ALL DE FR ES IT

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Problemesis

Solutions to No. 28, August 2002

R59 - Jean-Marie Chorein

WPb3 is from a2, needs 1 cap. WPd3 is from d2 WPa7 is from f2, needs 5 cap. WPg3 is from g2 That accounts for all 6 cap by white.

BPf6 is from e7, needs 1 cap. BPe4 is from h7 or b7, needs 3 cap either way. That accounts for all cap by black BPb6 didn't capture, came from b7. BPe4 is from h7.

BWB isn't on the board, so it was captured. The only capture by white on a white square was on b3, so it was captured there. BWB couldn't leave its home square until BPb7 advanced. BPb7 advanced after the WP reached a7, or the pawn couldn't have gotten there. WP reached a7 after BB reached b8, or the BB couldn't have gotten there. Therefore, BWB was captured on b3 after WPa7 had made all its captures and the BB was on b8.

BPa7 wasn't captured on its home square as a pawn (or a7 and c7 would prevent the BB from getting to b8). It wasn't captured as a pawn elsewhere because it couldn't get off its home file (no captures available), and no other captures occurred on the a a-file. Since it is not on the board, it promoted. It couldn't have promoted until after a2 captured (it was blocked), so it promoted after white had made all six captures; the promoted piece is still on the board. It is not the king, a pawn, or the BB (which was already there before the pawn promoted); it must be the rook. Therefore, black can not castle.

The WRa1 must have moved for the pawn to promote, so white can not castle queenside. White may castle kingside; horribly inefficient proof game follows.

1. d3 a5 2. g3 Nf6 3. Nh3 Ng4 4. Nf4 Ne3 5. fxe3 Nc6 6. Bd2 Nd4 7. Bb4 Ra6 8. Bd6 Rc6 9. Be5 Rc5 10. Bf6 exf6 11. exd4 Be7 12. Nd5 Rc6 13. Nb6 Bc5 14. Na8 Qe7 15. Bg2 Ra6 16. Be4 Ra7 17. Bg6 hxg6 18. Qd2 Ra6 19. Qg5 Ba7 20. Qf5 Bb8 21. Nc3 gxf5 22. Rb1 Qc5 23. dxc5 Ra7 24. Rc1 Rh4 25. Rb1 Rb4 26. Ne4 Rb6 27. Rc1 fxe4 28. cxb6 Kf8 29. bxa7 b6 30. Rd1 Ba6 31. Rb1 Bc4 32. Rc1 Bb3 33. axb3 a4 34. Rb1 a3 35. Rc1 a2 36. Rb1 a1=R 37. Rc1 Ra5 38. Rb1 Rh5 39. Ra1 Rh8 40. Rc1 Ke7 41. Ra1 Ke8 42. O-O

[Solution provided by  Mark Jeffrey Tilford <tilford(at)ugcs.caltech.edu>]


R60 - Andrew Buchanan

 1.  d2-d4  f7-f5    2. Ke1-d2 Ke8-f7    3. Kd2-e3 Kf7-e6 
 4. Ke3-f4 Ke6-d5    5. Kf4-g5  h7-h6+   6. Kg5-g6  e7-e5 
 7. Kg6-f7 Qd8-h4    8. Bc1-g5 Kd5-e4    9.  f2-f4 Qh4-e1 
10.  g2-g3 Ke4-e3   11. Bf1-g2 Ke3-f2   12. Bg2-c6  e5-e4 
13.  d4-d5  e4-e3   14. Qd1-d4 Qe1-d1   15. Qd4-f6 Kf2-e1 
16. Qf6-d8 Bf8-e7   17. Kf7-e8 

R61 - Andrew Buchanan

1.  d2-d4  e7-e5     2. Bc1-d2 Bf8-e7    3. Bd2-b4 Be7-g5 
4. Bb4-f8 Bg5-c1     5.  e2-e3  d7-d6    6. Bf1-a6 Bc8-h3 
7.  g2-g4  b7-b5     8. Ba6-c8 Bh3-f1

R62 - Andrew Buchanan

 1. Ng1-f3 Nb8-c6    2. Nf3-d4 Nc6-e5    3. Nd4-c6 Ng8-f6 
 4.  d2-d4 Nf6-e4    5. Bc1-h6 Ne4-d2    6. Nb1-c3 Nd2-b1 
 7. Qd1-d2 Ne5-f3+   8. Ke1-d1  e7-e5    9. Nc3-d5 Bf8-a3 
10. Nd5-e7 Nf3-g1   11. Ne7-g8 Qd8-e7   12. Nc6-b8

R63 - Andrew Buchanan

 1.  f2-f4  a7-a5    2.  f4-f5  a5-a4    3.  f5-f6  a4-a3 
 4.  f6xg7  a3xb2    5. g7xh8=R b2xa1=R  6. Bc1-b2 Bf8-g7 
 7. Bb2-c3 Bg7xc3    8.  d2xc3  h7-h5    9.  c3-c4  h5-h4 
10.  c4-c5 h4-h3    11.  c5-c6  h3xg2   12.  c6xb7  g2xh1=R 
13. b7xa8=R

R64 - Nicolas Dupont

 1.  h2-h4  d7-d6    2. Rh1-h3 Bc8xh3    3.  g2-g3 Bh3xf1 
 4.  c2-c4 Bf1xe2    5. Qd1-b3 Be2xc4    6. Qb3-a3 Bc4xa2 
 7. Nb1-c3