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Die Schwalbe

No. 235, February 2009

14001 - Roberto Osorio, Jorge Joaquin Lois

Die Schwalbe 235, February 2009

1. f4 g5 2. f5 g4 3. f6 g3 4. fxe7 f5 5. Nh3 Kf7 6. e8=Q+ Kf6 7. Qe3 Qe8 8. Qxa7 Ke5 9. Qxb8 Ra3 10. Qa7 Rf3 11. Qg1 Bc5 12. Nf2 Kd4 13. Nh3+ Qe3 14. Qf2 Bb6 15. Ng1 gxf2#

Double platzwechsel wPf2/wSg1


14002 - Valeri Liskovets

Die Schwalbe 235, February 2009

Of the pieces on c7/d8/e7, a maximum of 2 pieces can be volage: In case Sd8 is volage, it must've come from c7 or e7. So the idea is to force both black pawns to a white square, which proves that Sd8 isn't volage.

1. Kd4 c5 2. Ke5 c4(=w) 3. Qg6 e6(=w) 4. Sf7#

1... e5 2. Kc5 e4(=w) 3. Qg6 c6(=w) 4. Sf7#


14003 - Klaus Wenda

Die Schwalbe 235, February 2009

-1. Qb8-d8 Bd8-f6+ -2. Rf2xBd2[Ra1] Rf8-f2[=w] -3. Ke1xPf2[Ke1] Rh8-h6+ -4. g5xPf6ep[bPf2] and 1. ef5[bPf2] Rh1#


14004 - Bernd Gräfrath

Die Schwalbe 235, February 2009

For a mate, a bP must be added on f7. But how did the black king go to g7? It must've been through OO-f8-g7, which means that the black rook had to leave f8 after castling. So either d8 had to be empty, which can only be achieved by capturing the black queen on d8 or e8 after castling, and then replacing it with a Pronkin queen, or by capturing Rf8, and then replacing it with a Pronkin rook. So the second pawn must be white. It can't have left the g-file, on g5 it would be in the way of the mate, on g2 the white queen/king couldn't have left the first row, so it must be on g3 (this also proofs that the Pronkin-piece was a queen). #1 with 1. Qg5#

A possible proofgame: 1. Nc3 Nc6 2. Nd5 Na5 3. Nb4 Nb3 4. Nc6 Nxc1 5. Nxd8 Nb3 6. g3 Nxa1 7. Bg2 Nb3 8. Be4 g5 9. Nf3 Bh6 10. O-O Nf6 11. Qa1 O-O 12. Rd1 Rxd8 13. Kf1 Kf8 14. Kg2 Kg7 15. Nd4 Rh8 16. Kf3 Nh5 17. Kg4 Na5 18. Kf5 g4 19. Bg2 Nc4 20. Nf3 gxf3 21. Kg4 fxg2 22. Rf1 gxf1=Q 23. Qd1 Qh1 24. Qa1 Qc6 25. Qh1 Kf6 26. Qe4 Qa6 27. Kf3 Kg5 28. Kg2 Qg6 29. Kf3 Qg8 30. Qe6 Kf6 31. Qa6+ Kg7 32. Kg4 Nf6+ 33. Kf5 Ne5 34. Qa3 Neg4 35. Qc3 Qd8 36. Qe3 Rf8


14005 - Bernd Gräfrath

Die Schwalbe 235, February 2009

1. f3 Nf6 2. f4 Ne4 3. Nf3 Nxd2 4. Rg1 Nxf1 5. Rxf1 Nc6 6. Rf2 Nd4 7. Kf1 Nxe2 8. Qe1 Ng3+ 9. Kg1 Nf1 10. Rxf1


14006 - Andreas Witt

Die Schwalbe 235, February 2009

Dedicated to Herbert Ahues' birthday

1. c8=Q 2. f8=Q 3. g8=Q 4. Qc3 5. Qf4 6. Qg6

Now white has 25(Qc3)+25(Qf4)+23(Qg6)+14(Rd5)=87 moves, which is Herbert
Ahues' age.


14007 - Andreas Witt

Die Schwalbe 235, February 2009

Dedicated to Herbert Ahues' birthday

wX(#) means: After the first move X, white has # next moves. Idem bX(#). wKa2(7), wKb2(10), wKb3(9), wKb4(8), wKa4(6), wPa6(7), wPc5(7) bKh3(7), bKg3(10), bKg4(9), bKg5(9), bPe2(8).

In total (7+10+9+8+6+7+7)*(7+10+9+9+8)=2322 possible move combinations,
which is the birthday of Herbert Ahues (2.3.22)


14008 - Alain Brobecker

Die Schwalbe 235, February 2009

Which unique monochromatic proof game ends with 5...BxN#?

1. d4 g5 2. Kd2 Bg7 3. Ke3 Be5 4. Bd2 Bh2 5. f4 Bg1#


14009 - Claudius Gottstein

Die Schwalbe 235, February 2009

Construct a position with only white rooks, in which exactly 45 squares are observed. How many positions are there?

Five rooks are needed to guard 45 squares (e.g. Ra1, Rb2, Rb3, Rb4, Rb5). All these positions have the characteristic '4 rooks on the same file/row, and one unguarded rook). The unguarded rook can be placed on 64 squares, which leaves 14 rows/files for the 4-rook-cluster. There are binomial(7,4)=35 possible rook-clusters on a file. So in total there are 64*14*35=31360 positions in which exactly 45 squares are guarded by only white rooks.