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Karpov played on board six for the Armed Forces team, his 64/7 earned him a board prize and was also the absolute best result. Karpov's round-by-round results;

Karpov played on board six for the Armed Forces team, his 64/7 earned him a board prize and was also the absolute best result. Karpov's round-by-round results;

PR1 1 v. Lisenko

PR2 1 v. Gofstein

PR3 1 v. Markevsky

FR1 1 v. Steinberg

FR2 1 v. Peresipkin FR3 i v. A. Petrosian (FR4 = PR1) FR5 1 v. Veselovsky

1701 AK-MUchail Steinberg:

Sicilian

1 P-K4 P-QB4 2 N-KB3 P-Q3 3 P-Q.4 PxP 4 NxP N-KB3 5 N-QB3 P-K3 6 P-KN4 P-KR3 7 P-N5 PxP 8 BxP P-R3 9 Q-Q2 B-Q2 10 (MM) N-B3 11 P-KR4 Q-B2 12 B-K2 (MM) 13 P-B4 B-K2 14 P-R5 ± K-Nl 15 K-Nl P-Q416 P-K5N-K5 17N3xNPxN 18BxB NxB 19 Q-K3 N-B4 20 NxN PxN 21 KR-N1 KR-N1 22 P-N3?l 22 R-Q6 B-K3 23 Q-Q4 R-QB1 24 P-B3 ± 22 . . . B-K3 23 RxR+ QxR 24 R-Ql Q B2 25 R-Q6 P-KN3 26 PxP RxP 27 QrQ4 R-Nl = 28 K-N2 R-QB1 29 P-B3 R-Rl 30 P-R4 R-R7 31 Q-K3 Q-K2 32 R-Q2 B-Bl 33 Q-N6

Q-B2 34 Q-Q4 B-K3 35 Q K3 R-N7 36 Q Q4 R-R7 37 P-B4 R-R6 38B-Q1 (158)

39 P-B5± R-R5 40 P-N4 Q-B2 41 QrK3 R-Rl 42 Q.-QB3 R-QB1 43 R-Q.6 Q-K2 43 . . . P-R4! 44 K-R3 PxP+ 45 KxP Q-K2 46

K-R5! ± 44 Q-K3± R-Nl 45 B B2 Q-B2 46 Q-QB3 R-QB1 47 B-N3 BxB 48 KxB P-R4 49 Q-Q4 PxP 50 KxP Q-K2 51 Q-K3 R-B3 52 Q-Q2 K-R2 53 Q-Q4 Q-R5? 54 R xR Q-K8+ 55 K-B4 Q-K7+ 56 K-N3!±± PxR 57 Q-Q7 + K-Nl 58 Q-K8+ K-N2 59 QxKBP-f- K-Nl 60 Q-K8 + K-N2 61 Q-Q7+ K-Nl 62 Q-Q8 + K-N2 63 Q-N6 + K-Bl 64 Q x P+ K-Ql 65 Q-Q6+ K-Bl 66 P-B6 Q-B6 + 67 K-B4 Q-K7+ 68 K-Q5 1-0

1702 Peresipkin-AK:

FR2:

Sicilian

1 P-K4 P-QB4 2 N-KB3 P-K3 3 P-Q4 P x P 4 N x P N-QB3 5 N-N5 P-Q3 6 B-KB4 P-K4 7 B-K3 P-QR3 8 N5-B3 N-B3 9 B-KN5

Gufeld, in his notes to this game, in Informator 12 gives 9 B-QB4!? Against this Black should not play 9 . . . B-K2 10 N-Q5! NxN 11 BxN 0-0 12 N-B3± Fischer-Badilles, Manila 1967, but 9 ... N-QR4 10 B-Q5 B-K3 11 Q-Q3 R-B1 12 N-Q2 N-N5 and Black gains control of. . . QB5. 9 . . . B-K2 10 BxN BxB 11 B-B4± 11 N-Q5 is not so good: 11 . . . B-N4! 12 B-B4 0-0 13 0-0 B-K3 14 B-N3 B-R3 15 N1-B3 K-Rl 16 Q-Q3 R-Bl 17 QR-Q1 Q-R5 18 P-B3 N-Q5 and Black's counterplay was more than adequate compensation for the weakness of his Q4, Kapengut-Tal, 1965, or, in this, 13 N1-B3 B-K3 14 0-0 P-N4 15 B-N3 N-Q5 16 N-K2 NxB 17 RPxN Q-Nl = Kapen-gut-Furman, USSR Ch 1967. 11 ... N-R4 Or 11 ... O-O 12 O-O B-K3 13 B-N3 R-Bl ± - Gufeld 12 B-N3 0-0 12 . . . NxB 13 RPxB B-K3= is Fischer's opinion. 13 0-0 B-K3 14

N-Q5 B-N4 15 N1-B3 R-Bl 16 Q-K2 B-R3 17 KR-Q1 K-Rl 18 R-

Q3 N-B3 Gufeld assesses this position as equal. 19 Q-Ql N-Q5 20 N-K3 P-QN4 Kapengut-Furman would, almost certainly, have been discussed by Karpov with his trainer. Here 20 . . . Q-R5!? is Gufeld's suggestion. 21 N/B3-Q5 Q-R5 22 P-KB3 P-B4 23 PxP NxP/f5 24 N-KN4 R-B4 25 N-N6 25 P-B3 ! ? -Gufeld 25 . . .Q-K2 26 N xB If 26 BxB QxB 27 N-Q5 P-K5 28 PxP QxKP 29 NxB NxN 30 N-B3 Q-KR5 and White's K-side is probably untenable. 26 .. • P x N ^ (159)

27 B xB QxB 28 N-Q5 R-KN1 29 K-Rl P-KR4 30 Q-Q2 P-R5T

31 N-K3 P-R6 32 PxP No better is 32 P-KN3 P-K5 33 R-B3 P x P 34 NxN RxN 35 Q-Q4+ Q-K4T-

32 . . . N-Q5 33 R-KB1 QxKRP 34RxNPxR35QxP+ R-K4 36 QxP? 36 R-B2 R-N2 37 Q xP Q-K3 T would have prolonged the game. 36... QxRP+ 0-1

1703 AK-Arshak Petrosian:

FR3:

Alekhine

1 P-K4 N-KB3 2 P-K5 N-Q4 3 P-Q4 P-Q3 4 PxP BP xP 5 P-QB4 N-N3 6 P-KR3 P-KN3 7 B-

K2 B-N2 8 N-KB3 0-0 9 0-0 N-B3 10 N-B3 B-B4 11 B-B4 P-KR3! 12 B-K3 P-Q4 13 P-Q.N3 P xP 14 PxP R-Bl! Stronger than 14 . . . N-R4 of Karpov-Neckar. Now Black has fully sufficient counter-play - Suetin. 15 R-Bl N-R4 16 P-B5 N3-B5 17 B-B4 P-KN4 18 B-N3 Krogius, in an openings survey of the Team Championship in Shakhmatny Bulletin No. 10 1971, gives the continuation of the game as: 18 B-K3? Q-Q2 19 R-Kl P-N3 20 BxN NxB 21 Q-K2? PxP 22 P-Q5 BxN and Black has an extra pawn. 18 ... P-N3! 19 B xN N xB 20 Q-K2? P xPl (160)

21 P-Q5 or 21 QxN PxPiT 21... BxNT 22 RxB N-N3

1704 Veselovsky-AK:

FR5:

Sicilian

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