ODT070424

=**Close competition in club events**=


 * Problem:** White to play and mate in 3.

The first two tournaments of the Otago Chess Club season have shown there will be no easy games once the trophy events begin. Former club champion John Sutherland recently returned from overseas and started with 4/4 in the opening Swiss Rapid tournament. However, losses in the last two rounds to Duncan Watts and Richard Sutton allowed Quentin Johnson to slip through to first place on 4½/6, with Sutherland, Sutton and Hamish Gold finishing tied for second on 4.

Mostly the same names took the top places in the Swiss Standard tournament that followed. Johnson finishing first on 4½/5 followed by Sutton second on 4, Sutherland third on 3½, and Terry Duffield fourth on 3. The top three players were unbeaten since byes taken during the tournament meant that only Johnson and Sutton played each other – the game finishing in a draw.

Today's game from the standard tournament features some instructive mistakes and tactical motifs. Quentin Johnson is playing White against Geoff Aimers. media type="custom" key="26891232" In this offbeat line of the Modern Benoni White's loss of time with his bishop is compensated by the relatively poor placing of the black bishop on d7 interfering with his knight's development. White has been maneouvring towards the thematic Benoni advance f4 a followed by e5, and anticipates that the bishop will be more useful on b1 for the subsequent kingside attack. However, a brief glance at the actual position shows this to be a tactical blunder – 16 Be2 was necessary, defending c4 and f3. Taking one of the free pawns on offer – since 17 gxh3?? Nf3+ wins the queen. But White is able to exploit the exposed position of the black bishop on d4 to generate an attack on the dark squares. More disruptive to White's position was 16 ... Nc4! 17 Qd3 Nxb2 18 Qf3 Nc4 when the d4 bishop still defends the kingside and White's queenside is totally undermined. Stopping 19 Nxd4 since cxd4 wins the pinned knight. White now lays a trap to lure the two central black pieces away from the defence of the holes on f6 and h6. Stranding the knight on f2. If 20 ... Ne5? 21 Nxd4! cxd4 22 Bf6! forces Black to give up his queen by Qd8 to prevent 23 Qh6 and mate on g7. So 20 ... f6! was necessary when 21 Bf4 Ne5 is still better for Black. White's combination has a flaw. Instead 22 Kh2! will result in the eventual win of the trapped knight after 23 Be3. Missing the only other way to stave off mate on g7 – 22 ... Re5! 23 g4 (not 23 Qh6?? Rh5+) Nc7! and now 24 Qh6 would be met Ne8 defending g7 so White has nothing better than 24 Rf1 Bd4 25 Bxe5 winning back the exchange but still a pawn down with no more attack. Now White has a material advantage in addition to ongoing threats. Necessary was 25 ... Nc7 to meet 26 Bf6 with Ne8. Both sides thought that 26 Bf6!! forcing mate on g7 allowed perpetual check after 26 ... Qxe1+ 27 Kh2 Rh5+ 28 Nxh5 Qh4+ overlooking the fact that the bishop now defends the square it moved from 29 Bxh4 – a common calculation error. White instead forces the win of the black queen. The threat of mate by 30 Bf6 means the bishop on f5 is lost.
 * 1. d4 Nf6**
 * 2. c4 c5**
 * 3. d5 e6**
 * 4. Nc3 exd5**
 * 5. cxd5 d6**
 * 6. e4 g6**
 * 7. Bb5+ Bd7**
 * 8. Bd3 ---**
 * 8. --- Bg7**
 * 9. Nge2 O-O**
 * 10. O-O Re8**
 * 11. Ng3 Na6**
 * 12. Bg5 Qb6**
 * 13. Qd2 Ng4**
 * 14. Rae1 Bd4**
 * 15. h3 Ne5**
 * 16. Bb1? ---**
 * 16. --- Bxh3**
 * 17. Qc1 Bd7**
 * 18. Nge2 Rac8**
 * 19. Kh1! Ng4**
 * 20. f3 Nf2+?**
 * 21. Rxf2?! ---**
 * 21. --- Bxf2**
 * 22. Bf6 Bh4?**
 * 23. Bxh4 Re5**
 * 24. Ng3 c4**
 * 25. Qh6 Qf2?**
 * 26. Nf5 Qxe1+**
 * 27. Bxe1 Bxf5**
 * 28. f4 Ree8**
 * 29. Bh4 Resigns**


 * 1-0**


 * Solution:** 1 Qg6+! Bxg6 2 Ng5+ hxg5 3 hxg6#