Chess

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The Independent Online
The new Wimbledon-style knock-out format for the Fide world championship produced a Wimbledon-like result with two Englishmen meeting in the semi- finals. But which of Nigel Short and Michael Adams will turn out to be the Greg Rusedski and which the Tim Henman is still undecided at the moment of writing.

The Pete Sampras of the event, however, has undoubtedly been the Indian grandmaster Viswanathan Anand who has steered a faultless path through the precarious two-game knock-out matches and rapid play-offs. (Not that Short and Adams have been any less impressive, but Anand earns the Sampras role simply by virtue of having been the highest seeded player at the start.)

Anand will play Boris Gelfand, with the winner meeting Short or Adams. When only one man remains undefeated, the contest then moves from Groningen to Lausanne for a six-game match between the last survivor and Anatoly Karpov.

Here is the game that took Anand through to this stage. Playing against the imaginative Latvian grandmaster Alexei Shirov, Anand kept a firm control throughout the game, methodically denying space to Black's Q-side pieces while organising his own attack on the opposite wing.

The finishing combination is very neat. Just as Black seemed to have fought his way to some Q-side activity, Anand stifled his hopes with 36.h6! (see diagram). After 36...gxh6 37.Bxf6 White wins comfortably, but the alternative, as played in the game, did not last much longer.

White: Viswanathan Anand

Black: Alexei Shirov

1 e4 e5 22 d5 Bd8

2 Nf3 Nc6 23 Bd2 Qa6

3 Bb5 a6 24 Qa4 Ra8

4 Ba4 Nf6 25 Bc3 Nb7

5 0-0 b5 26 Qd1 Ba5

6 Bb3 Bc5 27 Bb2 Bb4

7 a4 Rb8 28 Nh4 Bg6

8 c3 d6 29 f4 Qa4

9 d4 Bb6 30 Qxa4 Rxa4

10 Na3 0-0 31 f5 Rfa8

11 axb5 axb5 32 Re7 Bh5

12 Nxb5 Bg4 33 g4 f6

13 Be3 exd4 34 gxh5 Rxa2

14 cxd4 Qe8 35 Rxa2 Rxa2

15 h3 Bd7 36 h6 Rxb2

16 Nc3 Nxe4 37 Rxg7+ Kh8

17 Re1 Nxc3 38 Rxb7 Bc3

18 bxc3 Qc8 39 Rd7 Kg8

19 c4 Bf5 40 Rd8+ Kf7

20 Re2 Na5 41 Rh8 Bd4+

21 Ba2 c5 42 Kf1 resigns

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