By Mikhail Botvinnik
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R-B3 White ignores the possibility of 3 8 R-N4, which would have forced the retreat 3 8 . . R-B2. His primary concern is to ensure advance of his Q-side pawns. 38 K-Ql ? Polugayevsky plays the ending weakly. Now the White pawns advance without hindrance, and it is unlikely that Black can save the game. By 38 . . N-N3 39 R-N4 R-B2 followed by 40 . . R-K4, Black could have hindered his opponent's plan, and consolidated his position. 39 P-B4 K-B2 41 P-N5 40 P-QN4 N-N3 The adjournment analysis of this position can not have brought Black any comfort.
40 THE SEMI-FINAL MATCH 28 B-N3 As was shown by Tai, the active continuation 28 Q-Q 8 + K-N2 29 Q-B7 Q xP 30 Q x B N x NP 3 1 B-N3 N-K6 leads to a win for Black. 28 . K-N2 29 B-K2 Most probably, White has more chances of saving the game in an ending. Therefore, he should have offered the exchange of queens by 29 Q-Q3. 29 B-B3 ? , when he is a clear pawn up, and in no particular danger at all. Tai gives the following interesting variations: 30 Q-Q6 PxP 3 1 B-K5 Q-KB+ 32 B-B l Q-K6 + 33 K-Rl Q x BP+ 34 B-N2 Q-87 35 Q xN+ Q x Q 36 B xQ+ K x B 37 P-K5 + K x P 38 B xB P-BS; 30 P-R3 N x KP; 30 B-Kl Q-N7 3 1 P-R3 P xP 32 RP xP B-B3, and in all cases, Black should win.
16 . . The correct continuation was 16 . . Q-R4 ! 17 QR-B l P XP! , when White has only one saving move-19 B-Q3 ( 19 P-B4 is bad because of 19 . . R xB, while 19 P-N3 P-KN4 20 BxNP fails to 20 . . B X N) 19 . . P-B5 20 N-Nl Q x Q 21 B xQ, buc afrer 2 1 . N-K4 Black has at least an equal game. This line was suggested by a listener ac one of my lectures; unfortunately, I don't know his name. The above line confirms our conclusion that White has not played the opening too well. After the move played, the position becomes blocked.