[Retros] Christmas Contest of Stuttgarter Zeitung

[Jens Guballa]

Hi all,

the solution for the Christmas contest was published in the last two
issues of the Stuttgarter Zeitung.

Here is a summary:

78 senders sent their solutions, 37 of them made it to the drawing lot
(Problem A and B solved correctly). The winners are:

  Martin Kummer, Karlsruhe (cordless telephone)
  Martin Pfleiderer, Korntal (micro wave)
  Werner Munk, Stuttgart (colored TV)
  
Winners for the expert contest (correct solutions for problem A - E) are
the following 8 persons:
Dietmar Fauth, Norbert Geisler, Rolf Heimerdinger, Olli Heimo, Martin
Kummer, Klaus Wenda, Nils Empacher and Rolf Schreiber.

The following solvers have missed a penny: Silvio Baier, Jens Gballa,
Markus Knapp, Herbert Lang, Hanno Lehto, Kurt Lorenz, Erhard Mozer, Karl
Roscher, Klaus Bernhardt, Peter Lau Thomas Lutzius and Bernd Schuh. They
also receive a book price.

Solutions:

A: Not 1. d5:e6 e.p#, because e7-e5 must not have been the last move,
   possibe is also Kf8:Bf7, and before that Bg8:f7+. So, after 1. B:g7
   it takes two moves to mate the black king (2. Rf8#).

B: Two solutions:

   a) Black pawn f7 becomes white. Bf8 is a promoted piece, and now the
   only legal last move is g7-g5. Therefore 1. f5:g6 e.p.# works.
  
   b) White pawn h6 becomes black. Black does not have a last move here,
   so he can checkmate with 1...Nf6#.
  
C: Solution: wKh1, wQg8, bKa1, bPb4. 1. Qg8-b3 stalemate.

   The quadrangle here has a size of 30 squares.
  
D: 1.Rc2!! Rf3: 2. Rb2 Rf1+ 3. Kc2 Rf2+ 4. Kc3 Rb2: 5. Kb2: 
   Most solvers stopped here, but white must play carefully to secure
   the win (avoiding stalemate and knight forks). E.g.:
   5... g5 ( 5...a4 6.Kc3! Nd6 7.Kb4 g5 8.Kc5 Nc8 9.Kd4 g6 10.Ke5 )
   6.Kc3 Nd6 7.a4  preventing any knight forks 7...e5 8.Kd2  ( 8.g4!? )
   8...g4 9.Kd3 g6 10.Ke2 h5 11.Kf2 g5 12.Ke3 e4 13.Kd4 h4 14.Kd5 Nc8
   15.Kxe4 h3 16.Ke5 Nd6  stalemate? 17.Ke6  No! White wins.  

E: Sample solution from Klaus Wenda: White cannot checkmate in one move.
   1. Bh1-f1! (treathens 2. Kf1-e1 and forward Bg2#):
     1... Kg1-h1? 2. Kd1-e1 (or anything else), Rf1#.
     1... Bg1-h2? (Bg1:Ph2?) 2. Rh5-a5, Bg2#
     1... Bg3-h2? 2. Ra1-a5, 0-0-0#.
     
   The bBh2 has been promoted on g1 (pawn c7). This pawn captured both
   knights, the queen and another white piece, that has been promoted by
   the a-pawn without any captures (pawn c7 captured 4 times). The white
   pawns also captured 4 times (both knights, queen and rook from a8). 
   So the 0-0-0 is legal. The following chessmen could never been
   captured by pawns: Rh8, Bf8, Pa7 (Pa2 has promotes without any
   captures), Bc1, Ph2. Therefore the retraction 1. Bh3:Qf1? Qg2-f1+ 2.
   Kf1-e1 and forward Bg2# is illegal.
   But black still have an ace: 1.... Bg3:Ph2!! Now the promoted bishop
   must have moved from g1 to g3 via f2, so the white King has moved and
   0-0-0 is not legal, therefore white cannot checkmate!




Best regards,
Jens