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Die Schwalbe

No. 239, October 2009

14241 - Andrei Frolkin

Die Schwalbe 239, October 2009

1. ...Sd7xSb8! 2-6. Sh8-b3 Ka7-a8 7. h7-h8=S Ka8-a7 8-10. h4-h7 11-15. Sa1-h5 16. S~xPh5 Ka7-a8 17. b7-b8=S h6-h5 18. Rb8-c8 h7-h6 19. Kc8-c7 etc. Schnoebelen-knight on b8.


14242 - Andrei Frolkin

Die Schwalbe 239, October 2009

1....Se7xBc8! 2-4. Qh8-a5 Ka7-b8 5. h7-h8=Q Kb8-a7 6. h6-h7 Ka7-b8 c7-c8=B h7xPg6 Kc8-d7 etc. Schnoebelen-bishop on c8.


14243 - Andrei Frolkin

Die Schwalbe 239, October 2009

1...Qa8xRb8 2. e3-e4 Kc6-d6 3. e2-e3 Bd6-c5 4. b7-b8=R Bb8-d6 5. a6xSb7. Schnoebelen-rook on b8.


14244 - Dmitri Baibikov

Die Schwalbe 239, October 2009

-1. b2-b3! Bg1-f2 (-1...Bg1xSf2? -2. Bh1-g2 & 1. Rd1#; -1...Rh1xQh2? -2. Qg1-h2 & 1. Ree2#) -2. Bh1-g2 Bf2xBg1 -3. Bg2-h1 Rh1-h2 -4. Bh2-g1 Rg1-h1 -5. Bh1-g2 Rg2-g1 -6. Bg1-h2 Rh2-g2 -7. Bg2-h1 Rh1-h2 -8. Bh2-g1 and now Rg1-h1 is illegal because it would forcibly lead to a third repetition. So -8...Bg1xSf2 -9. Sd1-f2 Bf2-g1 -10. Sc3-d1 & 1. Rd1#


14245 - Unto Heinonen

Die Schwalbe 239, October 2009

1. e4 b5 2. Bxb5 Nf6 3. Bxd7+ Qxd7 4. d3 Qxd3 5. e5 Qxc2 6. exf6 Qxb1 7. fxg7 Qxb2 8. gxh8=B Bh6 9. Bhxb2 Bxc1 10. Bxc1. Homebase


14246 - Alfred Pfeiffer

Die Schwalbe 239, October 2009

1. OO! Castling shows that Rb1, Kc1, Rg1 and Qh1 never moved. Ba1 (which never moved) shows that on a8 there was a black bishop, which must've been captured by one of the knights. So black can't castle. 1... b5/b6 2. Sfd7; 1... e5/e6/Kd8 2. S(x)e6; 1... Rxg6 2. Sxg6; 1... hxg6 Sh7.


14247 - Bernd Gräfrath

Die Schwalbe 239, October 2009

1. e4 e5 2. Bc4 d5 3. Bxd5 Qxd5 4. f4 Qxe4+ 5. Ne2 Qe3 6. Rf1 Qxd2+ 7. Kf2 Qd8 8. Kg1. Artificial castling. There's a unique 7.0 proofgame 6. d3 Qxd3 7. OO Qd8.


14248 - Ivan Antonov

Die Schwalbe 239, October 2009

1. a4 b5 2. Sa3 bxa4-a5 3. b4 axb4-b5 4. c4 bxc4-c5 5. d4 cxd4-d5 6. e4 dxe4-e5 7. f4 exf4-f5 8. g4 fxg4-g5 9. h4 gxh4-h5 10. Qg4 hxg4-b4 11. Rb1 bxa3-c4 12. Bxc4-c3 h6 13. Bxh6-h5

Cooked: 1. e3 h5 2. Qg4 hxg4-d1=S 3. Ba6 Sxb2-b4 4. Bxb7-b6 Sxa2-a3 5. Bb2 Sxb1-a3 6. Rb1 Sxc2-c4 7. Bc3 Sxe3-e4 8. Kd1 Sxd2-d3 9. h3 Sxf2-f4 10. Bf2 Sxh3-h4 11. Ke1 Sxg2-g3 12. Bxg3-h5


14249 - Werner Keym

Die Schwalbe 239, October 2009

History in which the maximum number of visited squares is reached:

Last moves were 1. OOO and e.g. Kg1xSh1. A white knight captured Sb8, Bc8, Qd8, then black castled queenside, the white knight captured Rd8, and then Kc8-h1. The black queen is a promoted pawn after a2xQb1=Q (a white knight on d1 interferes with the check).

Maximum number of squares visited only once:

wK: 1 (c1)
wR: 1 (d1)
bK: 35 (c8, d8, e8, f7, all squares on 6th-4th row, f3, g3, h3, f2, g2, h2, f1, g1, h1)
bQ: 46 (c1, b1, a2, all squares on the 6th-3rd row, h1, h2, g1, g2, g8, f7, a8-e8)


14250 - Per Grevlund

Die Schwalbe 239, October 2009

bernd ellinghoven zum Geburtstag

One solution is: 1. Sd4 2. Se6 3. Kd4 4. Ke5 5. Kf6 6. d4 7. d5 8. d6 9. d7#
All the three routes go over d4.

1) Begin with 1. Sd4 2. Se6. Now the king has 3 moves to f6, and the pawn 3 moves to d6 (d7# is the last move). Without collision this is (6 over 3) = 6!/(3!*3!) = 20. Both the king and pawn have 2 moves to reach their destination from d4, this results in (2 over 1) * (4 over 2) = 12 routes in which the routes collide. So in this case there are 20-12=8 possibilities

2) Begin with 1. Kd4 2. Ke5. Now the knight and pawn collide. Using the method in 1), this results in (5 over 3) - (2 over 1)*(3 over 1) = 4 possibilities. The move Kf6 can be anywhere from move 3 to 8, so in total there are 4*6=24 possibilities.

3) Begin with 1. d4 2. d5. Similar to 2), this results in 24 possibilities.

So in total 8+24+24=56 ways to deliver a series-mate in 9.


14251 - Bernd Schwarzkopf

Die Schwalbe 239, October 2009

Kc1 Ra1 Rd1 Rd2 - Kg1


14252 - Stephan Dietrich

Die Schwalbe 239, October 2009

A white knight is placed on a1. On the board are further seven white rooks. (8+0)

a) How many positions like this exist with no piece guarding another?

3720

b) Generalize for an nxn board with (n-1) rooks (n >= 3)

(n-1) rooks can be placed in (n-1)! non-attacking ways. This includes positions with Rc2 ((n-2)!) and Rb3 (also (n-2)!). Now the position with Rc2 *and* Rb3 is counted double, this is (n-3)! positions. So the generalised formula is (n-1)!-2*(n-2)!+(n-3)!.

This formula is known as '2nd differences of factorial numbers', see
http://oeis.org/A001564