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StrateGems

No. 49, Jan-Mar 2010

R0154 - Kevin Begley

StrateGems 49, Jan-Mar 2010

[n3k2r/8/8/8/8/8/8/R3K(RH)2]

3+3. h#3.5 (3 solutions: 3.1.1...) (AP) (Circe Parrain) (Rosehopper f1)

[n3k2r/8/8/8/8/8/8/R3K(RH)2]

Solution

There is no solution possible without some immediate rebirth into the diagram. In Circe Parrain, no previous capture may be assumed (no unit awaits rebirth) without proof by retrograde analysis. White's castling rights are dependent upon black having previously captured a white unit. In order to prove a rebirth, A.P. must be applied to mutual castling rights - long known possible in orthodox problems (see diagram A) - to restrict black's last capture on the a8 square (leaving only 5 retractions to be analyzed as retro-variants). Two of these retractions (-1.SxQa8 and -1.SxRa8) lead to "self-invalidating" rebirths: e.g. 1...0-0-0(+wQ/wRb8)+ prevents black from ever castling.

This leaves three unique retro-variants to be considered, each with a unique solution:

i) 1…0-0-0(+wSb8)! 2.0-0! Sc6 3.Kf7 Rd7+ 4.Ke8 Re7# (2.Rh6? Sc6 3.Rf6 Rd7 4.Rf8 Re7#, but fails to validate rebirth)

ii) 1…0-0-0(+wBb8)! 2.0-0! Be5 3.Rf7 Bf6 4.Kf8 Rd8# (2.Rh7? Be5 3.Rf7 Bf6 4.Kf8 Rd8#, but fails to validate rebirth)

iii) 1…0-0-0(+wRHb8)! 2.0-0! Rd7 3.Kh8 RHe5 4.Rg8 Rf7# (2.Rg8? Rd6 3.Sc7 RHb5 4.Kf8 Rd7#, but fails to validate rebirth)

A.Nikita Plaksin & Valery Liskovets Die Schwalbe, 1986

[r3kS2/p4pp1/3B1pp1/2PP4/2PpP2B/1P1B1S2/1p2P3/n3K2R]

13+10. h#2 A.P.

1.dxc3 e.p. 0-0! 2.0-0-0! Ba6#. If both sides can castle, white's last move must have been -1.c2-c4, thus, by A.P. reasoning, 1.dxc3 e.p. is legal, providing both sides castle in the solution.


R0155 - Klaus Wenda

StrateGems 49, Jan-Mar 2010

[k7/1qP2p2/4PpB1/6Br/p5PP/1P5b/6R1/4K3]

9+7. -12 & #1, Proca retractor, Anticirce

[k7/1qP2p2/4PpB1/6Br/p5PP/1P5b/6R1/4K3]

Solution

Retract: -1.Kh7xBg8(Ke1)! Qc8-b7+ -2.Bf5-g6 Rh6-h5 -3.Bg6-f5 Rh5-h6 -4.Bf5-g6 Rh6-h5 -5.Bg6-f5 a5-a4 [forced by Zugzwang to avoid the 3rd repetition of the position] -6.b2-b3! [threatens -7.c5xb6(Pb2) e.p. b7-b5 -8.Ra2-g2 & 1.Rxa5(Ra1)#] a6-a5 -7.Bf4-g5 Rh5-h6 -8.Bg5-f4 Rh6-h5 -9.Bf4-g5 Rh5-h6 -10.Bg5-f4 a7-a6 [forced again by Zugzwang] -11.c5xb6(Pb2) e.p. b7-b5 -12.Ra2-g2 & 1.Rxa7(Ra1)#

The goal of the 2 draw pendulums is the decoy of pawn a4 to a7, where it can be captured. Both pendulums work without self-check. White moves like -2.Be4-g6? or -7.Be3-g5? would be refuted by -2...,g6xf5(Pf7) or -7.g5xf4(Pf7). The comparison with Dittmann's R0153 (SG47) might be interesting for the solvers, as well as for the judge (A).


R0156 - Nikolai I. Beluhov

StrateGems 49, Jan-Mar 2010

[8/B2p2pp/ppp1p3/5p2/4PPPP/3PRKq1/2PR1NrN/3bbBk1]

13+13. Add the necessary pieces and Release the position

[8/B2p2pp/ppp1p3/5p2/4PPPP/3PRKq1/2PR1NrN/3bbBk1]

Solution

Add wQe2. Retract: -1.Qh3xSg3+ Sh1-f2 -2.Bf2-e1 Bb8-a7 -3.Be1-f2 Sh5-g3 -4.Bg3-e1+ Qe1-e2 -5.Re2-g2+ Bg2xSf1 -6.Bf2xSg3 Sf6-h5 -7.a7-a6 Sh5-g3 -8.Sg3-f1+ Sf1-h2+ -9.Qh2-h3 (The captures in the problem are completely determined. The two remaining captures are a6xRb7 and b6xSc7). This is a Chinese box type of problem. Presented is the change of parity theme, and multiple uncaptures of knights, which then help an empty square enter the box along the path f6-h5-g3-f1-h2 in order to unlock it.(A)

Unfortunately, a cook was reported by Ryan McCracken: Add bSe2, retract -1.Qh3xSg3+ Sh1-f2 -2.Sc1xSe2 Sc3-e2 -3.Re2-g2+.Sf2-h1, then Bg2, Bh1, Qf1, h3, Sh5, Kg3 etc. The composer offers the simple correction: +wQe2 (14+13): The new stipulation is: Release the position


R0157 - Nikolai I. Beluhov

StrateGems 49, Jan-Mar 2010

[3bQrRq/2ppNBBQ/1nPp1Knq/p1PP1p1k/7p/5P2/5PPP/8]

14+13. Release the position

[3bQrRq/2ppNBBQ/1nPp1Knq/p1PP1p1k/7p/5P2/5PPP/8]

Solution

Retract: -1.Sa4-b6 c4-c5 -2.Sc5-a4 c3-c4 -3.Se6-c5 c5-c6 -4.Qc1-h6 Bh6-g7+ -5.Sg7-e6+ c4-c5 -6.Qb1-c1 c2-c3 -7.b2-b1Q c3-c4 -8.b3-b2 b2xRc3 -9.Re3-c3 Sc6-e7 -10.Re7-e3+ Se5-c6 -11.Sf4-g6 Sg6-e5+ -12.Sh3-f4 d4-d5 -13.Sg5-h3 Bc4-f7 -14.Sf7-g5+ Se5-g6 -15.g6xRf5 Rf4-f5+ -16.b4-b3 Re4-f4 -17.b5-b4 Re1-e4 -18.b6-b5 Rh1-e1 -19.a6-a5 Bf1-c4 -20.a7-a6 e2xPf3

Multiple unchecks (A).


R0158 - Nikolai I. Beluhov

StrateGems 49, Jan-Mar 2010

[6Nn/p3pprR/P5pr/3p2Pk/2p4p/3PnP1K/1PPRBPP1/3bNQq1]

15+14. Last 11 single moves?

[6Nn/p3pprR/P5pr/3p2Pk/2p4p/3PnP1K/1PPRBPP1/3bNQq1]

Solution

Retract: -1.Sg4-e3 a5-a6 -2.Sh2-g4 a4-a5 -3.Qh1-g1 Qg1-f1 -4.c5-c4 Bf1-e2 -5.Be2-d1 Rd1-d2 -6.c6-c5. But why should the black c and d pawns retract these exact moves? The answer is given as the retractions go further: -6…Ra1-d1 -7.Bd1-e2 Be2-f1 -8.c7-c6 Qf1-g1 -9.Qg1-h1 Rc1-a1 -10.Sg4-h2 Rb1-c1 -11.Se5-g4 Rc1-b1 -12.Sc4-e5 Rb1-c1 -13.Sd2-c4 Rc1-b1 -14.Sb3-d2 Rb1-c1 -15.Sa1-b3 Rc1-b1 -16.a2-a1S Rb1-c1 -17.a3-a2 Rc1-b1 -18.b4xBa3 Rb1-c1 -19.b5-b4 Bd6-a3 -20.b6-b5 Bh2-d6 -21.Qh1-g1 Bf4-h2 -22.Qg1-h1+ Bc1-f4 -23.d6-d5 d2-d3 -24.b7-b6 Ba6-e2 -25.Be2-d1. Any other retraction by the black pawns would not allow the move -19...Bd6, making it one move too slow.

A Chinese box is formed in the South-East cage. (A)


R0159 - Dragan Lj. Petrovic

StrateGems 49, Jan-Mar 2010

[8/pp3P2/p7/RP3P2/QPPnkP2/r1Kppp1P/pB1rp1bR/bqBN1NnB]

16+16. Release the position (Circe)

[8/pp3P2/p7/RP3P2/QPPnkP2/r1Kppp1P/pB1rp1bR/bqBN1NnB]

Solution

Retract: -1.Sb3xBd4(+Bc1)+ Bc1-b2 -2.Rb2-d2+ Bb6-d4 -3.Rd2xBb2(+Bc1) Bc1-b2 -4.Rb2-d2+ Bd8xPb6(+Pb7) .... -6....Bh4-d8 .... -8....Be1-h4 .... -10....Bd2-e1 -11.c7xBb6(+Bc1) Bc1-d2 .... -13....Bc5-b6 .... -15....Be7xPc5(+Pc7) .... -17.Bh4-e7 .... -19.Be1-h4 .... -21....Bd2-e1 -22.d6xBc5(+Bc1) Bc1-d2 .... -24....Bd4-c5 -25.d7-d6 Bf6xPd4(+Pd7) -26.d5-d4+ Bh4-f6 -27.d6-d5 Be1-h4 .... -29....Bd2-e1 -30.e7xBd6(+Bc1) Bc1-d2 .... -32....Be5-d6 .... -34....Bf6xPe5(+Pe7) .... -36....Bh4-f6 .... -38....Be1-h4 .... -40....Bd2-e1 -41.f6xBe5(+Bc1) Bc1-d2 .... -43....Bd6-e5 .... -45....Bf8-d6 .... -47....Bh6-f8 .... -49....Bg5-h6 -50.g7xBf6(+Bc1) Bh4xPg5(+Pg7) -51.g6-g5 Bg3-h4 -52.g7-g6 Be1xPg3(+Pg7) -53.g4-g3 Bd2-e1 -54.g5-g4 Bc1-d2 .... -56....Bh8-f6 .... -58....h7-h8B .... -60....h6-h7 .... -62....h5-h6 -63.h6xBg5(+Bc1) Bh4-g5 64.h7-h6 g4xPh5(+Ph7) -65.h7-h5 Be1xPh4(+Ph7) -66.h5-h4 Bd2-e1 -67.h6-h5 Bc1-d2 -68.Rd2xBb2(+Bc1) Bc1-b2 -69.Rb2-d2+ g3-g4 -70.g7xBh6(+Bc1) f2xPg3(+Pg7) -71.g4xPf3(+Pf2) f2-f3+ etc.

For economy, the repeated mechanism -(n).Rd2xBb2(+Bc1) Bc1-b2 -(n+1).Rb2-d2+ is omitted The white dark-squared bishops switch roles 4 times. Excelsior (A)


P0259 - Bernd Gräfrath

StrateGems 49, Jan-Mar 2010

[rnb2r2/pppp1p1p/7n/8/4p3/7N/PPP1P1P1/RN3B2]

9+12. Shortest proof game in 10.0 moves (Losing chess) (C+)

[rnb2r2/pppp1p1p/7n/8/4p3/7N/PPP1P1P1/RN3B2]

Solution

1.h4 e5 2.h5 e4 3.Rh4 Qxh4 4.h6 Qxf2 5.hxg7 Qxe1 6.gxf8B Qxd1 7.Sh3 Qxd2 8.Bxd2 Kxf8 9.Bb4 Sh6 10.Bxf8 Rxf8.

Schnoebelen bishop combined with captured anti-Pronkin bishop.


P0260 - James Soliman

StrateGems 49, Jan-Mar 2010

[1nq2bnr/1k3ppp/1pp2r2/p7/1P1PP1P1/5b2/P1P2P1P/RNBQKBNR]

16+14. Shortest proof game in 11.5 moves (C+)

[1nq2bnr/1k3ppp/1pp2r2/p7/1P1PP1P1/5b2/P1P2P1P/RNBQKBNR]

Solution

1.d4 a5 2.Bg5 Ra6 3.Bxe7 Rf6 4.Ba3 b6 5.b4 Bb7 6.Bc1 Bf3 7.e4 Qc8 8.Bb5 Kd8 9.Bxd7 c6 10.Bh3 Kc7 11.g4 Kb7 12.Bf1.

Circuits by the white bishops are motivated by the missing central black pawns.


P0261 - Guy Sobrecases

StrateGems 49, Jan-Mar 2010

[rnB1k2r/p1Qpnb1p/p3p3/6p1/1B1R1P1q/b4N2/PPP1P2P/2K2N2]

13+14. Shortest proof game in 17.5 moves (C+)

[rnB1k2r/p1Qpnb1p/p3p3/6p1/1B1R1P1q/b4N2/PPP1P2P/2K2N2]

Solution

1.d4 f5 2.d5 f4 3.d6 f3 4.dxc7 fxg2 5.f4 gxf1B!! 6.Sf3 Bg2! 7.Rg1 Bh3! 8.Rg6 Bf1! 9.Ra6 bxa6 10.Qd6 Bb7 11.Sbd2 Bd5 12.Sxf1 Bf7 13.Bd2 e6 14.0-0-0! Qh4 15.c8B g5 16.Qc7 Ba3 17.Bb4 Se7 18.Rd4. The knight promotion on f1 fails: 5….gxf1S? 6.Sf3 Se3 7.Rg1 Sd5 8.Rg6 Sb4 (but 16…Ba3??) or 7….Sf5!? 8.Rg6 Sd4 (but 10.Qd6??) or 8…Sd6!!? (but 9.Ra6??).

Three tempo moves showing a switchback and sacrifice of the promoted black bishop on its promotion square. Without the promoted black bishop’s maneuver, black could find two tempo moves…but not three. wB Phoenix (A).


P0262 - Satoshi Hashimoto

StrateGems 49, Jan-Mar 2010

[2b5/2knnppp/5p2/p2p4/1q1r1b1P/2r4P/PP1P1P2/RNBQKBNR]

14+14. Shortest proof game in 24.5 moves

[2b5/2knnppp/5p2/p2p4/1q1r1b1P/2r4P/PP1P1P2/RNBQKBNR]

Solution

1.Sc3 d5 2.Se4 Bh3 3.Sf6+ exf6 4.c4 Bd6 5.c5 Bf4 6.c6 Qd6 7.cxb7 c5 8.e4 c4 9.e5 c3 10.e6 c2 11.e7 Kd7 12.e8S Se7 13.Sc7 Rc8 14.Sb5 Rc3 15.Sa3 Kc7 16.Sb1 cxb1B 17.gxh3 Bf5 18.h4 Bc8 19.h3 Sd7 20.b8S a5 21.Sc6 Rb8 22.Sd4 Rb4 23.Sc2 Rd4 24.Sa3 Qb4 25.Sb1

Pronkin promotions. The first (Sb1) is captured by the second (Bc8)'s promotion. The captured knight is restored by a third Pronkin (Sb1 again). (A)