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Probleemblad

Oct-Dec 2009

R358 - Unto Heinonen

Probleemblad, Oct-Dec 2009

[rnbq2nr/ppkp3p/8/8/8/1R1P4/PP1KPPP1/RNBQ2N1]

13+11. SPG 13.0 (C+)

[rnbq2nr/ppkp3p/8/8/8/1R1P4/PP1KPPP1/RNBQ2N1]

Solution

1. c4 e5 2. c5 Bd6 3. cxd6 g5 4. dxc7 g4 5. cxd8=N g3 6. Nxf7 gxh2 7. Nxe5 hxg1=Q 8. Rh3 Kd8 9. Rb3 Kc7 10. d3 Qxf1+ 11. Kd2 Qh1 12. Nf3 Qh4 13. Ng1

Qd8 Pronkins on d8 (promoted on g1) and g1 (promoted on d8).


R359 - Unto Heinonen

Probleemblad, Oct-Dec 2009

[r2q2k1/1ppnnp1p/1pB2p2/8/1P1Q2b1/2PK3P/1PPBPP1P/R5N1]

14+12. SPG 20.0 (C+)

[r2q2k1/1ppnnp1p/1pB2p2/8/1P1Q2b1/2PK3P/1PPBPP1P/R5N1]

Solution

1. Nf3 g5 2. Ne5 g4 3. Nxd7 g3 4. Nf6+ exf6 5. a3 Bb4 6. axb4 Qd7 7. Ra6 Qh3 8. Rb6 axb6 9. gxh3 Ra3 10. Bg2 Rc3 11. dxc3 Ne7 12. Qd4 O-O 13. Bc6 g2 14. Kd2 g1=Q 15. Kd3 Qg7 16. Nd2 Bg4 17. Nf3 Nd7 18. Bd2 Ra8 19. Ra1 Qf8 20. Ng1 Qd8

'Pronkin-siblings' (original captured, sibling replaces) on a1, a8, g1 and Pronkin on d8.


R360 - Ivan Denkovski

Probleemblad, Oct-Dec 2009

[3q2nr/p1kr1ppp/5p2/2n5/3p4/2b3Pb/2PPPPP1/RNBKRBNQ]

14+14. SPG 21.0 (C+)

[3q2nr/p1kr1ppp/5p2/2n5/3p4/2b3Pb/2PPPPP1/RNBKRBNQ]

Solution

1. a4 c5 2. a5 Qc7 3. a6 Qg3 4. hxg3 c4 5. Rh6 c3 6. Rf6 cxb2 7. Nc3 b1=Q 8. Ba3 Qb6 9. Qb1 d5 10. Qb4 Nd7 11. Qh4 Nc5 12. Qh1 Bh3 13. O-O-O O-O-O 14. Re1 Rd7 15. Kd1 Qd8 16. axb7+ Kc7 17. b8=R exf6 18. Rb1 Bd6 19. Ra1 Be5 20. Nb1 Bc3 21. Bc1

d4 Pronkins (a1, d8), switchbacks (b1, c1)


R361 - Gianni Donati

Probleemblad, Oct-Dec 2009

[1n2k3/1pb4p/1ppq2pr/n2p3r/5p1P/2P2P2/PPP1PP2/R2QK2R]

12+14. SPG 22.5

[1n2k3/1pb4p/1ppq2pr/n2p3r/5p1P/2P2P2/PPP1PP2/R2QK2R]

Solution

1. Nf3 e5 2. Nd4 exd4 3. Nc3 dxc3 4. dxc3 f5 5. Be3 f4 6. Bb6 axb6 7. h4 Ra5 8. Rh3 Rh5 9. Rg3 d5 10. Rg6 Bg4 11. Rd6 Bf3 12. Rd7 Bd6 13. gxf3 Ne7 14. Bh3 Rf8 15. Bf5 Rf6 16. Bd3 Rfh6 17. Bb5 g6 18. Bc6 Nexc6 19. Rf7 Na5 20. Rf5 c6 21. Rg5 Bc7 22. Rg1 Qd6 23. Rh1

Long journey of white rook for shielding


R362 - Guus Rol

Probleemblad, Oct-Dec 2009

[r3k3/8/8/1KpP4/P1P5/8/QP6/8]

6+3. h#2.5, (Circe) (AP) (Two solutions)

[r3k3/8/8/1KpP4/P1P5/8/QP6/8]

Solution

I) 1...Kc6! 2. Rxa4 Qxa4[Ra8] 3. OOO Qa8# II) 1...dxc6ep[Pc7] 2.OOO c5 3. Rg8 Qxg8[Ra8]#

In II), the ep capture has to be justified by castling. A try would be 1...dxc6ep[Pc7] 2. Rxa4? Qxa4[Ra8] 3. OOO Qa8# because in this scenario, black's last move could've been with the rook, so the ep-capture isn't justified.


R363 - Guus Rol

Probleemblad, Oct-Dec 2009

[r3k1b1/1p5p/3P3p/PPPKBpP1/pQrPRP2/Npp1p3/8/8]

12+12. h#2.5 (Circe) (AP)

[r3k1b1/1p5p/3P3p/PPPKBpP1/pQrPRP2/Npp1p3/8/8]

Solution

1... gxf6ep[Pf7] 2. Ra6 bxa6[Ra8] 3. OOO Qxb7# After the first move, direct castling is illegal (black's move before f7-f5 must've been with either the king or rook). But black has to castle because he must prove the last move wasn't Kf7-e8. In the solution, black castles with a reborn rook, proving his king didn't move.


R364 - René J. Millour

Probleemblad, Oct-Dec 2009

This problem was later found to be cooked.

[1nbkqbnr/ppppp3/R1r5/8/8/8/1PPPP1B1/1NB3N1]   [8/8/5p2/5p2/P5KP/5PP1/8/8]
A: 9+13.

[1nbkqbnr/ppppp3/R1r5/8/8/8/1PPPP1B1/1NB3N1]

  B: 5+2.

[8/8/5p2/5p2/P5KP/5PP1/8/8]

SPG 19.0 (Alice chess)

Solution

No solution available, the problem was later found to be cooked.


R365 - Wolfgang Dittmann

Probleemblad, Oct-Dec 2009

[nB3brk/p2p2p1/1P6/3p4/1r6/8/8/4K3]

3+9. -6 & #1 (Proca, Anticirce Cheylan)

[nB3brk/p2p2p1/1P6/3p4/1r6/8/8/4K3]

Solution

Main plan: -x. b7-b8=B & 1. ba8=R[Rh1] but black can defend with Rh4. So the 4th row must be obstructed. But this isn't as easy as it seems: -1. Kc5xBd4[Ke1]? e7xXd6[Pd7] -2. b7-b8=B & 1. ba8=R[Rh1] but 1... Ba1! -1. Kb2xPc3[Ke1]? b7xXa6[Pa7]! -1. Ke4xPd4[Ke1]? Bh4xPe7[Bf8]!

Solution: -1. Kc2xBb1[Ke1] Ba2-b1 -2. Kb1-c2 Bb3-a2 -3. Kc2-b1 Ba2-b3 -4. Kb1-c2 Bb3-a2 -5. Kc2-b1 Bc4-b3 (Ba2-b3?? illegal) -6. b7-b8=B & 1. ba8=R[Rh1]#

Thematic tries: -1. Kc2xBb3[Ke1]? Ba2-b3 ... -5. Kc2-b1?? illegal. -1. Kb1xBa2[Ke1]? Bb3-a2 ... -5. Kb1-c2?? illegal.


R366c - Klaus Wenda

Probleemblad, Oct-Dec 2009

The original problem was later found to be cooked. This is the corrected version, published in Probleemblad 2/2010:

[R7/pb4k1/1P1P4/8/KB6/rPp5/B1p3P1/8]

7+7. -9 & #1 (Proca, Anticirce)

[R7/pb4k1/1P1P4/8/KB6/rPp5/B1p3P1/8]

Solution

-1. Rb8-a8! Ba8-b7 -2. Ka5-a4 Kh8-g7 -3. b7-b8=R (threat: -4. f5xg6ep[Pg2] g7-g5 -5. b2-b3 & 1. ba8=R[Rh1]#) c4-d3 -4. Bc5-b4 Ra4-a3 -5. Bb4-c5 Ra3-a4 -6. Bc5-b4 Ra4-a3 -7. Bb4-c5 c5-c4 (Ra3-a4?? illegal) -8. f5xg6ep[Pg2] g7-g5 -9. b2-b3 & 1. ba8=R[Rh1]# -7... c3-c2 -8. hg6ep[Pg2]! g7-g5 -9. Bb1-a2 & 1. b8=Q# Try: -4. Bd2-b4? Ra4-a3 -5. Bb4-d2 Ra3-a4 -6. Bd2-b4 c5-c4! and now: -7. f5xg6ep[Pg2] g7-g5 -8. b2-b3 & 1. ba8=R[Ra1] Rh3! or -7. Bb4-d2 c6-c5! or -7. Be3-d2 Ra4-a3!