Chess

Jon Speelman
Tuesday 08 June 1999 23:02 BST
Comments

THE DUTCH Championship in Rotterdam is moving towards its conclusion. After a splendid start of 3/3, Predrag Nikolic still looks very much the favourite and with three rounds to go after the rest day on Monday he led by a clear point with 6.5/8 ahead of Reinderman and Piket 5.5, Van Wely 5, Van der Wiel 4.5, Van der Sterren 4, Ivan Sokolov (who has at least now won a game against the bottom marker Ruud Janssen but is still in the doldrums), Van den Doel and Cifuentes 3.5, Nijboer 2.5, Peng Zhaoqin 2.5 and Janssen just 1.5.

The surprise of the tournament, apart from Sokolov's execrable form, has been the excellent showing of Dmitri Reinderman.

Reinderman, who will be 27 in August and, incidentally, doesn't play chess full time since he has a "day job" with the Netherlands subdivision of Demon Internet, was only a name to me until I met him in tournaments a couple of times recently, in Leewarden at the end of 1997 and the Andorra zonal last November, from which he was one of the six qualifiers.

He is currently rated 2,541, and has a confident approach and pleasingly fluid attacking style which work well to get him out of the positional scrapes in which he sometimes finds himself. Once he gets going he can be absolutely lethal, as you can see from his beautiful win against the usually extremely solid Paul Van der Sterren

Despite starting as a Caro Kann, they transposed into one of the main lines of the Nimzo-Indian (1 d4 Nf6 2 c4 e6 3 Nc3 Bb4 4 e3 0-0 5 Bd3 d5 6 Nf3 c5 7 0-0 cxd4 8 exd4 dxc4 9 Bxc4 b6).

The paradoxical 14 ...Qxd7 may be possible probably inducing 15 Bxf6 gxf6 after which the reduced material makes it hard for White to attack - though the thematic d5 break 15 d5!? Bxd5 17 Rad1 is possible.

Although 20 ...e5 may look logical it does unshackle the b3 bishop. Reinderman reacted splendidly by sacrificing his c pawn to defend the bishop with tempo. He quickly developed an enormous attack and, of course, spurned the exchange starting with 25 Rbd1!.

26 ...Qc6 is also met by 27 Rg3!. The position after 27 ...Kh8, of course, deserves a diagram but I'm afraid I've run out of space.

28 Rxd7! then prepared a beautiful queen sacrifice. At the end 33 ...Rf7 34 h3 is simplest and if Rcf8, 35 Rxb7.

White: Dmitri Reinderman

Black: Paul Van der Sterren

Nimzo-Indian Defence

1 e4 c6

2 d4 d5

3 exd5 cxd5

4 c4 Nf6

5 Nc3 e6

6 Nf3 Bb4

7 Bd3 dxc4

8 Bxc4 0-0

9 0-0 b6

10 Bg5 Bb7

11 Ne5 Nbd7

12 Qe2 Bxc3

13 bxc3 Qc7

14 Nxd7 Nxd7

15 Rac1 h6

16 Bh4 Rac8

17 Bb3 Rfe8

18 a4 Qd6

19 Rfe1 Qa3

20 Qd1 e5

21 Ra1! Qb2

22 Rb1 Qxc3

23 Qh5 Rf8

24 Be7 Qxd4

25 Rbd1! Qc3

26 Re3 Qc7

27 Rg3 Kh8

28 Rxd7! Qxd7

29 Qxh6+! gxh6

30 Bf6+ Kh7

31 Rg7+ Kh8

32 Rxf7+ Kg8

33 Rxd7+

1-0

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