Chess

By PHILIDOR No. 283 W. I. KENNARD (Boston Transcript,

1957).

WHITE to play and mate in two moves; solution next week. Solution to No. 282 (Anderson): P—

B 4 1, no threat.

• . . . Kt—B 3 or Kt—Kt 6 ; 2 K- B 31, PxB=Kt mate. I. . . K— Kt 2 ; 2 Kt—Q 3, PxB=Q. BLACK (7 men) Kt x R ; 2 K—Q 4 !, P x B= Q. Last variation replaces the set play (W P on B 2). x . . Kt x R ; Z K—Q 2, P x B= Q. Beautiful and ingenious problem.

World championship chess is very much a psycho- logical as well as a technical struggle : this shows very clearly in the seventh game—the first six having been drawn—of the Petrosian/Spassky title match. Spassky, normally a free attacking player given to King's Pawn games, is quite unrecognisable and seems to have hypnotised by Petrosian into playing, very

White, SPASSKY Black, PETROSIAN Opening, QUEEN'S PAWN (Moscow, April 1966)

1 P—Q 4 Kt—K B 3 2 ICI—K B 3 . . . Much more passive than usual for him.

— P—K3

3 1—Kt 5 . . . A line which is (rightly) little played. P—Q 4 4 QKI— B—K 2 5 P—K 3O Q Kr—Q a 6 B—Q 3 P—B 4 Black already has the initiative. 7 P—B 3 P—Q Kt 3

8 0-0 2

9 Kt—K g? . . . This would be fine if Black had castled king's, as after the exchange White would have good king's side attacking chances; as, however, he hasn't, it merely weakens White's pawns and his own king's side. I would have thought it better to try to force P—K 4, though I prefer Black's game.

Kt xKg ro P x Kt Kt—Q 2

sr 13—K B 4 . — If zr BxB, QxB; 12 P—K B 4, Black can castle queen's side and disrupt the White position by P—B 3 andjor PB —K Kt 4.

Q— a

is Kt—B 3 P—K R 3! The combined attack on the king's side and on the weak king pawn is probably decisive; incredible that a really great player like Spassky vivid get a lost position in such style in 12 moves (less—he may have been lost after his ninth) with the White pieces. 13 P—Q Kg 4 . . . He has no good move. P—K Kt 4 P—K R 41 15 P—K R 4 • • • If t5 P—K R 3, Black will force the position at his leisure with P—Kt 5—probably playing 0-0--0 and Q R—Kt t first. .73 . — Kt P x PI Even stronger than P—Kt 5. the doubled pawns are not weak, as Petrosian convincingly shows:

indifferently, a version of Petrosian's own tortuous style.

:6 B—K B 4 0-0-0 17 P—R 4 P—B 5

18 B—K 2 P—R 31 Now he has no difficulty in blocking the queen's side and can proceed undisturbed with his own attack.

zg K—R z Q R—Kt r

TO R—K Kt r R—Kt 5

continuation I haven't fathomed.

aa P—R g P—Kt 4

R (R r)—Kt r sot: at on c e l o o ls

better unless (as is quite likely) Black has looks

K—

sr Q—Q 2 . ..

23 Q R—Q r B—B r

24 Kt—R 2? . . . Too late: now the exchange sacrifice wins easily. 24 Q—Q 4 looks best, though after 24 . • • B—Kt 2 or R 3 I think Petrosian would have had enough pressure to win.

4 Kt x PI 30 R—Q 2

2.22 946 RP--. _42— K 44

P •

28 B xB PP —x BKt 41

—K Kt—Q 2 B—Q 3 s Q—K 3 QxB . 25 Kt x R

3z P x9_12 P—B 32 VC 4 Kt-0 3 33 B 5 ch K—Kt r B—B r 333 5 46RQ P----BKKg3zr P—Kt 6 dzil B—B r P—R 6 R—R r

Petrosian plays a not yen/ t attack most incisively.

38 PxP B x P 39 K—Kt r B xB P—K gl

4; KQ_Qx B 1 Kg—Kt 5! 41 P x P?, P—B 6. . . . rapidly decisive (or 42 P x P, P—B 61°).therwise (2.--R 3 will be fa P x Kt . . .

P—B 6

43 R—K Kt 2 . . . Of 43 K—Kt r, Q—R 3. 44 Resign& P. .X .R ch K--Kt 1, Q—B 5; 46 44 K x P, R—R 7 ch; 45 play badly! 4, Q—B 7 chi; 47 QxQ,PxQ ch;Queer how often Petrotuan's opponents

48 K x R, P x 11..=Q.