This is the move which I had not considered. I thought that Rubinstein would have to play B - Kt 2, when I had in mind the following winning combination: 16 B - Kt 2, Kt - K 4 ! 17 Kt - B 4 (if R - B 1, Q × R !! Q × Q, B × P ch wins), Kt - Kt 5; 18 P - K R 3 (if Kt - R 3, B × P ch wins the exchange), Kt × P; 19 R × Kt, B × R ch; 20 K × B, P - K Kt 4, and Black should win. It is curious that this combination has been overlooked. It has been taken for granted that I did not see the 17th move Q - B 1.
16. ........
Q R - Q 1
After White's last move there was nothing for me to do but submit to the inevitable.
17. Q - B 1 !
P × Kt
18. Q × B
Q - Q 7
19. Q - Kt 5
Kt - Q 5
20. Q - Q 3
Q × Q
21. P × Q
K R - K 1
22. B - Kt 4
This gives Black a chance. He should have played K R - K 1. If then Kt - B 7; R × R ch, R × R; R - Q B 1, R - K 7; K - B 1, Kt - Q 5 (if R - Q 7; B - K 6 ch, K - B 1; B × P would win); R - B 8 ch, K - B 2; R - B 7 ch, R - K 2; R - B 5 wins.
