Wednesday, December 31, 2008

Happy New Year

Here is some fireworks for the new year:

Salaske-Leisebein, corr 1988

1.b4 e5 2.Bb2 Bxb4 3.Bxe5 Nf6 4.c4 0–0 5.Nf3 Nc6 6.Bb2 Re8 7.e3 d5 8.cxd5 Nxd5 9.Be2 Rxe3!? 10.fxe3 Nxe3 11.Qc1 (Dia)

In earlier entries I have had a look at 11.Qb3. I assume I will have to return to 11.Qa4 some day.

11...Nxg2+

This is natural and probably more promising than 11...Bf5, e.g. 12.Kf2 Nc2 13.Rd1 Qe7 14.Nc3 Rd8 15.Na4 Nxa1 16.Bxa1 Re8 17.Re1 +/- Bungo-Le.Karlsson, corr 1990.

12.Kf2 Bh3

Again this seems to be the critical try. After 12...Nf4 13.d4 Nxe2 14.Kxe2 Bg4 15.Qf4 Bxf3+ 16.Kxf3 Qd5+ 17.Qe4 Qh5+ 18.Kf2 f5 19.Qd5+ White had a clear advantage in Engelhardt-Rollwitz, Berlin 1995.

13.Rg1! Qe7 14.d4 (Dia)

White may also try 14.Ba3 Re8 15.Bb5 Ne5 16.Rxg2 when Black has these options:

a) 16...Bxg2 17.Kxg2 Nxf3 18.Bxe8 Qe2+ 19.Kg3 h5 (19...Ne1 20.Bxf7+ Kh8 looks winning for Black) 20.Bxf7+ Kxf7 21.Qxc7+ Kg6 22.Qxb7 (22.Qd7 Kh6 23.h4 Nxh4 is no better) 22...Qxh2+ 0–1

b) 16...Qf6!? 17.Be2 Ng4+ 18.Rxg4 Bxg4 19.Bd1 Bxf3 20.Bxf3 Qxa1 21.Bxb4 Qd4+ 22.Kg2 Qxb4 –+.

Tilp-Hammerschmidt, corr 1988.


14...Re8 15.Bd3

15.Bb5 is met by 15...a6

15...Nh4

This looks sufficient for at least equality. Alternatives are:

a) 15...Be1+? 16.Rxe1 Nxe1 17.Qxe1 Qd7 18.Qd1 Nb4 19.Nc3 Nxd3+ 20.Qxd3 Qg4 21.Rg1 1–0 Woelfelschneider-Guenther, corr 1990.

b) 15...Ne3 16.Nc3 (16.d5 Nf5 17.dxc6 Bc5+ 0–1 Milligan-J.Svensson, corr 1995) 16...Ng4+ 17.Rxg4 Bxg4 18.Qf4 Bxf3 19.Kxf3 Bd6 20.Nd5 Bxf4 21.Nxe7+ Nxe7 22.Kxf4 Nd5+ 23.Kg4 g6 = Schuehler-Salaske, corr 1989.

c) 15...Qf6 16.Qg5? (16.Nbd2 Nf4 17.Bb5 Qf5 18.Bc4 Bg4 19.Rg3 h5 unclear) 16...Be1+ 17.Rxe1 Nxe1 18.Bxh7+ Kxh7 19.Qh5+ Kg8 –+ Frenzel-Nebe, corr 1989.

16.Qg5

16.Qh6!? Ng6 17.Qg5 may be an improvement.

16...f6 (Dia)


17.Qf4?!

This may hand Black the advantage, so White should investigate line c) below.

a) 17.Qxg7+? Qxg7 18.Rxg7+ Kxg7 19.Nxh4 Be1+ 20.Kf3 Bxh4 –+.

b) 17.Qxh4? Qe3+ 18.Kg3 Qxd3! 19.Qxh3 Bd6+ 20.Kh4 (20.Kg2 Re2+ 21.Kh1 Qxf3+! 22.Qxf3 Rxh2#) 20...Re4+ 21.Rg4 (21.Kh5 g6+ –+) 21...g5+ 22.Kh5 Rxg4 0–1 Dziel-Zarebski, corr 1993.

c) 17.Bc4+ Kh8 18.Qxh4 Qe3+ 19.Kg3 Bd6+ 20.Kxh3 Qxf3+ 21.Rg3 Bxg3 22.Nd2 and the position is unclear.

17...Nxf3 18.Kxf3 g5 19.Nc3 Qg7

Rybka's 19...Kh8! looks like a worthwhile improvement.

20.Nd5 gxf4 21.Nxf6+ Kf8 22.Rxg7 Re3+ 23.Kxf4 Rxd3 24.Rxh7 Bd2+ 25.Ke4 Re3+ (Dia)

The position is totally bewildering. I would be more worried with White than with Black. However, this was a correspondence game (as most other examples in this line) and it seems the players were able to handle it correctly.

26.Kf4 Rb3+ 27.Ke4 Bg2+ 28.Kf5 Ne7+ 29.Ke6 Rb6+ 30.Ke5 Nc6+ 31.Kf5 Rxb2 32.Rh8+ 1/2–1/2


Addendum January 27th
There are three quite interesting articles by Tim Harding on the Sokolsky and even this 9...Rxe3 variation on Chess Cafe:

How Sokolsky Played the Sokolsky
Significant Games in the Sokolsky Opening
Goodbye to the Friendly Orang-Utan

1 comment:

M.Nieuweboer said...

Only recently I have found this article. An addition:
15...Qf6 (Frenzel-Nebe, corr 1989) 16.Nbd2 Re3 is good for a draw after 17.Be2 Qe7 18.Bc4 Bxd2 19.Qxd2 Rxf3+.