[Retros] Orbit 39

[raosorio at fibertel.com.ar]

Hi,
These are the retro problems published in Orbit 39; solutions included at the end.

(I) Leonid Borodatov
(r3k3/2K5/8/3PPP2/2NpPpP1/3N 1P1P/4PB1p/R6R)   H#2 * AP  (14+5)  
         
(II) Leonid Borodatov
(4k2r/8/1N5P/BPP3NP/qpP3P1/rp3B1P/RK4PR/8) H#2 AP  (15+6)

(III) Sergio Orce
(Bn6/k7/P2P4/8/1K6/8/8/8)   Retractor type Proca - 3 & #1 (4 + 2)

(IV) Itamar Faybish
(2R1kbnr/p1ppp1pp/8/8/8/PQ6/q2P1PPP/1NB1K1NR) PG 10.5  (12+11) C+     

(V) Paul Raican
(1n1qkbnr/rppppp1p/5P2/4pN1B/1P5P/NP2PR2/2PPK2B/8) PG 28.0 (13+14)  C-

(VI) Satoshi Hashimoto  
(6R1/5N2/1QP2P2/pppppppp/2PRB3/1P3P2/1P1KPPk1/2B2N2) PG 24.5 (16+9) C-

(VII) Nicolas Dupont
(3B1kB1/1b1np2p/1p4r1/3n4/2p2pq1/p2P2Pr/PP1bPK1P/RN1Q2NR) PG 21.5 (14+14)  
 Antecedent (rnkq2nr/1pp2ppp/6P1/6QP/3PR1P1/4B2N/P3P1BN/1KR5) R. Aschwanden  PG 20.0

Best,
Roberto



SOLUTIONS - Comments, Roberto Osorio (RO), Valery Liskovets (VL), Jorge Lois (JL)
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I) 1.-, Kd6  2. Kd8,  RxRa8# (set play)
1. f4xg3 e.p.! (retroplay 1. g2-g4, g3xQh2), Kc6 2. O-O-O! (argument for e.p.), Ra8#
There is no waiting move and after ep, castling should be done to justify ep! 
WPs d, e and f captured 10 men, so 10+5+Ph7=16 and then white could not
capture in his last move . h4xg3xh2 was impossible because the captured wB 
was the light squares one.. So, indeed, unless 000 is destroyed, the last moves 
could be only g2-g4 and g3xh2 (VL) / 1….,Kd6?  2.Kd8 would not prove the 
e.p. capture right, a typical try (RO).

II) 1.b4xc3 e.p.!, Bxc3 2. O-O!, h7# (Retroplay: 1. c2-c4!,  c4xQb3  2.Qd3-b3, Rc3-a3)
Again, 10 black men were captured and the balance is equivalent to the previous 
problem. The only one white retraction capable to keep the black castling right is
1. c2-c4! followed by the explained retroplay. A thematic couple of problems showing
The e.p. & castling rights dependency (RO) / A pity that it doesn't contain a thematic 
try: an (illegal) mate without 00 after ep (VL).

III) -1.c5xd6 e.p., d7-d5 -2. Ka5xb4, b5-b4  -3. Bc6xRa8! (.-3.Bc6?, Ka8xRa7!! shows that 
b7-b5 was not necessarily the last black move) and, 1. c5xb6 e.p.# (A) / A nice 
miniature that starts  with an e.p. retraction and ends with an e.p. mate (RO).

IV) 1.Pc2-c4   Pf7-f5   2.Pc4-c5   Pf5-f4   3.Pc5-c6   Pf4-f3   4.Pc6xb7   Pf3xe2   
5.Pb7xc8=R Pe2xf1=B 6.Rc8xb8   Bf1-a6   7.Rb8xa8   Qd8-b8   8.Pa2-a3   Ba6-c8   9.Ra1-a2   
Qb8xb2   10.Qd1-b3   Qb2xa2   11.Ra8xc8
Many Tourney themes in a short sequence: an officer captures an officer / Gianni Donatti 50th JT 
(switchback of a promoted piece from and to its promotion square) / Kostas Prentos 40th JT 
(a promoted piece is captured by an officer) / Messigny 2008 (a promoted piece is captured by 
another promoted piece) (RO).

V) COOKED 1.Ph2-h4   Pa7-a5   2.Rh1-h3   Pa5-a4   3.Rh3-f3   Pa4-a3   4.Rf3-f6   Pg7xf6   5.Pb2-b4   
Bf8-h6   6.Bc1-b2   Ke8-f8   7.Bb2-e5   Kf8-g7   8.Be5-h2   Bh6-f4   9.Pe2-e3   Sg8-h6   10.Qd1-h5   
Qd8-g8   11.Qh5-e5   Pf6xe5   12.Bf1-e2   Kg7-f6   13.Be2-h5   Qg8xg2   14.Ke1-e2   Qg2-d5   
15.Sg1-f3   Qd5-b3   16.Pa2xb3   Pa3-a2   17.Sb1-a3   Ra8-a7   18.Ra1-g1   Pa2-a1=Q 19.Rg1-g8   
Qa1-g1   20.Rg8xc8   Qg1-g7 21.Rg8  Qf8  22.Rg3  Sg8 23.Sd4  Bh6  24.f4  Qd8  25.Rf3  Kg7  26.f5  
Kf8 27.f6  Ke8  28.Sf5  Bf8.   
Queen-sarcophagus (A) / bQ Pronkin plus bK, bN and bB switchbacks; quite a term 
(sarcophagus) to refer to a Phenix piece standing on a square “entombed” by its
own army (RO).

VI) 1. d4 a5 2. d5 Ra6 3. Qd4 Rb6 4. Sd2 Rb3 5. axb3 b5 6. Qb6 c5 7. Ra4 Sc6 8. dxc6 d5 9. Rd4 Bg4 
10. c4 Bf3 11. gxf3 e5 12. Bh3 Ke7 13. Bf5 Kf6 14. Be4 Kg5 15. Sh3+ Kh4 
16. Sg5 f5 17. Sf7 Kh3 18. Sf1 Kg2 19. h4 Qg5 20. hxg5 Sf6 21. gxf6 g5 22. Kd2 Bh6 23. Rxh6 Rg8 
24. Rg6 h5 25. Rxg8. Black pawn double steps in alphabetical order; 
all black officers moved to be captured (A) / Nice added value (alphabetical order) on a well 
known task (displaced pawn line) (JL).

VII) 1. Pf2-f4   Pb7-b5  2. Pf4-f5   Fc8-b7  3. Pf5-f6   Dd8-c8 4. Pf6xg7   Pf7-f5   5. Pc2-c4   Cg8-f6   
6. Pg7-g8=F Ff8-h6 7. Pc4-c5   Re8-f8  8. Fg8-b3  Th8-g8  9. Pc5-c6  Tg8-g6 10.Pc6xd7 Pc7-c5 11. Pd7-d8=F 
Pc5-c4  12. Fd8-b6   Pa7xb6 13. Re1-f2   Ta8-a3    14. Fb3-a4  Ta3-h3  15. Pd2-d3   Pb5xa4 16. Fc1-f4   
Pa4-a3    17. Ff4-c7   Fh6-d2   18. Pg2-g3  Pf5-f4 19. Ff1-g2  Dc8-g4   20. Fg2-d5 Cb8-d7 21. Fd5-g8  
Cf6-d5  22. Fc7-d8.  Two anti-Pronkin bishops from the same side. Cross captures (A) 
/ See the R. Aschwanden’s antecedent: (1. h4 d5 2. h5 d4 3. Th4 d3 4. Sh3 dxc2 5. d4 a5 6. Dd2 a4 7. Dg5 a3 
8. Le3 c1=L 9. f4 Ld2+ 10. Sxd2 axb2 11. Sf3 b1=L 12. Sh2 Lg6 13. f5 e5 14. fxg6 Lf5 15. 0-0-0 Lb1 16.Kxb1 e4 
17. Txe4+ Kd7 18. g4 La3 19. Lg2 Lc1 20. Txc1 Kc8); the motivation here is that the (a) and (d) bP had to 
promote in order to be captured and after that there is no other square to capture the original bishops than 
the promotion ones. In the Dupont’s’ problem the (a) and (b) bP had to make a cross capture in order to allow 
the bR to reach h3 in two moves (see “Selected problems” to complete the discussion on this theme) (RO).