Joint leaders at Simpsons
The third round of the 2008 Staunton Memorial saw the last 100% scores
disappear. Mickey Adams faced the ultra-solid Erwin L'Ami with the black pieces,
and a theoretical 4.Qc2 Nimzo soon simplified to a drawn ending. The other
co-leader, Jan Timman, came much closer to maintaining his winning run, but saw
half a point slipped through his fingers:
Timman,Jan H - Cherniaev,Alexander [E18]
Staunton Memorial (3), 08.2008
1.d4 Nf6 2.c4 e6 3.Nf3 b6 4.g3 Bb7 5.Bg2 Be7 6.Nc3 Na6
The favourite line of Russian-born, Dutch-resident GM Sergey
Tiviakov.
7.0–0 0–0 8.Ne5 Bxg2 9.Kxg2 Qb8 10.e4 Qb7 11.Qf3 Bb4
12.Bg5!?
Definitely the critical move, and a typical Timman choice.
The safer move 12.Re1 would maintain a safe space advantage, whereas the text
provokes a much sharper and tactical battle.
12...Bxc3 13.bxc3 Nxe4 14.Be7
14.Nxd7 is the alternative, with a similarly unclear position
14...d6 15.Bxf8 dxe5!?
The solid 15...Rxf8 would also offer Black reasonable
compensation for the exchange.
16.Ba3 f5 17.Rfe1?!
Neither player liked this after the game. Timman was hoping
to organise g3-g4, undermining the e4 knight, but he never succeeds in making
this work, with the result that Rfe1 turns out to be a waste of time.17.Rad1 is
stronger.
17...Qc6 18.dxe5 Qxc4!?
Played after long thought, but perhaps not best. Chernaiev
had wanted to play 18...Qa4, but rejected it after calculating the line 19.Rxe4
dxe4 20.Qxe4 Re8 21.Bc1 Nc5 22.Qg4 Qc2 23.Bh6, but he had missed 23...Qe4+, when
Black seems OK.
19.Rad1 Qa4 20.c4 Re8?
And this definitely looks like a serious mistake. 20...Qxc4
21.Rd7 Qc6 would be a better try.
21.Qb3!
Now the black queen is driven back and White assumes a
decisive initiative.
21...Qc6 22.Qb5 Qa8 23.f3 Nec5
24.Re2?
Missing the strength of Black's counterplay. Instead, 24.Bc1!
would put a stop to it before it started, and the plan of doubling rooks on the
d-file should then win easily.
24...g5 25.Red2 Kf7! 26.Rd7+ Nxd7 27.Rxd7+ Kg6 28.Qxa6 g4
White as won a piece, but his out-of-play queen makes it
difficult to meet Black's threats on the kingside.
29.Bc1
29.Rd3 is the obvious reply, but after 29...gxf3+ 30.Kf2 Qe4,
Black has strong play.
29...Qxf3+ 30.Kg1 Qe2! 31.Qa3 ½–½
The day's only winners were Jan Smeets and Nigel Short. The
former triumphed in a tough Ruy Lopez against compatriot Ivan Sokolov. This was
a game which was perhaps lent additional significance by the fact that Smeets
won the 2008 Dutch Championship earlier this year. However, the top-rated Dutch
players Sokolov and van Wely were both absent, and both may therefore feel they
have something to prove against Smeets at this event. Sokolov played a somewhat
discredited line of the Zaitsev (theory considers 12...Nb8 to be superior), but
emerged in the middlegame with a reasonable position, only for a pawn to drop
off in the complications running up to the time-control.
Nigel Short, after his trauma of the day before, faced Bob
Wade with the black pieces. The former played solidly, but unambitiously in the
opening, and Short gradually ground out a win with a standard Minority Attack.
He clinched the game with an attractive combination:
Short demolished his opponent's position with 39...Nxc3! 40.Nxc3 Rbc8
41.Ra3 Bd6! 42.Rha1 42.Rb3 is answered in the same fashion: 42...Bb4!
43.Rxb4 Rxc3+ and 44...Rxf3. 42...Bb4! 43.Kd3 Rxc3+ 44.Rxc3 Rxc3+
45.Ke2 Bd6 and Black won in a few more moves.
A cheerful Nigel Short, who has obviously just heard the cricket score...
The two drawn games between Werle-Speelman and van Wely-Wells
both saw the English players obtain the better of the game, despite the black
pieces, but both were unable to convert. Although a draw with black against a
higher-rated opponent is rarely a bad result, one cannot help fearing that these
various dropped half points may come back to haunt the British in the team
event.
After three rounds, the scores are therefore as follows:
Adams, Timman 2.5, Van Wely 2, L'Ami, Wells, Short, Smeets, Werle 1.5, Cherniaev,
Speelman, Sokolov 1, Wade 0.
