CHESS by Phitidor

No. 303 J. HARTONG

(Good Companions,

1920)

WHITE to play and mate in two moves ; solution next week. Solution to No. 302 (Barry) : Q - Kt 4 !. threat Q4. K Kt 4 ; 2 Kt - B 7. . . . K - Q 4; 2 Kr- Kr 5. I ... K - Q 2; 2 Kt x Kt P. Very elegant.

wurrE

The outstanding (6 men)

features of this year's Piatagorsky tournament at Los Angeles were, first, the very poor performance of the world champion, Petrosian, who only succeeded in scoring 50 per cent (3W, 3L, raD), being badly beaten twice by Larsen, and, second, the wonderful recovery

in the second half of the tournament by Bobby Fischer; in the first half Fischer scored 3f out of 9 (07, 3L, 5D)—in the second he scored 71 (6W, 3D) and finished second, half a point behind Spassky (HD; and a point ahead of Larsen (ro). Here is the start of the Fischer landslide.

White, FISCHER Black, Ivicov Opening, StaLtax

(Round ro, Los Angeles, 1966)

I P-K 4 P -QB 4 2 Kt-K B3 P-K3 3 P-Q. — * It is interesting to see Fischer play a close form of the Sicilian; perhaps he chose it because after 3 P -9 4, P XP; 4 Kt XP, Kt -K B 3 one can easily get into a drawish variation.

- Q B3

4 P-5(I5 QKt-Qa B- Q3 03 P-34t 6 B-Kta KKt-K2 This formation (B on Q 3, Kt on

K 2) may be all right—it is too difficult for me to judge adequately--but B on K st, Kt on B3 looks sounder; it puts more pressure on the White centre and is better defensively.

7 0 - 0 0 - 0

8 Kt-Rat P-QKt3 P-KB4 PxP zo PxP B-R3

zz R-Kz P-Bs And I don't at all like this—it Is too optimistic a move. Black cannot occupy Q 6, and his advanced pawnis only a liability. If then 12 P-B 3(12 Ch-R 5, Kt -Q5), Kt - Kt 3 and although White stands better, Black has better defensive chances than in the game.

12 P-B 3 Kt-R4 13 P-K 5 B-B ch 14 K-Rz Kt-Q 14...B- Kt 22; is B x B, Kt x B;

16 Kt XP.

15 Kt- K 4 B - Kt 1 z6 Q-Rs Kt-Ka . . . 13- K 2 is better. 17 P-KKI4 BxKt If this is best, Black must be lost— and I think this may well be the case. White threatens P - B 5 and if 17 . . Q - B 2 tO St051 this, then i8 Kt - B 3 is very strong. But I would have tried Q - B 2 rather than B x Kt.

r8 BxB P-K13

19 Q-R6 Kr Q 4 20 P-B5 R- Kr Otherwise P - B 6 wins.

sz B PxKt P B PxP 22 la XP . Obvious and decisive.

Q2 22 ...Pxict; 2.2QxP ch,K - B 1; 24R-B z ch,x- K2; 25 R - B 7 17121e.

23 Kt-B QR-Qz

24 Kt-R5 K-R r 25 Kt - B 6 Kt x Kt 26 PXKt R-K Kt r 27 B-B RXP

28 QR-Qz QR-KKtz 28 QxR; 29 QxP mate.

Now, of course, 29 R xf,17? is met by 29. . . R - Kt & ch, 3o RXR. R xR mate. 29 P- B 7! Resigns 29 Q x P; 30 B - K 5 ch, R (9) - Kt 2; 3t Q X P mate.