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Die Schwalbe

No. 241, February 2010

14375 - Werner Keym

Die Schwalbe 241, February 2010

[r3k2r/1pp1p1p1/N1p1P1p1/4P2N/6Q1/4P3/1P2P1PB/R3K2R]

13+9. #3 RS

[r3k2r/1pp1p1p1/N1p1P1p1/4P2N/6Q1/4P3/1P2P1PB/R3K2R]

Solution

There are two different ways to reach the position:

Tries: 1. 000/Rd1? 00!, 1. Rf1? 000!

Solution: 1. 00! [2. Qd4/Rd1] Rf8 2. Sg7 Kd8 3. Rf8# 1... Kd8 2. Qd4 Kc8 3.


14376 - Nikolai Beluchov

Die Schwalbe 241, February 2010

[8/3p4/1p2p2p/8/PPPPPp1p/KBrQp2B/1RrnRP1P/q1bkb3]

13+14. Last 31 half moves?

[8/3p4/1p2p2p/8/PPPPPp1p/KBrQp2B/1RrnRP1P/q1bkb3]

Solution

1... Qb1-a1 2. Bf5-h3 h5-h4 3. Bh7-f5 f5-f4 4. Bg8-h7 f6-f5 5. g7-g8=B f7-f6 6. g6-g7 g7xSh6 7. Sf5-h6 h6-h5 8. Sg3-f5 h7-h6 9. Sf1-g3 Sf3-d2 10. Sd2-f1 Se5-f3 11. g5-g6 Sc6-e5 12. g4-g5 Sa5-c6 13. Ba2-b3 Sb3-a5 14. g3-g4 Qa1-b1 15. Rb1-b2 Rb2-c2 16. g2-g3 Rc2-c3


14377 - Gianni Donati, Olli Heimo

Die Schwalbe 241, February 2010

[4n1k1/pppnR1P1/4p1r1/5p2/8/P3P1P1/P1PrPPbp/1N1qKBNR]

14+13. Shortest proof game in 24.0 moves

[4n1k1/pppnR1P1/4p1r1/5p2/8/P3P1P1/P1PrPPbp/1N1qKBNR]

Solution

1. g3 d5 2. Bg2 Bh3 3. Kf1 e6 4. Qe1 Ba3 5. bxa3 f5 6. Bb2 Nf6 7. Bd4 O-O 8. Be3 d4 9. Nc3 dxe3 10. dxe3 Qd2 11. Rd1 Qc1 12. Rd7 Rd8 13. Re7 Rd2 14. Qd1 Nbd7 15. Ke1 Rf8 16. Bf1 Bg2 17. h4 Ne8 18. h5 Rf6 19. h6 Rg6 20. hxg7 h5 21. Rh3 h4 22. Rh2 h3 23. Rh1 h2 24. Nb1 Qxd1


14378 - Dragan Petrovic

Die Schwalbe 241, February 2010

[8/5p2/4p1P1/2P2PP1/1PrkPKPR/BRqbrpQ1/1pppppNb/4NnnB]

15+16. Shortest resolution?

[8/5p2/4p1P1/2P2PP1/1PrkPKPR/BRqbrpQ1/1pppppNb/4NnnB]

Solution

The author's intention required 192 moves, but the position can be resolved much quicker. Starting position, with black to move:

[5Q2/3p4/2B1R1p1/2P2P1P/1PrkPKPP/BRqbrpP1/1pppppSb/4Sss1]

1... g5 2. hg5 d6 3. Sh4 g6 4. hg6 d5 5. Qf6 gf6 6. Rf6 Sg3 7. Re6 Sf1 8. g3 Sg3 9. Bd5 Sh1 10. g3 Sg3 11. Bc6 Sh1 12. g3 d6 13. Seg2 f6 14. Qf1 d5 15. Se1 Sg3 16. Rf6 Sh1 17. g3 Sg3 18. Re6 Sh1 19. g3 f6 20. Qh3 Sg3 21. Rf6 Sh1 22. g3 Sg3 23. Re6 Sh1 24. g3 Sg3 25. Bd5 Sh1 26. g3 Sg3 27. Bc6 Sh1 28. g3 d6 29. Shg2 d5 30. Qh7 f6 31. Sh4 Sg3 32. Rf6 Sf1 33. g3 Sg3 34. Bd5 Sf1 35. g3 Sg3 36. Rc6 Sf1 37. g3 Sg3 38. Rc8 Sf1 39. g3 Sg3 40. Be6 Sh1 41. g3 de6 42. Bh3 Sg3 43. Re8 Sf1 44. g3 e5 45. Re5 f6 46. Bg2 e6 47. Bh1 Sg3 48. Re6 Sf1 49. g3 Sg3 50. Rf6 Sf1 51. g3 Sg3 52. Rc6 Sf1 53. g3 Sg3 54. Rc8 Sf1 55. g3 Sg3 56. Re8 Sf1 57. g3 e6 58. Shg2 e5 59. Re5 e6 60. Qh3 f6 61. Sh4 Sg3 62. Re6 Sf1 63. g3 Sg3 64. Rf6 Sf1 65. g3 Sg3 66. Rc6 Sf1 67. g3 Sg3 68. Rc8 Sf1 69. g3 Sg3 70. Rh8 Sf1 71. g3 e6 72. Shg2 Sg3 73. Rh4 Sf1 74. Qg3


14379 - Bernd Gräfrath

Die Schwalbe 241, February 2010

[rnb1Qbn1/pppp1qp1/5k2/5p2/5P2/8/PPPP3P/RNB1KBNR]

14+13. Shortest proof game in 9.5 moves (Duellist)

[rnb1Qbn1/pppp1qp1/5k2/5p2/5P2/8/PPPP3P/RNB1KBNR]

Solution

1. g4 h5 2. gxh5 Rxh5 3. f4 Rf5 4. e4 Rd5 5. exd5 f5 6. d6 Kf7 7. dxe7 Kf6 8. e8=Q Qxe8+ 9. Qe2 Qf7 10. Qe8


14380 - Bernd Gräfrath

Die Schwalbe 241, February 2010

Dedicated to John Beasley's 70th birthday

[5r1q/p1p1p1b1/2n2pk1/8/b7/8/PPPPPP1P/RNBQKBNR]

15+10. Shortest proof game in 11.5 moves (White must capture)

[5r1q/p1p1p1b1/2n2pk1/8/b7/8/PPPPPP1P/RNBQKBNR]

Solution

1. g4 f6 2. Bg2 Kf7 3. Bxb7 d5 4. Bxd5+ Kg6 5. Bxg8 h6 6. g5 Bd7 7. gxh6 Ba4 8. hxg7 Nc6 9. gxh8=R Bg7 10. Be6 Qxh8 11. Bh3 Rf8 12. Bf1


14381 - Wolfgang Dittmann

Die Schwalbe 241, February 2010

[n2n2Bb/pp6/k3P3/p7/4P3/8/1P1P4/r3Kb2]

6+9. -19 & #1 (Proca retractor, Anticirce)

[n2n2Bb/pp6/k3P3/p7/4P3/8/1P1P4/r3Kb2]

Solution

Master plan: c5xBb6[Pb2] & Bg8*X[Bf1]#.

1. Kf2xBg1[Ke1] Bh2-g1 2. e5xd6ep! d7-d5 3. Kg3-f2 Bg1-h2 4. Kf2-g3 Bh2-g1 5. Kg3-f2 Bg1-h2 6. Kf2-g3 g2-g1=B 7. Ke2-f2 f2-f1=B 8. Ke1-e2 f3-f2 9. Kf2xBg1[Ke1] Bh2-g1 10. Kf1-f2 g3-g2 11. Kf2-f1 g4-g3 (11... Sf7-d8? and short solution) 12. Kg3-f2 Bg1-h2 13. Kf2-g3 Bh2-g1 14. Kg3-f2 Bg1-h2 15. Kf2-g3 g2-g1=B 16. Kf1-f2 g3-g2 17. Kf2-f1 Sf7-d8 18. Kg2-f2 f4-f3/Bg7-h8 19. c5xBb6[Pb2] & 1. Bf7[Bf1]#


14382 - Günther Weeth

Die Schwalbe 241, February 2010

[8/bP3k2/1P6/pP4P1/3N4/3p4/4P3/3bK3]

7+5. -6 & s#1 (Proca retractor, Anticirce Cheylan, magic square type II a4)

[8/bP3k2/1P6/pP4P1/3N4/3p4/4P3/3bK3]

Solution

1. Kb8xBc8[Ke1] Kf8-f7 2. Bc2xRa4=bB[Bc8] a6-a5/d2-d1=B 3. Ra1-a4=bR d2-d1=B/a6-a5. Now white wants to play Rh1xBh6[Ra1] but black can defend with Rb4xPa4=wR[Rh1]!. So white first needs to block the magic square: 4. Ra1xPa3[Ra1] a4-a3 5. Rh1xBh6 Bg7-h6 6. Re7xQf7[Rh1] & 1. Re8 Qxe8[Qd8]#


14383 - Günther Weeth, Klaus Wenda

Die Schwalbe 241, February 2010

[8/7k/8/8/8/8/8/2K5]

1+1. Add the minimum amount of material for a correct -2 & #1 (Proca retractor, Anticrce, magic square type II d1)

[8/7k/8/8/8/8/8/2K5]

Solution

Add bRa8 with solution 1. Rd8xRd1=bR[Ra8] Kh8-h7 2. 000[bR] & 1. Ra7#


14384 - Per Grevlund

Die Schwalbe 241, February 2010

New year's greeting

[1k3K2/1P5p/7p/6pr/6Pp/7P/3p4/3b4]

1+1. ser-=12, how many solutions?

[1k3K2/1P5p/7p/6pr/6Pp/7P/3p4/3b4]

Solution

All solutions are of the form 6. Kd2 7. Kd1 12. Kb6=. Using Pascal's triangle, it's easy to see that there are 67 ways to play 6. Kd2, 1 way to play 7. Kd1 and 30 ways to play Kb6=. So in total there are 67*1*30=2010 solutions.


14385 - Andreas Witt

Die Schwalbe 241, February 2010

[8/8/1P3R2/8/8/1Q3P2/8/8]

4+0. The centers of the squares on which the white pieces stand form a rectangle. Play 3 white moves to change this rectangle into a different one with the same surface.

Zeroposition: a) wBf3 b) wSf3 c) wBb6 d) wSb6

Original: Die Mittelpunkte der Standfelder der wei?#376;en Steine sind die Eckpunkte eines Rechtecks. Bilde mit 3 Zügen ein neues Rechteck mit gleichem Flächeninhalt an einer anderen Stelle des Bretts!

[8/8/1P3R2/8/8/1Q3P2/8/8]

Solution

The rectangle has a surface of 4*3=12 units.

a) 1. Qb2 2. Re6 3. Be2 (3x4=12)
b) 1. Qb4 2. Rh6 3. Sh4 (6x2=12)
c) 1. Qc3 2. Rf7 3. Bc7 (3x4=12)
d) 1. Qd1 2. Rc6 3. Sa4 (2*sqrt(2)x3*sqrt(2)=12)


14386 - Stephan Dietrich

Die Schwalbe 241, February 2010

The white king is on e1. The eight light officers (B+N) are on the first two rows of the board, and each type of both colors is once on a white, once on a black square.

a) How many positions exist in which the white king is not in check?

b) What is the result of a) if the pieces are on the first three rows?

Original: Auf einem Schahbrett steht auf e1 der wei?#376;e König. Die acht Leichtfiguren des Partiesatzes befinden sich auf den ersten beiden Reihen des Brettes, wobei jede Figurenart jeder Partei jeweils einmal auf einem wei?#376;en und einmal auf einem schwarzen Feld steht.

a) Wie viele solcher Stellungen gibt es, bei denen der wei?#376;e König nicht im Schach steht?

b) Welches Resultat ergibt sich bei a), wenn die Läufer und die Springer auf die ersten 3 Reihen verteilt werden?

Solution

a) First the pieces on the white squares: bS can't be on c2 or g2, so 6 squares left. For the second piece, there are 7 squares left, for the third 6, for the fourth 5. In total 6x7x6x5=1260 possibilities.

Then the pieces on the black squares: bB can't be on d2 or f2, so 5 squares left (e1 is already occupied). For the second piece there are 6 squares left, for the third 5, for the fourth 4. In total 5x6x5x4=600 possibilities.

So in total there are 1260x600=756000 possibilities

b) White squares: Black knight can't be on c2, d3, f3, g2, so 8 possibilities. 11 for second piece, 10 for third, 9 for fourth. In total 8x11x10x9=7920 possibilities.

Black squares: Black bishop can't be on d2 or f2. First the case that the bishop isn't on c3 or g3: 7x10x9x8=5040 possibilities.

Next the case that the bishop is on c3 or g3. In this case, a piece must be on d2 or f2, and there are 3 possible pieces for this. So in total 2x3x9x8=432 possibilities. So in total 5040+432=5472 possibilities.

So in total there are 7920x5472=43338240 possibilities


14387 - Werner Keym

Die Schwalbe 241, February 2010

Construct an illegal cluster with wK, wR, bK. The distance of the 1) white 2) black king to the initial square is a) minimal b) maximal (4 positions: a1, a2, b1, b2)

Original: Konstruiere jeweils ein Illegal Cluster mit wK, wT, sK. Der Abstand des 1) wei?#376;en, 2) schwarzen König zu seinem Partieanfangsfeld ist a) minimal, b) maximal (4 Stellungen).

Solution

1a) wKg1, wRh1, bKh3
2a) wKh7, wRh8, bKg8
1b) wKa7, wRa8, bKc8
2b) wKa2, wRa1, bKc1


14388 - Bernd Schwarzkopf

Die Schwalbe 241, February 2010

Illegal cluster with wK, 4wS, bK. The white king is as close as possible to his initial square, and 3 knights are on border squares.

Original: Illegal Cluster mit wK, 4wS, sK. Der wei?#376;e König steht möglichst nahe an seinem Ursprungsfeld, 3 wei?#376;e Springer auf Randfeldern.

Solution

wKe2, wSa1, wSa2, wSb1, wSb2, bKc2. Last move must've been Sb3xXa1, but neither king nor this uncaptured piece have a retromove.