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Die Schwalbe

No. 237, June 2009

14123 - Unto Heinonen

Die Schwalbe 237, June 2009

1. Nf3 b5 2. Nd4 Bb7 3. Nxb5 Bxg2 4. Nxa7 c6 5. Nxc6 Ra3 6. h3 Rxh3 7. Rxh3 Bxh3 8. Nxb8 Bxf1 9. Kxf1 Qxb8 10. Ke1 Qd8

Cook: 1. Nf3 c6 2. Ne5 Qc7 3. Nxc6 Qxh2 4. Nxa7 Qxg2 5. Rh2 Qxh2 6. Bg2 Qc7 7. Bxb7 Nc6 8. Bxc8 Rxc8 9. Nxc6 Rd8 10. Nxd8 Qxd8


14124 - Alexei Gasparjan

Die Schwalbe 237, June 2009

1. f4 Nh6 2. f5 Rg8 3. f6 gxf6 4. Kf2 Rg3 5. Qe1 Rd3 6. exd3 Na6 7. Qe6 dxe6 8. g4 Kd7 9. g5 Kd6 10. g6 Ke5 11. g7 Qd6 12. g8=Q Bd7 13. Qg2 Bg7 14. Qxb7 Rh8 15. Bh3 Be8 16. Bf5 Kxf5 17. Nh3 Kg6 18. Rf1 Nf5 19. Kg1 h6

Cook: 1. g4 Na6 2. Bg2 Rb8 3. Bxb7 Nh6 4. Bd5 Rb3 5. Be6 Rd3 6. exd3 dxe6 7. f4 Qd6 8. f5 Bd7 9. f6 gxf6 10. g5 Nf5 11. Nh3 Bg7 12. O-O O-O 13. Qf3 h6 14. Qb7 Kh7 15. g6+ Kxg6 16. Kh1 Rh8 17. Kg1 Be8


14125 - Bernd Gräfrath

Die Schwalbe 237, June 2009

1. b3 b5 2. Bb2 b4 3. a4 ba3 4. Qc1 ab2 5. Kd1 ba1=Q 6. Qa3 d5 7. Kc1 Kd7 8. Kb2 Kd6 9. Ka1

Queen Schnoebelen


14126 - Henryk Grudzinski

Die Schwalbe 237, June 2009

1. f4 g5 2. fg5=S b5 3. Sh7=B b4 4. Bg8=R b3 5. Rg7=B ba2=S 6. Bc3=S Sc1=B 7. Ra7=Q Bb2=R 8. Qa8 Rb5=B 9. Qa1=R Bc6=S 10. Sa2=P Sa7=P

Cook: 1. f4 g5 2. fg5=S a5 3. b3 a4 4. Sh7=B ab3=S 5. Ba3=S Ra6=B 6. Bg8=R Bb5=S 7. Sb5=B Sa1=B 8. Bc6=S bc6=S 9. Rg7=B Sa7=P 10. Ba1=R


14127 - René Jean Millour

Die Schwalbe 237, June 2009

Without Rd1: -1. f6-f7 and 1. a8=Q#. So how to get rid of the rook?

In Mars-Circe, if two pawns of opposite colour are 'past each other' on the same file, at least one of the two must be on the second (black) or seventh (white) row, otherwise the pawns couldn't have gotten past each other (sequence is: wP->7th row, black captures on the 6th row, or black pawn to second row, white captures on third)

-1. b7-b8=R a2-a1=S -2. Kd5-c6 f2-f1=Q -3. Ke6-d5 b2-b1=B -4. Kf6-e6 h2-h1=S -5. Kg7-f6 g2-g1=B -6. Kg8-g7 d2-d1=R -7. a6-a7 (a3-a2, f3-f2 and h3-h2 are now illegal retractions, see text above) d3-d2 -8. f6-f7 & 1. b8=Q#


14128 - Valentin Blacker

Die Schwalbe 237, June 2009

ELM: RxS


14129 - Werner Keym

Die Schwalbe 237, June 2009

In an illegal cluster with (a) 2 white pieces and 2 black pieces (b) 3 white pieces and 1 black piece should the sum of the smallest distances between the diagram position and the origin square (either original or promotion) be as large as possible (example: Distance from origin square for a wBf4 is not sqrt(18) (distance to c1) but 4 (distance to f8).

Original text: In einem Illegal Cluster mit (a) zwei wei?#376;en Steinen und zwei schwarzen, (b) drei wei?#376;en Steinen und einem schwarze soll die Summe der kleinsten Abstände der Standfelder der (vier) Steine zu inhren Partieanfangs- oder Umwandlungsfeldern möglichst gro?#376; sein. (Beispiel: Für den wLf4 beträgt der kleinste Abstand 4, nämlich vron f4 bis f8, und nicht ca. 4,2 von f4 bis c1).

- - -

a) wKb8, wBa2; bKb1, bBa7 (distance of the kings is sqrt(58), distance of bishops is sqrt(26).

b) wKa8, wQa2, wQh1; bKa1 (distance of the kings is sqrt(65), distance of the queens is 6 (Qa2<-->a8)+4 (Qh1<-->d1), not 7 (Qh1<-->h8)+sqrt(10) (Qa2<-->d1).