ODT120110

=**Impressive field for Queenstown**=




 * Problem:** Black to play and mate in 4.

Central Otago will once again the focus of chess attention in New Zealand when the Queenstown Chess Classic starts on Sunday at the Millennium Hotel in Queenstown. Incorporating the 119th New Zealand Championship, the nine-round Swiss Open has attracted the strongest ever field in a New Zealand tournament. Eleven grandmasters, nine IMs and four WGMs head the more than 120 players from 19 countries.

The top three seeds all look to be in excellent touch. Second seeded English GM Gawain Jones, who played in the previous Queenstown event in 2009, will be looking to continue the convincing form he showed in winning the recent British Rapidplay championship in Leeds with a score of 10½/11. But he will face stiff competition, as the first and third seeds, young Chinese GM Li Chao B and Indian GM Surya Ganguly shared first place on 7/9 ahead on a strong field in the 2011 Indonesian Open Chess Championship at Jakarta in October. Readers interested in following the Queenstown results and games should visit the New Zealand Chess Federation website www.newzealandchess.co.nz.

Today's game features Queenstown top seed GM Li Chao punishing some rash opening play in a recent game from the Chinese Chess League. He is playing White against another Chinese GM, Wang Rui.

media type="custom" key="25615918" In the Catalan White seizes the long diagonal in preparation for long-term pressure on the light squares. Black frequently responds by capturing the pawn on c4, as the bishop is removed from its protection. Tactical tricks down the diagonal are common. White doesn't hurry to recapture his pawn and Black must be careful. Here 7...Nxd4? 8 e3 loses a piece as 8 ... Nc6 9 Bxc6+ bxc6 10 Qxd8+ Kxd8 11 Nxf7+ wins. Also after 7...Nxe5 8 dxe5 Qxd1 9 Rxd1 White will easily regain the pawn on c4. Aiming to exchange off a couple of White's active minor pieces. The drawbacks are the loosening of the queenside pawns, weakening the c6 square and the fact that castling is still two moves away. Another ambitious idea, defending b5 with the rook. The reason can be seen after the more natural 13...Rc8 14 Rc1 Bc5 15 Qd3 a6 16 a4! and Black can't defend the pawns (16...Qb6? 17 b4! wins), though after 16...Nd5! he seems to have a reasonable game. This move is a stretch too far: Placing the knight on the edge undefended while neglecting development is sure to be punished. Black had to relinquish the c-file 14...Rd5!? (14...Rxc1? 15 Qxc1 just brings the queen more quickly to c6, and 14...Be7 15 Rxc5 Bxc5 16 Qc2 Qd5 17 b4! is similar) and submit to an attack: 15 Qc2! Bc5 (or 15... Qb6 16 Qc7 Qxc7 17 Rxc7 and the rook on the seventh gives White the advantage) 16 b4! Be7 17 Qc6+ Nd7! 18 Nxd7 Rxd7 19 Qxb5 O-O 20 a3 and White has his pawn back but not too much more. Giving Black a reprieve. Correct was the c-file breakthrough 15 b4! Rxc1 16 Qxc1 Bxb4 (16....Nxf4 17 Qc6+ Ke7 18 gxf4 is no better) 17 Qc6+ Kf8 18 Qxb5 Be7 19 Qb7! Nxf4 20 gxf4 f6 21 Rc1! fxe5 22 Rc8 winning the queen. Presumably not keen on either 15...Nxf4 16 Rxc5! Nxe2+ 17 Kg2 Bxc5 18 Qxb5+ Kf8 19 Qxc5+ Kg8 20 Kf3! d3 21 Ke3! trapping the knight or 15...Qb6 16 Rxc5 Bxc5 17 Qf3 Nxf4 18 Qa8+ Ke7 19 Qxh8, Black simply loses a piece. But the counter-intuitive 15...Rd5! still would have put the onus on White to justify his pawn minus. There is no escaping the coming fork: 16...Nxf4 17 Qf3! Bxc5 18 Qc6+ Kf8 19 Qxc5+ followed by taking the knight. The game is over as 17...Nxf4 transposes to the previous note – a tactical exploitation of the Catalan diagonal. It takes a couple more moves to sink in.
 * 1.d4 Nf6**
 * 2.c4 e6**
 * 3.g3 ---**
 * 3.--- d5**
 * 4.Bg2 dxc4**
 * 5.Nf3 c5**
 * 6.O-O Nc6**
 * 7.Ne5 Bd7**
 * 8.Na3 Rc8**
 * 9.Naxc4 cxd4**
 * 10.Bf4 b5!?**
 * 11.Nxc6 Bxc6**
 * 12.Bxc6+ Rxc6**
 * 13.Ne5 Rc5!?**
 * 14.Rc1! Nh5?**
 * 15.Qb3?! ---**
 * 15.--- a6?**
 * 16.Rxc5 Bxc5**
 * 17.Qf3! Qd5**
 * 18.Qxh5 O-O**
 * 19.Qf3 Resigns**


 * 1-0**


 * Solution:** 1 ... Qh3!! (1...Qg4 2 Ng3 Qh3!! also works) 2 gxh3 (2 gxf3 Nf4 & 3...Qg2#) Nf4 3 Ng5 Bxg5 and 4 ... Nh3#.