[Retros] Hans Gruber 50 jubilee

[Joost de Heer]

The award of the Hans Gruber 50 jubilee tourney was published in Die Schwalbe 
Heft 244. Several retros ended up in the award. The theme was 'Anything that 
shows 'hg' in the solution'.

1st prize - Itamar Faybish
rs2k2S/p1p5/4p3/3p4/3S1P2/1P2QRKP/2PPP1p1/RrB5 (13+9)
SPG 18.0
1. Nf3 d5 2. Nd4 Bh3 3. gxh3 e6 4. Bg2 Bd6 5. O-O Bg3 6. hxg3 h5 7. Kh2 h4 8. 
gxh4 Qg5 9. hxg5 Nh6 10. gxh6 f5 11. hxg7 f4 12. gxh8=N f3 13. Kg3 fxg2 14. f4 
b5 15. Rf3 b4 16. Qg1 b3 17. Qe3 bxa2 18. b3 axb1=R

2nd prize - Cornel Pacurar, Ion Murarasu
A: 3k3K/8/6b1/6p1/6p1/6p1/6p1/6b1 (1+7)
B: 3k2K1/8/7b/7p/7p/7p/7p/7b (1+7)
ser-a=>b50, PWC
1.Kh8-g7   2.Kg7*g6 [+bBg7]   3.Kg6*g5 [+bPg6]   4.Kg5*g4 [+bPg5]   5.Kg4*g3 
[+bPg4]   6.Kg3*g2 [+bPg3]   7.Kg2-h1   8.Kh1*g1 [+bBh1]   9.Kg1-f1  10.Kf1-e2 
  11.Ke2-d3  12.Kd3-c4  13.Kc4-c5  14.Kc5-d6  15.Kd6-e6  16.Ke6-f7  17.Kf7-g8 
18.Kg8*g7 [+bBg8]  19.Kg7*g6 [+bPg7]  20.Kg6*g5 [+bPg6]  21.Kg5*g4 [+bPg5] 
22.Kg4-h3  23.Kh3*g3 [+bPh3]  24.Kg3-h2  25.Kh2*h3 [+bPh2]  26.Kh3-g4 
27.Kg4*g5 [+bPg4]  28.Kg5-h4  29.Kh4*g4 [+bPh4]  30.Kg4-h3  31.Kh3*h4 [+bPh3] 
32.Kh4-g5  33.Kg5*g6 [+bPg5]  34.Kg6-h5  35.Kh5*g5 [+bPh5]  36.Kg5-h4 
37.Kh4*h5 [+bPh4]  38.Kh5-g6  39.Kg6*g7 [+bPg6]  40.Kg7-h6  41.Kh6*g6 [+bPh6] 
42.Kg6-h5  43.Kh5*h6 [+bPh5]  44.Kh6-g7  45.Kg7*g8 [+bBg7]  46.Kg8-h7 
47.Kh7*g7 [+bBh7]  48.Kg7-h6  49.Kh6*h7 [+bBh6]  50.Kh7-g8 a=>b

4th prize - Klaus Wenda
kb1B4/PrP2P2/1P4pp/8/8/7R/8/R3K2b (8+6)
-5 & =1, Proca retractor Anticirce, Magic square Type II h2
1. Kg7xRh7[Ke1] Rh8-h7 2. Rg2xPh2=bR[Rh8] zz h7-h6 3. g5xPh6[Ph2=bP] zz h2-h1=B 
4. Rf8xQh8[Ra1] zz Qg8-h8 -5. Rh5-h3 & 1. Rh2=bR[Rh8]=

5th prize - Andrei Frolkin
rs1q2sr/ppp2pp1/P2ppk2/6P1/3P4/8/1PPP1PPb/1SB4K (11+14)
SPG 15.5
1. a4 d6 2. a5 Bf5 3. Ra4 Bd3 4. exd3 e6 5. Qh5 Be7 6. Rg4 Bg5 7. d4 Ke7 8. Bd3 
Kf6 9. Bg6 hxg6 10. h4 gxh5 11. Nh3 hxg4 12. O-O gxh3 13. Kh1 h2 14. Rg1 hxg1=B 
15. a6 Bh2 16. hxg5+

1st HM - Andrei Frolkin, Joaquim Crusats
1k6/pP5p/pPK3P1/BPPPPpP1/2pr1p2/8/8/8 (10+8)
-4 & #1 Proca retractor
Main plan: 1. b4-b5 ~ 2. b5-b6 & 1. Bc7#, but black defends with 1. b5xXc4!
Try: 1. h4xPg5? h6xRg5!
Solution: 1. h5xPg6! ~ 2. h4xPg5! ~ 3. b4-b5 (now b5xXc4 is illegal) ~ 4. b5-b6 
& 1. Bc7#

2nd HM - Itamar Faybish
rsb3k1/ppppp3/5pPs/6p1/5q2/1P6/P1PPPPP1/RS1QKBSR (15+13)
SPG 11.0
1. b3 g6 2. Bb2 Bh6 3. Bxh8 f6 4. Bg7 Kf7 5. Bxh6 Qf8 6. Bg5 Qh6 7. Bh4 Qf4 8. 
Bg5 h6 9. h4 hxg5 10. h5 Nh6 11. hxg6+ Kg8

3rd HM - Andrei Frolkin, Nikolai Beluhov
5sQq/3pPpRB/1p1pRKp1/4p1p1/5kPb/5PB1/1P3PP1/b5S1 (13+12)
Release the position b) wSg7

a) 1. Bh2xSg3 h6xSg5 2. h3xRg4 a2-a1=B 3. Se4-g5 Rg5-g4 4. Sc3-e4 a3-a2 5. 
Sb5-c3 a4-a3 6. Sa7-b5 a5-a4 7. Sc8-a7 a6-a5 8. c7-c8=S a7-a6 9. c6-c7 c7xSd6 
10. Se4/f5-d6 S~-g3 11. Sg3-e4/f5 S~ 12. d6xRe7 etc
b) 1. Bh2xRg3 h6xSg5 2. h3xSg4 Se3-g4 3. Se4-g5 Rg5-g3 4. Sg3-e4 Sf5-e3 5. 
Se8-g7 Sg7-f5 6. Sc7-e8 a2-a1=B 7. Sb5-c7 a3-a2 8. Sa7-b5 a4-a3 9. Sc8-a7 a5-a4 
10. c7-c8=S a6-a5 11. c6-c7 c7xRd6 12. R~-d6 a7-a6 13. d6xRe7 etc

6th HM - Andrei Frolkin
rsbqkb1R/pppppp2/7r/6rB/7r/8/PPPPPP2/RSBQKBS1 (15+15)
SPG 14.0
1. h4 g5 2. Rh3 gxh4 3. Rg3 h3 4. Rg7 h2 5. g4 h1=R 6. g5 Rh5 7. g6 Rg5 8. Rxg8 
h5 9. g7 Rh6 10. Rh8 h4 11. g8=B h3 12. Bh7 h2 13. Bg6 h1=R 14. Bh5 Rh4

7th HM - Andrei Kornilov
8/8/P1p1PPP1/K1k1B1P1/BSS1R1P1/p1R3P1/P1Q1ppp1/8 (16+6)
#1 How many solutions?
Capture balance: bxa (1), cxdxexfxg (4), dxexfxg (3), hxg(1), and to get the 
black pawns past the white pawns two more captures are needed. That is one 
capture too many, so the position needs to be rotated 180 degrees.
Now black has no last move, so there aren't 8 solutions (Be3, Be5, Sd6, Se3, 
Se7, Sg3, Sg7, Sh6) but only one: 1...hxg5#

Comm. - Bernd Gräfrath
6s1/7P/q5p1/7k/7p/P7/5BP1/4K2R (6+5)
Add a piece for a correct and legal ser-s#6, Monochrom
Try: bRg5 or bRc3, but then the position is illegal if white still may castle: 
Qa6 is promoted and needed 4 captures, the black g- and h-pawn needed another 
two, and Rc3/g5 would be promoted, needing another 4 captures including Sg1. 
But this rook couldn't leave g1 without destroying white's castling rights
Solution: +bQg5, 1. hg8=Q 2. Qc4 3. OO 4. Kh2 5. Bg1 6. g4 hg3ep#

Comm. - Unto Heinonen
1s1kqb2/1pps4/1p1rpp2/1b1prp2/4P3/1BRP4/P1P2PP1/RS1KQ1S1 (13+15)
1. b4 f5 2. b5 Nf6 3. b6 axb6 4. e4 Ra5 5. Bc4 Re5 6. Bb3 d5 7. Qh5+ Kd7 8. 
Qxh7 Qe8 9. Qh5 Kd8 10. Qe2 Rxh2 11. Kd1 Rh6 12. Rh3 Bd7 13. Rc3 Bb5 14. d3 
Nfd7 15. Bg5 Rd6 16. Bf6 gxf6 17. Qe1 e6