April 15 1934 Los Angeles Times Chess by Herman Steiner

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Chess by Herman Steiner 15 Apr 1934, Sun The Los Angeles Times (Los Angeles, California) Newspapers.com

The Los Angeles Times Chess by Herman Steiner Sunday April 15, 1934 Los Angeles, California L.A. Times Problem No. 454....

Posted by Bobby Fischer's True History on Tuesday, June 15, 2021

L.A. Times Problem No. 454. From the International Two-Mover Tourney of the Budapest Chess Club, 1933. Seventh and eighth honorable mention. By O. Nagy. White mates in two.
FEN 6bB/1P3Q2/K3pn1r/R5p1/3Nkpn1/1N3p1B/2PP4/6r1 w - - 0 1
Key: NxKP/Nxe6

The Los Angeles Times Chess by Herman Steiner Sunday April 15, 1934 Los Angeles, California L.A. Times Problem No. 455....

Posted by Bobby Fischer's True History on Tuesday, June 15, 2021

L.A. Times Problem No. 455. “La Liberte” by N. De Teretschenko. White mates in three.
FEN 6q1/2P1r2N/2R2b1r/1Q4p1/3pkB2/Np4P1/3R4/1KnB4 w - - 0 1
Key: B-N4/Bg4
The very strong tries, RxB? will be defeated by N-K7! or P-B8(N) by B-R! Bl. Additional variations, see April 29, 1934.

Game Department
There have been many inquiries about correspondence chess. We are therefore publishing a theoretically interesting game between H. Opsahl of Canada and A. G. Pearsall of Chula Vista, Cal. The game was played under the auspices of the Correspondence Chess League of America.

Haakon Opsahl, Canada (white) vs. Allen G. Pearsall, USA (black)
Slav Defense: Alapin Variation

Descriptive 1. P-Q4 P-Q4 2. N-KB3 N-KB3 3. P-B4 P-B3 4. N-B3 PxP 5. P-QR4 N-Q4 6. P-K4 NxN 7. PxN P-QN4 8. B-K2 P-K3 9. O-O B-N2 10. R-QN B-R3 11. Q-B2 N-Q2 12. PxP PxP 13. Q-R2 Q-B 14. N-K B-K2 15. P-B4 O-O 16. B-B3 N-N3 17. P-K5 R-N 18. B-K4 N-R5 19. Q-QB2 P-N3 20. R-B3 Q-Q2 21. R-R B-N2 22. BxB RxB 23. R-KR3 N-N3 24. Q-B2 P-B4 25. N-B2 N-Q4 26. B-Q2 B-Q 27. R-B3 P-QR4 28. R-N B-K2 29. N-K3 NxN 30. QxN P-N5 31. B-B R-QB 32. P-KN4 Q-B3 33. PxBP KPxP 34. R-B P-N6 35. B-N2 P-R5 36. Q-B R(B)-N 37. R-B2 Q-K5 38. R-R P-R6 39. RxP BxR 40. BxB R-R 41. B-N2 R-R7 42. Q-B Q-Q6 43. QxQ PxQ 44. K-N2 RxB 45. RxR P-Q7 46. R-N P-N7 47. K-B2 P-Q8(Q) 48. RxQ P-N8(Q) 49. RxQ RxR 50. K-K2 K-B2 51. K-Q3 K-K3 52. K-B4 P-N4 53. P-Q5ch K-Q2 54. PxP P-B5 55. K-K3 R-K8+! White resigns 0-1

Algebraic 1. d4 d5 2. Nf3 Nf6 3. c4 c6 4. Nc3 dxc4 5. a4 Nd5(a) 6. e4(b) Nxc3 7. bxc3 b5 8. Be2(c) e6 9. 0-0 Bb7(d) 10. Rb1 Ba6(e) 11. Qc2(f) Nd7(g) 12. axb5(h) cxb5 13. Qa2 Qc8 14. Ne1 Be7 15. f4 0-0 16. Bf3(i) Nb6 17. e5 Rb8 18. Be4 Na4 19. Qc2 g6 20. Rf3 Qd7 21. Ra1(j) Bb7 22. Bxb7 Rxb7 23. Rh3 Nb6 24. Qf2 f5 25. Nc2 Nd5 26. Bd2 Bd8 27. Rf3 a5(k) 28. Rb1 Be7 29. Ne3 Nxe3 30. Qxe3 b4 31. Bc1 Rc8(1) 32. g4(m) Qc6 33. gxf5 exf5 34. Rf1 b3 35. Bb2 a4 36. Qc1 Rcb8 37. Rf2 Qe4(n) 38. Ra1 a3 39. Rxa3(o) Bxa3 40. Bxa3 Ra8 41. Bb2 Ra2 42. Qf1 Qd3 43. Qxd3 cxd3 44. Kg2 Rxb2(p) 45. Rxb2 d2 46. Rb1 b2 47. Kf2 d1=Q 48. Rxd1 b1=Q 49. Rxb1 Rxb1 50. Ke2 Kf7 51. Kd3 Ke6 52. Kc4 g5 53. d5+ Kd7 54. fxg5 f4 55. Kd3 Rd1+ White resigns(q) 0-1

Annotation by Herman Steiner

(a) Fundamentally wrong! The object is to hold the gambit-pawn, but if White should play correctly, this cannot be done.
(b) This is just what Black wants. N-K5 would have regained the pawn with a good game, for example if 6 … NxN; 7. PxN, P-QN4?; 8. PxP, followed by Q-B3!
(c) N-K5 or B-R3 should have been considered.
(d) P-QR3 was necessary as is proved by White's next move.
(e) The only way to hold the gambit pawn.
(f) Too passive, N-K5 (threats NxQBP) would still be troublesome for Black, because of his inferior development.
(g) B-K2 would have been better. (See the following note:)
(h) This is bad because it gives up all chance to regain the gambit pawn. P-Q5 would have won back the pawn with a good game as follows: 12. P-Q5! KPxP; 13. KPxP, PxQP; 14. BxBP! etc.
(i) Why not P-B5, with possibilities?
(j) The strong-looking P-B5 would be defeated as shown: 21. P-B5? PxP; 22. BxP PxB; 23. B-R6 K-R!; 24. RxBP NxP; 25. BxR RxB; 26. R-R5 QxPch; 27. K-R, Q-K5!
(k) The correct way to win, the rest being a matter of technique.
(l) P-N6 would have been quicker.
(m) Desperation!
(n) P-R6 would have won the exchange as follows: 37. P-R6; 38. BxP P-N7; 39. BxP B-R6; 40. R-B2 R-N6; followed by Q-N2 or N4.
(o) Forced! The game should have been resigned at this point.
(p) The quickest way to victory.
(q) A very interesting game, which might have resulted differently if White had taken the many opportunities which occurred during this encounter.