A
Chess 23 Sep 1934, Sun Buffalo Courier Express (Buffalo, New York) Newspapers.com
OCR Text
Chess Problem.
Attack la ine White Horowits White 16 KtKt3 17 PxP PxB Bx R-K2 20 Kt-B 21 Kt-K3 22 0-03 23 Kt-B3 34 R-K3 25K-R 26 R-B3 37 K-Kt 38 BxKt Resigns Q-R4 Kt-B2 P-B5 Q-R5 Kt-B3 RxP R-Kt4ch K-K PxR Res lens RKt4-Kt3 7BxP 8 Castles 9Q-K3 I0R-Q 11 P-K4 12 K-Kt3 13 P-R3 14Q-B 15B-K3 16 P-KU 17 P-K15 18 P-R4 19 P-Q5 ine Black P-KR4 B-Q3 Kt-Kt4 K-B3(a R-R Q-R5 Kt-R4 R-R3 R-K13 R-R Kt-Kt6ch Kt-R8ch Kt-K7eh PXP B-B4 Kt-R3 Kt-K P-K3 B-K3 CMtlea B-Kt5 Q-R4 BxKt QR-R-Q2 KR-Q Kt-K P-QB4 P-K4 : ine White 1 P-O4 2-QB4 P-QB3 3 KI-QB3 Kt-B3 !
In problem No 90 we again return to the two-mover and trust that it may bring a good response from the solvers.
It is by C S Kipping of England Black— 7 pieces ine Horowitz White Black 20 P-R5 B-Q3 21 K-Kt3 P-QR3 33 R-R Kt-Q6 33 P-KU P-B5 34 B-B3 Kt-B5eh zp nxs4 36 P-K5 ST R-nA 38 PxBPcbKxP 29QR-K BxKt 30 B-K6chK-B 31 PXB 33 R-R4 33 RxP 34K-B 35RxKt 36 QxBP 37 P-R6 White— 7 pieces White to move end mate in lo The principal idea behind last week’s end-game was to endeavor to trade off the two Black pieces which could give mate the queen and the pawn leaving Black with the two knights which cannot win.
Therefore the play is as follows: 1— BxPch QxB 3— R-Q4ch K-K4 3— R-K4ch KxR 4— Kt-B5eh loalns qtieen drew 1— BxPch QxB: 3— R-Q4ch K-B3 3— B-B4ch QxR: 4— Kt-R5 come result 1— BxPch QxB: 3— R-Q4ch K-B3 3— R-B4ch KtxR 4— Kt-Q4ch and came result Correct solution to chess problem No 89 was received from W C Daly Herkimer N Y and Harold Armstrong Niagara Falls.
Two Match Games Divided Here are the first two games of the match between Reuben ine and I A Horowits both of New York and played in Philadelphia Dutch Defense P-KB4 Kt-KB3 I B-K3 CfeStlM p3 Kt-B3 P-K4 3 P-KKU P-K3 5 Castles ?P-B4 Kt-B3 IQ-B3 SP-QR3 P-Q5 II P-QKt4 12Kt-Q3 13P-B3 14 P-K4 15 P-K14 (a1 — The bevlnnln? of an which Black times his moves with ex cellent Judgment Queen’s Gambit Declined Horowits Black P-Q4
Attack la ine White Horowits White 16 KtKt3 17 PxP PxB Bx R-K2 20 Kt-B 21 Kt-K3 22 0-03 23 Kt-B3 34 R-K3 25K-R 26 R-B3 37 K-Kt 38 BxKt Resigns Q-R4 Kt-B2 P-B5 Q-R5 Kt-B3 RxP R-Kt4ch K-K PxR Res lens RKt4-Kt3 7BxP 8 Castles 9Q-K3 I0R-Q 11 P-K4 12 K-Kt3 13 P-R3 14Q-B 15B-K3 16 P-KU 17 P-K15 18 P-R4 19 P-Q5 ine Black P-KR4 B-Q3 Kt-Kt4 K-B3(a R-R Q-R5 Kt-R4 R-R3 R-K13 R-R Kt-Kt6ch Kt-R8ch Kt-K7eh PXP B-B4 Kt-R3 Kt-K P-K3 B-K3 CMtlea B-Kt5 Q-R4 BxKt QR-R-Q2 KR-Q Kt-K P-QB4 P-K4 : ine White 1 P-O4 2-QB4 P-QB3 3 KI-QB3 Kt-B3 !
In problem No 90 we again return to the two-mover and trust that it may bring a good response from the solvers.
It is by C S Kipping of England Black— 7 pieces ine Horowitz White Black 20 P-R5 B-Q3 21 K-Kt3 P-QR3 33 R-R Kt-Q6 33 P-KU P-B5 34 B-B3 Kt-B5eh zp nxs4 36 P-K5 ST R-nA 38 PxBPcbKxP 29QR-K BxKt 30 B-K6chK-B 31 PXB 33 R-R4 33 RxP 34K-B 35RxKt 36 QxBP 37 P-R6 White— 7 pieces White to move end mate in lo The principal idea behind last week’s end-game was to endeavor to trade off the two Black pieces which could give mate the queen and the pawn leaving Black with the two knights which cannot win.
Therefore the play is as follows: 1— BxPch QxB 3— R-Q4ch K-K4 3— R-K4ch KxR 4— Kt-B5eh loalns qtieen drew 1— BxPch QxB: 3— R-Q4ch K-B3 3— B-B4ch QxR: 4— Kt-R5 come result 1— BxPch QxB: 3— R-Q4ch K-B3 3— R-B4ch KtxR 4— Kt-Q4ch and came result Correct solution to chess problem No 89 was received from W C Daly Herkimer N Y and Harold Armstrong Niagara Falls.
Two Match Games Divided Here are the first two games of the match between Reuben ine and I A Horowits both of New York and played in Philadelphia Dutch Defense P-KB4 Kt-KB3 I B-K3 CfeStlM p3 Kt-B3 P-K4 3 P-KKU P-K3 5 Castles ?P-B4 Kt-B3 IQ-B3 SP-QR3 P-Q5 II P-QKt4 12Kt-Q3 13P-B3 14 P-K4 15 P-K14 (a1 — The bevlnnln? of an which Black times his moves with ex cellent Judgment Queen’s Gambit Declined Horowits Black P-Q4
