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David Pritchard Maximummer: 1 Bxa5 Rh8 2 Bd8 Rh1 3 Bh4 Ra1 4 Be1 Rxa6. The rook tour is expected, that of the bishop perhaps is not. Stalemate. It seems natural to drive the Black king towards the edge, but in fact he must be piloted the other way: 1 Ne1+ Ke3 2 Bc2 Kd2 3 Kf2! White, far from being able to profit from Black's inability to reach the squares "off the board", needs to make use of one of them himself.

Augsberg Chess (A)  1.Rc6 2.(R+S)e7 3.(R+B+S)e3 or    1.Bd7 2.(R+B)c6 3.(R+B+S)f3 or    1.Se7 2.(B+S)d6 3.(R+B+S)g3. (B) set: 1....(R+B)c2† 2.Kh1 Be4 play: 1.Bh2 (R+B)g6† 2.Kh1 Be4 or         1.Bf2 Be4† 2.Kg1 Rh1. (C)  1.e2 e7 2.e1=(B+B) e8(R+B) 3.(B+B)c3 Re2 4.Bd4 Bb5. (D)  (a) 1.Bc6 d8=(B+S) 2.(B+S)b5 (B+S)b6.        (b) 1.Sb4 d8=(R+S) 2.(R+S)b5 (R+S)c6. (E) (a) 1.Kh8 2.f5 3.(S+P)e7 4.(B+S+P)g6 (R+B)ืg6.       (b) 1.Kh7 2.Se7 3.(B+S)g8 4.Sf6 Qืf6. (F) 5.b1=(B+B+S+S) f8=(B+B+S+S)       6.(B+B+S+S)h7 (B+S)ืh7 =.

Spirits of the Knight A, A', B: In text. C: 1 Bืd6$; 2 $Bf7; 3 Ba2$c1; 4 $Bf4; 5 Bg6$e7; 6 gBb1; 7 e5$c6 $Bืe5‡. D: 1 bืa6$ b4$; 2 d5$ f4$; 3 Kh5$ Bd1‡ (3 ... Be8?; 4 $Kg7!). E: 1 Sc6† Kd5; 2 Sd6 Kืc6$; 3 b5† Kd5$f4‡. 2 ... Kืd6$; 3 (b5$??) h7! Kf7$g5‡. F. Black in irreal check unless wBf1 already inspired, so: 1 Kg1 Bg3$e2; Kh1 Kf1$g3‡ G: 1 Sb5! (2 B~) Bืb5$/Cืb5$ 2 Bf6/Be7‡. H: 1 Kf3 Ne2; 2 Kืe2$ Kb4; 3 $Kh8 Kb5$‡. I: If wK inspired, 1 Kf1 $Kd3; 2 Ke1$† Ke3=. If bK inspired, 1 $Ke2 Kc3$†; 2 Kd1$† Kd3=.

Eureka Problem Solutions: (A) “Gitterschach” is Grid Chess. (R Queck, Problemkiste , 1990). And don't blame me for not showing the grid - Popeye didn't display it either! (B) Circe. (J C van Gool, The Problemist , 1984) (C) Camelrider. (Stefan Klebes, Feenschach , 1986) (D) It isn't mate! White may have to move straight forwards or backwards next move, but then so must Black; he cannot play Sc2×e1, and so White is not even in check at the moment!

Solutions to ‘Two's Company' (A) Set 1...Nh4 2.h1B Ng2=; 1.h1B Nh4 2.Ba8 Nb7=. (B) (a) 1...b4 2.Kc7 b5 3.Kd6 b6 4.Kc5 b7 5.Kb4 b8Q† 6.Ka3 Qb3‡      (b) 1.Kd7 b4 2.Kc6 b5† 3.Kb5(Pb7) b8Q† 4.Ka6 Qb6‡      (c) 1.Kc7 b4 2.Kb6 b5 3.Ka5 b6 4.Kb6(Pc7) c8Q 5.Ka7 Qb7‡      (d) 1...b3 2.Kb3(Pf7) f8Q 3.Ka4 Qb4. (C) 1.Ra4 = wS Sc5 = bR 2.Rh5 = wB Bd1=bQ‡ (D) 1.Kf6(B) Kd7(Q) 2.Bb2 (S) Qd4‡ 1.Kf5(B) Kd8(Q) 2.Bb1 (S) Qd3‡ (E) (a) Retract Kb6×Q 1.Kc6 2.Kb5 3.Kc4 4.Kd5 5.Ke5 Qd4‡      (b) Retract Kf4×Q 1.Kg4 2.Kf3 3.Kg2 4.Kf1 5.Kg1 Qa1‡ (F) (a) 1.Kc7 c3 2.Kb8 c4 3.Ka7 c5 4.Ka8 c6 =/=.      (b) 1.Kd7 c3 2.Kc8 c4 3.Kb7 c5 4.Ka7 c6 5.Ka8 c7 =/=. (G) Set 1...h3 2.Cg4 h2 3.Ch1 =; 1.Ce3C h3 2.Ce2 h2 3.Ch1 =. (H) (a) 1.Sg5 2.Sf3 3.Sg1 Ke3 =/= (b)1.Sf3 2.Sg5 3.Sh3 Ke3 =/= (I) Partial Retro-Analysis: if BK entered from b1 1.Qb1; if from b2 1.Qb2; and if from a2 then 1.Qc1, not 1.Qa2? Kb2! - a trap for the unwary. (J) 1.Kh6 Kh8 2.Kh7 Kg8 3.Kh8 Kh7 4.Kg8 Kh6 ... 9.Kb8 Kg1 10.Ka7 ... 15.Ka2 Ka1 16.Kb1 Ka2 17.Kc1 Kb1=. (K) 1.Kf4 e3† 2.Kf5 e4† 3.Kf6 e5† 4.Kf7 e6† 5.Kf8 e7† 6.Kf7 e8Q† 7.Ke8(Qd1) Qa4† 8.Kd8 Qd7‡ (L) 5.h8R 6.Rb8 7.Rb2=P 12.b8Q 13.Qb1‡

SOLUTIONS to Progressive Fischer Random: (3) Santoni v Pfeiffer: 6. Re6, Rd6, Rd3, R×h3, R×h2, R×h1‡ (4) Honkela v Hansen: 5.Kd3, Nd5, Nf4, Rf1, Ne6‡ (5) Blazsik v Cassano: 4. f×e4, e3, Qf7, Q×f2† 5. Q×f2† 6. Kg8, Bh4, B×g3, B×h2, Rf8, R×f2‡ (also mates with 6. e×f2 etc) (6) Sasata v Castelli: 8. e4, e3, e×d2, d1Q, Kg6, Bf4, Bd2, Qc1‡

PUZZLE Answers for Space article 1: (1, -1, 2) + (-1, 2, -1) = (0, 1, 1). 2: (2, -1, 2) + (1, 2, -2) + (-3, 0, 0) = (0, 1, 0).

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