10th Match Game, Candidates 1965
3 P-Q4 PxP
13 B xN P-QN4?! More active (and correct) would have been 13 .. P-K4! 14 P-N5 B-N5!, e.g. 15 Q-N2 P xB 16 P xN PxN 17 PxB PxP+ 18 K-Nl BxR 19P xR=Q+ Q xQ 20B-B4 B-R4 21 Q-R3 P-KN3 22 Q-Q7, when White should probably regain most of the sacrificed material but he cannot hope for any advantage— Nikitin.
15 . .. B-N2 would be too slow— White replies 16 P-QR3! and Black has little counterplay.
Assessments of the merit of this sacrifice have varied between ! ? (Nikitin), ! (Shamkovich) and !! (Chess Review). Three years after the game was played there was still some controversy over what should have been the correct result. Now it appears that Tal's idea qualifies for a half point at the very least but of course in a variation as complex as this it is necessary to take into consideration the immense practical difficulties facing the defending player.
Acceptance is obligatory because 16... B-Ql is refuted by 17 N-B6 + !. e.g. 17 ... PxN 18 PxP BxP 19 KR-N1 + K-Rl 20 P-K5 B-KN2 21 RxB! KxR 22 Q-N4+ K-Rl 23 R-Nl and mate next move.
The Q3-KR7 diagonal has now been opened up for White's bishop (and queen). The immediate threat is 18 Q-K4 winning a piece. Black cannot play 17 ... B-Ql because of 18 BxKRP +! KxB 19 Q-R5 + K-Nl 20 BxP! KxB 21 Q-R6 + K-Nl 22 P-N6 N-B3 23 KR-N1 B-B4 24 P-N7! winning.
Considerable controversy centred around the question of whether 17 ... P-N3 would have been a better defensive move as was claimed by Larsen after the game. In his original notes to the game in Shakh-matny Bulletin (number 6, 1966) Shamkovich 'refuted' 17 .. . P-N3. Almost two years later however, Nikitin (SB number 3, 1968) retaliated with some analysis that 'refuted'
the refutation. Not to be deterred, Shamkovich came back two months later in the same learned journal presenting the refutation of the refutation of the refutation. This last refutation was deemed to be a good thing and the correspondence was closed! Let us see what all the fuss was about.
After 17 ... P-N3, White has two likely looking continuations. A 18 P-KR4 B 18 QR-K1!
Tal's suggestion of 18 Q-R3 (to which he appended an exclamation mark) is met by 18 ... N-B3 19 Q-R6 N-R4, and if 20 B-K2 R-Kl! 21 BxN B-B1++, or 20 P-B5 BxBP 21 BxB R-K1+ + . A 18 P-KR4
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