Chess
The Guardian, Chess, Thursday, November 19, 1959, Greater London, England Problem No. 551 By the Rev. V. T. Ducker...
Posted by Bobby Fischer's True History on Friday, May 20, 2022
Problem No. 551 By the Rev. V. T. Ducker (Buxton)
Black (10)
White (10)
White mates two moves.
FEN 2K1N3/1np3p1/1qNpkBBb/3p2R1/6P1/1n1Pp3/2Q5/5R2 w - - 0 1
Solution: 1. Qc5 dxc5 2. Re5#
England v. Holland
Five losses, fifteen draws, and not a single win is an undistinguished result, but the English selectors can take comfort from the fact that most of the losses in the match against Holland came from the younger team members overreaching themselves through premature attacks. More disturbing was the inability of the experienced players to make any impression on the solid Dutch positional style, which is based on the teachings of their former world champion Dr. Euwe. Here is a game from the match which illustrates the power of the two bishops on an open board.
Jan Hein Donner (white) vs. Peter Hugh Clarke (black) Nimzo-Indian Defense: Normal Variation, Bernstein Defense
- 1. The latest idea in what has become one of the standard positions in this fashionable modern opening. The extra finesse of this move compared with the natural 11. B-Q3 is that after 11. B-N5 P-QR3; 12. Q-B3 P-K4; 13. Q-B2 R-K1; 14. PxKP NxP; 15. NxN QxN; 16. P-B3, followed by P-K4, Black's pawn formation is a little weaker than in the variations where the QRP will eventually require protection by … P-QN3, to which White can reply with the promising advance P-QR4-5.
Such is the theory behind 11. B-N5, but a game between Petrosian and Averbakh at Portoroz in 1958 casts doubt on the whole plan. It continued (after 18. P-B3) P-QN4; 17. R-K1 B-N2; 18. P-K4 P-B5; 19. B-B1 N-Q4; 20. B-Q2 N-B5; 21. B-K3 N-K3; 22. QR-Q1 N-B4 with a draw. White's two bishops are of little value with his centre pawns so immobile. - 2. This move is weaker than either 11. … P-QR3 or 11. … B-Q3, since Black's fianchettoed bishop cannot prevent the advance of White's KP.
- 3. White could already play 13. P-K4 here, since the combination 13. … PxP; 14. PxP NxP; 15. NxN QB6 (15. … P-K4; 16. Q-B2); 16. B-K3 P-K4 falls against 17. N-K2.
- 4. At this stage the central advance is a slight inaccuracy; White should prepare it by 14. Q-B2.
- 5. Black in turn fails to find the correct line. He can free himself by 14. … P-K4; for if then 15. BxN (15. P-Q5 N-K2; 16. P-B4 N-N3; and … N-K1-Q3), BxB; 16. NxP BxP and 17. N-N4 is met by 17. … NxN threatening the KRP.
- 6. Apparently Black has the makings of a promising king's side attack but White's next few moves emphasize that the vital factors are his pawn centre and greater control of the board.
- 7. Threatening 17. R-K4, followed by R-R4.
- 8. A forced pawn sacrifice, since otherwise White rolls up the centre by P-B4 and P-Q5.
- 9. Again White's hold on the centre nips any counterplay in the bud. He threatens 23. B-N5 while if 22. … NxP; 23. NxN QxN; 24. B-N2 Q-B4; 25. Q-Q2 K-N1 (otherwise 26. Q-R6ch): 26. Q-N5 R-Q2; 27. QR-B1 Q-K2; 28. Q-N6, with a winning attack.
- 10. Black is short of time and so facilitates the win. Rather better is the immediate 25. … Q-K1.
- 11. White now decides the game with a few powerful strokes which reveal the fatal weakness of the enemy king.
- 12. For if 32. … K-R2 White has the pleasant choice between 33. RxR NxQ; 34. R-R3 mate and 33. Q-N5.
Hastings congress
The definite entries for the Premier at Hastings so far are Golombek, Haygarth, and Winsor (England), Gligoric (Yugoslavia), Uhlmann (East Germany), Bobotsov (Bulgaria), and either Heemsoth or Rautenberg of West Germany. A reply is awaited from Anderson of Canada, the highest scorer on second board in the Munich team tournament. Petrosian may be the Russian entry. Other possible competitors are Pomar (Spain), Larsen (Denmark), and Feuerstein (United States).
PENROSE LIKELY TO KEEP CHESS TITLE
By our Chess Correspondent
Jonathan Penrose is almost certain of retaining his title of British chess champion. In the play-off tournament yesterday he beat Golombek in 57 moves. Penrose can now only be caught if Haygarth wins all his three games. However, since Haygarth has a probably lost adjourned ending against Golombek, Penrose is virtually certain of winning the title for the second year running. The scores are: Penrose 3, Golombek and Haygarth 0 and 1 adjourned.
Penrose opened with the King's Indian Defence in his second game against Golombek, who, tried to build up an attacking position on the queen's wing. Penrose maintained aggressive posts for his minor pieces, and then opened the centre and switched his attack to the king's side. Golombek had to sacrifice a pawn and transpose into an ending which Penrose won comfortably.
CHESS CHAMPION FOR SECOND YEAR
Penrose's convincing style
By our Chess Correspondent
Jonathan Penrose retained the British chess championship when his two opponents, Golombek and Haygarth agreed a draw in their adjourned game on Saturday morning after 47 moves. Final scores were: Penrose, 3 out of 3, Haygarth ½ out of 2, and Golombek ½ out of 3. The remaining games were abandoned.
Penrose's overwhelming victory, was achieved in a convincing style which leaves no doubt that he has taken over Alexander's mantle as our outstanding master. He is now expected to participate in next year's zonal tournament in Madrid; this will be the first stage in a series of eliminating events to find a challenger for the world title in 1963.