CHESS by Philidor

No. 293.

A. ELLERMAN BLACK (8 men) (Caissa, 1943) WHITE to play and mate in two moves; solution next week. Solution to No. 292 (Godfrey) : B—Q 2! n, threat. z . . . K—K 5 ; 2 Kt— B 5. z. .K—B 5; 2 KtxB. 1 . . . K—B 3; 2 Kt x B. . . . P—B 3 c.h ; Kt (B 7)—Q 6.

Kt (Kt 7)—Q 6. I. . . B moves ; 2 Kt (B 7)--Q 8. z . . P—K B 4 ; fine and unexpected key.

In this year's Clare Benedict team tournament we had a strong team (Penrose, Kottnauer, Hindle and Keene) which was expected to be amongst the leaders ; in fact, we did inexplicably badly, finishing fifth out of six with a game score of z win (!), 9 losses, Jo draws. WHITE (8 men) 2 Kt—Kt 5. Very The eighteen-year-old Keene scored the one win on bottom board and, at one moment, had a chance to win the following exciting if unsound game as well. Let us hope our players have worked off all their bad patches simultaneously before the world team tourna- ment in Cuba later this year. Incidentally, the final order at Brunnen was : Holland, Spain, W. Germany, Austria, England, Switzerland.

White, HECHT (W. Germany) Black, KEENE (England) Opening, CARO-KANN e P—K 4 P—Q B 3 a P—Q 4 P—Q 4

3 Kt—Q B 3 P x P 4 KtxP Kt—Q2 5 B-0 B 4 K Kt—B 3

6 Kt—Kt 5 . . . A good alternative, intro- duced by Keres, is 6 Kt x Kt, Kt x Kt; 7 P—Q B 3. Black cannot reply 7 ... B—B 4? because of 8 Q—Kt 3, and after 7 ... P—K 3 White 122E some positional advantage. The text is also good.

6... P—K 3 8 7 Q— 2 Kt—Kt 3 B—QKz. P—K R 3 9 Kt (5) B 3 P—B 4

iv PxP B x P zr Kt—K5 0-0

1.2 Kt (r)—B 3 Q K1—Q 4 13 0-0 P—Q Kt 3 14_P—Q R 3 P—Q R 4 Otherwise is P—Q B 4, Kt—K 2; x6 P--Q Kt 4, B—Q 3; 17 13—Kt 2 leaves White with a clear advantage.

15 P—B 4 Kt—K 2

z6 R—Q r Q—K z An unsavoury move to have to play. The alternative is 16 ... Q—B 2 and if 17 B—B 4, then

17 . — Kt—R 4, e.g. 18 Q—K 4, P—B 4 ; 19 Q x R, B—Kt a. If Q—B z cannot be played, then I think Black's opening strategy is not wholly satisfactory.

17 B—Q 2 P—R r8 B—B 3 Kt—B 4? He should play B—Kt 1,

but White's position is preferable.

r9 P—K Kt 4! Kt—Q 3 20 P—Kt 5 P x P 22 Kt x Kt P B—Kt 2 22 B—B 2! . . . Threat 23 R x Kt, BxR;

24 Q—Q 3, R—Q r; 25 Q—R 3 with a winning attack.

P—K t3 Or 22 R—QB I; 23 Kt—Kt 4 and now (a ) 23 . . . Kt x Kt; 24 Q x Q— 3; 25 B—R 7 ch, K—R 1; 26 B—K 4! (b; 23 . . . Q—B 3; 24 Kt x Kt ch, P x Kt; 25 R—Q! and the threat of Q—R 5 is too strong, e.g. 25. P x R; 26 Q—R 5, K R—K ; 27 BxBP or z5. P x Kt??; 26 R x P mate or 25 . . . K—Kt z; 26 Q—Kt 4. 'A complicated and fascinating position. 23 Kt—Q 71? . . . Ingenious but unsound. 23 Kt—Kt 4! Kt x Kt; 24 Q x Kt wins (24 • • Q—B 3; 25 R—Q 5 P x R; 26 Q—R 4). Q xIO—. R 51, P x Q; 25 B—R 7 mate.! 23 . . . Kt x Kt?; 24 Q

24 B x Kt Q—B 3

R—Q 5 . . . The point of White's com- bination ...

P x . . . but Black could have escaped by 25 . Kt—K t!; z6 B—B 3, Kt—Kt z! and, though very complicated, the position looks strong for Black. I leave readers to amuse themselves with the analysis. 26 Q—R 5! B x P ch z6 . . P x QT; 27 B—R 7 Mite. 27 K xB Kt—K5 ch

28 Kt x Kt P x Q 24 R—Kt i eh K—R a

30 Kt—B5 dia.ek. Revives 3o K—R 3; 3t I3--Kt 7 name. Highly entertaining.