Chess: (Published 1975) (2025)

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Chess: (Published 1975) (1)

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March 11, 1975

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This is a digitized version of an article from The Times’s print archive, before the start of online publication in 1996. To preserve these articles as they originally appeared, The Times does not alter, edit or update them.

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No matter how alluring a tactical or strategical operation is, it must always be evaluated in terms of the time it takes to carry it out. Thus, a pinning operation may well succeed in its limited objective of producing double pawns in the enemy camp or even winning a pawn and still turn out a failure when the opponent is able to use the time to develop an attack.

There are no general rules telling when to embark on such an expedition; each concrete situation must be judged on its individual merits. Neither pessimism nor optimism, but diligence in ferreting out the opponent's resources, is mandatory.

In the game between Alexander Beljaysy, co‐winner with Mikhail Tal of the 42d championship of the Soviet Union, and Alburt, a surprising pawn sacrifice by White derailed an imposing pinning maneuver.

Potent Surprise

Alburt's reinforcing of his pin with 6 . . . N‐K5 and 7 . . . Q‐R4 seemed reasonable in that 8 N‐N5, P‐QR3; QxN, PxN; 10 B‐Q2, PxP; 11 QxBP, Q‐Q would yield Black a powerful game after 12 P‐K4, P‐Q4! Yet Beljaysky's 8 N‐B2! proved a remarkable resource that could not have been refuted by 8 ... BxNch; 9 PxB, QxPch; 10 QxQ, NxQ, since, after 11 B‐QN2, N‐R5; 12 BxP, R‐N1; 13 B‐R6, White's bishop pair will be more significant than his isolated QBP in the end game.

Alburt proceeded to set up the gain of a pawn by 9 ... NxRP, but, after Beljaysky's 12 B‐N2, White led in development while Black was bound to have trouble guarding his black squares, particularly Q3. Alburt could not play 12 . . . Q‐Q because 13 B‐R5, Q‐B4; 14 P‐QN4, Q‐K4; 15 P‐N5 shackles all three pieces on Black's queenside.

Perhaps Alburt should have daired 14 . . . NxPch; 15 K‐R1, N‐Q5 to force Beljavsky to make up a two‐pawn deficit. In any case, after 14 . . . P‐Q3, Beljaysky had everything his own way; Alburt could not defend his QP by 17 . . . R‐Q1 because of 18 P‐B5!! winning a piece no matter how Black plays.

When Beljaysky recovered his pawn by 21 RxP, his powerful bishops and tremendous lead in mobility already guaranteed him victory. Accordingly, he had no need to grab a pawn at the expense of bishops of opposite color by 23 BxN?, nor was it necessary to grant Alburt unnecessary chances. by 25 R‐Q7, BxR; 26 RxB, PxP; 27 RxQ, RxQ; 28 RxN, R‐R8; 29 K‐B1, R‐B1, etc.

Alburt still coud not develop by 29 . . . B‐K3, since 30 BxBch, RxB; 31 R‐Q7, Q‐B1; 32 Q‐R2, Q‐K1; 33 R‐Q8!, RxR; 34 RxR, QxR; 35 QxRch costs Black a piece.

Beljavksy launched the concluding attack with the breakthrough 31 P‐B6!, the point being that, after 33 R‐Q8, Alburt had to lose piece, since 33 . . . B‐K3; 34 RxRch, RxR; 35 BxB does not allow 35 ... RxB; 36 R‐Q8ch, and it is mate on the next move.

After Beljaysky's 37 B‐B8, Alburt tendered his somewhat overdue resignation.

ENGLISH OPENING

While Beljavsky

Black Alburt

White Beljavsky

Black Alburt

White Beljavsky

Black Alburt

1 P‐Q4

N‐KB3

14 O‐O

P‐Q3

27 P‐N5

P‐K4

2 P‐QB4

P‐B4

15 P‐K3

N‐N6

28 B‐N4

R‐K1

3 N‐KB3

PxP

16 B‐QB3

O‐O

29 B‐Q5ch

K‐R1

4 NxP

P‐K3

17 R‐Q1

N‐B4

30 P‐N6

Q‐Q2

5 N‐QB3

B‐N5

18 P‐QN4

N‐Q2

31 P‐B6

PxP

6 P‐KN3

N‐K5

19 R/2‐Q2

Q‐B2

32 B‐B4

Q‐N2

7 Q‐Q3

Q‐R4

20 Q‐R1

P‐B3

33 R‐Q8

B‐Q2

8 N‐B2

NxN

21 RxP

N‐N3

34 R/1xB

QxP

9 NxB

NxRP

22 P‐B5

N‐Q4

35 RxQR

RxR

10 RxN

QxNch

23 B‐K1

N‐K2

36 B‐Q3

N‐R3

11 B‐Q2

Q‐N3

24 Q‐R4

P‐QR4

37 B‐B8

Resigns

12 B‐N2

N‐B3

25 Q‐R3

N‐B4

13 Q‐R3

N‐Q5

26 R/6‐Q3

P‐R5

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Chess: (Published 1975) (2025)

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