg7 4. gxh7 Rh8 5. e5 b3 6. exd6) 4. e5 Ra8 5. exd6 (5. Rc4 ? dxe5 6. Rc7 b3
7. Rb7 b2 8. Rb8+ Kg7) 5... Rxa7 6. Rc4 b3 (6... Rf7 7. Rc7 Kg7 8. d7) 7. Rc7
b2 8. d7 Ra8 9. Rc8+ Kg7 10. Rxa8 b1=Q 11. d8=Q Qb2+ (11... Qc2+ 12. Kg3 f4+
13. Kh3) 12. Kf3 Qb7+ 13. Kg3 f4+ 14. Kh2 Qb2+ 15. Kh3 1-0

But what after: 1. ... Rxg5!?

For instance: 2.a7 Rg8 3.e5 Ra8 4.Rxb4 Rxa7 5.f6 h5 6.exd6 Rd7, I see no win.

Harold van der Heijden writes in the guestbook:

In my view the cook is not correct: 1…Rxg5 2.a7 Rg8 3.e5 Ra8 and now 4.exd6! Rxa7 5.Rc4! exf5 6.Rc7 wins.

So the study is correct!

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Gulbrandsen, A. 1924. [+4400.22a2h8]

White plays and wins!?

Solution author: 1. Qf2 Qc4+ (1... Rxe6 2. Qf8+ Kh7 3. Qc8 Rg6 (3... Re4 4. Qc2) 4. Rh3+) (1...
Qg6 2. Qf7) (1... Kh7 2. Qf7) 2. Kb1 Rxe6 3. Qf7 Re4 4. Rh3+ Rh4 5. Qxc4 1-0

But what after: 2. ... Qg4!

Black threatens Qd1+ to draw and square h3 is covered.

Harold van der Heijden writes in the guestbook: in addition, it looks to me like 1.Qf1! wins!

So may be the first move was entered wrong. The study is correct but starting with 1. Qf1!
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Nio_Bertholee, R. 1979. [+0014.11c2f3]

White plays and wins.

Author-solution: 1. d7 (1. Nxg3 ? Nxd6) 1... g2 (1... Nd4+ 2. Kd3 g2
3. Nh2+ Kf2 4. Bb6) (1... Ng7 2. Nxg3 Ne6 3. Bc7) 2. Nd2+ ! (2. Nh2+ ? Ke2 3.
Bb6 Ne3+ 4. Bxe3 Kxe3 5. d8=B g1=Q 6. Bb6+ Kf4 ! 7. Bxg1 Kg3) 2... Ke2 3. Bb6
Ne3+ 4. Bxe3 Kxe3 5. d8=B ! (5. d8=Q ? g1=Q 6. Qb6+ Ke2 7. Qxg1) 5... g1=Q (
5... Kf2 6. Bb6+ Kg3 7. Bg1) 6. Bb6+ 1-0.

But better: 1... Nd4+ 2. Kd3 Nc6! 3. Bc7 g2 4.Nd2+ Kg4 5. Bh2 g1=Q 6. Bxg1 Ne5+

or 4. Nh2+ Kf2 5. Bb6+ Kg3 6. Bg1 Ne5+ and the white pawn falls.

After 2. ... Nc6

Harold van der Heijden responds in the guestbook:

The study is correct: 1…Sd4+ 2.Kd3 Sc6 3.Bc7 g2 and now 4.Sh2+! Kf2 5.Ke4 g1Q 6.Bb6+ Kg2 7.Bxg1 Kxg1 8.Sf3+ wins (EGTB).

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Matous, M. 1967

White plays and wins.

Author-solution: 1. Rb3! Bxb3 2. c4! Bxc4 3. Bh7 Bd3
4. Be4 Bc2 5. Bb7 Bf5 6. Ba6 Bd3 7. Bc8+ Bf5 8. Bxf5# 1-0

Possibly a dual at move 1 with: 1. Be6+ Bxe6 2. Re8 Bd5 3. Re3 and white has a winning position?

After 1. Be6+

Not True! Marcel van Herck points out that the black bishop stays on the diagonal a8-d5

and white can't make any progess without stalemating black.

Study is correct!

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Hajek, K. 1924 [+0302.10f6e1]

White plays en wins!?

Solution author: 1. g7 Rh6+ 2. Kf5 Rh5+ 3. Kf4 Rh4+ 4. Kf3 Rh3+ 5. Kg2 Rh4 6. g8=R (6. g8=Q
Rg4+ 7. Qxg4=) 1-0

Position after 6.g8=R.

Black can play 5... Rh6 or Rh7 and now 6... Rg4. The king is already in a stalemate!

May be wrongly entered in the database by Harold ...?!

Reaction by Martin van Essen, 22-2-2005:

I think white can win, but not in the way the author had in mind:

1. g7, Rh6+ 2. Kf5, Rh5+ 3. Kf4, RTh4+ and now 4. Kg3! after that black must play 4. ..., Th1.

But then follows 5. Ne2!, Kxe2 6. Kg2 and wins. There is no stalemate in then.

This variation is also a possible dual for repairs. With Knigts on c3 or a3 the author-solution might work.

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Veurman, J - Marwitz, J. 1937. [+0331.53a8e3]

White plays and wins.

Solution in database: 1. c6 Rg1 2. c7 Rc1 3. Nd1+ Kd3 4. Nc3 1-0

Marwitz also gives in his Arves-book "Eindspelkunst" at page 13: 1... Rg6 2.Ng4+ and 3.Nf6.

(By the way: he does not mention the name Veurman,J there)

And also after 1... Rg5 2. Ng4+ K any 3. Ne5 Beautiful Novotny's!

but now after 2.... Ke2 3. Ne5 Bxe5 4. c7 Bxc7 5. h8=Q Bf4 6. Qf6 Bd2
7. Qxf7 Rxa5. Does black have a fortress?

If so than it is only a draw. Black can play h5 and try to walk with the king to the a-pawn also.

Marcel Van Herck has found that white should play 3.Nxh6! and then white still wins.

That makes at least that the study is correct, although the author-solution with three beautiful interferences is reduced to two.

But 1... Rg5! After 7... Rxa5.

The fortress found might be worthwile to make a study for...?

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Nio_Bertholee, R. [+0332.23c1a1]

White plays and wins.

Author-solution: 1. Nf5 (1. Nxh5 Rxe3 2. Na4 Re1+ 3. Kxc2 Re7) 1... Bg6 (1... Rd8 2. Nd4 Rxd4 3.
exd4 b5 4. Nd3 Bf7 5. Nb4 Bb3 6. d5) (1... Bf7 2. Nd4 Rc8 3. Nc4 ! Rxc4 (3...
Bxc4 4. Nxc2#) 4. Nb3#) 2. Nd4 Rxe3 3. Nd3 ! Rxd3 (3... Bxd3 4. Nb3#) 4. Nxc2# 1-0

But what if black plays: 1... Rf8!? 2.Nc4 Rd8 3.gxh5 b5 4.Na5 Rd1+ 5.Kxc2 Rb1 6.Nd4 Rb2+

Black must try to get his king out of the corner, this seems to work this way.

This variation is refuted by Marcel Van Herck.

He points out that white should play 3.Nd4! (threathening Nb3#) Rxd4 4.exd4 Bxg4 (b5 5.Na3) 5.Ne3 +- and white wins.

SO this study is still CORRECT !

After 1... Rf8!?

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Pogosiants, E. 1961 [+0041.11h6h4]

White plays and wins.

Author-solution: 1. Bf1 Bb5 ! (1... Bd7 2. Ng5 g2 3. Bxg2 Kg3 4. Bf1 Bc6 5. Be2 Bd5 6. Kg6 Kf4
7. Kh5 Bc6 8. Bd1 Bd5 9. Kh4 Bc6 10. Nh3+) 2. Bg2 ! Bf1 ! (2... Be2 3. Kg7 ?
(3. Bxf1 g2 ! 4. Ng3 ! g1=Q (4... gxf1=Q 5. Nxf1) (4... Kxg3 5. Bxg2
Kxg2 6. f4) 5. Nf5# 1-0

The study is correct but the variation 2... Be2 3. Kg7 is wrong. There follows 3... Bxf3! 4. Bxf3 Kh3=

Winning is: 3. f4! Kg4 4. f5 Kxf5 5. Nxg3+ Kf4 6. Nxe2+ Ke3 7. Nc3)

3... Lxf3! = 3.f4! +-

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Roycroft, John 1994 [=0133.22b1d3]

White plays and Draws.

Author-solution: 1. f3 (1. Rg8 ? Bxd7 2. Rd8 Nf6 3. Rb8 b5 4. Rb6 Ne4) 1... Bf5 2. Rf8 Bxd7 3.
Rd8 Be6 (3... Nf6 4. Rb8 (4. Rf8 ? Ne8 5. Rf7 Bc6) 4... b5 5. Rb6 Ne8 6. Rb7
Bc6 7. Rb6 Bd7 (7... Bxf3 8. Rxb5 a4 9. Rb4 Bd1 10. Kb2 Nd6 11. Ka3 Nc4+ 12. Rxc4)

8. Rb7) (3... Bc6 4. Rd6 Bb7 5. Rd7 Ba8 6. Rd8) 4. Rd6 Bf7
5. Rd7 Bg8 6. Rd8 Bh7 7. Rd7 (7. Rxd5+ ? Kc4+) 7... Bg6 8. Rd6 Bf7 9. Rd7
1/2-1/2

In the underlined subvariation the proposed move: 9.Rb4? is very risky because after Bd1 10. Kb2 Nd6 11. Ka3 there follows strongly Kc3!

And then black can move the pawn, because 12. Rxa4? Nb5+ .

Better is 9.Ra5 and the draw is reached. Some beauty goes with it.

For instance: 9. Ra5 Bc6 (9... Bd1? 10. Rd5+ Ke2 11. Re5+) 10. Ka2 Nd6 11. Ra6 Bd5+ 12. Ka3

After 11... Kc3! -+ Better: 9.Ra5! =

Harold van der Heijden in the guestbook:

The problem is not in move 9. After 3…Sf6 (which is actually a main line) 4.Rb8 b5 5.Rb6 Se8 6.Rb7 Bc6 7.Rb6 Bxf3 8.Rxb5 a4 9.Rb4 Bd1 10.Kb2 Sd6, White should play 11.Rf4(Rh4) and Black can’t make progress (11…Sc4+ 12.Rxc4).

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In EBUR/1990/2/Pag.17-18 Julien Vandiest and Rene Olthof analyse the following position:

Seletsky, A. 1933 [+4044.10f2f7]

White wins.

Solution according to the database by Harold van der Heijden:

1. Qg5 ! Ke6+ ! (1... Bxd7 2. Nf4 !) 2. Kg1 ! (2. Ke1 ? Kxd7 3. Nc5+ Kc8
4. Qe5 Qh6 !) 2... Kxd7 3. Nc5+ Kc8 (3... Kd6 4. Qg3+ Ke7 (4... Kd5 5. Bc4+
Kxc4 6. Qb3+ Kxc5 7. Qa3+ Kb6 8. Qxf8) 5. Qe5+ Kf7 6. Bc4+ Kg6 7. Bd3+) 4. Ba6+
Kb8 5. Qg3+ Ka8 6. Bb7+ ! Bxb7 7. Nd7 ! Qd8 8. Qb8+!! Qxb8 9. Nb6# 1-0

It's all about move: 3. ... Kd6.

The underlined variation ends with a position with Q-B+K about which Vandiest thought it was a draw, but is winning according to theory by A.Cheron for the Q which means that the studie is correct. So you may see that theory-knowledge is not as easy.

After 4. Qg3+ Kd5 in EBUR after 5.Qd3+ this looks also winning, meaning it is correct anyway.

The rest of the analyses: 5... Ke5 6.Qe3+ Kd5 7. Bf3+ Kd6 8.Ne4+ Ke5 9.Ng5+ Kf5 10.Bxc6 Nxc6 11.Qf2+

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Peter Boll.

Homepage ARVES.