CHESS by Philidor

No. 287.

C. BECKER

(Prablemist,

Nov. 1965) WRITE to play and mate in two moves ; solution next week. Solution to No. 286 (Meredith) : Q— Q 2 I, threat Q— K R 2. I . . . K—Q 3; 2 Kt- 13 4. ...QxQ; 2 Kt—B 4. I . . . Q x B ; 2 Q—Q 4. BLACK (8 men) WHITE (8 men)

. * Smyslov again demonstrated what a formidable player he is by winning the x966 Mar del Plata

. . . K—Q 3 ; 2

tournament with 105 (7 wins, 8 draws), the other Soviet grandmaster Stein being second with to}, followed by Portisch 9i and the veteran US grand- master Sammy Reshevsky with 9. The following fine game demonstrates Smyslov's skill in dealing with positionally unsound play ; he maintains an absolutely sound position structurally and from this demolishes with ease the tactical threats raised by his opponent.

White, SMYSLOV Black, PELIKAN Opening, Dural (Mar del Plata, 1966)

P—Q 4 P—K B 4 1 P--Q B 4 Kt—K B 3 3 Kt—QB3

4 Kt—B3 11:-.723Kt 3 The Leningrad variation; lively but slightly dubious.

5 P—K Kt 3 B—Kt

6 B—Kt a0--0 7 0-0 P—Q R4 Th's does not seem to fit7in with the Leningrad: an immediate 7 . . . Kt—B 3; 8 P—Q 5, Kt—l( 4 seems better.

8B—KC 5 1(1-83 9P.-Q5 Kt—K 4

so Kt X la PxKt

srP—K

V2-12 ii . . . This far-sighted centralising move is positionally correct and later shown to be tactically very strong. Kt—K1 ç I think 52 . . . P—K 4 is better, offering a more effective challenge in the centre. • 13 P—B 3 1(1-1( 4 14 P—Kt 3 P—R 51? Ingenious but not quite sound: however, after 14. . . P xP; Is P xP Black's pawn on K 2 is an unpleasant object. 15 BPxP R PxP n5 R PxP R—R 6 17 10-10 5! R x P z8 Kt—Q 4! R—Kt 3 And not 111 . . Kt xP?; 59 Q—R 21 winning (z9 R—Kt 5; 20 Kt—B 2 or ig R—B 6; ib Kt—Kt 5). • P—K 5 R—R 3 • PxP R—R Or 20 .. P x P; 21 ICt—K 61, B xKt; 22 P x B, Q x Q; 23 R x Q and Black's K P or Q Kt I'will fall. White's position is completely co-ordinsted—Black's

all over the place. The text threatens R x Kt; 22 Q x R, ics —B 6 c.h, but White stops this without being diverted from he own plans. 21 Kt—KS B x Kt • B PxB RxR ch

- 23 R x R P—Kt 3 Or 23 ... R—Q 5; 24 Q—R 59 threatening P—Q 6.

. 24 P—Q6! B PxP

25 PxQ P R—Q 5 Same idea again, which White now deals with effectively and entertainingly. 26 P! Q—Kz 26 . . . RxQ; 27 $ R x P; 28 P—K 7, R—K 3; 29 B—Q 5. 27 Q x RI . . . Allowing the famous combina- tion

Kt—B 6 ch . R x Kt BxQch 29 P—K 3 B—Kt

jo R-8 71 Resigns He has no defence against the threat of P—Q 7, except 30 Q x R, which leaves him • Piece

down.