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Die Schwalbe

No. 233, October 2008

13879 - Werner Keym

Die Schwalbe, October 2008

Each castling is legal in itself, but the following castlings are mutually exclusive: (1) both white (2) both black (3) both long (4) both short. The short retrogenesis is: wSxf8, sPd3xSc2-c1=B, Rb7 and Ra5 appeared through promotion on h1 and a8, or through promotion on a1 and h8.

In the first case the solution is: 1. OO? OOO!; 1. Rf1! [2. Qc6] Kd8 2. Qc6 Kc8 3. Qc7#

In the second case the solution is: 1. OOO? OO!; 1. Rd1! [2. Qg6] Kf8 2. Qg6 Kg8 3. Qg7#


13880 - Thierry le Gleuher

Die Schwalbe, October 2008

-1. Qa7-b8 f4-f3 -2. Kh3-h2 h2-h1=B -3. Kg4-h3 h3-h2 -4. Bh2-g3 g3-g2 -5. Kh5-g4 g4-g3 -6. Bg3-h4 h4-h3 -7. Kh6-h5 h5-h4 -8. Qh4-g5 g5-g4 -9. Sf8-g6 g6-g5 -10. Rg5-f5 f5-f4 -11. Qf7-f6 f6-f5 -12. Qg8-f7 f7-f6 -13. Rf6-e6 e6-e5 -14. Bd8-e7 e7-e6 -15. Re6-d6 d6-d5 -16. Sc5-d7 Ba8-b7 -17. Sb7-c5 Kd7-c8 -18. Sh7-f8 Kc8-d7


13881 - Gianni Donati, Olli Heimo

Die Schwalbe, October 2008

Dedicated to Günter Lauinger

1. d4 e5 2. Bh6 Qg5 3. d5 Qc1 4. d6 g5 5. dxc7 d5 6. Bg7 Bd7 7. c8=N Bd6 8. Nb6 axb6 9. Bf6 Ra3 10. Bd8 Rg3 11. f3 Ba4 12. Kf2 b5 13. Ba5 Ne7 14. Bd2 O-O 15. c3 Rc8 16. Qc2 Qxf1+ 17. Ke3 Qf2+ 18. Kd3 Qb6 19. Bc1 Qd8

Rundlauf and Platzwechsel by Bc1 and Qd8.


13882 - Bernd Gräfrath

Die Schwalbe, October 2008

a) 1. d3 Sc6 2. Bf4 Sd4 3. Kd2 Se2 4. Be2 c5 5. Qf1 Qb6 6. Bc7 Qa5

b) 1. d3 Sc6 2. Bd2 Sd4 3. Bf4 Se2 4. Be2 c6 5. Kd2 c5 6. Qf1 Qa5 7. Bc7


13883 - Bernd Gräfrath

Die Schwalbe, October 2008

1. h4 Sh6 2. h5 Sg4 3. h6 Sf2 4. hg7 Sh1 5. gh8=K h6 6. g4 e5 7. Bg2 Qf6 8. Bh1 Qh8

King-Schnoebelen


13884 - Bernd Schwarzkopf

Die Schwalbe, October 2008

White and black castling are mutually exclusive.

1. OOO? and black can't castle, so his longest moves are Ba4/Bg4 and no
mate in 3.

1. Rd1! OOO 2. Rgd4 Bh3 3. Rd8#


13885 - Tadashi Wakashima

Die Schwalbe, October 2008

-1. Ke5xBh8! Kg7xSe8 -2. Sf6-e8 and 1. e8=S#


13886 - Günther Weeth, Klaus Wenda

Die Schwalbe, October 2008

Main plan: -1. Bc8xSh3[Bf1] Kb8-c7 -2. e5xf6[Pf2]ep f7-f5 -3. c7-c8=B and 1. Rd1[Rh1] Kc7[Ke8] but 2. Rh3[Rh1].

Solution: -1. h2xRg3[Pg2] Rh3-g3 -2. Bf1xRg2[Bf1] Rh8-h2 and now the main plan: -3. Bc8xSh3[Bf1] Kb8-c7 -2. e5xf6[Pf2]ep f7-f5 -3. c7-c8=B and 1. Rd1[Rh1] Kc7[Ke8]#

The key is a double Antizielelement: h3 is blocked, and white gains a
flight square g2.


13887 - Mario Richter

Die Schwalbe, October 2008

Last moves when black has the move: -1. Rb1xBc1 Bb2-c1 -2. Qc1xSc2 S~
-3. c3-c2 etc
Last moves when white has the move: -1...Rb2xBa2 -2. Bb1-a2 Ra2xSb2 -3.
S~ Rb2-a2 SxQ[RBP] etc.


13888 - Bernd Schwarzkopf, Werner Keym

Die Schwalbe, October 2008

If black moved last, it must've been Qg6(x-)h7. If nothing was captured, then the check to the black king would be illegal. So the ELM is of the type QxX. If white moved last, it was Qg6xXh6. With X=Q/B, the white king is in an illegal check. With X=R/S, black has no last move. So the equal last moves were Qg6xPh6/Qg6xPh7, so type QxP.


13889 - Werner Keym, Bernd Schwarzkopf

Die Schwalbe, October 2008

Retracting g2xXh3 leads to an impossible K/Q swap. Any removal leads to either a different last move, or a hole through which the K/Q can escape.


13890 - Werner Keym

Die Schwalbe, October 2008

a) 8/8/7k/8/8/8/8/2B1K2R (sum of distance to home squares is ~3.6)

b) BK5Q/8/8/8/8/8/8/7k (sum of distance to home squares is ~31.9)