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5 problem(s) found in 1829 milliseconds (displaying 5 problem(s)). [A='Stojni��' and FIRSTNAME='Dragan'] [download as LaTeX]

1 - P1075235
Dragan Stojnic
Best Problems 2002
Lob
Garofalo-50-Jubiläumsturnier
P1075235
(6+11) C+
h#2
2.1...
1) 1. Dc2 Ld3 2. Tb4 Sc4#
2) 1. Te5 Le4 2. Td6 Sd5#
play all play one stop play next play all

Genre: h#
Computer test: (Popeye WINDOWS-32Bit V4.37 (1496444 KB))
FEN: 8/8/2b2pB1/1pkr2RK/2qr2p1/4N3/1p1p4/1nR3B1
Input: hpr, 2007-10-09
Last update: hpr, 2007-10-09 more...
2 - P1080355
Dragan Stojnic
R397v The Problemist 09/2008
P1080355
(13+12)
#2

Genre: Retro
FEN: 4k2r/2p1ppnb/1p2p2Q/bP2N1N1/2P2pp1/2P2P2/2PR1PP1/R3K3
Input: Gerd Wilts, 2009-01-03
Last update: Gerd Wilts, 2009-01-03 more...
3 - P1088951
Dragan Stojnic
R411 The Problemist 07/2009
P1088951
(14+12)
#2
Paulo Roque: Solution Part(1):

Apparently 1. Dxf7+ Kd8 2. Dxd7# ,but the position of diagram is illegal then problem insoluble.

Demonstration the illegality of position:

a) Logical elements in the diagram (LED):
(a.1) The light square sL was captured in original square, note square c8.
(a.2) captured two white pieces ( two bishops)
(a.3) captured four black pieces ( one knigth, one light square bishop, one queen and one unknown ).
The piece unknown: (uk1) is one pawn, or (uk2) is one pawn-promoted, or (uk3) is one piece original equal the pawn-promoted which is in the diagram. (2009-07-21)
Paulo Roque: Solution Part(2):
b) Development:
(b.1) The white pieces were captured in: cxb(one capture) and g3xf2(one capture) = two captures.
(b.2) The black pieces were captured in: b2xa3(one capture) + e2xd3(one capture)+ h2xg3(one capture) + bishop in square original xLc8(one capture) = four captures.
(b.3) Then by (b.1) light square wL captured in -b5- and dark square wL captured in f2.
(b.4) By (a.2) + (b.1) the pawn-h black not captured ( not sBh x ).
(b.5) By (b.2) the pawn-h black was not captured ( not x sBh ).
(b.6) Then (b.4) + (b.5) = pawn-h black was promotion in h1. Also not (uk1) in (a.3).
(b.7) But if pawn-h black promotion in h1 then first occurred Bh2xXg3.
But if Bh2xXg3 then first occurred Bg3xLf2.
But if .Bg3xLf2 then first occurred wBf3 and Bb2xXa3(freeing the path for wLc1 square f2)
But if wBf3 then first occurred Be2xXd3(freeing the path for wLf1 square b5)
(b.8) It is impossible because: have three captures of black pieces preceding the promotion pawn-h black (b.7) + one capture black bishop(a.1). Paradox !

c) Conclusion: The position is illegal then problem insoluble. (2009-07-21)
Henrik Juel: The position seems legal to me. One possible sequence of events is (forward): e2xSd3, c6xLb5, f2-f3, b2xDa3, g3xLf2, Th8-g3, h2xTg3, h2-h1=T, etc., preserving the black castling. But even if Black could not castle, 1.Dxf7+ etc. is the only mate in 2. So what is the point of the problem? (2009-07-21)
Paulo Roque: .
Henrik you are right, the position is legal. I missed the step (b.8), because it is impossible (Uk2) but possible (uk3).
1) Example in retroplay:
0... e4-e3 -1. Db5-h5 e5-e4 -2. Sg5-h7 Th1-h8 -3. Te4-d4 h2-h1=T -4. Te1-e4 h3-h2 -5. h2xTg3 Tg4-g3 -6. Th1-e1 g3xLf2 -7. Tc6-f6 Lb2-c1 -8. Sh6-g8 Ld4-b2 -9. Tc3-c6 Lc5-d4 -10. Kc1-b1 Lf8-c5 -11. Tb3-c3 Th4-g4 -12. Sf5-h6 Th8-h4 -13. Tb1-b3 Sh6-f7 -14. Kd1-c1 Sg8-h6 -15. Ld4-f2 h4-h3 -16. Ta1-b1 h5-h4 -17. Se4-g5 h7-h5 -18. Ke1-d1 g4-g3 -19. Lb2-d4 a5-a4 -20. Lc1-b2 a7-a5 -21. Sc3-e4 e6-e5 -22. De5-b5 b5-b4 -23. f2-f3 c6xLb5 -24. Lc4-b5 g5-g4 -25. b2xDa3 De7-a3 -26. Ld5-c4 Dd8-e7 -27. Dc7-e5 g7-g5 -28. Dc8-c7 e7-e6 -29. Sd4-f5 f5-f4 -30. Dc7xLc8 Sh6-g8 -31. De5-c7 c7-c6 -32. De2-e5 Sg8-h6 -33. Dd1-e2 Sh6-g8 -34. Lf3-d5 Sg8-h6 -35. Le2-f3 Sh6-g8 -36. Lf1-e2 Sg8-h6 -37. e2xSd3 Sb4-d3 -38. Sf3-d4 Sc6-b4 -39. Sg1-f3 Sb8-c6 -40. Sf3-g1 f6-f5 -41. Sb1-c3 f7-f6 -42. Sg1-f3 (2009-07-21)
Paulo Roque: .
Diagram Error! The correct position is with sSh8 (not sTh8).
I think it should be deleted, because the true position is P1192187. (2011-04-11)
comment

Genre: Retro
FEN: r3k1Nr/1p1p1n1N/5R2/7Q/pp1R1p2/P2PpPP1/P1PP1pP1/1Kb5
Input: Gerd Wilts, 2009-07-17
Last update: Gerd Wilts, 2009-07-17 more...
4 - P1293762
Dragan Stojnic
The Macedonian Problemist 2014
P1293762
(12+1) C+
ser-h=35*
*) 1. ... h8=L=
1) 1. Kxg7 2. Kf8 3. Ke8 4. Kd8 5. Kc8 6. Kb7 7. Kb6 8. Ka5 9. Kb4 10. Kc3 11. Kd2 12. Kxe1 13. Kd2 14. Kc3 15. Kb4 16. Ka5 17. Kb6 18. Kb7 19. Kc8 20. Kd8 21. Ke8 22. Kf8 23. Kg7
24. Kh6 25. Kh5 26. Kxh4 27. Kg3 28. Kxg2 29. Kxg1 30. Kf2 31. Ke3 32. Kxd4 33. Kxe5 34. Kf6 35. Kf7 h8=D=
play all play one stop play next play all
Keywords: Series mover
Genre: Fairies
Computer test: Popeye Windows-32Bit v4.51 (100000 KB)
FEN: 8/6NP/3P3k/3PN3/3P3P/3B4/6P1/K3R1B1
Input: hpr, 2015-01-01
Last update: Olaf Jenkner, 2018-01-01 more...
5 - P1293763
Dragan Stojnic
The Macedonian Problemist 2014
P1293763
(14+1) C+
ser-h=36*
*) 1. ... e8=T=
1) 1. Kxg8 2. Kf7 3. Ke8 4. Kd7 5. Kxd6 6. Kc7 7. Kb6 8. Kb5 9. Ka4 10. Ka3 11. Ka2 12. Kb1 13. Kc1 14. Kd1 15. Ke1 16. Kf2 17. Kxe3 18. Kf2 19. Kg1 20. Kxh1 21. Kh2 22. Kxg3 23. Kxf4
24. Ke3 25. Kxe2 26. Kd1 27. Kc1 28. Kb1 29. Ka2 30. Ka3 31. Ka4 32. Kb5 33. Kb6 34. Kc7 35. Kd6 36. Kd5 e8=S=
play all play one stop play next play all
Keywords: Series mover
Genre: Fairies
Computer test: Popeye Windows-32Bit v4.51 (100000 KB)
FEN: 6R1/4PkB1/3P4/N4P2/1P3N2/2KPR1P1/4P3/7B
Input: hpr, 2015-01-01
Last update: Olaf Jenkner, 2018-01-01 more...
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The problems of this query have been registered by the following contributors:

hpr (3)
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