[Retros] Die Schwalbe Heft 241, februari 2010

[Joost de Heer]

Die Schwalbe Heft 241 arrived yesterday. In it an article by Werner Keym 
about partial retroanalysis and retro-strategy in the 2009 Codex. Otto 
already published an English version of this article on the retro corner.

And of course lots of originals.

14375 - Werner Keym
r3k2r/1pp1p1p1/S1p1P1p1/4P2S/6Q1/4P3/1P2P1PB/R3K2R (13+9)
#3 RS

14376 - Nikolai Beluchov
8/3p4/1p2p2p/8/PPPPPp1p/KBrQp2B/1RrsRP1P/q1bkb3 (13+14)
Last 31 halfmoves?

14377 - Gianni Donati, Olli Heimo
4s1k1/pppsR1P1/4p1r1/5p2/8/P3P1P1/P1PrPPbp/1S1qKBSR (14+13)
SPG 24.0

14378 - Dragan Petrović
8/5p2/4p1P1/2P2PP1/1PrkPKPR/BRqbrpQ1/1pppppSb/4SssB (15+16)
Shortest resolution?

14379 - Bernd Gräfrath
rsb1Qbs1/pppp1qp1/5k2/5p2/5P2/8/PPPP3P/RSB1KBSR (14+13)
SPG 9.5 Duellist

14380 - Bernd Gräfrath
Dedicated to John Beasley's 70th birthday
5r1q/p1p1p1b1/2s2pk1/8/b7/8/PPPPPP1P/RSBQKBSR (15+10)
SPG 11.5 White must capture

14381 - Wolfgang Dittmann
s2s2Bb/pp6/k3P3/p7/4P3/8/1P1P4/r3Kb2 (6+9)
-19 & #1, Proca retractor, Anticirce

14382 - Günther Weeth
8/bP3k2/1P6/pP4P1/3S4/3p4/4P3/3bK3 (7+5)
-6 & s#1, Proca retractor, Anticirce Cheylan, magic square type II a4

14383 - Günther Weeth, Klaus Wenda
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

14384 - Per Grevlund
New year's greeting
1k3K2/1P5p/7p/6pr/6Pp/7P/3p4/3b4
ser-=12, how many solutions?

14385 - Andreas Witt
8/8/1P3R2/8/8/1Q3P2/8/8 (4+0)
The centres 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 weißen Steine sind die 
Eckpunkte eines Rechtecks. Bilde mit 3 Zügen ein neues Rechteck mit 
gleichem Flächeninhalt an einer anderen Stelle des Bretts!

14386 - Stephan Dietrich
The white king is on e1. The eight light officers (B+S) are on the first 
two rows of the board, and each type of both colours 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 weiße König. Die acht 
Leichtfiguren des Partiesatzes befinden sich auf den ersten beiden 
Reihen des Brettes, wobei jede Figurenart jeder Partei jeweils einmal 
auf einem weißen und einmal auf einem schwarzen Feld steht.
a) Wie viele solcher Stellungen gibt es, bei denen der weiße König nicht 
im Schach steht?
b) Welches Resultat ergbit sich bei a), wenn die Läufer und die Springer 
auf die ersten 3 Reihen verteilt werden?

14387 - Werner Keym
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) weißen, 2) schwarzen König zu seinem Partieanfangsfeld 
ist a) minimal, b) maximal (4 Stellungen).

14388 - Bernd Schwarzkopf
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 weiße König steht 
möglichst nahe an seinem Ursprungsfeld, 3 weiße Springer auf Randfeldern.

CORRECTIONS
-----------
14186 is cooked (see below), correction: bRg6 instead of bPg6.

SOLUTIONS HEFT 238
------------------
14182c - A. Yarosh
The version printed in the magazine is an incorrect version. The author 
had sent the correction before publication, but due to an error of the 
editor, the incorrect version was printed.

Correct diagram: 4SBBk/1pP1PRRS/p1pppKP1/P3Ppp1/6b1/5sr1/6P1/q7 (13+13)
Add a black knight and another black officer, one of them on h6, and 
release the position.
+bRh6, +bSf2 and -1...Rh1-h6 -2. a4-a5 Rb1-h1 -3. a3-a4 Rb6-b1 -4. a2-a3 
Rb1xb6 -5. b5-b6 Sh1-f2 -6. b4-b5 h2-h1=S -7. b3-b4 h3-h2 -8. b2-b3 
h4-h3 -9. e4-e5 h5-h4 -10. e3-e4 h6-h5 -11. h5xSg6 etc

14183 - B. Schwarzkopf
Try: -1. e7-e8=S and 1. e8=R/Q#, but 9 white pawns
Try: -1. d2xSc3 & d4#. White needs 6 captures for the pawn structure, 
and with Bc8 this means all missing black pieces. So black needs 6 pawn 
captures too for his pawn structure, since g2 can't have come from g7 
without a capture. But Bc1 isn't available for capture, so this results 
in an illegal position.
Solution: -1. f2xSg3 & 1. f4#

14184 - G. & I. Denkovski
1. e4 Nf6 2. Qh5 Rg8 3. Qh6 gxh6 4. a4 Rg3 5. Ra2 Ra3 6. g4 Bg7 7. g5 
Ng4 8. g6 Bc3 9. g7 Ba5 10. b4 d6 11. Bb2 Be6 12. g8=Q+ Kd7 13. Bh8 Bb3 
14. Qg7 Ke6 15. Qa1 Nd7 16. Nc3 Ndf6 17. Bc4+ Ke5 18. Bd5 Kf4 19. Nge2+ 
Kf3 20. O-O Qg8 21. Rb1 Qg5 22. Rbb2 Rg8 23. Qd1 Rg6 24. Nb1 Ng8 25. Bc3 
Rf6 26. Kh1
Pronkin queen, sibling knight, switchback

*14185 - P. Harris
Intentions: -1. bSe4xwSc5 and 1. Rf8 Kc8 2. Sd6 e6#; -1. Qe4xQb7 and 1. 
Qb1 Qf3 2. Qb8 Qf4#
But the first one doesn't work (the retraction is illegal), plus this is 
horribly cooked, e.g. -1. Sg1xQf3 & 1. Sh3 Sf5 2. Ra2 Qg3#

*14186 - G. Ettl
-1. Kf3-g2 Rf7-g7 -2. Ke3-f3 Re7-f7 -3. Kd3-e3 Rd7-e7 -4. Kc3-d3 Rc7-d7 
-5. Be8xSa4[Bf1] Sb2-a4 -6. Kd3-c3 Rd7-c7 -7. Ke3-d3 Re7-d7 -8. Kf3-e3 
Rf7-e7 -9. Rg2-f3 Rg7-f7 -10. Rh5-e5 & 1. Rh8#
Cook, found by Mario Richter: -1. Rh5-e5!! ~ 2. Be8xXf7[Bf1] & 1. 
Rh5-h8#. Nothing black does on his first move can avoid this short solution.

14187 - B. Gräfrath
a) Possible proofgame: 1. Nf3 Nf6 2. Nd4 Rg8 3. Ne6 Rh8 4. h4 Rg8 5. Rh3 
Rh8 6. Rg3 Ne4 7. Rg6 hxg6 8. Nxf8 Rh6 9. h5 Rh7 10. h6 Rh8 11. h7 Rg8 
12. h8=R Ng3 13. Rh4 Rh8 14. Ne6 Nh5 15. Nd4 Kf8 16. Nf3 Kg8 17. Ng1 Kh7 
18. Rh3 Rf8 19. Rh1 Kh8
b) 1. h4 Sf6 2. h5 Sh8[g8=R] 3. Rf8[-] Kf8[Rh6] 4. Rh7[Pg6] Kg8 5. 
Rh8[Rf8] Kh8[-]

14188 - M. Richter
1. a3 b5 2. c4 bc4 3. Qb3 cb3 4. Ra2 ba2 5. Sc3 a1=K 6. b3 Bb7 7. Bb2 
Bg2 8. Ba1
Fastest King-Schnoebelen in losing chess

14189 - B. Gräfrath
1. e4 d5 2. ed5 Sc6 3. dc6 Qd5 4. cb7 Qa2 5. bc8=S! Qe6 6. Se2 Qe3 7. h3 
Qd2 8. Sd2 c5 9. Sc4 Rc8 10. Sb6 Rb8 11. Sc8
Schnoebelen-knight and antipronkin knight on c8.

14190 - Osorio, Lois
1. h4 g5 2. hg5[Pg7] b6 3. Rh6 Ba6 4. Rb6[Pb7] h5 5. e4 Rh6 6. Qh5[Ph7] 
Rc6 7. Sf3 Rc3 8. Bc4 d5 9. d3 Qd7 10. Sbd2 Qa4 11. Sb3 Sc6 12. Bd2 OOO 
13. Qh3 f5 14. ef6[Pf7] Kb8 15. fe7 Ka8 16. e8=R Re8[Rh1] 17. OO Bc5 18. 
ed5[Pd7] Re1 19. Kh1 Rf1
All four special moves (doublestep, en passant, promotion, castling) 
performed by Ph2.

14191 - S. Dietrich
A bishop has maximum 13 moves, and a knight maximum 8. The wanted number 
of moves is one less than the maximum, so a knight must guard the bishop.
So on the board is a bishop that has 13 moves, one knight that has 8 and 
one that has 7 moves.
The bishop is on one of the center squares. Take e.g. e5. There are only 
6 squares where the '8-knight' can be: d5, e4, f5, e6, c5, e3. With the 
first four squares, the second knight must be on c6, c4, d3 or f3. With 
the other two squares, the second knight must be on c6, (d3 or c4) and 
f3. In total that is 4*4+2*3=22. The same number is reached for the 
other three central squares where the bishop can be. So the number of 
positions is 88.

14192 - S. Dietrich
The maximum is reached when there are two major officers on b1 and g1, 
and the knights have 8 moves without blocking white lines. With a rook 
on g1, the different positions are:
Qb1, Rc1: knights on d6, e3, e6, f4 (except the combination e6/f4): 5 
positions
Qb1, Rd1: knights on c4, c5, e6, f4 (except the combinations e6/f4 and 
c5/e6): 4 positions
Qb1, Re1: knights on c4, c5, d6, f4 (except the combination c4/d6): 5 
positions
Qb1, Rf1: knights on c4, c5, d6, e6 (except the combinations c4/d6 and 
c5/e6): 4 positions
Qc1, Rb1: knights on d6, e6, f5 (except the combination d6/f5): 2 positions
Qd1, Rb1: knights on c4, c5, e6, f4, f5 (except the combinations c5/e6 
and e6/f4): 8 positions
All these positions can be mirrored, so the total is 2*(5+4+5+4+2+8)=56