[Retros] Solutions Die Schwalbe 239
Joost de Heer
joost at sanguis.xs4all.nl
Sat May 1 13:28:37 EDT 2010
I forgot the solutions from Heft 239 in my last post, so here they are.
But first the corrections in 'Bemerkungen und Berichtigungen'
Heft 215, 12832 (G. Weeth) is cooked by Paul Raican: -1. Ke1xRd1 Rd8-d1
(Rd2..7-d1 leads to short solutions, and Qh8-h2 -2. Sf4-g6 & 1. Se2#)
-2. Ke1xPf2 f3-f2 -3. Ke1xPf2 Qh8-h2 -4. Ke2xSd1 f4-f3 -5. Ke3-e2 f5-f4
-6. Sf4-g6 & 1. Se2#
Correction:
12832v - Günther Weeth, Klaus Wenda
2bs5/6k1/4B1P1/8/8/b1q3s1/p6p/K6R (4+8)
-9 & #1
-1. Rh1xRb1 Rb8-b1 (Kh8-g7? -2. Re7xXh7 & 1. g7#) -2. Rh1xRb1 Rb7-b1...
-7. Rh1xRb1 Rb2-b1 -8. Re1xRb1! (-8. Rh1xRb1? Kh8-g7 and there are 16
black pieces on the board, so white can't uncapture anything on h7)
Kh8-g7 -9. Bg8-e6 & 1. g7#
Heft 229, 13636 - Kostas Prentos
Version Silvio Baier
r3rkb1/2p1p2q/sbp2p2/p1s4p/3P4/8/1PP1PP2/RSBQKBSR (13+14)
SPG 18.0
1. h4 Nf6 2. h5 Ne4 3. h6 f6 4. hxg7 h5 5. g8=N Bh6 6. g4 Be3 7. g5 Bb6
8. g6 Nc5 9. g7 Kf7 10. a4 Qxg8 11. a5 Qh7 12. a6 Re8 13. axb7 a5 14.
g8=N Nba6 15. b8=N Kf8 16. Nc6 dxc6 17. d3 Be6 18. d4 Bxg8
CF-knight on b8, two Schnoebelen-knights on g8.
Heft 240, 14322 (A. Thoma)
The correct stipulation is: "After 5 moves, all pieces are on their home
square. White starts. Circe + Anticirce"
SOLUTIONS HEFT 239
------------------
14241 - Frolkin
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 - Frolkin
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 - Frolkin
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 - Baibikov
-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 - Heinonen
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 - Pfeiffer
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 - Gräfrath
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 - Antonov
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 - Keym
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 - Grevlund
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 - Schwarzkopf
Kc1 Ra1 Rd1 Rd2 - Kg1
14252 - Dietrich
a) 3720
b) (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://www.research.att.com/~njas/sequences/A001564
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